Calculo Thomas (Solutions).pdf

resposta cálculos Thomas

1. to accompany
Thomas’ Calculus
Eleventh Edition
and
Thomas’ Calculus: Early Transcendentals
Eleventh Edition
INSTRUCTOR’S
SOLUTIONS MANUAL
PART TWO
WILLIAM ARDIS
JOSEPH E. BORZELLINO
LINDA BUCHANAN
ALEXIS T. MOGILL
PATRICIA NELSON
Weir •Hass •Giordano
Instructor’s Solutions Manual Part Two
Thomas’ Calculus, Eleventh Edition, and Thomas’ Calculus: Early Transcendentals, Eleventh Edition
Weir
•
Hass
•
Giordano
,!7IA3C1-ccgfab!:t;K;k;K;k
ISBN 0-321-22650-X
ISM
This work is
protected by US copyright laws
and is for instructors’ use only.
weir_22650_ISM_CVR.qxd 10/15/04 11:01 AM Page 1

2. to accompany
THOMAS’ CALCULUS
ELEVENTHEDITION
BASED ON THEORIGINALWORK BY
George B. Thomas, Jr.
Massachusetts Institute of Technology
ASREVISED BY
Maurice D. Weir
Naval Postgraduate School
Joel Hass
University of California, Davis
Frank R. Giordano
Naval Postgraduate School
INSTRUCTOR’S
SOLUTIONSMANUAL
PARTONE
ARDIS• BORZELLINO• BUCHANAN• MOGILL• NELSON

3. Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors.
Copyright © 2005 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted,
in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior
written permission of the publisher. Printed in the United States of America.
ISBN 0-321-22653-4
1 2 3 4 5 6 BB 07 06 05 04
weir_22646_SSM_TTL_CPY.qxd 9/27/04 9:59 AM Page 2

4. PREFACE TO THE INSTRUCTOR
This Instructor's Solutions Manual contains the solutions to every exercise in the 11th Edition of THOMAS' CALCULUS
by Maurice Weir, Joel Hass and Frank Giordano, including the Computer Algebra System (CAS) exercises. The
corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the
CAS exercises (because the CAS command templates would give them all away).
In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or
rewritten every solution which appeared in previous solutions manuals to ensure that each solution
conforms exactly to the methods, procedures and steps presented in the textì
is mathematically correctì
includes all of the steps necessary so a typical calculus student can follow the logical argument and algebraì
includes a graph or figure whenever called for by the exercise, or if needed to help with the explanationì
is formatted in an appropriate style to aid in its understandingì
Every CAS exercise is solved in both the MAPLE and computer algebra systems. A template showingMATHEMATICA
an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within
the text grouping require a change only in the input function or other numerical input parameters associated with the problem
(such as the interval endpoints or the number of iterations).
Acknowledgments
Solutions Writers
William Ardis, Collin County Community College-Preston Ridge Campus
Joseph Borzellino, California Polytechnic State University
Linda Buchanana, Howard College
Tim Mogill
Patricia Nelson, University of Wisconsin-La Crosse
Accuracy Checkers
Karl Kattchee, University of Wisconsin-La Crosse
Marie Vanisko, California State University, Stanislaus
Tom Weigleitner, VISTA Information Technologies
Thanks to Rachel Reeve, Christine O'Brien, Sheila Spinney, Elka Block, and Joe Vetere for all their guidance and help at
every step.

5. CHAPTER 1 PRELIMINARIES
1.1 REAL NUMBERS AND THE REAL LINE
1. Executing long division, 0.1, 0.2, 0.3, 0.8, 0.9
"
99999
2389
œœœœœ
2. Executing long division, 0.09, 0.18, 0.27, 0.81, 0.99
"
11 11 11 11 11
23911
œœœœœ
3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.
a) NNT. 5 is a counter example.
b) NT. 2 < x < 6 2 2 < x 2 < 6 2 0 < x 2 < 2.??
c) NT. 2 < x < 6 2/2 < x/2 < 6/2 1 < x < 3.ÊÊ
d) NT. 2 < x < 6 1/2 > 1/x > 1/6 1/6 < 1/x < 1/2.ÊÊ
e) NT. 2 < x < 6 1/2 > 1/x > 1/6 1/6 < 1/x < 1/2 6(1/6) < 6(1/x) < 6(1/2) 1 < 6/x < 3.ÊÊÊ Ê
f) NT. 2 < x < 6 x < 6 (x 4) < 2 and 2 < x < 6 x > 2 x < 2 x + 4 < 2 (x 4) < 2.?? ??? ?
The pair of inequalities (x 4) < 2 and (x 4) < 2 | x 4 | < 2.?
g) NT. 2 < x < 6 2 > x > 6 6 < x < 2. But 2 < 2. So 6 < x < 2 < 2 or 6 < x < 2.ÊÊ
h) NT. 2 < x < 6 1(2) > 1(x) < 1(6) 6 < x < 2??
4. NT = necessarily true, NNT = Not necessarily true. Given: 1 < y 5 < 1.
a) NT. 1 < y 5 < 1 1 + 5 < y 5 + 5 < 1 + 5 4 < y < 6. Ê Ê
b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6)‘‘
c) NT. From a), 1 < y 5 < 1, 4 < y < 6 y > 4. ? ?
d) NT. From a), 1 < y 5 < 1, 4 < y < 6 y < 6. ? ?
e) NT. 1 < y 5 < 1 1 + 1 < y 5 + 1 < 1 + 1 0 < y 4 < 2. Ê Ê
f) NT. 1 < y 5 < 1 (1/2)( 1 + 5) < (1/2)(y 5 + 5) < (1/2)(1 + 5) 2 < y/2 < 3. Ê Ê
g) NT. From a), 4 < y < 6 1/4 > 1/y > 1/6 1/6 < 1/y < 1/4.ÊÊ
h) NT. 1 < y 5 < 1 y 5 > 1 y > 4 y < 4 y + 5 < 1 (y 5) < 1. Ê Ê Ê Ê Ê
Also, 1 < y 5 < 1 y 5 < 1. The pair of inequalities (y 5) < 1 and (y 5) < 1 | y 5 | < 1. Ê Ê
5. 2x 4 x 2 l?
6. 8 3x 5 3x 3 x 1 x
1
  Ê   Ê Ÿ ïïïïïïïïïñqqqqqqqqp
7. 5x 3x 8x 10 x „ ? „ ? „
5
4
8. 3(2 x) 2(3 x) 6 3x 6 2x Ê
0 5x 0 x x
0
? k ? k ‹‹‹‹‹‹‹‹‹aaaaaaaab
9. 2x 7x 5x  Ê 
""
##
77
66
x or x Ê   
""
56 3
10
ˆ‰
10. 12 2x 12x 16
6x 3x4
42

?
28 14x 2 x x
2
? ‘ ? ‘ aaaaaaaaa‹‹‹‹‹‹‹‹

6. 2 Chapter 1 Preliminaries
11. (x 2) (x 6) 12(x 2) 5(x 6)
4
53
Ê
"
12x 24 5x 30 7x 6 or x Ê Ê
6
7
12. (4x 20) 24 6x„ ?„
x 5 12 3x
24
‘‘
44 10x x x
22/5
Ê Ÿ Ê Ÿ qqqqqqqqqñïïïïïïïïî

22
5
13. y 3 or y 3??
14. y 3 7 or y 3 7 y 10 or y 4œ œÊ œ œ
15. 2t 5 4 or 2t 4 2t 1 or 2t 9 t or t œ &œ Ê œ œ Ê œ œ
"
##
9
16. 1 t 1 or 1 t 1 t or t 2 t 0 or t 2œ œ Êœ! œ Ê œ œ
17. 8 3s or 8 3s 3s or 3s s or sœ œÊœ œ Êœ œ
99725725
2 66###
18. 1 1 or 1 1 2 or s 4 or s 0
ss ss
## ##
œ œÊ œ œ!Ê œ œ
19. 2 x 2; solution interval ( 2 2) ß
20. 2 x 2; solution interval [ 2 2] x
22
Ÿ Ÿ ß qqqqñïïïïïïïïñqqqqp

21. 3 t 1 3 2 t 4; solution interval [ 2 4] Ÿ Ÿ ÊŸŸ ß
22. 1 t 2 1 3 t 1;‘ ‘ ?‘‘
solution interval ( 3 1) t
31
ß qqqqðïïïïïïïïðqqqqp

23. 3y 7 4 3 3y 11 1 y ;% Ê Ê
11
3
solution interval 1 ˆ‰ß
11
3
24. 1 2y 5 6 2y 4 3 y 2; "Ê Ê
solution interval ( 3 2) y
32
ß qqqqðïïïïïïïïðqqqqp

25. 1 1 1 0 2 0 z 10;„„ ? „ „ ? „„
zz
55
solution interval [0 10] ß
26. 2 1 2 1 3 z 2; Ÿ Ÿ Ê Ÿ Ÿ Ê Ÿ Ÿ
3z 3z 2
3##
solution interval 2 z
2/3 2
‘ ß qqqqñïïïïïïïïñqqqqp

2
3
27. 3 Ê Ê
""" " "
### ###xxx
7575
x ; solution interval ?
22 22
75 75
ˆ‰
28. 3 4 3 1 1??ll
22x
xx 7 #
"
2 x x 2; solution interval 2 x
2/7 2
? k k ? ‘ ‘ aaaa‹‹‹‹‹‹‹‹aaaab
22 2
77 7
ˆ‰

7. Section 1.1 Real Numbers and the Real Line 3
29. 2s 4 or 2s 4 s 2 or s 2;¼¼?¼ „
solution intervals ( 2] [2 ) _ß ß _
30. s 3 or (s 3) s or s    Ê    
""
####
57
s or s ;?¼ „
57
##
solution intervals s
7/2 5/2
‘‰?_ß ß _ ïïïïïïñqqqqqqñïïïïïïî

75
##
31. 1 x 1 or ( x) 1 x 0 or x 2 " Ê x 0 or x 2; solution intervals ( ) (2 ) Ê _ß!ß_
32. 2 3x 5 or (2 3x) 5 3x 3 or 3x 7 Ê
x 1 or x ;? l
7
3
solution intervals ( 1) x
1 7/3
_ß ß _ ïïïïïïðqqqqqqðïïïïïïî

ˆ‰
7
3
33. 1 or 1 r 1 2 or r 1 2
rr1‘" ‘
##
    Ê  Ÿˆ‰
r 1 or r 3; solution intervals ( 3] [1 ) Ê  Ÿ _ßß_
34. or
3r 2 3r 2
555 5
" " ˆ‰
or r or r 1Ê Ê
3r 7 3r 3 7
55 5 5 3
solution intervals ( ) r
1 7/3
_ß " ß _ ïïïïïïðqqqqqqðïïïïïïî

ˆ‰
7
3
35. x x 2 2 x 2 ;
#
# Ê Ê kk
ÈÈ È
solution interval 2 2 xŠ‹
ÈÈ
ÈÈ
ß qqqqqqðïïïïïïðqqqqqqp
# #
36. 4 x 2 x x 2 or x 2;„?„ ?¼ „
#
kk
solution interval ( 2] [2 ) r
22
_ß ß _ ïïïïïïñqqqqqqñïïïïïïî

37. 4 x 9 2 x 3 2 x 3 or 2 x 3‘‘?‘ ‘?‘‘ ‘‘
#
kk
2 x 3 or 3 x 2;Ê
solution intervals ( 3 2) (2 3) x
3223
ß ß qqqqðïïïïðqqqqðïïïïðqqqp

38. x x x or x
""" """" "#
###943 3 3
‘‘?‘ ‘?‘‘ ‘‘kk
x or x ;Ê
""" "
##33
solution intervals x
1/2 1/3 1/3 1/2
ˆ‰ˆ‰ ß ß qqqqðïïïïðqqqqðïïïïðqqqp

"" ""
##33
39. (x 1) 4 x 1 2 2 x 1 2 Ê Ê
#
kk
1 x 3; solution interval ( ) x
13
Ê "ß $ qqqqqqðïïïïïïïïðqqqqp

40. (x 3) x 3 2?
#
kk
È
2 x 3 2 or 3 2 x 3 2 ;Ê
ÈÈÈ È
solution interval 3 2 3 2 x
33
Š‹
ÈÈ
ÈÈ
ß qqqqqqðïïïïïïïïðqqqqp
# #

8. 4 Chapter 1 Preliminaries
41. x x 0 x x + < x < x < < x < 0 < x < 1.
##
Ê Ê Ê Ê Ê
11 1 1 1 1 1 11
44 2 4 2 2 2 22
2
ˆ‰ ¹¹
So the solution is the interval (0 1)ß

42. x x 2 0 x x + x x or x x 2 or x 1.
##
  Ê   Ê   Ê     Ê   Ÿ
19 1 3 13 1 3
44 2 2 22 2 2
¹¹ ˆ‰
The solution interval is ( 1] [2 ) _ß ß _

43. True if a 0; False if a 0.È‘
44. x 1 1 x (x 1) 1 x 1 x 0 x 1kk k k? ?¼„
45. (1) a b (a b) or a b (a b);kk kkœ œ
both squared equal (a b)
#
(2) ab ab a b Ÿœkk kkkk
(3) a a or a a, so a a ; likewise, b bkk kk kk kk??? ?
## ##
(4) x y implies x y or x y for all nonnegative real numbers x and y. Let x a b and
## ##
ŸŸŸ œ
È È kk
y a b so that a b a b a b a b .œ Ÿ ÊŸkkkk kka b kkkkkk kk kk
##
46. If a 0 and b 0, then ab 0 and ab ab a b .     œœ kk kkkk
If a 0 and b 0, then ab 0 and ab ab ( a)( b) a b .‘ ‘ k ? ?? kk kkkk
If a 0 and b 0, then ab 0 and ab (ab) (a)( b) a b .¼ „ ??? kk kkkk
If a 0 and b 0, then ab 0 and ab (ab) ( a)(b) a b .¼ „ ??? kk kkkk
47. 3 x 3 and x x 3.Š Š k ? ‘ Š
""
##
48. Graph of x y 1 is the interiorkk kkŸ
of “diamond-shaped" region.

49. Let be a real number > 0 and f(x) = 2x + 1. Suppose that | x 1 | < . Then | x 1 | < 2| x 1 | < 2 $$$$ ??
| 2x | < 2 | (2x + 1) 3 | < 2 | f(x) f(1) | < 2# Ê Ê $$$
50. Let > 0 be any positive number and f(x) = 2x + 3. Suppose that | x 0 | < /2. Then 2| x 0 | < and%%%
| 2x + 3 3 | < . But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) f(0) | < .%%
51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | a by definition. Now, a > 0 a < 0. Let a = b. By definition, | b | b. Since b = a,???
| a | ( a) a and | a | | a | a.?? ??
ii) For a < 0, | a | a. Now, a < 0 a > 0. Let a b. By definition, | b | b and thus | a| a. So againœ Ê œ œ œ
| a | | a|.?
iii) By definition | 0 | 0 and since 0 0, | 0 | 0. Thus, by i), ii), and iii) | a | | a | for any real number.??? ?

9. Section 1.2 Lines, Circles and Parabolas 5
52. i) Prove | x | > 0 x > a or x < a for any positive number, a.?
For x 0, | x | x. | x | > a x > a. œ Ê
For x < 0, | x | x. | x | > a x > a x < a.œ Ê Ê
ii) Prove x > a or x < a | x | > 0 for any positive number, a.?
a > 0 and x > a | x | x. So x > a | x | > a.Êœ Ê
For a > 0, a < 0 and x < a x < 0 | x | x. So x < a x > a | x | > a.?ʜ??
53. a) 1 = 1 | 1 | = 1 b bÊÊœÊ œÊœÊœ¹ ¹ ¹¹ ¹¹ ¹¹ ††
"" "ll ll
ll llbb bb b
bb
b b
b b b b
¹¹¹¹ ¹¹
¹¹ ¹¹¹¹ ¹¹
†
†

b "
b) a a a
ll ll
ll ll ll
""aa
bb b bb
œœ œ œ¹ ¹ ¹¹ ¹¹ ¹¹†† †
"
54. Prove S a a for any real number a and any positive integer n.n
n nœœkk kk
a a a, so S is true. Now, assume that S a a is true form some positive integer .k k kk kk ¸¸
""
"
œœ œœ 5 k
kk
Since a a and a a , we have a a a a a a a a . Thus,k k kk kk k k kk kk kk¸¸ ¸ ¸ ¸ ¸ ¸¸
"" ‘" " " " "
œœ œ†œœœ
kk k k k k k+
S a a is also true. Thus by the Principle of Mathematical Induction, S a ak n
kk+ nn
‘"
‘" "œ œ œl lœl l¸¸ kk
is true for all n positive integers.
1.2 LINES, CIRCLES, AND PARABOLAS
1. x 1 ( 3) 2, y 2 2 4; d ( x) ( y) 4 16 2 5?? ??œ œ œ œ œ œ œ È È È
##
2. x ( 1) 2, y 2 ( 2) 4; d ( 2) 4 2 5??œ$ œ œ œ œ œ È È
##
3. x 8.1 ( 3.2) 4.9, y 2 ( 2) 0; d ( 4.9) 0 4.9??œ œ œ œ œ œ È
##
4. x 0 2 2, y 1.5 4 2.5; d 2 ( 2.5) 8.25??œ œ œ œ œ œ
ÈÈ È
ÊŠ‹
È
#
#
5. Circle with center ( ) and radius 1. 6. Circle with center ( ) and radius 2.!ß ! !ß !
È
7. Disk (i.e., circle together with its interior points) with center ( ) and radius 3.!ß !
È
8. The origin (a single point).
9. m 3 10. m?? ? ?? ?
? ?
? ?
y y
x2(1) x2(2) 4
12 3 #"

perpendicular slope perpendicular slope? ?
"
3 3
4

10. 6 Chapter 1 Preliminaries
11. m 0 12. m ; no slopeœœ œ œœ
? ?
? ?
y y
x12 x ()
33 0 #
# #
perpendicular slope does not exist perpendicular slope 0 œ

13. (a) x 1 14. (a) x 2 15. (a) x 0 16. (a) x? ? ? ?
È
1
(b) y (b) y 1.3 (b) y 2 (b) y 0????
4
3 È
17. P( 1 1), m 1 y 1 1 x ( 1) y x ß œ Ê œ Ê œ ab
18. P(2 3), m y ( 3) (x 2) y x 4ß œ Ê œ Ê œ
"""
###
19. P(3 4), Q( 2 5) m y 4 (x 3) y xß ß Ê œ œ œ Ê œ Ê œ
?
?
y
x23 5 5 5 5
54 23" " "

20. P( 8 0), Q( 1 3) m y 0 x ( 8) y xß ß Ê œ œ œ Ê œ Ê œ
?
?
y
x1(8)7 7 7 7
30 3 3 3 24

ab
21. m , b 6 y x 6 22. m , b 3 y x 3œ œ Ê œ œ œ Ê œ
55
44
""
##
23. m 0, P( 12 9) y 9 24. No slope, P x?ʜ ʜ ˆ‰
""
33
25. a 1, b 4 (0 4) and ( 0) are on the line m 4 y 4x 4œ œ Ê ß "ß Ê œ œ œ Ê œ
?
?
y
x10
04

26. a 2, b 6 (2 0) and ( 6) are on the line m 3 y 3x 6??? ʜ? ?ʜ
?
?
y
x02
60

27. P(5 1), L: 2x 5y 15 m parallel line is y ( 1) (x 5) y x 1ß œ Ê œ Ê œ Ê œ
L
222
555
28. P 2 2 , L: 2x 5y 3 m parallel line is y 2 x 2 y xŠ‹ Š ‹
ÈÈ È È
Š‹ß œ Ê œÊ œ Êœ
L
ÈÈÈ
222
5555
8
29. P(4 10), L: 6x 3y 5 m 2 m perpendicular line is y 10 (x 4) y x 12ß œÊ œÊ œÊ œ Êœ
L ¼
"""
###
30. P( 1), L: 8x 13y 13 m m perpendicular line is y x 1!ß œ Ê œ Ê œ Ê œ L
813 13
13 8 8
¼

11. Section 1.2 Lines, Circles and Parabolas 7
31. x-intercept 4, y-intercept 3 32. x-intercept 4, y-intercept 2? ? ? ?

33. x-intercept 3, y-intercept 2 34. x-intercept 2, y-intercept 3?? ??
È È

35. Ax By C y x and Bx Ay C y x . Since 1 is the œ Íœ œ Íœ œ"#
ABAB
BB AA BA
CC
‘ ˆ‰ˆ‰
product of the slopes, the lines are perpendicular.
36. Ax By C y x and Ax By C y x . Since the lines have the sameœ Íœ œ Íœ"#
AA
BB BB
CC

slope , they are parallel.
A
B
37. New position x x y y ( 3 ( 6)) ( 3).œ ß œ#&ß œ$ßab
old old??
38. New position x x y y (6 ( 6) 0 0) (0 0).œ ß œßœßab
old old??
39. x 5, y 6, B(3 3). Let A (x y). Then x x x 5 3 x x 2 and?? ??? ? ?ʜʜ
#"
y y y 6 3 y y 9. Therefore, A ( 9).?? ʜʜ ?#"
40. x , y??œ""œ! œ!!œ!

12. 8 Chapter 1 Preliminaries
41. C( 2), a 2 x (y 2) 4 42. C( 0), a 3 (x 3) y 9!ß œ Ê œ $ß œ Ê œ
## ##

43. C( 1 5), a 10 (x 1) (y 5) 10ß œ Ê œ
È
##

44. C( ), a 2 (x 1) (y 1) 2"ß " œ Ê œ
È
##
x 0 (0 1) (y 1) 2 (y 1) 1œÊ œÊ œ
## #
y 1 1 y 0 or y 2.??ʜ ?
Similarly, y 0 x 0 or x 2œÊœ œ

45. C 3 2 , a 2 x 3 (y 2) 4,Š‹ Š‹
ÈÈ
ß œÊ œ
#
#
x 0 0 3 (y 2) 4 (y 2) 1œÊ œÊ œŠ‹
È
#
##
y 2 1 y 1 or y 3. Also, y 0??ʜ ? ?
x 3 (0 2) 4 x 3 0Ê œÊ œŠ‹ Š‹
ÈÈ
##
#
x3ʜ
È

46. C 3 , a 5 (x 3) y 25, soˆ‰ ˆ ‰???
""
##
#
#
x 0 (0 3) y 25œÊ œ
# "
#
#ˆ‰
y 16 y 4 y? ???ʜˆ‰
""
###
#
9
or y . Also, y 0 (x 3) 0 25œ œ Ê œ
7
##
# "
#
ˆ‰
(x 3) x 3? ? ??
#
#
99
4
311
È
x3Êœ„
311
È
#

13. Section 1.2 Lines, Circles and Parabolas 9
47. x y 4x 4y 0
##
?
x y 4y 4??
##
x 4x4y 4y44??
##
(x 2) (y 2) 4 C ( 2 2), a 2.? ?ʜ ?
##

48. x y 8x 4y 16 0
##
?
x 8x y 4y 16??
##
x 8x16y 4y44Ê œ
##
(x 4) (y 2) 4? ?
##
C ( 2), a 2.ʜ ?

49. x y 3y 4 0 x y 3y 4
## ##
œÊ œ
xy3y??
## 925
44
xy C0,? ? ʜ
#
##
#ˆ‰ ˆ‰
325 3
4
a.œ
5
#

50. x y 4x 0
##
?
9 4
x4xy??
## 9
4
x 4x4y??
## 25
4
(x 2) y??
## 25
4
C(20), a .Êœß œ
5
#

51. x y 4x 4y 0
##
?
x 4x y 4y 0??
##
x 4x4y 4y48??
##
(x 2) (y 2) 8Ê œ
##
C(2 2), a 8.??
È

14. 10 Chapter 1 Preliminaries
52. x y 2x 3
##
œ
x 2x1y 4ʜ
##
(x 1) y 4ʜ
##
C ( 1 0), a 2.ʜ ?

53. x 1? ? ?
b2
a2(1)#

y (1) 2(1) 3 4Êœ œ
#
V ( 4). If x 0 then y 3.ʜ ? ?
Also, y 0 x 2x 3 0œÊ œ
#
(x 3)(x 1) 0 x 3 orÊ œÊœ
x 1. Axis of parabola is x 1.? ?

54. x 2? ? ?
b4
a2(1)#
y ( 2) 4( 2) 3 1Ê œ œ
#
V ( 2 1). If x 0 then y 3.ʜ ? ?
Also, y 0 x 4x 3 0œÊ œ
#
(x 1)(x 3) 0 x 1 orÊ œÊœ
x 3. Axis of parabola is x 2.? ?

55. x 2? ? ?
b4
a2(1)#
y (2) 4(2) 4Êœ œ
#
V (2 4). If x 0 then y 0.Êœß œ œ
Also, y0 x 4x0?? ?
#
x(x 4) 0 x 4 or x 0.Ê œ Ê œ œ
Axis of parabola is x 2.œ

56. x 2? ? ?
b4
a2(1)#
y (2) 4(2) 5 1Êœ œ
#
V (2 1). If x 0 then y 5.ʜ ? ?
Also, y 0 x 4x 5 0œÊ œ
# x 4x 5 0 x??ʜ
# „
#
44
È
no x intercepts. Axis of parabola is x 2.ʜ

15. Section 1.2 Lines, Circles and Parabolas 11
57. x 3? ? ?
b6
a2(1)#

y ( 3) 6( 3) 5 4Ê œ œ
#
V ( 3 ). If x 0 then y 5.ʜ ? ?
Also, y 0 x 6x 5 0œÊ œ
#
(x 5)(x 1) 0 x 5 orÊ œÊœ
x 1. Axis of parabola is x 3.? ?

58. x? ? ?
b1
a2(2)4#
"
y2 3Êœ œˆ‰
""
#
44 8
23
V . If x 0 then y 3.Êœß œ œˆ‰
"
48
23
Also, y 0 2x x 3 0œÊ œ
#
x no x intercepts.Êœ Ê
123
4
„
È
Axis of parabola is x .œ
"
4

59. x 1? ? ?
b1
a2(1/2)#
y ( 1) ( 1) 4Êœ œ
"
#
# 7
2
V . If x 0 then y 4.ʜ ? ?ˆ‰
7
2
Also, y0 x x40œÊ œ
"
#
#
x no x intercepts.Êœ Ê
„ 17
1
È
Axis of parabola is x 1.?

60. x 4? ? ?
b2
a2(1/4)#
y (4) 2(4) 4 8Êœ œ
" #
4
V (4 8) . If x 0 then y 4.Êœß œ œ
Also, y 0 x 2x 4 0œÊ œ
"#
4
x 4 4 2.Êœ œ„
„

28
1/2
È È
Axis of parabola is x 4.œ

61. The points that lie outside the circle with center ( 0) and radius 7.!ß È
62. The points that lie inside the circle with center ( 0) and radius 5.!ß È
63. The points that lie on or inside the circle with center ( 0) and radius 2."ß
64. The points lying on or outside the circle with center ( 2) and radius 2.!ß
65. The points lying outside the circle with center ( 0) and radius 1, but inside the circle with center ( 0),!ß !ß
and radius 2 (i.e., a washer).

16. 12 Chapter 1 Preliminaries
66. The points on or inside the circle centered
at ( ) with radius 2 and on or inside the!ß !
circle centered at ( 2 0) with radius 2.?

67. x y 6y 0 x (y 3) 9.
## # #
Ê
The interior points of the circle centered at
( 3) with radius 3, but above the line!ß
y3.œ

68. x y 4x 2y 4 (x 2) (y 1) 9.
## # #
Ê
The points exterior to the circle centered at
(2 1) with radius 3 and to the right of the?
line x 2.œ

69. (x 2) (y 1) 6 70. (x 4) (y 2) 16
## ##
71. x y 2, x 1 72. x y 4, (x 1) (y 3) 10
## ## # #
Ÿ  
73. x y 1 and y 2x 1 x 4x 5x
## # # #
œ œ Êœ œ
x and y or x and y .Êœ œ œ œŠ‹Š ‹
""
ÈÈ È È
55 5 5
22
Thus, A , B are theŠ‹Š ‹
""
ÈÈ È È
55 5 5
22
??
points of intersection.

17. Section 1.2 Lines, Circles, and Parabolas 13
74. x y 1 and (x 1) y 1œ œ
##
1 ( y) y 2yÊœ œ
## #
y and x orÊœ œ"Š‹
""
ÈÈ
22
y and x 1 . Thus,Š‹œ œ
""
ÈÈ
22
A and B 1Š‹Š ‹" ß ß
"" " "
ÈÈ È È
22 2 2
are intersection points.

75. y x 1 and y x x x 1œ œ Ê œ
##
xx10 x .ʜʜ
# „
#
15
È
If x, then yx1.œ œœ
15 35
##
ÈÈ
If x, then yx1.œ œœ
15 35
##
ÈÈ
Thus, A and BŠ‹Š‹
1535 1535
## ##
ÈÈ ÈÈ
ßß
are the intersection points.

76. y x and (x 1) (x 1) xœ Cœ Ê œ
##
x 3x 0 x . IfÊ"œÊœ
# „
#
35
È
x , then y x . Ifœœœ
35 53
##
ÈÈ
x , then y x .œœœ
35 35
##
ÈÈ
Thus, A and BŠ‹Š ‹
35 53 35 35
## # #
ÈÈ È È
??
are the intersection points.

77. y 2x 1 x 3x 1œœÊ œ
###
x and y or x and y .Ê œ œ œ œ
""" "
ÈÈ
33 33
Thus, A and B are theŠ‹Š ‹
"" ""
ÈÈ
33 33
ß ß
intersection points.

18. 14 Chapter 1 Preliminaries
78. y (x 1) 0 2x 1œœ Êœ
x3x
44

#
0 3x 8x 4 (3x 2)(x 2)Êœ œ
#
x 2 and y 1, or x andÊœ œ œ œ
x2
43

y . Thus, A(2 1) and Bœœ ß ß
x2
49 39

""
ˆ‰
are the intersection points.

79. x y 1 (x 1) y
## ##
œœ
x (x 1) x 2x 1Êœœ
###
0 2x 1 x . Henceʜʜ
"
#
y x or y . Thus,
##
#
œ" œ œ „
3
4
3
È
A and B are theŠ‹ Š ‹
""
## # #
??
ÈÈ
33
intersection points.

80. x y 1 x y y y
## # #
œœÊ œ
y(y 1) 0 y 0 or y 1.ʜʜ œ
If y 1, then x y 0 or x 0.œœ"œœ
##
If y 0, then x 1 y 1 or x 1.œœœœ„
##
Thus, A(0 1), B( 0), and C( 1 0) are theß"ß ß
intersection points.

81. (a) A (69° 0 in), B (68° .4 in) m 2.5°/in.¸ß ¸ß Êœ ¸
68° 69°
.4 0


(b) A (68° .4 in), B (10° 4 in) m 16.1°/in.¸ß ¸ß Êœ ¸
10° 68°
4 .4


(c) A (10° 4 in), B (5° 4.6 in) m 8.3°/in.¸ß ¸ß Êœ ¸
5° 10°
4.6 4


82. The time rate of heat transfer across a material, , is directly proportional to the cross-sectional area, A, of the mate
?
?
U
>
rial,
to the temperature gradient across the material, (the slopes from the previous problem), and to a constant characteristic
?
?
X
B
of the material. -kA k = . Note that and are of opposite sign because heat flow is toward lower
??
?? ??
??UU
>B >B
XX Î
œÊ





U
Ê
œ
A
temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good
insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are not changing), we may define another constant, K, characteristics of the material: K Using the values of fromœ Þ
"X
B


Ê
œ
?
?
the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the
poorest insulator, with K 0.4.œ
83. p kd 1 and p 10.94 at d 100 k 0.0994. Then p 0.0994d 1 is the diver'sœ œ œ Êœ œ œ
10.94
100
"
pressure equation so that d 50 p (0.0994)(50) 1 5.97 atmospheres.œÊœ œ
84. The line of incidence passes through ( 1) and ( 0) The line of reflection passes through ( 0) and ( )!ß "ß Ê "ß #ß "
m 1 y 0 1(x 1) y x 1 is the line of reflection.Ê œ œ Ê œ Ê œ
10
1

#

19. Section 1.2 Lines, Circles, and Parabolas 15
85. C (F 32) and C F F F F or F 40 ° gives the same numerical reading.œ œÊœÊœ œ
5 5 160 4 160
99999

86. m x . Therefore, distance between first and last rows is (14) 40.25 ft.œœÊœ ¸
37.1 14 14 14
100 x .371 .371?
? É ˆ‰
#
#
87. length AB (5 1) (5 2) 16 9 5œœ œÈ È
##
length AC (4 1) ( ) 9 16 5œ##œœÈ È
##
length BC (4 5) ( 5) 1 49 50 5 2 5œ#œœœÁÈ È È È
##
88. length AB (1 0) 3 0 1 3 2œ œœÊ Š‹
È È
#
#
length AC (2 0) (0 0) 4 0 2œœœÈ È
##
length BC (2 1) 0 3 1 3 2œ œœÊ Š‹
È È
#
#
89. Length AB ( x) ( y) 1 4 17 and length BC ( x) ( y) 4 1 17.œœœ œœœÈÈ ÈÈ ÈÈ
?? ??
#### ####
Also, slope AB and slope BC , so AB BC. Thus, the points are vertices of a square. The coordinateœœ¼
4
14
"
increments from the fourth vertex D(x y) to A must equal the increments from C to B 2 x x 4 and? ?
1 y y x 2 and y 2. Thus D( 2) is the fourth vertex. œ œ" Ê œ œ #ß?
90. Let A (x 2) and C (9 y) B (x y). Then 9 x AD and 2 y DC 2(9 x) 2(2 y) 56œß œß Ê œß œ œ Ê œ kk kk
and 9 x 3(2 y) 2(3(2 y)) 2(2 y) 56 y 5 9 x 3(2 ( 5)) x 12.œ Ê œ Ê œÊ œ Ê œ
Therefore, A ( 12 2), C (9 5), and B ( 12 5).X ? X ? X ?
91. Let A( ), B( ), and C(2 ) denote the points."ß " #ß $ ß !
Since BC is vertical and has length BC 3, letkkœ
D ( 4) be located vertically upward from A and
" ?
D ( 2) be located vertically downward from A so
#"ß
that BC AD AD 3. Denote the pointkkkkkkœœœ
"#
D (x y). Since the slope of AB equals the slope of$ß
CD we have 3y 9 x 2 or$


"y3
x2 3 œ Ê œ
x 3y 11. Likewise, the slope of AC equals the slopeœ
of BD so that 3y 2x 4 or 2x 3y 4.$


y 0
x 2 3
2 œÊ œ œ

Solving the system of equations we find x 5 and y 2 yielding the vertex D (5 ).
x3y
2x 3y 4

œ""
œ
œœ ß#
$
92. Let x, y , x and/or y be a point on the coordinate plane. The slope, m, of the segment to x, y is . A 90ab ababÁ! Á! !ß!
y
x
‰
rotation gives a segment with slope m . If this segment has length equal to the original segment, its endpoint
w "
œ œ
my
x
will be y, x or y, x , the first of these corresponds to a counter-clockwise rotation, the latter to a clockwiseabab
rotation.
(a) ( 4); (b) (3 2); (c) (5 2); (d) (0 x);"ß ß ß ß

20. 16 Chapter 1 Preliminaries
(e) ( y 0); (f) ( y x); (g) (3 10)ß ß ß
93. 2x ky 3 has slope and 4x y 1 has slope 4. The lines are perpendicular when ( 4) 1 orœ œ œ
2 2
k k
k 8 and parallel when 4 or k .œ œ œ
2
k
"
#
94. At the point of intersection, 2x 4y 6 and 2x 3y 1. Subtracting these equations we find 7y 7 orœ œ œ
y 1. Substitution into either equation gives x 1 (1 1) is the intersection point. The line through (1 1)œœÊß ß
and ( ) is vertical with equation x 1."ß # œ
95. Let M(a b) be the midpoint. Since the two trianglesß
shown in the figure are congruent, the value a must
lie midway between x and x , so a ."#

# œ
xx
Similarly, b .œ
yy
#

96. (a) L has slope 1 so M is the line through P(2 1) with slope 1; or the line y x 3. At the intersection? X
point, Q, we have equal y-values, y x 2 x 3. Thus, 2x 1 or x . Hence Q has coordinatesœœ œ œ
"
#
. The distance from P to L the distance from P to Q .ˆ‰ ˆ‰ˆ‰ É É
"
## ## #
##
?XXXX
5 3318
4
32
È
(b) L has slope so M has slope and M has the equation 4y 3x 12. We can rewrite the equations ofœ
43
34
the lines as L: x y 3 and M: y 4. Adding these we get y 7 so y . Substitutionœ Bœ œ œ
3 4 25 84
4 3 12 25
into either equation gives x 4 so that Q is the point of intersection. The distanceXX ?
484 12 1284
325 25 2525
ˆ‰ ˆ ‰
from P to L 4 6 .œœÉˆ‰ˆ‰
12 84 22
25 25 5
##
(c) M is a horizontal line with equation y b. The intersection point of L and M is Q( b). Thus, theX‰?
distance from P to L is (a 1) 0 a 1 .È kkœ
##
(d) If B 0 and A 0, then the distance from P to L is x as in (c). Similarly, if A 0 and B 0, theœÁ œÁ ¸¸
C
A
!
distance is y . If both A and B are 0 then L has slope so M has slope . Thus,¸¸
CAB
BBA
Á!
L: Ax By C and M: Bx Ay Bx Ay . Solving these equations simultaneously we find theœ œ !!
point of intersection Q(x y) with x and y . The distance fromßœ œ
AC B Ay Bx BC A Ay Bx
AB AB


ab ab ŸŸ ŸŸ

P to Q equals ( x) ( y) , where ( x)È Š‹?? ?
## #

#
œ
xA B ACABy Bx
ABŸŸŸ

ab
, and ( y) .œœ œ
AAxByC yAB BCAyABx BAxByC
AB AB AB

ŸŸ Ÿ Ÿ Ÿ ŸŸ



ab ab ab
ab ab


#

#
? Š‹
Thus, ( x) ( y) .È É??
##




œ œ
ab kk
È
Ax By C
AB
Ax By C
ABŸŸ


ŸŸ

1.3 FUNCTIONS AND THEIR GRAPHS
1. domain ( ); range [1 ) 2. domain [0 ); range ( 1]X?X? X?X?
3. domain ( ); y in range y , t 0 y and y y can be any positive real numberX!? > X B > X B!>
"" #
Èt t
range ( ).Ê œ !ß _

21. Section 1.3 Functions and Their Graphs 17
4. domain [0 ); y in range y , t 0. If t 0, then y 1 and as t increases, y becomes a smallerX? > X B X X
"
1tÈ
and smaller positive real number range (0 1].Êœß
5. 4 z (2 z)(2 z) 0 z [ 2 2] domain. Largest value is g(0) 4 2 and smallest value isœ  Í−ßœ œ œ
# È
g( 2) g(2) 0 0 range [0 2].X X X > X?
È
6. domain ( 2 2) from Exercise 5; smallest value is g(0) and as 0 z increases to 2, g(z) gets larger andX? X U
"
#
larger (also true as z 0 decreases to 2) range .? ‰
"
#
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.
9. y x 1 and x . So,œ "Ê " !Ê Ÿ !Ɉ‰
""
xx
(a) No (x ; (b) No; division by undefined;!Ñ !
(c) No; if x , ; (d)  " "Ê "! Ð!ß"Ó
""
xx
10. y x x x and x . x x and x x So, x .œ # Ê#  !Ê  ! Ÿ#  !Ê  ! Ÿ# Ê Ÿ%Þ !Ÿ Ÿ%É ÈÈÈ ÈÈ È
(a) No; (b) No; (c) Ò!ß %Ó
11. base x; (height) x height x; area is a(x) (base)(height) (x) x x ;œœÊœ œ œ œ
## #
#####
#
""ˆ‰ Š‹
x 333
4
ÈÈÈ
perimeter is p(x) x x x 3x.œœ
12. s side length s s d s ; and area is a s a dœÊœÊœ œÊœ
## # # # "
#
d
2
È
13. Let D diagonal of a face of the cube and the length of an edge. Then D d and (by Exercise 10)œjœjœ
###
D 2 3 d . The surface area is 6 2d and the volume is .
## ## # # $
$Î#
œj Ê jœ Êjœ jœ œ jœ œ
d6d dd
3 3333È È
 
Š‹
14. The coordinates of P are x x so the slope of the line joining P to the origin is m (x 0). Thus,ˆ‰È?
È
È
x
x x
"
x, x , .ˆ‰ˆ‰Èœ
""
mm

15. The domain is . 16. The domain is .ab ab_ß _ _ß _

22. 18 Chapter 1 Preliminaries
17. The domain is . 18. The domain is .ab? ? !

19. The domain is . 20. The domain is .abab abab_ß! !ß_ _ß! !ß_

21. Neither graph passes the vertical line test
(a) (b)
22. Neither graph passes the vertical line test
(a) (b)
x y 1
xy y1x
or or
xy y x
kk
ÚÞÚÞ
ÛßÛß
ÜàÜà
œÍ Í
œ" œ
œ" œ"

23. Section 1.3 Functions and Their Graphs 19
23. x012 24. x012
y010 y100
25. y 26. y
3 x, x 1
2x, 1 x
, x 0
x, 0 x
œœ
Ÿ
#
#
Ÿ
œœ
"
x
27. (a) Line through and : y x ab ab!ß ! "ß " œ
Line through and : y x 2ab ab"ß " #ß ! œ
f(x)
x, 0 x 1
x 2, 1 x 2
œ
ŸŸ
Ÿ
œ
(b) f(x)
2, x
x
2x
x
œ
$& #Ñ
$Ð Ñ & # %
Ð%&#
!ß $ Ÿ Ÿ %
Ú
Ý
Ý
Û
Ý
Ý
Ü
28. (a) Line through 2 and : y x 2 ab ab!ß #ß ! œ
Line through 2 and : m , so y x 2 xab ab a bß " &ß ! œ œ œ œ " œ
!" " " " " &
&# $ $ $ $ $


f(x)
x, 0x
x, x
œ
# Ÿ#
#Ÿ&
œ
"&
$$
(b) Line through and : m , so y x abab"ß ! !ß $ œ œ $ œ $ $
$!
!Ð"Ñ
Line through and : m , so y xab a b!ß $ #ß " œ œ œ # œ # $
"$ %
#! #
f(x)
x, x
x, x
œ
$ $ " Ÿ !
# $ ! Ÿ #
œ
29. (a) Line through and : y x abab"ß " !ß ! œ
Line through and : yab ab!ß " "ß " œ "
Line through and : m , so y x xab ab a b"ß" $ß! œ œ œ œ " "œ
!" " " " " $
$" # # # # #
f(x)
xx
x
xx
œ
"Ÿ!
Ñ$#&Ñ
"$
Ú
Û
Ü
"$
## (b) Line through and : y x abab#ß" !ß! œ
"
#
Line through and : y xab ab!ß # "ß ! œ # #
Line through and : yabab"ß " $ß " œ "

24. 20 Chapter 1 Preliminaries
f(x)
xx
xx
x
œ
# Ÿ Ÿ !
# # ! Ÿ "
" " Ÿ $
Ú
Û
Ü
"
#
30. (a) Line through and T : m , so y x 0 x ˆ‰ ˆ ‰ab
TT
TT T T T#Î##
"! # # #
ß! ß" œ œ œ œ "
ab
f(x)
, 0 x
x, xT
œ
!ŸŸ
" Ÿ

T
T
T
#
#
#
(b) f(x)
A, x
AxT
A T x
AxT
œ
$& #
ß Ÿ
Ð&#
ß Ÿ Ÿ#
Ú
Ý
Ý
Ý
Û
Ý
Ý
Ý
Ü
T
T
T
T
#
#
$
#
$
#
31. (a) From the graph, 1 x ( 2 0) ( )
x4
x#
Ê −ß %ß_
(b) 1 1 0
x4x4
xx##
Ê
x 0: 1 0 0 0Ê Ê
x4 x2x8
x2x x
(x 4)(x 2)
##


x 4 since x is positive;Ê
x 0: 1 0 0 0Ê Ê
x4 x2x8
2x 2x x
(x 4)(x 2)


#
x 2 since x is negative;Ê
sign of (x 4)(x 2)

2
ïïïïïðïïïïïðïïïïî
ÎÎ


%
Solution interval: ( 0) ( )#ß %ß _

32. (a) From the graph, x ( 5) ( 1 1)
32
x1 x1
Ê−_ßß
(b) x 1: 2Case Ê
32
x1 x1 x1
3(x 1)


3x 3 2x 2 x 5.ÊÊ
Thus, x ( 5) solves the inequality.− _ß
1x1: 2Case Ê
32
x1 x1 x1
3(x 1)


3x 3 2x 2 x 5 which is trueÊÊ
if x 1. Thus, x ( 1 1) solves the − ß
inequality.
1 x: 3x 3 2x 2 x 5CaseÊÊ
32
x1 x1
which is never true if 1 x, so no solution#
here.

In conclusion, x ( 5) ( 1 1).−_ß ß
33. (a) x 0 for x [0 1) (b) x 0 for x ( 1 0]? Þ Ð ? ÞÐ
34. x x only when x is an integer.ÚÛœÜÝ
35. For any real number x, n x n , where n is an integer. Now: n x n n x n. ByŸ Ÿ " Ÿ Ÿ "ÊÐ "ÑŸ Ÿ
definition: x n and x n x n. So x x for all x . ? ? * ? ? Þ

25. Section 1.3 Functions and Their Graphs 21
36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.

37. v f(x) x 2x 22 2x x 72x x; x 7œ œ Ð"% ÑÐ Ñ œ % $!) ! Þ
$#
38. (a) Let h height of the triangle. Since the triangle is isosceles, AB AB 2 AB 2 So,?Î?*?
### È
h 2 h B is at slope of AB The equation of AB is
##
#
ÎÑ ? * ?Ñ* $ÐÑ * ?Ñ*Š‹
È
ab
y f(x) ; x .œ œ B " − Ò!ß "Ó
(b) A x 2x y 2x x 2x x; x . ? ? ÎÑ? Î% Þ$ÐÑ
#
39. (a) Because the circumference of the original circle was and a piece of length x was removed.)1
(b) r??)
)
##
1
11
xx
(c) h r 16œ "' œ "' % œ "' œ œ œ
È É ˆ‰ ˆ ‰ ÉÉÉ
#
#%%%%#
#
"' "' x4xx4xxxx xx
11111111
1 1


È
(d) V r hœœ † œ
"")
$$# # #%
#
# "' ) "'
11 ˆ‰
1
11 1
1 11x x x xx xÈ ab
È



40. (a) Note that 2 mi = 10,560 ft, so there are x feet of river cable at $180 per foot and x feet of land
È
ab)!! "!ß &'!
##
cable at $100 per foot. The cost is C x x x .ab a b
È
œ ")! )!! "!! "!ß &'!
##
(b) C $ab! œ "ß #!!ß !!!
C$ ab&!! ¸ "ß "(&ß )"#
C$ ab"!!! ¸ "ß ")'ß &"#
C$ ab"&!! ¸ "ß #"#ß !!!
C$ ab#!!! ¸ "ß #%$ß ($#
C$ ab#&!! ¸ "ß #()ß %(*
C$ ab$!!! ¸ "ß $"%ß )(!
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the
point P.
41. A curve symmetric about the x-axis will not pass the vertical line test because the points x, y and x, y lie on the samab a b e
vertical line. The graph of the function y f x is the x-axis, a horizontal line for which there is a single y-value, ,œœ! !ab
for any x.
42. Pick 11, for example: , the original number.""&œ"'Ä#†"'œ$#Ä$#'œ#'Ä œ"$Ä"$#œ""
#'
#
f x x, the number you started with.ab?%?
# &'
#
abx

26. 22 Chapter 1 Preliminaries
1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS
1. (a) linear, polynomial of degree 1, algebraic. (b) power, algebraic.
(c) rational, algebraic. (d) exponential.
2. (a) polynomial of degree 4, algebraic. (b) exponential.
(c) algebraic. (d) power, algebraic.
3. (a) rational, algebraic. (b) algebraic.
(c) trigonometric. (d) logarithmic.
4. (a) logarithmic. (b) algebraic.
(c) exponential. (d) trigonometric.
5. (a) Graph h because it is an even function and rises less rapidly than does Graph g.
(b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
6. (a) Graph f because it is linear.
(b) Graph g because it contains .ab!ß "
(c) Graph h because it is a nonlinear odd function.
7. Symmetric about the origin 8. Symmetric about the y-axis
Dec: x Dec: x_ _ _ !
Inc: nowhere Inc: x ! _

9. Symmetric about the origin 10. Symmetric about the y-axis
Dec: nowhere Dec: x ! _
Inc: x Inc: x_ ! _ !
x ! _

27. Section 1.4 Identifying Functions; Mathematical Models 23
11. Symmetric about the y-axis 12. No symmetry
Dec: x Dec: x_ Ÿ ! _ Ÿ !
Inc: x Inc: nowhere! _

13. Symmetric about the origin 14. No symmetry
Dec: nowhere Dec: x !Ÿ _
Inc: x Inc: nowhere_ _

15. No symmetry 16. No symmetry
Dec: x Dec: x!Ÿ _ _ Ÿ!
Inc: nowhere Inc: nowhere

28. 24 Chapter 1 Preliminaries
17. Symmetric about the y-axis 18. Symmetric about the y-axis
Dec: x Dec: x_Ÿ! !Ÿ_
Inc: x Inc: x! _ _ !

19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the
function is even.
20. f x x and f x x f x . Thus the function is odd.ab abab ab ˆ‰œ œ œ œ œ œ
& """ &
xx x
ŸŸ Ÿ
ab
21. Since f x x x f x . The function is even.ab a b abœ "œ "œ
# #
22. Since f x x x f x x x and f x x x f x x x the function is neither even norÒ œ ÓÁÒ œ Ó Ò œ ÓÁÒ œ Óab a b a b ab ab ab
## ##
odd.
23. Since g x x x, g x x x x x g x . So the function is odd.ab a b a b abœ œœœ
$$$
24. g x x x x g x thus the function is even.ab ab ab abœ $ "œ $B "œ ß
%# %#25. g x g x . Thus the function is even.ab a b?? ?
""
" "x x

ab
26. g x ; g x g x . So the function is odd.ab ab ab???
xx
x x

" "
27. h t ; h t ; h t . Since h t h t and h t h t , the function is neither even nor odd.ab a b ab ab ab ab a b??? & &
"""
" " "t t t
28. Since t | t |, h t h t and the function is even.'?' ?
$ $
ab ab ab
29. h t 2t , h t 2t . So h t h t . h t 2t , so h t h t . The function is neither even norab a b ab a b ab ab abœ " œ " Á œ " Á
odd.
30. h t 2 t and h t 2 t 2 t . So h t h t and the function is even.ab a b ab a bœ l l" œ l l"œ l l" œ
31. (a) The graph support the assumption that y is proportional to x.=
The constant of proportionality is estimated from the slope of the
regression line, which is 0.166.

29. Section 1.4 Identifying Functions; Mathematical Models 25
(b) The graph support the assumption that y is proportional to x .=
"Î#
The constant of proportionality is estimated from the slope of the
regression line, which is 2.03.
32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the
regression line.

The graphs support the assumption that y is proportional to . The constant of proportionality is estimated from the$
x
slope of the regression line, which is 5.00.
(b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from
the slope of the regression line, which is 2.99.

33. (a) The scatterplot of y reaction distance versus x speed isœœ

Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is
approximately 1.1.

30. 26 Chapter 1 Preliminaries
(b) Calculate x speed squared. The scatterplot of x versus y braking distance is:
ww
œœ

Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which
is approximately 0.059.
34. Kepler's 3rd Law is T days R , R in millions of miles. "Quaoar" is 4 miles from Earth, or aboutabœ!Þ%" ‚"!
$Î# *
4 miles from the sun. Let R 4000 (millions of miles) and) Ñ$ Î $ ) Ñ$ " ) ) Ñ$ ?
*'*
T days days.œ !Þ%" %!!! ¸ "!$ß (#$aba b
$Î#
35. (a)
The hypothesis is reasonable.
(b) The constant of proportionality is the slope of the line in./unit mass in./unit mass.¸ œ !Þ)(%
)Þ(%" !
"!!


(c) y(in.) in./unit mass unit mass in.œ !Þ)( "$ œ ""Þ$"abab
36. (a) (b)
Graph (b) suggests that y k x is the better model. This graph is more linear than is graph (a).œ
$
1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. D : x , D : x 1 D D : x 1. R : y , R : y 0, R : y 1, R : y 0
f g fg fg f g fg fg_ _   Ê œ   _ _      

2. D : x 1 0 x 1, D : x 1 0 x 1. Therefore D D : x 1.
fg fgfg  Ê     Ê   œ  

R R : y 0, R : y 2, R : y 0
fg fg fgœ     

È

31. Section 1.5 Combining Functions; Shifting and Scaling Graphs 27
3. D : x , D : x D : x since g(x) 0 for any x; D : x
fg fg gf_ _ _ _ Ê _ _ Á _ _

since f(x) 0 for any x. R : y 2, R : y 1, R : 0 y 2, R : yÁœ Ÿ 
fgfg gfÎ
"
#
4. D : x , D : x 0 D : x 0 since g(x) 0 for any x 0; D : x 0 since f(x) 0
fgfg gf_ _   Ê   Á     Á

for any x 0. R : y 1, R : y 1, R : 0 y 1, R : y œ Ÿ  "
fgfg gfÎ
5. (a) f(g(0)) f( 3) 2œ œ
(b) g(f(0)) g(5) 22œœ
(c) f(g(x)) f(x 3) x 3 5 x 2œœœ
## #
(d) g(f(x)) g(x 5) (x 5) 3 x 10x 22œ œ œ
##
(e) f(f( 5)) f(0) 5œ œ
(f) g(g(2)) g(1) 2œœ
(g) f(f(x)) f(x 5) (x 5) 5 x 10œ œœ
(h) g(g(x)) g(x 3) (x 3) 3 x 6x 6œœœ
###%#
6. (a) f g fˆ‰ˆ‰ˆ‰
""
#
œœ
2
33
(b) g f g 2ˆ‰ˆ‰ˆ‰
""
##
œœ
(c) f(g(x)) f 1œœœˆ‰
""
__ _x1 x1 x1
x
(d) g(f(x)) g(x 1)œœ œ
""
(x 1) 1 x
(e) f(f(2)) f(1) 0œœ
(f) g(g(2)) gœœœˆ‰
""
34
3
4
3
(g) f(f(x)) f(x 1) (x 1) 1 x 2œ œœ
(h) g(g(x)) g (x 1 and x 2)œœœÁÁˆ‰
""_"
__# _x1 x 1
x

_
x1
7. (a) u(v(f(x))) u v u 4 5 5œœœœˆ‰ˆ‰ˆ‰ˆ‰
"""
#
xxx x
4
ll
(b) u(f(v(x))) u f x u 4 5 5œœœœabab ˆ‰ ˆ‰
# ""
xx x
4
ll l
(c) v(u(f(x))) v u v 4 5 5œœœˆ‰ˆ ‰ˆ ‰ˆ‰ ˆ‰
""
#
xxx
4
(d) v(f(u(x))) v(f(4x 5)) vœœ œ ˆ‰ˆ‰
""

#
4x 5 4x 5
(e) f(u(v(x))) f u x f 4 x 5œœœaba bab ab
## "
4x 5
l
(f) f(v(u(x))) f(v(4x 5)) f (4x 5)œœœ ab
# "
(4x 5)
l
8. (a) h(g(f(x))) h g x h 4 8 x 8œœœœˆ‰ˆ‰ÈÈ Š‹ Š‹
ÈÈxx
44
(b) h(f(g(x))) h f h 4 8 2 x 8œœœœˆ‰ˆ‰ˆ‰ ÈÈ È
xxx
444
(c) g(h(f(x))) g h x g 4 x 8 x 2œœœœˆ‰ˆ ‰ˆ‰ÈÈ È
4x 8
4
È
(d) g(f(h(x))) g(f(4x 8)) g 4x 8œœ œ œ Š‹
È
ÈÈ
4x 8
4
x2
#
(e) f(g(h(x))) f(g(4x 8)) f f(x 2) x 2œœ œœ ˆ‰ È
4x 8
4

(f) f(h(g(x))) f h f 4 8 f(x 8) x 8œœœœˆ‰ˆ ‰ˆ‰ ˆ‰ È
xx
44
9. (a) y f(g(x)) (b) y j(g(x))œœ
(c) y g(g(x)) (d) y j(j(x))œœ
(e) y g(h(f(x))) (f) y h(j(f(x)))œœ
10. (a) y f(j(x)) (b) y h(g(x)) g(h(x))œœœ
(c) y h(h(x)) (d) y f(f(x))œœ
(e) y j(g(f(x))) (f) y g(f(h(x)))œœ

32. 28 Chapter 1 Preliminaries
11. g(x) f(x) (f g)(x) ‰
(a) x 7 x x 7 È È
(b) x2 3x 3(x2)3x6œ
(c) x x 5 x 5
# # È È

(d) x
xx x
x1 x1 1 x(x1)
x
x1
x
x1
Ê
Ê
œœ
(e) 1 x
""
x1 x

(f) x
""
xx
12. (a) f g x g x .a bab ab‰œllœ
"
l"lx
(b) f g x so g x x .a bab ab‰ œ œÊ"œÊ"œÊœß œ"
gx
gx x gx x x gx x gx
xxxab
ab ab ab ab
"
_" _" _" _"
""""
(c) Since f g x g x x , g x x .a bab ab abȉœ œllœ
#
(d) Since f g x f x x , f x x . (Note that the domain of the composite is .)a bab abˆ‰È‰œ œllœ Ò!ß_Ñ
#
The completed table is shown. Note that the absolute value sign in part (d) is optional.
g x f x f g x
x
x
xxx
xx x
ab ab a bab
È
È
‰
ll
"
ll
ll
""
" l "l
"
_"
#
#
xx
xx
xx13. (a) f g x 1ababÉÉœœ
11x
xx
_
gfx ababœ
1
x1È_
(b) Domain f g : 0, , domain g f : 1, ab ab‰ Ð _Ñ ‰ Ð _Ñ
(c) Range f g : 1, , range g f : 0, ab ab‰ Ð _Ñ ‰ Ð _Ñ
14. (a) f g x 1 2 x xabab Ȝ
gfx 1 x ab kkabœ
(b) Domain f g : 0, , domain g f : 0, ab ab‰ Ð _Ñ ‰ Ð _Ñ
(c) Range f g : 0, , range g f : , 1ab ab‰ Ð _Ñ ‰ Ð_ Ñ
15. (a) y (x 7) (b) y (x 4)œ œ
# #
16. (a) y x 3 (b) y x 5œ œ
# #
17. (a) Position 4 (b) Position 1 (c) Position 2 (d) Position 3
18. (a) y (x 1) 4 (b) y (x 2) 3 (c) y (x 4) 1 (d) y (x 2)œ œ œ œ
## # #

33. Section 1.5 Combining Functions; Shifting and Scaling Graphs 29
19. 20.
21. 22.
23. 24.
25. 26.

34. 30 Chapter 1 Preliminaries
27. 28.
29. 30.
31. 32.
33. 34.
35. 36.

35. Section 1.5 Combining Functions; Shifting and Scaling Graphs 31
37. 38.
39. 40.
41. 42.
43. 44.

36. 32 Chapter 1 Preliminaries
45. 46.
47. 48.
49. (a) domain: [0 2]; range: [ ] (b) domain: [0 2]; range: [ 1 0]ß#ß$ ßß

(c) domain: [0 2]; range: [0 2] (d) domain: [0 2]; range: [ 1 0]ßß ßß

(e) domain: [ 2 0]; range: [ 1] (f) domain: [1 3]; range: [ ]ß !ß ß !ß"

37. Section 1.5 Combining Functions; Shifting and Scaling Graphs 33
(g) domain: [ 2 0]; range: [ ] (h) domain: [ 1 1]; range: [ ]ß !ß" ß !ß"

50. (a) domain: [0 4]; range: [ 3 0] (b) domain: [ 4 0]; range: [ ]ßß ß!ß$

(c) domain: [ 4 0]; range: [ ] (d) domain: [ 4 0]; range: [ ]ß !ß$ ß "ß%

(e) domain: [ 4]; range: [ 3 0] (f) domain: [ 2 2]; range: [ 3 0]#ß ß ß ß

38. 34 Chapter 1 Preliminaries
(g) domain: [ 5]; range: [ 3 0] (h) domain: [0 4]; range: [0 3]"ß ß ß ß

51. y 3x 3œ
#
52. y 2x 1 x 1œœ%ab
# #
53. yœ" œ
""""
###
ˆ‰
xx
ßß
54. y 1 1œ œ
"*
Î$abx x
ß ß
55. y x 1œ% È
56. y 3 x 1œ È
57. y 16 xœ% œ É ˆ‰ È
x
##
#
" #
58. y xœ%
"
$
#È
59. y 3x 27xœ" œ"ab
$ $60. yœ" œ"ˆ‰
xx
#)
$
#
61. Let y x f x and let g x x , h x x , i x x , andœ # "œ œ œ œ # È ab ab ab ab ˆ‰ ˆ‰ È
"Î# ""
##
"Î#" Î#
j x x f . The graph of h x is the graph of g x shifted left unit; the graph of i x is the graphab ab ab ab ab’“
Ȉ‰œ # œ B
" "
# #
"Î#
of h x stretched vertically by a factor of ; and the graph of j x f x is the graph of i x reflected across the x-axis.ab ab ab ab
È
#œ

39. Section 1.5 Combining Functions; Shifting and Scaling Graphs 35
62. Let y f x Let g x x , h x x , and i x x f xœ " œ Þ œ œ # œ # œ " œ ÞÈ Èab ab a b ab a b ab a b ab
x x
# #
"Î#" Î#" Î#"
#
È
The graph of g x is the graph of y x reflected across the x-axis. The graph of h x is the graph of g x shifted rightab ab ab Ȝ
two units. And the graph of i x is the graph of h x compressed vertically by a factor of .ab ab
È
#

63. y f x x . Shift f x one unit right followed by a shift two units up to get g x x .œœ œ" #ab ab ab a b
$ 3

64. y x f x . Let g x x , h x x , i x x , andœ "B #œÒ " # Óœ œ œ " œ " #a b a b a b ab ab ab a b ab a b a b
$$ $$ $
j x x . The graph of h x is the graph of g x shifted right one unit; the graph of i x is the graph ofab a b a b ab ab abœÒ " #Ó
$
h x shifted down two units; and the graph of f x is the graph of i x reflected across the x-axis.ab ab ab

40. 36 Chapter 1 Preliminaries
65. Compress the graph of f x horizontally by a factor of 2 to get g x . Then shift g x vertically down 1 unit toab ab abœœ
""
#xx
get h x .abœ"
"
#x

66. Let f x and g x Since , we see that the graph ofab ab
È
œ œ " œ " œ " œ "Þ # ¸ "Þ%
"#"" "
Î#" Î#B
xx
x
ßß
$
ß
ß
ßß
Š‹ Š‹ ’Š‹“
ÈÈ
f x stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of g x .ab ab

67. Reflect the graph of y f x x across the x-axis to get g x x.œœ œab abÈÈ
##

68. y f x x x x x . So the graph of f x is the graph of g x xœ œ # œ Ò " # Ó œ " # œ # œab a b a bab a b a b a b ab ab
#Î$# Î$# Î$# Î$#Î$ #Î$
compressed horizontally by a factor of 2.

41. Section 1.5 Combining Functions; Shifting and Scaling Graphs 37
69. 70.
71. x y 72. x y* #& œ ##& Ê œ " "' ( œ ""# Ê œ "
## ##
&$ %
(
x xy y
* *
** *
* *
*
Š‹
È

73. x y 74. x y$ # œ$Ê œ" " # œ%Ê œ"
# ##*# #
" #
$ #
"
ab ab
x y x y
*
* *
*
* *
*
* ab
Š‹ Š‹
È È
* ‘ab

75. x y 76. x y$"##œ' ' * œ&%abab ˆ‰ˆ‰
## $"
##
##
Ê œ" Ê œ"
ab
Š‹ Š‹ Š‹
ÈÈ
* ‘ab
’“ˆ‰
ˆ‰
È
x y
x
y "
#$ '
**#
**
$
*
*
** *
* "
*
*
*
#
*
*

42. 38 Chapter 1 Preliminaries
77. has its center at . Shiftinig 4 units left and 3 units up gives the center at h, k . So the
x y
* *
"' *
œ " !ß ! œ %ß $ab aba b
equation is . Center, C, is , and major axis, AB, is the segment
* ‘ab ab ab abx 4
43 43
y 3 x y ** *œ%*$
*
** **
***
œ"Êœ" %ß$ ab
from to .abab)ß $ !ß $

78. The ellipse has center h, k . Shifting the ellipse 3 units right and 2 units down produces an ellipse
x y
* *
%#&
œ " œ !ß ! abab
with center at h, k and an equation . Center, C, is 3 , and AB, the segment fromabab abœ $ß# œ" ß#
ab * ‘abx 3 y *
%#&
**#
*
*
to is the major axis.aba b$ß $ $ß (

79. (a) (fg)( x) f( x)g( x) f(x)( g(x)) (fg)(x), oddœ œ œ
(b) ( x) (x), oddŠ‹ Š‹
ff
gg(x)g(x)g
f( x) f(x)
œœœ
*
**
(c) ( x) (x), oddˆ‰ ˆ‰
gg(x)g(x)g
ff(x)f(x)f
œœœ
**
*
(d) f ( x) f( x)f( x) f(x)f(x) f (x), even
##
œ œ œ
(e) g ( x) (g( x)) ( g(x)) g (x), even
####
œ œ œ
(f) (f g)( x) f(g( x)) f( g(x)) f(g(x)) (f g)(x), even‰œ œ œ œ‰
(g) (g f)( x) g(f( x)) g(f(x)) (g f)(x), even‰œ œ œ‰
(h) (f f)( x) f(f( x)) f(f(x)) (f f)(x), even‰œ œ œ‰
(i) (g g)( x) g(g( x)) g( g(x)) g(g(x)) (g g)(x), odd‰ œ œ œ œ ‰
80. Yes, f(x) 0 is both even and odd since f( x) 0 f(x) and f( x) 0 f(x).œœœœœ

43. Section 1.6 Trigonometric Functions 39
81. (a) (b)
(c) (d)
82.
1.6 TRIGONOMETRIC FUNCTIONS
1. (a) s r (10) 8 m (b) s r (10)(110°) mœœ œ œœ œ œ)1 )ˆ‰ ˆ ‰
4 110 55
5 180° 18 9
1 111
2. radians and 225 °)œœ œ œ
s 10 5 5 180 °
r8 4 4
11 1
1
ˆ‰
3. 80° 80 ° s (6) 8.4 in. (since the diameter 12 in. radius 6 in.)))œÊœ œÊœ œ œ Ê œˆ‰ ˆ‰
11 1
180° 99
44

44. 40 Chapter 1 Preliminaries
4. d 1 meter r 50 cm 0.6 rad or 0.6 34 °œÊœÊœœœ ¸ )
s 30 180 °
r50
ˆ‰
1
5. 0 6
sin 0 0
cos 1 0
tan 0 3 0 und.
cot und. und. 0 1
sec 1 und. 2
csc und. und. 2
)1
)
)
)
)
)
)

"
"
"

#"
"
23
34
3
2
2
3
2
3
111
#
#
"
""
#
"
È
È
È
È
È
È
È
È
.
sin
cos
tan und. 3
cot 3 3
sec und. 2
csc
)
)
)
)
)
)
)

"
!
"
! "
#
" #
3
3
3
2
33
2
33
3
22
33
2
3
11 1 1 1
#'%'
&
## #
"" "
""
## #
""
"
È
È
ÈÈ
È
ÈÈ
È
ÈÈ
È
È
ÈÈ
È
È
2 #
7. cos x , tan x 8. sin x , cos xœ œ œ œ
43 2
54 55
ÈÈ
"
9. sin x , tan x 8 10. sin x , tan xœ œ œ œ
È
8
3 13 5
12 12
È
11. sin x , cos x 12. cos x , tan xœ œ œ œ
" "
#ÈÈ È
È
55 3
2 3
13. 14.
period period 4œœ11
15. 16.
period 2 period 4œœ
17. 18.
period 6 period 1œœ

45. Section 1.6 Trigonometric Functions 41
19. 20.
period 2 period 2œœ11
21. 22.
period 2 period 2œœ11
23. period , symmetric about the origin 24. period 1, symmetric about the originœœ
1
#

25. period 4, symmetric about the y-axis 26. period 4 , symmetric about the originœœ 1

27. (a) Cos x and sec x are positive in QI and QIV and
negative in QII and QIII. Sec x is undefined when
cos x is 0. The range of sec x is ( 1] [ );_ß "ß_
the range of cos x is [ 1]."ß

46. 42 Chapter 1 Preliminaries
(b) Sin x and csc x are positive in QI and QII and
negative in QIII and QIV. Csc x is undefined when
sin x is 0. The range of csc x is ( 1] [1 );_ß ß _
the range of sin x is [ ]."ß "

28. Since cot x , cot x is undefined when tan x 0œœ
"
tan x
and is zero when tan x is undefined. As tan x approaches
zero through positive values, cot x approaches infinity.
Also, cot x approaches negative infinity as tan x
approaches zero through negative values.

29. D: x ; R: y 1, 0, 1 30. D: x ; R: y 1, 0, 1_ _ œ _ _ œ

31. cos x cos x cos sin x sin (cos x)(0) (sin x)( 1) sin xˆ ‰ ˆ‰ ˆ‰œ œ œ
111
###
32. cos x cos x cos sin x sin (cos x)(0) (sin x)(1) sin xˆ ‰ ˆ‰ ˆ‰œ œ œ
111
###
33. sin x sin x cos cos x sin (sin x)(0) (cos x)(1) cos xˆ ‰ ˆ‰ ˆ‰œœœ
111
###
34. sin x sin x cos cos x sin (sin x)(0) (cos x)( 1) cos xˆ ‰ ˆ‰ ˆ‰œ œ œ
111
###
35. cos (A B) cos (A ( B)) cos A cos ( B) sin A sin ( B) cos A cos B sin A ( sin B)œ œ œ
cos A cos B sin A sin Bœ
36. sin (A B) sin (A ( B)) sin A cos ( B) cos A sin ( B) sin A cos B cos A ( sin B)œ œ œ
sin A cos B cos A sin Bœ
37. If B A, A B 0 cos (A B) cos 0 1. Also cos (A B) cos (A A) cos A cos A sin A sin AœœÊ œ œ œ œ
cos A sin A. Therefore, cos A sin A 1.œ œ
## ##
38. If B 2 , then cos (A 2 ) cos A cos 2 sin A sin 2 (cos A)(1) (sin A)(0) cos A andœœœœ1111
sin (A 2 ) sin A cos 2 cos A sin 2 (sin A)(1) (cos A)(0) sin A. The result agrees with theœœœ111
fact that the cosine and sine functions have period 2 .1
39. cos ( x) cos cos sin sin x ( 1)(cos x) (0)(sin x) cos x111œ B œ œ

47. Section 1.6 Trigonometric Functions 43
40. sin (2 x) sin 2 cos ( x) cos (2 ) sin ( x) (0)(cos ( x)) (1)(sin ( x)) sin x11 1œ œ œ
41. sin x sin cos ( x) cos sin ( x) ( 1)(cos x) (0)(sin ( x)) cos xˆ ‰ ˆ‰ ˆ‰
33 311 1
## #
œ œ œ
42. cos x cos cos x sin sin x (0)(cos x) ( 1)(sin x) sin xˆ ‰ ˆ‰ ˆ‰
33 311 1
## #
œ œ œ
43. sin sin sin cos cos sin
7
1434343 4
22 362111 1111
#####
"
œœœ œˆ‰ ˆ‰ Š‹ Š‹Š‹
ÈÈÈÈ È
44. cos cos cos cos sin sin
11 2 2 2
1434343 4
22 326111 1111
#####
"
œœ œ œˆ‰ ˆ‰ Š‹ Š‹Š‹
ÈÈÈÈ È
45. cos cos cos cos sin sin
1111111
12 3 4 3 4 3 4
22 313
22
œœ œ œˆ ‰ ˆ‰ ˆ‰ˆ‰ Š‹Š‹Š ‹
"
## # #

ÈÈÈ È
È
46. sin sin sin cos cos sin
52 2 2
134 3 4 34
313 22
22
111 1 1 11
# ## # #
"
œœœœˆ ‰ ˆ‰ ˆ ‰ ˆ‰ ˆ ‰ ˆ ‰ Š‹Š‹ Š ‹
ÈÈ È È
È
47. cos 48. cos
# #

## ###
1 1
84 1 4
1cos 1 1cos 1 22 23
œœœ œœœ
ˆ‰ ˆ‰È È
2 2
8 1
2 3


49. sin 50. sin
# #
## # # #
œœ œœ œ œ1 1
14 84
1cos 1 1cos 1 23 22
œœœ œœœ
ˆ‰ ˆ‰È È
2 2
1 8
3 2


51. tan (A B)œ œ œ
sin (A B)
cos (A B) cos A cos B sin A sin B
sin A cos B cos A cos B


sin A cos B cos A sin B
cos A cos B cos A cos B
cos A cos B sin A sin B
cos A cos B cos A cos B
œ

œ
œ
tan A tan B
1tan A tan B
52. tan (A B)œ œ œ
sin (A B)
cos (A B) cos A cos B sin A sin B
sin A cos B cos A cos Bœ

œ œ
sin A cos B cos A sin B
cos A cos B cos A cos B
cos A cos B sin A sin B
cos A cos B cos A cos B

œ

œ
tan A tan B
1tan A tan B
53. According to the figure in the text, we have the following: By the law of cosines, c a b 2ab cos
###
œ )
1 1 2 cos (A B) 2 2 cos (A B). By distance formula, c (cos A cos B) (sin A sin B)œ œ œ
## ###
cos A 2 cos A cos B cos B sin A 2 sin A sin B sin B 2 2(cos A cos B sin A sin B). Thusœ œ
####
c 2 2 cos (A B) 2 2(cos A cos B sin A sin B) cos (A B) cos A cos B sin A sin B.
#
œ œ Ê œ
54. (a) cos A B cos A cos B sin A sin Babœ
sin cos and cos sin))))œ œˆ‰ ˆ‰
11
##
Let A B)œ
sin A B cos A B cos A B cos A cos B sin A sin Bab ab ’“’ “ ˆ‰ ˆ‰ ˆ‰œ œ œ
1111
####
sin A cos B cos A sin Bœ
(b) cos A B cos A cos B sin A sin Babœ
cos A B cos A cos B sin A sin Ba b ab abab œ
cos A B cos A cos B sin A sin B cos A cos B sin A sin BÊœ œ a b ab ab a b
cos A cos B sin A sin Bœ
Because the cosine function is even and the sine functions is odd.
55. c a b 2ab cos C 2 3 2(2)(3) cos (60°) 4 9 12 cos (60°) 13 12 7.
### ## "
#
œ œ œ œ œ ˆ‰
Thus, c 7 2.65.œ¸
È
56. c a b 2ab cos C 2 3 2(2)(3) cos (40°) 13 12 cos (40°). Thus, c 13 12 cos 40° 1.951.
### ##
œ œ œ œ ¸
È

48. 44 Chapter 1 Preliminaries
57. From the figures in the text, we see that sin B . If C is an acute angle, then sin C . On the other hand,œœ
hh
cb
if C is obtuse (as in the figure on the right), then sin C sin ( C) . Thus, in either case,œœ1
h
b
h b sin C c sin B ah ab sin C ac sin B.œœ Êœ œ
By the law of cosines, cos C and cos B . Moreover, since the sum of theœœ
abc acb
2ab 2ac


interior angles of a triangle is , we have sin A sin ( (B C)) sin (B C) sin B cos C cos B sin C11 œœœ
2a b c c b ah bc sin A.œœœÊœˆ‰ ˆ‰ ˆ ‰’“’“ ab
h abc acb h h ah
c 2ab 2ac b 2abc bc

#####
Combining our results we have ah ab sin C, ah ac sin B, and ah bc sin A. Dividing by abc givesœœ œ
.
h sin A sin C sin B
bc a c b
œœœ
ðóóóóóóóñóóóóóóóò
law of sines
58. By the law of sines, . By Exercise 55 we know that c 7.
sin A sin B
3c
3/2
#
œœ œ
È
È
Thus sin B 0.982.Ϧ
33
27
È
È
59. From the figure at the right and the law of cosines,
b a 2 2(2a) cos B
###
œ
a44a a2a4.œ œ
## "
# ˆ‰
Applying the law of sines to the figure,
sin A sin B
ab
œ
b a. Thus, combining results,ʜʜ
ÈÈ
2/2
ab
3/2 3
É
#
a 2a 4 b a 0 a 2a 4
####
##
"
œœ Êœ
3

0 a 4a 8. From the quadratic formula and the fact that a 0, we haveʜ
#
a 1.464.œœ¶

##
œ4 4 4(1)( 8) 434
ÈÈ

60. (a) The graphs of y sin x and y x nearly coincide when x is near the origin (when the calculatorœœ
is in radians mode).
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The
curves look like intersecting straight lines near the origin when the calculator is in degree mode.
61. A 2, B 2 , C , D 1œ œ œ œ11

62. A , B 2, C 1, Dœœœœ
""
##

49. Section 1.6 Trigonometric Functions 45
63. A , B 4, C 0, Dœ œ œ œ
2
11
"

64. A , B L, C 0, D 0œœœœ
L
21

65. (a) amplitude A 37 (b) period B 365œœ œœkk kk
(c) right horizontal shift C 101 (d) upward vertical shift D 25œœ œœ
66. (a) It is highest when the value of the sine is 1 at f(101) 37 sin (0) 25 62°F.œœ
The lowest mean daily temp is 37( 1) 25 12 °F. œ
(b) The average of the highest and lowest mean daily temperatures 25°F.œœ
62°( 12)°
#
The average of the sine function is its horizontal axis, y 25.œ
67-70. Example CAS commands:
Maple
f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#67 (Section 1.6)" ); Mathematica
Clear[a, b, c, d, f, x] f[x_]:=a Sin[2 /b (x c)] + d1
Plot[f[x]/.{a 3, b 1, c 0, d 0}, {x, 4 , 4 }]ÄÄÄÄ 11
67. (a) The graph stretches horizontally.

50. 46 Chapter 1 Preliminaries
(b) The period remains the same: period B . The graph has a horizontal shift of period.œl l
"
#

68. (a) The graph is shifted right C units.

(b) The graph is shifted left C units.
(c) A shift of one period will produce no apparent shift. C „ l lœ'
69. The graph shifts upwards D units for D and down D units for Dll ! ll !Þ

70. (a) The graph stretches A units.ll

(b) For A , the graph is inverted.!
1.7 GRAPHING WITH CALCULATORS AND COMPUTERS
1-4. The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and
has little unused space.

51. Section 1.7 Graphing with Calculators and Computers 47
1. d. 2. c.

3. d. 4. b.

5-30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5 30
are not unique in appearance.
5. 2 5 by 15 40 6. 4 4 by 4 4Ò ß Ó Ò ß Ó Ò ß Ó Ò ß Ó

7. 2 6 by 250 50 8. 1 5 by 5 30Ò ß Ó Ò ß Ó Ò ß Ó Ò ß Ó

52. 48 Chapter 1 Preliminaries
9. 4 4 by 5 5 10. 2 2 by 2 8Ò ß Ó Ò ß Ó Ò ß Ó Ò ß Ó

11. 2 6 by 5 4 12. 4 4 by 8 8Ò ß Ó Ò ß Ó Ò ß Ó Ò ß Ó

13. by 14. by Ò"ß 'Ó Ò"ß %Ó Ò"ß 'Ó Ò"ß &Ó

15. 3 3 by 16. by Ò ß Ó Ò!ß "!Ó Ò"ß #Ó Ò!ß "Ó

53. Section 1.7 Graphing with Calculators and Computers 49
17. by 18. by Ò&ß "Ó Ò&ß &Ó Ò&ß "Ó Ò#ß %Ó

19. by 20. by Ò%ß %Ó Ò!ß $Ó Ò&ß &Ó Ò#ß #Ó

21. by 22. by Ò"!ß "!Ó Ò'ß 'Ó Ò&ß &Ó Ò#ß #Ó

23. by 24. by Ò'ß "!Ó Ò'ß 'Ó Ò$ß &Ó Ò#ß "!Ó

54. 50 Chapter 1 Preliminaries
25. by 26. by Ò!Þ!$ß !Þ!$Ó Ò"Þ#&ß "Þ#&Ó Ò!Þ"ß !Þ"Ó Ò$ß $Ó

27. by 28. by Ò$!!ß $!!Ó Ò"Þ#&ß "Þ#&Ó Ò&!ß &!Ó Ò!Þ"ß !Þ"Ó

29. by 30. by Ò!Þ#&ß !Þ#&Ó Ò!Þ$ß !Þ$Ó Ò!Þ"&ß !Þ"&Ó Ò!Þ!#ß !Þ!&Ó

31. x x y y y x x .
## #
# œ%% Ê œ#„ # )
È
The lower half is produced by graphing
yxx.œ# # )
È
#

32. y x y x . The upper branch
## #
"' œ"Ê œ „ ""'
È
is produced by graphing y x .œ""'
È
#

55. Section 1.7 Graphing with Calculators and Computers 51
33. 34.
35. 36.
37. 38 Þ
39. 40.
41. (a) y xœ "!&*Þ"% #!(%*(#
(b) m dollars/year, which is the yearly increase in compensation.œ "!&*Þ"%

56. 52 Chapter 1 Preliminaries
(c)
(d) Answers may vary slightly. y 14 $ 899œ "!&*Þ #!"! #!(%*(# œ &$ßabab
42. (a) Let C cost and x year.œœ
Cx œ (*'!Þ(" "Þ' ‚ "!ab
(
(b) Slope represents increase in cost per year
(c) C xœ #'$(Þ"% &Þ# ‚ "!ab
'
(d) The median price is rising faster in the northease (the slope is larger).
43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is
dxx.œ !Þ!)'' "Þ*( &!Þ"
#
(b)
(c) From the graph in part (b), the stopping distance is about feet when the vehicle is mph and it is about feet$(! (# &#&
when the speed is mph.)&
Algebraically: d ft.
quadraticab ab ab(# œ !Þ!)'' (# "Þ*( (# &!Þ" œ $'(Þ'
#
d ft. quadraticab ab ab)& œ !Þ!)'' )& "Þ*( )& &!Þ" œ &##Þ)
#
(d) The linear regression function is d x d ft andœ 'Þ)* "%!Þ% Ê (# œ 'Þ)* (# "%!Þ% œ $&&Þ( linearab ab
d ft. The linear regression line is shown on the graph in part (b). The quadratic
linearab ab)& œ 'Þ)* )& "%!Þ% œ %%&Þ#
regression curve clearly gives the better fit.

44. (a) The power regression function is y x .œ %Þ%%'%(
!Þ&""%"%

57. Chapter 1 Practice Exercises 53
(b)
(c) 15 2 km/hÞ
(d) The linear regression function is y x and it is shown on the graph in part (b). The linearœ !Þ*"$'(& %Þ")**('
regession function gives a speed of km/h when y m. The power regression curve in part (a) better fits the"%Þ# œ ""
data.
CHAPTER 1 PRACTICE EXERCISES
1. 2x x x (  $ Ê #  %Ê  #
2. 3x x x "! Ê qqqqqqqqðïïïïïïïî

"!
$ "!
$
3. x x x x
"" &%
abab abab" #Ê%"&#
xx xÊ% %& "!Ê'
4. x x x
xx$ %
#$ "
&
  Ê $ $   # % qqqqqqqqñïïïïïïïîab ab
xxxxÊ$* )#Ê& "Ê  
"
&
5. x x or x x or xl "lœ(Ê "œ( " œ(Ê œ' œ)ab
6. y y yl $l%Ê % $%Ê" (
7. or or x or x¹¹" Ê " " Ê Ê & "
xxxxx
## ## ## ####
$$$&"
x or xÊ& "
8. 1 x 1 22 x 8 x¹¹
#( #(
$$
xx
Ÿ&Ê &Ÿ Ÿ&Ê &Ÿ# (Ÿ &Ê Ÿ# Ÿ Ê""Ÿ Ÿ%
9. Since the particle moved to the y-axis, x x 2. Since y 3 x 6, the new coordinates# œ ! Ê œ œ œ?? ??
are (x x y y) ( ) (0 11). ß œ ##ß&' œ ß??
10. (a)

58. 54 Chapter 1 Preliminaries
(b) line slope
AB
10 1 9 3
28 6
Ò
ÒÒ #
œœ
BC
10 6 4 2
2(4) 6 3
Ò
ÒÒ
œœ
CD
6(3)
26
93ÒÒ
Ò% Ò Ò #
œœ
DA
1(3)
82 6 3
42ÒÒ
Ò
œœ
CE 0
66Ò
Ò% Ò
14
3œ
BD is vertical and has no slope
(c) Yes; A, B, C and D form a parallelogram. (d) Yes. The line AB has equation y 1 (x 8). Replacing x by gives y 8œ œ "
314314
33##
ˆ‰
1 5 1 6. Thus, E 6 lies on the line AB and the points A, B and E are collinear.œ œœ ß
310 14
33#
ˆ‰ ˆ ‰
(e) The line CD has equation y 3 (x 2) or y x. Thus the line passes through the origin.œ œ
33
##
11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are
53, 72 and 65, respectively. The slopes of AB, BC and AC are , 1 and , respectively.
ÈÈ È
7
8#
"

12. P(x 3x 1) is a point on the line y 3x 1. If the distance from P to ( 0) equals the distance from P toß œ !ß
( ), then x (3x 1) (x 3) (3 3x) x 9x 6x 1 x 6x 9 9 18x 9x$ß% œ Ê œ
###### # #
18x 17 or x y 3x 1 3 1 . Thus the point is P .Êœ œÊœœ œ ß
17 17 23 17 23
18 18 6 18 6
ˆ‰ ˆ ‰
13. y x y xœ$ " ' Ê œ$ *abab
14. y x y xœ " #Ê œ
""$
###
ab
15. xœ!
16. m y x y xœ œ œ#Ê œ# $ 'Ê œ#
Ò#Ò' Ò&
"Ò Ò$ %ab
ab
17. yœ#
18. m y x y xœ œ œ Ê œ $ $Ê œ
&Ò$ # # # # #"
Ò#Ò$ Ò& & & & &
ab
19. y xœ$ $
20. Since x y is equivalent to y x , the slope of the given line (and hence the slope of the desired line) is 2.# œ# œ##
yx yxœ# " "Ê œ# &ab
21. Since x y is equivalent to y x , the slope of the given line (and hence the slope of the desired line) is%$œ"# œ %
%
$
. y x 4 2 y xœ"Êœ
%% %#!
$$ $$
ab
22. Since x y is equivalent to y x , the slope of the given line is and the slope of the perpendicular line is$&œ" œ
$" $
&& &
. y x y xœ#$Êœ
55 5
$$ $$
"*
ab
23. Since x y is equivalent to y x , the slope of the given line is and the slope of the perpendicular line
"" $ $
#$ # #
œ" œ$
is . y x y x
## #) $$ $$
œ"#Êœab
24. The line passes through and . m y xabab!ß & $ß ! œ œ Ê œ &
!Ò Ò&
$Ò! $ $
&&ab

59. Chapter 1 Practice Exercises 55
25. The area is A r and the circumference is C r. Thus, r A .œœ#œÊœœ111
#
##%
#
CCC
111 ˆ‰
Ò
26. The surface area is S r r . The volume is V r r . Substitution into the formula forœ% Ê œ œ Ê œ11
#$
%$%
"Î#
%$ ˆ‰ É
SV
11
surface area gives S r .œ% œ%11
# $
%
#Î$ˆ‰
V
1
27. The coordinates of a point on the parabola are x x . The angle of inclination joining this point to the origin satisfiesabß
#
)
the equation tan x. Thus the point has coordinates x x tan tan .)))œœ ß œ ß
x
x
Ò
aba b
##
28. tan h tan ft.))œ œ Ê œ &!!
rise h
run &!!
29. 30.
Symmetric about the origin. Symmetric about the y-axis.
31. 32. Neither Symmetric about the y-axis.
33. y x x x y x . Even.abab abœ "œ "œ
# #
34. y x x x x x x x y x . Odd.abab ab ab abœœœ
&$ &$
35. y x cos x cos x y x . Even.ab ab abœ" œ" œ36. y x sec x tan x sec x tan x y x . Odd.ab abab ab œ œ œ œ œ
sin x
cos x cos x
sin xab
ab
Ò
Ò
Ò
ÒÒ
37. y x y x . Odd.ab abœ œ œ œ
ab
ab ab
"
ÒÒ#Ò
" "
# #
x
xx
xx
xx xx
&

&&

38. y x sin x sin x. Neither even nor odd.ab abœ" œ"
39. y x x cos x x cos x. Neither even nor odd.ab abœ œ
40. y x x x y x . Even.ab ab abÉ È
œ "œ "œ
%
%

60. 56 Chapter 1 Preliminaries
41. (a) The function is defined for all values of x, so the domain is .ab_ß _
(b) Since x attains all nonnegative values, the range is .l l Ò#ß _Ñ
42. (a) Since the square root requires x , the domain is ."   ! Ð_ß "Ó
#
(b) Since x attains all nonnegative values, the range is .
È
" Ò#ß _Ñ
43. (a) Since the square root requires x , the domain is ."'   ! Ò%ß %Ó
#
(b) For values of x in the domain, x , so x . The range is .! Ÿ "' Ÿ "' ! Ÿ "' Ÿ % Ò!ß %Ó
# # È
44. (a) The function is defined for all values of x, so the domain is .ab_ß _
(b) Since attains all positive values, the range is .$"ß_
#x
ab
45. (a) The function is defined for all values of x, so the domain is .ab_ß _
(b) Since e attains all positive values, the range is .# $ß _
x
ab
46. (a) The function is equivalent to y tan x, so we require x for odd integers k. The domain is given by x forœ# #Á Á
kk11
#%
odd integers k.
(b) Since the tangent function attains all values, the range is .ab_ß _
47. (a) The function is defined for all values of x, so the domain is .ab_ß _
(b) The sine function attains values from to , so sin x and hence sin x . The" " #Ÿ# $ Ÿ# $Ÿ# $ "Ÿ"ab ab11
range is 3 1 .Ò ß Ó
48. (a) The function is defined for all values of x, so the domain is .ab_ß _
(b) The function is equivalent to y x , which attains all nonnegative values. The range is .œ Ò!ß _Ñ
È

#
49. (a) The logarithm requires x , so the domain is . $ ! $ß _ ab
(b) The logarithm attains all real values, so the range is .ab_ß _
50. (a) The function is defined for all values of x, so the domain is .ab_ß _
(b) The cube root attains all real values, so the range is .ab_ß _
51. (a) The function is defined for x , so the domain is .% Ÿ Ÿ % Ò%ß %Ó
(b) The function is equivalent to y x , x , which attains values from to for x in the domain. Theœll%ŸŸ% !#È
range is .Ò!ß #Ó
52. (a) The function is defined for x , so the domain is .# Ÿ Ÿ # Ò#ß #Ó
(b) The range is .Ò"ß "Ó
53. First piece: Line through and . m y x xab ab!ß" "ß! œ œ œ"Ê œ "œ"
!" "
"! "
Second piece: Line through and . m y x x xab ab a b"ß" #ß! œ œ œ"Ê œ " "œ #œ#
!" "
#" "
fx
x, x
x, x
abœœ
" !Ÿ "
# "Ÿ Ÿ#
54. First piece: Line through and 2 5 . m y xab ab!ß ! ß œ œ Ê œ
55 5
22 2
!
!
Second piece: Line through 2 5 and 4 . m y x 2 5 x 10 10ab ab a bß ß! œ œ œ Ê œ œ œ
!

55 5 5 5 5x
42 2 2 2 2 2
f x (Note: x 2 can be included on either piece.)
x, x 2
10 ,2x4
ab

œœ
!Ÿ
ŸŸ
5
2
5x
2

61. Chapter 1 Practice Exercises 57
55. (a) f g f g f fabab a b abab Š‹‰ "œ " œ œ "œ œ"
""
"# "È
(b) g f g f g or abab a b ab ˆ‰ ɉ#œ #œ œ œ
"" " #
Ò# #Þ& &2
É
È
%
!
(c) f f x f f x f x, xabab a bab ˆ‰‰œ œœœÁ!
""
"Îxx
(d) g g x g g x gabab a b ab Š‹‰œ œ œ œ
""
Ò# Ò#
Ò#
lÒ# Ò#
È
ÉÉ
È
Èx
x
x
%
"!
Ò

x
56. (a) f g f g f fabab a b abab ˆ‰È‰ "œ " œ ""œ !œ#!œ#
Þ
(b) g f f g g ga bab a b a b abab ȉ#œ #œ##œ!œ !"œ"
Þ
(c) f f x f f x f x x xabababab abab‰ œ œ # œ# # œ
(d) g g x g g x g x xabab a b ab ˆ‰ÈÈ É‰ œ œ " œ ""
ÞÞ Þ
57. (a) f g x f g x f x x x, x .abab a bab ˆ‰ˆ‰Èȉ œ œ # œ# # œ  #
#
gfxfgxgx x xabababababab È È
‰œ œ#œ##œ%
# ##
(b) Domain of f g: ‰ Ò#ß _ÑÞ
Domain of g f: ‰ Ò#ß #ÓÞ
(b) Range of f g: ‰ Ð_ß #ÓÞ
Range of g f: ‰ Ò!ß #ÓÞ
58. (a) f g x f g x f x x x.abab a bab Š‹
ÈÈÈ É‰œ œ"œ"œ"
Ò
gf x fgx g x xabab a b ab ˆ‰ÈÈ É‰œ œ œ"
(b) Domain of f g: ‰ Ð_ß "ÓÞ
Domain of g f: ‰ Ò!ß "ÓÞ
(b) Range of f g: ‰ Ò!ß _ÑÞ
Range of g f: ‰ Ò!ß "ÓÞ
59. 60.
The graph of f (x) f x is the same as the The graph of f (x) f x is the same as the
#" #"œœab abkk kk
graph of f (x) to the right of the y-axis. The graph of f (x) to the right of the y-axis. The
" "
graph of f (x) to the left of the y-axis is the graph of f (x) to the left of the y-axis is the# #
reflection of y f (x), x 0 across the y-axis. reflection of y f (x), x 0 across the y-axis.œ  œ  " "

62. 58 Chapter 1 Preliminaries
61. 62.
It does not change the graph. The graph of f (x) f x is the same as the
#"œabkk
graph of f (x) to the right of the y-axis. The
"
graph of f (x) to the left of the y-axis is the #
reflection of y f (x), x 0 across the y-axis. œ  "
63. 64.
The graph of f (x) f x is the same as the The graph of f (x) f x is the same as the
#" #"œœab abkk kk
graph of f (x) to the right of the y-axis. The graph of f (x) to the right of the y-axis. The
" "
graph of f (x) to the left of the y-axis is the graph of f (x) to the left of the y-axis is the# #
reflection of y f (x), x 0 across the y-axis. reflection of y f (x), x 0 across the y-axis.œ  œ  " "
65. 66.
Whenever g (x) is positive, the graph of y g (x) It does not change the graph.
"# œ
g (x) is the same as the graph of y g (x).œœkk
""
When g (x) is negative, the graph of y g (x) is"# œ
the reflection of the graph of y g (x) across theœ
"
x-axis.
67. Whenever g (x) is positive, the graph of y g (x) g (x) is"#" œœ kk
the same as the graph of y g (x). When g (x) is negative, theœ
""
graph of y g (x) is the reflection of the graph of y g (x)œœ #"
across the x-axis.

63. Chapter 1 Practice Exercises 59
68. Whenever g (x) is positive, the graph of y g (x) g (x) is"#" œœ kk
the same as the graph of y g (x). When g (x) is negative, theœ
""
graph of y g (x) is the reflection of the graph of y g (x)œœ #"
across the x-axis.
69. 70.
period period 4œœ11
71. 72.
period 2 period 4œœ
73. 74.
period 2 period 2œœ11
75. (a) sin B sin b 2 sin 2 3. By the theorem of Pythagoras,œœœÊœ œ œ
11
3c 3
bb 3
##
Š‹
È
È
abc a cb 431.
### ##
œ Êœ œ œ
È È
(b) sin B sin c . Thus, a c b (2) .œœœÊœœœ œœ œœ
1
3cc sin 3
b2 2 2 4 4 4 2
333

3
3 Š‹
ÈÈÈ

!
È
ÊŠ‹ É## #
#
76. (a) sin A a c sin A (b) tan A a b tan AœÊœ œÊœ
a a
c b
77. (a) tan B a (b) sin A cœÊœ œÊœ
bb aa
atan B csin A

64. 60 Chapter 1 Preliminaries
78. (a) sin A (c) sin Aœœœ
a a
c cc
cb
È
ßß

79. Let h height of vertical pole, and let b and c denote theœ
distances of points B and C from the base of the pole,
measured along the flatground, respectively. Then,
tan 50° , tan 35° , and b c 10.œœœ
hh
cb
Thus, h c tan 50° and h b tan 35° (c 10) tan 35°œœœ
c tan 50° (c 10) tan 35°Êœ
c (tan 50° tan 35°) 10 tan 35°Êœ
c h c tan 50°Êœ Êœ
10 tan 35°
tan 50° tan 35°
16.98 m.œ¸
10 tan 35° tan 50°
tan 50° tan 35°

80. Let h height of balloon above ground. From the figure atœ
the right, tan 40° , tan 70° , and a b 2. Thus,œœœ
hh
ab
h b tan 70° h (2 a) tan 70° and h a tan 40°œÊœ œ
(2 a) tan 70° a tan 40° a(tan 40° tan 70°)Ê œ Ê
2 tan 70° a h a tan 40°œÊœ Êœ
2 tan 70°
tan 40° tan 70°Ò
1.3 km.œ¸
2 tan 70° tan 40°
tan 40° tan 70°Ò

81. (a)
(b) The period appears to be 4 .1
(c) f(x 4 ) sin (x 4 ) cos sin (x 2 ) cos 2 sin x cos œ œ œ 11 1 1 ˆ‰ ˆ ‰
x4 x xÒ
###
1
since the period of sine and cosine is 2 . Thus, f(x) has period 4 .11
82. (a)
(b) D ( 0) ( ); R [ 1 1]œ_ß !ß_ œß
(c) f is not periodic. For suppose f has period p. Then f kp f sin 2 0 for allˆ‰ˆ‰
""
##11
œ œ œ 1
integers k. Choose k so large that kp 0 . But then
"" "
#Ò11 1
Ê
(1/2 ) kp
1
f kp sin 0 which is a contradiction. Thus f has no period, as claimed.ˆ‰ Š‹
""
##Ò11
œ
(1/ ) kp

65. Chapter 1 Additional and Advanced Exercises 61
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
1. (a) The given graph is reflected about the y-axis. (b) The given graph is reflected about the x-axis.

(c) The given graph is shifted left 1 unit, stretched (d) The given graph is shifted right 2 units, stretched
vertically by a factor of 2, reflected about the vertically by a factor of 3, and then shifted
x-axis, and then shifted upward 1 unit. downward 2 units.

2. (a) (b)
3. There are (infinitely) many such function pairs. For example, f(x) 3x and g(x) 4x satisfyœœ
f(g(x)) f(4x) 3(4x) 12x 4(3x) g(3x) g(f(x)).œœœœœœ
4. Yes, there are many such function pairs. For example, if g(x) (2x 3) and f(x) x , thenœ œ
$ "Î$
(f g)(x) f(g(x)) f (2x 3) (2x 3) 2x 3.‰œ œ œ œ abab
$$
"Î$
5. If f is odd and defined at x, then f( x) f(x). Thus g( x) f( x) 2 f(x) 2 whereasœ œœ
g(x) (f(x) 2) f(x) 2. Then g cannot be odd because g( x) g(x) f(x) 2 f(x) 2 œ œ œ Ê œ
4 0, which is a contradiction. Also, g(x) is not even unless f(x) 0 for all x. On the other hand, if f isÊœ œ
even, then g(x) f(x) 2 is also even: g( x) f( x) 2 f(x) 2 g(x).œ œœœ
6. If g is odd and g(0) is defined, then g(0) g( 0) g(0). Therefore, 2g(0) 0 g(0) 0.œœ œ Ê œ

66. 62 Chapter 1 Preliminaries
7. For (x y) in the 1st quadrant, x y 1 xߜ kk kk
x y 1 x y 1. For (x y) in the 2ndÍ œ Í œ ß
quadrant, x y x1 xyx1kk kk œ Íœ
y 2x 1. In the 3rd quadrant, x y x 1Íœ œ kk kk
x y x 1 y 2x 1. In the 4thÍœ Í œ
quadrant, x y x 1 x ( y) x 1kk kkœÍœ
y 1. The graph is given at the right.͜

8. We use reasoning similar to Exercise 7.
(1) 1st quadrant: y y x xœkk kk
2y 2x y x.͜͜
(2) 2nd quadrant: y y x xœkk kk
2y x ( x) 0 y 0.ÍœœÍœ
(3) 3rd quadrant: y y x xœkk kk
y ( y) x ( x) 0 0͜͜
points in the 3rd quadrantÊall
satisfy the equation.
(4) 4th quadrant: y y x xœkk kk
y ( y) 2x 0 x. Combining͜ ͜
these results we have the graph given at the
right:

9. By the law of sines, b 2.
sin
3
sin A sin B
abb sin(/3)
sin 3 sin ( /4)
3

3 4
2
3
È
È
È
Š‹
œœœ Êœ œ œ
1
1


!
!
È
10. By the law of sines, sin B sin .
sin
4ab3 444 8
sin A sin B sin B 3 3 232

4
œœœ Ê œ œ œ
1
Š‹
ÈÈ
#
11. By the law of cosines, a b c 2bc cos A cos A .
###
œ Ê œ œ œ
bca 232 3
2bc 2(2)(3) 4
!!! ! ! !
12. By the law of cosines, c a b 2ab cos C 2 3 (2)(2)(3) cos 4 9 12
### ##
#
œ œ œ
1
4
2
Š‹
È
13 6 2 c 13 6 2 , since c 0.œ Êœ
ÈÈ É
13. By the law of cosines, b a c 2ac cos B cos B
###
#
œ Ê œ œ œ
a c b 2 4 3 4169
ac (2)(2)(4) 16
!!! !!!
. Since 0 B , sin B 1 cos B 1 .œœœœœ
11 121
16 256 16 16
135 3 15
1
È
É
#
ÈÈ
14. By the law of cosines, c a b 2ab cos C cos C
###
œ Ê œ œ œ
a b c 2 4 5 41625
2ab (2)(2)(4) 16
!!! !!!
. Since 0 C , sin C 1 cos C 1 .œ œ œ œ
525
16 256 16
231
1
È
É
#
È
15. (a) sin x cos x 1 sin x 1 cos x (1 cos x)(1 cos x) (1 cos x)
## # #
Ò
œÊ œœ Êœ
sin x
1cos x
!
ʜ
1cos x sin x
sin x 1 cos x

Ò
(b) Using the definition of the tangent function and the double angle formulas, we have
tan .
#
#Ò
ˆ‰
x1cos xsin
cos 1cos x
œœ œ
!
!!
!
!
%&
!
%"
!
!ˆ‰
ˆ‰
x
x
cos 2
x
cos 2
x
%!%!
%!%!

67. Chapter 1 Additional and Advanced Exercises 63
16. The angles labeled in the accompanying figure are#
equal since both angles subtend arc CD. Similarly, the
two angles labeled are equal since they both subtend!
arc AB. Thus, triangles AED and BEC are similar which
implies
ac 2a cos b
bac

Ò
œ
)
(a c)(a c) b(2a cos b)Ê œ )
a c 2ab cos bʜ
## #
)
c a b 2ab cos .Ê œ
###
)

17. As in the proof of the law of sines of Section P.5, Exercise 57, ah bc sin A ab sin C ac sin Bœœœ
the area of ABC (base)(height) ah bc sin A ab sin C ac sin B.Êœœœœœ
"""""
#####
18. As in Section P.5, Exercise 57, (Area of ABC) (base) (height) a h a b sin C
# # ######"""
œœœ
444
a b cos C . By the law of cosines, c a b 2ab cos C cos C .œ " œ Ê œ
" ## # # # #
4 2ab
abc
ab
ßßß
Thus, (area of ABC) a b cos C a b
### # ##""
#
#

œ"œ " œ"
44ab44ab
abc ab abc
ab Œ _Š‹ Š ‹
ßßß ßß ßßß
ß
ßß ab
4a b a b c 2ab a b c 2ab a b cœ œ
"" ## ### ### ###
#
16 16
Š‹ ab cdabababab
(a b) c c (a b) ((a b) c)((a b) c)(c (a b))(c (a b))œ œ
"" ### #
16 16
cdc dabab
s(s a)(s b)(s c), where s .œœœß ‘ˆ‰ˆ ‰ˆ‰ˆ‰
abc abc abc abc abc
#### #
Therefore, the area of ABC equals s(s a)(s b)(s c) .È
19. 1. b c (a c) b a, which is positive since a b. Thus, a c b c. œ
2. b c (a c) b a, which is positive since a b. Thus, a c b c. œ
3. c 0 and a b c 0 c and b a are positive (b a)c bc ac is positive ac bc.Êœ Êœ Ê
4. a b and c 0 b a and c are positive (b a)( c) ac bc is positive bc ac.Ê Êœ Ê
5. Since a 0, a and are positive 0.Ê
""
aa
6. Since 0 a b, both and are positive. By (3), a b and 0 a b or 1 Ê
"" " " "
ab a a a a
b
ˆ‰ ˆ‰
1 by (3) since 0 .Ê Êˆ‰ ˆ‰
"" " ""
bab b ba
b
7. a b 0 and are both negative, i.e., 0 and 0. By (4), a b and 0 b a Ê Ê
"" " " " " "
ab a b a a a
ˆ‰ ˆ‰
1 1 by (4) since 0 .ÊÊ Ê
bb
abab bba
ˆ‰ ˆ‰
"" " ""
20. (a) If a 0, then 0 a b b 0 0 a b . Since a a a a a andœœÍÁÍœ œœœkk kk kk kk kk kkkk k k
## # ##
b b we obtain a b . If a 0 then a 0 and a b a b . On the other hand,kk kk kk kk
# ### ##
œÁÊ
if a b then a a b b 0 b a b a b a . Since b a 0
## # # ## ##
œœÊœ kk kk kk kk abababkk kk kk kk kk kk
and the product b a b a is positive, we must have b a 0 b a . Thusabab ab kk kkkk kk kk kk kk kk Ê
ab ab.kk kkÍ
##
(b) ab ab ab 2 ab by Exercise 19(4) above a 2ab b a 2 a b b , sinceŸÊ  Ê  kk kk kk kkkk kk
## ##
a a and b b . Factoring both sides, (a b) a b a b a b , by part (a).kk kk ab kkk kkk kk kk kk
## ### #
œœ  Ê 
21. The fact that a a a a a a holds for n 1 is obvious. It also holds fork k kk kk kk
"# " #á Ÿ á œ
nn
n 2 by the triangle inequality. We now show it holds for all positive integers n, by induction.œ
Suppose it holds for n k 1: a a a a a a (this is the inductionœ  á Ÿ ák k kk kk kk
"# " # kk
hypothesis). Then a a a a a a a a a a a akkk kkkkk ab "# "# "#á œ á Ÿ á
kk1 k k1 k k1
(by the triangle inequality) a a a a (by the induction hypothesis) and theŸákk kk kk k k "# kk1
inequality holds for n k 1. Hence it holds for all n by induction.œ

68. 64 Chapter 1 Preliminaries
22. The fact that a a a a a a holds for n 1 is obvious. It holds for n 2k k kk kk kk "# " #á   á œ œ
nn
by Exercise 21(b), since a a a ( a ) a a a a a a .k k k k k k k k kk kkkk k k kk kk "# " # " # "# "#œ  œ  
We now show it holds for all positive integers n by induction.
Suppose the inequality holds for n k 1. Then a a a a a a (this isœ  á   ák k kk kk kk
"# " # kk
the induction hypothesis). Thus a a a a a akkk k abab ""á œ á
kk1 k k1
a a a (by Exercise 21(b)) a a a a a a  á œ á   á kkkkkkkkkkkk kkkkab """ kk1 kk1 kk1
a a a a (by the induction hypothesis). Hence the inequality holds for all ákk kk kk k k"# kk1
n by induction.
23. If f is even and odd, then f( x) f(x) and f( x) f(x) f(x) f(x) for all x in the domain of f.œ œ Ê œ
Thus 2f(x) 0 f(x) 0.œÊ œ24. (a) As suggested, let E(x) E( x) E(x) E is anœÊœ œœÊ
f(x) f( x) f( x) f( ( x)) f(x) f( x)
###
even function. Define O(x) f(x) E(x) f(x) . Thenœ œ œ
f(x) f( x) f(x) f( x)
##
O( x) O(x) O is an odd functionœ œ œ œ Ê
f( x) f( ( x)) f( x) f(x) f(x) f( x)
## #
Š‹
f(x) E(x) O(x) is the sum of an even and an odd function.ʜ
(b) Part (a) shows that f(x) E(x) O(x) is the sum of an even and an odd function. If alsoœ
f(x) E (x) O (x), where E is even and O is odd, then f(x) f(x) 0 E (x) O (x)œ œœ
"" " " "" ab
(E(x) O(x)). Thus, E(x) E (x) O (x) O(x) for all x in the domain of f (which is the same as the œ
""
domain of E E and O O ). Now (E E )( x) E( x) E ( x) E(x) E (x) (since E and E are œœ "" " " " "
even) (E E )(x) E E is even. Likewise, (O O)( x) O ( x) O( x) O (x) ( O(x))œ Ê œ œ "" " " "
(since O and O are odd) (O (x) O(x)) (O O)(x) O O is odd. Therefore, E E and"" "" " œ œ Ê
O O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is,
"
E E and O O, so the decomposition of f found in part (a) is unique.
""œœ
25. y ax bx c a x x c a x cœ œ œ
##
#
Š‹ ˆ‰
bb b b b
a4a 4a 2a 4a
ßß ß
ß
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift
of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward.
Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward.
(b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the
graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the
right.
If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward
to the right. If b 0, decreasing b shifts the graph upward to the left.
(c) Changing c (for fixed a and b) by c shifts the graph upward c units if c 0, and downward c????
units if c 0.?
26. (a) If a 0, the graph rises to the right of the vertical line x b and falls to the left. If a 0, the graphœ
falls to the right of the line x b and rises to the left. If a 0, the graph reduces to the horizontalœ œ
line y c. As a increases, the slope at any given point x x increases in magnitude and the graphœœ kk
!
becomes steeper. As a decreases, the slope at x decreases in magnitude and the graph rises or fallskk !
more gradually.
(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right.
(c) Increasing c shifts the graph upward; decreasing c shifts it downward.
27. If m 0, the x-intercept of y mx 2 must be negative. If m 0, then the x-intercept exceeds œ
"
#
0 mx 2 and x x 0 m 4.Êœ ÊœÊ
""
##
2
m

69. Chapter 1 Additional and Advanced Exercises 65
28. Each of the triangles pictured has the same base
b v t v(1 sec). Moreover, the height of eachœœ?
triangle is the same value h. Thus (base)(height) bh
""
##
œ
A A A . In conclusion, the object sweepsœœœœá
"#$
out equal areas in each one second interval.

29. (a) By Exercise #55 of Section P.2, the coordinates of P are . Thus the slopeˆ‰ˆ‰
a0 b0 a bÒÒ
## ##
ßœß
of OP . œœœ
?
?
y
xa/2a
b/2 b
(b) The slope of AB . The line segments AB and OP are perpendicular when the productœœ
b0 b
0a a


of their slopes is . Thus, b a a b (since both are positive). Therefore, AB" œ œ œ Ê œˆ‰ˆ ‰
bb b
aa a
ß
ß
##
is perpendicular to OP when a b.œ

70. 66 Chapter 1 Preliminaries
NOTES:

71. CHAPTER 2 LIMITS AND CONTINUITY
2.1 RATE OF CHANGE AND LIMITS
1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x)
approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x 1.Ä
(b) 1
(c) 0
2. (a) 0
(b) 1
(c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t)
approaches 1. There is no single number L that f(t) gets arbitrarily close to as t 0.Ä
3. (a) True (b) True (c) False
(d) False (e) False (f) True
4. (a) False (b) False (c) True
(d) True (e) True
5. lim does not exist because 1 if x 0 and 1 if x 0. As x approaches 0 from the left,
x0Ä
xxxxx
xxxxxkk kk kk
œœ œ œ

approaches 1. As x approaches 0 from the right, approaches 1. There is no single number L that all
xx
xxkk kk

the function values get arbitrarily close to as x 0.Ä
6. As x approaches 1 from the left, the values of become increasingly large and negative. As x approaches 1
"
x1
from the right, the values become increasingly large and positive. There is no one number L that all the
function values get arbitrarily close to as x 1, so lim does not exist.Ä
x1Ä
"
x1
7. Nothing can be said about f(x) because the existence of a limit as x x does not depend on how the functionÄ !
is defined at x . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when!
x is close enough to x . That is, the existence of a limit depends on the values of f(x) for x x , not on the! ! near
definition of f(x) at x itself.!
8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of
x0Ä
the value f(0) itself.
9. No, the definition does not require that f be defined at x 1 in order for a limiting value to exist there. If f(1)œ
is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) 5.
x1Ä
œ
10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If
lim f(x) exists, its value may be some number other than f(1) 5. We can conclude nothing about lim f(x),
x1 x1Ä Ä
œ
whether it exists or what its value is if it does exist, from knowing the value of f(1) alone.

72. 68 Chapter 2 Limits and Continuity
11. (a) f(x) x /(x 3)œ*ab
#
x 3.1 3.01 3.001 3.0001 3.00001 3.000001
f(x) 6.1 6.01 6.001 6.0001 6.00001 6.000001


x 2.9 2.99 2.999 2.9999 2.99999 2.999999
f(x) 5.9 5.99 5.999 5.9999 5.99999 5.999999


The estimate is lim f(x) 6.
xÄ$
œ
(b)
(c) f(x) x 3 if x 3, and lim (x 3) 3 3 6.œ œ œ Á œ œ
x9
x3 x3
(x 3)(x 3)


**

xÄ$
12. (a) g(x) x / x 2œ# ab Š‹
È
#
x 1.4 1.41 1.414 1.4142 1.41421 1.414213
g(x) 2.81421 2.82421 2.82821 2.828413 2.828423 2.828426
(b)
(c) g(x) x 2 if x 2, and lim x 2 2 2 2 2." " "' Î ' "'"
x2
x2
x2x2
x2





È
Š‹Š‹
ÈÈ
Š‹
È ÈÈ ÈÈÈÈ
Š‹
xÄ#È
13. (a) G(x) (x 6)/ x 4x 12œ ab
# x 5.9 5.99 5.999 5.9999 5.99999 5.999999
G(x) .126582 .1251564 .1250156 .1250015 .1250001 .1250000


x 6.1 6.01 6.001 6.0001 6.00001 6.000001
G(x) .123456 .124843 .124984 .124998 .124999 .124999

73. Section 2.1 Rates of Change and Limits 69
(b)
(c) G(x) if x 6, and lim 0.125.œ œ œ Á œ œ œ
x6 x6
x 4x12 (x6)(x2) x x2 2 8
**œ œœœ
# 'ab

xÄ'
14. (a) h(x) x 2x 3 / x 4x 3œ abab
##
x 2.9 2.99 2.999 2.9999 2.99999 2.999999
h(x) 2.052631 2.005025 2.000500 2.000050 2.000005 2.0000005
x 3.1 3.01 3.001 3.0001 3.00001 3.000001
h(x) 1.952380 1.995024 1.999500 1.999950 1.999995 1.999999
(b)
(c) h(x) if x 3, and lim 2.œœ œÁ œœœ
x 2x3 x1 x1 31 4
x 4x3 (x3)(x1) x1 x1 31
(x 3)(x 1)



#

xÄ$
15. (a) f(x) x 1 / x 1œ aba bkk
#
x 1.1 1.01 1.001 1.0001 1.00001 1.000001
f(x) 2.1 2.01 2.001 2.0001 2.00001 2.000001

x .9 .99 .999 .9999 .99999 .999999
f(x) 1.9 1.99 1.999 1.9999 1.99999 1.999999

(b)

74. 70 Chapter 2 Limits and Continuity
(c) f(x) , and lim (1 x) 1 ( 1) 2.
x 1, x 0 and x 1
1 x, x 0 and x 1
œœ œœ
"' Ñ Î
œ Á
x
x1
(x 1)(x 1)
x1
(x 1)(x 1)
(x 1)

"





kk

x1Ä
16. (a) F(x) x 3x 2 / 2 xœ abab kk
#
x 2.1 2.01 2.001 2.0001 2.00001 2.000001
F(x) 1.1 1.01 1.001 1.0001 1.00001 1.000001


x 1.9 1.99 1.999 1.9999 1.99999 1.999999
F(x) .9 .99 .999 .9999 .99999 .999999


(b)
(c) F(x) , and lim (x 1) 2 1
, x 0
x 1, x 0 and x 2
œœ œœ
 
œ Á
x3x2
2x
(x 2)(x 1)
x
(x 2)(x )
2x

**

**
#
**œ
*
kk

xÄ#
1.
17. (a) g( ) (sin )/)))œ
.1 .01 .001 .0001 .00001 .000001
g( ) .998334 .999983 .999999 .999999 .999999 .999999
)
)
.1 .01 .001 .0001 .00001 .000001
g( ) .998334 .999983 .999999 .999999 .999999 .999999
)
)

lim g( ) 1
)Ä!
)œ
(b)
18. (a) G(t) (1 cos t)/tœ
#
t .1 .01 .001 .0001 .00001 .000001
G(t) .499583 .499995 .499999 .5 .5 .5
t .1 .01 .001 .0001 .00001 .000001
G(t) .499583 .499995 .499999 .5 .5 .5

lim G(t) 0.5
tÄ!
œ

75. Section 2.1 Rates of Change and Limits 71
(b)
Graph is NOT TO SCALE
19. (a) f(x) xœ
"ÎÐ" Ñx
x .9 .99 .999 .9999 .99999 .999999
f(x) .348678 .366032 .367695 .367861 .367877 .367879
x 1.1 1.01 1.001 1.0001 1.00001 1.000001
f(x) .385543 .369711 .368063 .367897 .367881 .367878
lim f(x) 0.36788
x1Ä
¸
(b)
Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point
(1 2.71820).ß
20. (a) f(x) 3 1 /xœab
x
x .1 .01 .001 .0001 .00001 .000001
f(x) 1.161231 1.104669 1.099215 1.098672 1.098618 1.098612
x .1 .01 .001 .0001 .00001 .000001
f(x) 1.040415 1.092599 1.098009 1.098551 1.098606 1.098611

lim f(x) 1.0986
xÄ!
¸
(b)
21. lim 2x 2(2) 4 22. lim 2x 2(0) 0
xxÄ# Ä!
œœ œœ
23. lim (3x 1) 3 1 0 24. lim
x x1Ä Ä

œ
œ œ œ œˆ‰
"""
#3 3x 1 3(1) 1
1

76. 72 Chapter 2 Limits and Continuity
25. lim 3x(2x 1) 3( 1)(2( 1) 1) 9 26. lim 1
x x1Ä Ä"
œ œ œ œ œ
3x 3
2x1 2(1)1 3
3( 1)



27. lim x sin x sin 28. lim
x
x
Ä
Ä

!
œœ œœœ
111 1
1111###
" "
1
cos x cos
111 1
29. (a) 19 (b) 1
??
??
f289 f 20
x3 1 x1(1)
f(3) f(2) f(1) f( )
œœœ œ œœ
"
# #

30. (a) 0 (b) 2
??
??
g g(1) g( 1) g g(0) g( 2)
x1(1) 2 x0(2)
11 0 4
œœœ œœœ

#

31. (a) (b)
??
?1?1
h114 h
t t
hh hh 03 33
œœœ œœœ
ˆ‰ ˆ‰ ÈȈ‰ ˆ‰
3
44
3
44
6
63









32. (a) (b) 0
?1 ?11
?1 1 1 ?11 1
g g( ) g(0) (2 1) (2 1) g g( ) g( ) (2 1) (2 )
t0 0 t()
2
œœ œ œ œ œ
"
#
33. 1
?
?)
R3R(2) R(0)
20
81 1
œœ œœ

##
"
È È
34. 2 2 0
?
?)
P P(2) P(1) (8 16 10) ( )
21 1
œ œ œœ
"%&

35. (a) Q Slope of PQ œ
?
?
p
t
Q (10 225) 42.5 m/sec"

ߜ
650 225
20 10
Q (14 375) 45.83 m/sec#

ߜ
650 375
20 14
Q (16.5 475) 50.00 m/sec$

ߜ
650 475
20 16.5
Q (18 550) 50.00 m/sec%

ߜ
650 550
20 18
(b) At t 20, the Cobra was traveling approximately 50 m/sec or 180 km/h.œ
36. (a) Q Slope of PQ œ
?
?
p
t
Q (5 20) 12 m/sec"

ߜ
80 20
10 5
Q (7 39) 13.7 m/sec#

ߜ
80 39
10 7
Q (8.5 58) 14.7 m/sec$

ߜ
80 58
10 8.5
Q (9.5 72) 16 m/sec%

ߜ
80 72
10 9.5
(b) Approximately 16 m/sec
37. (a)
(b) 56 thousand dollars per year
?
?
p
t 1994 1992
174 62 112
œœœ

#
(c) The average rate of change from 1991 to 1992 is 35 thousand dollars per year.
?
?
p
t 1992 1991
62 27
œœ


The average rate of change from 1992 to 1993 is 49 thousand dollars per year.
?
?
p
t 1993 1992
111 62
œœ


So, the rate at which profits were changing in 1992 is approximatley 35 49 42 thousand dollars per year.
" #
ab'"

77. Section 2.1 Rates of Change and Limits 73
38. (a) F(x) (x 2)/(x 2)œ
x 1.2 1.1 1.01 1.001 1.0001 1
F(x) 4.0 3.4 3.04 3.004 3.0004 3
5.0; 4.4;
??
??
FF
x1.21 x1.11
4.0 ( 3) 3.4 ( 3)
œœ œœ


4.04; 4. ;
??
??
FF
x1.011 x1.0011
3.04 ( 3) 3.004 ( 3)
œ œ œ œ !!%


4. ;
?
?
F
x 1.0001 1
3. ( 3)
œ œ !!!%
!!!%

(b) The rate of change of F(x) at x 1 is 4.œ
39. (a) 0.414213 0.449489
??
??
g g(2) g(1) 2 g g(1.5) g(1)
x21 1 x1.51 0.5
1.5
œœ¸ œ œ¸
"
#
"
ÈÈ

?
?
g g(1 h) g(1)
x(1h)1 h
1h
œœ


"È
(b) g(x) xœÈ
1 h 1.1 1.01 1.001 1.0001 1.00001 1.000001
1 h 1.04880 1.004987 1.0004998 1.0000499 1.000005 1.0000005
1 h 1 /h 0.4880 0.4987 0.4998 0.499 0.5
'
'

È
Š‹
È 0.5
(c) The rate of change of g(x) at x 1 is 0.5.œ
(d) The calculator gives lim .
hÄ!
œ
È1h
h
" "
#
40. (a) i)
f(3) f(2)
32 1 1 6


"
œœœ
"" !"

36
ii) , T 2
f(T) f(2)
TTTT(T2)T(2T)T
2T 2T
# # # # # #
"
œœ œ œ œÁ
œœ

TTT
2T
(b) T 2.1 2.01 2.001 2.0001 2.00001 2.000001
f(T) 0.476190 0.497512 0.499750 0.4999750 0.499997 0.499999
f(T) f(2) / T 2 0.2381 0.2488 0.2abab 500 0.2500 0.2500 0.2500
(c) The table indicates the rate of change is 0.25 at t 2.œ
(d) lim
TÄ#
ˆ‰
""
#T4
œ
41-46. Example CAS commands:
:Maple
f := x -> (x^4 16)/(x 2);
x0 := 2;
plot( f(x), x x0-1..x0+1, color black,œœ
title "Section 2.1, #41(a)" );œ
limit( f(x), x x0 );œ
In Exercise 43, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be
overcome in Maple by entering the function as f := x -> (surd(x+1, 3) 1)/x.
: (assigned function and values for x0 and h may vary)Mathematica
Clear[f, x]
f[x_]:=(x x 5x 3)/(x 1)
32 2

x0= 1; h = 0.1;
Plot[f[x],{x, x0 h, x0 h}]
Limit[f[x], x x0]Ä

78. 74 Chapter 2 Limits and Continuity
2.2 CALCULATING LIMITS USING THE LIMIT LAWS
1. lim (2x 5) 2( 7) 5 14 5 9 2. lim (10 3x) 10 3(12) 10 36 26
x x1Ä( Ä#
œœœ œ œœ
3. lim x 5x2 (2) 5(2)2 41024
xÄ#
ab œ œ œ
##
4. lim x 2x 4x8 (2) 2(2) 4(2)8 8888 16
xÄ#
ab
$# $ #
œ œ œ
5. lim 8(t 5)(t 7) 8(6 5)(6 7) 8 6. lim 3s(2s 1) 3 2 1 2 1
t sÄ' Ä
œ œ œ œ œ
!
œ
ˆ‰œ ‘ˆ ‰ˆ‰
22 4 2
33 3 3
7. lim 8. lim 2
xxÄ# Ä&
x3 23 5 4 4 4
x6 26 8 x7 57
**
#
œœ œœœ
9. lim
yÄ&
y(5)
5y 5(5) 10
25 5

#

œœœ
10. lim
yÄ#
y2
y5y6 (2)5()6 4106 0 5
22 4 4*
# #
*œ
œœœœ
11. lim 3(2x 1) 3(2( 1) 1) 3( 3) 27
xÄ"
œ œœ
###
12. lim (x 3) ( 4 3) ( 1) 1
xÄ%
œœœ
"*)% "*)% "*)%
13. lim (5 y) [5 ( 3)] (8) (8) 2 16
yÄ$
œœ œ œœ
%Î$ %Î$ %Î$ "Î$ %
% ˆ‰
14. lim (2z 8) (2(0) 8) ( 8) 2
zÄ!
œœœ
"Î$ "Î$ "Î$
15. lim
hÄ!
3333
3h11 3(0) 1 1 11 2ÈÈ È** ** *
œœœ
16. lim
hÄ!
5555
5h42 5(0) 4 2 4 4ÈÈ È** ** #
œœœ
17. lim lim lim lim lim
h0 h0 h0 h0 h0ÄÄ ÄÄÄ
ÈÈÈ
È È
ab
Š‹ Š‹
ÈÈ
3h1 3h1 3h11 3h 1
hh 3h11 3h 1h 3h1 h 3h1
3h 3" " "
** **œ**œ **œ
œ†œœœ
œœ
3
È
œ*œ
$
#
18. lim lim lim lim lim
h0 h0 h0 h0 h0ÄÄ ÄÄÄ
ÈÈÈ
È È
ab
Š‹ Š‹
ÈÈ
5h42 5h42 5h42 5h4 4
hh 5h42 5h42h 5h42 h 5h42
5h 5
** **** **
œ†œœœ
œœ
55
42 4È
*
19. lim lim lim
xx xÄ& Ä& Ä&
x5 x5 1
x 25 (x5)(x5) x5 55 10
""

œœœœ
20. lim lim lim
xx xÄ$Ä $Ä $
x3 x3 1
x 4x3 (x3)(x1) x1 31 2
** œœ

œœœœ
21. lim lim lim (x 2) 7
xx xÄ&Ä &Ä &
x3x 0
x5 x5
(x 5)(x 2)

"
**

œœœ&#œ

79. Section 2.2 Calculating Limits Using the Limit Laws 75
22. lim lim lim (x 5) 2 5 3
xx xÄ# Ä# Ä#
x7x 0
xx2
(x 5)(x 2)

"
#

œ œ œœ
23. lim lim lim
tt tÄ" Ä" Ä"
tt2 t2123
t 1 (t 1)(t 1) t 1 1 1
(t 2)(t 1)



#

œœœœ
24. lim lim lim
tt tÄ"Ä "Ä "
t 3t2 t2 12 1
t t2 (t2)(t1) t2 12 3
(t 2)(t 1)




**
œœœœ
25. lim lim lim
xx xÄ#Ä #Ä #

**
2x 4 2 2 1
x2x x(x2) x 4 2
2(x 2)
œœœœ
26. lim lim lim
y0 yyÄ Ä! Ä!
5y 8y y (5y 8) 5y 8
3y 16y y3y16 3y16 16
8


***
#
"
œœœœ
ab
27. lim lim lim
u1 u1 u1ÄÄ Ä
u 4
u 1 u u1(u1) u u1 111 3
u(u1)(u1) u(u1) (1 1)(1 1)
*


"

" " **
œœœœ
ab ab
ab
28. lim lim lim
vv vÄ# Ä# Ä#
v 8 v 2v4 444 12 3
v16
(v 2) v 2v 4
(v 2)(v 2) v 4 (v 2) v 4 (4)(8) 32 8
#
*






œœœœœ
ab
ab ab
29. lim lim lim
xx xÄ* Ä* Ä*
ÈÈ
ˆ‰ˆ‰ÈÈ È È
x3 x3
x9 6 x3 x3 x3 93


"""
*
œœœœ
30. lim lim lim lim x 2 x 4(2 2) 16
xx x xÄ% Ä% Ä% Ä%
4x x
2x 2x 2x
x(4 x) x2 x 2 x



ÈÈ È
ˆ‰ˆ‰ÈÈ
"" "'"'" ˆ‰È
31. lim lim lim lim x 3
x1 x1 x1 x1ÄÄ Ä Ä
x1
x32 x3 x3
(x1) x32 (x1) x3
(x 3) 4

# #
#
ÈÈÈ
ˆ‰ ˆ‰ÈÈ
ˆ‰ˆ‰
"""''Ð Š‹È
424 "'"
È
32. lim lim lim
xx x1Ä Ä Ä""
œœ
È Š‹Š‹
ÈÈ
Š‹ Š‹
ÈÈ
abx83
x1
x8 x8
(x 1) x 8 (x 1) x 8
x8




*
$ $
***" ***"
*
lim lim œœœœ
x1 x1Ä Ä
(x 1)(x 1)
(x 1) x
x1 2
x 33 3

* *%*"
"
*%*" *
Š‹
È È

33. lim lim lim
x2 x2 x2ÄÄ Ä
È Š‹Š‹
ÈÈ
Š‹ Š‹
ÈÈ
abx124
x2
x124 x124
(x 2) x 12 4 (x 2) x 12 4
x1216








œœ
lim lim œœœœ
x2 x2ÄÄ
(x 2)(x 2)
(x 2) x 12 4
x2 4
x124 16 4 2


*œ
** *Š‹
È ÈÈ

34. lim lim lim
x2 x2 x2Ä Ä Ä
x2
x53
x2 x 53 x2 x 53
x53 x53
x59
*

*** ***

È
ab abŠ‹ Š‹
ÈÈ
Š‹Š‹
ÈÈ ab




œœ
lim lim œœœœ
x2 x2Ä Ä
ab Š‹
È
ÈÈx2 x 53
(x 2)(x 2) x 2 4 2
x53 93 3
***

** *


35. lim lim lim
x3 x3 x3Ä Ä Ä
2x5
x3
2x52x5
(x 3) 2 x 5 (x 3) 2 x 5
4x5
*


È Š‹Š‹
ÈÈ
Š‹ Š‹
ÈÈ
ab




œœ
lim lim lim œœœœœ
x3 x3 x3Ä Ä Ä
9x 3x 6 3
(x 3) 2 x 5 (x 3) 2 x 5
(3 x)(3 x)
2x5 24 2



*


Š‹ Š‹
ÈÈ ÈÈ

80. 76 Chapter 2 Limits and Continuity
36. lim lim lim
x4 x4 x4ÄÄ Ä
4x
5x9
4x5 x 9 4x5 x 9
5x95x9
25 x 9




È
ab abŠ‹ Š‹
ÈÈ
Š‹Š‹
ÈÈ ab




œœ
lim lim lim œœ œœœ
x4 x4 x4ÄÄÄ
ab abŠ‹ Š‹
ÈÈ
ÈÈ4x5 x 9 4x5 x 9
16 x (4 x)(4 x) 4 x 8 4
5x9 525 5


** *



37. (a) quotient rule
(b) difference and power rules
(c) sum and constant multiple rules
38. (a) quotient rule
(b) power and product rules
(c) difference and constant multiple rules
39. (a) lim f(x) g(x) lim f(x) lim g(x) (5)( 2) 10
xc xc xcÄÄÄ
œœœ’“’ “
(b) lim 2f(x) g(x) 2 lim f(x) lim g(x) 2(5)( 2) 20
xc xc xcÄÄÄ
œœœ’“’ “
(c) lim [f(x) 3g(x)] lim f(x) 3 lim g(x) 5 3( 2) 1
xc xc xcÄÄÄ
œ œœ
(d) lim
xcÄ
f(x)
f(x) g(x) lim f(x) lim g(x) 5 ( 2) 7
lim f(x)
55

œœœ
xc
xc xc#
##
40. (a) lim [g(x) 3] lim g(x) lim 3
xxxÄ% Ä% Ä%
œ œ$$œ!
(b) lim xf(x) lim x lim f(x) (4)(0) 0
xxxÄ% Ä% Ä%
œœœ †
(c) lim [g(x)] lim g(x) [ 3] 9
xxÄ% Ä%
##
#
œœœ’“
(d) lim 3
xÄ%
g(x)
f(x) 1 lim f(x) lim 1 0 1
lim g(x)
3


œœœ
x
xx#*

41. (a) lim [f(x) g(x)] lim f(x) lim g(x) 7 ( 3) 4
xb xb xbÄÄÄ
œ œœ
(b) lim f(x) g(x) lim f(x) lim g(x) (7)( 3) 21
xb xb xbÄÄÄ
†œœœ’“’ “
(c) lim 4g(x) lim 4 lim g(x) (4)( 3) 12
xb xb xbÄÄÄ
œœœ’“’ “
(d) lim f(x)/g(x) lim f(x)/ lim g(x)
xb xb xbÄÄÄ
œœœ
77
33
42. (a) lim [p(x) r(x) s(x)] lim p(x) lim r(x) lim s(x) 4 0 ( 3) 1
xxxxÄ#Ä #Ä #Ä #
œ œœ
(b) lim p(x) r(x) s(x) lim p(x) lim r(x) lim s(x) (4)(0)( 3) 0
xxxxÄ#Ä #Ä #Ä #
†† œœœ’“’“’“
(c) lim [ 4p(x) 5r(x)]/s(x) 4 lim p(x) 5 lim r(x) lim s(x) [ 4(4) 5(0)]/ 3
xxxxÄ#Ä #Ä #Ä #
œ œœ ’“ ‚
"6
3
43. lim lim lim lim (2 h) 2
hh hhÄ! Ä! Ä! Ä!
(1 h) 1 h(2 h)
hhh
12hh 1

"""'"
44. lim lim lim lim (h 4) 4
hhhhÄ! Ä! Ä! Ä!
(2 h) (2) h(h 4)
hhh
44hh 4

œœœœ
45. lim lim 3
hhÄ! Ä!
[3(2 h) 4] [3(2) 4]
hh
3h
œœ
46. lim lim lim lim
hhhhÄ! Ä! Ä! Ä!
ˆ‰ˆ‰
"" !

h2h
2
"
#
"
h 2h 2h( h) h(4 2h) 4
2(2h) h
œœ œ œ

81. Section 2.2 Calculating Limits Using the Limit Laws 77
47. lim lim lim
hh hÄ! Ä! Ä!
È È Š‹Š‹
ÈÈ ÈÈ
Š‹ Š‹
ÈÈ ÈÈ
7h 7
h
7h 7 7h 7
h7h 7 h7h 7
(7 h) 7

** **

œœ
lim lim œœœ
hhÄ! Ä!
h
h7h7 7h 7 7Š‹
È È È ÈÈ
**
""
#
48. lim lim lim
hh hÄ! Ä! Ä!
ÈÈ Š‹Š‹
ÈÈ
Š‹ Š ‹
ÈÈ
3(0 h) 1 3(0) 1
h
3h 1 3h 1
h 3h1 h 3h11
(3h 1)
" "
**œ * *
"
œœ
lim lim œœœ
hhÄ! Ä!
3h 3 3
h3h1 3h11Š‹
È È
**œ ** #
49. lim 5 2x 5 2(0) 5 and lim 5 x 5 (0) 5; by the sandwich theorem,
xxÄ! Ä!
ÈÈ ÈÈ ÈÈ
œ œ œ œ## ##
lim f(x) 5
xÄ!
œ
È
50. lim 2 x 2 0 2 and lim 2 cos x 2(1) 2; by the sandwich theorem, lim g(x) 2
xx xÄ! Ä! Ä!
ab œœ œ œ œ
#
51. (a) lim 1 1 1 and lim 1 1; by the sandwich theorem, lim 1
xx xÄ! Ä! Ä!
Š‹œœ œ œ
x0 x sin x
66 22 cos x


(b) For x 0, y (x sin x)/(2 2 cos x)Áœ
lies between the other two graphs in the
figure, and the graphs converge as x 0.Ä

52. (a) lim lim lim 0 and lim ; by the sandwich theorem,
xxx xÄ! Ä! Ä! Ä!
Š‹
"""""
#######
œ œœ œ
x1x
24 4

lim .
xÄ!
1cos x
x
"
#
œ
(b) For all x 0, the graph of f(x) (1 cos x)/xÁœ
#
lies between the line y and the parabolaœ
"
#
y x /24, and the graphs converge as x 0." ?
"
#
#

53. lim f(x) exists at those points c where lim x lim x . Thus, c c c 1 c 0
xc xc xcÄÄÄ
%#%###
œœÊœ ab
c 0, 1, or 1. Moreover, lim f(x) lim x 0 and lim f(x) lim f(x) 1.Êœ œ œ œ œ
xx x1 x1Ä! Ä! Ä Ä
#
54. Nothing can be concluded about the values of f, g, and h at x 2. Yes, f(2) could be 0. Since theœ
conditions of the sandwich theorem are satisfied, lim f(x) 5 0.
xÄ#
œ Á
55. 1 lim lim f(x) 5 2(1) lim f(x) 2 5 7.œ œ œ Ê œ Ê œœ
xxxÄ% Ä% Ä%
f(x) 5
x 2 lim x lim 2
lim f(x) lim 5 lim f(x) 5

%#

xx x
xx#

82. 78 Chapter 2 Limits and Continuity
56. (a) 1 lim lim f(x) 4.œœœÊ œ
xxÄ#Ä #
f(x)
x lim x
lim f(x) lim f(x)

xx
x#'! #'!
#'!
%
(b) 1 lim lim lim lim lim 2.œœ œ Ê œ
xxxx xÄ#Ä #Ä #Ä #Ä #
f(x) f(x) f(x) f(x)
xxxx x
’“’“’“ ˆ‰
""
#
57. (a) 0 3 0 lim lim (x 2) lim (x 2) lim [f(x) 5] lim f(x) 5œ œ œ œ œ †’“’ “’ “ Š‹
xx x x xÄ# Ä# Ä# Ä# Ä#
f(x) 5 f(x) 5
xx

# #
lim f(x) 5.ʜ
xÄ#
(b) 0 4 0 lim lim (x 2) lim f(x) 5 as in part (a).œœ Ê œ†’“’ “
xx xÄ# Ä# Ä#
f(x) 5
x

#
58. (a) 0 1 0 lim lim x lim lim x lim x lim f(x). That is, lim f(xœœ œ œ œ††’“’“’“’“’“
xx xxx x xÄ! Ä! Ä! Ä! Ä! Ä! Ä!
f(x) f(x) f(x)
xxx

#
##
)0.œ
(b) 0 1 0 lim lim x lim x lim . That is, lim 0.œœ œ œ œ††’“’“’“
xxx x xÄ! Ä! Ä! Ä! Ä!
f(x) f(x) f(x) f(x)
xxxx

59. (a) lim x sin 0
xÄ!
"
x
œ

(b) 1 sin 1 for x 0:Ÿ Ÿ Á
"
x
x 0 x x sin x lim x sin 0 by the sandwich theorem;ÊŸ ŸÊ œ
""
xx
xÄ!
x 0 x x sin x lim x sin 0 by the sandwich theorem.Ê   Ê œ
""
xx
xÄ!
60. (a) lim x cos 0
xÄ!
# "ˆ‰
x
#œ

(b) 1 cos 1 for x 0 x x cos x lim x cos 0 by the sandwichŸ Ÿ Á Ê Ÿ Ÿ Ê œˆ‰ ˆ‰ ˆ‰
""" ## # #
xxx
###
xÄ!
theorem since lim x 0.
xÄ!
#
œ
2.3 PRECISE DEFINITION OF A LIMIT
1.
Step 1: x5 x5 5x 5kk Ê Ê$$ $$ $
Step 2: 5 7 2, or 5 1 4.$$$$œ Ê œ œ Ê œ
The value of which assures x 5 1 x 7 is the smaller value, 2.$$ $ kkÊ œ

83. Section 2.3 Precise Definition of a Limit 79
2.
Step 1: x2 x2 x 2kk Ê Ê # $$ $$ $
Step 2: 2 1 1, or 2 7 5.œÊœ œÊœ$$$$
The value of which assures x 2 1 x 7 is the smaller value, 1.$$ $ kkÊ œ
3.
Step 1: x ( 3) x 3 x 3kkÊ$Ê$$ $$ $
Step 2: 3 , or . œ Ê œ $œ Ê œ$$$$
75
## ##
""
The value of which assures x ( 3) x is the smaller value, .$$ $ kk Ê œ
7
## #
""
4.
Step 1: x x x¸¸ˆ‰ Ê Ê
3333
####
$$ $$ $
Step 2: , or 1. œ Ê œ# œ Ê œ$$$$
37 3
## ##
"
The value of which assures x x is the smaller value, .$$ $ ¸¸ˆ‰ Ê œ"
37
###
"
5.
Step 1: x x x¸¸ÊÊ
""""
####
$$ $$ $
Step 2: , or .œÊœ œÊœ$$$$
""""
##
44
918 714
The value of which assures x x is the smaller value, .$$ $ ¸¸Ê œ
""
#
44
97 18
6.
Step 1: x3 x3 3x 3kk Ê Ê$$ $$ $
Step 2: 2.7591 0.2409, or 3.2391 0.2391.$œ Ê œ $œ Ê œ$$$$
The value of which assures x 3 2.7591 x 3.2391 is the smaller value, 0.2391.$$ $ kkÊ œ
7. Step 1: x 5 x 5 5 x 5kk Ê Ê$$ $$ $
Step 2: From the graph, 5 4.9 0.1, or 5 5.1 0.1; thus 0.1 in either case.œÊœ œÊœ œ$$$$$
8. Step 1: x ( 3) x 3 3 x 3kk Ê Ê$$ $$ $
Step 2: From the graph, 3 3.1 0.1, or 3 2.9 0.1; thus 0.1. œ Ê œ œ Ê œ œ$$$$$
9. Step 1: x 1 x 1 1 x 1kk Ê Ê$$ $$ $
Step 2: From the graph, 1 , or 1 ; thus .œÊœ œÊœ œ$$$$$
97 259 7
16 16 16 16 16
10. Step 1: x 3 x 3 3 x 3kk Ê Ê$$ $$ $
Step 2: From the graph, 3 2.61 0.39, or 3 3.41 0.41; thus 0.39.œÊœ œÊœ œ$$$$$
11. Step 1: x 2 x 2 2 x 2kk Ê Ê$$ $$ $
Step 2: From the graph, 2 3 2 3 0.2679, or 2 5 5 2 0.2361;œÊœ¸ œÊœ¸$$ $$
ÈÈ ÈÈ
thus 5 2. $œ
È

84. 80 Chapter 2 Limits and Continuity
12. Step 1: x ( 1) x 1 1 x 1kk Ê Ê$$ $$ $
Step 2: From the graph, 1 0.1180, or 1 0.1340; œ Ê œ ¸ œ Ê œ ¸$$ $$
ÈÈ È È
552 323
## ##

thus . $œ
È
52
#
13. Step 1: x ( 1) x 1 1 x 1kk Ê Ê$$ $$ $
Step 2: From the graph, 1 0.77, or 1 0.36; thus 0.36. œ Ê œ ¸ œ Ê œ œ œ$$$ $
16 7 16 9 9
9 9 25 25 25
14. Step 1: x x x¸¸ÊÊ
""""
####
$$ $$ $
Step 2: From the graph, 0.00248, or 0.00251;œÊœ¸ œÊœ¸$$ $$
""" "
## # #
11 1 1
2.01 2 .01 1.99 1.99
thus 0.00248.$œ
15. Step 1: (x 1) 5 0.01 x 4 0.01 0.01 x 4 0.01 3.99 x 4.01kk kk Ê Ê Ê
Step 2: x 4 x 4 4 x 4 0.01.kk Ê Ê Ê œ$$ $$ $ $
16. Step 1: (2x 2) ( 6) 0.02 2x 4 0.02 0.02 2x 4 0.02 4.02 2x 3.98kkkk Ê Ê Ê
2.01 x 1.99Ê
Step 2: x ( 2) x 2 2 x 2 0.01.kk Ê Ê Ê œ$$ $$ $ $
17. Step 1: x 1 0.1 0.1 x 1 0.1 0.9 x 1 1.1 0.81 x 1 1.21¹¹ÈÈÈ"Ê"ÊÊ
0.19 x 0.21Ê
Step 2: x 0 x . Then, or ; thus, 0.19.kk Ê œ !Þ"* Ê œ !Þ"* œ !Þ#" œ$$ $ $ $ $ $
18. Step 1: x 0.1 0.1 x 0.1 0.4 x 0.6 0.16 x 0.36¸¸ÈÈÈ Ê Ê Ê
""
##
Step 2: x x .¸¸ Ê Ê B
""""
4444
$$ $$ $
Then, 0.16 0.09 or 0.36 0.11; thus 0.09.œÊœ œÊœ œ$$$$$
""
44
19. Step 1: 19 x 19 x 1 2 19 x 4 19 x 16¹¹
ÈÈÈ
$"Ê" $ Ê %Ê
x 19 16 15 x 3 or 3 x 15Ê% Ê
Step 2: x 10 x 10 10 x 10.kkÊÊ$$ $$ $
Then 10 3 7, or 10 15 5; thus 5. œ Ê œ œ Ê œ œ$$$ $$
20. Step 1: x 7 4 1 x 7 1 3 x 7 5 9 x 7 25 16 x 32¹¹
ÈÈÈÊ" %Ê Ê Ê
Step 2: x 23 x 23 23 x 23.kkÊÊ$$ $$ $
Then 23 16 7, or 23 32 9; thus 7.œÊœ œÊœ œ$$$$$
21. Step 1: 0.05 0.05 0.05 0.2 0.3 x or x 5.¸¸
"" "" "
#x4 x4 x 3 3
10 10 10
Ê Ê Ê
Step 2: x 4 x 4 4 x 4.kk Ê Ê$$ $$ $
Then or , or 4 5 or 1; thus .%œ œ œ œ œ$$$$$
10 2 2
33 3
22. Step 1: x 3 .1 0.1 x 3 0.1 2.9 x 3.1 2.9 x 3.1kk
ÈÈ
###
!Ê Ê Ê
Step 2: x 3 x 3 3 x 3.¹¹
ÈÈÈÈ
ÊÊ$$ $$ $
Then 3 2.9 3 2.9 0.0291, or 3 3.1 3.1 3 0.0286;œÊœ¸ œÊœ¸$$ $$
ÈÈ ÈÈ ÈÈ È È
thus 0.0286.$œ

85. Section 2.3 Precise Definition of a Limit 81
23. Step 1: x 4 0.5 0.5 x 4 0.5 3.5 x 4.5 3.5 x 4.5 4.5 x 3.5,kk kk
ÈÈÈ È
###
Ê Ê Ê Ê
for x near 2.
Step 2: x ( 2) x 2 x 2.kk Ê Ê#$$ $$ $
Then 4.5 4.5 0.1213, or 3.5 3.5 0.1292; #œ Ê œ #¸ #œ Ê œ# ¸$$ $$
ÈÈ È È
thus 4.5 2 0.12.$œ¸
È
24. Step 1: ( 1) 0.1 0.1 1 0.1 x or x .¸¸
"""
x x 10x10119911
11 910101010
Ê Ê Ê
Step 2: x ( 1) x 1 x .kk Ê Ê " "$$ $$ $
Then , or ; thus . "œ Ê œ "œ Ê œ œ$$$$$
10 10
9 9 11 11 11
"""
25. Step 1: x 5 11 x 16 1 x 16 1 15 x 17 15 x 17.kkkkab
ÈÈ
## ##
"Ê Ê" Ê Ê
Step 2: x4 x4 x .kk Ê Ê % %$$ $$ $
Then 15 15 0.1270, or 17 17 0.1231; %œ Ê œ% ¸ %œ Ê œ %¸$$ $$
ÈÈ ÈÈ
thus 17 4 0.12.$œ¸
È
26. Step 1: 5 1 4 6 30 x 20 or 20 x 30.¸¸
120 120 120 x
x x x 4 120 6
"Ê" & Ê Ê Ê
""
Step 2: x 24 x 24 24 x 24.kkÊÊ$$ $$ $
Then 24 20 4, or 24 30 6; thus 4.œÊœ œÊœ Êœ$$$$$
27. Step 1: mx 2m 0.03 0.03 mx 2m 0.03 0.03 2m mx 0.03 2m kk Ê Ê Ê
2x2.
0.03 0.03
mm
Step 2: x2 x2 x .kk Ê Ê # #$$ $$ $
Then , or . In either case, .#œ# Ê œ #œ# Ê œ œ$$$$ $
0.03 0.03 0.03 0.03 0.03
mm mm m
28. Step 1: mx 3m c c mx 3m c c 3m mx c 3m 3 x 3kk ÊÊ Ê
cc
mm
Step 2: x3 x3 .kk Ê Ê $B $$$ $$ $
Then , or . In either case, .$œ$ Ê œ $œ$ Ê œ œ$$$$ $
cc cc c
mm mm m
29. Step 1: (mx b) b c mx c c mx c x .¸¸ ˆ‰ -Ê Ê Ê
mmmmcc
mm######
""
Step 2: x x x.¸¸ÊÊ
""""
####
$$ $$ $
Then , or . In either case, .œ Ê œ œ Ê œ œ$$$$ $
"" ""
## ##
cc cc c
mm mm m
30. Step 1: (mx b) (m b) 0.05 0.05 mx m 0.05 0.05 m mx 0.05 mkk Ê Ê
1x . Ê "
0.05 0.05
mm
Step 2: x1 x1 x .kk Ê Ê " "$$ $$ $
Then , or . In either case, ."œ" Ê œ "œ" Ê œ œ$$$$ $
0.05 0.05 0.05 0.05 0.05
mm mm m
31. lim (3 2x) 3 2(3) 3
x3Ä
œ œ
Step 1: 3 2x ( 3) 0.02 0.02 6 2x 0.02 6.02 2x 5.98 3.01 x 2.99 orkkab Ê Ê Ê
2.99 x 3.01.##
Step 2: 0 x 3 x 3 x . Ê Ê $ $kk $$ $$ $
Then 2.99 0.01, or 3.01 0.01; thus 0.01.$œ Ê œ $œ Ê œ œ$$$$$
32. lim ( 3x ) ( 3)( 1) 2 1
x1Ä
#œœ
Step 1: ( 3x 2) 1 0.03 0.03 3x 3 0.03 0.01 x 1 0.01 1.01 x 0.99.kk Ê Ê Ê
Step 2: x ( 1) x 1 x 1.kk Ê Ê"$$ $$ $

86. 82 Chapter 2 Limits and Continuity
Then 1.01 0.01, or 0.99 0.01; thus 0.01. "œ Ê œ "œ Ê œ œ$$$$$
33. lim lim lim (x 2) 4, x 2
xx xÄ# Ä# Ä#
x4
x(x2)
(x 2)(x 2)


#

""'"Ð'Ð"Î
Step 1: 4 0.05 0.05 0.05 3.95 x 2 4.05, x 2¹¹Š‹
x4
x2 (x2)
(x 2)(x 2)




Ê % Ê Á
1.95 x 2.05, x 2.Ê Á
Step 2: x2 x2 x 2.kk Ê Ê # $$ $$ $
Then 1.95 0.05, or 2.05 0.05; thus 0.05.#œ Ê œ #œ Ê œ œ$$$$$
34. lim lim lim (x 1) 4, x 5.
xx xÄ&Ä &Ä &
x6x5
x5 (x5)
(x 5)(x 1)

**
**
**
œœœÁ
Step 1: ( 4) 0.05 0.05 4 0.05 4.05 x 1 3.95, x 5¹¹Š‹
x6x5
x5 (x5)
(x 5)(x )

**
**
**œ
Ê Ê Á
5.05 x 4.95, x 5.Ê Á
Step 2: x ( 5) x 5 x .kk Ê Ê & &$$ $$ $
Then 5.05 0.05, or 4.95 0.05; thus 0.05. &œ Ê œ &œ Ê œ œ$$$$$
35. lim 1 5x 1 5( 3) 16 4
xÄ$
ÈÈ Èœ œ œ
Step 1: 1 5x 4 0.5 0.5 1 5x 4 0.5 3.5 1 5x 4.5 12.25 1 5x 20.25¹¹
ÈÈÈ
ÊÊÊ
11.25 5x 19.25 3.85 x 2.25.ÊÊ
Step 2: x ( 3) x 3 x .kk Ê Ê $ $$$ $$ $
Then 3.85 0.85, or 2.25 0.75; thus 0.75. $œ Ê œ $œ Ê œ$$$ $
36. lim 2
xÄ#
44
x
œœ
#
Step 1: 2 0.4 0.4 2 0.4 1.6 2.4 x or x .¸¸
4 4 4 10x10 10 105 5
x x x 16424 4 6 3 2
Ê Ê Ê Ê
Step 2: x2 x2 x .kk Ê Ê # #$$ $$ $
Then , or ; thus .#œ Ê œ #œ Ê œ œ$$$$$
55
33 3
"""
##
37. Step 1: (9 x) 5 4 x 4 x 4 x 4 x 4 .kk ÊÊÊ%Ê%%% %% % % % % %
Step 2: x4 x4 x .kk Ê Ê % %$$ $$ $
Then 4 4 , or . Thus choose . œ Ê œ %œ %Ê œ œ$ % $% $ % $% $%
38. Step 1: (3x 7) 2 3x 9 9 3x 3 x 3 .kkÊ Ê *Ê%% % % %
%%
33
Step 2: x 3 x 3 3 x 3.kk Ê Ê$$ $$ $
Then 3 , or 3 3 . Thus choose .œ$ Ê œ œ Ê œ œ$$$$ $
%% %% %
33 33 3
39. Step 1: x 5 2 x 5 x 5 ( ) x 5 ( )¹¹
ÈÈ È Ê # Ê# # Ê# #%% % % % % %
##
() x()5.Ê#&#%%
##
Step 2: x 9 x 9 9 x 9.kk Ê Ê$$ $$ $
Then , or . Thus choose *œ % * Ê œ% *œ % * Ê œ% $%% $%%$%% $%%
####
the smaller distance, . $%%œ%
#
40. Step 1: 4 x 2 4 x 4 x ( ) x ( )¹¹
ÈÈ È
Ê # Ê# # Ê# %#%% % % % % %
##
()x4() () x().Ê # # Ê # % # %%%%%
####
Step 2: x 0 x .kkÊ$$ $
Then ( ) 4 , or ( ) 4 4 . Thus choose œ# œ % Ê œ% œ# œ $% %%$%%$%%%
## # # #
the smaller distance, 4 .$%%œ
#

87. Section 2.3 Precise Definition of a Limit 83
41. Step 1: For x 1, x 1 x x 1 x 1Á Ê "Ê" "Ê kk kk
È È
## #
%% % % % % %
x 1 near .Ê " Bœ"
È È%%
Step 2: x1 x1 x .kk Ê Ê " "$$ $$ $
Then 1 1 , or 1 1. Choose"œ Êœ" œ"Êœ"$%$%$%$%
ÈÈ ÈÈ
min 1 1 , that is, the smaller of the two distances.$%%œ"ß"š›
È È
42. Step 1: For x 2, x 4 x 4 4 x 4 4 x 4Á Ê Ê Ê kk kk
È È
## #
%% % % % % %
4 x 4 near 2.Ê BœÈ È
%%
Step 2: x ( 2) x 2 2 x 2.kk Ê Ê$$ $$ $
Then 2 , or . Choose œ % Ê œ % # #œ % Ê œ# %$%$%$%$%ÈÈ ÈÈ
min . $% %œ % #ß# %š›È È
43. Step 1: 1 x .¸¸
"" """
xx x11
Ê " Ê" " Ê %% % % %
%%
Step 2: x1 x1 x .kk Ê Ê" "$$ $ $ $
Then , or ." œ Ê œ" œ " œ Ê œ "œ$$ $$
"" ""
" " " " " "%%%%%%
%%
Choose , the smaller of the two distances.$œ
%
%œ*
44. Step 1: x¸¸
"" "" " " " " $
"$ "$
#
x3 x3 3 x 3 3 x 3
13 1 3 3
ÊÊÊ Ê %% % % %
%%
%%
x , or x for x near .Ê $ÉÉ kkÉÉ È
3333
1$ "$ "$ "$%%%%
Step 2: x3 x3 3 x3.¹¹
ÈÈÈÈ
ÊÊ$$ $ $ $
Then 3 3 , or 3 3.
ÈÈÈ È ÉÉ ÉÉœ Ê œ œ Ê œ $$ $$
3333
"$ "$ "$ "$%%%%
Choose min 3 3 .$œßš›
ÈÈÉ É
33
"$ "$%%
45. Step 1: ( 6) (x 3) 6 , x 3 x 3 x .¹¹Š‹
x
x3

*
*
Ê Á Ê Ê $ $%% % % %% %
Step 2: x ( 3) x 3 x 3.kk Ê Ê$$$ $$ $
Then , or . Choose . $œ $ Ê œ $œ $ Ê œ œ$ % $% $ % $% $%
46. Step 1: 2 (x 1) 2 , x 1 x .¹¹Š‹
x1
x1



Ê ÁÊ""%% % % %
Step 2: x1 x1 x .kk Ê Ê" "$$ $ $ $
Then , or . Choose ." œ" Ê œ " œ" Ê œ œ$ % $% $ % $% $%
47. Step 1: x 1: (4 2x) 2 2 2x since x 1 Thus, 1 x ;l lÊ! Þ !%%
%
#
x 1: (6x 4) 2 6x 6 since x 1. Thus, x 1 . llÊ!Ÿ   "Ÿ%%
%
6
Step 2: x1 x1 x1 .kk Ê Ê"$$ $ $ $
Then 1 , or 1 . Choose . œ" Ê œ " œ Ê œ œ$$$$ $
%% %% %
## 66 6
48. Step 1: x : 2x 0 2x x 0;! Ê !Ê kk %%
%
#
x0: x . !Ê!Ÿ#¸¸
x
#
%%
Step 2: x 0 x .kkÊ$$ $
Then , or . Choose .œ Ê œ œ# Ê œ# œ$ $ $% $% $
%% %
## #
49. By the figure, x x sin x for all x 0 and x x sin x for x 0. Since lim ( x) lim x 0,Ÿ Ÿ     œ œ
""
xx
xxÄ! Ä!

88. 84 Chapter 2 Limits and Continuity
then by the sandwich theorem, in either case, lim x sin 0.
xÄ!
"
x
œ
50. By the figure, x x sin x for all x except possibly at x 0. Since lim x lim x 0, thenŸ Ÿ œ œ œ
## # # # "
x
xxÄ! Ä!
ab
by the sandwich theorem, lim x sin 0.
xÄ!
# "
x
œ
51. As x approaches the value 0, the values of g(x) approach k. Thus for every number 0, there exists a %$!!*
such that x 0 g(x) k .!Ê kk k k$%
52. Write x h c. Then x c x c , x c h c c , h c cœ !l l Í Á Í Á$$ $ $ $ ab
h, h h .Í Á!Í!l !l$$ $
Thus, f x L for any , there exists such that f x L whenever x clim
xcÄ
ab abœ Í ! ! l l !l l%$ % $
f h c L whenever h f h c L.Íl l !l !l Í œab ab %$ lim
hÄ!
53. Let f(x) x . The function values do get closer to 1 as x approaches 0, but lim f(x) 0, not 1. Theœœ
#
xÄ!
function f(x) x never gets to 1 for x near 0.œ
#
arbitrarily
close
54. Let f(x) sin x, L , and x 0. There exists a value of x (namely, x ) for which sin x for anyœœ œ œ
" "
# #
!
1
6
¸¸ %
given 0. However, lim sin x 0, not . The wrong statement does not require x to be arbitrarily close to%!"
xÄ!
"
#
x . As another example, let g(x) sin , L , and x 0. We can choose infinitely many values of x near 0!!
""
# œœ œ
x
such that sin as you can see from the accompanying figure. However, lim sin fails to exist. The
"" "
#x x
œ
xÄ!
wrong statement does not require values of x arbitrarily close to x 0 to lie within 0 of L . Againall !
"
#"!" %
you can see from the figure that there are also infinitely many values of x near 0 such that sin 0. If we
"
x
œ
choose we cannot satisfy the inequality sin for all values of x sufficiently near x 0.%%œ
"""
#
!
4x
¸¸

55. A 0.01 0.01 9 0.01 8.99 9.01 (8.99) x (9.01)kk ˆ‰*ŸÊŸ ŸÊŸŸÊ ŸŸ 1
xx44
4#
#
#1
11

2 x 2 or 3.384 x 3.387. To be safe, the left endpoint was rounded up and the rightÊŸŸ ŸŸÉÉ
8.99 9.01
11
endpoint was rounded down.
56. V RI I 5 0.1 0.1 5 0.1 4.9 5.1 œÊœÊŸ ÊŸŸ Ê ŸŸ Ê  Ê
V V 120 120 10 R 10
R R R R 49 1 0 51
¸¸
#
R 23.53 R 24.48.
(120)(10) (120)(10)
51 49
ŸŸ Ê ŸŸ
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a) x 1 0 x 1 f(x) x. Then f(x) 2 x 2 2 x 2 1 1. That is, Ê" Ê œ œ œœ$$ kkkk
f(x) 2 1 no matter how small is taken when x 1 lim f(x) 2.kk   "Ê Á
"
#
$$
x1Ä

89. Section 2.3 Precise Definition of a Limit 85
(b) 0 x 1 x f(x) x 1. Then f(x) 1 (x 1) 1 x x 1. That is, Ê"" Ê œ œ œ œ$$ kkk kkk
f(x) 1 1 no matter how small is taken when x lim f(x) 1.kk  ""Ê Á $$
x1Ä
(c) x 1 x 1 f(x) x. Then f(x) 1.5 x 1.5 1.5 x 1.5 1 0.5.!Ê"Ê œ œ œœ$$ kkkk
Also, x 1 1 x f(x) x 1. Then f(x) 1.5 (x 1) 1.5 x 0.5! Ê " Ê œ œ œ $$ kkk kkk
x 0.5 0.5 0.5. Thus, no matter how small is taken, there exists a value of x such thatœ " œ $ x 1 but f(x) 1.5 lim f(x) 1.5.   Ê Á$$ kk
"
#
x1Ä
58. (a) For 2 x 2 f(x) 2 f(x) 4 2. Thus for 2, f(x) 4 whenever 2 x 2 no Ê œ Ê œ   $%%$ kk kk
matter how small we choose 0 lim f(x) 4.$Ê Á
xÄ#
(b) For 2 x 2 f(x) 2 f(x) 3 1. Thus for 1, f(x) 3 whenever 2 x 2 no Ê œ Ê œ   $%%$ kk kk
matter how small we choose 0 lim f(x) 3.$Ê Á
xÄ#
(c) For 2 x 2 f(x) x so f(x) 2 x 2 . No matter how small 0 is chosen, x is close to 4 Ê œ œ $$
## #
kkkk
when x is near 2 and to the left on the real line x 2 will be close to 2. Thus if 1, f(x) 2Ê  kk k k
#
%%
whenever 2 x 2 no mater how small we choose 0 lim f(x) 2. Ê Á$$
xÄ#
59. (a) For 3 x 3 f(x) 4.8 f(x) 4 0.8. Thus for 0.8, f(x) 4 wheneverÊ Ê    $%% kk kk
3 x 3 no matter how small we choose 0 lim f(x) 4. Ê Á$$
xÄ$
(b) For 3 x 3 f(x) 3 f(x) 4.8 1.8. Thus for 1.8, f(x) 4.8 whenever 3 x 3 Ê Ê     $%%$ kk kk
no matter how small we choose 0 lim f(x) 4.8.$Ê Á
xÄ$
(c) For 3 x 3 f(x) 4.8 f(x) 3 1.8. Again, for 1.8, f(x) 3 whenever x 3 Ê Ê     $$%%$ kk kk
no matter how small we choose 0 lim f(x) 3.$Ê Á
xÄ$
60. (a) No matter how small we choose 0, for x near 1 satisfying x , the values of g(x) are$$$""
near 1 g(x) 2 is near 1. Then, for we have g(x) 2 for some x satisfyingÊ œ  kk kk %
""
##
x , or x 1 lim g(x) 2."" !Ê Á$$ $ kk
x1Ä
(b) Yes, lim g(x) 1 because from the graph we can find a such that g(x) 1 if x ( 1) .
x1Ä
œ!! $%$ kk k k
61-66. Example CAS commands (values of del may vary for a specified eps):
:Maple
f := x -> (x^4-81)/(x-3);x0 := 3;
plot( f(x), x=x0-1..x0+1, color=black, # (a)
title="Section 2.3, #61(a)" );
L := limit( f(x), x=x0 ); # (b)
epsilon := 0.2; # (c)
plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01,
color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" );
q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d)
delta := abs(x0-q);
plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" );
s in [0.1, 0.005, 0.001 ] do # (e)for ep
q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 );
delta := abs(x0-q);
head := sprintf("Section 2.3, #61(e) epsilon = %5f, delta = %5f", eps, delta );
print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta,
color=black, linestyle=[1,3,3], title=head ));

90. 86 Chapter 2 Limits and Continuity
end do:
(assigned function and values for x0, eps and del may vary):Mathematica
Clear[f, x]
y1: L eps; y2: L eps; x0 1;œ œ œ
f[x_]: (3x (7x 1)Sqrt[x] 5)/(x 1)œ
2
Plot[f[x], {x, x0 0.2, x0 0.2}]
L: Limit[f[x], x x0]π
eps 0.1; del 0.2;œœ
Plot[{f[x], y1, y2},{x, x0 del, x0 del}, PlotRange {L 2eps, L 2eps}]' ? '
2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY
1. (a) True (b) True (c) False (d) True
(e) True (f) True (g) False (h) False
(i) False (j) False (k) True (l) False
2. (a) True (b) False (c) False (d) True
(e) True (f) True (g) True (h) True
(i) True (j) False (k) True
3. (a) lim f(x) , lim f(x)
x xÄ# Ä#
Á '
œ "œ# œ$#œ"
2
#
(b) No, lim f(x) does not exist because lim f(x) lim f(x)
xx xÄ# Ä# Ä#
Á '
Á
(c) lim f(x) 1 3, lim f(x)
x xÄ% Ä%
' Á
"'" "'$"Û
44
##
(d) Yes, lim f(x) 3 because 3 lim f(x) lim f(x)
xx xÄ% Ä% Ä%
œœœ
' Á
4. (a) lim f(x) 1, lim f(x) , f(2) 2
x xÄ# Ä#
Á '
œ œ œ$#œ" œ
2
#
(b) Yes, lim f(x) 1 because lim f(x) lim f(x)
xx xÄ# Ä# Ä#
œ"œ œ
Á '
(c) lim f(x) 3 ( 1) 4, lim f(x) 3 ( 1) 4
x xÄ" Ä"
' Á
œœ œœ
(d) Yes, lim f(x) 4 because 4 lim f(x) lim f(x)
xx xÄ"Ä " Ä"
œœ œ
' Á 5. (a) No, lim f(x) does not exist since sin does not approach any single value as x approaches 0
xÄ!
Á
ˆ‰
"
x
(b) lim f(x) lim 0 0
xxÄ! Ä!
''
œœ
(c) lim f(x) does not exist because lim f(x) does not exist
x xÄ! Ä!
Á
6. (a) Yes, lim g(x) 0 by the sandwich theorem since x g(x) x when x 0Yes, lim g(x) 0
x xÄ!!Ä
Á Á
œ œŸŸ ÈÈ
(b) No, lim g(x) does not exist since x is not defined for x 0
xÄ!
'
È #
(c) No, lim g(x) does not exist since lim g(x) does not exist
xxÄ! Ä!
'

91. Section 2.4 One-Sided Limits and Limits at Infinity 87
7. (a Ñ (b) lim f(x) lim f(x)
x1 x1Ä Ä
' Î
œ"œ
(c) Yes, lim f(x) 1 since the right-hand and left-hand
x1Ä
œ
limits exist and equal 1
8. (a) (b) lim f(x) 0 lim f(x)
x1 x1Ä Ä
Î '
œœ
(c) Yes, lim f(x) 0 since the right-hand and left-hand
x1Ä
œ
limits exist and equal 0
9. (a) domain: 0 x 2ŸŸ
range: 0 y 1 and y 2Ÿ œ
(b) lim f(x) exists for c belonging to
xcÄ
(0 1) ( )ß "ß #
(c) x 2œ
(d) x 0œ

10. (a) domain: x_ _
range: y 1" Ÿ Ÿ
(b) lim f(x) exists for c belonging to
xcÄ
( 1)( )( )_ß "ß " "ß _
(c) none
(d) none

11. lim 3 12. lim 0
x x1Ä!Þ& Ä
' Î
É ÉÉ É ÈÈ
É
x2 0.52 3/2 x 11
x1 0.51 1/2 x
"
# "#
œœœ œœœ!
13. lim (2) 1
xÄ#
Î
ˆ‰ˆ ‰ˆ ‰ ˆ‰ Š‹
x2x5 2
x1 x x ( ) (2)
2( 2) 5
#" # #
"
œœœ
14. lim 1
x1Ä
'
ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰
" " "
#x1 x 7 11 1 7 1 7
x6 3x 16 31 7 2
œœœ
15. lim lim
hhÄ! Ä!
ÎÎ
ÈÈÈ ÈÈÈ
È È
h4h5 5 h4h5 5 h4h5 5
hh h4h5 5


***
œ Š‹Š‹
lim lim œœœœ
hhÄ! Ä!
ÎÎ
ab
Š‹Š‹
ÈÈ ÈÈ ÈÈ È
h4h55
hh4h5 5 hh4h5 5
h(h 4) 04 2
55 5



*** ***
* *
*

92. 88 Chapter 2 Limits and Continuity
16. lim lim
hhÄ! Ä!
''
ÈÈÈÈÈÈ
È È
6 5h 11h 6 6 5h 11h 6 6 5h 11h 6
hh 6 5h 11h 6

***

œ Š‹Š‹ lim lim œœœœ
hhÄ! Ä!
''
6 5h 11h 6
h 6 5h 11h 6 h 6 5h 11h 6
h(5h 11) (0 11)
66 26
11
*** ***

*
ab
Š‹Š‹
ÈÈÈÈ ÈÈ È


17. (a) lim (x 3) lim (x 3) x 2 x 2 for x 2
xxÄ#Ä #
ÎÎ
œ œ
kkx2
x2 (x )
(x 2)*
#
*
abkk
lim (x 3) ( 2) 3 1 œœœ
xÄ#
Î (b) lim (x 3) lim (x 3) x 2 (x 2) for x 2
xxÄ#Ä #
''
œ œ
kkx2
x2 (x )
(x 2)*
#

’“ abkk
lim (x 3)( 1) ( 2 3) 1 œœœ
xÄ#
'
18. (a) lim lim x 1 x 1 for x 1
x1 x1ÄÄ
ÎÎ
ÈÈ
kk
2x (x 1) 2x (x 1)
x1 (x1)


œœ abkk
lim 2x 2 œœ
x1Ä
Î
ÈÈ
(b) lim lim x 1 (x 1) for x 1
x1 x1ÄÄ
''
ÈÈ
kk
2x (x 1) 2x (x 1)
x1 (x1)


œœ abkk
lim 2x 2 œœ
x1Ä
'
ÈÈ
19. (a) lim 1 (b) lim
) )Ä$ Ä$
Î '
ÚÛ ÚÛ) )
) )
œœ œ
3 2
3 3
20. (a) lim t t 4 4 0 (b) lim t t 4 3 1
t tÄ% Ä%
Î '
ab abÚÛœœ ÚÛœœ
21. lim lim 1 (where x 2 )
)Ä! Ä!
sin 2
2
sin x
x
È
È
)
)
œœ œ
x
È
)
22. lim lim lim k lim k 1 k (where kt)
ttÄ! Ä! Ä! Ä!
sin kt k sin kt k sin sin
tkt
œœœ œœ œ
))
))
))
† )
23. lim lim lim lim (where 3y)
yy yÄ! Ä! Ä! Ä!
sin 3y 3 sin 3y sin 3y
4y 4 3y 4 3y 4 4
33sin 3
œœœœ œ
"
)
)
)
)
24. lim lim lim 1 (where 3h)
hh hÄ! Ä! Ä!
'' '
#Ð
h3h
sin 3h 3 sin 3h 3 3 3 3 lim
œœœœœœˆ‰
Π!
"""""""
††
ˆ‰
sin 3h
3h
sin
"

"
" )
25. lim lim lim lim lim 1 2 2
xx x x xÄ! Ä! Ä! Ä! Ä!
tan 2x sin 2x 2 sin 2x
x x x cos 2x cos 2x x
œœ œ œœ
ˆ‰
sin 2x
cos 2x
Š‹Š ‹
"
#
†
26. lim 2 lim 2 lim 2 lim cos t 2 2
tt t tÄ! Ä! Ä! Ä!
2t t t cos t
tan t sin t lim
œœœ œ""œ
ˆ‰
sin t sin t
cos t t Š‹ Œ !
"
t#Ð
††
27. lim lim lim lim 1 (1)
xx xxÄ! Ä! Ä! Ä!
x csc 2x x 2x
cos 5x sin 2x cos 5x sin 2x cos 5x
œœ œœˆ‰ ˆ‰ Š‹Š‹††
"" "" "
###
28. lim 6x (cot x)(csc 2x) lim lim 3 cos x 3 1 3
xxxÄ! Ä! Ä!
#
œœ œ"œ
6x cos x x 2x
sin x sin 2x sin x sin 2x

ˆ‰ †† ††
29. lim lim lim lim
xx x xÄ! Ä! Ä! Ä!
x x cos x x x cos x x x
sin x cos x sin x cos x sin x cos x sin x cos x sin x
*œ
"'" 'ˆ‰ˆ‰ †
lim lim lim (1)(1) 1 2"'"'"
xx xÄ! Ä! Ä!
Š‹ Š‹ ˆ‰
"" "
sin x sin x
xx†
cos x
30. lim lim 0 (1) 0
xxÄ! Ä!
x x sin x x sin x
xx

"" ""
# ### ##
œ œœˆ‰ ˆ‰

93. Section 2.4 One-Sided Limits and Limits at Infinity 89
31. lim lim 1 since 1 cos t 0 as t 0
tÄ! Ä!
sin(1 cos t)
1cos t
sin

"""??
)
)
)
)
32. lim lim 1 since sin h 0 as h 0
hÄ! Ä!
sin (sin h)
sin h
sin
œœœÄÄ
)
)
)
)
33. lim lim lim 1 1
)) )Ä! Ä! Ä!
sin sin 2 sin 2
sin 2 sin 2 sin 2
))) ))
))) ))
œœ œœˆ‰ ˆ ‰††††
## # #
"""
34. lim lim lim 1 1
xx xÄ! Ä! Ä!
sin 5x sin 5x 4x 5 5 sin 5x 4x 5 5
sin 4x sin 4x 5x 4 4 5x sin 4x 4 4
œœ œœˆ‰ˆ‰†† † ††
35. lim lim lim
xx xÄ! Ä! Ä!
tan 3x sin 3x sin 3x 8x 3
sin 8x cos 3x sin 8x cos 3x sin 8x 3x 8
œœ ˆ‰ˆ ‰††††
""
lim 1 1 1œœœ
3 sin 3x 8x 3 3
8 cos 3x 3x sin 8x 8 8
xÄ!
ˆ‰ˆ‰ˆ‰
"
†††
36. lim lim lim
yy yÄ! Ä! Ä!
sin 3y cot 5y sin 3y sin 4y cos 5y sin 3y sin 4y cos 5y 3
y cot 4y y cos 4y sin 5y y cos 4y sin 5y
œœ Š‹Š‹Š‹Š‹
††
††
45y
345y
lim 1111œœœ
yÄ!
Š‹Š‹Š‹Š‹ ˆ‰
sin 3y sin 4y 5y cos 5y
3y 4y sin 5y cos 4y 5 5 5
3 4 12 12†
††††
Note: In these exercises we use the result lim 0 whenever 0. This result follows immediately from
xĄ_
"
x
m
n
mnÎ"!
Example 6 and the power rule in Theorem 8: lim lim lim 0 0.
xx xĄ_ Ą_ Ą_
ˆ‰ ˆ‰ Š‹
"" "
xx x
mn"œœ œœ
mn
mn
mn


37. (a) 3 (b) 3
38. (a) (b) 11
39. (a) (b)
" "
# #
40. (a) (b)
" "
8 8
41. (a) (b)
5 5
3 3
42. (a) (b)
3 3
4 4
43. lim 0 by the Sandwich TheoremŸ Ÿ Ê œ
""
xxx x
sin 2x sin 2x
xÄ_
44. lim 0 by the Sandwich TheoremŸ Ÿ Ê œ
""
333 3
cos cos
))) )
))
)Ä_
45. lim lim 1
ttÄ_ Ä_
2tsin t 010
tcos t 10
1
1

**

*
œœœ
2sin t
tt
cos t
t
ˆ‰
ˆ‰
46. lim lim lim
rr rÄ_ Ä_ Ä_
rsin r 10
2r75 sin r 200
1
25
**œ
#
*

œœœ
ˆ‰
ˆ‰
sin r
r
7sin r
rr
47. (a) lim lim (b) (same process as part (a))
xxÄ_ Ä_
2x 3 2 2
5x 7 5 5
2
5
*
*
*
*
œœ
3
x
7
x
48. (a) lim lim 2
xxÄ_ Ä_
2x 7
xxx7
2
1
#
#
#
""

*

*

œœ
Š‹
7
x
xxx
7
(b) 2 (same process as part (a))

94. 90 Chapter 2 Limits and Continuity
49. (a) lim lim 0 (b) 0 (same process as part (a))
xxÄ_ Ä_
x1
x3 1
*
*
*
*

""

œœ
xx
3
x
50. (a) lim lim 0 (b) 0 (same process as part (a))
xxÄ_ Ä_
3x 7
x2 1
*

*



œœ
37
xx
2
x
51. (a) lim lim (b) 7 (same process as part (a))
xxÄ_ Ä_
7x 7
x3x6x 1
#



œœ(
36
xx
52. (a) lim lim (b) 0 (same process as part (a))
xxÄ_ Ä_
"
x4x1 1
#"
"
#
œœ!
x
4
xx
53. (a) lim lim 0
xxÄ_ Ä_
10xx31
x1
#*

"

**
**
œœ
10 31
x xx
(b) 0 (same process as part (a))
54. (a) lim lim
xxÄ_ Ä_
9x x 9
2x 5x x 6
9
2
*

"
#

"
*
*

*
#
œœ
x
56
xx x
(b) (same process as part (a))
9
#
55. (a) lim lim
xxÄ_ Ä_




2x 2x 3 2
3x 3x 5x 3
2
3
#



œœ
23
xx
35
xx
(b) (same process as part (a))
2
3
56. (a) lim lim 1
xxÄ_ Ä_
"

x
x7x7x9 1
*

*
œœ
77 9
xx x
(b) 1 (same process as part (a))
57. lim lim 0 58. lim lim 1
xx xxÄ_ Ä_ Ä_ Ä_
2x x 2 x
3x 7 3 2x
1
È ÈŠ‹Š‹ Š‹
È
Š‹
* *

* *œ


!"

"


œœ œœ
2 2
x xx
7
x
2
x
59. lim lim lim 1
xx xÄ_Ä _Ä _
# Ð
#Ð
%"ÎÐ& !%"Î#&
%"ÎÐ& !%"Î#&
"

"

ÈÈ
ÈÈ
Š‹
Š‹
xx
xx
1x
1x
1
1

*

*

*
œœœ
x
x60. lim lim
xxÄ_ Ä_
xx
xx
x
1


"

"
*

*

œœ_
x
x
61. lim lim
xxÄ_ Ä_
2xx7
x3xx 1
2x
ÐÎ# "Î#
$ÎÐ
"Î"Ð "
"*Î"Ð $ÎÐ
#ÎÐ ""Î")
"

** * *

È
œœ_
xx
7
3
xx
62. lim lim
xxÄ_Ä _
#

"

"
"Î#
Èx5x3
2x x 4
5
2
5


#
œœ
x
3
x
x
4
x
63. Yes. If lim f(x) L lim f(x), then lim f(x) L. If lim f(x) lim f(x), then lim f(x) does not exi
xa xa xa xa xa xa
ÄÄ ÄÄ ÄÄ
Î Î
' '
œœ œ Á st.
64. Since lim f(x) L if and only if lim f(x) L and lim f(x) L, then lim f(x) can be found by calculating
xc xc
xc xcÄ Ä
Ä Ä
œœœ
Î
'
lim f(x).
xcÄ
Î
65. If f is an odd function of x, then f( x) f(x). Given lim f(x) 3, then lim f(x) .œ œ œ$
x xÄ! Ä!
Î '
66. If f is an even function of x, then f( x) f(x). Given lim f(x) 7 then lim f(x) 7. However, nothingœ œ œ
x xÄ# Ä#
' Î
can be said about lim f(x) because we don't know lim f(x).
x xÄ# Ä#
' Î

95. Section 2.4 One-Sided Limits and Limits at Infinity 91
67. Yes. If lim 2 then the ratio of the polynomials' leading coefficients is 2, so lim 2 as well.
x xÄ_ Ä_
f(x) f(x)
g(x) g(x)
œœ
68. Yes, it can have a horizontal or oblique asymptote.
69. At most 1 horizontal asymptote: If lim L, then the ratio of the polynomials' leading coefficients is L, so
xÄ_
f(x)
g(x)
œ
lim L as well.
xÄ_
f(x)
g(x)
œ
70. lim x x x x lim x x x x lim
xx xÄ_ Ä_ Ä_
ÈÈ ÈÈ
’“’“## ##


œ œ †
È È
ÈÈ ÈÈ
ababxx xx
xx xx xx xx
xx xx


lim lim 1œœœœ
xxÄ_ Ä_
2x 2 2
xx xx 11
11È È
ÉÉ
""

*
xx
71. For any 0, take N 1. Then for all x N we have that f(x) k k k 0 .%%œ œœ kkkk
72. For any 0, take N 1. Then for all y N we have that f(x) k k k 0 .% % œ œœ kkkk
73. I (5 5 ) 5 x . Also, x 5 x 5 x . Choose œß Ê & Ê Ê & œ$$%%%$%
È
## #
lim x 5 0.ʜ
xÄ&
Î
È
74. I ( ) x 4. Also, x x x . Choose œ % ß% Ê % % Ê % Ê % œ$$ % % %$%
È
## #
lim x 0.Ê%œ
xÄ%
'
È
75. As x 0 the number x is always negative. Thus, ( 1) 1 0 which is always?#'##


¹¹ ¸¸
xx
xxkk
%%%
true independent of the value of x. Hence we can choose any 0 with x lim 1.$$!Ê œ
xÄ!
'
x
xkk
76. Since x we have x 2 and x 2 x 2. Then, 0Ä# œ " œ " Ê
*

kk ¹¹ ¸¸
x2 x2
x2 x2kk
%%
which is always true so long as x . Hence we can choose any , and thus x!Ð !* Ð# #Ð' $$
. Thus, lim 1.Ê" œ¹¹
x2 x2
x2 x2

kk kk
%
xÄ#
Î
77. (a) lim x 400. Just observe that if 400 x 401, then x 400. Thus if we choose , we have for any
xÄ %!!
Î
ÚÛœ ÚÛœ œ" $
number that 400 x 400 x 400 400 400 .%$ %! ÊlÚ Û lœl lœ!
(b) lim x 399. Just observe that if 399 x 400 then x 399. Thus if we choose , we have for any
xÄ %!!
'
ÚÛœ ÚÛœ œ" $
number that 400 x 400 x 399 399 399 .%$ %! ÊlÚ Û lœl lœ!
(c) Since lim x lim x we conclude that lim x does not exist.
x xxÄ %!! Ä %!! Ä %!!
Î '
ÚÛÁ ÚÛ ÚÛ
78. (a) lim f(x) lim x 0 0; x 0 x x for x positive. Choose
xxÄ! Ä!
ÎÎ
œ œ œ Ê Ê! œÈÈ ÈÈ ¸¸ %% % % $%
##
lim f(x) 0.ʜ
xÄ!
Î
(b) lim f(x) lim x sin 0 by the sandwich theorem since x x sin x for all x 0.
xxÄ! Ä!
''
œœ ŸŸÁ
####""ˆ‰ ˆ‰
xx
Since x 0 x 0 x whenever x , we choose and obtain x sin 0kkk k kk Èȸ¸ ˆ‰
### # "
œœ œ %%$% %
x
if x 0. $
(c) The function f has limit 0 at x 0 since both the right-hand and left-hand limits exist and equal 0.
!œ
79. lim x sin lim sin 1, 80. lim lim 1,
x 0
xÄ„_ ÄÄ!
Ä_
"" " " "
* *xx 11x
cos
1
cos
œœœ œœœœ
))
) )
)
)) )ˆ‰ ˆ‰
"
"
x
x
'

96. 92 Chapter 2 Limits and Continuity
81. lim lim lim , t
xx t0Ä„_ Ä„_ Ä
3x 4 3 4t 3
2x 5 2 5t x
3
2
**œ
#
*

œœœœ
4
x
5
x
ˆ‰
82. lim lim z 1, z
x
z
Ä_
Ä!
ˆ‰ ˆ ‰
""
"Î
xx
x
z
œœœ
Î
83. lim 3 cos lim (3 2 )(cos ) (3)(1) 3,
x 0Ä„_ Ä
ˆ‰ˆ‰ ˆ‰'"'"""
2
xx x
""
)
)) )
84. lim cos 1 sin lim 3 cos (1 sin ) (0 1)(1 0) 1,
xÄ_
Ä!
ˆ‰ˆ‰ ˆ‰ ab
3
xx x x
œ œœœ
"" "#
)
Î
)) ) )
2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES
1. lim 2. lim
x xÄ! Ä!
Î '
"
3x positive 2x negative
positive positive 5
œ_ œ_ Š‹ Š ‹
3. lim 4. lim
x xÄ# Ä$
' Î
3
x 2 negative x 3 positive
positive positive

"
œ_ œ_ Š‹ Š‹
5. lim 6. lim
x xÄ) Ä&
Î '
2x 3x
x 8 positive 2x 10 negative
negative negative
* *
œ_ œ_ Š‹ Š‹
7. lim 8. lim
x xÄ( Ä!
4
(x 7) positive x (x 1) positive positive
positive negative
*
"
œ_ œ_ Š‹ Š ‹
†
9. (a) lim (b) lim
x xÄ! Ä!
Î '
2 2
3x 3x
"Î# "Î#œ_ œ_
10. (a) lim (b) lim
x xÄ! Ä!
Î '
2 2
x x
"ÎÐ "ÎÐœ_ œ_
11. lim lim 12. lim lim
xx xxÄ! Ä! Ä! Ä!
44
x xx x
ÎÐ
"ÎÐ "Î#
œœ_ œœ_
ab ab
""
13. lim tan x 14. lim sec x
x xÄ Ä
ˆ‰ ˆ‰

!
'
'
!
Î
œ_ œ_
15. lim (1 csc )
)Ä!
'
œ_)
16. lim (2 cot ) and lim (2 cot ) , so the limit does not exist
) )Ä! Ä!
Î '
œ_ œ_))
17. (a) lim lim
xxÄ# Ä#
ÎÎ
"" "
x 4 (x 2)(x 2) positive positive
œœ_ Š‹
†
(b) lim lim
xxÄ# Ä#
''
"" "
x 4 (x 2)(x 2) positive negative
œœ_ Š‹
†
(c) lim lim
xxÄ#Ä #
ÎÎ
"" "
x 4 (x 2)(x 2) positive negative
œœ_ Š‹
†
(d) lim lim
xxÄ#Ä #
''
"" "
x 4 (x 2)(x 2) negative negative
œœ_ Š‹
†
18. (a) lim lim
xxÄ" Ä"
ÎÎ
xx
x 1 (x 1)(x 1) positive positive
positive


œœ_ Š‹
†
(b) lim lim
xxÄ" Ä"
''
xx
x 1 (x 1)(x 1) positive negative
positive


œœ_ Š‹
†
(c) lim lim
xxÄ"Ä "
ÎÎ
xx
x 1 (x 1)(x 1) positive negative
negative


œœ_ Š‹
†

97. Section 2.5 Infinite Limits and Vertical Asymptotes 93
(d) lim lim
xxÄ" Ä"
œœ
xx
x 1 (x 1)(x 1) negative negative
negative
œ

œœ_ Š‹
†
19. (a) lim 0 lim
xxÄ! Ä!

x
x x negative
œ
_œ
"" "
œ œ_ Š‹
(b) lim 0 lim
xxÄ! Ä!
œœ
x
x x positive
œ
_œ
"" "
œ œ_ Š‹
(c) lim 2 2 0
x2Ä
_È
x2
x
œœ

##
""
#
Ä!#$ Ä!#$
œœœ
(d) lim
x1Ä
x3
x1
œ
__œ_
"" "
œ œ ˆ‰
20. (a) lim (b) lim
x xÄ# Ä#
œ
x1 x1
2x 4 positive 2x 4 negative
positive positive
œ œ
œœ

œ_ œ_ Š‹ Š ‹
(c) lim lim 0
xxÄ" Ä"

x1 20
2x 4 2x 4 4
(x 1)(x 1)
œ
œ
#

œœœ
†
(d) lim
xÄ!
œ
x1
2x 4 4
œ
ÄÄ!

œ
21. (a) lim lim
xxÄ! Ä!

x3x2
x 2x x (x 2) positive negative
(x 2)(x 1) negative negative
œ
_œ œ

œœ
œœ
œœ_ Š‹
†
†
(b) lim lim lim , x 2
xx xÄ# Ä# Ä#

x 3x2 x1
x2x x(x2) x 4
(x 2)(x 1)
œ
_œ œ œ
"
œœ
œœ
œœœÁ
(c) lim lim lim , x 2
xx xÄ# Ä# Ä#
œœ œ
x 3x2 x1
x2x x(x2) x 4
(x 2)(x 1)
œ
_œ œ œ
"
œœ
œœ
œœœÁ
(d) lim lim lim , x 2
xx xÄ# Ä# Ä#
x 3x2 x1
x2x x(x2) x 4
(x 2)(x 1)
œ
_œ œ œ
"
œœ
œœ
œœœÁ
(e) lim lim
xxÄ! Ä!
x3x2
x 2x x (x 2) positive negative
(x 2)(x 1) negative negative
œ
_œ œ

œœ
œœ
œœ_ Š‹
†
†
22. (a) lim lim lim
xx xÄ# Ä# Ä#

x3x2
x 4x x(x )(x 2) x(x ) (4) 8
(x 2)(x ) (x 1)
œ
_
" "
###
ÄÄ! Ä
œœœœ
(b) lim lim lim
xx xÄ# Ä# Ä#

x3x2
x 4x x(x )(x 2) x(x ) negative positive
(x 2)(x ) (x 1) negative
œ
_

##
ÄÄ! Ä
œœœ_ Š‹
†
(c) lim lim lim
x0 xxÄ Ä! Ä!
œ œœ
x3x2
x 4x x(x )(x 2) x(x ) negative positive
(x 2)(x ) (x 1) negative
œ
_

##
ÄÄ! Ä
œœœ_ Š‹
†
(d) lim lim lim 0
xx xÄ" Ä" Ä"

x3x2 0
x 4x x(x )(x 2) x(x ) (1)(3)
(x 2)(x ) (x 1)
œ
_

##
ÄÄ! Ä
œœœœ
(e) lim
xÄ!

x
x(x ) positive positive
negativeÄ!
#
œ_ Š‹
†
and lim
xÄ!
œ
x
x(x ) negative positive
negativeÄ!
#
œ_ Š‹
†
so the function has no limit as x 0.Ä
23. (a) lim 2 (b) lim 2
t tÄ! Ä!
œ
œ ‘ œ ‘œ_ œ_
3 3
t t

24. (a) lim 7 (b) lim 7
t tÄ! Ä!
œ
œ ‘ œ ‘
" "
t t
œ_ œ_
25. (a) lim (b) lim
x xÄ! Ä!
œ
’“ ’“
" "
œœx(x1) x(x1)
2 2
œ !Ä" !Ä"
(c) lim (d) lim
x xÄ" Ä"
œ
’“ ’“
" "
œœx(x1) x(x1)
2 2
œ !Ä" !Ä"
26. (a) lim (b) lim
x xÄ! Ä!
œ
’“ ’“
" "
œœx(x1) x(x1)
1 1
œ_ œ_
(c) lim (d) lim
x xÄ" Ä"
œ
’“ ’“
" "
œœx(x1) x(x1)
1 1
œ_ œ_

98. 94 Chapter 2 Limits and Continuity
27. y 28. yœœ
" "
œ x1 x1

29. y 30. yœœ
" œ
# œx4 x3
3

31. y 1 32. yœœ œœ#
x3 2x 2
x2 x x1 x1
"
#

33. y x 1 34. y xœœ œ œ"
x x
x x x1 x1
œ œ
" "
""#

99. Section 2.5 Infinite Limits and Vertical Asymptotes 95
35. y x 36. y xœœ" œœ"
x x
xx x x
œ
% $ " " $
" " # % # # %
2

37. y x 38. y xœœ œœ
x1 x1
xx x x
œ #
œœ
" "

39. Here is one possibility. 40. Here is one possibility.

41. Here is one possibility. 42. Here is one possibility.

100. 96 Chapter 2 Limits and Continuity
43. Here is one possibility. 44. Here is one possibility.

45. Here is one possibility. 46. Here is one possibility.


47. For every real number B 0, we must find a 0 such that for all x, 0 x 0 B. Now, Ê $$ kk
Ä!
x
Ä
B B 0 x x . Choose , then 0 x x !Í Í Í œ Ê
"" """ " #
xx B BB B
ÄÄ kk kk kk
ÈÈ È
$$
B so that lim .Ê œ_
Ä! !
xx
ÄÄ
xÄ!
48. For every real number B 0, we must find a 0 such that for all x, x 0 B. Now, ! Ê $$ kk
"
llx
B x . Choose . Then x 0 x B so that lim .
""" "" "
ll ll llxBB Bx x
!Íl l œ ! Êl l Ê œ_$$ kk
xÄ!
49. For every real number B 0, we must find a 0 such that for all x, 0 x 3 B. Ê $$ kk
Ä
Ä
2
(x 3)
Ä
Now, B B 0 (x 3) . Choose
Ä!
ÄÄ
Ä #22 2 2
(x 3) (x 3) 2 B B B
(x 3)
ÄÄ
Ä ! Í Í Í Í ! B$ kk É
, then 0 x 3 B 0 so that lim .$$œ Ê œ_É kk
222
B (x3) (x3)
ÄÄ
ÄÄ
ÄÄ
xÄ$
50. For every real number B 0, we must find a 0 such that for all x, 0 x ( 5) B. Ê $$ kk
1
(x 5)
Ä
Now, B (x 5) x 5 . Choose . Then 0 x ( 5)
1
(x 5) B BB
# "" "
Ä!Í Í œ kk k k
ÈÈ
$$
x 5 B so that lim .Ê Ê œ_kk
"" "
È
B (x 5) (x 5)
ÄÄ
xÄ&
51. (a) We say that f(x) approaches infinity as x approaches x from the left, and write lim f(x) , if!
xxÄ
!
Ä
œ_
for every positive number B, there exists a corresponding number 0 such that for all x,$
x x x f(x) B.
!! Ê $
(b) We say that f(x) approaches minus infinity as x approaches x from the right, and write lim f(x) ,
!
xxÄ
!

œ_
if for every positive number B (or negative number B) there exists a corresponding number 0 such $
that for all x, x x x f(x) B.
!! Ê $

101. Section 2.5 Infinite Limits and Vertical Asymptotes 97
(c) We say that f(x) approaches minus infinity as x approaches x from the left, and write lim f(x) ,!
xxÄ
!
Ä
œ_
if for every positive number B (or negative number B) there exists a corresponding number 0 such $
that for all x, x x x f(x) B.
!! Ê $ 52. For B 0, B 0 x . Choose . Then x 0 x B so that lim .Í œ ! Ê Ê œ_
""" "" "
xBB Bx x
$$
xÄ!

53. For B 0, B 0 B 0 x x. Choose . Then x Í Í Í œ !
"" "" "
xx BBB
$$
x B so that lim .Ê Ê œ_
"" "
Bx x
xÄ!
Ä
54. For B , B B (x 2) x 2 x 2 . Choose . Then! Í Í Í Í œ
"" """"
Ä" Ä"xx BBBB
$
2 x 2 x 2 x 2 0 B 0 so that lim . Ê !Ê Ê œ_$$
"" "
Ä" Ä "Bx x
xÄ#
Ä
55. For B 0, B x 2 . Choose . Then x x x 2 Í! œ # # Ê! # Ê!
""" "
Ä"xBB B
$$$
B so that lim .Ê ! œ_
""
Ä" Ä"xx
xÄ#

56. For B 0 and x 1, B 1 x ( x)( x) . Now 1 since x 1. Choose! Í Í""
"" "
Ä #
#
1x B B
1x
Ä
. Then x x 1 0 x ( x)( x)$$$ $ " "Ê Ê" Ê " "
" """
# ##B BBB
1x
ˆ‰
B for x 1 and x near 1 lim .Ê! Ê œ_
""
Ä!Ä1x x
Ä Ä
xÄ"
Ä
57. y sec x 58. y sec xœ œ
" "
x x
Ä

102. 98 Chapter 2 Limits and Continuity
59. y tan x 60. y tan xœ œ
""
xx
Ä

61. y 62. yœœ
x
4x 4x
È È
Ä Ä
Ä!
Ä Ä

63. y x 64. y sinÄ! Ä
#Î$ "
x x1
Ä
ˆ‰
1

2.6 CONTINUITY
1. No, discontinuous at x 2, not defined at x 2œœ
2. No, discontinuous at x 3, lim g(x) g(3) 1.5œ"œ Á œ
xÄ$
Ä

103. Section 2.6 Continuity 99
3. Continuous on [ 1 3]ß
4. No, discontinuous at x 1, 1.5 lim k(x) lim k(x)œœÁœ!
x xÄ" Ä"
ß
5. (a) Yes (b) Yes, lim f(x) 0
x?B
ß
œ
(c) Yes (d) Yes
6. (a) Yes, f(1) 1 (b) Yes, lim f(x) 2œœ
x1Ä
(c) No (d) No
7. (a) No (b) No
8. [ )()()()"ß! !ß" "ß# #ß$
9. f(2) 0, since lim f(x) 2(2) 4 0 lim f(x)œœœœ
x xÄ# Ä#
ß
10. f(1) should be changed to 2 lim f(x)œ
x1Ä
11. Nonremovable discontinuity at x 1 because lim f(x) fails to exist ( lim f(x) 1 and lim f(x) 0).œœœ
x1 x xÄ Ä" Ä"
ß
Removable discontinuity at x 0 by assigning the number lim f(x) 0 to be the value of f(0) rather thanœœ
xÄ!
f(0) 1.œ
12. Nonremovable discontinuity at x 1 because lim f(x) fails to exist ( lim f(x) 2 and lim f(x) 1).œœœ
x1 x xÄ Ä" Ä"
ß
Removable discontinuity at x 2 by assigning the number lim f(x) 1 to be the value of f(2) rather thanœœ
xÄ#
f(2) 2.œ
13. Discontinuous only when x 2 0 x 2 14. Discontinuous only when (x 2) 0 x 2œÊœ œÊœ
#
15. Discontinuous only when x x (x 3)(x 1) 0 x 3 or x 1
#
% $œ! Ê œ Ê œ œ
16. Discontinuous only when x 3x 10 0 (x 5)(x 2) 0 x 5 or x 2
#
œÊ œÊœ œ
17. Continuous everywhere. ( x 1 sin x defined for all x; limits exist and are equal to function values.)kk
18. Continuous everywhere. ( x 0 for all x; limits exist and are equal to function values.)kk"Á
19. Discontinuous only at x 0œ
20. Discontinuous at odd integer multiples of , i.e., x = (2n ) , n an integer, but continuous at all other x.
11
##
"
21. Discontinuous when 2x is an integer multiple of , i.e., 2x n , n an integer x , n an integer, but11 œÊœ
n1
#
continuous at all other x.
22. Discontinuous when is an odd integer multiple of , i.e., (2n 1) , n an integer x 2n 1, n an
1111xx
####
œ Êœ
integer (i.e., x is an odd integer). Continuous everywhere else.

104. 100 Chapter 2 Limits and Continuity
23. Discontinuous at odd integer multiples of , i.e., x = (2n 1) , n an integer, but continuous at all other x.
11
##

24. Continuous everywhere since x 1 1 and sin x 1 0 sin x 1 1 sin x 1; limits exist
%##
  "Ÿ Ÿ Ê Ÿ Ÿ Ê  
and are equal to the function values.
25. Discontinuous when 2x 3 0 or x continuous on the interval . Ê ß_
33
##
‰
26. Discontinuous when 3x 1 0 or x continuous on the interval . Ê ß_
""
33
‰
27. Continuous everywhere: (2x 1) is defined for all x; limits exist and are equal to function values.
"Î$
28. Continuous everywhere: (2 x) is defined for all x; limits exist and are equal to function values.
"Î&
29. lim sin (x sin x) sin ( sin ) sin ( 0) sin 0, and function continuous at x .
xÄ1
œœœœ œ11 1 1 1
30. lim sin cos (tan t) sin cos (tan (0)) sin cos (0) sin 1, and function continuous at t .
tÄ!
ˆ‰ˆ ‰ˆ‰ˆ‰
11 11
## ##
œœœœ œ!
31. lim sec y sec y tan y 1 lim sec y sec y sec y lim sec (y 1) sec y sec ( ) sec 1
y1 y1 y1ÄÄÄ
a ba babab
## ## # #
œ œ œ""
sec 0 1, , and function continuous at y .œœ œ"
32. lim tan cos sin x tan cos (sin(0)) tan cos (0) tan 1, and function continuous at x .
xÄ!
%%#?#?ˆ‰
1111
4444
"Î$
œœœœ œ!
33. lim cos cos cos cos , and function continuous at t .
tÄ!
’“’“
1111
ÈÈÈ
È
19 3 sec 2t 19 3 sec 0 16 4
2
#
œœœœ œ!
34. lim csc x 5 3 tan x csc 5 3 tan 4 5 3 9 3, and function continuous at
xÄ

!
É ÈÈÈÈ É ˆ‰ ˆ‰
Ê Š‹## "
Êa Ê aÊ aa
11
66 3
È
x.œ
1
'
35. g(x) x 3, x 3 g(3) lim (x 3) 6aa aÊbŠa Êa
x9
x3 (x3)
(x 3)(x 3)




xÄ$
36. h(t) t 5, t h(2) lim (t 5) 7œœ œÁ#Êœœ
t3t10
tt
(t 5)(t 2)


# #

tÄ#
37. f(s) , s 1 f(1) lim œœ œ ÁÊœ œ
sss ss13
s 1 (s 1)(s 1) s 1 s 1
ss1(s1)



" "
#
ab
s1Ä
Š‹
38. g(x) , x 4 g(4) lim œœ œÁÊœ œ
x16 x4 x4 8
x 3x4 (x4)(x1) x1 x1 5
(x 4)(x 4)





xÄ%
ˆ‰
39. As defined, lim f(x) (3) 1 8 and lim (2a)(3) 6a. For f(x) to be continuous we must have
x xÄ$ Ä$
#
œœ œ
#
6a 8 a .œÊœ
4
3
40. As defined, lim g(x) 2 and lim g(x) b( 2) 4b. For g(x) to be continuous we must have
x x?ˆ ?ˆ
#
œ œ œ
#
4b 2 b .œ Ê œ
"
#

105. Section 2.6 Continuity 101
41. The function can be extended: f(0) 2.3. 42. The function cannot be extended to be continuous at¸
x 0. If f(0) 2.3, it will be continuous from the œ¸
right. Or if f(0) 2.3, it will be continuous from the ¸
left.

43. The function cannot be extended to be continuous 44. The function can be extended: f(0) 7.39.¸
at x 0. If f(0) 1, it will be continuous fromœœ
the right. Or if f(0) 1, it will be continuousœ
from the left.

45. f(x) is continuous on [ ] and f(0) 0, f(1) 0‘ˆ ’ ¹ š
by the Intermediate Value Theorem f(x) takesÊ
on every value between f(0) and f(1) theÊ
equation f(x) 0 has at least one solution betweenœ
x 0 and x 1.œœ

46. cos x x (cos x) x 0. If x , cos 0. If x , cos 0. Thus cos x x 0œÊ œ œ œ œ
111 111
### ###
ˆ‰ˆ‰ ˆ‰
for some x between and according to the Intermediate Value Theorem.
11
##
47. Let f(x) x 15x 1 which is continuous on [ 4 4]. Then f( 4) 3, f( 1) 15, f(1) 13, and f(4) 5.œ ß œ œ œ œ
$
By the Intermediate Value Theorem, f(x) 0 for some x in each of the intervals x 1, x 1, andœ%"
x 4. That is, x 15x 1 0 has three solutions in [ 4]. Since a polynomial of degree 3 can have at most 3" œ %ß
$
solutions, these are the only solutions.
48. Without loss of generality, assume that a b. Then F(x) (x a) (x b) x is continuous for all values ofœ
##
x, so it is continuous on the interval [a b]. Moreover F(a) a and F(b) b. By the Intermediate Valueßœœ
Theorem, since a b, there is a number c between a and b such that F(x) .¹¹ a
ab abß ß
# #

106. 102 Chapter 2 Limits and Continuity
49. Answers may vary. Note that f is continuous for every value of x.
(a) f(0) 10, f(1) 1 8(1) 10 3. Since 10, by the Intermediate Value Theorem, there exists a cœœœ $
$
1
so that c 1 and f(c) .‘¹ ¹ a 1 (b) f(0) 10, f( 4) ( 4) 8( 4) 10 22. Since 22 3 10, by the Intermediate Valueœ œ œ
$ È
Theorem, there exists a c so that 4 c 0 and f(c) 3. œ
È
(c) f(0) 10, f(1000) (1000) 8(1000) 10 999,992,010. Since 10 5,000,000 999,992,010, by theœœœ
$
Intermediate Value Theorem, there exists a c so that c 1000 and f(c) 5,000,000.‘¹ ¹ a
50. All five statements ask for the same information because of the intermediate value property of continuous
functions.
(a) A root of f(x) x 3x 1 is a point c where f(c) 0.œ œ
$
(b) The points where y x crosses y 3x 1 have the same y-coordinate, or y x 3x 1œœ œœ
$ $
f(x) x 3x 1 0.Êœœ
$
(c) x 3x 1 x 3x 1 0. The solutions to the equation are the roots of f(x) x 3x 1.
$$ $
œÊ œ œ
(d) The points where y x 3x crosses y 1 have common y-coordinates, or y x 3x 1œ œ œœ
$ $
f(x) x 3x 1 .Êœœ!
$
(e) The solutions of x 3x 1 0 are those points where f(x) x 3x 1 has value 0.
$$
œ œ
51. Answers may vary. For example, f(x) is discontinuous at x 2 because it is not defined there.œœ
sin (x 2)
x2


However, the discontinuity can be removed because f has a limit (namely 1) as x 2.Ä
52. Answers may vary. For example, g(x) has a discontinuity at x 1 because lim g(x) does not exist.œœ
"
ßx1
x?B
lim g(x) and lim g(x) .Š‹
x x?B ?B
ß
œ_ œ_
53. (a) Suppose x is rational f(x ) 1. Choose . For any 0 there is an irrational number x (actually!!
"
# Ša a š %$
infinitely many) in the interval (x x ) f(x) 0. Then 0 x x but f(x) f(x )
!! ! !ß Ê œ $$ $ kk k k
1 , so lim f(x) fails to exist f is discontinuous at x rational.aša Š
"
#
!
%
xxÄ Á
On the other hand, x irrational f(x ) 0 and there is a rational number x in (x x ) f(x)!! !! Êœ ßÊ $$
1. Again lim f(x) fails to exist f is discontinuous at x irrational. That is, f is discontinuous atœÊ
xxÄ Á
!
every point.
(b) f is neither right-continuous nor left-continuous at any point x because in every interval (x x ) or
!!! ß$
(x x ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and
!!ß $
xxÄ
Á

lim f(x) exist by the same arguments used in part (a).
xxÄ
Á
ß
54. Yes. Both f(x) x and g(x) x are continuous on [ ]. However is undefined at x sinceœœ !ß" œ
" "
# #
f(x)
g(x)
g 0 is discontinuous at x .ˆ‰
""
##
œÊ œ
f(x)
g(x)
55. No. For instance, if f(x) 0, g(x) x , then h(x) 0 x 0 is continuous at x 0 and g(x) is not.œœÜÝ œÜÝœ œ ab
56. Let f(x) and g(x) x 1. Both functions are continuous at x 0. The composition f g f(g(x))œœ œ ‰œ
"
x1
is discontinuous at x 0, since it is not defined there. Theorem 10 requires that f(x) beœœ œ
""
(x 1) 1 x
continuous at g(0), which is not the case here since g(0) 1 and f is undefined at 1.œ
57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to
equal zero at some point between a and b since f is continuous on [a b].ß

107. Section 2.7 Tangents and Derivatives 103
58. Let f(x) be the new position of point x and let d(x) f(x) x. The displacement function d is negative if x isœ
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the
Intermediate Value Theorem, d(x) 0 for some point in between. That is, f(x) x for some point x, which isœœ
then in its original position.
59. If f(0) 0 or f(1) 1, we are done (i.e., c 0 or c 1 in those cases). Then let f(0) a 0 and f(1) b 1aa aa aš a¹
because 0 f(x) 1. Define g(x) f(x) x g is continuous on [0 1]. Moreover, g(0) f(0) 0 a 0 andŸŸ œÊ ß œœ
g(1) f(1) 1 b 1 0 by the Intermediate Value Theorem there is a number c in ( ) such thatœ œ Ê !ß"
g(c) 0 f(c) c 0 or f(c) c.œÊ œ œ
60. Let 0. Since f is continuous at x c there is a 0 such that x c f(x) f(c)%$$%œ œ Ê
kkf(c)
#
kk k k
f(c) f(x) f(c) .Ê %%
If f(c) 0, then f(c) f(c) f(x) f(c) f(x) 0 on the interval (c c ).œÊÊ ß%$$
""
## #
3
If f(c) 0, then f(c) f(c) f(x) f(c) f(x) 0 on the interval (c c ).œÊÊ ß%$$
""
## #
3
61. By Exercises 52 in Section 2.3, we have lim f x L lim f c h L.
xc h0Ä Ä
ab a bœÍ œ
Thus, f x is continuous at x c lim f x f c lim f c h f c .ab ab ab a b abœÍ œ Í œ
xc h0Ä Ä
62. By Exercise 61, it suffices to show that lim sin c h sin c and lim cos c h cos c.
h0 h0ÄÄ
ab abÊa Êa
Now lim sin c h lim sin c cos h cos c sin h sin c lim cos h cos c lim sin h
h0 h0 h0 h0ÄÄ Ä Ä
a b aba ba babab a b‘ Š‹Š‹Êa Ê a Ê
By Example 6 Section 2.2, lim cos h and lim sin h . So lim sin c h sin c and thus f x sin x is
h0 h0 h0ÄÄÄ
a’ a‘ Ê a a ab ab
continuous at x c. Similarly,œ
lim cos c h lim cos c cos h sin c sin h cos c lim cos h sin c lim sin h cos c.
h0 h0 h0 h0ÄÄ Ä Ä
a b ababababab ab‘ Š‹Š‹œ œ œ
Thus, g x cos x is continuous at x c.abœœ
63. x 1.8794, 1.5321, 0.3473 64. x 1.4516, 0.8547, 0.4030¸ ¸
65. x 1.7549 66. x 1.5596¸¸
67. x 3.5156 68. x 3.9058, 3.8392, 0.0667¸¸
69. x 0.7391 70. x 1.8955, 0, 1.8955¸¸
2.7 TANGENTS AND DERIVATIVES
1. P : m 1, P : m 5 2. P : m 2, P : m 0
"" ## "" ##œœ œœ

108. 104 Chapter 2 Limits and Continuity
3. P : m , P : m 4. P : m 3, P : m 3"" ## "" ##
##
" œœ œœ
5
5. m lim œ
hÄ!
cdab4( h) 4(1)
h
"

lim lim 2;œœœ
hhÄ! Ä!
#ab12hh 1
hh
h( h)

at ( ): y (x ( 1)) y 2x 5,"ß $ œ $ # Ê œ
tangent line

6. m lim lim œœ
hhÄ! Ä!
cdcd(1h1) 1 ( ) 1
hh
h ""

lim h 0; at ( ): y 1 0(x 1) y 1,œœ"ß"œÊœ
hÄ!
tangent line

7. m lim lim œœ
hhÄ! Ä!
21h21 21h2 21h2
hh 21 h
ÈÈÈ È
È

ßß&
†
lim lim 1;œœœ
hhÄ! Ä!
4(1 h) 4
2h 1h1
2
1h1

ßß ßߊ‹
È È
at ( ): y 2 1(x 1) y x 1, tangent line"ß # œ Ê œ

8. m lim lim œœ
hhÄ! Ä!
œœ
Á! Áœ



(1 h) ( )


hh(1h)
1(1h)
lim lim 2;œœœ
hhÄ! Ä!


ab2h h
h(1h) (1h)
2h


at ( ): y 1 2(x ( 1)) y 2x 3,"ß " œ Ê œ
tangent line

109. Section 2.7 Tangents and Derivatives 105
9. m lim lim œœ
hhÄ! Ä!
(2 h) (2)
hh
8 12h 6h h 8
ˆˆ
lim 12 6h h 12;œœ
hÄ!
ab
#
at ( 2 8): y 8 12(x ( 2)) y 12x 16,ß œ Ê œ
tangent line

10. m lim lim œœ
hhÄ! Ä!
aa

ˆˆ
ˆ
ˆ
(h) ()

#
#h 8h( h)
8( h)
lim lim œœ
hhÄ! Ä!

#
ab12h 6h h
8h( h) 8( 2 h)
12 6h h
ˆ
ˆˆ

;œœ
"

23
8( 8) 16
at : y (x ( 2))ˆ‰#ß œ
""
8816
3
y x , tangent lineʜ
3
16
"
#

11. m lim lim lim ;œœœœ%
hhhÄ! Ä! Ä!
cd ab(2 h) 1 5
hhh
54hh 5 h(4 h) ˆ

at (2 5): y 5 4(x 2), tangent lineߜ
12. m lim lim lim 3;œœœœ
hhhÄ! Ä! Ä!
cd ab(h)2(1h) (1)
hhh
1h24h2h 1 h( 3 2h)"

at ( ): y 1 3(x 1), tangent line"ß " œ
13. m lim lim lim 2;œœ œœ
hh hÄ! Ä! Ä!
3h
(3 h) 2‘
‘b


ˆˆ

3
h h(h 1) h(h 1)
(3 h) 3(h 1) 2h
at ( ): y 3 2(x 3), tangent line$ß $ œ
14. m lim lim lim lim 2;œœ œ œ œœ
hh h hÄ! Ä! Ä! Ä!
8
(2 h)
‘






ˆˆˆ

2
h h(2 h) h(2 h) h(2 h) 4
8 2(2 h) 2h(4 h) 8244hh 8ab
at (2 2): y 2 2(x 2)ߜ
15. m lim lim lim 12;œœ œ œ
hh hÄ! Ä! Ä!
(2 h) 8
hh h
8 12h 6h h 8 h 12 6h h
ˆ
ab ab
at (2 ): y 8 12(t 2), tangent lineß) œ
16. m lim lim lim 6;œœ œœ
hh hÄ! Ä! Ä!
cd abab(1 h) 3(1 h) 4
hhh
13h3h h 33h 4 h63hh
ˆ
at ( ): y 4 6(t 1), tangent line"ß % œ
17. m lim lim lim lim œœ œœœ
hh h hÄ! Ä! Ä! Ä!
ÈÈÈ
ÈÈ
Š‹ Š‹
ÈÈ
4h2 4h2 4h2
hh 4h2
(4 h) 4
h4h h4h
h
4

ˆˆ

ˆˆ# ˆˆ#
"
ˆ#
†
; at ( ): y 2 (x 4), tangent lineœ%ß#œ
""
44
18. m lim lim lim lim œœ œœ
hh hhÄ! Ä! Ä! Ä!
ÈÈÈ
È
Š‹ Š‹
ÈÈ
(8 h) 1 3
hh
9h3 9h3
9h3
(9 h) 9
h 9h3 h 9h3
h
ˆˆ

ˆˆ ˆˆ
†
; at (8 3): y 3 (x 8), tangent lineœœ ßœ
"" "
ˆ
È
93 66
19. At x 1, y 5 m lim lim lim 10, slopeœ œ Ê œ œ œ œ
hh hÄ! Ä! Ä!
5( h) 5 5h( 2 h)
hhh
51 2h h 5"

ab

110. 106 Chapter 2 Limits and Continuity
20. At x 2, y 3 m lim lim lim 4, slopeœœÊœ œ œ œ
hhhÄ! Ä! Ä!
cd ab1(2h) (3)
hhh
144h h 3 h(4 h)

21. At x 3, y m lim lim lim , slopeœœÊœ œ œ œ
" "
#ˆˆ


hhhÄ! Ä! Ä!
aa

(3 h) 1
h 2h(2 h) 2h(2 h) 4
2(2h) h
22. At x 0, y 1 m lim lim lim 2, slopeœœÊœ œ œ œ
hh hÄ! Ä! Ä!
h1
h1b
‘
"
ˆˆ
(1)
h h(h 1) h(h 1)
(h 1) (h ) 2h
23. At a horizontal tangent the slope m 0 0 m lim œÊœœ
hÄ!
cdab(x h) 4(x h) 1 x 4x 1
h


lim lim lim (2x h 4) 2x 4;œœœœ
hhhÄ! Ä! Ä!
abababx2xhh4x4h1 x4x1 2xhh4h
hh


2x 4 0 x 2. Then f( 2) 4 8 1 5 ( 2 5) is the point on the graph where there is aœ Ê œ œœÊß
horizontal tangent.
24. 0 m lim lim œœ œ
hhÄ! Ä!
cdab abab(x h) 3(x h) x 3x
hh
x3xh3xhh3x3h x3x
ˆˆ
lim lim 3x 3xh h 3 3x 3; 3x 3 0 x 1 or x 1. ThenœœœœÊœœ
hhÄ! Ä!
3x h 3xh h 3h
h

####
ab
f( 1) 2 and f(1) 2 ( 2) and ( 2) are the points on the graph where a horizontal tangent exists. œ œ Ê "ß "ß
25. 1 m lim lim lim œ œ œ œ œ
hh hÄ! Ä! Ä!
aa
‘b b

(x h) 1 x 1



"
h h(x 1)(x h 1) h(x 1)(x h 1) (x 1)
(x 1) (x h 1) h
(x 1) 1 x 2x 0 x(x 2) 0 x 0 or x 2. If x 0, then y 1 and m 1Ê œÊ œÊ œÊœ œ œ œ œ
##
y 1 (x 0) (x 1). If x 2, then y 1 and m 1 y 1 (x 2) (x 3).Êœœ œ œ œÊœœ
26. m lim lim lim
"
ˆˆ

ˆˆ
4hh
xh x xh x xh x
xh x
(x h) x
hxh x
œœ œ œ
hh hÄ! Ä! Ä!
ÈÈÈ ÈÈÈ
È È
Š‹
È È
† lim . Thus, x 2 x 4 y 2. The tangent line isœ œ œ Ê œÊœÊœ
hÄ!
h
hxh x
xx 4
Š‹
È È
ÈÈ
ˆˆ
"""
## È
y 2 (x 4) 1.œ œ
"
44
x
27. lim lim lim
hh hÄ! Ä! Ä!
f(2 h) f(2)
hh h
100 4.9( h) 100 4.9(2) 4.9 4 4h h 4.9(4) #
œœ
abab ab

lim ( 19.6 4.9h) 19.6. The minus sign indicates the object is falling at a speed ofœœ
hÄ!
downward
19.6 m/sec.
28. lim lim lim 60 ft/sec.
hh hÄ! Ä! Ä!
f(10 h) f(10) 3(10 h) 3(10)
hhh
3 20h h ˆ
œœœ

ab
29. lim lim lim lim (6 h) 6
hh h hÄ! Ä! Ä! Ä!
f(3h)f(3) (3h) (3)
hh h
96hh 9
œœ œœ
11 1

cd
11
30. lim lim lim lim 12 6h h 16
hh h hÄ! Ä! Ä! Ä!
f(2 h) f(2)
hh h3
(2 h) (2) 12h 6h h 4
#
œœœœ
44 4
33 3
cd 1
cd 1
31. Slope at origin lim lim lim h sin 0 yes, f(x) does have a tangent atœœœœÊ
hhhÄ! Ä! Ä!
f(0 h) f(0)
hh h
h sin "
aˆ‰
h ˆ‰
the origin with slope 0.
32. lim lim lim sin . Since lim sin does not exist, f(x) has no tangent at
hhhhÄ! Ä! Ä! Ä!
g(0 h) g(0)
hhhh
h sin ""
œœ
ˆ‰
a
h
the origin.

111. Section 2.7 Tangents and Derivatives 107
33. lim lim , and lim lim . Therefore,
hh hhÄ! Ä! Ä! Ä!
ßß
f(0 h) f(0) f(0 h) f(0)
hh hh
10 10
œœ_ œœ_
lim yes, the graph of f has a vertical tangent at the origin.
hÄ!
f(0 h) f(0)
h

œ_ Ê
34. lim lim , and lim lim 0 no, the graph of f
hh hhÄ! Ä! Ä! Ä!
ßß
U(0 h) U(0) U(0 h) U(0)
hh hh
01 11
œœ_ œœÊ
does not have a vertical tangent at ( ) because the limit does not exist.!ß "
35. (a) The graph appears to have a cusp at x 0.œ

(b) lim lim lim and lim limit does not exist
hhh hÄ! Ä! Ä! Ä!
ß
f(0 h) f(0)
hh
h0
hh
" "
œœœ_ œ_Ê
&_
#&_ #&_
the graph of y x does not have a vertical tangent at x 0.Êœ œ
#Î&
36. (a) The graph appears to have a cusp at x 0.œ

(b) lim lim lim and lim limit does not exist
hhh hÄ! Ä! Ä! Ä!
ß
f(0 h) f(0)
hh
h0
hh
" "
œœœ_ œ_Ê

!&_ !&_
y x does not have a vertical tangent at x 0.Êœ œ
%Î&
37. (a) The graph appears to have a vertical tangent at x .œ!

(b) lim lim lim y x has a vertical tangent at x 0.
hhhÄ! Ä! Ä!
f(0 h) f(0)
hh
h0
h
" "Î&
œœœ_Êœ œ
!&_

38. (a) The graph appears to have a vertical tangent at x 0.œ

(b) lim lim lim the graph of y x has a vertical tangent
hhhÄ! Ä! Ä!
f(0 h) f(0)
hh
h0
h
" $Î&
œœœ_Ê œ
#&_

at x 0.œ

112. 108 Chapter 2 Limits and Continuity
39. (a) The graph appears to have a cusp at x 0.œ

(b) lim lim lim 2 and lim
hhh hÄ! Ä! Ä! Ä!
ß
f(0 h) f(0)
hh
4h 2h 4 4
hh

œœœ_#œ_
&_
#&_ #&_
limit does not exist the graph of y 4x 2x does not have a vertical tangent at x 0.ÊÊœ œ
#Î&

40. (a) The graph appears to have a cusp at x 0.œ

(b) lim lim lim h 0 lim does not exist the graph of
hhh hÄ! Ä! Ä! Ä!
f(0 h) f(0)
hh
h5h 5 5
hh
#Î$
œœœ Ê

œ"ß œ"ß
y x 5x does not have a vertical tangent at x .œ œ!
&Î$# Î$
41. (a) The graph appears to have a vertical tangent at x 1œ
and a cusp at x 0.œ

(b) x 1: lim lim œœœ_
hhÄ! Ä!
(1h) (1h1) (1h) h
hh
" "

y x (x 1) has a vertical tangent at x 1;Êœ œ
#Î$" Î$
x 0: lim lim lim œœ œ
hh hÄ! Ä! Ä!
f(0 h) f(0) h (h 1) ( 1) (h )
hh hh h
" ""

œ"ß
’“
does not exist y x (x 1) does not have a vertical tangent at x 0.Êœ œ
#Î$" Î$

113. Section 2.7 Tangents and Derivatives 109
42. (a) The graph appears to have vertical tangents at x 0 andœ
x1. œ

(b) x 0: lim lim y x (x 1) has aœœ œ_Êœ
hhÄ! Ä!
f(0 h) f(0) h (h 1) ( )
hh
" "Î$" Î$
œ"ß œ"ß œ"ß
vertical tangent at x 0; œ
x 1: lim lim y x (x 1) has aœœ œ_Êœ
hhÄ! Ä!
f(1h)f(1) (1h) ( h1) 1
hh
" "Î$" Î$
œ"ß œ"ß
vertical tangent at x . œ"
43. (a) The graph appears to have a vertical tangent at x 0.œ

(b) lim lim lim ;
hx hÄ! Ä! Ä!
ßß
f(0 h) f(0)
hh
h0
h
"
œœœ_
È
È
lim lim lim lim
hhhhÄ! Ä! Ä! Ä!

f(0 h) f(0)
hhh
h0 h
h


"
œœœœ_
ÈÈkk kk
kk Èkk
y has a vertical tangent at x 0.ʜ
44. (a) The graph appears to have a cusp at x 4.œ

(b) lim lim lim lim ;
hh hhÄ! Ä! Ä! Ä!
ßß ßß
f(4 h) f(4)
hhh
4(4h) 0 h
h
"
œœœœ_
ÈÈkk kk
È
lim lim lim lim
hh hhÄ! Ä! Ä! Ä!

f(4 h) f(4)
hhh
4(4h) h
h

ll
"
œ œœœ_
ÈÈkk kk
Èkk
y x does not have a vertical tangent at x 4.Êœ% œ
È
45-48. Example CAS commands:
:Maple
f := x -> x^3 + 2*x;x0 := 0;
plot( f(x), x=x0-1/2..x0+3, color=black, # part (a)
title="Section 2.7, #45(a)" );
q := unapply( (f(x0+h)-f(x0))/h, h ); # part (b)
L := limit( q(h), h=0 ); # part (c)
sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 ); # part (d)
tan_line := f(x0) + L*(x-x0);
plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black,

114. 110 Chapter 2 Limits and Continuity
linestyle=[1,2,5,6,7], title="Section 2.7, #45(d)",
legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)",
"Secant line (h=2)","Secant line (h=3)"] );
: (function and value for x0 may change)Mathematica
Clear[f,m,x,h]
x0 p;œ
f[x_]: Cos[x] 4Sin[2x]œ
Plot[f[x], {x, x0 1, x0 3}]
dq[h_]: (f[x0+h] f[x0])/hœ
m Limit[dq[h], h 0]π
ytan: f[x0] m(x x0)œ
y1: f[x0] dq[1](x x0)œ
y2: f[x0] dq[2](x x0)œ
y3: f[x0] dq[3](x x0)œ
Plot[{f[x], ytan, y1, y2, y3}, {x, x0 1, x0 3}]
CHAPTER 2 PRACTICE EXERCISES
1. At x 1: lim f(x) lim f(x) 1œ œ œ
x x?B ?B
ß
lim f(x) 1 f( 1)Êœœ
x1?
f is continuous at x 1.ʜ
At x 0: lim f(x) lim f(x) 0 lim f(x) 0.œœœÊœ
xx xÄ! Ä! Ä!
ß
But f(0) 1 lim f(x)œÁ
xÄ!
f is discontinuous at x 0.ʜ
If we define f , then the discontinuity at x isab!œ! œ!
removable.
At x 1: lim f(x) 1 and lim f(x) 1œœ œ
x xÄ" Ä"
ß
lim f(x) does not existÊ
x1Ä
f is discontinuous at x 1.ʜ

2. At x 1: lim f(x) 0 and lim f(x) 1œ œ œ
x x?B ?B
ß
lim f(x) does not existÊ
"
x?
f is discontinuous at x 1.ʜ
At x 0: lim f(x) and lim f(x)œœ_ œ_
x xÄ! Ä!
ß
lim f(x) does not existÊ
xÄ!
f is discontinuous at x 0.ʜ
At x 1: lim f(x) lim f(x) 1 lim f(x) 1.œœœÊœ
x x x1Ä" Ä" Ä
ß
But f(1) 0 lim f(x)œÁ
x1Ä
f is discontinuous at x 1.ʜ
If we define f , then the discontinuity at x isab"œ" œ"
removable.

3. (a) lim 3f t 3 lim f t 3( 7) 21
tt ttÄÄÁÁ
ab ababœ œœ
(b) lim f t lim f t 49
tt ttÄÄÁÁ
ab ab abab Š‹
##
#
œœ(œ

115. Chapter 2 Practice Exercises 111
(c) lim f t g t lim f t lim g t ( 7)(0) 0
tt tt ttÄÄÄbbb
a b ab abab ab††œœœ
(d) lim 1
ttÄ b
‘‘
‘‘‘
ft
g(x) 7 lim g t 7 lim g t lim 7 0 7
lim f t lim f t
7ab
ab ab
ab abab

œœ œœ
tt tt
tt tt tt

(e) lim cos g t cos lim g t cos 1
tt ttÄÄbb
ab abab Š‹œœ!œ
(f) lim f t lim f t 7 7
tt ttÄÄbb
kk ab kkab¹¹œœœ
(g) lim f t g t lim f t lim g t 7 0 7
tt tt ttÄÄÄbbb
a b ab abab abœ œœ
(h) lim
ttÄ b ‘
Š‹
"""
ft lim ft 7 7
1
ab ab
œœœ
tt
4. (a) lim g(x) lim g(x) 2
xxÄ! Ä!
œ œ
È
(b) lim g(x) f(x) lim g(x) lim f(x) 2
xxxÄ! Ä! Ä!
ab Š‹
Ȉ‰††œœœ
"
##
È
2
(c) lim f(x) g(x) lim f(x) lim g(x) 2
xxxÄ! Ä! Ä!
ab
È
Êa Ê aÊ
"
#
(d) lim 2
xÄ!
"""
f(x) lim f(x)
œœœ
x‘b
a

(e) lim x f(x) lim x lim f(x) 0
xxxÄ! Ä! Ä!
abÊa Ê aÊa
""
##
(f) lim
xÄ!
f(x) cos x
x 1 lim x lim 1 0 1
lim f(x) lim cos x
(1)†
†
#
"
œœœ
xx
xx‘b ‘b
‘b ‘b
ˆ‰
a

5. Since lim x 0 we must have that lim (4 g(x)) 0. Otherwise, if lim ( g(x)) is a finite positive
xx xÄ! Ä! Ä!
œœ%
number, we would have lim and lim so the limit could not equal 1 as
x xÄ! Ä!
ˆ
’“ ’“
4 g(x) 4 g(x)
xx

œ_ œ_
x 0. Similar reasoning holds if lim (4 g(x)) is a finite negative number. We conclude that lim g(x) 4.Ä œ
xxÄ! Ä!
6. 2 lim x lim g(x) lim x lim lim g(x) 4 lim lim g(x) 4 lim g(x)œ œ œ œ
xxx x xxxx?‘ ?‘ ?‘ ?‘Ä! Ä! Ä! Ä!
’ “ ’“ ’“ † (since lim g(x) is a constant) lim g(x) .
xxÄ! Ä!
Êœœ
2
% #
"
7. (a) lim f x lim x c f c for every real number c f is continuous on .
xc xcÄÄ
ab ab a bœœœ Ê _ß_
"Î$" Î$
(b) lim g x lim x c g c for every nonnegative real number c g is continuous on .
xc xcÄÄ
ab abœœœ Ê Ò!ß_Ñ
$Î% $Î%
(c) lim h x lim x h c for every nonzero real number c h is continuous on and .
xc xcÄÄ
ab ab a b a bœ œ œ Ê _ß ! _ß _
#Î$ "
c
’ˆ
(d) lim k x lim x k c for every positive real number c k is continuous on
xc xcÄÄ
ab ab a bœ œ œ Ê !ß _
"Î' "
c
a’È
8. (a) n n , where I the set of all integers.
nI
-
−
ß œˆ‰ˆ‰ˆ‰
""
##
11
(b) n n 1 , where I the set of all integers.
nI
-
−
ß œabab11
(c) abab_ß ß _11
(d) abab_ß! !ß_
9. (a) lim lim lim , x 2; the limit does not exist because
xxxÄ! Ä! Ä!
x 4x4 x2
x 5x 14x x(x 7)(x 2) x(x 7)
(x 2)(x 2)





œœÁ
lim and lim
x xÄ! Ä!
ˆ
x2 x2
x(x 7) x(x 7)

ˆˆ
œ_ œ_
(b) lim lim lim , x 2, and lim 0
xxx xÄ# Ä# Ä# Ä#
x 4x4 x2 x2 0
x 5x 14x x(x 7)(x ) x(x 7) x(x 7) 2(9)
(x 2)(x 2)



#

œœÁ œœ
10. (a) lim lim lim lim , x 0 and x 1.
xx x x0Ä! Ä! Ä! Ä
xx x1
x2xx
x(x 1)
x x 2x 1 x (x 1)(x 1) x (x 1)

“Ɉ
ˆˆ"
ˆˆ
ˆ
ˆˆ ˆ ˆ ˆ
œœœÁÁ
ab
Now lim and lim lim .
xx xÄ! Ä! Ä!
ˆ
11xx
x(x 1) x(x 1) x2xx


“Ɉ
ˆˆ
ˆ
ˆˆ
œ_ œ_ Ê œ_

116. 112 Chapter 2 Limits and Continuity
(b) lim lim lim , x 0 and x 1. The limit does not
xx x?" ?" ?"
xx 1
x2xx
x(x 1)
x x 2x 1 x (x 1)

“Ɉ
ˆ
ˆˆ
ˆ
ˆˆ ˆ
œœÁÁ
ab
exist because lim and lim .
x x?" ?"
ˆ
""
ˆˆx(x 1) x(x 1)
œ_ œ_
11. lim lim lim
x1 x1 x1ÄÄ Ä
1x x
1x 1x1x 1x
"
#
""
ÈÈ
ˆ‰ˆ‰ÈÈ È
œœœ
12. lim lim lim
xa xa xaÄÄ Ä
xa
xa
xa
xaxa xa a

ÉÉ


""


#
œœœ
ab
abab
13. lim lim lim (2x h) 2x
hh hÄ! Ä! Ä!
(x h) x
hh
x 2hx h x

œœœ
ab
14. lim lim lim (2x h) h
xx xÄ! Ä! Ä!
(x h) x
hh
x 2hx h x

œœœ
ab
15. lim lim lim
xx xÄ! Ä! Ä!
aa

x

#ˆ ˆ#
" "
x 2x( x) 4 x 4
2(2x)
œœœ
16. lim lim lim x 6x 12 12
xx xÄ! Ä! Ä!
(x)8
xx
x6x12x88# #
ˆ
œœœ
ab
ab
17. lim [4 g(x)] 2 lim 4 g(x) 2 lim 4 g(x) 8, since 2 8. Then lim g(x) 2.
xxx xÄ! Ä! Ä! Ä!
ˆˆˆ ˆ
"Î$ $
"Î$
œÊ œÊ œ œ œ’“18. lim 2 lim (x g(x)) 5 lim g(x) lim g(x) 5
xx x5 x5Ä& Ä& ÄÄÈÈ ÈÈ
""""
ˆ###xg(x)
œÊ œÊ œÊ œ
ÈÈ
19. lim lim g(x) 0 since lim 3x 1 4
x1 x1 x1ÄÄÄ
3x 1
g(x)

ˆ #
aŒ Š a Ê a ab
20. lim 0 lim g(x) since lim 5 x 1
xx x?ˆ ?ˆ ?ˆ
5x
g(x)
#

È
œÊ œ_ œ ab
21. lim lim . lim lim
xx x x?“ ?“ ?“ ?“
#ˆ$ #ˆ ! # # ˆ$ #ˆ ! #
&ˆ(& ˆ!& & ˆ(& ˆ!&
#ˆ
&ˆ &ˆ
#ˆ
x x
x x
œœœ ## œ œœ
ˆ



ˆ


x x
x x
23. lim lim
xx?“ ?“
xx
xxxx


%ˆ) " %)
$$$$
œ œ!!!œ!ˆ‰
24. lim lim
xxÄ_ Ä_
" !
(" " !ˆ!" xx

a

Êa
œœœ!
x
xx
25. lim lim . lim lim
xx x x?“ ?“ ?“ ?“
xx x xx x
x1 x
Ɉ
a ˆ
ˆ
( ( ˆˆ"
ˆ "# ˆ "#)"ˆ "# ˆ
œ œ _ #' œ œ _
x x27. lim lim since int x as x lim .
xx xÄ_ Ä_ Ä_
ll ll "sin x sin x
xx xgh gh gh
Ÿ œ! Ä_ Ä_Ê œ!
28. lim lim lim .
)))Ä_ Ä_ Ä_
l "ll #ll "lcos cos ))
)) )
Ÿœ!Ê œ!
29. lim lim
xxÄ_ Ä_
xsin x x
xsin x
ˆˆ#
ˆ"ˆ !
"ˆ ˆ
"ˆ
"ˆ!ˆ!
È
œœœ"
sin x
x x
sin x
x

30. lim lim
xxÄ_ Ä_
xx x
xcosx
’ˆ ba b“’ˆ


ˆ"ˆ"ˆ !
ˆ "ˆ
"ˆ!
œœœ"_¸
cos x
x

117. Chapter 2 Practice Exercises 113
31. At x 1: lim f(x) lim œ œ
xx?" ?"

xx 1
x1
ab
kk



lim lim x 1, andœœœ
xx?" ?"

xx 1
x1
ab




lim f(x) lim lim
xx x?" ?" ?"
ˆˆ ˆ
œœ
xx 1 xx 1
x1 x
ab ab
kk ab



"
lim ( x) ( 1) 1. Sinceœœœ
x1?
lim f(x) lim f(x)
x x?" ?"
ˆ
Á
lim f(x) does not exist, the function f beÊ
x1?
cannot
extended to a continuous function at x 1.œ
At x 1: lim f(x) lim lim lim ( x) 1, andœœœ œœ
xx x xÄ" Ä" Ä" Ä"

xx 1 xx 1
x1 x1
ab ab
kk ab



lim f(x) lim lim lim x 1. Again lim f(x) does not exist so f
xx x x1 x1Ä" Ä" Ä" Ä Ä
ˆˆˆˆ
œœœœ
xx 1 xx 1
x1 x
ab ab
kk



"
be extended to a continuous function at x 1 either.cannot
œ
32. The discontinuity at x 0 of f(x) sin is nonremovable because lim sin does not exist.œœ ˆ‰
""
xx
xÄ!
33. Yes, f does have a continuous extension to a 1:œ
define f(1) lim .œœ
x1Ä
x4
xx 3
"

É
È

34. Yes, g does have a continuous extension to a :œ
1
#
g lim .ˆ‰
1)
)1##
œœ
)Ä

’
5 cos 5
44

35. From the graph we see that lim h(t) lim h(t)
t tÄ! Ä!
ˆ
Á
so h be extended to a continuous functioncannot
at a 0.œ

118. 114 Chapter 2 Limits and Continuity
36. From the graph we see that lim k(x) lim k(x)
x xÄ! Ä!
ß
Á
so k be extended to a continuous functioncannot
at a 0.œ

37. (a) f( 1) 1 and f(2) 5 f has a root between 1 and 2 by the Intermediate Value Theorem.œ œ Ê
(b), (c) root is 1.32471795724
38. (a) f( 2) 2 and f(0) 2 f has a root between 2 and 0 by the Intermediate Value Theorem.œ œ Ê
(b), (c) root is 1.76929235424
CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES
1. (a) x 0.1 0.01 0.001 0.0001 0.00001
x 0.7943 0.9550 0.9931 0.9991 0.9999
x
Apparently, lim x 1
xÄ!
ß
x
œ
(b)
2. (a) x 10 100 1000
0.3679 0.3679 0.3679ˆ‰
"
"ÎÑ
x
_ln x
Apparently, lim 0.3678
xÄ_
ˆ‰
""
"ÎÐ Ñ
xe
ln x
œœ
(b)

119. Chapter 2 Additional and Advanced Exercises 115
3. lim L lim L L 1 L 1 0
vc vcÄÄ

‘œ"œœœ!! !ÉÉ É
vc
ccc
lim v



vc
ˆ
a
The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move fasterš
than the speed of light).
4. 1 0.2 0.2 1 0.2 0.8 1.2 1.6 x 2.4 2.56 x 5.76.¹¹ È
ÈÈÈxxx
###
ÊÊÊÊ
1 0.1 0.1 1 0.1 0.9 1.1 1.8 x 2.2 3.24 x 4.84.¹¹ È
ÈÈÈxxx
###
ÊÊÊÊ
5. 10 (t 70) 10 10 0.0005 (t 70) 10 0.0005 0.0005 (t 70) 10 0.0005kkkk‚Ê‚Ê‚
% % %
5 t 70 5 65° t 75° Within 5° F.Ê Ê Ê
6. We want to know in what interval to hold values of h to make V satisfy the inequality
V h . To find out, we solve the inequality:l "!!!l œ l$' "!!!l Ÿ "!1
hhhhl$' "!!!l Ÿ "! Ê "! Ÿ $' "!!! Ÿ "! Ê **! Ÿ $' Ÿ "!"! Ê Ÿ Ÿ111
**! "!"!
$' $'11
h . where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe.Ê )Þ) Ÿ Ÿ )Þ*
The interval in which we should hold h is about cm wide (1 mm). With stripes 1 mm wide, we can expect)Þ* )Þ) œ !Þ"
to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f(x) lim x 7 f(1).
x1 x1ÄÄ
œœ'œab
#
Step 1: x76 x1 1 x1 1 x 1.kkab
È È
###
ÊÊ Ê %% % % % % %
Step 2: x1 x1 x .kk Ê Ê " "$$ $$ $
Then 1 or 1 . Choose min 1 1 1 1 , then "œ "œ œ ß $%$%$ %%
ÈÈ ÈÈ š›
0 x 1 x 6 and lim f(x) 6. By the continuity test, f(x) is continuous at x 1.Ê ( œ œkk k k ab$%
#
x1Ä
8. Show lim g(x) lim 2 g .
xxÄÄ
ÉÉ

œœœ
""
2x 4
ˆ‰
Step 1: 2 x .¸¸
"" """
## ###xx x44
Ê # Ê# # Ê %% % % %
%%
Step 2: x x .¸¸B Ê Ê
""""
4444
$$ $$ $
Then , or .œÊœœ œÊœœ$$ $$
"" "" "" ""
# # # # 44 44 4(2) 44 4 44(2)%%%%%%
%%
Choose , the smaller of the two values. Then 0 x 2 and lim 2.$$%œÊœ
%
%4( ) 4 x x#ˆ ##
"" "
¸¸ ¸ ¸
xÄ
É

By the continuity test, g(x) is continuous at x .œ
"
4
9. Show lim h(x) lim 2x 3 h(2).
xxÄ# Ä#
œœ"œ
È
Step 1: 2x 3 1 2x 3 2x 3 x .¹¹
ÈÈ È
Ê " Ê" " Ê %% % % %
(1 ) ( ) 3$ "
##
%%

Step 2: x 2 x 2 or x .kk Ê ##$$ $$ $
Then , or #œ Ê œ# œ œ #œ$$ %$
() () (1) ()" $ " $ " " $
#####
%%% % %

. Choose , the smaller of the two values . Then,Êœ #œ œ œ$%$%
(("ˆÑˆ$ "ˆ Ñ"
### #
%% %%

x 2 2x 3 , so lim 2x 3 1. By the continuity test, h(x) is continuous at x 2.!Ê " œ œkk ¹¹
ÈÈ
$%
xÄ#
10. Show lim F(x) lim 9 x F(5).
xxÄ& Ä&
œœ#œ
È
Step 1: 9x2 9x 9(2 ) x ( ).¹¹
ÈÈ
Ê # Ê *#%% % % %
##
Step 2: 0x5 x x. Ê & Ê & &kk $$ $$ $
Then () () , or () () . &œ*# Ê œ# %œ # &œ*# Ê œ%# œ #$%$%%%$%$%%%
### # ##

120. 116 Chapter 2 Limits and Continuity
Choose , the smaller of the two values. Then, x 5 9 x , so$% % $ %œ# !Ê #
#
kk ¹¹
È
lim 9 x . By the continuity test, F(x) is continuous at x 5.
xÄ&
È
œ# œ
11. Suppose L and L are two different limits. Without loss of generality assume L L . Let (L L )."# #" #"
" œ %
3
Since lim f(x) L there is a 0 such that 0 x x f(x) L f(x) L
xxÄ b
œÊÊ"" ! "" " $$%%% kk k k
(L L ) L f(x) (L L ) L 4L L 3f(x) 2L L . Likewise, lim f(x) LÊ Ê œ
""
#" " #" " "# "# #
33
xxÄ b
so there is a such that 0 x x f(x) L f(x) L$$%%% # ! ## # Ê Ê kk k k
(L L ) L f(x) (L L ) L 2L L 3f(x) 4L LÊ Ê
""
#" # #" # #" #"
33
L 4L 3f(x) 2L L . If min both inequalities must hold for 0 x x :Ê œ ß "# #" "# ! $$$ $ef k k
5(L L ) 0 L L . That is, L L 0 L L 0,
4L L 3f(x) 2L L
LL 3f(x)2LL
Í
"# "#
"# #"
"# "# "# "#
%
Ê and
a contradiction.
12. Suppose lim f(x) L. If k , then lim kf(x) lim 0 lim f(x) and we are done.
xc xc xc xcÄÄÄÄ
œ œ! œ œ!œ!†
If k 0, then given any , there is a so that x c f x L k f x LÁ ! ! !l l Êl l Êl ll l%$ $ % ab ab
%
l5l
k f x L | kf x kL . Thus, lim kf(x) kL k lim f(x) .Êl Êl l œ œab ababab ab Š‹%%
xc xcÄÄ
13. (a) Since x 0 , 0 x x 1 x x 0 lim f x x lim f(y) B where y x x.ÄÊÄÊ œ œ œ
$ $ $ $
ab ab
x yÄ! Ä!
ˆ
(b) Since x 0 , 1 x x 0 x x 0 lim f x x lim f(y) A where y x x.ÄÊÄÊ œ œ œ
$$ $ $
ab ab
x yÄ! Ä!
ˆ
(c) Since x 0 , 0 x x 1 x x 0 lim f x x lim f(y) A where y x x .Ä Ê Ä Ê œ œ œ
ˆ % ## % ˆ# % # %
ab ab
xyÄ! Ä!
ˆˆ
(d) Since x 0 , 1 x 0 x x 1 x x 0 lim f x x A as in part (c).Ä Ê!Ê Ä Ê œ
% ## % ˆ# %
ab ab
xÄ!
ˆ
14. (a) True, because if lim (f(x) g(x)) exists then lim (f(x) g(x)) lim f(x) lim [(f(x) g(x)) f(x)]
xa xa xa xaÄÄÄÄ
œ
lim g(x) exists, contrary to assumption.œ
xaÄ
(b) False; for example take f(x) and g(x) . Then neither lim f(x) nor lim g(x) exists, butœœ
""
xx
xxÄ! Ä!
lim (f(x) g(x)) lim lim 0 0 exists.
xxxÄ! Ä! Ä!
œ œ œ ˆ‰
""
xx
(c) True, because g(x) x is continuous g(f(x)) f(x) is continuous (it is the composite of continuousœÊœkk k k
functions).
(d) False; for example let f(x) f(x) is discontinuous at x 0. However f(x) 1 is
1, x 0
1, x 0
œÊ œ œ
Ÿ
š
œ kk
continuous at x 0.œ
15. Show lim f(x) lim lim , x 1.
x1 x1 x1? ? ?
œœ œ#Á
x
x1 (x1)
(x 1)(x )

"
ˆˆ
"
Define the continuous extension of f(x) as F(x) . We now prove the limit of f(x) as x 1
, x
2 , x 1
π
Á"
œ
œ
x1
x1


ˆ
exists and has the correct value. Step 1: ( ) (x 1) , x x .¹¹
x
x1 (x1)
(x 1)(x )

"
ˆˆ
"
# Ê # Ê # Á"Ê" "%% %% % % %
Step 2: x ( 1) x 1 x .kk Ê Ê " "$$ $$ $
Then , or . Choose . Then x ( 1) "œ " Ê œ "œ " Ê œ œ ! $ % $% $ % $% $% $ kk
lim F(x) 2. Since the conditions of the continuity test are met by F(x), then f(x) has aÊ#Ê œ¹¹ ab
x
x1

"
ˆ
%
x1?
continuous extension to F(x) at x 1.œ

121. Chapter 2 Additional and Advanced Exercises 117
16. Show lim g(x) lim lim , x 3.
xx xÄ$ Ä$ Ä$
œœ œ#Á
x2x3
2x 6 2(x 3)
(x 3)(x )



"
Define the continuous extension of g(x) as G(x) . We now prove the limit of g(x) as
, x 3
2 , x 3
œ
Á
œ
œ
x2x3
2x 6



x 3 exists and has the correct value.Ä Step 1: 2 , x x .¹¹
x2x3 x
x6 2(x3)
(x 3)(x )

"
# #
"
Ê # Ê # Á$Ê$# $#%% %% % % %
Step 2: x3 x3 x .kk Ê Ê$ $$$ $ $ $
Then, , or . Choose . Then x 3$ œ$# Ê œ# $œ$# Ê œ# œ# ! $ % $%$ % $% $% $ kk
2 lim 2. Since the conditions of the continuity test hold for G(x),ÊÊ œ¹¹
x2x3
2x 6 (x 3)
(x 3)(x )


#
"
%
xÄ$
g(x) can be continuously extended to G(x) at 3.Bœ
17. (a) Let be given. If x is rational, then f(x) x f(x) 0 x 0 x 0 ; i.e., choose%%%! œÊ œÍ k kkk kk
. Then x 0 f(x) 0 for x rational. If x is irrational, then f(x) 0 f(x) 0$% $ % %œ Ê œÊ kk kk kk
which is true no matter how close irrational x is to 0, so again we can choose . In either case,Í! œ%$%
given there is a such that x 0 f(x) 0 . Therefore, f is continuous at%$% $ %! œ! !Ê kk k k
x0. œ
(b) Choose x c . Then within any interval (c c ) there are both rational and irrational numbers.œ! ß $$
If c is rational, pick . No matter how small we choose there is an irrational number x in%$aš‘
c
#
(c c ) f(x) f(c) 0 c c . That is, f is not continuous at any rational c 0. Onß Ê œ œ œ $$ % kkkk
c
#
the other hand, suppose c is irrational f(c) 0. Again pick . No matter how small we choose Ša a š‘ %$
c
#
there is a rational number x in (c c ) with x c x . Then f(x) f(c) x 0ß œ Í œ $$ % kk k kkk
cc3c
###
x f is not continuous at any irrational c 0.ašaŠ škk
c
#
%
If x c 0, repeat the argument picking . Therefore f fails to be continuous at anya¹ a a %
kkc c
##

nonzero value x c.œ
18. (a) Let c be a rational number in [0 1] reduced to lowest terms f(c) . Pick . No matter howœßÊœœ
m
n nn
""
#
%
small is taken, there is an irrational number x in the interval (c c ) f(x) f(c) 0$$$! ß Ê œ kk ¸¸
"
n
. Therefore f is discontinuous at x c, a rational number.aš a a
""
#nn
%
(b) Now suppose c is an irrational number f(c) 0. Let 0 be given. Notice that is the only rationalŠa š %
"
#
number reduced to lowest terms with denominator 2 and belonging to [0 1]; and the only rationals withß
"
33
2
denominator 3 belonging to [0 1]; and with denominator 4 in [0 1]; , , and with denominator 5 inßß
""
4 4 555 5
3234
[0 1]; etc. In general, choose N so that there exist only finitely many rationals in [ ] havingˆ¹Š ‘ˆ’
"
N
%
denominator N, say r , r , , r . Let min c r : i 1 p . Then the interval (c c )Ÿáœœßáß ß
"# pi $$$ efkk
contains no rational numbers with denominator N. Thus, 0 x c f(x) f(c) f(x) 0ŸÊœ kk k kk k$
f(x) f is continuous at x c irrational.ad¹Š akk
"
N
%

122. 118 Chapter 2 Limits and Continuity
(c) The graph looks like the markings on a typical ruler
when the points (x f(x)) on the graph of f(x) areß
connected to the x-axis with vertical lines.

19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the
zero point, 0, on the equator 0 R represents the midnight point (at the same exact time). Suppose xÊ 1
"
is a point on the equator “just after" noon x R is simultaneously “just after" midnight. It seemsÊ "1
reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically
opposite point just after midnight: That is, T(x ) T(x R) 0. At exactly the same moment in time
"" 1
pick x to be a point just before midnight x R is just before noon. Then T(x ) T(x R) 0.
#### Ê 11
Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate
Value Theorem says there is a point c between 0 (noon) and R (simultaneously midnight) such that1
T(c) T(c R) 0; i.e., there is always a pair of antipodal points on the earth's equator where the œ 1
temperatures are the same.
20. lim f(x)g(x) lim f(x) g(x) f(x) g(x) lim f(x) g(x) lim f(x) g(x)
xc xc xc xcÄÄ Ä Ä
œœ
""
%%
##
##
’“’ “abab ab ab Š‹Š‹
. œ$" œ#
"
%
# #
ˆ‰ab
21. (a) At x 0: lim r (a) lim lim œœ œ
aa aÄ! Ä! Ä!
ß
" " "
"
ÈÈÈ
È
1a 1a 1a
aa 1a Š‹Š‹
lim œœœ
aÄ!
1( a)
a1a 10
1"
" "
"
#ˆ‰ÈÈ
At x 1: lim r (a) lim lim 1œ œ œ œ œ
aa a1?B ?B ?
ßß
ß

"
"
"
1(1a)
a1 1a a 1a
a
0ˆ‰ ˆ‰ÈÈÈ
(b) At x 0: lim r (a) lim lim œœ œ
aa aÄ! Ä! Ä!

" " "
"
ÈÈÈ
È
1a 1a 1a
aa 1a Š‹Š‹
lim lim lim (because theœœœœ_
aaaÄ! Ä! Ä!

1( a)
a1a a11a 1a
a"
" "
"
ˆ‰ ˆ‰ÈÈÈ
denominator is always negative); lim r (a) lim (because the denominator
aaÄ! Ä!
ßß

"
" œœ_
È1a
is always positive). Therefore, lim r (a) does not exist.
aÄ!

At x 1: lim r (a) lim lim 1œœœœ
aa a1?B ?B ?
ßß ß

"
"
11a
a 1a
È
È

123. Chapter 2 Additional and Advanced Exercises 119
(c)
(d)
22. f(x) x 2 cos x f(0) 0 2 cos 0 2 0 and f( ) 2 cos ( ) 0. Since f(x) isœ Ê œ œ œ œ # 11 11
continuous on [ ], by the Intermediate Value Theorem, f(x) must take on every value between [ ].ß! #ß#1 1
Thus there is some number c in [ ] such that f(c) 0; i.e., c is a solution to x 2 cos x 0.ß! œ œ1
23. (a) The function f is bounded on D if f(x) M and f(x) N for all x in D. This means M f(x) N for all x Ÿ ŸŸ
in D. Choose B to be max M N . Then f(x) B. On the other hand, if f(x) B, thene f kk kkkkkkߟ Ÿ
B f(x) B f(x) B and f(x) B f(x) is bounded on D with N B an upper bound andŸ ŸÊ   ŸÊ œ
M B a lower bound.œ
(b) Assume f(x) N for all x and that L N. Let . Since lim f(x) L there is a such thatdša aš‘ %$
LN
#
xxÄ Á
0 x x f(x) L L f(x) L L f(x) L Ê Í Í kk k k !

##$%%%
LN LN
f(x) . But L N N N f(x) contrary to the boundedness assumptionÍ ÊÊ
LN 3LN LN
## #
f(x) N. This contradiction proves L N.ŸŸ
(c) Assume M f(x) for all x and that L M. Let . As in part (b), 0 x xŸœ %$
ML
#
!
kk
L f(x) L f(x) M, a contradiction.Ê Í
ML ML 3LM ML
####
24. (a) If a b, then a b 0 a b a b max (a b) a.   Ê œÊ ßœ œ œ œkk
ab ab ab 2aab
## ###
kk
If a b, then a b 0 a b (a b) b a max (a b)Ÿ Ÿ Ê œ œ Ê ß œ œ kk
ab ab baab
## ##
kk
b. œœ
2b
#
(b) Let min (a b) .ߜ
ab abß
##
kk

124. 120 Chapter 2 Limits and Continuity
25. lim lim lim lim
x0 x0 x0 x0ÄÄ ÄÄ
œœ ††œ †
sin cos x sin cos x sin cos x
x cos x x cos x cos x x co
cos x cos x cos xab ab ab
ab
" " "
" " " "
" " "

s x x cos x
sin x
œ"† lim
x0Ä

abBß
lim .œ†œ"†œ!
x0Ä
sin x sin x
xcos xBß &
!
ˆ‰
26. lim lim lim lim x .
x0 x0 x0 x0ÄÄ ÄÄ
ßß ßß
sin x sin x x
sin x sin x x
x
ÈÈÈ
È
Š‹
œ † † œ"† † œ"†!†!œ!
B
" sin x
x

È
27. lim lim lim lim .
x0 x0 x0 x0ÄÄ ÄÄ
sin sin x sin sin x sin sin x
xsin xxsin xx
sin x sin xab ab ab
œ†œ†œ"†"œ"
28. lim lim x lim lim x
x0 x0 x0 x0ÄÄ ÄÄ
sin x x sin x x sin x x
xxx xx
ab ab ab


ßß ß
ßß
a $ Ê’ a $ Ê’ a’$’a’ab ab
29. lim lim x 2 lim lim x 2
x2 x2 x2 x2ÄÄ ÄÄ
sin x sin x sin x
x2 x x
ab ab ab


% % %
% %
a $ Ê a $ Ê a’$‹a‹ab ab
30. lim lim lim lim
x9 x9 x9 x9ÄÄ ÄÄ
sin x sin x sin x
x9 xx x x
ˆ‰ ˆ‰ ˆ‰ÈÈ È
ÈÈ È È
$ $ $
''$ $ $ $
""""
œ†œ†œ"†œ

125. CHAPTER 3 DIFFERENTIATION
3.1 THE DERIVATIVE OF A FUNCTION
1. Step 1: f(x) 4 x and f(x h) 4 (x h)œ œ
##
Step 2:
f(x h) f(x) h( 2x h)
hh h hh
4(xh) 4x 4 x 2xh h 4 x 2xh h ÈÈ ÈÈ
œœ œœ
cdab ab
œœ œœœ œ
2x h œ
Step 3: f (x) lim ( 2x h) 2x; f ( ) 6, f (0) 0, f (1) 2
wwww
œœ$œœœ
hÄ!
2. F(x) (x 1) 1 and F(x h) (x h 1) F (x) lim œ œ "Ê œ
##w
hÄ!
cdcd(xh1) 1 (x1) 1
h
œœ
lim lim lim (2x h 2)œœœ
hhhÄ! Ä! Ä!
ababx 2xhh 2x2h11 x 2x11
hh
2xh h 2h
œœ œ œ

2(x 1); F ( 1) 4, F (0) 2, F (2) 2œœœœ
www
3. Step 1: g(t) and g(t h)œœ
""
œt(th)
œœ
Step 2:
g(t h) g(t)
h h h (th)th (th)th
tt2thh 2th h
œœ
ÈÈ
œœ œ œ
œœ
È
ŠŠ
ŠŠ
‹È
È
ŠŠ
(t h) t
t(th)
(t h) t
œ ?œ
ab
œ
œœ œ
œœ œœ
œ
††
œœ
h( 2t h)
(t h) t h (t h) t
2t hÈÈ
œœ
ÈÈ
œœ œœ
Step 3: g (t) lim ; g ( 1) 2, g (2) , g 3
wwww ÈÈ È È ‹
œ
œœœœœœ
hÄ!
2t h 2t 2 2
(t h) t t t t 4 33
œœ œ œ È
†
Š‹
È
È
4. k(z) and k(z h) k (z) lim œœÊœ
1z
z2(zh) h
1(zh)È
#œ
w
hÄ!
Š‹

ŠÈ Š
œ$(z h)
(z h) z
z
œ
lim lim lim lim œœ œœ
hh hhÄ! Ä! Ä! Ä!
(1 z h)z ( z)(z h)
(z h)zh 2(z h)zh 2(z h)zh (z h)z
zz zhzhz zh h "
#œ œ œ #œ
"
œœ
; k ( ) , k (1) , k 2œ"œ œ œ
È‹ ‹ ‹ ‹www
##2z 4
œ Š‹
È
5. Step 1: p( ) 3 and p( h) 3( h)))) )œœ
È È
Step 2:
p( h) p( ) (3 3h) 3
hh h
3( h) 3
33h 3 33h 3
33h 3 h 33h 3
)) ))))
))))
)) ))


œœ œœ
œœ œ
È È Š‹Š‹
ÈÈ ÈÈ
Š‹Š‹
ÈÈ ÈÈ
† œœ
3h 3
h33h 3 33h 3Š‹
È È È È
)) ))œœ œœ
Step 3: p ( ) lim ; p (1) , p (3) , p
wwww
œœ œ
"
# #
)œœœœœœ
hÄ!
3333 23
33h 3 3 3 23 23 3 2È ÈÈÈ È È È
)))))
ˆ‰
6. r(s) 2s 1 and r(s h) 2(s h) 1 r (s) lim œ œ ÊœÈ È
w
hÄ!
È È2s 2h 1 2s 1
h
lim lim œœ
hhÄ! Ä!
Š‹Š ‹
ÈÈ ÈÈ
Š‹Š‹
ÈÈ ÈÈ
2s h 1 2s 1 2s 2h 1 2s 1
h
2s 2h 1 2s 1 h 2s 2h 1 2s 1
(2s 2h 1) (2s 1)

œœœ œ œœœ œ

†
lim lim œœœœ
hhÄ! Ä!
2h 2 2 2
h2s2h1 2s1 2s 2h 1 2s 1 2s1 2s1 22s1Š‹
È È ÈÈÈÈ È
œœœ œ œœœ œ œœ œ œ
; r (0) 1, r (1) , rœœœœ
""""
œ
ww w
#ÈÈÈ2s 1 3 2
ˆ‰
7. y f(x) 2x and f(x h) 2(x h) lim lim œœ œÊœ œ
$$ dy 2(x h) 2x
dx h h
2 x 3x h 3xh h 2x
hhÄ! Ä!
ÈÈ
ab
lim lim lim 6x 6xh 2h 6xœœ œœ
hh hÄ! Ä! Ä!
6x h 6xh 2h
hh
h 6x 6xh 2h
œœ
œœ œœ ###ab
ab

126. 122 Chapter 3 Differentiation
8. r 1 lim lim œÊ œ œ
sdr
ds h h
11
(s h) 2 s 2
È ÈÈ
##

"
hhÄ! Ä!
”• ’“
cdcd
(s h) sÈ
”
ŠŠ
”
lim lim lim 3s 3sh h sœœœœ
" " "
####
œœ ###
hhhÄ! Ä! Ä!
s3sh3shh2s2 3
hh
h3s 3sh h
È œœ cd
ab
9. s r(t) and r(t h) lim œœ œ Êœ
tthds
2t 1 2(t h) 1 dt hœœœ
œ
È
hÄ!
Š‹ ˆ‰
th t
2(t h) 1 2t 1È
ÈÈ È
lim lim œœ
hhÄ! Ä!
Š‹
(t h)(2t 1) t(2t 2h 1)
(2t 2h 1)(2t 1)ÈÈ ‹ÈÈ
ÈÈ È
h (2t 2h 1)(2t 1)h
(t h)(2t 1) t(2t 2h 1)
œœ œ
lim lim lim œœœ
hhhÄ! Ä! Ä!
2t t 2ht h 2t 2ht t h
(2t 2h 1)(2t 1)h (2t 2h 1)(2t 1)h (2t 2h 1)(2t 1)
œœ
"
œœ œ œœ œ œœ œ
œœ
""
œœ œ(2t 1)(2t 1) (2t 1)
œ
10. lim lim lim
dv
dt h h h
(t h) t h
œœœ
hhhÄ! Ä! Ä!
’“ Š ‹ ˆ‰
œ
È È
œ
È
È‹ÈÈ
th (th)t th
h(th)tt(th)
Š
Š
t
t
lim lim 1œœœœ
hhÄ! Ä!
ht h t h t ht 1 t 1
h(t h)t (t h)t t t
œœ œ œ
œœ
œœ œœ œ $
œœ
11. p f(q) and f(q h) lim œœ œ Êœ
""
œ œœ
È
ÈÈ
Š‹Š‹
q1 (q h) 1
dp
dq h
hÄ!
œœ
ÈÈ Èœœ
(q h) 1 q1
lim lim œœ
hhÄ! Ä!
?
È È
È È
œ œ
œ œq1 qh1
qh1 q1È‹ÈÈ
ÈÈ È
h
q1 qh1
hqh1q1

œœ œ
lim lim œœ
hhÄ! Ä!
ˆ‰ˆ‰ÈÈ ÈÈ
ÈÈÈÈ ÈÈ ÈÈ ˆ‰ ˆ‰
q1 qh1 q1 qh1
hqh1q1 q1 qh1 hqh1q1 q1 qh1
(q 1) (q h 1)
œœ œ œœ œœ œœ œ œœ œœ

†
lim lim œœ
hhÄ! Ä!
ÈÈ‹
œœ œ œœ œœ œœ œ œœ œœ
h
hqh1q1 q1 qh1 qh1q1 q1 qh1ÈÈÈÈ ÈÈ ÈȈ‰ ˆ‰
œœ
È‹ È‹
œœ œœœ œ œÈÈ È È È ˆ‰q1q1 q1 q1 2(q1)q1
12. lim lim
dz
dw h
3w 2 3w 3h 2
h3w 3h 2 3w 2
œœ
hhÄ! Ä!
Š‹ È È
È È
œœ
È‹ ‹
œœ
3(w h) 2 3w 2
œ


lim œ
hÄ!
Š‹Š‹
ÈÈ ÈÈ
È È
Š‹
È È
3w 2 3w 3h 2 3w 2 3w 3h 2
h3w 3h 2 3w 2 3w 2 3w 3h 2


†
lim œ
hÄ!
(3w 2) (3w 3h 2)
h 3w 3h 2 3w 2 3w 2 3w 3h 2

ÈÈ ÈÈ
Š‹
lim œœ
hÄ!
ÈÈ

33
3w 3h 2 3w 2 3w 2 3w 3h 2 3w 2 3w 2 3w 2 3w 2ÈÈ ÈÈ ÈÈÈ È
Š‹Š‹
œ
È
ÈÈ
3
2(3w 2) 3w 2
È
13. f(x) x and f(x h) (x h) œ œ Ê œ
99
x(xh)hh
f(x h) f(x)
(x h) x
œ

’“’“
99
(x h) x
È
œœ œ
x(x h) 9x x (x h) 9(x h)
x(x h)h x(x h)h x(x h)h
x 2xhxh 9xx xh9x9h xhxh 9h
œœœ

œœ
; f (x) lim 1 ; m f ( 3) 0œœ œ œœœœ
h(x xh 9)
x(x h)h x(x h) x(x h) x x
xxh9 xxh9 x9 9
œ œœœ
œœ

œœ œ
ww
hÄ!
14. k(x) and k(x h) k (x) lim lim œœÊœ œ
""
#œ œ œ
w
x2(xh) h h
k(x h) k(x)
hhÄ! Ä!
Š‹
œœ
ŠÈ È ŠÈ
xh x
œ
lim lim lim ;œœœœ
hhhÄ! Ä! Ä!
( x)(2xh)
h(2x)(2xh) h(2x)(2xh) (2x)( xh) (2x)
h#
œ œœ œ œœ œ #œœ œ
ÈÈ‹È‹
œ
k(2)
w "
œ
16
15. lim lim
ds
dt h h
(t h) (t h) t t t3th3thh t2thh tt
œœ
hhÄ! Ä!
cdab abab
È
lim lim lim 3t 3th h 2t hœœœ
hhhÄ! Ä! Ä!
3th3thh2thh
hh
h3t 3th h 2t h
œœ œœ
##ab
ab

127. Section 3.1 The Derivative of a Function 123
3t 2t; m 5œ œ œ
#
•È‹ ¸
ds
dt
t
16. lim lim
dy (xh1) (x1) (x1) 3(x )h3(x1)h h (x1)
dx h h
œœ
hhÄ! Ä!
"
ÈÈ È
lim 3(x 1) 3(x 1)h h 3(x 1) ; m 3œ œ œ œ
hÄ!
cd ¹
### dy
dx
x=
17. f(x) and f(x h) œœÊœ
88
x2 (x h) 2
f(x h) f(x)
hhÈÈ
È

88
(x h) 2 x2
œœÈ‹ ‹
œ
œœ
8 x2 xh2 x2 xh2
hxh2x2 x2 xh2 hxh2x2 x2 xh2
8[(x 2) (x h 2)]
Š‹Š‹
ÈÈ ÈÈ
È È
Š‹ Š‹
ÈÈÈ ÈÈ È



†
f (x) lim œÊœ
ÈÈ

w8h 8
hxh2x2 x2 xh2 xh2x2 x2 xh2ÈÈ ÈÈ ÈÈ ÈÈ
Š‹ Š‹
hÄ!
; m f (6) the equation of the tangentœœœœœÊ
ÈÈÈ‹
ÈÈ
w
#
844
x2x2 x2 x2 (x 2) x 2 4 4ÈÈ È È
Š‹
ÈÈ
line at (6 4) is y 4 (x 6) y x y x .ß œ Ê œ $%Ê œ (
"" "
## #
18. g (z) lim lim
w


œœ
hhÄ! Ä!
ˆ‰È Š‹Š‹Š‹
ÈÈÈ ÈÈ
Š‹
È È
14(zh)14z 4zh4z 4zh4z
hh
4zh 4z
† lim lim lim ;œœœœ
hhhÄ! Ä! Ä!
(4 z h) (4 z)
h 4zh 4z h 4zh 4z 4zh 4z
h
24 z
ÈÈÈÈ

ÈÈ‹È‹
ÈŠ‹Š‹Š‹
ÈÈÈ ÈÈÈ È
m g (3) the equation of the tangent line at ( ) is w 2 (z 3)œœ œÊ $ß#œ
w È‹ ‹ "
È # #24 3
È
wz wz.ʜ#ʜ
"$ "(
## ##
19. s f(t) 1 3t and f(t h) 1 3(t h) 1 3t 6th 3h lim œœ œœ Êœ
#### ds
dt h
f(t h) f(t)
hÄ!
lim lim ( 6t 3h) 6t 6œœœÊœ
hhÄ! Ä!
abab13t 6th3h 13t
hdt
dsÈÈ È ÈÈ
œœœ
¸
t=
20. y f(x) and f(x h) 1 lim lim œœ" œ Êœ œ
""
œ

ÈÈÈ
xxhdxh h
dy f(x h) f(x)
11
hhÄ! Ä!
Š‹Š‹
œœ
È
xh x
lim lim lim œœœœÊœ
hhhÄ! Ä! Ä!
œœ
È
xxh
È
œœ
"" "
h x(x h)h x(x h) x dx 3
h dy
œ ¹
x= 3
œ
21. r f( ) and f( h) lim lim œœ œ Êœ œ))
22dr
4 4( h) dh h
f( h) f( )
ÈÈ
È

È
) ) )
))
hhÄ! Ä!
22
4h 4
œœ
‹‹ ‹œœ
lim lim œœ
hhÄ! Ä!
24 24 h 24 h
h4 4 h h4 4 h
224h
24 4 h
ÈÈ ÈÈ
ÈÈ ÈÈ
Š‹
ÈÈ
Š‹
ÈÈ
ÈÈ ÈÈ ÈÈ# “ÈÈ
ÈÈÈ ÈÈÈ
%
#
)) ))
)) ))
))
))
†
lim lim œœ
hhÄ! Ä!
4( ) 4( h)
2h44h4 4h 44h4 h
2“È È “È È
%
))
)) ) ) )) ) )
ÈÈ È È ÈÈ È È
Š‹ Š‹
œœÊœ
2dr
(4 ) 2 4 (4 ) 4 d8
ÈÈ
""
ÈÈ œ!
)) )) ))
Š‹
È È
¸
22. w f(z) z z and f(z h) (z h) z h lim œ œ œ Ê œÈ È
dw
dz h
f(z h) f(z)
hÄ!

lim lim lim 1œœœ
hhhÄ! Ä! Ä!
Š‹ Š‹
È Èˆ‰È ÈÈÈ ÈÈ
Š‹
È È
zh zh z z zh z
hhh
hzhz zhz
zh z
œœ

œœ
–—
†
1 lim 1 lim œ œ œ" Ê œ
hhÄ! Ä!
(z h) z
hzh z zh z 2z
dw 5
dz 4

œœ
""
œœŠ‹
È È
ÈÈ È
¸
z4‹
23. f x lim lim lim lim lim
w ##
# # # # #
È
ÈÈ‹
abœœœœœ
zx zx zx zx zxÄÄÄÄÄ
fz fx x z
z x zxzxzxzxzxz
xzab ab a b a b
ababab ababab abab
œœ
ÈŠ ÈŠ
zx
xœ# œ
È‹
œ#abx
œ

128. 124 Chapter 3 Differentiation
24. f x lim lim lim lim
w ÈÈ‹ÈÈ‹
ÈÈ
È
ÈÈ‹È‹
Ò" "ÓÒ"
abœœ œ œ
zx zx zx zxÄÄ Ä Ä
fz fx x z
zx z x zxz x
xzxzab ab a b a b
ababab
abababab
œœ
$œ $œ
ŠŠ
ÈŠ È Š
zx
œœ
œœ
È‹ k
ÈÈ‹È‹abababzxz x
œœ
lim limœœœœœ
zx zxÄÄ
abab ab abab
ababab abab ab ab ab
xzxz2 xz2 x x
zxzx zx x xx
" " # # # "
ÈÈ‹È‹ È‹È‹ È‹ È‹ È‹
È#
œœ œœ È‹‹
25. g x lim lim lim lim lim
w ÈÈ‹ÈÈ‹
È È È È‹ È‹ È È‹ È‹
È
"
abœœœœœ
zx zx zx zx zxÄÄÄÄÄ
gz gx zx xz
z x z x z x z x z x z x
zxab ab a b a b
ababab ababab
zx
zx
‹œ$œ
ababzxÈ‹ È‹ œ
È‹
È‹abx
œ
26. g x lim lim lim lim
w È
ÈÈÈ
" "

È
abœœ œ†œ
zx zx zx zxÄÄ Ä Ä
gz gx
z x z x z x
zx zx zx
zx z xzx
z xab ab ˆ ‰ˆ ‰ ÈÈ ÈÈÈÈ
ÈÈ ÈÈ ab ˆ‰
limœœ
zxÄ
""
œ#ÈÈ Èzx x
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x 0),œ
then positive the slope is always increasing which matches (b).Ê
28. Note that the slope of the tangent line is never negative. For x negative, f (x) is positive but decreasing as x
w
#
increases. When x 0, the slope of the tangent line to x is 0. For x 0, f (x) is positive and increasing. Thisœ
w
#
graph matches (a).
29. f (x) is an oscillating function like the cosine. Everywhere that the graph of f has a horizontal tangent we
$ $
expect f to be zero, and (d) matches this condition.
w
$
30. The graph matches with (c).
31. (a) f is not defined at x 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree.
w
œ
For example, lim slope of line joining ( 0) and ( ) but lim slope of
x xÄ! Ä!
œ ”
f(x) f(0) f(x) f(0)
x0 x0
È È
È#È
"
œ %ß !ß # œ œ
line joining (0 2) and ( 2) 4. Since these values are not equal, f (0) lim does not exist.ß"ßœ œ
w È
È
xÄ!
f(x) f(0)
x0
(b)
32. (a) (b) Shift the graph in (a) down 3 units

129. Section 3.1 The Derivative of a Function 125
33.
34. (a) (b) The fastest is between the 20 and 30 days;
th th
slowest is between the 40 and 50 days.
th th
35. Left-hand derivative: For h 0, f(0 h) f(h) h (using y x curve) lim œœ œ Ê
##
hÄ!
œ
f(0 h) f(0)
h
lim lim h 0;œœœ
hhÄ! Ä!
œœ
h0
h
œ
È
Right-hand derivative: For h 0, f(0 h) f(h) h (using y x curve) lim œœ œ Ê
hÄ!
”
f(0 h) f(0)
h

lim lim 1 1;œœœ
hhÄ! Ä!
””
h0
h
È
Then lim lim the derivative f (0) does not exist.
h hÄ! Ä!
œ ”
f(0 h) f(0) f(0 h) f(0)
hh
w
ÁÊ
36. Left-hand derivative: When h , 1 h 1 f(1 h) 2 lim lim ! Ê œ Ê œ
hhÄ! Ä!
œœ
f(1 h) f(1)
hh
22 È
lim 0 0;œœ
hÄ!
œ
Right-hand derivative: When h , 1 h 1 f(1 h) 2(1 h) 2 2h lim ! Ê œ œ Ê
hÄ!
”
f(1 h) f(1)
h

lim lim lim 2 2;œœœœ
hhhÄ! Ä! Ä!
”””
(2 2h) 2
hh
2h
Then lim lim the derivative f (1) does not exist.
h hÄ! Ä!
œ ”
f(1 h) f(1) f(1 h) f(1)
hh
w
ÁÊ
37. Left-hand derivative: When h 0, 1 h 1 f(1 h) 1 h lim ÊœÊ È
hÄ!
œ
f( h) f(1)
h
"
lim lim lim lim ;œœ œ œœ
hh h hÄ! Ä! Ä! Ä!
œœ œ œ
È Š‹Š‹
ÈÈ
Š‹ Š‹
ÈÈ È
1h
hh
1h 1h
1h1 h 1h
(1 h)
1h1
"
" "
œœ œœ$
" ""
œœ #
†
Right-hand derivative: When h 0, 1 h 1 f(1 h) 2(1 h) 1 2h 1 lim ÊœœÊ
hÄ!
”
f( h) f(1)
h
"
lim lim 2 2;œœœ
hhÄ! Ä!
””
(2h 1)
h
"
Then lim lim the derivative f (1) does not exist.
h hÄ! Ä!
œ ”
f(1 h) f(1) f(1 h) f(1)
hh
w
ÁÊ
38. Left-hand derivative: lim lim lim 1 1;
hhhÄ! Ä! Ä!
œœœ
f(1 h) f( ) (1 h)
hh
" "
œœœ
Right-hand derivative: lim lim lim
hhhÄ! Ä! Ä!
”””
f(1 h) f( )
hhh
"
È‹
œœ
Š‹ Š ‹
œ
ÈÈ
‹È
1h 1h
1(1h)
lim lim 1;œœœ
hhÄ! Ä!
””
ÈÈ‹
œœ
h
h(1 h) 1 h
Then lim lim the derivative f (1) does not exist.
h hÄ! Ä!
œ ”
f(1 h) f(1) f(1 h) f(1)
hh
w
ÁÊ

130. 126 Chapter 3 Differentiation
39. (a) The function is differentiable on its domain x 2 (it is smooth)$ Ÿ Ÿ
(b) none
(c) none
40. (a) The function is differentiable on its domain x 3 (it is smooth)# Ÿ Ÿ
(b) none
(c) none
41. (a) The function is differentiable on x 0 and x 3$ Ÿ ! Ÿ
(b) none
(c) The function is neither continuous nor differentiable at x 0 since lim f(x) lim f(x)œÁ
x xÄ! Ä!
œ "
42. (a) f is differentiable on x 1, x 0, 0 x 2, and 2 x 3# Ÿ " Ÿ
(b) f is continuous but not differentiable at x 1: lim f(x) 0 exists but there is a corner at x 1 sinceœ œ œ
x1Ĝ
lim 3 and lim 3 f ( 1) does not exist
h hÄ! Ä!
œ "
f( 1 h) f( ) f( h) f( 1)
hh
" " w
œ œ Ê
(c) f is neither continuous nor differentiable at x 0 and x 2:œœ
at x 0, lim f(x) 3 but lim f(x) 0 lim f(x) does not exist;œœ œÊ
x x x0Ä! Ä! Ä
œ "
at x 2, lim f(x) exists but lim f(x) f(2)œÁ
xxÄ# Ä#
43. (a) f is differentiable on x 0 and 0 x 2"Ÿ Ÿ
(b) f is continuous but not differentiable at x 0: lim f(x) 0 exists but there is a cusp at x 0, soœœ œ
xÄ!
f (0) lim does not exist
w
œ
hÄ!
f(0 h) f(0)
h
(c) none
44. (a) f is differentiable on x 2, 2 x 2, and 2 x 3$Ÿ Ÿ
(b) f is continuous but not differentiable at x 2 and x 2: there are corners at those pointsœ œ
(c) none
45. (a) f (x) lim lim lim lim ( 2x h) 2x
w
œœ œ œœ
hh h hÄ! Ä! Ä! Ä!
f(x h) f(x)
hh h
(x h) x x 2xh h x
œœ œœœ
ab
(b)
(c) y 2x is positive for x 0, y is zero when x 0, y is negative when x 0
www
œ œ
(d) y x is increasing for x 0 and decreasing for x ; the function is increasing on intervalsœ _ ! _
#
where y 0 and decreasing on intervals where y 0
ww
&Ð
46. (a) f (x) lim lim lim lim
w
È
œœ
""
œœœœœ
hhhhÄ! Ä! Ä! Ä!
f(x h) f(x) x (x h)
h h x(x h)h x(x h) x
Š‹
$œ$
È
xh x
1
œ

131. Section 3.1 The Derivative of a Function 127
(b)
(c) y is positive for all x 0, y is never 0, y is never negative
www
Á
(d) y is increasing for x 0 and xœ _ ! _
"
x
47. (a) Using the alternate formula for calculating derivatives: f (x) lim lim
w È
ÈÈ
È
œœ
zx zxÄÄ
f(z) f(x)
z x z x
Š‹
zx
33
””
lim lim lim x f (x) xœœ œ œÊœ
zx zx zxÄÄ Ä
z x z zx x
3(z x) 3(z x) 3
(z x) z zx x
ÈÈ œœœœ

ÈÈ
#w #ab
(b)
(c) y is positive for all x 0, and y 0 when x 0; y is never negative
www
Áœ œ
(d) y is increasing for all x 0 (the graph is horizontal at x 0) because y is increasing where y 0; y isœÁ œ
x
3
È
w
never decreasing
48. (a) Using the alternate form for calculating derivatives: f (x) lim lim
w È
ÈÈ
È
œœ
zx zxÄÄ
f(z) f(x)
z x z x
?
zx
44
••
lim lim lim x f (x) xœœ œ œÊœ
zx zx zxÄÄ Ä
z x z xz x z x
4(z x) 4(z x) 4
(z x) z xz x z x
‹‹

ÈÈ
$w $ab
(b)
(c) y is positive for x 0, y is zero for x 0, y is negative for x 0
www
œ
(d) y is increasing on 0 x and decreasing on x 0œ__
x
4
‹
49. y lim lim lim lim x xc c 3c .
w ###È
ÈÈ È
È
œœœ œœ
xc xc xc xcÄÄÄ Ä
f(x) f(c)
xc xc xc
xc (x c) x xc c
ÈÈ œœ ab
ab
The slope of the curve y x at x c is y 3c . Notice that 3c 0 for all c y x never has a negativeœœœ   Êœ
$w# # $
slope.
50. Horizontal tangents occur where y 0. Thus, y lim
ww

œœ
hÄ!
2xh2x
h
È È
lim lim lim .œœœœ
hhhÄ! Ä! Ä!
2xh x xh x
h
xh x h xh x
2((x h) x)) 2
xh x x
Š‹Š‹
ÈÈ ÈÈ
Š‹ Š‹
ÈÈ ÈÈ
ÈÈ È

œœ œœ

œœ
"
†

132. 128 Chapter 3 Differentiation
Then y 0 when 0 which is never true the curve has no horizontal tangents.
w "
œœ Ê
Èx
51. y lim lim
w
œœ
hhÄ! Ä!
abab2(x h) 13(x h) 5 2x 13x 5
hh
2x 4xh 2h 13x 13h 5 2x 13x 5
œœ œœ œ
lim lim (4x 2h 13) 4x 13, slope at x. The slope is 1 when 4x 13œœœ œ"
hhÄ! Ä!
4xh 2h 13h
h

œ
4x 12 x 3 y 2 3 13 3 5 16. Thus the tangent line is y 16 ( 1)(x 3)Ê œ ʜʜ œ œ ††
#
y x and the point of tangency is (3 16).Êœ"$ ß
52. For the curve y x, we have y lim lim œœ œÈ
w

œœ œœ

hhÄ! Ä!
Š‹Š‹
ÈÈ ÈÈ
Š‹ Š‹
ÈÈ ÈÈ
xh x xh x
h
xh x xh xh
(x h) x
†
lim . Suppose a is the point of tangency of such a line and ( ) is the pointœœ+ß "ß!
hÄ!
""
œœ #ÈÈ Èxh x x
ˆ‰È
on the line where it crosses the x-axis. Then the slope of the line is which must also equal
ÈÈa0 a
a(1) a1
È

œ
; using the derivative formula at x a 2a a 1 a 1. Thus such a line does
""
œ #2a a
a
a1ÈÈ
È
œÊ œ Ê œÊœ
exist: its point of tangency is ( ), its slope is ; and an equation of the line is y 1 (x 1)"ß " œ œ
"" "
# ##Èa
yx.ʜ
""
##
53. No. Derivatives of functions have the intermediate value property. The function f(x) x satisfies f(0) 0œÚ Û œ
and f(1) 1 but does not take on the value anywhere in [ ] f does not have the intermediate valueœ!ß"Ê
"
#
property. Thus f cannot be the derivative of any function on [ ] f cannot be the derivative of any function!ß " Ê
on ( )._ß _
54. The graphs are the same. So we know that
for f(x) x , we have f (x) .œœkk
w kkx
x

55. Yes; the derivative of f is f so that f (x ) exists f (x ) exists as well.ÈÈ •È
ww w
!!
56. Yes; the derivative of 3g is 3g so that g (7) exists 3g (7) exists as well.
ww w
Ê
57. Yes, lim can exist but it need not equal zero. For example, let g(t) mt and h(t) t. Then g(0) h(0)
tÄ!
g(t)
h(t)
œœ œ
0, but lim lim lim m m, which need not be zero.œœœœ
tttÄ! Ä! Ä!
g(t)
h(t) t
mt
58. (a) Suppose f(x) x for x 1. Then f(0) 0 f(0) 0. Then f (0) lim kk kkŸ"ŸŸ ŸÊœ œ
##w
hÄ!
f(0 h) f(0)
h
lim lim . For h 1, h f(h) h h h f (0) lim 0œœ ŸŸŸÊŸŸÊœœ
hh hÄ! Ä! Ä!
f(h) 0 f(h) f(h) f(h)
hh h h
È ## w
kk
by the Sandwich Theorem for limits.
(b) Note that for x 0, f(x) x sin x sin x x 1 x (since sin x 1). By part (a),Áœ œ Ÿœ "ŸŸk k kkk k kk¸¸
## ##"
x
†
f is differentiable at x 0 and f (0) 0.œœ
w
59. The graphs are shown below for h 1, 0.5, 0.1. The function y is the derivative of the functionœœ
"
2xÈ
y x so that lim . The graphs reveal that y gets closer to yœœ œ œÈ
" "
# #

È È
ÈÈ ÈÈ
x x
xh x xh x
hh
hÄ!

133. Section 3.1 The Derivative of a Function 129
as h gets smaller and smaller.

60. The graphs are shown below for h 2, 1, 0.5. The function y 3x is the derivative of the function y x soœœ œ
#$
that 3x lim . The graphs reveal that y gets closer to y 3x as h
# #
œœœ
hÄ!
(x h) x (x h) x
hh

gets smaller and smaller.

61. Weierstrass's nowhere differentiable continuous function.

62-67. Example CAS commands:
:Maple
f := x -> x^3 + x^2 - x;

134. 130 Chapter 3 Differentiation
x0 := 1;
plot( f(x), x=x0-5..x0+2, color=black,
title="Section 3_1, #62(a)" );
q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b)
L := limit( q(x,h), h=0 ); # (c)
m := eval( L, x=x0 );
tan_line := f(x0) + m*(x-x0);
plot( [f(x),tan_line], x=x0-2..x0+3, color=black,
linestyle=[1,7], title="Section 3.1 #62(d)",
legend=["y=f(x)","Tangent line at x=1"] );
Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ): # (e)
Yvals := map( f, Xvals ):
evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >);
plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" );
: (functions and x0 may vary) (see section 2.5 re. RealOnly ):Mathematica
<<Miscellaneous`RealOnly`
Clear[f, m, x, y, h]
x0= /4;1
f[x_]:=x Cos[x]
2
Plot[f[x], {x, x0 3, x0 3}]
q[x_, h_]:=(f[x h] f[x])/h
m[x_]:=Limit[q[x, h], h 0]Ä
ytan:=f[x0] m[x0] (x x0)
Plot[{f[x], ytan},{x, x0 3, x0 3}]
m[x0 1]//N
m[x0 1]//Nw
Plot[{f[x], m[x]},{x, x0 3, x0 3}]
3.2 DIFFERENTIATION RULES
1. y x 3 x (3) 2x 0 2œ Ê œ œ œ#B Ê œ
## dy d y
dx dx dx dx
dd
ab
œ
œ
2. y x x 8 2x 1 0 2x 1 œ Ê œ œ Ê œ#
# dy d y
dx dx
œ
œ
3. s 5t 3t 5t 3t 15t 15t 15t 15t 30t 60tœÊœœÊœ œ
$& $ & # % # % $ds d d d s d d
dt dt dt dt dt dt
ab ab ab ab
œ
œ
4. w 3z 7z 21z 21z 21z 42z 126z 42z 42œ Êœ Ê œ
($ # ' # & dw d w
dz dz
œ
œ
5. y x x 4x 1 8xœÊœÊœ
4
3dx dx
dy d y$#
œ
œ
6. y x x 2x 1 0 2x 1œÊ œÊ œ œ
xxx
34dx 4dx
dy d y
œ
œ
#
# "
7. w 3z z 6z z 18z 2zœÊœœÊœ œ
# " $ # % $ "dw 6 d w 18 2
dz z z dz z z

œ
$
8. s 2t 4t 2t 8t 4t 24tœ Ê œ œ Ê œ œ
" # # $ $ % ds 2 8 d s 4 24
dt t t dt t t

œ
$
9. y 6x 10x 5x 12x 10 10x 12x 10 12 0 30x 12œ Ê œ œ Ê œ œ
## $ % dy d y
dx x dx
10 30
x

œ
$

135. Section 3.2 Differentiation Rules 131
10. y 4 2x x 3x 0 12xœ Ê œ# œ# Ê œ œ
È” È“ È Èdy d y
dx dx x
312
x
‹
œ
œ"
11. r s s s s 2s 5sœ Êœ œÊœœ
‹ÈÈ# È‹ È” È# È“ È”
##3 ds 3 3s 2s ds s
5dr2525dr 25
s

œ
‹
12. r 12 4 12 12 4 24 48 20œÊœœÊœ))) ) )) ) ) )
È‹ È” È“ È# È“ È È” È È Èdr 12 12 4 d r
dd)))) )
œ "œ‹
œ
œ
24 48 20
)))
È”•
13. (a) y 3 x x x 1 y 3 x x x 1 x x 1 3 xœ Ê œ abababababab
#$ w # $ $ #
††
dd
dx dx
3 x 3x 1 x x 1 ( 2x) 5x 12x 2x 3œ œ aba ba b
## $ % #
(b) y x 4x x 3x 3 y 5x 12x 2x 3œ Êœ
&$# w % #
14. (a) y (x 1) x x 1 y (x 1)(2x 1) x x 1 ( ) 3xœ Ê œ "œab ab
#w # #
(b) y (x 1) x x 1 x 1 y 3xœ œÊ œab
#$w#
15. (a) y x 1 x 5 y x 1 x 5 x 5 x 1œ Ê œ ab ab abˆ‰ ˆ‰ˆ‰
#w# # """
xdxxxdx
dd
††
x 1 1 x x 5 x (2x) x 1 1 x 2x 10x 2 3x 10x 2œ œ œ aba ba b a ba b
#È# È‹# È## # "
x
œ
(b) y x 5x 2x 5 y 3x 10x 2œ Ê œ
$# w # ""
xx
œ
16. y x x 1œ ˆ‰ˆ ‰
""
xx
(a) y x x 1 x x x 1 1 x 2x 1
ŠÈ‹È# È‹ È# "
œ œababa bab†
xx
2
œ È
(b) y x x y 2x 1œ Ê œ
#w "" "
xx x x
2

17. y ; use the quotient rule: u 2x 5 and v 3x 2 u 2 and v 3 yœœœÊœœÊœ
2x 5 vu uv
3x 2 v
œ È
È
ww w
’’
œ
œœœ
(3x 2)(2) (2x 5)(3)
(3x 2) (3x ) (3x 2)
6x46x15 19
ÈÈ#È
ÈÈ È È
œœœ
18. z œÊœ œ œ œ
2x 1 dz 2x 2 4x 2x 2x 2x 2
x1 dx
x 1 (2) (2x 1)(2x) 2 x x 1
x1 x1 x1 x1

È

ÈÈÈÈ
œ
œ œ
œœœœ
œœœœ
œœ œ
ab a b
ab ab ab ab
19. g(x) ; use the quotient rule: u x 4 and v x 0.5 u 2x and v 1 g (x)œœœÊœœÊœ
x4 vu uv
x0.5 v
œ ’’
œ
È È
œ
#www
œœœ
(x 0.5)(2x) x 4 ( )
(x 0.5) (x 0.5) (x 0.5)
2x xx 4 x x4"
œœœ
ab
œ
œœœ
œœ œ
20. f(t) , t f (t)œœ œÁ"Êœ œ œ
tt tt
tt2 t t t2
tt t t
t2 t2 t2
œ
œ
" " # " "
# "
" " # " " " w
œœœ
abab abababab
abab ab ab ab
222
21. v (1 t) 1 t œ œ Ê œ œ œab
#
È‹ " "
œ
"
œœœ
1 t dv t 2t 2t t 2t
1t dt
1 t ( ) (1 t)(2t)
1t 1t 1t
œ
œ
œœœ
œœœ
œœœ
ab
ab ab ab
22. w wœÊœ œ œ
x5 2x72x10 17
2x 7 (2x 7) (2x 7) (2x 7)
(2x 7)(1) (x 5)(2)
ÈÈÈÈ
w
œœœ
23. f(s) f (s)œÊœ œ œ
Ȉ‰ˆ‰
È
ˆ‰ ˆ‰ÈÈ Š‹ Š‹
ˆ‰ ˆ‰ ˆ‰ÈÈÈÈÈ
ÈÈs
s1
ss1
s1 2 s s 1 s s 1
ss1È‹
œ
w
"
œœœ
" "
œœ
ŠŠœœ
ss
œœœ
NOTE: s from Example 2 in Section 2.1
d
ds s
ˆ‰Èœ
"
#È
24. u œÊœ œ
5x du 5x 1
x dx 4x
2 x (5) (5x 1)
4x
"
#

È
ˆ‰È Š‹
œ
œ
x
È%œ
25. v vœÊœ œ
1x4x 2x
xxx
x1 1x4x
"
w

ÈÈ Š‹ ˆ‰ È
2
x
œ
œœ

136. 132 Chapter 3 Differentiation
26. r 2 r 2œÊœ œŠ‹
È

""" w
È
È
È
Š‹
)
)
) ))
)
(0) 1
Š
œ

œ
œ
"
#
È
)
27. y ; use the quotient rule: u 1 and v x 1 x x 1 u 0 andœœœÊœ
"

## w
aba bx1xx1
œœ aba b
v x1(2x1)xx1(2x)2xx2x12x2x2x4x3x1
w# # $# $# $#
œ œœab a b
Êœ œ œ
dy
dx v
vu uv 4x 3x 10 1 4x 3x 1
x1xx1 x1xx1

œ

œœ œœ
œœ œœ


ab
aba b aba b
28. y yœœÊœ œ
(x 1)(x 2)
(x1)(x ) x 3x2 (x1)(x2) (x1)(x2)
x 3x 2 6x 12 x 3x 2 (2x 3) x 3x 2 (2x 3)œœ
#
w
œ œ
œœœœœ
œœ abab
œ
ÈÈ
ÈÈ
6x 2
(x 1) (x 2)
ab
œ
œœ
29. y x x x y 2x 3x 1 y 6x 3 y 12x y 12 y 0 for all n 5œ Ê œ Ê œ Ê œ Ê œ Ê œ  
"
##
% # w $ ww # www Ð%Ñ Ð Ñ3 n
30. y x y x y x y x y x y 1 y 0 for all n 6œ Êœ ʜʜʜʜʜ  
""""
## #
&w %ww$www#Ð%Ñ Ð&Ñ ÐÑ
10 4 6 n
31. y x 7x 2x 7x x 2 14xœ œ Ê œ œ# Ê œ œ#
x7
xdx xdx x
dy d y
È

œ
œß$% #È‹ È# È”
32. s 1 1 5t t 0 5t 2t 5t 2t œ œ œ Ê œ œ œ
t5t1 5 ds
ttt dt tt
œ
œœ
" È #È‹ È# È# È” È# È”
10t 6tÊœ œ
ds
dt t t
œ
‹
È” È“ "! '
33. r 1 0 12œœœ"œÊœ$œ$œÊœœ
() 1 dr d r
dd
)))
))) ) ) )
)
)
" È ‹ ‹ ” È‹# È” È“ È“ Èab
œ
ÈÈÈ œ"
È œ
‹
))) )
34. u 1 1 xœ œ œ œ œ œ
aba b a b abxxxx1 x(x1)xx xx1
xxxxx
xx x
œœ œ È
‹‹‹‹‹
‹
" œ È”
0 3x 3x 12xÊœ œ œÊ œ œ
du d u
dx dx x x
È“ È“ È È” ‹#
‹
œ
œ"
35. w (3 z) z 1 (3 z) z 3 z z z z 0 1 z 1œ œ œ œ Ê œœˆ‰ ˆ ‰
13z 8dw
3z 3 3 3 dz
œ$ $ È‹ È‹ È‹ È# È#
2z 0 2zœ"Ê œ œ œ
È‹ # È” È”
zdz z
dw
œœ È
œ
36. w (z 1)(z 1) z 1 z 1 z 1 z 1 4z 0 4z 12zœ œ œÊ œ œ Ê œababab
###% $$ # dw d w
dz dz
œ
œ
37. p q q q q q q qœœœÊœœŠ‹Š‹
q3 q1 qq3q3 dp
12q q 1 1 4 4 dq 6 6 6 6q q12q
œ$ Êœ$
È È”‹
"" "" "" """
##
#È# È“ È”È
q5qÊœ œ
dp
dq 6 6 q q
œ
œ ‹
"" " " &
#
È“ È
#
6
38. p qœœ œœœœ
q3 q3 q3 q3
(q 1) (q 1) q 3q 3q 1 q 3q 3q 1 2q 6q 2q q 3 q
œœœœ

œœœœ
# #
"" È‹
abab ab q qÊœ œÊ œ œ
dp d p
dq q dq q
"" "
##
È# È”

œ
39. u(0) 5, u (0) 3, v(0) 1, v (0) 2œœœœ
ww
(a) (uv) uv vu (uv) u(0)v (0) v(0)u (0) 5 2 ( 1)( 3) 13
dd
dx dx
œ Ê œ œ œ
ww w w ¸
x=0
†
(b) 7
d u vu uv d u
dx v v dx v (v(0)) ( 1)
v(0)u (0) u(0)v (0) ( )( 3) (5)(2)
ˆ‰ ¸ ˆ‰œÊ œ œ œ
’’
œœœ
’’
È ÈÈ‹ÈÈ
È
x=0
(c)
d v uv vu d v 7
dx u u dx u (u(0)) (5) 25
u(0)v (0) v(0)u (0) (5)(2) ( 1)( 3)
ˆ‰ ¸ ˆ‰œÊ œ œ œ
’’
œœœ
’’
È ÈÈÈÈ
x=0
(d) (7v 2u) 7v 2u (7v 2u) 7v (0) 2u (0) 7 2 2( 3) 20
dd
dx dx
œ Ê œ œ œ
ww w w ¸
x=0
†

137. Section 3.2 Differentiation Rules 133
40. u(1) 2, u (1) 0, v(1) 5, v (1) 1œœœœ
ww
(a) (uv) u(1)v (1) v(1)u (1) 2 ( 1) 5 0 2¸
d
dx
x=1
œœœ
ww
††
(b) ¸ˆ‰
du 2
dx v (v(1)) (5) 25
v(1)u ( ) u(1)v (1) 5 0 2 ( 1)
x=1
œœœ
ßß
œœ
‹È È È ††
(c) ¸ˆ‰
dv 1
dx u (u(1)) (2) 2
u(1)v ( ) v(1)u (1) 2 ( 1) 5 0
x=1
œœœ
ßß
œœ
‹È È È ††
(d) (7v 2u) 7v (1) 2u (1) 7 ( 1) 2 0 7¸
d
dx
œœœ
x=1
ww
††
41. y x 4x 1. Note that ( ) is on the curve: 1 2 4(2) 1œ #ß" œ
$$
(a) Slope of the tangent at (x y) is y 3x 4 slope of the tangent at ( ) is y (2) 3(2) 4 8. ThusßœÊ #ß" œœ
w# w #
the slope of the line perpendicular to the tangent at ( ) is the equation of the line perpendicular to‰’ ” È •
"
8
to the tangent line at ( ) is y 1 (x 2) or y .#ß " œ œ
"
884
x5
(b) The slope of the curve at x is m 3x 4 and the smallest value for m is 4 when x 0 and y 1.œ œ œ
#
(c) We want the slope of the curve to be 8 y 8 3x 4 8 3x 12 x 4 x 2. WhenÊ œÊ œÊ œ Ê œÊœ„
w# ##
x 2, y 1 and the tangent line has equation y 1 8(x 2) or y 8x 15; when x 2,œœ œ œ œ
y ( 2) 4( 2) 1 1, and the tangent line has equation y 1 8(x 2) or y 8x 17.œ œ œ œ
$
42. (a) y x 3x 2 y 3x 3. For the tangent to be horizontal, we need m y 0 0 3x 3œÊ œ œœÊœ
$w# w #
3x 3 x 1. When x 1, y 0 the tangent line has equation y 0. The lineʜʜ„ œœÊ œ
#
perpendicular to this line at ( ) is x 1. When x 1, y 4 the tangent line has equation"ß ! œ œ œ Ê
y 4. The line perpendicular to this line at ( ) is x 1.œ "ß % œ
(b) The smallest value of y is 3, and this occurs when x 0 and y 2. The tangent to the curve at ( 2)
w
œ œ !ß
has slope 3 the line perpendicular to the tangent at ( 2) has slope y 2 (x 0) or Ê !ß Ê œ
""
33
y x 2 is an equation of the perpendicular line.œ
"
3
43. y . When x 0, y 0 and yœÊœ œ œ œœ œ
4x 4x 4 8x
x1 dx 1
dy 4(0 1)x 1 (4) (4x)(2x) 4 x
x1 x1 x1
œ
œœ
œœœ
œœœ
œœ
œ
"
œœœ
w œab a b
ab ab ab
, so the tangent to the curve at ( ) is the line y 4x. When x 1, y 2 y 0, so the tangent to theœ% !ß! œ œ œ Ê œ
w
curve at ( 2) is the line y 2."ß œ
44. y y . When x 2, y 1 and y , so the tangentœÊœ œ œœ œ œ
8 16x
x4
x 4 (0) 8(2x)
x4 x4 24
16(2)
œ
œ
œœ œ
œœ œ
œ #
w w
œœ œ
È‹ Èab
ab ab ab
line to the curve at (2 ) has the equation y 1 (x 2), or y 2.ß" œ œ
"
##
x
45. y ax bx c passes through ( ) 0 a(0) b(0) c c 0; y ax bx passes through ( )œ !ß!Êœ Êœ œ "ß#
# #
2 a b; y 2ax b and since the curve is tangent to y x at the origin, its slope is 1 at x 0Êœ œ œ œ
w
y 1 when x 0 1 2a(0) b b 1. Then a b 2 a 1. In summary a b 1 and c 0 soÊœ œÊœ Êœ œÊœ œœ œ
w
the curve is y x x.œ
#
46. y cx x passes through ( ) 0 c(1) 1 c 1 the curve is y x x . For this curve,œ "ß ! Ê œ Ê œ Ê œ
# #
y 1 2x and x 1 y 1. Since y x x and y x ax b have common tangents at x 0,
ww##
œ œ Ê œ œ œ œ
y x ax b must also have slope 1 at x 1. Thus y 2x a 1 2 1 a a 3œ œ œ Êœ Êœ
#w
†
y x 3x b. Since this last curve passes through ( ), we have 0 1 3 b b 2. In summary,Êœ "ß! œÊœ
#
a 3, b 2 and c 1 so the curves are y x 3x 2 and y x x .œœœ œœ
##
47. (a) y x x y 3x 1. When x 1, y 0 and y 2 the tangent line to the curve at ( ) isœÊ œ œ œ œÊ "ß!
$w# w
y 2(x 1) or y 2x 2.œ œ

138. 134 Chapter 3 Differentiation
(b)
(c) x x 2x 2 x 3x 2 (x 2)(x 1) 0 x 2 or x 1. Since
yx x
y2x2
w
œ
œ
ʜʜ œÊœ œ
$
$$ #
y 2 2 2 6; the other intersection point is (2 6)œœ ßab
48. (a) y x 6x 5x y 3x 12x 5. When x 0, y 0 and y 5 the tangent line to the curve atœ Ê œ œ œ œÊ
$# w # w
(0 0) is y 5x.ߜ
(b)
(c) x 6x 5x 5x x 6x 0 x (x 6) 0 x 0 or x 6.
y x 6x 5x
y 5x
w
œ
œ
Ê œ Ê œÊ œÊœ œ
$#
$# $# #
Since y 5 6 , the other intersection point is (6 30).œ œ $! ßab
49. P x a x a x a x a x a P x na x n a x a x aab ab a bœ â Ê œ " â#
nn n n
nn n n
" # " ! " # "
" # w " #
50. R M M M , where C is a constant CM Mœœ Êœ
##$ #
##
"ˆ‰
CM C dR
33 dM
51. Let c be a constant 0 (u c) u c u 0 c c . Thus when one of theÊœÊ œ œ œ
dc d dc du du du
dx dx dx dx dx dx
††† †
functions is a constant, the Product Rule is just the Constant Multiple Rule the Constant Multiple Rule isÊ
a special case of the Product Rule.
52. (a) We use the Quotient rule to derive the Reciprocal Rule (with u 1): œœœ
d
dx v v v
v0 1
ˆ‰
"
"†† †
dv dv
dx dx
œœ
. œ
"
vdx
dv
Ϡ
(b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule: u
du d
dx v dx v
ˆ‰ ˆ ‰œ †
"
u (Product Rule) u (Reciprocal Rule) œ œ Êœ†† †
d du 1 dv du d u
dx v v dx v dx v dx dx v v
u v
ˆ‰ ˆ ‰ ˆ‰
"" "

œ œ
dv du
dx dx
, the Quotient Rule.œ
v u
v
du dv
dx dx

œ
53. (a) (uvw) ((uv) w) (uv) w (uv) uv w u v uv wu wv
dd dwd dwdvdudwdvdu
dx dx dx dx dx dx dx dx dx dx
œœœœ†† ˆ‰
uvw uv w u vwœ
ww w
(b) uuuu uuu u uuu u uuu uuuu
dd dd
dx dx dx dx dx
du
a ba bab ab a bab"#$% "#$ % "#$ % "#$ "#$%œœÊ
$
uuu u uu uu uu (using (a) a bove)œ "#$%"#$"$#
du du du du
dx dx dx dx
$ ˆ‰
uuuu uuu uuu uuu uuu ʜ
d
dx dx dx dx dx
du du du du
ab"#$% "#$ "#% "$% #$%

uuuu uuuu uuuu uuuuœ "#$ "# % " $% #$%
wwww
%$#"
(c) Generalizing (a) and (b) above, u u u u u u u u u u u u u u
d
dx
ab" "#" "## #
www
" "✠â â á â
nn nn n nn

139. Section 3.3 The Derivative as a Rate of Change 135
54. In this problem we don't know the Power Rule works with fractional powers so we can't use it. Remember
x (from Example 2 in Section 2.1)
d
dx x
ˆ‰Èœ
"
#È
(a) x x x x x x (x) x x 1 x x
dd d d 3
dx dx dx dx x
x3x
ˆ‰ ˆ ‰ ˆ‰ ÈÈ È È
$Î# "Î# "Î#"
# ###
œœœœœœ†† † †
È
ÈÈ
(b) x x x x x x x x x 2x x 2x x
dd d d 5
dx dx dx dx x
ˆ‰ ˆ ‰ ˆ‰ ÈÈ È ab Š‹
&Î# # "Î# # # # $Î# $Î# $Î# ""
# ##
œœœœœ†††
È (c) x x x x x x x x x 3x x 3x x
dd d d 7
dx dx dx dx x
ˆ‰ ˆ ‰ ˆ‰ ÈÈ È ab Š‹
(Î# $ "Î# $ $ $ # &Î# &Î# &Î# ""
# ##
œœœœœ†††
È (d) We have x x , x x , x x so it appears that x x
d3d5d7 dn
dx dx dx dx
ˆ‰ ˆ‰ ˆ‰ ab
$Î# "Î# &Î# $Î# (Î# &Î# Ð Î#Ñ"
### #
œœœ œ
nn•È
whenever n is an odd positive integer 3. 
55. p . We are holding T constant, and a, b, n, R are also constant so their derivatives are zeroœ
nRT an
Vnb VÈ
œ
œ
Êœ œ
dP nRT 2an
dV (V nb) (V nb) V
(V nb) 0 (nRT)(1) V(0) an (2V)
V
ÈÈ
ÈÈ
È È†
œœ È
œœ
œ
œ
œ
ab
ab
56. A q cm km q cm q km q km qab ab ab ab ˆ‰ ˆ‰œ œ Ê œ œ Ê œ# œ
km h dA h km h d A km
qdqqdtq
hq
## ##
È‹ È# È” #
œœ È
œ
3.3 THE DERIVATIVE AS A RATE OF CHANGE
1. s t t , 0 tœ$#ŸŸ#
# (a) displacement s s( ) s(0) m m m, v m/secœ œ # œ! # œ# œ œ œ"?
av
?
?
s
t
È#
#
(b) v t v(0) m/sec and v( ) 1 m/sec;œ œ# $ Ê œl$lœ$ # œ
ds
dt
kk kk
a a(0) m/sec and a( ) m/secœœ#Ê œ# #œ#
ds
dt
œ
œ
##
(c) v 0 t 0 t . v is negative in the interval t and v is positive when t the bodyœ Ê# $œ Ê œ ! #Ê
$$$
###
changes direction at t .œ
$
#
2. s t t , tœ' !Ÿ Ÿ'
#
(a) displacement s s( ) s(0) m, v m/secœ œ ' œ! œ œ œ!?
av
?
?
s
t
!
'
(b) v v(0) m/sec and v( ) m/sec;œ œ'#> Ê œl'lœ' ' œl'lœ'
ds
dt
kk kk
a a(0) m/sec and a( ) m/secœ œ # Ê œ # ' œ #
ds
dt
œ
œ
##
(c) v 0 t 0 t . v is positive in the interval t and v is negative when t the bodyœ Ê'#œ Êœ$ !$ $'Ê
changes direction at t .œ$
3. s t 3t 3t, 0 t 3œ Ÿ Ÿ
$#
(a) displacement s s(3) s(0) 9 m, v 3 m/secœœœ œœœ?
av
?
?
s9
t3
È
(b) v 3t 6t 3 v(0) 3 3 m/sec and v(3) 12 12 m/sec; a 6t 6œœÊ œœ œ œ œ œ
ds ds
dt dt
#
kkkk kkkk
œ
œ
a(0) 6 m/sec and a(3) 12 m/secÊœ œ
##
(c) v 0 3t 6t 3 0 t 2t 1 0 (t 1) 0 t 1. For all other values of t in theœ Ê œ Ê œ Ê œ Êœ
## #
interval the velocity v is negative (the graph of v 3t 6t 3 is a parabola with vertex at t 1 whichœ œ
#
opens downward the body never changes direction).Ê
4. s t t , 0 tœ ŸŸ$
t
4
‹
$#
(a) s s( ) s(0) m, v m/sec?œ$ œ œ œ œ
*$
%$%
av
?
?
s
t

‹
(b) v t 3t 2t v(0) 0 m/sec and v( ) m/sec; a 3t 6t 2 a(0) 2 m/sec andœ Ê œ $œ' œ Ê œ
$# ##
kk kk
a( ) m/sec$œ""
#
(c) v 0 t 3t 2t 0 t(t 2)(t 1) 0 t 0, 1, 2 v t(t 2)(t 1) is positive in theœÊ œÊ œÊœ Êœ
$#
interval for 0 t 1 and v is negative for 1 t 2 and v is positive for t the body changes direction atÐÐ ÐÐ ÒÐÐ"œ

140. 136 Chapter 3 Differentiation
t 1 and at t .œœ#
5. s , 1 t 5œ ŸŸ
25 5
tt
œ
(a) s s(5) s(1) 20 m, v 5 m/sec?œœ œœ
av
È20
4
(b) v v(1) 45 m/sec and v(5) m/sec; a a(1) 140 m/sec andœÊ œ œ œÊ œ
È‹ #50 5 150 10
tt 5 t t
È œ È‹ kk kk
a(5) m/secœ
4
25
#
(c) v 0 0 50 5t 0 t 10 the body does not change direction in the intervalœÊ œÊœÊœ Ê
50 5t
t
È
6. s , t 0œ%ŸŸ
25
t5œ
(a) s s(0) s( 4) 20 m, v 5 m/sec?œœ œœ
av
20
4
(b) v v( 4) 25 m/sec and v(0) m/sec; a a( 4) 50 m/sec andœÊœ œ" œÊœ
È
œ œ
#25 50
(t 5) (t 5)
œ È kk kk
a(0) m/secœ
2
5
#
(c) v 0 0 v is never 0 the body never changes directionœÊ œÊ Ê
È
œ
25
(t 5)
œ
7. s t 6t 9t and let the positive direction be to the right on the s-axis.œ
$#
(a) v 3t 12t 9 so that v 0 t 4t 3 (t 3)(t 1) 0 t 1 or 3; a 6t 12 a(1)œ œÊœ œÊœ œÊ
##
6 m/sec and a(3) 6 m/sec . Thus the body is motionless but being accelerated left when t 1, andœ œ œ
##
motionless but being accelerated right when t 3.œ
(b) a 0 6t 12 0 t 2 with speed v(2) 12 24 9 3 m/secœÊ œÊœ œ œ kkk k
(c) The body moves to the right or forward on 0 t 1, and to the left or backward on 1 t 2. TheÎÐ ÐÐ
positions are s(0) 0, s(1) 4 and s(2) 2 total distance s(1) s(0) s(2) s(1) 4 2œ œ œÊ œœ kkkkkkkk
6 m.œ
8. v t 4t 3 a 2t 4œÊœ
#
(a) v 0 t 4t 3 0 t 1 or 3 a(1) 2 m/sec and a(3) 2 m/secœÊ œÊœ Ê œ œ
###
(b) v 0 (t 3)(t 1) 0 0 t 1 or t 3 and the body is moving forward; v 0 (t 3)(t 1) 0Ê ÊŸ Ê
t 3 and the body is moving backwardœ$Ð Ð
(c) velocity increasing a 0 2t 4 0 t 2; velocity decreasing a 0 2t 4 0 t 2ÊÊ Ê ÊÊ Ê!Ÿ
9. s 1.86t v 3.72t and solving 3.72t 27.8 t 7.5 sec on Mars; s 11.44t v 22.88t and
mm jjœÊœ œÊ¸ œ Êœ
# #
solving 22.88t 27.8 t 1.2 sec on Jupiter.œÊ¸
10. (a) v(t) s (t) 24 1.6t m/sec, and a(t) v (t) s (t) 1.6 m/secœœ œœ œ
www# w
(b) Solve v(t) 0 24 1.6t 0 t 15 secœÊ œÊœ
(c) s(15) 24(15) .8(15) 180 mœ œ
#
(d) Solve s(t) 90 24t .8t 90 t 4.39 sec going up and 25.6 sec going downœÊ œÊœ ¸
# „
#
30 15 2
È
(e) Twice the time it took to reach its highest point or 30 sec
11. s 15t g t v 15 g t so that v 0 15 g t 0 g . Therefore g 0.75 m/secœ Êœ œÊœÊœ œœœ
"
#
# #
ss ss s
15 15 3
t204
12. Solving s 832t 2.6t 0 t(832 2.6t) 0 t 0 or 320 320 sec on the moon; solving
mœ œÊ œÊœ Ê
#
s 832t 16t 0 t(832 16t) 0 t 0 or 52 52 sec on the earth. Also, v 832 5.2t 0
e mœœÊ œÊœ Ê œœ
#
t 160 and s (160) 66,560 ft, the height it reaches above the moon's surface; v 832 32t 0Êœ œ œ œ
m e
t 26 and s (26) 10,816 ft, the height it reaches above the earth's surface.Êœ œ
e
13. (a) s 179 16t v 32t speed v 32t ft/sec and a 32 ft/secœ ÊœÊ œœ œ
# #
kk

141. Section 3.3 The Derivative as a Rate of Change 137
(b) s 0 179 16t 0 t 3.3 secœÊ œÊœ ¸
#
É
179
16
(c) When t , v 32 8 179 107.0 ft/secœœœ¸ÉÉ È
179 179
16 16
14. (a) lim v lim 9.8(sin )t 9.8t so we expect v 9.8t m/sec in free fall
))ÄÄ
œœ

œœ œ )
(b) a 9.8 m/secœœ
dv
dt
#
15. (a) at 2 and 7 seconds (b) between 3 and 6 seconds: t 6 $Ÿ Ÿ
(c) (d)
16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standingÐÐ ÐÐ ÐÐ
still when 1 t 2 or 3 t 5ÐÐ ÐÐ
(b)
17. (a) 190 ft/sec (b) 2 sec
(c) at 8 sec, 0 ft/sec (d) 10.8 sec, 90 ft/sec
(e) From t 8 until t 10.8 sec, a total of 2.8 secœœ
(f) Greatest acceleration happens 2 sec after launch
(g) From t 2 to t 10.8 sec; during this period, a 32 ft/secœœ œ ¸
v(10.8) v(2)
10.8 2


#
18. (a) Forward: 0 t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6;ÎÐ ÐÐ ÐÐ ÐÐ ÐÐ
Slows down: 0 t 1, 3 t 5, and 6 t 7ÎÐ ÐÐ ÐÐ
(b) Positive: 3 t 6; negative: 0 t 2 and 6 t 7; zero: 2 t 3 and 7 t 9ÐÐ ÎÐ ÐÐ ÐÐ ÐÐ
(c) t 0 and 2 t 3œŸŸ
(d) 7 t 9ŸŸ
19. s 490t v 980t a 980œÊœÊœ
#
(a) Solving 160 490t t sec. The average velocity was 280 cm/sec.œÊœ œ
# 4
74/7
s(4/7) s(0)
(b) At the 160 cm mark the balls are falling at v(4/7) 560 cm/sec. The acceleration at the 160 cm markœ
was 980 cm/sec .
#
(c) The light was flashing at a rate of 29.75 flashes per second.
17
4/7
œ

142. 138 Chapter 3 Differentiation
20. (a)
(b)
21. C position, A velocity, and B acceleration. Neither A nor C can be the derivative of B because B'sœœ œ
derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while
C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position.
Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That
leaves B for acceleration.
22. C position, B velocity, and A acceleration. Curve C cannot be the derivative of either A or B becauseœœ œ
C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C
has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is
negative where B has negative slopes and positive where B has positive slopes.
23. (a) c(100) 11,000 c $110œÊœœ
av
11,000
100
(b) c(x) 2000 100x .1x c (x) 100 .2x. Marginal cost c (x) the marginal cost of producing 100œÊ œ œÊ
#w w
machines is c (100) $80
w
œ
(c) The cost of producing the 101 machine is c(101) c(100) 100 $79.90
st
œœ
201
10
24. (a) r(x) 20000 1 r (x) , which is marginal revenue.œÊœˆ‰
" w
xx
20000
œ
(b) r $
w
ab"!! œ œ #Þ
20000
100
œ
(c) lim r (x) lim 0. The increase in revenue as the number of items increases without bound
xxÄ_ Ä_
w
œœ
20000
x
œ
will approach zero.
25. b(t) 10 10 t 10 t b (t) 10 (2) 10 t 10 (10 2t)œ Ê œ œ
'% $# w % $ $
ab
(a) b (0) 10 bacteria/hr (b) b (5) 0 bacteria/hr
w% w
œœ
(c) b (10) 10 bacteria/hr
w%
œ
26. Q(t) 200(30 t) 200 900 60t t Q (t) 200( 60 2t) Q (10) 8,000 gallons/min is the rateœœ ʜʜ
##w w
ab
the water is running at the end of 10 min. Then 10,000 gallons/min is the average rate the
Q(10) Q(0)
10

œ
water flows during the first 10 min. The negative signs indicate water is the tank.leaving

143. Section 3.3 The Derivative as a Rate of Change 139
27. (a) y 6 1 6 1 1œ œ Êœˆ‰ Š‹
ttt t
1 6 144 dt 12
dy
#
#
œ
(b) The largest value of is 0 m/h when t 12 and the fluid level is falling the slowest at that time. The
dy
dt
œ
smallest value of is 1 m/h, when t 0, and the fluid level is falling the fastest at that time.
dy
dt
œ
(c) In this situation, 0 the graph of y is
dy
dt
ŸÊ
always decreasing. As increases in value,
dy
dt
the slope of the graph of y increases from 1
to 0 over the interval 0 t 12.ŸŸ

28. (a) V r 4 r 4 (2) 16 ft /ftœÊœÊ œ œ
4dV dV
3dr dr
11 11
$# #$ ¸
r=2
(b) When r 2, 16 so that when r changes by 1 unit, we expect V to change by approximately 16 .œœ
dV
dr
11
Therefore when r changes by 0.2 units V changes by approximately (16 )(0.2) 3.2 10.05 ft . Note11 œ¸
$
that V(2.2) V(2) 11.09 ft .¸
$
29. 200 km/hr 55 m/sec m/sec, and D t V t. Thus V t t 25 sec. Whenœœ œÊœ œÊœÊœ
5 500 10 20 500 20 500
99 9 9 999
#
t 25, D (25) mœœ œ
10 6250
99
#
30. s v t 16t v v 32t; v 0 t ; 1900 v t 16t so that t 1900œ Êœ œÊœ œ œÊ œ!! !
# #
#
vv
32 32 3 64
vv

œœ
v (64)(1900) 80 19 ft/sec and, finally, 238 mph.Êœ œ ¸!
È È
80 19 ft
sec 1 min 1 hr 5280 ft
60 sec 60 min 1 mi
È
†† †
31.
(a) v 0 when t 6.25 secœœ
(b) v 0 when 0 t 6.25 body moves up; v 0 when 6.25 t 12.5 body moves down' %"( " "%(
(c) body changes direction at t 6.25 secœ
(d) body speeds up on (6.25 12.5] and slows down on [0 6.25)ßß
(e) The body is moving fastest at the endpoints t 0 and t 12.5 when it is traveling 200 ft/sec. It'sœœ
moving slowest at t 6.25 when the speed is 0.œ
(f) When t 6.25 the body is s 625 m from the origin and farthest away.œœ

144. 140 Chapter 3 Differentiation
32.
(a) v 0 when t secœœ
3
#
(b) v 0 when 0 t 1.5 body moves down; v 0 when 1.5 t 5 body moves up"%"( ' "%(
(c) body changes direction at t secœ
3
#
(d) body speeds up on and slows down on ‘‰ˆ œ
33
##
ß& !ß
(e) body is moving fastest at t 5 when the speed v(5) 7 units/sec; it is moving slowest atœœœ kk
t when the speed is 0œ
3
#
(f) When t 5 the body is s 12 units from the origin and farthest away.œœ
33.
(a) v 0 when t secœœ
615
3
„
È
(b) v 0 when t body moves left; v 0 when 0 t or t 4"""( '%" "%
615 615 615 615
33 33

ÈÈ ÈÈ
body moves rightÊ
(c) body changes direction at t secœ
615
3
„
È
(d) body speeds up on and slows down on 0 .Š‹ “ ‹Š‹ Š’
615 615 615 615
33 33

ÈÈ ÈÈ
ß# ß% ß #ß
(e) The body is moving fastest at t 0 and t 4 when it is moving 7 units/sec and slowest at t secœœ œ
615
3
„
È
(f) When t the body is at position s 6.303 units and farthest from the origin.œ¸
615
3

È

145. Section 3.4 Derivatives of Trigonometric Functions 141
34.
(a) v 0 when tœœ
615
3
„
È
(b) v 0 when 0 t or t 4 body is moving left; v 0 when"%" "%( '
615 615
33

ÈÈ
t body is moving right
615 615
33

ÈÈ
"" (
(c) body changes direction at t secœ
615
3
„
È
(d) body speeds up on and slows down on Š‹ “ ‹Š‹ Š’
615 615 615 615
33 33

ÈÈ ÈÈ
ß# ß% !ß #ß
(e) The body is moving fastest at 7 units/sec when t 0 and t 4; it is moving slowest and stationary atœœ
t œ
615
3
„
È
(f) When t the position is s 10.303 units and the body is farthest from the origin.œ¸
615
3

È
35. (a) It takes 135 seconds.
(b) Average speed furlongs/sec.œ œ œ ¸ !Þ!')
?
?
F
t
&! &
Ê$œ! Ê$
(c) Using a symmetric difference quotient, the horse's speed is approximately furlongs/sec.
?
?
F
t
œ œ ¸ !Þ!((
Ÿœ# #
&*$$ #'
(d) The horse is running the tastest during the last furlong (between the 9th and 10th furlong markers). This furlong takes
only 11 seconds to run, which is the least amount of time for a furlong.
(e) The horse accelerates the fastest during the first furlong (between markers 0 and 1).
3.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
1. y 10x 3 cos x 10 3 (cos x) 10 3 sin xœ Ê œ œ
dy
dx dx
d
2. y 5 sin x 5 (sin x) 5 cos xœ Ê œ œ
33d3
xdxxdxx
dy œœ
œœ
3. y csc x 4 x 7 csc x cot x 0 csc x cot xœ Ê œ œ È
dy
dx
42
xx#ÈÈ
4. y x cot x x (cot x) cot x x x csc x (cot x)(2x)œÊœ œ
## ### "
xdx dx dx x x
dy dd2 2
œ && †ab
x csc x 2x cot xœ
## 2
x
&
5. y (sec x tan x)(sec x tan x) (sec x tan x) (sec x tan x) (sec x tan x) (sec x tan x)œ Êœ
dy
dx dx dx
dd
(sec x tan x) sec x tan x sec x (sec x tan x) sec x tan x sec xœ abab
##
sec x tan x sec x tan x sec x sec x tan x sec x tan x sec x tan x sec x tan x sec x 0.œœabab
# #$# # #$ #
Note also that y sec x tan x tan x 1 tan x 1 0.Š‹ abœœ œÊœ
## # # dy
dx

146. 142 Chapter 3 Differentiation
6. y (sin x cos x) sec x (sin x cos x) (sec x) sec x (sin x cos x)œ Êœ
dy
dx dx dx
dd
(sin x cos x)(sec x tan x) (sec x)(cos x sin x)œ œ
(sin x cos x) sin x
cos x cos x
cos x sin x œ
œ
sec xœœœ
sin x cos x sin x cos x cos x sin x
cos x cos x
œœ
œœ
" #
Note also that y sin x sec x cos x sec x tan x 1 sec x.Š‹ œ œÊœ
dy
dx
#
7. y œÊœ œ
cot x
1 cot x dx (1 cot x) (1 cot x)
dy (1 cot x) (cot x) (cot x) (1 cot x) (1 cot x) csc x (cot x) csc x


dd
dx dx
œœ
œœ
ab ab
œœ


csc x csc x cot x csc x cot x csc x
(1 cot x) (1 cot x)
œœ œ œ
œœ
8. y œÊœ œ
cos x
1 sin x dx (1 sin x) (1 sin x)
dy (1 sin x) (cos x) (cos x) (1 sin x) (1 sin x) sin x (cos x) cos x


dd
dx dx
œœ
ab ab
œœœœ
"

sin xsinxcosx sin x1
(1 sin x) (1 sin x) (1 sin x) 1 sin x
(1 sin x)
œœ
œœœ
9. y 4 sec x cot x 4 sec x tan x csc xœœ Êœ
4
cos x tan x dx
dy" #
10. y œÊœœ
cos x x x sin x cos x cos x x sin x
x cos x dx x cos x x cos x
dy x( sin x) (cos x)(1) (cos x)(1) x( sin x)œœ œœ
œœœœ
11. y x sin x 2x cos x 2 sin x x cos x (sin x)(2x) (2x)( sin x) (cos x)(2) 2 cos xœ Êœ
## dy
dx
aba b
x cos x 2x sin x 2x sin x 2 cos x 2 cos x x cos xœ œ
##
12. y x cos x 2x sin x 2 cos x x ( sin x) (cos x)(2x) 2x cos x (sin x)(2) 2( sin x)œÊœ
## dy
dx
abab
x sin x 2x cos x 2x cos x 2 sin x 2 sin x x sin xœœ
##
13. s tan t t (tan t) 1 sec t 1 tan tœ Ê œ œ œ
ds d
dt dt
##
14. s t sec t 1 2t (sec t) 2t sec t tan tœ Ê œ œ
# ds d
dt dt
15. s œÊœ
1csc t ds
1csc t dt (1csc t)
(1 csc t)( csc t cot t) ( csc t)(csc t cot t)
œœ
"
œ
œœ

œœ
csc t cot t csc t cot t csc t cot t csc t cot t 2 csc t cot t
(1 csc t) (1 csc t)
œœ
œœ
16. s œÊœ œ œ œ
sin t ds cos t cos t sin t cos t
1 cos t dt (1 cos t) (1 cos t) (1 cos t) 1 cos t
(1 cos t)(cos t) (sin t)(sin t)
œœœœœ
œœ " "
œœœ
œœ
œ
"
œcos t 1
17. r 4 sin (sin ) (sin )(2 ) cos 2 sin ( cos sin )œ Ê œ œ œ #)) ) ) )) ) ))) ))) )
## # dr d
dd)) ˆ‰ ab
18. r sin cos ( cos (sin )(1)) sin cos œÊœ œ)) ) ) ) ) )) )
dr
d)
19. r sec csc (sec )( csc cot ) (csc )(sec tan )œÊœ)) ) )) ) ))
dr
d)
sec cscœœœˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰
" " " " " " ##
cos sin sin sin cos cos sin cos
cos sin
))) ))) ) )
))
œœ ))
20. r (1 sec ) sin ( sec ) cos (sin )(sec tan ) (cos ) tan cos secœ Ê œ" œ " œ )) ) ) ) )) ) ) ) )
dr
d)
##
21. p 5 tan q sec qœ& œ Ê œ
" #
cot q dq
dp
22. p (1 csc q) cos q (1 csc q)( sin q) (cos q)( csc q cot q) ( sin q 1) cot q sin q csc qœ Ê œ œ œ
dp
dq
##

147. Section 3.4 Derivatives of Trigonometric Functions 143
23. p œÊœ
sin q cos q dp (cos q)(cos q sin q) (sin q cos q)( sin q)
cos q dq cos q

œ
sec qœœœ
cos q cos q sin q sin q cos q sin q
cos q cos q
œœ
œœ
" #
24. p œÊœ œ œ
tan q dp sec q tan q sec q tan q sec q sec q
1 tan q dq (1 tan q) (1 tan q) (1 tan q)
(1 tan q) sec q (tan q) sec q

ab ab
œœ
œœœ
œœœ œ
25. (a) y csc x y csc x cot x y (csc x) csc x (cot x)( csc x cot x) csc x csc x cot xœÊœ Êœ œ
www# $#
ab ab
(csc x) csc x cot x (csc x) csc x csc x 1 2 csc x csc xœœœaba b
## ## $
(b) y sec x y sec x tan x y (sec x) sec x (tan x)(sec x tan x) sec x sec x tan xœÊœ Êœ œ
www# $#
ab
(sec x) sec x tan x (sec x) sec x sec x 1 2 sec x sec xœœœaba b
## ## $
26. (a) y 2 sin x y 2 cos x y 2( sin x) 2 sin x y 2 cos x y 2 sin xœ Ê œ Ê œ œ Ê œ Ê œ
w ww www Ð%Ñ
(b) y 9 cos x y 9 sin x y 9 cos x y 9( sin x) 9 sin x y 9 cos xœ Ê œ Ê œ Ê œ œ Ê œ
w ww www Ð%Ñ
27. y sin x y cos x slope of tangent atœÊœ Ê
w
x is y ( ) cos ( ) ; slope ofœ œ œ"11 1
w
tangent at x 0 is y (0) cos (0) 1; andœœœ
w
slope of tangent at x is y cos œœ
33 311 1
## #
w
ˆ‰
0. The tangent at ( ) is y 0 1(x ),œß!œ 11
or y x ; the tangent at (0 0) isœ ß1
y 0 1(x 0), or y x; and the tangent atœ œ
1 is y 1.ˆ‰
31
#
ß œ

28. y tan x y sec x slope of tangent at xœÊœ Ê œ
w# 1
3
is sec 4; slope of tangent at x 0 is sec (0) 1;
##ˆ‰œ œ œ
1
3
and slope of tangent at x is sec 4. The tangentœœ
11
33
#
ˆ‰
at tan 3 is y 3 4 x ;ˆ‰ ˆ‰ˆ‰ Š‹
ÈÈ
ß œß œ
11 1 1
33 3 3
the tangent at (0 0) is y x; and the tangent at tanßœ ß ˆ‰ˆ‰
11
33
3 is y 3 4x .œß œ Š‹
ÈÈ ˆ‰
11
33

29. y sec x y sec x tan x slope of tangent atœÊœ Ê
w
x is sec tan 2 3 ; slope of tangentœ œ
111
333
ˆ‰ˆ‰ È
at x is sec tan 2 . The tangent at the pointœœ
111
444
ˆ‰ ˆ‰ È
sec is y 2 3 x ;ˆ‰ˆ‰ ˆ‰ˆ‰ È
ß œß# œ#
11 1 1
33 3 3
the tangent at the point sec 2 is y 2ˆ‰ˆ‰ Š‹
ÈÈ
11 1
44 4
ßœß
2x .œ
Ȉ‰
1
4

30. y 1 cos x y sin x slope of tangent atœ Ê œ Ê
w
x is sin ; slope of tangent at xœ œ œ
11 1
33
3 3
ˆ‰
È
##
is sin 1. The tangent at the pointœˆ‰
31
#
cosˆ‰ˆ‰ ˆ‰ß" œß
111
333
3
#
is y x ; the tangent at the pointœ
3 3
3##
È
ˆ‰
1
cos 1 is y 1 xˆ‰ˆ‰ ˆ‰
333 3111 1
### #
ß" œ ß œ

148. 144 Chapter 3 Differentiation
31. Yes, y x sin x y cos x; horizontal tangent occurs where 1 cos x 0 cos x 1œ Ê œ" œ Ê œ
w
xʜ 1
32. No, y 2x sin x y 2 cos x; horizontal tangent occurs where 2 cos x 0 cos x . But thereœ Êœ œÊ œ#
w
are no x-values for which cos x .œ#
33. No, y x cot x y 1 csc x; horizontal tangent occurs where 1 csc x 0 csc x 1. But thereœ Ê œ œ Ê œ
w# # #
are no x-values for which csc x 1.
#
œ
34. Yes, y x 2 cos x y 1 2 sin x; horizontal tangent occurs where 1 2 sin x 0 1 2 sin xœ Ê œ œ Ê œ
w
sin x x or xÊœ Êœ œ
"
#
11
66
5
35. We want all points on the curve where the tangent
line has slope 2. Thus, y tan x y sec x soœÊœ
w#
that y 2 sec x 2 sec x 2
w#
œÊ œÊ œ„
È
x . Then the tangent line at hasÊœ„ ß"
11
44
ˆ‰
equation y 1 2 x ; the tangent line atœ ˆ‰
1
4
has equation y 1 2 x .ˆ‰ ˆ‰ß" œ
11
44

36. We want all points on the curve y cot x whereœ
the tangent line has slope 1. Thus y cot xœ
y csc x so that y 1 csc x 1Êœ œÊ œ
w# w #
csc x 1 csc x 1 x . Theʜʜ„Êœ
#
#
1
tangent line at is y x .ˆ‰
11
#
ß! œ
2

37. y 4 cot x 2 csc x y csc x 2 csc x cot xœ Ê œ œ
w# " ˆ‰ˆ ‰
sin x sin x
12 cos x
(a) When x , then y 1; the tangent line is y x 2.œœ œ
11
##
w
(b) To find the location of the horizontal tangent set y 0 1 2 cos x 0 x radians. When x ,
w
œÊ œÊœ œ
11
33
then y 3 is the horizontal tangent.œ%
È
38. y 1 2 csc x cot x y 2 csc x cot x csc xœ Ê œ œ
ÈÈ ˆ‰Š‹
w# "
sin x sin x
2 cos x 1
È
(a) If x , then y 4; the tangent line is y 4x 4.œœ œ
1
4
w
1
(b) To find the location of the horizontal tangent set y 0 2 cos x 1 0 x radians. When
w
œÊ œÊœ
È
3
4
1
x , then y 2 is the horizontal tangent.œœ
3
4
1
39. lim sin sin sin 0 0
x2Ä
ˆ‰ ˆ‰
"" ""
###x
œ œ œ
40. lim 1 cos ( csc x) 1 cos csc 1 cos ( 2) 2
xÄ
œ
6
ÈÈ É ˆ‰ È
œ œœ11 1
1
6
†
41. lim sec cos x tan 1 sec cos 0 tan 1 sec 1 tan 1 sec 1
xÄ!
œ ‘ œ ‘ œ ‘ˆ‰ ˆ‰ ˆ‰œœœœ1111
111
4 sec x 4 sec 0 4

149. Section 3.4 Derivatives of Trigonometric Functions 145
42. lim sin sin sin 1
xÄ!
ˆ‰ˆ‰ˆ‰
11 1
œœ#
tan x tan 0
tan x 2 sec x tan 0 2 sec 0
œœœ
43. lim tan 1 tan 1 lim tan (1 1) 0
ttÄ! Ä!
ˆ‰ Š‹œ œ œ
sin t sin t
tt
44. lim cos cos lim cos cos 1
))Ä! Ä!
ˆ‰ ˆ ‰ Š‹ Œ
1) )
))sin sin 1 lim
œœœœ111 ††
""

sin œ
œ
45. s sin t v 2 cos t a 2 sin t j 2 cos t. Therefore, velocity vœ## Ê œ œ Ê œ œ Ê œ œ œ
ds dv da
dt dt dt 4
ˆ‰
1
2 m/sec; speed v 2 m/sec; acceleration a 2 m/sec ; jerk j 2 m/sec .œ œ œ œ œ œ œ
ÈÈ ÈÈ ¸ ¸ ˆ‰ ˆ‰ˆ‰
111
444
#$
46. s sin t cos t v cos t sin t a sin t cos t j cos t sin t. Thereforeœ Êœœ Êœœ Êœœ
ds dv da
dt dt dt
velocity v 0 m/sec; speed v 0 m/sec; acceleration a 2 m/sec ;œœ œ œ œœˆ‰ ¸ ¸ ˆ‰ ˆ‰ È
11 1
44 4
#
jerk j 0 m/sec .œœˆ‰
1
4
$
47. lim f(x) lim lim 9 9 so that f is continuous at x 0 lim f(x) f(0)
xx x xÄ! Ä! Ä! Ä!
œœ œ œÊœ
sin 3x sin 3x sin 3x
x3x3x
œ
œ
ˆ‰ˆ‰
9c.ʜ
48. lim g(x) lim (x b) b and lim g(x) lim cos x 1 so that g is continuous at x 0 lim g(x)
xx xxxÄ! Ä! Ä!Ä! Ä!
&& &ÊÊ
œœ œ œ œÊ
lim g(x) b 1. Now g is not differentiable at x 0: At x 0, the left-hand derivative isœÊœ œœ
xÄ!
Ê
(x b) 1, but the right-hand derivative is (cos x) sin 0 0. The left- and right-hand¸¸
dd
dx dx
œ œœ
x=0 x=0
derivatives can never agree at x 0, so g is not differentiable at x 0 for any value of b (including b 1).œœ œ
49. (cos x) sin x because (cos x) cos x the derivative of cos x any number of times that is a
dd
dx dx
ÊÊÊ Ÿ
ÊÊÊ Ÿ
œœÊ multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 249 4 3 (cos x)œÊ†
d
dx
ÊÊÊ
ÊÊÊ
(cos x) (cos x) sin x.œœœ
dd d
dx dxdx

&& œŸÊ Ÿ
’“
œ
œ
50. (a) y sec x sec x tan xœœÊœ œœ œ
"" œœ
cos x dx (cos x) cos x cos x cos x
dy (cos x)(0) (1)( sin x) sin x sin x
œœ
ˆ‰ˆ‰
(sec x) sec x tan xʜ
d
dx
(b) y csc x csc x cot xœœÊœ œ œ œ
"" œ
sin x dx (sin x) sin x sin x sin x
dy (sin x)(0) (1)(cos x) cos x cos x
œœ
ˆ‰ˆ‰
(csc x) csc x cot xʜ
d
dx
(c) y cot x csc xœœÊœ œ œœ
cos x sin x cos x
sin x dx (sin x) sin x sin x
dy (sin x)( sin x) (cos x)(cos x)œœ " #
œœœ
œœ
(cot x) csc xʜ
d
dx
#
51.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y getœ
sin (x h) sin x
h

closer and closer to the black curve y cos x because (sin x) lim cos x. The sameœœœ
d
dx h
sin (x h) sin x
hÄ!

is true as h takes on the values of 1, 0.5, 0.3 and 0.1.

150. 146 Chapter 3 Differentiation
52.
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y getœ
cos (x h) cos x
h

closer and closer to the black curve y sin x because (cos x) lim sin x. Theœ œ œ
d
dx h
cos (x h) cos x
hÄ!

same is true as h takes on the values of 1, 0.5, 0.3, and 0.1.
53. (a)
The dashed curves of y are closer to the black curve y cos x than the corresponding dashedœœ
sin x h sin x h
h
ab ab
#
curves in Exercise 51 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
(b)
The dashed curves of y are closer to the black curve y sin x than the corresponding dashedœœ
cos x h cos x h
h
ab ab
#
curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
54. lim lim lim 0 0 the limits of the centered difference quotient exists even
hh xÄ! Ä!Ä!
kkkk kkkk0h 0h h h
2h 2h

œœœÊ
though the derivative of f(x) x does not exist at x 0.œœkk
55. y tan x y sec x, so the smallest valueœÊœ
w#
y sec x takes on is y 1 when x 0;
w# w
œœœ
y has no maximum value since sec x has no
w#
largest value on ; y is never negativeˆ‰ß
11
##
w
since sec x 1.
#
 

151. Section 3.4 Derivatives of Trigonometric Functions 147
56. y cot x y csc x so y has no smallestœÊœ
w#w
value since csc x has no minimum value on
#
( ); the largest value of y is 1, when x ;!ß œ1
w
#
1
the slope is never positive since the largest
value y csc x takes on is 1.
w
œ
2

57. y appears to cross the y-axis at y 1, sinceœœ
sin x
x
lim 1; y appears to cross the y-axis
xÄ!
sin x sin 2x
xx
œœ
at y 2, since lim 2; y appears to
x
œœœ
Ä!
sin 2x sin 4x
xx
cross the y-axis at y 4, since lim 4.œœ
xÄ!
sin 4x
x
However, none of these graphs actually cross the y-axis
since x 0 is not in the domain of the functions. Also,œ
lim 5, lim 3, and lim
xx xÄ! Ä! Ä!
sin 5x sin kx
xx x
sin ( 3x)
œœ
œ
k the graphs of y , y , andœÊ œ œ
sin 5x
xx
sin ( 3x)œ
y approach 5, 3, and k, respectively, asœ
sin kx
x
x 0. However, the graphs do not actually cross theÄ
y-axis.

58. (a) h
1 .017452406 .99994923
0.01 .017453292 1
0.001 .017453292 1
0.0001 .017453292 1
sin h sin h 180
hh
ˆ‰ˆ‰
1
lim lim lim lim ( h )
hh xÄ! Ä!Ä! Ä!
sin h°
h h h 180 180
sin h sin h sin
œœ œœœ
ˆ‰ ˆ‰††
†

180 180 180
œ
œ
180
180
)
)
)
11
) †
(converting to radians)
(b) h
1 0.0001523
0.01 0.0000015
0.001 0.0000001
0.0001 0
cos h 1
h
œ



lim 0, whether h is measured in degrees or radians.
hÄ!
cos h 1
h
œ
œ
(c) In degrees, (sin x) lim lim
d
dx h h
sin (x h) sin x (sin x cos h cos x sin h) sin x
œœ
hhÄ! Ä!

lim sin x lim cos x (sin x) lim (cos x) lim œœ
hh h hÄ! Ä! Ä! Ä!
ˆ‰ˆ‰ ˆ‰ ˆ‰††† †
cos h 1 sin h cos h 1 sin h
hh h h
œœ
(sin x)(0) (cos x) cos xœ œ ˆ‰
11
180 180
(d) In degrees, (cos x) lim lim
d
dx h h
cos (x h) cos x (cos x cos h sin x sin h) cos x
œœ
hhÄ! Ä!

lim lim cos x lim sin xœœ
hhhÄ! Ä! Ä!
(cos x)(cos h 1) sin x sin h
hhh
cos h 1 sin hœœ œ
ˆ‰ˆ‰††
(cos x) lim (sin x) lim (cos x)(0) (sin x) sin xœœœ
hhÄ! Ä!
ˆ‰ ˆ‰ ˆ‰
cos h 1 sin h
h h 180 180
œ 11
(e) (sin x) cos x sin x; (sin x) sin x cos x;
dd dd
dx dx 180 180 dx dx 180 180
œ &
œ &
œœ œœˆ ‰ ˆ‰ ˆ‰ ˆ‰ Š‹
11 1 1
##$
(cos x) sin x cos x; (cos x) cos x sin x
dd dd
dx dx 180 180 dx dx 180 180
œ &
œ &
œ œ œ œˆ ‰ ˆ‰ ˆ‰ ˆ‰ Š‹
11 1 1
##$

152. 148 Chapter 3 Differentiation
3.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS
1. f(u) 6u 9 f (u) 6 f (g(x)) 6; g(x) x g (x) 2x ; therefore f (g(x))g (x)œÊ œÊ œ œ Ê œ œ
ww %w$ ww "
#
dy
dx
6 2x 12xœœ†
$$
2. f(u) 2u f (u) 6u f (g(x)) 6(8x 1) ; g(x) 8x 1 g (x) 8; therefore f (g(x))g (x)œÊ œÊ œ œÊ œ œ
$w #w # w w w dy
dx
6(8x 1) 8 48(8x 1)œ œ
##
†
3. f(u) sin u f (u) cos u f (g(x)) cos (3x 1); g(x) 3x 1 g (x) 3; therefore f (g(x))g (x)œÊœ Ê œ œÊœ œ
ww w ww dy
dx
(cos (3x 1))(3) 3 cos (3x 1)œœ
4. f(u) cos u f (u) sin u f (g(x)) sin ; g(x) g (x) ; therefore f (g(x))g (x)œÊœÊ œ œÊœ œ
ww w ww " ˆ‰
xx
33 3dx
dy
sin sinœ œˆ‰ˆ‰ ˆ‰
""xx
333 3
†
5. f(u) cos u f (u) sin u f (g(x)) sin (sin x); g(x) sin x g (x) cos x; thereforeœÊœÊ œ œÊœ
ww w
f (g(x))g (x) (sin (sin x)) cos x
dy
dx
œœ
ww
6. f(u) sin u f (u) cos u f (g(x)) cos (x cos x); g(x) x cos x g (x) 1 sin x; thereforeœÊœ Ê œ œÊœ
ww w
f (g(x))g (x) (cos (x cos x))(1 sin x)
dy
dx
œœ
ww
7. f(u) tan u f (u) sec u f (g(x)) sec (10x 5); g(x) 10x 5 g (x) 10; thereforeœÊœ Ê œ œÊœ
w#w # w
f (g(x))g (x) sec (10x 5) (10) 10 sec (10x 5)
dy
dx
œœœ
ww # #
ab
8. f(u) sec u f (u) sec u tan u f (g(x)) sec x 7x tan x 7x ; g(x) x 7xœ Ê œ Ê œ œ
ww###
abab
g (x) 2x 7; therefore f (g(x))g (x) (2x 7) sec x 7x tan x 7xÊœ œ œ
www## dy
dx
abab
9. With u (2x 1), y u : 5u 2 10(2x 1)œ œ œ œ œ
&%%dy dy
dx du dx
du
†
10. With u (4 3x), y u : 9u ( 3) 27(4 3x)œ œ œ œ œ
*) )dy dy
dx du dx
du
†
11. With u 1 , y : 7uœ œ? œ œ œ"ˆ‰ ˆ‰ˆ‰
xdu x
7dxdudx 77
dy dyœÊ œ# "
œ#
†
12. With u 1 , y : 10u 5 1œ œ? œ œ œ ˆ‰ ˆ‰ ˆ‰
xdux dy dy
dx du dx###
"! "" "
""
†
13. With u x , y : 4u 1 4 x 1œ œ? œ œ œ Š‹ Š‹ ˆ‰ ˆ‰
xduxxx
8 x dx du dx 4 x 8 x 4 x
dy dy
œ œ
œœ
"""" %$
$
†
14. With u , y : 5u 1œ œ? œ œ œ ˆ‰ ˆ ‰ˆ‰ˆ‰
xdu1x
55x dx dudx 55x 55x x
dy dy"""" &%
%
†
œœ
15. With u tan x, y sec u: (sec u tan u) sec x (sec (tan x) tan (tan x)) sec xœœ œœ œ
dy dy
dx du dx
du
ab
##
16. With u , y cot u: csc u cscœ œ œ œ œ 11
"""" ##
xdxdudx xxx
dy dy du
ab ˆ‰ ˆ ‰
œœ
17. With u sin x, y u : 3u cos x 3 sin x (cos x)œœœœ œ
$##dy dy
dx du dx
du
ab
18. With u cos x, y 5u : 20u ( sin x) 20 cos x (sin x)œœ œœœ
% & &dy dy
dx du dx
du
ab ab

153. Section 3.5 The Chain Rule and Parametric Equations 149
19. p 3 t (3 t) (3 t) (3 t) (3 t)œœ Êœ œ œ
È
"Î# "Î# "Î#"""
## œ
dp
dt dt
d
23 t
†
È
20. q 2r r 2r r 2r r 2r r 2r r (2 2r)œœ Êœ œ œ
È
ab ab abab
# ####
"Î# "Î# "Î#"""
##
œ
dq
dr dr
dr
2r r
†
È œ
21. s sin 3t cos 5t cos 3t (3t) ( sin 5t) (5t) cos 3t sin 5tœ Êœ œ
44 ds4d4 d44
3 5 dt 3 dt 5 dt11 1 1 11
††
(cos 3t sin 5t)œ
4
1
22. s sin cos cos sin cos sinœ Êœ œ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
3t 3t ds 3t d 3t 3t d 3t 3 3t 3 3t
dt dt dt 2 2
1 1 11 111111
# # ## ## # #
††
cos sin œ
33t3t
2
11 1
ˆ‰
##
23. r (csc cot ) (csc cot ) (csc cot )œ Êœ œ œ)) )) ))
" #

dr d csc cot csc
d d (csc cot ) (csc cot )
csc (cot csc )
))))))
)) ) )) )
œ
œœ
œ
csc
csc cot
)
))
24. r (sec tan ) (sec tan ) (sec tan )œ Ê œ œ œ)) )) ))
" #

dr d sec tan sec
d d (sec tan ) (sec tan )
sec (tan sec )
)) ))))
)) ) )) )
œ
œœ
œ
sec
sec tan
)
))
25. y x sin x x cos x x sin x sin x x x cos x cos x (x)œ Êœ
# Ÿ œ# # Ÿ Ÿ # œ# œ# dy
dx dx dx dx dx
ddd d
ab ab a b ††
x 4 sin x (sin x) 2x sin x x 2 cos x (cos x) cos xœ
#$ Ÿ œ$ œ#ˆ‰ˆ ‰
dd
dx dx
†
x 4 sin x cos x 2x sin x x 2 cos x ( sin x) cos xœ
#$ Ÿ œ$ œ#
ab a b ab
4x sin x cos x 2x sin x 2x sin x cos x cos xœ
#$Ÿœ$œ#
26. y sin x cos x y sin x sin x cos x cos xœÊœ
"""& $ w & & $ $
x 3 xdx dx x 3dx dx 3
xd dxd dx
ab ab ˆ‰ ˆ‰††
5 sin x cos x sin x 3 cos x ( sin x) cos xœ
"" " ' & # $
xx33
x
ababa bab ˆ‰ ˆ‰ ab
œ
sin x cos x sin x x cos x sin x cos xœ
5
xx 3
' & # $ ""
œ
27. y (3x 2) 4 (3x 2) (3x 2) ( 1) 4 4œ Êœ
"" "" ('
###
" #
21 x dx 21 dx x dx x
dy 7d d
ˆ‰ ˆ‰ˆ‰
œ œœ ††
(3x 2) 3 ( 1) 4 (3x 2)œ œ
7
21 x x
x4
'' "" "
#
œ#
œ
† ˆ‰ˆ‰
œ &
&
œ
Š‹
œ


x
28. y (5 2x) 1 3(5 2x) ( 2) 1 6(5 2x) 1œ Ê œ œ
œ$ œŸ œŸ" "
%$$
8x dx 8x x x x
24222 dy ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ‰
œœ
œ
6
(5 2x)
1
xœ

Ÿ
&
œ
Š‹
2
x
29. y (4x 3) (x 1) (4x 3) ( 3)(x 1) (x 1) (x 1) (4)(4x 3) (4x 3)œ Êœ
Ÿ œ$ Ÿ œŸ œ$ $ dy
dx dx dx
dd
††
(4x 3) ( 3)(x 1) (1) (x 1) (4)(4x 3) (4) 3(4x 3) (x 1) 16(4x 3) (x 1)œ œ
ŸœŸ œ$$ ŸœŸ $œ$
3(4x 3) 16(x 1)œœ
(4x 3) (4x 3) (4x 7)
(x 1) (x 1)


&&
ŸŸ
cd
30. y (2x 5) x 5x (2x 5) (6) x 5x (2x 5) x 5x ( 1)(2x 5) (2)œ Êœ
" # " # # #
'&'
ab ab ab
dy
dx
6x 5xœab
#
& œ
œ
2x 5x
(2x 5)
ab
œ
¸
œ
31. h(x) x tan 2 x 7 h (x) x tan 2x tan 2x (x) 0œÊœ ˆ‰ ˆ ‰ ˆ‰È ˆ‰
w "Î# "Î#dd
dx dx
†
x sec 2x 2x tan 2x x sec 2 x tan 2 x x sec 2 x tan 2 xœœœ
# "Î# "Î# "Î# # # "ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ÈÈÈÈȆ†
d
dx xÈ

154. 150 Chapter 3 Differentiation
32. k(x) x sec k (x) x sec sec x x sec tan 2x secœÊœ œ
#w# ##"""""""ˆ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ab
x dx x xdx x xdxx x
dd d
††
x sec tan 2x sec 2x sec sec tanœ œ
# "" " " " ""ˆ‰ ˆ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
xx x x x xx
†
œ
33. f( ) f ( ) 2))œÊœ œˆ‰ ˆ‰ˆ‰
sin sin d sin 2 sin
1cos 1cos d 1cos 1cos (1cos )
(1 cos )(cos ) (sin )( sin )))))
))))))
)) ) )

#
w
††
œ
œœœ
(2 sin ) cos cos sin
(1cos ) (1cos ) (1cos )
(2 sin ) (cos 1) 2 sin )) ) )
)))
)) )ab


œœ

34. g(t) g (t)œ Ê œ œˆ‰ ˆ‰ˆ‰
1cos t 1cos t d 1cos t sint
sin t sin t dt sin t (1 cos t) (sin t)
(sin t)( sin t) ( cos t)(cos t)
" #
w

"
††
œ
œœ
œœ
œœ œ œ

"absin t cos t cos t
(1 cos t) 1 cos t
œœ
œ
35. r sin cos (2 ) sin ( sin 2 ) (2 ) cos (2 ) cosœÊœab ab a b ab ab)) ) )) )))
## ## dr d d
dd d)) )
†
sin ( sin 2 )(2) (cos 2 ) cos (2 ) 2 sin sin ( ) 2 cos (2 ) cosœ œ #ab a b ab ab ab)) ))) )))))
####
36. r sec tan sec sec tan sec tan œÊœ Š‹ Š‹ Š ‹Š‹
ÈÈ ÈȈ‰ ˆ ‰ˆ ‰ ˆ‰)) ))
""""" #
#)) )) ) )
dr
d
œ È
sec sec tan sec tan sec œ œ
"""" #
#))) )
œ
ÈÈÈȈ‰ ˆ‰ Š‹ ”•))))
È
tan tan sec
È ˆ‰ ˆ‰
È
)
) )
œœ

œœ
#
37. q sin cos cosœÊœ œŠ‹ Š‹Š‹ Š‹
ttdtt
t1 t1 t1 t1
dq
dt dt
t1(1)t t1
t1
ÈÈÈÈ
ÈÈ ˆ‰
ˆ‰È


††
†
d
dt
œ
cos cos cosœœœŠ‹ Š‹Š ‹Š ‹Š‹
ttt2t
t1 t1 t1
t1
t1
2(t 1) t
2(t 1) 2(t 1)ÈÈ È
È






†
t
2t 1
œ'

38. q cot csc cscœÊœ œˆ‰ ˆ‰ ˆ‰ˆ ‰ˆ ‰ ˆ‰
sin t sin t d sin t sin t t cos t sin t
tdt tdtt t t
dq ## œ
†
œ
39. y sin ( t 2) 2 sin ( t 2) sin ( t 2) 2 sin ( t 2) cos ( t 2) ( t 2)œÊœ œ
#
111111
dy
dt dt dt
dd
†††
2 sin(t2) cos(t2)œ11 1
40. y sec t (2 sec t) (sec t) (2 sec t)(sec t tan t) ( t) 2 sec t tan tœÊœ œ œ
# #
1 1 1 1 11 1 111
dy
dt dt dt
dd
††
41. y (1 cos 2t) 4(1 cos 2t) (1 cos 2t) 4(1 cos 2t) ( sin 2t) (2t)œ Ê œ œ œ
% & &

dy
dt dt dt (1 cos 2t)
d d 8 sin 2t
††

42. y 1 cot 2 1 cot 1 cot 2 1 cot cscœ Êœ œ ˆ‰ ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
ttdtttdt dy
dt dt dt######
œ# œ$ œ$
#
†††
œ
csc
cot


ˆ‰
ˆ‰
t
t
ˆ‰1
&
43. y sin cos (2t 5) cos (cos (2t 5)) cos (2t 5) cos (cos (2t 5)) ( sin (2t 5)) (2t 5)œÊœ œab
dy
dt dt dt
dd
†††
2 cos (cos (2t 5))(sin (2t 5))œ
44. y cos 5 sin sin 5 sin 5 sin sin 5 sin 5 cosœ Ê œ œˆ‰ ˆ‰ˆ‰ˆ‰ˆ ‰ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
ttdtttdt
3dt 3dt 3 3 3dt3
dy
††
sin 5 sin cosœ
5tt
333
ˆ‰ˆ‰ˆ‰ ˆ‰
45. y 1 tan 3 1 tan 1 tan 3 1 tan 4 tan tanœ Ê œ œ œ ‘ œ ‘ œ ‘œ ‘œ ‘ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
%%%%$
######
$# #
ttdtttdt
1dt 1dt 1 1 1dt1
dy
††
12 1 tan tan sec 1 tan tan secœ œœ ‘œ ‘œ ‘œ ‘ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
%$# %$#
#### ###
##
"ttt ttt
1111 111
†
46. y 1 cos (7t) 1 cos (7t) 2 cos (7t)( sin (7t))(7) 7 1 cos (7t) (cos (7t) sin (7t))œ Êœ œ
" ## #
$# #
6dt6
dy 3
cd cd cd †

155. Section 3.5 The Chain Rule and Parametric Equations 151
47. y 1 cos t 1 cos t 1 cos t 1 cos t sin t tœ Ê œ œ ab abababababab ab ab ab ˆ‰
######
"Î# "Î# "Î#""
##
dy
dt dt dt
dd
††
1 cos t sin t 2tœ œ
"
#
##
"Î#

ababab ab †
t sin t
1cost
ab
È ab
œ
œ
48. y 4 sin 1 t 4 cos 1 t 1 t 4 cos 1 t 1 tœÊœ œ Œ Œ Œ Œ
ÉÉÉÉÈÈÈÈÈ ˆ‰
dy
dt dt dt
dd
1t
†††
"
#ÉÈ
œœ
2 cos 1 t cos 1 t
1t2t tt
Œ œ Œ œ
ÉÉÈÈ
ÉÉÈÈ È

†
49. y 1 y 3 1 1 y 1 1œ Êœ œ Ê œ ˆ‰ ˆ‰ˆ‰ ˆ‰ ˆ‰ˆ‰ˆ‰ˆ‰
"""" ""
$# # ##
www
x x x x x x dx x x dx x
33dd3
œœ œ œ ††
21 11111œ œœ ˆ‰ˆ ‰ˆ‰ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ ‰ˆ‰ˆ‰
36666
x xx x x xxxxxx x x
œœ &&& Ÿ
"" """"""
##
11œ
62
xx x
&
ˆ‰ˆ‰
"
50. y 1 x y 1 x x 1 x xœ Êœ œ ˆ‰ ˆ‰ˆ ‰ˆ‰ÈÈ È
" # #
w "Î# "Î# ""
##
y 1 x x x ( 2) 1 x xʜ
ww $Î# "Î# "Î#"" "
## #
œ# œ$
’“ˆ‰ˆ ‰ ˆ‰ˆ ‰ÈÈ
x1x x1x x1x x1x1œœ
"" " "
## # #
$Î# " " "Î#
œ# œ$ œ$
’“ ˆ‰ ˆ‰ ˆ‰ œ ‘ÈÈ È È ˆ‰
1x 1 1xœ œ
"""" "
####
œ$ œ$
##xx xx
3
ˆ‰ ˆ‰ÈÈ Š‹ Š‹
ÈÈ
51. y cot (3x 1) y csc (3x 1)(3) csc (3x 1) y (csc (3x 1) csc (3x 1))œÊœ œ Êœ
""" w# # ww
993 3dx
2d
ˆ‰ †
csc (3x 1)( csc (3x 1) cot (3x 1) (3x 1)) 2 csc (3x 1) cot (3x 1)œ œ
2d
3dx
†
#
52. y 9 tan y 9 sec 3 sec y 3 2 sec sec tan 2 sec tanœÊœ œ Êœ œˆ‰ ˆ ‰ˆ‰ ˆ‰ ˆ‰ˆ ‰ˆ‰ ˆ‰ ˆ‰ˆ‰ ˆ‰ ˆ‰
x x x x xx xx
3333 333333
w# # ww # " "
†
53. g(x) x g (x) g(1) 1 and g (1) ; f(u) u 1 f (u) 5u f (g(1)) f (1) 5;œÊ œ Ê œ œ œÊ œÊ œœÈ
ww&w%ww""
# #Èx
therefore, (f g) (1) f (g(1)) g (1) 5‰œ œœ
ww w "
##
††
5
54. g(x) (1 x) g (x) (1 x) ( 1) g( 1) and g ( 1) ; f(u) 1œ Ê œ œ Êœ œ œ
" w # w """"
œ#(1 x) 4 u
œ
f (u) f (g( 1)) f 4; therefore, (f g) ( 1) f (g( 1))g ( 1) 4 1ʜʜ œ ‰œœœ
www www"" "
#u 4
œ
ˆ‰ †
55. g(x) 5 x g (x) g(1) 5 and g (1) ; f(u) cot f (u) cscœÊœÊœ œ œ ÊœÈ ˆ‰ ˆ‰ˆ‰
ww w#
# #
55uu
x 10 10 10È
111
csc f (g(1)) f (5) csc ; therefore, (f g) (1) f (g(1))g (1)œÊœœœ ‰œœ
œ #ww # www
# #
11 1 1 1 1
10 10 10 10 10
u 5
ˆ‰ ˆ‰ †
=
1
4
56. g(x) x g (x) g and g ; f(u) u sec u f (u) 1 2 sec u sec u tan uœÊ œÊ œ œ œ Ê œ11 1
ww#w "" ˆ‰ ˆ‰
44 4
1
†
1 2 sec u tan u f g f 1 2 sec tan 5; therefore, (f g) f g g 5œ Ê œ œ œ ‰ œ œ
#ww # www " """ ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰ˆ‰ ˆ‰
44 44 4 44
111
1
57. g(x) 10x x 1 g (x) 20x 1 g(0) 1 and g (0) 1; f(u) f (u)œÊ œÊœ œ œ Ê œ
#w w w



2u
u1
u 1 (2) (2u)(2u)
u1
œ
œ
œ
œ
ab
ab
f (g(0)) f (1) 0; therefore, (f g) (0) f (g(0))g (0) 0 1 0œÊœœ ‰œ œœ


ww www2u 2
u1
œ
œ
œ
ab
†
58. g(x) 1 g (x) g( 1) 0 and g ( 1) 2; f(u) f (u) 2 œ Ê œ Ê œ œ œ Ê œ
" œœœwww

#
x x u1 u1 du u1
2u1u1du1
œ &
ˆ‰ ˆ‰ˆ‰
2 f (g( 1)) f (0) 4; therefore,œœœÊœœˆ‰
u1
u 1 (u 1) (u 1) (u 1)
(u 1)(1) (u 1)(1) 2(u 1)(2) 4(u 1)œ

ww
†
œ &&
(f g) ( 1) f (g( 1))g ( 1) ( 4)(2) 8‰œ œ œ
www

156. 152 Chapter 3 Differentiation
59. (a) y 2f(x) 2f (x) 2f (2) 2œÊœ Ê œœœ
dy dy
dx dx 3 3
2ww "
¹ ˆ‰
x=2
(b) y f(x) g(x) f (x) g (x) f (3) g (3) 2 5œ Êœ Ê œ œ
dy dy
dx dx
ww ww
¹
x=3
1
(c) y f(x) g(x) f(x)g (x) g(x)f (x) f(3)g (3) g(3)f (3) 3 5 ( 4)(2 ) 15 8œÊœÊœœœ††
dy dy
dx dx
ww ww
¹
x=3
11
(d) y œÊœ Ê œ œ œ
f(x) dy g(x)f (x) f(x)g (x) dy g(2)f (2) f(2)g (2)
g(x) dx [g(x)] dx [g(2)] 6
(2) (8)( 3) 37
&& &&
œœœ

œœ œœ
#
¹
x=2
ˆ‰
3
(e) y f(g(x)) f (g(x))g (x) f (g(2))g (2) f (2)( 3) ( 3) 1œ Ê œ Ê œ œ œ œ
dy dy
dx dx 3
ww ww w "
¹
x=2
(f) y (f(x)) (f(x)) f (x) œ Êœ œ Ê œ œœœ œ
"Î# "Î# w " ""
# ## # #
dy f (x) dy f (2) 2
dx dx 24 f(x) f(2) 868 12
†

ÈÈÈÈÈ
ˆ‰ È
¹
x=2
3
(g) y (g(x)) 2(g(x)) g (x) 2(g(3)) g (3) 2( 4) 5œÊœ Êœ œœ
œ# œ$ ß œ$ ß œ$
#
dy dy
dx dx 3
5
†† ¹
x=3
(h) y (f(x)) (g(x)) (f(x)) (g(x)) 2f(x) f (x) 2g(x) g (x)œ Êœ ababa b
## ## w w
"Î# "Î# "
#
dy
dx
††
(f(2)) (g(2)) 2f(2)f (2) 2g(2)g (2) 8 2 2 8 2 2 ( 3)Êœ œ ¹ aba bab ˆ‰
dy
dx 3
x=2
"""
##
## w w ##
"Î# "Î#
†† ††
œ
5
317
È
60. (a) y 5f(x) g(x) 5f (x) g (x) 5f (1) g (1) 5 1œÊœ Ê œ œœ
dy dy
dx dx 3 3
8ww ww "
¹ ˆ‰ˆ‰
x=1
(b) y f(x)(g(x)) f(x) 3(g(x)) g (x) (g(x)) f (x) f(0)(g(0)) g (0) (g(0)) f (0)œÊœ Êœ$
$#w$w #w$wdy dy
dx dx
ab ¹
x=0
3(1)(1) (1) (5) 6œœ
#$"ˆ‰
3
(c) y œÊœ Êœ
f(x) dy (g(x) 1)f (x) f(x) g (x) dy (g(1) 1)f (1) f(1)g (1)
g(x) 1 dx (g(x) 1) dx (g(1) 1)

&& &&
œœ
¹
x=1
1 œœ
(4 ) (3)
(41)
"

ˆ‰ ˆ‰

œ
33
8
(d) y f(g(x)) f (g(x))g (x) f (g(0))g (0) f (1)œÊœ Êœ œ œœ
dy dy
dx dx 3 3 3 9
ww ww w """"
¹ ˆ‰ ˆ ‰ˆ‰
x=0
(e) y g(f(x)) g (f(x))f (x) g (f(0))f (0) g (1)(5) (5)œÊœ Êœ œ œœ
dy dy
dx dx 3 3
840ww ww w
¹ ˆ‰
x=0
(f) y x f(x) 2 x f(x) 11x f (x) 2(1 f(1)) 11 f (1)œ Êœ Ê œ ab aba b ab ¹
"" "" "! w $ w
œ# œ$dy dy
dx dx
x=1
2(1 3) 11œ œ œ
œ$ ""ˆ‰ˆ‰ˆ‰
3433
232
&
(g) y f(x g(x)) f (x g(x)) 1 g (x) f (0 g(0)) 1 g (0) f (1) 1œÊœÊœœ
dy dy
dx dx 3
ww www "
ab ab ¹ ˆ‰
x=0
œ œˆ‰ˆ‰
"
33 9
44
61. : s cos sin sin 1 so that 1 5 5
ds ds d ds ds 3 ds ds d
dt d dt d d dt d dt
œœÊœÊœœ œœœ
))) )
)1)
††† )) ¸ˆ‰
=
3
2œ
#
62. : y x 7x 5 2x 7 9 so that 9 3
dy dy dy dy dy dy
dt dx dt dx dx dt dx dt 3
dx dx
œœÊœÊœ œœœ†††
# "
¹
x=1
63. With y x, we should get 1 for both (a) and (b):œœ
dy
dx
(a) y 7 ; u 5x 35 5; therefore, 5 1, as expectedœÊœ œÊœ œ œ œ
ududu
5 du 5 dx dx du dx 5
dy dy dy""
††
(b) y 1 ; u (x 1) (x 1) (1) ; therefore œ Ê œ œ Ê œ œ œ
"" " " #
œu du u dx (x 1) dx du dx
dy dy dydu du
œ œ †
(x 1) 1, again as expectedœœ œœ
" " " " "
œœœ œ
#
u (x1) (x1) (x1)(x 1)
œœ œ œ

œ†††
ab
64. With y x , we should get x for both (a) and (b):œœ
$Î# "Î#
#
dy
dx
3
(a) y u 3u ; u x ; therefore, 3u 3 x x,œÊœ œ Êœ œ œ œ œ
$# # """
###
#
#
dy dy dy
du dx dx du dx
du du 3
xxx ÈÈÈ ˆ‰
ÈÈÈ
†† †
as expected.
(b) y u ; u x 3x ; therefore, 3x 3x x ,œÊœ œÊœ œ œ œ œÈ
dy dy dy
du dx dx du dxuu
du du 3
x
"""
##
$ # # # "Î#
#
#ÈÈÈ
†† †
&
again as expected.

157. Section 3.5 The Chain Rule and Parametric Equations 153
65. y 2 tan 2 sec secœÊœ œˆ‰ ˆ ‰ˆ‰
11111xxx
4dx 44 4
dy ##
#
(a) sec slope of tangent is 2; thus, y(1) 2 tan 2 and y (1) tangent line is¹ ˆ‰ ˆ‰
dy
dx 4 4
x=1
œœÊ œœœÊ
11 1
#
# w
11
given by y 2 (x 1) y x 2œ Êœ111
(b) y sec and the smallest value the secant function can have in x 2 is 1 the minimum
w#
#
œ#Ê
11
ˆ‰
x
4
value of y is and that occurs when sec 1 sec 1 sec x 0.
w##
###
111111
œÊœÊ„œÊœˆ‰ ˆ‰ ˆ‰
xx x
44 4
66. (a) y sin 2x y 2 cos 2x y (0) 2 cos (0) 2 tangent to y sin 2x at the origin is y 2x;œÊœ Êœ œÊ œ œ
ww
y sin y cos y (0) cos 0 tangent to y sin at the origin isœ Ê œ Ê œ œ Ê œˆ‰ ˆ‰ ˆ‰
xx x
### ## #
ww """
y x. The tangents are perpendicular to each other at the origin since the product of their slopes isœ
"
#
1.
(b) y sin (mx) y m cos (mx) y (0) m cos 0 m; y sin y cosœÊœ ÊœœœÊœ
ww w " ˆ‰ ˆ‰
xx
mmm
y (0) cos (0) . Since m 1, the tangent lines are perpendicular at the origin.Êœ œ œ
w "" "
mm m
†ˆ‰
(c) y sin (mx) y m cos (mx). The largest value cos (mx) can attain is 1 at x 0 the largest valueœÊœ œÊ
w
y can attain is m because y m cos (mx) m cos mx m 1 m . Also, y sin
ww
kk kk k k kkk k kk kk ˆ‰œœŸœœ †
x
m
y cos y cos cos the largest value y can attain is .Êœ Ê œ Ÿ Ÿ Ê
ww w"""" "
mm m m m m m m
xxx ˆ‰ ¸ ¸ ¸¸¸ ¸ ¸¸kk ˆ‰ ˆ‰
kk
(d) y sin (mx) y m cos (mx) y (0) m slope of curve at the origin is m. Also, sin (mx) completesœÊœ ÊœÊ
ww
m periods on [0 2 ]. Therefore the slope of the curve y sin (mx) at the origin is the same as the numberߜ1
of periods it completes on [0 2 ]. In particular, for large m, we can think of “compressing" the graph ofß1
y sin x horizontally which gives more periods completed on [0 2 ], but also increases the slope of theϧ 1
graph at the origin.
67. x cos 2t, y sin 2t, 0 t 68. x cos ( t), y sin ( t), 0 tœ œ ŸŸ œ œ ŸŸ 1111
cos 2t sin 2t 1 x y 1 cos ( t) sin ( t) 1ʜʜ Ê œ
## ## # #
11
x y 1, y Êœ  !
##

69. x 4 cos t, y 2 sin t, 0 t 2 70. x 4 sin t, y 5 cos t, 0 t 2œ œ ŸŸ œ œ ŸŸ 11
1 1 1 1Ê œÊ œ Ê œÊ œ
16 cos t 4 sin t x 16 sin t 25 cos t x
16 4 16 4 16 25 16 5
y y
œœ œ œ œ œ œ œ
#

158. 154 Chapter 3 Differentiation
71. x 3t, y 9t , t y x 72. x t , y t, t 0 x yœœ__Êœ œ œ Êœ
## È È
or y x , x 0 œŸ
#
73. x 2t 5, y 4t 7, t 74. x 3 3t, y 2t, 0 t 1 tœ œ__ œ œ ŸŸÊœ
y
#
x 5 2t 2(x 5) 4t x 3 3 2x 6 3yÊœ Ê œ Êœ Ê œ ˆ‰
y
#
y 2(x 5) 7 y 2x 3 y 2 x, xÊœ Êœ Êœ !ŸŸ$
2
3

75. x t, y 1 t , 1 t 0 76. x t 1, y t, t 0œœ ŸŸ œ œ  
È È È
#
y 1 x y t x y 1, y 0Êœ Ê œÊœ  
È È
# ##

77. x sec t 1, y tan t, t 78. x sec t, y tan t, tœœ œ œ
#
## ##
11 11
sec t 1 tan t x y sec t tan t 1 x y 1ʜʜ ʜʜ
## # ## ##

159. Section 3.5 The Chain Rule and Parametric Equations 155
79. (a) x a cos t, y a sin t, 0 t 2 80. (a) x a sin t, y b cos t, tœœŸŸ œœ ŸŸ 1
11
##
5
(b) x a cos t, y a sin t, 0 t 2 (b) x a cos t, y b sin t, 0 t 2œœŸŸ œœŸŸ 11
(c) x a cos t, y a sin t, 0 t 4 (c) x a sin t, y b cos t, tœœŸŸ œœ ŸŸ 1
11
##
9
(d) x a cos t, y a sin t, 0 t 4 (d) x a cos t, y b sin t, 0 t 4œœŸŸ œœŸŸ 11
81. Using we create the parametric equations x at and y bt, representing a line which goesab"ß $ œ " œ $
through at t . We determine a and b so that the line goes through when t .ab ab"ß$ œ! %ß" œ"
Since a a . Since b b . Therefore, one possible parameterization is x t,%œ" Ê œ& "œ$ Ê œ% œ"&
y t, 0 t .œ$% Ÿ Ÿ"

82. Using we create the parametric equations x at and y bt, representing a line which goes throughab"ß$ œ" œ$
at t . We determine a and b so that the line goes through when t . Since a a .ab ab"ß $ œ ! $ß # œ " $ œ " Ê œ %
Since b b . Therefore, one possible parameterization is x t, y t, 0 t .#œ$ Ê œ& œ"% œ$& Ÿ Ÿ"
83. The lower half of the parabola is given by x y for y . Substituting t for y, we obtain one possibleœ" Ÿ!
#
parameterization x t , y t, t 0œ"œ ŸÞ
#
84. The vertex of the parabola is at , so the left half of the parabola is given by y x x for x . Substitutingab"ß " œ # Ÿ "
#
t for x, we obtain one possible parametrization: x t, y t t, t .œœ#Ÿ"
#
85. For simplicity, we assume that x and y are linear functions of t and that the point x, y starts at for t and passesab a b #ß $ œ !
through at t . Then x f t , where f and f .a b ab ab ab"ß " œ " œ ! œ # " œ "
Since slope , x f t t t. Also, y g t , where g and g .œ œ œ$ œ œ$ #œ#$ œ ! œ$ " œ"
?
?
x
t
"#
"!
ab ab a b a b
Since slope 4. y g t t t.œ œ œ œ œ% $œ$%
?
?
y
t
3"
"!
ab
One possible parameterization is: x t, y t, t .œ#$ œ$%  !
86. For simplicity, we assume that x and y are linear functions of t and that the point x, y starts at for t andab a b "ß # œ !
passes through at t . Then x f t , where f and f .a b ab ab ab!ß! œ" œ !œ" "œ!
Since slope , x f t t t. Also, y g t , where g and g .œ œ œ" œ œ" " œ" œ ! œ# " œ!
?
?
x
t
! "
"!
ab
ab a b ab a b a b
Since slope . y g t t t.œ œ œ# œ œ# #œ##
?
?
y
t
!œ#
"!
ab
One possible parameterization is: x t, y t, t .œ" œ##  !
87. t x 2 cos 2, y 2 sin 2; 2 sin t, 2 cos t cot tœ Ê œ œ œ œ œ œ Ê œ œ œ
11 1
4 4 4 dt dt dx dx/dt 2 sin t
dx 2 cos t dy dy dy/dt ÈÈ
œ
cot 1; tangent line is y 2 1 x 2 or y x 2 2 ; csc tÊœœ œ œ œ¹Š‹
ÈÈ È
dy dy
dx 4 dt
tŸ
œ
4
1

#
2Êœ œ œ Ê œ
dy dy/dt dy
dx dx/dt 2 sin t 2 sin t dx
csc t
œ& œ

œ
œ
"
¹
È
tŸ
œ
4
88. t x cos , y 3 cos ; sin t, 3 sin t 3œ Ê œ œ œ œ œ œ Ê œ œ
22 2dx
33 3dtdt dxsin t
33 sin t dy dy11 1 "
## œ
œ ÈÈÈ
ÈÈ
3 ; tangent line is y 3 x or y 3 x; 0 0Êœ œ œ œÊœœ¹Š‹
ÈÈÈ œ ‘ˆ‰
dy dy d y
dx dt dx sin t
3 0
tŸ
2
3œ
È
## œ
" &œ
œ
0Êœ¹
dy
dx
œ
œ
tŸ
2
3œ
89. t x , y ; 1, 1; tangent line isœÊœ œ œ œ Ê œ œ Ê œ œ
11dx 1
4 4 dt dt dx dx/dt dx
dy dy dy/dt dy
t2t
"" "
# #
#
ÈÈ
É
¹
tŸ
1
4
4 y 1x or yx ; t t 2œœœÊœœÊœ
"" "" "
#
œ$# œ$#
†ˆ‰ ¹
4 4 dt 4 dx dx/dt 4 dx
dy d y dy /dt d y
&œ& œ
œœ
tŸ
1
4

160. 156 Chapter 3 Differentiation
90. t 3 x 3 1 2, y 3(3) 3; (t 1) , (3t) œÊœ œ œ œ œ œ Ê œÈ È
dx 3
dt dt dx
dy dy (3t)
(t 1)
"
##
"Î# "Î#

ˆ‰
ˆ‰
3
œ


œ

2; tangent line is y 3 2[x ( 2)] or y 2x 1;œ œ œ œ œ œ
3t1 33 1
3t
dy
dx 3(3)
ÈÈ
ÈÈ

¹
t3Ÿ

dy d y
dt 3t dx
3t (t 1) 3 t 1 (3t) 33
2t 3t t 1 t 3t
œ
œœ

œ
œœÊœœ
È ‘ ‘È
ÈÈÈ

33 Š‹
Š‹
3
2t 3t t 1
1
2t 1
œœ
œ
&
Ê
&
Êœ¹
dy
dx 3
œ
œ
t3Ÿ
"
91. t 1 x 5, y 1; 4t, 4t t ( 1) 1; tangent line isœ Ê œ œ œ œ Ê œ œ œ Ê œ œ
dx 4t
dt dt dx dx/dt 4t dx
dy dy dy/dt dy$##
&
¹
t1Ÿ&
y 1 1 (x 5) or y x 4; 2t œ œ œ Ê œ œ œ Ê œ†
dy d y dy /dt d y
dt dx dx/dt 4t dx
2t
&œ& œ
œœ
""
##
¹
t1Ÿ&
92. t x sin , y 1 cos 1 ; 1 cos t, sin t œ Ê œ œ œ œœ œ œ Ê œ
1111 1
3 3 3 3 3 dt dt dx dx/dt
3 dx dy dy dy/dt
È
###
""
3 ; tangent line is y 3 xœÊœ œœ œ
sin t
1cos t dx 3
dy sin
1cos
3
œ ##œ
"
¹Š‹
ÈÈ
tŸ
œ
3
ˆ‰
ˆ‰
Š‹
ˆ‰
È
œ
œ
3
3
3
œ
œ

œ
1
y3x 2; Êœ œ œ Ê œ œ
È
1
È ˆ‰
3
3 dt (1 cos t) 1 cos t dx dx/dt 1 cos t
dy (1 cos t)(cos t) (sin t)(sin t) d y dy /dt1
œ&
œœ


œœ
œœ œ
œ
1
1 cos t
4œÊœ
œ
œ
1
(1 cos t) dx
dy
œœ
œ ¹
tŸ
œ
3
93. t x cos 0, y 1 sin 2; sin t, cos t cot tœÊœ œ œ œ œ œ Ê œ œ
11 1
22 2dtdtdxsin t
dx cos t dy dy
œ
cot 0; tangent line is y 2; csc t csc t 1Êœœ œœÊœœÊ œ¹ ¹
dy dy d y d y
dx dt dx sin t dx
csc t
t tŸ Ÿ
œ œ
2 2
1
#œ
#$
&œ œ
œœ
œ
94. t x sec 1 1, y tan 1; 2 sec t tan t, sec tœ Ê œ œ œ œ œ œ
11 1
44 4dtdt
dx dy###
ˆ‰ ˆ‰
cot t cot ; tangent line isÊœ œ œ Ê œ œ
dy dy
dx 2 sec t tan t 2 tan t dx 4
sec t
œ
œ
"" " "
###
¹ ˆ‰
tŸ&
œ
4
1
y ( 1) (x 1) or y x ; csc t cot t œ œ œ Ê œ œ
"""" "
####
#$
œdy d y
dt dx 2 sec t tan t 4
csc t
&œ
œœ

œ
œ
Êœ¹
dy
dx 4
œ
œ
tŸ&
œ
4
"
95. s A cos (2 bt) v A sin (2 bt)(2 b) 2 bA sin (2 bt). If we replace b with 2b to double theœÊœœ œ11111
ds
dt
frequency, the velocity formula gives v 4 bA sin (4 bt) doubling the frequency causes the velocity toœ Ê11
double. Also v bA sin (2 bt) a 4 b A cos (2 bt). If we replace b with 2b in theœ# Ê œ œ11 1 1
dv
dt
##
acceleration formula, we get a 16 b A cos (4 bt) doubling the frequency causes the acceleration toœ Ê11
##
quadruple. Finally, a 4 b A cos (2 bt) j 8 b A sin (2 bt). If we replace b with 2b in the jerkœ Ê œ œ11 11
## $$ da
dt
formula, we get j 64 b A sin (4 bt) doubling the frequency multiplies the jerk by a factor of 8.œÊ11
$$
96. (a) y 37 sin (x 101) 25 y 37 cos (x 101) cos (x 101) .œÊœ œ œ ‘ œ ‘ˆ ‰ œ ‘
222742
365 365 365 365 365
11111 w
The temperature is increasing the fastest when y is as large as possible. The largest value of
w
cos (x 101) is 1 and occurs when (x 101) 0 x 101 on day 101 of the yearœ ‘
22
365 365
11
œÊœÊ
( April 11), the temperature is increasing the fastest.µ
(b) y (101) cos (101 101) cos (0) 0.64 °F/day
w
œœœ¸
74 2 74 74
365 365 365 365
11 1 1
œ ‘
97. s ( 4t) v (1 4t) (4) 2(1 4t) v(6) 2( 6) m/sec;œ" Ê œ œ œ Ê œ "% œ
"Î# "Î# "Î# "Î#"
#
ds 2
dt 5
†
v 2( 4t) a 2(1 4t) (4) 4(1 4t) a(6) 4(1 4 6) m/secœ"Êœœ œÊœ œ
"Î# $Î# $Î# $Î# # "
# #
dv 4
dt 15
††

161. Section 3.5 The Chain Rule and Parametric Equations 157
98. We need to show a is constant: a and k s a vœœœœœÊœœ
dv dv dv ds dv d k dv ds dv
dt dt ds dt ds ds ds dt ds 2s
††† ˆ‰È
È
k s which is a constant.œœ
kk
2sÈ
†È
œ
#
99. v proportional to v for some constant k . Thus, a v
"
ÈÈss
k dv k dv dv ds dv
ds dt ds dt ds2s
Êœ Ê œ œ œ œ
††
acceleration is a constant times so a is inversely proportional to s .œ œ Ê
kk k
2s s ss
&
œ
œœ†
È #
"" #
ˆ‰
100. Let f(x). Then, a f(x) f(x) (f(x)) f(x) f (x)f(x), as required.
dx dv dv dx dv d dx d
dt dt dx dt dx dx dt dx
œœœœœœ œ †† † † ˆ‰
w
101. T 2 2 . Therefore, kL 2 kœÊœ œœ œœ œœ11 1É É
LdT dTdTdL L
g dL g du dL du g
g
gL gL
kL
g
†† † † †
"" "
#
#
ÉÉ
ÈÈÈ
È
LL
gg
11 1 1
, as required.œ
kT
2
102. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f g is‰
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so
there is no contradiction.
103. The graph of y (f g)(x) has a horizontal tangent at x 1 provided that (f g) (1) 0 f (g(1))g (1) 0œ‰ œ ‰ œ Ê œ
www
either f (g(1)) 0 or g (1) 0 (or both) either the graph of f has a horizontal tangent at u g(1), or theÊœœÊ œ
ww
graph of g has a horizontal tangent at x 1 (or both).œ
104. (f g) ( 5) 0 f (g( 5)) g ( 5) 0 f (g( 5)) and g ( 5) are both nonzero and have opposite signs.‰ Ê Ê
wwwww
†
That is, either f (g( 5)) 0 and g ( 5) 0 or f (g( 5)) 0 and g ( 5) 0 .cdcd
wwww

105. As h 0, the graph of yĜ
sin 2(x h) sin 2x
h

approaches the graph of y 2 cos 2x becauseœ
lim (sin 2x) 2 cos 2x.
hÄ!
sin 2(x h) sin 2x
hdx
d
œœ

106. As h 0, the graph of yĜ
cos (x h) cos x
h
cd ab
œœ
approaches the graph of y 2x sin x becauseœ ab
#
lim cos x 2x sin x .
hÄ!
cos (x h) cos x
hdx
dcd ab ##
œœ
œœcd abab

162. 158 Chapter 3 Differentiation
107. cos t and 2 cos 2t ; then 0 0
dx 2 cos 2t
dt dt dx dx/dt cos t cos t dx cos t
dy dy dy/dt dy 22 cost 1 22 cost 1
œœÊœœœ œÊœ
ab ab
œœ
œœ
2 cos t 1 0 cos t t , , , . In the 1st quadrant: t x sin andʜʜ„Êœ œÊœœ
# "
#È
È
2 4444 4 4
357 21111 1 1
y sin 2 1 1 is the point where the tangent line is horizontal. At the origin: x 0 and y 0œœÊß œœˆ‰ Š‹
1
4
2
È
#
sin t 0 t 0 or t and sin 2t 0 t 0, , , ; thus t 0 and t give the tangent lines atÊ œ Êœ œ œ Êœ œ œ 111
11
##
3
the origin. Tangents at origin: 2 y 2x and 2 y 2x¹¹
dy dy
dx dx
t0 t%%
œÊœ œÊœ
œ
108. 2 cos 2t and 3 cos 3t
dx 3 cos 3t
dt dt dx dx/dt 2 cos 2t 2 2 cos t 1
dy dy dy/dt 3(cos 2t cos t sin 2t sin t)
œœÊœœœ
œ
œab
œ
œœœ
3 2 cos t 1 (cos t) 2 sin t cos t sin t
2 2 cos t 1 2 2 cos t 1 2 2 cos t
(3 cos t) 2 cos t 1 2 sin t (3 cos t) 4 cos t 3cdab
ab ab ab
abab
œ
œœœ
œœ œ
œœ
œœ
œœ œ
œ1
; then
0 0 3 cos t 0 or 4 cos t 3 0: 3 cos t 0 t , and
dy
dx 2 2 cos t 1
(3 cos t) 4 cos t 3 3
œÊ œÊ œ œ œÊœ
ab
ab
œ
œ
œ
œ ##
# 11
4 cos t 3 0 cos t t , , , . In the 1st quadrant: t x sin 2
#
# #
œ Ê œ„ Êœ œ Ê œ œ
È È
3 3
66 6 6 6 6
5711111 1 1 1
ˆ‰
and y sin 3 1 1 is the point where the graph has a horizontal tangent. At the origin: x 0œœÊß œˆ‰ Š‹
1
6
3
È
#
and y 0 sin 2t 0 and sin 3t 0 t 0, , , and t 0, , , , , t 0 and t giveœÊ œ œÊœ œ Êœ œ
1 1 11 11
##
11 1
3245
33 3 3
the tangent lines at the origin. Tangents at the origin: y x, and ¹¹
dy dy
dx 2 cos 0 dx
3 cos 0 3 3
t0 t%%
œœÊœ
##
œ
yxœœÊœ
3 cos (3 )
2 cos (2 )
331
1 ##
109. From the power rule, with y x , we get x . From the chain rule, y xœœ œ
"Î% $Î% "dy
dx 4
ÉÈ
x x , in agreement.Êœ œ œ
dy
dx dx 4
xx
d
x
""""
##
#
œ$Ÿ
ÉÉÈÈ
È
††ˆ‰È
110. From the power rule, with y x , we get x . From the chain rule, y x xœœ œ
$Î% "Î% dy
dx 4
3
ÉÈ
xx x x xÊœ Êœ œ œ
dy dy
dx dx dx
xx xx xx 4xx
d 3
x
3x""""
###
# #
ÉÉÉÉÈÈÈÈ
È
È
††††ˆ‰ ˆ ‰ÈÈÈ Š‹
x , in agreement.œœ
3x
4x x
3
4
È
ÈÈÉ
"Î%
111. (a)
(b) 1.27324 sin 2t 0.42444 sin 6t 0.2546 sin 10t 0.18186 sin 14t
df
dt
œ
(c) The curve of y approximates yœœ
df
dt dt
dg
the best when t is not , , 0, , nor .11
11
##

163. Section 3.5 The Chain Rule and Parametric Equations 159
112. (a)
(b) 2.5464 cos (2t) 2.5464 cos (6t) 2.5465 cos (10t) 2.54646 cos (14t) 2.54646 cos (18t)
dh
dt
œ
(c)
111-116. Example CAS commands:
:Maple
f := t -> 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t)
- 0.02546*cos(10*t) - 0.01299*cos(14*t);
g := t -> piecewise( t<-Pi/2, t+Pi, t<0, -t, t<Pi/2, t, Pi-t );
plot( [f(t),g(t)], t=-Pi..Pi );
Df := D(f);
Dg := D(g);
plot( [Df(t),Dg(t)], t=-Pi..Pi );
: (functions, domains, and value for t0 may change):Mathematica
To see the relationship between f[t] and f'[t] in 111 and h[t] in 112
Clear[t, f]
f[t_] = 0.78540 0.63662 Cos[2t] 0.07074 Cos[6t] 0.02546 Cos[10t] 0.01299 Cos[14t]
f'[t]
Plot[{f[t], f'[t]},{t, , }]11
For the parametric equations in 113 - 116, do the following. Do NOT use the colon when defining tanline.
Clear[x, y, t]
t0 = p/4;
x[t_]:=1 Cos[t]
y[t_]:=1 Sin[t]
p1=ParametricPlot[{x[t], y[t]},{t, , }]11
yp[t_]:=y'[t]/x'[t]
ypp[t_]:=yp'[t]/x'[t]
yp[t0]//N
ypp[t0]//N
tanline[x_]=y[t0] yp[t0] (x x[t0])
p2=Plot[tanline[x], {x, 0, 1}]
Show[p1, p2]

164. 160 Chapter 3 Differentiation
3.6 IMPLICIT DIFFERENTIATION
1. y x x 2. y x x*Î* * Î*%
*Î% &Î% $Î& )Î&dy dy
dx 4 dx 5
9 3
3. y 2x (2x) (2x) 2 4. y 5x (5x) (5x) 5œœ Êœ œ œœ Êœ œ
$ %"Î$ #Î$ "Î% $Î%" "È È
dy dy
dx 3 dx 4
2 5
3x 4x
††
*Î% *Î&
%Î&
5. y 7 x 6 7(x 6) (x 6)*&*& Î*& *È
'Ê! "'Ê!
#
dy
dx
77
2x 6È
6. y 2 x 1 2(x 1) 1(x 1)*% %*%% Î *%% *%
È
'Ê! "'Ê! "
"
dy
dx x1
È
7. y (2x 5) (2x 5) 2 (2x 5)*& Î*% & *%&
"'Ê! "†Ê! "†Ê! "
#
dy
dx
†
8. y ( 6x) (1 6x) ( 6) 4(1 6x)œ" Ê œ œ
!ʆ "'ʆ "'ʆdy
dx 3
2
9. y x x 1 y x x 1 x x 1 x 1 x xœ Êœ† #†"œ "œab ababab aba b
#w# # ###
'Ê! "'Ê! 'Ê! "'Ê!"
#
2x 1
x1
"
"
È
10. y x x 1 y x x 1 x x 1 x 1 x xœ Ê œ † # †"œ " œab ababab aba b ˆ‰
#w# # ###
"'Ê! "†Ê! "'Ê! "†Ê! " "
#
abx1
"

11. s t t t 12. r **Î* * * Î*%
( %# "†#Î( &Î( $Î% (Î%ÈÈ ds 2 dr 3
dt 7 d4
)) )
)
13. y sin (2t 5) cos (2t 5) (2t 5) 2 (2t 5) cos (2t 5)*& Î* & %& *%& &ˆ‰ ˆ‰ˆ‰ ˆ‰
#Î$ #Î$ &Î$ &Î$ #Î$dy
dt 3 3
24
††
14. z cos ( 6t) sin ( 6t) (1 6t) ( ) 4(1 6t) sin (1 6t)œ" Êœ" 'œ ˆ‰ ˆ‰ ˆ‰
!ʆ !ʆ "'ʆ "'ʆ !ʆdz 2
dt 3
†
15. f(x) 1 x 1 x f (x) 1 x x*%*% Î * % % * *É È ˆ‰ ˆ‰ˆ ‰
'Ê! ß 'Ê! "'Ê!
'Ê! "'Ê!
' ' "' "'
##
" "41 xx 4x1 xŠ‹ÉÉÈÈ ˆ‰È
16. g(x) 2 2x 1 g (x) 2x 1 ( 1)x 2x 1 x*&Î*%&%*&ˆ‰ ˆ‰ ˆ‰
"'Ê! ß "'Ê! "†Ê! "'Ê! "†Ê!
"Î$ %Î$ %Î$
22
33
†
17. h( ) 1 cos (2 ) (1 cos 2 ) h ( ) (1 cos 2 ) ( sin 2 ) 2 (sin 2 )(1 cos 2 )))))))))*& *& Î *& % *% &
$ 'ʆ ß "!ʆ "!ʆ"
È
33
2
††
18. k( ) (sin ( 5)) k ( ) (sin ( 5)) cos ( 5) cos ( 5)(sin ( 5)))) ) ) ) ) )*&Î* & &* & &
&Î% w "Î% "Î% 55
44
†
19. x y xy 6:
##
&*
Step 1: x y 2x x 2y y 1 0Š‹Š ‹
##dy dy
dx dx
&& &*†† †
Step 2: x 2xy 2xy y
##dy dy
dx dx
&*%%
Step 3: x 2xy 2xy y
dy dx
ab
##
&*%%
Step 4:
dy 2xy y dx x 2xy
œ
""

"
"
20. x y 18xy 3x 3y 18y 18x 3y 18x 18y 3x
$$ # # # # "
"
&* Î & *& Î % *% Î*
dy dy dy dy 6y x
dx dx dx dx y 6x
ab
"
"
21. 2xy y x y:&*&
#
Step 1: 2x 2y 2y 1Š‹
dy dy dy
dx dx dx
&& *&

165. Section 3.6 Implicit Differentiation 161
Step 2: 2x 2y 1 2y
dy dy dy
dx dx dx
&%*%
Step 3: (2x2y1) 2y
dy dx
œ"
Step 4:
dy 1 2y dx 2x 2y 1
œ
"
Á"
22. x xy y 1 3x y x 3y 0 3y x y 3x
$$ # # # # "
"
%&*Î %% & *Î % *% Î *
dy dy dy dy y 3x
dx dx dx dx 3y x
ab
"
"
23. x (x y) x y :
####
%*%
Step 1: x 2(x y) 1 (x y) (2x) 2x 2y
##
’“ Š‹%%&% *%
dy dy
dx dx
Step 2: 2x (x y) 2y 2x 2x (x y) 2x(x y)%%& *% %%%
### dy dy
dx dx
Step 3: 2x (x y) 2y 2x 1 x(x y) (x y)
dy dx
cdc d% %&* %%%%
##
Step 4:
dy
dx 2x (x y) 2y y x (x y) x y x y
2x 1 x(x y) (x y) x 1 x(x y) (x y) x 1 x xy x 2xy y
œœœ
cdcd ab" """ " """
""Á "" "Á
"Á "Á "
""
""" %
"" "
œ
x2x 3xyxy
xy x y
"Á "
"Á


24. (3xy 7) 6y 2(3xy 7) 3x 3y 6 2(3xy 7)(3x) 6 6y(3xy 7)&*Î& &*Î& %*%&
#
†Š‹
dy dy dy dy
dx dx dx dx
[6x(3xy 7) 6] 6y(3xy 7) Î&%*%&Î*%*
dy dy y(3xy 7) 3xy 7y
dx dx x(3xy 7) 1 1 3x y 7x

Á" " "
"
"
25. y 2y
# "' '

Á""
œÊ œ œ Êœ
x2
x 1 dx (x 1) (x 1) dx y(x 1)
dy (x 1) (x 1) dy
"" "
26. x x x y x y 3x 2xy x y 1 y x 1 y 1 3x 2xy y
#$# ##ww#w#w" ""

* Î & *% Î & & *% Î & *% % Î *
xy 1 3x 2xy
xy x1
ab
"
"
27. x tan y 1 sec y cos yœÊœ Êœœ ab
## "dy dy
dx dx sec y
"
28. xy cot xy x y csc (xy) x y x x csc (xy) y csc (xy) y*Î&*% &Î& *% %ab Š‹
dy dy dy dy
dx dx dx dx
###
x x csc (xy) y csc (xy)Ê œ " Ê œ œ
dy dy y
dx dx x
y csc (xy)
x csc (xy)
* ‘ * ‘
##
"Á'
"
* ‘
* ‘
"
"
29. x tan (xy) 1 sec (xy) y x 0 x sec (xy) 1 y sec (xy) &*#Î& &*Î *%% Î*cd Š‹
### "' "dy dy dy y sec (xy)
dx dx dx x sec (xy)
"
"
*%*%*
" """1
x sec (xy) x x x x
y cos (xy) y cos (xy) y
"
""
30. x sin y xy 1 (cos y) y x (cos y x) y 1 & * Î & *& Î % *% Î *
dy dy dy dy y 1
dx dx dx dx cos y x
"
"
31. y sin 1 xy y cos ( 1) sin x y Š‹ ’ “ Š‹ Š‹
""""
y y y dx y dx dx
dy dy dy
*% Î % & *% % Î †† †
"
cos sin x y
dy dy y y
dx y y y dx
cos sin x y sin cos xy
’“ Š‹ Š‹%&&*%Î* *
"" " ""
"ÁÁ "Á
** * * *
"
yy y y y
Š‹ Š‹ Š‹ Š‹
32. y cos 2x 2y y sin ( 1) cos 2y 2 2
## """"
Š‹ ’ “ Š‹ Š‹
y y y dx y dx dx
dy dy dy
*& Î % % & *& Î †† †
"
sin 2y cos 2 2
dy dy
dx y y dx
2
sin 2y cos
’“Š‹ Š‹
""
Á"!
&%*Î*
Š‹ Š‹
**
yy
33. r 1 r 0 ))
'Ê! 'Ê! "'Ê! "'Ê! '' '"'
## # #
& * Î & * Î * Î *% *% †
dr dr dr
dd d r
2r r
2)) ) )))
’“
ÈÈ È È
ÈÈ

166. 162 Chapter 3 Differentiation
34. r 2 %* & Î%*& Î*&&
È
)) ) ))) )))
3 4 dr dr
3d d#
#Î$ $Î% "Î# "Î$ "Î% "Î# "Î$ "Î%
))
35. sin (r ) [cos (r )] r 0 [ cos (r )] r cos (r ) ,))) )))*Î & *Î *% Î * *%
"
#
"
ˆ‰
dr dr dr r
dd d cos(r)
r cos (r )
)) ))))
)
cos (r ) 0)Á
36. cos r cot r ( sin r) csc r [ sin r ] r csc &*Î% % *&Î%%*& Î*%)) ) ) ) )
dr dr dr dr r csc
ddd dsin r))) ))
)##

"
37. x y 1 2x 2yy 0 2yy 2x y ; now to find , y
## w w w w
&*Î& *Î *%Î**% * %
dy d y
dx y dx dx dx y
xddx
"
"
ab Š‹
y since y yÎ* * *%Î ** * *
ww w ww"Á ""
"Á "
""'" "'y( 1) xy d y y x
yy ydx yyy
yx
x yy
†"""

"" Š‹
ab
x
y
38. x y 1 x y 0 y x y ;
!ʆ !ʆ "'ʆ "'ʆ "'ʆ "'ʆ ß
"Î$
& * Î & * Î *% Î * *% *%
22 2 2 x
3 3 dx dx 3 3 dx x
dy dy dy y
y
œ ‘ˆ‰
)*Î%
)*Î%
Differentiating again, y
ww
"Á
""Á
œœ
xyyyx
xx
xy yx
œœ

œœ
*Î%
*Î%
†
†
ˆ‰ˆ‰
ˆ‰ ˆ‰
Œ Ê
33
33
y
x
xy yxÎ* & * &
dy y
dx 3 3 3x 3y x
" *Î%
"
"" "#Î$ "Î$ "Î$ %Î$
39. y x 2x 2yy 2x 2 y ; then y
## w w ww "Á
"Á
*& Î *&Î * * * *
2x 2 x 1
2y y y y
y(x1)y
y(x1)
†
""
' Š‹
x1
y
yÊœœ
dy y (x 1)
dx y
"""

ww "Á
40. y 2x 1 2y 2y y 2 2y y (2y 2) 2 y (y 1) ; then y (y 1) y
!ßßßß"'ßß"!ß "

% *% Î %*% Î &*Î * *& *%&† †
y1
(y 1) (y 1) y*% & & Î * *
"! "' ßß "'

dy
dx (y 1)
"
" %
41. 2 y x y y y 1 y y y 1 1 y ; we canÈ ˆ‰*% Î *% Î & * Î * * *
"'Ê! ß ß ß "'Ê! ß "

dy
dx y 1 y1
y
#œÊ"
È
È
differentiate the equation y y 1 1 again to find y : y y y y 1 y 0
ß "'Ê! ßß ß "†Ê! ß "'Ê! ßß "
#ˆ‰ ˆ ‰ˆ‰&* % & & *
y1yyy yÎ&* Î** * *ˆ‰ cd
"'Ê! ßß ß "†Ê! ßß"""
#
#
#
dy
dx
y
y1 2y y 1 1y
"
"
œ
"
"

#œÊ"

%% Œ Ê
ab ab ˆ‰È
œ
Î*%&
"
y142. xy y 1 xy y 2yy 0 xy 2yy y y (x 2y) y y ; y&*Î&& *Î& *%Î &*%Î* *
#wwww w w ww "

ydy
(x 2y) dx
"
"
œœ œ
"Á Á Á
ÁÁ Á
"Á Á Á ÁÁÁ"
(x 2y)y y(1 2y )
(x 2y) (x 2y) (x 2y)
(x 2y) y 1 2 y(x 2y) y(x 2y) 2y
††
"" "
" ’“’ “ Š‹ cd
ÊÊ
""
œ
"yy
(x 2y) (x 2y) (x 2y)
œœœ
2y(x 2y) 2y 2y 2xy 2y(x y)
(x 2y) (x 2y) (x 2y)
Á" Á Á

""
%%%
43. x y 16 3x 3y y 0 3y y 3x y ; we differentiate y y x to find y :
$$ # #w #w # w #w # ww
&* Î & *Î *% Î *% *%
x
y
"
"
y y y 2y y 2x y y 2x 2y y y
#ww w w #ww w ww #
"" " ""
&*%Î*%%Î* *cd cd†
2x 2y 2x
yy
Š‹
x2x
yy
&$
&)
"
""
2*Î**%
"" ""2xy 2x d y
ydx 32
32 32

!"
¹
(2 2)œ
44. xy y 1 xy y 2yy 0 y (x 2y) y y y ;&*Î && *Î & *%Î * Î *
#www w ww "

Á"""Áy
(x 2y) (x 2y)
(x 2y) y ( y) 1 2yab a b
††
"
since y we obtain ykk
www ""
#
""
(0 1) (0 1)œÊœÊ
*% * *%
( 2) (1)(0)
44
ˆ‰
œ
"
45. y x y 2x at ( ) and ( 1) 2y 2x 4y 2 2y 4y 2 2x
## % $ $
œ #ß " #ß Ê œ Ê œ
dy dy dy dy
dx dx dx dx

167. Section 3.6 Implicit Differentiation 163
2y 4y 2 2x 1 and 1Î % *% % Î * Î *% *
dy dy dy dy
dx dx y y dx dx
x
ab ¹¹
$ "
!"
%
(21) (2 1)Êœ ÊœÊ
46. x y (x y) at( ) and ( 1) 2 x y 2x 2y 2(x y) 1ab ab Š‹ Š‹
## # ##
#
œ "ß ! "ß Ê œ
dy dy
dx dx
2y x y (x y) 2x x y (x y) 1Î & & % *% & & % Î * Î *%
dy dy dy
dx dx 2y x y (x y) dx
2x x y (x y)
cdabab ¹
## ## "ÁÁ"
ÁÁ"
ab
ab
""
""
(1 0)œ and 1¹
dy
dx
(1 1)œÊ
œ
47. x xy y 1 2x y xy 2yy 0 (x 2y)y 2x y y ;
## ww w w
"
&%*Î && % *Î % *%%Î*
2x y
2y x
(a) the slope of the tangent line m y the tangent line is y 3 (x 2) y x**Î %*%Î*%k
w "
#
(2 3)œ
777
444
(b) the normal line is y 3 (x 2) y x%*% % Î *% &
4429
777
48. x y 25 2x 2yy 0 y ;
## w w
&* Î & *Î *%
x y
(a) the slope of the tangent line m y the tangent line is y 4 (x 3)**%*Î &*%k ¹
w
(3 4)
(3 4)œÊ
œÊ
x3 3
y4 4
yxÎ* %
325
44
(b) the normal line is y 4 (x 3) y x&*% % Î *%
44
33
49. x y 9 2xy 2x yy 0 x yy xy y ;
## # # w # w # w
*Î & *Î *% Î *%
y
x
(a) the slope of the tangent line m y 3 the tangent line is y 3 3(x 1)**%*Î %*&k ¸
w
(13) (13)ʜ ʜ
y
x
y3x6Î* &
(b) the normal line is y 3 (x 1) y x%*% & Î *% &
""
333
8
50. y 2x 4y 2yy 2 4y 0 2(y 2)y 2 y ;
#wwww "
"!
"œ!Ê œÊ œÊœ
y
(a) the slope of the tangent line m y 1 the tangent line is y 1 1(x 2) y x 1**%Î %*%&Î*%%k
w
(21)ʜ
(b) the normal line is y 1 1(x 2) y x 3%* & Î *&
51. 6x 3xy 2y 17y 6 0 12x 3y 3xy 4yy 17y 0 y (3x 4y 17) 12x 3y
# # www w
&&&%*Î&& & & *Î &&*%%
y;ʜ
w ""

12x 3y
3x 4y 17
(a) the slope of the tangent line m y the tangent line is y 0 (x 1)** *Î %*&k ¹
w "' "

(10)
(10)ʜ
ʜ
2x 3y
3x 4y 17 7 7
66
yxÎ* &
66
77
(b) the normal line is y 0 (x 1) y x%*% & Î *% %
777
666
52. x 3xy 2y 5 2x 3xy 3y 4yy 0 y 4y 3x 3y 2x y ;
## www w "
"
%&*Î% %&*Î % *%Î*
ÈÈÈ ÈÈ
Š‹
È
È
3y 2x
4y 3x
(a) the slope of the tangent line m y 0 the tangent line is y 2œœ œÊ œk ¹
w "
"
œÊ
%
œÊ
%32
32œ
œ
È
È
3y 2x
4y 3x (b) the normal line is x 3œ
È
53. 2xy sin y 2 2xy 2y (cos y)y 0 y (2x cos y) 2y y ;&*Î&& *Î& *%Î*11 1 1
www w "

2y
2x cos y1
(a) the slope of the tangent line m y the tangent line is** *%Îk ¹
w "
#

1
1œ
œ
œ
œ
2
2
2y
2x cos y1
1
y(x1) yx%*% % Î *% &
11 1
## #
1
(b) the normal line is y (x 1) y x%* % Î * %&
11
111##
222

168. 164 Chapter 3 Differentiation
54. x sin 2y y cos 2x x(cos 2y)2y sin 2y 2y sin 2x y cos 2x y (2x cos 2y cos 2x)*Î &*%& Î %
www
sin 2y 2y sin 2x y ;*% % Î *
w
"
sin 2y 2y sin 2x
cos 2x 2x cos 2y
(a) the slope of the tangent line m y 2 the tangent line isœœ œœÊk ¹
w
"


œœ
œœ
42
42
œ
œ
sin 2y 2y sin 2x
cos 2x 2x cos 2y
1
œ
"
y2x y2x%* % Î *
11
#
ˆ‰
4
(b) the normal line is y x y x%*% % Î *% &
11 1
## #
""
ˆ‰
48
5
55. y 2 sin ( x y) y 2 [cos ( x y)] y y [1 2 cos ( x y)] 2 cos ( x y)*%Î* %%Î&%* %111 111
www
†ab
y;ʜ
w "
Á! "
2 cos(x y)
1 cos(xy)
11
1
(a) the slope of the tangent line m y 2 the tangent line isœœ œÊk ¹
w "
Á"
(1 0)
(1 0)œ
œ
2 cos(x y)
12 cos(xy)
11
1
1
y02(x1) y2x2%* % Î * %111
(b) the normal line is y 0 (x 1) y%*% % Î *% &
""
##111
x
2
56. x cos y sin y 0 x (2 cos y)( sin y)y 2x cos y y cos y 0 y 2x cos y sin y cos y
## # w # w w #
%*Î % & % *Î% % cd
2x cos y y ;*% Î *
#w

2x cos y
2x cos y sin y cos y
"
"
(a) the slope of the tangent line m y 0 the tangent line is yœœ œÊ œk ¹
w

(0 )
(0 )œ
œ
œ
œ
2x cos y
2x cos y sin y cos y
"
"
1
(b) the normal line is x 0œ57. Solving x xy y 7 and y 0 x 7 x 7 7 and 7 are the points where the
## #
&&* *Î *Î*( Î%w# w#
ÈÈ È
Š‹Š‹
curve crosses the x-axis. Now x xy y 7 2x y xy 2yy 0 (x 2y)y 2x y
## ww w
&&*Î && & *Î & *%% y m the slope at 7 is m 2 and the slope at 7 isÎ *% Î *% Î % w# *% *% w#
w

"
"
2x y 2x y
x2y x2y
27
7
Š‹ Š‹
ÈÈ
È
È
m 2. Since the slope is 2 in each case, the corresponding tangents must be parallel.*% *% %
27
7
È
È
58. x xy y 7 2x y x 2y 0 (x 2y) 2x y and ;
## "" Á
Á""
&&*Î && & *Î & *%%Î * *
dy dy dy dy 2x y x 2y
dx dx dx dx x 2y dy 2x y
dx
(a) Solving 0 2x y 0 y 2x and substitution into the original equation gives
dy
dx
*Î%%*Î*%
x x( 2x) ( 2x) 7 3x 7 x and y 2 when the tangents are parallel to the
###
&% &% *Î *Î*( *' ÉÉ
77
33
x-axis.
(b) Solving 0 x 2y 0 y and substitution gives x x 7 7
dx x x x 3x dy 4
*Î& *Î*% &% &% *Î *
###
#
#
ˆ‰ˆ‰
"
x 2 and y when the tangents are parallel to the y-axis.Êœ„ œ… ÉÉ
77
33
59. y y x 4y y 2yy 2x 2 2y y y 2x y ; the slope of the tangent line at
%## $w w $ w w
"
*% Î * % Î % *% Î * ab
x
y2y
%
is 1; the slope of the tangent line at Š‹ ¹ Š‹
ÈÈ È
33 3
4y2y 3 4
x
w****% w
!" !" !
"
"
""
% œ
"
œ
"
†#
œœ33
42
œ
œ
œœ
3
44
363
8
3
4
is 3 ¹
È
x
y2y 4 2
23
"" "
% œ
"
†#
œ3
42
1
œ
œœœ
œ3
4
2
8
È
60. y (2 x) x 2yy (2 x) y ( 1) 3x y ; the slope of the tangent line is
#$w##w
"
%* Î %& %* Î *
y3x
2y(2 x)
""
m 2 the tangent line is y 1 2(x 1) y 2x 1; the normal line is***Î %*%Î*%¹
y3x
2y(2 x)
4
""

"!
(1 1)œ
y 1 (x 1) y x%*% % Î *% &
""
###
3
61. y 4y x 9x 4y y 8yy 4x 18x y 4y 8y 4x 18x y
%#%# $w w $ w$ $ w ""
""
%*%Î %*%Î %*%Î* * ab
4x 18x 2x 9x
4y 8y 2y 4y
%%
%%

169. Section 3.6 Implicit Differentiation 165
m; ( 3 2): m ; ( ): m ; (3 ): m ; (3 ): m* *%w * *%%„w%)* w)* w%)*%
x2x 9
y2y 4 2(8 4) 8 8 8 8
( 3)(18 9) 27 27 27 27ab
ab
"
"
"
""
""
62. x y 9xy 0 3x 3y y 9xy 9y 0 y 3y 9x 9y 3x y
$$ # #w w w# # w ""
""
&% *Î & % %*Î % *% Î* * ab
9y 3x 3y x
3y 9x y 3x
""
"" (a) y and y ;kk
ww
(4 2) (2 4)œœ
œœ
54
45
(b) y 0 0 3y x 0 y x 9x 0 x 54x 0
w#$'$ "
"
$
*Î *Î %*Î* Î & % *Î % *
3y x
y3x 3 3 3
xxx
"
"
"""
Š‹ Š‹
x x 54 0 x 0 or x 54 3 2 there is a horizontal tangent at x 3 2 . To find theÎ%*Î*** Î *
$$ $$ $
ab
È ÈÈ
corresponding y-value, we will use part (c).
(c) 0 0 y 3x 0 y 3x ; y 3x x 3x 9x 3x 0
dx
dy 3y x
y3x
*Î *Î % *Î*( * Î & % *
"
"
"
"
#$
$ ÈÈ È È
Š‹
x 6 3 x 0 x x 6 3 0 x 0 or x 6 3 x 0 or x 108 3 4 .Î% *Î % *Î * * Î* * *
$ $Î# $Î# $Î# $Î# $Î# $$ÈÈ ÈÈ
Š‹
È
Since the equation x y 9xy 0 is symmetric in x and y, the graph is symmetric about the line y x.
$$
&% * *
That is, if (a b) is a point on the folium, then so is (b a). Moreover, if y m, then y .ßßœœ kk
ww "
(a b) (b a)œœ m
Thus, if the folium has a horizontal tangent at (a b), it has a vertical tangent at (b a) so one might expectßß
that with a horizontal tangent at x 54 and a vertical tangent at x 3 4, the points of tangency areœœ
$$È È
54 3 4 and 3 4 54 , respectively. One can check that these points do satisfy the equationŠ‹Š‹
ÈÈ ÈÈ
$$ $$
ßß
x y 9xy 0.
$$
&% *
63. x 2tx 2t 4 2x 2x 2t 4t 0 (2x 2t) 2x 4t ;
## ""
""
%&*Î %% &*Î % *%Î* *
dx dx dx dx 2x 4t x 2t
dt dt dt dt 2x 2t x t
2y 3t 4 6y 6t 0 ; thus ; t 2
$# # "
"
%*Î %*Î** * * * *
dy dy dy dy/dt t(x t)
dt dt 6y y dx dx/dt y (x 2t)
6t t
"" #"
"
#
Š‹
ˆ‰
t
y
x2t
xt
x 2(2)x 2(2) 4 x 4x 4 0 (x 2) 0 x 2; t 2 2y 3(2) 4Î% & *Î%&*Î%*Î* *Î % *
### # $#
2y 16 y 8 y 2; therefore 0ʜʜʜ œ œ
$$ "
"
¹
dy 2(2 2)
dx (2) (2 2(2))
t2%
"
64. x 5 t 5 t t ; y(t 1) t y (t 1) t*%Î* % % *% %*Î&% *É ÈÈ È ˆ‰ˆ ‰
dx
dt dt
4t 5 t
dy""" "
## #
"'Ê!
"'Ê! "'Ê!
"ÈÈÉ
t 1 y ; thus Î% * %Î * * * * * $ab
dy dy dy
dt dt t 1 dxttt2t
y
yt

"
!!"
"
"
'"!
ÈÈÈ ab
È
œ
&&Î
*Î&
ʜ
Ê
œœœ
œ
œœÊ
ttt2t
yt
4t 5 t
dy
dt
dx
dt
*Î%
%Î*
Ê
ʜ
yt
tt
4t 5 tÈ
Èab
ÈÈÉ
; t 4 x 5 4 3; t 4 y(3) 4 2**Î*%**Î**
%*Î% &Î
œÊ
ˆ‰ ÈÈÉyt t
t
É ÈÈÈ
therefore, ¹
dy
dx 3
14
t4%
œœ
2224 4
4
Š‹ab
ÈÈÉ*Î &Î
œÊ
65. x 2x t t 3x 2t 1 1 3x 2t 1 ; y t 1 2t y 4& *&Î & *&Î & *&Î * && *
$Î# # "Î# "Î#

dx dx dx dx 2t 1
dt dt dt dt 13x ˆ‰ È È
œÊ"
t1y (t1) 2y2t y 0 t1 2y 0Î && & && *Î && && *
dy dy dy y dy
dt dt dt y dt 2t 1
tÈÈ ˆ‰ ˆ ‰ ÈÈ Š‹ ""
##
"'Ê! "'Ê!
ÈÈ
t 1 2 y ; thusÎ&& * %Î* *Š‹È È
t
ydt dt
dy y dy
2t1ÈÈ
"

Š‹ È
Š‹È
È È
È È
Ê
"y
2t 1
t
y
œ
œ
Ê
""
ÊÊ "
"" "
2y
t1
yy4yt 1
2 y (t 1) 2t t 1
; t 0 x 2x 0 x 1 2x 0 x 0; t 0
dy dy/dt
dx dx/dt
* * *Î& *Î & *Î* *
Œ œ
Š‹
ÊÊ "
"" "
"
"
œ%&yy 4yt 1
2 y (t 1) 2t t 1
2t 1
13xœ œ
œ œ
$Î# "Î# ˆ‰
y 0 1 2(0) y 4 y 4; therefore 6Î&& *Î* * *%È È ¹
dy
dx
t0%
Œ Ê
Œ Ê
ÊÊ "
"" "
"
"
*%&44 4(4)0 1
2 4(0 1) 2(0) 0 1
2(0) 1
13(0)
œ œ
œ œ

170. 166 Chapter 3 Differentiation
66. x sin t 2x t sin t x cos t 2 1 (sin t 2) 1 x cos t ;&*Î & & *Î & *% Î *
dx dx dx dx 1 x cos t
dt dt dt dt sin t 2
"

t sin t 2t y sin t t cos t 2 ; thus ; t x sin 2x%*Î & %* * *Î & *
dy dy
dt dx sin t t cos t 2"Ê
ˆ‰
1x cos t
sin t 2Ê
" 111
x ; therefore 4Î* * * *%
11111
1#
Á"""
¹
dy
dx 2
sin cos 2 4 8
t%œ
–—
1cos
sin 2Ê
&
"
%&
œ
œ
œ
67. (a) if f(x) x 3, then f (x) x and f (x) x so the claim f (x) x is false*% * *% *
3
3#
#Î$ w "Î$ ww %Î$ ww "Î$ "
(b) if f(x) x 7, then f (x) x and f (x) x is true*% * *
93
10
&Î$ w #Î$ ww "Î$
#
(c) f (x) x f (x) x is true
ww "Î$ www %Î$ "
*Î*%
3
(d) if f (x) x 6, then f (x) x is true
ß !ʆ ßß "'ʆ
#
*& *
3
68. 2x 3y 5 4x 6yy 0 y y and y ;
## w w w w
&*Î& *Î*%Î *% *% *% *
2x 2x 2 2x 2
3y 3y 3 3y 3
kk ¹¹
(1 1) (1 1)
(1 1) (1 1)œœ Ê
œœÊ
also, y x 2yy 3x y y and y . Therefore
#$ w # w w w
##
*Î * Î* Î * * * *%
3x 3x 3 3x 3
2y 2y 2y
"" "
kk ¹¹
(1 1) (1 1)
(1 1) (1 1)œœ Ê
œœÊ
the tangents to the curves are perpendicular at (1 1) and (1 1) (i.e., the curves are orthogonal at these twoww%
points of intersection).
69. x 2xy 3y 0 2x 2xy 2y 6yy 0 y (2x 6y) 2x 2y y the slope of the
## www w
"
&%*Î& &% *Î %*%%Î* Î
xy
3y x
tangent line m y 1 the equation of the normal line at (1 1) is y 1 1(x 1)** *Î w%*%%k ¹
w
"
(1 1)
(1 1)œ
œ
xy
3y x
y x 2. To find where the normal line intersects the curve we substitute into its equation:Î*%&
x 2x(2 x) 3(2 x) 0 x 4x 2x 3 4 4x x 0 4x 16x 12 0
######
&%%%*Î&%%%&*Î%&%* ab
x 4x 3 0 (x 3)(x 1) 0 x 3 and y x 2 1. Therefore, the normal to the curveÎ % &*Î % % * Î * *%&*%
#
at (1 1) intersects the curve at the point (3 1). Note that it also intersects the curve at (1 1).ww% w
70. xy 2x y 0 x y 2 0 ; the slope of the line 2x y 0 is 2. In order to be& %*Î &&% *Î * &* %
dy dy dy y 2
dx dx dx 1 x

"
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope of%
the tangent is . Therefore, 2y 4 1 x x 3 2y. Substituting in the original equation,
""
!"!
y2
1x
* Î &*% Î *%%
y( 3 2y) 2( 3 2y) y 0 y 4y 3 0 y 3 or y 1. If y 3, then x 3 and% % & % % % * Î & & * Î *% *% *% *
#
y 3 2(x 3) y 2x 3. If y 1, then x 1 and y 1 2(x 1) y 2x 3.& *% % Î *% & *% *% & *% & Î *% %
71. y x . If a normal is drawn from (a 0) to (x y ) on the curve its slope satisfies 2y
# "
# "
"" "
"
*Î * w w *%
dy y0
dx y xa œ
œ
y 2y (x a) or a x . Since x 0 on the curve, we must have that a . By symmetry, theÎ*% % *& j j""" " "
""
##
two points on the parabola are x x and x x . For the normal to be perpendicular,ˆ‰ˆ ‰ÈÈ "" " "ww%
1 1 x (a x ) x x x x and y .Š‹Š‹ ˆ‰
ÈÈxx
xa ax (ax) 4
x
œœ
œœ œ
œ
"
"" " ! !
"""""" "
# """
#
*% Î * Î * % Î * & % Î * * (
Therefore, and a .ˆ‰
""
#44
3
ß„ œ
72. Ex. 6b.) y x has no derivative at x 0 because the slope of the graph becomes vertical at x 0.œœ œ
"Î#
Ex. 7a.) y 1 x has a derivative only on ( ) because the function is defined only on [ ] andœ "ß " "ß "ab
#
"Î%
the slope of the tangent becomes vertical at both x 1 and x 1.*% *
73. xy x y 6 x 3y y x 2xy 0 3xy x y 2xy
$# # $# ## $ ""

&*Î && &*Î &*%% Î*Š‹ ab
dy dy dy dy y 2xy
dx dx dx dx 3xy x
%
""
; also, xy x y 6 x 3y y x y 2x 0 y 2xy 3xy x*% & * Î & & & * Î & *% %
y 2xy
3xy x dy dy dy
dx dx dx
%
""


$# # $ # $ ##
ab a b Š‹
; thus appears to equal . The two different treatments view the graphs as functionsÎ*%
dx dx
dy y 2xy dy
3xy x
""
%


"
dy
dx

171. Section 3.6 Implicit Differentiation 167
symmetric across the line y x, so their slopes are reciprocals of one another at the corresponding pointsœ
(a b) and (b a).ßß
74. x y sin y 3x 2y (2 sin y)(cos y) (2y 2 sin y cos y) 3x
$# # # # "
"
&* Î & * Î % *% Î *
dy dy dy dy
dx dx dx dx 2y 2 sin y cos y
3x
"
; also, x y sin y 3x 2y 2 sin y cos y ; thus *&*Î&*Î*
3x dx dx dx
2 sin y cos y 2y dy dy 3x dy
2 sin y cos y 2y
"
"
"
$# # # "
appears to equal . The two different treatments view the graphs as functions symmetric across the line
"
dy
dx
y x so their slopes are reciprocals of one another at the corresponding points (a b) and (b a).œßß
75. x 4y 1:
%#
&*
(a) y y 1 x
# "'
#
%
*Î*(%
1x
4
&
È
1 x 4x ;Î*( % % *
dy
dx 4
x
1x
"„ %$
"'Ê!
"
abab
%
&
œÊ"
ab
differentiating implicitly, we find, 4x 8y 0
$
&*
dy
dx
. Êœ œ œ
dy
dx 8y
4x 4x x
81x 1x
"" $
$" "
%% %
œ
"
& &
œÊ"
Š‹
È ab
(b)
76. (x 2) y 4:%&*
##
(a) y 4 (x 2)*( % %È
#
4(x2) (2(x2))Î*( %% %%
dy
dx
"
#
#
"'Ê!
ab
; differentiating implicitly,œ
$"
""
(x 2)
4(x2)cd
"
œÊ"
2(x 2) 2y 0 %& *Î *
dy dy 2(x 2)
dx dx 2y
""
. œœ œ
"" "" $"
$" " " "!
(x 2) (x 2) (x 2)
y
4(x2) 4(x )cdcd
""
œÊ" œÊ"
(b)
77-84. Example CAS commands:
:Maple
q1 := x^3-x*y+y^3 = 7;
pt := [x=2,y=1];
p1 := implicitplot( q1, x=-3..3, y=-3..3 ):
p1;

172. 168 Chapter 3 Differentiation
eval( q1, pt );
q2 := implicitdiff( q1, y, x );
m := eval( q2, pt );
tan_line := y = 1 + m*(x-2);
p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ):
p3 := pointplot( eval([x,y],pt), color=blue ):
display( [p1,p2,p3], ="Section 3.6 #77(c)" );
: (functions and x0 may vary):Mathematica
Note use of double equal sign (logic statement) in definition of eqn and tanline.
<<Graphics`ImplicitPlot`
Clear[x, y]
{x0, y0}={1, /4};1
eqn=x + Tan[y/x]==2;
ImplicitPlot[eqn,{ x, x0 3, x0 3},{y, y0 3, y0 3}]%& %&
eqn/.{x x0, y y0}ÄÄ
eqn/.{ y y[x]}Ä
D[%, x]
Solve[%, y'[x]]
slope=y'[x]/.First[%]
m=slope/.{x x0, y[x] y0}ÄÄ
tanline=y==y0 m (x x0)&%
ImplicitPlot[{eqn, tanline}, {x, x0 3, x0 3},{y, y0 3, y0 + 3}]%& %
3.7 RELATED RATES
1. A r 2 r œÊœ11
# dA dr
dt dt
2. S 4 r 8 r œÊœ11
# dS dr
dt dt
3. (a) V r h r (b) V r h 2 rh œÊœ œÊœ11 1 1
## # dV dh dV dr
dt dt dt dt
(c) V r h r 2 rh *Î* &111
## dV dh dr
dt dt dt
4. (a) V r h r (b) V r h rh œÊœ œÊœ
"" "## #
3dt3dt 3dt3dt
dV dh dV 2 dr
11 11
(c) r rh
dV dh 2 dr
dt 3 dt 3 dt
*&
"#
11
5. (a) 1 volt/sec (b) amp/sec
dV dI
dt dt 3
**%
"
(c) R I R
dV dI dR dR dV dI dR dV V dI
dt dt dt dt I dt dt dt I dt I dt
*& Î*%Î*%ˆ‰ ˆ ‰ ˆ ‰ ˆ ‰
""
(d) 1 (3) ohms/sec, R is increasing
dR 12 3
dt 3
*%%* *
"""
## # #
œ ‘ˆ‰ˆ‰
6. (a) P RI I 2RI *Î* &
## dP dR dI
dt dt dt
(b) P RI 0 I 2RI * Î * * & Î *% *% *%
## dP dR dI dR 2RI dI dI 2P dI
dt dt dt dt I dt I dt I dt
2
"" %
ˆ‰
P
I
7. (a) s x y x y *&*& Î*È ab
## ##
"Î#

ds x dx
dt dtxyÈ
""
(b) s x y x y *&*& Î* &È ab
## ##
"Î#

ds x dx
dt dt dtxy xy
ydy
ÈÈ
"" ""
(c) s x y s x y 2s 2x 2y 2s 0 2x 2y *&Î*&Î * & Î * & Î*%È
## ### ds dx dx dx
dt dt dt dt dt dt x dt
dy dy y dy
†
8. (a) s x y z s x y z 2s 2x 2y 2z * &&Î*&&Î * & &È
### #### ds dx dz
dt dt dt dt
dy

173. Section 3.7 Related Rates 169
Î* & &
ds x dx z dz
dt dt dt dtxyz xyz xyz
ydy
ÈÈÈ
""" """ """
ÁÁ ÁÁ ÁÁ
(b) From part (a) with 0
dx ds z dz
dt dt dt dt
ydy
xyz xyz
*Î * &
ÈÈ""" """
ÁÁ ÁÁ
(c) From part (a) with 0 0 2x 2y 2z 0
ds dx dz dx z dz
dt dt dt dt dt x dt x dt
dy y dy
*Î* & & Î & & *
9. (a) A ab sin ab cos (b) A ab sin ab cos b sin *Î* *Î* &
"" """
## ###
)) )))
dA d dA d da
dt dt dt dt dt
) )
(c) A ab sin ab cos b sin a sin *Î* & &
""""
####
))))
dA d da db
dt dt dt dt
)
10. Given A r , 0.01 cm/sec, and r 50 cm. Since 2 r , then 2 (50)œœ œ œ œ111
# "dr dA dr dA
dt dt dt dt 100 ¸ˆ‰
r=50
cm /min.œ1
#
11. Given 2 cm/sec, 2 cm/sec, 12 cm and w 5 cm.
ddw
dt dt
j
*% * Þ* *
(a) A w w 12(2) 5( 2) 14 cm /sec, increasing*ÞÎ*Þ& Î* &%*
dA dw d dA
dt dt dt dt
j #
(b) P 2 2w 2 2 2( 2) 2(2) 0 cm/sec, constant*Þ& Î * & * % & *
dP d dw
dt dt dt
j
(c) D w w w 2w 2 * &Þ * &Þ Î * &Þ & Þ Î *
È
ab ab ˆ‰
## ## ##
'Ê! "'Ê! Á   "j
# j
dD dw d dD
dt dt dt dt
w
w
dw d
dt dt
""
È
cm/sec, decreasing**%
(5)(2) (12)( 2)
25 144
14
13
Á"
È
12. (a) V xyz yz xz xy (3)(2)(1) (4)(2)( 2) (4)(3)(1) 2 m /sec*Î* & & Î * & %& *
dV dx dz dV
dt dt dt dt dt
dy
¸
(432)œœ
$
(b) S 2xy 2xz 2yz (2y 2z) (2x 2z) (2x 2y) * && Î*& && &&
dS dx dz
dt dt dt dt
dy
(10)(1) (12)( 2) (14)(1) 0 m /secÎ*&%&*¸
dS
dt
(432)œœ
#
(c) xyz xyz Þ* &&* && Î * & &È ab
### ###
"Î# j
ÁÁ ÁÁ ÁÁ
dxdx zdz
dt dt dt dtxyz xyz xyz
ydy
ÈÈÈ
""" """ """
(1) ( 2) (1) 0 m/secÎ* &%& *¸ Š‹ Š‹ Š‹
d43 2
dt 29 29 29
j
(432)œœ ÈÈ È
13. Given: 5 ft/sec, the ladder is 13 ft long, and x 12, y 5 at the instant of time
dx
dt
œœœ
(a) Since x y 169 (5) 12 ft/sec, the ladder is sliding down the wall
##
&* Î *% *% *%
dy
dt y dt 5
xdx 12
ˆ‰
(b) The area of the triangle formed by the ladder and walls is A xy x y . The area*Î* &
""
##
dA dx
dt dt dt
dy
ˆ‰Š‹
is changing at [12( 12) 5(5)] 59.5 ft /sec.
"
##
#
% & *% *%
119
(c) cos sin (5) 1 rad/sec))*Î% * Î*% *% *%
xddxd dx
13 dt 13 dt dt 13 sin dt 5
))
)
"""
†† ˆ‰
14. s y x 2s 2x 2y x y [5( 442) 12( 481)]
### ""
*& Î * & Î * & Î * % & %
ds dx ds dx ds
dt dt dt dt s dt dt dt
dy dy
169
Š‹
È
614 knots*%
15. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the
girl and kite s (300) x 20 ft/sec.Î* &Î* * *
### ds x dx
dt s dt 500
400(25)
16. When the diameter is 3.8 in., the radius is 1.9 in. and in/min. Also V 6 r 12 r
dr dV dr
dt 3000 dt dt
œœÊœ
" #
11
12 (1.9) 0.0076 . The volume is changing at about 0.0239 in /min.Êœ œ
dV
dt 3000
11ˆ‰
" $
17. V r h, h (2r) r V h œœœÊœÊœ œÊœ
""#
#
3 8 4 3 3 3 27 dt 9 dt
3 3r 4h 4h 16 h dV 16 h dh
11 ˆ‰
11
% "
(a) (10) 0.1119 m/sec 11.19 cm/sec¸ˆ‰
dh 9 90
dt 16 4 256h=4
œœ¸ œ
11
"
(b) r 0.1492 m/sec 14.92 cm/secœÊœ œ œ ¸ œ
4h dr 4 dh 4 90 15
3 dt 3 dt 3 256 32
ˆ‰
11

174. 170 Chapter 3 Differentiation
18. (a) V r h and r V h œœÊœœÊœÊœœ
"" "#
##
# "
3 3 4 dt 4 dt dt 225 (5) 225
15h 15h 75 h dV 225 h dh dh 8 4( 50)
11 ˆ‰ ¸
11
11
% "
"
h=5
0.0113 m/min 1.13 cm/min†% *% (b) r 0.0849 m/sec 8.49 cm/sec* Î * Î * * †% *%
15h dr 15 dh dr 15 8 4
dt dt dt 225 15## #
""
¸ˆ‰ˆ‰
h=5 11
19. (a) V y (3R y) 2y(3R y) y ( 1) 6Ry 3y at R 13 and*%Î* %&%Î*% Î*
11 1
3 dt 3 dt dt 3 dt
dV dVdy dy###
"'
cdab œ ‘
y 8 we have ( 6) m/min**%*
dy
dt 144 24
'"'
11
(b) The hemisphere is on the circle r (13 y) 169 r 26y y m
## #
&%* Î* % È
(c) r 26y y 26y y (26 2y) * % Î* % % Î* Î *ab ab ¸ˆ‰
##
'Ê! "'Ê! " ""'
# #
"
" "
dr dr dr 13 8
dt dt dt dt dt 4
dy 13 y dy
26y y 26 8 64
ÈÈ
" y=8 † 1
m/minœ
"5
2881
20. If V r , S 4 r , and kS 4k r , then 4 r 4k r 4 r k, a constant.œœ œœ œÊœÊœ
4dV dVdrdrdr
3dt dtdtdtdt
11 1 1 11
$# # # ##
Therefore, the radius is increasing at a constant rate.
21. If V r , r 5, and 100 ft /min, then 4 r 1 ft/min. Then S 4 r œœ œ œ Êœ œÊ
4dV dVdrdr dS
3dt dtdtdt dt
111 1
$$# #
8 r 8 (5)(1) 40 ft /min, the rate at which the surface area is increasing.œœ œ11 1
dr
dt
#
22. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.
(a) We have s x 36 . Therefore, the boat is approaching the dock at
##
"
*& Î * *
dx s ds s ds
dt x dt dt
s36
È
" ( 2) 2.5 ft/sec.¸
dx 10
dt
10 36
s=10
*%*%
È "
"
(b) cos sin . Thus, r 10, x 8, and sin )) )*Î% *% Î * * * *
6d6drd6dr 8
r dt r dt dt r sin dt 10
))
)
""
( 2) rad/secÎ* %*%
d6 3
dt 2010
)
"ˆ‰
8
10†
23. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal
distance between the balloon and the bicycle. The relationship between the variables is s h x
###
*&
h x [68(1) 51(17)] 11 ft/sec.Î* & Î* & *
ds dh dx ds
dt s dt dt dt 85
""
ˆ‰
24. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is
V 9 h 9 the rate the coffee is rising is in/min.œÊœ Ê œ œ11
dV dh dh dV 10
dt dt dt 9 dt 9
"
11
(b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter r V r hœÊ œ
h
3#
"#
1
, the volume of the filter. The rate the coffee is falling is ( 10) in/min.***%*%
1
11 1
h dh 4 dV 4 8
1 dt h dt 5 5
%
"
# #
25. y QD D QD (0) ( 2) L/min increasing about 0.2772 L/min*Î* % *%%* Î
"' "' "! "dy
dt dt dt 41 (41) 1681
dQ dD 233 466
"
26. (a) 3x 12x 15 3(2) 12(2) 15 (0.1) 0.3, 9 9(0.1) 0.9, 0.9 0.3 0.6
dc dx dr dx
dt dt dt dt dt
dp
* % & * % & * * * * *%*aba b
##
(b) 3x 12x 45x 3(1.5) 12(1.5) 45(1.5) (0.05) = 1.5625, 70 70(0.05) 3.5,
dc dx dr dx
dt dt dt dt
*%% * % % % * * *aba b
!"! ! "!
3.5 ( 1.5625) 5.0625
dp
dt
*%% *
27. Let P(x y) represent a point on the curve y x and the angle of inclination of a line containing P and theߜ
#
)
origin. Consequently, tan tan x sec cos . Since 10 m/sec)) ) )œÊ œœÊ œ Ê œ œ
y
x x dt dt dt dt dt
xddxddxdx
"
##))
and cos , we have 1 rad/sec.k ¸
#

"
)
x=3 x=3
œœœ œ
x3 d
yx 93 10 dt
""
"" ""
)
28. y ( x) and tan tan sec *% * Î * Î *
"Î# # " """"
)) )
y(x)
xxdtxdt
ddx ( x) ( 1)x ( x) (1)
œÊ" œ
"
#œÊ" œÊ"
"
)
ˆ‰

175. Section 3.7 Related Rates 171
cos . Now, tan cos cos . ThenÎ* **%Î *% Î *
ddx224
dt x dt 4 5
x
5
)
Œ Êab ˆ‰
Ê
Êx
2x
œ
""
# #
"!
"
È
È
" )) ) )
( 8) rad/sec.
d42
dt 16 5 5
2)
*%*Š‹ ˆ‰
4
4
"
29. The distance from the origin is s x y and we wish to find *&È ¸
## ds
dt
(5 12)œ
x y 2x 2y 5 m/sec*& & * *%
"
#
##
"'Ê! "Á "

ab ¹Š‹
dx
dt dt
dy (5)( 1) (12)( 5)
25 144
(5 12)œ
È
30. When s represents the length of the shadow and x the distance of the man from the streetlight, then s x.œ
3
5
(a) If I represents the distance of the tip of the shadow from the streetlight, then I s x *& Î * &
dI ds dx
dt dt dt
(which is velocity not speed) 5 8 ft/sec, the speed the tip of theÎ* &* *%*¸¸ ¸ ¸ ¸¸¸¸ kk
dI 3 dx dx 8 dx 8
dt 5 dt dt 5 dt 5
shadow is moving along the ground.
(b) ( 5) 3 ft/sec, so the length of the shadow is at a rate of 3 ft/sec.
ds 3 dx 3
dt 5 dt 5
**%*% decreasing
31. Let s 16t represent the distance the ball has fallen,œ
#
h the distance between the ball and the ground, and I
the distance between the shadow and the point directly
beneath the ball. Accordingly, s h 50 and since&*
the triangle LOQ and triangle PRQ are similar we have
I h 50 16t and I*Î*% *
30h
50 h 50 50 16t
30 50 16t
"""
# "ab
ab
"
"
30 1500 ft/sec.*%Î*% Î *%
1500 dI 1500 dI
16t dt 8t dt
" %
¸
t=
1
2

32. Let s distance of car from foot of perpendicular in the textbook diagram tan sec œÊœÊœ ))
sdds
13 dt 13 dt##
# ")
; 264 and 0 2 rad/sec. A half second later the car has traveled 132 ftÎ * *% * Î *%
d cos ds ds d
dt 132 dt dt dt
)) )
"
)
right of the perpendicular , cos , and 264 (since s increases) (264) 1 rad/sec.Êœ œ œ Êœ œkk))
1)
4 dt dt 132
ds d# "
#
ˆ‰
œ
"
33. The volume of the ice is V r 4 4 r in./min when 10 in /min, the*% Î* Î * *%
4 4 dV dr dr 5 dV
3 3 dt dt dt 72 dt
11 1
$$ # $ " ¸
r=6 1
thickness of the ice is decreasing at in/min. The surface area is S 4 r 8 r 48
5dSdrdS5
72 dt dt dt 721 1
œÊœ Ê œ11 1
# " ¸ˆ‰
r=6
in /min, the outer surface area of the ice is decreasing at in /min.*%
10 10
3 3
# #
34. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between
the car and plane 9 s r ( 160) 200 mphÎ&* Î * Î * % *%
##
"
ds r dr ds 5
dt dt dt
r9 16
ÈÈ
"
¸
r=5
speed of plane speed of car 200 mph the speed of the car is 80 mph.Î&*Î
35. When x represents the length of the shadow, then tan sec .))*Î *% Î*
80 d 80 dx dx x sec d
x dt x dt dt 80 dt
# ")))
"
""
We are given that 0.27° rad/min. At x 60, cos
d3 3
dt 000 5
)1
œœ œ œÊ
#
)
ft/min 0.589 ft/min 7.1 in./min.¸¸ ¹¹¹
dx x sec d 3
dt 80 dt 16
œœ¸¸
"
""
)) 1
œÊ
d3 5
dt 2000 3œÊ
= and sec =)
36. Let A represent the side opposite and B represent the side adjacent . tan sec )))) *Î * %
AddAAdB
BdtBdtBdt
# ")
"
t at A 10 m and B 20 m we have cos and ( 2) (1)Î* * ** * %% )
20 2 d 10 4
10 5 5 dt 0 400 5ÈÈ
)
œ ‘ˆ ‰ˆ‰ ˆ ‰
"
#
rad/sec /sec 6 °/sec* % *% *% †%ˆ‰ˆ‰
"' ' '
10 40 5 10
418 °
1
37. Let x represent distance of the player from second base and s the distance to third base. Then 16 ft/sec
dx
dt
*%
(a) s x 8100 2s 2x . When the player is 30 ft from first base, x 60
##
*& Î * Î * *
ds dx ds x dx
dt dt dt s dt

176. 172 Chapter 3 Differentiation
s 30 13 and ( 16) 8.875 ft/secÎ* * % * †%
È
ds 60 32
dt 30 13 13
ÈÈ
"
(b) cos sin . Therefore, x 60 and s 30 13)) ""*Î% *% Î * * * *
90 90 ds 90 ds 90 ds
s dt s dt dt s sin dt sx dt
dd))
) œœ
""
œ
†††
È
rad/sec; sin cos Î* * *Î *% Î*
d dd
dt 65 s dt s dt dt s cos dt
90 32 8 90 90 ds 90 ds
30 13 (60) 13
) ))
)œ ""
""
"
Š‹
È È
†††Š‹
"" "
##
))
. Therefore, x 60 and s 30 13 rad/sec.œœœÊœ
"90 ds 8
sx dt dt 65
d
†
È
)"
(c) lim
d d
dt s sin dt s dt s dt x 8100 dt dt
90 ds 90 x dx 90 dx 90 dx
s
) )
)œ œ
"""
œ
"
œœ œœ ʆ††
ˆ‰†
x
s
ˆ‰ˆ‰ ˆ‰ˆ‰ ˆ ‰

xÄ!
lim ( 15) rad/sec; *%*%***
xÄ!
ˆ ‰ ˆ‰ˆ‰ ˆ ‰ˆ‰ Š‹
90 90 ds 90 x dx 90 dx
x 8100 6 dt s cos dt s s dt s dt
d
""""
"
"

'"" " )
)
†
†
x
s
lim rad/secœÊœˆ‰
"'

90 dx
x 8100 dt dt 6
d
"
"
xÄ!
)
38. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and
D the distance between the ships. By the Law of Cosines, D a b 2ab cos 120°
###
*&%
2a 2b a b . When a 5, 14, b 3, and 21, then Î* & & & * * * * *
dD da db db da da db dD 413
dt D dt dt dt dt dt dt dt 2D
"
#
œ ‘
where D 7. The ships are moving 29.5 knots apart.œœ
dD
dt
3.8 LINEARIZATION AND DIFFERENTIALS
1. f(x) x 2x 3 f (x) 3x 2 L(x) f (2)(x 2) f(2) 10(x 2) 7 L(x) 10x 13 at x 2* % &Î * %Î * %& * %&Î * % *
$w# w 2. f(x) x 9 x 9 f (x) x 9 (2x) L(x) f ( 4)(x 4) f( 4)*&*& Î * & * Î *%&&%
È
ab ab ˆ‰
# #w# w
'Ê! "'Ê! "
#
x
x9
È
"
(x 4) 5 L(x) x at x 4*% & & Î *% & *%
449
555
3. f(x) x f (x) 1 x L(x) f(1) f (1)(x 1) (x 1)*& Î *% Î * & % *)&#% *)
" ß"! ß
x
4. f(x) x f (x) L(x) f ( 8) x 8 f 8 (x 8) 2 L(x) x*Î *Î *%%%&%*&%Î *%
"Î$ w w """
$ ##x 113
4
"Î% ababab
5. f(x) x 2x f (x) 2x 2 L(x) f (0)(x 0) f(0) 2(x 0) 0 L(x) 2x at x 0*& Î *&Î * %& * %&Î * *
#w w
6. f(x) x f (x) x L(x) f (1)(x 1) f(1) ( 1)(x 1) 1 L(x) x 2 at x 1* Î *% Î * %& *% %&Î *%& *
"' ß "! ß
7. f(x) 2x 4x 3 f (x) 4x 4 L(x) f ( 1)(x 1) f( 1) 0(x 1) ( 5) L(x) 5 at x 1*&%Î*&Î*%&&%*&&%Î*%*%
#w w
8. f(x) 1 x f (x) 1 L(x) f (8)(x 8) f(8) 1(x 8) 9 L(x) x 1 at x 8*& Î * Î * % & * % & Î *& *
ww
9. f(x) x x f (x) x L(x) f (8)(x 8) f(8) (x 8) 2 L(x) x at x 8**Î * Î * %&*%&Î *& *
$ 'ʆ ß "!ʆ ß """
##
È ˆ‰
3113
4
10. f(x) f (x) L(x) f (1)(x 1) f(1) (x 1)*Î * * Î* %&*%&
x
x 1 (x 1) (x 1) 4
(1)(x 1) ( )(x)
#
ww Á"' """
""
L(x) x at x 1Î*& *
""
44

177. Section 3.8 Linearization and Differentials 173
11. f(x) sin x f (x) cos xœÊœ
w
(a) L(x) f (0)(x 0) f(0) 1(x 0) 0*%&*%&
w
L(x) x at x 0Êœ œ
(b) L(x) f ( )(x ) f( ) ( 1)(x ) 0*%&*%%&
w
111 1
L(x) x at xÎ*% * 11

12. f(x) cos x f (x) sin x*Î*%
w
(a) L(x) f (0)(x 0) f(0) 0(x 0) 1*%&*%&
w
L(x) 1 at x 0Êœ œ
(b) L(x) f x f*% &&%
w
## #ˆ‰ˆ ‰ ˆ‰
11 1
( 1) x 0 L(x) x*%% & & Î * &ˆ‰
11
# 2
at x *%
1
#

13. f(x) sec x f (x) sec x tan xœÊœ
w
(a) L(x) f (0)(x 0) f(0) 0(x 0) 1*%&*%&
w
L(x) 1 at x 0Êœ œ
(b) L(x) f x f*% &&%
wˆ‰ˆ ‰ ˆ‰
11 1
33 3
2 3 x 2 L(x) 2 2 3 x*% & & Î * % &
ÈȈ‰ ˆ‰
11
33
at x *%
1
3

14. f(x) tan x f (x) sec xœÊœ
w#
(a) L(x) f (0)(x 0) f(0) 1(x 0) 0 x*%&*%&*
w
L(x) x at x 0Êœ œ
(b) L(x) f x f 2 x 1*%&*%&
wˆ‰ˆ ‰ ˆ‰ ˆ ‰
111 1
444 4
L(x) 1 2 x at xÎ*&% * ˆ‰
11
44

15. f x k x . We have f and f k. L x f f x k x kx
www "'
ab a b ab ab ab ab aba b a bœ " ! œ" ! œ œ ! ! ! œ" ! œ"
k
16. (a) f x x x x xaba b ab abœ ‘œ " œ " ¸"' œ"'
' '
(b) f x x x xab ab ababœ ‘ œ ‘œ œ# " ¸# " " œ##
#
'"
"'
x
(c) f x x xab a b ˆ‰œ " ¸" œ"
"'Ê! "
##
x
(d) f x xab
È ÈÈÈ
Š‹ Š ‹ Š‹œ " œ #" ¸ #" œ #"
#
###%
"Î#
"xxx
"""
(e) f x xab a b ˆ‰ ˆ ‰ ˆ‰œ %$ œ% " ¸% " œ% "
"Î$ "Î$ "Î$ "Î$$"$
%$%%
"Î$
xxx

178. 174 Chapter 3 Differentiation
(f) f xabˆ ‰ ˆ‰ ˆ‰ ’“œ " œ " ¸" œ"
""#"#
# # $ # '$
Î$
Î$
xx xx
2
2
17. (a) (1.0002) (1 0.0002) 1 50(0.0002) 1 .01 1.01
&! &!
* & †& *& *
(b) 1.009 (1 0.009) 1 (0.009) 1 0.003 1.003
$ "Î$ "È ˆ‰* & †& *& *
3
18. f(x) x 1 sin x (x 1) sin x f (x) (x 1) cos x L (x) f (0)(x 0) f(0)*&& *& & Î * & & Î * %&È ˆ‰
'Ê! ß "'Ê! ß "
#
f
(x 0) 1 L (x) x 1, the linearization of f(x); g(x) x 1 (x 1) g (x)*%&Î *& *&*& Î
33
##
"Î# w
f
È
(x 1) L (x) g (0)(x 0) g(0) (x 0) 1 L (x) x 1, the linearization of g(x);*& Î* %&*%&Î*&ˆ‰
"""
###
"'Ê! ß
g g
h(x) sin x h (x) cos x L (x) h (0)(x 0) h(0) (1)(x 0) 0 L (x) x, the linearization of*Î* Î * %&*%&Î *
ww
h h
h(x). L (x) L (x) L (x) implies that the linearization of a sum is equal to the sum of the linearizations.
fgh*&
19. y x 3 x x 3x dy 3x x dx dy 3x dx*% *% Î * % Î * %
† † 'Ê! ! "'Ê! !
#È ˆ‰ Š‹
33
2xÈ
20. y x 1 x x 1 x dy (1) 1 x (x) 1 x ( 2x) dx*%*% Î* % & % %
È
ab ab ab ’“ ˆ‰
# ###
'Ê! 'Ê! "'Ê! "
#
1 x 1 x x dx dx*% % % *abc d ab
###
"'Ê! "
"
ab
È
12x
1x
"
"
21. y dy dx dxœÊœ œ
2x 2 2x
1x
(2) 1 x (2x)(2x)
1x 1x
Á"

"
"
"
""
""
" Š‹
ab
ab ab
22. y dy dx dxœœ Êœ œ
2x
31 x
2x 3x 3 3
31 x
x31x 2x x
91 x 91 x
È
ˆ‰È ab
ˆ‰ˆ‰ˆ‰
ab ab
Á"

Á" œÊ" #œÊ"
œÊ"
#œÊ" œÊ" œÊ" #œÊ"
"
œÊ" œÊ"
""
Š‹
3
dy dxʜ
"
3x1 xÈȈ‰
"
23. 2y xy x 0 3y dy y dx x dy dx 0 3y x dy (1 y) dx dy dx
$Î# "Î# "Î# "

& %*Î & & % *Î & *% Î * ˆ‰
1y
3yxÈ
24. xy 4x y 0 y dx 2xy dy 6x dx dy 0 (2xy 1) dy 6x y dx
# $Î# # "Î# "Î# #
%%*Î & % %*Î%* % ˆ‰
dy dxʜ
6x y
2xy 1
È"
"
"
25. y sin 5 x sin 5x dy cos 5x x dx dy dxœ œ Êœ Êœˆ‰ ˆ‰ ˆ ‰ˆ ‰È ˆ‰
'Ê! 'Ê! "'Ê!
#
5 5 cos 5 x
2x
ˆ‰È
È
26. y cos x dy sin x (2x) dx 2x sin x dx*Î*% *%ab c d ab ab
## #
27. y 4 tan dy 4 sec x dx dy 4x sec dxœÊœ ÊœŠ‹ Š ‹ Š‹ Š‹ab
xx x
33 3
%% %
## ##
28. y sec x 1 dy sec x 1 tan x 1 (2x) dx 2x sec x 1 tan x 1 dx*%Î* % % * % %abcdcd abab abab
### ##
29. y 3 csc 1 2 x 3 csc 1 2x dy 3 csc 1 2x cot 1 2x x dx*%*%Î*%% %%ˆ‰ˆ‰ ˆ ‰ˆ‰ˆ‰È ˆ‰
'Ê! 'Ê! 'Ê! "'Ê!
dy csc 1 2 x cot 1 2 x dxÎ* % %
3
xÈ
ˆ‰ˆ‰ÈÈ
30. y 2 cot 2 cot x dy 2 csc x x dx dy csc dx* * Î*% % Î*Š‹ Š‹ˆ‰ ˆ‰ˆ‰ˆ‰
"""" "'Ê! ! "'Ê! "†Ê! !
#È ÈÈx xx
%
31. f(x) x 2x, x 1, dx 0.1 f (x) 2x 2*& * * Î *&
#w
!
(a) f f(x dx) f(x ) f(1.1) f(1) 3.41 3 0.41?*&% * %* %* !!
(b) df f (x ) dx [2(1) 2](0.1) 0.4**&*
w
!

179. Section 3.8 Linearization and Differentials 175
(c) f df 0.41 0.4 0.01kkk k?%* % *
32. f(x) 2x 4x 3, x 1, dx 0.1 f (x) 4x 4*&% *% * Î *&
#w
!
(a) f f(x dx) f(x ) f( .9) f( 1) .02?*&% *%%%* !!
(b) df f (x ) dx [4( 1) 4](.1) 0**%&*
w
!
(c) f df .02 0 .02kkkk?%* %*
33. f(x) x x, x 1, dx 0.1 f (x) 3x 1*% * * Î * %
$w#
!
(a) f f(x dx) f(x ) f(1.1) f(1) .231?*&% * %* !!
(b) df f (x ) dx [3(1) 1](.1) .2**%*
w#
!
(c) f df .231 .2 .031kkk k?%* %*
34. f(x) x , x 1, dx 0.1 f (x) 4xœœœÊœ
%w$
!
(a) f f(x dx) f(x ) f(1.1) f(1) .4641?*&% * %* !!
(b) df f (x ) dx 4(1) (.1) .4œœœ
w$
!
(c) f df .4641 .4 .0641kkk k?%* %*
35. f(x) x , x 0.5, dx 0.1 f (x) x***Î*%
"' ß "!
!
(a) f f(x dx) f(x ) f(.6) f(.5)?*&% *%*% !!
"
3
(b) df f (x ) dx ( 4)**%*%
w
!
" ˆ‰
10 5
2
(c) f dfkk ¸¸?%*%&*
""
35 15
2
36. f(x) x 2x 3, x 2, dx 0.1 f (x) 3x 2*%& * * Î * %
$w#
!
(a) f f(x dx) f(x ) f(2.1) f(2) 1.061?*&% * %* !!
(b) df f (x ) dx (10)(0.10) 1œœ œ
w
!
(c) f df 1.061 1 .061kkk k?%* %*
37. V r dV 4 r dr 38. V x dV 3x dxœÊœ œÊœ
4
3
11
$# $#
! !
39. S 6x dS 12x dxœÊœ
#
!
40. S r r h r r h , h constant r h r r r h*&*& Î*&& &11 1 1
È
ab ab ab
## ## ## ##
'Ê! 'Ê! "'Ê! dS
dr
†
dS dr, h constantʜ ʜ
dS
dr
rh r
rh
2r h
rh
11 1ab
È
ab
É
"" "
""
""

"

"




41. V r h, height constant dV 2 r h dr 42. S 2 rh dS 2 r dhœÊœ œÊœ1111
#
!
43. Given r 2 m, dr .02 mœœ
(a) A r dA 2 r dr 2 (2)(.02) .08 mœÊœ œ œ1111
##
(b) (100%) 2%ˆ‰
.08
4
1
1
œ
44. C 2 r and dC 2 in. dC 2 dr dr the diameter grew about in.; A r dA 2 r drœœÊœÊœÊ œÊœ11 11
" #
11
2
2 (5) 10 in.œœ1ˆ‰
" #
1
45. The volume of a cylinder is V r h. When h is held fixed, we have rh, and so dV rh dr. For h in.,œœ#œ#œ$!111
# dV
dr
r in., and dr in., the volume of the material in the shell is approximately dV rh drœ' œ!Þ& œ# œ# ' $! !Þ& 11 aba ba b
in .œ ")! ¸ &'&Þ&1
$

180. 176 Chapter 3 Differentiation
46. Let angle of elevation and h height of building. Then h tan , so dh sec d . We want dh h,))))œ œ œ$! œ$! l l!Þ!%
#
which gives: sec d tan d d sin cos d sin cos l$! l!Þ!%l$! lÊ l l Êl l!Þ!% Êl l!Þ!%
# "!Þ!% &&
"# "#
)) ) ) ) ) ) )
cos cos
sin
"
))
)11
radian. The angle should be measured with an error of less than radian (or approximatley degrees),œ !Þ!" !Þ!" !Þ&(
which is a percentage error of approximately %.!Þ('
47. V h dV 3 h dh; recall that V dV. Then V (1%)(V) dVœÊœ ¸ Ÿ œ Ê Ÿ11 ? ?
$#
kk kk
(1) h (1) h
100 100
ab ab11
%%
3 h dh dh h % h. Therefore the greatest tolerated error in the measurementÊŸÊŸœkk kk ˆ‰1
# ""(1) h
100 300 3
ab1
%
of h is %.
"
3
48. (a) Let D represent the inside diameter. Then V r h h and h 10 V
i œœ œ œÊœ Ê11
#
##
# ˆ‰
D Dh 5 D
4ii i
11
""
dV 5 D dD . Recall that V dV. We want V (1%)(V) dVœ¸ŸÊŸœ1? ?
ii kk kk ˆ‰Š‹
"
#100 40
5D D1 1
" "
i i
5 D dD 200. The inside diameter must be measured to within 0.5%.ÊŸÊŸ1
ii
1D
40 D
dD
"
ii
i
(b) Let D represent the exterior diameter, h the height and S the area of the painted surface. S D h dS hdD
e ee œÊœ11
. Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameterʜ
dS
SD
dD e
e
is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to
within 5%, the tanks's exterior diameter must be measured to within 5%.
49. V r h, h is constant dV 2 rh dr; recall that V dV. We want V V dVœÊœ ¸ ŸÊŸ11??
# "
kk kk
1000 1000
rh1
"
2 rh dr dr (.05%)r a .05% variation in the radius can be tolerated.ÊŸÊŸœÊkk kk1
1rh r
1000 000
"
#
50. Volume (x x) x 3x ( x) 3x( x) ( x)*&*&&&????
$$ # # $

51. W a a bg dW bg dg 37.87, so a change of*& *& Î *% *% Î * * *
b 32
g g dW 5.2
b dg dW"' "!
#
"
moon
earth
Š‹
Š‹Ê
Ê
b dg
(5.2)
b dg
(32)
&
&
ˆ‰
gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth.
52. (a) T 2 dT 2 L g dg L g dg*Î*%*%11 1Š‹
ÈȈ‰
L
g
"Î#
"
#
"†Ê! "†Ê!
(b) If g increases, then dg 0 dT 0. The period T decreases and the clock ticks more frequently. BothÊ
the pendulum speed and clock speed increase.
(c) 0.001 100 980 dg dg 0.977 cm/sec the new g 979 cm/sec*% Î †% Î †1
È ˆ‰
"†Ê! ! !
53. The error in measurement dx (1%)(10) 0.1 cm; V x dV 3x dx 3(10) (0.1) 30 cm theœœ œÊœœ œÊ
$## $
percentage error in the volume calculation is (100%) 3%ˆ‰
30
1000
œ

181. Section 3.8 Linearization and Differentials 177
54. A s dA 2s ds; recall that A dA. Then A (2%)A dA 2s dsœÊ œ ¸ Ÿ œœÊ ŸÊ Ÿ
#
?? kk kk k k
2s s s s
100 50 50 50
"" " "
ds (1%) s the error must be no more than 1% of the true value.ÊŸ œœ Êkk
ss
(2s)(50) 100
"
55. Given D 100 cm, dD 1 cm, V dV D dD (100) (1) . Then (100%)œœœœÊœœœ
4D D 10 dV
36 V
1ˆ‰
####
$
##111 1
% &
10 % % 3%œœœ”• ”•ab
10 10
10 10
66
†'
&&
''
œœ
œœ
#
56. V r dV dD; recall that V dV. Then V (3%)Vœœ œÊœ ¸ Ÿ œ
44DD D 3D
33 6 100 6
11 ? ?
$
##
$ ˆ‰ ˆ ‰ kk Š‹
11 1
% " %
dV dD dD (1%) D the allowable percentage error inœÊ ŸÊ ŸÊ Ÿœ Ê
1111DDDDD
200 200 00 100

kk kk ¹¹
##
measuring the diameter is 1%.
57. A 5% error in measuring t dt (5%)t . Then s 16t ds 32t dt 32t sÊœ œ œ Êœ œ œœœ
t t 32t 16t
20 20 20 10 10
# "
ˆ‰ ˆ‰
""
(10%)s a 10% error in the calculation of s.œÊ
58. From Example 8 we have 4 . An increase of 12.5% in r will give a 50% increase in V.
dV dr
Vr
œ
59. lim 1 60. lim lim (1)(1) 1
x0 x0 x0ÄÄÄ
ÈÈ1x 10
1 xxcos x1
tan x sin x

"
x 0
" "
œœ œ œœ ˆ‰ˆ‰
61. E(x) f(x) g(x) E(x) f(x) m(x a) c. Then E(a) 0 f(a) m(a a) c 0 c f(a). Next*% Î *%%% *Î %%%*Î*
we calculate m: lim 0 lim 0 lim m 0 (since c f(a))
xa xa xaÄÄ Ä
E(x) f(x) m(x a) c f(x) f(a)
xa xa xa"" "
""" "
*Î *Î % * * ’“
f (a) m 0 m f (a). Therefore, g(x) m(x a) c f (a)(x a) f(a) is the linear approximation,Î%*Î* *%&* %&
ww w
as claimed.
62. (a) i. Q a f a implies that b f a .ab ab abœœ
!
ii. Since Q x b b x a , Q a f a implies that b f a .
wwww
"# "
ab a b ab ab ab*&) % * *
iii. Since Q x b , Q a f a implies that b .
ww ww ww
#"
#
ab ab abœ# œ œ
fa
††
ab
In summary, b f a , b f a , and b .!" "
w
#œœ œab ab
fa
††
ab
(b) f x xab a bœ"
"'
fx x x
w "! "!
ab abababœ"" " œ "
fx x x
ww "† "†
ab abababœ#" " œ#" Since f , f , and f , the coefficients are b , b , b . The quadraticab ab ab!œ" !œ" !œ# œ" œ" œ œ"
www
!"#
#
#
approximation is Q x x x .abœ"
#
(c) As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(d) g x xabœ
"'
gx x
ß"!
abœ"
gx x
ßß "†
abœ#
Since g , g , and g , the coefficients are b , b , b . The quadraticab ab ab" œ" " œ" " œ# œ" œ" œ œ"
www
!" #
#
#

182. 178 Chapter 3 Differentiation
approximation is Q x x x .ab ababœ" " "
#
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(e) h x xab a bœ"
"Î#
hx x
w "
#
"'Ê!
ab a bœ"
hx x
ww "
%
"†Ê!
ab a bœ "
Since h , h , and h , the coefficients are b , b , b . The quadraticab ab ab! œ" ! œ ! œ œ" œ œ œ
www "" ""
#% #
!" #
"
œ
&
28
approximation is Q x .abœ"
xx
8#
"
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f) The linearization of any differentiable function u x at x a is L x u a u a x a b b x a , whereab ab ab aba b a b**&%*&%
w
!"
b and b are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization!"
for f x at x is x; the linearization for g x at x is x or x; and the linearization for h x atab ab a b abœ! " œ" " " #
x is .œ! "
x
#
63. (a) x 1 œ

(b) x 1; m 2.5, e 2.7 x 0; m 1, e 1 x 1; m 0.3, e 0.4** † *** *%* †
10 1 "

183. Section 3.8 Linearization and Differentials 179
64. If f has a horizontal tangent at x a, then f (a) 0 and the linearization of f at x a isœœ œ
w
L(x) f(a) f (a)(x a) f(a) 0 (x a) f(a). The linearization is a constant.*& %*& %*
w
†
65. Find v when m m . m m m v cll œ"Þ!" œ Ê "œ Ê "œ Ê"œ Êœ "!!
'"
##mm vvv
ccmcmm
mm
ßß ß
"
"
""
"""" "
" ""
É
v
c
ÉÉ Š‹
v c dv c dm, dm m dv . m m ,Ê l l œ " Ê œ † " œ !Þ!" Ê œ œÉ Š‹Š‹ ˆ‰
m
mmm
mm c m
m
m
ßßß ßß
""" "

% " "!"
# "!! "!!
"'Ê!
#
!!
'"Ê
m
m

&
&
dv 0.69c. Body at rest v and v v dv**†Î*#*&
cm m 1†
'"
"!!
!!!
'Ä' ' "
!!

"
%
œ!œ
#
œ!! œ!œ
)# &

&
œ!!
&
&
œ!!
&
œ!œ
&
m
3
m
m
Í
Í
Í
Ì
Ê ˆ‰
v 0.69c.ʜ
66. (a) The successive square roots of 2 appear to converge to the number 1. For tenth roots the convergence is more rapid.
(b) Successive square roots of 0.5 also converge to 1. In fact, successive square roots of any positive number converge
to 1.
A graph indicates what is going on:

Starting on the line y x, the succesSive square roots are found by moving to the graph of y x and then across toœœ È
the line y x again. From any positive starting value x, the iterates converge to 1.œ
67-70. Example CAS commands:
:Maple
with(plots):
a:= 1: f:=x -> x 3 x 2 2*x;••
plot(f(x), x= 1..2);%
diff(f(x),x);
fp := unapply ( ,x);
ww
L:=x -> f(a) fp(a)*(x a);&%
plot({f(x), L(x)}, x= 1..2);%
err:=x -> abs(f(x) L(x));%
plot(err(x), x= 1..2, title = absolute error function );% ##
err( 1);%
: (function, x1, x2, and a may vary):Mathematica
Clear[f, x]
{x1, x2} = { 1, 2}; a = 1;%
f[x_]:=x x 2x
32
&%
Plot[f[x], {x, x1, x2}]
lin[x_]=f[a] f'[a](x a)&%
Plot[{f[x], lin[x]}, {x, x1, x2}]
err[x_]=Abs[f[x] lin[x]]%

184. 180 Chapter 3 Differentiation
Plot[err[x], {x, x1,x 2}]
err//N
After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del)
eps = 0.5; del = 0.4
Plot[{err[x], eps},{x, a del, a del}]%&
CHAPTER 3 PRACTICE EXERCISES
1. y x 0.125x 0.25x 5x 0.25x 0.25*% & Î * % &
&# % dy
dx
2. y 3 0.7x 0.3x 2.1x 2.1x*% & Î *% &
$( #' dy
dx
3. y x 3 x 3x 3(2x 0) 3x 6x 3x(x 2)*% & Î * % &* %* %
$## # #
ab1
dy
dx
4. y x 7x 7x 7*& % Î * &
(' "
ÈÈ
11dx
dy
5. y (x 1) x 2x (x 1) (2x 2) x 2x (2(x 1)) 2(x 1) (x 1) x(x 2)*& & Î *& && & & * & & & &
## # # #
ab ab c d
dy
dx
2(x1)2x 4x1*& && ab
#
6. y (2x 5)(4 x) (2x 5)( 1)(4 x) ( 1) (4 x) (2) (4 x) (2x 5) 2(4 x)*% % Î*%%% %&% *% %&%
"' "! "' "!dy
dx
cd
3(4 x)*%
"!
7. y sec 1 3 sec 1 (2 sec tan )*& & Î* & & &abab)) )) )))
##
$# dy
d)
8. y 1 2 1 1 (csc cot )*%% % Î * %% % % *%% % %Š‹Š‹ Š‹ ˆ‰
csc csc csc cot csc
4d 4 4
dy)) )) ))) ))
)#####
#
"" "
)))
9. s œÊœ œ œ
Ȉ‰
È
ˆ‰ÈÈ Š‹
ˆ‰ ˆ‰ ˆ‰ÈÈÈÈÈ
ÈÈ
t
1t
ds
dt
1t t
1t 2t1t t1t
1t t

Á"
#
Á" "
†
œœ
""
"""
œœ
tt
10. s œÊœ œ
'"'
"
""
""
È
ˆ‰È Š‹
ˆ‰ ˆ‰ÈÈÈt1
ds
dt
t1(0)1
t1 2t t1
œ
"
""
œ
t
11. y 2 tan x sec x (4 tan x) sec x (2 sec x)(sec x tan x) 2 sec x tan x*%Î* % *
## # # dy
dx
ab
12. y csc x 2 csc x (2 csc x)( csc x cot x) 2( csc x cot x) (2 csc x cot x)(1 csc x)*%* % Î* % %% * %
" #
sin x sin x dx
2 dy
"
13. s cos (1 2t) 4 cos (1 2t)( sin (1 2t))( 2) 8 cos (1 2t) sin (1 2t)* % Î* %% % %* % %
%$ $ ds
dt
14. s cot 3 cot csc cot csc*Î* % *
$#### "ˆ‰ ˆ‰ˆ ‰ˆ ‰ ˆ‰ ˆ‰ ˆ‰
2ds 2 2262 2
tdt t ttt t t
""
15. s (sec t tan t) 5(sec t tan t) sec t tan t sec t 5(sec t)(sec t tan t)*& Î* & & * &
&%# &ds
dt
ab
16. s csc 1 t 3t 5 csc 1 t 3t csc 1 t 3t cot 1 t 3t ( 1 6t)* %& Î * %& % %& %& %&
&# %# # #
ab aba b abab
ds
dt
5(6t 1) csc 1 t 3t cot 1 t 3t*% % % & % &
&# #
abab
17. r 2 sin (2 sin ) (2 sin ) ( cos 2 sin )** Î* )&*
È
)) )) )) ) ) )
'Ê! "'Ê! "
#
dr cos sin
d 2 sin )
)) )
))
È

185. Chapter 3 Practice Exercises 181
18. r 2 cos 2 (cos ) 2 (cos ) ( sin ) 2(cos ) 2 cos ** Î* %& *&)))) ) ) ) ) )
È Èˆ‰
'Ê! "'Ê! 'Ê! '"
#
dr sin
d cos )
))
)
È
œ
2 cos sin
cos
)) )
)
"
È
19. r sin 2 sin (2 ) cos (2 ) (2 ) (2)œœ Êœ œ
È ˆ‰)) ) )
'Ê! 'Ê! "'Ê! "
#
dr
d
cos 2
2)
)
)
È
È
20. r sin 1 cos 1 1 cos 1*&&Î* &&& * &&Š‹ Š‹Š‹ Š‹
ÈÈ È)) )) ))
dr
d 21
21
) ))
)"
#"
"
ÈÈ
È
21. y x csc x csc cot csc 2x csc cot x csc *Î*% & *&
''"'
## #
## 22222222
xdx xxx x xx x
dy
ˆ‰ˆ‰ˆ‰ˆ‰
" †
22. y 2 x sin x 2 x cos x sin x cos x*Î* & *&ÈÈ È È È È ˆ‰ ˆ‰Š‹ Š‹
dy
dx 2x 2x x
2 sin x"
ÈÈÈ
È
23. y x sec (2x) x sec (2x) tan (2x) (2(2x) 2) sec (2x) x*Î* &%
"'Ê! ! "'Ê! ! ! ! "†Ê! "
#
dy
dx
† ˆ‰
8x sec (2x) tan (2x) x sec (2x) x sec (2x) 16 tan (2x) x or sec x 16x tan 2xœœ #"
'Ê! ! ! "†Ê! ! 'Ê! ! ! "! !"" "
## #
#
c d ab ab œ ‘
x
2
%Ê"
24. y x csc (x 1) x csc (x 1)*&*&È
$ "Î# $
x csc (x 1) cot (x 1) 3(x 1) csc (x 1) xÎ* % & & & & &
dy
dx
'Ê! † † ! † "'Ê! "
#
abab ˆ‰
3 x (x 1) csc (x 1) cot (x 1) x csc (x 1) 6(x 1) cot (x 1)*% & & & & * & % & &ÈÈ œ ‘
#$$ $ #$ ""
#
csc (x 1)
2x x
%
È
or csc(x 1) 1 6x(x 1) cot (x 1)
"
#
$#$
Èx
&%& &cd
25. y 5 cot x 5 csc x (2x) 10x csc x*Î*% *%
#####dy
dx
ab ab
26. y x cot 5x x csc 5x (5) (cot 5x)(2x) 5x csc 5x 2x cot 5x*Î*% & *% &
### ## dy
dx
ab
27. y x sin 2x x 2 sin 2x cos 2x (4x) sin 2x (2x) 8x sin 2x cos 2x 2x sin 2x*Î* & * &
## # # # # # # $ # # # #
ab a ba b ab ab ab abab ab
dy
dx
28. y x sin x x 2 sin x cos x 3x sin x 2x 6 sin x cos x 2x sin x*Î* &%* %
"! ! † "! † † ! ! † "† † † "† ! †
ab a ba ba b aba b ab ab abab ab
dy
dx
29. s 2 2* Î *% *% *%ˆ‰ ˆ‰ ˆ‰ Š‹
4t ds 4t 4t 4
t 1 dt t 1 (t 1) t 1 (t 1) 8t
(t 1)(4) (4t)(1) (t 1)

"! "† "† Á" Á
"" %
30. s (15t 1) ( 3)(15t 1) (15)**%%Î*%%%*
"' ' '
"
$ %
"15(15t 1) 15 dt 15
ds 3
(15t 1)
% &
31. y 2œÊœ œœŠ‹ Š‹
ÈÈ Š‹ ˆ‰È
xx
x 1 dx x 1 (x 1) (x 1) (x 1)
dy (x 1) 2x
(x 1) x (1)
1x

# Á"
Á" "
†
œ
"

œ
x
32. y 2 œÊœ œœŠ‹ Š‹
%&
2x 2x
2x 1 2x 1
dy
dx
2x 1 2x 4x
2x 1 2x 1 2x 1
4
ÈÈ
ÈÈ
ˆ‰ ˆ‰ÈÈ ÈŠ‹ Š‹ Š‹
ˆ‰ ˆ‰ˆ‰ÈÈÈ
# Á"

œœ œ

œœ œ
xx x
33. y 1 1**&Î*&%*%É ˆ‰ ˆ‰ˆ‰
xx
xxdxxx
dy
x1
"
" "
"
" """ "
'Ê! "'Ê!
#
#É
œ
x34. y 4x x x 4x x x 4x x x 1 x x x (4)*&*& Î* & & &&É È ˆ‰ ˆ‰ˆ‰ˆ ‰ˆ‰
'Ê! 'Ê! "'Ê! 'Ê!
'Ê! "'Ê! 'Ê!
""
##
dy
dx
xx 2x1 4xx xx 2xx4x4x*& & & & *& & && *ˆ‰ ˆ‰ˆ‰ˆ ‰ÈÈÈÈÈ’“Š‹
"'Ê! "'Ê!
"
#


È
È
É Èx
6x 5 x
xx

186. 182 Chapter 3 Differentiation
35. r 2œÊœˆ‰ ˆ‰ ’“
sin dr sin
cos 1 d cos 1 (cos 1)
(cos 1)(cos ) (sin )( sin )))
))) )
))))
"""
# """
"
2œœœˆ‰ Š‹
sin cos cos sin 2 sin
cos (cos ) (cos 1) (cos )
(2 sin ) (1 cos ))))) )
)) ) )
))
"' "' " "'
"Á " "
""
"
36. r 2œÊœˆ‰ ˆ‰ ’“
sin 1 dr sin 1
1 cos d 1 cos (1 cos )
(1 cos )(cos ) (sin )(sin )))
)) ) )
)) ) )
"""
# ""Á'
"
cos cos sin sin *%%%*
2(sin ) 2(sin 1)(cos sin 1)
(1 cos ) (1 c os )
))))
))
Á' Á " "
""
##
% %ab))))
37. y (2x 1) 2x 1 (2x 1) (2x 1) (2) 3 2x 1*& &*& Î* & * &ÈÈ
$Î# "Î#
#
dy
dx
3
38. y 20(3x 4) (3x 4) 20(3x 4) 20 (3x 4) (3)*% % *% Î* % *
"Î% "Î& "Î#! "*Î#! "
"
dy
dx 20
3
(3x 4) ˆ‰
œ„Ê"ß
39. y 3 5x sin 2x 3 5x sin 2x [10x (cos 2x)(2)]*& Î*% & & *ab ab ˆ‰
##
$Î# &Î#
#
"Á

dy 9(5x cos 2x)
dx
3
5x sin 2xab
"
!Ê"
40. y 3 cos 3x 3 cos 3x 3 cos 3x ( sin 3x)(3)*& Î *% & % *ab abab
$$#
"Î$ %Î$ "

dy
dx 3
3 cos 3x sin 3x
3 cos 3x
"
%
&Î%
ab
41. xy 2x 3y 1 xy y 2 3y 0 xy 3y 2 y y (x 3) 2 y y&&*Î &&& *Î & *%%Î &*%%Î*%ab
wwww w w

y2
x3
42. x xy y 5x 2 2x x y 2y x 2y 5 2x y (x 2y)
##
&&%*Î & && %â*#Î & *%%Î &Š‹
dy dy dy dy dy
dx dx dx dx dx
52xy *% % Î *
dy 52xy
dx x 2y
""

43. x 4xy 3y 2x 3x 4x 4y 4y 2 4x 4y 2 3x 4y
$ %Î$ # "Î$ "Î$ #
&% *Î & &% *Î % *%% Š‹
dy dy dy dy
dx dx dx dx
4x 4y 2 3x 4y Î%*%%Î*
dy dy 2 3x 4y
dx dx 4x 4y
ˆ‰
"Î$ # ""
"
"
*Î%
44. 5x 10y 15 4x 12y 0 12y 4x x y
%Î& 'Î& "Î& "Î& "Î& "Î& "Î& "Î& ""
&*Î & *Î *%Î*% *%
dy dy dy
dx dx dx 3 3(xy)
œÊ!
45. (xy) 1 (xy) x y 0 x y x y x y
'Ê! "'Ê! 'Ê! "'Ê! "'Ê! 'Ê! "'"
#
*Î & *Î *% Î *% Î *%Š‹
dy dy dy dy y
dx dx dx dx x
46. x y 1 x 2y y (2x) 0 2x y 2xy
## # # # #
*Î & *Î *% Î *%Š‹
dy dy dy y
dx dx dx x
47. y 2y
#
#
Á" "
œÊ œ Êœ
x
x 1 dx (x 1) dx y(x 1)
dy (x 1)(1) (x)(1) dy
""
48. y y 4y
#%$ " "
"" '" "
"Î# ""Á%"'
œÊœÊœ Êœˆ‰
1x x
1x 1x dx ( x) dx 2y(1x)
dy dy(1 x)(1) (1 x) )
"
49. p 4pq 3q 2 3p 4 p q 6q 0 3p 4q 6q 4p 3p 4q 6q 4p
$## # #
& % *Î & & %*Î & *% Î & *%
dp dp dp dp dp
dq dq dq dq dq
Š‹ ab
ʜ
dp 6q 4p
dq 3p 4q
"

"
50. q 5p 2p 1 5p 2p 10p 2 5p 2p (10p 2)*& Î*%& &Î%&* &ab ab ab Š‹
## #
$Î# &Î# &Î#
#
32 dp dp dp
dq dq 3 dq
Î*%
dp
dq 3(5p 1)
5p 2pab
"
!Ê"

187. Chapter 3 Practice Exercises 183
51. r cos 2s sin s r( sin 2s)(2) (cos 2s) 2 sin s cos s 0 (cos 2s) 2r sin 2s 2 sin s cos s&*Î% & & *Î * %
#
1 ˆ‰
dr dr
ds ds
(2r 1)(tan 2s)Î* * *%
dr 2r sin 2s sin 2s
ds cos 2s cos 2s
(2r 1)(sin 2s)" "
52. 2rs r s s 3 2 r s 1 2s 0 (2s 1) 1 2s 2r %%& *% Î & % % & * Î % * % % Î *
# '" "
" ˆ‰
dr dr dr dr 2s 2r
ds ds ds ds 2s 1
53. (a) x y 1 3x 3y 0
$$ # #
"""
&*Î & *Î *% Î *
dy dy d y
dx dx y dx
x
y ( 2x) x 2y
y
"
""
"
""
& abŠ‹
dy
dx
Êœ œ œ
d y 2xy 2x
dx y
2xy 2yx
yy
2xy
" %&
"
""
&&
" "Á " ""
""
abŠ‹
x
yy
2x
&
&
†
(b) y 1 2y yx yx y(2x) x
#### " "' "!
*%Î *Î* Î* Î *% &
22
x dx x dx yx dx dx dx
dy dy dy d y dy
"" "
" ab ab ’“
Êœ œ
d y 2xy 1
dx
2xy x
yx yx
""
"
"
"&%& ""
""
Š‹
œ
&
yx
54. (a) x y 1 2x 2y 0 2y 2x
##
%*Î % *Î% *% Î *
dy dy dy
dx dx dx y
x
(b) (since y x 1)
dy d y y x
dx y dx y y y y
x
y(1) x
yx
*Î * * * * %*%
"""
"" " %% "
"
" "' ##
dy
dx
x
y
Š‹
55. (a) Let h(x) 6f(x) g(x) h (x) 6f (x) g (x) h (1) 6f (1) g (1) 6œÊ œ Ê œ œ %œ(
www www "
# ˆ‰ab
(b) Let h(x) f(x)g (x) h (x) f(x) g(x) g (x) g (x)f (x) h (0) f(0)g(0)g (0) g (0)f (0)*Î*)& Î*) &
# w w#w w w#w
ab
(1)(1) (1) ( )*) & %„ *%)ˆ‰
"
#
#
(c) Let h(x) h (x) h (1)œÊœ Êœ
f(x) (g(x) 1)f (x) f(x)g (x) (g(1) )f (1) f(1)g (1)
g(x) 1 (g(x) 1) (g(1) 1)ÁÁ Á
ww Á" Á'"
†† ††
""
œœ
(1) 3
(1)
& %
& "#
&
ˆ‰ ab
œ
"
"
(d) Let h(x) f(g(x)) h (x) f (g(x))g (x) h (0) f (g(0))g (0) f (1)œÊœ Êœ œ œ œ
www wwww """"
###% ˆ‰ ˆ‰ˆ‰
(e) Let h(x) g(f(x)) h (x) g (f(x))f (x) h (0) g (f(0))f (0) g (1)f (0) ( )œÊœ Êœ œ œ%$œ"#
www wwwww
ab
(f) Let h(x) (x f(x)) h (x) (x f(x)) 1 f (x) h (1) (1 f(1)) 1 f (1)*& Î*& & Î*& &
$Î# w "Î# w w "Î# w
##
33
ab ab
(1 3) 1*& &*
3
###
"Î# "*
ˆ‰
(g) Let h(x) f(x g(x)) h (x) f (x g(x)) 1 g (x) h (0) f (g(0)) 1 g (0)*&Î*& & Î* &
ww w ww w
ab ab f (1) 1*&* *
w ""$$
###%ˆ ‰ ˆ‰ˆ‰
56. (a) Let h(x) x f(x) h (x) x f (x) f(x) h (1) 1 f (1) f(1) ( 3)*Î*& Î*& *&%*%ÈÈ È ˆ‰
ww ww """"
# # #
††
ÈÈx 1 510
13
(b) Let h(x) (f(x)) h (x) (f(x)) f (x) h (0) (f(0)) f (0) (9) ( 2)*Î* Î* *%*%
'Ê! ß "'Ê! ß ß "'Ê! ß "'Ê!""""
###
ab
3
(c) Let h(x) f x h (x) f x h (1) f 1œÊœ Êœ œœˆ‰ ˆ‰ÈÈ Š‹
È
ww ww """""
# # #
†††
ÈÈx 1 510
(d) Let h(x) f(1 5 tan x) h (x) f (1 5 tan x) 5 sec x h (0) f (1 5 tan 0) 5 sec 0*% Î*% % Î*% %
ww # ww #
ab ab
f (1)( 5) ( 5) 1*%*%*%
w "
5
(e) Let h(x) h (x) h (0)*Î* Î* **%
f(x) (2 cos x)f (x) f(x)( sin x) (2 1)f (0) f(0)(0) 3( 2)
2 cos x (2 cos x) (2 1) 9 3
2
ÁÁ Á
ww Á"" Á" "
††
""
(f) Let h(x) 10 sin f (x) h (x) 10 sin 2f(x)f (x) f (x) 10 cos*Î* &ˆ‰ ˆ‰ ˆ ‰ˆ‰ ab ˆ‰
11 11xx x
## ##
#w w#
h (1) 10 sin 2f(1)f (1) f (1) 10 cos 20( 3) 12Î* & *%&#*%
ww#
###
" ˆ‰ ˆ ‰ˆ‰ ˆ‰ab ˆ‰
111
5
57. x t 2t; y 3 sin 2x 3(cos 2x)(2) 6 cos 2x 6 cos 2t 2 6 cos 2t ; thus,*& Î * * Î * * * & *
# ##
11
dx
dt dx
dy
abab
6 cos 2t 2t 6 cos (0) 0 0
dy dy dy
dt dx dt dt
dx
œœ Êœ œ†† † ab ¹
#
t=0
58. t u 2u u 2u (2u 2) u 2u (u 1); s t 5t 2t 5*& Î* & &* & & *&Î*&ab ab ab
## ##
'ʆ "!ʆ "!ʆ"dt 2 ds
du 3 3 dt
2 u 2u 5; thus 2 u 2u 5 u 2u (u 1)*& & * * & & & &ab ab ab ’“ ˆ‰
###
'ʆ 'ʆ "!ʆds ds dt 2
du dt du 3
†

188. 184 Chapter 3 Differentiation
2 2 2(2) 5 2 2(2) (2 1) 2 2 8 5 8 2(2 2 5)Î*& & & &* & *&*¸ˆ‰ˆ‰ˆ‰ˆ‰’“ab ab
ds 2 9
du 3 4
u=2
! ! 'ʆ "!ʆ
'ʆ "!ʆ "
#
††
59. r 8 sin s 8 cos s ; w sin r 2 cos r 2*&Î* &* %Î* %ˆ‰ ˆ‰ ˆ ‰ ˆ ‰ ÈÈ Š‹
11
6ds 6 dr
dr dw
r
"
#È
; thus, 8 cos s*** &
cos 8 sin s 2
2 8 sin s 8 sin s
cos 8 sin s 2ÉŠ‹ˆ‰
ÉÉ ˆ‰ ˆ ‰
É ˆ‰"Ê
#
"Ê
œ
œœ
œ
6
66
6 dw dw dr
ds dr ds 6
†† œ ‘ˆ‰
1
3Êœ œ œ¸ È
dw
ds
s=0
cos 8 sin 2 8 cos (cos 0)(8)
28 sin 24
Š‹ Š‹É ˆ‰ ˆ‰
É ˆ‰
È
œœ
œ
66
6
3
ʆ
œ
&
60. t 1 t 2 0 (2 t 1) ; r 7)) ) ) ) ) )
## # # "

"Î$
&* Î & & * Î & *% Î * * &ˆ‰ ˆ‰ ab
dd d d
dt dt dt dt 2 t 1
)) ) ))
)
"
7 (2 ) 7 ; now t 0 and t 1 1 so that 1Î* & * & * &*Î* **%
dr 2 d1
d3 3 dt 1)
)" "## #
"!ʆ "!ʆ
ab ab ¸)))) )))
t=0, =1œ
and (1 7) ( 1)¸¸¸¸ˆ‰
dr 2 dr dr d
d3 6dtddt 6 6))
)
œ=1 t=0 t=0 t=0
*& *Î * * %*%
"!ʆ """
†
61. y y 2 cos x 3y 2 sin x 3y 1 2 sin x
$# # "

& * Î & *% Î & *% Î * Î
dy dy dy dy dy
dx dx dx dx 3y 1 dx
2 sin x
ab ¹
"
(0 1)œ
0; œœœ
"

Á" ""

2 sin (0) d y
31 dx
3y 1 ( 2 cos x) ( 2 sin x) 6y
3y 1
"
"
"
"
" ab Š‹
ab
dy
dx
Î* *%¹
d y (3 1)( 2 cos 0) ( 2 sin 0)(6 0)
dx (3 1)
"
""
(0 1)œ
Á" ""
#
"†
62. x y 4 x y 0 1;
'ʆ 'ʆ "!ʆ "!ʆ "" "
&*Î & *Î*%Î *% *
3 3 dx dx dx dx
dy dy y dy dy y
xx
"

¹
(8 8)œ
Êœ Ê œ
dy dy
dx dx
xy yx
x
88(1)88
8
" "
" "
"

"

&Î% ˆ‰ ˆ ‰ˆ ‰Š‹
ab
ˆ‰œ ‘ˆ ‰ˆ ‰"""
""Á
22
3dx 3
dy
22
33
¹
(8 8)œ
†† † œœœ
œœ
"Î%
33 3
2
"
8 46
63. f(t) and f(t h) *&*Î**
""
Á!ÁÁ ÁÁÁ
Á" Á" Á Á
"
2t 1 (t h) 1 h h (2t 2h 1)(2t 1)h
f(th)f(t) 2t1(2t2h1)
œœ

(t h) 1 t 1
f (t) lim lim œœÊœœ
"" "
ÁÁ Á ÁÁ Á ÁÁ!Á
w Á"2h 2 2
(2t 2h 1)(2t 1)h (2t 2h 1)(2t 1) h (2t 2h 1)( t 1)
f(t h) f(t)
hhÄ! Ä!
œ
"!
(2t 1)
"
64. g(x) 2x 1 and g(x h) 2(x h) 1 2x 4xh 2h 1 *& &*&&*&&&Î
#### Á"g(x h) g(x)
h
4x 2h g (x) lim lim (4x 2h)***&Î**&
abab2x 4xh 2h 1 2x 1
hh h
4xh 2h g(x h) g(x)
""" "
ÁÁÁ" Á w Á"
hhÄ! Ä!
4xœ
65. (a)
(b) lim f(x) lim x 0 and lim f(x) lim x 0 lim f(x) 0. Since lim f(x) 0 f(0) it
xx x x xxÄ! Ä! Ä! Ä!Ä! Ä!
ÊÊ %%
** *%*Î * **
##
follows that f is continuous at x 0.œ
(c) lim f (x) lim (2x) 0 and lim f (x) lim ( 2x) 0 lim f (x) 0. Since this limit exists, it
xx x xxÄ! Ä! Ä!Ä! Ä!
ÊÊ %%
ww w
** *%*Î*
follows that f is differentiable at x 0.œ

189. Chapter 3 Practice Exercises 185
66. (a)
(b) lim f(x) lim x 0 and lim f(x) lim tan x 0 lim f(x) 0. Since lim f(x) 0 f(0), it
xx x x xxÄ! Ä! Ä! Ä!Ä! Ä!
ÊÊ %%
œœ œ œÊ œ œœ
follows that f is continuous at x 0.œ
(c) lim f (x) lim 1 1 and lim f (x) lim sec x 1 lim f (x) 1. Since this limit exists it
xx x xxÄ! Ä! Ä!Ä! Ä!
ÊÊ %%
ww#w
œœ œ œÊ œ
follows that f is differentiable at x 0.œ
67. (a)
(b) lim f(x) lim x 1 and lim f(x) lim (2 x) 1 lim f(x) 1. Since lim f(x) 1 f(1), it
xx x x xxÄ" Ä" Ä" Ä"Ä" Ä"
ÊÊ %%
** * %*Î * **
follows that f is continuous at x 1.œ
(c) lim f (x) lim 1 1 and lim f (x) lim 1 1 lim f (x) lim f (x), so lim f (x) does
xx x xx x x1Ä" Ä" Ä"Ä" Ä" Ä" Ä
ÊÊ Ê%% %
ww www
œœ œœÊ Á
not exist f is not differentiable at x 1.ʜ
68. (a) lim f(x) lim sin 2x 0 and lim f(x) lim mx 0 lim f(x) 0, independent of m; since
xx x xxÄ! Ä! Ä!Ä! Ä!
ÊÊ %%
œœ œœÊœ
f(0) 0 lim f(x) it follows that f is continuous at x 0 for all values of m.œœ œ
xÄ!
(b) lim f (x) lim (sin 2x) lim 2 cos 2x 2 and lim f (x) lim (mx) lim m m f is
xx x xx xÄ! Ä! Ä! Ä! Ä! Ä!
ÊÊ Ê %% %
ww ww
œœœ œœœÊ
differentiable at x 0 provided that lim f (x) lim f (x) m 2.œœÊœ
x xÄ! Ä!
Ê %
ww
69. y x (2x 4) 2(2x 4) ; the slope of the tangent is *& * & % Î *% % % Î%
x 33
x4 dx
dy
!!" ! ! ! !
"" " "' "!
2(2x 4) 2 2(2x 4) 1 (2x 4) 1 4x 16x 16 1*% % Î%*% % Î * Î % * Î % & *
""
!"
"! "! ! !
(2x 4)
"
4x 16x 15 0 (2x 5)(2x 3) 0 x or x and are points on theÎ%&*Î% %*Î* *Îw w%
#
### #
"53593
44 ˆ‰ ˆ ‰
curve where the slope is .%
3
#
70. y x 1 1 ; the slope of the tangent is 3 3 1 2 x*% Î *& *& Î *& Î * Î *
"" """
###
#
2x dx (2x) x x x 4
dy 2
"" ""
x and are points on the curve where the slope is 3.Î*( Î w% %w
""" ""
### ##
ˆ‰ˆ‰
71. y 2x 3x 12x 20 6x 6x 12; the tangent is parallel to the x-axis when 0*%%&Î*%% *
$# # dy dy
dx dx
6x 6x 12 0 x x 2 0 (x 2)(x 1) 0 x 2 or x 1 ( ) and ( 7) areÊ œ Ê œ Ê œ Ê œ œ Ê #ß! "ß#
##
points on the curve where the tangent is parallel to the x-axis.
72. y x 3x 12; an equation of the tangent line at ( ) is y 8 12(x 2)œ Ê œ Ê œ #ß ) œ
$# dy dy
dx dx
¹
(2 8)ÊœÊ
y 12x 16; x-intercept: 0 12x 16 x ; y-intercept: y 12(0) 16 16 (0 16)Î* & * & Î*%Î%w# * & * Îw
44
33
ˆ‰

190. 186 Chapter 3 Differentiation
73. y 2x 3x 12x 20 6x 6x 12*%%&Î*%%
$# # dy
dx
(a) The tangent is perpendicular to the line y 1 when 24; 6x 6x 12 24*% *% * % % *
x
24 dx
dy
Š‹
"
"
#
ˆ‰
œ
"
4
x x 2 4 x x 6 0 (x 3)(x 2) 0 x 2 or x 3 ( 16) and ( 11) areÎ%%*Î%%*Î% &*Î*% *Î%)w „w
##
points where the tangent is perpendicular to y 1 .*%
x
24
(b) The tangent is parallel to the line y 2 12x when 12 6x 6x 12 12 x x 0*% *%Î %%*%Î%*
È
dy
dx
##
x(x 1) 0 x 0 or x 1 ( 20) and ( ) are points where the tangent is parallel toʜʜ œÊ!ß "ß(
y 2 12x.*%
È
74. y m 1 and m 1.*Î* Î***% * *
111 11
11
sin x
x dx x dx dx
dy x( cos x) ( sin x)(1) dy dy

"
"#
"
"""
"" ¹¹
x= x=œœ Ê
Since m the tangents intersect at right angles."
"*%
m"
75. y tan x, x sec x; now the slopeœÊœ
11
##
#dy
dx
of y is the normal line is parallel to*% % Î
x
##
"
y when 2. Thus, sec x 2 2*% * * Î *
x dy
dx cos x#
# "
"
cos x cos x x and xÎ*Î*Î*% *
# "„"
# È
2 44
11
for x 1 and are points Êß ß"
11 1 1
##
ˆ‰ˆ‰
44
where the normal is parallel to y .*%
x
#

76. y 1 cos x sin x 1*& Î *% Î *%
dy dy
dx dx
¹
œÊ
œ
2
œ1
the tangent at 1 is the line y 1 xÎw%*%% ˆ‰ ˆ ‰
11
##
y x 1; the normal at 1 isÎ*%&& w
11
##
ˆ‰
y1(1)x yx 1%* % Î *%&ˆ‰
11
##

77. y x C 2x and y x 1; the parabola is tangent to y x when 2x 1 x y ;*&Î * *Î * * *Î*Î*
# ""
##
dy dy
dx dx
thus, C C
"" "
##
#
*&Î*ˆ‰
4
78. y x 3x 3a the tangent line at a a is y a 3a (x a). The tangent line*Î* Î * Î w %* %
$# # $$#dy dy
dx dx
¹ ab
x=a
intersects y x when x a 3a (x a) (x a) x xa a 3a (x a) (x a) x xa 2a 0* %*%Î%&&*%Î%&%*
$$$# # ## # #
ab a b
(x a) (x 2a) 0 x a or x 2a. Now 3( 2a) 12a 4 3a , so the slope atÎ% & *Î* *% *% * *
# ## #
¹ ab
dy
dx
x= 2aÊ
x 2a is 4 times as large as the slope at a a where x a.*% w * ab
$
79. The line through ( ) and (5 2) has slope m 1 the line through ( ) and ( 2) is#w „ w % * * % Î #w „ âw %
3(2)
05
""
"
y x 3; y , so the curve is tangent to y x 3 1*% & * Î * *% & Î *% *
cc c
x 1 dx (x 1) dx (x 1)
dy dy

" "
" "
(x 1) c, x 1. Moreover, y intersects y x 3 x 3, x 1Ê œ Á œ œ Ê œ Á
#

cc
x1 x1
c (x 1)( x 3), x 1. Thus c c (x 1) (x 1)( x 3) (x 1)[x 1 ( x 3)]Êœ Á œÊ œÊ
#
, x 1 (x 1)(2x 2) 0 x 1 (since x 1) c 4.œ! Á Ê œ Ê œ Á Ê œ

191. Chapter 3 Practice Exercises 187
80. Let b a b be a point on the circle x y a . Then x y a 2x 2y 0 Š‹
È
w( % & * & * Î & * Î *%
## ### ### dy dy
dx dx y
x
normal line through b a b has slope normal line isÎ* Î w(% ιŠ‹
È
dy
dx b
b
ab
ab
x=b
"
$"
## $"
È
È
""
""
yab (xb) yab xab y x%( % * % Î ' % * ' % Î *(Š‹
ÈÈÈ
## ## ## $" $" "
ÈÈÈ
ab ab ab
bbb
"" "" ""
which passes through the origin.
81. x 2y 9 2x 4y 0 the tangent line is y 2 (x 1)
## ""
&*Î& *Î*%Î *%Î *% %
dy dy dy
dx dx 2y dx 4 4
x
¹
(1 2)œ
x and the normal line is y 2 4(x 1) 4x 2.*% & * & % * %
"
44
9
82. x y 2 3x 2y 0 the tangent line is y 1 (x 1)
$# # ""
##
&*Î&*Î*Î *%Î *&%
dy dy dy
dx dx 2y dx
3x 3 3
"
¹
(1 1)œ
x and the normal line is y 1 (x 1) x .*% & * & % * &
35 2 2
333##
"
83. xy 2x 5y 2 x y 2 5 0 (x 5) y 2 2&%*Î &&% *Î %*%%Î * Î *Š‹ ¹
dy dy dy dy y 2 dy
dx dx dx dx x 5 dx
""
"
(3 2)œ
the tangent line is y 2 2(x 3) 2x 4 and the normal line is y 2 (x 3) x .Î*&%*% *&%*%&
"'
###
17
84. (y x) 2x 4 2(y x) 1 2 (y x) 1 (y x) %*&Î % %*Î% *&%Î* Î *
# Á"
"
Š‹ ¹
dy dy dy 1 y x dy
dx dx dx y x dx 4
3
(6 2)œ
the tangent line is y 2 (x 6) x and the normal line is y 2 (x 6) x 10.Î*&%*% *%%*%&
335 4 4
44 3 3 #
85. x xy 6 1 x y 0 x y 2 xy &*Î& &*Î&*% Î* Î *ÈÈ Š‹ ¹
" "
#
""
È
È
xy dx dx dx x dx 4
dy dy dy dy 2xy y
5
(4 1)œ
the tangent line is y 1 (x 4) = x 6 and the normal line is y (x 4) x .Ê œ œ"œ
55 4411
44 555
86. x 2y 17 x 3y 0 the tangent line is
$Î# $Î# "Î# "Î# "'
&*Î & *Î*Î *%Î
3x
2dxdxdx4
dy dy dy
2y
œÊ"
œÊ"
¹
(1 4)œ
y 4 (x 1) x and the normal line is y 4 4(x 1) 4x.*% % *% & *& % *
""
444
17
87. x y y x y x 3y y 3x 2y 1 3x y 2y 3x y
$$ # $ # $ # $# #$
œÊ œ Ê œ"’“Š‹ ab
dy dy dy dy dy dy
dx dx dx dx dx dx
3x y 2y 1 1 3x y , but is undefined.Î&%*%Î* Î*%
dy dy 1 3x y dy dy
dx dx 3x y 2y 1 dx 4 dx
2
ab ¹¹
$# #$ "
Á"
"%

(1 1) (1 1)œœ Ê
Therefore, the curve has slope at ( ) but the slope is undefined at ( 1). "ß " "ß
"
#
88. y sin (x sin x) [cos (x sin x)](1 cos x); y 0 sin (x sin x) 0 x sin x k ,*% Î* % % *Î %*Î%*
dy
dx
1
k 2, 1, 0, 1, 2 (for our interval) cos (x sin x) cos (k ) 1. Therefore, 0 and y 0 when*% % Î % * * ( * * 1
dy
dx
1 cos x 0 and x k . For x 2 , these equations hold when k 2, 0, and 2 (sinceœ œ #ŸŸ œ 111
cos ( ) cos 1). Thus the curve has horizontal tangents at the x-axis for the x-values 2 , 0, and 2%* *% %11 11
(which are even integer multiples of ) the curve has an infinite number of horizontal tangents.1Ê
89. x tan t, y sec t sin t sin ; tœœÊœœœœÊœœœ
""
## #
dy dy/dt dy
dx dx/dt sec t dx 3 3
sec t tan t
sec t
tan t 3
œ
"
œ
"
"
¹
t3%"œ
11
È
x tan and y sec 1 y x ; 2 cos t Î* * * *Î* & * * * Î
"" "
## # #
$11
3 3 4 dx dx/dt dx
33 dy dy/dt dy cos t
sec t
ÈÈ
" †"
""
"
œ
& ¸
t3%"œ
2 cosœœ
$ "ˆ‰
1
34
90. x , y t (2) 3; t 2 x 1 andœ" œ" Ê œ œ œ Ê œ œ œ Ê œ œ
" "
"
##t t dx dx/dt 2 dx 4
333 5dy dy/dt dy
" "
"
%
Š‹
Š‹
3
t
2
t
¹
t2%

192. 188 Chapter 3 Differentiation
y1 y 3x ; t (2) 6*%*% Î *% & * * * Î * *
3333
4 dx dx/dt 4 dx 4
dy dy/dt dy
##
"" "
"
$$
" †"
""
"
% ˆ‰
Š‹
3
2
t
¹
t2%
91. B graph of f, A graph of f . Curve B cannot be the derivative of A because A has only negative slopesœœ
w
while some of B's values are positive.
92. A graph of f, B graph of f . Curve A cannot be the derivative of B because B has only negative slopesœœ
w
while A has positive values for x 0.
93. 94.
95. (a) 0, 0 (b) largest 1700, smallest about 1400
96. rabbits/day and foxes/day
97. lim lim (1) 1
xxÄ! Ä!
sin x sin x
2x x x ( x 1) 1
"
"!""
""
***%’“ˆ‰ ˆ‰†
98. lim lim lim 1 1 2
xx xÄ! Ä! Ä!
3x tan 7x 3x sin 7x 3 sin 7x 3 7
x 2x 2x cos 7x cos 7x 7x 2
"''
## #
* % *% *% *%ˆ‰ ˆ‰ Š‹†† ††
ˆ‰
2
7
99. lim lim (1) lim (1)
rr rÄ! Ä! Ä!
sin r sin r 2r cos 2r
tan 2r r tan 2r 1
œœœœˆ ‰ ˆ‰ ˆ‰ ˆ‰††
"" " ""
## # # ˆ‰
sin 2r
2r
100. lim lim lim . Let x sin . Then x 0 as 0
)) )Ä! Ä! Ä!
sin (sin ) sin (sin ) sin (sin )
sin sin
sin )) )
))))
)
œœœÄÄŠ‹ ˆ‰ ))
lim lim 1Êœœ
)Ä! Ä!
sin (sin )
sin x
sin x)
)
x
101. lim lim 4
))ÄĈ‰ ˆ‰
œœ
22
ÊÊ
4 tan tan 1
tan (1 0)
4
(400)
"
"
œœ
"
"
))
)

&

"

œœœ
Š‹
Š‹
tan tan
5
tanœ œ
œ
102. lim lim
))Ä! Ä!
%%
12 cot 2
5 cot 7 cot 8 (5 0 0) 5
2
5
(0 2)"
"" ""
"
""
"
"
"
œ
"
"
)
))
***%
Š‹
Š‹
cot
78
cot cotœ
œ œ
103. lim lim lim lim
xx x xÄ! Ä! Ä! Ä!
x sin x x sin x x sin x sin x
2 2 cos x 2(1 cos x) x 22 sin sin""
œœ œ
ˆ‰ ˆ‰ˆ‰""
""
""xx
xx ’“
†
†
lim (1)(1)(1) 1œœœ
xÄ!
’“
ˆ‰ ˆ‰
ˆ‰ ˆ‰
xx
xx
""
""
sin sin
sin x
x
††
104. lim lim lim (1)(1)
)) )Ä! Ä! Ä!
1cos 2 sin sin sin"'''
###
)
))
""
"
"""
""œœ œœ
ˆ‰ ˆ‰ ˆ‰
ˆ‰ ˆ‰
œœœ
œœ
’“ ˆ‰††105. lim lim 1; let tan x 0 as x 0 lim g(x) lim
xx xxÄ! Ä! Ä! Ä!
tan x sin x
xcos xx tan x
tan (tan x)
œœœÊÄÄÊœˆ‰
"
† ))
lim 1. Therefore, to make g continuous at the origin, define g(0) 1.œœ œ
)Ä!
tan )
)

193. Chapter 3 Practice Exercises 189
106. lim f(x) lim lim 1 lim (us
xx x xÄ! Ä! Ä! Ä!
œœ œ
tan (tan x) tan (tan x)
sin (sin x) tan x sin (sin x) cos x sin (sin x)
sin x sin x
’“ †† †
"
ing the result of
#105); let sin x 0 as x 0 lim lim 1. Therefore, to make f))œÊÄ ÄÊ œ œ
xÄ! Ä!
sin x
sin (sin x) sin
)
)
)
continuous at the origin, define f(0) 1.œ
107. (a) S 2 r 2 rh and h constant 4 r 2 h (4 r 2 h) *& Î* & *&11 1 1 11
# dS dr dr dr
dt dt dt dt
(b) S 2 r 2 rh and r constant 2 r *& Î*11 1
# dS dh
dt dt
(c) S 2 r 2 rh 4 r r h (4 r 2 h) 2 r *& Î* &) & *& &11 1 1 11 1
# dS dr dh dr dr dh
dt dt dt dt dt dt ˆ‰
(d) S constant 0 0 (4 r 2 h) 2 r (2r h) r Î*Î* & & Î & *% Î*
dS dr dh dr dh dr r dh
dt dt dt dt dt dt 2r h dt
11 1
"

108. S r r h r r h ;*&Î* &&111
ÈÈ
## ##


dS dr
dt dt
r h
rh
†
ˆ‰
È
dr dh
dt dt
"" (a) h constant 0 r h r h Î*Î* & & * &&
dh dS dr r dr
dt dt dt dt
r
rh rh
1 1
"
"" ""
"dr
dt
ÈÈ

## ##
11
ÈÈ
’“
(b) r constant 0 ÊœÊ œ
dr dS rh dh
dt dt dt rh
1
È
""

(c) In general, r h
dS r dr rh dh
dt dt dt rh rh
*&& &’“
È
1
##

11
"
"" ""
ÈÈ
109. A r 2 r ; so r 10 and m/sec (2 )(10) 40 m /sec*Î* * *% Î* %*%11 1
# #dA dr dr 2 dA 2
dt dt dt dt 11 ˆ‰
110. V s 3s ; so s 20 and 1200 cm /min (1200) 1 cm/minœÊ œ Êœ œ œ Êœ œ
$# $ ""dV ds ds dV dV ds
dt dt dt 3s dt dt dt 3(20)
†
" "
111. 1 ohm/sec, 0.5 ohm/sec; and . Also,
dR dR dR dR
dt dt R R R R dt dt dt
dR
RRœ" œ"
œ"
" ""
œ"
*% * * & Î * %
''' "' "' '
R 75 ohms and R 50 ohms R 30 ohms. Therefore, from the derivative equation,"#
"""**Î*&Î*
R7550
( 1) (0.5) ( 900)
"' "' ' ' ' " '
(30) dt (75) (50) 5625 5000 dt 5625 5000 50(5625) 50
dR dR 5000 5625 9(625)
"" "*%% * % Î*% * * ˆ‰ ˆ‰
†
0.02 ohm/sec.œ
112. 3 ohms/sec and 2 ohms/sec; Z R X so that R 10 ohms and
dR dX dZ
dt dt dt
R X
RX
**%*&Î* *
È
##


dR dX
dt dt
È""
X 20 ohms 0.45 ohm/sec.*Î* *†%
dZ
dt
(10)(3) (20)( 2)
10 20 5
Á"

"'
ÈÈ
""
113. Given 10 m/sec and 5 m/sec, let D be the distance from the origin D x y 2D
dx dD
dt dt dt
dy
** Î*&Î
###
2x 2y D x y . When (x y) ( ), D and*& Î *& w*„w%Ð*„&%Ð*â
dx dD dx
dt dt dt dt dt
dy dy
É ab#
#
(5)(10) (12)(5) 22. Therefore, the particle is moving the origin at 22 m/secâ* & Î * *
dD dD 110
dt dt 5
away
from
(because the distance D is increasing).
114. Let D be the distance from the origin. We are given that 11 units/sec. Then D x y
dD
dt
**&
###
x x x x 2D 2x 3x x(2 3x) ; x 3 D 3 3 6*& *& Î * & * & *Î * &*
# $Î# # $ #
#
#$ˆ‰ È
dD dx dx dx
dt dt dt dt
and substitution in the derivative equation gives (2)(6)(11) (3)(2 9) 4 units/sec.*& Î*
dx dx
dt dt
115. (a) From the diagram we have r h.
10 4 2
hr 5
œÊœ
(b) V r h h h , so 5 and h 6 ft/min.** *Î* *%*Î*%
""#
#
3 3 5 75 dt 25 dt dt dt 144
2 4 h dV 4 h dh dV dh 125
11 ˆ‰
11
1
% "
116. From the sketch in the text, s r r . Also r 1.2 is constant 0*Î* & * Î*))
ds d dr dr
dt dt dt dt
)
r (1.2) . Therefore, 6 ft/sec and r 1.2 ft 5 rad/secÊœ œ œ œ Êœ
ds d d ds d
dt dt dt dt dt
)) )

194. 190 Chapter 3 Differentiation
117. (a) From the sketch in the text, 0.6 rad/sec and x tan . Also x tan sec ; at
ddxd
dt dt dt
) )
*% * * Î * )))
#
point A, x 0 0 sec 0 ( 0.6) 0.6. Therefore the speed of the light is 0.6 km/sec*Î*Î * % *% *)
dx 3
dt 5
ab
#
when it reaches point A.
(b) revs/min
(3/5) rad
sec 2 rad min
1 rev 60 sec 18
††
11
œ
118. From the figure, . We are given
ab a b
rBC r
br
œÊœ
È""
"
that r is constant. Differentiation gives,
. Then,
"
""
"rdt br
da
br (b)
†œ
Š‹ Š‹
È ˆ‰ ˆ‰
""
""
#
"" db b db
dt dt
br
œ
b 2r and 0.3r**%
db
dt
rʜ
da
dt (2r) r
(2r) r ( 0.3r) (2r)
Ô×
ÖÙ
ÕØÈ
%& "" #
""
#
""
"" "
"
2r( 0.3r)
(2r) r
Ê

m/sec. Since is positive,œœ œœ
È
ab ab
ÈÈÈ
3r ( 0.3r)
3r dt
3r ( 0.3r) 4r (0.3r)
33r 33 103
0.3r r da"
"
"
""
"
"Á
"Á
4r (0.3r)
3r
œ
the distance OA is increasing when OB 2r, and B is moving toward O at the rate of 0.3r m/sec.œ119. (a) If f(x) tan x and x , then f (x) sec x,**% *
1
4
w#
f 1 and f 2. The linearization ofˆ‰ ˆ‰%*% %*
11
44
w
f(x) is L(x) 2 x ( 1) 2x .*&&%*&ˆ‰
11
4
2"
#

(b) If f(x) sec x and x , then f (x) sec x tan x,**% *
1
4
w
f 2 and f 2. The linearizationˆ‰ ˆ‰ÈÈ
%* %*%
11
44
w
of f(x) is L(x) 2 x 2*% & &
ÈȈ‰
1
4
2x .*% &
È
È
2( )
4
%1

120. f(x) f (x) . The linearization at x 0 is L(x) f (0)(x 0) f(0) 1 x.*Î* * *%&*%
'"

ww
1tan x (1tan x)
sec x
"
"
121. f(x) x 1 sin x 0.5 (x 1) sin x 0.5 f (x) (x 1) cos x*&&%*&&%Î * & &È ˆ‰
'Ê! ß "'Ê! "
#
L(x) f (0)(x 0) f(0) 1.5(x 0) 0.5 L(x) 1.5x 0.5, the linearization of f(x).Î* %&*%&Î*&
w
122. f(x) 1 x 3.1 2(1 x) (1 x) 3.1 f (x) 2(1 x) ( 1) (1 x)*&&%*% && %Î *%% %&&
2
1x" #
"' 'Ê! ß "! "'Ê! "È
L(x) f (0)(x 0) f(0) 2.5x 0.1, the linearization of f(x).*& Î* %&*%
2
(1 x) 21 x"
"

w
" È
123. S r r h , r constant dS r r h h dh dh. Height changes from h to h dh*& Î*$&)* &11
È
ab
## "
#
##
"'Ê!

!!
1 r h
rh
È
""
dSʜ
1 r h dh
rh
"

"ab
É

195. Chapter 3 Additional and Advanced Exercises 191
124. (a) S 6r dS 12r dr. We want dS (2%) S 12r dr dr . The measurement of theœÊœ Ÿ Ê Ÿ Ê Ÿ
#
kk k k kk
12r r
100 100
"
edge r must have an error less than 1%.
(b) When V r , then dV 3r dr. The accuracy of the volume is (100%) (100%)œœ œ
$# ˆ‰ Š‹
dV 3r dr
Vr
"
%
(dr)(100%) (100%) 3%œœ œˆ‰ ˆ‰ˆ ‰
33r
r r 100
125. C 2 r r , S 4 r , and V r . It also follows that dr dC, dS dC andœÊœ œ œ œ œ œ œ111
CC4C 2C
23611 1 11
#$ "
#
" %
"
dV dC. Recall that C 10 cm and dC 0.4 cm.œœœ
C
2
"
"
1
(a) dr cm (100%) (100%) (.04)(100%) 4%œœ Ê œ œ œ
0.4 0.2 dr 0.2 2
2r 1011 1
1
ˆ‰ ˆ ‰ˆ ‰
(b) dS (0.4) cm (100%) (100%) 8%œœÊ œ œ
20 8 dS 8
S 10011 1
1
ˆ‰ ˆ‰ˆ ‰
(c) dV (0.4) cm (100%) (100%) 12%œœÊ œ œ
10 20 dV 20 6
2 V 1000
" "
"" "
11 1
1
ˆ‰ ˆ‰ Š‹
126. Similar triangles yield h 14 ft. The same triangles imply that h 120a 6
35 15 20 a a
h6 h 6
*Î* *Î* &
"'
dh 120a da da .0444 ft 0.53 inches.Î*% *% *% (*% (*(†( *(
"! " "#! "
#"&"#
120 120 2
aa1 45
"" "
ˆ‰ˆ‰ˆ‰ˆ‰
CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES
1. (a) sin 2 2 sin cos (sin 2 ) (2 sin cos ) 2 cos 2 2[(sin )( sin ) (cos )(cos )]))) ) )) ) ))))*Î* Î*%&
dd
dd))
cos 2 cos sinÎ*% )))
##
(b) cos 2 cos sin (cos 2 ) cos sin 2 sin 2 (2 cos )( sin ) (2 sin )(cos )))) ) )) ) )) ))*%Î * % Î% * %%
## ## dd
dd))
ab
sin 2 cos sin sin cos sin 2 2 sin cos Î* & Î*))))) ) ))
2. The derivative of sin (x a) sin x cos a cos x sin a with respect to x is&* &
cos (x a) cos x cos a sin x sin a, which is also an identity. This principle does not apply to the&* %
equation x 2x 8 0, since x 2x 8 0 is not an identity: it holds for 2 values of x ( 2 and 4), but not
##
%%* %%* %
for all x.
3. (a) f(x) cos x f (x) sin x f (x) cos x, and g(x) a bx cx g (x) b 2cx g (x) 2c;*Î*%Î*% *&&Î*&Î*
www #w ww
also, f(0) g(0) cos (0) a a 1; f (0) g (0) sin (0) b b 0; f (0) g (0)*Î *Î* * Î% *Î* *
w w ww ww
cos (0) 2c c . Therefore, g(x) 1 x .Î% * Î *% * %
""
##
#
(b) f(x) sin (x a) f (x) cos (x a), and g(x) b sin x c cos x g (x) b cos x c sin x; also,*&Î* & * & Î* %
ww
f(0) g(0) sin (a) b sin (0) c cos (0) c sin a; f (0) g (0) cos (a) b cos (0) c sin (0)*Î * & Î* * Î * %
ww
b cos a. Therefore, g(x) sin x cos a cos x sin a.Î* * &
(c) When f(x) cos x, f (x) sin x and f (x) cos x; when g(x) 1 x , g (x) 0 and g (x) 0.** * *%* *
www Ð%Ñ # www Ð%Ñ "
#
Thus f (0) 0 g (0) so the third derivatives agree at x 0. However, the fourth derivatives do not
www www
œœ œ
agree since f (0) 1 but g (0) 0. In case (b), when f(x) sin (x a) and g(x)
Ð%Ñ Ð%Ñ
** *&
sin x cos a cos x sin a, notice that f(x) g(x) for all x, not just x 0. Since this is an identity, we*& * *
have f (x) g (x) for any x and any positive integer n.
ÐÑ ÐÑnn
œ
4. (a) y sin x y cos x y sin x y y sin x sin x 0; y cos x y sin x* Î * Î *% Î & *% & * * Î *%
www ww w
y cos x y y cos x cos x 0; y a cos x b sin x y a sin x b cos xÎ*% Î&*% & * * & Î*% &
ww ww w
y a cos x b sin x y y ( a cos x b sin x) (a cos x b sin x) 0Î*% % Î&*% % & & *
ww ww
(b) y sin (2x) y 2 cos (2x) y 4 sin (2x) y 4y 4 sin (2x) 4 sin (2x) 0. Similarly,* Î * Î *% Î & *% & *
www ww
y cos (2x) and y a cos (2x) b sin (2x) satisfy the differential equation y 4y 0. In general,**& &*
ww
y cos (mx), y sin (mx) and y a cos (mx) b sin (mx) satisfy the differential equation y m y 0.** *& &*
ww #

196. 192 Chapter 3 Differentiation
5. If the circle (x h) (y k) a and y x 1 are tangent at ( ), then the slope of this tangent is œ œ "ß #
### #
m 2x 2 and the tangent line is y 2x. The line containing (h k) and ( ) is perpendicular tokœ œ œ ß "ß #
(1 2)œ
y 2x h 5 2k the location of the center is (5 2k k). Also, (x h) (y k) a*Î *%Î*%Î %w %&%*
k2
h1
"'
"!
###
x h (y k)y 0 1 y (y k)y 0 y . At the point ( ) we knowÊ œ Ê œ Ê œ "ß #
wwwww #w
"
ab
1y
ky
ab
†"
y 2 from the tangent line and that y 2 from the parabola. Since the second derivatives are equal at ( )
www
œ œ "ß #
we obtain 2 k . Then h 5 2k 4 the circle is (x 4) y a . Since ( )œ Ê œ œ œ Ê œ "ß #
1(2)
k
9 9
"! ! #
##
#
"
ˆ‰
lies on the circle we have that a .œ
55
2
È
6. The total revenue is the number of people times the price of the fare: r(x) xp x 3 , where** % ˆ‰
x
40
#
0 x 60. The marginal revenue is 3 2x 3 3 3ŸŸ œ Ê œ
dr x x dr x x 2x
dx 40 40 40 dx 40 40 40
ˆ‰ ˆ‰ˆ‰ ˆ‰ œ ‘ˆ‰
#
"
3 3 1 . Then 0 x 40 (since x 120 does not belong to the domain). When 40 people*% % *Î* *ˆ‰ˆ‰
xx dr
40 40 dx
are on the bus the marginal revenue is zero and the fare is p(40) 3 $4.00.*% * ¹ˆ‰
x
40
#
x=40
7. (a) y uv v u (0.04u)v u(0.05v) 0.09uv 0.09y the rate of growth of the total production is*Î*&*&**Î
dy
dt dt dt
du dv
9% per year.
(b) If 0.02u and 0.03v, then ( 0.02u)v (0.03v)u 0.01uv 0.01y, increasing at 1% per
du dv
dt dt dt
dy
*% * *%& **
year.
8. When x y 225, then y . The tangent
## w
&* *%
x
y
line to the balloon at (12 9) is y 9 (x 12)w% & * %
4
3
y x 25. The top of the gondola is 15 8Î* % &
4
3
23 ft below the center of the balloon. The inter-œ
section of y 23 and y x 25 is at the far*% * %
4
3
right edge of the gondola 23 x 25Î% * %
4
3
x . Thus the gondola is 2x 3 ft wide.Êœ œ
3
#

9. Answers will vary. Here is one possibility.

10. s(t) 10 cos t v(t) 10 sin t a(t) 10 cos t* & Î * *% & Î * * *% &ˆ‰ ˆ‰ ˆ‰
11 1
4 dt 4 dt dt 4
ds dv d s
"
"
(a) s(0) 10 cosœœ ˆ‰
1
4
10
2
È
(b) Left: 10, Right: 10%
(c) Solving 10 cos t 10 cos t 1 t when the particle is farthest to the left.ˆ‰ ˆ‰&*%Î &*%Î*
111
444
3
Solving 10 cos t 10 cos t 1 t , but t 0 t 2 when the particleˆ‰ ˆ‰& * Î & *Î*% jÎ* & *
111 11
444 44
7
1
"
is farthest to the right. Thus, v 0, v 0, a 10, and a 10.ˆ‰ ˆ‰ ˆ‰ ˆ‰
373 7
444 4
111 1
*** *%
(d) Solving 10 cos t 0 t v 10, v 10 and a .ˆ ‰ ˆ‰ ¸ ¸ ˆ‰ ˆ‰&*Î*Î *% * *#
1111 1
4444 4

197. Chapter 3 Additional and Advanced Exercises 193
11. (a) s(t) 64t 16t v(t) 64 32t 32(2 t). The maximum height is reached when v(t) 0*% Î **%* % *
# ds
dt
t 2 sec. The velocity when it leaves the hand is v(0) 64 ft/sec.Êœ œ
(b) s(t) 64t 2.6t v(t) 64 5.2t. The maximum height is reached when v(t) 0 t 12.31 sec.*% Π**% *Ά
# ds
dt
The maximum height is about s(12.31) 393.85 ft.œ
12. s 3t 12t 18t 5 and s t 9t 12t v 9t 24t 18 and v 3t 18t 12; v v
"#"#"#
$# $# # #*% && *%&% Î*%& *%&% *
9t 24t 18 3t 18t 12 2t 7t 5 0 (t 1)(2t 5) 0 t 1 sec and t 2.5 sec.Î % & *%& % Î %&*Î% %*Î* *
###
13. m v v k x x m 2v k 2x m k m kx . Thenabab ˆ‰ˆ ‰ ˆ‰ ˆ‰
## ##
!!
"
%* % Î *% Î *% Î *%
dv dx dv 2x dx dv dx
dt dt dt 2v dt dt v dt
substituting v m kx, as claimed.
dx dv
dt dt
*Î *%
14. (a) x At Bt C on t t v 2At B v 2A B A t t B is the*&& wÎ**&Î * &* &&
#
"# "#

##
cd a b ˆ‰ ˆ‰
dx
dt
tt tt œ "œ"
instantaneous velocity at the midpoint. The average velocity over the time interval is v
avœ
?
?
x
t
At t B.***&&
abab abc d abAtBtC AtBtC
tt tt
ttAtt B
""
"œ "œ
"œ "œ
"œ "œ
ÁÁ" ÁÁ
""
"ÁÁ
#"
ab
(b) On the graph of the parabola x At Bt C, the slope of the curve at the midpoint of the interval*&&
#
t t is the same as the average slope of the curve over the interval.cd "#ß15. (a) To be continuous at x requires that lim sin x lim (mx b) 0 m b m ;**&Î*&Î*%11
x xÄ Ä1 1
Ê
%
b
1
(b) If y is differentiable at x , then lim cos x m m 1 and b .
cos x, x
m, x
w
***Î*%*

 
œ
1
1
11
xÄ1
Ê
16. f x is continuous at because lim f . f (0) lim lim ab ab !œ!œ!œœ
xxxÄ! Ä! Ä!
'" w "
"
"cos x
xx0x
f(x) f(0) 0
1 cos x
x#
lim lim . Therefore f (0) exists with value .œœœ
xxÄ! Ä!
ˆ‰ˆ‰ ˆ‰ˆ‰
1cos x 1cos x sin x
x1cos x x1cos x
"Á '' '
# #
#
w
"
17. (a) For all a, b and for all x 2, f is differentiable at x. Next, f differentiable at x 2 f continuous at x 2ÁœÊœ
lim f(x) f(2) 2a 4a 2b 3 2a 2b 3 0. Also, f differentiable at x 2Ê œ Ê œÊ œ Á
x2Ä
Ê
f (x) . In order that f (2) exist we must have a 2a(2) b a 4a b 3a b.
a, x 2
2ax b, x 2
Î* *%Î*%Î*


ww
œ Then 2a 2b 3 0 and 3a b a and b .%&* *Î* *
39
44
(b) For x , the graph of f is a straight line having a slope of and passing through the origin; for x , the graph of f#  #
$
%
is a parabola. At x , the value of the y-coordinate on the parabola is which matches the y-coordinate of the pointœ#
$
#
on the straight line at x . In addition, the slope of the parabola at the match up point is which is equal to theœ#
$
%
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there.
18. (a) For any a, b and for any x 1, g is differentiable at x. Next, g differentiable at x 1 g continuous at% *% Î
x 1 lim g(x) g( 1) a 1 2b a b b 1. Also, g differentiable at x 1œ Ê œ Ê œ Ê œ Á
xĜ"
%
g (x) . In order that g ( 1) exist we must have a 3a( 1) 1 a 3a 1
a, x 1
3ax 1, x 1
Î* % *%&Î*&


ww#
#
œ
a.Î*%
"
#
(b) For x , the graph of f is a straight line having a slope of and a y-intercept of . For x , the graph of f isŸ" " "
"
#
a parabola. At x , the value of the y-coordinate on the parabola is which matches the y-coordinate of the pointœ"
$
#
on the straight line at x . In addition, the slope of the parabola at the match up point is which is equal to theœ"
"
#
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there.
19. f odd f( x) f(x) (f( x)) ( f(x)) f ( x)( 1) f (x) f ( x) f (x) f is even.Î%*% Î % * % Î %%*% Î %* Î
dd
dx dx
wwwww

198. 194 Chapter 3 Differentiation
20. f even f( x) f(x) (f( x)) (f(x)) f ( x)( 1) f (x) f ( x) f (x) f is odd.Î%* Î %* Î%%* Î%*% Î
dd
dx dx
wwwww
21. Let h(x) (fg)(x) f(x) g(x) h (x) lim lim œœ Êœ œ
w ""
""
xx xxÄÄ ††
h(x) h(x ) f(x) g(x) f(x ) g(x )
xx xx
lim lim f(x) lim g(x )**&
xx xx xxÄÄƆ†
f(x) g(x) f(x) g(x ) f(x) g(x ) f(x ) g(x ) g(x) g(x ) f(x) f(x )
xx xx xx
"Á" " "
"""
! ßßßß ß ß

’“’ “’“ ’“
f(x) lim g(x)f(x) 0 lim g(x)f(x) g(x)f(x), if g is*&*&* !!! !!!!
""
""
www
xx xxÄÄ ††
’“ ’“
g(x) g(x ) g(x) g(x )
xx xx

†
continuous at x . Therefore (fg)(x) is differentiable at x if f(x ) 0, and (fg) (x ) g(x ) f (x ).
!!!!!!
ww œœ
22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f(0) 0 and g is continuousœ
at 0.
(a) If f(x) sin x and g(x) x , then x sin x is differentiable because f (0) cos (0) 1, f(0) sin (0) 0œ œ œ œœœ kk kk
w
and g(x) x is continuous at x 0.œœkk
(b) If f(x) sin x and g(x) x , then x sin x is differentiable because f (0) cos (0) 1, f(0) sin (0) 0œ œ œ œœœ
#Î$ #Î$ w
and g(x) x is continuous at x 0.œœ
#Î$
(c) If f(x) 1 cos x and g(x) x, then x (1 cos x) is differentiable because f (0) sin (0) 0,*% * % * *
$$ w
ÈÈ
f(0) 1 cos (0) 0 and g(x) x is continuous at x 0.*% * * *
"Î$ (d) If f(x) x and g(x) x sin , then x sin is differentiable because f (0) 1, f(0) 0 andœœ œœ ˆ‰ ˆ‰
"" #w
xx
lim x sin lim lim 0 (so g is continuous at x 0).
xx tÄ! Ä! Ä_
ˆ‰
"
xt
sin sin t
œœœ œ
ˆ‰
œ
œ
x
x
23. If f(x) x and g(x) x sin , then x sin is differentiable at x 0 because f (0) 1, f(0) 0 andœœ œ œœ ˆ‰ ˆ‰
"" #w
xx
lim x sin lim lim 0 (so g is continuous at x 0). In fact, from Exercise 21,
xx tÄ! Ä! Ä_
ˆ‰
"
xt
sin sin t
œœœ œ
ˆ‰
œ
œ
x
x
h (0) g(0) f (0) 0. However, for x 0, h (x) x cos 2x sin . But
ww w# "" "
œœ Áœ œ ‘ˆ ‰ ˆ ‰ˆ‰
xx x
"
lim h (x) lim cos 2x sin does not exist because cos has no limit as x 0. Therefore,
xxÄ! Ä!
w "" "
*% & !œ ‘ˆ‰ˆ‰ ˆ‰
xx x
the derivative is not continuous at x 0 because it has no limit there.œ
24. From the given conditions we have f(x h) f(x) f(h), f(h) 1 hg(h) and lim g(h) 1. Therefore,&* %* *
hÄ!
f (x) lim lim lim f(x) f(x) lim g(h) f(x) 1 f(x)
w Á" " "
œœ œ œœœ
hh h hÄ! Ä! Ä! Ä!
f(x h) f(x) f(x) f(h) f(x) f(h) 1
hh h
’“ ’ “ †
f (x) f(x) and f x exists at every value of x.ʜ
ww
ab
25. Step 1: The formula holds for n 2 (a single product) since y u u u u .**Î*& "# # "
dy
dx dx dx
du du
œ "
Step 2: Assume the formula holds for n k:œ
yuuu uuuu uu uuu .œ â Ê œ â âá â"# #$ " $ "#kkkk-1
dy
dx dx dx dx
du du du
œ" k
If yuuuu uuuu, then u uuu œâ œ â œ â"# "# "#
â k k1 k k1 k1 k%% %ab
dy d(u u u )
dx dx dx
du œ" k k1 œ
uu u u u u uu u u uu u œ â âââ ∉
du du
dx dx dx dx
du duœ "
#$ " $ "# "#k k k1 k1 k Ê %
k k1 œ
uu u u u u uu u u uu u .œ â âââ â
du du
dx dx dx dx
du duœ "
#$ " $ "# "#k1 k1 k1 k1 k%% Î%
k k1 œ
Thus the original formula holds for n (k 1) whenever it holds for n k.*& *
26. Recall . Then m and ˆ‰ ˆ‰ ˆ‰ ˆ ‰
mm mm
k k! (m k)! 1 1! (m 1)! k k 1 k! (m k)! (k 1)! (m k 1)!
m! m! m! m!
***&*&
""Á"Á""
. Now, we proveœœœ œ
m! (k 1) m! (m k) m! (m 1) (m 1)!
(k 1)! (m k)! (k 1)! (m k)! (k 1)! ((m 1) (k 1))! k 1
m1ÁÁ " Á Á
Á" Á" Á Á"Á Á

ˆ‰
Leibniz's rule by mathematical induction.
Step 1: If n 1, then u v . Assume that the statement is true for n k, that is:**& *
d(uv)
dx dx dx
dv du
v k u .
d (uv)
dx dx dx dx 2 dx dx k 1 dv dx dx
du dudv dudv dudv dv kk
k
kk k k k k
kk k kk
œ á
Î% Î& Î%
Î% Î& Î%
ˆ‰ ˆ ‰
"
"
" Step 2: If n k 1, then v k k *& * * & & &
d (uv) d (uv)
dx dx dx dx dx dx dx dx dx dx
d dududv dudvdudv
kk
kkkkkk
kk k kœ%
*% *% Î%
*% Î%
Š‹ ’ “œ ‘
"
"

199. Chapter 3 Additional and Advanced Exercises 195
v á ’“’ “ˆ‰ ˆ‰ ˆ ‰ ˆ ‰
kk k k
2 dx dx 2 dx dx k 1 dx dx k 1 dx dx
d u dv d u dv du d v du du
kk k k
kk k kÎ% Î& Î%
Î% Î& Î%"

""
u v(k1) œ ᜠ‘ œ ‘ˆ‰ ˆ‰
dudv du du dudv dudv
dx dx dx dx dx dx 1 2 dx dx
kk
kk k k k
kk k k k œ%*% Î%
*% *% Î% "
" u v(k1) œ ᜠ‘ˆ‰ˆ‰ˆ‰
kk k1
k 1 k dx dx dx dx dx dx 2 dx dx
du dv d v d u du dv d u dv
"

kkk k k
kkk k k œ% Î%
*% *% Î% 'œ "
" u . &&ˆ‰
k1
kdxdx dx
du d v d v
kk
kk œ%
*%
Therefore the formula (c) holds for n (k 1) whenever it holds for n k.*& *
27. (a) T L L L 0.8156 ft
#
œÊœÊœ ʸ
4L
g4 4
Tg 1 sec 32.2 ft/sec1
11
" "
""
"" aba b
(b) T T L; dT dL dL; dT ft 0.00613 sec.
# ##"
# !Þ)"&'
œÊœ œ† œ œ !Þ!"¸
4L
g gg L Lg ft 32.2 ft/sec
11 1 1 1
"
"
ÈÈÈÈÈ aba b
È
ab
(c) Since there are 86,400 sec in a day, we have 0.00613 sec 86,400 sec/day 529.6 sec/day, or 8.83 min/day; theaba b ¸
clock will lose about 8.83 min/day.
28. v s s k s k. If s the initial length of the cube's side, then s s k* Î *„ *% " Î *%) * * %)
$##
!"!
dv ds ds
dt dt dt
ab
k s s . Let t the time it will take the ice cube to melt. Now, tÎ) * % * * * *!"
!"
"
ss
kss
v
vv

œ

œÎ%

*Î% %
&
*Î% ab
ab ˆ‰
hr.œ¸""
"
'"ˆ‰
%
&
*Î%

200. 196 Chapter 3 Differentiation
NOTES:

201. CHAPTER 4 APPLICATIONS OF DERIVATIVES
4.1 EXTREME VALUES OF FUNCTIONS
1. An absolute minimum at x c , an absolute maximum at x b. Theorem 1 guarantees the existence of suchœœ
#
extreme values because h is continuous on [a b].ß
2. An absolute minimum at x b, an absolute maximum at x c. Theorem 1 guarantees the existence of suchœœ
extreme values because f is continuous on [a b].ß
3. No absolute minimum. An absolute maximum at x c. Since the function's domain is an open interval, theœ
function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values.
4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill
the conclusions of Theorem 1.
5. An absolute minimum at x a and an absolute maximum at x c. Note that y g(x) is not continuous butœœœ
still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the
hypothesis is not satisfied, absolute extrema may or may not occur.
6. Absolute minimum at x c and an absolute maximum at x a. Note that y g(x) is not continuous but stillœœœ
has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when
the hypothesis is not satisfied, absolute extrema may or may not occur.
7. Local minimum at , local maximum at ab ab$? Î $? Î
8. Minima at and , maximum at a b ab ab&?Î &?Î Î?&
9. Maximum at . Note that there is no minimum since the endpoint is excluded from the graph.ab ab!ß & #ß !
10. Local maximum at , local minimum at , maximum at , minimum at a b ab ab a b*?Î &?Î $?& Î?$
11. Graph (c), since this the only graph that has positive slope at c.
12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c.
13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a.
14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b.

202. 198 Chapter 4 Applications of Derivatives
15. f(x) x 5 f (x) no critical points;?' ?'
22
33
w
f( 2) , f(3) 3 the absolute maximum? ?'
19
3
is 3 at x 3 and the absolute minimum is atœ
19
3
x2?

16. f(x) x 4 f (x) 1 no critical points;œ Ê œ Ê
w
f( 4) 0, f(1) 5 the absolute maximum is 0? ?'
at x 4 and the absolute minimum is 5 at xœ œ"

17. f(x) x 1 f (x) 2x a critical point at?' ? '
#w
x 0; f( 1) 0, f(0) 1, f(2) 3 the absolute?? ? ?'
maximum is 3 at x 2 and the absolute minimum is 1?
at x 0œ

18. f(x) x f (x) 2x a critical point at?( ' ? '
#w
x 0; f( 3) 5, f(0) 4, f(1) 3 the absolute?? ? ?'
maximum is 4 at x 0 and the absolute minimum is 5?
at x 3?

19. F(x) x F (x) 2x , however? ? ' ? ?
" # w $
xx
2

x 0 is not a critical point since 0 is not in the domain;œ
F(0.5) 4, F(2) 0.25 the absolute maximum is? ? '
0.25 at x 2 and the absolute minimum is 4 atœ
x 0.5œ

203. Section 4.1 Extreme Values of Functions 199
20. F(x) x F (x) x , however? ? ' ? ?
"" " w #
xx

x 0 is not a critical point since 0 is not in the domain;œ
F( 2) , F( 1) 1 the absolute maximum is 1 atœ œ Ê
"
#
x 1 and the absolute minimum is at x 2? ?
"
#

21. h(x) x x h (x) x a critical pointœœÊ œ Ê
$ "Î$ w #Î$ "
È
3
at x 0; h( 1) 1, h(0) 0, h(8) 2 the absolute?? ? ?'
maximum is 2 at x 8 and the absolute minimum is 1?
at x 1?

22. h(x) 3x h (x) x a critical point at? ' ?& '
#Î$ w "Î$
x 0; h( 1) 3, h(0) 0, h(1) 3 the absolute?? ? ?'
maximum is 0 at x 0 and the absolute minimum is 3?
at x 1 and at x 1??

23. g(x) 4 x 4 x??
È
ab
# #
"Î#
g (x) 4 x ( 2x)Êœ œ
w# "
#
"Î#

ab
x
4x
È

critical points at x 2 and x 0, but not at x 2'???
because 2 is not in the domain; g( 2) 0, g(0) 2,? ? g(1) 3 the absolute maximum is 2 at x 0 and theœÊ œ
È
absolute minimum is 0 at x 2?
24. g(x) 5 x x 5 x ( 2x)œ œ &
È
abab
# ##
"Î# "Î#
g (x) critical points at x 5'?? ' ?
w "
#
& ˆ‰ È
x
x
È
and x 0, but not at x 5 because 5 is not in theœœ ÈÈ
domain; f 5 0, f(0) 5Š‹
ÈÈ? ?
the absolute maximum is 0 at x 5 and the absolute'? È
minimum is 5 at x 0?È

25. f( ) sin f ( ) cos is a critical point,))) ))œÊœ Êœ
w
#
1
but is not a critical point because is not interior to)œ

##
11
the domain; f 1, f 1, fˆ‰ ˆ‰ ˆ‰
"
## #
111
? ? ?
5
6
the absolute maximum is 1 at and the absoluteʜ )
1
#
minimum is 1 at ?)

#
1

204. 200 Chapter 4 Applications of Derivatives
26. f( ) tan f ( ) sec f has no critical points in))) )œÊœ Ê
w#
. The extreme values therefore occur at theˆ‰
11
34
ß
endpoints: f 3 and f 1 the absoluteˆ‰ ˆ‰È
11
34
? ? '
maximum is 1 at and the absolute)œ
1
4
minimum is 3 at ?
È
)
1
3

27. g(x) csc x g (x) (csc x)(cot x) a critical point?'? '
w
at x ; g , g 1, g theœœ œ œÊ
11 1 1
##
ˆ‰ ˆ‰ ˆ ‰
33
222
33
ÈÈ
absolute maximum is at x and x , and the
22
3 33È
œœ
11
absolute minimum is 1 at xœ
1
#

28. g(x) sec x g (x) (sec x)(tan x) a critical point atœÊœ Ê
w
x 0; g 2, g(0) 1, g the absolute?? ? ?'ˆ‰ ˆ‰
11
36
2
3
È
maximum is 2 at x and the absolute minimum is 1?
1
3
at x 0œ

29. f(t) 2 t t t? ?& ?&kk a b
È
# #
"Î#
f (t) t (2t)'? ??
w# "
#
"Î#
ab
tt
t
tÈ kk

a critical point at t 0; f( 1) 1,'??
f(0) 2, f(3) 1 the absolute maximum is 2 at t 0??' ?
and the absolute minimum is 1 at t 3?

30. f(t) t 5 (t 5) (t 5) f (t)œ œ œ Êkk a bÈ
# #w
"Î#
(t 5) (2(t 5))œ œ œ
"
#
#
"Î#

ab
t5 t5
(t 5) t5È kk

a critical point at t 5; f(4) 1, f(5) 0, f(7) 2Ê œœœœ
the absolute maximum is 2 at t 7 and the absoluteʜ
minimum is 0 at t 5œ

31. f(x) x f (x) x a critical point at x 0; f( 1) 1, f(0) 0, f(8) 16 the absolute?'?' ????'
%Î$ w "Î$ 4
3
maximum is 16 at x 8 and the absolute minimum is 0 at x 0œœ
32. f(x) x f (x) x a critical point at x 0; f( 1) 1, f(0) 0, f(8) 32 the absolute?' ? ' ?? ? ?'
&Î$ w #Î$ 5
3
maximum is 32 at x 8 and the absolute minimum is 1 at x 1??
33. g( ) g ( ) a critical point at 0; g( 32) 8, g(0) 0, g(1) 1 the absolute)) ) ) )?' ? ' ?? ? ?'
$Î& w #Î& 3
5
maximum is 1 at 1 and the absolute minimum is 8 at 32))??

205. Section 4.1 Extreme Values of Functions 201
34. h( ) 3 h ( ) 2 a critical point at 0; h( 27) 27, h(0) 0, h(8) 12 the absolute)) )) )?'? ' ?? ? ?'
#Î$ w "Î$
maximum is 27 at 27 and the absolute minimum is 0 at 0))? ?
35. Minimum value is 1 at x .œ#

36. To find the exact values, note that y x ,
w#
œ$ #
which is zero when x . Local maximum atœ„É
#
$
; localŠ‹É ab ß % ¸ !Þ)"'ß &Þ!)*
#
$*
%'È
minimum at Š‹É ab
#
$*
%'
ß % ¸ !Þ)"'ß #Þ*""
È

37. To find the exact values, note that that y x x
w#
œ$ # )
x x , which is zero when x or x .œ $ % # œ# œabab
%
$
Local maximum at ; local minimum at ab ˆ‰#ß "( ß
%%"
$#(

38. Note that y x x x , which is zero at
w# #
œ$ ' $œ$ " ab
x . The graph shows that the function assumes lowerœ"
values to the left and higher values to the right of this point,
so the function has no local or global extreme values.

39. Minimum value is 0 when x or x .?$ ?$

206. 202 Chapter 4 Applications of Derivatives
40. The minimum value is 1 at x .œ!

41. The actual graph of the function has asymptotes at x ,œ„"
so there are no extrema near these values. (This is an
example of grapher failure.) There is a local minimum at
.ab!ß "

42. Maximum value is 2 at x ;œ"
minimum value is 0 at x and x .?$ ?*

43. Maximum value is at x
"
#
œ"à
minimum value is as x .?$
"
#

44. Maximum value is at x 0
" #
ψ
minimum value is as x 2.?
"
#

207. Section 4.1 Extreme Values of Functions 203
45. y x x x
w #Î$ "Î$ #& %
$ $
?$) )&?ab a b
x
xÈ
"
crit. pt. derivative extremum value
x local max
x undefined local min 0
œ ! "! œ "Þ!$%
œ!
%"#
&#&
"Î$

46. y x x x x
w #Î$ "Î$ # #))
$ $
œ# %œab a b
x
x

"
È
crit. pt. derivative extremum value
x minimum
x undefined local max 0
x minimum
œ" ! $
œ!
œ" ! $

47. y x x x
w "
#%
#
?&)$(
È
x
abab
È
œœ
%
% %
%#xx
xx
x



ab
ÈÈ
crit. pt. derivative extremum value
x undefined local max
x minimum
xmaximum
x undefined local min
?& Î
œ # ! #
œ# ! #
œ# !
È
È

48. y x 1 x x
w# "
#$
?)&*
È
x
ab È
œœ
% $
#$ #$
"#xxx
xx
_5x x

aba b
ÈÈ
crit. pt. derivative extremum value
x 0 minimum
x local max
x undefined minimum
œ! !
œ ! "& ¸ %Þ%'#
œ$ !
"# "%%
& "#&
"Î#

49. y
,x
,x
w
œ
# "
""
œ
crit. pt. derivative extremum value
x undefined minimumœ" #

208. 204 Chapter 4 Applications of Derivatives
50. y
,x
x, x
w
œ
" !
## !
œ
crit. pt. derivative extremum value
x undefined local min
x local max
œ! $
œ" ! %

51. y
2x 2, x 1
2x 6, x 1
w
œ


œ
crit. pt. derivative extremum value
x 1 maximum 5
x 1 undefined local min 1
x 3 maximum 5
œ !
œ
œ!

52. We begin by determining whether f x is defined at x , where f x
xx,x
xxx,x
w
"""&
%#%
#
$#
ab ab œœ" œ
Ÿ"
' ) " Clearly, f x x if x , and f h . Also, f x x x if x , and
www# ""
##
Ä!
ab a b abœ " " œ" œ$ "# ) "lim
h

f h . Since f is continuous at x , we have that f . Thus,lim
hÄ!
w w
"
ab ab" œ" œ" " œ"
fx
x , x
xx ,x
w
""
##
#
abœœ
Ÿ"
$"# ) "
Note that x when x , and x x when x . œ! œ" $ "# )œ! œ œ œ#„
""
## #$ ' $
# "#„"# %$) "#„%) #$
ÈÈÈ abab
ab

But , so the critical points occur at x and x .# ¸!Þ)%&" œ" œ# ¸$Þ"&&
#$ #$
$ $
È È
crit. pt. derivative extremum value
x local max 4
x local min
œ" !
¸ $Þ"&& ! ¸ $Þ!(*

53. (a) No, since f x x , which is undefined at x .
w #
$
"Î$
ab a b? ?
(b) The derivative is defined and nonzero for all x . Also, f and f x for all x .Á# # œ! ! Á#ab ab
(c) No, f x need not have a global maximum because its domain is all real numbers. Any restriction of f to a closedab
interval of the form a, b would have both a maximum value and minimum value on the interval.ÒÓ
(d) The answers are the same as (a) and (b) with 2 replaced by a.
54. Note that f x . Therefore, f x .
x x, x or x x , x or x
x x, x or x x , x or x
ab abœœœœ
* Ÿ $ ! Ÿ $ $ * $ ! $
* $ !  $ $ * $ ! $
$ $
$ $
w
(a) No, since the left- and right-hand derivatives at x , are and , respectively.œ! * *
(b) No, since the left- and right-hand derivatives at x , are and , respectively.œ$ ") ")

209. Section 4.1 Extreme Values of Functions 205
(c) No, since the left- and right-hand derivatives at x , are and , respectively.œ$ ") ")
(d) The critical points occur when f x (at x ) and when f x is undefined (at x and x ). The
ww
ab ab Èœ! œ „ $ œ! œ „$
minimum value is at x , at x , and at x ; local maxima occur at and .! œ$ œ! œ$ $ß' $ $ß' $ Š‹Š‹
ÈÈ ÈÈ
55.
(a) The construction cost is C x x x million dollars, where x miles. The following isab a b
È
œ !Þ$ "' !Þ# * ! Ÿ Ÿ *
#
a graph of C x .ab
Solving C x gives x miles, but only x miles is a critical point is
w !Þ$
"' "
)&
&
abœ !Þ# œ ! œ „ ¸ „ $Þ&) œ $Þ&)
x
x
È
È

the specified domain. Evaluating the costs at the critical and endpoints gives C $3 million, C $2.694ab Š‹!œ ¸
)&
&
È
million, and C $2.955 million. Thereform, to minimize the cost of construction, the pipeline should be placedab*¸
from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the
refinery.
(b) If the per mile cost of underwater construction is p, then C x p x x andab a b
È
œ "' !Þ# *
#
C x gives x , which minimizes the construction cost provided x . The value
w !Þ$! Þ)
"' " !Þ!%
abœ!Þ#œ! œ Ÿ*
x
x
c c
pÈÈ

of p that gives x miles is . Consequently, if the underwater construction costs $218,864 per mile or less,cœ * !Þ#"))'%
then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost
of construction.
In theory, p would have to infinite to justify running the pipe directly from the docking facility to point A (i.e., for x
c
to be zero). For all values of p there is always an x that will give a minimum value for C. This !Þ#"))'% − Ð!ß *Ñ c
is proved by looking at C x which is always positive for p .
ww "'
"' "
abc
p
x?
ab
c


56. There are two options to consider. The first is to build a new road straight from village A to Village B. THe second is to
build a new highway segment from village A to the Old Road, reconstruct a segment of Old Road, and build a new
highway segment from Old Road to village B, as shown in the figure. The cost of the first option is C million
"œ !Þ& "&!ab
dollars 75 million dollars.œ

210. 206 Chapter 4 Applications of Derivatives

The construction cost for the second option is C x x x million dollars for#
#ab a bŠ‹
È
œ !Þ& # #&!! !Þ$ "&! #
x miles. The following is a graph of C x .!Ÿ Ÿ(&
#ab
Solving C x give x miles, but only x miles is in the specified domain. In
w
#
#&!! "
abœ !Þ' œ ! œ „ $(Þ& œ $(Þ&
x
x
È

summary, C $75 million, C $95 million, C $85 million, and C $90.139 million. Consequently,"# # #œ ! œ $(Þ& œ (& œab a b a b
a new road straight from village A to village B is the least expensive option.
57.
The length of pipeline is L x x x for x . The following is a graph of L x .ab a b ab
È Éœ % #& "! ! Ÿ Ÿ "!
#
#
Setting the derivative of L x equal to zero gives L x . Note that cos andab ab
w
%" %"
"!
#&"!
? ? ?
xx
xx
x
x
AÈÈ
ab
É ab

) cos , therefore, L x when cos cos , or and ACP is similar to BDP. Use
"!
#&"!
wx
x
BABAB
É ab

œœ!œœ˜˜))))) ab
simple proportions to determine x as follows: x miles along the coast from town A to town B.
xx
2
œ Ê œ ¸ #Þ)&(
"! #!
&(
If the two towns were on opposite sides of the river, the obvious solution would be to place the pump station on a straight
line (the shortest distance) between two towns, again forcing . The shortest length of pipe is the same regardless of))
ABœ
whether the towns are on thee same or opposite sides of the river.

211. Section 4.1 Extreme Values of Functions 207
58.
(a) The length of guy wire is L x x x for x . The following is a graph ofab a b
È Éœ *!! #&!! "&! ! Ÿ Ÿ "&!
#
#
Lx. ab
Setting L x equal to zero gives L x . Note that cos and
ww
*!! " *!! "
"&!
#&!! " "&!
ab ab ? ? ?
xx
xx
x
x
AÈÈ
ab
É ab

) cos . Therefore, L x when cos cos , or and ACE is similar to ABD.
ab
É ab
"&!
#&!! " "&!
wx
x
BABAB

œœ!œœ˜˜))))) ab
Use simple proportions to determine x: x feet.
xx
$! &! %
"&! ## &
œ Ê œ œ &'Þ#&
(b) If the heights of the towers are h and h , and the horizontal distance between them is s, then
BC L x h x h s x and L x . However, cos andab a b abÉ Éœ œ œ
C B
xx
hx hx
sx
hsx
## #
# w
""
G
ÉÉ É
ab
ab
CC B

)
cos . Therefore, L x when cos cos , or and ACE is similar to ABD.
ab
É ab
sx
hsx
BCBCB


w
B

œœ!œœ˜˜))))) ab
Simple proportions can again be used to determine the optimum x: x s.
xsx
hh hh
h
cB Bc
c
œÊœ

"
Š‹
59. (a) V x x x xabœ "'! &# %
#$
Vx x x x x
w#
ab a ba bœ "'! "!% "# œ % # $ #!
The only critical point in the interval is at x . The maximum value of V x is 144 at x .ab ab!ß & œ # œ #
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x units.œ#
60. (a) P x x
w#
abœ # #!!
The only critical point in the interval is at x . The minimum value of P x is at x .ab ab!ß _ œ "! %! œ "!
(b) The smallest possible perimeter of the rectangel is 40 units and it occurs at x units which makes the rectangle aœ"!
10 by 10 square.
61. Let x represent the length of the base and x the height of the triangle. The area of the triangle is represented by
È
#&
#
A x x where x . Consequently, solving A x x . Sinceab ab
È
œ #& !Ÿ Ÿ& œ!Ê œ!Ê œ
x x
x
#
# w #&# &
##& #


ÈÈ A A , A x is maximized at x . The largest possible area is A cm .ab ab ab Š‹!œ &œ! œ œ
&& #&
## %
#
ÈÈ

212. 208 Chapter 4 Applications of Derivatives
62. (a) From the diagram the perimeter P x rœ# # œ%!!1
x r. The area A is 2rxÊ œ #!! 1
A r r r where r .Ê œ %!! # ! Ÿ Ÿab 1
# #!!
1
(b) A r r so the only critical point is r .
w "!!
abœ %!! % œ1
1
Since A r if r and x r , theab? ? ? ? 1
values r 31.83 m and x m theœ ¸ œ "!!
"!!
1
maximize
area over the interval r .!Ÿ Ÿ
#!!
1

63. s gt v t s gt v t . Now s t s t v 0 t 0 or t .œ Ê œ œ!Ê œ œ Í œ Í œ œ
"
#
#
!! !
ds
dt g 2 g
v2v
00
gt
ab ˆ‰
0
Thus s g v s s s is the height over the interval 0 t .Š‹ Š‹ Š‹
vvv 2v
ggg2g g
2
00 00
v
œ œ Ÿ Ÿ
"
#
2
0
maximum
64. sin t cos t, solving tan t t n where n is a nonnegative integer (in this exercise t is
dI dI
dt dt
œ# # œ!Ê œ"Ê œ
1
%
1 never negative) the peak current is amps.Ê##
È
65. Yes, since f(x) x x x f (x) x (2x) is not defined at x 0. Thus itœœ œ Ê œ œ œ œkk ab ab
È
# #w #
"Î#" Î#"
#
xx
x
x
ab
kk


is not required that f be zero at a local extreme point since f may be undefined there.
ww
66. If f(c) is a local maximum value of f, then f(x) f(c) for all x in some open interval (a b) containing c. SinceŸß
f is even, f( x) f(x) f(c) f( c) for all x in the open interval ( b a) containing c. That is, f assumesœ Ÿ œ ß
a local maximum at the point c. This is also clear from the graph of f because the graph of an even function
is symmetric about the y-axis.
67. If g(c) is a local minimum value of g, then g(x) g(c) for all x in some open interval (a b) containing c. Since ß
g is odd, g( x) g(x) g(c) g( c) for all x in the open interval ( b a) containing c. That is, gœ Ÿ œ ß
assumes a local maximum at the point c. This is also clear from the graph of g because the graph of an odd
function is symmetric about the origin.
68. If there are no boundary points or critical points the function will have no extreme values in its domain. Such
functions do indeed exist, for example f(x) x for x . (Any other linear function f(x) mx b? ?
with m 0 will do as well.)Á
69. (a) f x ax bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The
w#
abœ$ #
function f x x x has two critical points at x and x . The function f x x has one critical pointab ab? ? ? ?
$ $
at x The function f x x x has no critical points.œ!Þ œ ab
$
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the
cubic function has no extreme values.)

213. Section 4.1 Extreme Values of Functions 209
70. (a)
f is not a local extreme value because in any open interval containing x , there are infinitely many pointsab!œ! œ!
where f x and where f x .ab ab? ?
(b) One possible answer, on the interval :Ò!ß "Ó
fx
xcos , x
, x
abœ
ab
œ
" !Ÿ "
!œ"
"
"x
This function has no local extreme value at x . Note that it is continuous on .œ" Ò!ß"Ó
71. Maximum value is 11 at x ;œ&
minimum value is 5 on the interval ;Ò$ß #Ó
local maximum at ab&ß *

72. Maximum value is 4 on the interval ;Ò&ß (Ó
minimum value is on the interval .% Ò#ß "Ó

73. Maximum value is on the interval ;& Ò$ß _Ñ
minimum value is on the interval .& Ð_ß #Ó

214. 210 Chapter 4 Applications of Derivatives
74. Minimum value is 4 on the interval Ò"ß $Ó

75-80. Example CAS commands:
:Maple
with(student):
f := x -> x^4 - 8*x^2 + 4*x + 2;
domain := x=-20/25..64/25;
plot( f(x), domain, color=black, title="Section 4.1 #75(a)" );
Df := D(f);
plot( Df(x), domain, color=black, title="Section 4.1 # 75(b)" )
StatPt := fsolve( Df(x)=0, domain )
SingPt := NULL;
EndPt := op(rhs(domain));
Pts :=evalf([EndPt,StatPt,SingPt]);
Values := [seq( f(x), x=Pts )];
Maximum value is 2.7608 and occurs at x=2.56 (right endpoint).
Minimum value is -6.2680 and occurs at x=1.86081 (singular point).
$
%
: (functions may vary) (see section 2.5 re. RealsOnly ):Mathematica
<<Miscellaneous `RealOnly`
Clear[f,x]
a = 1; b = 10/3;
f[x_] =2 2x 3 x
2/3
f'[x]
Plot[{f[x], f'[x]}, {x, a, b}]
NSolve[f'[x]==0, x]
{f[a], f[0], f[x]/.%, f[b]//N
In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here)
is observed from the graph and the following command is used:
FindRoot[f'[x]==0,{x, 1.1}]
4.2 THE MEAN VALUE THEOREM
1. When f(x) x 2x 1 for 0 x 1, then f (c) 3 2c 2 c .œ ŸŸ œ Êœ Êœ
#w
#
"f(1) f(0)
10
2. When f(x) x for 0 x 1, then f (c) 1 c c .œŸŸ œÊœ Êœ
#Î$w" Î$
#
f(1) f(0)
10 3 7
28 ˆ‰
3. When f(x) x for x 2, then f (c) 0 c 1.œ ŸŸ œ Ê œ" Ê œ
"" "
#
w
x21/2 c
f(2) f(1/2)

4. When f(x) x 1 for 1 x 3, then f (c) c .œ ŸŸ œ Êœ Êœ
È
f(3) f(1) 2
31 c1
3
# #
w "
#
È
È
5. Does not; f(x) is not differentiable at x 0 in ( 8).?

215. Section 4.2 The Mean Value Theorem 211
6. Does; f(x) is continuous for every point of [0 1] and differentiable for every point in (0 1).ßß
7. Does; f(x) is continuous for every point of [0 1] and differentiable for every point in (0 1).ßß
8. Does not; f(x) is not continuous at x 0 because lim f(x) 1 0 f(0).œ œÁœ
xÄ!

9. Since f(x) is not continuous on 0 x 1, Rolle's Theorem does not apply: lim f(x) lim x 1 ? ?
x1 x1ÄÄ

0 f(1).Áœ
10. Since f(x) must be continuous at x 0 and x 1 we have lim f(x) a f(0) a 3 andœœ œœÊœ
xÄ!
"
lim f(x) lim f(x) 1 3 a m b 5 m b. Since f(x) must also be differentiable at
x1 x1Ä Ä
"
? ? ?
x 1 we have lim f (x) lim f (x) 2x 3 m 1 m. Therefore, a 3, m 1 and b 4.?? ?????
x1 x1Ä Ä
"
ww
kk
x=1 x=1
11. (a) i
ii
iii
iv
(b) Let r and r be zeros of the polynomial P(x) x a x a x a , then P(r ) P(r ) 0.
"# "! " # œ á œ œ
nn-1
n-1
Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem P (r) 0 for some r
w
œ
between r and r , where P (x) nx (n 1) a x a .
"# "
w œ á
n-1 n-2
n-1
12. With f both differentiable and continuous on [a b] and f(r ) f(r ) f(r ) 0 where r , r and r are in [a b],ß œœœ ß "#$ "#$
then by Rolle's Theorem there exists a c between r and r such that f (c ) 0 and a c between r and r""# " ##$
w œ
such that f (c ) 0. Since f is both differentiable and continuous on [a b], Rolle's Theorem again applies and
ww
#
ϧ
we have a c between c and c such that f (c ) 0. To generalize, if f has n 1 zeros in [a b] and f is
$"# $
w ÐÑw œ ß
n
continuous on [a b], then f has at least one zero between a and b.ß
ÐÑn
13. Since f exists throughout [a b] the derivative function f is continuous there. If f has more than one zero in
ww w w
ß
[a b], say f (r ) f (r ) 0 for r r , then by Rolle's Theorem there is a c between r and r such thatßœœÁ
ww
"# "# "#
f (c) 0, contrary to f 0 throughout [a b]. Therefore f has at most one zero in [a b]. The same argument
ww ww w
?
holds if f 0 throughout [a b].
ww
ß
14. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f (x) has three or more zeros,
w
f (x) has 2 or more zeros and f (x) has at least one zero. This is a contradiction since f (x) is a non-zero
ww www www
constant when f(x) is a cubic polynomial.
15. With f( 2) 11 0 and f( 1) 1 0 we conclude from the Intermediate Value Theorem thatœ œ
f(x) x 3x 1 has at least one zero between 2 and 1. Then 2 x 1 x 1œ Ê)
% $
32 4x 4 29 4x 3 1 f (x) 0 for 2 x 1 f(x) is decreasing on [ 1]Ê Ê Ê Ê #ß
$$w
f(x) 0 has exactly one solution in the interval ( 1).Êœ #ß
16. f(x) x 7 f (x) 3x 0 on ( 0) f(x) is increasing on ( 0). Also, f(x) 0 ifœ Ê œ _ß Ê _ß
$w#48
xx
!
x 2 and f(x) 0 if 2 x 0 f(x) has exactly one zero in ( ). Ê _ß !
17. g(t) t t 1 4 g (t) 0 g(t) is increasing for t in ( ); g(3) 3 2 0œ Ê œ Ê !ß_ œ È È È
w ""
#"ÈÈt2t1
and g(15) 15 0 g(t) has exactly one zero in ( )œÊ !ß_Þ
È

216. 212 Chapter 4 Applications of Derivatives
18. g(t) 1 t 3.1 g (t) 0 g(t) is increasing for t in ( 1 1);œ Ê œ Ê ß
"""
"œ "œ
w
t(t) 21tÈ
œ È
g( 0.99) 2.5 and g(0.99) 98.3 g(t) has exactly one zero in ( 1 1).œ œ Ê ß
19. r( ) sin 8 r ( ) 1 sin cos 1 sin 0 on ( ) r( ) is)) ) )œ Ê œ œ _ß_Ê
#w "ˆ‰ ˆ‰ ˆ‰ ˆ ‰
))))
333333
22
increasing on ( ); r(0) 8 and r(8) sin 0 r( ) has exactly one zero in ( )._ß _ œ œ Ê _ß _
#ˆ‰
8
3
)
20. r( ) 2 cos 2 r ( ) 2 2 sin cos 2 sin 2 0 on ( ) r( ) is increasing on)) ) ) )) ) )œ Ê œ œ _ß_Ê
#wÈ
( ); r( ) 4 cos ( ) 2 4 1 2 0 and r(2 ) 4 1 2 0 r( ) has_ß _ # œ # œ œ Ê11 1 1 11 )
ÈÈ È
exactly one zero in ( )._ß _
21. r( ) sec 5 r ( ) (sec )(tan ) 0 on r( ) is increasing on ;)) ) )) )œÊœ !ßÊ !ß
" w
##) )
11
w
3
ˆ‰ ˆ‰
r(0.1) 994 and r(1.57) 1260.5 r( ) has exactly one zero in .¸ ¸ Ê !ß ) ˆ‰
1
#
22. r( ) tan cot r ( ) sec csc 1 sec cot 0 on r( ) is increasing))))) )) )) )œ Ê œ œ !ß Ê
w## ##
# ˆ‰
1
on 0 ; r 0 and r(1.57) 1254.2 r( ) has exactly one zero in .ˆ‰ˆ‰ ˆ‰ßœ ¸Ê !ß
11 1 1
# #44
)
23. By Corollary 1, f (x) 0 for all x f(x) C, where C is a constant. Since f( 1) 3 we have C 3
w
œÊœ œ œ
f(x) 3 for all x.ʜ
24. g(x) 2x 5 g (x) 2 f (x) for all x. By Corollary 2, f(x) g(x) C for some constant C. ThenœÊ œœ œ
ww
f(0) g(0) C 5 5 C C 0 f(x) g(x) 2x 5 for all x.œÊœÊœÊ œœ
25. g(x) x g (x) 2x f (x) for all x. By Corollary 2, f(x) g(x) C.œÊ œœ œ
#w w
(a) f(0) 0 0 g(0) C 0 C C 0 f(x) x f(2) 4œÊœœÊœÊœÊœ
#
(b) f(1) 0 0 g(1) C 1 C C 1 f(x) x 1 f(2) 3œÊœ œÊ œÊ œÊ œ
#
(c) f( 2) 3 3 g( 2) C 3 4 C C 1 f(x) x 1 f(2) 3œÊœÊœÊ œÊ œÊ œ
#
26. g(x) mx g (x) m, a constant. If f (x) m, then by Corollary 2, f(x) g(x) b mx b whereœÊ œ œ œœ
ww
b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines
ymxb.œ
27. (a) y C (b) y C (c) y Cœ œ œ
xxx
34
œ w
#
28. (a) y x C (b) y x x C (c) y x x x Cœ œ œ
##$#
29. (a) y x y C (b) y x C (c) y 5x C
w# "" "
œ Ê œ œ œ
xx x
30. (a) y x y x C y x C (b) y 2 x C
w "Î# "Î#"
#
œÊœÊœ œ ÈÈ
(c) y 2x 2 x Cœ
#È
31. (a) y cos 2t C (b) y 2 sin Cœ œ
"
# #
t
(c) y cos 2t 2 sin Cœ
"
##
t
32. (a) y tan C (b) y y C (c) y tan Cœ œÊœ œ)))))
w "Î# $Î# $Î# 22
33
33. f(x) x x C; 0 f(0) 0 0 C C 0 f(x) x xœ œ œÊœÊ œ
## #

217. Section 4.2 The Mean Value Theorem 213
34. g(x) x C; 1 g( 1) ( 1) C C 1 g(x) x 1œ œ œ Ê œ Ê œ
"" "## #
œx1 x
35. r( ) 8 cot C; 0 r 8 cot C 0 2 1 C C 2 1)) ) 1 1œ œ œ Êœ Êœˆ‰ ˆ‰ ˆ‰
11 1
44 4
r( ) 8 cot 2 1ʜ)) )1
36. r(t) sec t t C; 0 r(0) sec (0) 0 C C 1 r(t) sec t t 1œœœ ʜʜ
37. v t s t t C; at s and t we have C s t tœ œ*Þ) &Ê œ%Þ* & œ"! œ! œ"!Ê œ%Þ* & "!
ds
dt
# #
38. v t s t t C; at s and t we have C s t tœ œ $# # Ê œ "' # œ % œ œ " Ê œ ' # "
ds
dt
## "
#
39. v sin t s cos t C; at s and t we have C sœ œ Ê œ œ! œ! œ Ê œ
ds
dt
cos t
ab ab11
"" "œ
111
1ab
40. v cos s sin C; at s and t we have C s sinœ œ Êœ œ" œ œ"Êœ "
ds 2 t t t
dt 11 1 1
ˆ‰ ˆ‰ ˆ‰
## # #
1
41. a v t C ; at v and t we have C v t s t t C ; at s andœ$#Ê œ$# œ#! œ! œ#!Ê œ$# #!Ê œ"' #! œ&
"" #
#
t we have C s t tœ! œ&Ê œ"' #! &#
#
42. a 9.8 v 9.8t C ; at v and t we have C v t s t t C ; at s andœ Ê œ œ$ œ! œ$Ê œ*Þ) $Ê œ%Þ* $ œ!"" #
#
t we have C s t tœ! œ!Ê œ%Þ* $#
#
43. a sin t v cos t C ; at v and t we have C v cos t s sin t C ; at sœ% # Ê œ# # œ# œ! œ!Ê œ# # Ê œ # œ$ab ab ab ab "" #
and t we have C s sin tœ! œ$Ê œ # $ # ab
44. a cos v sin C ; at v and t we have C v sin s cos C ; atœ Ê œ œ! œ! œ!Ê œ Ê œ
* $ $$ $$ $
"" #
1 1 11 11 1
œ
ˆ‰ ˆ‰ ˆ‰ ˆ‰
tt tt
s and t we have C s cosœ" œ! œ!Ê œ #
$
ˆ‰
t
1
45. If T(t) is the temperature of the thermometer at time t, then T(0) 19° C and T(14) 100° C. From theœ œ
Mean Value Theorem there exists a 0 t 14 such that 8.5° C/sec T (t ), the rate at which œ œ !!
œ
œ
wT(14) T(0)
14 0
the temperature was changing at t t as measured by the rising mercury on the thermometer.œ !
46. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that
speed at least once during the trip.
47. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that
speed at least once during the trip.
48. The runner's average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value Theorem, the
runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0
mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 11 mph
at least twice.
49. Let d(t) represent the distance the automobile traveled in time t. The average speed over 0 t 2 isŸŸ
. The Mean Value Theorem says that for some 0 t 2, d (t ) . The value d (t ) is
d(2) d(0) d(2) d(0)
0 0
œ œ
ʜ ʜ
!! !
ww
œ
the speed of the automobile at time t (which is read on the speedometer).
!
50. a t v t v t t C; at we have C v t t. When t , then v m/sec.ab ab ab a b ab a bœ œ "Þ' Ê œ "Þ' !ß ! œ ! Ê œ "Þ' œ $! $! œ %)
w

218. 214 Chapter 4 Applications of Derivatives
51. The conclusion of the Mean Value Theorem yields c a b c ab.
œœ
ba
œ
œ
"œ #
ba c ab
ab
œ Ê œ Ê œ
œ
ˆ‰ È
52. The conclusion of the Mean Value Theorem yields 2c c .
ba ab
ba
œœ

œÊ
œÊœ
53. f (x) [cos x sin (x 2) sin x cos (x 2)] 2 sin (x 1) cos (x 1) sin (x x 2) sin 2(x 1)
w
œœ
sin (2x 2) sin (2x 2) 0. Therefore, the function has the constant value f(0) sin 1 0.7081œœ œ¸
#
which explains why the graph is a horizontal line.
54. (a) f x x x x x x x x x is one possibility.abababababœ#" "#œ&%
&$
(b) Graphing f x x x x and f x x x on by we see that each x-intercept ofab abœ & % œ & "& % Ò$ß $Ó Ò(ß (Ó
&$ w % #
f x lies between a pair of x-intercepts of f x , as expected by Rolle's Theorem.
w
ab ab

(c) Yes, since sin is continuous and differentiable on .ab_ß_
55. f x must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f x is zero twiceab ab
between a and b. Then by the Mean Value Theorem, f x would have to be zero at least once between the two zeros of
w
ab
f x , but this can't be true since we are given that f x on this interval. Therefore, f x is zero once and only onceab ab ab
w
Á!
between a and b.
56. Consider the function k x f x g x . k x is continuousab ab ab abœ
and differentiable on a, b , and since k a f a g a andÒÓ œ ab ab ab
k b f b g b , by the Mean Value Theorem, there mustab ab abœ
be a point c in a, b where k c . But sinceab ab
w
œ!
k c f c g c , this means that f c g c , and c is a
www w w
ab ab ab ab abœ œ
point where the graphs of f and g have tangent lines with the
same slope, so these lines are either parallel or are the same
line.

57. Yes. By Corollary 2 we have f(x) g(x) c since f (x) g (x). If the graphs start at the same point x a,œ œ œ
ww
then f(a) g(a) c 0 f(x) g(x).œÊœÊœ
58. Let f(x) sin x for a x b. From the Mean Value Theorem there exists a c between a and b such thatœŸŸ
cos c 1 1 1 sin b sin a b a .
sin b sin a sin b sin a sin b sin a
ba ba ba
œœœ
œœœ
œ ÊŸ ŸÊ ŸÊ Ÿ ¸¸ kkkk
59. By the Mean Value Theorem we have f (c) for some point c between a and b. Since b a 0 and f(b) f(a),
f(b) f(a)
ba
œ
œ
w
œ
we have f(b) f(a) 0 f (c) 0. Ê
w
60. The condition is that f should be continuous over [a b]. The Mean Value Theorem then guarantees the
w
ß existence of a point c in (a b) such that f (c). If f is continuous, then it has a minimum andßœ
f(b) f(a)
ba
œ
œ
ww
maximum value on [a b], and min f f (c) max f , as required.ߟŸ
ww w

219. Section 4.3 Monotonic Functions and the First Derivative Test 215
61. f (x) 1 x cos x f (x) 1 x cos x 4x cos x x sin x
w% ww % $%
œ" œÊ
œ Ê œ ab aba b
x 1 x cos x (4 cos x x sin x) 0 for 0 x 0.1 f (x) is decreasing when 0 x 0.1œ Ÿ Ÿ Ê Ÿ Ÿ
$% w
œÊ
ab
min f 0.9999 and max f 1. Now we have 0.9999 1 0.09999 f(0.1) 1 0.1ʸ œ Ÿ ŸÊ ŸŸ
ww œ"f(0.1)
0.1
1.09999 f(0.1) 1.1.ÊŸŸ
62. f (x) 1 x f (x) 1 x 4x 0 for 0 x 0.1 f (x) is increasing when
w%ww %$ w
œ" œÊ
œ
œ Ê œ œ Ÿ Êab abab
4x
1x
w

w
ab
0 x 0.1 min f 1 and max f 1.0001. Now we have 1 1.0001ŸŸ Ê œ œ Ÿ Ÿ
ww œf(0.1) 2
0.1
0.1 f(0.1) 2 0.10001 2.1 f(0.1) 2.10001.ÊŸ Ÿ ÊŸ Ÿ
63. (a) Suppose x 1, then by the Mean Value Theorem 0 f(x) f(1). Suppose x 1, then by the&&#$$
f(x) f(1)
x1
œ
œ
Mean Value Theorem 0 f(x) f(1). Therefore f(x) 1 for all x since f(1) 1.
f(x) f(1)
x1
œ
œ
Ê   œ
(b) Yes. From part (a), lim 0 and lim 0. Since f (1) exists, these two one-sided
x1 x1Ä Ä
œ w
f(x) f(1) f(x) f(1)
x1 x1
œœ
œœ
w
Ÿ 
limits are equal and have the value f (1) f (1) 0 and f (1) 0 f (1) 0.
ww w w
ÊŸ  Êœ
64. From the Mean Value Theorem we have f (c) where c is between a and b. But f (c) 2pc q 0
f(b) f(a)
ba
œ
œ
ww
œœœ
has only one solution c . (Note: p 0 since f is a quadratic function.)œ Á
q
p#
4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST
1. (a) f (x) x(x 1) critical points at 0 and 1
w
œÊ
(b) f increasing on ( ) and ( ), decreasing on ( )
w
œ ± ± Ê _ß ! "ß _ !ß "
!"
(c) Local maximum at x 0 and a local minimum at x 1œœ
2. (a) f (x) (x 1)(x 2) critical points at 2 and 1
w
œ Ê
(b) f increasing on ( ) and ( ), decreasing on ( 2 )
w
œ ± ± Ê _ß # "ß _ ß "
# "
(c) Local maximum at x 2 and a local minimum at x 1œ œ
3. (a) f (x) (x 1) (x 2) critical points at 2 and 1
w#
œ Ê
(b) f increasing on ( 2 1) and ( ), decreasing on ( 2)
w
œ ± ± Ê ß "ß _ _ß
# "
(c) No local maximum and a local minimum at x 2œ
4. (a) f (x) (x 1) (x 2) critical points at 2 and 1
w##
œ Ê
(b) f increasing on ( 2) ( ) ( ), never decreasing
w
œ ± ± Ê _ß #ß " "ß _
# "
(c) No local extrema
5. (a) f (x) (x 1)(x 2)(x 3) critical points at 2, 1 and 3
w
œÊ
(b) f increasing on ( 2 1) and ( ), decreasing on ( 2) and ( )
w
œ ± ± ± Ê ß $ß _ _ß "ß $
# " $
(c) Local maximum at x 1, local minima at x 2 and x 3œœœ
6. (a) f (x) (x 7)(x 1)(x 5) critical points at 5, 1 and 7
w
œÊ
(b) f increasing on ( 5 1) and (7 ), decreasing on ( 5) and ( 7)
w
œ ± ± ± Ê ß ß _ _ß "ß
& " (
(c) Local maximum at x 1, local minima at x 5 and x 7œ œ œ

220. 216 Chapter 4 Applications of Derivatives
7. (a) f (x) x (x 2) critical points at 2 and 0
w "Î$
œÊ
(b) f )( increasing on ( 2) and (0 ), decreasing on ( 2 0)
w
œ ± Ê _ß ß _ ß
# !
(c) Local maximum at x 2, local minimum at x 0œ œ
8. (a) f (x) x (x 3) critical points at 0 and 3
w "Î#
œÊ
(b) f ( increasing on ( ), decreasing on (0 3)
w
œ ± Ê $ß _ ß
! $
(c) No local maximum and a local minimum at x 3œ
9. (a) g(t) t 3t 3 g (t) 2t 3 a critical point at t ; g , increasing onœ Ê œ Ê œ œ ±
$Î#
#w w
#
3
, decreasing on ˆ‰ ˆ‰_ß ß _
33
##
(b) local maximum value of g at tˆ‰œ œ
321 3
4##
(c) absolute maximum is at t
21 3
4
œ
#
(d)
10. (a) g(t) 3t 9t 5 g (t) 6t 9 a critical point at t ; g , increasing onœ Ê œ Ê œ œ ±
$Î#
#w w
#
3
, decreasing on ˆ‰ ˆ‰_ß ß _
33
2 #
(b) local maximum value of g at tˆ‰
347 3
4##
œœ
(c) absolute maximum is at t
47 3
4
œ
#
(d)
11. (a) h(x) x 2x h (x) 3x 4x x(4 3x) critical points at x 0, œ Ê œ œ Ê œ
$# w # 4
3
h , increasing on 0 , decreasing on ( ) and Ê œ ± ± ß _ß ! ß _
! %Î$
w ˆ‰ ˆ ‰
44
33
(b) local maximum value of h at x ; local minimum value of h(0) 0 at x 0ˆ‰
432 4
327 3
œœ œœ
(c) no absolute extrema

221. Section 4.3 Monotonic Functions and the First Derivative Test 217
(d)
12. (a) h(x) 2x 18x h (x) 6x 18 6 x 3 x 3 critical points at x 3œ Ê œœ Ê œ„
$w#
Š‹Š‹
ÈÈ È
h | | , increasing on 3 and , decreasing on 3Ê œ _ß $ß _ $ß
$ $
w
ÈÈ
Š‹Š‹ Š‹
ÈÈ ÈÈ
(b) a local maximum is h 3 12 3 at x 3; local minimum is h 3 12 3 at x 3Š‹ Š‹
ÈÈ È È ÈÈ
œ œ œ œ
(c) no absolute extrema
(d)
13. (a) f( ) 3 4 f ( ) 6 12 6 (1 2 ) critical points at 0, f ,))) )))) ) )œ Ê œ œ Ê œ Ê œ ± ±
! "Î#
#$ w # w "
#
increasing on 0 , decreasing on ( ) and ˆ‰ ˆ ‰ß _ß ! ß _
""
##
(b) a local maximum is f at , a local minimum is f(0) 0 at 0ˆ‰
"" "
##
œœ œœ
4
))
(c) no absolute extrema
(d)
14. (a) f( ) 6 f ( ) 6 3 3 2 2 critical points at 2 ))) ) ) ) ) )œ Ê œ œ Ê œ„ Ê
$w #
Š‹Š‹
ÈÈ È
f , increasing on 2 2 , decreasing on 2 and 2
w
œ ± ± ß _ß ß _
# #
ÈÈ
Š‹ Š ‹Š‹
ÈÈ È È
(b) a local maximum is f 2 4 2 at 2, a local minimum is f 2 2 at 2Š‹ Š ‹
ÈÈ È È È È
œœ œ%œ))
(c) no absolute extrema

222. 218 Chapter 4 Applications of Derivatives
(d)
15. (a) f(r) 3r 16r f (r) 9r 16 no critical points f , increasing on ( ), neverœ Ê œ Ê Ê œ _ß _
$w# w
decreasing
(b) no local extrema
(c) no absolute extrema
(d)
16. (a) h(r) (r 7) h (r) 3(r 7) a critical point at r 7 h , increasing onœ Ê œ Ê œ Ê œ ±
(
$w # w
( 7) ( ), never decreasing_ß (ß _
(b) no local extrema
(c) no absolute extrema
(d)
17. (a) f(x) x 8x 16 f (x) 4x 16x 4x(x 2)(x 2) critical points at x 0 and x 2œ Ê œ œ Ê œ œ„
%# w $
f , increasing on ( ) and ( ), decreasing on ( 2) and ( )Ê œ ± ± ± #ß ! #ß _ _ß !ß #
# ! #
w
(b) a local maximum is f(0) 16 at x 0, local minima are f 2 0 at x 2œœ „œœ„ ab
(c) no absolute maximum; absolute minimum is 0 at x 2œ„
(d)

223. Section 4.3 Monotonic Functions and the First Derivative Test 219
18. (a) g(x) x 4x 4x g (x) 4x 12x x 4x(x 2)(x 1) critical points at x 0, 1, 2œ Ê œ )œ Ê œ
%$# w $ #
g , increasing on (0 1) and ( ), decreasing on ( 0) and (1 )Ê œ ± ± ± ß #ß _ _ß ß #
!"#
w
(b) a local maximum is g(1) 1 at x 1, local minima are g(0) 0 at x 0 and g(2) 0 at x 2œœ œœ œœ
(c) no absolute maximum; absolute minimum is 0 at x 0, 2œ
(d)
19. (a) H(t) t t H (t) 6t 6t 6t (1 t)( t) critical points at t 0, 1œ Ê œœ "Ê œ„
3
#
%' w $ & $
H , increasing on ( 1) and (0 1), decreasing on ( 0) and ( )Ê œ ± ± ± _ß ß "ß "ß _
" ! "
w
(b) the local maxima are H( 1) at t 1 and H(1) at t 1, the local minimum is H(0) 0 at t 0œ œ œ œ œ œ
""
##
(c) absolute maximum is at t 1; no absolute minimum
" #
œ„
(d)
20. (a) K(t) 15t t K (t) 45t 5t 5t (3 t)(3 t) critical points at t 0, 3œÊ œœ Ê œ„
$& w # % #
K , increasing on ( 3 0) (0 3), decreasing on ( 3) and (3 )Ê œ ± ± ± ß ß _ß ß _
$ ! $
w
(b) a local maximum is K(3) 162 at t 3, a local minimum is K( 3) 162 at t 3œœ œœ
(c) no absolute extrema
(d)
21. (a) g(x) x 8 x x 8 x g (x) 8 x x x ( 2x)œœ Ê œ ) œ
È
ab ab ab ˆ‰
# #w # #
"ßÊ "ßÊ œ"ßÊ "
#


2(2 x)(2 x)
22 x 22 xÊŠ‹Š‹
ÈÈ
critical points at x 2, 2 2 g ( , increasing on ( ), decreasing onÊ œ „ „ Ê œ ± ± Ñ #ß #
# #
# #
##
È
È È
w
2 and 2Š‹Š‹
ÈÈ
# ß# #ß#

224. 220 Chapter 4 Applications of Derivatives
(b) local maxima are g(2) 4 at x 2 and g 2 2 0 at x 2 2, local minima are g( 2) 4 atœœ œœ œ Š‹
ÈÈ
x 2 and g 2 2 0 at x 2 2œ œ œŠ‹
ÈÈ
(c) absolute maximum is 4 at x 2; absolute minimum is 4 at x 2œœ
(d)
22. (a) g(x) x 5 x x (5 x) g (x) 2x(5 x) x (5 x) ( 1)œœÊœ œ
# # "Î# w "Î# # "Î# "
#
œ
œÈ ˆ‰
5x(4 x)
25x
È
critical points at x 0, 4 and 5 g , increasing on (0 4), decreasing on ( )Ê œ Ê œ ± ± Ñ ß _ß !
! % &
w
and ( )%ß &
(b) a local maximum is g(4) 16 at x 4, a local minimum is 0 at x 0 and x 5œœ œ œ
(c) no absolute maximum; absolute minimum is 0 at x 0, 5œ
(d)
23. (a) f(x) f (x) critical points at x 1, 3œÊœ œ Ê œ
x3
x(x2)(x)
2x(x 2) x 3 (1) (x 3)(x )
œ œ
œœ
œ
œÊ œ œÊ
w œœ œ œœ"ab
f )( , increasing on ( 1) and ( ), decreasing on ( ) and ( ),Ê œ ± ± _ß $ß _ "ß # #ß $
"$ #
w
discontinuous at x 2œ
(b) a local maximum is f(1) 2 at x 1, a local minimum is f(3) 6 at x 3œœ œœ
(c) no absolute extrema (d)
24. (a) f(x) f (x) a critical point at x 0œÊœ œ Ê œ
x
3x 1
3x 3x 1 x (6x) 3x x 1
3x 1 3x 1
w
œ

œœ
œœ

w

ab ab
ab ab
f , increasing on ( ) ( ), and never decreasingÊ œ ± _ß ! !ß _
!
w
(b) no local extrema
(c) no absolute extrema

225. Section 4.3 Monotonic Functions and the First Derivative Test 221
(d)
25. (a) f(x) x (x 8) x 8x f (x) x x critical points at x 0, 2œœÊœ œ Ê œ
"Î$ %Î$ "Î$ w "Î$ #Î$ 48
33
4(x 2)
3x
œ#w
f )( , increasing on ( ) ( ), decreasing on ( 2)Ê œ ± #ß ! !ß _ _ß
# !
w
(b) no local maximum, a local minimum is f( 2) 6 2 7.56 at x 2œ ¸ œ
$È
(c) no absolute maximum; absolute minimum is 6 2 at x 2œ
$È
(d)
26. (a) g(x) x (x 5) x 5x g (x) x x critical points at x 2 andœœÊœ œ Ê œ
#Î$ &Î$ #Î$ w #Î$ "Î$ 510
33
5(x 2)
3x
w
È
x 0 g )( , increasing on ( 2) and ( ), decreasing on ( 2 )œ Ê œ ± _ß !ß _ ß !
# !
w
(b) local maximum is g( 2) 3 4 4.762 at x 2, a local minimum is g(0) 0 at x 0œ ¸ œ œ œ
$È
(c) no absolute extrema
(d)
27. (a) h(x) x x 4 x 4x h (x) x x critical points atœœÊœœ Ê
"Î$ # (Î$ "Î$ w %Î$ #Î$
#
ab
74
33
7x 2 7x
3x
Š‹Š‹
ÈÈ
È
wœ
x 0, h )( , increasing on and , decreasing onœ Ê œ ± ± _ß ß _
#Î ( #Î (
!
„ œw2 22
7 77
È ÈÈ
ÈÈ
Š‹Š‹
and Š‹Š‹
œ22
77
ÈÈ
ß! !ß
(b) local maximum is h 3.12 at x , the local minimum is h 3.12Š‹ Š‹
œœ222
77 7
24 2 24 2
7 7ÈÈ È
È È
œ¸ œ œ¸
w w
$#Î $#Î
(c) no absolute extrema

226. 222 Chapter 4 Applications of Derivatives
(d)
28. (a) k(x) x x 4 x 4x k (x) x x critical points atœœÊœœ Ê
#Î$ # )Î$ #Î$ w &Î$ "Î$
ab
88
33
8(x 1)(x 1)
3x
w
È
x 0, 1 k )( , increasing on ( ) and ( ), decreasing on ( 1)œ „ Ê œ ± ± "ß ! "ß _ _ß
" "!
w
and ( 1)!ß
(b) local maximum is k(0) 0 at x 0, local minima are k 1 3 at x 1œœ „œœ„ ab
(c) no absolute maximum; absolute minimum is 3 at x 1œ„
(d)
29. (a) f(x) 2x x f (x) 2 2x 2(1 x) a critical point at x 1 f ] and f(1) 1,œ Ê œ œ Ê œ Ê œ ± œ
" #
#w w
f(2) 0 a local maximum is 1 at x 1, a local minimum is 0 at x 2œÊ œ œ
(b) absolute maximum is 1 at x 1; no absolute minimumœ
(c)
30. (a) f(x) (x 1) f (x) 2(x 1) a critical point at x 1 f ] and f( 1) 0, f(0) 1œ Ê œ Ê œ Ê œ ± œ œ
" !
#w w
a local maximum is 1 at x 0, a local minimum is 0 at x 1Êœ œ
(b) no absolute maximum; absolute minimum is 0 at x 1œ
(c)

227. Section 4.3 Monotonic Functions and the First Derivative Test 223
31. (a) g(x) x 4x 4 g (x) 2x 4 2(x 2) a critical point at x 2 g [ andœ Ê œ œ Ê œ Ê œ ±
" #
#w w
g(1) 1, g(2) 0 a local maximum is 1 at x 1, a local minimum is g(2) 0 at x 2œœÊ œ œœ
(b) no absolute maximum; absolute minimum is 0 at x 2œ
(c)
32. (a) g(x) x 6x 9 g (x) 2x 6 2(x 3) a critical point at x 3 g [ andœ Ê œ œ Ê œ Ê œ ±
% $
#w w
g( 4) 1, g( 3) 0 a local maximum is 0 at x 3, a local minimum is 1 at x 4œ œ Ê œ œ
(b) absolute maximum is 0 at x 3; no absolute minimumœ
(c)
33. (a) f(t) 12t t f (t) 12 3t 3(2 t)(2 t) critical points at t 2 f [œ Ê œ œ Ê œ „ Ê œ ± ±
$ # #
$w # w
and f( 3) 9, f( 2) 16, f(2) 16 local maxima are 9 at t 3 and 16 at t 2, a localœ œ œ Ê œ œ
minimum is 16 at t 2œ
(b) absolute maximum is 16 at t 2; no absolute minimumœ
(c)
34. (a) f(t) t 3t f (t) 3t 6t 3t(t 2) critical points at t 0 and t 2œ Ê œ œ Ê œ œ
$# w #
f ] and f(0) 0, f(2) 4, f(3) 0 a local maximum is 0 at t 0 and t 3, aÊ œ ± ± œ œ œ Ê œ œ
!# $
w
local minimum is 4 at t 2œ
(b) absolute maximum is 0 at t 0, 3; no absolute minimumœ

228. 224 Chapter 4 Applications of Derivatives
(c)
35. (a) h(x) 2x 4x h (x) x 4x 4 (x 2) a critical point at x 2 h [ andœ Ê œ œ Ê œ Ê œ ±
! #
x
3
œ
#w# # w
h(0) 0 no local maximum, a local minimum is 0 at x 0œÊ œ
(b) no absolute maximum; absolute minimum is 0 at x 0œ
(c)
36. (a) k(x) x 3x 3x 1 k (x) 3x 6x 3 3(x 1) a critical point at x 1œ Ê œ œ Ê œ
$# w # #
k ] and k( 1) 0, k(0) 1 a local maximum is 1 at x 0, no local minimumÊ œ ± œ œ Ê œ
" !
w
(b) absolute maximum is 1 at x 0; no absolute minimumœ
(c)
37. (a) f(x) 2 sin f (x) cos , f (x) 0 cos a critical point at xœ Ê œ œ Ê œ Ê œ
xx x x 2
3## ## ##
ww ""
ˆ‰ ˆ‰ ˆ‰
1
f [ ] and f(0) 0, f 3, f(2 ) local maxima are 0 at x 0 and Ê œ ± œ œ œ Ê œ
!##Î$
w
1 1
11 1ˆ‰ È
2
33
11
at x 2 , a local minimum is 3 at xœœ1
11
33
2È
(b) The graph of f rises when f 0, falls when f 0,
ww

and has a local minimum value at the point where f
w
changes from negative to positive.

229. Section 4.3 Monotonic Functions and the First Derivative Test 225
38. (a) f(x) 2 cos x cos x f (x) 2 sin x 2 cos x sin x 2(sin x)(1 cos x) critical points atœ Ê œ œ Ê
#w
x , 0, f [ ] and f( ) 1, f(0) 3, f( ) 1 a local maximum is 1 atœ Ê œ ± œ œ œ Ê
Î
!
11 1 1
11
w
x , a local minimum is 3 at x 0œ„ œ1
(b) The graph of f rises when f 0, falls when f 0,
ww

and has local extreme values where f 0. The
w
œ
function f has a local minimum value at x 0, whereœ
the values of f change from negative to positive.
w

39. (a) f(x) csc x 2 cot x f (x) 2(csc x)( csc x)(cot x) 2 csc x 2 csc x (cot x 1) a criticalœ Êœ œ Ê
#w ##
abab
point at x f ( ) and f 0 no local maximum, a local minimum is 0 at xœ Ê œ ± œ Ê œ
! Î%
11 1
44 4
w
1
1
ˆ‰
(b) The graph of f rises when f 0, falls when f 0,
ww

and has a local minimum value at the point where
f 0 and the values of f change from negative to
ww
œ
positive. The graph of f steepens as f (x) .
w
Ä„_

40. (a) f(x) sec x 2 tan x f (x) 2(sec x)(sec x)(tan x) 2 sec x 2 sec x (tan x 1) a critical pointœ Êœ œ Ê
#w ##
ab
at x f ( ) and f 0 no local maximum, a local minimum is 0 at xœ Ê œ ± œ Ê œ
Î# Î# Î%
11 1
44 4
w
11 1
ˆ‰
(b) The graph of f rises when f 0, falls when f 0,
ww

and has a local minimum value where f 0 and the
w
œ
values of f change from negative to positive.
w

41. h( ) 3 cos h ( ) sin h [ ] , ( ) and ( 3) a local maximum is 3 at 0,)) 1 )
1
œ Ê œ Ê œ !ß $ # ß Ê œ
!#
ˆ‰ ˆ‰
))
###
ww 3
a local minimum is 3 at 2œ)1
42. h( ) 5 sin h ( ) cos h [ ] , ( 0) and ( 5) a local maximum is 5 at , a local)) 1 )1
1
œ Ê œ Ê œ !ß ß Ê œ
!
ˆ‰ ˆ‰
))
###
ww 5
minimum is 0 at 0)œ

230. 226 Chapter 4 Applications of Derivatives
43. (a) (b) (c) (d)
44. (a) (b) (c) (d)
45. (a) (b)
46. (a) (b)
47. f(x) x 3x 2 f (x) 3x 3 3(x 1)(x 1) f rising for x c sinceœ Ê œ œ Ê œ ± ± Ê œ œ #
" "
$w# w
f (x) 0 for x c 2.
w
œœ
48. f(x) ax bx c a x x c a x x c a x , a parabola whoseœ œ œ œ
## #
#
œ ˆ‰ ˆ‰ Š‹
bbbbbb 4ac
a a 4a 4a 2a 4a
ÎÎ Î
Î
vertex is at x . Thus when a 0, f is increasing on and decreasing on ; when a 0,œ ß _ _ß
bbb
2a 2a a
ˆ‰ ˆ ‰
œœ
#
f is increasing on and decreasing on . Also note that f (x) 2ax b 2a x forˆ‰ ˆ‰ ˆ‰_ß ß _ œ œ Ê
œœ
## #
wbb b
aa a
a 0, f | ; for a 0, f . œ œ ±
Î Î
ww
b2a b2aÎ Î
4.4 CONCAVITY AND CURVE SKETCHING
1. y 2x y x x 2 (x 2)(x 1) y 2x 1 2 x . The graph is rising onœÊ œœ Ê œ œ
xx
33
œ Î
# #
" "w# ww
ˆ‰
( 1) and ( ), falling on ( ), concave up on and concave down on . Consequently,_ß #ß _ "ß # ß _ _ß ˆ‰ ˆ ‰
""
##
a local maximum is at x 1, a local minimum is 3 at x 2, and is a point of inflection.
3 3
4##
"
œ œ ß ˆ‰

231. Section 4.4 Concavity and Curve Sketching 227
2. y 2x 4 y x 4x x x 4 x(x 2)(x 2) y 3x 4 3x 2 3x 2 . Theœ Ê œ œ œ Ê œ œ
x
4
%
#w$ # ww#
ab Š‹Š‹
ÈÈ
graph is rising on ( 2 0) and ( ), falling on ( ) and ( ), concave up on and ß #ß _ _ß # !ß # _ß ß _ Š‹Š‹
22
33
ÈÈ
and concave down on . Consequently, a local maximum is 4 at x 0, local minima are 0 atŠ‹ß œ
22
33
ÈÈ
x 2, and and are points of inflection.œ„ ß ßŠ‹Š‹
216 216
3399ÈÈ
3. y x 1 y x 1 (2x) x x 1 , y ) ( )(œ Ê œ œ œ ±
" " !
332
443
ab ab ab ˆ‰ˆ‰
#w# #w
#Î$ "Î$ "Î$
the graph is rising on ( ) and ( ), falling on ( ) and ( ) a local maximum is at x 0, localÊ "ß! "ß_ _ß" !ß" Ê œ
3
4
minima are 0 at x 1; y x 1 (x) x 1 (2x) ,œ„ œ œ
ww # #
"Î$ %Î$ "œ
œ
ab ab ˆ‰
3
x3
3 x 1
Î
œ Î
%
Éab
y ) ( )( the graph is concave up on 3 and 3 , concave
ww
œ ± ± Ê _ß ß _
$ $
" "ÈÈ
Š‹Š‹
ÈÈ
down on 3 3 points of inflection at 3Š‹ Š ‹
ÈÈ È
ß Ê „ ß
$%
%
È
œ
4. y x x 7 y x x 7 x (2x) x x 1 , y )(œ Ê œ œ œ ± ±
" "!
9393
14 14 14
"Î$ # w #Î$ # "Î$ #Î$ # w
#
ab ab ab
the graph is rising on ( 1) and ( ), falling on ( 1 ) a local maximum is at x 1, a localÊ_ß"ß_ß"Ê œ
27
7
minimum is at x 1; y x x 1 3x 2x x x 2x 1 , œœ œœ
27
7
ww &Î$ # "Î$ "Î$ &Î$ &Î$ #
ab a b
y )( the graph is concave up on ( ), concave down on ( ) a point of inflection at
ww
œ Ê !ß _ _ß ! Ê
!
()!ß !
5. y x sin 2x y 1 2 cos 2x, y [ ] the graph is rising on , fallingœ Ê œ œ ± ± Ê ß
# Î$ # Î$Î$ Î$
ww
11 11
ˆ‰
11
33
on and local maxima are at x and at x , local minima areˆ‰ˆ‰ß ß Ê œ œ
#
##
11 11 1 1 1 1
33 33 3 3 3 3
222 33
ÈÈ
at x and at x ; y 4 sin 2x, y [ ] the œ œ œ œ ± ± ± Ê
# Î$ # Î$Î# Î# !
1111
3333
33 22
ÈÈ
##
ww ww
11 11
graph is concave up on and , concave down on and points of inflection atˆ‰ˆ‰ ˆ ‰ˆ‰ß! ß ß !ß Ê
111 111
## ##
22
33
, ( ), and ˆ‰ ˆ‰ß !ß! ß
11 11
## ##
6. y tan x 4x y sec x 4, y ( ) the graph is rising on andœ Ê œ œ ± ± Ê ß
Î# Î#Î$ Î$
w# w
11 11
ˆ‰
11
23
, falling on a local maximum is 3 at x , a local minimum is 3 at x ;ˆ‰ ˆ ‰ ÈÈ
11 11 1 1 1 1
333 33 33
44
ßßÊ œ œ
#
y 2(sec x)(sec x)(tan x) 2 sec x (tan x), y ( ) the graph is concave up on 0 ,
ww # ww
#
œ œ œ ± Ê ß
Î# Î# !
ab ˆ‰
11
1
concave down on 0 a point of inflection at (0 )ˆ‰ß Ê ß!
1
2
7. If x 0, sin x sin x and if x 0, sin x sin ( x) œ œkk kk
sin x. From the sketch the graph is rising onœ
, and , falling on 2 ,ˆ‰ˆ‰ˆ‰ ˆ ‰ß !ß ß# ß
33 311 1 1 1
## # # #
11
and ; local minima are 1 at xˆ‰ˆ‰ß! ß œ„
111 1
### #
33
and 0 at x ; local maxima are 1 at x andœ! œ „
1
#
0 at x ; concave up on ( ) and ( ), andœ„# #ß ß#11111
concavedown on ( 0) and ( ) points of inflection ß !ß Ê11
are ( ) and ( )ß! ß!11

232. 228 Chapter 4 Applications of Derivatives
8. y 2 cos x 2 x y 2 sin x 2, y [ ] rising onœ Ê œ œ ± ± ± Ê
Î
$ Î% Î% & Î% $Î#
ÈÈ
ww
1
111 1
and , falling on and local maxima are 2 2 at x , 2ˆ‰ˆ‰ ˆ‰ˆ‰ ÈÈ
ß ß ß ß Ê œ
353 3 5
44 4 4 44 4
211 11 1 11 1
#
111
È
at x and at x , and local minima are 2 at x and 2 at x ;œ œ œ œ
11 1 1 111
4 44 44
32 32 52 335
ÈÈÈ
## ÈÈ
y 2 cos x, y [ ] concave up on and , concave down on
ww ww
###
œ œ ± ± Ê ß ß
Î
Î# Î# $Î#
1
11 1
1ˆ‰ˆ‰
111 3
points of inflection at and ˆ‰ Š‹Š‹ß Ê ß ß
11 1 1 11
## # # # #
ÈÈ
22
9. When y x 4x 3, then y 2x 4 2(x 2) andœ œœ
#w
y 2. The curve rises on ( ) and falls on ( ).
ww
œ #ß_ _ß#
At x 2 there is a minimum. Since y 0, the curve isœ
ww
concave up for all x.

10. When y 2x x , then y 2x 2( x) andœ' œ# œ "
#w
y 2. The curve rises on ( 1) and falls on
ww
œ _ß
( 1 ). At x 1 there is a maximum. Since y 0, theß_ œ
ww
curve is concave down for all x.

11. When y x 3x 3, then y 3x 3 3(x 1)(x 1)œ œ œ
$w#
and y 6x. The curve rises on ( 1) ( ) and
ww
œ _ß "ß _
falls on ( 1 1). At x 1 there is a local maximum and atß œ
x 1 a local minimum. The curve is concave down onœ
( 0) and concave up on ( ). There is a point of_ß !ß _
inflection at x 0.œ

12. When y x(6 2x) , then y 4x(6 2x) ( 2x)œ œ '
#w #
12(3 x)( x) and y 12(3 x) 12( x)œ " œ "
ww
24(x 2). The curve rises on ( ) ( ) andœ _ß" $ß_
falls on ( ). The curve is concave down on ( ) and"ß $ _ß #
concave up on ( ). At x 2 there is a point of#ß _ œ
inflection.

233. Section 4.4 Concavity and Curve Sketching 229
13. When y 2x 6x 3, then y 6x 12xœ œ
$# w #
6x(x 2) and y 12x 12 12(x 1). Theœ œ œ
ww
curve rises on ( ) and falls on ( 0) and ( ).!ß # _ß #ß _
At x 0 there is a local minimum and at x 2 a localœœ
maximum. The curve is concave up on ( ) and_ß "
concave down on ( ). At x 1 there is a point of"ß _ œ
inflection.

14. When y 1 9x 6x x , then y 9 12x 3xœ œ
#$ w #
(x 3)( 1) and y 12 6x 6(x 2).œ$ B œ œ
ww
The curve rises on ( ) and falls on ( 3) and$ß " _ß
( ). At x 1 there is a local maximum and at"ß _ œ
x 3 a local minimum. The curve is concave up onœ
( 2) and concave down on ( ). At x 2_ß #ß _ œ
there is a point of inflection.

15. When y (x 2) 1, then y 3(x 2) andœ œ
$w#
y 6(x 2). The curve never falls and there are no
ww
œ
local extrema. The curve is concave down on ( )_ß #
and concave up on ( ). At x 2 there is a point#ß _ œ
of inflection.

16. When y 1 (x 1) , then y 3(x 1) andœ œ
$w #
y 6(x 1). The curve never rises and there are
ww
œ
no local extrema. The curve is concave up on ( 1)_ß
and concave down on ( ). At x 1 there is a"ß _ œ
point of inflection.

234. 230 Chapter 4 Applications of Derivatives
17. When y x 2x , then y 4x 4x 4x(x 1)(x 1)œ œ œ
%# w $
and y 12x 4 12 x x . The curve
ww # ""
œœ Š‹Š‹
ÈÈ
33
rises on ( ) and ( ) and falls on ( 1) and ( )."ß ! "ß _ _ß !ß "
At x 1 there are local minima and at x 0 a localœ„ œ
maximum. The curve is concave up on andŠ‹_ß
"
È
3
and concave down on . At xŠ‹ Š ‹
"""„"
ÈÈÈÈ
3333
ß_ ß œ
there are points of inflection.

18. When y x 6x 4, then y 4x 12xœ œ
%# w $
4x x 3 x 3 and y 12x 12œ œ Š‹Š‹
ÈÈ
ww #
12(x 1)(x 1). The curve rises on 3œ _ß Š‹
È
and 3 , and falls on 3 and 3 . AtŠ‹ Š ‹Š ‹
ÈÈÈ
!ß ß ! ß _
x 3there are local maxima and at x 0 a localœ„ œ
È
minimum. The curve is concave up on ( ) and concave"ß "
down on ( 1) and ( ). At x 1 there are points_ß "ß _ œ „
of inflection.

19. When y 4x x , then y 12x 4x 4x ( x) andœ œ œ$
$% w # $ #
y 24x 12x 12x(2 x). The curve rises on
ww #
œ œ _ß$ ab
and falls on . At x 3 there is a local maximum, butab$ß _ œ
there is no local minimum. The graph is concave up on
and concave down on and . There areab a b a b!ß # _ß ! #ß _
inflection points at x 0 and x 2.œœ

20. When y x 2x , then y 4x 6x 2x (2x 3) andœ œœ
%$ w $# #
y 12x 12x 12x(x 1). The curve rises on
ww #
œœ
and falls on . There is a localˆ‰ ˆ ‰ ß _ _ß
33
2#
minimum at x , but no local maximum. The curve isœ
3
#
concave up on ( 1) and ( ), and concave down on_ß !ß _
( 1 0). At x 1 and x 0 there are points of inflection.ß œ œ

235. Section 4.4 Concavity and Curve Sketching 231
21. When y x 5x , then y 5x 20x 5x (x 4) andœ œ œ
&% w % $ $
y 20x 60x 20x (x 3). The curve rises on
ww $ # #
œœ
( ) and ( ), and falls on ( ). There is a local_ß! %ß_ !ß%
maximum at x 0, and a local minimum at x 4. Theœœ
curve is concave down on ( 3) and concave up on*()
(3 ). At x 3 there is a point of inflection.ß_ œ

22. When y x 5 , then y 5 x(4) 5œ œ ˆ‰ ˆ‰ ˆ‰ˆ‰
xxx
####
%%$
w "
5 5 , and y 3 5 5œ œ ˆ‰ˆ ‰ ˆ‰ˆ‰ˆ ‰
x5x x 5x
## ###
$#
ww "
5 5 5 (x 4). The curve is rising œ ˆ‰ˆ‰ˆ‰
x5x
###
$#
on ( ) and (10 ), and falling on ( 10). There is a_ß# ß_ #ß
local maximum at x 2 and a local minimum at x 10.œœ
The curve is concave down on ( ) and concave up on_ß %
( ). At x 4 there is a point of inflection.%ß _ œ

23. When y x sin x, then y cos x and y sin x.œ œ" œ
www
The curve rises on ( 2 ). At x 0 there is a local and!ß œ1
absolute minimum and at x 2 there is a local and absoluteœ1
maximum. The curve is concave down on ( ) and concave!ß1
up on ( ). At x there is a point of inflection.11 1ß# œ

24. When y x sin x, then y cos x and y sin x.œ œ" œ
www
The curve rises on ( 2 ). At x 0 there is a local and!ß œ1
absolute minimum and at x 2 there is a local and absoluteœ1
maximum. The curve is concave up on ( ) and concave!ß1
down on ( ). At x there is a point of inflection.11 1ß# œ

25. When y x , then y x and y x .œœ œ
"?& w œ%?& ww œ?&"
525
4
The curve rises on ( ) and there are no extrema.*() (
The curve is concave up on ( ) and concave down*() '
on ( ). At x 0 there is a point of inflection.!ß _ œ

236. 232 Chapter 4 Applications of Derivatives
26. When y x , then y x and y x .œœ œ
$?& w œ#?& ww œ_?&36
525
The curve rises on ( ) and there are no extrema.*() (
The curve is concave up on ( ) and concave down*() '
on ( ). At x 0 there is a point of inflection.!ß _ œ

27. When y x , then y x and y x .œœ œ
#?& w œ$?& ww œß?&26
525
The curve is rising on (0 ) and falling on ( ). At)( *()'
x 0 there is a local and absolute minimum. There isœ
no local or absolute maximum. The curve is concave
down on ( ) and ( ). There are no points of*() ' ') (
inflection, but a cusp exists at x 0.œ

28. When y x , then y x and y x .œœ œ
%?& w œ"?& ww œ!?&44
525
The curve is rising on (0 ) and falling on ( ). At)( *()'
x 0 there is a local and absolute minimum. There isœ
no local or absolute maximum. The curve is concave
down on ( ) and ( ). There are no points of*() ' ') (
inflection, but a cusp exists at x 0.œ

29. When y 2x 3x , then y 2 2x andœ œ
#?$ w œ"?$
y x . The curve is rising on ( ) and
ww œ%?$
œ _ß !
2
3
( ), and falling on ( ). There is a local maximum"ß _ !ß "
at x 0 and a local minimum at x 1. The curve isœœ
concave up on ( ) and ( ). There are no*() ' ') (
points of inflection, but a cusp exists at x 0.œ

30. When y 5x 2x, then y 2x 2 2 x 1œ œ œ
#?& w œ$?& œ$?& ˆ‰
and y x . The curve is rising on (0 1) and
ww ϧ?&
œ ß
6
5
falling on ( 0) and ( ). There is a local minimum*() ) (
at x 0 and a local maximum at x 1. The curve isœœ
concave down on ( ) and ( ). There are no*() ' ') (
points of inflection, but a cusp exists at x 0.œ

237. Section 4.4 Concavity and Curve Sketching 233
31. When y x x x x , thenœœ
#Î$ #Î$ &Î$
##ˆ‰
55
y x x x (1 x) and
w œ"?$ #?$ œ"?$
œœ
555
333
y x x x (1 2x).
ww œ%?$ œ"?$ œ%?$
œ œ
510 5 99 9
The curve is rising on ( ) and falling on ( ) and') *() '
( ). There is a local minimum at x 0 and a local"ß _ œ
maximum at x 1. The curve is concave up on œ _ß ˆ‰
"
#
and concave down on and (0 ). There is a pointˆ‰*)' )(
"
#
of inflection at x and a cusp at x 0.œ œ
"
#

32. When y x (x 5) x 5x , thenœœ
#Î$ &Î$ #Î$
y x x x (x 2) and
w #?$ œ"?$ œ"?$
œ œ
510 5
33 3
y x x x (x 1). The curve
ww œ"?$ œ%?$ œ%?$
œœ
10 10 10
99 9
is rising on ( ) and ( ), and falling on ( )._ß! #ß_ !ß#
There is a local minimum at x 2 and a local maximumœ
at x 0. The curve is concave up on ( 0) and ( ),œ "ß !ß _
and concave down on ( 1). There is a point of*() *
inflection at x 1 and a cusp at x 0.œ œ
33. When y x 8 x x 8 x , thenœœ
È
ab
# #
"Î#
y 8 x (x) 8 x ( x)
w# #
"?# œ"?# "
#
œ #ab ab ˆ‰
8x 82x andœ œaba b
##
œ"?# œ%
%œ
2(2 x)(2 x)
22 x 22 xÊŠ‹Š‹
ÈÈ
y 8 x ( 2x) 8 2x 8 x ( 4x)
ww # # #"
#
œœ
œˆ‰ab a bab

__
. The curve is rising on ( ), and fallingœ #ß #
2x x 12
8x
ab
Éab
_
_
œ
œ
œ
on 2 2 2 and 2 2 . There are local minimaŠ‹Š‹
ÈÈ
ß #ß
x 2 and x 2 2, and local maxima at x 2 2 andœ œ œ
ÈÈ
x 2. The curve is concave up on 2 2 andœß! Š‹
È
concave down on 2 . There is a point of inflectionŠ‹
È
!ß #
at x 0.œ

34. When y 2 x , then y 2 x ( 2x)œ œ ab ab ˆ‰
#w #
$Î# "Î#
#
3
3x2x 3x 2x 2x andœ œ
È
ÊŠ‹Š‹
ÈÈ
#
y ( 3) 2 x ( 3x) 2 x ( 2x)
ww # #
"?# œ"?# "
#
œ ab ab ˆ‰
. The curve is rising onœ
œ"œ %
œ%
6( x)(1 x)
2x 2xÊŠ‹Š‹
ÈÈ
2 and falling on 2 . There is a localŠ‹ Š‹
ÈÈ
*)' ')
maximum at x 0, and local minima at x 2. Theœœ„
È
curve is concave down on ( ) and concave up on* )
2 and 2 . There are points of inflection atŠ‹Š‹
ÈÈ
*)* )
x1.œ„

238. 234 Chapter 4 Applications of Derivatives
35. When y , then yœœ
x3
x(x2)
2x(x 2) x 3 ( )
_ _
_
œ
œ# œ
w œœ œ "ab
andœ
(x 3)(x 1)
(x 2)
œœ
œ
_
y.
ww œœœœ%#œ
œ œ
œœ
(2x 4)(x 2) x 4x 3 (x 2)
(x 2)
2
(x 2)
__
ß œ
ab
The curve is rising on ( ) and ( ), and falling on_ß " $ß _
( ) and ( ). There is a local maximum at x 1 and a"ß # #ß $ œ
local minimum at x 3. The curve is concave down onœ
( ) and concave up on ( ). There are no points_ß # #ß _
of inflection because x 2 is not in the domain.œ

36. When y , then yœœ
x
3x 1
3x 3x 1 x (6x)
3x 1
œ
_
__ œ
_
_
%
w %œ
%
ab
ab andœ
3x x 1
3x 1
__
_
_
ab
ab
%
% y
ww
%%"œ%" %
%
œ
a babab ˆ‰
ab
12x 6x 3x 2 3x (6x) 3x 3x
3x 1
œ_ _ ß_
_
_
ß . The curve is rising on ( ) soœ _ß _
6x(1 x)( x)
3x 1
œ"%
%ab
_
œ
there are no local extrema. The curve is concave up on
( ) and ( ), and concave down on ( ) and*()* ') * )'
( ). There are points of inflection at x , x 0,"ß _ œ " œ
and x 1.œ

37. When y x 1 , then
x 1, x 1
1x, x 1
œœ
*
*
kk œ
kk
kk
#
#
#
y and y . The
2x, x 1 2, x
2x, x , x
www
œœ
"
" #"
œœ
kk kk
kk kk
curve rises on ( ) and ( ) and falls on ( 1)* ) ' ) ( *() *
and (0 1). There is a local maximum at x 0 and localߜ
minima at x 1. The curve is concave up on ( 1)œ„ _ß
and ( ), and concave down on ( ). There are no ) ( * )
points of inflection because y is not differentiable at x 1œ„
(so there is no tangent line at those points).

38. When y x 2x , then
x 2x, x 0
2x x , 0 x 2
x 2x, x 2
œœ
*
ŸŸ

kk
Ú
Û
Ü
#
#
#
#
y , and y .
2x 2, x 0 2, x 0
2 2x, 0 x 2 2, 0 x 2
2x 2, x 2 2, x 2
www
œœ

* *

ÚÚ
ÛÛ
ÜÜ
The curve is rising on ( 1) and ( ), and falling on!ß #ß _
( ) and ( ). There is a local maximum at x 1 and_ß ! "ß # œ
local minima at x 0 and x 2. The curve is concave upœœ
on ( ) and ( ), and concave down on ( )._ß! #ß_ !ß#
There are no points of inflection because y is not
differentiable at x 0 and x 2 (so there is no tangentœœ
at those points).

239. Section 4.4 Concavity and Curve Sketching 235
39. When y x , then
x , x 0
x, x 0
œœ
 
*
Èkk
œ
È
È
y and y .
, x 0
, x 0
, x 0
, x 0
www
"
#
œ"
œ
œ
œœ
œœ




Ú
Û
Ü
œ
È
È
x
2x
x
4
(x)
4
!œ%_
!œ%_
Since lim y and lim y there is a
x xÄ! Ä!
œ *
ww
œ_ œ_
cusp at x 0. There is a local minimum at x 0, but noœœ
local maximum. The curve is concave down on ( )*() '
and ( ). There are no points of inflection.!ß _

40. When y x 4 , then
x 4 , x 4
4 x , x 4
œœ
*
*
Èkk
œ
È
È
y and y .
, x 4
, x 4
, x 4
, x 4
www
"
œ
œ"
#œ
œœ
œœ
œœ




Ú
Û
Ü
œ
2x 4
4x
(x 4)
4
(4 x)
4
È
È
!œ%_
!œ%_
Since lim y and lim y there is a cusp
x4 x4Ä Ä
œ *
ww
œ_ œ_
at x 4. There is a local minimum at x 4, but no localœœ
maximum. The curve is concave down on ( )_ß %
and ( ). There are no points of inflection.%ß _

41. y 2 x x (1 x)( x), y
w# w
œ œ # œ ± ±
" #
rising on ( ), falling on ( 1) and ( )Ê "ß # _ß #ß _
there is a local maximum at x 2 and a local minimumʜ
at x 1; y 1 2x, yœ œ œ ±
"Î#
ww ww
concave up on , concave down on Ê_ß ß_ ˆ‰ ˆ‰
""
##
a point of inflection at xʜ
"
#

42. y x x 6 (x 3)(x 2), y
w# w
œ œ œ ± ±
# $
rising on ( ) and (3 ), falling on ( 2 3)Ê_ß#ß_ ß
there is a local maximum at x 2 and a localʜ
minimum at x 3; y 2x 1, yœ œ œ ±
"Î#
ww ww
concave up on , concave down on Êß_ _ß ˆ‰ ˆ ‰
""
##
a point of inflection at xʜ
"
#

43. y x(x 3) , y rising on
w#w
œ œ ± ± Ê
!$
( ), falling on ( ) no local maximum, but there!ß _ _ß ! Ê
is a local minimum at x 0; y (x 3) x(2)(x 3)œœ
ww #
3(x 3)(x 1), y concaveœ œ ± ± Ê
"$
ww
up on ( ) and ( ), concave down on ( ) _ß" $ß_ "ß$ Ê
points of inflection at x 1 and x 3œœ

240. 236 Chapter 4 Applications of Derivatives
44. y x (2 x), y rising on
w# w
œ œ ± ± Ê
!#
( ), falling on (2 ) there is a local maximum at_ß# ß_ Ê
x 2, but no local minimum; y 2x(2 x) x ( 1)œœ
ww #
x(4 3x), y concave upœ œ ± ± Ê
! %Î$
ww
on , concave down on ( ) and pointsˆ‰ ˆ ‰!ß _ß! ß_ Ê
44
33
of inflection at x 0 and xœœ
4
3

45. y x x 12 x x 2 3 x 2 3 ,
w#
œœ ab Š‹Š‹
ÈÈ
y rising on
w
œ ± ± ± Ê
# $ # $
!ÈÈ
2 3 and 3 , falling on 3Š‹Š‹ Š ‹
ÈÈ È
ß! # ß_ _ß#
and 3 a local maximum at x 0, local minimaŠ‹
È
!ß # Ê œ
at x 3 ; y (1) x 12 (x)(2x)œ„# œ
È
ab
ww #
3(x 2)(x 2), yœ œ ± ±
# #
ww
concave up on ( ) and ( ), concave down onÊ _ß # #ß _
( ) points of inflection at x 2#ß # Ê œ „

46. y (x 1) (2x 3), y
w# w
œ œ ± ±
$Î# "
rising on , falling on no localÊ ß_ _ß Êˆ‰ ˆ ‰
33
##
maximum, a local minimum at x ;œ
3
#
y 2(x 1)(2x 3) (x 1) (2) 2(x 1)(3x 2),
ww #
œ œ
y concave up on
ww
œ ± ± Ê
#Î$ "
and ( ), concave down on ˆ‰ ˆ‰*()* )( * )
22
33
points of inflection at x and x 1Êœœ
2
3

47. y 8x 5x (4 x) x(8 5x)( x) ,
w## #
œ œ%ab
y rising on ,
w
œ ± ± ± Ê !ß
! )Î& %
ˆ‰
8
5
falling on ( ) and a local maximum at_ß! ß_ ʈ‰
8
5
x , a local minimum at x 0;œœ
8
5
y (8 10x)(4 x) 8x 5x (2)( x)( 1)
ww # #
œ % ab
4(4 x) 5x 16x 8 ,œ ab
#
y concave up
ww
ßœ# ! ß%# !
&&
œ ± ± ± Ê
%
ÈÈ
on and , concave down onŠ‹Š‹_ß ß %
826 826
55
œ%
ÈÈ
and (4 ) points of inflection atŠ‹
826826
55
œ%
ÈÈ
ßß_Ê
x and x 4œœ
826
5
„
È

241. Section 4.4 Concavity and Curve Sketching 237
48. y x 2x (x 5) x(x 2)(x 5) ,
w# # #
œ œ ab
y rising on ( )
w
œ ± ± ± Ê _ß !
!#&
and ( ), falling on ( ) a local maximum at x 0,#ß _ !ß # Ê œ
a local minimum at x 2;œ
y (2x 2)(x 5) x 2x (2)(x 5)
ww # #
œ ab
2(x5)2x 8x5,œ ab
#
y concave up
ww
%œ ! %% !
##
œ ± ± ± Ê
&
ÈÈ
on and (5 ), concave down onŠ‹
4646
2
œ%
#
ÈÈ
ßß_
and points of inflection atŠ‹Š‹_ß ß & Ê
%œ %
#
ÈÈ
646
2
x and x 5œœ
46
2
„
È

49. y sec x, y ( ) rising on ,
w#w
##
œ œ Ê ß
Î# Î#11
ˆ‰
11
never falling no local extrema;Ê
y 2(sec x)(sec x)(tan x) 2 sec x (tan x),
ww #
œœ ab
y ( ) concave up on , ,
ww
#
œ ± Ê !
Î# Î# !11
ˆ‰
1
concave down on , is a opoint of inflection.ˆ‰*)' '
1
#

50. y tan x, y ( ) rising on 0 ,
ww
#
œ œ ± Ê ß
Î# Î# !11
ˆ‰
1
falling on no local maximum, a local minimumˆ‰ß! Ê
1
#
at x 0; y sec x, y ( ) concave upœ œ œ Ê
Î# Î#
ww # ww
11
on no points of inflectionˆ‰ß Ê
11
##

51. y cot , y ( ) rising on ( ),
ww
#
œ œ ± Ê !ß
!#
)
1 1
1
falling on ( ) a local maximum at , no local11 ) 1ß# Ê œ minimum; y csc , y ( ) never
ww # ww"
##
œ œ Ê
!#
)
1
concave up, concave down on ( ) no points of!ß # Ê1
inflection

52. y csc , y ( ) rising on ( 2 ), never
w#w
#
œ œ Ê !ß
!#
)
1
1
falling no local extrema;Ê y 2 csc csc cot
ww
####
"
œˆ‰ˆ ‰ˆ‰ˆ‰
)))
csc cot , y ( )œ œ ±
!#
ˆ‰ˆ‰
#ww
##
))
1 1
concave up on ( ), concave down on ( )Êß# !ß 11 1
a point of inflection at ʜ )1

242. 238 Chapter 4 Applications of Derivatives
53. y tan 1 (tan 1)(tan 1),
w#
œœ )))
y ( | ) rising on
w
œ ± Ê
Î# Î% Î# Î%11 1 1 and , falling on ˆ‰ˆ‰ ˆ‰*)* ) *)
1 1 11 11
##44 44
a local maximum at , a local minimum at ;Êœ œ ))
11
44
y 2 tan sec , y ( )
ww # ww
œ œ ±
Î# Î# !
))
11
concave up on , concave down on Ê!ß ß! ˆ‰ ˆ ‰
11
##
a point of inflection at 0ʜ )

54. y 1 cot ( cot )(1 cot ),
w#
œ œ" )))
y ( | ) rising on ,
w
œ ± Ê ß
! Î%$Î%1 1
1
ˆ‰
11
44
3
falling on 0 and a local maximum atˆ‰ ˆ ‰ßßÊ
11
44
3
1
, a local minimum at ;))œœ
3
44
11
y 2(cot ) csc , y ( )
ww # ww
œ œ ±
! Î#
))
1
1
ab
concave up on , concave down on Ê!ß ß ˆ‰ ˆ‰
11
##
1
a point of inflection at ʜ )
1
#

55. y cos t, y [ ] rising on
ww
œ œ ± ± Ê
!# Î# $ Î#11 1
and 2 , falling on local maxima atˆ‰ ˆ ‰ ˆ ‰!ß ß ß Ê
11 11
## ##
33
1
t and t 2 , local minima at t 0 and t ;œœ œœ
11
##
1
3
y sin t, y [ ]
ww ww
œ œ ±
!# 1 1
concave up on ( ), concave downÊß# 11
on ( ) a point of inflection at t!ß Ê œ11

56. y sin t, y [ ] rising on ( ),
ww
œ œ ± Ê !ß
!# 1 1
1
falling on ( 2 ) a local maximum at t , local11 1ßÊ œ
minima at t 0 and t 2 ; y cos t,œœœ 1
ww
y [ ] concave up on
ww
#
œ ± ± Ê !ß
!# Î# $ Î#11 1
ˆ‰
1
and , concave down on pointsˆ‰ ˆ‰
33111
###
ß# ß Ê1
of inflection at t and tœœ
11
##
3

57. y (x 1) , y ) ( rising on
w œ#?$ w
œ œ Ê
*
( ), never falling no local extrema;_ß _ Ê
y(x1), y )(
ww œ&?$ ww
œ œ
*
2
3
concave up on ( 1), concave down on ( )Ê _ß "ß _
a point of inflection and vertical tangent at x 1ʜ

243. Section 4.4 Concavity and Curve Sketching 239
58. y (x 2) , y )( rising on (2 ),
w "Î$ w
œ œ Ê ß _
#
falling on ( ) no local maximum, but a local_ß # Ê
minimum at x 2; y (x 2) ,œœ
ww %Î$1
3
y )( concave down on ( 2) and
ww
œ Ê _ß
#
( ) no points of inflection, but there is a cusp at#ß _ Ê
x2œ

59. y x (x 1), y )( rising on
w #Î$ w
œ œ ± Ê
! "
( ), falling on ( ) no local maximum, but a"ß _ _ß " Ê
local minimum at x 1; y x xww "
ww #Î$ &Î$"
33
2
x(x2), y )(œ œ ±
# !
"&Î$ ww
3
concave up on ( 2) and ( ), concave down onÊ_ß!ß_
( ) points of inflection at x 2 and x 0, and a#ß ! Ê œ œ
vertical tangent at x 0œ

60. y x (x 1), y )( rising on
w %Î& w
œ œ ± Ê
" !
( 0) and ( ), falling on ( ) no local"ß !ß _ _ß " Ê
maximum, but a local minimum at x 1;œ
y x x x (x 4),
ww %Î& *Î& *Î&""
œœ
55 5
4
y )( concave up on ( 0) and
ww
œ ± Ê _ß
! %
(4 ), concave down on (0 4) points of inflection atß_ ß Ê
x 0 and x 4, and a vertical tangent at x 0œœ œ

61. y , y rising on
x, x 0
2x, x 0
ww
w w """ & """ Î
# Ÿ

!
œ
( ) no local extrema; y ,
2, x 0
2, x 0
_ß _ Ê œ


ww
œ
y )( concave up on ( ), concave
ww
œ Ê !ß _
!
down on ( ) a point of inflection at x 0_ß ! Ê œ

62. y , y rising on
x, x 0
x , x 0
ww
#
#
œ œ ± Ê
Ÿ

!
œ
( ), falling on ( ) no local maximum, but a!ß _ _ß ! Ê
local minimum at x 0; y ,
2x, x 0
2x, x 0
œœ
Ÿ

ww
œ
y concave up on ( )
ww
œ ± Ê _ß _
!
no point of inflectionÊ

244. 240 Chapter 4 Applications of Derivatives
63. The graph of y f (x) the graph of y f(x) is concaveœÊ œ
ww
up on ( ), concave down on ( ) a point of!ß _ _ß ! Ê
inflection at x 0; the graph of y f (x)œœ
w
y the graph y f(x) hasÊ œ ± ± Ê œ
w
both a local maximum and a local minimum

64. The graph of y f (x) y theœ Ê œ ± Ê
ww ww
graph of y f(x) has a point of inflection, the graph ofœ
y f (x) y the graph ofœ Ê œ ± ± Ê
ww
y f(x) has both a local maximum and a local minimumœ

65. The graph of y f (x) yœ Ê œ ± ±
ww ww
the graph of y f(x) has two points of inflection, theʜ
graph of y f (x) y the graph ofœ Ê œ ± Ê
ww
y f(x) has a local minimumœ

66. The graph of y f (x) y theœ Ê œ ± Ê
ww ww
graph of y f(x) has a point of inflection; the graph ofœ
y f (x) y the graph ofœ Ê œ ± ± Ê
ww
y f(x) has both a local maximum and a local minimumœ
67. Point y y
P
Q
R
S
T
www

"*

!

245. Section 4.4 Concavity and Curve Sketching 241
68. 69.
70.
71. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here.
(a) The body is moving away from the origin when displacement is increasing as t increases, 0 t 2 andkk ÞÞ
6 t 9.5; the body is moving toward the origin when displacement is decreasing as t increases, 2 t 6ÞÞ ÞÞkk
and 9.5 t 15ÞÞ
(b) The velocity will be zero when the slope of the tangent line for y s(t) is horizontal. The velocity is zeroœ
when t is approximately 2, 6, or 9.5 sec.
(c) The acceleration will be zero at those values of t where the curve y s(t) has points of inflection. Theœ
acceleration is zero when t is approximately 4, 7.5, or 12.5 sec.
(d) The acceleration is positive when the concavity is up, 4 t 7.5 and 12.5 t 15; the acceleration isÞÞ ÞÞ
negative when the concavity is down, 0 t 4 and 7.5 t 12.5ÞÞ ÞÞ
72. (a) The body is moving away from the origin when displacement is increasing as t increases, 1.5 t 4,kk ÞÞ
10 t 12 and 13.5 t 16; the body is moving toward the origin when displacement is decreasing as tÞÞ ÞÞ kk
increases, 0 t 1.5, 4 t 10 and 12 t 13.5ÞÞ ÞÞ ÞÞ
(b) The velocity will be zero when the slope of the tangent line for y s(t) is horizontal. The velocity is zeroœ
when t is approximately 0, 4, 12 or 16 sec.
(c) The acceleration will be zero at those values of t where the curve y s(t) has points of inflection. Theœ
acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec.
(d) The acceleration is positive when the concavity is up, 0 t 1.5, 6 t 8 and 10 t 13.5, theÞÞ ÞÞ ÞÞ
acceleration is negative when the concavity is down, 1.5 t 6, 8 t 10 and 13.5 t 16.ÞÞ ÞÞ ÞÞ
73. The marginal cost is which changes from decreasing to increasing when its derivative is zero. This is a
dc dc
dx dx
w
w
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units.
74. The marginal revenue is and it is increasing when its derivative is positive the curve is concave up
dy d y
dx dx
w
w
Ê t 2 and 5 t 9; marginal revenue is decreasing when 0 the curve is concave downÎ*ÞÞ ÞÞ Þ Î
dy
dx
w
w
2 t 5 and 9 t 12.ÎÞÞ ÞÞ
75. When y (x 1) (x 2), then y 2(x 1)(x 2) (x 1) . The curve falls on ( 2) and rises on
w# ww #
œ œ _ß
( ). At x 2 there is a local minimum. There is no local maximum. The curve is concave upward on ( ) and#ß _ œ _ß "

246. 242 Chapter 4 Applications of Derivatives
, and concave downward on . At x 1 or x there are inflection points.ˆ‰ ˆ‰
555
333
ß_ "ß œ œ
76. When y (x 1) (x 2)(x 4), then y 2(x 1)(x 2)(x 4) (x 1) (x 4) (x 1) (x 2)
w# ww # #
œ œ
(x 1) 2 x 6x 8 x 5x 4 x 3x 2 2(x 1) 2x 10x 11 . The curve rises onœ œ cdabababab
### #
( 2) and ( ) and falls on ( ). At x 2 there is a local maximum and at x 4 a local minimum. The_ß %ß _ #ß % œ œ
curve is concave downward on ( ) and and concave upward on 1 and_ß " ß ßŠ‹ Š‹
5353 53
2

##
ÈÈ È
. At x 1, and there are inflection points.Š‹
53 53 53
###
ÈÈÈ
ß_ œ
77. The graph must be concave down for x 0 because
f (x) 0.
ww "
œ
x
w

78. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always
be concave up or concave down so it will have no inflection points and no cusps or corners.
79. The curve will have a point of inflection at x 1 if 1 is a solution of y 0; y x bx cx dwww"""
ww $ #
y 3x 2bx c y 6x 2b and 6(1) 2b 0 b 3.Êœ Ê œ œÊœ
w# w w
80. (a) True. If f(x) is a polynomial of even degree then f is of odd degree. Every polynomial of odd degree has
w
at least one real root f (x) 0 for some x r f has a horizontal tangent at x r.Êœ œÊ œ
w
(b) False. For example, f(x) x 1 is a polynomial of odd degree but f (x) 1 is never 0. As anotherœ œ
w
example, y x x x is a polynomial of odd degree, but y x 2x 1 (x 1) 0 for all x.w"" w""w"(
"$# w # #
3
81. (a) f(x) ax bx c a x x c a x c a x a parabolaœœ œ B œ
## #
#
#
ˆ‰ ˆ‰ Š‹
bbbbbb4ac
a a 4a 4a a 4a
ww w
w
whose vertex is at x the coordinates of the vertex are œ Ê ß
bbb4ac
2a 2a 4a
Š‹
w

(b) The second derivative, f (x) 2a, describes concavity when a 0 the parabola is concave up and
ww
œÊ
when a 0 the parabola is concave down.Þ
82. No, f (x) could be decreasing to zero at x c and then increase again so it would be concave up on every
ww
œ
interval even though f (x) 0. For example f(x) x is always concave up even though f (0) 0.
ww % ww
œœ œ
83. A quadratic curve never has an inflection point. If y ax bx c where a 0, then y 2ax b andw"" ) w"
#w
y 2a. Since 2a is a constant, it is not possible for y to change signs.
ww ww
œ
84. A cubic curve always has exactly one inflection point. If y ax bx cx d where a 0, thenw""" )
$#
y 3ax 2bx c and y 6ax 2b. Since is a solution of y 0, we have that y changes its sign
w # ww ww ww
w"" w" w
b
3a
at x and y exists everywhere (so there is a tangent at x ). Thus the curve has an inflectionœ œ
b b
3a 3a
w
point at x . There are no other inflection points because y changes sign only at this zero.œ
b
3a
ww

247. Section 4.4 Concavity and Curve Sketching 243
85. If y x 5x 240, then y 5x (x 4) andœ œ
&% w $
y 20x (x 3). The zeros of y are extrema, and
ww # w
œ
there is a point of inflection at xœ$Þ

86. If y x 12x , then y 3x(x 8) andœ œ
$# w
y 6(x 4). The zeros of y and y are
ww w ww
œ
extrema and points of inflection, respectively.

87. If y x 16x 25, then y 4x x 8 andœ œ
4
5
&# w $
ab
y 16 x 2 . The zeros of y and y are
ww $ w ww
w"ab
extrema and points of inflection, respectively.

88. If y 4x 12x 20, thenœ
xx
43

#
y x x x (x 3)(x 2)
w$# #
œ ) "#œ Þ
So y has a local minimum at x as its only extremeœ$
value. Also y x x (3x 4)(x 2) and there
ww #
œ$ # )œ
are inflection points at both zeros, and 2, of y .
%
$
ww

89. The graph of f falls where f 0, rises where f 0,
ww
Þ(
and has horizontal tangents where f 0. It has local
w
œ
minima at points where f changes from negative to
w
positive and local maxima where f changes from
w
positive to negative. The graph of f is concave down
where f 0 and concave up where f 0. It has an
ww ww
Þ(
inflection point each time f changes sign, provided a tangent
ww
line exists there.

248. 244 Chapter 4 Applications of Derivatives
90. The graph f is concave down where f 0, and concave
ww
Þ
up where f 0. It has an inflection point each time
ww

f changes sign, provided a tangent line exists there.
ww

91. (a) It appears to control the number and magnitude of the
local extrema. If k 0, there is a local maximum to theÞ
left of the origin and a local minimum to the right. The
larger the magnitude of k (k 0), the greater theÞ
magnitude of the extrema. If k 0, the graph has only
positive slopes and lies entirely in the first and third
quadrants with no local extrema. The graph becomes
increasingly steep and straight as k .Ä_

(b) f (x) 3x k the discriminant 0 4(3)(k) 12k is positive for k 0, zero for k 0, and
w# #
œÊ œ œ
negative for k 0; f has two zeros x when k 0, one zero x 0 when k 0 and no real zerosœ„œœ
w
É
k
3
when k 0; the sign of k controls the number of local extrema.
(c) As k , f (x) and the graph becomesÄ_ Ä_
w
increasingly steep and straight. As k , theÄ_
crest of the graph (local maximum) in the second
quadrant becomes increasingly high and the trough
(local minimum) in the fourth quadrant becomes
increasingly deep.

92. (a) It appears to control the concavity and the number of
local extrema.

249. Section 4.5 Applied Optimization Problems 245
(b) f(x) x kx 6x f (x) 4x 3kx 12xw" " Î w " "
%$# w $ #
f (x) 12x 6kx 12 the discriminant isÎw""Î
ww #
36k 4(12)(12) 36(k 4)(k 4), so the sign line
#
œ
of the discriminant is the ± ± Ê
% %
discriminant is positive when k 4, zero whenkk
k 4, and negative when k 4; f (x) 0 hasœ„ œ kk
ww
two zeros when k 4, one zero when k 4, andkkœ„
no real zeros for k 4; the value of k controls thekkÞ
number of possible points of inflection.

93. (a) If y x x 2 , then y x 2x 1 andœ œ
#Î$ # w "Î$ #
ab a b
4
3
y x 10x 1 . The curve rises on
ww %Î$ #
w"
4 9
ab
0 and and falls on Š‹Š‹ Š ‹ ß ß_ _ß
"" "
ÈÈ È
22 2
and . The curve is concave up on ( )Š‹!ß _ß !
"
È
2
and ( ).!ß _
(b) A cusp since lim y and lim y .
x xÄ! Ä!
œ
ww
œ_ œ_

94. (a) If y 9x (x 1), then y andœ œ
#Î$ w
15 x
x
ˆ‰
2
5
Î$"
y . The curve rises on ( ) and
ww

œ _ß !
10 x
x
ˆ‰
Ê

5
and falls on . The curve is concaveˆ‰ ˆ‰
22
55
ß_ !ß
down on and concave up on andˆ‰ ˆ‰_ß ß !
""
55
().!ß _
(b) A cusp since lim y and lim y .
x xÄ! Ä!
œ
ww
œ_ œ_

95. Yes: y x 3 sin 2x y 2x 6 cos 2x. The graphw" Î w"
#w
of y is zero near 3 and this indicates a horizontal tangent
w

near x 3.œ

4.5 APPLIED OPTIMIZATION PROBLEMS
1. Let and w represent the length and width of the rectangle, respectively. With an area of 16 in. , we havej
#
that ( )(w) 16 w 16 the perimeter is P 2 2w 2 32 and P ( ) 2 .jœÊœjÊ œjœjj jœœ
" " w
jj
j32 216
œœ
œ
ab
Solving P ( ) 0 0 4, 4. Since 0 for the length of a rectangle, must be 4 and
w j j
j
jœ Ê œ Êjœ j j
2( 4)( 4)
œ
w 4 the perimeter is 16 in., a minimum since P ( ) 0.œÊ jœ
ww
j
16
"
2. Let x represent the length of the rectangle in meters ( x ) Then the width is x and the area is! % %
A x x x x x . Since A x x, the critical point occurs at x . Since, A x for x andab a b ab abœ % œ% œ%# œ# ! ! #
#w w
A x for x , this critical point corresponds to the maximum area. The rectangle with the largest area measures
w
abÞ* #Þ Þj
m by m, so it is a square.#%#œ#

250. 246 Chapter 4 Applications of Derivatives
Graphical Support:
3. (a) The line containing point P also contains the points ( ) and ( ) the line containing P is y 1 x!ß " "ß ! Ê œ
a general point on that line is (x 1 x).Êß
(b) The area A(x) 2x(1 x), where 0 x 1.œ ŸŸ
(c) When A(x) 2x 2x , then A (x) 0 2 4x 0 x . Since A(0) 0 and A(1) 0, we concludeœ œÊœÊœ œ œ
#w "
#
that A sq units is the largest area. The dimensions are unit by unit.ˆ‰
"" "
## #
œ"
4. The area of the rectangle is A 2xy 2x 12 x ,œœ ab
#
where 0 x 12 . Solving A (x) 0 24 6x 0ŸŸ œ Ê œ
È
w#
x 2 or 2. Now 2 is not in the domain, and sinceʜ
A(0) 0 and A 12 0, we conclude that A(2) 32œœ œ Š‹
È
square units is the maximum area. The dimensions are 4 units
by 8 units.

5. The volume of the box is V(x) x(15 2x)(8 2x)œ
120x 46x 4x , where 0 x 4. Solving V (x) 0œ ŸŸ œ
#$ w
120 92x 12x 4(6 x)(5 3x) 0 xÊ œœÊœ
# 5
3
or 6, but 6 is not in the domain. Since V(0) V(4) 0,œœ
V in must be the maximum volume ofˆ‰
5
3
œ¸*"
#%&!
#(
$
the box with dimensions inches.
14 35 5
333
‚‚

6. The area of the triangle is A ba 400 b , whereœœ
"
##
#b È
0 b 20. Then 400 bŸŸ œ
dA b
db
2 400 b
"
#
#
È
w
w
È 0 the interior critical point is b 10 2.œœÊ œ
200 b
400 b


w
w
È
È
When b 0 or 20, the area is zero A 10 2 is theœÊ Š‹
È
maximum area. When a b 400 and b 10 2, the
##
"w w
È
value of a is also 10 2 the maximum area occurs when
È
Ê
ab.œ

7. The area is A(x) x(800 2x), where 0 x 400.œ ŸŸ
Solving A (x) 800 4x 0 x 200. With
w
œœÊœ
A(0) A(400) 0, the maximum area isœœ
A(200) 80,000 m . The dimensions are 200 m by 400 m.œ
#

251. Section 4.5 Applied Optimization Problems 247
8. The area is 2xy 216 y . The amount of fenceœÊœ
108
x
needed is P 4x 3y 4x 324x , where x;w" w" *Þ
"
4 0 x 81 0 the critical points are
dP 324
dx x
œ œ Ê œÊ
w
#
0 and 9, but 0 and 9 are not in the domain. Then„
P (9) 0 at x 9 there is a minimum the
ww
Ê œ Ê
dimensions of the outer rectangle are 18 m by 12 m
72 meters of fence will be needed.Ê

9. (a) We minimize the weight tS where S is the surface area, and t is the thickness of the steel walls of the tank. Theœ
surace area is S x xy where x is the length of a side of the square base of the tank, and y is its depth. Thew"j
#
volume of the tank must be ft y . Therefore, the weight of the tank is w x t x . Treating the&!! Ê œ œ
$ #&!! #!!!
xx
w abˆ‰
thickness as a constant gives w x t x for x . The critical value is at x . Since
w #!!!
abˆ‰œ# Þ! œ"!
x
w
w t , there is a minimum at x . Therefore, the optimum dimensions of the tank are ft on
ww %!!!
"!
ab ˆ‰"! œ # ! œ "! "!
"
the base edges and ft deep.&
(b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel
walls would likely be determned by other considerations such as structural requirements.
10. (a) The voluem of the tank being ft , we have that yx y . The cost of building the tank is""#& œ ""#& Ê œ
$# ""#&
x
w
c x x x , where x. Then c x x the critical points are and , but is notab ab ˆ‰œ& $! ! œ"! œ!Ê ! "& !
#w ""#& $$(&!
xx
w w
in the domain. Thus, c at x we have a minimum. The values of x ft and y ft will
ww
ab"& ! Ê œ "& œ "& œ &
minimize the cost.
(b) The cost function c x xy xy, can be separated into two items: (1) the cost of the materials and labor toœ& % "!ab
#
fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of the tanks is x xy ,ab
#
"j
it can be deduced that the unit cost to fabricate the tanks is $ /ft . Normally, excavation costs are per unit volume of&
#
excavated material. Consequently, the total excavation cost can be taken as xy x y . This suggests that the"! œˆ‰ab
"! #
x
unit cost of excavation is where x is the length of a side of the square base of the tank in feet. For the least
$ft
x
"!Î
w
expensive tank, the unit cost for the excavation is . The total cost of the least expensive tank is
$ft
ft ft yd
$$"!Î
"&
!Þ'( ")
w
""
œœ
$ , which is the sum of $ for fabrication and $ for the excavation.$$(& #'#& (&!
11. The area of the printing is (y 4)(x 8) 50.œ
Consequently, y 4. The area of the paper isw"ˆ‰
50
x8
A(x) x 4 , where 8 x. Thenw" Þˆ‰
50
x8
A (x) 4 x 0
w


œ œ œˆ‰ Š‹
50 50
x 8 (x 8) (x 8)
4(x 8) 400
ww
w
the critical points are 2 and 18, but 2 is not in theÊ
domain. Thus A (18) 0 at x 18 we have
ww
Ê œ
a minimum. Therefore the dimensions 18 by 9 inches
minimize the amount of paper.

12. The volume of the cone is V r h, where r x 9 y and h y 3 (from the figure in the text).œœœœ
"# #
3
1 È
Thus, V(y) 9 y (y 3) 27 9y 3y y V (y) 9 6y 3y (1 y)(3 y).œ œ Ê œœ
11 1
33 3
ab a b a b
##$w#
1
The critical points are 3 and 1, but 3 is not in the domain. Thus V (1) ( 6(1)) 0 at y 1 œ'Êœ
ww 1
3
we have a maximum volume of V(1) (8)(4) cubic units.œœ
11
33
32

252. 248 Chapter 4 Applications of Derivatives
13. The area of the triangle is A( ) , where 0 .))1wÞÞ
ab sin )
#
Solving A ( ) 0 0 . Since A ( )
www
##
)))œÊ œÊœ
ab cos )1
A 0, there is a maximum at .œ Ê œ
ab sin )1 1
## #
ww
ˆ‰ )
14. A volume V r h 1000 h . The amount ofœœ Êœ1
# 1000
r1
w
material is the surface area given by the sides and bottom of
the can S 2 rh r r , 0 r. ThenÎw " w " Þ11 1
##2000
r
2 r 0. The critical points
dS 2000 r 1000
dr r r
œ œ! Ê œ
ww
" 1
1
are 0 and , but 0 is not in the domain. Since
10
"
È1
0, we have a minimum surface area when
d S 4000
dr r
w
w"
w"#( 1 r cm and h cm. Comparing this result toœœœ
10 1000 10
r
"" w
ÈÈ11 1
the result found in Example 2, if we include both ends of the
can, then we have a minimum surface area when the can is
shorter-specifically, when the height of the can is the same as
its diameter.

15. With a volume of 1000 cm and V r h, then h . The amount of aluminum used per can isœœ1
# 1000
r1
w
A 8r 2 rh 8r . Then A (r) 16r 0 0 the critical points are 0 and 5,œ œ œ œÊ œÊ
## w
1
2000 2000 8r 1000
rrr
ww
"
but r 0 results in no can. Since A (r) 16 0 we have a minimum at r 5 h and h:r 8: .ww"(wÎww
ww 1000 40
r
"
1
1
16. (a) The base measures x in. by in., so the volume formula is V x x x x."! # œ œ # #& (&
"&#
##
"!# "&# $#x xx x
ab
abab
(b) We require x , x , and x . Combining these requirements, the domain is the interval . ! # "! # "& !ß & ab

(c) The maximum volume is approximately 66.02 in. when x in.
$
¸"Þ*'

(d) V x x x . The critical point occurs when V x , at x
w# w
&! „ &! % ' (&
#' "#
&! „ (!!
ab abœ' &! (& œ! œ œ
Éab abab
ab
È
w
, that is, x or x . We discard the larger value because it is not in the domain. Sinceœ ¸ "Þ*' ¸ 'Þ$(
#&„& (
'
È
V x x , which is negative when x , the critical point corresponds to the maximum volume. The
ww
abœ"# &! ¸"Þ*'
maximum volume occurs when x , which comfimrs the result in (c).œ ¸ "Þ*'
#&& (
'
È
17. (a) The" sides" of the suitcase will measure x in. by x in. and will be x in. apart, so the volume formula is#% # ") # #
Vxxx xxxx.abababœ # #% # ") # œ ) "') )'#
$#

253. Section 4.5 Applied Optimization Problems 249
(b) We require x , x , and x . Combining these requirements, the domain is the interval . ! # ") # #% !ß * ab

(c) The maximum volume is approximately 1309.95 in. when x 3 3 in.
$
¸Þ*

(d) V x x x x x . The critical point is at x
w# #
"% „ "% % " $'
#" #
"% „ &#
ab a bœ #% $$' )'% œ #% "% $' œ œ
Éab abab
ab
È

, that is, x or x . We discard the larger value because it is not in the domain. Sinceœ ( „ "$ ¸ $Þ$* ¸ "!Þ'"È
V x x which is negative when x , the critical point corresponds to the maximum volume. The
ww
ab a bœ #% # "% ¸ $Þ$* maximum value occurs at x , which confirms the results in (c).œ ( "$ ¸ $Þ$*È
(e) x x x 8 x x x x x x . Since is not in) "') )'# œ ""#! Ê #" "!) "%! œ ! Ê ) # & "% œ ! "%
$# $#
a b ababa b
the fomain, the possible values of x are x in. or x in.œ# œ&
(f) The dimensions of the resulting box are x in., x in., and x . Each of these measurements must be# #%# ")#ab ab
positive, so that gives the domain of .ab!ß *
18. If the upper right vertex of the rectangle is located at x cos x for x , then the rectangle has width x andabß % !Þ& ! # 1
height cos x, so the area is A x x cos x. Solving A x graphically for x , we find that%!Þ& œ) !Þ& œ! !ab ab
w
1
x . Evaluating x and cos x for x , the dimensions of the rectangle are approximately (width) by¸ #Þ#"% # % !Þ& ¸ #Þ#"% %Þ%$
(height), and the maximum area is approximately ."Þ(* (Þ*#$
19. Let the radius of the cylinder be r cm, r . Then the height is r and the volume is! "! # "!!
È
#
V r r r cm . Then, V r r r r rab ab a b a b
ÈÈ
Š‹ Š ‹œ # "!! œ # # # "!! #111
#$w# # #"
"!!
È
r

. The critical point for r occurs at r . Since V r forœ œ ! "! œ œ"! !
# % "!! # #!! $
"!! "!!
#!! #
$$
w11 1rr r r r
rr


ab a b
ÈÈ
ÉÉ ab
r and V r for r , the critical point corresponds to the maximum volume. The! "! ! "! "!ÉÉ ab
##
$$
w
dimensions are r cm and h cm, and the volume is cm .œ "! ¸ )Þ"' œ ¸ ""Þ&& ¸ #%")Þ%!É
# #! %!!!
$ $$$
$
ÈÈ
1
20. (a) From the diagram we have 4x 108 and V x .jœ œ j
#
The volume of the box is V(x) x (108 4x), whereœ
#
0 x 27. ThenŸ
V (x) 216x 12x 12x(18 x) 0
w#
*†* †*
the critical points are 0 and 18, but x 0 results inʜ
no box. Since V (x) 216 24x 0 at x 18 we
ww
*†! *
have a maximum. The dimensions of the box are
18 18 36 in.‚‚

254. 250 Chapter 4 Applications of Derivatives
(b) In terms of length, V( ) x . The graphjœ jœ j
# j
#ˆ‰
108
4
indicates that the maximum volume occurs near 36,jœ
which is consistent with the result of part (a).

21. (a) From the diagram we have 3h 2w 108 andœ
V h w V(h) h 54 h 54h h .œÊ œ œ
## #$
## ˆ‰
33
Then V (h) 108h h h(24 h) 0
w#
##
*†* †*
99
h 0 or h 24, but h 0 results in no box. SinceÊœ œ œ
V (h) 108 9h 0 at h 24, we have a maximum
ww
*†! *
volume at h 24 and w 54 h 18.œœœ
3
#

(b)
22. From the diagram the perimeter is P 2r 2h r,œ 1
where r is the radius of the semicircle and h is the
height of the rectangle. The amount of light transmitted
proportional to
A 2rh r r(P 2r r) rœ œ
""##
44
111
rP 2r r . Then P 4r r 0*† † *†† *
##
#
3dA 3
4dr
11
r 2h P .Êœ Ê œ œ
2P 4P 2 P
83 83 83 83
(4 )P
""""
"
1111
1 1
Therefore, gives the proportions that admit the
2r 8
h4
œ
"1
light since 4 0.most
dA 3
dr


*† † !
#
1

23. The fixed volume is V r h r h , where h is the height of the cylinder and r is the radiusœ Êœ11
#$ 2V2r
3r3 1

of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the
surface area of the hemisphere. Thus, we minimize C 2 rh 4 r 2 r 4 r r .œœ œ111 1 1
### ˆ‰
V2r 2V8
r3 r31

Then r 0 V r r . From the volume equation, h
dC 2V 16 8 3V V 2r
dr r 3 3 8 r 3
œ œ Ê œ Ê œ œ
11
$
"Î$ ˆ‰
11
. Since 0, theseœ œ œ œ
4V 23 V 3 24V 23 V 3V d C 4V 16
33 3 dr r 3


11 1 1††† ††
†† ††† ††
##

"Î$
ˆ‰ 1
dimensions do minimize the cost.
24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram
the area of the cross section is A( ) cos sin cos , 0 . Then A ( ) sin cos sin))))) ) )))œ œ
1
#
w##
2 sin sin 1 (2 sin 1)(sin 1) so A ( ) 0 sin or sin 1 becauseœ œ œ Ê œ œ Ê œab
#w "
#
)) ) ) ) ) ) )
1
6
sin 1 when 0 . Also, A ( ) 0 for 0 and A ( ) 0 for . Therefore, at )))))) )Á œ
11111
# #
ww
66 6
there is a maximum.

255. Section 4.5 Applied Optimization Problems 251
25. (a) From the diagram we have: AP x, RA L x ,œœ
È
#
PB 8.5 x, CH DR 11 RA 11 L x ,*† * *† *† †
È
#
QB x (8.5 x) , HQ 11 CH QBœ œÈ
##
11 11 L x x (8.5 x)œ ’“
È È
## #
L x x (8.5 x) , RQ RH HQœ œ
È È
## #
## #
(8.5) L x x (8.5 x) . Itœ
# ## #
#
Š‹
È È

follows that RP PQ RQ L x L x x (x 8.5) (8.5)
# ##
## # ## # #
#
œ Êœ Š‹
È È
L x L x 2 L x 17x (8.5) 17x (8.5) (8.5)ʜ
#### # # ## # È È
17 x 4 L x 17x (8.5) L xÊœ Êœ œ
## # # # # #

aba b
17 x 17x
4 17x (8.5) 17x (8.5)
'

cd
. œœ
17x 4x
17x
4x 17
''



ˆ‰
17
(b) If f(x) is minimized, then L is minimized. Now f (x) f (x) 0 when xœœÊ
4x 51
4x 17 (4x 17) 8
4x (8x 51)
'


#w w
and f (x) 0 when x . Thus L is minimized when x .
w#
œ
51 51
88
(c) When x , then L 11.0 in.œ¸
51
8

26. (a) From the figure in the text we have P 2x 2y y x. If P 36, then y 18 x. When theœ Êœ œ œ
P
#
cylinder is formed, x 2 r r and h y h 18 x. The volume of the cylinder is V r hœÊœ œÊœ œ11
x
#
#
1
V(x) . Solving V (x) 0 x 0 or 12; but when x 0, there is no cylinder.Êœ œ œÊœ œ
18x x
44
3x(12 x)
'
w
11
Then V (x) 3 V (12) 0 there is a maximum at x 12. The values of x 12 cm and
ww ww
#
œÊ Ê œ œ
3x
1
ˆ‰
y 6 cm give the largest volume.œ
(b) In this case V(x) x (18 x). Solving V (x) 3 x(12 x) 0 x 0 or 12; but x 0 would result inœ œœÊœ œ11
#w
no cylinder. Then V (x) 6 (6 x) V (12) 0 there is a maximum at x 12. The values of
ww ww
œÊ Ê œ1
x 12 cm and y 6 cm give the largest volume.œœ
27. Note that h r and so r h . Then the volume is given by V r h h h h h for
## ## $#
$$ $
œ$ œ $ œ œ $ œ
È
ab
11 1
1
h , and so r r . The critical point (for h ) occurs at h . Since for! $ œ œ " ! œ" !È ab
dV dV
dh dh
11 1
##
h , and for h , the critical point corresponds to the maximum volume. The cone of greatest! " ! " $
dV
dh È
volume has radius m, height m, and volume m .
È
#"
#
$
$1
28. (a) f(x) x f (x) x 2x a , so that f (x) 0 when x 2 implies a 16œÊ œ œ œ œ
#w#$ wa
x
ab
(b) f(x) x f (x) 2x x a , so that f (x) 0 when x 2 implies a 1œÊ œ œ œ œ
#ww$$ wwa
x
ab
29. If f(x) x , then f (x) 2x ax and f (x) 2 2ax . The critical points are 0 and , but x 0.œ œ œ Á
# w # ww $ $
#
a a
x
È
Now f 6 0 at x there is a local minimum. However, no local maximum exists for any a.
ww$$
##ˆ‰ÈÈ
aa
œ Ê œ
30. If f(x) x ax bx, then f (x) 3x 2ax b and f (x) 6x 2a.œ œ œ
$# w # ww
(a) A local maximum at x 1 and local minimum at x 3 f ( 1) 0 and f (3) 0 3 2a b 0 andœ œÊ œ œÊœ
ww
27 6a b 0 a 3 and b 9. œ Ê œ œ
(b) A local minimum at x 4 and a point of inflection at x 1 f (4) 0 and f (1) 0 48 8a b 0œœÊœœÊœ
www

256. 252 Chapter 4 Applications of Derivatives
and 6 2a 0 a 3 and b 24.œÊœ œ
31. (a) s t t t v t s t t . At t , the velocity is v ft/sec.ab ab ab a bœ "' *' ""# Ê œ œ $# *' œ ! ! œ *'
#w
(b) The maximum height ocurs when v t , when t . The maximum height is s ft and it occurs at tab a bœ! œ$ $ œ#&' œ$
sec.
(c) Note that s t t t t t , so s at t or t . Choosing the positive valueab ababœ"' *' ""#œ"' " ( œ! œ" œ(
#
of t, the velocity when s is v ft/sec.œ ! ( œ "#)ab
32.
Let x be the distance from the point on the shoreline nearest Jane's boat to the point where she lands her boat. Then she
needs to row x mi at 2 mph and walk x mi at 5 mph. The total amount of time to reach the village is
È
% '
#
f x hours ( x ). Then f x x . Solving f x , weab ab a b abœ !ŸŸ' œ #œ œ!
È
ÈÈ
%"
#& # & &
' " " " " ww
#%" #%"
x x x
xx

have: x x x x x x . We discard the negative
x
x#%"
" %
&
# ###
#"
È È

œ Ê & œ # % Ê #& œ % % Ê #" œ "' Ê œ „
È
ab
value of x because it is not in the domain. Checking the endpoints and critical point, we have f ,ab! œ #Þ#
f , and f . Jane should land her boat miles donw the shoreline from the pointŠ‹ ab
% %
#" #"
È È
¸#Þ"# ' ¸$Þ"' ¸!Þ)(
nearest her boat.
33. h and L x h x
) #"'
"#*
#
#
xx x
h
œÊœ) œ#( ab a bÉ
x when x . Note that L x isœ) #(  !Ɉ‰ ab ab
#"'
# #
x
minimized when f x x isab a bˆ‰œ) #(
#"'
# #
x
minimized. If f x , then
w
abœ!
x# ) # #( œ!ˆ‰ˆ‰ ab
#"' #"'
xx

x x (not acceptableÊ #( " œ ! Ê œ #(ab ˆ‰
"(#)
x
!

since distance is never negative or x . Then L ftœ "# "# œ #"*( ¸ %'Þ)( Þab
È
34. (a) From the diagram we have d 4r w . The strength of the beam is S kwd kw 4r w . When
### # ##
*† * * † ab
r 6, then S 144kw kw . Also, S (w) 144k 3kw 3k 48 w so S (w) 0 w 4 3 ;œœ œœ œÊœ„
$w # #w
ab
È
S 4 3 0 and 4 3 is not acceptable. Therefore S 4 3 is the maximum strength. The dimensions
ww
Š‹ Š‹
ÈÈ È
!†
of the strongest beam are 4 3 by 4 6 inches.
ÈÈ
(b) (c)
Both graphs indicate the same maximum value and are consistent with each other. Changing k does not
change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce
the strongest beam).

257. Section 4.5 Applied Optimization Problems 253
35. (a) From the situation we have w 144 d . The stiffness of the beam is S kwd kd 144 d ,
## $$#
"Î#
*† * * † ab
where 0 d 12. Also, S (d) critical points at 0, 12, and 6 3. Both d 0 andŸŸ œ Ê œ
w

4kd 108 d
144 d


ab
È È
d 12 cause S 0. The maximum occurs at d 6 3. The dimensions are 6 by 6 3 inches.œœ œ
ÈÈ
(b) (c)
Both graphs indicate the same maximum value and are consistent with each other. The changing of k has
no effect.
36. (a) s s sin t sin t sin t sin t cos sin cos t sin t sin t cos t tan t 3"#
"
##œÊ œ Ê œ Ê œ Ê œˆ‰ È
111
333
3
È
t or ʜ
11
33
4
(b) The distance between the particles is s(t) s s sin t sin t sin t 3 cos tœœ œ kk ¸¸ ˆ‰ ¹¹
È "#
"
#
1
3
s (t) since x critical times and endpointsÊœ œÊ
w


Š‹Š‹
ÈÈ
¹¹
È kk
sin t 3 cos t cos t 3 sin t
2 sin t 3 cos t
dx
dx x
kk
are 0, , , , , 2 ; then s(0) , s 0, s 1, s 0, s 1, s(2 ) the
111 1 1 1 1 1
36 3 6 3 6 3 6
5411 5 4 11 3 3
11 œ œœœ œœÊ
È È
# #
ˆ‰ ˆ‰ ˆ‰ ˆ ‰
greatest distance between the particles is 1.
(c) Since s (t) we can conclude that at t and , s (t) has cusps and
w w


œœ
Š‹Š‹
ÈÈ
¹¹
È
sin t 3 cos t cos t 3 sin t
2 sin t 3 cos t
33
411
the distance between the particles is changing the fastest near these points.
37. (a) s 10 cos ( t) v 10 sin ( t) speed 10 sin ( t) 10 sin ( t) the maximum speed isœÊœ Êœ œ Ê111 1111 kkkk
10 31.42 cm/sec since the maximum value of sin ( t) is 1; the cart is moving the fastest at t 0.5 sec,11¸ œkk 1.5 sec, 2.5 sec and 3.5 sec when sin ( t) is 1. At these times the distance is s 10 cos 0 cm andkk ˆ‰1 œœ
1
#
a 10 cos ( t) a 10 cos ( t) a 0 cm/secœ Êœ Êœ11 1 1
## #
kk k k kk
(b) a 10 cos ( t) is greatest at t 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times thekk k kœœ11
#
magnitude of the cart's position is s 10 cm from the rest position and the speed is 0 cm/sec.kkœ
38. (a) 2 sin t sin 2t 2 sin t 2 sin t cos t 0 (2 sin t)(1 cos t) 0 t k where k is a positiveœÊ œÊ œÊœ 1
integer
(b) The vertical distance between the masses is s(t) s s s s (sin 2t 2 sin t)*†* † * †kkab a bˆ‰
"# "#
#
"Î#
#
"Î#
s (t) (sin 2t 2 sin t) (2)(sin 2t 2 sin t)(2 cos 2t 2 cos t)ʜ
w# "
#
"Î#ˆ‰ab
critical times atœœ Ê
2(cos 2t cos t)(sin 2t 2 sin t) 4(2 cos t 1)(cos t )(sin t)(cos t 1)
sin 2t 2 sin t sin 2t 2 sin t
"
kk kk
0, , , , 2 ; then s(0) 0, s sin 2 sin , s( ) 0, s
24 2 4 2 4
33 3 3 3 3
3311 1 1 1 1
11 1 œœ œ œˆ‰ ¸ ¸ ˆ‰ˆ‰ ˆ‰
È
#
sin 2 sin , s(2 ) 0 the greatest distance is at t and œ œ œÊ œ¸¸ˆ‰ ˆ‰
84 24
332 33
33 3311 11
ÈÈ
1
#
39. (a) s (12 12t) (8t) (12 12t) 64tœœÈ ab
## ##
"Î#
(b) (12 12t) 64t [2(12 12t)( 12) 128t]
ds 208t 144
dt (12 12t) 64t
œ œ
"
#
##
"Î#

ab
È 12 knots and 8 knotsÊœ œ¸¸
ds ds
dt dt
t=0 t=1

258. 254 Chapter 4 Applications of Derivatives
(c) The graph indicates that the ships did not see
each other because s(t) 5 for all values of t.

(d) The graph supports the conclusions in parts (b)
and (c).

(e) lim lim lim 208 4 13
tt tÄ_ Ä_ Ä_
ds 208
dt 144( t) 64t 144 64
(208t 144)
208
144 1 64
œœœœœÉ
Ë
É ÈÈ

"






Š‹
Š‹
144
t
t

which equals the square root of the sums of the squares of the individual speeds.
40. The distance OT TB is minimized when OB is
a straight line. Hence .nœn Ê œ!"))
"#

41. If v kax kx , then v ka 2kx and v 2k, so v 0 x . At x there is a maximum sinceœ œ œ œÊœ œ
#w ww w
##
aa
v 2k 0. The maximum value of v is .
ww
#ˆ‰
aka
4
*† !

42. (a) According to the graph, y .
w
ab!œ!
(b) According to the graph, y L .
w
abœ!
(c) y , so d . Now y x ax bx c, so y implies that c . There fore, y x ax bx andab ab ab ab!œ! œ! œ$ # !œ! œ! œ
w# w $#
y x ax bx. Then y L aL bL H and y L aL bL , so we have two linear
w# $# w #
ab ab abœ$ # œ œ œ$ # œ!
equations in two unknowns a and b. The second equation gives b . Substituting into the first equation, we haveœ
$
#
aL
aL H, or H, so a . Therefore, b and the equation for y is œ œ œ# œ$
$$
##
aL aL H H
LL
!!

y x x x , or y x H .ab ab ’“ˆ‰ ˆ‰œ# $ œ # $
HH x x
LL L L

$#
$#
43. The profit is p nx nc n(x c) a(x c) b(100 x) (x c) a b(100 x)(x c)œœœœ cd
"
a (bc 100b)x 100bc bx . Then p (x) bc 100b 2bx and p (x) 2b. Solving p (x) 0 œ œ œ œ Ê
#w ww w
x 50. At x 50 there is a maximum profit since p (x) 2b 0 for all x.œ œ œ
cc
##
ww
44. Let x represent the number of people over 50. The profit is p(x) (50 x)(200 2x) 32(50 x) 6000œ
2x 68x 2400. Then p (x) 4x 68 and p 4. Solving p (x) 0 x 17. At x 17 there is aœ œ œ œ Ê œ œ
#wwww
maximum since p (17) 0. It would take 67 people to maximize the profit.
ww
!

259. Section 4.5 Applied Optimization Problems 255
45. (a) A(q) kmq cm q, where q 0 A (q) kmq and A (q) 2kmq . Theœ Êœœ œ
" w # ww $
##
hh hq 2km
2q


critical points are , 0, and , but only is in the domain. Then A 0 atÊÉÉ É É Š‹
2km 2km 2km 2km
hh h h
ww
q there is a minimum average weekly cost.σ
2km
h
(b) A(q) cm q kmq bm cm q, where q 0 A (q) 0 at q as in (a).œœ Êœœ
(k bq)m
q h
h h 2km"
##
" w
É
Also A (q) 2kmq 0 so the most economical quantity to order is still q which minimizes
ww $
œ œ É
2km
h
the average weekly cost.
46. We start with c x the cost of producing x items, x , and the average cost of producing x items, assumedabœ!œ
cx
x
ab
to be differentiable. If the average cost can be minimized, it will be at a production level at which
d
dx x
cx
Š‹
ab
œ!
(by the quotient rule) x c x c x (multiply both sides by x ) c x whereʜ! ʜ! ʜ
x c x c x cx
x x
"

ab ab ab w#w
ab ab ab
c x is the marginal cost. This concludes the proof. (Note: The theorem does not assure a production level that will give a
w
ab
minimum cost, but rather, it indicates where to look to see if there is one. Find the production levels where the average cost
equals the marginal cost, then check to see if any of them give a mimimum.)
47. The profit p(x) r(x) c(x) 6x x 6x 15x x 6x 9x, where x 0. Thenœ œ œ   ab
$# $#
p (x) 3x 12x 9 3(x 3)(x 1) and p (x) 6x 12. The critical points are 1 and 3. Thus
w# w w
œ œ œ
p (1) 6 0 at x 1 there is a local minimum, and p (3) 0 at x 3 there is a local maximum.
ww ww
œÊœ œ'Êœ
But p(3) 0 the best you can do is break even.œÊ
48. The average cost of producing x items is c x x x c x x x , theab abœ œ #! #!ß !!! Ê œ # #! œ ! Ê œ "!
cx
x
ab # w
only critical value. The average cost is c $ per item is a minimum cost because c .ab ab"! œ "*ß *!! "! œ # !
ww
49. (a) The artisan should order px units of material in order to have enough until the next delivery.
(b) The average number of units in storage until the next delivery is and so the cost of storing then is s per
px px
##
ˆ‰
day, and the total cost for x days is sx. When added to the delivery cost, the total cost for delivery and storageˆ‰
px
#
for each cycle is: cost per cycle d sx.œ
px
#
(c) The average cost per day for storage and delivery of materials is: average cost per day x.œ
ˆ‰dx
ps
#"
xx
d ps
#
To minimize the average cost per day, set the derivative equal to zero. d x x d x
d
dx
ps ps
Š‹ab ab
" #
##
œ œ!
x . Only the positive root makes sense in this context so that x . To verify that x gives aÊœ„ œÉÉ
# #‡‡d d
ps ps
minimum, check the second derivative d x a minimum.’“ˆ‰ab ºº
ddd
dx x
ps
œ œ !Ê
#
#
##
ÉÉ
""dd
ps ps
" "
Œ É
#d
ps
The amount to deliver is px .
‡ #
σ
pd
s
(d) The line and the hyperbola intersect when x. Solving for x gives x . For x ,
dd
xps
ps
intersection
œœ„!
#
#
É
x x . From this result, the average cost per day is minimized when the average daily cost ofintersection
d
ps œœÉ
# ‡
delivery is equal to the average daily cost of storage.
50. Average Cost: x x x . Check for a minimum:
cx cx
xx dxx x
dab ab
œ *' % Ê œ # œ ! Ê œ "!!
#!!! #!!! "Î# "Î#
Š‹
a minimum at x . At a production level of units,
d
dx x
cx
x

"
Š‹º
ab
œ"!!
%!!!
"!!
$Î#
œ "!! œ !Þ!!$ ! Ê œ "!! "!!ß !!!
the average cost will be minimized at $ per unit."&'

260. 256 Chapter 4 Applications of Derivatives
51. We have CM M . Solving C M M . Also, at M there is a
dR d R C d R C
dM dM dM
œ œ#œ!Êœ œ#!Ê œ
#
##
!

maximum.
52. (a) If v cr r cr , then v 2cr r 3cr cr 2r 3r and v 2cr 6cr 2c r 3r . The solution of*† *†* † *†* †
!!!!!
#$ w # ww ab ab
v 0 is r 0 or , but 0 is not in the domain. Also, v 0 for r and v 0 for r at
www
œœ Ê
2r 2r 2r
333) ))
r there is a maximum.œ
2r
3)
(b) The graph confirms the findings in (a).

53. If x 0, then (x 1) 0 x 1 2x 2. In particular if a, b, c and d are positive integers, Ê Ê 
## "x1
x

then 16.Š‹Š‹Š‹Š‹
a1 b1 c1 d
abcd

""""l
 
54. (a) f(x) f (x) 0œÊœ œ œ
xaxxa
ax
ax xax
ax
ax ax
È
ab ab
ab
ab ab







"
w
"

""
f(x) is an increasing function of xÊ
(b) g(x) g (x)œÊœ
dx
b(dx)
b (dx) (dx)b (dx)
b(dx)


w
È
ab ab




0 g(x) is a decreasing function of xœœÊ


ab
ab a b
b (dx) (dx)
b(dx) b (dx)
b




(c) Since c , c 0, the derivative is an increasing function of x (from part (a)) minus a decreasing"#
dt
dx
function of x (from part (b)): f(x) g(x) f (x) g (x) 0 since f (x) 0 and
dt d t
dx c c dx c c
œ Êœ
"" " " ww w



g (x) 0 is an increasing function of x.
w
Ê
dt
dx
55. At x c, the tangents to the curves are parallel. Justification: The vertical distance between the curves isœ
D(x) f(x) g(x), so D (x) f (x) g (x). The maximum value of D will occur at a point c where D 0. At*† * † *
www w
such a point, f (c) g (c) 0, or f (c) g (c).
ww w w
œ œ
56. (a) f(x) 3 4 cos x cos 2x is a periodic function with period 2œ 1
(b) No, f(x) 3 4 cos x cos 2x 3 4 cos x 2 cos x 1 2 1 2 cos x cos x 2(1 cos x) 0œœ œœ  aba b
###
f(x) is never negativeÊ
57. (a) If y cot x 2 csc x where 0 x , then y (csc x) 2 cot x csc x . Solving y 0*† !! * † *
ÈÈ
Š‹1
ww
cos x x . For 0 x we have y 0, and y 0 when x . Therefore, at xÊ œ Ê œ œ
" ww
È
2 44 4 4
11 1 1
1
there is a maximum value of y 1.œ

261. Section 4.5 Applied Optimization Problems 257
(b)
The graph confirms the findings in (a).
58. (a) If y tan x 3 cot x where 0 x , then y sec x 3 csc x. Solving y 0 tan x 3œ œ œÊ œ„
1
#
w# # w È
x , but is not in the domain. Also, y 2 sec x tan x 3 csc x cot x 0 for all 0 x .Êœ„ œ
11 1
33 2
ww # #
Therefore at x there is a minimum value of y 2 3.œœ
1
3 È
(b)
The graph confirms the findings in (a).
59. (a) The square of the distance is D x x x x x , so D x x and the criticalab abˆ‰ˆ ‰ Èœ !œ# œ##
$*
#%
# #
#w
point occurs at x . Since D x for x and D x for x , the critical point corresponds to theœ" ! " ! "
ww
ab ab
minimum distance. The minimum distance is D .Èab"œ
È&
#
(b)
The minimum distance is from the point to the point on the graph of y x, and this occurs at theˆ‰ ab È
$
#
ß! "ß" œ
value x where D x , the distance squared, has its minimum value.œ" ab
60. (a) Calculus Method:
The square of the distance from the point to x x is given byŠ‹Š ‹
È È
"ß $ ß "'
#
Dx x x x x x x x x.ab a b Š‹
ÈÈÈ Èœ " "' $ œ # ""' # %)$ $œ # #!# %)$
#
###
#
##
Then D x x . Solving D x we have: x x
w w#'
%)$ %)$
#
ab a b ab
È
œ # ' œ # œ! ' œ# %)$
ÈÈ
x x
x

262. 258 Chapter 4 Applications of Derivatives
x x x x x x . We discard x as an extraneous solution,Ê$'œ%%)$ Ê*œ%)$Ê"#œ%)Êœ„# œ#
#####
ab
leaving x . Since D x for x and D x for x , the critical point corresponds to theœ# ! % # ! # %
ww
ab ab minimum distance. The minimum distance is D .Èab#œ#
Geometry Method:
The semicircle is centered at the origin and has radius . The distance from the origin to is%"ß$ Š‹
È
. The shortest distance from the point to the semicircle is the distance along the radiusÊ Š‹
È" $ œ#
#
#
containing the point . That distance is .Š‹
È"ß $ % # œ #
(b)
The minimum distance is from the point to the point on the graph of y x , and thisŠ‹ Š ‹
ÈÈ È
"ß $ #ß # $ œ "'
#
occurs at the value x where D x , the distance squared, has its minimum value.œ# ab
61. (a) The base radius of the cone is r and so the height is h a r a . Therefore,œœœ
# #
##
## #
#
11
11
ax a x È É ˆ‰
Vx rh a .ab ˆ‰ ˆ‰É** †
111 1
11$$# #
# # #
##
#a x a x
(b) To simplify the calculations, we shall consider the volume as a function of r: volume f r r a r , where** †ab
È
1
$
# ##
ra. fr ra r r r a r r! œ œ † # # œ
w# #
$$ $
## ## "
#
#
ab a b a bŠ‹ Š‹
ÈÈ
”•”•
11 1d
dr
a r a r
r ra r
ÈÈ
ab


. The critical point occurs when r , which gives r a . Thenœœ œ œœ
1 1
$ $$$
#$ # #
$
#$ # '
”• É
ar r a
a r a r
ra r a
!


ÈÈ
ab È
h a r a . Using r and h , we may now find the values of r and hœœœœ œ œ
È ÉÉ
## # #
$$$ $ $
$$ 'aa aa a
ÈÈÈ
for the given values of a.
When a : r , hœ% œ œ à
%'
$$
%$
ÈÈ
When a : r , hœ& œ œ à
&'
$$
&$
ÈÈ
When a : r , hœ' œ# ' œ# $àÈÈ
When a : r , hœ) œ œ à
)'
$$
)$
ÈÈ
(c) Since r and h , the relationship is .œœ œ#
a a r
h
ÈÈ'
$$
$ È
62. (a) Let x represent the fixed value of x at the point P, so the P has the coordinates x a , and let m f x be the
! !!
w ab abߜ
slope of the line RT. Then the equation of the line RT is y m x x a. The y-intercept of this line isœab
!
m x a a mx , and the x-intercept is the solution of m x x a , or x . Let O designateab ab! œ œ! œ !! !
mx a
m
)
the origin. Then
(Area of triangle RST)

263. Section 4.5 Applied Optimization Problems 259
(Area of triangle ORT)œ#
(x-intercept of line RT)(y-intercept of line RT)œ#†
"
#
amxœ#†
"
#

!
ˆ‰ ab
mx a
m)
m œ ˆ‰ˆ‰
mx a mx a
mm))
m œ ˆ‰
mx a
m)#
mx*† †ˆ‰ !
#
a
m
Substituting x for x , f x for m, and f x for a, we have A x f x x .!
ww
#ab ab ab ab ”•*† †
fx
fx
ab
ab
"
(b) The domain is the open interval . To graph, let y f x , y f x NDER y , anda b ab ab a b É!ß "! œ œ & & " œ œ "#"
"!!
wx

y A x y x . The graph of the area function y A x is shown below.$# $
#œœ œab abŠ‹
y
y#


The vertical asymptotes at x and x correspond to horizontal or vertical tangent lines, which do not formœ! œ"!
triangles.
(c) Using our expression for the y-intercept of the tangent line, the height of the triangle is
amxfx fxx x x x œ † œ & "!! œ & "!! ab ab
ÈÈ
w ""
##
##
"!! "!!
xx
2 x 2 x
ÈÈ


We may use graphing methods or the analytic method in part (d) to find that the minimum value of A x occurs atab
x . Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the¸)Þ''
y-coordinate of the center of the ellipse.
(d) Part (a) remains unchanged. Assuming C B, the domain is C . To graph, not that !ß ab
f x B B B C x and f x x . Therefore we haveab ab a bÉ È
œ " œ œ #œ
xB
CC


## w BBx
C C x C C x
!#
!& &
ÈÈ
"" ""
Ax f x x x xab ab ӥ Π!#
*† † * † * †
w
#

#
#
fx
fx
Bx
CC x
ab
ab È
"

B C x Bx
CC x
BC B C x C x
Bx

#

#
B
C
Bx
CC xÈ
È
Š‹Š‹
ÈÈ""
#
""
#
""
"" ""

Bx BC B C x C x Bx BC C x B C xœœ
""
##BCx C x BCx C x
ÈÈ
"" ""”•”•Š‹Š‹
ÈÈ È
ab
# ##### ## ##
##
BC C C xœœ
"
##

BCx C x x C x
BC C C x
ÈÈ
Š‹
È
"" ""
""
"”•Š‹
È
##
#
Ax BC
w


abϠ
Š‹Š‹Š‹Š‹Š ‹ab
abx C x C C x C C x x C x
ÈÈ È È
ab
"" "" "" "" ##
"" ""
##
# "
xx
C x C x


xC x


x C Cx Cxœ #
BC C C x
xC x
Š‹
È
ab

#
""
"" "
–—
Š‹Š ‹
ÈÈ
# ## ##

x
C x


È
xCCxxCxœ#
BC C C x
xC x
Š‹
È
ab

#
""
"" "
”•
È
ab
####

##Cx
C x


È
CC x C Cx CC x C C xœœ
BC C C x BC C C x
xC x xC x
Š‹ Š‹
ÈÈ
ab ab

# #
"" ""
"" "
"" "

Š‹
ÈÈ
’“ ab
Cx
C x


È

## #######
xCCCxœ#
BC C C x
xC x
" ""
"" "

Š‹
È
ab

#
Š‹
È
## ##
To find the critical points for x C, we solve: x C C C x x C x C C C x! #œ Ê%% œ
## % ##% %## ## È
x C x x x C . The minimum value of A x for x C occurs at the critical pointÊ% $ œ!Ê % $ œ! !
%## ###
ab ab

264. 260 Chapter 4 Applications of Derivatives
x , or x . The corresponding triangle height isœœ
C C
È
$
#%
# $

amxfx fxxœ †ab ab
w
BCx œ
B
CÈ## Bx
CC
"
" #
"
)
É#
C
BCx œ
B
CÈ##
B
CC
Š‹
É
#
"
)
" #
"
)C
C
#
B œ
BC
C
ˆ‰
#
#
"
)
"
"BC
C
B œ
BB
##
$
B œ$
This shows that the traingle has minimum arrea when its height is B.$
4.6 INDETERMINATE FORMS AND L'HOPITAL'S RULE
^
1. l'Hopital: lim or lim lim lim ^
x2 x2 x2 x2ÄÄÄÄ
x2 x2 x2
x4 x 4 x4 x x x 4
x
" " ""
# ## #
œ#
œœ œ œ œ¹
abab
2. l'Hopital: lim 5 or lim 5 lim 5 1 5
5
^
x0 x0 x0ÄÄ Ä
sin 5x 5 cos 5x sin 5x sin 5x
x1 x 5x
x
œ œ œ œ†œ¹
œ!
3. l'Hopital: lim lim lim or lim lim ^
xxx xxÄ_ Ä_ Ä_ Ä_ Ä_
5x 3x 0x 3 10 5 5x 3x 5
7x 1 14x 14 7 7x 1 7
5
7


"
""

"
œœœ œœ
3
x
x

4. l'Hopital: lim lim or lim lim ^
x1 x1 x1 x1ÄÄ ÄÄ
x1 3x 3 x1
4x x 3 12x 1 11 4x x 3 x 4x + 4x + 3
xxx
"



"
" "
œœ œ
aba b
aba b
2
lim œœ
x1Ä
ab
ab
xx
4x + 4x + 3 11
3

""l
2
5. l'Hopital: lim lim lim or lim lim ^
xxx x x1Ä! Ä! Ä! Ä! Ä
1 cos x sin x cos x 1 cos x cos x
x2x2 x xco
cos x""
#l"
"
œœœ œ ”
ˆ‰
ab
2
s x•
lim lim œœ œ
x1 x1ÄÄ
sin x sin x sin x
x cos x x x cos x

2
abl" l" #
"" ”•
ˆ‰ˆ‰ˆ ‰
6. l'Hopital: lim lim lim or lim lim ^
xxxxxÄ_ Ä_ Ä_ Ä_ Ä_
#$ % % # !
"" $ "l ! "" l
"
l" "
xx x3 x3x
xx1 x x xx1


œ œ œ! œ œ œ!
"
"

"#
x
3
x
xx
7. lim lim 0
t0 t0ÄÄ
sin t 2t cos t
t1

œœ
8. lim lim 2
x
ÄÎ ÄÎ1)122
2x 2 2
cos x sin x

"
1
œœœ
9. lim lim
)1 )1ÄÄ
sin cos ))
1)""
"
œœœ"
10. lim lim lim
xx
x2 2 2ÄÎ ÄÎ ÄÎ111
1 sin x cos x sin x
1 cos 2x 2 sin 2x 4 cos 2x 4( 1) 4
""

œœœœ
11. lim lim
xxÄÎ% ÄÎ%11
sin x cos x cos x sin x
x

"##
##

)œ œœ#
ÈÈ
È
12. lim lim
xxÄÎ$ ÄÎ$11
cos x
x
sin x



"#
$

"
#

œœ
È

265. Section 4.6 Indeterminate Forms and L'Hopital's Rule 261^
13. lim x tan x lim lim
x2 x2 x2ÄÎ ÄÎ ÄÎ111
œ œ œ œ"ˆ‰
1
#"
"
"
ˆ‰ ˆ‰ abx sin x x cos x sin x
cos x sin x

""
14. lim lim lim
xxxÄ! Ä! Ä!
## %†!
"* "* l"
%
#†!(
x
xx 2x
x
ÈÈ
È
œœœœ!
%
"
x
15. lim lim lim
xx xÄ" Ä" Ä!
#$" #
" " "
#$ #
%
xx x
xx
xx x
xx



ab È
œœœ"

" "
x
16. lim lim lim
xx xÄ# Ä# Ä#
È abab
È
x5
x2x
x5 2x
2x 5

#




$
% '
" ""
"
œœœ
17. lim lim , where a 0.
x0 x0ÄÄ
Èab
ÈÈ
aa x a
x
aa
aax a

#" #
"
#
œœœ

18. lim lim lim lim
t0 t0 t0 t0ÄÄ ÄÄ
10(sin t t) 10(cos t )
ttt
sin t cos t 10"
$'''$
"! "! †" &
œœœœœ
ab
19. lim lim lim lim lim
x0 x0 x0 x0 x0ÄÄ Ä Ä Ä
x(cos x )
sin x x cos x sin x sin x
xsin x cos x xcos x sin x xcos x sin x xs"
"
" # #
œœœœ
in x cos x
cos x
"$ $
"
œœ$
20. lim lim
h0 h0ÄÄ
sin a h sin a cos a h cos a
h
ab ab
"
œœ!
21. lim lim an lim r an, where n is a positive integer.
r r1 r1Ä" ÄÄ
ar anr
r1
nab ab
nn
" †
"
"
œœœ
%#
22. lim lim lim x
xx xÄ! Ä! Ä!

Š‹ Š‹Š ‹ ˆ‰È
"" "
x x does not apply xx
x l'Hopital's rule 1
œ œ œ "†œ_
È
È
23. lim x x x lim x x x lim lim
xx xxÄ_ Ä_ Ä_ Ä_
Š‹Š‹Š‹
ÈÈ
œ œ œ
## ""
"" ""

"
xxx
xxx xxx
xxxÈ
ÈÈ
ab



x
x
xxx
x
x


"

" lim œœ
xÄ_
" "
l" l"
#
É

x
Š‹
l'Hopital's rule
is unnecessary
24. lim x tan lim lim lim sec sec
xxx xÄ_ Ä_ Ä_ Ä_
ˆ‰ ˆ‰
"" ##
xx
tan
œœ œ œ!œ"
Š‹



"
"

"
x
x
x x
x
#
#
secˆ‰
25. lim lim
xxĄ_ Ą_
$&
# # "
x3
xx 4x
œœ!
26. lim lim
x0 x0ÄÄ
sin x
tan x sec x
cos x((†"(
"" "" "" ""†" ""
((
œœœ
ab
ab

27. lim lim lim 9 3
xxxÄ_ Ä_ Ä_
È
È
9x 1
x1
9x 1 9
x1 1
"
"
"
"
œœœœÉÉ
È
28. lim 1
xÄ!

!#
È
È
x
sin x lim 1
œœœ
Ê
É
""
x

sin x
x
29. lim lim lim 1
x2 x2 x2ÄÎ ÄÎ ÄÎ11 1
## #
sec x cos x
tan x cos x sin x sin x
œœœ ˆ‰ˆ‰
""
30. lim lim lim cos x 1
xx xÄ! Ä! Ä!

cot x
csc x
œœœ
ˆ‰
Š‹
cos x
sin x
sin x

266. 262 Chapter 4 Applications of Derivatives
31. Part (b) is correct because part (a) is neither in the nor form and so l'Hopital's rule may not be used.^
0
0
_
_
32. Answers may vary.
(a) f x x ;g x xab abœ$ " œ
lim lim lim
xx xÄ_ Ä_ Ä_
f(x)
g(x) x
x
œœœ$
$"l $
"
(b) f x x ;g x xab abœ" œ
#
lim lim lim
xx xÄ_ Ä_ Ä_
f(x)
g(x) x x
x
œœœ!
"l l
#

(c) f x x ;g x xab abœœ"
#
lim lim lim
xx xÄ_ Ä_ Ä_
f(x)
g(x) x
xx
œœœ_

"l l
#
33. If f(x) is to be continuous at x 0, then lim f(x) f(0) c f(0) lim lim œœÊœœœ
x0 x0 x0ÄÄÄ
9x 3 sin 3x 9 9 cos 3x
5x 15x


lim lim .œœœ
x0 x0ÄÄ
27 sin 3x 81 cos 3x 27
30x 30 10
34. (a) For x 0, f (x) (x 2) 1 and g (x) (x 1) 1. Therefore, lim 1, while lim Áœœ œœ œœ
ww dd 1
dx dx g (x) 1 g(x)
f (x) f(x)
x0 x0ÄÄ
"
"
2.œœœ
x2 02
x1 01
""
""
(b) This does not contradict l'Hopital's rule because neither f nor g is differentiable at x 0^ œ
(as evidenced by the fact that neither is continuous at x 0), so l'Hopital's rule does not apply.^œ
35. The graph indicates a limit near 1. The limit leads to the
indeterminate form : lim
0
0x1
2x (3x 1) x 2
x1Ä



È
lim lim œœ
x1 x1ÄÄ
2x 3x x 2
x1 1
4x x x





9
!!

1œœœ
4
11
45
9

#

36. (a)
(b) The limit leads to the indeterminate form :__
lim x x x lim x x x lim lim
xx xxÄ_ Ä_ Ä_ Ä_
Š‹Š‹Š‹Š‹
ÈÈ
œ œ œ
## ""
"" "" ""
xxx
xxx xxx xxx
xxx x
È
ÈÈÈ
ab



lim œœœ
xÄ_
" " "
l" l"
" "! #
É
È

x
37. Graphing f x on th window by it appears that lim f x . However, we see that if we letab abœ Ò"ß "Ó Ò!Þ&ß "Ó œ !
"cos x
x
ß

x0Ä
u x , then lim f x lim lim lim .œœœœœ
' " "
###
x0 u0 u0 u0ÄÄ ÄÄ
ab
cos u sin u cos u
uu

38. (a) We seek c in so that . Since f c and g c c we have that a b ab ab#ß! œ œ œ œ" œ# œ
fc f f
gc g g c
"
"
ab ab a b
ab ab a b
! #
! # !% # # #
!# " " " ww
c.ʜ"

267. Section 4.7 Newton's Method 263
(b) We seek c in any open interval a b so that cabß œ œ œ œÊœÊœ
fc fb fa
gc gb ga b a baba ba c ba
ba ba ba
"
" ←←
ab ab ab
ab ab ab a ba b
←
# #
"""
(c) We seek c in so that c c c .ab!ß$ œ œ œ Ê œ Ê$ # "#œ!Ê œ
fc f f
gc g g c
c
"
"
←
ab ab ab
ab ab ab
È$ !
$ ! *! $ # $ $
$! " % " # " $(
(Note that c is not in the given interval .)œ!ß$
←" ← $
$
È
ab
39. (a) By similar triangles, where E is the point on AB such that CE AB :
PA CE
AB EB
œ¼
←→ ←→ ←→

Thus , since the coordinates of C are cos sin . Hence, x .
"← "←
← ←
"←xcos
sin sin
cos
))) ))
) ))
œß"œ ab))
ab
(b) lim x lim lim lim lim
)) ) ) )ÄÄ Ä Ä Ä00 0 0 0
ab" œ œ œ œ
))
)) ) )
)) ) )))) ))ab"←
←"←
" #cos
sin cos sin
sin cos cos sin sin cos sin
sin
)
)
lim lim œœœœ$
))ÄÄ00
)) ) )
))
)) )ab# $ !$
"
sin cos cos
cos cos
sin cos
(c) We have that lim x cos lim cos lim cos
)) )Ä_ Ä_ Ä_
# ‘abab ab ab ”•”•" " œ " œ " ")))
))
)) ))
)ab"←
←←
cos
sin sin
As cos oscillates between and , and so it is bounded. Since lim ,))Ä" !# "œ""œ!_
Ä_
, ab ˆ‰
)
)
))←sin
lim cos . Geometrically, this means that as , the distance between points P and D
)Ä_
ab ”•" " œ! Ä_))
)
))←sin
approaches 0.
40. Throughout this problem note that r y , r y and that both r and y as
##
#
œ " Ä_ Ä_ Ä Þ )
1
(a) lim r y lim
)1 )1ÄÎ ÄÎ22
œ œ!
"
"ry
(b) lim r y lim
)1 )1ÄÎ ÄÎ22
##
œ "œ"
(c) We have that r y r y r ry y y .
$$ # # "" "" $
"
œ œ œ œ$†aba b
r ryy y yyy y y
ry r r r
←←← ←←
Since lim y lim sin y we have that lim r y .
)1 )1 )1ÄÎ ÄÎ ÄÎ22 2
$† œ $ † œ_ œ_
y
r
)
$$
4 7 NEWTON'S METHODÞ
1. y x x 1 y 2x 1 x x ; x 1 x 1œÊ œÊ œ œÊ œ œ
#w
!"

#"
n1 n←
xx1
x1
"
nn
n

# 111 2
13
x x .61905; x 1 x 1 2Ê œ Ê œ œ œ ¸ œÊ œ œ## !"

"
"
# #
2 2 469 2 13 111
331293121 1
1
1
42
93
4
3
x 2 1.66667Ê œ œ ¸#
←←"

42 5
41 3
2. y x 3x 1 y 3x 3 x x ; x 0 x 0œÊ œ Ê œ œÊ œœ
$w#
!"
""
n1 n←
x3x1
3x 3
#
"
n n
n
←←
←
33
x 0.32222Êœ œœ¸#
"""
"3 3 90 90
11
3
29
#
←
#
7
3

268. 264 Chapter 4 Applications of Derivatives
3. y x x 3 y 4x 1 x x ; x 1 x 1œÊ œ Ê œ œÊ œ œ
%w$
!"

"
n1 n
xx3
4x 1
)
#
nn
n

113 6
41 5
x 1.16542; x 1 x 1Êœ œ œ œ ¸ œÊœ# !"

"
"
"
6 6 1296 750 1875 6 171 5763 1 3
5 5 4320 625 5 4945 4945 4 1
3
1
1296 6
625 5
864
125
2 x 2 1.64516œ Ê œ# œ œ ¸#

#
16 2 3 11 51
3 1 31 31
4. y 2x x 1 y 2 2x x x ; x 0 x 0œÊœ Ê œ œÊ œ œ
#w
!"
" "
# #
n1 n
2xx1
22xnn
n
#
"
00
0
x .41667; x 2 x 2 xÊœ œœ¸ œÊœ œÊœ#!"#
""" "
# # # # # # # # #
" 1 51
111 4 5
54455
#
4 4
25
2.41667œ œ œ ¸
5 20254 5 29
1112## ##
"
5. y x 2 y 4x x x ; x 1 x 1 xœÊœ Ê œ œÊœ œÊœ œ
%w$
!" #
"
n1 n
x2
4x
)
#
n
n
# 2 5 5 5 625 512
4 4 4 4 2000
2
625
256
125
16
1.1935œ œ œ ¸
5 113 2500 113 2387
4 2000 2000 000

#
6. From Exercise 5, x x ; x 1 x 1 1 x
n1 nœ œ Ê œ œ œ Ê œ
x2
4x
)
#
n
n
#
!" #
" "



255
444 4
2
625
256
125
16
1.1935œ œ ¸
5 625 512 5 113
4 2000 4 2000


7. f(x ) 0 and f (x ) x x gives x x x x x x for all n 0. That is, all of! ! "! #! !
wœ Á!Ê œ œÊœÊœ  
n1 n n
fx
fx
ab
abn
n

the approximations in Newton's method will be the root of f(x) 0.œ
8. It does matter. If you start too far away from x , the calculated values may approach some other root.œ
1
#
Starting with x 0.5, for instance, leads to x as the root, not x .!
##œ œ œ
11
9. If x h 0 x x h!"!œ Ê œ œ
f(x ) f(h)
f(x) f(h))
""
)
hhh2hh;œ œ œ
È
Š‹
h
#


h
Š‹Š ‹
ÈÈ
if x h 0 x x h!"!

œ Ê œ œ
f(x ) f( h)
f(x) f( h))
""
)
hhh2hh.œ œ œ
È
Š‹
h
%#
2h

Š‹Š ‹
ÈÈ

10. f(x) x f (x) x x xœÊ œ Êœ
"Î$ w #Î$ " ˆ‰
3
n1 n
x
x


n
3
n
ˆ‰
2x ; x 1 x 2, x 4, x 8, andœ œ Ê œ œ œ
n!"#$
x 16 and so forth. Since x 2 x we may conclude%œœll kk
nn1 #
that n x .Ä_Ê Ä_kk
n

11. i) is equivalent to solving x x .
$
$ "œ!
ii) is equivalent to solving x x .
$
$ "œ!
iii) is equivalent to solving x x .
$
$ "œ!
iv) is equivalent to solving x x .
$
$ "œ!
All four equations are equivalent.
12. f(x) x 1 0.5 sin x f (x) 1 0.5 cos x x x ; if x 1.5, thenœ Ê œ Ê œ œ
w
!
n1 n
x 1 0.5 sin x
1 0.5 cos xnn
n##
#
x 1.49870"œ

269. Section 4.7 Newton's Method 265
13. For x , the procedure converges to the root !œ !Þ$ !Þ$##")&$&ÞÞÞÞ
(a)
(b)
(c)
(d) Values for x will vary. One possible choice is x 1.
!œ!Þ
(e) Values for x will vary.
14. (a) f(x) x 3x 1 f (x) 3x 3 x x the two negative zeros are 1.53209œÊ œ Ê œ Ê
$w#
n1 nœ
x3x1
3x 3
œ

n n
nœœ
œ
and 0.34730
(b) The estimated solutions of x 3x 1 0 are
$
œ
1.53209, 0.34730, 1.87939.

(d) The estimated x-values where
g(x) 0.25x 1.5x x 5 has horizontal tangentsœ
%#
are the roots of g (x) x 3x 1, and these are
w$
œ
1.53209, 0.34730, 1.87939.

15. f(x) tan x 2x f (x) sec x 2 x x ; x 1 x 12920445œÊ œ Êœ œÊœ
w#
!"
n1 nœ
tan x 2x
sec x
ab
abnn
nœ

x 1.155327774 x x 1.165561185Êœ Ê œœ# 16 17
16. f(x) x 2x x 2x 2 f (x) 4x 6x 2x 2 x x ;œ Ê œ Ê œ
%$# w $#
n1 nœ
x 2xx2x2
4x 6x 2x 2


nnn n
nn n

œœœ
if x 0.5, then x 0.630115396; if x 2.5, then x 2.57327196!% !%œœ œœ

270. 266 Chapter 4 Applications of Derivatives
17. (a) The graph of f(x) sin 3x 0.99 x in the windowœ
#
2 x 2, 2 y 3 suggests three roots.Ÿ Ÿ Ÿ Ÿ
However, when you zoom in on the x-axis near x 1.2,œ
you can see that the graph lies above the axis there.
There are only two roots, one near x 1, the otherœ
near x 0.4.œ
(b) f(x) sin 3x 0.99 x f (x) 3 cos 3x 2xœÊœ
#w
x x and the solutionsʜ
n1 nœ
sin (3x ) 0.99 x
3 cos (3x ) 2xn n
nn


are approximately 0.35003501505249 and
1.0261731615301

18. (a) Yes, three times as indicted by the
graphs
(b) f(x) cos 3x x f (x)œÊ
w
3 sin 3x 1 xœ Ê
n1œ
x; atœ
n
cos 3x x
3 sin 3x 1
ab
ab nn
nœ
œœ
approximately 0.979367,
0.887726, and 0.39004 we have
cos 3x xœ

19. f(x) 2x 4x 1 f (x) 8x 8x x x ; if x 2, then x 1.30656296; ifœÊ œÊ œ œ œ
%# w $
!'
n1 nœ
2x 4x 1
8x 8x

œ
nn
n n
œ
x 0.5, then x 0.5411961; the roots are approximately 0.5411961 and 1.30656296 because f(x) is!$œ œ „ „
an even function.
20. f(x) tan x f (x) sec x x x ; x 3 x 3.13971 x 3.14159 and weœÊœ Êœ œÊœ Êœ
w#
!" #
n1 nœ
tan x
sec x
ab
abn
n

approximate to be 3.14159.1
21. From the graph we let x 0.5 and f(x) cos x 2x
!œœ
x x x .45063ʜ ʜ
n1 nœ
cos x 2x
sin x 2
ab
abnn
nœ
œœ
"
x .45018 at x 0.45 we have cos x 2x.Êœ Ê ¸ œ#

22. From the graph we let x 0.7 and f(x) cos x x
!œ œ
x x x .73944ʜ ʜ
n1 nœ
x cos x
1sinxnn
n
œ
ab
ab
"
x .73908 at x 0.74 we have cos x x.Ê œ Ê ¸ œ#

271. Section 4.7 Newton's Method 267
23. If f(x) x 2x 4, then f(1) 1 0 and f(2) 8 0 by the Intermediate Value Theorem the equationœ œ œÊ
$
x 2x 4 0 has a solution between 1 and 2. Consequently, f (x) 3x 2 and x x .
$ w#
œ œ œ
n1 nœ
x2x4
3x 2
œ

n n
n


Then x 1 x 1.2 x 1.17975 x 1.179509 x 1.1795090 the root is approximately!" # $ %œÊ œ Ê œ Ê œ Ê œ Ê
1.17951.
24. We wish to solve 8x 14x 9x 11x 1 0. Let f(x) 8x 14x 9x 11x 1, then
%$# %$#
œ œ
f (x) 32x 42x 18x 11 x x .
w$#
œÊœ
n1 nœ
8x 14x 9x 11x 1
3 x 42x 18x 11


nnn n
nn n

#
x approximation of corresponding root
1.0 0.976823589
0.1 0.100363332
0.6 0.642746671
2.0 1.983713587!

25. f(x) 4x 4x f (x) 16x 8x x x x . Iterations are performed using theœ Ê œ Ê œ œ
%# w $
i
œ1i i
fx
fx
xx
x
ab
abi
i
i i
i
%
œ

œ
ޜ!
procedure in problem 13 in this section.
(a) For x or x , x as i gets large.
!!œ# œ!Þ) Ä" i
(b) For x or x , x as i gets large.!!œ !Þ& œ !Þ#& Ä !i
(c) For x or x , x as i gets large.!!œ!Þ) œ# Ä" i
(d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.)
For x or x , Newton's method does not converge. The values of x alternate between!!œ œ
ÈÈ
21 21
77
i
x or x as i increases.!!œ œ
ÈÈ
21 21
77
26. (a) The distance can be represented by
D(x) (x 2) x , where x 0. Theœ  É ˆ‰
## "
#
#
distance D(x) is minimized when
f(x) (x 2) x is minimized. Ifœ
## "
#
#ˆ‰
f(x) (x 2) x , thenœ
## "
#
#ˆ‰
f (x) 4 x x 1 and f (x) 4 3x 1 0.
w$ w # w
œ œ ab ab
Now f (x) 0 x x 1 0 x x 1 1
w$ #
œÊ œÊ œ ab
x.ʜ
"
œx1
œ

(b) Let g(x) x x 1 x g (x) x 1 (2x) 1 1œœÊ œ œ
"
œ
#w#
" #
œx1
2x
x1
œ
œ
œ ab ab
ab
x x ; x 1 x 0.68233 to five decimal places.Êœ œÊœ
n1 nœ
Œ œ
ÎÑ
ÏÒ
"

Î
&

Î&

x1
n
n
2xn
x1 1
n
œx

!%
27. f(x) (x 1) f (x) 40(x 1) x x . With x 2, our computerœ Ê œ Ê œ œ œ
%! w $*
!
n1 nœ
ab
ab
x1
40 x 1
39x
40n
n
nœ
œ
"
*Ð
œ&
gave x x x x 1.11051, coming within 0.11051 of the root x 1.)( )) )* #!!œ œ œâœ œ œ
28. f(x) 4x 4x f (x) 16x 8x 8x 2x 1 x x ; if x .65, thenœ Ê œ œ Ê œ œ
%# w $ #
!
ab
n1 nœ
xx
22x 1nn
nab
ab


œ$
œ
x .000004, if x .7, then x 1.000004; if x .8, then x 1.000000. NOTE: .654654"# ! "# ! '¸ œ œ œ œ ¸
È
21
7
29. f(x) x 3.6x 36.4 f (x) 3x 7.2x x x ; x 2 x 2.5303œ Ê œ Ê œ œÊ œ
$# w #
!"
n1 nœ
x 3.6x 36.4
3x 7.2x


nn
n n

x 2.45418225 x 2.45238021 x 2.45237921 which is 2.45 to two decimal places. Recall thatʜ ʜ ʜ#$%

272. 268 Chapter 4 Applications of Derivatives
x 10 H O H O (x) 10 (2.45) 10 0.000245œÊœœ œ
% % %
$$
cdcd ab ab
30. Newton's method yields the following:
the initial value 2 i 3 i
the approached value 1 5.55931i 29.5815 17.0789i
È


4.8 ANTIDERIVATIVES
1. (a) x (b) (c) x x
# #xx
33


2. (a) 3x (b) (c) 3x 8x
# #xx
88
* *

3. (a) x (b) (c) x 3x
$ #

xx
33

4. (a) x (b) (c) x
#
#
xx xx
43 2

5. (a) (b) (c) 2x
"
xx x
55

6. (a) (b) (c)
"" "
#x4x4x
x
œ œ œ
(
7. (a) x (b) x (c) x 2 x
È ÈÈÈ
$ $2
3

8. (a) x (b) x (c) x x
%Î$ #Î$ %Î$ #Î$ "
##
33
4

9. (a) x (b) x (c) x
#Î$ "Î$ "Î$
10. (a) x (b) x (c) x
"Î# "Î# $Î#
11. (a) cos ( x) (b) 3 cos x (c) cos (3x)1
cos ( x)1
1
12. (a) sin ( x) (b) sin (c) sin sin x1 1ˆ‰ ˆ‰ ˆ‰
11
1
x2x
##

13. (a) tan x (b) 2 tan (c) tan ˆ‰ ˆ ‰
x23x
33

#
14. (a) cot x (b) cot (c) x 4 cot (2x) ˆ‰
3x
#
15. (a) csc x (b) csc (5x) (c) 2 csc
"
#5
x
ˆ‰
1
16. (a) sec x (b) sec (3x) (c) sec
42x
3 1
1
ˆ‰
#
17. (x 1) dx x C 18. (5 6x) dx 5x 3x C''œ œ
x
œ
#
#
19. 3t dt t C 20. 4t dt t C''ˆ‰ Š‹
#$ $ %
##
œ œ
tt t t
4 6
œ œ

273. Section 4.8 Antiderivatives 269
21. 2x 5x 7 dx x x 7x C 22. 1 x 3x dx x x x C''ab ab
$%# #&$' " ""
## #
œ œ
5
3
23. x dx x x dx x C C''ˆ‰ˆ ‰
"" " " "###
x 3 3 1 33 x33
xx xx
œ
œ œ œ
24. 2x dx 2x 2x dx x C x C'ˆ‰ˆ ‰ ' Š‹
""" " $ #
# #5x 5 5 xx
22x2x5
œ œ œ
œ
Þœ œ
25. x dx C x C 26. x dx C C''"Î$ #Î$ &Î%
#

œœ œ œ
x3 x 4
x
œ )Î&(
Î (
2
3 4
È
27. x x dx x x dx C x x C''ˆ‰ˆ‰ÈÈœœœ
$ "Î# "Î$ $Î# %Î$ xx 2 3
34

œ
34
3
28. dx x 2x dx 2 C x 4x C''Š ‹ Š‹ Š‹ ˆ‰
È
È
x 2xx
x 3## #
"" ""Î# "Î# $Î# "Î#
œ œ œ

œ
Î
œ
3
29. 8y dy 8y 2y dy 2 C 4y y C''Š‹ Š‹ ˆ‰œ œœ
2 8
y
8y y
3
Î&(

"Î% # $Î%
#
3
4
30. dy y dy y C C''Š‹ Š‹ ˆ‰
""" &Î%
777 7
1 4
y y
yy
œ œœ
Ð&( Î&(
)Î&( 1
4
31. 2x 1 x dx 2x 2x dx 2 C x C''ab a b Š‹œ œœ
$ # #
#
2x x 2
1x
œ )Î
32. x (x 1) dx x x dx C C''$ # $
# #
""
œ œ œab Š‹
xx
1xx
Þ# Þœ
œ
33. dt dt t t dt C 2 t C'' '
tt t
ttt
tt t t 2
t
ÈÈ
È
œ "Î# $Î#

œœœ

ÎÎ
œœ œœ œœŠ‹ Š‹ ˆ‰ È
34. dt dt 4t t dt 4 C C'' '
4t
ttt t
4t t t 2 2
3t
œ $ &Î#
#
È
œ

œ
œ œ œ œŠ‹ Š‹Š‹ ˆ‰
3
35. 2 cos t dt 2 sin t C 36. 5 sin t dt 5 cos t C''œ œ
37. 7 sin d 21 cos C 38. 3 cos 5 d sin 5 C''
))
33 5
3
))))œ œ
39. 3 csc x dx 3 cot x C 40. dx C''œ œ
# sec x tan x
33
œ
41. d csc C 42. sec tan d sec C''
csc cot 2 2
55
))
##
"
)) ))))œ œ
43. 4 sec x tan x 2 sec x dx 4 sec x 2 tan x C
'ab œ
#
44. csc x csc x cot x dx cot x csc x C'
"""#
##2
abœ
45. sin 2x csc x dx cos 2x cot x C 46. (2 cos 2x 3 sin 3x) dx sin 2x cos 3x C''ab œ œ
# "
#
47. dt cos 4t dt t C C''
1 cos 4t sin 4t t sin 4t
428
œ""""
### ##
œ œ œˆ‰ ˆ‰
48. dt cos 6t dt t C C''
1 cos 6t sin 6t t sin 6t
6212
""""
### ##
œ œ œˆ‰ ˆ‰

274. 270 Chapter 4 Applications of Derivatives
49. 1 tan d sec d tan C''abœ œ
##
)) )) )
50. 2 tan d 1 1 tan d 1 sec d tan C
'' 'aba babœœœ
###
)) )) )) ) )
51. cot x dx csc x 1 dx cot x x C
''##
œœab
52. 1 cot x dx 1 csc x 1 dx 2 csc x dx 2x cot x C
'' 'aba bab abœœœ
## #
53. cos (tan sec ) d (sin 1) d cos C')) )) ) ) ))œœ '
54. d d d d sec d tan C''csc csc sin
csc sin csc sin sin 1 sin cos
)))
)) )) ) ) )
"" #
)))))))œœœœœˆ‰ˆ‰ '''
œœ
55. C (7x 2)
d
dx 28 28
(7x 2) 4(7x 2) (7)
Š‹
$

œ œ
56. C (3x 5)
d
dx 3 3
(3 x 5) (3x 5) (3)
Š‹Š‹œ œ
œœ #
Þ# Þœ
57. tan (5x 1) C sec (5x 1) (5) sec (5x 1)
d
dx 5 5
ˆ‰ ab
"" ##
œ œ
58. 3 cot C 3 csc csc
dx x x
dx 3 3 3 3
ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ ˆ‰œ œ
" " " " ##
59. C ( 1)( 1)(x 1) 60. C
d dx
dx x 1 (x 1) dx x 1 (x 1) (x 1)
(x 1)( ) x(1)
ˆ‰ ˆ‰
" " "
œœœœœ
# "
œ œ œ œ
œ œœ
61. (a) Wrong: sin x C sin x cos x x sin x cos x x sin x
dx 2x x x
dx
Š‹
œœœ
### #
œœÁ
(b) Wrong: ( x cos x C) cos x x sin x x sin x
d
dx
œÁ
(c) Right: ( x cos x sin x C) cos x x sin x cos x x sin x
d
dx
œœ
62. (a) Wrong: C (sec tan ) sec tan tan sec
dsec 3 sec
d3 3)
))
Š‹

œ œ Á )) )) ) )
$#
(b) Right: tan C (2 tan ) sec tan sec
d
d)
ˆ‰
""
##
###
)))))œ œ
(c) Right: sec C (2 sec ) sec tan tan sec
d
d)
ˆ‰
""
##
##
))))))œ œ
63. (a) Wrong: C 2(2x 1) (2x 1)
d
dx 3 3
(2x 1) 3(2x 1) (2)
Š‹
œœ ##

œ œ Á
(b) Wrong: (2x 1) C 3(2x 1) (2) 6(2x 1) 3(2x 1)
d
dx
abœ œÁ
$###
(c) Right: (2x 1) C 6(2x 1)
d
dx
abœ
$#
64. (a) Wrong: x x C x x C (2x 1) 2x 1
d2x1
dx 2x xC
abab È
##
"Î# "Î#"œ
# œœ
œ œ Á
Ȝ
(b) Wrong: x x C x x (2x 1) 2x 1
d2x1
dx 2x x
Š‹ab ab È
##
"Î# "Î# "œ
# œ
œ œ Á
Ȝ
(c) Right: 2x 1 C (2x 1) C (2x 1) (2) 2x 1
dd3
dx 3 dx 3 6
Œ œŠ‹ÈÈ ˆ‰
""
$
$Î# "Î#
œ œ œ
65. Graph (b), because 2 y x C. Then y(1) 4 C 3.
dy
dx
œBÊ œ œ Ê œ
#
66. Graph (b), because y x C. Then y( 1) 1 C .
dy dx
3
œB Ê œ œ Ê œ
"
##
#

275. Section 4.8 Antiderivatives 271
67. 2x 7 y x 7x C; at x 2 and y 0 we have 0 2 7(2) C C 10 y x 7x 10
dy
dx
œÊœ œ œ œ Êœ Êœ
###
68. 10 x y 10x C; at x 0 and y 1 we have 1 10(0) C C 1
dy dx
x0
œÊœ œ œ œ Êœ
œ œ
##
y 10x 1ʜ
x
œ
#
69. x x x y x C; at x 2 and y 1 we have 1 2 C C
dy
dx x
x2
œœ Êœ œ œ œÊœ
" "# " "
###
œ
œ œ
y x or yÊœ œ
"
## ##
"""xx
x
œœ
70. 9x 4x 5 y 3x 2x 5x C; at x 1 and y 0 we have 0 3( 1) 2( 1) 5( 1) C
dy
dx
œÊœ œ œ œ
#$# $#
C 10 y 3x 2x 5x 10ʜ ʜ
$#
71. x y C ; at x 9x C; at x and y we have ( ) C C
dy
dx
x
œ$ Ê œ œ* œ œ" œ& &œ*" Ê œ%
#Î$ "Î$ "Î$ $

Î

y9xʜ %
"Î$
72. x y x C; at x 4 and y 0 we have 0 4 C C 2 y x 2
dy
dx x
œ œ Êœ œ œ œ Ê œÊœ
""
# #
"Î# "Î# "Î# "Î#
È
73. 1 cos t s t sin t C; at t 0 and s 4 we have 4 0 sin 0 C C 4 s t sin t 4
ds
dt
œ Ê œ œ œ œ Ê œ Ê œ
74. cos t sin t s sin t cos t C; at t and s 1 we have 1 sin cos C C 0
ds
dt
œÊœ œ œ œ Êœ 111
s sin t cos tʜ
75. sin r cos ( ) C; at r 0 and 0 we have 0 cos ( 0) C C r cos ( ) 1
dr
d)
œ Ê œ œ œ œ Ê œ" Ê œ 11) 1) ) 1 1)
76. cos r sin( ) C; at r 1 and 0 we have 1 sin ( 0) C C r sin ( ) 1
dr
d)1 1 1
œÊœœœ œÊœ"Êœ1) 1) ) 1 1)
"""
77. sec t tan t v sec t C; at v 1 and t 0 we have 1 sec (0) C C v sec t
dv
dt
œÊœœœœÊœÊœ
"" " """
## # ###
78. 8t csc t v 4t cot t C; at v 7 and t we have 7 4 cot C C 7
dv
dt
œ Ê œ œ œ œ Ê œ
## #
###
#
111 ˆ‰ ˆ‰ 1
v 4t cot t 7ʜ
##
1
79. 2 6x 2x 3x C ; at 4 and x 0 we have 4 2(0) 3(0) C C 4
d y dy dy
dx dx dx
œ
œ
œ Ê œ œ œ œ Ê œ
##
""" 2x 3x 4 y x x 4x C ; at y 1 and x 0 we have 1 0 0 4(0) C C 1Ê œ Êœ œ œ œ Ê œ
dy
dx
##$ #$
###
yxx4x1ʜ
#$
80. 0 C ; at 2 and x 0 we have C 2 2 y 2x C ; at y 0 and x 0 we
d y dy dy dy
dx dx dx dx
œ
œ
œÊ œ œ œ œÊ œÊœ œ œ"" #
have 0 2(0) C C 0 y 2xœÊœÊœ ##
81. 2t t C ; at 1 and t 1 we have 1 (1) C C 2 t 2
dr 2 dr dr dr
dt t dt dt dt
œ
œ
œ œ Ê œ œ œ œ Ê œ Ê œ
$ # # #
"""
r t 2t C ; at r 1 and t 1 we have 1 1 2(1) C C 2 r t 2t 2 orÊœ œ œ œ Ê œÊœ
" " "
###
r2t2œ
"
t
82. C ; at 3 and t 4 we have 3 C C 0 s C ; at
d s 3t ds 3t ds ds 3t t
dt 8 dt16 dt 16 dt16 16
3(4)
œœ
œ
œ
œÊœ œ œ œ Ê œÊœ Êœ"""#
s 4 and t 4 we have 4 C C 0 sœœ œÊœÊœ
4t
16 16

##

276. 272 Chapter 4 Applications of Derivatives
83. 6 6x C ; at 8 and x 0 we have 8 6(0) C C 8 6x 8
dy dy dy dy
dx dx dx dx
œ
œ
œÊ œ œ œ œ Ê œÊ œ""" 3x 8x C ; at 0 and x 0 we have 0 3(0) 8(0) C C 0 3x 8xÊœ œ œ œ Ê œÊœ
dy dy dy
dx dx dx
###
###
y x 4x C ; at y 5 and x 0 we have 5 0 4(0) C C 5 y x 4x 5Êœ œ œ œ Ê œÊœ
$# $ # $#
$$$
84. 0 C ; at 2 and t 0 we have 2 2t C ; at and t 0 we
ddd d d d
dt dt dt dt dt dt


))) ) ) )
œ Ê œ œ œ œ Ê œ œ œ" #
"
# have 2(0) C C 2t t t C ; at 2 and t 0 we have œ Ê œ Ê œ Ê œ œ œ
""""
####
## $
#d
dt
)
))
È
20 (0)C C 2 t t 2
ÈÈÈ
œ Ê œ Ê œ
##""
##
$$
)
85. y sin t cos t y cos t sin t C ; at y 7 and t 0 we have 7 cos (0) sin (0) C
Ð%Ñ www www
" "
œ Ê œ œ œ œ
C 6 y cos t sin t 6 y sin t cos t 6t C ; at y 1 and t 0 we haveʜʜ Êœ œ œ
" #
www ww ww
1 sin (0) cos (0) 6(0) C C 0 y sin t cos t 6t y cos t sin t 3t C ;œ Ê œ Ê œ Ê œ ## $
ww w #
at y 1 and t 0 we have 1 cos (0) sin (0) 3(0) C C 0 y cos t sin t 3t
w #w#
$$
œ œ œ Ê œ Ê œ
y sin t cos t t C ; at y 0 and t 0 we have 0 sin (0) cos (0) 0 C C 1Ê œ œ œ œ Ê œ
$ $
% %%
y sin t cos t t 1ʜ
$
86. y cos x 8 sin (2x) y sin x 4 cos (2x) C ; at y 0 and x 0 we have
Ð%Ñ www www
"
œ Ê œ œ œ
0 sin (0) cos (2(0)) C C 4 y sin x 4 cos (2x) 4 y cos x 2 sin (2x) 4x C ;œ % Ê œ Ê œ Ê œ
"" #
www ww
at y 1 and x 0 we have 1 cos (0) 2 sin (2(0)) 4(0) C C 0 y cos x 2 sin (2x) 4x
ww ww
##
œœ œ ʜʜ
y sin x cos (2x) 2x C ; at y 1 and x 0 we have 1 sin (0) cos (2(0)) 2(0) C C 0Êœ œ œ œ Ê œ
w#w #
$ $$
y sin x cos (2x) 2x y cos x sin (2x) x C ; at y 3 and x 0 we haveÊœ Êœ œ œ
w# $ "
#
%
2
3
3 cos (0) sin (2(0)) (0) C C 4 y cos x sin (2x) x 4œ Ê œ Ê œ
" "
# #
$$
%%
22
33
87. m y 3 x 3x y 2x C; at ( 4) we have 4 2(9) C C 50 y 2x 50œœœ Êœ*ß œ ʜʜ
w "Î# $Î# $Î# $Î#È
88. (a) 6x 3x C ; at y 0 and x 0 we have 0 3(0) C C 0 3x
dy dy dy
dx dx dx
œ
œ
œÊœ œ œ œ ʜʜ
#w # #
"""
y x C ; at y 1 and x 0 we have C 1 y x 1Êœ œ œ œÊœ
$$
##
(b) One, because any other possible function would differ from x 1 by a constant that must be zero because
$

of the initial conditions
89. 1 x y 1 x dx x x C; at (1 0.5) on the curve we have 0.5 1 1 C
dy
dx 3 3
44
œ Êœ œ ß œ
"Î$ "Î$ %Î$ %Î$'ˆ‰
C 0.5 y x xʜ ʜ
%Î$ "
#
90. x 1 y (x 1) dx x C; at ( 1 1) on the curve we have 1 ( 1) C
dy ()
dx
x
œ Ê œ œ ß œ '
œ œ
##
"
C y xʜʜ
""
###
x
œ
91. sin x cos x y (sin x cos x) dx cos x sin x C; at ( 1) on the curve we have
dy
dx
œ Êœ œ ß' 1
cos ( ) sin ( ) C C 2 y cos x sin x 2"œ Ê œÊ œ 11
92. sin x x sin x y x sin x dx x cos x C; at (1 ) on the
dy
dx x
œ œ Êœ œ ß#
"" "
# ##
"Î# "Î# "Î#
È
11 11 1 1 'ˆ‰
curve we have 2 1 cos (1) C C 0 y x cos xœ ʜʜ
"Î#
11 È
93. (a) 9.8t 3 s 4.9t 3t C; (i) at s 5 and t 0 we have C 5 s 4.9t 3t 5;
ds
dt
œÊœ œ œ œÊœ
# #
displacement s(3) s(1) ((4.9)(9) 9 5) (4.9 3 5) 33.2 units; (ii) at s 2 and t 0 we haveœœ œ œ œ
C 2 s 4.9t 3t 2; displacement s(3) s(1) ((4.9)(9) 9 2) (4.9 3 2) 33.2 units;œÊœ œœ œ
#
(iii) at s s and t 0 we have C s s 4.9t 3t s ; displacement s(3) s(1)œœ œÊœ œ !!!
#
((4.9)(9) 9 s ) (4.9 3 s ) 33.2 unitsœœ !!

277. Section 4.8 Antiderivatives 273
(b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is
s f(t) C for some constant C. Therefore, the displacement from t a to t b is (f(b) C) (f(a) C)œ œ œ
f(b) f(a). Thus we can find the displacement from any antiderivative f as the numerical differenceœ
f(b) f(a) without knowing the exact values of C and s.
94. a(t) v (t) 20 v(t) 20t C; at (0 ) we have C 0 v(t) 20t. When t 60, then v(60) 20(60)œœÊœ ß! œÊœ œ œ
w
1200 m/sec.œ
95. Step 1: k kt C ; at 88 and t 0 we have C 88 kt 88
d s ds ds ds
dt dt dt dt
œ
œ
œ Ê œ œ œ œ Ê œ Ê"" s k 88t C ; at s 0 and t 0 we have C 0 s 88tœ œ œ œ Ê œ Š‹
t kt
œ œ
# #
##
Step 2: 0 0 kt 88 t
ds 88
dt k
œÊœ Êœ
Step 3: 242 88 242 242 k 16œ Êœ Êœ Êœ

#
k 88
k2kk2k
(88) (88) (88)
ˆ‰
88
k
œ
œœ œ
ˆ‰
96. k k dt kt C; at 44 when t 0 we have 44 k(0) C C 44
d s ds ds
dt dt dt
œ
œ
œ Ê œ œ œ œ œ Ê œ'
kt 44 s 44t C ; at s 0 when t 0 we have 0 44(0) C C 0ʜʜ œ œ œ Ê œ
ds kt
dt
k(0)
œ œ
##
"""
s 44t. Then 0 kt 44 0 t and s 44 45Êœ œÊ œÊœ œ œ
kt ds 44 44 44
dt k k k
k
œ
œ
# #
ˆ‰ ˆ‰
ˆ‰
44
k
45 45 k 21.5 sec.Ê œ Ê œ Ê œ ¸
968 1936 968 968
kk k 45
97. (a) v a dt 15t 3t dt 10t 6t C; (1) 4 4 10(1) 6(1) C C 0œœ œ œÊœ Êœ' 'ˆ‰
"Î# "Î# $Î# "Î# $Î# "Î# ds
dt
v 10t 6tʜ
$Î# "Î#
(b) s v dt 10t 6t dt 4t 4t C; s(1) 0 0 4(1) 4(1) C C 0œœ œ œÊœ Êœ' 'ˆ‰
$Î# "Î# &Î# $Î# &Î# $Î#
s4t 4tʜ
&Î# $Î#
98. 5.2 5.2t C ; at 0 and t 0 we have C 0 5.2t s 2.6t C ; at s 4
d s ds ds ds
dt dt dt dt
œ
œ
œ Ê œ œ œ œ Ê œ Ê œ œ"" #
#
and t 0 we have C 4 s 2.6t 4. Then s 0 0 2.6t 4 t 1.24 sec, since t 0œœÊœœÊœÊœ¸ #
## É
4
2.6
99. a a dt at C; v when t 0 C v at v s v t C ; s s
d s ds ds ds at
dt dt dt dt
œ œ
œ
œÊ œ œ œ œÊ œ Ê œ Êœ œ'
!!!!"!
# when t 0 s v (0) C C s s v t sœÊ œ Ê œ Êœ !!""! !!
##
a(0) at
œ œ
100. The appropriate initial value problem is: Differential Equation: g with Initial Conditions: v and
ds ds
dt dt
œ
œ
œ œ ! s s when t 0. Thus, g dt gt C ; (0) v v ( g)(0) C C vœœ œœœÊœÊœ!"!!""!
ds ds
dt dt
'
gt v . Thus s gt v dt gt v t C ; s(0) s (g)(0) v (0) C C sÊœ œ œ œ œ Ê œ
ds
dt
!!!#!!##!
""
##
## 'ab
Thus s gt v t sœ
"
#
#
!! .
101. (a) f(x) dx 1 x C x C (b) g(x) dx x 2 C x C'' œ œ œ œÈÈ ""
(c) f(x) dx 1 x C x C (d) g(x) dx (x 2) C x C'' œ œ œ œ ˆ‰ÈÈ ""
(e) [f(x) g(x)] dx 1 x (x 2) C x x C' œ œ ˆ‰ÈÈ "
(f) [f(x) g(x)] dx 1 x (x 2) C x x C' œ œ ˆ‰ÈÈ "
102. Yes. If F(x) and G(x) both solve the initial value problem on an interval I then they both have the same first
derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that
F(x) G(x) C for all x. In particular, F(x ) G(x ) C, so C F(x ) G(x ) 0. Hence F(x) G(x)œ œ œ œ œ
!! !!
for all x.

278. 274 Chapter 4 Applications of Derivatives
103 106 Example CAS commands:
:Maple
with(student):
f := x -> cos(x)^2 + sin(x);
ic := [x=Pi,y=1];
F := unapply( int( f(x), x ) + C, x );
eq := eval( y=F(x), ic );
solnC := solve( eq, {C} );
Y := unapply( eval( F(x), solnC ), x );
DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=1]],
color=black, linecolor=black, stepsize=0.05, title="Section 4.8 #103" );
: (functions and values may vary)Mathematica
The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution
of the initial value problems for exercises 103 - 105.
Clear[x, y, yprime]
yprime[x_] = Cos[x] Sin[x];
2

initxvalue = ; inityvalue = 1;1
y[x_] = Integrate[yprime[t], {t, initxvalue, x}] inityvalue
If the solution satisfies the differential equation and initial condition, the following yield True
yprime[x]==D[y[x], x] //Simplify
y[initxvalue]==inityvalue
Since exercise 106 is a second order differential equation, two integrations will be required.
Clear[x, y, yprime]
y2prime[x_] = 3 Exp[x/2] 1;
initxval = 0; inityval = 4; inityprimeval = 1;
yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] inityprimeval
y[x_] = Integrate[yprime[t], {t, initxval, x}] inityval
Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue).
y2prime[x]==D[y[x], {x, 2}]//Simplify
y[initxval]==inityval
yprime[initxval]==inityprimeval
Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle {RGBColor[1,0,0], RGBColor[0,0,1]}] Ä
CHAPTER 4 PRACTICE EXERCISES
1. No, since f(x) x 2x tan x f (x) 3x 2 sec x 0 f(x) is always increasing on its domainœ Ê œ Ê
$w##
2. No, since g(x) csc x 2 cot x g (x) csc x cot x 2 csc x (cos x 2) 0œ Ê œ œœ
w# "cos x 2
sin x sin x sin x
œœ œ
g(x) is always decreasing on its domainÊ
3. No absolute minimum because lim (7 x)(11 3x) . Next f (x)
xÄ_
œ_ œ
"Î$ w
(11 3x) (7 x)(11 3x) x 1 and x are critical points. œ œ Êœ œ
"Î$ #Î$

(11 3x) (7 x) 4(1 x)
(11 3x) (11 3x)
11
3
œ
Since f 0 if x 1 and f 0 if x 1, f(1) 16 is the absolute maximum.
ww
œ
4. f(x) f (x) ; f (3) 0 ( a b a) a b .œ Ê œ œ œ Ê * ' œ!Ê& $ œ!
ax b
x1
a x 1 2x(ax b) ax 2bx a
x1 x1
œ "
'%
ww

œ
œœ
œœ
œœ
ab a b
ab ab
We require also that f(3) 1. Thus 3a b . Solving both equations yields a 6 and b 10. Now,œ"œÊœ) œœ
3a b
8
œ
f (x) so that f . Thus f changes sign at x from
1131/3
ww w # $ " $

œ œ ± ± ± ± œ $

abab
ab
xx
x1
œ
œ
positive to negative so there is a local maximum at x which has a value f(3) 1.œ$ œ

279. Chapter 4 Practice Exercises 275
5. Yes, because at each point of [ except x 0, the function's value is a local minimum value as well as a!ß "Ñ œ
local maximum value. At x 0 the function's value, 0, is not a local minimum value because each openœ
interval around x 0 on the x-axis contains points to the left of 0 where f equals 1.œ
6. (a) The first derivative of the function f(x) x is zero at x 0 even though f has no local extreme value atœœ
$
x0. œ
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x c then f (c) 0. It does notœœ
w
assert the (false) reverse implication f (c) 0 f has a local extreme at x c.
w
œÊ œ
7. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is
continuous throughout a finite closed interval a x b then the existence of absolute extrema is guaranteed onŸŸ
that interval.
8. The absolute maximum is 1 1 and the absolute minimum is 0 0. This is not inconsistent with the Extreme Valuekk kkœ œ
Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that
interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half
closed, such as , so there is nothing to contradict.Ò"ß "Ñ
9. (a) There appear to be local minima at x 1.75œ
and 1.8. Points of inflection are indicated at
approximately x 0 and x 1.œœ„

(b) f (x) x 3x 5x 15x x x 3 x 5 . The pattern y
w (&% ### $ w
$
œ œ œ ± ± ± ±
$ & $
!
abab
ÈÈÈ
indicates a local maximum at x 5 and local minima at x 3 .œœ„
$È È
(c)
10. (a) The graph does not indicate any local
extremum. Points of inflection are indicated at
approximately x and x .œ œ"
$
%

(b) f (x) x 2x 5 x x 2 x 5 . The pattern f )( indicates
w(% $$ ( w
( $
œ œ œ ± ±
!
& #
10
x
abab
ÈÈ

280. 276 Chapter 4 Applications of Derivatives
a local maximum at x 5 and a local minimum at x 2 .œœ
($È È
(c)
11. (a) g(t) sin t 3t g (t) 2 sin t cos t 3 sin (2t) 3 g 0 g(t) is always falling and hence mustœÊœ œ ÊÊ
#w w
decrease on every interval in its domain.
(b) One, since sin t 3t 5 0 and sin t 3t 5 have the same solutions: f(t) sin t 3t 5 has the same
## #
œ œ œ
derivative as g(t) in part (a) and is always decreasing with f( 3) 0 and f(0) 0. The Intermediate Value
Theorem guarantees the continuous function f has a root in [ 0].$ß
12. (a) y tan sec 0 y tan is always rising on its domain y tan increases on everyœ Ê œ Êœ Êœ)) ) )
dy
d)
#
interval in its domain
(b) The interval is not in the tangent's domain because tan is undefined at . Thus the tangent‘
1 1
4
ߜ1))
#
need not increase on this interval.
13. (a) f(x) x 2x 2 f (x) 4x 4x. Since f(0) 2 0, f(1) 1 0 and f (x) 0 for 0 x 1, weœ Ê œ œ œ   ŸŸ
%# w $ w
may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 x 1.ŸŸ
(b) x 0 x 3 1 and x 0 x .7320508076 .8555996772
##„
#
œÊœ Ê¸ ¸
248È ÈÈ
14. (a) y y 0, for all x in the domain of y is increasing in every interval inœÊœ Êœ
xxx
x1 (x1) x1 x1œœ œœ
w "
œ
its domain
(b) y x 2x y 3x 2 0 for all x the graph of y x 2x is always increasing and can neverœ Ê œ Ê œ
$w# $
have a local maximum or minimum
15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) a be the initialœ
!
amount and V(1440) a (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoirœ !
after the rain, where 24 hr 1440 min. Assume that V(t) is continuous on [ 1440] and differentiable onœ!ß
( 1440). The Mean Value Theorem says that for some t in ( 1440) we have V (t )!ß !ß œ !!
w

V(1440) V(0)
1440 0
316,778 gal/min. Therefore at t the reservoir's volumeœœœ
a (1400)(43,560)(7.48) a 456,160,320 gal
1440 1440 min
!
was increasing at a rate in excess of 225,000 gal/min.
16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the difference 3x g(x) is a constant K because g (x) 3 (3x). Thus g(x) 3x K, the same form as F(x).œœœ
w d
dx
17. No, 1 differs from by the constant 1. Both functions have the same derivative
x1x 1
x1 x1 x1 x1œœœ œ

œ Ê
.
dx d
dx x 1 (x 1) (x 1) dx x 1
(x 1) x(1)
ˆ‰ ˆ‰
œœœ œ
""
œœœ
œœ
18. f (x) g (x) f(x) g(x) C for some constant C the graphs differ by a vertical shift.
ww
œ
œœ Êœ Ê
2x
x1ab
œ
œ
19. The global minimum value of occurs at x .
"
#
œ#

281. Chapter 4 Practice Exercises 277
20. (a) The function is increasing on the intervals and .Ò$ß #Ó Ò"ß #Ó
(b) The function is decreasing on the intervals and .Ò#ß !Ñ Ð!ß "Ó
(c) The local maximum values occur only at x , and at x ; local minimum values occur at x and at xœ# œ# œ$ œ"
provided f is continuous at x .œ!
21. (a) t 0, 6, 12 (b) t 3, 9 (c) 6 t 12 (d) 0 t 6, 12 t 14œ œ
22. (a) t 4 (b) at no time (c) 0 t 4 (d) 4 t 8œ
23. 24.
25. 26.
27. 28.
29. 30.

282. 278 Chapter 4 Applications of Derivatives
31. 32.
33. (a) y 16 x y the curve is rising on ( ), falling on ( 4) and ( )
w#w
œ Ê œ ± ± Ê %ß % _ß %ß _
% %
a local maximum at x 4 and a local minimum at x 4; y 2x y the curveÊ œ œ œ Ê œ± Ê
!
ww ww
is concave up on ( ), concave down on ( ) a point of inflection at x 0_ß ! !ß _ Ê œ
(b)
34. (a) y x x 6 (x )(x 2) y the curve is rising on ( 2) and ( ),
w# w
œ œ $ Ê œ ± ± Ê _ß $ß _
# $
falling on ( ) local maximum at x 2 and a local minimum at x 3; y 2x 1#ß $ Ê œ œ œ
ww
y concave up on , concave down on a point of inflection at xÊ œ ± Ê ß _ _ß Ê œ
"Î#
ww "" "
## # ˆ‰ ˆ ‰
(b)
35. (a) y 6x(x 1)(x 2) 6x 6x 12x y the graph is rising on ( )
w$#w
œ œ Ê œ ± ± ± Ê "ß !
" ! #
and ( ), falling on ( 1) and ( ) a local maximum at x 0, local minima at x 1 and#ß _ _ß !ß # Ê œ œ
x 2; y 18x 12x 12 6 3x 2x 2 6 x x œœœ œ Ê
ww # #
ab Š‹Š‹
17 17
33
ÈÈ
y the curve is concave up on and , concave down
ww
" ( " (
$$

œ ± ± Ê _ß ß _
ÈÈ
ÈÈ
Š‹Š‹
17 17
33
on points of inflection at xŠ‹
1717 17
33 3
„
ÈÈ È
ßÊ œ
(b)

283. Chapter 4 Practice Exercises 279
36. (a) y x (6 4x) 6x 4x y the curve is rising on , falling on
w# # $ w
##
œ œ Ê œ ± ± Ê _ß ß _
! $Î#
ˆ‰ ˆ‰
33
a local maximum at x ; y 12x 12x 12x( x) y concave up onÊ œ œ œ " Ê œ ± ± Ê
!"
3
#
ww # ww
( ), concave down on ( ) and ( ) points of inflection at x 0 and x 1!ß " _ß ! "ß _ Ê œ œ
(b)
37. (a) y x 2x x x 2 y the curve is rising on 2 and
w% ### w
œ œ Ê œ ± ± ± Ê _ß
# #
!
ab
ÈÈ
Š‹
È
2 , falling on 2 2 a local maximum at x 2 and a local minimum at x 2 ;Š‹ Š ‹
ÈÈÈ È È
ß_ ß Ê œ œ
y 4x 4x 4x(x 1)(x 1) y concave up on ( 0) and ( ),
ww $ ww
œ œ Ê œ ± ± ± Ê "ß "ß _
" ! "
concave down on ( 1) and (0 1) points of inflection at x 0 and x 1_ß ß Ê œ œ „
(b)
38. (a) y 4x x x 4 x y the curve is rising on ( 2 0) and (0 2),
w#%# # w
œ œ Ê œ ± ± ± Ê ß ß
# ! #
ab
falling on ( 2) and ( ) a local maximum at x 2, a local minimum at x 2; y 8x 4x_ß #ß _ Ê œ œ œ
ww $
4x 2 x y concave up on 2 and 0 2 , concaveœ Ê œ ± ± ± Ê _ß ß
# #
!
ab
ÈÈ
Š‹Š‹
ÈÈ
#ww
down on 2 0 and 2 points of inflection at x 0 and x 2Š‹Š‹
ÈÈ È
ß ß_Ê œ œ„
(b)
39. The values of the first derivative indicate that the curve is rising on ( ) and falling on ( 0). The slope!ß _ _ß
of the curve approaches as x , and approaches as x 0 and x 1. The curve should therefore_ Ä ! _ Ä Ä

have a cusp and local minimum at x 0, and a vertical tangent at x 1.œœ

284. 280 Chapter 4 Applications of Derivatives

40. The values of the first derivative indicate that the curve is rising on and ( ), and falling on ( )ˆ‰!ß "ß _ _ß !
"
#
and . The derivative changes from positive to negative at x , indicating a local maximum there. Theˆ‰
" "
# #
ß" œ
slope of the curve approaches as x 0 and x 1 , and approaches as x 0 and as x 1 ,†% ( ( % ( (
!! ß ß
indicating cusps and local minima at both x 0 and x 1.œœ

41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches _
as x 0 and as x 1, indicating vertical tangents at both x 0 and x 1.ÄÄ œœ

42. The graph of the first derivative indicates that the curve is rising on and , fallingŠ‹Š ‹!ß ß _
17 33 17 33
16 16
!ß
ÈÈ
on ( ) and a local maximum at x , a local minimum at_ß ! ß Ê œŠ‹
17 33 17 33 17 33
16 16 16
!ß !
ÈÈ È
x . The derivative approaches as x 0 and x 1, and approaches as x 0 ,!†%((%(
17 33
16
ß !ß
È
indicating a cusp and local minimum at x 0 and a vertical tangent at x 1.œœ

285. Chapter 4 Practice Exercises 281

43. y 1 44. y 2!!Î !!†
x 1 4 2x 10
x 3x 3 x 5x 5
ß
!! ßß

45. y x 46. y x 1!!Î ! !†Î
x 1 x x 1
xx x x
! !
ß" !ß "

286. 282 Chapter 4 Applications of Derivatives
47. y 48. y x!!Î !!†
x 2 x x 1
xx x x
ß! "
!!
ß" ! "
##
#

49. y 1 50. y 1!!† !!Î
x 4 x4
x 3x 3 x 4x 4
! !
!! !!
!"
!! !!

51. lim lim
xxÄ" Ä"
x x x
x
!
ß" !_ # ß"
!" "
œœ&
52. lim lim
xxÄ" Ä"
xaxa
xbx b
aa
bb
!"
!"
œœ
%†
%†
53. lim
xÄ1
tan x tan
x
œœ!
!
!
54. lim lim
xxÄ! Ä!
tan x sec x
xsin x cos xß "ß "ß" #
""
œœœ
!
55. lim lim lim lim
xx x xÄ! Ä! Ä! Ä!
sin x sin x cos x
tan x x sec x x sec x x sec x tan x x
sin x cos x
!
!!!!! !!!
ab ab ab a b
ab ab
ab ab
œœœ
#†
####†#
### sec x#!#†"
#
!!
ab
œœ"
56. lim lim
xxÄ! Ä!
sin mx m cos mx
sin nx n cos nx n
mab ab
ab ab
œœ
57. lim sec x cos x lim lim
xxxÄÎ# ÄÎ# ÄÎ#111
!!!
ab ab($œ œ œ
cos x sin x
cos x sin x
ab ab
ab ab
"!""
Ä!ÄÄÄ
$
58. lim x sec x lim
xxÄ! Ä!
ßß
È œœœ!
Èx
cos x
!
"
59. lim csc x cot x lim lim
xxxÄ! Ä! Ä!
abœ œ œœ!
"! œ
"
cos x sin x
sin x cos x

287. Chapter 4 Practice Exercises 283
60. lim lim lim x lim x lim
xxx xxÄ! Ä! Ä! Ä! Ä!
ˆ‰ Š‹ ab ab
"" "! " " ##
xx x xx
x
$$ $ $!
!†! !$†„!$†! !$„%!%
61. lim x x x x lim x x x x
xxÄÄ__
ÎÎ$† † ! ÎÎ$† † „Š‹Š‹
ÈÈ ÈÈ
## ## ßß"ß !
ß ß"ß !
È È
È È
x x x x
x x x x
!!
!!
lim œ
_
xÄ
##"
ß ß"ß !
x
x x x x
È È
!!
Notice that x x for x so this is equivalent toœ!
È
#
lim lim œœ œœ"
__
xxÄÄ
!ß" "
!!
ßß" _
!!
$$ $
!x
xx
x x x x
xx xxx
ÉÉ É É!
ß!
"! ! ! "_


#
"# "
ÈÈ
62. lim lim lim lim lim
xx xxxÄÄ ÄÄÄ__ ___
†! ! ! !Š‹
xx x x
x x x x x
xx xx
x
## ß!
!! !! ß
ß! ß!
$
!" ß" !" ß" _
ß" ! !" #'"#
!"
abab
abab
x
x"#
!
lim lim œœœ!
__
xxÄÄ
"# "
#% #xx
63. (a) Maximize f(x) x 36 x x (36 x) where 0 x 36!† †!†† œœÈ È
"Î# "Î#
f (x) x (36 x) ( 1) derivative fails to exist at 0 and 36; f(0) 6,'! †† †! ' !†
& !" # !" #""
##
!ß
#!
È È
ÈÈ
36 x x
x 36 x
and f(36) 6 the numbers are 0 and 36œÊ
(b) Maximize g(x) x 36 x x (36 x) where 0 x 36!Î †!Ά œœÈ È
"Î# "Î#
g (x) x (36 x) ( 1) critical points at 0, 18 and 36; g(0) 6,'! Ά †! ' !
& !" # !" #""
##
!!
#!
È È
ÈÈ
36 x x
x 36 x
g(18) 2 18 6 2 and g(36) 6 the numbers are 18 and 18œœ œÊ
È È
64. (a) Maximize f(x) x (20 x) 20x x where 0 x 20 f (x) 10x x!†!† œœ'! †È
" # " # & !" # " #
#
3
0 x 0 and x are critical points; f(0) f(20) 0 and f 20!!'! ! !! ! †
20 3x 20 20 20 20
x 3333
!
#È
ˆ‰ ˆ ‰É
the numbers are and .œÊ
40 20
33
20 40
33
È
È
(b) Maximize g(x) x 20 x x (20 x) where 0 x 20 g (x) 0!Î †!Î † œœ ' ! !
È
"Î# w !!
#!
2 20 x 1
20 x
È
È
20 x x . The critical points are x and x 20. Since g and g(20) 20,'†!'! ! ! ! !
È ˆ‰
"
#
79 79 79 81
4444
the numbers must be and .
79
44
"
65. A(x) (2x) 27 x for 0 x 27!†œœ
"
#
#
ab
È
A (x) 3(3 x)(3 x) and A (x) 6x.'!Ά !†
ww w
The critical points are 3 and 3, but 3 is not in the†† domain. Since A (3) 18 0 and A 27 0,
ww
!† * ! Š‹
È
the maximum occurs at x 3 the largest area isœÊ
A(3) 54 sq units.œ

66. The volume is V x h 32 h . TheœœÊœ
# 32
x
!
surface area is S(x) x 4x x ,œ œ
## ˆ‰
32 128
xx
!
where x 0 S (x)‡' !
w !ßß2(x 4) x 4x 16
x
ab
!
!
the critical points are 0 and 4, but 0 is not in theÊ
domain. Now S (4) 2 0 at x 4 there
ww
!Î ‡ ' !
256
4
#
is a minimum. The dimensions 4 ft by 4 ft by 2 ft
minimize the surface area.

288. 284 Chapter 4 Applications of Derivatives
67. From the diagram we have r 3ˆ‰ Š‹
È
h
#
#
#
#
œ
r . The volume of the cylinder isʜ
# !12 h
4
!
V r h h 12h h , where!! ! †11
#$ !
Š‹ ab
12 h
44
!
1
0 h 2 3 . Then V (h) (2 h)(2 h)œœ ! Î †
È
w 3
4
1
the critical points are 2 and 2, but 2 is not in'††
the domain. At h 2 there is a maximum sinceœ
V (2) 3 0. The dimensions of the largest
ww
!† *1
cylinder are radius 2 and height 2.œœ
È

68. From the diagram we have x radius andœ
y height 12 2x and V(x) x (12 2x), where!!† ! †
"#
3
1
0 x 6 V (x) 2 x(4 x) and V (4) 8 . Theœœ ' ! † !†
ww w
11
critical points are 0 and 4; V(0) V(6) 0 x 4œœÊœ
gives the maximum. Thus the values of r 4 andœ
h 4 yield the largest volume for the smaller cone.œ

69. The profit P 2px py 2px p , where p is the profit on grade B tires and 0 x 4. Thusœ œ ŸŸ ˆ‰
40 10x
5 x
!
!
P (x) x 10x 20 the critical points are 5 5 , 5, and 5 5 , but only 5 5 is in
w#
!
!†Î' † Î †
2p
(5 x)
!ab Š‹Š‹ Š‹
ÈÈ È
the domain. Now P (x) 0 for 0 x 5 5 and P (x) 0 for 5 5 x 4 at x 5 5 there
ww
‡**† * †**'!†Š‹ Š‹ Š‹
ÈÈ È
is a local maximum. Also P(0) 8p, P 5 5 4p 5 5 11p, and P(4) 8p at x 5 5 thereœœ¸ œÊœŠ‹Š‹ Š‹
ÈÈ È
is an absolute maximum. The maximum occurs when x 5 5 and y 2 5 5 , the units are!† ! †Š‹ Š‹
ÈÈ
hundreds of tires, i.e., x 276 tires and y 553 tires.¸¸
70. (a) The distance between the particles is f t where f t cos t cos t . Then, f t sin t sin t .ll œ œ ab ab ab ˆ‰ ˆ‰
11
%%
w
Solving f t graphically, we obtain t , t , and so on.
w
abœ! ¸"Þ"() ¸%Þ$#!

Alternatively, f t may be solved analytically as follows. f t sin t sin t
ww
)) ))
ab ab ’“’“ˆ‰ ˆ‰œ! œ
11 11
sin t cos cos t sin sin t cos cos t sin sin cos tœ œ# ’“’“ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
11 11 11 11 1 1
)) )) )) )) ) )
so the critical points occur when cos t , or t k . At each of these values, f t cosˆ‰ ab œ! œ œ„
11 1
)) )
$$
1
units, so the maximum distance between the particles is units.¸ „ !Þ('& !Þ('&

289. Chapter 4 Practice Exercises 285
(b) Solving cos t cos t graphically, we obtain t , t , and so on.œ ¸ #Þ(%* ¸ &Þ)*!ˆ‰
1
%

Alternatively, this problem can be solved analytically as follows.
cos t cos t œ ˆ‰
1
%
cos t cos t ’“’“ˆ‰ ˆ‰ ($
11 11
)) ))
cos t cos sin t sin cos t cos sin t sinˆ‰ ˆ‰ ˆ‰ ˆ‰ $ (
11 11 11 11
)) )) )) ))
sin t sin # œ!ˆ‰
11
))
sin t ˆ‰ œ!
1
)
tk œ
(
)
1
1
The particles collide when t . (plus multiples of if they keep going.)œ ¸ #Þ(%*
(
)
1
1
71. The dimensions will be x in. by x in. by x in., so V x x x x x x x for"! # "' # œ "! # "' # œ % &# "'!ababab
$#
x . Then V x x x x x , so the critical point in the correct domain is x .! & œ "# "!% "'! œ % # $ #! œ #
w#
ab a ba b
This critical point corresponds to the maximum possible volume because V x for x and V x for
ww
ab ab! ! # !
2 x . The box of laargest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.**w
$
Graphical support:

72. The length of the ladder is d d 8 sec 6 csc . We
"# œ ))
wish to maximize I( ) 8 sec 6 csc I ( )))))œ Ê
w
8 sec tan 6 csc cot . Then I ( ) 0$( $)) )) )
w
8 sin 6 cos 0 tan '†!'!'
$$
#
)) )
#È
6
d 4 4 36 and d 36 4 36"#
$$$œ œ ÉÉ ÈÈÈ
the length of the ladder is aboutÊ
436436436 ft.Š‹ Š‹
ÈÈ ÈÉ œ ¸ "*Þ(
$$ $
$Î#

73. g(x) 3x x 4 g(2) 2 0 and g(3) 14 0 g(x) 0 in the interval [ 3] by the Intermediateœ Ê œ œÊ œ #ß
$
Value Theorem. Then g (x) 3 3x x x ; x 2 x 2.22 x 2.196215, and
w#
!" #
!† ' !† !'! '!
n1 n#
3x x 4
33xn n
n_!
_
(
!
so forth to x 2.195823345.&œ

290. 286 Chapter 4 Applications of Derivatives
74. g(x) x x 75 g(3) 21 0 and g(4) 117 0 g(x) 0 in the interval [ ] by the Intermediateœ Ê œ œ Ê œ $ß%
%$
Value Theorem. Then g (x) 4x 3x x x ; x 3 x 3.259259
w$#
!"
!† ' !† !'!
n1 n#
x x 75
4x 3x
$(

nn
nn__
_
x 3.229050, and so forth to x 3.22857729.Êœ œ #&
75. x 5x 7 dx 7x C'ab
$
#
( $ (
x5x
4
"!
76. 8t t dt C 2t C'Š‹
$%
###
( $ ( $ (
t 8ttt tt
46 6
! "ß! ß!
77. 3 t dt 3t 4t dt C 2t C''ˆ‰ˆ ‰È $ $ $(
43t4t4
t1t
!

" # !# " #
!
Š‹
3
!
78. dt t 3t dt C t C''Š‹ ˆ‰
Œ!
È
"" " "
# ##!
!" # !_
Èt
3t3t
t (3) t
($ ( $ ( $
"

# "
!
79. Let u r 5 du drœ Ê œ
u du C u C C''dr du u
r 5 u1 r 5ab ab#
!# !"
!ß
"
! !
%†$$ $ $( $( '
80. Let u r 2 du dr!† ' !
È
6 6 6 u du 6 C 3u C C'''6 dr dr du u 3
r 2 r 2 r2
u
Š‹ Š‹ Š‹
ÈÈ È
!! !
!" !#
!#
## !#
_!$$$$ $( $( ' Š‹
81. Let u 1 du 2 d du dœ Ê œ Ê œ)))))
# "
#
3 1 d u du u du C u C 1 C'')) ) )
È Èˆ‰
Œ! ab
#
## #
"Î# $Î# #
$Î#
œ œ œ œ œ
33 3u '
#œ!
3
!
82. Let u 7 du 2 d du dœ Ê œ Ê œ)))))
2 "
#
d du u du C u C 7 C'')
)
ÈÈ
7 u
u 2
#
"" " "
## #
!" # " #
2
))œœœ œ œ ˆ‰
Œ!
È '

"
!
83. Let u 1 x du 4x dx du x dxœ Ê œ Ê œ
%$ $ "
4
x 1 x dx u du u du C u C 1 x C'" _ !" _ !" _ " _ _
!" _ $Î%
"" " " "
ab ab ˆ‰
Œ! œ œ œ œ œ ''
44 4 3 3
u
#$"
3
4
84. Let u 2 x du dx du dx!† ' !† '† !
(2 x) dx u ( du) u du C u C (2 x) C'($($( $( $( $((
$Î& $Î& $Î& )Î& )Î&''
u5 5
88
ÄœÊ
Š‹
8
5
85. Let u du ds 10 du dsœÊœ Ê œ
s
10 10
"
sec ds sec u (10 du) 10 sec u du 10 tan u C 10 tan C' ## #s s
10 10
œœœ œ ''ab
86. Let u s du ds du dsœÊœ Ê œ11
"
1
csc s ds csc u du csc u du cot u C cot s C' ## # "" " "
11$$$( $( ''ab ˆ‰
11 1 1
87. Let u 2 du 2 d du dœÊœ Ê œ
ÈÈ
)))
"
È
2
csc 2 cot 2 d (csc u cot u) du ( csc u) C csc 2 C''ÈÈ È
Š‹))) ) $$( $( "" "
ÈÈ È
22 2

291. Chapter 4 Additional and Advanced Exercises 287
88. Let u du d 3 du dœÊ œ Ê œ
)
33
"
))
sec tan d (sec u tan u)(3 du) 3 sec u C 3 sec C' )) )
33 3
)œœ œ '
89. Let u du dx 4 du dxœÊ œ Ê œ
x
44
"
sin dx sin u (4 du) 4 du 2 (1 cos 2u) du 2 u C'' '## !
##
x 1 cos 2u sin 2u
4
!! !†!†Îab ˆ‰ ˆ ‰ '
2u sin 2u C 2 sin 2 C sin C!† Î! † Î!† Î ˆ‰ ˆ‰
xxxx
44 ##
90. Let u du dx 2 du dxœÊ œ Ê œ
x
##
"
cos dx cos u (2 du) 2 du (1 cos 2u) du u C'' '##
## #
ßx 1 cos 2u sin 2u
œœ œ œ ab ˆ‰ '
sin x Cœ
x
##
"
91. y dx 1 x dx x x C x C; y 1 when x 1 1 C 1!!Î!†Î!†Î!†!'†Î!†''x 1
xx1
!
!
ß" " !# !"
ab C 1 y x 1'!†'!††
"
x
92. y x dx x 2 dx x 2 x dx 2x x C 2x C;! Î ! ÎÎ ! ÎÎ !Ά Î!ΆÎ'' 'ˆ‰ ˆ ‰ ab
"" "
#
# # !# !"
xx 33x
xx
!
ßß
y 1 when x 1 2 C 1 C y 2x!!'ΆÎ!'!†'!Ά†
""""
31 3 3 x3
1x
ß
93. 15 t dt 15t 3t dt 10t 6t C; 8 when t 1
dr 3 dr
dt dtt
œ œ œ œœ''Š‹È ˆ‰
È
" # !" # " # " #
10(1) 6(1) C 8 C 8. Thus 10t 6t 8 r 10t 6t 8 dt' Î Î!'!† ! Ά'! Ά
$Î# "Î# $Î# "Î# $Î# "Î# dr
dt 'ˆ‰
4t 4t 8t C; r 0 when t 1 4(1) 4(1) 8(1) C 0 C 0. Therefore,!ΆÎ! !' Î †Î!'!
&Î# $Î# &Î# $Î#
""
r4t4t8t!Ά
&Î# $Î#
94. cos t dt sin t C; r 0 when t 0 sin 0 C 0 C 0. Thus, sin t
dr dr
dt dt
! !
! !
!† !† Î ! !'† Î!' ! !†' ww
sin t dt cos t C ; r 0 when t 0 1 C 0 C 1. Then cos t 1'!† ! Î ! !'Î!' !† ! †
dr dr
dt dt'
"""
w
r (cos t 1) dt sin t t C ; r 1 when t 0 0 0 C 1 C 1. Therefore,' ! † ! † Î !† ! ' † Î !† ' !†'
###
rsin tt1!††
CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES
1. If M and m are the maximum and minimum values, respectively, then m f(x) M for all x I. If m MŸŸ − œ
then f is constant on I.
2. No, the function f(x) has an absolute minimum value of 0 at x 2 and an absolute
3x 6, 2 x 0
9 x , 0 x 2
!!†
Άœ*
†œœ
œ
#
maximum value of 9 at x 0, but it is discontinuous at x 0.œœ
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical
point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the
closed endpoint. Extreme values occur only where f 0, f does not exist, or at the endpoints of the interval.
ww
œ
Thus the extreme points will not be at the ends of an open interval.
4. The pattern f indicates a local maximum at x 1 and a local
w
œ ± ± ± ± œ
"#$ %
minimum at x 3.œ

292. 288 Chapter 4 Applications of Derivatives
5. (a) If y 6(x 1)(x 2) , then y 0 for x 1 and y 0 for x 1. The sign pattern is
w#w w
!Î † * *† ‡ ‡†
f f has a local minimum at x 1. Also y 6(x 2) 12(x 1)(x 2)
w ww #
œ ± ± Ê œ œ
" #
6(x 2)(3x) y 0 for x 0 or x 2, while y 0 for 0 x 2. Therefore f has points of inflection!† '‡ * ‡ * **
wwww
at x 0 and x 2. There is no local maximum.œœ
(b) If y 6x(x 1)(x 2), then y 0 for x 1 and 0 x 2; y 0 for x 0 and x 2. The sign
ww w
!Ά * *† **‡†$** ‡
sign pattern is y . Therefore f has a local maximum at x 0 and
w
œ ± ± ± œ
" ! #
local minima at x 1 and x 2. Also, y x x , so y 0 forœ œ œ ")
ww ww !ß
$$
’“’“Š‹ Š‹
17 17
ÈÈ
x and y 0 for all other x f has points of inflection at x .
17 17 17!ß „
$$ $
ww
ÈÈ È
** ‡ ' !
6. The Mean Value Theorem indicates that f (c) 2 for some c in (0 6). Then f(6) f(0) 12
f(6) f(0)
60
!
!
w
œŸ ß Ÿ
indicates the most that f can increase is 12.
7. If f is continuous on [a c) and f (x) 0 on [a c), then by the Mean Value Theorem for all x [a c) we haveßŸß −ß
w
0 f(c) f(x) 0 f(x) f(c). Also if f is continuous on (c b] and f (x) 0 on (c b], then for
f(c) f(x)
cx
!
!
w
ŸÊ ŸÊ   ß   ß
all x (c b] we have 0 f(x) f(c) 0 f(x) f(c). Therefore f(x) f(c) for all x [a b].−ß   Ê   Ê     −ß
f(x) f(c)
xc
!
!
8. (a) For all x, (x 1) 0 (x 1) 1 x 2x 1 x .†Î œœ † '† Î œ œ Î '†œ œ
### # ""
## #
ab ab
x
1x
!
(b) There exists c (a b) such that , from part (a)−ß œ Ê œ Ÿ
cc
1c ba ba 1c
f(b) f(a) f(b) f(a)
ß! ! ß#
!! "
!! ¹¹ ¸¸
f(b) f(a) b a .'†œ†kkkk
"
#
9. No. Corollary 1 requires that f (x) 0 for x in some interval I, not f (x) 0 at a single point in I.
ww
œœ all
10. (a) h(x) f(x)g(x) h (x) f (x)g(x) f(x)g (x) which changes signs at x a since f (x), g (x) 0 when!'!Î ! ‡
ww w ww
x a, f (x), g (x) 0 when x a and f(x), g(x) 0 for all x. Therefore h(x) does have a local maximum**‡ ‡
ww
at x a.œ
(b) No, let f(x) g(x) x which have points of inflection at x 0, but h(x) x has no point of inflectionœœ œ œ
$'
(it has a local minimum at x 0).œ
11. From (ii), f( 1) 0 a 1; from (iii), either 1 lim f(x) or 1 lim f(x). In either case,†! ! ' ! ! !
!" ß
!ß#
a
bc
xxÄ_ Ä!_
lim f(x) lim lim b 0 and c . For if b , then
xx xĄ_ Ą_ Ą_
œ œ œ"Ê œ œ" œ"
x
bx cx
#"
###
!
1
bx c
!
!!
"
x
2
x lim and if c 0, then lim lim . Thus a 1, b 0, and c 1.
xxxĄ_ Ą_ Ą_
111
xc bx
!!!
!! !
"""
xxx
222
xxx
œ! œ œ œ „_ œ œ œ
12. 3x 2kx 3 0 x x has only one value when 4k 36 0 k 9 or
dy
dx 6
2k 4k 36
!ÎÎ!'! ' †!'!
# ##!† !
È
!
k3.œ„
13. The area of the ABC is A(x) (2) 1 x 1 x ,? $($(
"
#
# #
"Î#È
ab
where 0 x 1. Thus A (x) 0 and 1 areŸŸ œ Ê „
w !
!
x
1x
È
!
critical points. Also A 1 0 so A(0) 1 is theab„œ œ
maximum. When x 0 the ABC is isosceles sinceœ ?
AC BC 2 .œœ
È

293. Chapter 4 Additional and Advanced Exercises 289
14. lim f (c) for f (c) 0 there exists a 0 such that 0 h
h0Ä
f(c h) f(c)
h

ß! ww ww "
#
!'! ‡ ‡ **%$$kk kk
f (c) f (c) . Then f (c) 0 f (c) f (c) f (c)'†* !'†*†*¹¹ kk kk kk
f(c h) f(c) f(c h)
h h

ß! #ww ww w ww ww ww"""
###
f (c) f (c) f (c) f (c) . If f (c) 0, then f (c) f (c)'† * *Î * !†
ww ww ww ww ww ww ww""
##
#
kk kk kk
f(c h)
h

f (c) f (c) 0; likewise if f (c) 0, then 0 f (c) f (c).'*** ‡ ***
3 3f(c h) f(c h)
h h## ##
ww ww ww ww ww# #""

(a) If f (c) 0, then h 0 f (c h) 0 and 0 h f (c h) 0. Therefore, f(c) is a local
ww w w
*†**'· **'Î*$$
maximum.
(b) If f (c) 0, then h 0 f (c h) 0 and 0 h f (c h) 0. Therefore, f(c) is a local
ww w w
‡†**'Î* **'·$$
minimum.
15. The time it would take the water to hit the ground from height y is , where g is the acceleration ofÉ
2y
g
gravity. The product of time and exit velocity (rate) yields the distance the water travels:
D(y) 64(h y) 8 hy y , 0 y h D (y) 4 hy y (h 2y) 0, and h!†!†œœ'!†††'ÉÈ ÉÉab ab
2y
gg g
22h #w#
" # !" #
#
are critical points. Now D(0) 0, D and D(h) 0 the best place to drill the hole is at y .œœ œÊ œˆ‰
h8h h
g# #È
16. From the figure in the text, tan ( ) ; tan ( ) ; and tan . These equations") ") ) œ œ œ
ba a
h 1 tan tan h
tan tan # #
!
")
")
give . Solving for tan gives tan or
ba bh
h h a tan h a(b a)
h tan a# #
!ßß
œœ œ
tan
1 tan
"
"
!
_
a
h
a
h
"
"
""
!
h a(b a) tan bh. Differentiating both sides with respect to h givesab
#
( $ "
2h tan h a(b a) sec b. Then 0 2h tan b 2h b"" " œ œÊ œÊ œab Š‹
##
##
dd
dh dh h a(b a)
bh""
!
2bh bh ab(b a) h a(b a) h a(a b) .ʜ ʜ ʜ
## # È
17. The surface area of the cylinder is S 2 r 2 rh. Fromœ 11
#
the diagram we have h and
rHh RHrH
RH R
œÊœ
!!
S(r) 2 r(r h) 2 r r H r$ $ (11 ˆ‰
H
R
2 1 r 2 Hr, where 0 r R.!†Î œœ11ˆ‰
H
R
#
Case 1: H R S(r) is a quadratic equation containing*'
the origin and concave upward S(r) is maximum atÊ
rR. œ
Case 2: H R S(r) is a linear equation containing theœÊ
origin with a positive slope S(r) is maximum atÊ
rR. œ

Case 3: H R S(r) is a quadratic equation containing the origin and concave downward. Then‡'
4 1 r 2 H and 0 4 1 r 2 H 0 r . For simplification
dS H dS H RH
dr R dr R 2(H R)
!†Î !' †Î!'!11 11ˆ‰ ˆ‰
!
we let r .
‡
!
œ
RH
2(H R)
(a) If R H 2R, then 0 H 2R H 2(H R) R which is impossible.** ‡† ' ‡ † ' ‡
RH
2(H R)!
(b) If H 2R, then r R S(r) is maximum at r R.œœœÊ œ
‡ 2R
2R
!
(c) If H 2R, then 2R H 2H H 2(H R) 1 R r R. Therefore,‡ Î*'*†'*'*'*
HRH
2(H R) 2(H R)!!
‡
S(r) is a maximum at r r .œœ
‡
!
RH
2(H R)
: If H (0 R] or H 2R, then the maximum surface area is at r R. If H (R 2R), then r RConclusion
−ß œ œ − ß
which is not possible. If H (2R ), then the maximum is at r r .−ß_ œœ
‡
!
RH
2(H R)
18. f(x) mx 1 f (x) m and f (x) 0 when x 0. Then f (x) 0 x yields a!†Î' !† !‡ ‡ !'!
"" "ww w w
xxx
2
m
! # È
minimum. If f 0, then m 1 m 2 m 1 0 m . Thus the smallest acceptable valueŠ‹ ÈÈÈ
" "
Èm 4
  œ Ê 

294. 290 Chapter 4 Applications of Derivatives
for m is .
"
4
19. (a) lim lim lim
xxxÄ! Ä! Ä!
#& #& &
$$&$$ &
"! "! "!sin x sin x sin x
xx x
ab ab ab
ab ab
œœ œ†"œ
#
Ê
(b) lim sin x cot x lim lim
xxxÄ! Ä! Ä!
ab ab&$œ œ œ
sin x cos x sin x sin x cos x cos x
sin x cos x
ab ab ab ab ab ab
ab ab
$" !"$"ß$$"
$$$$
&
(c) lim x csc x lim lim lim lim
xxxxxÄ! Ä! Ä! Ä! Ä!
#
#
#
##È
#œ œ œ œ x
sin x
x
sin x
!È
È
Š‹
È
"
##
!! !
!
"
!
!
!sin x cos x
x
x
x!!
!
!
!
cos xŠ‹
È
lim œœ
xÄ!
""
## †#
#
cos xŠ‹
È
(d) lim sec x tan x lim lim
x/2 x/2 x/2ÄÄÄ111
abœ œ œ!Þ
"! !
!
sin x cos x
cos x sin x
(e) lim lim lim lim lim li
xxxxxÄ! Ä! Ä! Ä! Ä!
x sin x cos x cos x cos x sin x
x tan x sec x tan x tan x tan x sec x
!"! "! !"!
!"! ! #
œœœœ œ
!!! !
m
xÄ!

_sin x

!
(sin x
cos xœ lim $$(
xÄ!
cos x

#
!# #
"
(f) lim lim lim
xx xÄ! Ä! Ä!
sin x x cos x x sin x cos x
xsin x xcos x sin x xsin x cos x
ab ab a b ab ab
!!!!!
œœ œœ"
#!#ß#
ß!ß##
#
(g) lim lim lim
xx xÄ! Ä! Ä!
sec x sec x tan x sec x tan x sec x
xx
!" ß "ßœ "
####
!
ß!œœ œœ
(h) lim lim lim
xx xÄ# Ä# Ä#
xxx
xxx x
xxx
ß!
!
!
!( ß# ß_ _ß_ß_
!_ !# ß# ß# _
!# ß# ß_
œœœœ$
aba b
abab
20. (a) lim lim lim
xxxÄ_ Ä_ Ä_
È
È
x
x
#
#
"
"
œœ œœ"
!
!
!
!
!x
x
x
x
x
xßÄ
ßÄ
Ä
Ä
É"!
"!
(b) lim lim lim
xxxÄ_ Ä_ Ä_
##
Ꭰ"#$
x
xxÈ
œœ œœ#
!
ßÊ "x
x
xx
x x
!
#
"(É
21. (a) The profit function is P x c ex x a bx ex c b x a. P x ex c bab a b a b a b abœ œ œ# œ!
#w
x . P x e if e so that the profit function is maximized at x .Êœ œ#! ! œ
cb cb
e e
! !
# #
ww
ab
(b) The price therefore that corresponds to a production level yeilding a maximum profit is
p c e dollars.¹ ˆ‰
x
cb cb
e
œ
!ß
##
cb
e_
!
$( $
(c) The weekly profit at this production level is P x e c b a a.ab a bˆ‰ ˆ‰$( ( ( $ (
cb cb
eee
cb!!
##%
# !ab
!
(d) The tax increases cost to the new profit function is F x c ex x a bx tx ex c b t x a.ab a b a b a b$( ( $( (( (
#
Now F x ex c b t when x . Since F x e if e , F is maximized
w wwß! !!
!# #
ab abœ# œ! œ œ œ# ! !
tbc cbt
ee
when x units per week. Thus the price per unit is p c e dollars. Thus, such a tax$$($
cbt cbt cbt
e e
!! !! ßß
# ##
ˆ‰
increases the cost per unit by dollars if units are priced to maximize profit.
cbt cb t## #
###
($
22. (a)
The x-intercept occurs when x .
"" "
$xx
$œ!Ê œ$Ê œ
(b) By Newton's method, x x . Here f x x . So x x x xnn n nn n
fx
fx xnn#" #"
&!# # "$( $( $ $( $ (
ab
abn
n n
ab Š‹
_"
_œ
x
n
xn
x
n
!
"
_"
!
xx x x xx x.œ $ œ# $ œ #$nn n n n nn
## ab

295. Chapter 4 Additional and Advanced Exercises 291
23. x x x x so that x is a weighted average of x"! ! ! "
!" "$( $( $ $ $
fx
fx q q
qab
ab


xa qxxa xq a
qx qx qx x
a


_" _" _" _"
qqqq
qq q q
___ _"_ ab
Š‹ Š‹ !
and with weights m and m .
a
x

_"
q !"
!" "œœ
q
qq
In the case where x we have x a and x .!" !
!" !" ""œœœ œ œ
a
x
q qq
qq qq

%†
q
aaa a
xxx x

_" _" _" _"
qqq qŠ‹ Š‹ Š ‹
24. We have that x h y h r and so x h y h and y h hold.abab abab ab œ # # œ! # # # œ!
## # dy dy d y
dx dx dx
!
!
Thus x y h h , by the former. Solving for h, we obtain h . Substituting this into the second# # œ# # œ
dy dy
dx dx xy!
"!
dy
dx
dy
dx equation yields y . Dividing by 2 results in y .# # # # œ! " œ!
dy d y dy d y
dx dx dx dx
! !
! !
Π! Π!
xy xy!!
"! "!
dy dy
dx dx
dy dy
dx dx
25. (a) a t s t k k s t kt C , where s C s t kt . Soab ab a b ab a b abœ œ ! Ê œ ! œ))Ê œ))Ê œ ))
ww w w w
""
s t t C where s C so s t t. Now s t whenab a b ab abœ )) ! œ ! Ê œ ! œ )) œ "!!
!!
##
##
kt kt
! !
t . Solving for t we obtain t . At such t we want s t , thus
!
#
(( † (( ! #œœ wkt k
k
! !
)) œ "!! œ œ !
È
ab
k or k . In either case we obtain k )) œ ! )) œ ! )) #!! œ !Š‹ Š‹
(( ß (( ! #œœ (( ! (( ! #œœ #
ÈÈ
!!
kk
kk
so that k ft/sec .œ ¸ $)Þ(#
))
#!!
#
!
(b) The initial condition that s ft/sec implies that s t kt and s t t where k is as above.
ww !
#
a b ab ab! œ%% œ %% œ %%
kt
!
The car is stopped at a time t such that s t kt t . At this time the car has traveled a distance
w %%
abœ %%œ!Ê œ
k
s feet. Thus halving the initial velocity quartersˆ‰ ˆ‰ ˆ‰ ˆ ‰
%% %% %% %% *') #!!
##))
#
kk kkk
k
œ %% œ œ œ *') œ #&
!
!
stopping distance.
26. h x f x g x h x f x f x g x g x f x f x g x g x f x g x g x f xab ab ab ab ab ab ab ab ab ab ab ab abab aba b# ‘# ‘abœ Êœ# # œ# œ#
## w w w w w
. Thus h x c, a constant. Since h , h x for all x in the domain of h. Thus h .œ#†!œ! œ ! œ& œ& "! œ&ab ab ab a b
27. Yes. The curve y x satisfies all three conditions since everywhere, when x , y , and everywhere.œœ"œ!œ!œ!
dy d y
dx dx
!
!
28. y x for all x y x x C where C C y x x .
w# $ $ $
œ$ # Ê œ # "œ" #†" Ê œ%Ê œ # %
29. s t a t v s t C. We seek v s C. We know that s t b for some t and s is at a
ww # w w ‡ ‡ !
$
!
ab ab a b a bœœÊœ œ œ !œ œ
t
#
maximum for this t . Since s t Ct k and s we have that s t Ct and also s t so that
‡ w‡!!
ab a b ab a bœ ! œ! œ œ!
tt
12 12
" "
t C . So C C b C C b C b C
‡ "Î$ %Î$"Î$ "Î$ "Î$ "Î$!" $$ %
"# % $
!& Î & !'& † !'& !'& !ab ab ab ab ˆ‰ ˆ‰
! ‘abC
12
CC b
($#"
C . Thus v s b .Êœ œ!œ œ
ab ab È%%
$$$
!
w $Î% ##bb
#$" #$"
ab
30. (a) s t t t v t s t t t k where v k v t t t
&& " # !" # & " # " # " # " # #%#%
$$$$
ab ab ab a b abœÊœœ# !œœÊœ# Þ
(b) s t t t t k where s k . Thus s t t t t .ab a b abœ !œœ œ
%%% % %%%%
"&$$ "& "&$$"&
&Î# $Î# &Î# $Î#
##
31. The graph of f x ax bx c with a is a parabola opening upwards. Thus f x for all x if f x for at mostab ab abœ !  ! œ!
#
one real value of x. The solutions to f x are, by the quadratic equation . Thus we requireabœ!
!# † # ! _
#
bbac
a
Éab
!
bac bac.ab#%Ÿ!ÊŸ!
# #
32. (a) Clearly f x a x b a x b for all x. Expanding we seeabab abœ ÞÞÞ  ! ""
##
nn
fx ax abx b ax abx babababœ # ÞÞÞ #
## # ## #
"" "" nn nn
a a a x ab ab ab x b b b .œ ÞÞÞ # ÞÞÞ ÞÞÞ  !aba ba b
## ## ## #
" "# #"" ##n nnn
Thus a b a b a b a a a b b b by Exercise 31.ababab "" ##
# ## ### #
""## ÞÞÞ ÞÞÞ ÞÞÞ Ÿ!nn nn

296. 292 Chapter 4 Applications of Derivatives
Thus ab ab ab a a a b b b .ababab "" ##
# ## ### #
""## ÞÞÞ Ÿ ÞÞÞ ÞÞÞ nn nn
(b) Referring to Exercise 31: It is clear that f x for some real x b ac , by quadratic formula.abœ! Í % œ!
#
Now notice that this implies that
fx ax b ax babab abœ ÞÞÞ
""
##
nn
a a a x ab ab ab x b b bœ ÞÞÞ # ÞÞÞ ÞÞÞ œ!aba ba b
## ## ## #
" "# #"" ##n nnn
ab ab ab a a a b b bÍ ÞÞÞ ÞÞÞ ÞÞÞ œ!ababab "" ##
# ## ### #
""##nn nn
ab ab ab a a a b b bÍ ÞÞÞ œ ÞÞÞ ÞÞÞ ababab "" ##
# ## ### #
""##nn nn
But now f x a x b for all i n a x b for all i n.abœ!Í œ! œ"ß#ßÞÞÞß Í œ œ! œ"ß#ßÞÞÞßii i i

297. CHAPTER 5 INTEGRATION
5.1 ESTIMATING WITH FINITE SUMS
1. f x xabœ
#

Since f is increasing on , we use left endpoints to obtainÒ!ß "Ó
lower sums and right endpoints to obtain upper sums.
(a) x and x i x a lower sum is ?? ? ??? ? ? ? ! ?
"! " " " " "
## # ### # )
œ!
"
#
#
#
i
ii
i !ˆ‰ ˆ‰ Š‹
(b) x and x i x a lower sum is ˜œ œ œ˜œ Ê † œ ! œ † œ
"! " " " " " $ " ( (
%% % # %)$#
œ!
$
#
#
###
i
ii
i
444 4 4 !ˆ‰ ˆ‰ ˆ‰ ˆ‰Š‹
(c) x and x i x an upper sum is +1˜œ œ œ˜œ Ê † œ œ
"! " " " " &
## # #### )
œ
# #
#
i
ii
i1
2 !ˆ‰ ˆ‰ Š‹
(d) x and x i x an upper sum is +1˜œ œ œ˜œ Ê † œ œ † œ
"! " " " " " $ " $! "&
%% % # %"'$#
œ"
%
# ###
#
i
ii
i
4444 4 !ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ ‰Š‹
2. f x xabœ
$

Since f is increasing on , we use left endpoints to obtainÒ!ß "Ó
lower sums and right endpoints to obtain upper sums.
(a) x and x i x a lower sum is ?? ? ??? ? ? ? ! ?
"! " " " " "
## # ### # "'
œ!
"
$
$
$
i
ii
i !ˆ‰ ˆ‰ Š‹
(b) x and x i x a lower sum is ˜œ œ œ˜œ Ê † œ ! œ œ
"! " " " " " $ $' *
% % % # #&' '%
œ!
$
$
$
$$$
i
ii
i
444 4 4 !ˆ‰ ˆ‰ ˆ‰ ˆ‰Š‹
(c) x and x i x an upper sum is +1˜œ œ œ˜œ Ê † œ œ † œ
"! " " " " " * *
# # # # ### #)"'
œ
$ $
$
i
ii
i1
2 !ˆ‰ ˆ‰ Š‹
(d) x and x i x an upper sum is +1˜œ œ œ˜œ Ê † œ œœ œ
" ! " " " " " $ "!! #&
% % % # #&' '%
œ"
%
$ $$$
$
i
ii
i
4444 4 !ˆ‰ ˆ‰ ˆ‰ ˆ‰Š‹

298. 294 Chapter 5 Integration
3. f xabœ
"
x

Since f is decreasing on , we use left endpoints to obtainÒ!ß "Ó
upper sums and right endpoints to obtain lower sums.
(a) x and x i x i a lower sum is ˜ œ œ# œ" ˜ œ"# Ê †#œ# œ
&" " " " "'
# $& "&
œ"
#
i
i
x ! ˆ‰
i
(b) x and x i x i a lower sum is ˜ œ œ% œ" ˜ œ" Ê †"œ" œ
&" " """" ((
% #$%& '!
œ"
%
i
i
x ! ˆ‰
i
(c) x and x i x i an upper sum is ? ? ? ?" ? ?" ? ?? " ?
&" "")
# $$
œ!
"
i
i
x ! ˆ‰
i
(d) x and x i x i an upper sum is ˜ œ œ% œ" ˜ œ" Ê †"œ" " œ
&" " " " " #&
% #$% "#
œ!
$
i
i
x ! ˆ‰
i
4. f x xab?
#

Since f is increasing on and decreasing on , we use?œ!? !? ?!? !?
left endpoints on and right endpoints on to obtain?œ!? !? ?!? !?
lower sums and use right endpoints on and left endpoints?? !?
on to obtain upper sums.Ò!ß #Ó
(a) x and x i x i a lower sum is ˜ œ œ# œ# ˜ œ## Ê #† % # #† %# œ!
# # #
#
#ab
i
ˆ‰ab a b
(b) x and x i x i a lower sum is x x?? ?" ?œ!??œ!? )œ ?" )œ ?"
# # ##
%
œ! œ$
"%
ab
iii
ii
!!ˆ‰ ˆ‰ab ab
œ" % # % " %" %# œ'ˆ‰ˆ‰ˆ‰ab ab a ba b
## ##
(c) x and x i x i a upper sum is ˜ œ œ# œ# ˜ œ## Ê #† % ! #† %! œ"'
# # #
#
#ab
i
ˆ‰ab a b
(d) x and x i x i a upper sum is x x?? ?" ?œ!??œ!? )œ ?" )œ ?"
# # ##
%
œ" œ#
#$
ab
iii
ii
!!ˆ‰ ˆ‰ab ab
œ" %" %! %! %" œ"%ˆ‰ˆ‰abababab
# ###
5. f x xabœ
#

Using 2 rectangles x f fʘ œ œ Ê
"! " " " $
###% %
ˆ‰ˆ‰ ˆ‰
???
"" $ "! &
#% % $# "'
##
Š‹ˆ‰ ˆ‰
Using 4 rectangles xʘ œ œ
"! "
%%
Ê
""$&( %))))
ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ffff
œ œ
""$&( #"
%)))) '%
## # #
Š‹ˆ‰ ˆ‰ ˆ‰ ˆ‰

299. Section 5.1 Estimating with Finite Sums 295
6. f x xabœ
$

Using 2 rectangles x f fʘ œ œ Ê
"! " " " $
###% %
ˆ‰ˆ‰ ˆ‰
???
"" $ #) (
#% % #†'% $#
$$
Š‹ˆ‰ ˆ‰
Using 4 rectangles xʘ œ œ
"! "
%%
Ê
""$&( %))))
ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ffff
œœœœ
" " )' )& )$ ¸*( "#¸ '"
% ) % † ) ) "#)
Š‹


7. f xabœ
"
x

Using 2 rectangles x f fʘ œ œ#Ê# # %
&"
#
abab ab
œ# œˆ‰
"" $
#% #
Using 4 rectangles xʘ œ œ"
&"
%
Ê" ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ffff
$&(*
####
œ" œ œ œˆ‰
#### "%)) %*' %*'
$&(* $†&†(†* &†(†* $"&
8. f x xab?
#

Using 2 rectangles x f fʘ œ œ#Ê# " "
# #
#
ab
abab ab
? ?"ab
Using 4 rectangles xʘ œ œ"
# #
%
ab
Ê" ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ffff
$ ""$
# ###
œ" % % % %Š‹Š‹Š‹Š‹Š‹ˆ‰ ˆ‰ ˆ‰ ˆ‰
$ ""$
# ###
# ###
œ "' † # † # œ "' œ ""ˆ‰
*" "!
%% #
9. (a) D (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) 87 inches?
(b) D (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) 87 inches¸ œ
10. (a) D (1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300)¸
(1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) 5220 meters (NOTE: 5 minutes 300 seconds)? ?
(b) D (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300)¸
(1.8)(300) (1.5)(300) (1.2)(300) (0)(300) 4920 meters (NOTE: 5 minutes 300 seconds)? ?
11. (a) D (0)(10) (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10)¸
(35)(10) (44)(10) (30)(10) 3490 feet 0.66 miles? %
(b) D (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10)¸
(44)(10) (30)(10) (35)(10) 3840 feet 0.73 miles? %
12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the
midpoints of each time interval to approximate this area using rectangles. Thus,
D (20)(0.001) (50)(0.001) (72)(0.001) (90)(0.001) (102)(0.001) (112)(0.001) (120)(0.001)¸
(128)(0.001) (134)(0.001) (139)(0.001) 0.967 miles¸
(b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours 22.7 sec. At 22.7œ
sec, the velocity was approximately 120 mi/hr.

300. 296 Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing
acceleration t. Thus, t 1 and speed [32.00 19.41 11.77 7.14 4.33](1) 74.65 ft/sec†?? œ ¸ œ
(b) Using right end-points we obtain a lower estimate: speed [19.41 11.77 7.14 4.33 2.63](1)¸
45.28 ft/secœ
(c) Upper estimates for the speed at each second are:
t0 12345
v 0 32.00 51.41 63.18 70.32 74.65
Thus, the distance fallen when t 3 seconds is s [32.00 51.41 63.18](1) 146.59 ft.?%?
14. (a) The speed is a decreasing function of time right end-points give an lower estimate for the height (distance)Ê
attained. Also
t012345
v 400 368 336 304 272 240
gives the time-velocity table by subtracting the constant g 32 from the speed at each time incrementœ
t 1 sec. Thus, the speed 240 ft/sec after 5 seconds.?œ¸
(b) A lower estimate for height attained is h [368 336 304 272 240](1) 1520 ft.¸ œ
15. Partition [ ] into the four subintervals [0 0.5], [0.5 1], [1 1.5], and [1.5 2]. The midpoints of these!ß # ß ß ß ß
subintervals are m 0.25, m 0.75, m 1.25, and m 1.75. The heights of the four approximating
"#$ %œœœ œ
rectangles are f(m ) (0.25) , f(m ) (0.75) , f(m ) (1.25) , and f(m ) (1.75)"#$ %
$$$ $œœ œœ œœ œœ
1 27 125 343
64 64 64 64
Notice that the average value is approximated by
""""""$" #####"'
$$$ $
’“ˆ‰ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ‰
4444
357
?
. We use this observation in solving the next several exercises.
approximate area under
curve f(x) x
œ
œ
"
!ß# $length of [ ]
†”•
16. Partition [1 9] into the four subintervals [ ], [3 ], [ ], and [ ]. The midpoints of these subintervals areß "ß$ ß& &ß( (ß* m 2, m 4, m 6, and m 8. The heights of the four approximating rectangles are f(m ) ,"#$ % "
"
#œœœ œ œ
f(m ) , f(m ) , and f(m ) . The width of each rectangle is x 2. Thus,#$ %
"" "œœ œ œ
46 8
?
Area2222 average value .%?? % ??ˆ‰ ˆ‰ ˆ‰ ˆ‰
""""
##"ß* 4 6 8 1 length of [ ] 8 96
25 area 25
ˆ‰
25
12
17. Partition [0 2] into the four subintervals [0 0.5], [0.5 1], [1 1.5], and [1.5 2]. The midpoints of the subintervalsßßßßß
are m 0.25, m 0.75, m 1.25, and m 1.75. The heights of the four approximating rectangles are
"#$ %œœœ œ
f(m ) sin 1, f(m ) sin 1, f(m ) sin "# $
""" " "" " ""
######
## #
#? ?? ? ?? ? ?œ
111
424 24
35
2
Š‹
È
1, and f(m ) sin 1. The width of each rectangle is x . Thus,?? ? ?œ ? ?
"" " " " "
## # #
%
#
#
24
7
2
1
Š‹
È
?
Area(1111) 2 average value 1.¸ œ Ê ¸ œ œˆ‰
"
#ß#
area 2
length of [0 2]
18. Partition [0 4] into the four subintervals [0 1], [1 2 ], [2 3], and [3 4]. The midpoints of the subintervalsßßßßßß
are m , m , m , and m . The heights of the four approximating rectangles are"#$ %
"
### #œœœ œ
35 7
f(m ) 1 cos 1 cos 0.27145 (to 5 decimal places),"
%
%?œ ?œ ?Š‹Š‹ ˆ‰ˆ‰
1 1
ˆ‰
#
%
48
f(m ) 1 cos 1 cos 0.97855, f(m ) 1 cos 1 cos#
%
%
%
%?œ ?œ ? ?œ ?œŠ‹ Š‹Š‹ Š‹ ˆ‰ ˆ‰ˆ‰ ˆ‰
1 11 1
ˆ‰ ˆ‰
3 5
% %
48 48
3 5
3
0.97855, and f(m ) 1 cos 1 cos 0.27145. The width of each rectangle is??œ?œ? %
%
% Š‹Š‹ ˆ‰ˆ‰
1 1
ˆ‰
7
%
48
7
x . Thus, Area (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) 2.5 average??"%??
value .¸œœ
area 2.5 5
length of [0 4] 4 8ß

301. Section 5.1 Estimating with Finite Sums 297
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left
endpoints:
(a) upper estimate (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) 758 gal,??
lower estimate (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) 543 gal.œ œ
(b) upper estimate (70 97 136 190 265 369 516 720) 2363 gal,? ?
lower estimate (50 70 97 136 190 265 369 516) 1693 gal.? ?
(c) worst case: 2363 720t 25,000 t 31.4 hrs;? ?%
best case: 1693 720t 25,000 t 32.4 hrs? ?%
20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate
uses left endpoints:
(a) upper estimate (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) 60.9 tons??
lower estimate (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) 46.8 tons??
(b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) 126.6 tons,?
so near the end of September 125 tons of pollutants will have been released.
21. (a) The diagonal of the square has length 2, so the side length is . Area
ÈÈ
Š‹#œ#œ#
#
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring
.
#
"' )
11
œ
Area sin cos sin œ"' œ% œ# #¸#Þ)#)ˆ‰ˆ ‰ˆ ‰ È
"
#) ) %
11 1
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring
.
#
$# "'
11
œ
Area sin cos sin œ$# œ) œ# #¸$Þ!'"ˆ‰ˆ ‰ˆ ‰ È
"
#"' "' )
11 1
(d) Each area is less than the area of the circle, . As n increases, the area approaches .11
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring
. The area of each isosceles triangle is A sin cos sin .
#""#
###
11 1 1 1
nn n n n
T
œœ#œ ˆ‰ˆ ‰ˆ ‰
(b) The area of the polygon is A nA sin , so sin PT
nn
nn
nn
sin œœ œ † œ
##
##
Ä_ Ä_
11
lim lim 11
)
)


n
n
ˆ‰
(c) Multiply each area by r .
#
Arsin T
nœ
"#
#
# 1
Arsin P
n
nœ
#
# #1
Ar lim
n
P
Ä_
#
œ1
23-26. Example CAS commands:
:Maple
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;
plot( f(x), x=a..b, title="#23(a) (Section 5.1)" );
N := [ 100, 200, 1000 ]; # (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do # (c)
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));

302. 298 Chapter 5 Integration
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot); # (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
: (assigned function and values for a and b may vary):Mathematica
Symbols for , , powers, roots, fractions, etc. are available in Palettes (under File).1Ä
Never insert a space between the name of a function and its argument.
Clear[x]
f[x_]:=x Sin[1/x]
{a,b}={ /4, }11
Plot[f[x],{x, a, b}]
The following code computes the value of the function for each interval midpoint and then finds the average. Each
sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell.
n =100; dx = (b a) /n;œ
values = Table[N[f[x]], {x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =200; dx = (b a) /n;œ
values = Table[N[f[x]],{x, a + dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =1000; dx = (b a) /n;œ
values = Table[N[f[x]],{x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
FindRoot[f[x] == average,{x, a}]
5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS
1. 7!
2
k1

6k 6 12
k1 11 21 2 3
6(1) 6(2)

???
2. 0!
3
k1

k1 11 21 31 1 2 7
k 123 236

???
3. cos k cos(1)cos(2)cos(3)cos(4) 11110!
4
k1

11111???
4. sin k sin(1)sin(2)sin(3)sin(4)sin(5)000000!
5
k1

111111?? ?
5. ( 1) sin ( 1) sin ( 1) sin ( ) sin 0 1!
3
k1
k1

#
?"??
1 111
k1 3
332"" #" $"
###

ÈÈ
6. ( 1) cos k ( 1) cos (1 ) ( 1) cos (2 ) ( 1) cos (3 ) ( 1) cos (4 )!
4
k1
k

?11111
"#$%
(1) 1 (1) 1 4œ œ

303. Section 5.2 Sigma Notation and Limits of Finite Sums 299
7. (a) 2222222 12481632!
6
k1
k1

%
??
"" #" $" %" &" '"
(b) 2222222 12481632!
5
k0
k

??
!"#$%&
(c) 22 22222 12481632!
4
k
k1
)œ
#
œ œ
"" !" "" #" $" %"
All of them represent 12481632
8. (a) (2) (2) (2) (2) (2) (2) (2) 12481632!
6
k1
k1

%
œ œ
"" #" $" %" &" '"
(b) (1)2 (1)2 )2 (1)2 (1)2 (1)2 (1)2 12481632!
5
k0
kk

œ Ð" œ
!! "" ## $$ %% &&
(c) ( 1) 2 ) 2 ( ) 2 ( ) 2 ( 1) 2 ( ) 2!
3
k2
k1 k2
%
##
œÐ" " " "
#" ## "" "# !" !# "" "# #" ##
(1)2 12481632; œ
$" $#
(a) and (b) represent 12481632; (c) is not equivalent to the other two
9. (a) 1!
4
k2

()(1)()()
k1 21 31 41 3
" " "
#
""
k
??
%
(b) 1!
2
k0

() (1)()()
k1 01 11 21 3
" " "
) ))) #
""
k
??
$#%
(c) 1!
1
k
)œ
() (1) () ()
k2 12 02 12 3
" " "
#
""
k
??
œ!(!
(a) and (c) are equivalent; (b) is not equivalent to the other two.
10. (a) (k 1) (1 1) (2 1) (3 1) (4 1) 0149!
4
k1

??
# ####
(b) (k1) (11) (01) (11) (21) (31) 014916!
3
k1
%
œ œ
# #####
(c) k ( 3) ( 2) ( 1) 9 4 1!
)œ

k3
# ###
œ œ
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
11. k 12. k 13. !!!
6 44
k1 k1 k1

# "
#
k
14. 2k 15. ( 1) 16. ( 1)!!!
5 55
k1 k1 k1
k1 k

#
œœ
"
k5
k
17. (a) 3a 3 a 3( 5) 15!!
nn
k1 k1
kk

? ??
(b) b (6) 1!!
nn
k1 k1
k

b
66 6
k
œœœ
""
(c) (a b ) a b 5 6 1!!!
nnn
k1 k1 k1
kk k k

œ œœ
(d) (a b ) a b 5 6 11!!!
nnn
k1 k1 k1
kk k k

œ? œ ?œœ?œ
(e) (b 2a ) b 2 a 6 2( 5) 16!!!
nnn
k1 k1 k1
kk k k

œ? œ ?œœ?

304. 300 Chapter 5 Integration
18. (a) 8a 8 a 8(0) 0 (b) 250b 250 b 250(1) 250!! ! !
nn n n
k1 k1 k1 k1
kk kk

œœœ œ œœ
(c) (a 1) a 1 0 n n (d) (b 1) b 1 n!!! !!!
nnn n nn
k1 k1 k1 k1 k1 k1
kk k k

œ œœ œ œ"
19. (a) k 55 (b) k 385!!
10 10
k1 k1

œœ œ œ
10(10 1) 10(10 1)(2(10) 1)
6

#
#
(c) k 55 3025! ’“
10
k1

$#
#
#
œœœ
10(10 1)
20. (a) k 91 (b) k 819!!
13 13
k1 k1

œœ œ œ
13(13 1) 13(13 1)(2(13) 1)
6

#
#
(c) k 91 8281! ’“
13
k1

$#
#
#
œœœ
13(13 1)
21. 2k 2 k 2 56 22. k!! !! Š‹ Š‹
77 55
k1 k1 k1 k1

? ? ? ? ? ?
7(7 ) 5(5 1) k
15 15 15
"
# #
11 1
1
23. 3 k 3 k 3(6) 73!!!ab
666
k1 k1 k1

œ? œ ? œ ?œ
## )" )6(6 )(2(6) 1)
6
24. k 5 k 5 5(6) 61!!!ab
666
k1 k1 k1

## )" )
œ? œ ? œ ?
6(6 )(2(6) 1)
6
25. k(3k 5) 3k 5k 3 k 5 k 3 5 240!! !! ab Š‹Š‹
55 55
k1 k1 k1 k1

œ œ œ œ
## )) )
#
5(5 1)(2(5) 1) 5(5 1)
6
26. k(2k 1) 2k k 2 k k 2 308!! !! ab Š‹
77 77
k1 k1 k1 k1

œ œ œ œ
## )) )
#
7(7 1)(2(7) 1) 7(7 1)
6
27. k k k 3376!! !!
Œ Œ Š‹Š‹
55 55
k1 k1 k1 k1

k
225 2 5 25
5(5 1) 5(5 1)

œœ œ
$$
""
####
$
#$
28. k k k 588Œ Œ
!! ! ! Š‹Š‹
77 7 7
k1 k1 k1 k1

##
"" $
##
##
œ? œ ? œ ?
k
44 4
7(7 1) 7(7 1)

29. (a) (b) (c)

305. Section 5.2 Sigma Notation and Limits of Finite Sums 301
30. (a) (b) (c)

31. (a) (b) (c)

32. (a) (b) (c)

33. x x 1.2 0 1.2, x x 1.5 1.2 0.3, x x 2.3 1.5 0.8, x x 2.6 2.3 0.3,kkkkkkk kkkk kkkk k
"! #" $# %$œ?œ? œ?œ? œ?œ? œ?œ?
and x x 3 2.6 0.4; the largest is P 1.2.kkkk ll
&%? ? ?
34. x x 1.6 ( 2) 0.4, x x 0.5 ( 1.6) 1.1, x x 0 ( 0.5) 0.5,kkk kkkk kkkk k
"! #" $#œ?œœœ? œ?œœœ ? œ?œœ ?
x x 0.8 0 0.8, and x x 1 0.8 0.2; the largest is P 1.1.kkkk kkkk ll
%$ &%œ? œ? œ?œ ? ?
35. f x xab?"
#

Since f is decreasing on , 1 we use left endpoints to obtainÒ! Ó
upper sums. x and x i x . So an upper sum˜œ œ œ˜œ
"! "
nn n
i
i
is x n i!! !ab a b Š‹ˆ‰
ii i
nn n
inn n n
i
œ! œ! œ!
" " "
###"" "
#
"œ ? "œ ? œ

? ?" ?"
nnnn
nn n n
i
n
nnn
%

"#$
œ!
# " # " "
''!i
aba bab
?"
#
'
#
%
nn
. Thus,
xlim lim
nn
i
n
in
Ä_ Ä_
œ!
"
#""#
#
'$$
!ab Π" ? " ?" ?
#
%
nn

306. 302 Chapter 5 Integration
36. f x xabœ#

Since f is increasing on , we use right endpoints to obtainÒ! $Ó
upper sums. x and x i x . So an upper˜œ œ œ˜œ
$! $ $
nn n
i
i
sum is x i!! !ˆ‰
ii i
nn n
i
nnnn n n
inn nn
œ" œ" œ"
$'$")") ** "
#
#œ †œ œ† œ
)) )
)
ab
Thus, .lim lim lim
nnn
i
n
inn
nn n n
Ä_ Ä_ Ä_
œ"
'$ ** *
! ˆ‰?? ? * ?*
)
)
37. f x xab?"
#

Since f is increasing on , we use right endpoints to obtainÒ! $Ó
upper sums. x and x i x . So an upper˜œ œ œ˜œ
$! $ $
nn n
i
i
sum is x!! !ab Š‹ Š‹ˆ‰
ii i
nn n
i nnnnn
ii
œ" œ" œ"
# $$$$*
#
" œ " œ "
)
)
œ†œ $
#( $ #(
œ"
# )" # )"
'nnn
i
n
nn n! Š‹in

aba b
??
*# )' )
##
"% ) )abnnn
n
)

)%&
)
n n
Thus,
x .lim lim
nn
i
n
i n
Ä_ Ä_
œ"
# $
"% ) )
#
!ab Œ " œ $ œ*$œ"#
)%&
)
n n
38. f x xabœ$
#

Since f is increasing on , we use right endpoints to obtainÒ! "Ó
upper sums. x and x i x . So an upper sum˜œ œ œ˜œ
"! "
nn n
i
i
is x i!! !ˆ‰ ˆ‰ˆ‰ Š‹
ii i
nn n
innnnn
i nn n
œ" œ" œ"
##""$$
# )" # )"
'
$œ$ œ œ†

aba b
œœ $
#$ "
##
#) )
Ä_
œ"
#nnn
nn
n
i
n
i
)


)
nn
. Thus, xlim!ˆ‰
œœœ"lim
nÄ_
#) )
##
#
Œ
!
)
nn
.
39. f x x x x xab a b? ? "
#

Since f is increasing on , we use right endpoints to obtainÒ! "Ó
upper sums. x and x i x . So an upper sum˜œ œ œ˜œ
"! "
nn n
i
i
is x x i i!! !!ab Š‹ˆ‰
ii ii
nn nn
i innnnn
ii
n
œ" œ" œ" œ"
##""""
#
œ œ
)
? ?
"" )#)'))" )" # )"
#'#'nnn
nn nn n
n
nn n nn

Š‹Š ‹
ab aba b
œ
"
#'
#) )
Ä_
œ"
#"
!
)
nn n
. Thus, x xlim
n
i
n
i in
!ab
œ œœlim
nÄ_
"
#'#''
#) )
"# &
”•Š‹ Œ
!
)
nn n
.
40. f x x xabœ$ #
#

Since f is increasing on , we use right endpoints to obtainÒ! "Ó
upper sums. x and x i x . So an upper sum˜œ œ œ˜œ
"! "
nn n
i
i
is x x i i!! !!ab Š‹ ˆ‰
ii ii
nn nn
i innnnn
ii
n
œ" œ" œ" œ"
##"$ "$#
#
$# œ # œ
)
? ?
$# $$#$")" )" # )"
#'#$nnn
nn nn n
n
nn nn
)))
))Š‹Š ‹
ab aba b
? (!
$
#$
#) )
Ä_
œ"
#"
!
)
nn n
. Thus, x xlim
n
i
n
i in
!ab
œ œœlim
nÄ_
$
#$#$'
#) )
$# "$
”•Š‹ Œ
!
)
nn n
.

307. Section 5.3 The Definite Integral 303
5.3 THE DEFINITE INTEGRAL
1. x dx 2. 2x dx 3. x 3x dx
'''
0
2
#$#
" (
!&
abœ
4. dx 5. dx 6. 4 x dx'''
"#
%$"
""

#
x1x 0
È
œ
7. (sec x) dx 8. (tan x) dx
''
%'$
'$


!
0
9. (a) g(x) dx 0 (b) g(x) dx g(x) dx 8'''
# &"
"&2
???
(c) 3f(x) dx 3 f(x) dx 3( 4) 12 (d) f(x) dx f(x) dx f(x) dx 6 ( 4) 10
'' '''
"" #""
&&22 2
? ?œ?œ ? œ ?œœ?
(e) [f(x) g(x)] dx f(x) dx g(x) dx 6 8 2
'''
"""
&&&
œ ? œ ?œ?œ
(f) [4f(x) g(x)] dx 4 f(x) dx g(x) dx 4(6) 8 16
'''
"""
&&&
œ? œ ?œ?
10. (a) 2f(x) dx 2 f(x) dx 2( 1) 2
''
""
**
? ??
(b) [f(x) h(x)] dx f(x) dx h(x) dx 5 4 9
'''
(((
***
œ œœ
(c) [2f(x) 3h(x)] dx 2 f(x) dx 3 h(x) dx 2(5) 3(4) 2
'''
(((
***
œ ? œ ?œ?œ
(d) f(x) dx f(x) dx ( 1) 1
''
*"
"*
? ? ?
(e) f(x) dx f(x) dx f(x) dx 1 5 6
'''
"" (
(* *
???
(f) [h(x) f(x)] dx [f(x) h(x)] dx f(x) dx h(x) dx 5 4 1
''''
*( ((
(* **
œ ? œ ? œ ?œ?
11. (a) f(u) du f(x) dx 5 (b) 3 f(z) dz 3 f(z) dz 5 3'' ' '
"" " "
22 2 2
œœ œ œ
ÈÈ È
(c) f(t) dt f(t) dt 5 (d) [ f(x)] dx f(x) dx 5
'' ' '
#" " "
"
?œ ?œ œ ?œ ?œ
222
12. (a) g(t) dt g(t) dt 2 (b) g(u) du g(t) dt 2'' ''
!$ $$
$ ! ! !
? ? ? ?
ÈÈ
(c) [ g(x)] dx g(x) dx 2 (d) dr g(t) dt 2 1'' ''
$ $ $ $
!! !!
""
? ? ? ? ?
ÈÈ
Š‹Š ‹
g(r)
22 2
ÈÈ È
13. (a) f(z) dz f(z) dz f(z) dz 7 3 4'''
$!!
%%$
? œ ?œ?
(b) f(t) dt f(t) dt 4
''
%$
$%
? ?
14. (a) h(r) dr h(r) dr h(r) dr 6 0 6
'''
"""
$$ "
? œ ?œ?
(b) h(u) du h(u) du h(u) du 6? ? ?
'''
$""
"$$
Œ

308. 304 Chapter 5 Integration
15. The area of the trapezoid is A (B b)h?
"
#
(5 2)(6) 21 3 dxœ œÊ
"
## #
% 'ˆ‰
x
21 square unitsœ

16. The area of the trapezoid is A (B b)h?
"
#
(3 1)(1) 2 ( 2x 4) dx? ?? œ
"
# "Î#
$Î# '
2 square unitsœ

17. The area of the semicircle is A r (3)œœ
""
##
##
11
9 x dx square units?? œ ?
99
## $
$
#
11'È

18. The graph of the quarter circle is A r (4)œœ
""##
44
11
4 16 x dx 4 square units?? œ ?11 '
%
!
#
È

19. The area of the triangle on the left is A bh (2)(2)œœ
""
##
2. The area of the triangle on the right is A bhœœ
"
#
(1)(1) . Then, the total area is 2.5œœ
""
##
x dx 2.5 square unitsʜ'
#
"
kk

309. Section 5.3 The Definite Integral 305
20. The area of the triangle is A bh (2)(1) 1œœ œ
""
##
1 x dx 1 square unit??'
"
"
abkk

21. The area of the triangular peak is A bh (2)(1) 1.œœ œ
""
##
The area of the rectangular base is S w (2)(1) 2.œj œ œ
Then the total area is 3 2 x dx 3 square units??
'
"
"
abkk

22. y 1 1 x y 1 1 xœ Ê œ
ÈÈ
##
(y 1) 1 x x (y 1) 1, a circle with? ??Þ ?
### #
center ( ) and radius of 1 y 1 1 x is the!ß " Ê œ
È
#
upper semicircle. The area of this semicircle is
A r (1) . The area of the rectangular baseœœ œ
""
## #
##
11
1
is A w (2)(1) 2. Then the total area is 2?! ? ? Þ
1
#
1 1 x dx 2 square units?Þ?Þ'
"
"
#
#
Š‹
È
1

23. dx (b)( ) 24. 4x dx b(4b) 2b''
! !
" "
# #
#
b b
xbb
224
œœ œœ

310. 306 Chapter 5 Integration
25. 2s ds b(2b) a(2a) b a 26. 3t dt b(3b) a(3a) b a''
a a
b b
3
?? ??
"" ""
## ###
## ##
ab

27. x dx 28. x dx 3''
" !Þ&
# #Þ&
### ##
"
È
Š‹
È
œ œ œœ
2
(1) (2.5) (0.5)


29. d 30. r dr 24''
1
1
1 11
# &#
### # # #
))œœ œ œ
(2 ) 3
52 2


È
È
Š‹ Š‹
ÈÈ
31. x dx 32. s ds 0.009''
" !
( !Þ$
# #
È
Š‹
È
j
j
j
j
œœ œœ
7
33 3
7 (0.3)
33. t dt 34. d''
! !
"Î# Î#
# #"
#
œœ œœ
ˆ‰ ˆ‰
Þ

j j
j
324 3 4
1
1
))

35. x dx 36. x dx a''
a a
a a
(2a) a3a a
3a
# $
### # #
#
œœ œ œ


È
Š‹
È
37. x dx 38. x dx 9b''
! !
# #$
$
È
Š‹
È
j
j
j
j
b b
b
33 3
b (3b)
œœ œœ
39. 7 dx 7(1 3) 14 40. 2 dx 2 ( ) 2 2''
$ !
"
œœ œ #!œ
2
ÈÈ È
41. 5x dx 5 x dx 5 10 42. dx x dx 1'' ''
!! $$ ## ##
&&
""
22
20 x 53 16
88 8 16
œ œ œ œ œ œœ’“ ’“

43. (2t 3) dt 2 t dt 3 dt 2 3(2 0) 4 6 2'''
!"!
"
##
22
20
œ œ œœ ’“

44. t 2 dt t dt 2 dt 2 2 0 1 2 1'''
!!! ##
ÈÈÈ
Š‹
È222
2
0
Š‹ ’ “
ÈÈÈÈ
–—
œ œ œœ


45. 1 dz 1 dz dz 1 dz z dz 1[1 2]'''''
####"
""""#
### #####
"""
ˆ‰ ˆ‰’“œ œ œœ"œ
zz 2137
4

46. (2z 3) dz 2z dz 3 dz 2 z dz 3 dz 2 3[0 3] 9 9 0'''''
$$$!$
!!!$!
##
œ œ œœœ ’“
30

47. 3u du 3 u du 3 u du u du 3 3 3 7''''
""!!
###"
## ## "
œ œ œ œ œ œ”• Š‹’“’“’“ ˆ‰
20 0 21 7
33 33 33 3
jj jj jj

311. Section 5.3 The Definite Integral 307
48. 24u du 24 u du 24 u du u du 24 24 7''''
"Î# "Î# ! !
" " " "Î#
## ##
?????
–— ”• ’“
1
33 3
j
Þ

j ˆ‰ ˆ ‰
7
8
49. 3x x 5 dx 3 x dx x dx 5 dx 3 5[2 0] (8 2) 10 0''''
!!!!
####
##
##
ab ’“’“ œ œ œœ
20 20
33
jj
50. 3x x 5 dx 3x x 5 dx 3 x dx x dx 5 dx'' '''
"! !!!
!" """
###
ab ab ”• œ œ
3 5(1 0) 5œ œœ’“Š‹Š‹ ˆ‰
10 10 3 7
33
jj
## # #
51. Let x and let x 0, x x,??œœ œ œ
b0 b
nn

!"
x 2 x x (n 1) x, x n x b.#?? ? ????
nn
Let the c 's be the right end-points of the subintervals
k
c x , c x , and so on. The rectanglesÊœ œ ""##
defined have areas:
f(c ) x f( x) x 3( x) x 3( x)
"
#$??????œœ œ
f(c)xf(2x)x3(2x)x3(2)(x)
#
##$??? ?? ?œœ œ
f(c)xf(3x)x3(3x)x3(3)(x)
$
##$??? ?? ?œœ œ
ã
f(c)xf(nx)x3(nx)x3(n)(x)
n??? ?? ?œœ œ
##$
Then S f(c ) x 3k ( x)
nk
nn
k1 k1œœ!!
ÞÞ
??
#$
3( x) k 3œœ?
$# áá! Š‹Š ‹
n
k1
Þ
b
n6
n(n1)(2n1)
j
j

2 3x dx lim 2 b .? ÞÞ? ? ÞÞ?
b3 b3
nn nn
b
j j

##
""
!
#$
ˆ‰ ˆ‰ '
nÄ_
52. Let x and let x 0, x x,??œœ œ œ
b0 b
nn

!"
x 2 x x (n 1) x, x n x b.#?? ? ????
nn
Let the c 's be the right end-points of the subintervals
k
c x , c x , and so on. The rectanglesÊœ œ ""##
defined have areas:
f(c) x f(x) x (x) x (x)
"
#$???1??1?œœ œ
f(c ) x f(2 x) x (2 x) x (2) ( x)
#
##$???1??1?œœ œ
f(c ) x f(3 x) x (3 x) x (3) ( x)
$
##$???1??1?œœ œ
ã
f(c ) x f(n x) x (n x) x (n) ( x)
n???1??1?œœ œ
##$
Then S f(c ) x k ( x)
nk
nn
k1 k1œœ!!
ÞÞ
?1?
#$
(x) kœœ1? 1
$# áá! Š‹Š ‹
n
k1
Þ
b
n6
n(n1)(2n1)
j
j

2 x dx lim 2 .? ÞÞ? ? ÞÞ?
111b3 b3 b
6nn 6nn3
b
j jj

ˆ‰ ˆ‰
""
!
#'
1
nÄ_

312. 308 Chapter 5 Integration
53. Let x and let x 0, x x,??œœ œ œ
b0 b
nn

!"
x 2 x x (n 1) x, x n x b.#?((? ? ????
nn
Let the c 's be the right end-points of the subintervals
k
c x , c x , and so on. The rectanglesÊœ œ ""##
defined have areas:
f(c) x f(x) x 2(x)(x) 2(x)
"
#??? ?? ?œœ œ
f(c ) x f(2 x) x 2(2 x)( x) 2(2)( x)
#
#??? ?? ?œœ œ
f(c ) x f(3 x) x 2(3 x)( x) 2(3)( x)
$
#??? ?? ?œœ œ
ã
f(c ) x f(n x) x 2(n x)( x) 2(n)( x)
n??? ?? ?œœ œ
#
Then S f(c ) x 2k( x)
nk
nn
k1 k1œœ!!
ÞÞ
??
#
2( x) k 2œœ?
# á! Š‹Š ‹
n
k1
Þ
b
n2
n(n 1)



b 1 2x dx lim b 1 b .?Þ? ? Þ?
### ""
!ˆ‰ ˆ‰
nn
b
'
nÄ_
54. Let x and let x 0, x x,??œœ œ œ
b0 b
nn

!"
x 2 x x (n 1) x, x n x b.#?((? ? ????
nn
Let the c 's be the right end-points of the subintervals
k
c x , c x , and so on. The rectanglesÊœ œ ""##
defined have areas:
f(c) x f(x) x 1(x) (x) x"
##
" #??? ? ????Þ?Þ ˆ‰
?x
f(c ) x f(2 x) x 1 ( x) (2)( x) x#
##
" #??? ? ????Þ?Þ ˆ‰
2x?
f(c ) x f(3 x) x 1 ( x) (3)( x) x$
##
" #??? ? ????Þ?Þ ˆ‰
3x?
ã
f(c ) x f(n x) x 1 ( x) (n)( x) xn
nx??? ? ????Þ?Þ ˆ‰
?
##
" #

Then S f(c ) x k( x) x ( x) k x 1 (n)
nk
nn nn
k1 k1 k1 k1?? Þ? Þ? Þ!! !! ˆ‰ ˆ‰ Š‹Š ‹
ÞÞ ÞÞ
?????
"" "
## #
## ább
n2 n
n(n 1)


b 1 b 1 dx lim b 1 b b b.?ÞÞ?Þ? ÞÞ?Þ
""""###
!#4n 4n 4
1x
b
ˆ‰ ˆ‰ ˆ ‰ ˆ‰ '
nÄ_
55. av(f) x 1 dx?Š‹ ab
"
!
$
#
È
È
30'
x dx 1 dx?
""
!!
$$
#
ÈÈ
ÈÈ
33''
30 110.œ œœ
""
ÈÈ
Š‹
È
33
3
3

Š‹
È
j

56. av(f) dx x dx??ˆ‰ ˆ‰Š‹
"""
# #!!
$$
#
30 3
x''

; .œ œ œ
"
## #63
33x3
Š‹
j

313. Section 5.3 The Definite Integral 309
57. av(f) 3x 1 dx??ˆ‰ ab
"
!
"
#
10'
3 x dx 1 dx 3 (1 0)œ œ ''
!!
""
#
Š‹
1
3
j
.?Î

58. av(f) 3x 3 dx??ˆ‰ ab
"
!
"
#
10'
3 x dx 3 dx 3 3(1 0)??''
!!
""
#
Š‹
1
3
j
.?Î

59. av(f) (t 1) dt?ˆ‰
"
!
$
#
30'
t dt t dt 1 dt?Þ
""
!!!
$$$
#
333
2'''
(3 0) 1.?Þ?
""
##33 3 3
3230
Š‹ Š ‹
j

60. av(f) t t dt?Š‹ ab
1
1(2) #
"
#'
t dt t dt?
""
# #
""
#
33''
t dt t dtœ
"" "
!!
"#
##
##

33 3
1 (2)''
Š‹

.? Þ?
"" "
##33 3 3
13 (2)
Š‹ Š ‹
j j

61. (a) av(g) x 1 dx?Š‹ abkk
"
"
"
1(1)'
( x 1) dx (x 1) dx?Þ
""
##" !
!"''
x dx 1 dx x dx 1 dxœ
"" ""
## ##" " ! !
!!""''''
(0 ( 1)) (1 0)œ
""""
## # # ## # #

Š‹ Š‹
010(1)

. ?
"
#

314. 310 Chapter 5 Integration
(b) av(g) x 1 dx (x 1) dx??ˆ‰ abkk
""
# ""
$$
31''
x dx 1 dx (3 1)? ?
"" " "
## ####""
$$''
Š‹
31

1. œ

(c) av(g) x 1 dx?Š‹ abkk
"
"
$
3(1)'
x 1 dx x 1 dx?Þ
""
" "
"$
44''
ab abkk kk
( 1 2) (see parts (a) and (b) above).?Þ?
""
44

62. (a) av(h) x dx ( x) dx??Š‹ kk
"
" "0(1)
00''
x dx .???'
" ## #
"
0
0 (1)


(b) av(h) x dx x dx??ˆ‰ kk
"

""
10 00''
. œ œŠ‹
""
## #

0

(c) av(h) x dx?Š‹ kk
"
"
"
1(1)'
x dx x dx?Þ
"
# "
"
Œ Þkk kk
''
0
0
(see parts (a) and (b)?Þ ?
"" " "
## # #
ˆ‰ ˆ‰
above).

63. To find where x x 0, let x x 0 x(1 x) 0 x 0 or x 1. If 0 x 1, then 0 x x a 0  œÊ œÊœ œ Êœ
## #
and b 1 maximize the integral.œ

315. Section 5.3 The Definite Integral 311
64. To find where x 2x 0, let x 2x 0 x x 2 0 x 0 or x 2. By the sign graph,
%# %# ##
Ÿ œÊ œÊœ œ„ ab
È
0 0 0 , we can see that x 2x 0 on 2 2 a 2 and b 2 Ÿ ß Ê œ œ
# #
!ÈÈ
’“
ÈÈ È È
%#
minimize the integral.
65. f(x) is decreasing on [0 1] maximum value of f occurs at 0 max f f(0) 1; minimum value of fœßÊ Êœœ
"
á1x

occurs at 1 min f f(1) . Therefore, (1 0) min f dx (1 0) max fÊœœœ Ÿ Ÿ
"" "
á# á
"
11 1x 0

'
dx 1. That is, an upper bound 1 and a lower bound .ÊŸ Ÿ œ œ
"" "
#á #
"'
01x

66. See Exercise 39 above. On [0 0.5], max f 1, min f 0.8. Thereforeßœœœœ
""
áá10 1(0.5)

(0.5 0) min f f(x) dx (0.5 0) max f dx . On [0.5 1], max f 0.8 and* * ?* * ( ??''
00
0.5 0.5
2
51x 1(0.5)
"" "
á# á

min f 0.5. Therefore (1 0.5) min f dx (1 0.5) max f dx .œœ Ÿ Ÿ ÊŸ Ÿ
"""
ááá
""
11 1x 4 1x 5 0.5 0.5
12

''
Then dx dx dx .
"" "" "
áá# á
""
4 5 1x 1x 5 20 1x 10
22139
00.5 0
0.5
Þ* Þ *Þ ? * *'' '

67. 1 sin x 1 for all x (1 0)( 1) sin x dx (1 0)(1) or sin x dx 1 sin x dx cannotŸ Ÿ Ê Ÿ Ÿ Ÿ Êab ab
####
""" '''
000
equal 2.
68. f(x) x 8 is increasing on [ ] max f f(1) 1 8 3 and min f f(0) 0 8 2 2 .?Þ &($? ??Þ? ??Þ?ÈÈÈ È
Therefore, (1 0) min f x 8 dx (1 0) max f 2 2 x 8 dx 3.*Þ* ?*Þ* ''
00
""
ÈÈ È
69. If f(x) 0 on [a b], then min f 0 and max f 0 on [a b]. Now, (b a) min f f(x) dx (b a) max f. ß    ß Ÿ Ÿ
'
a
b
Then b a b a 0 (b a) min f 0 f(x) dx 0. Ê Ê  Ê   '
a
b
70. If f(x) 0 on [a b], then min f 0 and max f 0. Now, (b a) min f f(x) dx (b a) max f. ThenŸß Ÿ Ÿ Ÿ Ÿ '
a
b
b a b a 0 (b a) max f 0 f(x) dx 0. Ê Ê ŸÊ Ÿ '
a
b
71. sin x x for x 0 sin x x 0 for x 0 (sin x x) dx 0 (see Exercise 70) sin x dx x dxŸ ÊŸ Ê Ÿ Ê '''
000
"""
0 sin x dx x dx sin x dx sin x dx . Thus an upper bound is .ŸÊ Ÿ Ê Ÿ Ê Ÿ''' '
000 0
10
""" "
## # #
""
Š‹

72. sec x 1 on sec x 1 0 on sec x 1 dx 0 (see  ß Ê   ß Ê  
xx x
0

### # ## #
"
ˆ‰ ˆ‰ Š‹ ’ “ Š‹
11 11 '
Exercise 69) since [0 1] is contained in sec x dx 1 dx 0 sec x dx((?Þœ? ˆ‰ Š‹
11
## #
"" "'' '
00 0
x

1 dx sec x dx 1 dx x dx sec x dx (1 0) sec x dx .œÞ ? œ Þ ? œÞ ? œ''''' '
00000 0
x 17
36
""""" "
## #
"" #
Š‹ Š‹
j
Thus a lower bound is .
7
6
73. Yes, for the following reasons: av(f) f(x) dx is a constant K. Thus av(f) dx K dxœœ
"
ba aaa
bbb'''
K(b a) av(f) dx (b a)K (b a) f(x) dx f(x) dx.?? ?? ? '''
aaa
bbb
ba
†
"

316. 312 Chapter 5 Integration
74. All three rules hold. The reasons: On any interval [a b] on which f and g are integrable, we have:ß
(a) av(f g) [f(x) g(x)] dx f(x) dx g(x) dx f(x) dx g(x) dxÞ? Þ ? Þ ? Þ
"" ""
ba ba ba baaaaaa
bbbbb'''''
”•
av(f) av(g)?Þ
(b) av(kf) kf(x) dx k f(x) dx k f(x) dx k av(f)œœœœ
"" "
ba ba baaaa
bbb'''
”•” •
(c) av(f) f(x) dx g(x) dx since f(x) g(x) on [a b], and g(x) dx av(g).œŸ Ÿß œ
"" "
ba ba baaa a
bb b'' '
Therefore, av(f) av(g).Ÿ
75. Consider the partition P that subdivides the interval a, b into n subintervals of width x and let c be the rightÒÓ ˜œ
ba
n
k

endpoint of each subinterval. So the partition is P a, a , a , . . . , a and c a .? Þ Þ Þ ? Þ
ba
nn n n
ba nba kba
k
# ab ab ab
We get the Riemann sum f c x c n c b a . As n and P!! !ab a b
kk k
nn n
k
ba
nn n
cb a cb a
œ" œ" œ"

˜ œ † œ "œ † œ Ä_ m mÄ!
ab ab
this expression remains c b a . Thus, c dx c b a .ab ab? '
a
b
76. Consider the partition P that subdivides the interval a, b into n subintervals of width x and let c be the rightÒÓ ˜œ
ba
n
k

endpoint of each subinterval. So the partition is P a, a , a , . . ., a and c a .? Þ Þ Þ ? Þ
ba
nn n n
ba nba kba
k
# ab ab ab
We get the Riemann sum f c x c a a!! ! !ab ˆ‰ Š‹Š ‹
kk k k
nn n n
k k
ba ba ba
nn n n n n
kb a akb a k b a
œ" œ" œ" œ"
## #
#
? ? Þ ? Þ Þ
ab ab ab


akkna?ÞÞ ?ÞÞ
ba ba
nnnnn n
kkk
nnn
aba ba aba nn ba nn n
œ" œ" œ"
#### # " "#"
#'
Œ Þ
!!!
ab ab ab abab aba b
j

baa aba baa abaœ † † œ † †ab ab ab ab
## #"#" # á"
'"'"
"á #á á
n
nn
ba n n baababa b ab
j j

Þ !Þ

nn n
As n and P this expression has value b a a a b aÄ_ m mÄ! †" †# ab ab
# #
'
abba
j
ba a ab a b a b b a ba a . Thus, x dx .œ # $$ œ œ
#$##$ $# #$ # "
$$$$$
ab
ba ba
a
b
jj jj '
77. (a) U max x max x max x where max f(x ), max f(x ), , max f(x ) since f is?ÞÞÞ ? ?? " # ""##?? ?
n nn
increasing on [a b]; L min x min x min x where min f(x ), min f(x ), ,ßœ á œ œ á " # "!#"?? ?
n
min f(x ) since f is increasing on [a b]. Therefore
nn1ϧ

U L (max min ) x (max min ) x (max min ) x?ÞÞÞ "" ## ?? ?
nn
(f(x ) f(x )) x (f(x ) f(x )) x (f(x ) f(x )) x (f(x ) f(x )) x (f(b) f(a)) x.? Þ ÞÞ ? ?"! #" !?? ? ??
nn1 n
(b) U max x max x max x where max f(x ), max f(x ), , max f(x ) since f?ÞÞÞ ? ?? "" ## " " # #?? ?
nn nn
is increasing on[a b]; L min x min x min x where(? Þ ÞÞ "" ##?? ?
nn
min f(x ), min f(x ), , min f(x ) since f is increasing on [a b]. Therefore"!#"œœáœ ß
nn1
U L (max min ) x (max min ) x (max min ) x?ÞÞÞ """ ###?? ?
nnn
(f(x ) f(x )) x (f(x ) f(x )) x (f(x ) f(x )) x? Þ ÞÞ "!"#"# ?? ?
nn1n
(f(x ) f(x )) x (f(x ) f(x )) x (f(x ) f(x )) x . ThenŸ á "! #" ?? ?
max max n n 1 max
U L (f(x ) f(x )) x (f(b) f(a)) x f(b) f(a) x since f(b) f(a). ThusŸ œ œ  
n max max max!??? kk
lim (U L) lim (f(b) f(a)) x 0, since x P .
ll llP0 P0ÄÄ
œ œ œ ??
max max ll

317. Section 5.3 The Definite Integral 313
78. (a) U max x max x max x where? "#?? ?
n
max f(x ), max f(x ), , max f(x )"!#"œœáœ
nn j
since f is decreasing on [a b];ß
L min x min x min x where?
"#?? ?
n
min f(x ), min f(x ) , min f(x )""##œœßáœ
nn
since f is decreasing on [a b]. Thereforeß
U L (max min ) x (max min ) xœ
"" ## ??
(max min ) xá
nn ?
(f(x ) f(x )) x (f(x ) f(x )) x? #
!" "# ??
(f(x ) f(x )) x (f(x ) f(x )) xá œ
nn n ??
(f(a) f(b)) x.? ?

(b) U max x max x max x where max f(x ), max f(x ), , max f(x ) since? ? ??
"" ## " ! # "?? ?
nn nn
f is decreasing on[a b]; L min x min x min x where? # ## "" ##?? ?
nn
min f(x ), min f(x ), , min f(x ) since f is decreasing on [a b]. Therefore""##œœáœ ß
nn
U L (max min ) x (max min ) x (max min ) x? """ ###?? ?
nnn
(f(x ) f(x )) x (f(x ) f(x )) x (f(x ) f(x )) xœ á !"""## ?? ?
nnn
(f(x ) f(x )) x (f(a) f(b) x f(b) f(a) x since f(b) f(a). ThusŸ œ œ Ÿ! n max max max?? ? kk
lim (U L) lim f(b) f(a) x 0, since x P .
ll llP0 P0ÄÄ
œ œ œ kk ll ??
max max
79. (a) Partition 0 into n subintervals, each of length x with points x 0, x x,$‘ßœœœ
11
##
!"
??
n
x 2 x, , x n x . Since sin x is increasing on 0 , the upper sum U is the sum of the areas#
##œáœœ ß??
n
11
$‘
of the circumscribed rectangles of areas f(x ) x (sin x) x, f(x ) x (sin 2 x) x, , f(x ) x
"#???? ?? ?œœá
n (sin n x) x. Then U (sin x sin 2 x sin n x) x x????? ? ? ?? ? ”•
cos cos n x
sin

x
x

j


#
ˆ‰ˆ‰ ?
œœœ”•
ˆ‰
cos cos n cos cos cos cos
sin 4n sin
jj jjjjjj
jj j
j
4n 2n 4n 4n 4n 4n
4n 4n
sin
4n
4n

#
ˆ‰ ˆ ‰ ˆ‰ˆ‰ ˆ ‰
Š‹
j

1
#n
1
(b) The area is sin x dx lim 1.'
!

ß"
œœœ
nÄ_
cos cos
jjj
j
j
4n 4n
sin
4n
4n
ˆ‰
Š‹

1cos
1


80. (a) The area of the shaded region is x m which is equal to L.!
i
n
ii
œ"
˜†
(b) The area of the shaded region is x M which is equal to U.!
i
n
ii
œ"
˜†
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure
and the first part of the figure. Thus this area is U L.
81. By Exercise 80, U L x M x m where M max f x on the ith subinterval andœ ˜† ˜† œ Ö ×!! ab
ii
nn
ii ii i
œ" œ"
m min f x on the ith subinterval . Thus U L M m x x provided x for eachi iii i i
ii
nnœ Ö × œ ˜ †˜ ˜ab a b !!
œ" œ"
%$
i , n. Since x x b a the result, U L b a follows.œ"ßÞÞÞ †˜ œ ˜ œ !! ab ab
ii
nn
ii
œ" œ"
%% % %

318. 314 Chapter 5 Integration
82. The car drove the first 150 miles in 5 hours and the
second 150 miles in 3 hours, which means it drove 300
miles in 8 hours, for an average of mi/hr
300
8
37.5 mi/hr. In terms of average values of functions,œ
the function whose average value we seek is
v(t) , and the average value is
30, 0 t 5
50, 5 1 8
œ
ŸŸ
Ÿ
œ
37.5.
(30)(5) (50)(3)
8
á
œ

83-88. Example CAS commands:
:Maple
with( plots );
with( Student[Calculus1] );
f := x -> 1-x;
a := 0;
b := 1;
N :=[ 4, 10, 20, 50 ];
P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]:
display( P, insequence=true );
89-92. Example CAS commands:
:Maple
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;
plot( f(x), x=a..b, title="#23(a) (Section 5.1)" );
N := [ 100, 200, 1000 ]; # (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do # (c)
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot); # (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
83-92. Example CAS commands:
: (assigned function and values for a, b, and n may vary)Mathematica
Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands
Clear[x, f, a, b, n]

319. Section 5.4 The Fundamental Theorem of Calculus 315
{a, b}={0, }; n =10; dx = (b a)/n;1
f = Sin[x] ;
2
xvals =Table[N[x], {x, a, b dx, dx}];
yvals = f /.x xvals;Ä
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}];Þ
Plot[f, {x, a, b}, Epilog boxes];Ä
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b}={0, }; n =10; dx = (b a)/n;1
f = Sin[x] ;
2
xvals =Table[N[x], {x, a dx, b, dx}];Þ
yvals = f /.x xvals;Ä
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx,xvals, yvals}];
Plot[f, {x, a, b}, Epilog boxes];Ä
Sum[yvals[[i]] dx, {i, 1,Length[yvals]}]//N
Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b}={0, }; n =10; dx = (b a)/n;1
f = Sin[x] ;
2
xvals =Table[N[x], {x, a dx/2, b dx/2, dx}];
yvals = f /.x xvals;Ä
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx/2, xvals dx/2, yvals}];
Plot[f, {x, a, b},Epilog boxes];Ä
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
5.4 THE FUNDAMENTAL THEOREM OF CALCULUS
1. (2x 5) dx x 5x 0 5(0) ( 2) 5( 2) 6
'
2
0
Þ?Þ ?ÞÞ?cd aba b
## #
!
#
2. 5 dx 5x 5(4) 5( 3)'
3
4
ˆ‰ ’“Š ‹Š ‹?? ?
x x 4 133
44 44
(3)
#
%
$


3. 3x dx 8'
0
4
Š‹’ “Š ‹Š ‹???
x3xx 4
4161616
3(4) 3(0) (0)
j
###
%
!
4. x 2x 3 dx x 3x 2 3(2) ( 2) 3( 2) 12'
2
2
ab ’“Š ‹Š ‹
$## #
#
#

œ œ œ
x2
44 4
(2)
!! !
5. x x dx x 0 1'
0
1
ˆ‰ ˆ‰È ’“
# $Î#
"
!
"
Þ?Þ ?Þ?
x2 2
33 33
j
6. x dx x (5) 0 2(5) 10 5'
0
5
$Î# &Î# &Î# $Î#
&
!
????! ‘ È
22
55
7. x dx 5x ( 5)'
1
32
'Î& "Î&
$#
" ##
œ œ œ! ‘ˆ‰
55
8. dx 2x dx 2x 1''

22
11
222
x1
? ? ??
# "
"
#

#
cd ˆ‰ˆ‰

320. 316 Chapter 5 Integration
9. sin x dx [ cos x] ( cos ) ( cos 0) ( 1) ( 1) 2'
0

œ œ œ œ
1
!
1
10. (1 cos x) dx [x sin x] ( sin ) (0 sin 0)
'
0

Þ?Þ?ÞÞ?
1
!
11 1
11. 2 sec x dx [2 tan x] 2 tan (2 tan 0) 2 3 0 2 3'
0
3
ß
# Î$
!
????
1 1
ˆ‰ˆ‰ ÈÈ
3
12. csc x dx [ cot x] cot cot 3 3 2 3'


ß
ß6
56
# &Î'
Î'
œ œ œ œ
1
1
11
ˆ‰ˆ‰ˆ‰ ˆ‰ Š‹Š‹
ÈÈÈ
5
66
13. csc cot d [ csc ] csc csc 2 2 0'


ß
ß4
34
))) )œ œ œ œ
$Î%
Î%
1
1
11
ˆ‰ˆ‰ˆ‰ ˆ‰ ÈÈ
Š‹
3
44
14. 4 sec u tan u du [4 sec u] 4 sec 4 sec 0 4(2) 4(1) 4'
0
3
ß
œ œ œœ
1 1Î$
!
ˆ‰
3
15. dt cos 2t dt t sin 2t (0) sin 2(0) sin 2''
ßß22
00
"á " " " " " " " "
### # # ###
!
Î#
cos 2t
444
œ œ œ ˆ‰ ! ‘ˆ ‰ˆ ‰ ˆ‰ ˆ‰
1
11
?
1
4
16. dt cos 2t dt t sin 2t''

ßß

33
33
" " " " "
####
Î$
Î$
cos 2t
4
? ?ˆ‰ ! ‘
1
1
sin 2 sin 2 sin sinœ œ œˆ‰ˆ ‰ ˆ‰ˆ‰ ˆ‰ ˆ ‰ ˆ ‰
"" " " " "
##
111 111111
34 3 34 3 64 364 3 34
22 3
È
17. 8y sin y dy cos y cos cos'
ß"
ß"


ab ’“ Œ ÞŒ Þ
ˆ‰
#
Î#
Î#
##

Þ? ??
8y
333 3
88 2
j

jj
j 1
1
111
ˆ‰ ˆ ‰
18. 4 sec t dt 4 sec t t dt 4 tan t''

ß# ß#

33

###
Î%
Î$
ˆ‰ ! ‘abœ œ
11
1
1tt
1
4 tan 4 tan (4( 1) 4) 4 3 3 4 3 3œ œœŠ‹Š‹Š‹ˆ‰ ˆ‰ Š‹
ÈÈ
11 11
43 ˆ‰ ˆ‰

43
19. (r 1) dr r 2r 1 dr r r ( 1) ( 1) 1 1''
""
" "
## # # #
"
"

œ œ œ œab ’“Š ‹Š ‹
r18
33 33
(1)
j jj
20. (t 1) t 4 dt t t 4t 4 dt 2t 4t''

#$# #
$
$


33
33
ÞÞ? ÞÞÞ?ÞÞÞab a b ’“
tt
43
! j
È
È
2343 2343 103œ œ

Š‹ Š‹Š‹
ÈÈ È È È
Š‹ Š‹ Š ‹ Š ‹
ÈÈ È È
33 3 3
43 4 3
!j ! j
##
21. du u du''

22
""
##
"""" &
"
#
Š‹ Š ‹’ “Š ‹

uuu1 3
u1616164 4u 4(1)
2
42
ßß""
# !! !
"
œ œœ œ
È
Š‹
È
Š‹
È
22. dv v v dv''
12 12ßß
""
"" " " " " " $ %
"
"Î#
ˆ‰ ! ‘ab Š‹ Œ Þ
v 2v 3v 2(1) 3(1) 6v
15
23
j ! ÞÞ

œ œ œ œ
ˆ‰ ˆ‰
23. ds 1 s ds s 2 1 2 2 1''
11
22

ss
s
222
s
2
1


á
$Î# $Î%
#
"
È
ÈÈ
È
ÉÈ
?Þ ???Þˆ‰ ’“ Š ‹

ÈÈ
281? Þ
È È
%

321. Section 5.4 The Fundamental Theorem of Calculus 317
24. du u 1 du 2 u u 2 4 4 2 9 9 3''
99
44
1u
u

"Î#
%
*
È
È
????ˆ‰ ! ‘È Š‹Š‹
È È
25. x dx x dx x dx x dx x dx''' ''

!!
44 4
kk kk kk ’“ ’“Š ‹Š ‹?Þ?Þ?Þ?ÞÞ
xx 0 40 (4)

## ####
!%
% !

16œ
26. cos x cos x dx (cos x cos x) dx (cos x cos x) dx cos x dx [sin x]''''
!!ß"!
ß" ß"

"""
###
Î#
!
ab kkœ œ œ
1
sin sin 0 1??
1
#
27. (a) cos t dt [sin t] sin x sin 0 sin x cos t dt sin x cos x x''
! !

x x
x
???? ? ?
È
!
"
#
"Î#
ÈÈ ÈÈ
Œ Þ
ˆ‰ ˆ‰
dd
dx dx
œ
cos x
2x
È
È
(b) cos t dt cos x x cos x x
dd
dx dx
cos x
2x
Œ Þ
ˆ‰ˆ ‰ˆ ‰ˆ‰ÈÈ Èˆ‰
'
!

x
œœœ
"
#
"Î#
È
È
28. (a) 3t dt t sin x 1 3t dt sin x 1 3 sin x cos x''
11
sin x sin x
sin x
#$ $ # $ #
"
œœÊ œ œcd abŒ Þ
dd
dx dx
(b) 3t dt 3 sin x (sin x) 3 sin x cos x
dd
dx dx
Œ Þab ˆ‰
'
1
sin x
## #
œœ
29. (a) u du u du u t 0 t u du t 4t'' '
!! !
tt t
t

ÈÈ ! ‘ˆ‰ab Œ Þ????? ??
"Î# $Î# % ' ' &
!
$Î#
22 2d d2
33 3dt dt3
!
(b) u du t t t 4t 4t
dd
dt dt
Œ Þ
È Èˆ‰ab a b
'
!
t

œœœ
% %#$&
30. (a) sec y dy [tan y] tan (tan ) 0 tan (tan ) sec y dy (tan (tan ))''
! !
!
tan tan
tan j j
j
# #
œœ œ Ê œ )) )
dd
dd))
Œ Þ
sec (tan ) secœab
##
))
(b) sec y dy sec (tan ) (tan ) sec (tan ) sec
dd
dd))
Œ Þab ab ˆ‰
'
!
tan j
## ##
œœ )) ))
31. y 1 t dt 1 x 32. y dt , x 0? Þ ʜ Þ ? ʜ ''
!
x x
1
ÈÈ ## ""dy dy
dx t dx x
33. y sin t dt sin x x (sin x) xœÊœ œ œ'
!

x
# "Î#
#
"
#
dy
dx dx
dsin x
2x
Š‹ˆ‰ˆ ‰ ˆ ‰ÈÈ ˆ‰
È
34. y cos t dt cos x x 2x cos xœÊœ œ'
!
x
j
È Š‹
È ˆ‰ab kk
dy
dx dx
d# #
35. y , x (sin x) (cos x) 1 since xœÊœ œœœœ'
!
sin x
dt d cos x cos x
1t 1sinx cosx
dy
dx dx cos x cos xÈÈÈ kk

# #
""

kk kkˆ‰
1 1
36. y (tan x) sec x 1œÊœ œ œ'
!
tan x
dt d
1t dx 1tanx dx secx
dy
áá
"" #

ˆ‰ˆ ‰ˆ‰ ab

322. 318 Chapter 5 Integration
37. x 2x 0 x(x 2) 0 x 0 or x 2; Area ?? Þ?ʜ ?
#
x 2x dx x 2x dx x 2x dxœ '''
$ # !
# ! #
###
ababab
xxxœ ’“’“’“
xxx
333
jjj
###
# ! #
$ # !
(2) (3)?Š‹Š‹Š‹
(2) (3)
33
##
jj
0(2) Š‹Š‹Š ‹
0
33
(2)
j j
##
20 œŠ‹Š‹Š‹
2028
333
jj
##

38. 3x 3 0 x 1 x 1; because of symmetry about
##
? ? ? ? ?)
the y-axis, Area 2 3x 3 dx 3x 3 dxœ Œ Þab ab
''
!"
"#
##
2 x 3x x 3x 2 1 3(1) 0 3(0)Š‹cdcdc ababab œ
$$ $ $
"#
!"
2 3(2) 1 3(1) 2(6) 12 œœ daa b a b
$$

39. x 3x 2x 0 x x 3x 2 0
$# #
œÊ œ ab
x(x 2)(x 1) 0 x 0, 1, or 2;??ʜ
Area x 3x 2x dx x 3x 2x dx?ÞÞ
''
!"
"#
$# $#
abab
xx xxœ ’“’“
xx
44
!!
$# $#
"#
!"
11 00?ÞÞŠ‹Š‹
10
44
!!
$# $#
22 11ÞÞ?’“Š‹Š‹
21
44
!!
$# $# "
#

40. x 4x 0 x x 4 0 x(x 2)(x 2) 0
$#
œÊ œÊ œab
x 0, 2, or 2. Area x 4x dx x 4x dxÊœ œ
''
2
!#
$$
!
ab ab
2x 2x 2(0)œ œ ’“’“Š ‹
xx0
444
!!!
## #
!#
# !
2( 2) 2(2) 2(0) 8 œŠ‹’ “ Š‹Š‹
(2)
444
20 ###
! !!

41. x 0 x 0; Area x dx x dx
"Î$ "Î$ "Î$
!)
!
?ʜ ? Þ ''
j
xx? Þ! ‘ ! ‘
33
44
%Î$ %Î$
!)
" !
(0) ( 1) (8) (0)œ ˆ ‰ˆ ‰ˆ‰ˆ‰
3333
4444
%Î$ %Î$ %Î$ %Î$
œ
51
4

323. Section 5.4 The Fundamental Theorem of Calculus 319
42. x x 0 x 1 x 0 x 0 or
"Î$ "Î$ #Î$ "Î$
œ Ê œ Ê œˆ‰
1 x 0 x 0 or 1 x x 0 or?ʜ ? ʜ
#Î$ #Î$
1 x x 0 or 1;œÊœ „
#
Area xxdx xxdx xxdxœ '''
j
!" )
"Î$ "Î$ "Î$
!"
ˆ‰ ˆ‰ ˆ‰
xxxœ ’“’“’“
3x 3x3x
444
%Î$ %Î$ %Î$
###
!")
"!"

(0) ( 1)œ ’“Š‹Š ‹
303
44
(1)%Î$ %Î$
##


(1) (0)’“Š‹Š‹
3130
44
%Î$ %Î$
##

(8) (1)’“Š‹Š‹
3831
44
%Î$ %Î$
##

2?Þ&Þ ?
"" $"
#44 4 4
83
ˆ‰

43. The area of the rectangle bounded by the lines y 2, y 0, x , and x 0 is 2 . The area under the curveœœœ œ 11
y 1 cos x on [0 ] is (1 cos x) dx [x sin x] ( sin ) (0 sin 0) . Therefore the area ofœ ß œ œ œ1111
'
!
!

1
the shaded region is 2 .111?
44. The area of the rectangle bounded by the lines x , x , y sin sin , and y 0 isœœ œ œœ œ
11 1 1
66 6 6
55 "
#
. The area under the curve y sin x on is sin x dx [ cos x]
" #
&Î'
Î'
ˆ‰ ! ‘
5 5
66 3 66
11 1 11 1
1
? ? ( ? '


ß
ß6
56
cos cos 3. Therefore the area of the shaded region is 3 .œ œ œ ˆ‰ˆ‰ Š‹
ÈÈ
5
66 3
3311 1
ÈÈ
##
45. On 0 : The area of the rectangle bounded by the lines y 2, y 0, 0, and is 2! ‘ ˆ‰ÈÈ
( ? ? ? ?
1 11
4 44
))
. The area between the curve y sec tan and y 0 is sec tan d [ sec ]????
1
1
È
2
4
)) ))) ) '
ß
!
4
!
Î%
( sec 0) sec 2 1. Therefore the area of the shaded region on is 2 1 .œ œ ß! ˆ‰ ! ‘ˆ‰ ÈÈ
Š‹
1 1 1
4 44
2
È
On 0 : The area of the rectangle bounded by , 0, y 2, and y 0 is 2 . The area!‘ ˆ‰ÈÈ
ßœœœœœ
111 1
4444
2
))
È
under the curve y sec tan is sec tan d [sec ] sec sec 0 2 1. Therefore the area????)) ))) ) '
!
ß
4
1 1Î%
! 4
È
of the shaded region on is 2 1 . Thus, the area of the total shaded region is!‘ Š‹
È
!ß
1 1
44
2
È
21 21 .Š‹Š‹
ÈÈ
111
ÈÈÈ
222
44
œ
#
46. The area of the rectangle bounded by the lines y 2, y 0, t , and t 1 is 2 1 2 . Theœœœ œ œ
111
44
ˆ‰ˆ‰
#
area under the curve y sec t on is sec t dt [tan t] tan 0 tan 1. The area?(& ? ??
##!
Î%! ‘ˆ‰
11
144'
ß
!
4
under the curve y 1 t on [ ] is 1 t dt t 1 0 . Thus, the totalœ !ß " œ œ œ
##
!
"
! '
j
ab ’“Š‹Š‹
t102
3333
jjj
area under the curves on is 1 . Therefore the area of the shaded region is 2 .! ‘ˆ‰ß" œ œ
1 11
433 33
25 5
##
"
47. y dt 3 and y( ) dt 3 0 3 3 (d) is a solution to this problem.œ Ê œ œ œœÊ''


x
"""
tdxx t
dy
1
48. y sec t dt 4 sec x and y( 1) sec t dt 4 0 4 4 (c) is a solution to this problem.œ Ê œ œ œœ Ê''

1 1
x 1
dy
dx
49. y sec t dt 4 sec x and y(0) sec t dt 4 0 4 4 (b) is a solution to this problem.? Þ ? ? ? Þ?Þ? ?''
!!
!
x
dy
dx

324. 320 Chapter 5 Integration
50. y dt 3 and y(1) dt 3 0 3 3 (a) is a solution to this problem.œ Ê œ œ œœÊ''
jj
j
x
"""
tdxx t
dy
51. y sec t dt 3 52. y 1 t dt 2?Þ ?Þ''
" j
x x
È #
53. s f(x) dx s 54. v g(x) dx v?Þ ? Þ''
t t
t tÞ Þ
! !
55. Area h x dx hx? ?'
ß
ß
ß
b2
b2
b2
b2
ˆ‰ˆ‰ ’“
4h 4hx
b3b

j
#
hhœ Œ ÞŒ Þ
ˆ‰ ˆ ‰
bb 4h 4h
3b 3b##
ˆ‰ ˆ ‰
bb

jj

bh bh?Þ??ˆ‰ˆ ‰
bh bh bh bh bh 2
6633##

56. r 2 dx 2 1 dx 2 x 2 3 0œ œ œ œ ''
!!
áá
Š‹ Š‹ ’ “ ! ‘ˆ‰ Š‹Š‹
21
(x 1) (x 1) x 1 (3 1) (0 1)ááááá
"""
$
!

2 3 1 2 2 4.5 or $4500?? ?! ‘ˆ‰
""
44
57. x c t dt t x
dc
dx x
œœ Êœ œ œ
"" "
# ##
"Î# "Î# "Î#
È '
!
x
x
0
!‘È
c(100) c(1) 100 1 $9.00œ œ
È È
58. By Exercise 57, c(400) c(100) 400 100 20 10 $10.00 œ œœ
ÈÈ
59. (a) v f(x) dx f(t) v(5) f(5) 2 m/secœœ œ Ê œ œ
ds d
dt dt'
!
t
(b) a is negative since the slope of the tangent line at t 5 is negativeœœ
df
dt
(c) s f(x) dx (3)(3) m since the integral is the area of the triangle formed by y f(x), the x-axis,œœœ œ'
!
3
"
##
9
and x 3 œ
(d) t 6 since from t 6 to t 9, the region lies below the x-axisœœœ
(e) At t 4 and t 7, since there are horizontal tangents thereœœ
(f) Toward the origin between t 6 and t 9 since the velocity is negative on this interval. Away from theœœ
origin between t 0 and t 6 since the velocity is positive there.œœ
(g) Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the
x-axis than below it.
60. (a) v g(x) dx g(t) v(3) g(3) 0 m/sec.œœ œ Ê œ œ
dg
dt dt
d'
!
t
(b) a is positive, since the slope of the tangent line at t 3 is positiveœœ
df
dt
(c) At t 3, the particle's position is g(x) dx (3)( 6) 9??? '
!
á
"
#
(d) The particle passes through the origin at t 6 because s(6) g(x) dx 0œœœ '
!

(e) At t 7, since there is a horizontal tangent thereœ
(f) The particle starts at the origin and moves away to the left for 0 t 3. It moves back toward the origin
for 3 t 6, passes through the origin at t 6, and moves away to the right for t 6.'' ?
(g) Right side, since its position at t 9 is positive, there being more area above the x-axis than below it at t .œœ*

325. Section 5.4 The Fundamental Theorem of Calculus 321
61. k 0 one arch of y sin kx will occur over the interval 0 the area sin kx dx cos kx? ? ( ? ? ? !‘ ! ‘
1
1
kk '
!
ß
k
k
"
Î
!
cos k cos (0)œ œ
""
kk k k
2
ˆ‰ˆ ‰ˆ‰
1
62. dt .lim lim lim lim
xxxx
xx
t
tx
Ä! Ä! Ä! Ä!
""
á" " á"
jj

! !
'
!
x
œœœœ_
'
Þ

!
ßj

!
ßj

xt
t
x
x
dt
x$
ab
63. f(t) dt x 2x 1 f(x) f(t) dt x 2x 1 2x 2''
11
xx
œÊ œ œ œ
## dd
dx dx
ab
64. f(t) dt x cos x f(x) f(t) dt cos x x sin x f(4) cos (4) (4) sin (4) 1''
!!
xx
?ʜ ? ʜ ?1111111
d
dx
65. f(x) 2 dt f (x) f (1) 3; f(1) 2 dt 2 0 2;? ? ? ? ? ? ? ??''
" "
 j j j
x
999 9
1t 1(x1) x2 1táááá á
ww
L(x) 3(x 1) f(1) 3(x 1) 2 3x 5œ œ œ
66. g(x) 3 sec (t 1) dt g (x) sec x 1 (2x) 2x sec x 1 g ( 1) 2( 1) sec ( 1) 1œ Ê œ œ Ê œ
'
1
x
j
w# #w #
ab ab abab
2; g( 1) 3 sec (t 1) dt 3 sec (t 1) dt 3 0 3; L(x) 2(x ( 1)) g( 1)œ œ œ œœ œ
''
11
ab" "
j
2(x 1) 3 2x 1? Þ Þ ? Þ
67. (a) True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus.
(b) True: g is continuous because it is differentiable.
(c) True, since g (1) f(1) 0.
w
œœ
(d) False, since g (1) f (1) 0.
ww w
?
(e) True, since g (1) 0 and g (1) f (1) 0.
wwww
??
(f) False: g (x) f (x) 0, so g never changes sign.
ww w ww
?
(g) True, since g (1) f(1) 0 and g (x) f(x) is an increasing function of x (because f (x) 0).
ww w
?? ?
68. (a) True: by Part 1 of the Fundamental Theorem of Calculus, h (x) f(x). Since f is differentiable for all x,
w
œ
h has a second derivative for all x.
(b) True: they are continuous because they are differentiable.
(c) True, since h (1) f(1) 0.
w
œœ
(d) True, since h (1) 0 and h (1) f (1) 0.
wwww
œœ
(e) False, since h (1) f (1) 0.
ww w
œ
(f) False, since h (x) f (x) 0 never changes sign.
ww w
œ
(g) True, since h (1) f(1) 0 and h (x) f(x) is a decreasing function of x (because f (x) 0).
ww w
?? ? '
69. 70. The limit is 3x
#

326. 322 Chapter 5 Integration
71-74. Example CAS commands:
:Maple
p:=x^2*cos(x);
with( plots );
f := x -> x^3-4*x^2+3*x;
a := 0;
b := 4;
F := unapply( int(f(t),t=a..x), x ); # (a)
p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#71(a) (Section 5.4)" ):
p1;
dF := D(F); # (b)
q1 := solve( dF(x)=0, x );
pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ];
p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ):
display( [p1,p2], title="71(b) (Section 5.4)" );
incr := solve( dF(x)>0, x ); # (c)
decr := solve( dF(x)<0, x );
df := D(f); # (d)
p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#71(d) (Section 5.4)" ):
p3;
q2 := solve( df(x)=0, x );
pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ];
p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '(x)=0" ):
display( [p3,p4], title="71(d) (Section 5.4)" );
75-78. Example CAS commands:
:Maple
a := 1;
u := x -> x^2;
f := x -> sqrt(1-x^2);
F := unapply( int( f(t), t=a..u(x) ), x );
dF := D(F); # (b)
cp := solve( dF(x)=0, x );
solve( dF(x)>0, x );
solve( dF(x)<0, x );
d2F := D(dF); # (c)
solve( d2F(x)=0, x );
plot( F(x), x=-1..1, title="#75(d) (Section 5.4)" );
79. Example CAS commands:
:Maple
f := `f`;
q1 := Diff( Int( f(t), t=a..u(x) ), x );
d1 := value( q1 );
80. Example CAS commands:
:Maple
f := `f`;
q2 := Diff( Int( f(t), t=a..u(x) ), x,x );

327. Section 5.5 Indefinite Integrals and the Substitution Rule 323
value( q2 );
71-80. Example CAS commands:
: (assigned function and values for a, and b may vary)Mathematica
For transcendental functions the FindRoot is needed instead of the Solve command.
The Map command executes FindRoot over a set of initial guesses
Initial guesses will vary as the functions vary.
Clear[x, f, F]
{a, b}= {0, 2 }; f[x_] = Sin[2x] Cos[x/3]1
F[x_] = Integrate[f[t], {t, a, x}]
Plot[{f[x], F[x]},{x, a, b}]
x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}]
x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}]
Slightly alter above commands for 75 - 80.
Clear[x, f, F, u]
a=0; f[x_] = x 2x 3
2

u[x_] = 1 x
2
F[x_] = Integrate[f[t], {t, a, u(x)}]
x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}]
x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}]
After determining an appropriate value for b, the following can be entered
b = 4;
Plot[{F[x], {x, a, b}]
5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE
1. Let u 3x du 3 dx du dxœÊœ Ê œ
"
3
sin 3x dx sin u du cos u C cos 3x C '' œœœ
"" " 33 3
2. Let u 2x du 4x dx du x dxœÊœ Ê œ
# "
4
x sin 2x dx sin u du cos u C cos 2x C''ab
## "" "
œœœ
44 4
3. Let u 2t du 2 dt du dtœÊ œ Ê œ
"
#
sec 2t tan 2t dt sec u tan u du sec u C sec 2t C'' &&Î&Î
""" ###
4. Let u 1 cos du sin dt 2 du sin dtœ Ê œ Ê œ
tt t
22
"
##
1 cos sin dt 2u du u C 1 cos C''ˆ‰ˆ‰ ˆ‰œœœ
tt 2 2 t
33## #
#$
#$
5. Let u 7x 2 du 7 dx du dxœÊ œ Ê œ
"
7
28(7x 2) dx (28)u du 4u du u C (7x 2) C'''œ œ œœ
& & & % %"
7
6. Let u x du 4x dx du x dxœ"Ê œ Ê œ
%$$ "
4
x x 1 dx u du C x 1 C ''$% # %
#$
""
##
ab abœ œœ
411
u

328. 324 Chapter 5 Integration
7. Let u 1 r du 3r dr 3 du 9r drœ Ê œ Ê œ
$# #
3u du 3(2)u C 6 1 r C'9r dr
1r
&

È

"Î# "Î# $
"Î#
œ œ œ ' ab
8. Let u y 4y 1 du 4y 8y dy 3 du 12 y 2y dy&Î Î% & Î % & Î
%# $ $
ab ab
12y4y1y2y dy 3u duuCy4y1 C
''abab ab
%# $ # $ %#
#$
ÎÎ Î & &Î&ÎÎÎ
9. Let u x 1 du x dx du x dxœÊœ Ê œ
$Î# "Î#
#
32
3 È
x sin x 1 dx sin u du sin 2u C x 1 sin 2x 2 C''È ˆ‰ ˆ ‰ ˆ‰ˆ ‰
# $Î# # $Î# $Î#
#
"""
œ œ œ
22u
334 3 6
10. Let u du dxœ Ê œ
""
xx
&
cos dx cos u du cos u du sin 2u C sin C'''"" " ""## #
#xx 4 2x4x
u2
&
ˆ‰ ˆ ‰ ˆ ‰ ab abœœ œ œ
sin Cœ
""
2x 4 x
2
ˆ‰
11. (a) Let u cot 2 du 2 csc 2 d du csc 2 dœÊœ Êœ))) ))
## "
#
csc 2 cot 2 d u du C C cot 2 C''# #"" "
###
))) ) œ œ œ œ Š‹
uu
44
&&
(b) Let u csc 2 du 2 csc 2 cot 2 d du csc 2 cot 2 dœÊœ Êœ)))) )))
"
#
csc 2 cot 2 d u du C C csc 2 C''# #"" "
###
))) ) œ œ œ œ Š‹
uu
44
&&
12. (a) Let u 5x 8 du 5 dx du dx&Î% & % &
"
5
du u du 2u C u C 5x 8 C'' 'dx 22
5x 8 55 5 55uÈÈ"
"" " " "Î# "Î# "Î#
&&&Î&Î&ÎΊ‹ ˆ‰ È
(b) Let u 5x 8 du (5x 8) (5) dx du&Î%&Î % &È
"
#
"Î#
"
2dx
5 5x 8È
du uC 5x8C''dx 2 2 2
5x 8 55 5È"
&&Î&ÎÎ È
13. Let u 3 2s du 2 ds du dsœ Ê œ Ê œ
"
#
3 2s ds u du u du u C (3 2s) C'' 'È Èˆ‰ ˆ‰ˆ‰œ œ œ œ "" " "
## #
"Î# $Î# $Î# 2
33
14. Let u 2x 1 du 2 dx du dx&Î% & % &
"
#
(2x 1) dx u du u du C (2x 1) C'''Î& & & Î&ÎÎ
$$ $ % "" " "
## # ˆ‰ ˆ‰ Š‹
u
48
%
15. Let u 5s 4 du 5 ds du ds&Î% & % &
"
5
ds du u du 2u C 5s 4 C'' '"""" "
"
"Î# "Î#
ÈÈ5s 4 u55 5 5
2
&&&Î&ÎΈ‰ ˆ‰ˆ ‰ È
16. Let u 2 x du dx du dxœ Ê œ Ê œ
dx 3 u du 3 C C'''3u3
(2 x) u 1 2 x
3( du)

#
&&
œœ œœ Š‹
17. Let u 1 du 2 d du dœ Ê œ Ê œ)))))
# "
#
1 d u du u du u C 1 C'' '))) )
%
È Èˆ‰ ˆ‰ˆ‰ abœ œ œ œ # % "" "
## #
"Î% &Î% #
&Î%42
55
18. Let u 1 du 2 d 4 du 8 dœÊ œ Ê œ)))))
#
8 1 d u (4 du) 4 u du 4 u C 3 1 C''')) ) )

È È ˆ‰ ab# $ "Î$ %Î$ #
%Î$
œ œ œ œ
3
4

329. Section 5.5 Indefinite Integrals and the Substitution Rule 325
19. Let u 7 3y du 6y dy du 3y dyœ Ê œ Ê œ
# "
#
3y 7 3y dy u du u du u C 7 3y C'' 'È Èˆ‰ ˆ‰ˆ‰ abœ œ œ œ # "" " "
## #
"Î# $Î# #
$Î#2
33
20. Let u 2y 1 du 4y dy&Î%&
#
du u du 2u C 2 2y 1 C'4y dy
2y 1 uÈÈ
&
"
" "Î# "Î# #
&& &Î&ÎÎ'' È
21. Let u 1 x du dx 2 du dx&Î % & % &È
""
2x xÈÈ
dx C C' "
"" "ÈȈ‰ È
xx
2 du 2 2
uu 1x
& &œœœ'
22. Let u 1 x du dx 2 du dx&Î % & % &È
""
2x xÈÈ
dx u (2 du) 2 u C 1 x C'
ˆ‰È
È
1x
x 4
"
$% ""
#
%

&&Î&ÎÎ' ˆ‰ ˆ ‰ È
23. Let u 3z 4 du 3 dz du dz&Î% & % &
"
3
cos (3z 4) dz (cos u) du cos u du sin u C sin (3z 4) C'' 'Î& & & Î& ÎÎ ˆ‰
"" " "
33 3 3
24. Let u 8z 5 du 8 dz du dzœÊ œ Ê œ
"
8
sin (8z 5) dz (sin u) du sin u du ( cos u) C cos (8z 5) C'' 'œ œ œœ ˆ‰
"" " "
88 8 8
25. Let u 3x 2 du 3 dx du dx&Î% & % &
"
3
sec (3x 2) dx sec u du sec u du tan u C tan (3x 2) C'' '### "" " "
Î& & & Î& ÎÎab ˆ‰
33 3 3
26. Let u tan x du sec x dxœÊœ
#
tan x sec x dx u du u C tan x C''## # $ $ ""
&&Î&Î
33
27. Let u sin du cos dx 3 du cos dxœÊœ Êœˆ‰ ˆ‰ ˆ‰
xx x
333 3
"
sin cos dx u (3 du) 3 u C sin C''&&'' ""
#ˆ‰ ˆ‰ ˆ ‰ ˆ‰
xx x
33 6 3
&&Î&Î
28. Let u tan du sec dx 2 du sec dxœÊœ Êœˆ‰ ˆ‰ ˆ‰
xx x
### #
" ##
tan sec dx u (2 du) 2 u C tan C''(# ( ) )
## #
""ˆ‰ ˆ‰ ˆ ‰ ˆ‰
xx x
84
&&Î&Î
29. Let u 1 du dr 6 du r drœÊ œ Ê œ
rr
18 6
&
#
r 1 dr u (6 du) 6 u du 6 C 1 C'''#&&
& '
Š‹ Š‹ Š‹
rur
18 6 18

œ œ œ œ
30. Let u 7 du r dr 2 du r drœ Ê œ Ê œ
r
10
(
"
#
%%
r 7 dr u ( 2 du) 2 u du 2 C 7 C'''%$$
$ %
"
#
Š‹ Š‹ Š‹œœ œœ
rur
10 4 10
( %(
31. Let u x 1 du x dx du x dx&Î%& % &
$Î# "Î# "Î#
#
32
3
x sin x 1 dx (sin u) du sin u du ( cos u) C cos x 1 C'''"Î# $Î# $Î#ˆ‰ ˆ‰ ˆ‰œ œ œœ
22 2 2
33 3 3

330. 326 Chapter 5 Integration
32. Let u x 8 du x dx du x dxœÊœ Ê œ
%Î$ "Î$ "Î$ 43
34
x sin x 8 dx (sin u) du sin u du ( cos u) C cos x 8 C'''"Î$ %Î$ %Î$ˆ‰ ˆ‰ ˆ‰œ œ œœ
33 3 3
44 4 4
33. Let u sec v du sec v tan v dv&Î%&ΠΈ‰ ˆ‰ˆ‰
111
###
sec v tan v dv du u C sec v C''ˆ‰ˆ‰ ˆ‰ÎÎ&&Î&ÎÎ
11 1
## #
34. Let u csc du csc cot dv 2 du csc cot dvœÊœ Êœˆ‰ ˆ‰ˆ‰ ˆ‰ˆ‰
vvv vv "
# ### ##
111 11
csc cot dv 2 du 2u C 2 csc C''ˆ‰ˆ‰ ˆ‰
vv v
## #
11 1
œ œœ
35. Let u cos (2t 1) du 2 sin (2t 1) dt du sin (2t 1) dtœÊœ Êœ
"
#
dt C C''
sin (2t 1)
cos (2t 1) u u cos (2t 1)
du"
"###"
"" "
&& œ œœ
36. Let u 2 sin t du cos t dt&Î % &
dt du 6 u du 6 C 3(2 sin t) C'''6 cos t 6 u
(2 sin t) u#
$ #

Î&œœ œœ Š‹
37. Let u cot y du csc y dy du csc y dyœÊœ Êœ
##
cot y csc y dy u ( du) u du u C (cot y) C cot y C'''È È ab
# "Î# $Î# $Î# $
"Î#
œ œ œ œ œ
22 2
33 3
38. Let u sec z du sec z tan z dzœÊœ
dz du u du 2u C 2 sec z C'''sec z tan z
sec z uÈÈ
&& &Î&Î
" "Î# "Î# È
39. Let u 1 t 1 du t dt du dtœœ Ê œ Ê œ
"" " #
tt
&
cos 1 dt (cos u)( du) cos u du sin u C sin 1 C'''"" "
tt t
&
ˆ‰ ˆ‰œ œ œœ
40. Let u t 3 t 3 du t dt 2 du dt&Î&Î%& % &È
"Î# "Î# ""
# Èt
cos t 3 dt (cos u)(2 du) 2 cos u du 2 sin u C 2 sin t 3 C'''"
Èt
ˆ‰ ˆ‰ÈÈÎ& & & Î& ÎÎ
41. Let u sin du cos d du cos dœÊœ Êœ
""" ""
))) ))
ˆ‰ˆ‰
&&))
sin cos d u du u C sin C''""" " " "
##
##
))) )
& )œ œ œ
42. Let u csc du csc cot d 2 du cot csc dœÊœ Êœ
ÈÈÈ ÈÈ
Š‹Š‹)))) )))
""
#
ÈÈ
))
d cot csc d 2 du 2u C 2 csc C C'' '
cos
sin sin
2
È
ÈÈ È È
)
)) ) )
&
)))) )œœœœœ
" ÈÈ È
43. Let u s 2s 5s 5 du 3s 4s 5 dsœ Ê œ
$# #
ab
s 2s 5s 5 3s 4s 5 ds u du C C''abab
$# #
##

œ œœ
u s2s5s5
&
& ab
44. Let u 2 8 2 du 4 4 8 d du 2 dœ Ê œ Ê œ ))) )) ) )) )
%# $ $ "
ab ab
4
2 8 2 2 d u du u du C C'''abab ˆ‰ Š‹))) )) )
%# $ "" "
#

œ œ œ œ
44 4 8
u 282
&
%&
& ˆ‰)))

331. Section 5.5 Indefinite Integrals and the Substitution Rule 327
45. Let u 1 t du 4t dt du t dt&Î % & % &
%$ $ "
4
t 1 t dt u du u C 1 t C''$% $ % %
$%
""" "
ab ab ˆ‰ ˆ‰Î& & Î&ÎÎ
444 16
46. Let u 1 du dxœ Ê œ
""
xx
&
dx dx 1 dx u du u du u C 1 C'' ' ''ÉÉ É È ˆ‰
x1 x1 2 2
xxxxx 3 3x
""" " "Î# $Î#
$Î#
(& &œœœœœœ
47. Let u x . Then du xdx and du xdx and x u . Thus x x dx u u duœ" œ# œ œ" "œ "
##$ " "
# #
# ''È
ab È
uudu u u Cu uCx x Cœœœœ""
" "## "" " "
# #&$ &$ & $
$Î# "Î# &Î# $Î# &Î# $Î# # #
&Î# $Î#'ab
’“ ab ab
48. Let u x du x dx and x u . So x dx u u du u u duœ"Ê œ$ œ" $B "œ " œ
$ # $ & $Î# "Î# $ ''' abÈ
abÈ
uuCx x Cœœ""
## # #
&$ & $
&Î# $Î# $ $
&Î# $Î#
ab ab
49. (a) Let u tan x du sec x dx; v u dv 3u du 6 dv 18u du; w 2 v dw dv&%& &%& %& &Î%&
#$ # #
dx du 6 w dw 6w C C'''''18 tan x sec x 18u 6 dv 6 dw 6
2tanx 2u (2 v) w v
&& &

&& &&
ab ab"" "#"
# "
œœœœœœ
CC œ œ
66
2u 2tanx""

(b) Let u tan x du 3 tan x sec x dx 6 du 18 tan x sec x dx; v 2 u dv du&%& %& &Î%&
$## ##
dx C C C'''18 tan x sec x 6 du 6 dv 6 6 6
2tanx (2 u) v v 2 u tan x
&&

&
ab" ""#"
œ œ œ œ œ
(c) Let u 2 tan x du 3 tan x sec x dx 6 du 18 tan x sec x dx&Î % & % &
$## ##
dx C C''18 tan x sec x 6 du 6 6
2tanx uu 2tanx
&&

&
ab" "
œœœ
50. (a) Let u x 1 du dx; v sin u dv cos u du; w 1 v dw 2v dv dw v dvœ Ê œ œ Ê œ œ Ê œ Ê œ
# "
#
1 sin (x 1) sin (x 1) cos (x 1) dx 1 sin u sin u cos u du v 1 v dv''È ÈÈ ' œ œ ###
w dwwC1v C1sinu C1sin(x1) Cœ œ œ œ œ '"" " " "
#
$Î# # # #
$Î# $Î# $Î#
È ab a b a b
33 3 3
(b) Let u sin (x 1) du cos (x 1) dx; v 1 u dv 2u du dv u duœÊœ œÊœ Êœ
# "
#
1 sin (x 1) sin (x 1) cos (x 1) dx u 1 u du v dv v dv''''È È È œ œ œ## ""
##
"Î#
v C v C 1u C 1sin(x1) Cœœœœˆ‰ˆ‰ ab a b
""" "
#
$Î# $Î# # #
$Î# $Î#2
333 3
(c) Let u 1 sin (x 1) du 2 sin (x 1) cos (x 1) dx du sin (x 1) cos (x 1) dxœ Ê œ Ê œ
# "
#
1 sin (x 1) sin (x 1) cos (x 1) dx u du u du u C'''È È ˆ‰ œ œ œ # """
###
"Î# $Î# 2
3
1 sin (x 1) Cœ
" #
$Î#
3
ab
51. Let u 3(2r 1) 6 du 6(2r 1)(2) dr du (2r 1) dr; v u dv du dvœÊœ Ê œ œ Êœ Ê
# """
# #16 u È
È
duœ
"
#1uÈ
dr du (cos v) dv sin v C sin u C'
(2r 1) cos 3(2r 1) 6
3(2r 1) 6
cos u
u 1666


""""
#
ÈÈ
ÈÈ
&
&
&&&Î&Î''Š‹ ˆ‰ ˆ‰ È
sin 3(2r 1) 6 Cœ
" #
6È
52. Let u cos du sin d 2 du dœÊœ Êœ
ÈÈ
Š‹Š‹))) )
"
#
ÈÈ
È
))
)sin
d d 2 u du 2 2u C C''sin sin
cos cos
2 du 4
u u
ÈÈ
ÉÉ ÈÈÈ
È
))
)) ) )

)&
))œ œ œ œ œ '' $Î# "Î#
ˆ‰

332. 328 Chapter 5 Integration
C &Î
4
cos É È
)
53. Let u 3t 1 du 6t dt 2 du 12t dtœÊ œ Ê œ
#
s 12t 3t 1 dt u (2 du) 2 u C u C 3t 1 C;œ œ œ œ œ ''ab ab ˆ‰
#$%%#
$ %"""
##4
s 3 when t 1 3 (3 1) C 3 8 C C 5 s 3t 1 5œ œÊœ ÊœÊ œÊœ
""
##
%#
%
ab
54. Let u x 8 du 2x dx 2 du 4x dx&Î% & % &
#
y 4x x 8 dx u (2 du) 2 u C 3u C 3 x 8 C;& Î & & Î& Î& Î Î''ab ab ˆ‰
# "Î$ #Î$ #Î$ #
"Î$ #Î$
#
3
y 0 when x 0 0 3(8) C C 12 y 3 x 8 12œ œÊœ Ê œ Êœ
#Î$ #
#Î$
ab
55. Let u t du dt&Î % &
1
1#
s 8 sin t dt 8 sin u du 8 sin 2u C 4 t 2 sin 2t C;œœ œœ'' ##
## #
"ˆ‰ ˆ ‰ ˆ‰ ˆ ‰
1 11
1416
u
s 8 when t 0 8 4 2 sin C C 8 1 9œ œ Ê œ Ê œœˆ‰ ˆ‰
11 1 1
16 3 3#
s 4 t 2 sin 2t 9 4t 2 sin 2t 9Êœ œ ˆ‰ ˆ‰ ˆ‰
111 1
163 6#
56. Let u du dœÊ œ
1
4
))
r 3 cos d 3 cos u du 3 sin 2u C sin 2 C;œ œ œ œ '' ##
###
"ˆ‰ ˆ ‰ ˆ‰ ˆ ‰
1 11
4444
u33
)) ) )
r when 0 sin C C r sin 2œœÊœÊœÊœ
11111 111
8884 444 4
33 3 3 3 3
)))
## # ##
ˆ‰ ˆ ‰
r sin2 r cos 2ʜ ʜ
33 3 33 3
4842484##
)) ))ˆ‰
11 1
57. Let u 2t du 2 dt 2 du 4 dtœÊ œ Ê œ
1
#
4 sin 2t dt (sin u)( 2 du) 2 cos u C 2 cos 2t C ;
ds
dt
œ œ œ œ '' ˆ‰ ˆ‰
1 1
# #
""
at t 0 and 100 we have 100 2 cos C C 100 2 cos 2t 100œœ œÊœÊœ
ds ds
dt dt
ˆ‰ ˆ ‰
11
##
""
s 2 cos 2t 100 dt (cos u 50) du sin u 50u C sin 2t 50 2t C ;Êœ œ œ œ ''ˆ ‰ ˆ‰ˆ‰ˆ‰
1 11
# ##
##
at t 0 and s 0 we have 0 sin 50 C C 1 25œœ œÊœ ˆ‰ ˆ‰
11
##
##
1
s sin 2t 100t 25 (1 25 ) s sin 2t 100t 1Êœ Êœ ˆ‰ ˆ‰
11
##
11
58. Let u tan 2x du 2 sec 2x dx 2 du 4 sec 2x dx; v 2x dv 2 dx dv dxœÊœ Êœ œÊœÊœ
## "
#
4 sec 2x tan 2x dx u(2 du) u C tan 2x C ;
dy
dx
&&&Î&Î'' ###
""
at x 0 and 4 we have 4 0 C C 4 tan 2x 4 sec 2x 1 4 sec 2x 3œ œ œ Ê œ Ê œ œ œ
dy dy
dx dx
""
## #
ab
y sec 2x 3 dx sec v 3 dv tan v v C tan 2x 3x C ;%& Î & Î & Î Î & Î Î''abab ˆ‰
## "" "
## # #
##
3
at x 0 and y 1 we have 1 (0) 0 C C 1 y tan 2x 3x 1œœ œÊœÊœ
""
##
##
59. Let u 2t du 2 dt 3 du 6 dtœÊ œ Ê œ
s 6 sin 2t dt (sin u)(3 du) 3 cos u C 3 cos 2t C;œœ œœ
''
at t 0 and s 0 we have 0 3 cos 0 C C 3 s 3 3 cos 2t s 3 3 cos ( ) 6 mœœ œÊœÊœ Êœ œ ˆ‰
1
#
1
60. Let u t du dt du dtœÊ œ Ê œ1111
#
v cos t dt (cos u)( du) sin u C sin ( t) C ;&& &Î&Î''11 1 1 11
#
""
at t 0 and v 8 we have 8 (0) C C 8 v sin ( t) 8 s ( sin ( t) 8) dt&& &Î%&%&& Î%& Î 11111 ""
ds
dt
'
sin u du 8t C cos ( t) 8t C ; at t 0 and s 0 we have 0 1 C C 1œœœœ œÊœ'
## ## 1

333. Section 5.6 Substitution and Area Between Curves 329
s 8t cos ( t) 1 s(1) 8 cos 1 10 mÊœ Ê œ œ11
61. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on
the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
sinxC1cosxC C1C; also cosxC C CC C .
## #
""#" # #$#"
## # #
"""
œ Ê œ œ Ê œœ
cos 2x
62. Both integrations are correct. In each case, the derivative of the function on the right is the integrand on the
left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
CCC
tan x sec x 1 sec x
&& &
####
"
œ œ
ðñò
ˆ‰
a constant
63. (a) V sin 120 t dt 60 V cos (120 t) [cos 2 cos 0]Š‹ ‘ˆ‰
""

"Î'!
! #
"
60
0 120
V
'
0
160
max max

111œ œ
11
max
[1 1] 0œ œ
V
max
#1
(b) V 2 V 2 (240) 339 volts
max rmsœœ ¸
ÈÈ
(c) V sin 120 t dt V dt (1 cos 240 t) dt'''
000
160 160 160
max max

ab ab ˆ‰
###
##
11œœ
1 cos 240 t V1 ab max
&
t sin 240 t sin (4 ) 0 sin (0)œ œ œ
ab ab abVV V
240 60 240 40 1 0
max max max
&& &
## ##
""""
"Î'!
!
‘ ‘ˆ‰ ˆ ‰ˆ ‰ ˆ‰ ˆ‰
111
11
5.6 SUBSTITUTION AND AREA BETWEEN CURVES
1. (a) Let u y 1 du dy; y 0 u 1, y 3 u 4&Î % & & % & & % &
y 1 dy u du u (4) (1) (8) (1)''
01
34
È ‘ˆ‰ ˆ‰ ˆ‰ ˆ‰œ œ œ œœ
"Î# $Î# $Î# $Î#
%
"
22 2 2214
33 3 333
(b) Use the same substitution for u as in part (a); y 1 u 0, y 0 u 1œ Ê œ œ Ê œ
y 1 dy u du u (1) 0''
&10
01
È ‘ˆ‰œ œ œ œ
"Î# $Î# $Î#
"
!
22 2
33 3
2. (a) Let u 1 r du 2r dr du r dr; r 0 u 1, r 1 u 0œ Ê œ Ê œ œ Ê œ œ Ê œ
# "
#
r 1 r dr u du u 0 (1)''
01
10
È È ‘ˆ‰œ œ œ œ# "" ""
#
$Î# $Î#
!
"333
(b) Use the same substitution for u as in part (a); r 1 u 0, r 1 u 0œ Ê œ œ Ê œ
r 1 r dr u du 0''
&10
10
È Èœ œ# "
#
3. (a) Let u tan x du sec x dx; x 0 u 0, x u 1œÊœ œÊœœÊœ
# 1
4
tan x sec x dx u du 0''
00
41

#
## #
"
!
"
œœœœ ’“
u1

&&
(b) Use the same substitution as in part (a); x u 1, x 0 u 0œÊœ œÊœ
1
4
tan x sec x dx u du 0''
&41
00
#
###
!
"
""
œœœœ ’“
u
&
4. (a) Let u cos x du sin x dx du sin x dx; x 0 u 1, x u 1œ Ê œ Ê œ œ Ê œ œ Ê œ 1
3 cos x sin x dx 3u du u ( 1) (1) 2
''
01
1

##$$$

"
"
œ œ œœcd ab
(b) Use the same substitution as in part (a); x 2 u 1, x 3 u 1œÊœ œÊœ11
3 cos x sin x dx 3u du 2
''
21
31

##

œ œ

334. 330 Chapter 5 Integration
5. (a) u 1 t du 4t dt du t dt; t 0 u 1, t 1 u 2&Î % & % & & % & & % &
%$ $ "
4
t 1 t dt u du''
01
12
$% $
$
"
#
"
ab ’“ œ œ œœ
416161616
u2115
%%%
(b) Use the same substitution as in part (a); t 1 u 2, t 1 u 2œ Ê œ œ Ê œ
t 1 t dt u du 0''
&12
12
$% $
$
"
abÎ& &
4
6. (a) Let u t 1 du 2t dt du t dt; t 0 u 1, t 7 u 8&Î% & % & & % & & % &
# "
# È
t t 1 dt u du u (8) (1)''
01
78

ab ‘ˆ‰ ˆ‰ˆ‰ˆ‰
# "Î$ %Î$ %Î$ %Î$
"Î$ ""
##
)
"
œ œ œœ
33345
4888
(b) Use the same substitution as in part (a); t 7 u 8, t 0 u 1œ Ê œ œ Ê œ
È
t t 1 dt u du u du'''
&781
018
ab
# "Î$ "Î$
"Î$ ""
##
œ œ œ
45
8
7. (a) Let u 4 r du 2r dr du r dr; r 1 u 5, r 1 u 5œ Ê œ Ê œ œÊ œ œ Ê œ
# "
#
dr 5 u du 0''
"
"
#
#15
15
5r
4rab
&
&
œœ
(b) Use the same substitution as in part (a); r 0 u 4, r 1 u 5œÊœ œÊœ
dr 5 u du 5 u 5 (5) 5 (4)''
04
15
5r
4r 8ab"
"""""
####
# " " "
&
%
&
&
œœœœ ‘ˆ ‰ˆ ‰
8. (a) Let u 1 v du v dv du 10 v dv; v 0 u 1, v 1 u 2&Î % & % & & % & & % &
$Î# "Î#
#
320
3 È
dv du u du'' '
01 1
12 2
10 v
1v
u3 3 3u 3 1 3
20 20 20 20 1 10
È
ab"
""" #
#
" #
)&
& &
œ œ œ œ œˆ‰ ‘ ‘
(b) Use the same substitution as in part (a); v 1 u 2, v 4 u 1 4 9&%& &%&Î &
$Î#
dv du''
12
49
10 v
1v
u3 3u 39 2 3 18 7
20 20 20 1 20 7 70
È
ab"
"""
*
# #
)&
& &
œœœœœˆ‰ ‘ˆ‰ˆ‰
9. (a) Let u x 1 du 2x dx 2 du 4x dx; x 0 u 1, x 3 u 4&Î% & % & &%& & %&
# È
dx du 2u du 4u 4(4) 4(1) 4'''
011
344

4x 2
x1 uÈÈ
&
"
"Î# "Î# "Î# "Î#
%
"
œ œ œ œœ ‘
(b) Use the same substitution as in part (a); x 3 u 4, x 3 u 4œ Ê œ œ Ê œ
ÈÈ
dx du 0''
&

34
34
4x 2
x1 uÈÈ
&
"
œœ
10. (a) Let u x 9 du 4x dx du x dx; x 0 u 9, x 1 u 10&Î% & % & &%& &%&
%$$ "
4
dx u du (2)u (10) (9)''
09
110
x
x9 44
10 3

%
È
È
"
"" """Î# "Î# "Î# "Î#
"!
* ## #

œœœœ ‘
(b) Use the same substitution as in part (a); x 1 u 10, x 0 u 9œ Ê œ œ Ê œ
dx u du u du'' '
&110 9
09 10
x
x9 44
310

%
È
È
"
"""Î# "Î#
#
œœœ
11. (a) Let u 1 cos 3t du 3 sin 3t dt du sin 3t dt; t 0 u 0, t u 1 cos 1œ Ê œ Ê œ œ Ê œ œ Ê œ œ
"
#36
11
(1 cos 3t) sin 3t dt u du (1) (0)''
00
61

œœœœ
"" """
#
"
!
##
33 666
u
’“Š‹
&
(b) Use the same substitution as in part (a); t u 1, t u 1 cos 2œÊœ œÊœ œ
11
63
1
(1 cos 3t) sin 3t dt u du (2) (1)''



61
32
œœœœ
"" """
#
#
"
##
33 662
u
’“Š‹
&

335. Section 5.6 Substitution and Area Between Curves 331
12. (a) Let u 2 tan du sec dt 2 du sec dt; t u 2 tan 1, t 0 u 2&Î % & % & & % &Î & & % &
tt t
4### ##
"## 11
ˆ‰
2 tan sec dt u (2 du) u 2 1 3''
&21
02
ˆ‰ cd œ œ œœ
tt
##
####
#
"
(b) Use the same substitution as in part (a); t u 1, t u 3œÊœœÊœ

##
11
2 tan sec dt 2 u du u 3 1 8''
&


21
23
ˆ‰ cd œ œ œœ
tt
##
####
$
"
13. (a) Let u 4 3 sin z du 3 cos z dz du cos z dz; z 0 u 4, z 2 u 4&Î % & % & & % & & % &
"
3
1
dz du 0''
94
24

cos z
43 sin z u3ÈÈ"
""
œœ ˆ‰
(b) Use the same substitution as in part (a); z u 4 3 sin ( ) 4, z u 4œÊœ œ œÊœ111
dz du 0''
&

cos z
43 sin z u3ÈÈ"
""
œœ
4
4
ˆ‰
14. (a) Let u 3 2 cos w du 2 sin w dw du sin w dw; w u 3, w 0 u 5œ Ê œ Ê œ œ Ê œ œ Ê œ
"
##
1
dw u du u''
&23
05
sin w
(3 2 cos w) 5 3 15"###
# """ """" &
$
&œœœœˆ‰ ˆ‰ cd
(b) Use the same substitution as in part (a); w 0 u 5, w u 3œÊœ œ Êœ
1
#
dw u du u du''
%

23
5
5
3
sin w
(3 2 cos w) 15"##
# #"" "
&œœ œˆ‰ '
15. Let u t 2t du 5t 2 dt; t 0 u 0, t 1 u 3&Î % & Î & % & & % &
&%
ab
t 2t 5t 2 dt u du u (3) (0) 2 3''
00
13
È
ab ‘ È& % "Î# $Î# $Î# $Î#
$
!
œ œ œœ
222
333
16. Let u 1 y du ; y 1 u 2, y 4 u 3&Î % & & % & & % &È
dy
2yÈ
du u du u'''
122
433
dy
2y1 y
u326
11
ÈȈ‰"
"" # "
$
#
& &œœ œœœ cd ˆ‰ˆ‰
17. Let u cos 2 du 2 sin 2 d du sin 2 d ; 0 u 1, u cos 2œÊœ Êœ œÊœœÊœ œ))) ))))
" "
# #
11
66
ˆ‰
cos 2 sin 2 d u du u du'''
%

61212
11
$ $ $ "" " ""
## #
"Î#
"
))) œœ œ œœˆ‰ ’“Š‹
24(1)4
u3
4
Î&
&
& &
ˆ‰
1
18. Let u tan du sec d 6 du sec d ; u tan , œÊœ Êœ œÊœœœÊˆ‰ ˆ‰ ˆ‰ ˆ‰
)) ) 11
666 6 6 3
3" "##
#
)))1 )
È
utan 1œœ
1
4
cot sec d u (6 du) 6 12''


32 1
13

&# &

"
"Î $
"
"Î $
#ˆ‰ ˆ‰ ‘’“Š‹
&%
))
66 4
u333
2u 2(1)
)œœœœœ

Î%
%%%
"
È
È
Š‹
3
19. Let u 5 4 cos t du 4 sin t dt du sin t dt; t 0 u 5 4 cos 0 1, t œ Ê œ Ê œ œ Ê œ œ œ Ê
"
4
1
u 5 4 cos 9œ œ 1
5 (5 4 cos t) sin t dt 5u du u du u 9 1'''
%

œœœœœ$"
"Î% "Î% "Î% &Î% &Î% &Î# "
*
"
11
99
ˆ‰ ‘ˆ‰
44 45
554
20. Let u 1 sin 2t du 2 cos 2t dt du cos 2t dt; t 0 u 1, t u 0œ Ê œ Ê œ œ Ê œ œ Ê œ
"
#
1
4
(1 sin 2t) cos 2t dt u du u (0) (1)''
%

40
1
œœœœ
$Î# $Î# &Î# &Î# &Î# "" "
#
!
" ‘ˆ ‰ˆ ‰ˆ‰
25 5 5 5
211
21. Let u 4y y 4y 1 du 4 2y 12y dy; y 0 u 1, y 1 u 4(1) (1) 4(1) 1 8œ Ê œ œÊœ œÊœ œ
#$ # # $
ab
4y y 4y 1 12y 2y 4 dy u du 3u 3(8) 3(1) 3
''
%
1 8
1
abab ‘ œ œ œœ
# $ # #Î$ "Î$ "Î$ "Î$
#Î$ )
"

336. 332 Chapter 5 Integration
22. Let u y 6y 12y 9 du 3y 12y 12 dy du y 4y 4 dy; y 0 u 9, y 1œÊœ Ê œ œÊœœ
$# # # "
ab ab
3
u4ʜ
y 6y 12y 9 y 4y 4 dy u du 2u (4) (9) (2 3)''
%
1 4
9
abab ‘ˆ‰
$ # # "Î# "Î# "Î# "Î#
"Î# ""
%
*
œ œ œœ
33 333
22 2 œ
2
3
23. Let u du d du d ; 0 u 0, uœÊœ Ê œ œÊœœ Êœ)))))) )11
$Î# "Î#
#
$ #32
3 È È
cos d cos u du sin 2u sin 2 (0)''
%%
&

È ˆ‰ ˆ ‰ ‘ˆ ‰ˆ‰))) 1# $Î# #
##
""
!
œœœœ
22u 2 2
334 34 33
1
11
24. Let u 1 du t dt; t 1 u 0, t u 1œ Ê œ œÊ œ œ Ê œ
"" #
#t
t sin 1 dt sin u du sin 2u sin ( 2) sin 0''
& %
1
12 1
# # # """""
"
! ## ˆ‰ ‘ˆ ‰ˆ ‰ˆ‰ ’“œ œ œ
t2444
u0
sin 2œ
""
# 4
25. Let u 4 x du 2x dx du x dx; x 2 u 0, x 0 u 4, x 2 u 0œ Ê œ Ê œ œÊœ œÊœ œÊœ
# "
#
A x 4 x dx x 4 x dx u du u du 2 u du u duœ œ œ œ'' ''''
& %% %%2 4
02 4044
ÈÈ ## """
###
"Î# "Î# "Î# "Î#
u (4) (0)œ œœ ‘
22216
3333
$Î# $Î# $Î#
%
!
26. Let u 1 cos x du sin x dx; x 0 u 0, x u 2œ Ê œ œ Ê œ œ Ê œ 1
(1 cos x) sin x dx u du 2''
%%

œ œ œœ
2
’“
u20
2
&&&
#
!
##
27. Let u 1 cos x du sin x dx du sin x dx; x u 1 cos ( ) 0, x 0œ Ê œ Ê œ œÊ œ œ œ 11
u1cos 02%&Î &
A 3 (sin x) 1 cos x dx 3u ( du) 3 u du 2u 2(2) 2(0) 2œ œ œ œ œœ'''
& %%
022
È ‘ "Î# "Î# $Î# $Î# $Î# &Î#
#
!
28. Let u sin x du cos x dx du cos x dx; x u sin 0, x 0 uœ Ê œ Ê œ œÊœ œ œÊœ11 1 11 1
"
##1
11
ˆ‰
Because of symmetry about x , A 2 (cos x) (sin ( sin x)) dx 2 (sin u) duœ œ œ
11 1
1## #
" ''
&
2
0
11 ˆ‰
sin u du [ cos u] ( cos ) ( cos 0) 2œœœœ
'
%

1
!
1
29. For the sketch given, a 0, b ; f(x) g(x) 1 cos x sin x ;œœ œ œ œ1
##
#
1 cos 2x
A dx (1 cos 2x) dx x [( 0) (0 0)]œ œ œ œ œ''
%%

( cos 2x) sin 2x"
## ### #
"""
!
‘
1
1
1
30. For the sketch given, a , b ; f(t) g(t) sec t 4 sin t sec t 4 sin t;œ œ œ œ
11
33
""
##
####
ab
A sec t 4 sin t dt sec t dt 4 sin t dt sec t dt 4 dt&Î&Î&Î'''''
&


33333
33333
ˆ‰
"" "
## # #
## # # # "( cos 2t)
sec t dt 2 (1 cos 2t) dt [tan t] 2[t ] 3 4 3œ œ œ œ
""
###
#
Î$
Î$
Î$
Î$''
&


33
33
1
1
1
1
11sin 2t 4
33
ÈÈ
†
31. For the sketch given, a 2, b 2; f(x) g(x) 2x x 2x 4x x ;œ œ œ œ
#% # #%
ab
A 4x x dxœ œ œ œœ œ'
&2
2
ab ’“ ˆ‰ ‘ˆ‰
#%
#
#
4x x 32 32 32 32 64 64 320 192 128
3 5 35 3 5 35 15 15
(

337. Section 5.6 Substitution and Area Between Curves 333
32. For the sketch given, c 0, d 1; f(y) g(y) y y ;œœ œ
#$
A y y dy y dy y dyœ œ œ œ œœ'''
%%%
111
ab ’“ ’“
#$ # $
""
!!
" " "" "
#
yy(0)(0)
34 34341
%
33. For the sketch given, c 0, d 1; f(y) g(y) 12y 12y 2y 2y 10y 12y 2y;œœ œ œ abab
#$ # #$
A 10y 12y 2y dy 10y dy 12y dy 2y dy y y yœœ œ' '''
% %%%
1 111
ab ‘‘‘
#$ # $ $ % #
"""
!! ! #
10 12 2
34
0 (30)(10)œ œˆ‰
10 4
33
34. For the sketch given, a 1, b 1; f(x) g(x) x 2x x 2x ;œ œ œ œ
# %#%
ab
A x 2x dxœ œ œ œœ œ'
&1
1
ab ’“ ˆ‰ ‘ˆ‰
#%
"
"
"" "x 2x 2 2 24101222
3 5 35 3 5 35 15 15
(
35. We want the area between the line y 1, 0 x 2, and the curve y , the area of a triangleœ Ÿ Ÿ œ 738?=
x
4
&
(formed by y x and y 1) with base 1 and height 1. Thus, A 1 dx (1)(1) xœœ œœ '
%
2
Š‹ ’“
xx
41
&
""
###
#
!
22œ œœˆ‰
825
136## #
""
36. We want the area between the x-axis and the curve y x , 0 x 1 the area of a triangle (formed by x 1,œŸŸ:6?= œ
#
x y 2, and the x-axis) with base 1 and height 1. Thus, A x dx (1)(1)Î& & Î & Î&Î& '
%
1
# """"
###
"
!
’“
x5
336

37. AREA A1 A2&Î
A1: For the sketch given, a 3 and we find b by solving the equations y x 4 and y x 2xœ œ œ
##
simultaneously for x: x 4 x 2x 2x 2x 4 0 2(x 2)(x 1) x 2 or x 1 so
## #
œ Ê œ Ê Ê œ œ
b 2: f(x) g(x) x 4 x 2x 2x 2x 4 A1 2x 2x 4 dxœ œ œ Ê œ aba b a b
### # '
&
&3
2
4x 4 8 ( 18 9 12) 9 ;œ œœœ’“ ˆ‰
2x 2x 16 16 11
33 33
&
#
#
$
A2: For the sketch given, a 2 and b 1: f(x) g(x) x 2x x 4 2x 2x 4œ œ œ œ abab
## #
A2 2x 2x 4 dx x 4x 1 4 4 8Êœ œ œ'
&2
1
ab ’“ ˆ‰ˆ ‰
##
"
#
2x 2 16
333

14 489;œ œ
216
33
Therefore, AREA A1 A2 9&Î&Î&
11 38
33
38. AREA A1 A2&Î
A1: For the sketch given, a 2 and b 0: f(x) g(x) 2x x 5x x 3x 2x 8xœ œ œ œ abab
$# # $ A1 2x 8x dx 0 (8 16) 8;Êœ œ œœ'
&2
0
ab ’“
$
#
!
#
2x 8x
4
% &
A2: For the sketch given, a 0 and b 2: f(x) g(x) x 3x 2x x 5x 8x 2xœœœœ aba b
#$# $
A2 8x 2x dx (16 8) 8;Êœ œ œœ'
0
2
ab ’“
$
#
!
8x 2x
24
& %
Therefore, AREA A1 A2 16&Î&
39. AREA A1 A2 A3&ÎÎ
A1: For the sketch given, a 2 and b 1: f(x) g(x) ( x 2) 4 x x x 2œ œ œ œ ab
##
A1 x x 2 dx 2x 2 4 ;Ê œ œ œ œœ œ'
&
&2
1
ab ’“ ˆ‰ˆ‰
#
## #
"
#
"" " "xx 84 7 1431
3332366
&
A2: For the sketch given, a 1 and b 2: f(x) g(x) 4 x ( x 2) x x 2œ œ œ œ ab a b
##
A2 x x 2 dx 2x 4 2 3 8 ;Êœ œ œœœ'
&1
2
ab ’“ ˆ‰ˆ ‰
#
## ##
#
"
"x x 84 11 9
3332
&

338. 334 Chapter 5 Integration
A3: For the sketch given, a 2 and b 3: f(x) g(x) ( x 2) 4 x x x 2œœœœ ab
##
A3 x x 2 dx 2x 6 4 9 ;Ê œ œ œ œ'
2
3
ab ’“ ˆ‰ˆ‰
#
## #
$
#
xx 279 84 98
33323
&
Therefore, AREA A1 A2 A3 9 9œœœœ
11 9 9 8 5 49
6366##
ˆ‰
40. AREA A1 A2 A3&ÎÎ
A1: For the sketch given, a 2 and b 0: f(x) g(x) x x x 4xœ œ œ œ œ Š‹ ab
xxx4
33333

"$
A1 x 4x dx 2x 0 (4 8) ;Êœ œ œœ
"" " $#
!
#
33433
x4'
&2
0
ab ’“
%
A2: For the sketch given, a 0 and we find b by solving the equations y x and y simultaneouslyœœœ
xx
33

for x: x x 0 (x 2)(x 2) 0 x 2, x 0, or x 2 so b 2:
xxx4 x
3333 3

œÊ œÊ œÊœ œ œ œ
f(x) g(x) x x 4x A2 x 4x dx 4x x 2x œ œ Ê œ œ œ
xx x
33 3 3 3 3 4
Š‹ ’ “ ab ab ab
%
""""$$$#
#
! ''
00
22
(8 4) ;œœ
"
33
4
A3: For the sketch given, a 2 and b 3: f(x) g(x) x x 4xœœœœ Š‹ ab
xx
333

"$
A3 x 4x dx 2x 2 9 8 14 ;Êœ œ œ œ œ
""" "$#
$
#
3343443412
x81168125'
2
3
ab ’“ ‘ˆ ‰ˆ‰ˆ‰
%
†
Therefore, AREA A1 A2 A3&ÎÎ&ÎÎ& &
4425322519
3312 1 4
"
#
41. a 2, b 2;œ œ
f(x) g(x) 2 x 2 4 xœœ ab
## A4xdx4x 8 8Êœ œ œ'
&2
2
ab ’“ ˆ‰ˆ ‰
#
#
#
x88
333

2œœ†ˆ‰
24 8 32
33 3

42. a 1, b 3;œ œ
f(x) g(x) 2x x ( 3) 2x x 3œœab
##
A 2x x 3 dx x 3xÊœ œ '
&1
3
ab ’“
##
$
"
x
3

991311œœœˆ‰ˆ‰
27 1 32
3333
"

43. a 0, b 2;œœ
f(x) g(x) 8x x A 8x x dxœÊœ
%% '
0
2
ab
16œ œœ œ’“
8x x 32803248
25 555
& (
#
!


44. Limits of integration: x 2x x x 3x
##
œÊ œ
x(x 3) 0 a 0 and b 3;ʜʜ œ
f(x) g(x) x x 2x 3x xœœ ab
##
A 3x x dxÊœ œ '
0
3
ab ’“
#
$
!
3x x
23
&
9œœ œ
27 27 18 9
###

339. Section 5.6 Substitution and Area Between Curves 335
45. Limits of integration: x x 4x 2x 4x 0
## #
œ Ê œ
2x(x 2) 0 a 0 and b 2;ʜʜ œ
f(x) g(x) x 4x x 2x 4xœœab
###
A 2x 4x dxÊœ œ '
0
2
ab ’“
#
#
!
2x 4x
32
œ
œ œ œ
16 16 32 48 8
363#


46. Limits of integration: 7 2x x 4 3x 3 0œÊ œ
## #
3(x 1)(x 1) 0 a 1 and b 1;ʜʜ œ
f(x) g(x) 7 2x x 4 3 3xœœabab
## #
A 3 3x dx 3 xÊœ œ '
œ1
1
ab ’“
#
"
"
x
3

31 1 6 4œœ œ ‘ˆ‰ˆ‰ˆ ‰
""
333
2

47. Limits of integration: x 4x 4 x
%# #
œ
x5x40 x4x10Ê œÊ œ
%# # #
abab
(x 2)(x 2)(x 1)(x 1) x 2, 1, 1, 2;ʜ!ʜ
f(x) g(x) x 4x 4 x x 5x 4 and œ œ ab
%# #%#
g(x) f(x) x x 4x 4 x 5x 4œœ
#% # % #
ab
A x 5x 4 dx x 5x 4 dxʜ
''
œœ
œ21
11
abab
%# %#
x5x4dx'
1
2
ab
%#
4x 4x 4xœ ’“’“’“
x5x x5x x5x
53 53 5 3
$
" " #
# " "


4844 84œ ˆ‰ˆ ‰ˆ‰ˆ ‰ˆ ‰ˆ ‰
""" "
53 5 3 53 53 5 3 53
5 3240 5 5 3240 5
8œ œ œ
60 60 300 180
53 15

48. Limits of integration: x a x 0 x 0 or
È
##
œÊœ
a x 0 x 0 or a x 0 x a, 0, a;
È
## ##
œÊœ œÊœ
A x a x dx x a x dxœ ''
œa0
0a
ÈÈ## ##
ax axœ
""
##
## ##
$Î# $Î#
!
!
’“’“ab ab
22
33
a
a
aaœ œ
""##
$Î# $Î#
333
2a
ab ab’“


49. Limits of integration: y x and
x, x 0
x, x 0
œœ
Ÿ
 
Èkk
Î
È
È
5y x 6 or y ; for x 0: xœ œ Ÿ œ
x6 x6
55 55 È
5 x x 6 25( x) x 12x 36ʜʜÈ
#
x 37x 36 0 (x 1)(x 36) 0%Î Î&%Î Î &
#
x 1, 36 (but x 36 is not a solution);Êœ œ
for x 0: 5 x x 6 25x x 12x 36 œÊœÈ
#
x 13x 36 0 (x 4)(x 9) 0Ê œÊ œ
#
x 4, 9; there are three intersection points andʜ

A x dx x dx x dxœ'''
œ104
049
ˆ ‰ˆ‰ˆ‰ÈÈÈ
x6 x6 x6
55 5
"" "

340. 336 Chapter 5 Integration
(x) x xœ ’ “’“’“
(x 6) (x 6) (x 6)
10 3 10 3 3 10
222"" " $Î# $Î# $Î#
!%*
"!%
&& &
409 4œ œœˆ‰ˆ ‰ˆ ‰
36 25 2 100 2 36 2 225 2 100 50 20 5
10 10 3 10 3 10 3 10 3 10 10 3 3
†††
$Î# $Î# $Î#
50. Limits of integration:
yx4
x 4, x 2 or x 2
4 x , 2 x 2
œœ
Ÿ  
ŸŸ
kk œ
#
#
#
for x 2 and x 2: x 4 4Ÿ   œ
# x
2
&
2x 8 x 8 x 16 x 4;ʜʜʜ„
## #
for 2 x 2: 4 x 4 8 2x x 8Ÿ Ÿ œ Ê œ
###
#
x
&
x 0 x 0; by symmetry of the graph,ʜʜ
#

A2 44xdx2 4x4dx2 28xœœ''
02
24
’“’“’“’“Š‹ Š‹ ab ab
xxxx
2226
&&
##
#%
!#
20232 16 40œ œœˆ‰ˆ ‰
8 64 8 56 64
66 33#
51. Limits of integration: c 0 and d 3;œœ
f(y) g(y) 2y 0 2yœœ
##
A 2y dy 2 9 18Êœ œ œ œ'
0
3
#
$
!
’“
2y
3

†

52. Limits of integration: y y 2 (y 1)(y 2) 0
#
œ Ê œ
c 1 and d 2; f(y) g(y) (y 2) yÊœ œ œ
#
A y 2 y dy 2yÊœ œ '
&1
2
ab ’“
#
#
#
"
yy
3
&
4262œœœˆ‰ˆ‰
48 8 9
3333## ##
"" ""

53. Limits of integration: 4x y 4 and 4x 16 yœ œ
#
y 4 16 y y y 20 0 ʜʜÊ
##
(y 5)(y 4) 0 c 4 and d 5;œÊœ œ
f(y) g(y)œ œˆ‰ Š‹
16 y y 4 y y 20
44 4

&&
A y y 20 dyʜ
" #
4'
&4
5
ab
20yœ
"
#
&
%
43
yy
’“
&
100 80œ
""
#432 43
125 25 64 16
ˆ‰ˆ‰
180œ œ
"
432 8
189 9 243
ˆ‰

341. Section 5.6 Substitution and Area Between Curves 337
54. Limits of integration: x y and x 3 2yœœ
##
y 3 2y 3y 3 3(y 1)(y 1) 0Êœ Ê œÊ œ
###
c 1 and d 1; f(y) g(y) 3 2y yÊœ œ œ ab
##
33y 31y A3 1y dyœ œ Ê œ
## #
ab ab '
&1
1
3y 31 3 1œ œ’“ ˆ‰ˆ ‰
y
333
"
"
""
321 4œœ†ˆ‰
"
3

55. Limits of integration: x y and x 2 3yœ œ
##
y 2 3y 2y 2 0Ê œ Ê œ
###
2(y 1)(y 1) 0 c 1 and d 1;ʜʜ œ
f(y) g(y) 2 3y y 2 2y 2 1 yœœœabab ab
## # #
A 2 1 y dy 2 yÊœ œ '
&1
1
ab ’“
#
"
"
y
3

21 2 1 4œœ œˆ‰ˆ ‰ˆ‰
""
3333
28
56. Limits of integration: x y and x 2 yœœ
#Î$ %
y 2 y c 1 and d 1;ʜʜ œ
#Î$ %
f(y) g(y) 2 y yœ ab
% #Î$
A 2 y y dyʜ '
&1
1
ˆ‰
% #Î$
2y yœ’“
y
55
3
(
&Î$
"
"
22œˆ‰ˆ ‰
""
55 55
33
22œœˆ‰
"
55 5
312

57. Limits of integration: x y 1 and x y 1 yœ œ
# #
kkÈ
y1y1y y2y1y1yÊœ Ê œ
#%### #
kk abÈ
y2y1yy 2y3y10Ê œÊ œ
%# #% %#
2y 1 y 1 0 2y 1 0 or y 1 0Ê œÊœ œabab
## # #
y or y 1 y or y 1.Ê œ œ Ê œ„ œ„
##"
##
È
2
Substitution shows that are not solutions y 1;
„
#
È
2
Êœ„
for 1 y 0, f(x) g(x) y 1 y y 1Ÿ Ÿ œ È ab
# #
1 y y 1 y , and by symmetry of the graph,œ
##
"Î#
ab
A2 1y y1y dyœ
'
&1
0
’“ ab
##
"Î#

2 1 y dy 2 y 1 y dyœ
''
&&11
00
ab ab
##
"Î#
2y 2 2( 0) 1 0 2œ œ!œ’“ ˆ‰ ‘ˆ ‰
”•
ˆ‰
y
33 33
21 y 2
&
)&!
"
""
#

!
"
ab

342. 338 Chapter 5 Integration
58. AREA A1 A2&Î
Limits of integration: x 2y and x y y œœÊ
$#
y y 2y y y y 2 y(y 1)(y 2) 0
$# #
œ Ê œ œab
y 1, 0, 2:ʜ
for 1 y 0, f(y) g(y) y y 2yŸ Ÿ œ
$#
A1 y y 2y dy yÊœ œ'
&1
0
ab ’“
$# #
!
"
yy
43
%
01;œ œˆ‰
""
43 12
5
for 0 y 2, f(y) g(y) 2y y yŸŸ œ
$#
A2 2y y y dy yÊœ œ'
!
$# #
#
!
2
ab ’“
yy
43
%
40;Êœˆ‰
16 8 8
43 3
Therefore, A1 A2Î&Î&
5837
12 3 12

59. Limits of integration: y 4x 4 and y x 1œ œ
#%
x 1 4x 4 x 4x 5 0ÊœÊ œ
%#%#
x 5 (x 1)(x 1) 0 a 1 and b 1;Ê œÊœ œab
#
f(x) g(x) 4x 4 x 1 4x x 5 œ œ
#% #%
A 4x x 5 dx 5xÊœ œ'
&1
1
ab ’“
#%
"
"
4x x
35
(
5525œœœˆ‰ˆ‰ˆ‰
4 4 4 104
35 35 35 15
"" "

60. Limits of integration: y x and y 3x 4œœ
$#
x3x40 xx2(x2)0Ê œÊ œ
$# #
ab
(x 1)(x 2) 0 a 1 and b 2;Ê œÊœ œ
#
f(x) g(x) x 3x 4 x 3x 4œœ
$# $#
ab
A x 3x 4 dx 4xÊœ œ'
&1
2
ab ’“
$#
#
"
x3x
43
%
84œ"œˆ‰ˆ‰
16 24 1 27
43 4 4

61. Limits of integration: x 4 4y and x 1 yœ œ
#%
44y1y y4y30Ê œ Ê œ
#%%#
y 3 y 3 (y 1)(y 1) 0 c 1Ê œÊœŠ‹Š‹
ÈÈ
and d 1 since x 0; f(y) g(y) 4 4y 1 yœ œ abab
#%
3 4y y A 3 4y y dyœ Ê œ
#% #% '
&1
1
ab
3y 2 3œ œœ’“ ˆ‰
4y y
35 35 15
456
( "
"
"

62. Limits of integration: x 3 y and xœ œ
# y
4
&
3 y 3 0 (y 2)(y 2) 0Ê œ Ê œÊ œ
# y3y
44 4
3
&&
c 2 and d 2; f(y) g(y) 3 yÊœ œ œ ab Š‹
# y
4
&
3 1 A 3 1 dy 3 yœÊœ œŠ‹ Š‹ ’“
yyy
441

'
&2
2 #
#
#
32 2 34 12 4 8œœœœ ‘ˆ ‰ˆ‰ˆ ‰
88 16
12 12 12

343. Section 5.6 Substitution and Area Between Curves 339
63. a 0, b ; f(x) g(x) 2 sin x sin 2xœœ œ 1
A (2 sin x sin 2x) dx 2 cos xÊœ œ ' ‘

0
cos 2x
2
1
!
2( 1) 2 1 4œ œ ‘ˆ ‰
""
##
†

64. a , b ; f(x) g(x) 8 cos x sec xœ œ œ
11
33
#
A 8 cos x sec x dx [8 sin x tan x]Êœ œ '
&


3
3
ab
# Î$
Î$
1
1
838363œœŠ‹Š ‹
ÈÈÈ
††
ÈÈ
33
##

65. a 1, b 1; f(x) g(x) 1 x cosœ œ œ ab ˆ‰
#
#
1x
A 1 x cos dx x sinÊœ œ'
&1
1
‘ˆ‰ˆ‰ ’“
#
##
"
"
11
1
xx2x
3

112œœ œˆ‰ˆ ‰ˆ‰
""
3333
222244
1111

66. A A1 A2&Î
a 1, b 0 and a 0, b 1;
"" ##œ œ œ œ
f (x) g (x) x sin and f (x) g (x) sin x"" ##
##œ œ ˆ‰ ˆ‰
11xx
by symmetry about the origin,Ê
A A 2A A 2 sin x dx"# "
#œ Êœ '
0
1
‘ˆ‰
1x
2 cos 2 0 10œ œ ’“ ˆ‰ ‘ˆ‰ˆ‰
2xx 2 2
111
1
## #
"
!
"
&
††
22œœ œˆ‰ˆ‰
244
2111
11"
#

67. a , b ; f(x) g(x) sec x tan xœ œ œ
11
44
##
A sec x tan x dxʜ '
&


4
4
ab
##
sec x sec x 1 dxœ'
&


4
4
cd ab
##
1dx [x]œœœœ' ˆ‰



4
4
†
1
1
111Î%
Î% #44

68. c , d ; f(y) g(y) tan y tan y 2 tan yœ œ œ œ
11
44
###
ab
2 sec y 1 A 2 sec y 1 dyœÊœ ab ab
## '
&


4
4
2[tan y y] 2 1 1œœ
1
1
11Î%
Î%
‘ˆ‰ˆ ‰
44
41 4œœˆ‰
1
4
1

344. 340 Chapter 5 Integration
69. c 0, d ; f(y) g(y) 3 sin y cos y 0 3 sin y cos yœœ œ œ
1
#
ÈÈ
A 3 sin y cos y dy 3 (cos y)Êœ œ '
0
2

È ‘
2
3
$Î#
Î#
!
1
2(0 1) 2œ œ

70. a 1, b 1; f(x) g(x) sec xœ œ œ
# "Î$ˆ‰
1x
3
A sec x dx tan xÊœ œ '
&1
1
‘ ‘ˆ‰ ˆ‰
# "Î$ %Î$
"
"
11
1
x3x3
334
33œœŠ‹’ “
ÈÈ
Š‹
333 3
44
63
11 1
È

71. A A A&Î
"#
Limits of integration: x y and x y y yœœÊœ
$$
y y 0 y(y 1)(y 1) 0 c 1, d 0Ê œÊ œÊ œ œ
$
""
and c 0, d 1; f (y) g (y) y y and##""
$œœ œ
f (y) g (y) y y by symmetry about the origin,
##
$œÊ
A A 2A A 2 y y dy 2"# #
$
#
"
!œ Êœ œ '
0
1
ab ’“
yy
4
& %
2œœˆ‰
"" "
##4

72. A A A&Î
"#
Limits of integration: y x and y x x xœœÊœ
$&$&
x x 0 x (x 1)(x 1) 0 a 1, b 0Ê œÊ œÊ œ œ
&$ $
""
and a 0, b 1; f (x) g (x) x x and##""
$&œœ œ
f (x) g (x) x x by symmetry about the origin,
##
&$œÊ
A A 2A A 2 x x dx 2"# #
$&
"
!œ Êœ œ '
0
1
ab ’“
xx
46
% '
2œœˆ‰
"" "
46 6

73. A A A&Î
"#
Limits of integration: y x and y x , x 0œœÊœÁ
""
xx
&&
x 1 x 1 , f (x) g (x) x 0 xÊ œ Ê œ œœ
$
""
A x dx ; f (x) g (x) 0Êœ œ œ œ"##
"
!
""
#
'
0
1
’“
x
2x
&
&
x A x dx 1 ;œÊœ œ œœ
# #
#
" " "
#
" ## '
1
2
‘
x
AA A 1& Î &Î&"#
""
##

345. Section 5.6 Substitution and Area Between Curves 341
74. Limits of integration: sin x cos x x a 0œÊœÊœ
1
4
and b ; f(x) g(x) cos x sin xœœ
1
4
A (cos x sin x) dx [sin x cos x]Êœ œ '
0
4

1Î%
!
(0 1) 2 1œœŠ‹
È
ÈÈ
22
##

75. (a) The coordinates of the points of intersection of the
line and parabola are c x x c and y cœÊœ„ œ
# È
(b) f(y) g(y) y y 2 y the area of theœœ ÊÈÈȈ‰
lower section is, A [f(y) g(y)] dy
L
0
cœ'
2 y dy 2 y c . The area of theœœœ '
0
c
c
È ‘
24
33
$Î# $Î#
!
entire shaded region can be found by setting c 4: A 4 . Since we want c to divide the regionœœ œœˆ‰
44832
333
$Î# †
into subsections of equal area we have A 2A 2 c c 4œÊœ Êœ
L
32 4
33
ˆ‰
$Î# #Î$
(c) f(x) g(x) c x A [f(x) g(x)] dx c x dx cx 2 cœÊœ œ œ œ
# # $Î#
L
cc
cc
c
c
''
&&
&



ab ’“ ’ “
xc
33
)&
È
c . Again, the area of the whole shaded region can be found by setting c 4 A . From theœœÊœ
4 32
3 3
$Î#
condition A 2A , we get c c 4 as in part (b).œœÊœ
L
432
33
$Î# #Î$
76. (a) Limits of integration: y 3 x and y 1œ œ
#
3 x 1 x 4 a 2 and b 2;Ê œÊ œÊœ œ
##
f(x) g(x) 3 x ( 1) 4 xœœab
##
A 4 x dx 4xÊœ œ '
&1
2
ab ’“
#
#
#
x
3

8816œœœˆ‰ˆ ‰
8 8 16 32
3333
(b) Limits of integration: let x 0 in y 3 xœœ
#
y 3; f(y) g(y) 3 y 3 yÊœ œ ÈÈ ˆ‰
2(3 y)œ
"Î#

A 2 (3 y) dy 2 (3 y) ( 1) dy ( 2) 0 (3 1)Êœ œ œ œ ''
&&11
33
"Î# "Î# $Î#
$
"
’“ ˆ‰ ‘
2(3 y)
33
4
)&
(8)œœˆ‰
432
33
77. Limits of integration: y 1 x and y&Î &È
2
xÈ
1 x , x 0 x x 2 x (2 x)Ê œ ÁÊ œÊœÈÈ
2
xÈ
#
x 4 4x x x 5x 4 0Êœ Ê œ
##
(x 4)(x 1) 0 x 1, 4 (but x 4 does notÊ œÊœ œ
satisfy the equation); y and y œœÊœ
2x2x
xx 44ÈÈ
8 x x 64 x x 4.Êœ Ê œ ÊœÈ
$
Therefore, AREA A A : f (x) g (x) 1 xœ œ "#" "
"Î#
ˆ‰
x
4
A 1 x dx x xÊœ œ "
"Î# $Î#
"
!
'
0
1
ˆ‰ ’“
x2x
438
&

1 0 ; f (x) g (x) 2x A 2x dx 4xœœ œ Ê œ œ ˆ‰ ˆ ‰ ’“
237 x x x
38 24 4 4 8
"
## #
"Î# "Î# "Î#
%
" '
1
4
&
4 2 4 4 ; Therefore, AREA A Aœ œœ œ œœ œœˆ‰ˆ‰†
16 15 17 37 17 37 51 88 11
8888 24824243
" "
"#

346. 342 Chapter 5 Integration
78. Limits of integration: (y 1) 3 y y 2y 1œÊ
##
3 y y y 2 0 (y 2)(y 1) 0œ Ê œ Ê œ
#
y 2 since y 0; also, 2 y 3 yÊœ œ È
4y 9 6y y y 10y 9 0ÊœÊ œ
##
(y 9)(y 1) 0 y 1 since y 9 does notÊ œÊœ œ
satisfy the equation;
AREA A A&Î
"#
f (y) g (y) 2 y 0 2y""
"Î#œœ È

A 2 y dy 2 ; f (y) g (y) (3 y) (y 1)Ê œ œ œ œ"##
"Î# #
"
!
'
0
1
’“
2y
33
4
)&
A 3y(y1) dy 3y y (y1) 62 3 0 1 ;Ê œ œ œ œ œ#
##$ "" " " ""
###
#
"
'
1
2
cd ‘ˆ ‰ˆ ‰
3336
7
Therefore, A A"#Î &Î& &
47 15 5
3662
79. Area between parabola and y a : A 2 a x dx 2 a x x 2 a 0 ;œœ œ œœ
####$$ "
! '
0
a
a
ab ‘ Š‹
333
a4a

Area of triangle AOC: (2a) a a ; limit of ratio lim which is independent of a.
"
#
#$
abœœœ
aÄ!
Î
a


Š‹
4a
3
3
4
80. A 2f(x) dx f(x) dx 2 f(x) dx f(x) dx f(x) dx 4œœœœ'''''
aa aaa
bb bbb
81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the
region's upper and lower bounding curves at x 0. The area of the shaded region is actuallyœ
A [ x (x)] dx [x ( x)] dx 2x dx 2x dx .œ œ œ#
''''
&&10 10
01 01
82. It is sometimes true. It is true if f(x) g(x) for all x between a and b. Otherwise it is false. If the graph of f 
lies below the graph of g for a portion of the interval of integration, the integral over that portion will be
negative and the integral over [a b] will be less than the area between the curves (see Exercise 53).ß
83. Let u 2x du 2 dx du dx; x 1 u 2, x 3 u 6œÊœ Ê œ œÊœ œÊœ
"
#
dx du du F(u) F(6) F(2)'' '
12 2
36 6
sin 2x sin u sin u
xu
œœœœ
ˆ‰
u
&
ˆ‰ cd
"
#
'
#
84. Let u 1 x du dx du dx; x 0 u 1, x 1 u 0œ Ê œ Ê œ œ Ê œ œ Ê œ
f(1 x) dx f(u) ( du) f(u) du f(u) du f(x) dx
' ' '''
0 1 100
1 0 011
œ œ œ œ
85. (a) Let u x du dx; x 1 u 1, x 0 u 0œ Ê œ œ Ê œ œ Ê œ
f odd f( x) f(x). Then f(x) dx f( u) ( du) f(u) ( du) f(u) du f(u) duÊœ œ œ œ œ
'' ' ' '
&11 1 1 0
00 0 0 1
3 œ
(b) Let u x du dx; x 1 u 1, x 0 u 0œ Ê œ œ Ê œ œ Ê œ
f even f( x) f(x). Then f(x) dx f( u) ( du) f(u) du f(u) du 3Êœ œ œ œ œ
'' ''
&11 1 0
00 01
86. (a) Consider f(x) dx when f is odd. Let u x du dx du dx and x a u a and x'
&a
0
œ Ê œ Ê œ œ Ê œ œ!
u . Thus f(x) dx f( u) du f(u) du f(u) du f(x) dx.Êœ! œ œ œ œ
'' ' ' '
&aa a 0 0
00 0 a a
Thus f(x) dx f(x) dx f(x) dx f(x) dx f(x) dx .''' ''
&&aa0 00
a0a aa
œœœ!

347. Section 5.6 Substitution and Area Between Curves 343
(b) sin x dx [ cos x] cos cos .'
œ
/2
/2
œ œ œ!!œ!
1
1
11Î#
Î# ##
ˆ‰ ˆ ‰
87. Let u a x du dx; x 0 u a, x a u 0œ Ê œ œ Ê œ œ Ê œ
I ( du)œœ œœ'' ''
0a 00
a0 aa
f(x) dx f(a u) f(a u) du f(a x) dx
f(x) f(a x) f(a u) f(u) f(u) f(a u) f(x) f(a x)

I I dx dx [x] a 0 a.Êœœ œœœœ''' '
00 0 0
aa a a
a
f(x) dx f(a x) dx f(x) f(a x)
f(x) f(a x) f(x) f(a x) f(x) f(a x)

!
Therefore, 2I a I .œÊœ
a
#
88. Let u du dt du dt du dt; t x u y, t xy u 1. Therefore,œÊœ Ê œ Ê œ œÊœœÊœ
xy xy
ttxytut
t
œ
"""
dt du du du dt'' '''
xy y11
xy 1 1 y y
" " """
tu uut
œ œ œ œ
89. Let u x c du dx; x a c u a, x b c u bœ Ê œ œ Ê œ œ Ê œ
f(x c) dx f(u) du f(x) dx
'''
ac a a
bc b bœ
œ
Î& &
90. (a) (b) (c)
91-94. Example CAS commands:
:Maple
f := x -> x^3/3-x^2/2-2*x+1/3;
g := x -> x-1;
plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" );
q1 := [ -5, -2, 1, 4 ]; # (b)
q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )];
for i from 1 to nops(q2)-1 do # (c)
area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] );
end do;
add( area[i], i=1..nops(q2)-1 ); # (d)
: (assigned functions may vary)Mathematica
Clear[x, f, g]
f[x_] = x Cos[x]
2
g[x_] = x x
3

Plot[{f[x], g[x]}, {x, 2, 2}]
After examining the plots, the initial guesses for FindRoot can be determined.
pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, { 1, 0, 1}]
i1=NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}]
i2=NIntegrate[f[x] g[x], {x, pts[[2]], pts[[3]]}]
i1 i2Î

348. 344 Chapter 5 Integration
CHAPTER 5 PRACTICE EXERCISES
1. (a) Each time subinterval is of length t 0.4 sec. The distance traveled over each subinterval, using the?œ
midpoint rule, is h v v t, where v is the velocity at the left endpoint and v the velocity at???
"
#
ab
ii1 i i1
the right endpoint of the subinterval. We then add h to the height attained so far at the left endpoint v to?
i
arrive at the height associated with velocity v at the right endpoint. Using this methodology we build
i1
the following table based on the figure in the text:
t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0
v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65
h (ft) 0 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2
t (sec) 6.4 6.8 7.2 7.6 8.0
v (fps) 50 37 25 12 0
h (ft) 643.2 660.6 672 679.4 681.8
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph.
Remember that some shifting of the graph occurs in the printing process.
The total height attained is about 680 ft.
(b) The graph is based on the table in part (a).

2. (a) Each time subinterval is of length t 1 sec. The distance traveled over each subinterval, using the?œ
midpoint rule, is s v v t, where v is the velocity at the left, and v the velocity at the???
"
#
ab
ii1 i i1
right, endpoint of the subinterval. We then add s to the distance attained so far at the left endpoint v?
i
to arrive at the distance associated with velocity v at the right endpoint. Using this methodology we
i1
build the table given below based on the figure in the text, obtaining approximately 26 m for the total
distance traveled:
t (sec) 0 1 2 3 4 5 6 7 8 9 10
v (m/sec) 0 0.5 1.2 2 3.4 4.5 4.8 4.5 3.5 2 0
s (m) 0 0.25 1.1 2.7 5.4 9.35 14 18.65 22.65 25.4 26.4
(b) The graph shows the distance traveled by the
moving body as a function of time for
0 t 10.ŸŸ

3. (a) a ( 2) (b) (b 3a ) b 3 a 25 3( 2) 31!! ! !!
10 10 10 10 10
k1 k1 k1 k1 k1œœ œ œœ
a
44 4
k
? ? *?* * ? * ? **?
"""
#
kkkkk
(c) (a b 1) a b 2 25 (1)(10) 13!!!!
10 10 10 10
k1 k1 k1 k1œœœœ
kk k k*?*$?**?

349. Chapter 5 Practice Exercises 345
(d) b b (10) 25 0!!!ˆ‰
10 10 10
k1 k1 k1œœœ
555
###
*? * ? *?
kk
4. (a) 3a 3 a 3(0) 0 (b) (a b ) a b 0 7 7!! ! !!
20 20 20 20 20
k1 k1 k1 k1 k1œœ œ œœ
kk kkkkœ œ œ œ œœ
(c) b (20) (7) 8!!!ˆ‰
20 20 20
k1 k1 k1œœœ
"""
###
*? * ? * ?
2b
77 7
22
k
k
(d) a 2 a 2 0 2(20) 40!!!ab
20 20 20
k1 k1 k1œœœ
kk*? * ?* ?*
5. Let u 2x 1 du 2 dx du dx; x 1 u 1, x 5 u 9?*% ? % ? ?%? ?%?
"
#
(2x 1) dx u du u 3 1 2''
11
59
* ? ? ?*?
"Î# "Î# "Î# "
#
*
" ˆ‰ ‘
6. Let u x 1 du 2x dx du x dx; x 1 u 0, x 3 u 8?*% ? % ? ?%? ?%?
# "
#
x x 1 dx u du u (16 0) 6''
10
38
ab ˆ‰ ‘
# "Î$ %Î$
"Î$ "
#
)
!
*? ? ?*?
33
88
7. Let u 2 du dx; x u , x 0 u 0?% ? ?*%?* ?%?
x
2
1
1
#
cos dx (cos u)(2 du) [2 sin u] 2 sin 0 2 sin 2(0 ( 1)) 2''
ÎÎ*
00
2
ˆ‰ ˆ ‰
x
# #
!
Î#
???**?**?
1
1
8. Let u sin x du cos x dx; x 0 u 0, x u 1œÊœ œÊœœÊœ
1
#
(sin x)(cos x) dx u du''
00
21
*
œœœ ’“
u
2

"
!
"
#
9. (a) f(x) dx 3 f(x) dx (12) 4 (b) f(x) dx f(x) dx f(x) dx 6 4 2'' '''
ÎÎ ÎÎ22 222
22 552
? ? ? ? * ?*?
""
33
(c) g(x) dx g(x) dx 2 (d) ( g(x)) dx g(x) dx (2) 2'' ' '
52 2 2
25 5 5
Î
ÎÎÎ
?* ?* * ?* ?* ?* 11 11
(e) dx f(x) dx g(x) dx (6) (2)'''
ÎÎÎ222
555
Š‹
f(x) g(x)
55 5 555
8! "" ""
œ œœ
10. (a) g(x) dx 7 g(x) dx (7) 1 (b) g(x) dx g(x) dx g(x) dx 1 2 1'' '''
00 100
22 221
? ? ? ? * ?*?*
""
77
(c) f(x) dx f(x) dx (d) 2 f(x) dx 2 f(x) dx 2 ( ) 2'' ' '
20 0 0
02 2 2
?* ?* ? ? ? 111
ÈÈ ÈÈ
(e) [g(x) 3 f(x)] dx g(x) dx 3 f(x) dx 1 3
'''
000
222
*? * ?* 1
11. x 4x 3 0 (x 3)(x 1) 0 x 3 or x 1;
#
œÊ œÊœ œ
Area x 4x 3 dx x 4x 3 dx?***
''
01
13
abab
## 2x 3x 2x 3x?* **’“’“
xx
33
ÎÎ
##
"$
!"
2(1) 3(1) 0?* *’“Š‹
" #
Î
3
2(3) 3(3) 2(1) 3(1) ’“Š‹Š‹
31
33
ÎÎ
##
10 1?** ?ˆ‰ ‘ˆ‰
""
333
8

350. 346 Chapter 5 Integration
12. 1 0 4 x 0 x 2;*?%**%?)
x
4

#
Area 1 dx 1 dx?***''
Î22
23
Š‹ Š‹
xx
44

xx?* **’“’“
xx
12 12
ÎÎ
#$
# #
22 32?**** ****’“’“Š‹Š ‹Š‹Š‹
232
12 12 12 12
(2)
ÎÎÎ Î

?** **? ‘ˆ ‰ˆ‰
443413
33434

13. 5 5x 0 1 x 0 x 1;*?%*?%?)
#Î$ #Î$
Area 55x dx 55x dx?* **''
Î11
18
ˆ‰ˆ‰
#Î$ #Î$
5x 3x 5x 3x?* ** ‘ ‘
&Î$ &Î$
")
" "
5(1) 3(1) 5( 1) 3( 1)? * * *** ‘ˆ‰ˆ ‰
&Î$ &Î$
5(8) 3(8) 5(1) 3(1)** ** ‘ˆ‰ˆ‰
&Î$ &Î$
[2 ( 2)] [(40 96) 2] 62?*** * *?

14. 1 x 0 x 1;*?%?È
Area 1 x dx 1 x dx?* **''
01
14
ˆ‰ ˆ‰ÈÈ
xx xx?* ** ‘ ‘
22
33
$Î# $Î#
"%
!"
1 (1) 0 4 (4) 1 (1)?* *** ** ‘ ‘ˆ‰ˆ‰ˆ‰
222
333
$Î# $Î# $Î#
42?* * * ?
""
333
16
‘ˆ‰

15. f(x) x, g(x) , a 1, b 2 A [f(x) g(x)] dx????%? *
"
x

'
a
b
x dx 11œ œ œœ'
1
2
ˆ‰ ˆ‰ˆ‰ ’“
""""
####
#
"
xx
x4



16. f(x) x, g(x) , a 1, b 2 A [f(x) g(x)] dx????%? *
"
Èx '
a
b
xdx 2x?* ?*'
1
2
Š‹’ “ È
"
#
#
"
Èx
x

22 2?* **?Š‹
È ˆ‰
4
2
74 2"
##

È

17. f(x) 1 x , g(x) 0, a 0, b 1 A [f(x) g(x)] dx 1 x dx 1 2 x x dx?* ???%? * ?* ?*ˆ‰ ˆ‰ ˆ ‰ÈÈÈ
# # '''
a00
b11
12x x dx x x 1 (683)?*?*?*?*?'
0
1
ˆ‰ ’“
"Î# $Î#
##
"
!
"" "4x 4
3366

18. f(x) 1 x , g(x) 0, a 0, b 1 A [f(x) g(x)] dx 1 x dx 1 2x x dxœ œ œ œÊ œ œ œ ab ab a b
$ $$'
# # '''
a00
b11
x1?* ?*?’“
xx 9
7714
*$
##
"
!
""

351. Chapter 5 Practice Exercises 347
19. f(y) 2y , g(y) 0, c 0, d 3œœœœ
#
A [f(y) g(y)] dy 2y 0 dy%? * ? *''
c0
d3
ab
#
2 y dy y 18œœœ'
0
3
#$
$
!
2
3
cd

20. f(y) 4 y , g(y) 0, c 2, d 2?* ? ?* ?
#
A [f(y) g(y)] dy 4 y dy%? * ? *''
c2
d2 Î
ab
#
4y 2 8?* ? *?’“ ˆ‰
y
333
832
Î#
#

21. Let us find the intersection points:
yy2
44

œ
!
y y 2 0 (y 2)(y 1) 0 y 1ÊœÊ œÊœ
#
or y 2 c 1, d 2; f(y) , g(y)?%?* ? ? ?
y2 y
44
!

A [f(y) g(y)] dy dy%? * ? *''
c1
d2 Î
Š‹
y2 y
44
!

y 2 y dy 2y?*?*
"" #
#
#
"
443
yy'
Î1
2
ab ’“
Î
42œ œ
"""
##43 38
48 9
‘ˆ‰ˆ‰

22. Let us find the intersection points:
y4 y16
44


œ
y y 20 0 (y 5)(y 4) 0 y 4ÊœÊ œÊœ
#
or y 5 c 4, d 5; f(y) , g(y)?%?* ? ? ?
y16 y 4
44


A [f(y) g(y)] dy dy%? * ? *''
c4
d5 Î
Š‹
y16 y 4
44


y 20 y dy 20y?*?*
"" #
#
&
%
443
yy'
Î4
5
ab ’“
Î
100 80?***
""
##43 3
25 125 6 64
‘ˆ‰ˆ‰
180 63 117 (9 234)?*? ??
"""
##4488
9 9 243
ˆ‰ˆ‰

23. f(x) x, g(x) sin x, a 0, bœœ œœ
1
4
A [f(x) g(x)] dx (x sin x) dx%? * ? *''
a0
b4
*
cos x 1œ œ ’“Š‹
x
3
2

###
Î%
!
1
1
È

24. f(x) 1, g(x) sin x , a , b?? ?*? kk
11
# 2
A [f(x) g(x)] dx 1 sin x dx%? * ? *''
a2
b2 Î*
*


abkk
(1 sin x) dx (1 sin x) dxœ
''
Î*
*

20
02
2 (1 sin x) dx 2[x cos x]?*?'
0
2
*
1Î#
!
21 2?*?*ˆ‰
1
#
1

352. 348 Chapter 5 Integration
25. a 0, b , f(x) g(x) 2 sin x sin 2x?? *? *1
A (2 sin x sin 2x) dx 2 cos xÊœ œ '
0

‘
cos 2x
#!
1
2(1) 21 4œ œ ‘ˆ ‰††
""
##

26. a , b , f(x) g(x) 8 cos x sec x?* ? * ? *
11
33
#
A 8 cos x sec x dx [8 sin x tan x]%? * ? *'
Ÿ*
*

3
3
ab
# Î$
Î$
1
1
838363?***?Š‹Š ‹
ÈÈÈ
††
ÈÈ
33
##

27. f(y) y, g(y) 2 y, c 1, d 2??*??È
A [f(y) g(y)] dy y (2 y) dy%? * ? **''
c1
d2
‘È
y 2 y dy y 2y?*?*'
1
2
ˆ‰È ’“
2
3
y$Î#
#
#
"

242 2 2?***?*?Š‹
ÈÈ ˆ‰
4247
33366
827"
#

È

28. f(y) 6 y, g(y) y , c 1, d 2?* ? ? ?
#
A [f(y) g(y)] dy 6 y y dy%? * ? **''
c1
d2
ab
#
6y 12 2 6?** ?*****’“ ˆ‰ˆ‰
yy
333
8
Ÿ
##
#
"
""
4?*? ?
7 24143 13
366
"
#

29. f(x) x 3x x (x 3) f (x) 3x 6x 3x(x 2) fœ œ Ê œ œ Ê œ ± ±
!#
$## w # w
f(0) 0 is a maximum and f(2) 4 is a minimum. A x 3x dx x% ? ?* ?* * ?* * '
0
3
ab ’“
$# $
$
!
x
4
*
27?* * ?ˆ‰
81 27
44
30. A a x dx a 2 a x x dx ax a x a a a aœ œ œ œ ''
00
aa
a
0
ˆ‰ˆ ‰ ÈÈÈÈ ’“
"Î# "Î# "Î# $Î# #
#
##
4x4 a
33

†
a1 (683)?*?*?
# "
#ˆ‰
4a a
36 6

31. The area above the x-axis is A y y dy"
#Î$?*'
0
1
ˆ‰
; the area below the x-axis is?*?’“
3y y
510

#
"
!
"
Ayy dy#
#Î$
#
!
"?*?*?'
Ÿ1
0
ˆ‰ ’“
3y y
510
11

the total area is A A%? "#
6
5

353. Chapter 5 Practice Exercises 349
32. A (cos x sin x) dx (sin x cos x) dx?**''
04
454


**
*
(cos x sin x) dx [sin x cos x]*?'
54
32

*
*
1Î%
!
[ cos x sin x] [sin x cos x]
&Î% $Î#
Î% & Î%
11
11
(0 1)?****’“’ “Š‹ Š‹Š ‹
ÈÈ ÈÈ ÈÈ
22 22 22
## ## ##
(1 0) 2 4 2 2* ** * ? *? *’“ Š‹
È
ÈÈ È
22 82
## #

33. y x dt 2x 2 ; y(1) 1 dt 1 and y (1) 2 1 3œ Ê œ Ê œ œ œ œœ
# w""""''
1 1
x 1tdx xdx x t
dy d y


34. y 1 2 sec t dt 1 2 sec x 2 (sec x) (sec x tan x) sec x (tan x);? %? %? ?'
0
x
ˆ‰ ˆ‰È ÈÈ
dy d y
dx dx


"
#
"Î#
x 0 y 1 2 sec t dt 0 and x 0 1 2 sec 0 3œÊœ œ œÊ œ œ'
0
0
ˆ‰È È
dy
dx
35. y dt 3 ; x 5 y dt 3 3?*%??%?*?*''
5 5
x 5
sin t sin x sin t
tdxx t
dy
36. y 2 sin t dt 2 so that 2 sin x; x 1 y 2 sin t dt 2 2?* ?* ?*%? *?''
Ÿ Ÿ
Ÿ1 1
x 1
ÈÈ È ## # dy
dx
37. Let u cos x du sin x dx du sin x dx?%?* %*?
2(cos x) sin x dx 2u ( du) 2 u du 2 C 4u C''' "Î# "Î# "Î# "Î#
œ œ œ œ Š‹
u

±

4(cos x) Cœ
"Î#
38. Let u tan x du sec x dxœÊœ
#
(tan x) sec x dx u du C 2u C C'' $Î# # $Î# "Î#


???*?
u2
(tan x)

±


ˆ‰
39. Let u 2 1 du 2 d du d?% ? % ?)))
"
#
[2 1 2 cos (2 1)] d (u 2 cos u) du sin u C sin (2 1) C''
))) ) œ œ œ ˆ‰
"
#
""
!u
44
(2 1)

)
sin (2 1) C, where C C is still an arbitrary constantœ œ )) )
#
"
"
4
40. Let u 2 du 2 d du d?*% ? % ?)1 ) )
"
#
2 sec ( ) d 2 sec u du u 2 sec u du'
Š‹Š‹ ˆ‰ ˆ ‰
""""

# # "Î# #
##ÈÈ
2 u)1
#œ œ )1 ) ''
(2 tan u) C u tan u C (2 ) tan (2 ) Cœ œœ
""
##
"Î# "Î#
Š‹
u

±

)1 )1
41. t t dt t dt t 4t dt 4 C C'''ˆ‰ˆ‰ ˆ ‰ ab Š‹œœ œ œ
22 4 t t t4
tt t 3 1 3t
###


Ÿ!± Ÿ
42. dt dt dt t 2t dt 2 C C''' '
(t 1) 1
tt
t2t 2 t t
tt (1) tt
!" ""# $
#

**


?????**ˆ‰ ab Š‹
43. Let u t du t dt du t dtœ# Ê œ$ Ê œ
$Î# "
$ ÈÈ
t sin t dt sin u du cos u C cos t C''È ˆ‰ ˆ‰&? ?*?*&$Î# $Î#"""
$$$

354. 350 Chapter 5 Integration
44. Let u sec du sec tan d sec tan sec d u du u Cœ" Ê œ Ê " œ œ )))))))) '' È "Î# $Î##
$
sec Cœ"
#
$
$Î#
ab )
45. 3x 4x 7 dx x 2x 7x 1 2(1) 7(1) ( 1) 2( 1) 7( 1) 6 ( 10) 16
'
Î1
1
abc dc dc d
#$#$#$#
"
"
œ œ œ œ
46. 8s 12s 5 ds 2s 4s 5s 2(1) 4(1) 5(1) 0 3
'
0
1
abcdc d
$# %$ % $
"
!
œœ œ
47. dv 4v dv 4v 2''
11
22
444
v1
? ?* ?*?
# "
#
"

#
cd ˆ‰ˆ‰
48. x dx 3x 3(27) 3(1) 3 3(1) 2'
1
27
%Î$ "Î$ "Î$ "Î$
#(
"
"
œ œ œ œ ‘ˆ‰ˆ‰
3
49. t dt 2t 1'''
11 1
44 4
dt dt 2
tt t 41
(2)
ÈÈÈ
?? ?* ?*?
Î
$Î# "Î#
%
"
‘
50. Let x 1 u dx u du 2 dx ; u 1 x 2, u 4 x 3? % ? % ? ? % ? ? % ?È
"
#
"Î# du
uÈ
du x (2 dx) 2 x 3 2 4 3 2 3 3 2 2'
1
4
2
3
ˆ‰È
È
1u
u
244 84
333 33
!
"Î# $Î# $Î# $Î#
$
#

? ? ?*?*?*' ‘ˆ‰ˆ‰ˆ‰ ÈÈ ÈÈ
Š‹
51. Let u 2x 1 du 2 dx 18 du 36 dx; x 0 u 1, x 1 u 3?% ? % ? ?%? ?%?
18u du 8''
01
13
36 dx 8u 9 9 9
(2x 1) 2 u 3 1
$ "
$
"
$
"
Î
? ? ? ?*? ’“ ‘ˆ‰ˆ‰
52. Let u 7 5r du 5 dr du dr; r 0 u 7, r 1 u 2?* % ?* %* ? ? % ? ? % ?
"
5
(7 5r) dr u du 3u 7 2'' '
00 7
11 2
dr 3
(7 5r) 55 5
Î È
#Î$ #Î$ "Î$ ""
#
(
$$
?* ? *?* ? * ˆ‰ ‘ Š‹
È È
53. Let u 1 x du x dx du x dx; x u 1 ,?* % ?* %* ? ? % ?* ?
#Î$ "Î$ "Î$
#
""
#Î$
23 3
3884 ˆ‰
x1 u11 0?%?* ?
#Î$ x 1 x dx u du u (0)''
18 34
10**
"Î$ #Î$ $Î# &Î# &Î#
$Î# !
#
!
$Î%
$Î%
&Î#ˆ‰ ˆ‰ˆ‰ ‘ˆ‰ˆ‰’“ Š‹*?*?* ?*?***
33u 3 3 33
25554


5
œ
27 3
160
È
54. Let u 1 9x du 36x dx du x dx; x 0 u 1, x u 1 9? % ? % ? ? % ? ? % ? ?
%$ $ """
##
%
36 16
25 ˆ‰
x 1 9x dx u du u''
01
1 2 25 16
**
$ % $Î# "Î#
$Î#
"" "

#&Î"'
"
#&Î"'
"
ab ˆ‰ ‘’“Š‹? ? ?*
36 36 18
u

!

(1)?* * * ?
"""
"Î#
"Î#
18 16 18 90
25
ˆ‰ ˆ ‰
55. Let u 5r du 5 dr du dr; r 0 u 0, r u 5œÊ œ Ê œ œÊœ œÊœ
"
5
11
sin 5r dr sin u du 0''
00
5

## ""
&
! ## #
??*?***?ab ˆ‰ ‘ˆ ‰ˆ ‰
5524 0 20
u sin 2u sin 10 sin 0
1
11 1
56. Let u 4t du 4 dt du dt; t 0 u , t u?*% ? % ? ?%?* ? %?
1111
44 444
3"
cos 4t dt cos u du''
04
34


*$*
Î*
## "" " "
$Î%
Î%
ˆ‰ ˆ‰ ‘ab Š‹Š ‹œ œ œ
1 11
1
144424484484
usin 2u 3 sin sinˆ‰ ˆ ‰
3

œ œ
11
81616 8
""

355. Chapter 5 Practice Exercises 351
57. sec d [tan ] tan tan 0 3'
0
*Ê
# Î$
!
)) )??*?
1 1
3
È
58. csc x dx [ cot x] cot cot 2'


*
*4
34
# $Î%
Î%
?* ? * ** ?
1
1
11
ˆ‰ˆ‰
3
44
59. Let u du dx 6 du dx; x u , x 3 uœÊ œ Ê œ œÊœ œ Êœ
x
66 6
"
#
11
11
cot dx 6 cot u du 6 csc u 1 du [6( cot u u)] 6 cot 6 cot '' '


32 2
66
### Î#
Î' ##
x
6 66
??*?**?*****
**
**
ab ˆ‰ˆ‰
1
1
11 11
63 2?*
È
1
60. Let u du d 3 du d ; 0 u 0, uœÊ œ Ê œ œÊœ œÊœ
) 1
33 3
"
))) )1
tan d sec 1 d 3 sec u 1 du [3 tan u 3u]'' '
00 0
3

## # Î$
!
)) 1
33
))?*? *?*ˆ‰ ab
*
3 tan 3 (3 tan 0 0) 3 3?***?* ‘ˆ‰ È
11
33
1
61. sec x tan x dx [sec x] sec 0 sec 1 2 1'
Ÿ*3
0
? ? * * ?*?*
!
Î$1
1 ˆ‰
3
62. csc z cot z dz [ csc z] csc csc 2 2 0'


*
*4
34
œ œ œ œ
$Î%
Î%
1
1
11
ˆ‰ˆ‰ ÈÈ
3
44
63. Let u sin x du cos x dx; x 0 u 0, x u 1œÊœ œÊœœÊœ
1
#
5(sin x) cos x dx 5u du 5 u 2u 2(1) 2(0) 2'
0
2
0
1
*
$Î# $Î# &Î# &Î# &Î# &Î#
""
! !
? ? ? ?*?' ‘ ‘ˆ‰
2
5
64. Let u 1 x du 2x dx du 2x dx; x 1 u 0, x 1 u 0?* % ?* %* ? ?*% ? ? % ?
#
2x sin 1 x dx sin u du 0''
Ÿ10
10
ab*?* ?
#
65. Let u sin 3x du 3 cos 3x dx du cos 3x dx; x u sin 1, x u sin?%? %? ?*%?*??%?
"
####3
331111
ˆ‰ ˆ‰
1?*
15 sin 3x cos 3x dx 15u du 5u du u ( 1) (1) 2'''
Ÿ*
*ÎÎ

211
211
%%%&&& " "
"
????**?*ˆ‰ cd
3
66. Let u cos du sin dx 2 du sin dx; x 0 u cos 1, x u cos? % ?* %* ? ?%? ? ? %?ˆ‰ ˆ‰ ˆ‰ ˆ‰ Š‹
xx x 02
3### # # #
" 1
2
3
œ
"
#
cos sin dx u ( 2 du) 2 (1) (8 1)''
01
23 12
**
% % $
## #
"Î#
"
"
$ˆ‰ ˆ‰ ˆ‰ ’“Š‹
xx u 2 2 2 14
33 33 3
?*?* ?*?*?
!Ÿ
67. Let u 1 3 sin x du 6 sin x cos x dx du 3 sin x cos x dx; x 0 u 1, x? % ? % ? ? % ? ?
# "
##
1
u 1 3 sin 4%? ?
#
#
1
dx du u du u 4 1 1'''
011
244
*
3 sin x cos x u
13 sinx u 2ÈÈ
!
"" " "
##
"Î# "Î# "Î# "Î#
%
"
%
"


±

? ? ? ? ?*?ˆ‰ ‘’“Š‹
68. Let u 1 7 tan x du 7 sec x dx du sec x dx; x 0 u 1 7 tan 0 1,? % ? % ? ? % ? ?
## "
7
x u 1 7 tan 8?%? ?
11
44
dx du u du u (8) (1)'''
011
488
*
sec x u 3 3 3 3
(1 7 tan x) u 77 7 7777
±„Ÿ
±
! "
"" " " #Î$ "Î$ "Î$ "Î$
)
"
)
? ? ? ? ?*?ˆ‰ ‘’“Š‹
3

356. 352 Chapter 5 Integration
69. Let u sec du sec tan d ; 0 u sec 0 1, u sec 2œ Ê œ œÊœ œ œ Êœ œ))))) )
11
33
d d d du u du'' ' ' '
00 0 1 1
33 3 2 2
** *
tan sec tan sec tan
2 sec sec 2 sec 2 (sec ) 2 u 2
)))))
))) )
ÈÈÈÈÈ
)))œœœœ
Î
"" $Î#
21??*?***?*
"

##
" "
ÈÈÈȈ‰
2
u222
2u 2(2) 2(1)
’“’ “ Š ‹
È

!

70. Let u sin t du cos t t dt dt 2 du dt; t u sin ,œÊœ œ Êœ œÊœœÈÈ ˆ‰ˆ‰
" "
# #
"Î# cos t cos t
2t t 36 6
ÈÈ
ÈÈ
11

t usin 1œÊœ œ
11

4 #
dt (2 du) 2 u du 4 u 4 1 4 2 2 2'''




***
*36 12 12
41 1
cos t
t sin t
u
È
É È
È
????*?*
"" "Î#
"
"Î# #
‘È ÈÈ É Š‹
71. (a) av(f) (mx b) dx bx b(1) b( 1) (2b) b???**??
""" "
# # # #
"
"

1(1) 2 2
mx m(1) m( 1)'
Î1
1
’“’ “ Š‹Š ‹

(b) av(f) (mx b) dx bx b(k) b( k)???**
"""
# # #

k(k) k 2 k 2
mx m(k) m( k)'
Î
Îk
k
k
k
’“’ “ Š‹Š ‹

(2bk) bœœ
"
#k
72. (a) y 3x dx 3 x dx x (3) (0) 2 3 2
av
00
33?? ??*??
""

"Î# $Î# $Î# $Î#
$
!30 3 3 3 3 3 3 3
33 3222''ÈÈ È ‘ ‘ Š‹
ÈÈ È
(b) y ax dx a x dx x (a) (0) a a a
av
00
aa
a????*??
""

"Î# $Î# $Î# $Î#
!a0 a a 3 a 3 3 a 3 3
aa a222 22''ÈÈ È ‘ˆ ‰ˆ‰
ÈÈ È
73. f axf (x) dx [f(x)] [f(b) f(a)] so the average value of f over [a b] is the
ww w"""


av a
a
b
b
???*? w
b a ba ba b a
f(b) f(a)
'È
slope of the secant line joining the points (a f(a)) and (b f(b)), which is the average rate of change of f over [a b].ßß ß
74. Yes, because the average value of f on [a b] is f(x) dx. If the length of the interval is 2, then b a 2w*?
"
ba'
a
b
and the average value of the function is f(x) dx.
"
#'
a
b
75. We want to evaluate
f(x) dx sin x dx sin x dx dx
""# $(# #&
$'& ! $'& $'& $'& $'& $'&'' ' '
)) ) )
ʱ! ʱ! ʱ! ʱ!
œ $( "!" #& œ "!" Œ Δ• ”•ab ab
11
Notice that the period of y sin x is and that we are integrating this function over an iterval of? * $$ ? '†Þ”•ab
##
$'&
11

Î&%

length 365. Thus the value of sin x dx dx is .
$( # #& $( #&
$'& $'& $'& $'& $'&''
))
ʱ! ʱ!
”•ab
1
"!" † ! † $'& œ #&
76. T T dT T
" " #' "Þ)(
'(&#! '&& #†"! $†"!#
& #
#!
'(&'
)
±#!
ab ab ”•)Þ#( "! #' "Þ)( œ )Þ#(
TT
Î
%%
œ )Þ#( '(& )Þ#( #! ¸ $(#%Þ%% "'&Þ%!
" "
'&& #†"! $†"! #†"! $†"! '&&
#' '(& "Þ)( '(& #' #! "Þ)( #!
Œ Δ•”•ab ab a b
ab ab ab ab

%% %%
the average value of C on [20, 675]. To find the temperature T at which C , solveœ&Þ%$œ œ&Þ%$ vv
T T for T. We obtain T T&Þ%$ œ )Þ#( "! #' "Þ)( "Þ)( #' #)%!!! œ !
& # #
ab
T . So T or T . Only T lies in the% ? ? ? *'‰&á‰& ? '†á(& ? '†á(&
#' „ #' % "Þ)( #)%!!!
# "Þ)( $Þ(%
#' „ #"#%**'
Éab a ba b
ab
È

interval [20, 675], so T C.œ $*'Þ(#
‰
77. cos x
dy
dx
?&
È
$

357. Chapter 5 Practice Exercises 353
78. cos x x x cos x
dy
dx dx
d
?&(„(?$&(ÈÈ ab ab ab
$# $# #
79. dt
dy
dx dx
d
x
tx
?* ?*Œ Ÿ'
1
''
"! "!
**
80. dt dt sec x
dy
dx dx t dx t sec x dx sec x
d d d sec x tan x
sec x
sec x
??*?*?*Π٠Π٠ab''
#
"""
!" !" !" "! #

81. Yes. The function f, being differentiable on [a b], is then continuous on [a b]. The Fundamental Theorem ofßß
Calculus says that every continuous function on [a b] is the derivative of a function on [a b].ßß
82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on
[a b], then f(x) dx F(b) F(a). In particular, if F(x) is an antiderivaitve of 1 x on [0 1], thenßœ ß '
a
b
È %
1 x dx F(1) F(0).'
0
1
È
?* %
83. y 1 t dt 1 t dt 1 t dt 1 t dt 1 xœ œ Ê œ œ œ '' ' '
11 1 1
xx x x
dy
dx dx dx
dd
ÈÈ È ÈÈ
”•”•## # ##
84. y dt dt dt dt??* %?* ?*'' ' '
cos x 0 0 0
0 cos x cos x cos x
"" " "
1t 1t dx dx 1t dx 1t
dy dd
”•”•
(cos x) ( sin x) csc x?* ?* * ? ?ˆ‰ˆ ‰ˆ‰
"""
1 cos x dx sin x sin x
d

85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each
interval by averaging the widths at top and bottom. This gives the estimate
A¸"&† ˆ‰
!$' $'&% &%&" &"%*Þ& %*Þ&&% &%'%Þ% '%Þ%'(Þ& '(Þ&%#
########
A ft . The cost is Area ($2.10/ft ) 5961 ft $2.10/ft $12,518.10 the job cannot be done for $11,000.¸ &*'" ¸ œ Ê
####
† aba b
86. (a) Before the chute opens for A, a 32 ft/sec . Since the helicopter is hovering, v 0 ft/sec?* ?
#
!
v 32 dt 32t v 32t. Then s 6400 ft s 32t dt 16t s 16t 6400.Ê œ œ œ œ Ê œ œ œ ''
!! !
##
At t 4 sec, s 16(4) 6400 6144 ft when A's chute opens;??*?
#
(b) For B, s 7000 ft, v 0, a 32 ft/sec v 32 dt 32t v 32t s 32t dt!! !
#œ œ œ Ê œ œ œ Ê œ ''
16t s 16t 7000. At t 13 sec, s 16(13) 7000 4296 ft when B's chute opens;œ œ œ œ œ
## #
!
(c) After the chutes open, v 16 ft/sec s 16 dt 16t s . For A, s 6144 ft and for B,œ Ê œ œ œ '
!!
s 4296 ft. Therefore, for A, s 16t 6144 and for B, s 16t 4296. When they hit the ground,!??*?*
s 0 for A, 0 16t 6144 t 384 seconds, and for B, 0 16t 4296 tœ Ê œ Ê œ œ œ Ê œ
6144 4296
16 16
268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opensœ
B hits the ground first.Ê
87. av(I) (1200 40t) dt 1200t 20t (1200(30) 20(30) 1200(0) 20(0)?*?*? ***
""" ###
$!
!30 30 30'
0
30
cdc d abab
(18,000) 600; Average Daily Holding Cost (600)($0.03) $18œœ œ œ
"
30
88. av(I) (600 600t) dt 600t 300t 600(14) 300(14) 0 4800; Average Daily???*?
""" ##
"%
!14 14 14'
0
14
cdc d
Holding Cost (4800)($0.04) $192œœ

358. 354 Chapter 5 Integration
89. av(I) 450 dt 450t 450(30) 0 300; Average Daily Holding Cost?*?*?**?
"""
#
$!
!
30 30 6 30 6
tt 30'
0
30
Š‹’ “’ “
ÎÎ
(300)($0.02) $6œœ
90. av(I) 600 20 15t dt 600 20 15 t dt 600t 20 15 t?*?* ?*
"" " "Î# $Î#
'!
!
60 60 60 3
2''
00
60 60
Š‹Š ‹’ “
ÈÈÈ ˆ‰
600(60) (60) 0 36,000 15 200; Average Daily Holding Cost?* *?*?
"" $Î# #
60 3 60 3
40 15 320
’“ ˆ‰ ˆ‰
È
(200)($0.005) $1.00œœ
CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES
1. (a) Yes, because f(x) dx 7f(x) dx (7) 1''
00
11
œœœ
""
77
(b) No. For example, 8x dx 4x 4, but 8x dx 2 2 1 0''
00
11
?? ? ? *cd
È
’“
È
Š‹ ˆ‰
# $Î# $Î#
"
!
"
!
x 42
3
Î

3
È
4 œÁ
42
3
È È
2. (a) True: f(x) dx f(x) dx 3
''
52
25
?* ?*
(b) True: [f(x) g(x)] dx f(x) dx g(x) dx f(x) dx f(x) dx g(x) dx
''''''
ÎÎÎÎ Î222222
555255
œ œ
4329??
(c) False: f(x) dx 4 3 7 2 g(x) dx [f(x) g(x)] dx 0 [g(x) f(x)] dx 0.
''''
ÎÎÎÎ2222
5555
??? % * % *
On the other hand, f(x) g(x) [g(x) f(x)] 0 [g(x) f(x)] dx 0 which is a contradiction.ŸÊ  Ê  
'
Î2
5
3. y f(t) sin a(x t) dt f(t) sin ax cos at dt f(t) cos ax sin at dt?*? *
"" "
aa a'' '
00 0
xx x
f(t) cos at dt f(t) sin at dt cos ax f(t) cos at dt?* %?
sin ax cos ax
aadx
dy'' '
00 0
xx x
Œ Î
f(t) cos at dt sin ax f(t) sin at dt f(t) sin at dt
sin ax d cos ax d
adx adx
Œ Î Œ Î
'' '
00 0
xx x
cos ax f(t) cos at dt (f(x) cos ax) sin ax f(t) sin at dt (f(x) sin ax)?* ''
0 0
x x
sin ax cos ax
aa
cos ax f(t) cos at dt sin ax f(t) sin at dt. Next,ʜ
dy
dx ''
00
xx
a sin ax f(t) cos at dt (cos ax) f(t) cos at dt a cos ax f(t) sin at dt
dy
dx dx
d


œ '''
000
xxx
Œ Î
(sin ax) f(t) sin at dt a sin ax f(t) cos at dt (cos ax)f(x) cos ax?* Œ Î
d
dx''
00
xx
a cos ax f(t) sin at dt (sin ax)f(x) sin ax a sin ax f(t) cos at dt a cos ax f(t) sin at dt f(x).œ '''
0 00
x xx
Therefore, y a y a cos ax f(t) sin at dt a sin ax f(t) cos at dt f(x)
ww #
œ ''
00
xx
a f(t) cos at dt f(t) sin at dt f(x). Note also that y (0) y(0) 0.*???
# w
Œ Î
sin ax cos ax
aa''
00
xx
4. x dt (x) dt dt from the chain ruleœÊœ œ'''
000
yyy
"""
!!!
ÈÈÈ
14t 14t 14t
dd d
dx dx dy dx
dy
”• Š‹
1 1 4y . Then 1 4y 1 4yÊœ Ê œ œ œ
"
!
###
È14y
dy dy d y dy
dx dx dx dx dy dx
dd


Š‹ Š‹ ÈÈÈ ˆ‰ˆ‰
1 4y (8y) 4y. Thus 4y, and the constant of? ? ? ? ?
"
#
#
"Î#
!!
!
ab Š‹
dy d y
dx dx
4y
14y 14y
4y 1 4y
Š‹
ÈÈ
ˆ‰È
dy
dx

359. Chapter 5 Additional and Advanced Exercises 355
proportionality is 4.
5. (a) f(t) dt x cos x f(t) dt cos x x sin x f x (2x) cos x x sin x''
00
xx

?% ?* % ?*1111111
d
dx
ab
#
f x . Thus, x 2 f(4)Êœ œÊœ œab
# "cos x x sin x cos 2 2 sin 2
2x 4 4
11 1 11 1
(b) t dt (f(x)) (f(x)) x cos x (f(x)) 3x cos x f(x) 3x cos x'
0
f(x)
f(x)
#$$$
!
"" $
œœ Ê œ Ê œ Êœ’“
È
t
33 3
Î
11 1
f(4) 3(4) cos 4 12Êœ œ
$$
È È
1
6. f(x) dx sin a cos a. Let F(a) f(t) dt f(a) F (a). Now F(a) sin a cos a''
0 0
a a
œ œ Ê œ œ
aa aa

## # ## #
w1 1
f(a) F (a) a sin a cos a sin a f sin cos sin Êœ œ Ê œ œœ
w """"
# # # # ## # # ## # ### #
a 111111111 ˆ‰
ˆ‰


7. f(x) dx b 1 2 f(b) f(x) dx b 1 (2b) f(x)''
11
bb
œÊœ œ œ Êœ
ÈÈ
ab
# "
#
#
"Î#
!!
dbx
db b1 x1
ÈÈ

8. The derivative of the left side of the equation is: f(t) dt du f(t) dt; the derivative of the right
d
dx
”•”•
'' '
00 0
xu x
œ
side of the equation is: f(u)(x u) du f(u) x du u f(u) du
ddd
dx dx dx
”•
'''
000
xxx
*? *
x f(u) du u f(u) du f(u) du x f(u) du xf(x) f(u) du xf(x) xf(x)?*? *?*
dd d
dx dx dx
”• ” •
''' ' '
000 0 0
xxx x x
f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0œ'
0
x
when x 0, the constant must be 0. Therefore, f(t) dt du f(u)(x u) du.??* '' '
00 0
xu x
”•
9. 3x 2 y 3x 2 dx x 2x C. Then (1 1) on the curve 1 2(1) C 1 C 4
dy
dx
œÊœ œ ß Ê œÊœ
##$ $ 'ab
y x 2x 4%? *
$
10. The acceleration due to gravity downward is 32 ft/sec v 32 dt 32t v , where v is the initial*%?*?*
#
!! '
velocity v 32t 32 s ( 32t 32) dt 16t 32t C. If the release point, at t , is s 0, thenÊœ Êœ œ œ! œ ' #
C 0 s 16t 32t. Then s 17 17 16t 32t 16t 32t 17 0. The discriminant of thisœÊ œ œ Ê œ Ê œ
###
quadratic equation is 64 which says there is no real time when s 17 ft. You had better duck.*?
11. f(x) dx x dx 4 dx
'''
ÎÎ880
303
?*
#Î$
x [ 4x]?* ‘
3
5
&Î$ $
!
) !
0 ( 8) ( 4(3) 0) 12œ œˆ‰
396
55
&Î$
œ
36
5

12. f(x) dx x dx x 4 dx'' '
ÎÎ44 0
30 3
?* *È ab
#
(x) 4xœ ‘ ’“
2x
33
$Î#
!
%
$
!
Î
0 (4) 4(3) 0œ ‘ˆ‰ ’“Š‹
23
33
$Î#
Î
3?*?
16 7
33

360. 356 Chapter 5 Integration
13. g(t) dt t dt sin t dt'''
001
212
? 1
cos t?*’“ ‘
t
2

"
!
"
#
"1
1 0 cos 2 cos ?** **ˆ‰ ‘ˆ‰
"" "
# 11
11
?*
"
#
2
1

14. h(z) dz 1 z dz (7z 6) dz'' '
00 1
21 2
?**
È
"Î$
(1 z) (7z 6)œ ‘ ‘
23
314
$Î# #Î$
"#
! "
(1 1) (1 0)?* * ** * ‘ˆ‰
22
33
$Î# $Î#
(7(2) 6) (7(1) 6) ‘
33
14 14
#Î$ #Î$
? * ?
263 55
3 7 14 42
ˆ‰

15. f(x) dx dx 1 x dx 2 dx
''' '
ÎÎÎ
Î221 1
211 2
?* ab
#
[x] x [2x]œ
" #
#
"
"
"
’“
x
3
Î
1 ( 2) 1 1 2(2) 2(1)œ ab ”•Š‹Š ‹ ’“
1
33
(1)
Î Î

1 42œ œ
22 13
33 3
ˆ‰

16. h(r) dr r dr 1 r dr dr
''' '
ÎÎ110 1
201 2
?* ab
#
r [r]?*’“ ’ “
rr
23
Î
!"
" !
#
"
01021œ Š‹Š ‹ Š‹ ab
(1) 1
3

#
Î
1œ œ
"
#
27
36
17. Ave. value f(x) dx f(x) dx x dx (x 1) dx x?? ?*?*
""" ""
# # ##
"#
!"
ba 0 2 2
xx''''
a001
b212
”• ’“ ’ “

021? **** ?
""
#####
’“Š‹Š‹Š‹
121

18. Ave. value f(x) dx f(x) dx dx 0 dx dx [ 10032]œ œ œ œ œ
""" "
ba 30 3 3 3
2'''''
a0012
b3123
”•
19. f(x) dt f (x) x? % ? * ?* * ??'
1/x
x
""""""""w
t x dx dx x x x x x x
dx d 2
ˆ‰ ˆ ‰ ˆ ‰Š‹ ˆ‰
!
x
20. f(x) dt f (x) (sin x) (cos x)?%? * ?'
cos x
sin x
"" " "

w
t 1 t 1 sin x dx 1 cos x dx cos x sin x
ddcos xsin x

ˆ‰ˆ ‰ˆ‰ˆ ‰
?
""
cos x sin x
21. g(y) sin t dt g (y) sin 2 y 2 y sin y y?%? * ?*'

y
2y
#w
##
Š‹Š‹Š‹Š‹ˆ‰ ˆ‰ ˆ‰ ˆ‰ÈÈ ÈÈ
dd
dy dy y 2 y
sin 4y sin y
ÈÈ
22. f(x) t(5 t) dt f (x) (x 3)( (x 3)) (x 3) x(5 x) (x 3)(2 x) x(5 x)œÊœ& œ'
x
x3

w ˆ‰ ˆ‰
ddx
dx dx
6 x x 5x x 6 6x. Thus f (x) 0 6 6x 0 x 1. Also, f (x) 6 x 1 gives aœ œ œ Ê œ Ê œ œ!Ê œ
## w ww

361. Chapter 5 Additional and Advanced Exercises 357
maximum.
23. Let f(x) x on [0 1]. Partition [0 1] into n subintervals with x . Then , , , are theœß ß œœ á
& " "
?
10 2 n
nn nn n
right-hand endpoints of the subintervals. Since f is increasing on [0 1], U is the upper sum forßœ !Š‹ˆ‰


j1
j
nn
&
"
f(x) x on [0 1] lim lim lim œßÊ œ áœ
&
&
" " " ! !Þ!
& & &
nn nÄ_ Ä_ Ä_
!Š‹ ’ “ ’ “ˆ‰ ˆ‰ ˆ‰ ˆ‰


j1
j
nn nn n n n
2n 12n
%% %

x dxœœœ'
0
1
&
"
!
"
’“
x
66

24. Let f(x) x on [0 1]. Partition [0 1] into n subintervals with x . Then , , , are theœß ß œœ á
$ " "
?
10 2 n
nn nn n
right-hand endpoints of the subintervals. Since f is increasing on [0 1], U is the upper sum forßœ !Š‹ˆ‰


j1
j
nn
$
"
f(x) x on [0 1] lim lim lim œßÊ œ áœ
$
$
" " " ! !Þ!
$ $ $
nn nÄ_ Ä_ Ä_
!Š‹ ’ “ ’ “ˆ‰ ˆ‰ ˆ‰ ˆ‰


j1
j
nn nn n n
2n 12n
n
ÎÎ Î
* x dxœœœ'
0
1
$
"
!
"
’“
x
44
*
25. Let y f(x) on [0 1]. Partition [0 1] into n subintervals with x . Then , , , are theœß ß œœ á ?
10 2 n
nn nn n
" "
right-hand endpoints of the subintervals. Since f is continuous on [ 1], f is a Riemann sum of!ß!Š‹ˆ‰


j1
j
nn
"
y f(x) on [0 1] lim f lim f f f f(x) dxœßÊ œ áœ
nnÄ_ Ä_
!Š‹ˆ‰ ‘ˆ‰ ˆ‰ ˆ‰


j1
0
1
j
nn n n n n
2n""" '
26. (a) lim [2 4 6 2n] lim 2x dx x 1, where f(x) 2x
nnÄ_ Ä_
"" #
"
!n nnnn n
246 2n
á œ á œ œ œ œ ‘ cd '
0
1
on [0 1] (see Exercise 25)ß
(b) lim 1 2 n lim x dx , where
nnÄ_ Ä_
"" ""& "& "& "&
"& "& "&
"
!
nnnnn1616
12 n x
!&
!&cd ’“’“ˆ‰ ˆ‰ ˆ‰á œ á œ œ œ '
0
1
f(x) x on [0 1] (see Exercise 25)ϧ
"&
(c) lim sin sin sin sin n dx cos x cos cos 0
nÄ_
""""
"
!nn n n
2n
‘ ‘ˆ‰
11 1
111
? ?* ?* ** '
0
1
111
, where f(x) sin x on [0 1] (see Exercise 25)œœß
2
1
1
(d) lim 1 2 n lim lim 1 2 n lim x dx
nnn nÄ_ Ä_ Ä_ Ä_
""" ""& "& "& "& "& "& "&
n nn n
!$ !&cd cd Š‹Š ‹Š‹á œ á œ '
0
1
0 0 (see part (b) above)œœˆ‰
"
16
(e) lim 1 2 n lim 1 2 n
nnÄ_ Ä_
""& "& "& "& "& "&
nn
n
!% !&cdcdá œ á
lim n lim 1 2 n lim n x dx (see part (b) above)???Š‹Š ‹Š‹ cd
nn nÄ_ Ä_ Ä_
""& "& "& "&
n
!&
'
0
1
27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and
the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal
to r, the radius of the circle) and a vertex angle of where . The area of each triangle is))
nn œ
2
n
1
A r sin the area of the polygon is A nA sin sin .
nn nnœÊ œœœ
"
###
#
))
nr nr 2
n

1
(b) lim A lim sin lim sin lim r r lim r
nn n n
2n 0
Ä_ Ä_ Ä_ Ä_
ÎÄ
œœ œ œ œ
nr 2 n r 2
n2n
sin sin

#
## #111
1
ab ab11 1
ˆ‰ ˆ‰
ˆ‰ ˆ‰
22
nn
22
nn

1
28. y sin x cos 2t dt sin x cos 2t dt y cos x cos 2x ; when x we haveœ "œ "Ê œ œ''
x
x


w
ab 1
y cos cos 2 . And y sin x 2sin 2x ; when x , y sin cos 2t dt
www
œ œ""œ# œ œ œ "11 1 1ab ab '
x
1
.?$?$

362. 358 Chapter 5 Integration
29. (a) g f t dtab ab"œ œ! '
1
1
(b) g f t dtab ab abab' ? ?* & $ ?*$'
1
Ê
"
#
(c) g f t dt f t dta b ab ab a b*$ ? ? * ? * & ? *''
11
11


"
%
#
11
(d) g x f x x , , and the sign chart for g x f x is . So g has a
31 3
ww
ab ab ab abœ œ ! Ê œ $ " $ œ ± ± ±
*
relative maximum at x .œ"
(e) g f is the slope and g f t dt , by (c). Thus the equation is y x
w

ab ab ab ab a b"œ "œ# "œ œ œ# " '
1
1
11
yx .œ# # 1
(f) g x f x at x and g x f x is negative on and positive on so there is an
ww w ww w
ab ab ab ab a b a bœ œ ! œ " œ $ß " "ß "
inflection point for g at x . We notice that g x f x for x on and g x f x for x on?*$ ? *$w& ?
ww w ww w
ab ab a b ab ab
, even though g does not exist, g has a tangent line at x , so there is an inflection point at x .ab ab#ß% # œ# œ#
ww
(g) g is continuous on and so it attains its absolute maximum and minimum values on this interval. We saw in (d)Ò$ß %Ó
that g x x , , . We have that
w
ab?% ?*'$'
g f t dt f t dta b ab ab*' ? ? * ? * ? *&''
1
$ "
$
#
#
1

1
gft dtab ab"œ œ!
'
1
1
gft dtab ab'? ?*$ '
1
$
gft dtab ab ? ?*$ „$„$?*'
1
%
""
##
Thus, the absolute minimum is and the absolute maximum is . Thus, the range is .# ! Ò# ß !Ó11

363. Chapter 5 Additional and Advanced Exercises 359
NOTES:

364. 360 Chapter 5 Integration
NOTES:

365. CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS
1. (a) A (radius) and radius 1 x A(x) 1 xœœÊœ11
## # È
ab
(b) A width height, width height 2 1 x A(x) 4 1 xœœœÊœ†
È
ab
# #
(c) A (side) and diagonal 2(side) A ; diagonal 2 1 x A(x) 2 1 xœœÊœœÊœ
# #
#
# È È
ab
(diagonal)
œ
(d) A (side) and side 2 1 x A(x) 3 1 xœœÊœ
È
3
4
## #
È È
ab
2. (a) A (radius) and radius x A(x) xœœÊœ11
# È
(b) A width height, width height 2 x A(x) 4xœœœÊœ† È
(c) A (side) and diagonal 2(side) A ; diagonal 2 x A(x) 2xœœÊœœÊœ
#
# È È
(diagonal)
œ
(d) A (side) and side 2 x A(x) 3xœœÊœ
È
3
4
#
È È
3. A(x) 2x (see Exercise 1c); a 0, b 4;œœ œ œœ
(diagonal) xx
œ
œ
##
œœˆ‰ÈȈ‰
V A(x) dx 2x dx x 16œœœœ''
a0
b4
cd
#
%
!
4. A(x) 1 2x x ; a 1, b 1;œœ œ œœœ
1 11(diameter)
44 4
2x x 21x
œ œœ œ
œœ
cdcdab abœœ œ #%
1ab
V A(x) dx 1 2x x dx x x 2 1œœœœœ''
a1
b1 œ
11 1ab ’“ ˆ‰
#% $
"
π
"2x 2 16
35 3515
%
1
5. A(x) (edge) 1 x 1 x 2 1 x 4 1 x ; a 1, b 1;œœœœœœ
# ### #
##
’“Š‹
ÈÈ È
Š‹ ab
V A(x) dx 4 1 x dx 4 x 8 1œœœœœ''
a1
b1 œ
ab ’“ ˆ‰
#
"
π
"x16
333
Ê
6. A(x) 2 1 x (see Exercise 1c); a 1, b 1;œœ œ œ œœ
(diagonal)
1x 1x 21x
œ
œœ œ
œœ
## #
œœœ œ œ
#
’“Š‹
ÈÈ È
Š‹
ab
V A(x) dx 2 1 x dx 2 x 4 1œœœœœ''
a1
b1 œ
ab ’“ ˆ‰
#
"
π
"x8
333
Ê
7. (a) STEP 1) A(x) (side) (side) sin 2 sin x 2 sin x sin 3 sin xœœ œ
""
##
†† † †ˆ‰ ˆ‰ Š‹Š‹
ÈÈ È
11
33
STEP 2) a 0, bœœ 1
STEP 3) V A(x) dx 3 sin x dx 3 cos x 3(1 1) 2 3œœ œ œœ ''
a0
b
ÈÈÈÈ
’“
œ 1
!
(b) STEP 1) A(x) (side) 2 sin x 2 sin x 4 sin xœœ œ
#
Š‹Š‹
ÈÈ
STEP 2) a 0, bœœ 1
STEP 3) V A(x) dx 4 sin x dx 4 cos x 8œœ œœ
''
a0
b
œ
cd
1
!
8. (a) STEP 1) A(x) (sec x tan x) sec x tan x 2 sec x tan xœœœ
1 11(diameter)
44 4
œ
###
ab
sec x sec x 1 2 œ
1
4cosx
sin x
œ ‘ab
##
œ
STEP 2) a , bœ œ
11
33
STEP 3) V A(x) dx 2 sec x 1 dx 2 tan x x 2œœ œ ''
a3
b3 œ%
%
œ
œ
11
1
14cosx4 cos x
2 sin x
ˆ‰ œ ‘ˆ‰
# "
Î$
Î$
œ

366. 362 Chapter 6 Applications of Definite Integrals
23 2 23 2 43œœ
11 1 1 1
43 3 4 3
2
’“Š‹
ÈÈ È
Š‹Š ‹ Š‹
""
ˆ‰ ˆ‰
$$
œœ
(b) STEP 1) A(x) (edge) (sec x tan x) 2 sec x 1 2 œœœ
### ˆ‰
sin x
cos x
œ
STEP 2) a , bœ œ
11
33
STEP 3) V A(x) dx 2 sec x 1 dx 2 2 3 4 3œœ œœ ''
a3
b3 œ%
%
œ
œ
ˆ‰ Š‹
ÈÈ
# 2 sin x 2
cos x 3 3
œ
11
9. A(y) (diameter) 5y 0 y ;œœœ
11 1
44 4
5## %
#
Š‹
È
c 0, d 2; V A(y) dy y dyœœ œ œ ''
c0
d2
5
4
1%
208œœœ’“ˆ‰Š‹ ab
5
45 4
y11
%#
!
&
1

10. A(y) (leg)(leg) 1 y 1 y 2 1 y 2 1 y ; c 1, d 1;œœœœœœ
"" "
## #
## #
##
#
œ ‘ˆ ‰ÈÈ È ˆ‰ ab
V A(y) dy 2 1 y dy 2 y 4 1œœœœœ''
c1
d1 œ
ab ’“ ˆ‰
#
"
π
"y
333
8
Ê
11. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right
prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) (side length) s ;œœ
##
STEP 2) a 0, b h; STEP 3) V A(x) dx s dx s hœœ œ œ œ ''
a0
bh
##
(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the
prism described above, regardless of the number of turns V s hʜ
#
12. 1) The solid and the cone have the same altitude of 12.
2) The cross sections of the solid are disks of diameter
x . If we place the vertex of the cone at theœˆ‰
xx
##
origin of the coordinate system and make its axis of
symmetry coincide with the x-axis then the cone's cross
sections will be circular disks of diameter
(see accompanying figure).
xxx
44
œˆ‰
#
3) The solid and the cone have equal altitudes and identical
parallel cross sections. From Cavalieri's Principle we
conclude that the solid and the cone have the same
volume.

13. R(x) y 1 V [R(x)] dx 1 dx 1 x dx xœœ Ê œ œ œ œ
xxxxx
412## #
#
#
#
! '''
00 0
22 2
11 1 1 ˆ‰ Š‹’ “
œœ Ê
2œœ1ˆ‰
48 2
212 3
1
14. R(y) x V [R(y)] dy dy y dy y 8 6œœ Ê œ œ œ œ œ œ
3y 3y 933
444##
##$
# #
! '''
000
222
11 1111 ˆ‰ œ‘ ††
15. R(x) tan y ; u y du dy 4 du dy; y 0 u 0, y 1 u ;œœÊœÊœœÊœœÊœˆ‰
11 1 1
44 4 4
1
V [R(y)] dy tan y dy 4 tan u du 4 1 sec u du 4[ u tan u]œœ œœœ'' ''
00 00
11 44
11
###
# Î%
! œ ‘ˆ‰ ab
1 1
4
œœ%%

367. Section 6.1 Volumes by Slicing and Rotation About an Axis 363
4104œœˆ‰
1
4
1
16. R(x) sin x cos x; R(x) 0 a 0 and b are the limits of integration; V [R(x)] dxœœÊœœ œ
1
#
# '
0
2
œ%
1
(sin x cos x) dx dx; u 2x du 2 dx ; x 0 u 0,œœœÊœÊœœÊœ11''
00
22
œœ%%
# (sin 2x)
484
du dx
œ
œ
x u V sin u du sin 2u 0 0‘ œ ‘ œ ‘ˆ‰œÊœ Äœ œ œ œ
11111
1
###
""#
!
11 '
0
œ
8848 16
u
œ
17. R(x) x V [R(x)] dx x dxœÊœ œ
###
# ''
00
22
11 ab
x dxœœœ11 '
0
2
%
#
!
’“
x32
55
%
1

18. R(x) x V [R(x)] dx x dxœÊœ œ
$#$
# ''
00
22
11 ab
x dxœœœ11 '
0
2
'
#
!
’“
x 128
77
"
1

19. R(x) 9 x V [R(x)] dx 9 x dxœÊœ œ
È
ab
# ## ''
œœ33
33
11
9x 2 9(3) 2 18 36œ œ œ œ11 11’“ œ ‘
x27
33
!
$
$
††

20. R(x) x x V [R(x)] dx x x dxœ Ê œ œ
###
# ''
00
11
11 ab
x2xx dxœœ11'
0
1
ab ’“
#$%
"
!
x2xx
345
! Î%
(10156)œœ œ1ˆ‰
1
3 5 30 30
""
#
11

21. R(x) cos x V [R(x)] dx cos x dxœÊœ œÈ ''
00
22
œœ%%
11
#
sin x (1 0)œœœ111cd
1Î#
!

368. 364 Chapter 6 Applications of Definite Integrals
22. R(x) sec x V [R(x)] dx sec x dxœÊœ œ ''
œ
%%
œœ
œœ44
44
11
##
tan x [1 ( 1)] 2œœœ11 1cd
1
1
Î%
Î%

23. R(x) 2 sec x tan x V [R(x)] dxœ Êœ
È '
0
4
œ%
1
#
2 sec x tan x dxœ1 '
0
4
œ%
Š‹
È
#
2 2 2 sec x tan x sec x tan x dxœ 1 '
0
4
œ%
Š‹
È
##
2 dx 2 2 sec x tan x dx (tan x) sec x dxœ 1Œ %
È '' '
00 0
44 4
œœ œ%% %
##
[2x] 2 2 [sec x]œ 1Œ %
È
’“
11
1
Î% Î%
!!
Î%
!
tan x
3 Ê
02221 10 22œ œ11’“Š‹ˆ‰ ÈÈ È
Š‹ ab
11
##
"$
33
11

24. R(x) 2 2 sin x 2(1 sin x) V [R(x)] dxœ œ Ê œ
'
0
2
œ%
1
#
4(1 sin x) dx 4 1 sin x 2 sin x dxœœ11 ''
00
22
œœ%%
##
ab
4 1 (1 cos 2x) 2 sin x dxœ1 '
0
2
œ%
œ ‘
"
#
4 2 sin xœ1 '
0
2
œ%
ˆ‰
3 cos 2x
2#
4 x 2 cos xœ1œ ‘
3sin 2x
4#
Î#
!
1
4 00 (002) (3 8)œœ111œ ‘ˆ‰
3
4
1

25. R(y) 5 y V [R(y)] dy 5y dyœÊœ œ
È
†
##% ''
œœ11
11
11
y[1(1)]2œœœ11 1cd
&
"
π

26. R(y) y V [R(y)] dy y dyœÊœ œ
$Î# # $''
00
22
11
4œœ11’“
y
4
##
!

27. R(y) 2 sin 2y V [R(y)] dyœÊœÈ '
0
2
œ%
1
#
2 sin 2y dy cos 2yœœ11'
0
2
œ%
cd
1Î#
!
[1 ( 1)] 2œœ11

369. Section 6.1 Volumes by Slicing and Rotation About an Axis 365
28. R(y) cos V [R(y)] dyœÊœÉ
1y
4 '
œ2
0
1
#
cos dy 4 sin 4[0 ( 1)] 4œœœœ1 '
œ2
0
ˆ‰ œ ‘
11yy
44
!
œ#

29. R(y) V [R(y)] dy 4 dyœÊœ œ
2
y1 (y1)
# " ''
00
33
11
œ
44(1)3œœœ11 1’“ œ ‘
œÄ Ä

$
!
y1 4

30. R(y) V [R(y)] dy 2y y 1 dy;œÊœ œ
È2y
y1
œ

##
œ#
''
00
11
11 ab
u y 1 du 2y dy; y 0 u 1, y 1 u 2cdœÊ œ œÊœ œÊœ
#
V u du ( 1)Äœ œ œœ111 '
1
2
œ# ""
#
" ## œ‘ œ ‘
u
1

31. For the sketch given, a , b ; R(x) 1, r(x) cos x; V [R(x)] [r(x)] dxœ œ œ œ œ
11
##
## È ab '
a
b
1
(1 cos x) dx 2 (1 cos x) dx 2 [x sin x] 2 1 2œœ œœœ''
œ%
%%
œ
œœ20
22
11 1111
1 1Î#
! #
#
ˆ‰
32. For the sketch given, c 0, d ; R(y) 1, r(y) tan y; V [R(y)] [r(y)] dyœœ œ œ œ
1
4 '
c
d
1ab
##
1 tan y dy 2 sec y dy [2y tan y] 1œ œ œ œœ11 111''
00
44
œœ%%
ab ab ˆ‰
## Î%
! ##
1 11
œ
33. r(x) x and R(x) 1 V [R(x)] [r(x)] dxœœÊœ '
0
1
1ab
##
1 x dx x 1 0œœœœ'
0
1
111ab ’“ œ ‘ˆ‰
#
"
!
"x2
333
Ê
1

34. r(x) 2 x and R(x) 2 V [R(x)] [r(x)] dxœœÊœÈ ab '
0
1
1
##
(4 4x) dx 4 x 4 1 2œœœœ1111'
0
1
’“ ˆ‰
x
œ
##
"
!
"

370. 366 Chapter 6 Applications of Definite Integrals
35. r(x) x 1 and R(x) x 3œ œ
#
V [R(x)] [r(x)] dxʜ '
œ1
2
1ab
##
(x 3) x 1 dxœ1 '
œ1
2
’“ ab
##
#
x6x9 x2x1 dxœ1 '
œ1
2
cdaba b
#%#
xx6x8 dxœ1 '
œ1
2
ab
%#
8xœ 1’“
xx6x
53
% ʜ
#
#
π

16 8 3 28 3 8œ œ œ œ111œ ‘ˆ ‰ˆ ‰ˆ‰ˆ‰
32 8 24 6 33 5 30 33 117
53 53 5 5 5##
ÄÄ œ † 1
36. r(x) 2 x and R(x) 4 xœ œ
#
V [R(x)] [r(x)] dxʜ '
œ1
2
1ab
##
4 x (2 x) dxœ1 '
œ1
2
’“ab
##
#
16 8x x 4 4x x dxœ1 '
œ1
2
cdabab
#% #
12 4x 9x x dxœ1 '
œ1
2
ab
#%
12x 2x 3xœ 1’“
#$
#
π
x
5
%

24824 1223 15œ œ œ11œ ‘ˆ ‰ˆ‰ˆ‰
32 33 108
5555
" 1
37. r(x) sec x and R(x) 2œœ
È
V [R(x)] [r(x)] dxʜ
'
œ%
%
œ
œ4
4
1ab
##
2 sec x dx [2x tan x]œ œ11 '
œ%
%
œ
œ4
4
ab
# Î%
Î%
1
1
11(2)œœ111œ ‘ˆ‰ˆ ‰
11
##

38. R(x) sec x and r(x) tan xœœ
V [R(x)] [r(x)] dxʜ
'
0
1
1ab
##
sec x tan x dx 1 dx [x]œœœœ1111''
00
11
ab
## "
!

39. r(y) 1 and R(y) 1 yœœ
V [R(y)] [r(y)] dyʜ
'
0
1
1ab
##
(1 y) 1 dy 1 2y y 1 dyœœ11''
00
11
cda b
##
2y y dy y 1œœœœ111'
0
1
ab ’“ ˆ‰
##
"
!
"y
333
4
Ê
1

371. Section 6.1 Volumes by Slicing and Rotation About an Axis 367
40. R(y) 1 and r(y) 1 y V [R(y)] [r(y)] dyœœÊœ '
0
1
1ab
##
1 (1 y) dy 1 1 2y y dyœœ11''
00
11
cdc d ab
##
2y y dy y 1œœœœ111'
0
1
ab ’“ ˆ‰
##
"
!
"y
333
2
Ê
1

41. R(y) 2 and r(y) yœœ È
V [R(y)] [r(y)] dyʜ
'
0
4
1ab
##
(4 y) dy 4y (16 8) 8œœœœ1111'
0
4
’“
y
2
œ%
!

42. R(y) 3 and r(y) 3 yœœ
È È
#
V [R(y)] [r(y)] dyʜ '
0
3
œ
1ab
##
3 3 y dy y dyœœ11 ''
00
33
œœ
cdab
##
3œœ11’“
È
y
3
Ê
È
$
!

43. R(y) 2 and r(y) 1 yœœ È
V [R(y)] [r(y)] dyʜ
'
0
1
1ab
##
41 y dyœ1 '
0
1
’“ˆ‰È
#
412yy dyœ1 '
0
1
ˆ‰ È
32yy dyœ1 '
0
1
ˆ‰È
3y yœ1’“
4
3
y$Î#
#
"
!
œ
3œœ œ11ˆ‰ˆ‰
418837
366
Äœœ
#
1

44. R(y) 2 y and r(y) 1œ œ
"Î$
V [R(y)] [r(y)] dyʜ '
0
1
1ab
##
2y 1 dyœ1 '
0
1
’“ˆ‰
"Î$
#
44yy1 dyœ1 '
0
1
ˆ‰
"Î$ #Î$
3 4y y dyœ1 '
0
1
ˆ‰
"Î$ #Î$

3y 3y 3 3œ œœ11’“ ˆ‰
%Î$
"
!
3y
555
33
%&Ê
1

372. 368 Chapter 6 Applications of Definite Integrals
45. (a) r(x) x and R(x) 2œœÈ
V [R(x)] [r(x)] dxʜ
'
0
4
1ab
##
(4 x) dx 4x (16 8) 8œœœœ1111 '
0
4
’“
x
œ
#
%
!
(b) r(y) 0 and R(y) yœœ
#
V [R(y)] [r(y)] dyʜ '
0
2
1ab
##
y dyœœœ11 '
0
2
%
#
!
’“
y
55
32
%
1

(c) r(x) 0 and R(x) 2 x V [R(x)] [r(x)] dx 2 x dxœœÊœ œ ÈÈ ab ˆ‰ ''
00
44
11
##
#
4 4 x x dx 4x 16œ œœœ111 '
0
4
ˆ‰ ˆ‰È ’“
8x x 64 16 8
333
Ê&œ œ
##
%
!
1
(d) r(y) 4 y and R(y) 4 V [R(y)] [r(y)] dy 16 4 y dyœ œ Ê œ œ
####
# ''
00
22
11ab ab ’“
16 16 8y y dy 8y y dy yœœ œœœ1111 ''
00
22
abab ’“ ˆ‰
#% #% $
#
!
8 64 32 224
35 3515
y
%
1
46. (a) r(y) 0 and R(y) 1œœ
y
#
V [R(y)] [r(y)] dyʜ '
0
2
1ab
##
1 dy 1 y dyœœ 11 ''
00
22
ˆ‰ Š‹
yy
4#
#
œ
y œ œ#œ11’“ ˆ‰
yy
12 2 12 3
48 2
œ Ê
#
#
!
1

(b) r(y) 1 and R(y) 2œœ
y
#
V [R(y)] [r(y)] dy 2 1 dy 4 2y 1 dyÊœ œ œ '''
000
222
11 1ab ’“Š ‹ˆ‰
##
#
#yy
4
œ
3 2y dy 3y y 6 4 2œœœœœ1111 '
0
2
Š‹’ “ ˆ‰ˆ‰
yy
4121233
828
œ Ê
#
#
!
1
47. (a) r(x) 0 and R(x) 1 xœœ
#
V [R(x)] [r(x)] dxʜ '
œ1
1
1ab
##
1 x dx 1 2x x dxœœ11 ''
œœ11
11
ab a b
##%
#
x21œ œ 11’“ ˆ‰
2x x 2 1
35 35
Ê %
"
π
2 œœ1ˆ‰
15 10 3 16
15 15
œ% 1

(b) r(x) 1 and R(x) 2 x V [R(x)] [r(x)] dx 2 x 1 dxœœÊœ œ
####
# ''
œœ11
11
11abab ’“
4 4x x 1 dx 3 4x x dx 3x x 2 3œœœœ1111 ''
œœ11
11
abab ’“ ˆ‰
#% #% $
"
π
4x 41
35 35
%
(45203)œœ
256
15 15
11
(c) r(x) 1 x and R(x) 2 V [R(x)] [r(x)] dx 4 1 x dxœ œ Ê œ œ
####
# ''
œœ11
11
11ab ab ’“
4 1 2x x dx 3 2x x dx 3x x 2 3œœœœ1111 ''
œœ11
11
abab ’“ ˆ‰
#% #% $
"
π
2x 21
35 35
%
(45103)œœ
264
15 15
11

373. Section 6.1 Volumes by Slicing and Rotation About an Axis 369
48. (a) r(x) 0 and R(x) x hœœ
h
b
V [R(x)] [r(x)] dxʜ '
0
b
1ab
##
x h dxœ1 '
0
b
ˆ‰
h
b
#
xxh dxœ1 '
0
b
Š‹
h2h
bb
œœ
œ
##
hxhbbœœœ11
##
!
’“ ˆ‰
xx b hb
3b b 3 3
Ê œœ
œ
b
1

(b) r(y) 0 and R(y) b 1 V [R(y)] [r(y)] dy b 1 dyœœÊœ œ ˆ‰ ˆ‰ ab
yy
hh ''
00
hh
11
## #
#
b 1 dy b y b h hœœœœ111
###
!'
0
h h
Š‹’ “ ˆ‰
2y y y y
hh h3h 3 3
hbh
œœ Ê
œœ
œ
1
49. R(y) b a y and r(y) b a yœ œ ÈÈ
## ##
V [R(y)] [r(y)] dyʜ '
œa
a
1ab
##
b a y b a y dyœ1 '
œa
a
’“ˆ‰ˆ‰ÈÈ
## ##
##
4b a y dy 4b a y dyœœ11 ''
œœaa
aa
ÈÈ## ##
4b area of semicircle of radius a 4b 2a bœœœ111††
1a
œ
#
##

50. (a) A cross section has radius r y and area r y. The volume is ydy y .œ # œ# # œ œ#&È cd11 1 1 1
##
&
! '
0
Ê
(b) V h A h dh, so A h . Therefore A h , so .ab ab ab abœœ œ†œ†œ†' dV dV dV dh dh dh dV
dh dt dh dt dt dt A h dt
"
ab
For h , the area is , so .œ% # % œ) œ †$ œ †11ab
dh units units
dt sec sec
"$
))11
ÊÊ
51. (a) R(y) a y V a y dy a y a h a aœÊœ œ œ È ab ’“’ “ Š‹
## ## # # $ $ œ
111'
œ
œ œ
œa
ha ha
a
y(ha)
333
a
ÊÊ Ê
a h h 3h a 3ha a a h h a haœ œ œ11’“Š‹ab
#$##$ # ##" œ
3333
ah h (3a h)
ÊÊ œ
1
(b) Given 0.2 m /sec and a 5 m, find . From part (a), V(h) 5 h
dV dh h
dt dt 3 3
h (15 h)
œœ œœ
$ #œ ¸
h4$
1 1
œ Ê
1
10 h h h(10 h) m/sec.Êœ Êœ œ Ê œ œ œ
dV dV dV dh dh dh 0.2
dh dt dh dt dt dt 4 (10 4) (20 )(6) 1 0
11 1
#
œ#
""
† ¸
h4$ 111
52. Suppose the solid is produced by revolving y 2 x aboutœ
the y-axis. Cast a shadow of the solid on a plane parallel to
the xy-plane.
Use an approximation such as the Trapezoid Rule, to
estimate R y dy y.'
a
b
11cdab !
Œ%
#
œ"
#
¸˜
k
n
d
k
^
#

53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius
h has been removed. Thus its area is A R h R h . The cross section of the hemisphere is a disk of
"
## ##œœ 111 ab
radius R h . Therefore its area is A R h R h . We can see that A A . The altitudes of
ÈÈ
Š‹ ab
## ##
#"#
#
##
œœœ 11
both solids are R. Applying Cavalieri's Principle we find
Volume of Hemisphere (Volume of Cylinder) (Volume of Cone) R R R R R .œœœ ab ab11 1
##$ "
33
2

374. 370 Chapter 6 Applications of Definite Integrals
54. R(x) V [R(x)] dx dx r h, the volume ofœÊœ œ œ œ œ
rx r x r x r h
hhh3h33''
00
hh h
11 1
##
!
"
œœ œ ÊœÊ
œœ œ
11
’“ Š ‹Š‹
a cone of radius r and height h.
55. R(y) 256 y V [R(y)] dy 256 y dy 256yœÊœ œ œ È ab ’“
# ##
œ¸
œÄ† ''
œœ
œœ16 16
77
11 1
y
3
Ê
(256)( 7) (256)( 16) 256(16 7) 1053 cm 3308 cmœœœ ¸111’“Š‹ Š‹
716716
3333
ÊÊÊÊ
$$
56. R(x) 36 x V [R(x)] dx 36 x dx 36x x dxœÊœ œ œ
xx
1 144 144#
# ###%È
ab a b '' '
00 0
66 6
11
œ
1
12x 12 6 12 cm . The plumb bob willœœ œœ œ
11 1 11
144 5 144 5 144 5 144 5 5
x 6 6 36 196 60 36 36
’“Š ‹ ˆ‰ˆ‰ˆ‰
$$ $
'
!
œ
%% Ê
†
†
weigh about W (8.5) 192 gm, to the nearest gram.œ¸ ˆ‰
36
5
1
57. (a) R(x) c sin x , so V [R(x)] dx (c sin x) dx c 2c sin x sin x dxœ œ œ œ kk ab11 1 '' '
00 0
œœ œ
####
c 2c sin x dx c 2c sin x dxœ œ11 ''
00
œœ
ˆ‰ˆ ‰
## π
###
1 cos 2x cos 2x
c x 2c cos x c 2c 0 (0 2c 0) c 4c . Letœ œ œ 11111œ ‘ œ ‘ˆ ‰ˆ‰ ˆ ‰
## #"
## # !
sin 2x
4
1
11
V(c) c 4c . We find the extreme values of V(c): (2c 4) 0 c is a criticalœ œœÊœ11 11ˆ‰
#
#
1
1
dV 2
dc
point, and V 4; Evaluate V at the endpoints: V(0) andˆ‰ ˆ ‰ ˆ ‰
24 8 4
11 1 1
111 1
œ œ œ œ11
### #
œ œ
V(1) 4 (4 ) . Now we see that the function's absolute minimum value is 4,œœ 11 11ˆ‰
3
## #
1 1
œ œ
taken on at the critical point c . (See also the accompanying graph.)œ
2
1
(b) From the discussion in part (a) we conclude that the function's absolute maximum value is , taken on at
1
œ
#
the endpoint c 0.œ
(c) The graph of the solid's volume as a function of c for
0 c 1 is given at the right. As c moves away fromŸŸ
[ ] the volume of the solid increases without bound.!ß "
If we approximate the solid as a set of solid disks, we
can see that the radius of a typical disk increases without
bounds as c moves away from [0 1].ß

58. (a) R(x) 1 V [R(x)] dxœ Ê œ
x
16
œ
'
œ4
4
1
#
1 dx 1 dxœœ11 ''
œœ44
44
Š‹ Š ‹
xxx
16 8 16
œœ #
œ
#
x24œ œ 11’“Š‹
xx 44
24 5 16 24 5 16
Ê %!%
œœ
††
%
%
2 4 (60 40 12) ftœœ œ1ˆ‰
84 2 64
3 5 15 15
11 $

(b) The helicopter will be able to fly (7.481)(2) 201 additional miles.ˆ‰
64
15
1
¸
6.2 VOLUME BY CYLINDRICAL SHELLS
1. For the sketch given, a 0, b 2;œœ
V 2 dx 2 x 1 dx 2 x dx 2 2œ œ œ œœ'''
a00
b22
11111ˆ‰ ˆ‰Š‹ Š‹ Š‹ ’ “
shell shell
radius height 4 4 16 16
xxxx416
œ Êœ#
##
#
!
236œœ11†
2. For the sketch given, a 0, b 2;œœ
V 2 dx 2 x 2 dx 2 2x dx 2 x 2 (4 1) 6œ œ œ œœœ'''
a00
b22
111111ˆ‰ Š‹ Š ‹ Š ‹ ’ “
shell shell
radius height 4 4 16
xxx
œ Ê#
#
#
!

375. Section 6.2 Volume by Cylindrical Shells 371
3. For the sketch given, c 0, d 2;œœ
È
V 2 dy 2 y y dy 2 y dy 2 2œœœœœ'''
c00
d22
11111ˆ‰ Š‹ ’“ ab
shell shell
radius height 4
y

†
#$
#
!

È
4. For the sketch given, c 0, d 3;œœ
È
V 2 dy 2 y 3 3 y dy 2 y dy 2œœ œœœ'' '
c0 0
d3 3
1111ˆ‰ Š‹ ’“ cdab
shell shell
radius height 4
y
3
9

†
#$
!
#

È
1
5. For the sketch given, a 0, b 3;œœ
È
V 2 dx 2 x x 1 dx;œœ ''
a0
b3
11ˆ‰ Š‹ Š ‹
È
shell shell
radius height

†
#
u x 1 du 2x dx; x 0 u 1, x 3 u 4’“
È
œÊœ œÊœ œ Êœ
#
V u du u 4 1 (8 1)Äœ œ œ œ œ11 '
1
4
"Î# $Î# $Î#
%
" ‘ˆ ‰ˆ‰
22 2 14
33 3 3
111
6. For the sketch given, a 0, b 3;œœ
V 2 dx 2 x dx;œœ ''
a0
b3
11ˆ‰ Š‹ Š ‹
shell shell
radius height
9x
x9
È


u x 9 du 3x dx 3 du 9x dx; x 0 u 9, x 3 u 36c dœÊœ Ê œ œÊœ œÊœ
$# #
V 2 3u du 6 2u 12 36 9 36Äœ œ œ œ111 1 '
9
36
"Î# "Î#
$'
* ‘ Š‹
ÈÈ
7. a 0, b 2;œœ
V 2 dx 2 x x dxœœ ''
a0
b2
11ˆ‰ ‘Š‹ ˆ‰
shell shell
radius height 2
x
2 x dx 3x dx x 8œœœœ''
00
22
11 11
##$
#
#
!
†
3
cd

8. a 0, b 1;œœ
V 2 dx 2 x 2x dxœœ ''
a0
b1
11ˆ‰ ˆ ‰Š‹
shell shell
radius height 2
x
2 dx 3x dx xœœœœ1111'
0
1 1
0
Š‹ ' cd
3x

#
#$
"
!

9. a 0, b 1;œœ
V 2 dx 2 x (2 x) x dxœœ
''
a0
b1
11ˆ‰ Š‹ cd
shell shell
radius height
#
2 2xx x dx2xœ œ 11 '
0
1
ab ’“
#$ #
"
!
xx
34

21 2œ œœœ11ˆ‰ˆ‰
""
3 4 12 12 6
1243 10 5 11

376. 372 Chapter 6 Applications of Definite Integrals
10. a 0, b 1;œœ
V 2 dx 2 x 2 x x dxœœ
''
a0
b1
11ˆ‰ Š‹ cdab
shell shell
radius height
##
2 x2 2x dx 4 x x dxœ œ 11 ''
00
11
ab ab
#$
44œ œ œ111’“ ˆ‰
xx
424

#
"
!
""

11. a 0, b 1;œœ
V 2 dx 2 x x (2x 1) dxœœ ''
a0
b1
11ˆ‰ ‘Š‹ È
shell shell
radius height
2 x 2x x dx 2 x x xœ œ 11 '
0
1
ˆ‰ ‘
$Î# # &Î# $ # "
#
"
!
22
53
22œ œœ11ˆ‰ˆ‰
2 2 12 20 15 7
5 3 30 15
"
#
1

12. a , b 4;œ" œ
V 2 dx 2 x x dxœœ ''
a1
b4
11ˆ‰ ˆ ‰Š‹
shell shell
radius height 2
3"Î#
3 x dx 3 x 2 4œœœ "111'
1
4
"Î# $Î# $Î#
%
" ‘ˆ ‰
2
3
2(8 1) 14œ œ11

13. (a) xf(x) xf(x) ; since sin 0 0 we have
x, 0x
x, x 0
sin x, 0 x
0, x 0
œÊœ œ
Ÿ
œ
Ÿ
œ
œœ
†
sin x
x
1 1
xf(x) xf(x) sin x, 0 x
sin x, 0 x
sin x, x 0
œÊœŸŸ
Ÿ
œ
œ
1
1
(b) V 2 dx 2 x f(x) dx and x f(x) sin x, 0 x by part (a)œœ œŸŸ
''
a0
b
11 1ˆ‰ Š‹
shell shell
radius height

††
V 2 sin x dx 2 [ cos x] 2 ( cos cos 0) 4Êœ œ œ œ11111
'
0

1
!
14. (a) xg(x) xg(x) ; since tan 0 0 we have
x, 0x
x 0, x 0
tan x, 0 x /4
0, x 0
œÊœ œ
Ÿ
œ
Ÿ
œ
œœ
†
†
tan x
x4

1 #
1
xg(x) xg(x) tan x, 0 x /4
tan x, 0 x /4
tan x, x 0
œÊœŸŸ
Ÿ
œ
œ
#
#
#
1
1
(b) V 2 dx 2 x g(x) dx and x g(x) tan x, 0 x /4 by part (a)œœ œŸŸ
''
a0
b4
11 1ˆ‰ Š‹
shell shell
radius height

††
#
V 2 tan x dx 2 sec x 1 dx 2 [tan x x] 2 1Êœ œ œ œ œ11 11 ''
00
44

## Î%
!

#
ab ˆ‰
1 111
4
4

377. Section 6.2 Volume by Cylindrical Shells 373
15. c 0, d 2;œœ
V 2 dy 2 y y ( y) dyœœ ''
c0
d2
11ˆ‰ ‘Š‹ È
shell shell
radius height
2 y y dy 2œ œ 11 '
0
2
ˆ‰ ’“
$Î# #
#
!
2y y
53

22 2 16œ œ œ 111”•Š‹ Š ‹ Š ‹
È
22 8
535353
82 2
&
"
ÈÈ
32 5œ
16
15
1
Š‹
È

16. c 0, d 2;œœ
V 2 dy 2 y y ( y) dyœœ
''
c0
d2
11ˆ‰ Š‹ cd
shell shell
radius height
#
2 y y dy 2 16œ œ œ 111'
0
2
ab ’“ ˆ‰
$#
#
!
"yy
43 43
2

16œœ1ˆ‰
540
63
1

17. c 0, d 2;œœ
V 2 dy 2 y 2y y dyœœ
''
c0
d2
11ˆ‰ Š‹ ab
shell shell
radius height
#
2 2y y dy 2 2œ œ œ 111'
0
2
ab ’“ ˆ‰
#$
#
!
"2y y
34 34
16 6

32œ œœ1ˆ‰
""
34 12 3
32 811

18. c 0, d 1;œœ
V 2 dy 2 y 2y y y dyœœ
''
c0
d1
11ˆ‰ Š‹ ab
shell shell
radius height
#
2 y y y dy 2 y y dyœ œ 11 ''
00
11
ab a b
##$
22œ œ œ11’“ ˆ‰
yy
34 34 6
1
"
!
" 1

19. c 0, d 1;œœ
V 2 dy 2 y[y ( y)]dyœœ
''
c0
d1
11ˆ‰ Š‹
shell shell
radius height
2 2y dy yœœœ1 '
0
1
#$
"
!
44
33
11
cd

20. c 0, d 2;œœ
V 2 dy 2 y y dyœœ ''
c0
d2
11ˆ‰ ˆ ‰Š‹
shell shell
radius height 2
y
2 dy yœœœ1 '
0
2
y
23 3
3
1
8
2
11
cd
!

378. 374 Chapter 6 Applications of Definite Integrals
21. c 0, d 2;œœ
V 2 dy 2 y (2 y) y dyœœ
''
c0
d2
11ˆ‰ Š‹ cd
shell shell
radius height
#
2 2y y y dy 2 yœ œ 11 '
0
2
ab ’“
#$ #
#
!
yy
34

24 (483248)œ œ œ1ˆ‰
816 16
34 6 3
11

22. c 0, d 1;œœ
V 2 dy 2 y (2 y) y dyœœ
''
c0
d1
11ˆ‰ Š‹ cd
shell shell
radius height
#
2 2yy y dy2yœ œ 11 '
0
1
ab ’“
#$ #
"
!
yy
34

21 (1243)œ œ œ1ˆ‰
11 5
34 6 6
11

23. (a) V 2 dy 2 y 12 y y dy 24 y y dy 24œœ œ œ '' '
c0 0
d1 1
1111ˆ‰ Š‹ ’ “ ab ab
shell shell
radius height 45
yy
†
#$ $%
"
!
 
24 œ œœ1ˆ‰
11 24 6
45 20 5
11
(b) V 2 dy 2 (1 y) 12 y y dy 24 (1 y) y y dyœœ œ '' '
c0 0
d1 1
11 1ˆ‰ Š‹ cd abab
shell shell
radius height
#$ #$
24 y 2y y dy 24 24 24œ œ œ œœ1111 '
0
1
ab ’“ ˆ‰ˆ‰
#$%
"
!
""yyy
325 325 30 5
11 4

1
(c) V 2 dy 2 y 12 y y dy 24 y y y dyœœ œ '' '
c0 0
d1 1
11 1ˆ‰ ˆ‰ ˆ‰Š‹ cd abab
shell shell
radius height 5 5
88 #$ #$
24 y y y dy 24 y y 24 (32 39 12)œ œ œ œ 111 '
0
1
ˆ‰ ˆ‰ ’“
8 13 8 13 8 13 1 24
5 5 15 20 5 15 20 5 60
y#$% $%
"
!
 
1
2 œœ
24
12
1
1
(d) V 2 dy 2 y 12 y y dy 24 y y y dyœœ œ '' '
c0 0
d1 1
11 1ˆ‰ ˆ‰ ˆ‰Š‹ cd abab
shell shell
radius height 5 5
22 #$ #$
24 y y y y dy 24 y y y dy 24 y yœ œ œ 111 ''
00
11
ˆ‰ˆ‰ ’“
$%#$ #$% $ %
"
!
22 23 2 3
55 55 1520 5
y
 
24 (8 9 12) 2œ œ œœ11ˆ‰
231 24 24
15 20 5 60 12
11
24. (a) V 2 dy 2 y dy 2 y y dy 2 y dyœœ œ œ '' ''
c0 00
d2 22
11 11ˆ‰ Š‹ ’ “ Š‹ Š‹ Š‹
shell shell
radius height 4 4 4
yyy y y

##
#$
2 2 32 32 32œ œ œ œ œœ11 111’“Š‹ ˆ‰ ˆ‰ ˆ‰
yy
424 424 42446243
22 4 2 8
#
!
""" 1
(b) V 2 dy 2 (2 y) dy 2 (2 y) y dyœœ œ '' '
c0 0
d2 2
11 1ˆ‰ Š‹ ’ “ Š ‹ Š‹
shell shell
radius height 4 4
yyy y

##
#
2 2y y dy 2 2œ œ œ œ111 '
0
2
Š‹’“ ˆ‰
#$
#
#
!
y y 2yyyy
2 4 3 104 4 310424 5
16 32 16 64 8

1
(c) V 2 dy 2 (5 y) dy 2 (5 y) y dyœœ œ '' '
c0 0
d2 2
11 1ˆ‰ Š‹ ’ “ Š ‹ Š‹
shell shell
radius height 4 4
yyy y

##
#
2 5y y y dy 2 2 8œ œ œ œ1111 '
0
2
Š‹’“ ˆ‰
#%$
#
#
!
5 40 160 16 64
44 32044320424
y 5y5yyy
 
(d) V 2 dy 2 y dy 2 y y dyœœ œ '' '
c0 0
d2 2
11 1ˆ‰ ˆ‰ ˆ‰Š‹ ’ “ Š ‹ Š‹
shell shell
radius height 8 4 8 4
55yyy y

##
#
2 y y y dy 2 2 4œ œ œ œ1111 '
0
2
Š‹’“ ˆ‰
$#%
##
#
!
y yy5y5y
4 8 32 4 4 4 160 4 24 24 160
5 5 16 64 40 160
 

379. Section 6.2 Volume by Cylindrical Shells 375
25. (a) About x-axis: V 2 dyœ '
c
d
1ˆ‰ Š‹
shell shell
radius height
2 y y y dy 2 y y dyœ œ ''
00
11
11ˆ‰ ˆ ‰È
$Î# #
2y y 2œ œ œ11 ‘ˆ‰
#" #"#
&$ &$"&
&Î# $
"
!
1
About y-axis: V 2 dxœ '
a
b
1ˆ‰ Š‹
shell shell
radius height
2 x x x dx 2 x x dxœ œ ''
00
11
11ab a b
# 23
22 œ œ œ11’“ ˆ‰
xx

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(b) About x-axis: R x x and r x x V R x r x dx x x dxab ab ab ab c d ‘œœÊœ œ
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About y-axis: R y y and r y y V R y r y dy y y dyab ab ab ab c dÈ ‘œœÊœ œ ''
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26. (a) V R x r x dx x dxœ œ # ''
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radius height
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27. (a) V 2 dy 2 y(y 1) dyœœ ''
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radius height
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2 x(2 x) dx 2 2x x dx 2 x 2 4 1œ œ œ œ ''
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radius height 3 3 3
10 20 16 #
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radius height 33
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1

380. 376 Chapter 6 Applications of Definite Integrals
28. (a) V 2 dy 2 y y 0 dyœœ ''
c0
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radius height
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(c) V 2 dx 2 (4 x) 2 x dx 2 8 4x 2x x dxœœ œ '' '
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29. (a) V 2 dy 2 y y y dyœœ ''
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radius height
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30. (a) V 2 dyœ '
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(b) Use the washer method:
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111 1cd ab a b ’“ ’ “
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381. Section 6.2 Volume by Cylindrical Shells 377
31. (a) V 2 dy 2 y 8y y dyœœ ''
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d2
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(b) V 2 dx 2 x x dx 2 x dx 2 xœœ œ œ '' '
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radius height 8 8 5 32
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35. (a) k: V V VH3= œ "#
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b

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382. 378 Chapter 6 Applications of Definite Integrals
V [R (x)] [r (x)] dx with R (x) and r (x) x; a 0 and b 1### # ###
## œ œœœœ'
a
b

1ab É È
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3
two integrals are requiredÊ
(c) : V 2 dy 2 y dy where shell height y 3y 2 2 2y ;W2/66 œ œ œ œ
''
cc
dd
11ˆ‰ Š‹ Š‹ ab
shell shell shell
radius height height
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c 0 and d 1. Only integral is required. It is, therefore preferable to use the method.œœ 98/ =2/66
However, whichever method you use, you will get V .œ1
36. (a) k: V V V VH3= œ
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i
1
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shell shell shell
radius height height
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However, whichever method you use, you will get V .œ
5
6
1
6.3 LENGTHS OF PLANE CURVES
1. 1 and 3 ( 1) (3) 10
dx dx
dt dt dt dt
dy dy
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#
#
##
Length 10 dt 10 t 10 10Êœ œ œ œ '


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1
1
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2
33
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2. sin t and 1 cos t ( sin t) (1 cos t) 2 2 cos t
dx dx
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383. Section 6.3 Lengths of Plane Curves 379
6. 8t cos t and 8t sin t (8t cos t) (8t sin t) 64t cos t 64t sin t
dx dx
dt dt dt dt
dy dy
œœÊ œ œ ʈ‰ Š‹ È È
#
#
######
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7. x 2 2x x 2 x
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dx 3
3
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3

8. x L 1 x dx; u 1 x
dy
dx 4 4
399
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#
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0
4
du dx du dx; x 0 u 1; x 4Êœ Ê œ œÊœ œ
94
49
u 10 L u du ud ˆ‰ ‘Êœ Äœ œ '
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10
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#
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33
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34 31 34 134 1 1 6
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9
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99
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9
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y
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332

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4œ œ œœŠ‹ ˆ‰
16 128 1 8 4 123
4 (16)(2) 4 8 32 4 8 32 32
""" """

384. 380 Chapter 6 Applications of Definite Integrals
12. y 2 y
dx dx
dy y dy 4
y
œ Êœ


##
""
#
% %
Š‹ ab
L 1 y 2 y dyʜ '
2
3
É ab
"% %
4
y2y dyœ '
2
3
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4
y y dy y y dyœ œ
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# #
# # #''
22
33
Éab ab

y6œ œ œ œ œ
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333333 4
27 8 26 8 13

13. x x x
dy dy
dx 4 dx 16
x
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#
#
Š‹

L 1 x dxʜ '
1
8
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x
16

x dxœ '
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8
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x
16

x x dx x x dxœ œ ''
11
88
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44
x x 2x xœ œ ‘ ‘
33 3
48 8
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3399
888
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14. x 2x 1 x 2x 1
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4
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#
#
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u 108 1 4 3 106 53

15. sec y 1 sec y 1
dx dx
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44
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16. 3x 1 3x 1
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22
11
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385. Section 6.3 Lengths of Plane Curves 381
17. (a) 2x 4x
dy dy
dx dx
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#
#
L 1 dxʜ '
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2
Ê Š‹
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20. (a)
dx dx
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yy
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386. 382 Chapter 6 Applications of Definite Integrals
22. (a) cos x - cos x + x sin x x sin x
dy dy
dx dx
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#
##
L 1 x sin x dxʜ '
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23. (a) tan x tan x
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dx dx
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#
#
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00
66
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È É #  sin x cos x
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sec x dxœœ ''
00
66
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dx
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(c) L 0.55¸
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24. (a) sec y 1 sec y 1
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"

3
4
È ab #
sec y dy sec y dyœœ ''

""

33
44
kk
(c) L 2.20¸
(b)
25. 2 x 1 dt, x 0 2 1 1 y f(x) x C where C is any
ÈÈ
ÊÊŠ‹ Š‹œ  ÊœÊœ„Êœœ„'
0
x
dy dy dy
dt dx dx
##
real number.
26. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see that
dy f (x ) x length of kth tangent fin isœ˜Ê
w
k1 k
(x) (dy) (x) [f(x)x].ÈÈ! ?! !
kkk1k
## #w #


(b) Length of curve lim (length of kth tangent fin) lim ( x ) [f (x ) x ]??
nnÄ_ Ä_
!! È
nn
k1 k1
kk1k
ÎÎ

#w #
lim 1 [f (x )] x 1 [f (x)] dx??
nÄ_
!ÈÈ
n
k1
k1 k
a
b
Î

w# w# '

387. Section 6.3 Lengths of Plane Curves 383
27. (a) correspondes to here, so take as . Then y x C and since ( ) lies on the curve, C 0.Š‹ È
dy dy
dx 4x dx x
#
""
#È
? $&$ ?
So y x from ( ) to (4 2).œ"ß"ßÈ
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
28. (a) correspondes to here, so take as . Then x C and, since ( ) lies on the curve, C 1Š‹
dx
dy dx y y y
dy
#
"""
" œ !ß" œ
So y . œ
"
"x
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
29. (a) 2 sin 2t and 2 cos 2t ( 2 sin 2t) (2 cos 2t) 2
dx dx
dt dt dt dt
dy dy
œ œ Ê œ œ ʈ‰ Š‹ È
#
#
##
Length 2 dt 2tÊœ œœ '
0
2
"
cd
1Î#
!
1
(b) cos t and sin t ( cos t) ( sin t)
dx dx
dt dt dt dt
dy dy
??Î??Î?11 11 11 11 1 ʈ‰ Š‹ È
#
#
##
Length dt tÊœ œœ '
"
"12
12
11 1cd
"Î#
"Î#
30. x a( sin ) a(1 cos ) a 1 2 cos cos and y a(1 cos )œ Êœ Ê œ œ)) ) ) ) )
dx dx
dd))
ˆ‰ ab
#
##
a sin a sin Length d 2a (1 cos ) dÊœ Ê œ Ê œ œ
dy dy dy
dd dd
dx
)) ))
)) ) ))Š‹ Š‹ ʈ‰ È
##
##
#
#
''
00
22

a 2 2 d 2a sin d 2a sin d 4a cos 8a????Î?
ÈÈ É ¸¸ ‘ '''
000
222

1cos
2

###
#
!
))))
1
)))
31-36. Example CAS commands:
:Maple
with( plots );
with( Student[Calculus1] );
with( student );
f := x -> sqrt(1-x^2);a := -1;
b := 1;
N := [2, 4, 8 ];
for n in N do
xx := [seq( a+i*(b-a)/n, i=0..n )];
pts := [seq([x,f(x)],x=xx)];
L := simplify(add( distance(pts[i+1],pts[i]), i=1..n )); # (b)
T := sprintf("#31(a) (Section 6.3)n=%3d L=%8.5f", n, L );
P[n] := plot( [f(x),pts], x=a..b, title=T ): # (a)
end do:
display( [seq(P[n],n=N)], insequence=true, scaling=constrained );
L := ArcLength( f(x), x=a..b, output=integral ):
L = evalf( L ); # (c)
37-40. Example CAS commands:
:Maple
with( plots );
with( student );
x := t -> t^3/3;
y := t -> t^2/2;
a := 0;

388. 384 Chapter 6 Applications of Definite Integrals
b := 1;
N := [2, 4, 8 ];
for n in N do
tt := [seq( a+i*(b-a)/n, i=0..n )];
pts := [seq([x(t),y(t)],t=tt)];
L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n )); # (b)
T := sprintf("#37(a) (Section 6.3)n=%3d L=%8.5f", n, L );
P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): # (a)
end do:
display( [seq(P[n],n=N)], insequence=true );
ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ): # (c)
L := Int( ds(t), t=a..b ):
L = evalf(L);
31-40. Example CAS commands:
: (assigned function and values for a, b, and n may vary)Mathematica
Clear[x, f]
{a, b} = { 1, 1}; f[x_] = Sqrt[1 x ]ÎÎ
2
p1 = Plot[f[x], {x, a, b}]
n = 8;
pts = Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / NÎ
Show[{p1,Graphics[{Line[pts]}]}]
Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]]) (pts[[i 1, 2]] pts[[i, 2]]) ], {i, 1, n}]
22
NIntegrate[ Sqrt[ 1 f'[x] ],{x, a, b}]
2
6.4 MOMENTS AND CENTERS OF MASS
1. Because the children are balanced, the moment of the system about the origin must be equal to zero:
5 80 x 100 x 4 ft, the distance of the 100-lb child from the fulcrum.††œÊœ
2. Suppose the log has length 2a. Align the log along the x-axis so the 100-lb end is placed at x a and the?
200-lb end at x a. Then the center of mass x satisfies x x . That is, x is locatedœœÊœ
100( a) 200(a)
300 3
a
at a distance a (2a) which is of the length of the log from the 200-lb (heavier) end (see figure)? ?
a2a
333 3
""
or of the way from the lighter end toward the heavier end.
2
3
(2a)
" 3
èëëéëëê
100 lbs. 200 lbsñïïïïïïïïïïïïïïñïïïïñïïïïïïñ
Î !
a xa/3 aœ
3. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point
masses located at the centers of the rods at coordinates and 0 . Therefore xˆ‰ ˆ‰
LL
##
ß! ß œ
m
my
and y is the center ofœœœœœœœÊß
xm xm
mm mm m mm mm
m0 m ym ym 0m




L
x2
L
† †
LLLL
4444
ˆ‰
mass location.
4. Let the rods have lengths x L and y 2L. The center of mass of each rod is in its center (see Example 1).œœ
The rod system is equivalent to two point masses located at the centers of the rods at coordinates andˆ‰
L
#
ß!
( L). Therefore x and y is the center of mass location.!ß œ œ œ œ Ê ß
L
!
†† ††m02m
m2m m2m
mL2m

!
L2LL2L
6363
ˆ‰

389. Section 6.4 Moments and Centers of Mass 385
5. M x 4 dx 4 4 8; M 4 dx [4x] 4 2 8 x 1!
##
#
!
#
!œœœœœœœœÊœœ''
00
22
†† †’“
x4 M
M

Î
6. M x 4 dx 4 (9 1) 16; M 4 dx [4x] 12 4 8 x 2!
##
$
"
$
"????????ʜ??''
1 1
3 3
† ’“
x4 16M
M8

Î
7. M x 1 dx x dx ; M 1 dx x!
##
$$
!!???????'' '
00 0
33 3
ˆ‰ ˆ ‰ ˆ‰ Š‹’ “ ’“
xxxx92715xx
33929 36
 
3 x?? ? ? ? ? ?
99 155
6M93
M
# Î
ˆ‰
ˆ‰
15
2
9
2
8. M x 2 dx 2x dx x 16 16 16 ;!
#
%
!???????''
00
44
ˆ‰ ˆ ‰ Š‹’“
xxx6416232
4 4 12 12 3 3 3
 
†
M 2 dx 2x 8 6 x????ʜ??'
0
4
ˆ‰ ’“
x x 16 32 16
488 M369
M

Î
%
!
†
9. M x 1 dx x x dx 8 ;!
" ¸ "Î#
###
%
"œ œ œ œ œœ œ''
11
44
Š‹ ’ “ ˆ‰ ˆ‰ˆ‰
Èx
x 2x 16 2 15 14 45 28 73
3 3 3 366

M 1x dx x2x (44)(12)5 xœ œ œœÊœ œ œ'
1
4
ˆ‰ ‘
"Î# "Î#
%
"
M
M530
73 Î
ˆ‰
73
6
10. M x 3 x x dx 3 x x dx 3 2x 3 (2 2) 2!
$Î# &Î# "Î# $Î# "Î#
"
"Î%
"
#????''
14 14
11¸¸
† †ˆ‰ˆ‰ ‘ ’“ Š‹
22
x
˜
ˆ‰
3(4 1) 9; M 3 x x dx 3 3 2 4 3 2œœ œ œ œ œ '
14
1¸
ˆ‰ ‘ ‘ˆ ‰ˆ‰ˆ ‰
$Î# &Î#
"
"Î%
22 2 16 14
x3x 33 3

6 14 20 x? ? ? ? ?
M
M20
9Î
11. M x(2 x) dx x x dx 2x x dx x dx 1!
##
#
"#
!"
""œ œ œ œ''' '
010 1
121 2
† ab ’“’“ ˆ‰ˆ‰
2x x x 8
33 333
  
3; M (2 x) dx x dx 2x 2 3 x 1œœ œ œ œœÊœ œ
9xx4
3 M
M ''
01
12
’“’“ ˆ‰ˆ‰

Î
## ###
"#
!"
""
12. M x(x 1) dx 2x dx x x dx 2x dx x (4 1)!
##
"
!
#
"
""? ? ??''''
0101
1212
ab cd ’“ ˆ‰
xx
32 32

3 ; M (x 1) dx 2 dx x 2x 1 (4 ) 2œœ œ œ œ #œœ
523 x 37
66 ''
01
12
’“ cd ˆ‰

## ##
"
!
#
"
"
xÊœ œ œ
M
M6721
23 2 23Έ‰ˆ‰
13. Since the plate is symmetric about the y-axis and its density is
constant, the distribution of mass is symmetric about the y-axis
and the center of mass lies on the y-axis. This means that
x 0. It remains to find y . We model the distribution ofœœ
M
Mx
mass with strips. The typical strip has center of mass:@/<>3-+6
( x y ) x , length: 4 x , width: dx, area:
µµ
&?& Ί‹
x4

 
#
#

dA 4 x dx, mass: dm dA 4 x dx. The moment of the strip about the x-axis is?Î ? ? Îab ab
##
$$ dm 4 x dx 16 x dx. The moment of the plate about the x-axis is M dm? Î ? Î ?
µ µ
Š‹ ab a b
x4

 
##
#%
$
$
x
'
16 x dx 16x 16 2 16 2 32 . The mass of the?Î?Î? ÎÎÎ?Î?'
2
2
$$ $ $ $
## # #
%
#
#
ab ’“’ “ Š‹Š ‹ ˆ‰
x 2 2 2 32 128
55555
"""
††
†
plate is M 4 x dx 4x 2 8 . Therefore y . The plate's center of???? ???'$$ $ab ’“ ˆ‰
#
#
#
x 8 32 Mx 12
333 M5
 
$
Š‹
Š‹
128
5
32
3

mass is the point (x y) .ßœ!߈‰
12
5

390. 386 Chapter 6 Applications of Definite Integrals
14. Applying the symmetry argument analogous to the one in
Exercise 13, we find x 0. To find y , we use theœœ
M
Mx
strips technique. The typical strip has center of@/<>3-+6
mass: ( x y ) x , length: 25 x , width: dx,
µµ
&?& Ί‹
25 x
#
#

area: dA 25 x dx, mass: dm dA 25 x dx.?Î ? ? Îab ab
##
$$
The moment of the strip about the x-axis is
y dm 25 x dx 25 x dx. The moment of the plate about the x-axis is M y dm
µ µ
?? ?Š‹ ab ab
25 x
##
##
#

$
$
x
'
25 x dx 625 50x x dx 625x x 2 625 5 5?Î? Î?Î? Î''
55
55
$$ $ $
## # #
##%$ $
#
&
&
ab a b ’“Š ‹
50 x 50 5
35 3 5
""
†††
625 5 1 625 . The mass of the plate is M dm 25 x dx 25x?? ???$$ $$†††ˆ‰ ˆ‰ ab ’“
10 8 x
33 3 ' '
5
5
#
&
&
 
2 5 5 . Therefore y 10. The plate's center of mass is the point (x y) ( 10).?Î? ?? ? &?'&$$Š‹
$$54
33 M
M
 
†
x
$
$
††
††
5
5
Î
 ˆ‰
ˆ‰
8
3
4
3
15. Intersection points: x x x 2x x 0Î?Î? Î?
##
x(2 x) 0 x 0 or x 2. The typical Ê œ Ê œ œ @/<>3-+6
strip has center of mass: ( x y ) x
µµ
ߜߊ‹
abx x ( x)
#

x , length: x x ( x) 2x x , width: dx,?&Î Î ÎÎ?Ί‹ ab
x

#
##
area: dA 2x x dx, mass: dm dA 2x x dx.?Î ? ? Îab ab
##
$$
The moment of the strip about the x-axis is

y dm 2x x dx; about the y-axis it is x dm x 2x x dx. Thus, M y dm
µµµ
?Î Î ? Î ?Š‹ ab ab
x

#
##
$$ †
x
'
x 2x x dx 2x x dx 2 2 1?Î Î ?Î Î ?Î Î ?Î Î ?Î Î''
00
22
ˆ‰ ˆ‰ab a b ’“ Š‹
$$$$$
######
## $% $ $
#
!
xx 2 4
55 5
¸" "
†
; M x dm x 2x x dx 2x x x 2 ;?Î ? ? Î ? Î ? Î ? Î ? ?
µ
4 2x 22 24
5 34 3413
$ $$
y
00
22
' ''††$$ $$ab a b ’“Š ‹
##$$
#
!
#
¸ ¸¸
†
M dm 2x x dx 2x x dx x 4 . Therefore, x?? Î? Î?Î?Î? ?' ''
00
22
$$ $$ab ab ’“ ˆ‰
###
#
!
x84
333 M
M
 
$ y
1 and y (x y) 1 is the center of mass.???????ˆ‰ˆ‰ ˆ ‰ˆ‰ ˆ ‰
43 43 3 3
34 M 54 5 5
M$$
$$ x
16. Intersection points: x 3 2x 3x 3 0
###
Î?Î ? Î?
3(x 1)(x 1) 0 x 1 or x 1. Applying the??ʜ ?
symmetry argument analogous to the one in Exercise 13, we
find x 0. The typical strip has center of mass:œ @/<>3-+6
(x y ) x x ,
µµ
ßœß œßŠ‹Š‹

##
2x x 3 x 3

ab
length: 2x x 3 3 1 x , width: dx,ÎÎÎ? Î
## #
abab
area: dA 3 1 x dx, mass: dm dA 3 1 x dx.?Î ? ? Îab ab
##
$$
The moment of the strip about the x-axis is

y dm x 3 1 x dx x 3x x 3 dx x 2x 3 dx; M y dm
µ µ
œ œ œ œ
33 3
## #
## %## %#
$$ $abab a b a b
x
'
x 2x 3 dx 3x 2 3 3 ;œœœ œ œ
3 3 x 2x 3 2 3 10 45 32
53 53 15 5###
%#
"
"
"
$$$$'
1
1
ab ’“ ˆ‰ˆ‰
"  
††
$
M dm 3 1 x dx 3 x 3 2 1 4 . Therefore, y?? Î ?Î? Î? ??Î?Î' $$$$'
1
1
ab ’“ ˆ‰
#
"
"
"x 32 8
33 M545
M
 
†
x $
$
†
††
(x y) 0 is the center of mass.?? ˆ‰
8
5

391. Section 6.4 Moments and Centers of Mass 387
17. The typical strip has center of mass:29<3D98>+6
( x y ) y , length: y y , width: dy,
µµ
&? & Ί‹
yy
#
$
 
area: dA y y dy, mass: dm dA y y dy.?Î ? ? Îab ab
$$
$$
The moment of the strip about the y-axis is
x dm y y dy y y dy
µ
?Î?Î$Š‹ ab ab
yy
##
$$
#
 
$
y 2y y dy; the moment about the x-axis is?Î
$
#
#%'
ab

y dm y y y dy y y dy. Thus, M y dm y y dy ;
µµ
?Î ?Î ? ? Î ?Î?Î?$$ $ $$ab ab ab ’“ ˆ‰
$#% #%
"
!
""
x
0
1
' '
yy
35 35 15
2
  $
$
M x dm y 2y y dy ; M dm
y
0
1??Î?Î?Î? ??
µ' '
$$$$$
####
#%'
"
!
"" 'ab ’“ ˆ‰ˆ‰
y2yy
357 357 357 105
2 354215 4
  $&
††
y y dy . Therefore, x and y?Î?Î?Î? ?? ? ??$$$'
0
1
ab ’“ ˆ ‰ ˆ ‰ˆ‰ ˆ ‰ˆ‰
$
##
"
!
""yy
4 4 4 M 105 105 M 15

M 44 16 24 M
"
$$ $
$$ y x
(x y) is the center of mass.œÊßœ ß
8168
15 105 15
ˆ‰
18. Intersection points: y y y y 2y 0?Î? Î?
##
y(y 2) 0 y 0 or y 2. The typical?Î?ʜ ?
strip has center of mass:29<3D98>+6
(x y ) y y ,
µµ
ßœ ߜߊ‹Š‹
abyyy
2
y


#
length: y y y 2y y , width: dy,ÎÎ?Îab
##
area: dA 2y y dy, mass: dm dA 2y y dy.?Î ? ? Îab ab
##
$$
The moment about the y-axis is x dm y 2y y dy
µ
?Î
$
#
##
†ab

2y y dy; the moment about the x-axis is y dm y 2y y dy 2y y dy. Thus,?Î ?Î?Î
µ
$
#
$% ##$
ab abab $$
M y dm 2y y dy (4 3) ; M x dm
x y
0
2?? Î?Î?Î?Î??
µ µ'' '$$$ab ’“ ˆ‰
#$
#
!
#
2y y
34 34 1 3
16 16 16 4
  "
$$
2y y dy 8 ; M dm 2y y dy?Î?Î?Î????Î''
0 0
2 2
$$$$$
####
$% #
#
!

ab ab ’“ ˆ‰ˆ‰
yy
25 5 5 5
32 40 32 4
"$
' $
y 4 . Therefore, x and y 1?Î?Î? ?? ? ?? ?$$’“ ˆ ‰ ˆ‰ˆ‰ ˆ‰ˆ‰
#
#
!
y
3 3 3 M 54 5 M 34
84 43 3 43 M M
 
$$ $
$$ y x
(x y) is the center of mass.Êßœß" ˆ‰
3
5
19. Applying the symmetry argument analogous to the one used
in Exercise 13, we find x 0. The typical strip hasœ @/<>3-+6
center of mass: ( x y ) x , length: cos x, width: dx,
µµ
ߜ߈‰
cos x
#
area: dA cos x dx, mass: dm dA cos x dx. Theœœœ $$
moment of the strip about the x-axis is y dm cos x dx
µ
œ$††
cos x
#
cos x dx dx (1 cos 2x) dx; thus,?? ?
$$ $
###
#  
ˆ‰
1 cos 2x
4
M y dm (1 cos 2x) dx x 0 ; M dm cos x dx
x
2 2
2 2œœ œ œœœœ
µ'' ''
"
" "


$$ $11$1
1
144 4 4
sin 2x
‘ ‘ˆ‰ˆ‰
###
Î#
Î#
$
[sin x] 2 . Therefore, y (x y) is the center of mass.œœ œœœÊßœ!ß$$
1
1
$1 1 1
$
Î#
Î# #
M
M4 8 8
x
†
ˆ‰
20. Applying the symmetry argument analogous to the one used
in Exercise 13, we find x 0. The typical vertical strip hasœ
center of mass: ( x y ) x , length: sec x, width: dx,
µµ
ߜߊ‹
sec x

#
#
area: dA sec x dx, mass: dm dA sec x dx. Theœœœ
##
$$
moment about the x-axis is y dm sec x dx
µ
œŠ‹ ab
sec x

#
#
$
sec x dx. M y dm sec x dxœœœ
µ
$$
##
%%
x
44
44
''

""

392. 388 Chapter 6 Applications of Definite Integrals
tan x 1 sec x dx (tan x) sec x dx sec x dx [tan x]œœ œ
$$$$$
1
1
1
1### #
## ## #
Î
Î
Î%
Î%'''

"""

444
444
a bab ab ’“
23
(tan x)
4
4
 
[1 ( 1)] ; M dm sec x dx [tan x] [1 ( 1)] 2 .?ÎÎÎÎ?? ? ? ? ?ÎÎ?
$$$$ 1
123 3 3 3
4 4
4
‘ˆ‰
""
#
# Î
Î
$$$$$ ' '
"
"

4
4
Therefore, y (x y) is the center of mass.œœ œÊßœ!ß
M
M323 3
42 2xˆ‰ˆ‰ ˆ ‰
$
$
"
21. Since the plate is symmetric about the line x 1 and itsœ
density is constant, the distribution of mass is symmetric
about this line and the center of mass lies on it. This means
that x 1. The typical strip has center of mass:œ @/<>3-+6
(x y ) x x ,
µµ
ßœß œßŠ‹Š‹
aba b2x x 2x 4x x2x
##


length: 2x x 2x 4x 3x 6x 3 2x x ,aba b abÎÎ Î ?Î? Î
## # #
width: dx, area: dA 3 2x x dx, mass: dm dA?Î ?ab
#
$
3 2x x dx. The moment about the x-axis is?Î$ab
#
y dm x 2x 2x x dx x 2x dx
µ
?Î Î?ÎÎ
33
##
## #
#
$$abab ab

x 4x 4x dx. Thus, M y dm x 4x 4x dx x xœ œ œ œ
µ
333x4
2253#
%$# %$# % $
#
!
$$$ab ab ’“
x
0
2
' '
$
22 21 2 ; Mdmœ œ œ œ œ
3 2 4 3 2 2 3 6 15 10 8
25 3 5 3 15 5
$$$Š‹ ˆ‰ ˆ‰
$
%$ % %
##

†† †
$ '
3 2x x dx 3 x 3 4 4 . Therefore, y?Î?Î?Î? ??Î?Î'
0
2
$$$$ab ’“ ˆ‰ ˆ‰ˆ‰
##
#
!
"x8 82
33 M545
M
 
x $
$
(x y) 1 is the center of mass.?&?&Î ˆ‰
2
5
22. (a) Since the plate is symmetric about the line x y andœ
its density is constant, the distribution of mass is
symmetric about this line. This means that x y. Theœ
typical strip has center of mass:@/<>3-+6
( x y ) x , length: 9 x , width: dx,
µµ
&?& Ί‹
È
È
9x
#
#
area: dA 9 x dx,?Î
È
#
mass: dm dA 9 x dx.??Î$$
È
#
The moment about the x-axis is

y dm 9 x dx 9 x dx. Thus, M y dm 9 x dx 9x
µµ
?Î?Î ??Î?Î$Š‹ ’ “
È
ab ab
È
9x x
3

## ##
# ##
$
!  
$$$
x
0
3
' '
(27 9) 9 ; M dm dA dA (Area of a quarter of a circle of radius 3) .?Î????? ??
$ 11$
#
$$$$ $'' ' ˆ‰
99
44
Therefore, y (9 ) (x y) is the center of mass.œœ œÊßœß
M
M9
44 44x
$ˆ‰ ˆ ‰
1$ 1 1 1
(b) Applying the symmetry argument analogous to the one
used in Exercise 13, we find that x 0. The typicalœ
vertical strip has the same parameters as in part (a).
Thus, M y dm 9 x dx
x
3
3??Î
µ' '

$
#
#
ab
9 x dx 2(9 ) 18 ;? Î ? ?'
0
3
$
#
#
ab $$
M dm dA dAœœ œ
'' ' $$

(Area of a semi-circle of radius 3) . Therefore, y (18 ) , the same yœœœœœœ$$ $ ˆ‰ ˆ ‰
99 24
22 M 9
M11$
1$ 1 x
as in part (a) (x y) 0 is the center of mass.Êßœß ˆ‰
4
1

393. Section 6.4 Moments and Centers of Mass 389
23. Since the plate is symmetric about the line x y and itsœ
density is constant, the distribution of mass is symmetric
about this line. This means that x y. The typical œ @/<>3-+6
strip has
center of mass: ( x y ) x ,
µµ
ߜߊ‹
39x
#
È

length: 3 9 x , width: dx,ÎÎ
È
#
area: dA 3 9 x dx,?ΠΊ‹
È
#
mass: dm dA 3 9 x dx.??$$ Š‹
È
#
The moment about the x-axis is

y dm dx 9 9 x dx dx. Thus, M dx x . The area
µ
??????$
Š‹Š‹
ÈÈ
39x39x
xx9
6

## # # #
#$
$
!


$$ $$$
cd cdab
x
0
3
'
equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 A 3ʜÎ
#19
4
(4 ) M A (4 ). Therefore, y (x y) is the?Îʜ? Î ?? ? ?&? &
99 94222
44 M9(4)444
M
1$ 1
$$
$1 1 1 1 xˆ‰ ˆ ‰’“
#
center of mass.
24. Applying the symmetry argument analogous to the one used
in Exercise 13, we find that y 0. The typical stripœ @/<>3-+6
has center of mass: ( x y ) x (x 0),
µµ
ßœß œßŒ ¸
˜˜
  
xx

#
length: , width: dx, area: dA dx,
""
xxx x
22
      ? ?ˆ‰
mass: dm dA dx. The moment about the y-axis isœœ$
2
x
$
 
x dm x dx dx. Thus, M x dm dx
µµ
œœ œœ†
22 2
xx x
$$ $
  y
1
a
' '
2 2 1 ; M dm dx 1 . Therefore,? Î ? Î? ? ? ?Î ?Î?$$ $$‘ˆ ‰ ‘ˆ ‰
"" ""
" "

xaa x xaa
2(a 1) 2 a1
a a
1
a $ $ $ ''
 

ab
x (x y) 0 . Also, lim x 2.œœ œ Êßœ ß œ
M
M a a 1 a1 a1
2(a 1) a2a 2ay
’“’ “ ˆ‰
$
$




ab aÄ_
25. M y dm dx
x
1
2œœ †
µ' '
Š‹
2
x


#
$†ˆ‰
2
x
x dx dx 2 x dxœœœ'''
111
222
ˆ‰ ˆ‰ab
" ##
xx x
22

2 x 2 ( 1) 2 1;?Î ? Î ÎÎ ? ?cd ‘ˆ‰ˆ‰
"
#
"
""
##
M x dm x dx
y
1
2œœ
µ' '††$ˆ‰
2
x

x x dx 2 x dx 2œœœ''
11
22
abˆ‰ ’“
#
#
#
"
2x
x



2 2 4 1 3; M dm dx x dx 2 dx 2[x] 2(2 1) 2. So? Î ?Î? ? ? ? ? ? ? Î ?ˆ ‰ ˆ‰ ˆ‰
"
#
##
" ''' '
11 1
22 2
$
22
xx

x and y (x y) is the center of mass.œœ œœÊßœß
M
MM
33 My x
####
""
ˆ‰
26. We use the strip approach:@/<>3-+6
M y dm x x dx
x
0
1?? Î
µ' '
abxx 
#
#

ab †$
x x 12x dx?
"
#
#%'
0
1
ab †
6 x x dx 6??'
0
1
ab ’“
$&
"
!
xx
46
¸!
61;???ˆ‰
"" "
#46 4
6

M x dm x x x dx x x 12x dx 12 x x dx 12 12
y
00 0
11 1??Î ?Î ? Î?Î?Î
µ' '' 'ab ab ab ’“ ˆ‰
##$ $%
"
!
""
††$
xx
45 45
¸"

394. 390 Chapter 6 Applications of Definite Integrals
; M dm x x dx 12 x x dx 12 12 1. So?? ? ? Î ? Î ? Î ? Î??
12 3 xx 12
05 34 34 12#
##$
"
!
"" ' 'ab a b ’“ ˆ‰
1
0
0
1
†$ '
  ¸
x and y is the center of mass.œœ œœÊß
M
M5 M 5
33 My x ""
##
ˆ‰
27. (a) We use the shell method: V 2 dx 2 x dx?? ''
a1
b4
11ˆ‰ Š‹ ’ “ Š‹
shell shell
radius height
44
xxÈÈ
16 dx 16 x dx 16 x 16 8 (8 1)?? ????11 11 ''
11
44
x 2 2 2 32 224
x 33333È
"Î# $Î#
%
"
‘ˆ‰ †
11
(b) Since the plate is symmetric about the x-axis and its density (x) is a function of x alone, the$œ
"
x
distribution of its mass is symmetric about the x-axis. This means that y 0. We use the vertical stripœ
approach to find x: M x dm x dx x dx 8 x dx
y
111
444?? ÎÎ ? ?
µ' '''††††’“Š‹
44 8
xx x xÈÈ È
$
" "Î#
8 2x 8(2 2 2) 16; M dm dx 8 dx 8 x dx????? ? ? ‘ˆ‰ ’“ Š‹Š‹
"Î# $Î#
%
"
""
†† ' '''
111
444
44
xx x xÈÈ È
$
8 2x 8[ 1 ( 2)] 8. So x 2 (x y) (2 0) is the center of mass.? Î ?ÎÎÎ ? ? ? ? ? & ?& ‘
"Î#
%
"
M
M8
16 y
(c)
28. (a) We use the disk method: V R (x) dx dx 4 x dx 4???? '' '
a1 1
b4 4
1111
## "
%
" ˆ‰ ‘
4
xx

4 ( 1) [ 1 4] 3???111 ‘
"
4
(b) We model the distribution of mass with vertical strips: M y dm dx x dx
x
11
44œœ œ
µ' ''
ˆ‰
2
x
2x x
22
†† †ˆ‰ È$

2 x dx 2 2[ 1 ( 2)] 2; M x dm x dx 2 x dx?????? ?
µ '''
1 11
4 44
y
$Î# "Î#
%
"
’“
22
x xÈ ' ††$
2 2 ; M dm dx 2 dx 2 x dx 2 2x????? ? ? ?’“ ‘ ‘
2x 16 2 28 2
3333 x x
x
 
%
"
"Î# "Î#
%
"
' '''
111
444
†$
È
2(4 2) 4. So x and y (x y) is the center of mass.?? ?? ? ?????
M
M43 M4 3
72 7 My
28
3 x
ˆ‰
""
##
ˆ‰
(c)
29. The mass of a horizontal strip is dm dA L dy, where L is the width of the triangle at a distance of y aboveœœ$$
its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have
L
bh
hy
œ

L (h y). Thus, M y dm y (h y) dy hy y dyʜ Î ? ? Î ? Î ? Î
µ
bbbb
hhhh3
hy y
x
00
hh h
' ''$ˆ‰ ab ’“
$$ #
#
!
 
bh ; M dm (h y) dy h y dy hy?Î? Î? ?? Î? Î?Î
$$ $$bh h bh b b b
h3 36 h h h2
y
Š‹ ’“ˆ‰ ˆ‰ ab
  
##
#""
!
$$ ' ''
00
hh h
h . So y the center of mass lies above the base of the?? ?? ??
$$ $
$
b h bh bh 2 h
h22 M6bh3
M
Š‹ Š‹ ˆ‰
#

x

395. Section 6.4 Moments and Centers of Mass 391
triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be
placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the
medians, as claimed.
30. From the symmetry about the y-axis it follows that x 0.œ
It also follows that the line through the points ( ) and!ß !
( ) is a median y (3 0) 1 (x y) ( ).'& ? ? Î ? ? & ? '& $
"
3

31. From the symmetry about the line x y it follows thatœ
x y. It also follows that the line through the points ( )œ!ß!
and is a median y x 0ˆ‰ ˆ ‰
"" " "
## #
&ʜ?Î?
2
33
†
(x y) .Êßœß ˆ‰
""
33
32. From the symmetry about the line x y it follows thatœ
x y. It also follows that the line through the point ( )œ!ß!
and is a median y x 0 aˆ‰ ˆ ‰
aa 2 a
33## #
"
&ʜ?Î?
(x y) .Êßœß ˆ‰
aa
33
33. The point of intersection of the median from the vertex (0 b)ß
to the opposite side has coordinates ˆ‰!ß
a
#
y (b 0) and xʜÎ ? ? Î' ?††
"
#33 33
ba2a
ˆ‰
(x y) .Êßœß ˆ‰
ab
33

34. From the symmetry about the line x it follows thatœ
a
#
x . It also follows that the line through the pointsœ
a
#
and b is a median y (b 0)ˆ‰ ˆ‰
aa b
33##
"
&' & ? ? Î ?
(x y) .Êßœß ˆ‰
ab
3#

35. y x dy x dxœÊœ
"Î# "Î# "
#
ds (dx) (dy) 1 dx ;Êœ œ È É
## "
4x
Mx1 dx
x
0
2?$'ÈÉ
"
4x
x dx x??$ '
0
2
É ’“ˆ‰
""
$Î#
#
!
43 4
2$
2?Î
2
34 4
$
’“ˆ‰ˆ‰
""
$Î# $Î#

?Î?Î?
29 227 13
34 4 38 8 6
$$$
’“ˆ‰ ˆ‰ ˆ ‰
$Î#
""
$Î#

396. 392 Chapter 6 Applications of Definite Integrals
36. y x dy 3x dxœÊœ
$#
dx (dx) 3x dx 1 9x dx;Êœ œ É ab
È
## %
#
M x 1 9x dx;
x
0
1?$'$ %È
[u 1 9x du 36x dx du x dx;? ? ? ? ?
%$ $ "
36
x 0 u 1, x 1 u 10]œÊœ œÊœ
M u du u 10 1? ? ? Î
x
1
10$'""Î# $Î# $Î#
"!
"36 36 3 54
2$$
‘ˆ ‰
37. From Example 6 we have M a(a sin )(k sin ) d a k sin d (1 cos 2 ) d
x
000???Î'''

))) )) ))
##
#
ak

; M a(a cos )(k sin ) d a k sin cos d sin 0;?Î ? ? ? ? ?
ak sin 2 ak ak

## # #!
##
!
‘ cd))))))))
)1
1 1
y
00
''

M ak sin d ak[ cos ] 2ak. Therefore, x 0 and y ??Î? ??????'&'
0


)) )
 
M
MM22ak44
M ak a ay x
Š‹ˆ‰ ˆ ‰

111"
is the center of mass.
38. M y dm (a sin ) a d
x
0œœ
µ' '

)$ )††
a sin 1 k cos d?
'
0

aba b kk
#
)))
a (sin )(1 k cos ) d?
#'
0
2
"
)))
a (sin )(1 k cos ) d
#'


"2
)))

a sin d a k sin cos d a sin d a k sin cos dœ
## ##'' ''
00 22
22


""
""
)) ) )) )) ) ))
a [ cos ] ak a[ cos ] akœ
#### Î#
! ##
Î#
!Î#
Î#
))
1 ))
11
1
1
1
’“ ’“
sin sin

a[0(1)]ak 0 a[(1)0]ak0 a aœ œ
### ### ""
#### ˆ‰ ˆ‰
ak ak

2a ak a(2 k);œœ
## #
M x dm (a cos ) a d a cos 1 k cos d
y
00œœ œ
µ' ''

)$ ) ) ) )†† aba b kk
#
a (cos )(1 k cos ) d a (cos )(1 k cos ) d?Î
##''
02
2


"
"
))) )))
a cos dak da cos dak dœ
## ##   
##'''
022
2
2
0



"
"
""
)) ) )) )'ˆ‰ ˆ‰
1cos 2 1cos 2))
a [sin ] a [sin ]?Î
## Î#
! ## ##
Î#
!Î# Î#
)) ))
1 ))
11
1
1 1
ak sin 2 ak sin 2
‘ ‘
a (1 0) 0 ( 0) a (0 1) ( 0) 0 a a 0;œ ! œ œ
## ##
## # #
ak ak ak ak
44

‘ ‘ˆ‰ ˆ‰
1111
1
M a d a (1 k cos ) d a (1 k cos ) d a (1 k cos ) d?? ?Î
'' ' '
00 0 2
2


$ ) ) ) )) ))† kk
"
"
a[ k sin ] a[ k sin ] a k 0 a ( 0) kœ œ )))) 1
1 1
1
11Î#
! Î# ##
‘ ‘ˆ‰ ˆ‰
ak a k a 2ak a( 2k). So x 0 and yœ œ œ œ œ œ œ œ
a M
MMa(k)k
M a(2 k) a(2 k)11
11##  #  #
  
ˆ‰ 11
y x

0 is the center of mass.Ê߈‰
2a ka
k
 
 #1
39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc
length we have that the length of a particular segment is ds (dx) (dy) . This implies that?È
##
M y ds, M x ds and M ds. If is constant, then x and
xyœœ œ œœœ'' '$$ $$
M
M length
x ds x ds
dsy
''
'
y.œœ œ
M
M length
y ds y ds
dsx
''
'
40. Applying the symmetry argument analogous to the one used in Exercise 13, we find that x 0. The typicalœ

397. Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 393
vertical strip has center of mass: ( x y ) x , length: a , width: dx, area: dA a dx,
µµ
&?& Î ?ÎŒ " Š‹
a
24p 4p
xx
 
x
4p


mass: dm dA a dx. Thus, M y dm a a dxœœ œ œ
µ
$$ $Š‹ Š‹Š‹
xxx
4p 4p 4p

x
2pa
2pa
' '


"
#
a dx a x 2 a x 2a pa?Î?Î?Î?Î
$$$
###
## # #
#'

2pa
2pa pa 2 pa
2pa 0



Š‹’“ ’“Š ‹ È
xx x
16p 80p 80p 80p
2pa pa
"$ $


È
È
† $
2a pa 1 2a pa 2a pa ; M dm a dx?Î? ? ???Î
### "
$$$ $ÈÈȈ‰ ˆ‰ ˆ‰ 'Š‹
16 80 6 64 x
80 80 80 5 4p
8a pa


$È
'
2pa
2pa
ax 2 ax 2 2a pa 4a pa 1 4a pa?Î ? Î ? Î ? Î?$$$ $$’“ ’“ Š ‹ ÈÈÈ ˆ‰ ˆ‰
xx 4124
12p 12p 12p 12 1
2pa pa
  
 
2pa
2pa
2pa


 
†
È

#
. So y a, as claimed.œœœ œ
8a pa 8a pa
3M5 5
M 33
8a pa
$$
$
ÈÈ
È
x
Š‹Š‹

41. Since the density is constant, its value will not affect our answers, so we can set .$œ"
A generalization of Example 6 yields M y dm a sin d a [ cos ]
x
2
2
2?? ?Î
µ' '
"



"
"!
"!

##
)) ) a cos cos a (sin sin ) 2a sin ; M dm a d a[ ]œ œ œ œ œ œ
###
## ‘ˆ‰ ˆ‰
11
!!!!! )) ' '
"



"
"!
"!

2
2
2
a 2a . Thus, y . Now s a(2 ) and a sin ?ÎÎ? ?? ? ? ? ‘ˆ‰ˆ‰
11 !!
!!## #
!!! ! !
M
M2a
2a sin a sin cx

c 2a sin . Then y , as claimed.Êœ œ œ !
a(2a sin )
2a s
ac!
!
42. (a) First, we note that y (distance from origin to AB) d a cos d d .?ʜʜ
a sin a(sin cos )!
!!
!! !
!

Moreover, h a a cos . The graphs below suggest that?Î ? ? ?!
dsin cos
h a( cos ) cos
a(sin cos )!! !
!! ! !! !
!! !


lim .
!Ä!
!
sin cos 2
cos 3
!! !
!! !


¸

(b) 0.2 0.4 0.6 0.8 1.0
f( ) 0.666222 0.664879 0.662615 0.659389 0.655145
!
!
6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS
1. (a) sec x sec x
dy dy
dx dx
œÊ œ
#%
#
Š‹
S 2 (tan x) 1 sec x dxʜ 1 '
0
4
"
È %
(c) S 3.84¸
(b)

398. 394 Chapter 6 Applications of Definite Integrals
2. (a) 2x 4x
dy dy
dx dx
œÊ œ Š‹
#
S 2 x 1 4x dxʜ 1 '
0
2
# #È
(c) S 53.23¸
(b)
3. (a) xy 1 x ?ʜ? ?Î ? ?
"" "
#
ydy y dy
dx dx
y
¸ Š‹
S 2 1 y dyʜ 1 '
1
2
" %
yÈ
(c) S 5.02¸
(b)
4. (a) cos y cos y
dx dx
dy dy
œÊ œ Š‹
#
#
S 2 (sin y) 1 cos y dyʜ 1 '
0

È #
(c) S 14.42¸
(b)
5. (a) x y 3 y 3 x
"Î# "Î# "Î#
#
?ʜÎ ˆ‰
23 x xʜ Î Î
dy
dx
ˆ‰ˆ ‰
"Î# "Î#"
#
13xʜΊ‹ ˆ‰
dy
dx
#
"Î#
#
S 2 3 x 1 1 3x dxʜ 1 '
1
4
ˆ‰ É ab
"Î#
#
"Î#
#
(c) S 63.37¸
(b)

399. Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 395
6. (a) 1 y 1 y
dx dx
dy dy
œ Ê œ
"Î# "Î#
#
#
Š‹ ˆ‰
S2 y2y 1 1y dxʜ 1 '
1
2
ˆ‰ÈÉ ab "Î#
#
(c) S 51.33¸
(b)
7. (a) tan y tan y
dx dx
dy dy
œÊ œ Š‹
#
#
S 2 tan t dt 1 tan y dyʜ 1 ''
00
3y

Š‹ È
#
2 tan t dt sec y dyœ1 ''
00
3y

Š‹
(c) S 2.08¸
(b)
8. (a) x 1 x 1
dy dy
dx dx
?? ?
È
Š‹
#
#
#
S 2 t 1 dt 1 x 1 dxʜ 1 ''
11
5x

Š‹
È È ab
##
2 t 1 dt x dx?1 ''
11
5x

Š‹
È
#
(c) S 8.55¸
(b)
9. y ; S 2 y 1 dx S 2 1 dx x dxœÊ œ œ Êœ œ
x xdy dy
dx dx 4
5
## # #
" "
# '''
a00
b44
11Ê Š‹ ˆ‰É
1
È
4 5; Geometry formula: base circumference 2 (2), slant height 4 2 2 5?? ? ??
1
È
5x
##
%
!
##
’“
È ÈÈ

11
Lateral surface area (4 ) 2 5 4 5 in agreement with the integral valueÊœœ
"
#
11Š‹
ÈÈ
10. y x 2y 2; S 2 x 1 dy 2 2y 1 2 dy 4 5 y dy 2 5 yœÊœ Ê œ œ œ œ œ
xdx dx
dy dy#
#
# #
#
! '' '
c0 0
d2 2
1111Ê Š‹
È ÈÈ cd†
2 5 4 8 5; Geometry formula: base circumference 2 (4), slant height 4 2 2 5?? ? ??11 1ÈÈ ÈÈ
†
##
Lateral surface area (8 ) 2 5 8 5 in agreement with the integral valueÊœœ
"
#
11Š‹
ÈÈ
11. ; S 2 y 1 dx 2 1 dx (x 1) dx x
dy dy (x 1)
dx dx
55 x
œœ œ œ œ
""
######
# $
Ÿ #
"'' '
a1 1
b3 3
11Ê Š‹ ’ “ É ˆ‰
11
ÈÈ

3 1 (4 2) 3 5; Geometry formula: r 1, r 2,œ œ œ œœ œœ
11
ÈÈ
559 3
# # # # ## ##
" "" "
"#
‘ˆ‰ˆ‰ È
1
slant height (2 1) (3 1) 5 Frustum surface area (r r ) slant height (1 2) 5??? ? ?È ÈÈ
##
"#
11
3 5 in agreement with the integral valueœ1
È

400. 396 Chapter 6 Applications of Definite Integrals
12. y x 2y 1 2; S 2 x 1 dy 2 (2y 1) 1 4 dy 2 5 (2y 1) dyœ Ê œ Ê œ œ œ œ
xdx dx
dy dy##
"
# '' '
c1 1
d2 2
111Ê Š‹ È È
2 5 y y 2 5 [(4 2) (1 1)] 4 5; Geometry formula: r 1, r 3,? Î ? ÎÎÎ ? ? ?11 1
ÈÈ È
cd
#
#
" "#
slant height (2 1) (3 1) 5 Frustum surface area (1 3) 5 4 5 in agreement with??? ? ?È ÈÈÈ
##
11
the integral value
13. S 1 dx;
dy dy
dx 3 dx 9 9 9
xx2xx
œÊ œÊœ
%Î ¸
Š‹ É
#
'
0
2
1
u 1 du x dx du dx;’? ? ? ? ?
x4 x
9949
%Î
$ "
x 0 u 1, x 2 u ‘œÊœ œÊ œ
25
9
S 2 u du uÄœ œ1 '
1
25 9

"Î# $Î#"
#
#&Î*
"
†
43
21
‘
1?? ?
11 1
3 27 3 27 81
125 125 27 98
ˆ‰ˆ‰

14. x
dy dy
dx dx 4x
œÊœ
""
#
"Î#
#
Š‹
S 2 x 1 dxʜ '
34
15 4

1ÈÉ
"
4x
2 x dx 2 x??11 '
34
15 4

É ’“ˆ‰
""
$Î#
"&Î%
$Î%
434
2
1?Î? Î
415 3 44
344 44 32
11
’“’“ˆ‰ˆ‰ ˆ‰
""
$Î# $Î# $
(8 1)??
428
33
11

15.
dy (2 2x) dy (1 x)
dx dx 2x x
2x x 2x x
1x
œœÊœ
"
#


#
ÈÈ



Š‹
S 2 2x x 1 dxʜ '
05
15¸
¸
1
È É
#

(1 x)
2x x


2 2x x dx?1 '
05
15¸
¸
È #

È
È
2x x 1 2x x
2x x


2dx2[x]2œœœ111'
05
15¸
¸
"Þ&
!Þ&

16.
dy dy
dx dx 4(x 1)2x 1
œÊœ
""
Ÿ
#
ŸÈ
Š‹
S 2 x 1 1 dxʜ '
1
5
1È
É
"
Ÿ4(x 1)
2 (x 1) dx 2 x dx??11 ''
11
55
ÉÉ
"
44
5
2x 5 1??1’“’ “ˆ‰ ˆ‰ˆ‰
25 4 5 5
34 3 4 4
$Î# $Î# $Î#
&
"
1
??
425 9 453
34 4 322
11
’“Š‹ˆ‰ ˆ‰
$Î# $Î# ÎÎ
ÎÎ
(125 27)???
111
663
98 49

17. y y S 1 y dy;
dx dx
dy dy 3
2y
œÊ œÊœ
#%
#
%
Š‹ È '
0
1
1
Î
u 1 y du 4y dy du y dy; y 0? ? ? ? ? ?
%$ $ "
4
u 1, y 1 u 2 S 2 u dud ˆ‰ ˆ ‰Êœ œÊœ Äœ '
1
2
1
"""Î#
34
u du u 8 1???
111
6639
2'
1
2
"Î# $Î#
#
" ‘ Š‹
È

401. Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 397
18. x y y 0, when 1 y 3. To get positiveœŸ ŸŸˆ‰
"$Î# "Î#
3
area, we take x y y?ΠΈ‰
"$Î# "Î#
3
y y y2yʜÎ Î ? ? Î
dx dx
dy dy 4
""
#
"Î# "Î# "
#
ˆ‰ Š‹ ab
S 2 y y 1 y 2 y dyʜ '
1
3
1ˆ‰ É ab
""$Î# "Î# "
34
2 y y y2y dyœ 1 '
1
3
ˆ‰ Éab
""$Î# "Î# "
34

2 y y dy y y 1 y dy y 1 (y 1) dyœ œ œ 111 '''
111
333
ˆ ‰ ˆ‰ ˆ‰ Š‹
""""$Î# "Î# "Î# "Î#
Ÿ
#333
yy
y
Éab



y y 1 dy y 3 1 3 1œ œ œ œ 111 1'
1
3
ˆ‰ ‘ˆ ‰’“ ˆ‰ˆ‰
" "" ""#
$
"
3 3 9 3 9 3 93 93
2279 yy
Î
(1813)œ œ
11
99
16
19. S 2 2 4 y 1 dy 4 (4 y) 1 dy
dx dx
dy dy 4 y 4 y4y
?? ?ʜ Î ? Î
" " "

#
È
Š‹ È É È ''
00
15 4 15 4

11†
4 5y dy 4 (5y) 5 5 5?Î?ÎÎ?ÎÎÎ?ÎÎ11 '
0
15 4

È ‘ˆ‰ ˆ‰ ’“’“
281585
33434
$Î# $Î# $Î#
"&Î%
!
$Î# $Î#
11
55?Î? ?
88
38383
55 40555 35 511 1
Š‹Š‹
È
ÈÈÈÈ

20. S 2 2y 1 1 dy 2 (2y 1) 1 dy
dx dx
dy dy 2y 1 2y 12y 1
?ʜʜ Î ? Î
"" "

#
È
Š‹ È É È ''
58 58
11
11
2 2 y dy 2 2 y 1 1???Î?Î11 '
58
1
ÈÈ ‘ˆ‰ ’“Š‹"Î# $Î# $Î#
"
&Î)
$Î#
25
3383
42 42 55
88
11
ÈÈÈ
È
16 2 5 5??Î
42
312
82 2 5 5
82 2
1 1
ÈÈ È
È
Š‹Š ‹
È È
†
†

21. ds dx dy y 1 dy y 1 dy y dyœ œ œ œ È ÊŠ‹ Š‹Š‹ ÊÊ
## $ ' ' """""
#
##4y 16y 16y
Î &&
y dy y dy; S 2 y ds 2 y y dy 2 y y dyœ œ œ œ œ ÊŠ‹Š‹ Š‹ ˆ‰
$ "" " "
#
$$%#
4y 4y 4y 4
ÎÎ Î
'' '
11 1
22 2
11 1
2 y 2 2 (8 31 5)œ œ œ œ œ11 1’“ ‘ˆ ‰ˆ‰ˆ‰
y
54 58 54 58 40 20
32 31 2 253
*
""""""
#
"
11
†
22. y x 2 dy x x 2 dx ds 1 2x x dx S 2 x 1 2x x dxœ Êœ Êœ Êœ
"#
$Î#
##% #%
3
ab a b
È È È
1 '
0
2

2 x x 1 dx 2 x x 1 dx 2 x x dx 2 2 4??????111111'''
000
222

Éab ab ab ’“ ˆ‰#
# #$
#
#
!
xx 42
442

È
23. y a x a x ( 2x) ?Îʜ Î Î? ? ?
È
ab Š‹
## "
#
##
"Î#

#
dy dy
dx dx a x
xx
ax
È ab



S 2 a x 1 dx 2 a x x dx 2 a dx 2 a[x]Êœ œ œ œ1111 '''
ÎÎÎ
Îaaa
aaa
a
a
È
É Èab## ## #

x
ax


ab
2 a[a ( a)] (2 a)(2a) 4 a?ÎÎ? ?111
#
24. y x S 2 x 1 dx 2 x dxœÊœÊ œÊœ œ
rr r rr rhr
hdxhdxh h h hh
dy dy
Š‹ ÉÉ
#
Ÿ



11''
00
hh
x dx h r h r r h r????
2r h r 2r x 2r h
hh h h
111
É ÈÈÈ
’“ Š‹



Ÿ ## ## ##
##'
0
h h
0
1
25. y cos x sin x sin x S 2 (cos x) 1 sin x dxœÊœÊ œ Êœ
dy dy
dx dx
Š‹
È
#
# #
1'
Î


2
2

402. 398 Chapter 6 Applications of Definite Integrals
26. y 1 x 1 x x 1?Î ? ? Î Î ?Î ? ? ? Έ‰ ˆ‰ˆ ‰ Š‹
#Î$ #Î$ "Î$
$Î# "Î#
#

#
"dy dy
dx 3 dx
32 1x 1x
xxx
ˆ‰
$Î



S 2 2 1 x 1 1 dx 4 1 x x dxÊœ œ ''
0 0
1 1
11ˆ‰ ˆ‰ ˆ‰É
È
#Î$ #Î$
$Î# $Î#
" #Î$
x
$Î
4 1 x x dx; u 1 x du x dx du x dx; ? ?ʜ ??1 '
0
1
ˆ‰
#Î$ "Î$ #Î$ "Î$ "Î$
$Î#
23
32
x 0 u 1, x 1 u 0 S 4 u du 6 u 6 0d ˆ‰ ‘ˆ‰?ʜ ?ʜ $? Î ?Î ?Î Î ? 111 '
1
0
$Î# &Î#
#
!
"
32 212
555
1
27. The area of the surface of one wok is S 2 x 1 dy. Now, x y 16 x 16 yœ œÊœ '
c
d
1Ê Š‹ È
dx
dy
#
## # ##
; S 2 16 y 1 dy 2 16 y y dyÊœ Ê œ œ œ
dx dx
dy dy 16 y 16 y
yy y
16 y


#

## ## #
È



Š‹ È É Èab ''
ÎÎ
ÎÎ16 16
77
11
2 16 dy 32 9 288 904.78 cm . The enamel needed to cover one surface of one wok isœœœ¸111
'
Î
Î16
7
†
#
V S 0.5 mm S 0.05 cm (904.78)(0.05) cm 45.24 cm . For 5000 woks, we needœœ œ œ††
$$
5000 V 5000 45.24 cm (5)(45.24)L 226.2L 226.2 liters of each color are needed.††œœœÊ
$
28. y r x ; S 2 r x 1 dxœÊœ œ Ê œ œ
ÈÈ
Š‹ É
## ##"
#

#
dy
dx dy r x r x
2x x dx x x
rx rx
ÈÈ




1'
a
ah
‚
2 r x x dx 2 r dx 2 rh, which is independent of a.???111''
aa
ah ah
‚‚
Èab## #
29. y R x ; S 2 R x 1 dxœÊœ œ Ê œ œ
ÈÈ
Š‹ É
## ##"
#

#
dy
dx dy R x R x
2x x dx x x
Rx Rx
ÈÈ




1'
a
ah
‚
2 R x x dx 2 R dx 2 Rh???111''
aa
ah ah
‚‚
Èab## #
30. (a) x y 45 x 45 y ;
## # ##

#

? ʜ Î ? ? ? ?È Š‹
dx dx
dy dy 45 y
yy
45 yÈ



S 2 45 y 1 dy 2 45 y y dy 2 45 dy?Î? Î?'''
Î%Î%Î%22 5 22 5 22 5
45 45 45
111È É Èab
## ## #

y
45 y


†
(2 )(45)(67.5) 6075 square feetœœ11
(b) 19,085 square feet
31. y x 1 1 S 2 x 1 1 dx 2 ( x) 2 dx 2 x 2 dxœÊ œÊ œÊœ œ Š‹ Š‹ kkÈ ÈÈ
dy dy
dx dx
#
111'''
ÎÎ110
202
22 22 22 0 22(2 0) 52œ œ œ
ÈÈÈÈÈ
’“ ’“ ˆ‰11111
xx

## #
!#
" !
"
32. by symmetry of the graph that S 2 2 1 dx; u 1
dy dy
dx 3 dx 9 99 9
xx xx x
œÊ œÊ œ œ
¸ Î% ¸
Š‹ Š ‹ ’ É
#
'
Î
3
0
1
du x dx du dx; x 3 u 2, x 0 u 1 S 4 u du “
È ˆ‰? ? ? Î ?Î ?Î ? ? ? ? ? $ ? Î
4x
949 4
$ "Î#" "
Î
1'
2
1
u du u 8 8 1 . If the absolute value bars are dropped the?Î ?Î ?Î Î ? Î111 '
2
1
"Î# $Î#
"
# ‘ Š‹Š‹
ÈÈ
2222
3333
1
integral for S 2 f(x) ds will equal zero since 2 1 dx is the integral of an odd function? ''
ÎÎ

33
33
11 Š‹É
xx
99
Î ¸
over the symmetric interval 3 x 3.ŸŸ
ÈÈ
33. sin t and cos t ( sin t) (cos t) 1 S 2 y ds
dx dx
dt dt dt dt
dy dy
œ œ Ê œ œ Ê œÊˆ‰ Š‹ È
#
#
##
'1
2 (2 sin t)(1) dt 2 2t cos t 2 [(4 1) (0 1)] 8œ œ œ œ
'
0
2

1111 1 cd
#
!
#1
34. t and t t t S 2 x ds
dx dx t 1
dt dt dt dt t
dy dy
?????ʜ
"Î# "Î#
#
#
" Ÿ
ʈ‰ Š‹
È É

'1

403. Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 399
2 t dt t t 1 dt; u t 1 du 2t dt; t 0 u 1,???ʜ?ʜ''
00
33

1ˆ‰ É È
c
2t 4
3t3
$Î# # Ÿ" #

1
t 3 u 4 u du u’“
È È ‘œÊœÄ œ œ '
1
4
2428
399
111 $Î#
%
"
Note: 2 t dt is an improper integral but lim f(t) exists and is equal to 0, where'
0
3

1ˆ‰ É
2t1
3t
$Î# Ÿ

tÄ!
‚
f(t) 2 t . Thus the discontinuity is removable: define F(t) f(t) for t 0 and F(0) 0???1ˆ‰ É
2t
3t
$Î# Ÿ"

F(t) dt .ʜ'
0
3

28
9
1
35. 1 and t 2 1 t 2 t 2 2 t 3 S 2 x ds
dx dx
dt dt dt dt
dy dy
?????ʜ
ÈÈÈ
Êʈ‰ Š‹ Š ‹ É
#
##
##
'1
2 t 2 t 2 2 t 3 dt; u t 2 2 t 3 du 2t 2 2 dt; t 2 u 1,??ʜ?ʜ'
Î

2
2
1Š‹ Š ‹
ÈÈ È È ÈÉ ’
# #
t 2 u 9 u du u (27 1)?ʜ$ ? ? Î?
È
“ È ‘ '
1
9
11
22 52
33 3
$Î#
*
"
11
36. a 1 cos t and a sin t a sin t a1 cos t
dx dx
dt dt dt dt
dy dy
œ œ Ê œ Îab c dab ʈ‰ Š‹ Éab
#
#
# #
a 2 a cos t a cos t a sin t 2a 2a cos t a 2 1 cos t S 2 y dsœ œ œ Êœ
È È ÈÈ
22 2222 22 '1
2 a 1 cos t a 2 1 cos t dt 2 2 a 1 cos t dt?Î*Î? Î''
0 0
2 2

11ab ab
ÈÈÈ
2 3/2
37. 2 and 1 2 1 5 S 2 y ds 2 (t 1) 5 dt
dx dx
dt dt dt dt
dy dy
?????ʜ ? ʈ‰ Š‹
È ÈÈ
#
#
##
'11'
0
1
2 5 t 3 5. Check: slant height is 5 Area is (1 2) 5 3 5 .œœ Ê œ11 11
ÈÈ È ÈÈ
’“
t
2

"
!
38. h and r h r S 2 y ds 2 rt h r dt
dx dx
dt dt dt dt
dy dy
œœÊœÊœ œ ʈ‰ Š‹
ÈÈ
#
#
## ##
'11'
0
1
2 r h r t dt 2 r h r r h r . Check: slant height is h r Area isœ œœ Ê111
ÈÈÈ È
’“
## ## ## ##
"
!'
0
1
t
2

rh r.1
È
##

39. (a) An equation of the tangent line segment is
(see figure) y f(m ) f (m )(x m ).? Î
kkk
w
When x x we haveœ
k1Î
r f(m ) f (m )(x m )"
w? Î
kk1k $Î
f(m) f(m) f(m) f(m) ;? Î?Î
kk kk
ww
##ˆ‰
??xx
kk
when x x we haveœ
k
r f(m ) f (m )(x m )#
w? Î
kk k Ä
f(m) f(m) ;?
kk
w
#
?x k
(b) L ( x ) (r r )
k
## #
#"
??
k
( x ) f (m ) f (m ) ? ÎÎ?
kk k
#w w
##
# ‘ˆ‰
??xx
kk
( x ) [f (m ) x ] L ( x ) [f (m ) x ] , as claimedœ Êœ ?? ??
kkk kkk
#w # #w #
k È
(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent
line segment about the x-axis is given by S (r r )L [2f(m )] x [f (m ) x ]?1 1 ? ?
kkkkkœ œ "#
#
w#
k
Éab
using parts (a) and (b) above. Thus, S 2 f(m ) 1 [f (m )] x .?1 ?
kk kk? È
w#
(d) S lim S lim 2 f(m ) 1 [f (m )] x 2 f(x) 1 [f (x)] dxœœ œ
nnÄ_ Ä_
!! ÈÈ
nn
k1 k1
kkkk
a
b
ŸŸ
?1 ?1
w# w# '

404. 400 Chapter 6 Applications of Definite Integrals
40. S 2 f(x) dx 2 dx x 3œœ œœœ''
a0
b3
3
11 1


†
x3
33 3
ÈÈ È
11
cd
È
#
&
41. The centroid of the square is located at ( ). The volume is V (2 ) y (A) (2 )(2)(8) 32 and the#ß # œ œ œ 111ab
surface area is S (2 ) y (L) (2 )(2) 4 8 32 2 (where 8 is the length of a side).œœ œ11 1ab Š‹
ÈÈ È
42. The midpoint of the hypotenuse of the triangle is 3ˆ‰
3
#
ß
y 2x is an equation of the median the lineÊœ Ê
y 2x contains the centroid. The point isœß$ ˆ‰
3
#
units from the origin the x-coordinate of the
35
È
#
Ê
centroid solves the equation x (2x 3)Ɉ‰
3
#
#
#
x 3x 4x 12x 9??Î Î?
È
5 95
44#
##
ˆ‰ ab
5x 15x 9 1??
#

x 3x 2 (x 2)(x 1) 0 x 1 since the centroid must lie inside the triangle y 2. By the?Î?Î Î?ʜ ʜ
#
Theorem of Pappus, the volume is V (distance traveled by the centroid)(area of the region) 2 5 x (3)(6)?? 1ab ‘
"
#
(2 )(4)(9) 72œœ11
43. The centroid is located at ( ) V (2 ) x (A) (2 )(2)( ) 4#ß ! Ê œ œ œ1111ab
#
44. We create the cone by revolving the triangle with vertices
(0 0), (h r) and (h 0) about the y-axis (see the accompanyingßß ß
figure). Thus, the cone has height h and base radius r. By
Theorem of Pappus, the lateral surface area swept out by the
hypotenuse L is given by S 2 yL 2 h rœœ 11 ˆ‰È
r
#
##
r r h . To calculate the volume we need the position?1
È
##
of the centroid of the triangle. From the diagram we see that
the centroid lies on the line y x. The x-coordinate of the centroid solves the equation (x h) x?
r rr
2h 2h
É ˆ‰ #
#
#
h x x 0 x or x , since the centroid must lieœÊ œÊœ Êœ
"ŸŸ# # Ÿ
344h 2h49 33 3
r 4h r 4h r r 2h 4h 2h 2r 4h
É Š‹Š‹


ab
inside the triangle y x . By the Theorem of Pappus, V 2 hr r h.Êœ œ œ œ
rr r
2h 3 3 3
‘ˆ ‰ˆ‰11
""
#
#
45. S 2 y L 4 a 2 y ( a) y , and by symmetry x 0œÊœ Êœ œ1111
#
ab
2a
1
46. S 2 L 2 a ( a) 2 a ( 2)??Î ? Î13 1 1 1 1 ‘ˆ‰
2a
1
#
47. V 2 yA ab 2 y y and by symmetry x 0œÊ œ Êœ œ111
4ab4b
33
#
#
abˆ‰
1
1
48. V 2 A V 2 a?ʜ ?13 1 ‘ˆ‰ Š‹
4a a
33
a(3 4)
1
1 11
Î
#
Ÿ
49. V 2 A (2 )(area of the region) (distance from the centroid to the line y x a). We must find the?? ?13 1 †
distance from 0 to y x a. The line containing the centroid and perpendicular to y x a has slopeˆ‰? ?
4a
31
1 and contains the point . This line is y x . The intersection of y x a and y x is? ?? ˆ‰
4a 4a 4a
33 311 1
the point . Thus, the distance from the centroid to the line y x a isˆ‰
4a 3a 4a 3a
66
11
11
?
V(2)Ɉ‰ˆ ‰ Š‹Š‹
4a 3a 4a 4a 3a a
6366 6 6 6
2 (4a 3a ) 2 (4a 3a ) 2 a (4 3 )Ÿ
## ŸŸŸ
#
11 1
1111 1 1
1111
? ʜ ?
ÈÈÈ
1
Î

405. Section 6.6 Work 401
50. The line perpendicular to y x a and passing through the centroid has equation y x . Theœ !ß œ ˆ‰
2a 2a
11
intersection of the two perpendicular lines occurs when x a x x y . Thus? ʜ ʜ
2a 2a a 2a a
2211 1
11
the distance from the centroid to the line y x a is 0 .œ œÉˆ‰ˆ ‰
2a a 2a a 2a
22
a(2 )
2

##
#
Ÿ11 1
1
È
Therefore, by the Theorem of Pappus the surface area is S 2 ( a) 2 a (2 ).??1111’“
È
a(2 )
2
Ÿ #1
1
È
51. From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is M y M
xœ
.œœˆ‰Š‹
4a a 2a
331
1
Î
#
6.6 WORK
1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) kx. Theœ
work done by F is W F(x) dx k x dx x . This work is equal to 1800 J k 1800œœœœ Êœ ''
00
33
k9k 9
## #
#
$
!
cd
k 400 N/mʜ
2. (a) We find the force constant from Hooke's Law: F kx k k 200 lb/in.œÊœÊœœ
F 800
x4
(b) The work done to stretch the spring 2 inches beyond its natural length is W kx dxœ '
0
2
200 x dx 200 200(2 0) 400 in lb 33.3 ft lb????? '
0
2
’“
x

#
#
!
††
(c) We substitute F 1600 into the equation F 200x to find 1600 200x x 8 in.œœœÊœ
3. We find the force constant from Hooke's law: F kx. A force of 2 N stretches the spring to 0.02 mœ
2 k (0.02) k 100 . The force of 4 N will stretch the rubber band y m, where F ky yÊœ Êœ œ Êœ†
N F
m k
y y 0.04 m 4 cm. The work done to stretch the rubber band 0.04 m is W kx dxÊœ Êœ œ œ
4N
100
N
m
'
0
004
¸
100 x dx 100 0.08 Jœœœœ '
0
004
¸
’“
x (100)(0.04)

##
!Þ!%
!
4. We find the force constant from Hooke's law: F kx k k k 90 . The work done toœÊœÊœÊœ
F90 N
x1 m
stretch the spring 5 m beyond its natural length is W kx dx 90 x dx 90 (90) 1125 Jœœ œœœ ''
00
55
’“ ˆ‰
x25

##
&
!
5. (a) We find the spring's constant from Hooke's law: F kx k k 7238 œÊœœ œ Êœ
Flb
x85 3 in
21,714 21,714

(b) The work done to compress the assembly the first half inch is W kx dx 7238 x dxœœ ''
00
05 05
¸¸
7238 (7238) 905 in lb. The work done to compress the assembly theœœœ¸’“
x (0.5) (7238)(0.25)

###
!Þ&
!
†
second half inch is: W kx dx 7238 x dx 7238 1 (0.5)?? ? ?? ''
05 05
10 10¸¸
¸¸
’“ cd
x 7238 (7238)(0.75)

## #
"Þ!
!Þ&
#
2714 in lb¸ †
6. First, we find the force constant from Hooke's law: F kx k 16 150 2,400 . If someoneœÊœœ œ œ
F 150 lb
xinˆ‰
‚
16
†
compresses the scale x in, he/she must weigh F kx 2,400 300 lb. The work done to compress theœœœœ
""
88
ˆ‰
scale this far is W kx dx 2400 18.75 lb in. ft lbœœ œœ œ '
0
18

’“
x 2400 25
264 16

#
"Î)
!
†
††
7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to
x, the length of the rope still hanging: F(x) 0.624x. The work done is: W F(x) dx 0.624x dxœœœ
''
00
50 50

406. 402 Chapter 6 Applications of Definite Integrals
0.624 780 Jœœ ’“
x

#
&!
!
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the
ground is F x x. The work done is: W F(x) dx x dx 144x 2x 1944 ft lbab a b c d? (†† Î † ? ? (†† Î † ? Î ?
''
a0
b18
#
")
!
†
9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) (4.5)(180 x) where x?
is the position of the car off the first floor. The work done is: W F(x) dx 4.5 (180 x) dx??
''
00
180 180
4.5 180x 4.5 180 72,900 ft lb????’“Š ‹
x 180 4.5 180

###
")!
!
# †
†
10. Since the force is acting the origin, it acts opposite to the positive x-direction. Thus F(x) . Thetoward ?
k
x

work done is W dx k dx k k?Î ? Î ? ? Î? ''
aa
bb
b
a
k
xxxbaab
k(a b)

""""
‘ˆ‰
11. The force against the piston is F pA. If V Ax, where x is the height of the cylinder, then dV A dxœœ œ
Work F dx pA dx p dV.Êœ œ œ
'' '
%Î
%ÎpV
pV
¸¸†
†
12. pV c, a constant p cV . If V 243 in and p 50 lb/in , then c (50)(243) 109,350 lb.
"Þ% "Þ% $ $ "Þ%
""
œÊœ œ œ œœ
Thus W 109,350V dV????'
243
32
"Þ%
$#
#%$
"" ""
## ‘ˆ‰ˆ‰
109,350 109,350 109,350
0.4V 3 43 0.4 0.4 4 9
#߸ #߸ #߸
37,968.75 in lb. Note that when a system is compressed, the work done by the system is negative.? ?
(109,350)(5)
(0.4)(36)
†
13. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant raœ te,
the amount of water in the bucket is proportional to x , the distance the bucket is being raised. The leakage rate ofabÞw Î
the water is 0.8 lb/ft raised and the weight of the water in the bucket is F 0.8 x . So:?ÞwÎab
W 0.8 x dx 0.8 20x 160 ft lb.?ÞwÎ? Î?*'
0
20
ab ’“
x

#
#!
!

14. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant raœ te,
the amount of water in the bucket is proportional to x , the distance the bucket is being raised. The leakage rate ofabÞw Î
the water is 2 lb/ft raised and the weight of the water in the bucket is F 2 x . So:?ÞwÎab
W 2 x dx 2 20x 400 ft lb.?ÞwÎ? Î? *'
0
20
ab ’“
x

#
#!
!
Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5
times as great.
15. We will use the coordinate system given.
(a) The typical slab between the planes at y and y y has?
a volume of V (10)(12) y 120 y ft . The force???œœ
$
F required to lift the slab is equal to its weight:
F 62.4 V 62.4 120 y lb. The distance throughœœ?? †
which F must act is about y ft, so the work done lifting
the slab is about W force distance?œ‚
62.4 120 y y ft lb. The work it takes to lift allœ ††† †?
the water is approximately W W¸!
20
0
?
62.4 120y y ft lb. This is a Riemann sum forœ!
20
0
†† †?

the function 62.4 120y over the interval 0 y 20. The work of pumping the tank empty is the limit of these sums:† ŸŸ

407. Section 6.6 Work 403
W 62.4 120y dy (62.4)(120) (62.4)(120) (62.4)(120)(200) 1,497,600 ft lbœœœœœ '
0
20
† †’“ ˆ‰
y 400

##
#!
!
(b) The time t it takes to empty the full tank with –hp motor is t 5990.4 secˆ‰
5W
11 250 250
1,497,600 ft lb
œœ œ
ft lb ft lb
sec sec
†
1.664 hr t 1 hr and 40 minœÊ¸
(c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is
W 62.4 120y dy (62.4)(120) (62.4)(120) 374,400 ft lb and the time is tœœ œœ œ'
0
10
†† ’“ ˆ‰
y 100 W
250

##
"!
!
ft lb
sec
1497.6 sec 0.416 hr 25 minœœ¸
(d) In a location where water weighs 62.26 :
lb
ft
Î
a) W (62.26)(24,000) 1,494,240 ft lb.œœ †
b) t 5976.96 sec 1.660 hr t 1 hr and 40 minœœ ¸ ʸ
1,494,240
250
In a location where water weighs 62.59
lb
ft
Î
a) W (62.59)(24,000) 1,502,160 ft lbœœ †
b) t 6008.64 sec 1.669 hr t 1 hr and 40.1 minœœ ¸ ʸ
1,502,160
250
16. We will use the coordinate system given.
(a) The typical slab between the planes at y and y y has?
a volume of V (20)(12) y 240 y ft . The force???œœ
$
F required to lift the slab is equal to its weight:
F 62.4 V 62.4 240 y lb. The distance throughœœ?? †
which F must act is about y ft, so the work done lifting
the slab is about W force distance?œ‚

62.4 240 y y ft lb. The work it takes to lift all the water is approximately W Wœ ¸††† †?? !
20
10
62.4 240y y ft lb. This is a Riemann sum for the function 62.4 240y over the intervalœ!
20
10
†† † †?
10 y 20. The work it takes to empty the cistern is the limit of these sums: W 62.4 240y dyŸŸ œ
'
10
20
†
(62.4)(240) (62.4)(240)(200 50) (62.4)(240)(150) 2,246,400 ft lb???? ’“
y

#
#!
"!
†
(b) t 8168.73 sec 2.27 hours 2 hr and 16.1 minœœ ¸ ¸ ¸
W
275
2,246,400 ft lb
275
ft lb
sec
†
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is
W 62.4 240y dy (62.4)(240) (62.4)(240) (62.4)(240) 936,000 ft.?? ???'
10
15
† ’“ ˆ‰ ˆ‰
y 225 100 125

####
"&
"!
Then the time is t 3403.64 sec 56.7 minœœ¸ ¸
W
275
936,000
75
ft lb
sec
#
(d) In a location where water weighs 62.26 :
lb
ft
Î
a) W (62.26)(240)(150) 2,241,360 ft lb.œœ †
b) t 8150.40 sec 2.264 hours 2 hr and 15.8 minœœ œ ¸
2,241,360
275
c) W (62.26)(240) 933,900 ft lb; t 3396 sec 0.94 hours 56.6 minœœœœ¸¸ ˆ‰
125 933,900
75##
†
In a location where water weighs 62.59
lb
ft
Î
a) W (62.59)(240)(150) 2,253,240 ft lb.œœ †
b) t 8193.60 sec 2.276 hours 2 hr and 16.56 minœœ œ ¸
2,253,240
275
c) W (62.59)(240) 938,850 ft lb; t 3414 sec 0.95 hours 56.9 minœœœ¸¸¸ ˆ‰
125 938,850
275#
†
17. The slab is a disk of area x , thickness y, and height below the top of the tank y . So the work to pump11
#
#
#
? ß (w Έ‰ ab
y
the oil in this slab, W, is 57 y . The work to pump all the oil to the top of the tank isß(wÎ ab ˆ‰1
y
#
#
W y y dy 11,875 ft lb 37,306 ft lb.?(wÎ? Î? *% *'
0
10
57 57
44
yy11
ab ’“
#$ "!
$%
"!
!
Î ¸
1

408. 404 Chapter 6 Applications of Definite Integrals
18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is y and since the tank isabab ˆ‰(† Î1
y
#
#
half full and the volume of the original cone is V r h ft , half the volume ft , andœœ&"!œ œ
" " #&! #&!
$$ $ '
##
11 abab
1133
with half the volume the cone is filled to a height y, y y ft. So W y y dy
#&! "
'$%
&!!
#$11
? ? ? ‚ww ? († Î1
y 57
4

Î
Î
È ab '
0
È
60,042 ft lb.?Î% *
57
4
yy1
’“
"%
$%
&!!
!
Î ¸
Î
È
19. The typical slab between the planes at y and and y y has a volume of V (radius) (thickness) y????1 1?
#
#
# ˆ‰
20
100 y ft . The force F required to lift the slab is equal to its weight: F 51.2 V 51.2 100 y lbœœœ1? ? 1?† †
$
F 5120 y lb. The distance through which F must act is about (30 y) ft. The work it takes to lift all theʜ Î 1?
kerosene is approximately W W 5120 (30 y) y ft lb which is a Riemann sum. The work to pump the%? Î!!
30 30
00
?1? †
tank dry is the limit of these sums: W 5120 (30 y) dy 5120 30y 5120 (5120)(450 )???? '
0
30
11 1 1 ’“ ˆ‰
y 900

##
$!
!
7,238,229.48 ft lb¸ †
20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires
all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3
additional feet. Thus pumping through the valve requires ft 6 ft lb/ft 14,115 ft lb less work and thusÈab a b$ % '#Þ% ¸1
33
†
less time.
21. (a) Follow all the steps of Example 5 but make the substitution of 64.5 for 57 . Then,
lb lb
ft ft
ÎÎ
W (10 y)y dy 8 2?Î?Î?Î? Î'
0
8
64.5 64.5 64.5 10 8 8 64.5 10
443443443
10y y1111 # $
)
!
’“ Š ‹ ˆ‰ ˆ ‰ab
Î ¸ Î%
†
21.5 8 34,582.65 ft lbœœ ¸
64.5 8
3
1†
Î
1††
$
(b) Exactly as done in Example 5 but change the distance through which F acts to distance (13 y) ft.%Î
Then W (13 y)y dy 8 2?Î?Î?Î? Î? '
0
8
57 57 57 13 8 8 57 13 57 8 7
44344344334
13y y1111 1 #$
)
!
’“Š ‹ ˆ‰ ˆ ‰ab
Î ¸ Î% Î
†††
†
(19 ) 8 (7)(2) 53.482.5 ft lbœ¸1ab
#
†
22. The typical slab between the planes of y and y y has a volume of about V (radius) (thickness)???1
#
y y xy y m . The force F(y) is equal to the slab's weight: F(y) 10,000 Vœœ œ1?? ?ˆ‰È
#
$ N
m
Ά
10,000y y N. The height of the tank is 4 16 m. The distance through which F(y) must act to liftœœ1?
#
the slab to the level of the top of the tank is about (16 y) m, so the work done lifting the slab is aboutÎ
W 10,000 y(16 y) y N m. The work done lifting all the slabs from y 0 to y 16 to the top is?1?? ?? †
approximately W 10,000 y(16 y) y. Taking the limit of these Riemann sums, we get%Î!
16
0
1?
W 10,000 y(16 y) dy 10,000 16y y dy 10,000 10,000????''
00
16 16
11 1 1 ab ’“ Š‹
#
##
"'
!
16y y
33
16 16
Î ÎÎ
21,446,605.9 Jœ¸
10,000 16
6
††1
Î
23. The typical slab between the planes at y and y y has a volume of about V (radius) (thickness)???1
#
25 y y m . The force F(y) required to lift this slab is equal to its weight: F(y) 9800 V? œ1? ?ˆ‰È
#
#
$
†
9800 25 y y 9800 25 y y N. The distance through which F(y) must act to lift the??1?1?ˆ‰È ab
#
#
#
slab to the level of 4 m above the top of the reservoir is about (4 y) m, so the work done is approximatelyÎ
W 9800 25 y (4 y) y N m. The work done lifting all the slabs from y 5 m to y 0 m is?1 ? ??ab
#
†
approximately W 9800 25 y (4 y) y N m. Taking the limit of these Riemann sums, we get%ÎÎ! ab
0
5
Î
1?
#
†

409. Section 6.6 Work 405
W 9800 25 y (4 y) dy 9800 100 25y 4y y dy 9800 100 y y?ÎÎ? ÎÎ?ÎÎ''
ÎÎ55
00
11 1ab a b ’“
##$#$
#
!
&
25 4
34
y
¸
9800 500 125 15,073,099.75 Jœ ¸1ˆ‰
25 25 4 625
34
†
#
†
24. The typical slab between the planes at y and y y has a volume of about V (radius) (thickness)???1
#
100 y y 100 y y ft . The force is F(y) V 56 100 y y lb. The?? ? ?1?1? ?1?ˆ‰È ab ab
#
#
#$ # 56 lb
ft
Ά
distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about
(12 y) ft, so the work done is W 56 100 y (12 y) y lb ft. The work done lifting all the slabsÎ%ÎÎ ?1 ? ab
#
†
from y 0 ft to y 10 ft is approximately W 56 100 y (12 y) y lb ft. Taking the limit of these?? % ÎÎ ! ab
10
0
1?
#
†
Riemann sums, we get W 56 100 y (12 y) dy 56 100 y (12 y) dy??
''
00
10 10
11ab ab
##
56 1200 100y 12y y dy 56 1200??11 '
0
10
ab ’“
#$
#
"!
!
100y 12y y
34
Î%
56 12,000 4 1000 (56 ) 12 5 4 (1000) 967,611 ft lb.œ œ ¸11ˆ‰ˆ‰
10,000 10,000
4
5
##
††
It would cost (0.5)(967,611) 483,805¢ $4838.05. Yes, you can afford to hire the firm.œœ
25. F m mv by the chain rule W mv dx m v dx m v (x)œœ Êœ œ œ
dv dv dv dv
dt dx dx dx ''
xx
xx
x
x
¸¸
¸

ˆ‰ ‘
"
#
#
m v (x ) v (x ) mv mv , as claimed.??
"""
###
## ##
#" # "
cd
26. weight 2 oz lb; mass slugs; W slugs (160 ft/sec) 50 ft lbœœ œ œœ œ ¸
2
16 32 3 56 56
weight
‚
8
## ##
""" #
ˆ‰ˆ ‰ †
27. 90 mph 132 ft/sec; m slugs;œœœœ
90 mi 1 hr 1 min 5280 ft 0.3125 lb 0.3125
1 hr 60 min 60 sec 1 mi 32 ft/sec 32
†††

W (132 ft/sec) 85.1 ft lbœ¸ˆ‰ˆ ‰
"
#
#0.3125 lb
32 ft/sec
†
28. weight 1.6 oz 0.1 lb m slugs; W slugs (280 ft/sec) 122.5 ft lbœœÊœ œ œ œ
0.1 lb
32 ft/sec 30 30

"""
###
#
ˆ‰ˆ ‰ †
29. weight 2 oz lb m slugs slugs; 124 mph 181.87 ft/sec;œœ Êœ œ œ ¸
""
#8 32 56 (60)(60)
(124)(5280)
‚
8
W slugs (181.87 ft/sec) 64.6 ft lbœ¸ˆ‰ˆ ‰
""
#
#
256
†
30. weight 14.5 oz lb m slugs; W slugs (88 ft/sec) 109.7 ft lbœœÊœ œ ¸
14.5 14.5 4.5
16 (16)(32) (16)(32)
ˆ‰Š‹
""
#
#
†
31. weight 6.5 oz lb m slugs; W slugs (132 ft/sec) 110.6 ft lbœœÊœ œ ¸
6.5 6.5 6.5
16 (16)(32) (16)(32)
ˆ‰Š‹
"
#
#
†
32. F (18 lb/ft)x W 18x dx 9x ft lb. Now W mv mv , where W ft lb,?ʜ ?? ? ? '
0
16

cd
###
"Î
!
""
## "
6 11
44
††
m slugs and v 0 ft/sec. Thus, ft lb. slugs v v 8 2 ft/sec. With v 0œœ œ œ Êœ œ
‚
8
32 56 4 56
1"""
###
"
#
† ˆ‰ˆ ‰ È
at the top of the bearing's path and v 8 2 32t t sec when the bearing is at the top of its path.?ʜ
È
È
2
4
The height the bearing reaches is s 8 2 t 16t at t the bearing reaches a height of?ʜ
È
#
È
2
4
8 2 (16) 2 ftŠ ‹Š‹ Š‹
È
ÈÈ
22
44
?
#

410. 406 Chapter 6 Applications of Definite Integrals
33. (a) From the diagram,
r y x y 325ab a b É?wÎ ?wÎ ‚w Î Î
#
#
for 325 y 375 ft.ŸŸ
(b) The volume of a horizontal slice of the funnel
is V r y y˜¸ ˜1‘ab
# y 325 y? w Î ‚w Î Î ß1”•
É ab
#
#
#
(c) The work required to lift the single slice of
water is W 62.4 V y˜ ¸ ˜ $(& ab
62.4 y y 325 y.? '‚ Î w Î ‚w Î Î ßab ab ”•
É1
#
#
#
The total work to pump our the funnel is W
62.4 375 y 50 y 325 dy? Î w Î Î Î'
325
375
ab ab ”•
É1
#
#
#
6.3358 10 ft lb.¸††
(

34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft lb. Thereform, the total work†
required to pump out the throat and the funnel is 1,353,869,354 63,358,000 1,417227,354 ft lb.? *
(b) In horsepower-hours, the work required to pump out the glory hole is 715.8. Therefore, it would take
1,417227,354
1.98 10†
6œ
0.7158 hours 43 minutes.
715.8 hp h
1000 hp
†
œ¸
35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a
partition of the interval [ ]. The typical slab between the planes at y and y y has a volume of about!ß ( ?
V (radius) (thickness) y in . The force F(y) required to lift this slab is equal to its?1 1 ?œœ
#$ Ÿ
# ˆ‰
y 17.5
14
weight: F(y) V y oz. The distance through which F(y) must act to lift this slab toœœ
44
9914
y 17.5
??
1
ˆ‰
Ÿ
#
the level of 1 inch above the top is about (8 y) in. The work done lifting the slab is aboutÎ
W (8 y) y in oz. The work done lifting all the slabs from y 0 to y 7 is??? ??ˆ‰
4
914
(y 17.5)1 Ÿ


†
approximately W (y 17.5) (8 y) y in oz which is a Riemann sum. The work is the limit of?!
7
0
4
914
1
†

#
? † these sums as the norm of the partition goes to zero: W (y 17.5) (8 y) dy? '
0
7
4
914
1
†

#
2450 26.25y 27y y dy 9y y 2450y??
4 4 26.25
914 914 4
y11
††

¸
'
0
7
ab ’“
#$ $ #
#
(
!
9 7 7 2450 7 91.32 in ozœ ¸
4 7 26.25
914 4
1
†

¸’“ †† † †
$#
#
36. We fill the pipe and the tank. To find the work required to fill the tank follow Example 6 with radius 10 ft. Thenœ
V 100 y ft . The force required will be F 62.4 V 62.4 100 y 6240 y lb. The distance through? 1 ? ? 1? 1?œœœœ†††
$
which F must act is y so the work done lifting the slab is about W 6240 y y lb ft. The work it takes to?1? "œ †† †
lift all the water into the tank is: W W 6240 y y lb ft. Taking the limit we end up with
""¸œ!!
385 385
360 360
?1? †† †
W 6240 y dy 6240 385 360 182,557,949 ft lb"
##
$)&
$'!
##???'
360
385
11 ’“ cd
y 6240

1
†
To find the work required to fill the pipe, do as above, but take the radius to be in ft.
4
6#
"
œ
Then V y ft and F 62.4 V y. Also take different limits of summation and?1? ? ?œœœ††
" $
36 36
62.41
integration: W W W y dy 352,864 ft lb.###
##
$'!
!¸Êœ œ œ ¸! ’“ Š ‹ˆ‰
360
0
0
360
?1 '62.4 62.4 62.4 360
36 36 36
y11

†
The total work is W W W 182,557,949 352,864 182,910,813 ft lb. The time it takes to fill theœ¸ ¸
"# †

411. Section 6.7 Fluid Pressures and Forces 407
tank and the pipe is Time 110,855 sec 31 hrœ¸ ¸ ¸
W
1650 1650
182,910,813
37. Work dr 1000 MG 1000 MG?? ? ''
6 370 000 6 370 000
35 780 000 35 780 000†† ††
†† ††
1000 MG dr
rrr

‘
"
$&ß()!ß!!!
'ß$(!ß!!!
(1000) 5.975 10 6.672 10 5.144 10 J?aba b Š‹††
#% "" "! ""
6,370,000 35,780,000
38. (a) Let be the x-coordinate of the second electron. Then r ( 1) W F( ) d3333
##
? ? ? '
Î1
0
d 23 10 1 11.5 10????'
Î1
0
ab23 10
(1)
23 10‚
" #
‚"
!
"
#* #*



33
3 ’“ ab ˆ‰
(b) W W W where W is the work done against the field of the first electron and W is the work done?
"# " #
against the field of the second electron. Let be the x-coordinate of the third electron. Then r ( 1)33
##
"
?
and r ( 1) W d d 23 10
## #*
# "
‚‚ "
" "
&
$
?ʜ??333 ''
33
55
23 10 23 10
r ()


‚

33
’“
23 10 10 , and W d d?Î ! Î ? ! ? ?ab ˆ‰
#* #*"" ‚ ‚
#Ÿ"
#
44 ()
23 23 10 23 10
r ''
33
55




33
3
23 10 23 10 (3 2) 10 . Therefore?Î ! ? Î ! Î ? Î ? !
#* #* #*"""‚
Ÿ"
&
$
’“ ab ˆ‰
3 6 4 12 12
23 10 23

W W W 10 10 10 7.67 10 Jœœ‚ ‚ œ‚ ¸ ‚"#
#* #* #* #*
ˆ‰ˆ‰
23 23 23
4123
6.7 FLUID PRESSURES AND FORCES
1. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's
right-hand edge: y x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then?
x 5 y and the total width is L(y) 2x 2(5 y). The depth of the strip is ( y). The force exerted by theœ œ œ
water against one side of the plate is therefore F w( y) L(y) dy 62.4 ( y) 2(5 y) dy?Î ? Î
''
ÎÎ
ÎÎ55
22
†††
124.8 5y y dy 124.8 y y 124.8 4 8 25 125???'
Î
Î5
2
ab ‘ ‘ˆ‰ˆ ‰
##$
###
"""
#
&
555
333
†† † †
(124.8) (124.8) 1684.8 lb?? ?ˆ‰ ˆ‰
105 117 315 234
36#

2. An equation for the line of the plate's right-hand edge is y x 3 x y 3. Thus the total width is? ? ?
L(y) 2x 2(y 3). The depth of the strip is (2 y). The force exerted by the water isœœ
F w(2 y)L(y) dy 62.4 (2 y) 2(3 y) dy 124.8 6 y y dy 124.8 6y?Î ? Î? ÎÎ? ÎÎ'' '
ÎÎ Î33 3
00 0
†† ab ’“
#
#
!
$
yy
3
Î
( 124.8) 18 9 ( 124.8) 1684.8 lbœ œ œˆ‰ˆ‰
927
##
3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is
y x 3 x y 3. Thus the total width is L(y) 2x 2(y 3). The depth of the strip changes to (4 y)œ Ê œ œ œ
F w(4 y)L(y) dy 62.4 (4 y) 2(y 3) dy 124.8 12 y y dyÊœ œ œ
'' '
ÎÎ Î33 3
00 0
†† ab
# 124.8 12y ( 124.8) 36 9 ( 124.8) 2808 lb????’“ ˆ‰ˆ‰
yy
3
945
Î
###
!
$
4. Using the coordinate system of Exercise 4, we see that the equation for the line of the plate's right-hand edge
remains the same: y x 3 x 3 y and L(y) 2x 2(y 3). The depth of the strip changes to ( y)œ Ê œ œ œ
F w( y)L(y) dy 62.4 ( y) 2(y 3) dy 124.8 y 3y dy 124.8 yÊœ œ œ œ '' '
ÎÎ Î33 3
00 0
†† ab ’“
##
#
!
$
y
3
3
Î
( 124.8) 561.6 lb?Î Î ? ?ˆ‰
27 27
36
( 124.8)(27)(2 3)
#

5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be
y 2x 4 x and L(y) 2x y 4. The depth of the strip is (1 y).?Îʜ ?? Î
y4Ÿ
#

412. 408 Chapter 6 Applications of Definite Integrals
(a) F w(1 y)L(y) dy 62.4 (1 y)(y 4) dy 62.4 4 3y y dy 62.4 4y?Î ? Î? ÎÎ? ÎÎ'' '
ÎÎ Î44 4
00 0
† ab ’“
#
#
!
%
3y y
3
Î
( 62.4) ( 4)(4) ( 62.4) 16 24 1164.8 lbœ œ œ œ’“ ˆ‰
(3)(16) ( 62.4)( 120 64)64 64
333#

(b) F ( 64.0) ( 4)(4) 1194.7 lbœ œ ¸’“
(3)(16) ( 64.0)( 120 64)64
33#

6. Using the coordinate system given, we find an equation for
the line of the plate's right-hand edge to be y 2x 4œ
x and L(y) 2x 4 y. The depth of theʜ ? ?
4y
#
strip is (1 y) F w(1 y)(4 y) dyÎʜ Î Î '
0
1
62.4 y 5y 4 dy 62.4 4y??'
0
1
ab ’“
#
#
"
!
y5y
3
Î

(62.4) 4 (62.4) 114.4 lb?? ??ˆ‰ˆ‰
"
#366
5 2 15 24 (62.4)(11)
7. Using the coordinate system given in the accompanying
figure, we see that the total width is L(y) 63 and the depthœ
of the strip is (33.5 y) F w(33.5 y)L(y) dyÎʜ Î
'
0
33 (33.5 y) 63 dy (63) (33.5 y) dy?Î? Î''
00
33 33
64 64
112#
ÎΆ† ˆ‰
(63) 33.5y (33.5)(33)?Î? Έ‰ ˆ ‰’“ ’ “
64 64 63 33
12 1
y
ÎÎ

## #
$$
!
†
1309 lbœœ
(64)(63)(33)(67 33)
()12

#ab
Î

8. (a) Use the coordinate system given in the accompanying
figure. The depth of the strip is y ftˆ‰
11
6
Î
F w y (width) dyʜ Î '
0
11 6

ˆ‰
11
6
(62.4)(width) y dy? '
0
11 6

ˆ‰
11
6
(62.4)(width) y? ’“
11
6
y

#
""Î'
!

(62.4)(width) F (62.4)(2) 209.73 lb and F (62.4)(4) 419.47 lbœ Êœ¸œ¸ ’“ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ‰
11 121 121
636 36
end side
#
"" "
## #
†
(b) Use the coordinate system given in the accompanying
figure. Find Y from the condition that the entire volume
of the water is conserved (no spilling): 2 4 2 2 Y
11
6
†† ††œ
Y ft. The depth of a typical strip is y ftʜ Î
11 11
33
ˆ‰
and the total width is L(y) 2 ft. Thus,œ
F w y L(y) dy? '
0
113
ˆ‰
11
3
(62.4) y 2 dy (62.4)(2) y (62.4)(2) 838.93 lb the fluid??? ??'
0
113
ˆ‰ ˆ‰ˆ‰ ’“ ’ “
11 11 11
33 39
y (62.4)(12 )
†

##
""Î$
!
"
# "
force doubles.
9. Using the coordinate system given in the accompanying
figure, we see that the right-hand edge is x 1 y?È
#
so the total width is L(y) 2x 2 1 y and the depth?? ÎÈ
#
of the strip is ( y). The force exerted by the water isÎ
therefore F w ( y) 2 1 y dy?Î Î '
Î1
0
†† È
#

413. Section 6.7 Fluid Pressures and Forces 409
62.4 1 y d 1 y 62.4 1 y (62.4) (1 0) 416 lb?ÎÎ?Î? Î?'
Î1
0
È ab ab ’“ ˆ‰ # ##
$Î#
!
"
22
33
10. Using the same coordinate system as in Exercise 15, the right-hand edge is x 3 y and the total width is?ÎÈ
##
L(y) 2x 2 9 y . The depth of the strip is ( y). The force exerted by the milk is therefore?? Î ÎÈ
#
F w ( y) 2 9 y dy 64.5 9 y d 9 y 64.5 9 y (64.5) (27 0)?ÎÎ? ÎÎ? Î ? Î''
ÎÎ33
00
†† ÈÈ ab ab ’“ ˆ‰
## ##
$Î#
!
$
22
33
(64.5)(18) 1161 lbœœ
11. The coordinate system is given in the text. The right-hand edge is x y and the total width is L(y) 2xœœÈ
2y.œÈ
(a) The depth of the strip is (2 y) so the force exerted by the liquid on the gate is F w(2 y)L(y) dyÎ?Î
'
0
1
50(2 y) 2 y dy 100 (2 y) y dy 100 2y y dy 100 y y?Î ? Î ? Î ? Î'''
000
111
†ÈÈ ˆ‰ ‘
"Î# $Î# $Î# &Î#
"
!
42
35
100 (20 6) 93.33 lb?Î? Î?ˆ‰ˆ‰
4 2 100
35 15
(b) We need to solve 160 w(H y) 2 y dy for h. 160 100 H 3 ft?Î ? Îʜ '
0
1
†È ˆ‰
2H 2
35
12. Use the coordinate system given in the accompanying figure. The total width is L(y) 1.œ
(a) The depth of the strip is (3 1) y (2 y) ft. The force exerted by the fluid in the window isÎÎ?Î
F w(2 y)L(y) dy 62.4 (2 y) 1 dy (62.4) 2y (62.4) 2 93.6 lb?Î ? Î ? Î? Î??''
00
11
† ’“ ˆ‰
y (62.4)(3)

###
"
!
"
(b) Suppose that H is the maximum height to which the
tank can be filled without exceeding its design
limitation. This means that the depth of a typical
strip is (H 1) y and the force isÎÎ
F w[(H 1) y]L(y) dy F , where?ÎÎ ?
'
0
1
max

F 312 lb. Thus, F w [(H 1) y] 1 dy (62.4) (H 1)y (62.4) H
max max
0
1??ÎÎ?ÎÎ?Î ' † ’“ ˆ‰
y 3

##
"
!
(2H 3) 93.6 62.4H. Then F 93.6 62.4H 312 93.6 62.4H H? Î?Î ?Î ? ?Î ʜˆ‰
62.4 405.6
62.4#
max
6.5 ftœ
13. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for
the line of the end plate's right-hand edge is y x x y. The total width is L(y) 2x y and theœÊœ œœ
52 4
55#
depth of the typical horizontal strip at level y is (h y). Then the force is F w(h y)L(y) dy F ,Î?Î? '
0
h
max
where F 6667 lb. Hence, F w (h y) y dy (62.4) hy y dy
max
0
h h
0??Î?Î max
44
55
' † ˆ‰'ab
#
(62.4) (62.4) (62.4) h (10.4) h h?Î?Î? ?ʜˆ‰ ˆ‰ ˆ‰ˆ‰ ˆ‰ ˆ‰ˆ ‰’“ Š‹ É
44hh445
5 3 5 3 5 6 5 4 10.4
hy y F
Î ÎÎ Î
##
"$$
h
0
max
9.288 ft. The volume of water which the tank can hold is V (Base)(Height) 30, whereœ¸ œ
$ "
#
Ɉ‰ˆ ‰
5 6667
4 10.4
†
Height h and (Base) h V h (30) 12h 12(9.288) 1035 ft .œœÊœœ¸¸
"
#
## #$22
55
ˆ‰
14. (a) After 9 hours of filling there are V 1000 9 9000 cubic feet of water in the pool. The level of the waterœœ †
is h , where Area 50 30 1500 h 6 ft. The depth of the typical horizontal strip atœœœÊœœ
V 9000
Area 1500
†
level y is then (6 y) for the coordinate system given in the text. An equation for the drain plate'sÎ
right-hand edge is y x total width is L(y) 2x 2y. Thus the force against the drain plate isœÊ œ œ
F w(6 y)L(y) dy 62.4 (6 y) 2y dy (62.4)(2) 6y y (62.4)(2)?Î ? Î ? Î? Î'' '
00 0
11 1
† ab ’“
#
#
"
!
6y y
3
Î
(124.8) 3 (124.8) 332.8 lb?Î? ?ˆ‰ ˆ‰
"
33
8

414. 410 Chapter 6 Applications of Definite Integrals
(b) Suppose that h is the maximum height. Then, the depth of a typical strip is (h y) and the forceÎ
F w(h y)L(y) dy F , where F 520 lb. Hence, F (62.4) (h y) 2y dy?Î ? ? ? Î
''
0 0
1 1
max max max
†
124.8 hy y dy (124.8) (124.8) (20.8)(3h 2) 3h 2????ʜ'
0
1
ab ’“ ˆ‰
#
##
"
!
"hy y
3 3 20.8
h 520
Î
h9 ftÊœ œ
27
3
15. The pressure at level y is p(y) w y the averageœÊ†
pressure is p p(y) dy w y dy wœœœ
"" "
#bb b
y''
00
bb b
0
† ’“

. This is the pressure at level , whichœœˆ‰Š‹
wb wb b
b

## #
is the pressure at the middle of the plate.

16. The force exerted by the fluid is F w(depth)(length) dy w y a dy (w a) y dy (w a)œœœœ '''
000
bbb b
0
†† † † ’“
y

#
w (ab) p Area, where p is the average value of the pressure (see Exercise 21).œœ œŠ‹ ˆ‰
ab wb

##
†
17. When the water reaches the top of the tank the force on the movable side is (62.4) 2 4 y ( y) dy'
Î2
0
ˆ‰ÈÎÎ #
(62.4) 4 y ( 2y) dy (62.4) 4 y (62.4) 4 332.8 ft lb. The force??? ?'
Î2
0
ab ab ’“ ˆ‰ˆ ‰
# # $Î#
"Î# $Î#
!
#
22
33
†
compressing the spring is F 100x, so when the tank is full we have 332.8 100x x 3.33 ft. ThereforeœœÊ¸
the movable end does not reach the required 5 ft to allow drainage the tank will overflow.Ê
18. (a) Using the given coordinate system we see that the total
width is L(y) 3 and the depth of the strip is (3 y).?
Thus, F w(3 y)L(y) dy (62.4)(3 y) 3 dy?Î ? Î
''
00
33
†
(62.4)(3) (3 y) dy (62.4)(3) 3y?? '
0
3
’“
y

#
$
!
(62.4)(3) 9 (62.4)(3) 842.4 lb?? ? ˆ‰ ˆ‰
99
##

(b) Find a new water level Y such that F (0.75)(842.4 lb) 631.8 lb. The new depth of the strip is
Yœœ
(Y y) and Y is the new upper limit of integration. Thus, F w(Y y)L(y) dy?
Y
0
Y
'
62.4 (Y y) 3 dy (62.4)(3) (Y y) dy (62.4)(3) Yy (62.4)(3) Y?Î? Î? Î? Î''
00
YY Y
0
† ’“ Š‹
y Y

##
#
(62.4)(3) . Therefore, Y 6.75 2.598 ft. So, Y 3 Y?????Š‹ É É È
Y 1263.6 2F
(62.4)(3) 187.2

#
Y
?
3 2.598 0.402 ft 4.8 in%Î % %
19. Use ac oordinate system with y 0 at the bottom of the carton and with L(y) 3.75 and the depth of a typical strip beingœœ
(7.75 y). Then F w(7.75 y)L(y) dy (3.75) (7.75 y) dy (3.75) 7.75yÎ? Î? Î? Î ''
00
775 775
¸¸
ˆ‰ ˆ‰ ’“
64.5 64.5
12 12
y
ÎÎ

#
(Þ(&
!
(3.75) 4.2 lbœ¸ˆ‰
64.5
12
(7.75)
Î

#
20. The force against the base is F pA whA w h (length)(width) (10)(5.75)(3.5) 6.64 lb.
baseœœ œ œ ¸ †† ˆ‰
57
12
Î
To find the fluid force against each side, use a coordinate system with y 0 at the bottom of the can, so that the depth of aœ
typical strip is (10 y): F w(10 y) dy 10yÎ? Î ? Î '
0
10
ˆ‰ˆ‰ˆ‰ ’“
width of width of
the side 12 the side
57 y
Î

#
"!
!
F (50)(3.5) 5.773 lb and F (50)(5.75) 9.484 lbœÊœ¸œ¸ˆ‰ˆ ‰ˆ‰ ˆ‰ ˆ‰
57 100 57 57
12 the side 12 12
width of
ÎÎ Î
#
end side

415. Chapter 6 Practice Exercises 411
21. (a) An equation of the right-hand edge is y x x y and L(y) 2x . The depth of the stripœÊœ œœ
32
33
4y
#
is (3 y) F w(3 y)L(y) dy (62.4)(3 y) y dy (62.4) 3y y dyÎʜ Î ? Î ? Î '' '
00 0
33 3
ˆ‰ ˆ‰ ab
44
33
†
#
(62.4) y (62.4) (62.4) 374.4 lb????ˆ‰ ˆ‰ ‘ˆ‰ˆ‰’“
43 427 27 427
33 33 36
y
##
#
$
!
Î
(b) We want to find a new water level Y such that F (374.4) 187.2 lb. The new depth of the strip is
Yœœ
"
#
(Y y), and Y is the new upper limit of integration. Thus, F w(Y y)L(y) dy? Y
'
0
Y
62.4 (Y y) y dy (62.4) Yy y dy (62.4) Y (62.4)?Î ? Î? Î? Î''
00
YY Y
ˆ ‰ ˆ‰ ˆ‰ ˆ‰ ab ’“ Š‹
44 4 4YY
33 33323
yy#
#
†
Î ÎÎ
&
(62.4) Y . Therefore Y Y 13.5 2.3811 ft. So,œœœÊœœ¸ˆ‰ É È
2
9 2 (62.4) 124.8 124.8
9F (9)(187.2) (9)(187.2)$$ $ Y
†
Î
Y 3 Y 3 2.3811 0.6189 ft 7.5 in. to the nearest half inch.??Î %Î % %
(c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depthe of the
water.
22. The area of a strip of the face of height y and parallel to the base is 100 y, where the factor of accounts for?? ˆ‰
26 26
24 24
† the
inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then:
F w(24 y) 100 dy y 1,946,880 lb.? Î ? w Þ† Î ? w Þ† Î ?'
0
24
abˆ‰ ’“ Š ‹
26
24
y

##
##%
#ß
&
CHAPTER 6 PRACTICE EXERCISES
1. A(x) (diameter) x x??
11
44
##
#
ˆ‰È
x 2 x x x ; a 0, b 1œ œœ
1
4
ˆ‰È†
#%
V A(x) dx x 2x x dxÊœ œ ''
a0
b1
1
4
ˆ‰
&Î# %
xœ œ
11
47 5475
x4 x 4
’“ ˆ‰
†
##
(Î#
"
!
""
(354014)??
11
4 70 280
9
†

2. A(x) (side) sin 2 x x??
"
#
#
#
ˆ‰ ˆ ‰ È
1
34
3
È
4x 4x x x ; a 0, b 4???
È
3
4
ˆ‰ È
#
V A(x) dx 4x 4x x dxÊœ œ ''
a0
b4
È
3
4
ˆ‰
$Î# #
2x x 32??
ÈÈ
33
453453
8x 83264
’“ ˆ‰
# &Î#
%
!
Î
†
1 (15 24 10)???
32 3 8 3 8 3
45315 15
82
ÈÈ È
ˆ‰

3. A(x) (diameter) (2 sin x 2 cos x)??
11
44
##
4 sin x 2 sin x cos x cos xœ
1
4
†ab
##
(1 sin 2x); a , b? ? ?1
11
44
5
V A(x) dx (1 sin 2x) dxʜ ? Î ''
a4
b54
1




x?1 ‘
cos 2x
#
&Î%
Î%
1
1
??11’“Š‹Š‹
5
44
cos cos 11
5

##
#

4. A(x) (edge) 6 x 0 6 x 36 24 6 x 36x 4 6 x x ;œœœœ
# $Î# #
#%
#
Œ ¸Š‹ Š‹
ÈÈÈÈÈÈÈ
a 0, b 6 V A(x) dx 36 24 6 x 36x 4 6 x x dxœœÊœ œ ''
a0
b6
Š‹
ÈÈÈ
$Î# #

416. 412 Chapter 6 Applications of Definite Integrals
36x 24 6 x 18x 4 6 x 216 16 6 6 6 18 6 6 6 6œ œ ’“
È È ÈÈ ÈÈ
†† ††††
22x 86
353 53
$Î# # &Î# # #
'
!
Î Î
216 576 648 72 360œ œ œ œ
1728 1728 1800 1728 72
5555

5. A(x) (diameter) 2 x 4x x ; a 0, b 4 V A(x) dx?????ʜ
11 1
444416
xx# &Î#
#
Š‹Š ‹È
¸
'
a
b
4x x dx 2x x 32 32 32???
111
41647516475
x2x 82'
0
4
Š‹’ “ ˆ‰
&Î# # (Î#
%
!
¸ †
†
††
1 (35 40 14)???
32 8 2 8 72
47535 35
11 1
ˆ‰
6. A(x) (edge) sin 2 x 2 x??
"
#
#
#
ˆ‰ ‘ÈȈ‰
1
34
3
È
4 x 4 3 x; a 0, b 1œœœœ
È
3
4
ˆ‰È È
#
V A(x) dx 4 3 x dx 2 3 xÊœ œ œ ''
a0
b1
ÈÈ
’“ #
"
!
23œ
È

7. (a) :.3=5 7/>29.
V R (x) dx 3x dx 9x dxœœ œ
'' '
a1 1
b1 1
111
#%)
#
ÎÎ
ab
x2 œœ11cd
*
"
"

(b) :=2/66 7/>29.
V 2 dx 2 x 3x dx 2 3 x dx 2 3œœœœœ '''
a00
b11
11111ˆ‰ Š‹ ’“ ab
shell shell
radius height 6
x%&
"
!
††
&
Note: The lower limit of integration is 0 rather than 1.Î
(c) :=2/66 7/>29.
V 2 dx 2 (1 x) 3x dx 2 2?????''
a1
b1
11 11ˆ‰ ‘Š‹ ’ “ ab ˆ‰ˆ ‰
shell shell
radius height 5 2 5 5 5
3x x 3 3 12
Î
%
"
"
""
##
† &
1
(d) :A+=2/< 7/>29.
R(x) 3, r(x) 3 3x 3 1 x V R (x) r (x) dx 9 9 1 x dx??Î?Îʜ Î ? ÎÎ
%% ## %
#
ab c d ab ’“ ''
a1
b1
11
Î
9 1 1 2x x dx 9 2x x dx 9 18?ÎÎ? Î?Î?Î??1111 ''
ÎÎ11
11
cdabab ’“ ‘
%) %)
"
"
"2x x 2 2 13 26
59 59 5 5
†
11†
8. (a) :A+=2/< 7/>29.
R(x) , r(x) V R (x) r (x) dx dx x??ʜ Î ? Î ?ÎÎ
4416x
xx54
Î Î
""
##
## &
## #
" ''
a1
b2
111cd ’“ˆ‰ ˆ‰ ‘
( 2 10 64 5)? ÎÎÎÎ ?ÎÎ? ÎÎ?11 ‘ˆ ‰ˆ‰ˆ ‰
" " "" "
## #
16 16 16 57
532 5 4 10 5 4 20 0†
11
(b) :=2/66 7/>29.
V 2 x dx 2 4x 2 1 4 2?????111 1 '
1
2
ˆ‰ ‘ˆ‰’“ ˆ‰ˆ‰
4x4 55
x4 44
Î

""
###
"
#
"
1
(c) :=2/66 7/>29.
V 2 dx 2 (2 x) dx 2 1 dx???11 1 '' '
a1 1
b2 2
ˆ‰ ˆ ‰ ˆ ‰Š‹
shell shell
radius height x x x
484x
ÎÎ
"
##
2 x 2(1221) 441œ œ œ11’“ ‘ˆ‰
44 x 3
xx 4 4


#
"
"
#
1

417. Chapter 6 Practice Exercises 413
(d) :A+=2/< 7/>29.
V R (x) r (x) dx?
'
a
b
1cd
##
4 dx?1 '
1
2
’“ˆ‰ ˆ ‰
74
x#
# #

16 1 2x x dxœ
49
4
1
1'
1
2
ab
$ '
16xx? $
49 x
45
1
1’“
#
#
"
$'
16 2 1 1œ
49
44535
1
1 ‘ˆ‰ˆ‰
"" "
#†
16 œ
49
4 4 160 5
1
1ˆ‰
"""
(40 1 32)œ œœ
49 16 49 71 103
4 160 4 10 20
11 11 1

9. (a) :.3=5 7/>29.
V x 1 dx (x 1) dx x???111 ''
11
55
Š‹ ’“
È
#&
#
"
x
"
51 48???111 ‘ˆ ‰ˆ‰ˆ‰
25 24
## #
"
(b) :A+=2/< 7/>29.
R(y) 5, r(y) y 1 V R (y) r (y) dy 25 y 1 dyœœÊœ œ
### #
# ''
c2
d2
11cd ab’“

25 y 2y 1 dy 24 y 2y dy 24y y 2 24 2 8œœ œœ 1111 ''
22
22
abab ’“ ˆ‰
%# %# $
#
#
y
53 53
2322
'
††
32 3 (45 6 5)œœ œ1ˆ‰
2 32 1088
5 3 15 15
" 11
(c) :.3=5 7/>29.
R(y) 5 y 1 4 y? $ ?ab
##
V R (y) dy 4 y dyÊœ œ ''
c2
d2
11
##
#

ab
16 8y y dy?$1
'
2
2
ab
#%
16y 2 32?$?$11’“ ˆ‰
8y y
35 35
64 32
' #
#
64 1 (15 10 3)œœ œ1ˆ‰
2 64 512
3 5 15 15
" 11

10. (a) :=2/66 7/>29.
V 2 dy 2 y y dy?? ''
c0
d4
11ˆ‰ Š‹ Š ‹
shell shell
radius height 4
y
"
2 y dy 2 2???111 '
0
4
Š‹ ’“ ˆ‰
#
%
!
yyy
431634
64 64
&
64 œœ
232
13
11
#
†

(b) :=2/66 7/>29.
V 2 dx 2 x 2 x x dx 2 2x x dx 2 x????'''
a00
b44
1111ˆ‰ ˆ ‰ ˆ ‰Š‹ ’ “ È
shell shell
radius height 53
4x$Î# # &Î#
%
!

232??1ˆ‰
4 64 128
5315
†
1
(c) :=2/66 7/>29.
V 2 dx 2 (4 x) 2 x x dx 2 8x 4x 2x x dx???$'' '
a0 0
b4 4
111ˆ‰ ˆ ‰ ˆ ‰Š‹ È
shell shell
radius height
"Î# $Î# #
2 x 2x x 2 8 32 32 64 1?$?$?$111’“ ˆ‰ˆ‰
16 4 x 16 4 64 4 4 2
353353353
$Î# # &Î#
%
!

††
64 1??1ˆ‰
464
55
1
(d) :=2/66 7/>29.
V 2 dy 2 (4 y) y dy 2 4y y y dyœ œ œ '' '
c0 0
d4 4
111ˆ‰ Š‹ Š ‹ Š ‹
shell shell
radius height 4 4
yy
"
##

418. 414 Chapter 6 Applications of Definite Integrals
2 4y2y dy22y y 232 6416 322 1?????1111 '
0
4
Š‹’ “ ˆ‰ˆ‰
##$
%
!
yy
43163 33
22 832
&
†
1
11. :.3=5 7/>29.
R(x) tan x, a 0, b V tan x dx sec x 1 dx [tan x x]œœœÊœ œ œœ
1 1
11
3 3
33
11 1''
00
33
$$
## Î$
!

ab
Š‹
È
12. :.3=5 7/>29.
V (2 sin x) dx 4 4 sin x sin x dx 4 4 sin x dxœ œ œ11 1'' '
00 0

##
#
ab ˆ‰
1 cos 2x
4x4 cos x 4 4 0 (0400) 8 (9 16)œ œ œ œ 111 11 ‘ ‘ˆ ‰ˆ‰
xsin 2x 9
4## ## !
1
111
13. (a) :.3=5 7/>29.
V x 2x dx x 4x 4x dx x x 16????11 1 1 ''
00
22
ab a b ’“ ˆ‰
# %$# % $
#
#
!
x 4 32 32
53 5 3
'
(6 15 10)??
16 16
15 15
11
(b) :A+=2/< 7/>29.
V 1 x 2x dx dx x dx??? ? ?'''
000
222
1111111’“ ’“ab ab
##
# %"
&&&
#
!
#)abx
'
1
(c) :=2/66 7/>29.
V 2 dx 2 (2 x) x 2x dx 2 (2 x) 2x x dx???
'' '
a0 0
b2 2
11 1ˆ‰ Š‹ cd abab
shell shell
radius height
##
2 4x 2x 2x x dx 2 x 4x 4x dx 2 x 2x 2 4 8œ œ œ œ 1111 ''
00
22
abab ’“ ˆ‰
##$ $# $#
#
!
x4 32
43 3
&
(36 32)??
28
33
11
(d) :A+=2/< 7/>29.
V 2 x 2x dx 2 dx 4 4 x 2x x 2x dx 8œ œ 111 1
'''
000
222
cd ababab ’“
####
##
4 4x 8x x 4x 4x dx 8 x 4x 8x 4 dx 8œ œ 1111 ''
0 0
2 2
abab
# %$# %$
x 4x 4x 8 16 16 8 8 (32 40) 8œ œ œ œ œ11111’“ ˆ‰
x32 724032
555555
'
%#
#
!
1111
14. :.3=5 7/>29.
V 2 4 tan x dx 8 sec x 1 dx 8 [tan x x] 2 (4 )????11 111
''
00
44
$$
## Î%
!
ab
1
15. The material removed from the sphere consists of a cylinder
and two "caps." From the diagram, the height of the cylinder is 2h, where h , i.e. h . Thus
##
#
$ œ# œ"Š‹
È
V h ft . To get the volume of a cap,cylœ# $ œ'abŠ‹
È11
#
$
use the disk method and x y : V x dy
## # #
"
? ? cap
'
2
1
ydy y? ?'
"
#
#
"
2
11ab ’“
y
3
3
8 ft . Therefore,??1 ‘ˆ‰ˆ‰
8
333
"& $1
VVV ft.removed cyl cap
33 ??? 1
"! #) $11

16. We rotate the region enclosed by the curve y 12 1 and the x-axis around the x-axis. To find the?Ɉ‰
4x
121
"
volume we use the method: V R (x) dx 12 1 dx 12 1 dx.3=5 œ œ œ '' '
a112 112
b112 112
11 1
#
#

$$
Š‹ Š‹Éˆ‰
4x 4x
121 121
" "

419. Chapter 6 Practice Exercises 415
12 1 dx 12 x 24 132 1??? ?111 1'
$
$11 2
11 2
Š‹ ’“ ’ “ ’ “ ˆ‰ˆ‰ ˆ‰ Š‹
4x 4x 11 4 11 4 11
121 363 2 363 363 4
" "
""Î#
""Î#
#
$
132 1 88 276 in???(11ˆ‰
" $
33
2641
17. y x x x 2 x L 1 2 x dxœÊœ Ê œ Êœ
"Î# "Î# "Î# " " "" ""
##
#
x
3dx dx4x 4x
dy dy
%"
Š‹ ˆ‰ ˆ‰ É '
1
4
L 2 x dx x x dx x x dx 2x x&? $$ ? $ ? $ ? $'' '
11 1
44 4
Ɉ‰ ˆ ‰ ‘Éab
"" " " " "Î# "Î#
#
##
"Î# "Î# "Î# $Î#
%
"4x 4 3
2
482 2œ œœ
""
##
‘ˆ ‰ˆ‰ˆ‰
2 2 14 10
33 33
†
18. x y x L 1 dy 1 dy?&? & ? &? $ ? $
#Î$ "Î$
##
dx 2 dx 4x dx 4
dy 3 dy 9 dy 9x
Š‹ Š‹ Ê É


''
11
88
dx 9x 4 x dx; u 9x 4 du 6y dy; x 1 u 13,??$ ?$&??&?''
11
88
È
9x 4
3x 3

" #Î$ "Î$ #Î$ "Î$È ˆ‰
x 8 u 40 L u du u 40 13 7.634d ' ‘ ‘œÊœ Äœ œ œ ¸
""" "Î# $Î# $Î# $Î#
%!
"$ #18 18 3 7
2
40
13
19. y x x x x x 2 x?&? & ?$
55
12 8 dx dx 4
dy dy'Î& %Î& "Î& "Î& #Î& #Î&"" "
##
#
Š‹ ˆ‰
L 1 x 2 x dx L x 2 x dx x x dxÊœ Êœ œ ''
11
32 32 32
1
ÉÉ ab abab 'É
"""#Î& #Î& #Î& #Î& "Î& "Î&
#
444
x x dx x x 2 2?$?$?$$?$'
1
32
"" " "
## # #
"Î& "Î& 'Î& %Î& ' %
$#
"
ˆ‰ ‘ ‘ˆ ‰ˆ‰ˆ‰
5 5 5 5 5 5 315 75
64 64 64 64
††
(1260 450)?$??
"
48 48 8
1710 285
20. x y y y L 1 y dyœÊœÊ œÊœ
"" "" """ """
###
$# %
#
%
1 y dy 4 y dy 16 16
dx dx
yy
" &&Š‹ Š ‹ Ê '
1
2
y dy y dy y dy y?$$? $?$?'''
111
222
É ÊŠ‹ Š‹’“
""" "" "" ""%#
# #
##
#$
"
16 4y 4y 1 yy
& ""
1??$?ˆ‰ˆ‰
8713
12 1 1 12
"" "
## ##
21. 5 sin t 5 sin 5t and 5 cos t 5 cos 5t
dx dx
dt dt dt dt
dy dy
œ œ Ê Êˆ‰ Š‹
#
#
5 sin t 5 sin 5t 5 cos t 5 cos 5tœ Éabab
##
5 sin 5t sin t sin 5t sin t cos t cos t cos 5t cos 5t sin t sin 5t cos t cos 5 t?Î$$Î$?wÎÎ$
È È ab
####
5 cos t 5 cos t sin t sin t sin t (since t )œ # " % œ % " % œ "! # œ "!l # l œ "! # ! Ÿ ŸÈab ab Ɉ‰ È
"
# #
# 1
Length sin t dt 5 cos t& ? !j Î ? Î ?w!w!?!j'
!
Î
Î#
!
1
1
2
cd abababab
22. 3t 12t and 3t 12t 3t 12t 3t 12t 288t 8t
dx dx
dt dt dt dt
22 dy dy 22 4
œ œÊ œ œ " ʈ‰ Š‹ Éabab
È
#
#
##
#
3 2 t 16 t Length 3 2 t 16 t dt 3 2 t 16 t dt; u 16 t du 2t dt?$&? $? $?$&?
ÈÈÈ
kk kk
ÈÈÈ
’
222 2 ''
!!
""
du t dt; t 0 u 16; t 1 u 17 ; u du u 17 16Êœ œÊœ œÊœ œ œ
"
#
"
“ È ‘ Š‹ab ab
32 32 32
22323316
7
2223/2
17
16
3/2 3/2ÈÈÈ'
17 64 2 17 64 8.617.œ† œ ¸
32
23
2 3/2 3/2È
Š‹Š‹ab ab
È
23. sin and cos sin cos sin cos
dx dx
dd dd
dy dy
)) ))
œ$ œ$ Ê œ $ $ œ $ œ$)) )))) ʈ‰ Š‹ Éababa b È
#
#
##
##
Length d d&?? ?j? ''
!!
$Î $Î
$*
##
11
11
22
)) ˆ‰

420. 416 Chapter 6 Applications of Definite Integrals
24. x t and y t, 3 t 3 2t and t Length 2t t dt??&? ?!&? $!
# #

##
#tdx
3dtdt
dy
3
3

ÈÈ Éab a b '
È
È
t t dt t 2t dt t dt t dt t?$Î$!?$$!? $!?$!?$'''

%# %# #
#

#

ÈÈ È
ÈÈ È
È
È È
È33 3
33 3
3
3
t
3
3
3
ÈÈ Éab ab ’“ '
3
43œ
È
25. Intersection points: 3 x 2x 3x 3 0œ Ê œ
## #
3(x 1)(x 1) 0 x 1 or x 1. Symmetry&$?&? ?
suggests that x 0. The typical strip hasœ @/<>3-+6
center of mass: ( x y ) x x ,
µµ
ßœß œßŠ‹Š‹
2x 3 x x3
"" "

##
ab
length: 3 x 2x 3 1 x , width: dx,ab ab œ
## #
area: dA 3 1 x dx, and mass: dm dA? ?ab
#
$†
3 1 x dx the moment about the x-axis is? &$ab
#

y dm x 3 1 x dx x 2x 3 dx M y dm x 2x 3 dx
µµ
œ œ Ê œ œ
33 3
## #
# # %# %#
"
$$ $abab a b a b
x
1
' '

3x 3 3 ( 3 10 45) ; M dm 3 1 x dxœ œœœ œ œ
3 x 2x 2 3 32
53 53 15 5#
"
"
"
"
#
$$ $’“ ˆ‰ ab
'
$$ ' '
1
3 x 6 1 4 y . Therefore, the centroid is (x y) .? ??&??? ?j$$$’“ ˆ‰ ˆ‰
x3288
33 M545 5
M

"
"
"
x $
$†
26. Symmetry suggests that x 0. The typical œ @/<>3-+6
strip has center of mass: ( x y ) x , length: x ,
µµ
ߜߊ‹
x
"
#
#
width: dx, area: dA x dx, mass: dm dA x dxœœœ
##
$$†
the moment about the x-axis is y dm x x dxʜ
µ
$
#
##
†
x dx M y dm x dx xœÊœœ œ
µ
$$$
##
%%&
#
#
x
2
2
' '
10
cd

2 ; M dm x dx 2 y . Therefore, theœœœœ œ œœÊœœœ
232 x216 3236
10 5 3 3 3 M 5 16 5
M$$ $$ $
$
ab ab ’“
&#$
#
# ' $$'
2
2

x ††
††
centroid is(x y) .ßœ!߈‰
6
5
27. The typical strip has: center of mass: ( x y )@/<>3-+6 ß
µµ
x , length: 4 , width: dx,ϧ Π$
4 x
4

#
x
4
"
"
area: dA 4 dx, mass: dm dA? ?Š‹
x
4
"
$†
4 dx the moment about the x-axis is? &$Š‹
x
4
"
y dm 4 dx 16 dx; the
µ
??$†
Š‹4
xx
416

##
x
4
"
" &
Š‹ Š ‹
$

moment about the y-axis is x dm 4 x dx 4x dx. Thus, M y dm 16 dx
µµ
œ œ œ œ $$Š‹ Š ‹ Š ‹
xx x
44 16
" &
†
x
0
4
'
$
#'
16x 64 ; M x dm 4x dx 2xœœœ œ œ œ
µ
$$$
2516 55 4 16
x 64 128 x x
’“ Š‹’“ ‘
'
†
% %
! !
#
#
y
0
4
' $$'
(32 16) 16 ; M dm 4 dx 4x 16œœ œ œ œ œ œ$$$ $$ ' '
0
4
Š‹ ’ “ ˆ‰
x x 64 32
41213
"
%
!
#
$
x and y . Therefore, the centroid is (x y) .Êœ œ œ œ œ œ ßœß
M
M32 2 532 5 5
16 3 3 128 3 12 3 12y †† †
†††
$$
$$ M
Mx †
#
ˆ‰

421. Chapter 6 Practice Exercises 417
28. A typical strip has:29<3D98>+6
center of mass: ( x y ) y , length: 2y y ,
µµ
ßœ ß Š‹
y2y
"

#
#
width: dy, area: dA 2y y dy, mass: dm dA? ?ab
#
$†
2y y dy; the moment about the x-axis is?$ab
#
y dm y 2y y dy 2y y ; the moment
µ
??$$††aba b
##$
about the y-axis is x dm 2y y dy
µ
?$††
aby2y
"

#
#
ab
4y y dy M y dm 2y y dyœ Êœ œ
µ
$
#
#% #$
ab ab
x
0
2
' $'
y 8 ; M x dm 4y y dy yœœ œœœœ œ œ
µ
$$$’“ ’“ˆ‰ˆ‰ ab
2 2 16 16 16 16 4 4
34 3 4 34123 35
y y$ #% $
# #
! !
##
& '
†
$$ $ $†
y
0
2
' '
; M dm 2y y dy y 4 x andœœ œœ œœœÊœœœ
$$ $$
$#
##
#
!
ˆ‰ ˆ‰ ab ’“
4 8 32 32 84 3238
3 5 15 3 3 3 M 15 4 5
y M† ††
†† ' $$$'
0
2

y
y 1. Therefore, the centroid is (x y) 1 .œœ œ ßœß
M
M34 5
43 8x ††
††
$
$
ˆ‰
29. A typical horizontal strip has: center of mass: ( x y )
µµ
ß
y , length: 2y y , width: dy,œß Š‹
y2y
"

#
#
area: dA 2y y dy, mass: dm dA? ?ab
#
$†
(1 y) 2y y dy the moment about theœ Êab
#
x-axis is y dm y(1 y) 2y y dy
µ
œ ab
#
2y 2y y y dy?$ab
# $$%
2y y y dy; the moment about the y-axis is?$ab
#$%

x dm (1 y) 2y y dy 4y y (1 y) dy 4y 4y y y dy
µ
?$?$?$Š‹ ab a b a b
y2y
"

###
# #% # $%&""
M y dm 2y y y dy y 16& ? ? $ ? $ ?$? $
µ
x
0
2
' 'ab ’“ ˆ‰ˆ‰
#$% $
#
!
""2 161632 2
3 45 345 345
yy
& '
(20 15 24) (11) ; M x dm 4y 4y y y dy y y?$? ? ? ? $? $
µ
16 4 44 4
60 15 15 356
yy
y
0
2
' '""
##
# $%& $%
#
!
ab ’“
' Î
2 4 2 42 ; M dm (1y)2yy dy?$?$????$
"
#
% #
Š‹ ˆ‰ˆ‰ ab
42 2 2 4 4 8 4 24
356356 55
†
'Î
' '
0
2
2y y y dy y 4 x and y?$?$?$?&?? ? ?'
0
2
ab ’“ ˆ‰ ˆ‰ˆ‰
#$ #
#
!
yy
34 34 3 M 58 5 M
8168 2439 M M
&
y x
. Therefore, the center of mass is (x y) .œœœ ߜ߈‰ˆ‰ ˆ ‰
44 3 44 11 9 11
15 8 40 10 5 10
30. A typical vertical strip has: center of mass: ( x y ) x , length: , width: dx,
µµ
ߜ߈‰
33
2x x

area: dA dx, mass: dm dA dx the moment about the x-axis isœœœÊ
33
xx
$$††
y dm dx dx; the moment about the y-axis is x dm x dx dx.
µµ
œœ œœ
33 9 3 3
xx xx2x#
††$$
$ $
(a) M dx ; M x dx 3 2x 12 ;
xy
11
99???? ??$$$$''"
###
*
"
"Î#
*
"
ˆ‰ ˆ ‰ ‘’“
99x20 3
x9 x

$"

$$
M dx 6 x 4 x 3 and y???&??????$$$ '
1
9
3125
x
M
M4 M 4 9
M
%"
‘
"Î#
*
" y x $
$$
ˆ‰
20
9
(b) M dx 4; M x dx 2x 52; M x dx
xy
11 1
99 9???? ???'' 'x9 9 3 3
xx xx##
"
*
"
# $Î#
*
"
ˆ‰ ‘ˆ‰ ‘ˆ‰

6 x 12 x and yœœÊœœœœ‘
"Î#
*
"
"M
M3 M3
13 M y x
31. S 2 y 1 dx; S 2 2x 1 1 dx?$ ?&?&?$$''
a 0
b 3
11Ê Š‹ Š‹ È É
dy dy dy
dx dx dx x 1 x 12x 1
##
"" "
# #È
2 2x 1 dx 2 2 x 1 dx 2 2 (x 1) 2 2 (8 1)œ œ œ œ œ1111''
00
33
ÈÈ É
ÈÈ È ‘ 2x 2 2 2
2x 1 3 3 3
28 2

$Î#
$
!
†
1
È
32. S 2 y 1 dx; x x S 2 1 x dx 1 x 4x dx? $ ?& ?&? $ ? $'''
a 00
b 11
11Ê Š‹ Š‹
ÈÈ
ab
dy dy dy
dx dx dx 3 6
x
##
#% $ %%
†

1

422. 418 Chapter 6 Applications of Definite Integrals
1x d1x 1x 221?$$?$?
111
6639
2'
0
1
È
ab ab ’“’“
È % %%
$Î#
"
!
33. S 2 x 1 dy; 1?$ ??&$? ?'
c
d
1Ê Š‹ Š‹
dx dx dx 4
dy dy dy 4y y 4y y
(4 2y)
4y y 4y y
2 y 4y y 4 4y y
##




ˆ‰
ÈÈ
!
"
""
""
""
S 2 4y y dy 4 dx 4Êœ œ œ''
11
22
111È É
#

4
4y y
"
34. S 2 x 1 dy; 1 1 S 2 y dy?$ ?&$?$?&?''
c 2
d 6
11Ê Š‹ Š‹ È
dx dx 1 dx
dy dy 2 y dy 4y 4y
4y 1 4y 1
4y
##
"
È È
È
†
4y 1 dy (4y 1) (125 27) (98)?$?$???1 '
2
6
È ‘
1111
43 6 6 3
249 $Î#
'
#
35. x and y 2t, 0 t 5 t and 2 Surface Area 2 (2t) t 4 dt 2 u du??&??& ? $?
tdx
dt dt
dy
"
#
# "Î#
È È ''
04
59

11
2 u , where u t 4 du 2t dt; t 0 u 4, t 5 u 9?? ?$&??&??&?1 ‘ È
276
33
$Î# #
*
%
1
36. x t and y 4 t , t 1 2t and ?$ ? & ? ?
# "" "
2t dt 2t dt 2
dx 2 dy
t È
ÈÈ
"
Surface Area 2 t 2t dt 2 t 2t dt&?$$?$$ ''
12 12
1 1$ $

11ˆ‰ˆ ‰ ˆ‰ˆ ‰Ê Š‹ É
##"" ""
###
##
#
t2t t t
2
t
" " È
2 t 2t dt 2 2t t dt 2 t t t?$$? $$?$111 ''
12 12
11$$

ˆ‰ˆ ‰ ˆ ‰ ‘
#$$%#"" " " "
##
"
"Î #2t 2t 4 2 8
33
" È
22?1Š‹
32
4
È
37. The equipment alone: the force required to lift the equipment is equal to its weight F (x) 100 N.ʜ "
The work done is W F (x) dx 100 dx [100x] 4000 J; the rope alone: the force required""
%!
!œœœœ''
a0
b40
to lift the rope is equal to the weight of the rope paid out at elevation x F (x) 0.8(40 x). The workʜ #
done is W F (x) dx 0.8(40 x) dx 0.8 40x 0.8 40 640 J;##
###
%!
!
#??????''
a0
b40
’“Š ‹
x40 (0.8)(1600)
""
the total work is W W W 4000 640 4640 J?$? $ ? "#
38. The force required to lift the water is equal to the water's weight, which varies steadily from 8 800 lb to†
8 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is†
F(x) 8 800 (6400) 1 lb. The work done is W F(x) dx?? ?††ˆ‰ ˆ‰
2 4750 x x
2 4750 9500
†
†
'
a
b
6400 1 dx 6400 x 6400 4750 (6400)(4750)????'
0
4750
ˆ‰ ˆ‰ ’“ Š ‹
x x 4750 3
9500 2 9500 4 4750 4
""
††
%(&!
!
22,800,000 ft lbœ †
39. Force constant: F kx 20 k 1 k 20 lb/ft; the work to stretch the spring 1 ft isœÊœ Êœ †
W kx dx k x dx 20 10 ft lb; the work to stretch the spring an additional foot isœœœœ''
00
11
’“
x
"
#
"
!
†
W kx dx k x dx 20 20 20 30 ft lb??????''
11
22
’“ ˆ‰ˆ‰
x4 3
"
####
#
"
"
†
40. Force constant: F kx 200 k(0.8) k 250 N/m; the 300 N force stretches the spring xœÊ œ Êœ œ
F
k
1.2 m; the work required to stretch the spring that far is then W F(x) dx 250x dxœœ œ œ
300
250 ''
00
12 12
''
[125x ] 125(1.2) 180 Jœœœ
#"Þ# #
!

423. Chapter 6 Practice Exercises 419
41. We imagine the water divided into thin slabs by planes
perpendicular to the y-axis at the points of a partition of the
interval [0 8]. The typical slab between the planes at y andß
y y has a volume of about V (radius) (thickness)$???1
#
y y y y ft . The force F(y) required toœœ1? ?ˆ‰
525
416
#
#$1
lift this slab is equal to its weight: F(y) 62.4 Vœ ?
y y lb. The distance through which F(y)œ
(62.4)(25)
16
1?
#
must act to lift this slab to the level 6 ft above the top is

about (6 8 y) ft, so the work done lifting the slab is about W y (14 y) y ft lb. The work done œ ?1?
(62.4)(25)
16
#
†
lifting all the slabs from y 0 to y 8 to the level 6 ft above the top is approximatelyœœ
W y (14 y) y ft lb so the work to pump the water is the limit of these Riemann sums as the norm of¸!
8
"
(62.4)(25)
16
1?
#
†
the partition goes to zero: W y (14 y) dy 14y y dy (62.4) yœœœ ''
00
88
(62.4)(25) (62.4)(25) y
(16) 16 16 3 4
25 14
1
##$$
)
!
1 1
ab ˆ‰’“
&
(62.4) 8 418,208.81 ft lb?( ˆ‰Š‹
25 14 8
16 3 4
1
††
$
&
42. The same as in Exercise 41, but change the distance through which F(y) must act to (8 y) rather than
(6 8 y). Also change the upper limit of integration from 8 to 5. The integral is:
W y (8 y) dy (62.4) 8y y dy (62.4) yœœ œ ''
00
55
(62.4)(25) y
16 16 16 3 4
25 25 81 11##$$
&
!
ˆ‰ ˆ‰ab ’“
&
(62.4) 5 54,241.56 ft lb?( ˆ‰Š‹
25 8 5
16 3 4
1
††
$
&
43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x y . A typicalœœ
5
10
y
#
horizontal slab has volume V (radius) (thickness) y y y. The force required to lift this?1 1? ?œœœ
##
#
# ˆ‰
y
4
1
slab is its weight: F(y) 60 y y. The distance through which F(y) must act is (2 10 y) ft, so the?$†
1
4
#
?
work to pump the liquid is W 60 (12 y) dy 15 22,500 ft lb; the time needed??? '
0
10
11 1 Š‹ ’ “
y 12y y
434
" "!
!
†
to empty the tank is 257 sec
22,500 ft lb
275 ft lb/sec
†
†
¸
44. A typical horizontal slab has volume about V (20)(2x) y (20) 2 16 y y and the force required to?? ??? ˆ‰È
#
lift this slab is its weight F(y) (57)(20) 2 16 y y. The distance through which F(y) must act is? ˆ‰È
#
?
(6 4 y) ft, so the work to pump the olive oil from the half-full tank is
W 57 (10 y)(20) 2 16 y dy 2880 10 16 y dy 1140 16 y ( 2y) dyœ œ '''
444
000
ˆ‰ÈÈ ab ## #
"Î#
22,800 (area of a quarter circle having radius 4) (1140) 16 y (22,800)(4 ) 48,640?$?$†
2
3
’“ab
#
$Î#
!
%
1
335,153.25 ft lbœ †
45. F W L(y) dy F 2 (62.4)(2 y)(2y) dy 249.6 2y y dy 249.6 y?&???'''
a00
b22
††Š‹ ’ “ ab
strip
depth 3
y##
#
!

(249.6) 4 (249.6) 332.8 lb?? ?ˆ‰ ˆ‰
84
33
46. F W L(y) dy F 75 y (2y 4) dy 75 y 2y 4y dy?&?$?$'''
a00
b5656
††Š‹ ˆ‰ ˆ ‰
strip
depth 6 3 3
5510
$$
#
75 y 2y dy 75 y y y (75)???'
0
56
$
ˆ‰ ‘ ‘ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ ‰
10 7 10 7 2 50 7 25 2 125
3 3 3 6 3 18 6 36 3 216
##$
&Î'
!
(75) (25 216 175 9 250 3) 118.63 lb.œœ œ¸ˆ‰ˆ‰
25 175 250 75
9 216 3 16 9 16 9 16
(75)(3075)
†† †## #
†††

424. 420 Chapter 6 Applications of Definite Integrals
47. F W L(y) dy F 62.4 (9 y) 2 dy 62.4 9y 3y dy?&??'''
a00
b44
†† †Š‹ Š ‹ ˆ‰
strip
depth 2
yÈ
"Î# $Î#
62.4 6y y (62.4) 6 8 32 (48 5 64) 2196.48 lbœœ†œ œœ ‘ˆ ‰ˆ‰
$Î# &Î#
%
!
2 2 62.4
5555
(62.4)(176)
††
48. Place the origin at the bottom of the tank. Then F W L(y) dy, h the height of the mercury column,œœ
'
0
h
††Š‹
strip
depth strip depth h y, L(y) 1 F 849(h y) dy (849) (h y) dy 849 hy 849 hœ œÊ œ " œ œ œ ''
00
hh h
’“ Š‹
y h
" "
##
!
#
h . Now solve h 40000 to get h 9.707 ft. The volume of the mercury is s h 1 9.707 9.707 ftœœ¸ œ†œÞ
849 849
22
22## $
49. F w (8 y)(2)(6 y) dy w (8 y)(2)(y 6) dy 2w 48 14y y dy 2w 48 2y y dyœœ "# " #
##

'' ''
060 6
60 6 0
abab
2w 48y 7y 2w 48y y 216w 360w?$$$?$"# "#
##
'!
!'’“’“
yy
33

50. (a) F 62.4 (10 y) 8 dy? '
0
6
‘ˆ‰ˆ‰
yy
66
240 34y y dy?$
62.4
3'ab
6
0
#
240y 17y (1440 612 72)?$?$
62.4 62.4
333
y
’“
#
'
!

18,720 lb.œ

(b) The centroid 3 of the parallelogram is located at the intersection of y x and y x . The centroid ofˆ‰
7 6636
2 755
??$
the triangle is located at (7 2). Therefore, F (62.4)(7)(36) (62.4)(8)(6) (300)(62.4) 18,720 lb?$??
CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES
1. V f(x) dx b ab f(t) dt x ax for all x a [f(x)] 2x a f(x)??&?&?&?11 1''
aa
bx
cd cd É
## ## # 2x a
1
2. V [f(x)] dx a a [f(t)] dt x x for all x a [f(x)] 2x 1 f(x)??$&?$&?$&?11 1''
00
ax
## ## #
É
2x 1
1
3. s(x) Cx 1 [f (t)] dt Cx 1 [f (x)] C f (x) C 1 for C 1œÊ œÊ œÊ œ  '
0
x
ÈÈ È w# w # # w
f(x) C 1 dt k. Then f(0) a a 0 k f(x) C 1 dt a f(x) x C 1 a,Êœ œÊœÊœ Êœ ''
0 0
x x
ÈÈÈ# ##
where C 1. 
4. (a) The graph of f(x) sin x traces out a path from ( ) to ( sin ) whose length is L 1 cos d .?jj?$ !! )) '
0
$
È #
The line segment from (0 0) to ( sin ) has length ( 0) (sin 0) sin . Since theßß œ!! ! ! ! ! È È
####
shortest distance between two points is the length of the straight line segment joining them, we have
immediately that 1 cos d sin if 0 .'
0
$
ÈÈ
Ÿ###
#
)) ! ! !
1
(b) In general, if y f(x) is continuously differentiable and f(0) 0, then 1 [f (t)] dt f ( )??$$ '
0
$
ÈÈ w# # #
!!
for 0.!
5. From the symmetry of y 1 x , n even, about the y-axis for 1 x 1, we have x 0. To find y , weœ ŸŸ œ œ
n M
Mx
use the vertical strips technique. The typical strip has center of mass: ( x y ) x , length: 1 x ,
µµ
ßœß ˆ‰
1x
2

n
n
width: dx, area: dA 1 x dx, mass: dm 1 dA 1 x dx. The moment of the strip about the? ? ?ab ab
nn
†
x-axis is y dm dx M dx 2 1 2x x dx x
µ
?&? ?$?$
ab ab1x 1x 2x x
n1 n1

### $#$
"
"
!
nn""
x
10
11
n2n
''

ab ‘
n1 2n1

425. Chapter 6 Additional and Advanced Exercises 421
1 .? $ ? ? ?
2 2n 3n14n2n1 2n
n 1 n 1 (n 1)( n 1) (n 1)( n 1) (n 1)( n 1)
(n 1)(2n 1) 2(2n ) (n 1)
$ #$ $ #$ $ #$ $ #$
" "
""
Also, M dA 1 x dx 2 1 x dx 2 x 2 1 . Therefore,œ œœ œ œœ'' '
11 0
11 1
nn
ab ab ‘ˆ ‰
x2n
n1 n1 n1
n1

"
!
"
y is the location of the centroid. As n , y soœ œ œ Ê !ß Ä _ Ä
M
M (n1)(2n1) 2n 2n1 n1
2n n n (n 1)x
"
$$ $ #$ #
"
† ˆ‰
the limiting position of the centroid is .ˆ‰!ß
"
#
6. Align the telephone pole along the x-axis as shown in the
accompanying figure. The slope of the top length of pole is
(14.5 9) . Thus,
ˆ‰
14.5 9
88
$$
""
40 8 40 8 40 8 80
5.5 11
???
111
††
††
y x 9 x is an equation of the?$ ? $
911 11
8880 8 8011 1†
"
ˆ‰
line representing the top of the pole. Then,
M x y dx x 9 x dxy??$''
a0
b40
†11
# "
# ‘ˆ‰
880
11
1
x 9 x dx; M y dx?$ ?
"
#
#
64 80
11
1''
0a
40 b
ˆ‰ 1

9 x dx 9 x dx. Thus, x 23.06 (using a calculator to compute?$?$ ?((1 ''
00
40 40
‘ˆ‰ˆ‰
""
##
880 64 80 M
11 11 M 129,700
5623.311 y
the integrals). By symmetry about the x-axis, y 0 so the center of mass is about 23 ft from the top of the pole.œ
7. (a) Consider a single vertical strip with center of mass ( x y ). If the plate lies to the right of the line, then
µµ
ß
the moment of this strip about the line x b is (x b) dm (x b) dA the plate's first moment?? &
µµ
$
about x b is the integral (x b) dA x dA b dA M b A.???
'' $$$ $ '
y
(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x b isœ
b x dm b x dA the plate's first moment about x b is (b x) dA b dA x dAababœ Ê œ œ
µµ
$$$$
'''
bA M.?$
y
8. (a) By symmetry of the plate about the x-axis, y 0. A typical vertical strip has center of mass:œ
( x y ) (x 0), length: 4 ax, width: dx, area: 4 ax dx, mass: dm dA kx 4 ax dx, for some
µµ
ßœß œ œ ÈÈ È $ †
proportionality constant k. The moment of the strip about the y-axis is M x dm 4kx ax dx
y
0
aœœ
µ' ' #È
4k a x dx 4k a x 4ka a . Also, M dm 4kx ax dxœœœœœœÈÈ È ‘ ''
0 0
a a
a
0
&Î# (Î# "Î# (Î# 2 2 8ka
777
†
&
'
4k a x dx 4k a x 4ka a . Thus, x aœœœœœœœÈÈ ‘ '
0
a
a
0
$Î# &Î# "Î# &Î# 2 2 8ka 8ka 5 5
5 5 5 M 7 8ka 7
M
††
&

y
(x y) 0 is the center of mass.Êßœ ß ˆ‰
5a
7
(b) A typical horizontal strip has center of mass: ( x y ) y y , length: a ,
µµ
ßœ ßœ ß Œ $Š‹
y
4a


#
a y4a y
8a 4a
"" "
width: dy, area: a dy, mass: dm dA y a dy. Thus, M y dmŠ‹ Š‹ kk???
µ
yy
4a 4a
""
$
x
'
y y a dy y a dy y a dy??$'''
2a 2a 0
2a 0 2a
kkŠ‹ Š‹ Š‹
yyy
4a 4a 4a
"""
##
ay dy ay dy y yœ œ ''
2a 0
02a
a
a
Š‹Š‹’ “’“
##$$
##
!#
# !
yyyy
4a 4a 3 0a 3 0a
aa
&& ''
0; M x dm y a dyœ œ œ œ
µ
8a 32a 8a 32a
3 20a 3 0a 8a 4a
y 4a y
&'&' "" "
#

y
2a
2a
' '

Š‹Š‹ kk
y y 4a dy y 16a y dyœ œ
"" ## %%
8a 4a 32a
4a y'

2a
2a 2a
2a
kkab kkabŠ‹ '
""
"
16a y y dy 16a y y dy 8a y 8a yœ œ
"""
# ##
%& %& %# %#
!#
# !
3a 32a 3a 6 3a 6
yy 1
""""
ÎÎ
''
2a 0
02a
a
a
ab ab ’“’“
8a 4a 8a 4a 32a 32a a ;?$???
"" ""%# %# ' ' %
32a 6 32a 6 16a 3 16a 3 3
64a 64a 32a 2 4
"" ""
ÎÎÎ’“’“Š‹ ab†† †
M dm y dy y 4a y dyœœ œ ' kk kk abŠ‹ ''
2a 2a
2a 2a
4a y
4a 4a
""
" ##

426. 422 Chapter 6 Applications of Definite Integrals
4a y y dy 4a y y dy 2a y 2a y?$'' $?$'' $
"""" #$ #$ ## ##
!#
$# %
4a 4a 4a 4 4a 4
yy''
$2a 0
02a
a
a
ab ab ’“’“
!!
2 2a 4a 8a 4a 2a . Therefore, x a and?$?$? ???††
"" "## % % $ %
#4a 4 a M 3 2a 3
16a 4 2a M
Š‹ ab ˆ‰ˆ‰
!
$
y
y 0 is the center of mass.œœ
M
M
x
9. (a) On [0 a] a typical strip has center of mass: ( x y ) x, ,ß @/<>3-+6 ß œ
µµ
Š‹
ÈÈ
bx ax"" ""
$ $
#
length: b x a x , width: dx, area: dA b x a x dx, mass: dm dA
ÈÈ ÈÈ
Š‹
## ## ## ##
$$ $ ? $$ $ ? $
b x a x dx. On [a b] a typical strip has center of mass:? $ $ $ /<>+)$Š‹
ÈÈ
## ##
( x y ) x , length: b x , width: dx, area: dA b x dx,
µµ
? $ ? $Š‹
ÈÈ
È
bx""
$
#
## ##
mass: dm dA b x dx. Thus, M y dm?? $ ?
µ
$$
È '
##
x
bx ax bx ax dx bx bx dx?$'$ $$$'$$''
0 a
a b
" "
# #
## ## ## ## ## ##
Š‹Š‹
ÈÈ ÈÈ ÈÈ
$$
b x a x dx b x dx b a dx b x dx?$$$'$?$'$
$$$$
####
## ## ## ## ##''''
0a0a
abab
cd ab ab ababab
bax bx baa b ba?$ ' $?$' $$$
$$$$
####
## # ## $ #
!
cd cdab ab ’“ ’ “ Š‹Š ‹
a
b xba
333
a
$$$
ab a b ab ; M x dm?$' $'?$? ?
µ
$$ $$
##
#$ $ # $
ab Š‹ Š‹ '
2abab a
33333
$$$ $$
$
y
x b x a x dx x b x dx?$$$'$''
0a
ab
$$Š‹
ÈÈ È
## ## ##
xb x dx xa x dx xb x dx?$$ $' $$$ $ '''
00a
aab
ab ab ab
## ## ##
"Î# "Î# "Î#
?'$
$
###
$$ $
!
$$$
”•”•”•
2b x 2a x 2b x
33 3
ab ab ab
"" "" ""
$)" $)" $)"
aa b
0a
ba b 0a 0ba M;?$ $ $ '$ $$$ ?$? ?
$$$$$ $
333333
ba ba
’“’“’“abab ab ab
## # # ##
$Î# $Î# $Î# $Î# $
$$ $$ ab
x
We calculate the mass geometrically: M A b a . Thus, x?? $ ? $ ?$$ $Š‹ Š‹ ab
11$1ba
444 M
M
""
## y
; likewiseœœœœ
$
$1 1 1 1
ab abab
ab
ba (ba)aabb 4aabb
3 b a 3 b a 3 (b a)(b a) 3 (a b)
44ba4
$$ "" ""
"" ""
$$


$
† Š‹
y. œœ
M
M3(ab)
4a ab bx ab
""

1
(b) lim (x y) is the limiting
baÄ
4 aabb 4 aaa 4 3a 2a 2a2a
3ab 3 aa 32a11 1111
Š‹Š‹Š‹ ˆ‰ ˆ‰ ˆ ‰
" " """ "
$$ $$

œœœÊßœß
position of the centroid as b a. This is the centroid of a circle of radius a (and we note the two circlesÄ
coincide when b a).œ
10. Since the area of the traingle is 36, the diagram may be
labeled as shown at the right. The centroid of the triangle is , . The shaded portion is 144 36 108. Writeˆ‰
a24
3a
$?
, for the centroid of the remaining region. The centroidabx
y
of the whole square is obviously 6, 6 . Think of the squareab
as a sheet of uniform density, so that the centroid of the
square is the average of the centroids of the two regions,
weighted by area:
and 'œ 'œ
$' "!) $' "!)
"%% "%%
ˆ‰ ˆ ‰ab ab
a24
3a
x
y
which we solve to get and . Setx y?$ ?
a a
a*
$"ab
7 in. (Given). It follows that a , whence x yœœ*œ
'%
*

7 in. The distances of the centroid , from the other sides are easily computed. (Note that if we set 7 in.œ œ
"
*
abx
y y
above, we will find 7 .)xœ
"
*

427. Chapter 6 Additional and Advanced Exercises 423
11. y 2 x ds 1 dx A 2 x 1 dx (1 x)?&?$&? $?$ ?ÈÈ ÉÉ ‘
"" $Î#
$
!xx33
428 '
0
3
12. This surface is a triangle having a base of 2 a and a height of 2 ak. Therefore the surface area is11
(2 a)(2 ak) 2 a k.
"
#
##
11 1 œ
13. F ma t a v C; v 0 when t 0 C 0 x C ;? ? & ?? &? ? $ ? ?& ?& ? &? $
#
"
dx t dx t dx t t
dt m dt 3m dt 3m 12m
""
"
x 0 when t 0 C 0 x . Then x h t (12mh) . The work done isœœÊœÊœ œÊœ "
"Î%t
12m
&
W F dx F(t) dt t dt (12mh)œœ œ œ œ' ’“ ˆ‰ ''
00
12mh 12mh 12mh)
0
&% &% &
$'" $'" $'"
††
dx t t
dt 3m 3m 6 18m
# 'Î% ""
Î
2 3mh 3mhœœ œ œ
(12mh)
18m 18m 3 3
12mh 12mh 2h 4h

†
È
†
ÈÈ
14. Converting to pounds and feet, 2 lb/in 24 lb/ft. Thus, F 24x W 24x dxœœ œÊœ
2 lb 12 in
1 in 1 ft
† '
0
12
$
12x 3 ft lb. Since W mv mv , where W 3 ft lb, m lb?? ? ? ?cd ˆ‰ˆ ‰
###
"Î#
!
"" " "
## #! "
††
10 3 ft/sec
"
slugs, and v 0 ft/sec, we have 3 v v 3 640. For the projectile height,œœ œÊœ
"""
"
##
##
!320 3 0
ˆ‰ˆ ‰
" †
s 16t v t (since s 0 at t 0) v 32t v . At the top of the ball's path, v 0 t? $ ? ? & ? ? $ ? & ?
#
!!
#
ds
dt 3
v
and the height is s 16 v 30 ft.? $ ? ? ?ˆ‰ ˆ‰
vv
3 3 64 64
v 3 640
"
##
#
!
†
15. The submerged triangular plate is depicted in the figure
at the right. The hypotenuse of the triangle has slope 1
y ( 2) (x 0) x (y 2) is an equationÊ œ Ê œ
of the hypotenuse. Using a typical horizontal strip, the fluid
pressure is F (62.4) dyœ
' ††Š‹Š ‹
strip strip
depth length
(62.4)( y)[ (y 2)] dy 62.4 y 2y dy?$? $''

66
22
ab
# 62.4 y (62.4) 4 36?$?$$’“ ‘ˆ‰ˆ ‰
y
333
8 216

#
#
'
(62.4) 32 2329.6 lb??(ˆ‰
208
33
(62.4)(112)

16. Consider a rectangular plate of length and width w.j
The length is parallel with the surface of the fluid of
weight density . The force on one side of the plate is=
F ( y)( ) dy . The?!?! ?== '
w
0
w
’“
y w
" "
##
!

j=
average force on one side of the plate is F ( y)dy
av
w
0?
=
w'

. Therefore the force ? ?
== =
w
y ww
’“
" "
## #
!

j
w

( w) (the average pressure up and down) (thearea of the plate).œjœˆ‰
=w
#
†
17. (a) We establish a coordinate system as shown. A typical
horizontal strip has: center of pressure: ( x y )
µµ
ß
y , length: L(y) b, width: dy, area: dAœß œˆ‰
b
#
b dy, pressure: dp y dA b y dyœœœ ==kk kk
F y dp y b y dy b y dy&? ? ?
µ
x
hh
00
' ''

†==kk
#
bb0 ;œ œ œ==’“ ’ “ Š‹
y
333
hbh
!


h
=
F dp y L(y) dy b y dy?? ? ' ''
hh
00
==kk

428. 424 Chapter 6 Applications of Definite Integrals
b b 0 . Thus, y the distance below the surface is h.œ œ œ œ œ œ Ê==’“ ’ “
y hbh 2h 2 F
F3 3
" ""
###
!


h
= x
Š‹
Š‹
$
'



bh
3
bh
(b) A typical horizontal strip has length L(y). By similar
triangles from the figure at the right,
L(y) y a
bh
œ

L(y) (y a). Thus, a typical strip has center&?$
b
h
of pressure: ( x y ) ( x y), length: L(y)
µµ µ
ßœß
(y a), width: dy, area: dA (y a) dy,? $ ? $
bb
hh
pressure: dp y dA ( y) (y a) dy??$==kk ˆ‰
b
h
y ay dy F y dp?$&?
µ
=b
h
ab
#
x
'
y y ay dy y ay dy?$?$ ''

ah ah
aa
†
==bb
hh
ab a b
#$#
?$
=b
h4 3
yay
’“
& a
ah
&!%

œ œ
==ba a b
h43 4 3 h 4 3
(a h) a(a h) a (a h) a a(a h)
’“’“Š‹Š ‹
&& &

3a a4ah6ah4ahh 4aaa3ah3ahhœ
=b
12h
c dabababab
%%$##$% %$# #$
12a h 12a h 4ah 12a h 18a h 12ah 3h 6a h 8ah 3hœ œ
= =b b
12h 12h
abab
$ ## $ $ ## $ % ## $ %
6a 8ah 3h ; F dp y L(y) dy y ay dy?$$?? ? $?$
## #

===bh b b
12 h h 3 2
yay
ab kk ab ’“ '' = '
&!%

ah
a a
ah
"
?$$ ? $
==baa b
h3 3 h 3 2
(ah) a(ah) (ah) a a a(ah)
’“’“Š‹Š ‹

##


23ah 3ah h 32ah ah?$?$$$
==ba3ah3ahha b
h3 6h
aa2ahah
’“ cdabab


#
##$ ##ab
6a h 6ah 2h 6a h 3ah 3ah 2h (3a 2h). Thus, y?$$?$?$ ?
===bbbh
6h 6h 6 F
F
abab
##$## #$ x
the distance below the surface isœœ Ê
ˆ‰
ˆ‰
$


bh
1
bh
6
ab6a 8ah 3h
(3a 2h)
6a 8ah 3h
3a 2h
""
""


"
#
ˆ‰Š‹
.
6a 8ah 3h
6a 4h
""

429. CHAPTER 7 TRANSCENDENTAL FUNCTIONS
7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES
1. Yes one-to-one, the graph passes the horizontal test.
2. Not one-to-one, the graph fails the horizontal test.
3. Not one-to-one since (for example) the horizontal line y intersects the graph twice.œ#
4. Not one-to-one, the graph fails the horizontal test.
5. Yes one-to-one, the graph passes the horizontal test
6. Yes one-to-one, the graph passes the horizontal test
7. Domain: 0 x 1, Range: 0 y 8. Domain: x 1, Range: y 0Ÿ Ÿ

9. Domain: 1 x 1, Range: y 10. Domain: x , Range: yŸŸ ŸŸ __ Ÿ
11 11
## ##

11. The graph is symmetric about y x.œ

(b) y 1 x y 1 x x 1 y x 1 y y 1 x f (x)? Î%?Î%?Î%? Î%? Î?
ÈÈ È
### #### "

430. 426 Chapter 7 Transcendental Functions
12. The graph is symmetric about y x.œ

y x y f(x)œÊœÊœœ
""" "
xyx
13. Step 1: y x 1 x y 1 x y 1?w% ?Î%? Î
## È
Step 2: y x 1 f (x)?Î?
È
"
14. Step 1: y x x y, since x .? % ?Î "
#
È
Step 2: y x f (x)?Î ?È
"
15. Step 1: y x 1 x y 1 x (y 1)?Î% ?w%?w
$ $ "Î$
Step 2: y x 1 f (x)?w?
$ "È
16. Step 1: y x 2x 1 y (x 1) y x 1, since x 1 x 1 y?Îw%?Î % ?Î !%?w
##
ÈÈ
Step 2: y1 xf(x)?w ?È
"
17. Step 1: y (x 1) y x 1, since x 1 x y 1?w % ?w !Î%? Î
#
ÈÈ
Step 2: y x 1 f (x)?Î?È
"
18. Step 1: y x x yœÊœ
#Î$ $Î#
Step 2: y x f (x)œœ
$Î# "
19. Step 1: y x x yœÊœ
& "Î&
Step 2: y x f (x);œœ
& "
È
Domain and Range of f : all reals;
"
f f (x) x x and f (f(x)) x xab ab ˆ‰
" "Î& " &
& "Î&
œœ œœ
20. Step 1: y x x yœÊœ
% "Î%
Step 2: y x f (x);œœ
% "
È
Domain of f : x 0, Range of f : y 0;
" "
  
f f (x) x x and f (f(x)) x xab ab ˆ‰
" "Î% " %
% "Î%
œœ œœ
21. Step 1: y x 1 x y 1 x (y 1)?w% ?Î%?Î
$ $ "Î$
Step 2: y x 1 f (x);?Î?
$ "È
Domain and Range of f : all reals;
"
f f (x) (x 1) 1 (x 1) 1 x and f (f(x)) x 1 1 x xab a babˆ‰ ab
" "Î$ " $ $
$ "Î$ "Î$
?Î w?Îw? ? wÎ ? ?

431. Section 7.1 Inverse Functions and Their Derivatives 427
22. Step 1: y x x y x 2y 7?Î% ?w%?w
""
## # #
77
Step 2: y 2x 7 f (x);?w?
"
Domain and Range of f : all reals;
"
f f (x) (2x 7) x x and f (f(x)) 2 x 7 (x 7) 7 xab ˆ‰ ˆ ‰
" " ""
#### ##
? wÎ?wÎ? ? Îw?Îw?
777 7
23. Step 1: y x xœÊœÊœ
""" #
xyy
È
Step 2: y f (x)œœ
" "
Èx
Domain of f : x 0, Range of f : y 0;
" "
$$
f f (x) x and f (f(x)) x since x 0ab
" " "" ""
??? ??? $
Š‹ Š‹ Š‹ É


x
xx

"

x
24. Step 1: y x xœÊœÊœ
""" $
xy y
$ "Î$
Step 2: y f (x);œœ œ
"" $ "
x x
"Î$É
Domain of f : x 0, Range of f : y 0;
" "
ÁÁ
f f (x) x and f (f(x)) xab ˆ‰ ˆ‰
" " "" " "
"Î$ "
œœœ œ œœ
abx
xxx
&"Î$
$ &" $
25. (a) y 2x 3 2x y 3?w% ?Î
x f (x)%?Î% ?Î
y 3x3
## ##
"
(c) 2, ¸ ¹
df df
dx dx
x1
x1

œœ
&"
"
#
(b)
26. (a) y x 7 x y 7?w% ?Î
""
55
x 5y 35 f (x) 5x 35%? Î % ? Î
"
(c) , 5¸ ¹
df df
dx 5 dx
x1
x

œœ
"
&"
(b)
27. (a) y 5 4x 4x 5 y?Î % ?Î
x f (x)%?Î% ?Î
55x
44 44
y "
(c) 4, ¸ ¹
df df
dx dx 4
x1
x3

?Î ?Î
&"
"
(b)

432. 428 Chapter 7 Transcendental Functions
28. (a) y 2x x yœÊœ
## "
#
x y f (x)Êœ Ê œ
" "
#È
2
x
È È
(c) 4x 20,¸ k
df
dx
x x5
œœ
x ¹¹
df
dx 0 2
&"
x0 x50
œœ
""
#
"Î#
#È
(b)
29. (a) f(g(x)) x x, g(f(x)) x xœœ œœˆ‰È È
$
$
$
$
(c) f (x) 3x f (1) 3, f ( 1) 3;
w#w w
? ??
g (x) x g (1) , g ( 1)
w #Î$ w w"""
???
333
(d) The line y 0 is tangent to f(x) x at ( );œ œ !ß !
$
the line x 0 is tangent to g(x) x at (0 0)œœß
$
È
(b)
30. (a) h(k(x)) (4x) x,œœ
" "Î$
$
4
ˆ‰
k(h(x)) 4 xœœŠ‹†
x
4
$
"Î$
(c) h (x) h (2) 3, h ( 2) 3;
www
?% ? Î?
3x
4

k (x) (4x) k (2) , k ( 2)
w #Î$ w w ""
???
4
333
(d) The line y 0 is tangent to h(x) at ( );œ œ !ß !
x
4
$
the line x 0 is tangent to k(x) (4x) atœœ
"Î$
() !ß !
(b)
31. 3x 6x 32. 2x 4
df df df df
dx dx 9 dx dx 6
? ? ? ? ? ?
# " "
¹¹ ºº
&" &"
xf(3) xf(5)
x3 x5

""
df df
dx dx
33. 3 34. ¹¹ ¹¹ ºº
df df
dx dx dx dx 2
dg dg
&" &"
"
&" &"
x4 xf(2) x0 xf(0)
x2 x0

œœœœ œœœ
""
df dg
dx dx
" "
ˆ‰
3
35. (a) y mx x y f (x) xœÊœ Ê œ
"" "
mm
(b) The graph of y f (x) is a line through the origin with slope .œ
" "
m
36. y mx b x f (x) x ; the graph of f (x) is a line with slope and y-intercept .?w%?Î% ? Î Î
y
mm m m m m
bb b" " ""
37. (a) y x 1 x y 1 f (x) x 1?w % ?Î % ?Î
"
(b) y x b x y b f (x) x b?w % ?Î % ?Î
"
(c) Their graphs will be parallel to one another and lie on
opposite sides of the line y x equidistant from thatœ
line.

433. Section 7.1 Inverse Functions and Their Derivatives 429
38. (a) y x 1 x y 1 f (x) 1 x;?Îw% ?Îw% ?Î
"
the lines intersect at a right angle
(b) y x b x y b f (x) b x;?Î w % ?Î w % ? Î
"
the lines intersect at a right angle
(c) Such a function is its own inverse.

39. Let x x be two numbers in the domain of an increasing function f. Then, either x x or
"# "#Á
x x which implies f(x ) f(x ) or f(x ) f(x ), since f(x) is increasing. In either case,
"# "#"#
f(x ) f(x ) and f is one-to-one. Similar arguments hold if f is decreasing.
"#Á
40. f(x) is increasing since x x x x ; 3#" # "
"" " "$% w$ w ?% ? ?
3636dx3 dx
5 5 df df
&"
"
ˆ‰
3
41. f(x) is increasing since x x 27x 27x ; y 27x x y f (x) x ;#"
$ $ $ "Î$ " "Î$
# "
""$% $ ? %? % ?
33
81x x
df df
dx dx 81x 9 9x
œÊœ œœ
# #Î$ """
&"

¸
1
3
x

42. f(x) is decreasing since x x 1 8x 1 8x ; y 1 8x x (1 y) f (x) (1 x) ;#"
$ $ $ "Î$ " "Î$
# "
""
##Ê œ Êœ Ê œ
24x (1 x)
df df
dx dx 24x 6 6( x)
?Î % ? ? ?Î Î
# #Î$ """
"
&"

¸
1
2
!Ÿ 1x

43. f(x) is decreasing since x x (1 x ) (1 x ) ; y (1 x) x 1 y f (x) 1 x ;#" # "
$ $ $ "Î$ " "Î$Ê œ Êœ Ê œ
3(1 x) x
df df
dx dx 3(1 x) 3 3x
?Î Î % ? ? ?Î
# #Î$ """

&"

¹
1x"

44. f(x) is increasing since x x x x ; y x x y f (x) x ;#"
&Î$ &Î$
# "
&Î$ $Î& " $Î&$% $ ? %? % ? x x
df 5 df 3 3
dx 3 dx 5 x 5x
œÊœ œœ
#Î$ #Î& "
&"

¹5
3
x
$"Î
45. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x x then"#Á
f(x ) f(x ), so f(x ) f(x ) and therefore g(x ) g(x ). Therefore g(x) is one-to-one as well.
"# " # " #'Î'Î '
46. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x x then
"#Á
f(x ) f(x ), so , and therefore h(x ) h(x )."# " #
""ÁÁ Á
f(x ) f(x )
47. The composite is one-to-one also. The reasoning: If x x then g(x ) g(x ) because g is one-to-one. Since"# " #ÁÁ
g(x ) g(x ), we also have f(g(x )) f(g(x )) because f is one-to-one. Thus, f g is one-to-one because
"# " #ÁÁ ‰
x x f(g(x )) f(g(x )).
"# " #ÁÊ Á
48. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x x in the domain of g
"#Á
with g(x ) g(x ). For these numbers we would also have f(g(x )) f(g(x )), contradicting the assumption
"# " #œœ
that f g is one-to-one.‰

434. 430 Chapter 7 Transcendental Functions
49. The first integral is the area between f(x) and the x-axis
over a x b. The second integral is the area betweenŸŸ
f(x) and the y-axis for f(a) y f(b). The sum of theŸŸ
integrals is the area of the larger rectangle with corners
at (0 0), (b 0), (b f(b)) and (0 f(b)) minus the area of theßßß ß
smaller rectangle with vertices at (0 0), (a 0), (a f(a)) andßßß
(0 f(a)). That is, the sum of the integrals is bf(b) af(a).)Î

50. f x . Thus if ad bc , f x is either always positive or always negative. Hence f x i
w w
!!

ab ab ab?? Î'
abab
ab ab
cx d a ax b c
cx d cx d
ad bc
s
either always increasing or always decreasing. If follows that f x is one-to-one if ad bc .ab Á!
51. (g f)(x) x g(f(x)) x g (f(x))f (x) 1‰œÊ œÊ œ
ww
52. W(a) f (y) a dy 0 2 x[f(a) f(x)] dx S(a); W (t) f (f(t)) a f (t)?????''
f(a) a
f(a) a
11 1’“ ’ “ab a b
" # w " # w
# #
t a f (t); also S(t) 2 f(t) x dx 2 xf(x) dx f(t)t f(t)a 2 xf(x) dx?Î ? Î ? Î Î111111ab c d
##w ## '' '
aa a
tt t
S (t) t f (t) 2 tf(t) a f (t) 2 tf(t) t a f (t) W (t) S (t). Therefore, W(t) S(t)%? w Î Î ?Î % ? ?
w#w #w ##w ww
111 11 ab
for all t [a b].−ß
53-60. Example CAS commands:
:Maple
with( plots );#53
f := x -> sqrt(3*x-2);
domain := 2/3 .. 4;
x0 := 3;
Df := D(f); # (a)
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"],
title="#53(a) (Section 7.1)" );
q1 := solve( y=f(x), x ); # (b)
g := unapply( q1, y );
m1 := Df(x0); # (c)
t1 := f(x0)+m1*(x-x0);
y=t1;
m2 := 1/Df(x0); # (d)
t2 := g(f(x0)) + m2*(x-f(x0));
y=t2;
domaing := map(f,domain); # (e)
p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ):
p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ):
p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ):
p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ):
p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ):
display( [p1,p2,p3,p4,p5], scaling=constrained, title="#53(e) (Section 7.1)" );
(assigned function and values for a, b, and x0 may vary)Mathematica:
If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica
to do this. See section 2.5 for details.
<<Miscellaneous `RealOnly`
Clear[x, y]

435. Section 7.1 Inverse Functions and Their Derivatives 431
{a,b} = { 2, 1}; x0 = 1/2 ;
f[x_] = (3x 2) / (2x 11)
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[y == f[x], x]
g[y_] = x /. solx[[1]]
y0 = f[x0]
ftan[x_] = y0 f'[x0] (x-x0)
gtan[y_] = x0 1/ f'[x0] (y y0)
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Line[{{x0, y0},{y0, x0}}], PlotRange {{a,b},{a,b}}, AspectRatio Automatic]ÄÄÄ
61-62. Example CAS commands:
:Maple
with( plots );
eq := cos(y) = x^(1/5);
domain := 0 .. 1;
x0 := 1/2;
f := unapply( solve( eq, y ), x ); # (a)
Df := D(f);
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"],
title="#62(a) (Section 7.1)" );
q1 := solve( eq, x ); # (b)
g := unapply( q1, y );
m1 := Df(x0); # (c)
t1 := f(x0)+m1*(x-x0);
y=t1;
m2 := 1/Df(x0); # (d)
t2 := g(f(x0)) + m2*(x-f(x0));
y=t2;
domaing := map(f,domain); # (e)
p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ):
p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ):
p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ):
p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ):
p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ):
display( [p1,p2,p3,p4,p5], scaling=constrained, title="#62(e) (Section 7.1)" );
(assigned function and values for a, b, and x0 may vary)Mathematica:
For problems 61 and 62, the code is just slightly altered. At times, different "parts" of solutions need to be used, as in the
definitions of f[x] and g[y]
Clear[x, y]
{a,b} = {0, 1}; x0 = 1/2 ;
eqn = Cos[y] == x
1/5
soly = Solve[eqn, y]
f[x_] = y /. soly[[2]]
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[eqn, x]
g[y_] = x /. solx[[1]]
y0 = f[x0]
ftan[x_] = y0 f'[x0] (x x0)

436. 432 Chapter 7 Transcendental Functions
gtan[y_] = x0 1/ f'[x0] (y y0)wÎ
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Line[{{x0, y0},{y0, x0}}], PlotRange {{a, b}, {a, b}}, AspectRatio Automatic]ÄÄÄ
7.2 NATURAL LOGARITHMS
1. (a) ln 0.75 ln ln 3 ln 4 ln 3 ln 2 ln 3 2 ln 2? ?? ?
3
4
#
(b) ln ln 4 ln 9 ln 2 ln 3 2 ln 2 2 ln 3
4
9
?Î?Î? Î
##
(c) ln ln 1 ln 2 ln 2 (d) ln 9 ln 9 ln 3 ln 3
" ""
#
$ #
?? ? ? ?
È
33 3
2
(e) ln 3 2 ln 3 ln 2 ln 3 ln 2
È
? ?
"Î# "
#
(f) ln 13.5 ln 13.5 ln ln 3 ln 2 (3 ln 3 ln 2)
È
ab????
""" "
#### #
$27
2. (a) ln ln 1 3 ln 5 3 ln 5 (b) ln 9.8 ln ln 7 ln 5 2 ln 7 ln 5
" #
125 5
49
?Î ?Î ? ? Î? Î
(c) ln 7 7 ln 7 ln 7 (d) ln 1225 ln 35 2 ln 35 2 ln 5 2 ln 7
È
?? ?? ?
$Î# #
#
3
(e) ln 0.056 ln ln 7 ln 5 ln 7 3 ln 5???
7
125
$
(f)
ln 35 ln
ln 25 ln 5
ln 5 ln 7 ln 7! "
##
"
7
œœ
3. (a) ln sin ln ln ln 5 (b) ln 3x 9x ln ln ln (x 3))Î? ? Îw? ?Έ‰ ˆ‰

ab Š‹
sin sin 3x 9x
5 3x 3x
))
Š‹
sin
5
# "

(c) ln 4t ln 2 ln 4t ln 2 ln 2t ln 2 ln ln t
"
# #
%## %
ab ab
È
Š‹Î? Î? Î? ?
2t

4. (a) ln sec ln cos ln [(sec )(cos )] ln 1 0)) ))? ??
(b) ln (8x 4) ln 2 ln (8x 4) ln 4 ln ln (2x 1)?? ?
# ! ˆ‰
8x 4
4
(c) 3 ln t 1 ln (t 1) 3 ln t 1 ln (t 1) 3 ln t 1 ln (t 1) ln
$
È
ab ab ˆ‰ Š‹# ##
"Î$ " "
!
ÎÎ w? Î Î w? ÎÎ w?
3(t 1)
(t 1)(t )
ln (t 1)?
5. y ln 3x y (3) 6. y ln kx y (k) xœÊœ œ œÊœ œ
w w""ˆ‰ ˆ‰
1
3x x kx
7. y ln t (2t) 8. y ln t tœÊœ œ œ Êœ œab ˆ‰ ˆ ‰ ˆ ‰ˆ ‰
# $Î# "Î#" "
#
dy dy
dt t t dt 2t
2 33
t

9. y ln ln 3x 3x?? %? Î ?Î
3
xdx3x x
dy" # ""
ˆ‰ab
&"
10. y ln ln 10x 10x?? %? Î ?Î
10
x dx 10x x
dy" # ""
ˆ‰ ab
&"
11. y ln ( 1) (1) 12. y ln (2 2) (2)?w%? ? ? w%? ?))
dy dy
d 1 1 d 2 1)) ) ) ) )
ˆ‰ ˆ ‰
"" " "
!! #!!
13. y ln x 3x 14. y (ln x) 3(ln x) (ln x)œÊœ œ œ Êœ œ
$# $# "dy dy 3(ln x)
dx x x dx dx x
3 d ˆ‰ab
$
†
15. y t(ln t) (ln t) 2t(ln t) (ln t) (ln t) (ln t) 2 ln t? %?w ?w?w
## # #dy
dt dt t
d 2t ln t
†
16. y t ln t t(ln t) (ln t) t(ln t) (ln t) (ln t)?? %? w ? w
È
"Î# "Î# "Î# "Î# "
##
dy t(ln t)
dt dt t
d
†

(ln t)?
"Î# "
#(ln t)

17. y ln x x ln x x ln x???
xx x4x
416dx 4x16
dy
ww w$
$$ "
†

437. Section 7.2 Natural Logarithms 433
18. y ln x x ln x x ln x?Î%?wÎ?
xx x3x
39dx 3x9
dy

## "
†
19. y œÊœ œ
ln t
tdt
dy
t (ln t)(1)
tt
1 ln t
ˆ‰

""
t

20. y ?%? ? ?Î
" " ln t 1 ln t ln t
tdt t t
dy
t ( ln t)(1)
t
ˆ‰

"
t
"

21. y yœÊœ œ œ
ln x
1 ln x x(1 ln x)
(1 ln x) (ln x)
(1 ln x) (1 ln x)!!
w
!
!!
"
ˆ‰ ˆ‰

xxxxx
ln x ln x


22. y y 1?%? ? ?Î
x ln x ln x
1 ln x (1 ln x) (1 ln x)
(1 ln x) (x ln x)
(1 ln x)
( ln x) ln x
!!!
w
!
!
"
ˆ‰ˆ‰ln x x†

xx



23. y ln (ln x) yœÊœ œ
w "" "ˆ‰ˆ‰
ln x x x ln x
24. y ln (ln (ln x)) y (ln (ln x)) (ln x)œÊœ œ œ
w """"
ln (ln x) dx ln (ln x) ln x dx x (ln x) ln (ln x)
dd
†††
25. y [sin (ln ) cos (ln )] [sin (ln ) cos (ln )] cos (ln ) sin (ln )?w%?ww Î)) ) ) )) ) )
dy
d)))
$ ‘††
""
sin (ln ) cos (ln ) cos (ln ) sin (ln ) 2 cos (ln )?wwÎ?)))) )
26. y ln (sec tan ) sec ?w%? ? ?)) )
dy sec (tan sec )
d sec tan tan sec
sec tan sec
))) ))
)) ) )) )!
!!
!

27. y ln ln x ln (x 1) y??ÎÎw%?ÎÎ?Î?Î
ŸŸ ŸŸŸ !
! ##!!!
w !!
xx 1 x x 1 2x(x 1) 2x(x 1)
2(x 1) x 3x 2
È
ˆ‰
28. y ln ln (1 x) ln (1 x) y ( 1)??wÎÎ%?ÎÎ? ?
" " " " " " "
## # #"
w1x 1x1x
1x 1x 1x (1x)( x) 1x
cd $ ‘ˆ‰ ’“

29. y œÊœ œ œ
1ln t 2
1 ln t dt (1 ln t) (1 ln t) t(1 ln t)
dy!

(1 ln t) (1 ln t) ˆ‰ ˆ ‰

t t tttt
ln t ln t

30. y ln t ln t ln t ln t ln t tœœ Êœ œÉÈ ˆ‰ ˆ‰ ˆ‰ˆ‰ ˆ‰
"Î# "Î# "Î# "Î# "Î#
"Î# "Î# "Î#
"""
##
dy
dt dt dt
dd
t
†††
"
ln t tœœ
""""
##
"Î# "Î#
"Î#
ˆ‰ ††
t
4t ln t
"
ÉÈ
31. y ln (sec (ln )) (sec (ln )) (ln )œÊœ œ œ)))
dy sec (ln ) tan (ln ) tan (ln )
dsec(ln )d sec(ln ) d
dd
))) )) )
)) )"
††
32. y ln (ln sin ln cos ) ln (1 2 ln ) ??wÎw%?ÎÎ
È
sin cos
1 2 ln d sin cos 1 ln
dy cos sin ))
)))))
))
!# # !#
""
)) ) ˆ‰
2

cot tan ?ÎÎ
"
#!
’“))
4
(1 2 ln )))
33. y ln 5 ln x 1 ln (1 x) y ( 1)??wÎÎ%?ÎÎ?wŠ‹ ab ˆ‰
ab
È
x1
1x
5 2x 10x
x1 1x x1 (1x)

%

!

#w """"
###
†
34. y ln [5 ln (x 1) 20 ln (x 2)] y??wÎw%?Î?É ˆ‰ ’“
(x 1) (x 2) 4(x 1)
(x 2) x 1 x (x 1)(x 2)
5205!
!# #!!##!!
"" w
%

?Î
53x2
(x 1)(x )#!!#
!
’“
35. y ln t dt ln x x ln 2x ln x x ln ?%? Î ?Î'
x2
x
&
&
&
È Š‹ Š‹Š‹
È
ab kk É
dy
dx dx dx
dxdx x
2
# #
##
††
kk
È

438. 434 Chapter 7 Transcendental Functions
36. y ln t dt ln x x ln x x ln x x ln x x?%? Î ? Î'

x
x
$
dy
dx dx dx 3
dd
ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ÈÈÈÈ È È
$$ $ ""#Î$ "Î#
#
††
?
ln x
3 x
ln x
2x
$
$

È
È
È
È
37. dx ln x ln 2 ln 3 ln 38. dx ln 3x 2 ln 2 ln 5 ln ''
" "
"3 1
2 0
" # !
$ "#x33x 5
23 2
? ?Î? ? Î ?Î?cd c dkk k k
39. dy ln y 25 C 40. dr ln 4r 5 C''
2y
y25 4r 5
8r


# #
?Îw ? Îwkk kk
41. dt ln 2 cos t ln 3 ln 1 ln 3; or let u 2 cos t du sin t dt with t 0'
0

sin t
2cos t !
? Î ?Î? ?Î %? ?cdkk
1
u 1 and t u 3 dt du ln u ln 3 ln 1 ln 3? ?? ? ? ??1 ''
01
3

sin t
2cos t u
" $
"
cdkk
42. d ln 1 4 cos ln 1 2 ln 3 ln ; or let u 1 4 cos du 4 sin d'
0
3
&
4 sin
14 cos 3
)
)
1

Î$
!
"
)) )))?Î ?Î?Î? ?Î %?cdkkkk
with 0 u 3 and u 1 d du ln u ln 3 ln )) )? % ?Î ? % ?Î % ? ? ?Î ?
1)
)314 cos u 3
4 sin ''
03
31
& "
"

"" "
$
cdkk
43. Let u ln x du dx; x 1 u 0 and x 2 u ln 2;œÊœ œÊœ œÊœ
"
x
dx 2u du u (ln 2)''
10
2ln 2
ln 2
0
2 ln x
x
œœœ cd
##
44. Let u ln x du dx; x 2 u ln 2 and x 4 u ln 4;œÊœ œÊœ œÊœ
"
x
du ln u ln (ln 4) ln (ln 2) ln ln ln ln 2''
2ln 2
4ln 4
ln 4
ln 2
dx ln 4 ln 2 2 ln 2
x ln x u ln 2 ln 2 ln 2
? ???? ? ?
"
cd ˆ‰ ˆ ‰Š‹

45. Let u ln x du dx; x 2 u ln 2 and x 4 u ln 4;œÊœ œÊœ œÊœ
"
x
u du''
2ln 2
4ln 4
ln 4
ln 2
dx
x(ln x) u ln 4 ln ln ln 2 2 ln ln ln 2 ln 4
???????
# """""""""
## ###$‘
46. Let u ln x du dx; x 2 u ln 2 and x 16 u ln 16;œÊœ œÊœ œÊœ
"
x
u du u ln 16 ln 2 4 ln 2 ln 2 2 ln 2 ln 2 ln 2''
2ln 2
16 ln 16
ln 16
ln 2
dx
2x ln x
È
? ? ?Î? Î?Î?
"
#
"Î# "Î#
$‘ È ÈÈ È ÈÈÈ
47. Let u 6 3 tan t du 3 sec t dt;?w % ?
#
dt ln u C ln 6 3 tan t C''3 sec t du
63 tan t u

!
?? w?w wkk k k
48. Let u 2 sec y du sec y tan y dy;?w % ?
dy ln u C ln 2 sec y C''
sec y tan y
sec y u
du
#!
?? w?w wkk k k
49. Let u cos du sin dx 2 du sin dx; x 0 u 1 and x u ;?%?Î %Î? ?%? ?%?
xx x
2### # #
" "1
È
tan dx dx 2 2 ln u 2 ln 2 ln 2 ln 2'' '
00 1
22 12
12
1
&& &
&
xdu sin
cos u 2#
"
??????
x
x




cdkk
È
È
50. Let u sin t du cos t dt; t u and t u 1;œÊœ œÊœ œÊœ
11
4 2
"
#È
cot t dt dt ln u ln ln 2'''


&&&
&&4412
221
?????
cos t du
sin t u 2

cdkk
È
"
"Î #
"
È È
51. Let u sin du cos d 6 du 2 cos d ; u and u ;œÊœ Êœ œÊœ œÊœ
)) )1
333 3
3""
## #
))))1
È
2 cot d d 6 6 ln u 6 ln ln 6 ln 3 ln 27'''


&& &
&
&
&2212
32
32
12
)
3u
2 cos
sin
du 3
))??????


3
3


cdkk Š‹
È
È
##
"

439. Section 7.2 Natural Logarithms 435
52. Let u cos 3x du 3 sin 3x dx 2 du 6 sin 3x dx; x 0 u 1 and x u ;?%?Î %Î? ?%??%?
1
1 2#
"
È
6 tan 3x dx dx 2 2 ln u 2 ln ln 1 2 ln 2 ln 2'' '
00 1
12 12 1 2
12
1
&& &
&
??????
6 sin 3x du
cos 3x u 2


cdkk
È
"
È
53. ; let u 1 x du dx; ln u C'' 'dx dx dx du
2x 2x 2x1 x x 2x1 x uÈÈÈ ÈÈÈ ˆ‰ ˆ‰!! # !
"
??? ?? È kk '
ln 1 x C ln 1 x C??¸¸ ˆ‰ÈÈ
54. Let u sec x tan x du sec x tan x sec x dx (sec x)(tan x sec x) dx sec x dx ;?w %? w ? w % ? ab
# du
u
(ln u) du 2(ln u) C 2 ln (sec x tan x) C' sec x dx du
ln (sec x tan x) uln u uÈÈ !
"Î# "Î#"
?? ? w? ww' ' Ȇ
55. y x(x 1) (x(x 1)) ln y ln (x(x 1)) 2 ln y ln (x) ln (x 1) ?w?w %? w% ?ww%?wÈ
"Î# " ""
# !
2y
yxx1

y x(x 1)%? w w ? ?
w ŸŸŸ !Ÿ
#!!
!!
!ˆ‰ ˆ ‰È
x x 1 2x(x 1)
x(x 1) (2x 1) 2x
2x(x 1)
È
È
56. y x 1 (x 1) ln y ln x 1 2 ln (x 1) ?wÎ%? ww Î%? wÈab c d ab ˆ‰
## ""
##
# y
yx1x1
2x 2


y x +1 (x 1) x 1 (x 1)%? Î w ? w Î ?
w ## # #

"
ÈÈab a b ˆ‰ ’“
xxxx1
x1 x1 x1(x1)
2x x1 x1
x1(x1)



ab
abkk
È
57. y ln y [ln t ln (t 1)] ?? %?Îw%?ÎÉ ˆ‰ ˆ ‰
tt
t1 t1 ydt t t1
dy
!! # # !
"Î#
"""""
%? Î ? ?
dy
dt t 1 t t 1 t 1 t(t 1)
tt
2t(t 1)
"""" " "
#! ! #! ! !
ÉɈ‰ ’“
È $
58. y [t(t 1)] ln y [ln t ln (t 1)] ????É ˆ‰
1
t(t 1) y dt t t 1
dy
!# #!
"Î# """""
? ?
dy
dt t(t 1) t(t 1)
12t 2t1
2t t
Ÿ!Ÿ!
#! !
!
É ’“
ab


59. y 3 (sin ) ( 3) sin ln y ln ( 3) ln (sin ) ?w ?w % ? ww % ? wÈ))) ) ) )
"Î# """
##! yd ( 3) sin
dy cos
)) )
)
3 (sin ) cot %? w w
dy
d2(3)))È
’“)) ) "
!
60. y (tan ) 2 1 (tan )(2 1) ln y ln (tan ) ln (2 1) ????)) )) ) )È ˆ‰ˆ ‰
"Î# """
###! yd tan 1
dy sec 2
)) )
)

(tan ) 2 1 sec 2 1%? w w ? ww
dy
dtan 1
sec tan
21)))
))
)
)) ))ÈÈ
Š‹ ab

"
#!
#
!È
61. y t(t 1)(t 2) ln y ln t ln (t 1) ln (t 2) ?w w% ? w ww w% ?w w
""""
!!#ydt t t1 t
dy
t(t 1)(t+2) t(t 1)(t 2) 3t 6t 2%?w w w ?w w ?ww
dy (t 1)(t 2) t(t 2) t(t 1)
dt t t 1 t t(t 1)(t 2)
ˆ‰ ’“
"" "
!!# !!
! !!!!! #
62. y ln y ln 1 ln t ln (t 1) ln (t 2) ? % ?ÎÎwÎw% ?ÎÎÎ
" """"
!! !!#t(t1)(t2) ydt t t1 t
dy
%? ÎÎ Î ?
dy (t 1)(t 2) t(t 2) t(t 1)
dt t(t 1)(t 2) t t 1 t t(t 1)(t ) t(t 1)(t 2)
"""" "
!! ! !# !!# !!
! !!!!!
$ ‘ ’“
?
3t 6t 2
t3t2t



!!
!!ab
63. y ln y ln ( 5) ln ln (cos ) ?%?wÎÎ %?Îw
) )
)) ) ) ) )
!ŸŸŸ
!
5 sin
cos y d 5 cos
dy
)))
tan %? Îw
dy
d cos 5
5
)))) )
)
ˆ‰ˆ ‰
!ŸŸ
!
)

440. 436 Chapter 7 Transcendental Functions
64. y ln y ln ln (sin ) ln (sec ) ?%?w Î %?wÎ
)) )
) )) ) )
)) sin cos
sec y d sin 2 sec
dy (sec )(tan )
È
)) )
"""
#
’“
cot tan %? w Î
dy
d
sin
sec ))
))
)
È
ˆ‰
""
#
))
65. y ln y ln x ln x 1 ln (x 1) ???
xx 1 y
(x 1)
2x2
3 y xx13(x1)
È

Î$


!
!
""
#!!
#
ab
y%? w Î
w !
!
"
!!
xx 1
(x 1) xx13(x1)
x2
È

Î$
’“
66. y ln y [10 ln (x 1) 5 ln (2x 1)] ???É
(x 1) y
(2x1) y x1 2x1
55!
!# !!
"
"!
%

y%? Î
w !
!! !
É ˆ‰
(x 1)
(2x1) x1 2x1
55
Ÿ!
%
67. y ln y ln x ln (x 2) ln x 1 ???
$


É cd ab ˆ‰
x(x 2) y
x1 3 y 3x x x1
2x
! #
"""" #
y %? w Î
w"""
#3x1xx x1
x(x 2) 2x
$

É ˆ‰
68. y ln y ln x ln (x 1) ln (x 2) ln x 1 ln (2x 3)? % ? w ww ÎÎ wÎ w
$

É cd ab
x(x 1)(x 2)
x1(2x3) 3

!!
" #
ab
y %? wwÎ Î
w""""
# 3 x 1 (2x 3) x x 1 x x 1 2x 3
x(x 1)(x 2) 2x 2
$

É ˆ‰
ab
69. (a) f(x) ln (cos x) f (x) tan x 0 x 0; f (x) 0 for x 0 and f (x) 0 forœ Ê œ œ œ Ê œ Ÿ
www sin x
cos x 4
1
0 x there is a relative maximum at x 0 with f(0) ln (cos 0) ln 1 0; f ln cos" % ? ? ? ? Î ? Î
1 11
3 44
ˆ‰ ˆ ‰ ˆ‰
ln ln 2 and f ln cos ln ln 2. Therefore, the absolute minimum occurs at?? ? ??Š‹ ˆ‰ ˆ ‰ ˆ‰
"" "
##È
2 33
11
x with f ln 2 and the absolute maximum occurs at x 0 with f(0) 0.?? ??
11
33
ˆ‰
(b) f(x) cos (ln x) f (x) 0 x 1; f (x) 0 for x 1 and f (x) 0 for 1 x 2œÊœœÊœŸ Ÿ
www "
#
sin (ln x)
x
there is a relative maximum at x 1 with f(1) cos (ln 1) cos 0 1; f cos lnÊœœœœœ ˆ‰ ˆ ‰ ˆ‰
""
##
cos ( ln 2) cos (ln 2) and f(2) cos (ln 2). Therefore, the absolute minimum occurs at x and? ? ? ?
"
#
x 2 with f f(2) cos (ln 2), and the absolute maximum occurs at x 1 with f(1) 1.œœœ œœˆ‰
"
#
70. (a) f(x) x ln x f (x) 1 ; if x 1, then f (x) 0 which means that f(x) is increasing?Î % ?Î $ $
ww "
x
(b) f(1) 1 ln 1 1 f(x) x ln x 0, if x 1 by part (a) x ln x if x 1?Î ? % ?Î $ $ % $ $
71. (ln 2x ln x) dx ( ln x ln 2 ln x) dx (ln 2) dx (ln 2)(5 1) ln 2 ln 16
'' '
11 1
55 5
?? ???
%
72. A tan x dx tan x dx dx dx ln cos x ln cos x?Î w ? Î ? Î''''
"&"&
&&

40 40
0303
!
Î%
Î$
!
sin x sin x
cos x cos x
cdcdkk kk
1
1
ln 1 ln ln ln 1 ln 2 ln 2 ln 2?Î ÎÎ? w?Š‹ ˆ‰ È
""
##È
2
3
73. V dy 4 dy 4 ln y 1 4 (ln 4 ln 1) 4 ln 4?????11111''
00
33
Š‹ cdkk
2
y1 y1È!
#
"
!
$
!
74. V cot x dx dx ln (sin x) ln 1 ln ln 2??? ??111 11''


&&
&&66
22
cos x
sin x
cd ˆ‰
1
1
Î#
Î'
"
#
75. V 2 x dx 2 dx 2 ln x 2 ln 2 ln 2 (2 ln 2) ln 2 ln 16???????1111111''
12 12
22&&
ˆ‰ ˆ ‰ cdkk
"" " #
"Î# #
%
xx

441. Section 7.2 Natural Logarithms 437
76. V dx 27 dx 27 ln x 9 27 (ln 36 ln 9)????1111''
00
33
Š‹ cdab
9x
x9
È
$
!
#
$
$
!
27 (ln 4 ln 9 ln 9) 27 ln 4 54 ln 2???111
77. (a) y ln x 1 y 1 1 L 1 y dx?Î %w ?wÎ ?w ? %? w
x x x4 x4
84x4x4x

ab ab ˆ‰ Š‹Š‹ É
w# "
#
##
w
# '
4
8
dx dx ln x (8 ln 8) (2 ln 4) 6 ln 2?????''
44
88
x4 x x
4x 4 x 8

!Ÿ
)
%
ˆ‰ ’“ kk
(b) x 2 ln 1 1 1?Î %?Î%w ?wÎ?w ?ˆ‰ ˆ‰ Š‹ Š‹ Š‹Š‹
yy y y y16y16
4 4 dy 8 y dy 8 y 8y 8y
dx 2 dx 2
#
####


L 1 dy dy dy 2 ln y (9 2 ln 12) (1 2 ln 4)%? w ? ? w ? w ?w Îw'''
444
12 12 12
Ê Š‹ Š ‹ ’ “
dx 2
dy 8y 8 y 16
y16 y y
#"#
!
%

8 2 ln 3 8 ln 9? ?
78. L 1 dx y ln x C ln x C since x 0 0 ln 1 C C 0 y ln x? w % ?%? w? w $%? w% ?%?'
1
2
É kk
""
xdxx
dy

79. (a) M x dx 1, M dx dx , M dx ln x ln 2
yx
111 1
222 2??? ??????''' 'ˆ‰ ˆ ‰ˆ‰ $‘ cdkk
""""""""
#
#
"
#
"x2xxx2x4x

x 1.44 and y 0.36Êœ œ ¸ œ œ ¸
M
Mln 2 Mln 2
My x4 "
ˆ‰
"
(b)
80. (a) M x dx x dx x 42; M dx dx
y x
11 1 1
16 16 16 16œœœœœ œ'' ' 'Š‹ Š ‹Š‹ $‘
" """"Î# $Î#
"'
"
"
##È ÈÈ
x x
2
3xx
ln x ln 4, M dx 2x 6 x 7 and yœœœœœÊœœœœ
""
#
"'
"
"Î#
"'
"
cdkk $ ‘ '
1
16Èx
M
MM6
M ln 4
y x
(b) M x dx 4 dx 60, M dx x dx
y x
111 1
16 16 16 16œœœœ œ#''' 'Š‹Š‹ Š ‹Š‹Š‹
"" "
#
$Î#
ÈÈ ÈÈ È
xx xx
44
x
4 x 3, M dx 4 dx 4 ln x 4 ln 16 x and?Î ? ? ? ? ? % ? ?$ ‘ Š‹Š‹ cdkk
"Î#
"'
"
"" "'
" ''
11
16 16ÈÈxx
4 15
xMln 16
M
y
y œœ
M
M4 ln 16
3x
81. 1 at ( 3) y x ln x C; y 3 at x 1 C 2 y x ln x 2
dy
dx x
?w Þ) %?w w ? ?% ?%?w w
"
kk kk
82. sec x tan x C and 1 tan 0 C tan x 1 y (tan x 1) dx
d y dy dy dx dx dx


? %?w ?w%?w%? w
# '
ln sec x x C and 0 ln sec 0 0 C C 0 y ln sec x x????kk kk kk
""" 83. (a) L(x) f(0) f (0) x, and f(x) ln (1 x) f (x) 1 L(x) ln 1 1 x L(x) x?w ? w% ? ?% ?w % ?
ww "
!
†† k ¸
x0
x0
1x
(b) Let f x ln x . Since f x on , the graph of f is concave down on this interval and theab a b abœ " œ ! Ò!ß !Þ"Ó
ww "
!Ÿabx

largest error in the linear approximation will occur when x . This error is ln to fiveœ !Þ" !Þ" "Þ" ¸ !Þ!!%'*ab
decimal places.

442. 438 Chapter 7 Transcendental Functions
(c) The approximation y x for ln (1 x) is best for smaller?
positive values of x; in particular for 0 x 0.1 in theŸŸ
graph. As x increases, so does the error x ln (1 x).Îw
From the graph an upper bound for the error is
0.5 ln (1 0.5) 0.095; i.e., E(x) 0.095 forÎwÑ " kk
0 x 0.5. Note from the graph that 0.1 ln (1 0.1)"" Î w
0.00469 estimates the error in replacing ln (1 x) byÑw
x over 0 x 0.1. This is consistent with the estimateŸŸ
given in part (b) above.

84. For all positive values of x, and 0 . Since and haveln ln ln a ln x ln a ln x
d1a1d 11
dx x x dx x x
a a
x x
cd c d?? ?? ÎÎ a
x 2
the same derivative, then ln ln a ln x C for some constant C. Since this equation holds for all positve values of x,
a
x
?
it must be true for x 1 ln ln 1 ln x C 0 ln x C ln ln x C. By part 3 we know that?% ? Î w?Î w% ?Î w
11
xx
ln ln x C 0 ln ln a ln x.
1a xx
?Î % ? % ? Î
85. y ln kx y ln x ln k; thus the graph of??
y ln kx is the graph of y ln x shifted verticallyœœ
by ln k, k 0.$

86. To turn the arches upside down we would use the
formula y ln sin x ln .? ?kk
"
kksin x

87. (a) (b) y . Since sin x and cos x are less than
w
!
œllll
cos x
asin x
or equal to 1, we have for a$Þ y for all x.
" "
" "
w
aa
ŸŸ
Thus, y for all x the graph of y lookslim
a‰!−
w
œ! Ê
more and more horizontal as a .Ðw&

443. Section 7.3 The Exponential Function 439
88. (a) The graph of y x ln x to be concave?È appears
upward for all x 0.$

(b) y x ln x y y 1 0 x 4 x 16.? Î % ? Î% ?Î w? Î w ?% ?%?ÈÈ Š‹
www"" " ""
#È
È
x xxx4 4x
x
$
Thus, y 0 if 0 x 16 and y 0 if x 16 so a point of inflection exists at x 16. The graph of
ww ww
œ
y x ln x closely resembles a straight line for x 10 and it is impossible to discuss the point of? È
inflection visually from the graph.
7.3 THE EXPONENTIAL FUNCTION
1. (a) e 7.2 (b) e (c) e e
ln 7.2 ln x ln x ln y ln x y
e xy
x
œœœœœ
ÐÎÑ ""


ln x
2. (a) e x y (b) e (c) e e
ln x y ln 0 3 ln x ln 2 ln x 2
e 0.3
xab


## Þ ÐÎÑ ""
#
? ? ? ? ?
ln 0 3
11 1
3. (a) 2 ln e 2 ln e (2) ln e 1 (b) ln ln e ln (e ln e) ln e 1È ˆ‰ abœœ œ œ œœ
"Î# "
#
e
(c) ln e x y ln e x y
ab ## ##xy

?Î Î ?Î Îab
4. (a) ln e (sec )(ln e) sec (b) ln e e (ln e) eˆ‰ ab
sec e x x)
œœ œœ))
ab
x
(c) ln e ln e ln x 2 ln xˆ‰ Š‹
2lnx lnx
œœœ

#
5. ln y 2t 4 e e y e 6. ln y t 5 e e y e?w% ? %? ?Îw% ? %?
lny 2t4 2t4 lny t5 t5
7. ln (y 40) 5t e e y 40 e y e 40? ???
ln y 40) 5t 5t 5tÐ
8. ln (1 2y) t e e 1 2y e 2y e 1 y? ????
ln 1 2y) t t t eÐ "
#
Š‹
t
9. ln (y 1) ln 2 x ln x ln (y 1) ln 2 ln x x ln x e e eÎÎ ?w % ÎÎ Î ?% ?% ? % ? ˆ‰
y1 y1
2x x
ln x x
#
ˆ‰
y1
2x&
y 1 2xe y 2xe 1%Î? %? w
xx
10. ln y 1 ln (y 1) ln (sin x) ln ln (sin x) ln (y 1) ln (sin x) e eab Š‹
# ÐÑ Ð Ñ"
!
ÎÎ w? % ? % Î? % ?
y
y1
ln y 1 ln sin x

y 1 sin x y sin x 1%Î? %? w
11. (a) e 4 ln e ln 4 2k ln e ln 2 2k 2 ln 2 k ln 2
2k 2k
œÊ œ Ê œ Ê œ Êœ
#
(b) 100e 200 e 2 ln e ln 2 10k ln e ln 2 10k ln 2 k
10k 10k 10k ln 2
10
œÊœÊ œÊ œÊœÊœ
(c) e a ln e ln a ln e ln a ln a k 1000 ln a
k 1000 k 1000 kk
1000 1000
ÎÎ
œÊ œÊ œÊ œÊœ
12. (a) e ln e ln 4 5k ln e ln 4 5k ln 4 k
5k 5k
4 5
ln 4
? % ? % ?Î % ?Î % ?Î
" "
(b) 80e 1 e 80 ln e ln 80 k ln e ln 80 k ln 80
kk k
?% ? % ? % ?Î %?Î
" "
(c) e 0.8 e 0.8 (0.8) 0.8 k 1
ÐÞÑ Þln 0 8 k ln 0 8 k
k
œÊ œÊ œÊœˆ‰

444. 440 Chapter 7 Transcendental Functions
13. (a) e 27 ln e ln 3 ( 0.3t) ln e 3 ln 3 0.3t 3 ln 3 t 10 ln 3
Þ Þ $03t 03t
?% ? %Î ? %Î? %?Î
(b) e ln e ln 2 kt ln e ln 2 t
kt kt ln 2
k
?% ? ? ?Î %?Î
"
#
"
(c) e 0.4 e 0.4 0.2 0.4 ln 0.2 ln 0.4 t ln 0.2 ln 0.4 t
ÐÞÑ Þln 0 2 t ln 0 2 t t
t
ln 0.4
ln 0.2
œÊ œÊ œÊ œ Ê œ Êœˆ‰
14. (a) e 1000 ln e ln 1000 ( 0.01t) ln e ln 1000 0.01t ln 1000 t 100 ln 1000
Þ Þ0 01t 0 01t
? ? ???
(b) e ln e ln 10 kt ln e ln 10 kt ln 10 t
kt kt
10 k
ln 10
? % ? ? ?Î % ?Î % ?Î
" "
(c) e e 2 2 2 t 1
Ð Ñ " ""
#
ln 2 t ln 2 t
t
? ???ˆ‰
15. e x ln e ln x t 2 ln x t 4(ln x)
ÈÈtt
œÊ œ Ê œ Êœ
## # È
16. e e e e e ln e ln e t x 2x 1
x 2x1 t x 2x1 t x 2x1 t

!!! !! #
?% ?% ? %?ww
17. y e y e ( 5x) y 5e ?%? Î%?Î
w w5x 5x 5x d
dx
18. y e y e y eœÊœ Êœ
2x 3 2x 3 2x 3 d2x 2
dx 3 3
ÎwÎ w Î ˆ‰
19. y e y e (5 7x) y 7e?%? Î%?Î
57x 57x 57x d
dx
w w
20. y e y e 4 x x y 2x e?%? w%?w
ˆ‰ ˆ‰ ˆ‰ÈÈ È
È
4xx 4xx 4xx d2
dx x
w #w

ˆ‰È Š‹
21. y xe e y e xe e xe ?Î%?w Î?
xx x x x x w
ab
22. y (1 2x) e y 2e (1 2x)e ( 2x) y 2e 2(1 2x) e 4xe?w % ? ww Î % ? Î w ?Î
w w 2x 2x 2x 2x 2x 2x d
dx
23. y x 2x 2 e y (2x 2)e x 2x 2 e x e?Îw %?Î wÎw ?ab ab
#w## xxxx
24. y 9x 6x 2 e y (18x 6)e 9x 6x 2 e (3x) y (18x 6)e 3 9x 6x 2 e?Îw %?Î wÎw %?Î w Îwab ab ab
#w # w # 3x 3x 3x 3x 3x d
dx
27x eœ
#3x
25. y e (sin cos ) y e (sin cos ) e (cos sin ) 2e cos ???
))))
)) )) )) )
w
26. y ln 3 e ln 3 ln ln e ln 3 ln 1????ˆ‰)) ))
"))
))
dy
d
27. y cos e sin e e sin e e 2 e sin e?%?Î ?Î Î?Š‹ Š‹Š‹Š ‹Š‹ Š‹ Š‹ ab
#)))))))
)) )
dy
dd d
dd
))
28. y e cos 5 3 e cos 5 cos 5 e ( 2 ) 5(sin 5 ) e?%? w ÎÎ)) ) ))) )))
$ # # # $ # $ #))) )
))
dy
dd
d
ab a bˆ‰ ˆ‰
e (3 cos 5 2 cos 5 5 sin 5 )?))))))
##)
29. y ln 3te ln 3 ln t ln e ln 3 ln t t 1?????ab
"tt dy
dt t t
1t
30. y ln 2e sin t ln 2 ln e ln sin t ln 2 t ln sin t 1 (sin t) 1???? ?ab ˆ‰
"tt dy
dt sin t dt sin t
dcos t
œ
cos t sin t
sin t

31. y ln ln e ln 1 e ln 1+e 1 1 e 1??Îw?Î %?Î w?Î?
e de
1e 1e 1e 1e
dy
dd


! !!!
"")) ) )
))
ˆ‰ ˆ‰ ˆ‰ˆ‰)

445. Section 7.3 The Exponential Function 441
32. y ln ln ln 1 1??Îw%? Î w
È
ÈÈÈ
)
))) )) )1 1
dy
dd d
dd
! !
""
ÈÈ È È
Š‹ Š‹Š‹Š‹Š‹)) ) )
?Î ???Š‹Š‹Š ‹Š‹
"" " " " "
#!#

!#!
#!ÈÈ ÈÈ
Š‹
ÈÈ
Š‹ Š‹
ÈÈ ab)) ))
))
)) ))
))1
1
21 1
1
"
33. y e e e te e te (cos t) (1 t sin t) e???%?w ?Î
Ä!¸cos t ln t cos t ln t cos t cos t cos t cos tdy
dt dt
d
34. y e ln t 1 e (cos t) ln t 1 e e ln t 1 (cos t)?w%? ww? ww
sin t sin t sin t sin tdy
dt t t
22
ab ab ab $ ‘
###
35. sin e dt y sin e (ln x)'
0
ln x
tlnx dsin x
dx x
Êœ œ
wˆ‰ †
36. y ln t dt y ln e e ln e e (2x) 2e 4 x e 4 x?%? Î ?Î'
e
e
2x 2x 4 x 4 x 2x 4 xdd d
dx dx dx
4x
2x
w
abab ab Š ‹Š‹ Š‹ ˆ‰ ˆ‰ÈȆ† †
ÈÈ È
4xe 4 x e 4xe 8e?Î ?Î
2x 4 x 2x 4 x 2
x È Š‹
ÈÈ
È
37. ln y e sin x y y e (sin x) e cos x y e sin x e cos x?%? w%Î ?
yyyyy
yy
Š‹ Š ‹ab
""ww w
y e cos x yʜʜ
ww

Š‹
1 ye sin x ye cos x
y 1 ye sin x
y
yy
y
38. ln xy e ln x ln y e y 1 y e y e e?%w?%w ?w % Î ?Î
xy xy xy xy xy
xy y x
www "" " "
Š‹ Š ‹ab
y yʜʜ
ww " "

Š‹
1ye
yx x1ye
xe yxe
xy xy xy
xy

ab
ab
39. e sin (x 3y) 2e 1 3y cos (x 3y) 1 3y 3y 1
2x 2x 2e 2e
cos (x 3y) cos (x 3y)
?w%?w w%w? %? Î ab
www
!!
2x 2x
yʜ
w
!
2e cos (x 3y)
3 cos (x 3y)
2x
40. tan y e ln x sec y y e y?w % ?w% ?
xx
xx
xe cos y
ab
#w w " !Ÿab
x
41. e 5e dx 5e C 42. 2e 3e dx 2e e C''ab a b
3x x x x 2x x 2xe 3
3
w?Îw Î ?ww

#
3x
43. e dx e e e 3 2 1 44. e dx e e e 1 2 1''
ln 2 ln 2
ln 3 ln 3
x x ln 3 ln 2 x x ln 2ln 3 0
ln 2 ln 2
? ?Î?Î? ?Î ?Îw?Îw?cd c d

!

45. 8e dx 8e C 46. 2e dx e C''ÐÑ ÐÑ Ð Ñ Ð Ñx1 x1 2x1 2x1
?w ?w
47. e dx 2e 2 e e 2 e e 2(3 2) 2
'
ln 4
ln 9
x 2 x 2 ln 9 2 ln 4) 2 ln 3 ln 2
ln 9
ln 4
ÎÎ ÐÑÎÐÎ
??Î?Î?Î?$‘ $ ‘ˆ ‰
48. e dx 4e 4 e e 4 e 1 4(2 1) 4
'
0
ln 16
x4 x4 ln16 4 0 ln2
ln 16
0
ÎÎ ÐÑÎ
??Î?Î?Î?$‘ˆ ‰ˆ‰
49. Let u r du r dr 2 du r dr;œÊœ Ê œ
"Î# "Î# "Î#"
#
dr e r dr 2 e du 2e C 2e C 2e C''e
r
ruurr
r
È
È
???w?w?w'
"
†
"Î#
50. Let u r du r dr 2 du r dr;?Î % ?Î % Î ?
"Î# "Î# "Î# "
#
dr e r dr 2 e du 2e C 2e C'' 'e
r
rurr
& r
È
È
??Î?Îw?Îw
"Î#
†

446. 442 Chapter 7 Transcendental Functions
51. Let u t du 2t dt du 2t dt;?Î % ?Î % Î ?
#
2te dt e du e C e C''tuu t

?Î ?Î w ?Î w
52. Let u t du 4t dt du t dt;œÊ œ Ê œ
%$ $ "
4
t e dt e du e C''$ ""tut
44
ww
??w
53. Let u du dx du dx;?% ?Î %Î?
"" "
xx x

dx e du e C e C''e
x
uu 1x
1xÎ

?Î ?Îw?Î w
Î
54. Let u x du 2x dx du x dx;?Î % ? % ?
# $ $ "
#
dx e x dx e du e C e C e C'' 'e
x
xuux1x
&Î

$
1x
???w?w?w
$ Î """ "
### #
†
55. Let u tan du sec d ; 0 u 0, u 1;œÊœ œÊœœÊœ)))) )
# 1
4
1 e sec d sec d e du tan e tan tan (0) e e'''
000
441
tan u u 4
0 4
11
) 1 1
ÎÎ
## "!Î "
!
ˆ‰ $ ‘cd cd a b ˆ‰w?w?w?ÎwÎ)) )) )
(1 0) (e 1) e?ÎwÎ?
56. Let u cot du csc d ; u 1, u 0;?%?Î ?%??%?)))))
# 11
42
1 e csc d csc d e du cot e cot cot e e'''
11
11
) 1
1
11
ÎÎ
ÎÎ
## !"Î
Î
!
"
441
220
cot u u 2
4 24
ˆ‰ $ ‘cdcd ab ˆ‰ ˆ‰w?Î?ÎÎ?ÎwÎÎ)) )) )
(0 1) (1 e) e?wÎÎ?
57. Let u sec t du sec t tan t dt sec t tan t dt;œÊœ Êœ1 111 11
du
1
e sec ( t) tan ( t) dt e du C C''sec t u eeÐÑ "1
111
11 ??w?w
usect "$"
58. Let u csc ( t) du csc ( t) cot ( t) dt;?w%?Îw w111
e csc ( t) cot ( t) dt e du e C e C
''csc t u u csc tÄ!¸ Ä!¸1 1
11w w ?Î ?Î w ?Î w
59. Let u e du e dv 2 du 2e dv; v ln u , v ln u ;œÊ œ Ê œ œ Êœ œ Êœ
vv v
66
1111
##
2e cos e dv 2 cos u du 2 sin u 2 sin sin 2 1 1''
ln 6 6
ln 2 2
vv 2
6 6ÐÎÑ Î
ÐÎÑ Î
Î
Î ##
"
11
11
1
1
11
???Î?Î? cd $ ‘ˆ ‰ˆ‰ ˆ‰
60. Let u e du 2xe dx; x 0 u 1, x ln u e ;œÊœ œÊœœ Êœ œ
xx ln

È
11 1
2xe cos e dx cos u du sin u sin ( ) sin (1) sin (1) 0.84147''
01
ln
xx
È
11
1

Š‹ cd? ? ? Î ?Î ÑÎ
"
1
61. Let u 1 e du e dr;?w % ?
rr dr du ln u C ln 1 e C''e
1e u
r
r
r
!
"
??w?ww kk a b
62. dx dx;''"
!!1e e 1
e
xx
xœ
&
&
let u e 1 du e dx du e dx;?w%?Î %Î?
xxx
dx du ln u C ln e 1 C''e
e1 u
x
&
&x
x
!
"
?Î ?Î w ?Î w wkk a b
63. e sin e 2 y e sin e 2 dt;
dy
dt
tt tt
?Î%? Îab ab '
let u e 2 du e dt y sin u du cos u C cos e 2 C; y(ln 2) 0?Î% ? %? ?Î w?Î Î w ?
tt t ' ab

447. Section 7.3 The Exponential Function 443
cos e 2 C 0 cos (2 2) C 0 C cos 0 1; thus, y 1 cos e 2%Î Î w ? %Î Î w ? % ? ? ? ΠΈ‰ ab
ln 2 t
64. e sec e y e sec e dt;
dy
dt
tt tt
œÊœ
# #
ab ab11 '
let u e du e dt du e dt y sec u du tan u C? % ?Î % Î ? % ?Î ?Î w11
# """tt t
111 '
tan e C; y(ln 4) tan e C tan C?Î w ? %Î w ? %Î w ?
""""
111111
ab ˆ‰ ˆ‰111
tln4 222
4
† (1) C C ; thus, y tan e%Î w ? % ? ? Î
""
11111
23 3 t
ab1
65. 2e 2e C; x 0 and 0 0 2e C C 2; thus 2e 2
d y dy dy dy
dx dx dx dx
xx x


? % ?Î w ? ? % ?Î w % ? ?Î w
!
y2e 2xC; x0 and y1 12e C C 1 y2e 2x12e x 1%? w w ? ?%? w % ?Î%? w Î? wÎ
!
"""
x xx
ab
66. 1 e t e C; t 1 and 0 0 1 e C C e 1; thus
d y dy dy
dt dt dt
2t 2t


?Î % ?Î w ? ? % ?Î w % ? Î
"""
###
##
t e e 1 y t e e 1 t C ; t 1 and y 1 e e 1 C
dy
dt 4 4
2t 2t
?Î wÎ%?Î w Îw ? ?Î%ÎÞ?ÎwÎw
"" "" " """
## # # # #
### ##
""
ˆ‰
C e y t e e 1 t e%?ÎÎ %?Î w ÎÎw"
"" " " " ""
####
## # #
44 4
2t
ˆ‰ˆ‰
67. f(x) e 2x f (x) e 2; f (x) 0 e 2 x ln 2; f(0) 1, the absolute maximum;?Î % ?Î ?% ?%? ?
xx x ww
f(ln 2) 2 2 ln 2 0.613706, the absolute minimum; f(1) e 2 0.71828, a relative or local maximum?Î Ñ ?ÎÑ
since f (x) e is always positive.
ww
œ
x
68. The function f(x) 2e has a maximum whenever sin 1 and a minimum whenever sin 1.???Î
sin x 2 xxÐÎÑ
##
Therefore the maximums occur at x 2k(2 ) and the minimums occur at x 3 2k(2 ), where k is any?w ? w11 11
integer. The maximum is 2e 5.43656 and the minimum is 0.73576.¸¸
2
e
69. f(x) x ln f (x) 2x ln x x 2x ln x x(2 ln x 1); f (x) 0 x 0 or?%?w Î?Î?Îw ?%?
#w ## w""""
xx x
Š‹ab"
x
ln x . Since x 0 is not in the domain of f, x e . Also, f (x) 0 for 0 x andœ œ œ œ
"""
#
"Î# w
ÈÈee
f (x) 0 for x . Therefore, f ln e ln e ln e is the absolute maximum value
w "Î#""""""
##
$ ? ? ??
ÈÈee ee ee
Š‹ È
of f assumed at x .œ
"
Èe
70. f(x) (x 3) e f (x) 2(x 3) e (x 3) e?Î % ? Î wÎ
#w #xxx
(x 3) e (2 x 3) (x 1)(x 3) e ; thus?Î wÎ?Î Î
xx
f (x) 0 for x 1 or x 3, and f (x) 0 for
ww

1 x 3 f(1) 4e 10.87 is a local maximum and % ? Ñ
f(3) 0 is a local minimum. Since f(x) 0 for all x,œ 
f(3) 0 is also an absolute minimum.œ

71. e e dx e e e 3 1 2 2'
0
ln 3
ab ’“Š ‹Š‹ ˆ‰ˆ‰
2x x x ln 3 eee9 8
ln 3
0
Î?Î?ÎÎÎ?ÎÎÎ?Î?
2x 2ln3
######
! "
!
72. e e dx 2e 2e 2e 2e 2e 2e (4 1) (2 2) 5 4 1'
0
2 ln 2
ˆ‰ $ ‘ˆ ‰ ab
x2 x2 x2 x2 ln2 ln2
2ln2
0
ÎÎ Î Î ! !
Î ? w ? w Î w ?wÎw?Î?
73. L 1 dx y e C; y(0) 0 0 e C C 1 y e 1? w % ? %? w ?%?w% ?Î%? Î'
0
1
x2
É
ee
4dx
dy x2 0
x x2 Î
#
Î
&

448. 444 Chapter 7 Transcendental Functions
74. S 2 1 dy 2 1 e 2 e dy?w? wÎw11 ''
00
ln 2 ln 2
ˆ‰ ˆ‰ ˆ‰É É ab
ee ee ee
4
2y 2y
yy yy yy
"
## #
#

&& &
2 dy 2 dy e 2 e???ww11 '
0
ln 2
ˆ‰ˆ‰ ˆ‰É '' ab
ee ee ee
00
ln 2 ln 2
2y 2y
yy yy yy
!! !
## # #
# #

&& &
1
dy e 2y e e 2 ln 2 e 0?wÎ ? wÎ ÎwÎ
11
## # # # # # #
"" " """
$ ‘ $ ‘ˆ‰ˆ‰
2y 2y 2ln2 2ln2
ln 2
0
4 2 ln 2 2 2 ln 2 ln 2?wÎ?Îw?w
11
## # #
""""
ˆ‰ˆ‰ˆ‰††
48 16
15
1
75. (a) (x ln x x C) x ln x 1 0 ln x
d
dx x
Îw ? w Îw?†
"
(b) average value ln x dx x ln x x [(e ln e e) (1 ln 1 1)]??Î?ÎÎÎ
"" "
e1 e1 e1
e
1'
1
e
cd
(ee1)?Îw?
""
e1 e 1
76. average value dx ln x ln 2 ln 1 ln 2? ? ?Î?
""

#
"21 x'
1
2
cdkk
77. (a) f(x) e f (x) e ; L(x) f(0) f (0)(x 0) L(x) 1 x?% ? ? w Î% ?w
xx ww
(b) f(0) 1 and L(0) 1 error 0; f(0.2) e 1.22140 and L(0.2) 1.2 error 0.02140œœÊœœ¸ œÊ¸
02
(c) Since y e 0, the tangent line
ww
?$
x
approximation always lies below the curve y e .œ
x
Thus L(x) x 1 never overestimates e .?w
x

78. (a) e e e e 1 e for all x; e e e e
xx xx x x x x xx
eee
eÐÑ! ""
œœœÊœ œ œ œ
x
x
xx
"


ˆ‰
(b) y e ln y x ln e x x x x e e y e e eœÊœ œœÊœÊœÊœab ab
xx ln yxxxxxxxx x
##""#
79. f(x) ln(x) 1 f (x) x x x x 2 ln (x ) . Then x 2?Î% ?% ?Î % ?Î ?
w "
!! Ÿ

x
n1 n n1 n n
ln (x ) 1 n
1
xn
ˆ‰
cd
x 2.61370564, x 2.71624393 and x 2.71828183, where we have used Newton's method.Êœ œ œ
#$ &
80. e x and ln e x for all x 0
ln x x
??$ ab
81. Note that y ln x and e x are the same curve; ln x dx area under the curve between 1 and a;œœ œ
y
1
a '
e dy area to the left of the curve between 0 and ln a. The sum of these areas is equal to the area of the rectangle'
0
ln a
y
œ
ln x dx e dy a ln a.%w?
''
10
aln a
y
82. (a) y e y e 0 for all x the graph of y e is always concave upward?% ?$ % ?
xx x ww
(b) area of the trapezoid ABCD e dx area of the trapezoid AEFD (AB CD)(ln b ln a) Ê '
ln a
ln b
x
"
#
e dx (ln b ln a). Now (AB CD) is the height of the midpoint '
ln a
ln b
x
Š‹
ee
ln a ln b
!Ÿ
##
M e since the curve containing the points B and C is linear e (ln b ln a)?%Î
!ß _ !ß _ln a ln b 2 ln a ln b 2
e dx (ln b ln a) '
ln a
ln b
x
Š‹
ee
ln a ln b
!
#
(c) e dx e e e b a, so part (b) implies that'
ln a
ln b
xxlnblna ln b
ln a
??Î?Îcd
e (ln b ln a) b a (ln b ln a) e
ÐÑÎ ÐÑÎ
##
lna lnb 2 lna lnb 2 ee ba ab
ln b ln a
Ê Š‹
ln a ln b
ee ee abÊÊÊ
ln a 2 ln b 2 ba ab ba ab ba ab
ln b ln a ln b ln a ln b ln a
ln a ln bÎÎ
# # #
†
ÈÈ È

449. Section 7.4 a and log x 445
x
a
7.4 a and log x
x
a
1. (a) 5 7 (b) 8 2 (c) 1.3 75
log 7 log 2 log 755 8 13
œœ œ
È %
(d) log 16 log 4 2 log 4 2 1 2 (e) log 3 log 3 log 3 1 0.5
44 4 3 3 3œœ œœ œ œ œœœ
# "Î# """
###
††
È
(f) log log 4 1 log 4 1 1 1
44 4
ˆ‰
" "
4
??Î?Î?Î †
2. (a) 2 3 (b) 10 (c) 7
log 3 log 1 2 log 72 10
œœœ
!_  "
#
1

(d) log 121 log 11 2 log 11 2 1 2
11 11 11œœ œœ
#
†
(e) log 11 log 121 log 121 1
121 121 121œœ œœ
"Î# """
###ˆ‰ ˆ‰ †
(f) log log 3 2 log 3 2 1 2
33 3
ˆ‰
" #
9
??Î?Î?Î †
3. (a) Let z log x 4 x 2 x 2 x 2 xœ Ê œÊ œÊ œÊ œ
4
z2z z z ab È
#
(b) Let z log x 3 x 3 x 3 x 9 xœ ÊœÊ œÊ œÊœ
3
zz 2zz ab
# ###
(c) log e log 2 sin x
22
ln 2 sin x sin xab
! 
œœ
4. (a) Let z log 3x 5 3x 25 9xœÊœÊœ
5
zzab
## %
(b) log e x
e
xabœ
(c) log 2 log 4
44
ee
ˆ‰
xx x
sin x sin x e sin x
œœ
"$ Î#
#
5. (a) (b)
log x log x
log x ln ln 3 ln ln x ln 2 log x ln ln 8 ln ln x ln
ln x ln x ln x ln 3 ln 3 ln x ln x ln x ln 8 3 ln 22 2
3 8
œƒœ œ œƒœ œ
## ##
††
2
œ3
(c) 2
log a
log a ln x ln x ln x ln a ln x
ln a ln a ln a ln x 2 ln xx
x



œƒ œ œ œ †
6. (a)
log x
log x ln 9 ln 3 2 ln 3 ln x 2
ln x ln x ln x ln 3 19
3
œƒœ œ †
(b)
log x
log x ln x ln 10
ln x ln x ln x ln 2
ln 10 ln 2 ln 10
ln 2
10
2
œƒœ œ
ÈÈ ˆ‰
ˆ‰
"

"
†
(c)
log b
log a ln a ln b ln a ln a ln a
ln b ln a ln b ln b ln ba
b
œƒœ œ † ˆ‰
#
7. 3 2 5 7 5 x x 12
log 7 log 5 log x32 5ÐÑ ÐÑ ÐÑ
w?%w?%?
8. 8 e x 7 3 5 x 3x 0 x 3x 2 (x 1)(x 2) x 1 or x 2
log 3 ln 5 log 3x87ÐÑ # Ð Ñ # #
Î ? Î % Î? Î % ? Î w? Î Î % ? ?
9. 3 5e 3 10 x 5x 6 x 5x 6 0 (x 2)(x 3) 0 x 2 or x 3
log x ln x log 2310ab

? Î % ? Î% Î w?% Î Î?%? ?†
ÐÑ # #
10. ln e 4 log 100 1 4 log 10 1 x (2) 1 0w ? %w ? %w ? %wÎ?
ÐÑ # # """"2 log x log x
xxxxx
10 10
2 44 ab
&

ˆ‰
x 2x 1 0 (x 1) 0 x 1%Îw?%Î?%?
##
11. y 2 y 2 ln 2 12. y 3 y 3 (ln 3)( 1) 3 ln 3?%? ? %? Î?Î
xx xx x w w "" "
13. y 5 5 (ln 5) s 5œÊœ œ
ss sdy
ds
ln 5
2s
ˆ‰ Š‹
"
#
"Î#
È
14. y 2 2 (ln 2)2s ln 2 s2 (ln 4)s2œÊœ œ œ
ss s sdy
ds

abŠ‹
#
15. y x y x 16. y t (1 e) t?%? ? %?Î
w
1
!Ÿ  Ÿ Ÿ11ee dy
dt
17. y (cos ) 2 (cos ) (sin )?%?Î)))
Î&

2 21 dy
d) È
"

450. 446 Chapter 7 Transcendental Functions
18. y (ln ) (ln )œÊœ œ)1)
dy (ln )
d)))
1) !Ÿ 1ˆ‰
"
Ä&_"1
19. y 7 ln 7 7 ln 7 (ln 7)(sec tan ) 7 (ln 7) (sec tan )œÊœ œ
sec sec sec"" " dy
d)
ab )) ))
#
20. y 3 ln 3 3 ln 3 (ln 3) sec 3 (ln 3) secœÊœ œ
tan tan tan"" " dy
d)
ab
###
))
21. y 2 2 ln 2 (cos 3t)(3) (3 cos 3t) 2 (ln 2)œÊœ œ
sin 3t sin 3t sin 3tdy
dt
ab ab
22. y 5 5 ln 5 (sin 2t)(2) (2 sin 2t) 5 (ln 5)œÊœ œ
"" "cos 2t cos 2t cos 2tdy
dt
ab ab
23. y log 5 (5)œœÊœ œ
2)
ln 5
ln d ln 5 ln
dy)
)))## #
"" "
ˆ‰ˆ‰
24. y log (1 ln 3) (ln 3)?w? %? ?
3 )
ln (1 ln 3) dy
ln 3 d ln 3 1 ln 3 1 ln 3
! "" "
!!
)
)) )
ˆ‰ˆ ‰
25. y 2 3 y?w ?w ? %?
ln x ln x ln x ln x ln x 3
ln 4 ln 4 ln 4 ln 4 ln 4 x ln 4

w
26. y (x ln x) y 1?Î?Î? Î %? Î?
x ln e ln x x ln x x 1
ln 5 2 ln 5 ln 5 2 ln 5 ln 5 ln 5 x 2x ln 5## # #
""" w
ˆ‰ ˆ‰ˆ ‰
27. y log r log r (2 ln r)œœœÊœ œ
24† ˆ‰ˆ‰ ˆ‰ ’“
ln r ln r ln r 2 ln r
ln ln 4 (ln 2)(ln 4) dr (ln 2)(ln 4) r r(ln 2)(ln 4)
dy
#
""

28. y log r log r (2 ln r)œœœÊœ œ
39† ˆ‰ˆ‰ ˆ‰ ’“
ln r ln r ln r 2 ln r
ln 3 ln 9 (ln 3)(ln 9) dr (ln 3)(ln 9) r r(ln 3)(ln 9)
dy

""
29. y log ln ln (x 1) ln (x 1)?????wÎÎ
3
ln 3Š‹ˆ‰ ˆ‰
x1 x1
x1 ln 3 ln 3 x1
(ln 3) ln
!!
lnˆ‰
x1
x1
ln 3 x1
x1
"

" Š‹
%?Î?
dy
dx x 1 x 1 (x 1)(x 1)
2""

30. y log log lnœœœœœ
55
ln 5 ln 5 2Ɉ‰ ˆ‰ ˆ‰ ˆ‰
”•
7x 7x ln 5 7x
3x 2 3x 2 ln 5 ln 5 3x!! # #!#
"
! _ ln lnˆ‰ ˆ‰
7x 7x
3x 2 3x 2
ln 5 2

$Î&
ln 7x ln (3x 2) ?Î w%?Î ? ?
"" "
## ! ! !
dy (3x 2) 3x
dx 2 7x 2 (3x 2) 2x(3x 2) x(3x 2)
73
††
31. y sin (log ) sin sin cos sin (log ) cos (log )??%?w ?w))) ) ) )
7 77
ˆ‰ ˆ‰ $ ‘ˆ ‰ˆ‰
ln ln ln
ln 7 d ln 7 ln 7 ln 7 ln 7
dy)))
))
""
32. y log œœ œ
7
ˆ‰
sin cos
e
ln (sin ) ln (cos ) ln e ln 2 ln (sin ) ln (cos ) ln 2
ln 7 ln 7
)) )) ))))
"
#



(cot tan 1 ln 2)%? Î ÎÎ? Î ÎÎ
dy
d (sin )(ln 7) (cos )(ln 7) ln 7 ln 7 ln 7
cos sin ln 2
)) )
)) ""
ˆ‰ ))
33. y log e yœœœÊœ
5
x
ln e x
ln 5 ln 5 ln 5
x
w "
34. y log œœ œ
2Š‹
xe
2x 1
ln x ln e ln 2 ln x 1
ln 2 ln 2
2 ln x 2 ln 2 ln (x 1)
"

È
È
!

y%? Î ? ?
w Ÿ!
#! ! !#
23x4
x ln 2 (ln 2)(x 1) 2x(x 1)(ln 2) 2x(x 1) ln
4(x 1) x
35. y 3 3 3 (ln 3) log 3 3œœ Êœ œ
log t ln t ln 2 ln t ln 2 log t
22 2 ! _!   ! _!  dy
dt t ln 2 t
cdab ˆ‰
""
36. y 3 log (log t) œœœÊœ œ
82
3 ln (log t) dy
ln 8 ln 8 dt ln 8 (ln t)/(ln 2) t ln t(ln t)(ln 8)
3 ln 33 2
ln t
ln 2
ˆ‰
ˆ‰ ˆ ‰’“
""
#
œ
"
#t(ln t)(ln )

451. Section 7.4 a and log x 447
x
a
37. y log 8t 3 ln t ??? ?w%?
2
ln 2ab
ln 8 ln t
ln ln dt t
3 ln 2 (ln 2)(ln t) dy!
##
! "
ˆ‰
ln 2
38. y t sin t sin t t cos t????%?w
t ln e
ln 3 ln 3 ln 3 dt
t ln 3 t(sin t)(ln 3) dy
Š‹ˆ‰
ˆ‰
ln 3
sin t
sin t
39. y (x 1) ln y ln (x 1) x ln (x 1) ln (x 1) x y (x 1) ln (x 1)?w % ? w ? w % ? ww % ?w w w
xx x y
y(x1) x1
x

†
"
!!
w
$ ‘
40. y x ln y ln x (x 1) ln x ln x (x 1) ln x 1?%? ?w %?ww ?ww
!߉  !߉ xx y
yxx

ˆ‰
""
y x 1 ln x%? ww
w " !ß x1ˆ‰
x
41. y t t t ln y ln t ln t (ln t)???%?? %? w ?wˆ ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰ˆ‰È
t t
tt
"Î# Î#
#####
"" " " &w ttln t
ydt t
dy
t%? w
dy
dt
ln t
ˆ‰ˆ ‰È
t
##
"
42. y t t ln y ln t t (ln t) t (ln t) t?? % ? ? % ? w ?
"$ "$tt t

ˆ‰ ˆ ‰ ˆ‰
"Î# "Î# "Î# ŸŸ Ÿ!
#ydt t
dy ln t 2
2tÈ
tʜ
dy
dt
ln t 2
2t
Š‹
!
È t
43. y (sin x) ln y ln (sin x) x ln (sin x) ln (sin x) x y (sin x) ln (sin x) x cot x?%? ? %? w %? w
xx x y
ysin x
cos x

ˆ‰ cd
w
44. y x ln y ln x (sin x)(ln x) (cos x)(ln x) (sin x)?%? ? %? w ?
sin x sin x y sin x x (ln x)(cos x)
yxx

ˆ‰
" !
yxʜ
w ! sin x
’“
sin x x(ln x)(cos x)
x
45. y x , x 0 ln y (ln x) 2(ln x) y x?$%? %? %?
ln x ln x #w "y
yx x
ln x

ˆ‰ abŠ‹
46. y (ln x) ln y (ln x) ln (ln x) ln (ln x) (ln x) (ln x)?%? %? w ?w
ln x yln(ln x)
yx ln xdx x x
d

ˆ‰ ˆ ‰
"" "
y (ln x)ʜ
w !Ÿ
Š‹
ln (ln x)
x ln x
47. 5 dx C 48. (1.3) dx C''x x
?w ? w
5
ln 5 ln (1.3)
(1.3)
x x
49. 2 d d''
00
11
""
))? ? ?Î?Î? ?ˆ‰
–—
""""
##
"
!
) Š‹
Š‹ Š‹ Š‹ Š‹

w
%%%%
wwww
%%
ww

ln ln ln ln
2(ln 1 ln 2) ln 2
50. 5 d d (1 25)''
"
"22
00
"
))???Î?Î??

"""
!
#

ˆ‰
–—
5 ln 1 ln 5 ln 5
ln ln ln ln
24 24
) Š‹ Š‹
Š‹ Š‹ Š‹ Š‹


55
5555
&
51. Let u x du 2x dx du x dx; x 1 u 1, x 2 u 2;œÊœ Ê œ œÊœ œ Êœ
# "
# È
x2 dx 2 du 2 2''
11
22
u

abx 2
ln 2 ln 2 ln

???Î?ˆ‰ $‘ˆ‰ ab
"" " "
### #
#
"
#"
u
52. Let u x du x dx 2 du ; x 1 u 1, x 4 u 2;œÊœ Ê œ œÊœœÊœ
"Î# "Î# "
#
dx
xÈ
dx 2 x dx 2 2 du 2 2'' '
11 1
44 2
xu
224
x ln ln ln
x
È
????Î?

†
"Î# $ #
## #
#
"
"
’“ ˆ‰ab
u1

452. 448 Chapter 7 Transcendental Functions
53. Let u cos t du sin t dt du sin t dt; t 0 u 1, t u 0;?%?Î %Î? ?%??%?
1
#
7 sin t dt 7 du 7 7''
01
20
cos t u
&
?Î ? Î ? Î ?$ ‘ˆ‰ ab
76
ln 7 ln 7 ln 7
u!
"
" !
54. Let u tan t du sec t dt; t 0 u 0, t u 1;œÊœ œÊœœÊœ
# 1
4
sec t dt du''
00
41
tan t u
&

ˆ‰ ˆ‰ ˆ ‰ˆ‰ ˆ‰
–—
’“
"" """#
"
!
!
3 3 ln 3 3 3 3 ln 3
ln
2
???ÎÎ?
Š‹
Š‹


3
3
u
55. Let u x ln u 2x ln x 2 ln x (2x) 2u(ln x 1) du x (1 ln x) dx;?% ? % ? w %? w% ? w
2x 2x "" "
#udx x dx
du du
ˆ‰
x 2 u 2 16, x 4 u 4 65,536;œÊœ œ œÊœ œ
%)
x (1 ln x) dx du u (65,536 16) 32,760''
216
4 65 536
2x
65 536
16
w? ? ? Î??
"""
### #


cd
65,520
56. Let u ln x du dx; x 1 u 0, x 2 u ln 2;œÊœ œÊœœÊœ
"
x
dx 2 du 2 2''
10
2ln 2
uln2
ln 2
0
22 2
x ln ln ln
ln x ln 2
???Î? $‘ˆ‰ ab
u
## #
"" !
57. 3x dx C 58. x dx C''
Î&

3 21
?w ?w
3x x
31 2
"$
31 2w
È È
!
"
59. 2 1 x dx x 3 60. x dx''
0 1
3 e
2 ln 2 122
e
1
Š‹ ’“
È $‘w? ? ????
Î& Î&

)Ä )Ä ! Ÿ
$
!
##
"xe121
ln ln 2 ln 2 ln
ln 2 ln 2 ln 2
61. dx dx; u ln x du dx'
log x
x ln 10 x x
ln x10
œœÊœ'ˆ‰ˆ‰ $ ‘
""
dx u du u C CÐ??w?w''ˆ‰ˆ‰ ˆ‰ˆ‰
ln x
ln 10 x ln 10 ln 10 2 ln 10
(ln x)"" ""
#
#

62. dx dx; u ln x du dx; x 1 u 0, x 4 u ln 4''
11
44
log x
xln x x
ln x
2
œœÊœœÊœœÊœˆ‰ˆ‰ $ ‘
#
""
dx u du u (ln 4) ln 4Ä œ œ œ œœœ''
10
4ln 4
ln 4
0
ˆ‰ˆ‰ ˆ‰ ˆ‰ $‘ˆ‰ $ ‘
ln x
ln x ln ln ln 2 ln ln 4
(ln 4) (ln 4)
# # ## ## #
" " "" "" ##

63. dx dx dx (ln x) (ln 4) (ln 1) (ln 4)'' '
11 1
44 4
ln 2 log x
xxln x
ln 2 ln x ln x
2
????Î?ˆ‰ˆ‰ $ ‘ cd
####
"" " ####
%
"
(2 ln 2) 2(ln 2)œœ
"
#
##
64. dx dx (ln x) (ln e) (ln 1) 1''
11
ee
e
1
2 ln 10 (log x) (ln 10)(2 ln x)
x (ln 10) x
10
???Î? ˆ‰ cd
" ###
65. dx ln (x 2) dx''
00
22
log (x 2) (ln (x 2)) (ln 4) (ln 2)
x ln x ln ln
2!!
!# # !# # # # # #
""" "
#
!
?w ? ?Îcd ˆ ‰ ˆ‰ ˆ‰ ’“ ’ “

ln 2?Î?ˆ‰’“
"
## # #ln
4(ln 2) (ln 2) 3

66. dx ln (10x) dx''
110 110
10 10&&
log (10x) (ln (10x)) (ln 100) (ln 1)
x ln 10 10x ln 10 0 ln 10 0
000
10
???Î
""" "
###
"!
"Î"!
cd ˆ‰ ˆ‰ ˆ‰ ’“ ’ “

ln 10œœ#ˆ‰’“
"
#
0
ln 10 0
4(ln 10)

67. dx ln (x 1) dx''
00
99
2 log (x 1) (ln (x 1)) (ln 10) (ln )
x 1 ln 10 x 1 ln 10 ln 10 2
222
10!!Ÿ
!!##
"
*
!
?w? ? Î ˆ‰ ˆ‰ ˆ‰ ’“ ’ “

ln 10œ
68. dx ln (x 1) dx ln 2''
22
33
2 log (x 1) (ln (x 1)) (ln 2) (ln )
x 1 ln 2 x 1 ln 2 ln 2 2
222
2"
##
"
$
#
?Î? ? Î? ˆ ‰ ˆ‰ ˆ‰ ’“ ’ “

453. Section 7.4 a and log x 449
x
a
69. dx (ln 10) dx; u ln x du dx'' 'dx ln 10
x log x ln x x ln x x x
10
œœ œÊœˆ ‰ˆ‰ ˆ ‰ˆ‰ $ ‘
""" "
(ln 10) dx (ln 10) du (ln 10) ln u C (ln 10) ln ln x CÐ??w?w ''ˆ‰ˆ‰ kk k k
"" "
ln x x u
70. (ln 8) dx (ln 8) C C'' 'dx dx
x (log x) x 1 ln x
x
(ln x) (ln x) (ln 8)
8

?? ? w?Îw
ˆ‰
ln x
ln 8
##

71. dt ln t ln ln x ln 1 ln (ln x), x 1'
1
ln x
"
t
ln x
1
??Î? $cd kkkk
72. dt ln t ln e ln 1 x ln e x'
1
e
e x
x
x
"
t 1
??Î??cdkk
73. dt ln t ln ln 1 ln 1 ln x ln 1 ln x, x 0'
1
1/x
x
1
""
tx
??Î?ÎÎ?Î$cd a bkk k k¸¸
Ä&
74. dt ln t log x, x 0
"" "
ln a t ln a ln a ln a
ln x ln 1'
1
x
x
1
a
? ?Î? $$ ‘kk
75. A dx 2 dx; u 1 x du 2x dx; x 0 u 1, x 2 u 5?? ?w%??%??%?''
"20
22
2x 2x
1x 1x!!
#
cd
A 2 du 2 ln u 2(ln 5 ln 1) 2 ln 5Ð? ? ? Î ?'
1
5
" &
"u
cdkk
76. A 2 dx 2 dx 2 2?? ??ÎÎ?ÎÎ?''

""
######
"
"11
11
1x
x
!Ÿ  ˆ‰ ˆ ‰ ˆ ‰ˆ ‰
–—
Š‹
Š‹

w
%
w
x
ln
2233
ln ln ln
77. Let H O x and solve the equations 7.37 log x and 7.44 log x. The solutions of these equationscd $
!??Î?Î
10 10
are 10 and 10 . Consequently, the bounds for H O are 10 10 .
(Þ$( (Þ%% (Þ%% (Þ$(
$
cdc d ß
78. pH log 4.8 10 (log 4.8) 8 7.32?Î ?Î w ?
10 10ab
)
79. Let O original sound level 10 log I 10 db from Equation (6) in the text. Solvingœœ‚
10ab
"#
O 10 10 log kI 10 for k 10 log I 10 10 10 log kI 10 log I 10 1w? % w? % w
10 10 10 10ab ab ab ab
"# "# "# "#
log kI 10 log I 10 1 log k log I 10 1 log k 1? % w? w %? %?
10 10 10 10 10a b ab ab
"# "# "# ln k
ln 10
ln k ln 10 k 10ʜ ʜ
80. Sound level with 10I 10 log 10I 10 10 log 10 log I 10 10 10 log I 10??w?w
10 10 10 10abc d ab ab
"# "# "#
original sound level 10 an increase of 10 db?w%
81. (a) If x H O and S x OH , then x(S x) 10 S x 1?Î? Î?%?w%?Îcd cd $
"% 10 dS 10
xdx x
&"w &"w

and 0 a minimum exists at x 10
dS 210
dx x


?$% ?
† (
(b) pH log 10 7?Î ?
10ab
(
(c) the ratio equals 1 at x 10
cd cd
cd cd
OH OH
HO x x HO
Sx 10
& &
$ $

&"w
œœ œ Ê œ
(
Š‹xx
x

10
x
ŸÊ!
82. Yes, it's true for all positive values of a and b: log b and log a log b
ab aœœÊœœ
ln b ln a ln b
ln a ln b log a ln a
"
b

454. 450 Chapter 7 Transcendental Functions
83. From zooming in on the graph at the right, we estimate
the third root to be x 0.76666ÑÎ

84. The functions f(x) x and g(x) 2 appear toœœ
ln 2 ln x
have identical graphs for x 0. This is no accident,$
because x e e 2 .
ln 2 ln 2 ln x ln 2 ln x ln x
œœ œ

ab
85. (a) f(x) 2 f (x) 2 ln 2; L(x) 2 ln 2 x 2 x ln 2 1 0.69x 1?% ? ? w? wÑ w
xx w!!
ab
(b)
86. (a) f(x) log x f (x) , and f(3) L(x) (x 3) 1?%? ?%? Îw?Îw
3
w """
x ln 3 ln 3 3 ln 3 ln 3 3 ln 3 ln 3
ln 3 ln 3 x
0.30x 0.09Ñw
(b)
87. (a) log 8 1.89279 (b) log 0.5 0.35621
3 7?Ñ ? ÑÎ
ln 8 ln 0.5
ln 3 ln 7
(c) log 17 0.94575 (d) log 7 2.80735
20 05?Ñ ? ÑÎ
ln 17 ln 7
ln 0 ln 0.5#

(e) ln x (log x)(ln 10) 2.3 ln 10 5.29595 (f) ln x (log x)(ln 2) 1.4 ln 2 0.97041œœ¸ œœ¸
10 2
(g) ln x (log x)(ln 2) 1.5 ln 2 1.03972 (h) ln x (log x)(ln 10) 0.7 ln 10 1.61181? ?Î ÑÎ ? ?Î ÑÎ
2 10
88. (a) log x log x (b) log x log x
ln 10 ln 10 ln x ln x ln a ln a ln x ln x
ln ln ln 10 ln ln b ln b ln a ln b###
†† ††
10 2 a bœœœ œœœ
89. x k x and ln x c .
dd
dx dx x
ˆ‰ abÎw?Î w?
""
#
#
Since x for any x , these two curves will have perpendicular tangent lines.Î ?ÎÞ '
"
x
90. e x for x and ln e x for all x
ln x x
? ? ab

455. Section 7.5 Exponential Growth and Decay 451
91. Using Newton's Method: f x ln x f x x x x x ln x .ab ab ab a b ’“?ÎÞ%?%?Î %?#Î
w "
!Ÿ !Ÿ
x
nn nn n ln xabn
x"


Then, x 2, x 2.61370564, x 2.71624393, and x 2.71828183. Many other methods may be used. For example,12 3œœ œ œ &
graph y ln x and determine the zero of y.?ÎÞ
92. (a) The point of tangency is p ln p and m since . The tangent line passes through theab abßœœ !ß!Ê tangent
pdxx
dy""
equation of the tangent line is y x. The tangent line also passes through p ln p ln p p p e, andœßÊœœ"Êœ
" "
p p
ab
the tangent line equation is y x.œ
"
e
(b) for x y ln x is concave downward over its domain. Therefore, y ln x lies below the graph of
dy
dx x


?Î '% ? ?
"
y x for all x , x e, and ln x for x , x e.œ !Á !Á
"
ee
x
(c) Multiplying by e, e ln x x or ln x x.
e
(d) Exponentiating both sides of ln x x, we have e e , or x e for all positive x e.
eln xxex
Á
e
(e) Let x to see that e . Therefore, e is bigger.?11
e 11
7.5 EXPONENTIAL GROWTH AND DECAY
1. (a) y y e 0.99y y e k 0.00001?% ? %?ÑÎ !!!
kt 1000k ln 0.99
1000
(b) 0.9 e ( 0.00001)t ln (0.9) t 10,536 years?%Î?%?Ñ
!Ÿ Á0 00001)t ln (0.9)
0.00001
(c) y y e y e y (0.82) 82%œ¸œÊ !!!
!−   ŸÁ20 000 k 0 2
2. (a) kp p p e where p 1013; 90 1013e k 0.121
dp ln (90) ln (1013)
dh 20
?%? ? ? %? ÑÎ!!

kh 20k
(b) p 1013e 2.389 millibarsœ¸
"'605
(c) 900 1013e 0.121h ln h 0.977 km?%Î?%? Ñ
!Ÿ Á  0121 h ˆ‰
900
1013 0.121
ln (1013) ln (900)
3. 0.6y y y e ; y 100 y 100e y 100e 54.88 grams when t 1 hr
dy
dt
?Î % ? ? % ? % ? Ñ ?!!
"' "' "'06t 06t 06
4. A A e 800 1000e k A 1000e , where A represents the amount ofœÊœ Êœ Êœ!
kt 10k ln (0 8 10 t ln (0.8)
10 !Á _ 
sugar that remains after time t. Thus after another 14 hrs, A 1000e 585.35 kgœ¸
!!Á _ ln 0 8 10 24
5. L(x) L e L e ln 18k k 0.0385 L(x) L e ; when the intensity?%? %?Î%?Ñ %?!! !
##
"
"" "'kx 18k 0 0385xL ln 2
18!
is one-tenth of the surface value, L e ln 10 0.0385x x 59.8 ft
L
10!
œÊœÊ¸!
"0 0385x
6. V(t) V e 0.1V V e when the voltage is 10% of its original value t 40 ln (0.1)?%? %?Î !!!
"& "&t40 t40
92.1 sec¸
7. y y e and y 1 y e at y 2 and t 0.5 we have 2 e ln 2 0.5k k ln 4.œ œÊœ Ê œ œ œ Ê œ Êœ œ!!
kt kt 0 5k ln 2
0.5
Therefore, y e y e 4 2.81474978 10 at the end of 24 hrsœÊœœœ ‚
! ln 4 t 24 ln 4 24 14
8. y y e and y(3) 10,000 10,000 y e ; also y(5) 40,000 y e . Therefore y e 4y eœœÊœ œœ œ!!!!!
kt 3k 5k 5k 3k
e 4e e 4 k ln 2. Thus, y y e 10,000 y e y e 10,000 8yÊœ ʜʜ œ Ê œ œ Ê œ
5k 3k 2k ln 2 t 3 ln 2 ln 8
!!!!
! 
y 1250Êœ œ !
10,000
8
9. (a) 10,000e 7500 e 0.75 k ln 0.75 and y 10,000e . Now 1000 10,000e
k 1 k ln0 75 t ln0 75 t!  !Á  !Á 
œÊœÊœ œ œ
ln 0.1 (ln 0.75)t t 8.00 years (to the nearest hundredth of a year)Êœ Êœ¸
ln 0.1
ln 0.75
(b) 1 10,000e ln 0.0001 (ln 0.75)t t 32.02 years (to the nearest hundredth of aœÊœÊœ¸
!Á ln 0 75 t ln 0.0001
ln 0.75
year)

456. 452 Chapter 7 Transcendental Functions
10. (a) There are (60)(60)(24)(365) 31,536,000 seconds in a year. Thus, assuming exponential growth,œ
P 257,313,431e and 257,313,432 257,313,431e lnœœÊœ
kt 14k 31 536 000
Š‹
257,313,432
257,313,431 31,536,000
14k
k 0.0087542ʸ
(b) P 257,313,431e 293,420,847 (to the nearest integer). Answers will vary considerablyœ¸
0.0087542ab"&
with the number of decimal places retained.
11. 0.9P P e k ln 0.9; when the well's output falls to one-fifth of its present value P 0.2P
!! !œÊœ œ
k
0.2P P e 0.2 e ln (0.2) (ln 0.9)t t 15.28 yrÊœ Êœ Ê œ Êœ¸!!
ln 0 9 t ln 0 9 t ln 0.2
ln 0.9
12. (a) p dx ln p x C p e e e C e ;
dp dp
dx 100 p 100 100
œ Êœ Êœ Êœ œ œ
"" "
" 001x C C 001x 001x
p(100) 20.09 20.09 C e C 20.09e 54.61 p(x) 54.61e (in dollars)œÊœ Êœ ¸Êœ ""
%Î001 100 001x
(b) p(10) 54.61e $49.41, and p(90) 54.61e $22.20œœ œœ
0 01 10 0 01 90
(c) r(x) xp(x) r (x) p(x) xp (x);œÊœ #
ww
p (x) .5461e r (x)
ww
œ Ê
%Î001x
(54.61 .5461x)e . Thus, r (x) 0œ œ
%Î001x w
54.61 .5461x x 100. Since r 0ʜ ʜ
w
for any x 100 and r 0 for x 100, then
w
r(x) must be a maximum at x 100.œ

13. (a) A e A e
!!
004 5 02
œ
(b) 2A A e ln 2 (0.04)t t 17.33 years; 3A A e ln 3 (0.04)t!! !!œÊœÊœ¸ œÊœ
004 t 004 t ln 2
0.04
t 27.47 yearsÊœ ¸
ln 3
0.04
14. (a) The amount of money invested A after t years is A(t) A e!! œ
t
(b) If A(t) 3A , then 3A A e ln 3 t or t 1.099 yearsœœÊœ¸ !!!
t
(c) At the beginning of a year the account balance is A e , while at the end of the year the balance is A e .!!
t t1
The amount earned is A e A e A e (e 1) 1.7 times the beginning amount.!!!
t1 t t
œ ¸
15. A(100) 90,000 90,000 1000e 90 e ln 90 100r r 0.0450 or 4.50%œÊœ ʜʜʜ¸
r 100 100r ln 90
100
16. A(100) 131,000 131,000 1000e ln 131 100r r 0.04875 or 4.875%œÊœ ʜʜ¸
100r ln 131
100
17. y y e represents the decay equation; solving (0.9)y y e t 0.585 daysœœÊœ¸ !!!

%Î %Î0 18t 018t ln (0.9)
0.18
18. A A e and A A e e k 0.00499; then 0.05A A eœœÊœÊœ¸ œ !!! !!
""
##
kt 139k 139k 0 00499t ln (0.5)
139 %Î
t 600 daysÊœ ¸
ln 0.05
0.00499
19. y y e y e y e (0.05)(y ) after three mean lifetimes less than 5% remainsœœ œœ $œÊ !! ! !
kt k 3 k 3 yy
e20
#
20. (a) A A e e k 0.262œÊœÊœ¸ !
"
##
%%Îkt 2 645k ln 2
.645
(b) 3.816 years
" k
¸
(c) (0.05)A A exp t ln 20 t t 11.431 yearsœ Ê œ Êœ ¸ˆ‰ ˆ‰
ln 2 ln 2 2.645 ln 20
2.645 2.645 ln #
21. T T (T T ) e , T 90°C, T 20°C, T 60°C 60 20 70e eœ œœœÊ œÊœ
ss s
kt 10k 10k!!
% %% 4
7
k 0.05596Êœ ¸
ln
10
ˆ‰
7
4

457. Section 7.6 Relative Rates of Growth 453
(a) 35 20 70e t 27.5 min is the total time it will take 27.5 10 17.5 minutes longer to reachœœÊ¸ Ê œœ
− 0 05596t
35 °C
(b) T T (T T ) e , T 90 °C, T 15°C 35 15 105e t 13.26 minœœ œ œœ œ Ê #œÊ¸
ss s
kt 0 05596t!!
− − 
22. T 65°(T 65°) e 35°65°(T 65°) e and 50°65°(T 65°) e . Solvingœ œ œ Ê œ œ œœ œ œ !! !
−− −kt 10k 20k
30 °(T 65°) e and 15°(T 65°) e simultaneously (T 65°) e 2(T 65 °)eœœ œœ œ œ Ê œ œ œ !! !!
−− −−10k 20k 10k 20k
e 2 k and 30 ° 30°e65 °60°5°ÊœÊœ œœÊ œ œ œ œ
10k 10 ln 2 ln 2
10 e
T65° œœ
10k
ln 2
10
œ ‘

œ Ê œ T65° T65 °30°e((
ˆ‰
23. T T (T T ) e 39 T (46 T ) e and 33 T (46 T ) e e andœœ œ Ê œœ œœ œ œ Êœ
ss ss ss
kt 10k 20k 10k!
œ
œ
−− −− 39 T
46 Ts
s
e e (33 T )(46 T ) (39 T ) 1518 79T T
33 T 33 T 39 T
46 T 46 T 46 T
œœœ
œœœ
#
#
##ss
ss
œœ Ê œ Ê œœ œ œ Ê
−−20k 10k
ss s s s
ab Š‹
s
s
1521 78T T T 3 T 3 °Cœ ÊœœÊ œ œ
sss s
#
24. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room
temperature the silver will be 120 min from now, and t the time the silver will be 10°C above room
!
temperature. We then have the following time-temperature table:
time in min. 0 20 (Now) 35 140 t
temperature T 70°T60°TxTyT10 ° !
s s sss# # ###
T T (T T ) e (60 T ) T (70 T ) T e 60 70e k lnœœ œ Ê œ Êœ Êœ œ
s s ss ss
kt 20k 20k!
"
−−−
cd ˆ‰ˆ‰
20 7
6
0.00771¸
(a) T T (T T ) e (T x) T (70 T ) T e x 70e 53.44 °Cœœ œ Ê œ Êœ ¸
ss ss ss
0 00771t 0 00771 35 0 26985!
−
cd
(b) T T (T T ) e (T y) T (70 T ) T e y 70e 23.79 °Cœœ œ Ê œ Êœ ¸
ss ss ss
0 00771t 0 00771 140 1 0794!

cd
(c) T T (T T ) e (T 10) T (70 T ) T e 10 70eœœ œ Ê œ Êœ
ss ss ss
0 00771t 0 00771 t 0 00771t!

cd
œœ
ln 0.00771t t ln 252.39 252.39 20 232 minutes from now theÊœ œ Êœ œ œÊ œ¸ˆ‰ ˆ ‰ ˆ‰
"""
!!
7 0.00771 7
silver will be 10°C above room temperature
25. From Example 5, the half-life of carbon-14 is 5700 yr c c e k 0.0001216Êœ Êœ¸
"
#
!! −œÄk 5700 ln 2
5700
c c e (0.445)c c e t 6659 yearsÊœ Ê œ Êœ ¸!!!
œ
−  − 0 0001216t 0 0001216t ln (0.445)
0.0001216
26. From Exercise 25, k 0.0001216 for carbon-14.¸
(a) c c e (0.17)c c e t 14,571.44 years 12,571 BCœÊœÊ¸ Ê
!!!
−  − 0 0001216t 0 0001216t
(b) (0.18)c c e t 14,101.41 years 12,101 BC!!œÊ¸ Ê
− 0 0001216t
(c) (0.16)c c e t 15,069.98 years 13,070 BC!!œÊ¸ Ê
− 0 0001216t
27. From Exercise 25, k 0.0001216 for carbon-14. Thus, c c e (0.995)c c e¸œÊœ !!!
−  − 0 0001216t 0 0001216t
t 41 years oldÊœ ¸
ln (0.995)
0.0001216œ
7.6 RELATIVE RATES OF GROWTH
1. (a) slower, lim lim 0
xxÄ_ Ä_
œœ
x3
ee
Ê"
xx
(b) slower, lim lim lim lim 0 by the
xx x xÄ_ Ä_ Ä_ Ä_
œœœœ
x sin x 3x 2 sin x cos x 6x 2 cos 2x 6 4 sin 2x
ee ee

ÊÊ Ê œ
xx xx
Sandwich Theorem because for all reals and lim 0 lim
xx
2 6 4 sin 2x 10 2 0
eee e e
xx x x xŸŸ œœ
Ä_ Ä_
œ "
(c) slower, lim lim lim lim 0
xxx xÄ_ Ä_ Ä_ Ä_
œœ œ œ
È Š‹
È
x
ee e
x
x
xe
xx x x
Ä_Ê

œ
#
"
# (d) faster, lim lim since 1
xxÄ_ Ä_
œœ_ Ê
444
eee
x
x
x
ˆ‰
(e) slower, lim lim 0 since 1
xxÄ_ Ä_
œœ $
Š‹
3
#
x
x
e2e2e
33
x
ˆ‰

458. 454 Chapter 7 Transcendental Functions
(f) slower, lim lim 0
xxÄ_ Ä_
œœ
e
e e
x2
x x2 
 
"
(g) same, lim lim
xxÄ_ Ä_
œœ
Š‹
e
x
#
e
x
""
##
(h) slower, lim lim lim lim 0
xxxxÄ_ Ä_ Ä_ Ä_
œœœ œ
log x
e (ln 10) e (ln 10) e (ln 10)xe
ln x10
xxxx

x "
2. (a) slower, lim lim lim lim lim 0
xxxxxÄ_ Ä_ Ä_ Ä_ Ä_
œœœœœ
10x 30x 1 40x 30 20x 240x 240
eeeee

"
xxxxx
(b) slower, lim lim lim lim lim
xx x x xÄ_ Ä_ Ä_ Ä_ Ä_
œœ œ œ
x ln x x ln x 1 1 ln x
ee e ee
x (ln x 1)
ln x1x


xx x xx
Š‹

x
lim lim 0
xx
œœœ
Ä_ Ä_
Š‹

x
exe
xx
"
(c) slower, lim lim lim lim lim lim
x x xx xxÄ_ Ä_ Ä_ Ä_ Ä_ Ä_
œœœœœ
È
1x
e e 2e 4e 8e 16e
1 x 4x 12x 24x 24 ß
x2x2x2x2x2xÉ ÉÉ ÉÉ
00 œœ
È
(d) slower, lim lim 0 since 1
xxÄ_ Ä_
œœ $
Š‹
5
#
x
x
ee2e
55
x
ˆ‰
#
(e) slower, lim lim 0
xxÄ_ Ä_
œœ
e
ee
−x
x2x
"
(f) faster, lim lim x
xxÄ_ Ä_
œœ_
xe
e
x
x
(g) slower, since for all reals we have 1 cos x 1 e e e and alsoŸŸÊŸŸÊŸŸ
" "cos x ee e
eee
%$ $
xxx
cos x
lim 0 lim , so by the Sandwich Theorem we conclude that lim 0
xx xÄ_ Ä_ Ä_
œœ œ
ee e
ee e
%$ $
xx x
cos x
(h) same, lim lim lim
xx xÄ_ Ä_ Ä_
œœœ
e
eee e
x1
x xx1−
(% ‚á
"""
3. (a) same, lim lim lim 1
xxxÄ_ Ä_ Ä_
œœœ
x4x 2x4 2
x2x



# (b) faster, lim lim x 1
xxÄ_ Ä_
œ œ_
xx
x


$
ab
(c) same, lim lim lim 1 1 1
xxxÄ_ Ä_ Ä_
œœ #œœ
È
xx
xx
xx
xß#

ß#
ß
"
É É ˆ‰ È
(d) same, lim lim lim 1
xxxÄ_ Ä_ Ä_
œœœ
(x 3) 2(x 3)
xx
2
##


(e) slower, lim lim lim 0
xxxÄ_ Ä_ Ä_
œœ œ
x ln x ln x
xx1

Š‹

x
(f) faster, lim lim lim
xx xÄ_ Ä_ Ä_
œœ œ_
2
x2x
(ln 2) 2 (ln 2) 2
x xx


#
(g) slower, lim lim lim 0
xxxÄ_ Ä_ Ä_
œœœ
xe x
xee
#%
x
xx
" (h) same, lim lim 8 8
xxÄ_ Ä_
œœ
8x
x


4. (a) same, lim lim 1 1
xxÄ_ Ä_
œ # œ
xx
x x


"
È
ˆ‰ (b) same, lim lim 10 10
xxÄ_ Ä_
œœ
"0x
x


(c) slower, lim lim 0
xxÄ_ Ä_
œœ
xe
xe
−
x
x
"
(d) slower, lim lim lim lim lim 0
xx x x xÄ_ Ä_ Ä_ Ä_ Ä_
œœ œ œ œ
log x
x x ln 10 x ln 10 2x ln 10 x
2 ln x 210

Š‹ Š‹
ln x
ln 10 x
#

"""
(e) faster, lim lim (x 1)
xxÄ_ Ä_
œ œ_
xx
x



(f) slower, lim lim 0
xxÄ_ Ä_
œœ
Š‹

10
x
x
x10x

"
(g) faster, lim lim lim
xx xÄ_ Ä_ Ä_
œœ œ_
(1.1) (ln 1.1)(1.1) (ln 1.1) (1.1)
xx
xx x


##
(h) same, lim lim 1 1
xxÄ_ Ä_
œ # œ
x 100x 100
xx



ˆ‰

459. Section 7.6 Relative Rates of Growth 455
5. (a) same, lim lim lim
xx xÄ_ Ä_ Ä_
œœœ
log x
ln x ln x ln 3 ln 33
Š‹
ln x
ln 3
""
(b) same, lim lim 1
xxÄ_ Ä_
œœ
ln 2x
ln x
ˆ‰
ˆ‰
2
x
x
#

(c) same, lim lim lim
xx xÄ_ Ä_ Ä_
œœœ
ln x
ln x ln x
ln xÈ Š‹

#
""
##
(d) faster, lim lim lim lim lim
xxx x xÄ_ Ä_ Ä_ Ä_ Ä_
œœ œ œœ_
È ÈŠ‹
Š‹
È
x x
ln x ln x
xx
x
x


#

x
# # (e) faster, lim lim lim x
xx xÄ_ Ä_ Ä_
œœœ_
x
ln x
"
Š‹

x
(f) same, lim lim 5 5
xxÄ_ Ä_
œœ
5 ln x
ln x
(g) slower, lim lim 0
xxÄ_ Ä_
œœ
Š‹

x
ln x x ln x
"
(h) faster, lim lim lim xe
xxxÄ_ Ä_ Ä_
œœœ_
ee
ln x
x
xx
x
ˆ‰
$
6. (a) same, lim lim lim lim lim 2
xx x x xÄ_ Ä_ Ä_ Ä_ Ä_
œœ œ œœ
log x
ln x ln x ln ln x ln ln x ln
ln x 2 ln x 22
Š‹
ln x
ln 2
#
"" "
## # ln #
(b) same, lim lim lim lim lim
xx x x xÄ_ Ä_ Ä_ Ä_
œœ œ œ
log 10x
ln x ln x ln 10 ln x ln 10 ln 10
ln 10x10
Š‹ Š‹
Š‹
ln 10x 0
ln 10 10x
x
"" "


Ä_
œ 1
"
ln 10
(c) slower, lim lim 0
xxÄ_ Ä_
œœ
Š‹
ˆ‰È


x
ln x x(ln x)
"
(d) slower, lim lim 0
xxÄ_ Ä_
œœ
Š‹

#
x
ln x x ln x
"

(e) faster, lim lim 2 lim 2 lim 2 lim x 2
xx x x xÄ_ Ä_ Ä_ Ä_ Ä_
œ œ œ œ œ_
x2 ln x x x
ln x ln x ln x
"
ˆ‰ Š‹ Š‹

Š‹

x
(f) slower, lim lim 0
xxÄ_ Ä_
œœ
e
ln x e ln x
−x
x
"
(g) slower, lim lim lim 0
xxxÄ_ Ä_ Ä_
œœœ
ln (ln x)
ln x ln x
Š‹
Š‹

/x
ln x
x
"
(h) same, lim lim lim lim lim 1 1
xxxxxÄ_ Ä_ Ä_ Ä_ Ä_
œœœœœ
ln (2x 5)
ln x 2x 5
2x 2
#
Š‹
Š‹
2
2x 5
x


7. lim lim e e grows faster than e ; since for x e we have ln x e and lim
xx xÄ_ Ä_ Ä_
œœ_Ê
e
e
x2 x x2 e (ln x)
e
x
x2
x
x
 
ÎÎ
lim (ln x) grows faster than e ; since x ln x for all x 0 and lim lim
x xx
œœ_Ê œ
Ä_ Ä_ Ä_
ˆ‰ ˆ‰
ln x xx
e (ln x) ln x
x
xx
x
x
x
x grows faster than (ln x) . Therefore, slowest to fastest are: e , e , (ln x) , x so the order is d, a, c, bœ_ Ê
xx x2xxx Î
8. lim lim lim lim (ln 2) 0
xx x xÄ_ Ä_ Ä_ Ä_
œœ œ œ
(ln 2) (ln (ln 2))(ln 2) (ln (ln 2)) (ln 2) (ln (ln 2))
xx
x
xx x


###
(ln 2) grows slower than x ; lim lim lim 0 x grows slower than 2 ;
xx x
ÊœœœÊ
Ä_ Ä_ Ä_
x xx2x 2
2 (ln )2 (ln 2)
# #
##


xx x
lim lim 0 2 grows slower than e . Therefore, the slowest to the fastest is: (ln 2) , x , 2
xxÄ_ Ä_
œœÊ
22
ee
x
xx xx
x
x
ˆ‰
#
and e so the order is c, b, a, d
x
9. (a) false; lim 1
xÄ_
œ
x
x
(b) false; lim 1
xÄ_
œœ
x
x5 1
"
(c) true; x x 5 1 if x 1 (or sufficiently large)$# Ê
x
x5
(d) true; x 2x 1 if x 1 (or sufficiently large)$ Ê
x
2x
(e) true; lim lim 0
x x0Ä_
œœ
Ä
e
ee
x
2x x
"
(f) true; 1 1 1 2 if x 1 (or sufficiently large)
x ln x ln x
xxx
x
x
"
œ#$# œ
È
È

460. 456 Chapter 7 Transcendental Functions
(g) false; lim lim lim 1 1
xx xÄ_ Ä_ Ä_
œœœ
ln x
ln 2x
Š‹
Š‹

#
x
2
x
(h) true; 1 6 if x 1 (or sufficiently large)
ÈÈ
x5
xxxx
(x 5) x5 5# #
# # #
$$ œ
10. (a) true; 1 if x 1 (or sufficiently large)
Š‹
Š‹

#

x3
x
œ
x
x3#
(b) true; 1 2 if x 1 (or sufficiently large)
Š‹
Š‹

#

xx
x

œ
"
x
(c) false; lim lim 1 1
xxÄ_ Ä_
œ œ
Š‹
Š‹

#

xx
x
#
ˆ‰
"
x
(d) true; 2 cos x 3 if x is sufficiently large# ŸÊ Ÿ
2cos x 3#
##
(e) true; 1 and 0 as x 1 2 if x is sufficiently large
ex x x x
eee e
x
xxx x
#
œ# Ä Ä_Ê #$
(f) true; lim lim lim 0
xxxÄ_ Ä_ Ä_
œœ œ
x ln x ln x
xx1
#
Š‹

x
(g) true; 1 if x is sufficiently large
ln (ln x)
ln x ln x
ln x
$ œ
(h) false; lim lim lim lim
xxxxÄ_ Ä_ Ä_ Ä_
œœœ # œ
ln x x
ln x 1 x xab
Š‹
Š‹
###
#
####
#" " " "

#
#
x
2x
x1
ˆ‰
11. If f(x) and g(x) grow at the same rate, then lim L 0 lim 0. Then
xxÄ_ Ä_
œÁ Ê œÁ
f(x) g(x)
g(x) f(x) L
"
L 1 if x is sufficiently large L 1 L 1 L 1 if x is sufficiently large¹¹ kk
f(x) f(x) f(x)
g(x) g(x) g(x)
Ê ÊŸ #
f O(g). Similarly, 1 g O(f).Êœ Ÿ #Êœ
g(x)
f(x) L
¸¸
"
12. When the degree of f is less than the degree of g since in that case lim 0.
xÄ_
œ
f(x)
g(x)
13. When the degree of f is less than or equal to the degree of g since lim 0 when the degree of f is smaller
xÄ_
œ
f(x)
g(x)
than the degree of g, and lim (the ratio of the leading coefficients) when the degrees are the same.
xÄ_
œ
f(x)
g(x) b
a
14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the
same degree grow at the same rate.
15. lim lim lim lim 1 and lim lim
xxxx x xÄ_ Ä_ Ä_ Ä_ Ä_ Ä_
œœœœ œ
ln (x ) ln (x 999)
ln x x 1 1 ln x
x#" #
#
"
Š‹ Š‹
Š‹ Š‹

# #

x1 x999
x x
lim 1
x
œœ
Ä_
x
x 999#
16. lim lim lim lim 1. Therefore, the relative rates are the same.
xxxxÄ_ Ä_ Ä_ Ä_
œœœœ
ln (x a)
ln x x a 1
x#
#
"
Š‹
Š‹

#

xa
x
17. lim lim 10 and lim lim 1 1. Since the growth rate
xx xxÄ_ Ä_ Ä_ Ä_
œœ œœœ
ÈÈ
ÈÈ
10x 1 x 1
xx
0x 1 x 1
xx
## "# #
ÉÉ
È È
is transitive, we conclude that 10x 1 and x 1 have the same growth rate that of x .ÈÈ ˆ‰ È##
18. lim lim 1 and lim lim 1. Since the growth rate is
xx x xÄ_ Ä_ Ä_ Ä_
œœ œ œ
ÈÈ
xx
xx
xx xx
xx
xxß
# #
ß ß#
ß ß
ß#
#
ÉÉ
transitive, we conclude that x x and x x have the same growth rate that of x .
È È
ab
%%$ #

19. lim lim lim 0 x o e for any non-negative integer n
xx xÄ_ Ä_ Ä_
œœáœœÊœ
xnx n!
ee e
nx
nn1
xx x −
ab

461. Section 7.6 Relative Rates of Growth 457
20. If p(x) a x a x a x a , then lim a lim a lim
xx x
œ ## á## œ ## á
Ä_ Ä_ Ä_ nn1 n n1
nn1 p(x)
ee e
xx
"!
xx x
nn1
−
a lim a lim where each limit is zero (from Exercise 19). Therefore, lim 0
xx x
## œ
Ä_ Ä_ Ä_"!
"x
ee e
p(x) xx x
e grows faster than any polynomial.Ê
x
21. (a) lim lim lim x ln x o x for any positive integer n
xx xÄ_ Ä_ Ä_
œœ œ_Êœ
xx
ln x n n
1n 1n
1n 1 n n
x  (%áÎ
$
ˆ‰
ˆ‰ ˆ ‰
" ÎÎ
(b) ln e 17,000,000 e e 24,154,952.75ab Š‹
17 000 000
110
ß ß "(‚"! "(
Î
œ $ œ¸


(c) x 3.430631121 10¸‚
"&
(d) In the interval 3.41 10 3.45 10 we havecd‚ß ‚
"& "&
ln x 10 ln (ln x). The graphs cross at aboutœ
3.4306311 10 .‚
"&

22. lim
xÄ_
œœ
ln x
ax a x ax a a
lim
x
lim a
x
lim
x
nn1 n
nn1
n á
Ä_
Ä_
á
Ä_−
−
Š‹
Š‹
–—
ln x
x
n
aa
n1
xx
a
x
n1 n
x
nx
n1
$
$
 

$

lim 0 ln x grows slower than any non-constant polynomial (n 1)
x
œœÊ  
Ä_
"
aba banx
n
n
23. (a) lim lim 0 n log n grows
nnÄ_ Ä_
œœÊ
n log n
nlog n log n
2 2
2 2
ab

"
slower than n (log n) ; lim lim
nn 2
n log n
nn
#
Ä_ Ä_
œ
2

Š‹
ln n
ln 2
lim lim 0
nn
œœœ
Ä_ Ä_
""
##ln ln n
2
n
Š‹
ˆ‰

n
$


n log n grows slower than n . Therefore, n log nÊ 22
$Î#
grows at the slowest rate the algorithm that takesÊ
O(n log n) steps is the most efficient in the long run.
2

24. (a) lim lim lim
nnnÄ_ Ä_ Ä_
œœ
(log n) (ln n)
n n n(ln 2)2


Š‹
ln n
ln 2
lim lim
nn
œœ
Ä_ Ä_
2(ln n)
(ln 2) (ln 2) n
2ln n
Š‹

n

lim 0 (log n) grows slower
n
œœÊ
Ä_
2
(ln 2) 1
2

Š‹

n #
than n; lim lim
nnÄ_ Ä_
œ
(log n) log n
n log n n22
2

ÈÈ
lim lim
nn
œœ
Ä_ Ä_
Š‹
ln n
ln 2
nn ln
ln n

"
#

lim lim 0 (log n) grows slower than n log n. Therefore (log n) grows
x n
œœœÊ
Ä_ Ä_
""
##
##
ln ln n
2
n
222
Š‹
ˆ‰

n
$


È
at the slowest rate the algorithm that takes O (log n) steps is the most efficient in the long run.Ê ab
2
#
25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because
2 524,288 1,000,000 1,048,576 2 .
"* #!
œ $$ œ
26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because
2 262,144 450,000 524,288 2 .
") "*
œ $$ œ

462. 458 Chapter 7 Transcendental Functions
7.7 INVERSE TRIGONOMETRIC FUNCTIONS
1. (a) (b) (c) 2. (a) (b) (c)
111 111
436 436

3. (a) (b) (c) 4. (a) (b) (c)
11 1 1 11
64 3 6 43
5. (a) (b) (c) 6. (a) (b) (c)
111 111
346 346
325
7. (a) (b) (c) 8. (a) (b) (c)
32 5
46 3 4 63
11 1 1 11
9. (a) (b) (c) 10. (a) (b) (c)
111 111
436 436

11. (a) (b) (c) 12. (a) (b) (c)
32 5
46 3 4 63
11 1 1 11
13. sin cos , tan , sec , csc , and cot ! !!!! !œ Êœœœœ œ
"ˆ‰
5 12 5 13 13 12
13 13 12 12 5 5
14. tan sin , cos , sec , csc , and cot ! !!!! !œ Êœœœœ œ
"ˆ‰
4 4355 3
3 5534 4
15. sec 5 sin , cos , tan 2, csc , and cot !!!!!!œ Êœ œ œ œœ
"
Š‹
È
21 1
55
5
22ÈÈ
È
16. sec sin , cos , tan , csc , and cot !!!!!!œ Êœ œ œ œœ
"
##
Š‹
È È
ÈÈ
13 13323 2
13 13 33
17. sin cos sin 18. sec cos sec 2Š‹ ˆ‰ ˆ ‰ ˆ‰
" "
# #
""
È
È
2
4 32
œœ œœ
1 1
19. tan sin tan 20. cot sin cotˆ ‰ ˆ‰ ˆ‰ˆ‰ Š‹Š‹
" """ "
# #
œ œ œ œ
1 1
6 33 3
3
È È
È
21. csc sec 2 cos tan 3 csc cos cos csc cosab Š‹Š‹
È ˆ ‰ ˆ‰ ˆ‰ ˆ‰ˆ‰
" " " ""
##

œ œ œ #œ
11 1
33 3
2
323
43
ÈÈ
È
22. tan sec 1 sin csc 2 tan cos sin sin tan (0) sin 0aba b ab ˆ‰ˆ ‰ ˆ‰ˆ‰ ˆ‰
" " " " """
##
œ œ œ œ
12 6
1 1
23. sin sin cos sin sin 1ˆ‰ˆ‰ˆ‰ˆ‰ ˆ‰
" """
## #
œ œœ
11 1
63
2
24. cot sin sec 2 cot cos cot cot 0ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ ˆ‰
" " """
###
œ œ œ œ
1111
663
25. sec tan 1 csc 1 sec sin sec sec 2ab ˆ‰ˆ‰ˆ‰ È
" " "
#
# œ # œ #œœ
1111
41 4 4
13
26. sec cot 3 csc ( 1) sec sin ( ) sec sec 2Š‹
È ˆ‰ˆ‰ˆ‰
" " " "
#
œ # œ œ œ
1 111 1
61 32 3
27. sec sec sec cos
" " "
#ˆ‰ˆ‰ Š‹ Š‹ œœœ
11
66
2
3
3
È
È
28. cot cot cot ( 1)
" "ˆ‰ˆ‰ œ œ
11
44
3

463. Section 7.7 Inverse Trigonometric Functions 459
29. tan indicates the diagram sec tan sec ! !œÊœœ
" "
# ##
x x x4 ˆ‰
È
30. tan 2x indicates the diagram sec tan 2x sec 4x 1! !œÊœœ #
" " #
ab
È
31. sec 3y indicates the diagram tan sec 3y tan 9y 1! !œÊœœ
" " #
ab È
32. sec indicates the diagram tan sec tan ! !œÊœœ
" " y y
5 55
y25 ˆ‰
È
33. sin x indicates the diagram cos sin x cos 1 x! !œÊœœ
" " #
ab
È
34. cos x indicates the diagram tan cos x tan ! !œÊœœ
" "
ab
È
1x
x
35. tan x 2x indicates the diagram sin tan x 2x!œ Ê
" "# #ÈÈ
Š‹
sin œœ!
È
x2x
x1


36. tan indicates the diagram sin tan sin ! !œÊœœ
" "

x xx
x1 x1 2x1
È ÈÈ

Š‹
37. sin indicates the diagram cos sin cos ! !œÊœœ
" " 2y 2y
3 33
94y ˆ‰
È
38. sin indicates the diagram cos sin cos ! !œÊœœ
" " y y
5 55
25 y ˆ‰
È
39. sec indicates the diagram sin sec sin ! !œÊœœ
" " x x
4 4x
x16 ˆ‰
È

464. 460 Chapter 7 Transcendental Functions
40. sec indicates the diagram sin sec sin ! !œÊœœ
œ" œ"Ê Ê
Ê
È È
È
x4 x4
x x
2
x4
Ê Ê
Ê
Š‹
41. lim sin x 42. lim cos x
x1 x1Ä $#
− ß
œ" œ"
#
œœ
1
1
43. lim tan x 44. lim tan x
xx$Î $#Î
œ" œ"
# #
œœ œ
1 1
45. lim sec x 46. lim sec x lim cos
xxx$Î $#Î $#Î
œ" œ" œ"
# #
"
œœœ
1 1
ˆ‰
x
47. lim csc x lim sin 0 48. lim csc x lim sin 0
xx x x$Î $Î $#Î $#Î
œ" œ" œ" œ" " "
œœ œ œ ˆ‰ ˆ‰
x x
49. y cos x 50. y cos sec x œÊœ œ œœœÊœ
" # " "
œ
œ""
œ œ
ab ˆ‰
dy dy
dx x dx
2x 2x
1x
1x x x 1É ab
È È
kk
Ê
Ê ß Ê
51. y sin 2t 52. y sin (1 t) œÊœ œ œ œÊœ œ
œ" œ"
œ
œ œ
œ" œ"
œœÈ
dy 2 2 dy
dt dt
12t
12t 2t t1(1t)
ÈÈ
ÊŠ ‹
È
ÈÈÈ
Ê Ê ÊÊ
53. y sec (2s 1) œ #Êœ œ œ
œ"
ÊÊœ ÊÊ ÊÊ
"dy
ds
22
2s 1 (2s 1) 1 2s 1 4s 4s 2s 1 s skk ÈÈÈ
kk kk
Ê ÊÊ
54. y sec 5s œÊœ œ
œ"
œ
"
œ
dy
ds
5
5s (5s) 1 s 25s 1kkÈÈ
kk
Ê Ê
55. y csc x 1 œ #Êœ œ œ
" #
ÊÊœ
œ
ÊÊ
ab
dy
dx
2x 2x
x1 x1 1
x1 x2x
kkabÉ ab
È
ÊÊ
Ê Ê
56. y csc œÊœ œ œœ
œ"
#
œ
œ" œ
œˆ‰
x2 dy
dx
1 x
x x 4
Š‹
ÉɈ‰ kk
kk
È
œ
#
##
#
Ä
¸¸
xx x4
4 Ê Ê
57. y sec cos t œœÊœ
œ" œ""œ"
œˆ‰
tdt
dy
1t
È Ê
58. y sin csc œœ Êœ œ œœ
œ" œ"
œ
œœ
œˆ‰ Š‹
3t 2t6
t3dt
dy
1
t
tt 9
Ê
Ê
Ê
Ê
ß
ˆ‰
¹¹ Š‹
2t
3
tt
33
t9
9
##
−
Ä
Ê
É
È
59. y cot t cot t œœ Êœ œ œ
" " "Î#
Ê
œ"
#È
dy
dt
t
1t t(1 t)
Š‹
ab
È
œ
#−Ä_Ê
Ä_Ê
Ê
60. y cot t 1 cot (t 1) œ œœ œ Êœ œ œœ
" " "Î#
œ
ʜ
œ" œ"
œÊœ œÈ
dy
dt
(t 1)
1(t1) 2 t 1 (1 t 1) 2t t 1
Š‹
cd
ÈÈ
œ
# −Ä_Ê
Ä_Ê
Ê
61. y ln tan x œÊœœab
œ" "
Ê
dy
dx tan x tan x 1 x
Š‹
abab
œ
Ê
#
1x
−
62. y tan (ln x) œÊœœ
œ"
ÊÊ
"dy
dx 1 (ln x) x 1 (ln x)
ˆ‰
œ
x
ÊÊ
cd
63. y csc e œÊœ œ œ
œ"
œ
œ"
œ
ab
t dy
dt
e
e e 1
e1
t
tt
2t
kk abÉ
È
Ê

465. Section 7.7 Inverse Trigonometric Functions 461
64. y cos e œÊœ œ
"


ab
t dy
dt
ee
1e
1e
−−
− −tt
t
2t
É ab
È65. y s 1 s cos s s 1 s cos s 1 s s 1 s ( 2s)œ œ Êœ
È
ab ab ab ˆ‰
# " # " # #
"Î# "Î# "Î# ""
#

dy
ds
1s
È

1s 1sœ œ œœ
ÈÈ
##

" s s1 1ss1 2s
1s 1s 1s 1s 1s


ÈÈ È È È
66. y s 1 sec s s 1 sec s s 1 (2s)œ œ Êœ œ
È
ab ab ˆ‰
# " # " #
"Î# "Î# """
#

dy
dx
s s1 s1 s s1
s
kk kk
ÈÈ È

œ
s s 1
s s 1
kk
kk
È



67. y tan x 1 csc x tan x 1 csc x œ œ Êœ
" " " # "#
"Î#


"
È
ab dy
dx
x 1 (2x)
1x1
x x 1
Š‹ab
’“ab
kk
È

#





0, for x 1œ œ
""
xx 1 x x 1
ÈÈ
kk

68. y cos tan x tan x tan x 0 0œ œ Êœ œ œ
" " " " "" """
# ˆ‰ ab
xdx1xx11x
dy x
1x
1
−
%$

ab
69. y x sin x 1 x x sin x 1 x sin x x 1 x ( 2x)œ œ Êœ
" " # " # #
"Î# "Î# ""

# È
ab ab Š‹ ˆ‰ dy
dx
1x
È
sin x sin xœ œ
" "

xx
1x 1x
ÈÈ

70. y ln x 4 x tan tan x tanœ Êœ œ ab ˆ‰ ˆ‰ ˆ‰
–—
#" " "
## #

x2xx 2xx2xdy
dx x4 x4 4x
1

Š‹
ˆ‰

#
#
x
tanœ
"
#ˆ‰
x
71. dx sin C' "

"
È
9x
x
3

œ #ˆ‰
72. dx dx , where u 2x and du 2 dx'' '"" "

## ÈÈ È
14x 1u
2du
1 (2x)

œœ œœ
sin u C sin (2x) Cœ #œ #
""
##
" "
73. dx dx tan C''"""


"
17 x
17 x
17 17
x

œœ #
Š‹
È
ÈÈ
74. dx dx tan C tan C''""" "


" "
93x 3 9
3x
33 3 3
xx 3

œœ #œ #
Š‹
È
ÈÈ È
È
Š‹ Š‹
75. , where u 5x and du 5 dx''dx du
x 25x 2 u u 2
ÈÈ


œœœ
sec C sec Cœ #œ #
"" " "
ÈÈ ÈÈ
22 22
u5x
¹¹ ¹¹
76. , where u 5x and du 5 dx''dx du
x5x 4 uu 4
ÈÈ


œœœ
ÈÈ
sec C sec Cœ #œ #
""
## # #
" "
¸¸ ¹¹
u 5x
È
77. 4 sin 4 sin sin 0 4 0'
0
1
4 ds s 2
4s
63È

" " "
##
"
!
"

œœ œ œ ‘ˆ ‰ˆ‰
11

466. 462 Chapter 7 Transcendental Functions
78. , where u 2s and du 2 ds; s 0 u 0, s u''
00
324 324

##
ds du
94s 9u
32 32
4ÈÈ
ÈÈ

"
# #

œœœœÊœœÊœ
sin sin sin 0 0œœ œ œ ‘ˆ‰ Š‹
"" "
####
" " "u
348
2
322
0

# È
11
79. , where u 2t and du 2 dt; t 0 u 0, t 2 u 2 2''
00
222
dt du
82t 8u 2
"
œœœœÊœœÊœ
È

ÈÈ È
tan tan tan 0 tan 1 tan 0 0œœ œ œ œ’“Š ‹ ab ˆ‰
"" " " " " " " " "
##
!
ÈÈ È È
È
È
2 88 8
u
444416
22
†
11
80. , where u 3t and du 3 dt; t 2 u 2 3, t 2 u 2 3''
−−223
223
dt du
43t 4u 3
"
œœœœ Êœ œÊœ
È

ÈÈ È È
tan tan 3 tan 3œœ œ œ’“’ “
ÈÈ
Š‹ ‘ˆ‰
"" " "
##
" " "
#$
# $
##
ÈÈ ÈÈ
È
È33 333
u
33
†
11 1
81. , where u 2y and du 2 dy; y 1 u 2, y u 2''
−−
−#%12
22 2

dy 2
y4y 1
du
uu 1
ÈÈ
È


#
œœœœ Êœ œ Êœ
È
sec u sec 2 sec 2œœ œœcd kkkk ¹¹
È
" " "
#
# #
È
11 1
43 1
82. , where u 3y and du 3 dy; y u 2, y u 2''
−#%
%# %23 2
23 2

dy 2
y9y 1
du 2
uu 1
33ÈÈ
È


œœœœ Êœ œ Êœ
È
sec u sec 2 sec 2œœ œœcd kkkk ¹¹
È
" " "
#
# #
È
11 1
43 1
83. , where u 2(r 1) and du 2 dr''3 dr 3 du
14(r1) 1u
ÈÈ #


œœ œ
sin u C sin 2(r 1) Cœ #œ
33
##
" "
84. 6 , where u r 1 and du dr''6 dr du
4(r1) 4u
ÈÈ

œœ # œ
6 sin C 6 sin Cœ #œ #
" "
##
ur1 ˆ‰
85. , where u x 1 and du dx''dx du
2(x1) 2u
œœ œ
tan C tan Cœ #œ #
""" "
ÈÈ È È
22 2 2
ux1
Š‹
86. , where u 3x 1 and du 3 dx''dx du
1 (3x 1) 3 1 u
"
œœ # œ
tan u C tan (3x 1) Cœ #œ ##
""" "
33
87. , where u 2x 1 and du 2 dx'' dx du
(2x 1) (2x 1) 4 uu 4
"
#

ÈÈ
œœ œ
sec C sec Cœ #œ #
"" "
## # #
" "
† ¸¸ ¸ ¸
u2x1
4
88. , where u x 3 and du dx'' dx du
(x 3) (x 3) 25 uu 25
ÈÈ
œœ # œ
sec C sec Cœ #œ #
""" "
55 5 5
ux3
¸¸ ¸ ¸
89. 2 , where u sin and du cos d ; u , u''
−#%
#

21
21
2 cos d du
1(sin ) 1u
)) 11
) # #
œœœœ Êœ "œÊœ")))) )
2 tan u 2 tan 1 tan ( 1) 2œœ œ œcda b ‘ˆ‰
" " "
"
"
11
44
1

467. Section 7.7 Inverse Trigonometric Functions 463
90. , where u cot x and du csc x dx; x u 3 , x u 1''


#
#63
41
csc x dx du
1(cot x) 1u 6 4



#
œ œœ œÊœ œÊœ

11 È
tan u tan 1 tan 3œ œ œ œcd
È
" " "
"
$ #
È
11 1
43 1
91. , where u e and du e dx; x 0 u 1, x ln 3 u 3''
01
ln 3 3
xx

e dx du
1e 1u
x
2x

œœœœÊœœÊœ

ÈÈ
tan u tan 3 tan 1œœ œœcd
È
" " "
$
" #
È
11 1
34 1
92. 4 , where u ln t and du dt; t 1 u 0, t e u''
10
e4
4

4 dt du
t1 lnt 1 u t 4ab
"
œœœœÊœœÊœ


#
#
1
4 tan u 4 tan tan 0 4 tan œœ œcd ˆ‰
" " " "
Î%
!
1 11
44
93. , where u y and du 2y dy''
y dy
1y
du
1u
ÈÈ
"
#

#
ß
œœœ
sin u C sin y Cœ #œ #
""
##
" " #
94. , where u tan y and du sec y dy''
sec y dy
1tany
du
1u


ÈÈ
#
œœœ
sin u C sin (tan y) Cœ #œ #
" "
95. sin (x 2) C'' 'dx dx dx
x4x3 1x4x4 1(x2)ÈÈ È ab
"

œœœ
96. sin (x 1) C'' 'dx dx dx
2x x 1x2x1 1(x1)ÈÈ È ab
"

œœœ
97. 6 6 6 sin'' '

"
#
!
"11 1
00 0
6 dt dt dt t 1
32tt 4t2t1 2(t1)ÈÈ È ab

œœœ ‘ˆ‰
6sin sin 0 6 0œ œ œ ‘ˆ ‰ˆ‰
" ""
#
1
6
1
98. 3 3 3 sin'' '
12 12 12
11 1## #
6 dt 2 dt 2 dt 2t 1
34t4t 44t4t1 2 (2t 1)ÈÈ È ab
"
#
"
"Î#

œœœ ‘ˆ‰
3sin sin 0 3 0œ œ œ ‘ˆ ‰ˆ‰
" ""
##
11
6
99. tan C'' '
dy dy dy y 1
y 2y5 4y 2y1 (y1)

# # #
" "
œœœ #ˆ‰
100. tan (y 3) C'' '
dy dy dy
y 6y10 1 y 6y9 1(y3)


"
œœœ ##
ab
101. 8 8 8 tan (x 1)'' '
11 1
22 2
8 dx dx dx
x 2x 2 1 x 2x 1 1 (x 1)


"
#
"
œœœ
ab
cd
8 tan 1 tan 0 8 0 2œ œ œab ˆ‰
" " 1
4
1
102. 2 2 2 tan (x 3)'' '
22 2
44 4
2 dx dx dx
x 6x10 1 x 6x9 1(x3)


"
%
#
œœœ
ab
cd
2 tan 1 tan ( 1) 2œ œ œcd ‘ˆ‰
" " 11
44
1
103. '' 'dx dx dx
(x1)x 2x (x1)x 2x11 (x 1) (x 1) 1 ÈÈ È

œœ
, where u x 1 and du dxœœ # œ'
du
uu 1
È


sec u C sec x 1 Cœ #œ ##
" "
kk k k

468. 464 Chapter 7 Transcendental Functions
104. '' 'dx dx dx
(x2)x 4x3 (x2)x 4x41 (x 2) (x 2) 1œœÊ œœÊœ œœœÈÈ È
ÊÊ Ê
œœ
du, where u x 2 and du dxœœ œ œ'
"
œuu 1
È
Ê
sec u C sec x 2 Cœ #œ
œ" œ"
kk k k
105. dx e du, where u sin x and du''edx
1x 1x
sin x
−Ä
Ê Ê
ÈÈ
œœ
œ"
œœœ
u
eCe Cœ#œ #
u sin x
−Ä
106. dx e du, where u cos x and du''edx
1x 1x
cos x
%$
Ê Ê
ÈÈ
œœ
œ" œ
œœ œœ
u
eC e Cœ œ
u cos x
−Ä
107. dx u du, where u sin x and du''
ab
ÈÈ
sin x
1x 1x
dx
−Ä
Ê
Ê Ê
œœ
#"
œœœ
CC œ#œ #
u
33
sin x
# %$
# ab
108. dx u du, where u tan x and du''
È
tan x
1x 1x
dx%$
Ê Ê
ÊÊ
"Î# "
œœœ
uC tanx C tanxCœ #œ #œ #
22 2
33 3
$Î# "
$Î#
œ"
$
ab ab É
109. dy dy du, where u tan y and du'' ""
Ê Ê
œ"
abab
Š‹
tan y 1 y tan y u 1 y
dy
− Ê œœ œ œ'
œ
Ê
#
1y
ln u C ln tan y Cœ #œ #kk k k
œ"
110. dy dy du, where u sin y and du'' ""
Ê œ
œ"
ab È È

sin y 1 y 1ysin y u
dy
−ÄÊ Ê%$ œœ œœ'
œ
Ä
#Ä
1y
ln u C ln sin y Cœ #œ #kk k k
œ"
111. dx sec u du, where u sec x and du ; x 2 u , x 2 u''
Ê
24
23
sec sec x
xx 1 xx 1
dx
43
Ê %$
Ê Ê
ab
È È
œ œ
#"
œœœœÊœœÊœ
œ
œ
#
#
È 11
tan u tan tan 3 1œœ œ œ œcd
È
1
1
11Î$
Î4 34
112. dx cos u du, where u sec x and du ; x u , x 2 u''
2
cos sec x
xx 1 xx 1
dx 2
3 63##
#
Ê36
23
ab
È ÈÈ
−Ä
Ê Ê
œ œ
œ"
œœœœÊœœÊœ
œ
œ
11
sin u sin sin œœ œ œcd
1
1
11Î$
Î'
œ"
#36
3
È
113. lim lim 5
x0 x0ÄÄ
sin 5x
x1
−Ä
œœ
Š‹
5
1 25x
œ
Ä
#
114. lim lim lim lim x x 1
x1 x1 x1 x1ÄÄ Ä Ä
ßß ß ß
È ab
Š‹ab
Œ Ê
x1
sec x sec x
x1
x 1 (2x)Ê
%$ %$
Ê
Ä_Ê
Ê

œ œ
œ
œœ œœ
œ
#
œ
#
Ä  
œ
x x 1
kk
115. lim x tan lim lim lim 2
xxxxÄ_ Ä_ Ä_ Ä_
œ"
œÊˆ‰
22
xx x14x
tan 2x
œœœœ
−$%$

ab
Š‹
Ä
$#
Ê
$#2x
14x
116. lim lim lim
x0 x0 x0ÄÄÄ
2 tan 3x 6 6
7x 14x 7 71 9x
−ÄÊ
Ê ß
œœ œ
Š‹
ab
12x
19x
Ê
−
Ê

469. Section 7.7 Inverse Trigonometric Functions 465
117. If y ln x ln 1 x C, then dy dxœ œ
""
#
#

ab
–—
tan x x
xx1xx
tan x
−$

%$ Š‹
x
1x

#
dx dx dx,œ œœŠ‹
""


x 1x x1x x x1x x
x tan x tan x x1 x x x tan x 1 x

%$ %$
ab ab
ab a bab
which verifies the formula
118. If y cos 5x dx, then dy x cos 5x dxœ # œ ##
x5x x55x
44 4 4
1 25x 1 25x 1 25x
ßß ß ß

" $ "

'
ÈÈÈ
’“ Š‹Š‹Š‹
x cos 5x dx, which verifies the formulaœab
$"
119. If y x sin x 2x 2 1 x sin x C, thenœ ab
È
" "
#
#
dy sin x 2 sin x 2 1 x dx sin x dx, which verifiesœ œ’“ab ab
È
Š‹
" " "
# #

" #
2x sin x
1x 1x 1x
2xab
ÈÈ È
−$

the formula
120. If y x ln a x 2x 2a tan C, then dy ln a x 2 dxœ œ ab ab ˆ‰
–—
## " ##


x2x2
aax
1


Š‹
x
a
#
#
ln a x 2 2 dx ln a x dx, which verifies the formulaœ œ #’“ab ab Š‹
## ##

ax
ax


121. dy y sin x C; x 0 and y 0 0 sin 0 C C 0 y sin x
dy
dx
1x 1x
dx
œÊœÊœ # œœÊœ #ʜʜ
"

" " "
ÈÈ

122. 1 dy 1 dx y tan (x) x C; x 0 and y 1 1 tan 0 0 C
dy
dx x 1 1 x
œ Êœ Êœ œœÊœ
""

" "

ˆ‰
C 1 y tan (x) x 1ʜʜ
"
123. dy y sec x C; x 2 and y sec 2 C C sec 2
dy
dx
xx 1 xx 1
dx
œÊœÊœ # œœÊœ #Êœ
"

" " "
ÈÈ

kk 11 1
ysec(x) , x1œœÊœ 1
11 1
33 3
22 "
124. dy dx y tan x 2 sin x C; x 0 and y 2
dy
dx 1x 1x
22
1x 1x
œ Êœ Êœ œœ
""


" "

ÈÈ
Š‹
2 tan 0 2 sin 0 C C 2 y tan x 2 sin x 2ʜ ʜʜ
" " " "
125. The angle is the large angle between the wall and the right end of the blackboard minus the small angle!
between the left end of the blackboard and the wall cot cot .ʜ !
" "ˆ‰ ˆ‰
xx
15 3
126. V 2 (sec y) dy 4y tan y 3œ œ œ 111 '
0
3
#
cdcd Š‹
È
## Î$
!
1 14
3
127. V r h (3 sin ) (3 cos ) 9 cos cos , where 0 9 (sin ) 1 3 cosœœ œ ŸŸ Ê œ ˆ‰ ˆ‰ ab ab
""## $ #
#33 d
dV
11))1)) ) 1) )
1
)
sin 0 or cos the critical points are: 0, cos , and cos ; butœ! Ê œ œ „ Ê ))
""" " "
ÈÈÈ
333
Š‹ Š ‹
cos is not in the domain. When 0, we have a minimum and when cos 54.7 °, we
" "" "
Š‹ Š‹ œœ¸
È È
3 3
))
have a maximum volume.
128. 65°(90°)(90°) 180° 65 ° 65°tan 65 °22.78°42.22° œÊœ œ ¸ ¸"! !"
"ˆ‰
21
50
129. Take each square as a unit square. From the diagram we have the following: the smallest angle has a!
tangent of 1 tan 1; the middle angle has a tangent of 2 tan 2; and the largest angle ʜ ʜ!"" #
" "
has a tangent of 3 tan 3. The sum of these three angles is Êœ Ê ## œ#1!"#1
"
tan 1 tan 2 tan 3 .Ê ## œ
" " "
1

470. 466 Chapter 7 Transcendental Functions
130. (a) From the symmetry of the diagram, we see that sec x is the vertical distance from the graph of1
"
y sec x to the line y and this distance is the same as the height of y sec x above the x-axis atœœ œ
" "
1
x; i.e., sec x sec ( x). œ 1
" "
(b) cos ( x) cos x, where 1 x 1 cos cos , where x 1 or x 1
" " " " ""
œ ŸŸ Ê œ  Ÿ11 ˆ‰ ˆ‰
xx
sec ( x) sec xÊ œ
" "
1
131. sin (1) cos (1) 0 ; sin (0) cos (0) 0 ; and sin ( 1) cos ( 1) .
" " " " " "
## ## # #
# œ#œ # œ#œ œ œ
11 11 1 1
1
If x ( 0) and x a, then sin (x) cos (x) sin ( a) cos ( a) sin a cos a−"ß œ œ œ
" " " " " "
ab1
sin a cos a from Equations (3) and (4) in the text.œ œœ11ab
" "
##
11
132. tan x and tan tan x tan .Êœ œÊœ #œ #!" !"
""
#
" "
xx
1
133. (a) Defined; there is an angle whose tangent is 2.
(b) Not defined; there is no angle whose cosine is 2.
134. (a) Not defined; there is no angle whose cosecant is .
"
#
(b) Defined; there is an angle whose cosecant is 2.
135. (a) Not defined; there is no angle whose secant is 0.
(b) Not defined; there is no angle whose sine is 2.
È
136. (a) Defined; there is an angle whose cotangent is .
"
#
(b) Not defined; there is no angle whose cosine is 5.
137. (x) cot cot , x 0 (x) ; solving!!œ Êœ # œ
" " w

ˆ‰ ˆ‰
xx 153
15 3 225 x 9 x 225 x 9 x
15 9 x 3 225 x


aba b
abab
(x) 0 135 15x 675 3x 0 x 3 5 ; (x) 0 when 0 x 3 5 and (x) 0 for!!!
w##w w
œÊ œÊœ
ÈÈ
x 3 5 there is a maximum at 3 5 ft from the front of the room Ê
ÈÈ
138. From the accompanying figure, , cot !") 1 !## œœ
x
1
and cot cot x cot (2 x)")1œÊœ
2x
1
" "
Êœ œ
d
dx 1x 1(2x) 1x 1(2x)
1(2x) 1x) ""




ab
abc d

; solving 0 x 1; 0 for x 1 and 0 for x 1œœÊœ !
44x d d d
1x 1(2x) dx dx dx

abc d

)) )
at x 1 there is a maximum cot 1 cot (2 1)Êœ œ œ œ)1 1
" "
#
11 1
44
139. Yes, sin x and cos x differ by the constant
" "
#

1
140. Yes, the derivatives of y cos x C and y cos ( x) C are both œ œ
" " "

È
1x

141. csc u sec u csc u sec u 0 , u 1
" " " "
##

œ Êœ œ œ
11 dd
dx dx
u u 1 u u 1
ab kk ˆ‰
du du
dx dx
kk kk
ÈÈ

471. Section 7.7 Inverse Trigonometric Functions 467
142. y tan x tan y x (tan y) (x)œÊœÊ œ
" dd
dx dx
sec y 1 ʜʜœab
# ""

dy dy
dx dx sec y
1x



Š‹
È
, as indicated by the triangleœ
"
1x


143. f(x) sec x f (x) sec x tan x .œÊœ Ê œ œ œ
w
œ
"" "
„"
df
dx sec sec b tan sec b
xb bb
%$

%$
%$ %$

¹
df
dx
xf b
¹Š‹
abab È

Since the slope of sec x is always positive, we the right sign by writing sec x .
" " "
ll "
d
dx
xx
œ
È
144. cot u tan u cot u tan u 0
" " " "
##
œ Êœ œ œ
11 dd
dx dx 1 u 1 u
ab ˆ‰
du du
dx dx

145. The functions f and g have the same derivative (for x 0), namely . The functions therefore differ 
"
Èx(x 1)
by a constant. To identify the constant we can set x equal to 0 in the equation f(x) g(x) C, obtainingœ #
sin ( 1) 2 tan (0) C 0 C C . For x 0, we have sin 2 tan x .
" " " "
## #

œ #Êœ#Êœ   œ
11 1
ˆ‰ È
x1
x1
146. The functions f and g have the same derivative for x 0, namely . The functions therefore differ by a
"
1x

constant for x 0. To identify the constant we can set x equal to 1 in the equation f(x) g(x) C, obtaining œ #
sin tan 1 C C C 0. For x 0, we have sin tan .
" " " "" ""

Š‹
È È
2 x144 x
œ #Êœ #Êœ œ
11

147. V dx dx tan x tan 3 tanœœœœ 1111''
−#%#
%#


33 33
33
3
33
Š‹ ’ “ cd
È
Š‹
""

#

" " "
È
È
1x 1x 3
3


œ œ1 ‘ˆ‰
111
36

#
148. y 1 x 1 x y 1 x ( 2x) 1 y ; L 1 y dxœ œ Êœ Ê# œœ #
È
ab ab ab ab ˆ‰ É
# w#w # w
"Î# "Î# ""
#
# #
1x

'
−#
#12
12
2 dx 2 sin x 2 0œœœ œ '
0
12
12
0
#
#
"

"
È
1x
63

cd ˆ‰
11
149. (a) A(x) (diameter) V A(x) dx œœ œÊœ œ
11 1 1
4 4 1x 1x 1x 1x
dx# ""

#

’“ Š‹
ÈÈ

''
a1
b1 −
tan x ( )(2)œœœ11cd ˆ‰
"
"
" #
11
4

(b) A(x) (edge) V A(x) dxœœ œÊœ œ
# ""

#

’“ Š‹
ÈÈ
1x 1x
44 dx
1x 1x

''
a1
b1 −
4 tan x 4 tan (1) tan ( 1) 4 2œœ œ œcd c d ‘ˆ‰
" " "
"
"
11
44
1
150. (a) A(x) (diameter) 0 V A(x) dxœœ œœÊœ
11 1 1
44 4
24
1x 1x 1x
#

#

Š‹Š‹
ß

È ÈÈ
'
a
b
dx sin x sin sinœœœ œ œ '
−#
#
#
%#


22
22
22
22
1 111
È
ÈÈ
1x
22
44

" " "
## #


11 1cd ’“Š‹ Š ‹ ‘ˆ‰
(b) A(x) 0 V A(x) dx dxœœ œÊœ œ
(diagonal)
22
12 2 2
1x 1x 1x

ß

Š‹
È ÈÈ

#

''
a22
b22 −#
#


2sin x 2 2œœœcd ˆ‰
"


22
22#
%# 1
4
† 1151. (a) sec 1.5 cos 0.84107 (b) csc ( 1.5) sin 0.72973
" " " " " "
œ¸ œ ¸
1.5 1.5
ˆ‰
(c) cot 2 tan 2 0.46365
" "
#
œ ¸
1
152. (a) sec ( 3) cos 1.91063 (b) csc 1.7 sin 0.62887
" " " " ""
œ ¸œ¸ˆ‰ ˆ‰
31.7
(c) cot ( 2) tan ( 2) 2.67795
" "
#
œ ¸
1

472. 468 Chapter 7 Transcendental Functions
153. (a) Domain: all real numbers except those having
the form k where k is an integer.
1
#
#1
Range: y
11
##

(b) Domain: x ; Range: y_$$_ _$$_
The graph of y tan (tan x) is periodic, theœ
"
graph of y tan tan x x for x .œœ _Ÿ $_ab
"

154. (a) Domain: x ; Range: y_$$_ ŸŸ
11
##

(b) Domain: x 1; Range: y 1"Ÿ Ÿ "Ÿ Ÿ
The graph of y sin (sin x) is periodic; theœ
"
graph of y sin sin x x for x 1.œœ "Ÿ Ÿab
"

155. (a) Domain: x ; Range: 0 y_$$_ŸŸ 1

(b) Domain: 1 x 1; Range: y 1ŸŸ "Ÿ Ÿ
The graph of y cos (cos x) is periodic; theœ
"
graph of y cos cos x x for x 1.œœ "Ÿ Ÿab
"

473. Section 7.7 Inverse Trigonometric Functions 469
156. Since the domain of sec x is ( 1] [ ), we
"
_ß "ß _
have sec sec x x for x 1. The graph ofab kk
"
œ  
y sec sec x is the line y x with the openœœab
"
line segment from ( ) to ( ) removed."ß""ß"

157. The graphs are identical for y 2 sin 2 tan xœ ab
"
4 sin tan x cos tan x 4œœcdc dabab Š‹Š‹
" "

"x
x1 x1
ÈÈ

from the triangle œ
4x
x1



158. The graphs are identical for y cos 2 sec xœ ab
"
cos sec x sin sec xœ œ
#" #" "
abab
xx
x1


from the triangle œ
2x
x




159. The values of f increase over the interval [ 1] because"ß
f 0, and the graph of f steepens as the values of f
ww

increase towards the ends of the interval. The graph of f
is concave down to the left of the origin where f 0,
ww
$
and concave up to the right of the origin where f 0.
ww

There is an inflection point at x 0 where f 0 andœœ
ww
f has a local minimum value.
w

160. The values of f increase throughout the interval ( )_ß _
because f 0, and they increase most rapidly near the
w

origin where the values of f are relatively large. The
w
graph of f is concave up to the left of the origin where
f 0, and concave down to the right of the origin
ww

where f 0. There is an inflection point at x 0
ww
$ œ
where f 0 and f has a local maximum value.
ww w
œ

474. 470 Chapter 7 Transcendental Functions
7.8 HYPERBOLIC FUNCTIONS
1. sinh x cosh x 1 sinh x 1 1 , tanh x ,œÊœ # œ œ #œœ œœœ
3 3 9 25 5 sinh x 3
4 4 16 16 4 cosh x 5È É ˆ‰ ÉÉ #
# ˆ‰
ˆ‰
3
4
5
4
coth x , sech x , and csch xœœ œœ œœ
"" "
tanh x 3 cosh x 5 sin x 3
54 4
2. sinh x cosh x 1 sinh x 1 , tanh x , coth x ,œÊ œ # œ #œœ œœœ œœ
4 16 25 5 sinh x 4 5
3 9 9 3 cosh x 5 tanh x 4È ÉÉ # "
ˆ‰
ˆ‰
4
3
5
3
sech x , and csch xœœ œœ
""
cosh x 5 sinh x 4
33
3. cosh x , x 0 sinh x cosh x 1 1 1 , tanh xœ Êœ œ œ œœ œœ
17 17 289 64 8 sinh x
15 15 225 225 15 cosh xÈ Éˆ‰ ÉÉ #
# ˆ‰
ˆ‰
8
15
17
15
, coth x , sech x , and csch xœœœœœ œœ
81715 15
17 tanh x 8 cosh x 17 sinh x 8
"" "
4. cosh x , x 0 sinh x cosh x 1 1 , tanh x ,œ Êœ œ œœ œœœ
13 169 144 12 sinh x 12
525255cosh x13 È
ÉÉ #
ˆ‰
ˆ‰
12
5
13
5
coth x , sech x , and csch xœœ œœ œœ
"" "
tanh x 12 cosh x 13 sinh x 12
13 5 5
5. 2 cosh (ln x) 2 e xœœ # œ#Š‹
ee ln x
e x
ln x ln x
ln x
""
#
%
6. sinh (2 ln x)œœœœ
ee ee x
x
x
2lnx 2lnx lnx lnx
x
"
####

% ß

$

Š‹
7. cosh 5x sinh 5x e# œ # œ
ee ee 5x
5x 5x 5x 5x

##
%%
8. cosh 3x sinh 3x e œ œ
ee ee 3x
3x 3x 3x 3x

##

%%
9. (sinh x cosh x) e e# œ # œœ
%
##
% %ˆ‰ ab
ee ee x4x
xx xx%%
10. ln (cosh x sinh x) ln (cosh x sinh x) ln cosh x sinh x ln 1 0 œ œœab
##
11. (a) sinh 2x sinh (x x) sinh x cosh x cosh x sinh x 2 sinh x cosh xœ #œ # œ
(b) cosh 2x cosh (x x) cosh x cosh x sinh x sin x cosh x sinh xœ #œ # œ #
##
12. cosh x sinh x e e e e e e e e
## "
##
##
œ œ ˆ‰ˆ‰ cdcdabababab
ee ee
4
xx xxxx xx
xx xx%%
2e 2e 4e (4) 1œœœœ
""" !
444
xx
aba b ab
13. y 6 sinh 6 cosh 2 cosh œÊœ œ
xxx
3dx 33 3
dy
ˆ‰ˆ‰
"
14. y sinh (2x 1) [cosh (2x 1)](2) cosh (2x 1)œ #Êœ # œ #
""
##
dy
dx
15. y 2 t tanh t 2t tanh t sech t t 2t tanh t tœœ Êœ #ÈÈ ‘ˆ ‰ˆ ‰ ˆ ‰ˆ ‰ˆ‰
"Î# "Î# # "Î# "Î# "Î# "Î# "Î# "
#
dy
dt
sech tœ #
#È
tanh t
t
È
È
16. y t tanh t tanh t sech t t t (2t) tanh t sech 2t tanh œœ Êœ œ
# # " # " # # " #" ""
tdt tt
dy
cdabababab
17. y ln (sinh z) coth z 18. y ln (cosh z) tanh zœÊœœ œ Êœœ
dy dy
dz sinh z dz cosh z
cosh z sinh z

475. Section 7.8 Hyperbolic Functions 471
19. y (sech )(1 ln sech ) (sech ) ( sech tanh )(1 ln sech )œ Êœ )) ))))
dy
dsech
sech tanh
))
))
ˆ‰

sech tanh (sech tanh )(1 ln sech ) (sech tanh )[1 (1 ln sech )]œ œ )) )) ) )) )
(sech tanh )(ln sech )œ )) )
20. y (csch )(1 ln csch ) (csch ) (1 ln csch )( csch coth )œ Êœ )) ) )))
dy
dcsch
csch coth
))
))
ˆ‰

csch coth (1 ln csch )(csch coth ) (csch coth )(1 1 ln csch ) (csch coth )(ln csch )œ œ œ)) ) )) )) ) )) )
21. y ln cosh v tanh v (2 tanh v) sech v tanh v (tanh v) sech vœ Êœ œ
""
##
### dy
dv cosh v
sinh v
ˆ‰ ab ab
(tanh v) 1 sech v (tanh v) tanh v tanh vœ œœabab
##$
22. y ln sinh v coth v (2 coth v) csch v coth v (coth v) csch vœ Êœ œ #
""
##
### dy
dv sinh v
cosh v
ˆ‰ ab ab
(coth v) 1 csch v (coth v) coth v coth vœ # œœabab
##$
23. y x 1 sech (ln x) x 1 x 1 x 1 2x 2œ # œ # œ # œ # œÊœab ab ab ab ˆ‰ ˆ‰ ˆ‰
## ##

222x
ee xx x1 dx
dy
ln x ln x− −
24. y 4x 1 csch (ln 2x) 4x 1 4x 1 4x 1œ œ œ œ ab ab ab ab ˆ‰ ˆ‰ Š‹
## ##

224x
ee 2x (2x) 4x 1
ln 2x ln 2x−
4x 4œÊœ
dy
dx
25. y sinh x sinh x œœ Êœœœ
" " "Î#

""
# #È ˆ‰
dy
dx
x
1x
x1x x(1 x)
Š‹
É ab
ÈÈÈ

#−


26. y cosh 2 x 1 cosh 2(x 1) œ #œ # Êœ œ œ
" " "Î#


""
È ˆ‰
dy
dx
(2) (x 1)
2(x 1) 1
x14x3 4x 7x 3
Š‹
Écd
ÈÈ È

# −


27. y (1 ) tanh (1 ) ( 1) tanh tanhœ Êœ œ )) ) ) )
" " " ""

dy
d1 1)) ) ˆ‰

28. y 2 tanh ( 1) 2 (2 2) tanh ( 1)œ ## Êœ ####ab ab ’“)) ) )) ) )
#" # " "

dy
d1(1)))

(2 2) tanh ( 1) (2 2) tanh ( 1) 1œ ## # œ
))
))




" "2
2
))))
29. y (1 t) coth t (1 t) coth t (1 t) ( 1) coth t coth tœ œ Êœ œ
" " "Î# " "Î# "

"
#È Èˆ‰ ˆ‰
–—
dy
dt
t
1t t
Š‹
ab
È

#−


30. y 1 t coth t 1 t ( 2t) coth t 1 2t coth tœ Êœ œab ab ˆ‰
# " # " " "

dy
dt 1 t

31. y cos x x sech x x (1) sech x sech xœ Êœ œ
" " " " " " " "

dy
dx
1x x1x 1x 1x
ÈÈ ÈÈ

’“Š‹
sech xœ
"
32. y ln x 1 x sech x ln x 1 x sech x œ œ Ê
È
ab
# " # "
"Î# dy
dx
1 x 1 x ( 2x) sech x sech x sech xœ œ œ
""" "" # # " " "
"Î# "Î#

#x xx
x1 x 1 x 1 x
xx
ab ab Š‹ ˆ‰
ÈÈÈ
33. y csch œÊœ œ œ
""
#


ˆ‰
)
)
dy ln (1) ln (2)
d
ln
1
11
ln 2
’“Š‹Š‹
Š‹ Š‹Ë ”•
ÊÊŠ‹ Š‹

##

##

##

476. 472 Chapter 7 Transcendental Functions
34. y csch 2 œÊœ œ
"



)
)
dy (ln 2) 2
d
212
ln 2
12



É ab
È

35. y sinh (tan x) sec xœÊœœœœœ
"

dy
dx sec x sec x
sec x sec x sec x
1 (tan x) sec x
sec x sec x


ÈÈ kk kk
kkkk
kk
36. y cosh (sec x) sec x, 0 xœ Êœœœœ $$
"

#
dy (sec x)(tan x) (sec x)(tan x) (sec x)(tan x)
dx tan x
sec x 1 tan x
ÈÈ kk

1
37. (a) If y tan (sinh x) C, then sech x, which verifies the formulaœ # œœœ
"

dy
dx 1 sinh x cosh x
cosh x cosh x

(b) If y sin (tanh x) C, then sech x, which verifies the formulaœ # œœœ
"

dy
dx sech x
sech x sech x
1tanhx


È
38. If y sech x 1 x C, then x sech x x sech x,œ œ ## œ
xx2x dy
dx
x1 x 41 x


## #
" " """ #

È
Š‹
ÈÈ
which verifies the formula
39. If y coth x C, then x coth x x coth x, which verifiesœ ## œ ## œ
xx x dy
dx 1 x


" " " "
## ##
" " "
Š‹ ˆ‰
the formula
40. If y x tanh x ln 1 x C, then tanh x x tanh x, which verifiesœ œ ## œ
" # " """"
##
ab ˆ‰ˆ‰
dy
dx 1x 1x
2x

the formula
41. sinh 2x dx sinh u du, where u 2x and du 2 dx'' œœœ
"
#
CC œ #œ #
cosh u cosh 2x
##
42. sinh dx 5 sinh u du, where u and du dx'' xx 555
œœœ
"
5 cosh u C 5 cosh Cœ #œ #
x
5
43. 6 cosh ln 3 dx 12 cosh u du, where u ln 3 and du dx'' ˆ‰
xx ###
"
œœ œ
12 sinh u C 12 sinh ln 3 Cœ #œ ˆ‰
x
#
44. 4 cosh (3x ln 2) dx cosh u du, where u 3x ln 2 and du 3 dx'' œœ œ
4
3
sinh u C sinh (3x ln 2) Cœ #œ
44
33
45. tanh dx 7 du, where u and du dx'' x sinh u x 7 cosh u 7 7
œœœ
"
7 ln cosh u C 7 ln cosh C 7 ln C 7 ln e e 7 ln 2 Cœ #œ #œ #œ kk ¸¸ ¸ ¸ ¹¹ "" " "

#
ÎÎxee
7
x7 x7
x7 x7  − 
7 ln e e Cœ ##kk
x7 x7# %#
46. coth d 3 du, where u and du'' )))
ÈÈÈ
333
cosh u d
sinh u
)œœœ
È
3 ln sinh u C 3 ln sinh C 3 ln Cœ #œ #œ #
ÈÈ È
kk ¹¹ ¹ ¹ "" "

#
)
È
3
ee
 #%Î#
3 ln e e 3 ln 2 C 3 ln e e Cœ œ
ÈÈÈ
¹¹ ¹¹
)) ))Î $ Î $ Î Î
"
ÈÈ ÈÈ
33
47. sech x dx sech u du, where u x and du dx'' ## ""
##ˆ‰ ˆ‰ œœ œ
tanh u C tanh x Cœ #œ ˆ‰
"
#

477. Section 7.8 Hyperbolic Functions 473
48. csch (5 x) dx csch u du, where u (5 x) and du dx'' ##
œ œ œ
( coth u) C coth u C coth (5 x) Cœ œ #œ
49. dt 2 sech u tanh u du, where u t t and du''
sech t tanh t
t 2t
dt
ÈÈ
È È
œœœœ È
"Î#
2( sech u) C 2 sech t Cœ œ È
50. dt csch u coth u du, where u ln t and du''
csch (ln t) coth (ln t)
t t
dt
œœœ
csch u C csch (ln t) Cœ œ
51. coth x dx dx du ln u ln ln ln ln ,'''
ln 2 ln 2 3 4
ln 4 ln 4 15 8
œœœœ œœ
cosh x 15 3 15 4 5
sinh x u 8 4 8 3
#
#
" "&Î)
$Î% #
cdkk ¸¸ ¸¸ ¸ ¸ †
where u sinh x, du cosh x dx, the lower limit is sinh (ln 2) and the upperœœ œœœ
ee 3
2
4
ln 2 ln 2

##

− Š‹

#
limit is sinh (ln 4)œœœ
ee 15
4
8
ln 4 ln 4

##

− Š‹

4
52. tanh 2x dx dx du ln u ln ln 1 ln , where'' '
00 1
ln 2 ln 2 17 8
œœœœ œ
sinh 2x 17 17
cosh 2x u 8 8
""" " "
### #
"(Î)
"
#
cdkk ‘ˆ‰
u cosh 2x, du 2 sinh (2x) dx, the lower limit is cosh 0 1 and the upper limit is cosh (2 ln 2) cosh (ln 4)œœ œ œ
œœœ
ee 17
4
8
ln 4 ln 4
#
##
#
− Š‹

4
53. 2e cosh d 2e d e 1 d'' '
−− −
−− −
−ln 4 ln 4 ln 4
ln 2 ln 2 ln 2
2
ln 2
ln 4
))
)) ) ) )œœ # œ #Š‹ ab ‘
ee


#
$
# e
2
#

ln 2 ln 4 ln 2 ln 4 ln 2 2 ln 2 ln 2œ œ œ œ#Š‹Š‹ ˆ‰ˆ ‰
ee 3 3
83232 3
−−2ln2 2ln4
## #
""
54. 4e sinh d 4e d 2 1 e d 2'' '
00 0
ln 2 ln 2 ln 2
2
ln 2
0
−− −## #
)) ) ) )œœ œ #Š‹ ’ “ ab
ee e


##
−− #
2 ln 2 0 2 ln 2 2 ln 2 1 ln 4œ œ œ œ ’“Š‹Š‹ ˆ‰
ee 3
844
−2ln2 0
## #
"" "
55. cosh (tan ) sec d cosh u du sinh u sinh (1) sinh ( 1)''
−#%
#

41
41
)))
# "
"

##
œœœ œ cd Š‹Š‹
ee e e
$%$ %$$
e e , where u tan , du sec d , the lower limit is tan 1 and the upperœœ œœ œ
eeee
4

#
" #
%$ %$
))) ˆ‰
1
limit is tan 1ˆ‰
1
4
œ
56. 2 sinh (sin ) cos d 2 sinh u du 2 cosh u 2(cosh 1 cosh 0) 2 1''
00
21
#
))) œœœ œ cd Š‹
"
!
#
#
ee
−$
e e 2, where u sin , du cos d , the lower limit is sin 0 0 and the upper limit is sin 1œ œœ œ œ
"
#
))) ˆ‰
1
57. dt cosh u du sinh u sinh (ln 2) sinh (0) 0 , where''
10
2ln 2
ln 2
0
cosh (ln t)
t 4
ee 3
2
œœœ œ œœcd
ln 2 ln 2

##

−

#
u ln t, du dt, the lower limit is ln 1 0 and the upper limit is ln 2œœ œ
"
t
58. dx 16 cosh u du 16 sinh u 16(sinh 2 sinh 1) 16''
11
42
8 cosh x
x
ee ee
È
È
œœœ œ cd ’“Š‹Š‹
#
"

##
#
8 e e e e , where u x x , du x dx , the lower limit is 1 1 and the upperœ œœ œ œ œab È È
# # " "Î# "Î# "
#
dx
2xÈ
limit is 4 2
È
œ
59. cosh dx dx (cosh x 1) dx sinh x x'''
−−−
−ln 2 ln 2 ln 2
000
0
ln 2
#
### #
#" " "ˆ‰ cd
x cosh x
œœ # œ #

478. 474 Chapter 7 Transcendental Functions
[(sinh 0 0) (sinh ( ln 2) ln 2)] (0 0) ln 2 ln 2œ œ œ
"""
#####

’“ Š‹
–—
ee
2
−ln 2 ln 2 Š‹

#
1 ln 2 ln 2 ln 2œ œ# œ#
"" "
##
ˆ‰ È
48 8
33
60. 4 sinh dx 4 dx 2 (cosh x 1) dx 2 sinh x x'' '
00 0
ln 10 ln 10 ln 10
ln 10
0
#
##
ˆ‰ ˆ ‰ cd
x cosh x 1
œœ œ
2[(sinh (ln 10) ln 10) (sinh 0 0)] e e 2 ln 10 10 2 ln 10 9.9 2 ln 10œ œ œ œ
ln 10 ln 10− "
10
61. sinh ln 1 ln 62. cosh ln 1 ln 3
" "
#ˆ ‰ ˆ‰ ˆ‰Š‹ Š‹É É
5525 2 5525
1 12 144 3 3 3 9
œ œœ œ
63. tanh ln 64. coth ln ln 9 ln 3
" """ ""
## # # #
ˆ‰ ˆ‰ Š‹ Š‹œœ œœœ
1 (1/2) (9/4)
1 (1/2) 4 (1/4)
ln 3 5
65. sech ln ln 3 66. csch ln 3 ln 3 2
" " "ˆ‰ Š‹ Š‹ Š‹

ÈÈ
3
5 (3/5)
1 1 (9/25)
3
4/3
1/ 3
œœ œ œ
È È
È
Š‹
È
67. (a) sinh sinh 3 sinh 0 sinh 3'
0
23
23
0


dx x
4x
È

" " "
#

œœ œ ‘ ÈÈ
(b) sinh 3 ln 3 3 1 ln 3 2
"ÈÈ È
Š‹Š‹Èœ ## œ #
68. (a) 2 , where u 3x, du 3 dx, a 1''
00
13 1
#
6 dx dx
19x a u
ÈÈ


œœœœ
2 sinh u 2 sinh 1 sinh 0 2 sinh 1œœ œcda b
" " " "
"
! (b) 2 sinh 1 2 ln 1 1 1 2 ln 1 2
" #
œ ## œ #Š‹Š‹
ÈÈ
69. (a) dx coth x coth 2 coth '
54
2#
"

" " "
#
&Î%1x 4
5
œœ cd
(b) coth 2 coth ln 3 ln ln
" " """
##
œ œ
59/4
41/43
‘ˆ‰
70. (a) dx tanh x tanh tanh 0 tanh '
0
12
#
"""
##
" " " "
"Î#
!1x
œœ œcd
(b) tanh ln ln 3
""" "
## #

œœŠ‹
1 (1/2)
1 (1/2)
71. (a) , where u 4x, du 4 dx, a 1''
15 45
313 1213##
##
dx du
x1 16x ua u
ÈÈ


œœœœ
sech u sech sech œ œcd
" " "
12 13
45#
# 12 4
13 5 (b) sech sech ln ln œ
" " 12 4
13 5 (12/13) (4/5)
1 1 (12/13) 1 1 (4/5)
Š‹Š‹
ÈÈ
ln ln ln ln ln 2 ln œ œ œ Š‹Š‹ ˆ‰ ˆ ‰
13 169 144 5 25 16
14412
53 135 3
# #

ÈÈ
ln 2 ln œœˆ‰†
24
33
72. (a) csch csch 1 csch csch csch 1'
1
2
dx x
x4 x
È

"" """
### ###
" " " " "
#
"

œ œ œ ‘ˆ ‰ˆ ‰¸¸
(b) csch csch 1 ln 2 ln 1 2 ln
"" " "
## # #
" "

ˆ‰ ’“Š‹Š‹Š‹
È
œ œ
ÈÈ
È
5/4
(1/2)
25
12
73. (a) dx du sinh u sinh 0 sinh 0 0, where u sin x, du cos x dx''
00
0

cos x
1sinx 1u
ÈÈ

" " " "
!
!

œœœ œœœcd
(b) sinh 0 sinh 0 ln 0 0 1 ln 0 0 1 0
" "
œ œŠ‹Š‹ÈÈ

479. Section 7.8 Hyperbolic Functions 475
74. (a) , where u ln x, du dx, a 1''
10
e1
dx du
x 1 (ln x) au xÈÈÊ Ê
"
Ê ÊÊ
œœœœ
sinh u sinh 1 sinh 0 sinh 1œœ œ œcd
œ" œ" œ" œ"
"
!
(b) sinh 1 sinh 0 ln 1 1 1 ln 0 0 1 ln 1 2
œ" œ" ##
œ œ œ #Š‹Š‹Š‹
ÈÈÈ
75. (a) Let E(x) and O(x) . Then E(x) O(x)œœ # œ #
f(x) f( x) f(x) f( x) f(x) f( x) f(x) f( x)Êœ œœ Êœ œœ
## ##
f(x). Also, E( x) E(x) E(x) is even, andœœ œœœœÊ
2f(x) f( x) f( ( x)) f(x) f( x)
###
œÊœœ Êœ
O( x) O(x) O(x) is odd. Consequently, f(x) can be written asœœœ œ œœ Ê
f( x) f( ( x)) f(x) f( x)œœœœ œœ
##
a sum of an even and an odd function.
(b) f(x) because 0 if f is even and f(x) because 0 if f is odd.œœœœ
f(x) f( x) f(x) f( x) f(x) f( x) f(x) f( x)Êœ œœ œœ Êœ
## ##
Thus, if f is even f(x) 0 and if f is odd, f(x) 0œ # œ#
2f(x) 2f(x)
##
76. y sinh x x sinh y x 2x e 2xe e 1 e 2xe 1 0œÊœÊœÊœ œÊœ œÊ œœ œ
œ" œ"
#
ee y
e
yy
y −
y2y 2y y
e e x x 1 sinh x y ln x x 1 . Since e 0, we cannotÊœ Êœ ## Êœœ
yy y 2x 4x 4„
#
## œ"
È
Ê
ÈÈ
Š‹
choose e x x 1 because x x 1 0.
y
œ
ÈÈ
##
77. (a) v tanh t sech t g sech t .œÊœ œÉÉ Œ Ä Œ ÄŒ Ä Œ ÄÉÉÉÉ ”•
mg gk mg gk gk gk
kmdtk mm m
dv ##
Thus m mg sech t mg tanh t mg kv . Also, since tanh x when x , v
dv
dt m m
gk gk
œœ" œ œ œ œ! œ! œ!
###
Œ Ä Œ ÄÉÉ Œ Ä
when t . œ!
(b) lim v lim tanh t lim tanh t (1)
tt tÄ_ Ä_ Ä_
œœœœÉÉ ÉÉŒ Ä Œ ÄÉÉ
mg kg mg kg mg mg
kmk mkk
(c) 80 5 178.89 ft/secÉ É È
160 400
0.005 5
160,000
5
œœœ¸
È
78. (a) s(t) a cos kt b sin kt ak sin kt bk cos kt ak cos kt bk sin ktœ # Êœ Êœ œœ
ds d s
dt dt
Ê
Ê
##
k (a cos kt b sin kt) k s(t) acceleration is proportional to s. The negative constant kœ œœ Ê œ
## #
implies that the acceleration is directed toward the origin. (b) s(t) a cosh kt b sinh kt ak sinh kt bk cosh kt ak cosh kt bk sinh ktœ # Êœ # Êœ #
ds d s
dt dt
Ê
Ê
##
k (a cosh kt b sinh kt) k s(t) acceleration is proportional to s. The positive constant k impliesœ # œÊ
## #
that the acceleration is directed away from the origin.
79. y dx dx y sech (x) 1 x C; x 1 and
dy
dx
x1x 1x x1x 1x
xx
œ # Êœ # Êœ œ
œ" œ"
œœ œ œ
œ" #
ÈÈ È È
ÊÊ Ê Ê
'' È
y0 C0 ysech(x) 1xœÊ œÊœ œœ
œ" # È
80. To find the length of the curve: y cosh ax y sinh ax L 1 (sinh ax) dxœÊœÊœ #
" w #
a '
0
b
È
L cosh ax dx sinh ax sinh ab. Then the area under the curve is A cosh ax dxÊœ œ œ œ''
0 0
b b
b
0
œ ‘
"" "
aa a
sinh ax sinh ab sinh ab which is the area of the rectangle of height and length Lœœœœ ‘ˆ‰ˆ‰
"""" "
aaaa a
ÊÊ
b
0
as claimed.
81. V cosh x sinh x dx 1 dx 2 œ œ œœ111
''
00
22
ab
##
82. V 2 sech x dx 2 tanh x 2œœœœ1111 '
0
ln 3
ln 3
0
Ê
Ê
#
œ
Ê
cd
–—
ÈÈ
Š‹
ÈÈ
Š‹
31/3
31/3

480. 476 Chapter 7 Transcendental Functions
83. (a) y cosh 2x y sinh 2x L 1 (sinh 2x) dx cosh 2x dx sinh 2xœÊœÊœ # œœ
" "
# #
w # ''
00
ln 5 ln 5
ln 5
0


È ‘
5 œœ œ’“Š‹ ˆ‰
" " "
##
ee 6
455
2x 2x−
ln 5
0
(b) y cosh ax 1 y 1 sinh ax cosh ax L cosh ax dx cosh ax dxœÊ # œ# ÊÊœ œ œ
" w###
#
a a
sin ax
ab
È ‘ ''
00
bb
b
0
œ
sinh ab
a
84. (a) Let the point located at (cosh u 0) be called T. Then A(u) area of the triangle OTP minus the areaߜ ?
under the curve y x 1 from A to T A(u) cosh u sinh u x 1 dx.œ Êœ
ÈÈ
# #"
# '
1
cosh u
(b) A(u) cosh u sinh u x 1 dx A (u) cosh u sinh u cosh u 1 (sinh u)œ Êœ
""
##
## w## '
1
cosh u
È
ab Š‹
È
cosh u sinh u sinh u cosh u sinh u (1)œ œ œœ
"" " ""
## # ##
### ##
ab ˆ‰
(c) A (u) A(u) C, and from part (a) we have A(0) 0 C 0 A(u) u 2A
w "
## #
œÊ œ # œÊ œÊ œÊœ
u u
85. y 4 cosh 1 1 sinh cosh ; the surface area is S 2 y 1 dxœÊ # œ# œœ #
xxx
4dx 4 4 dx
dy dy
Š‹ Š‹ ˆ‰ ˆ‰ Ê
# #
##
'
−ln 16
ln 81
1
8 cosh dx 4 1 cosh dx 4 x 2 sinh œœ # œ #11 1''
−− −ln 16 ln 16
ln 81 ln 81
ln 81
ln 16
#
##ˆ‰ ˆ ‰ ‘
xxx
4
4 ln 81 2 sinh ln 16 2 sinh 4 ln (81 16) 2 sinh (ln 9) 2 sinh (ln 4)œ œ ##11 ‘ˆ‰ˆ ‰ ˆ‰ ˆ ‰ cd
ln 81 ln 16
##

†
4 ln (9 4) e e e e 4 2 ln 36 9 4 4 4 ln 6œ œ œ ##111cd abab ‘ˆ ‰ˆ‰ˆ‰†
# "" ln 9 ln 9 ln 4 ln 4−−
94 94
80 15
4 4 ln 6 16 ln 6œ # œ #11ˆ‰
320 135 455
36 9
1
86. (a) y cosh x ds (dx) (dy) (dx) sinh x (dx) cosh x dx; M y dsœÊœ # œ # œœÈ È ab
## # ##
x
ln 2
ln 2
'
−
cosh x ds cosh x dx (cosh 2x 1) dx x e e ln 2œœ œ # œ # œ '' '
−−
−ln 2 ln 2 0
ln 2 ln 2 ln 2
ln 2
0
ln 4 ln 4
#
#
" ‘ ab
sinh 2x
4
ln 2; M 2 1 sinh x dx 2 cosh x dx 2 sinh x e e 2 .œ# œ # œœœ œœ
15 3
16 ''
00
ln 2 ln 2
ln 2
0
ln 2 ln 2
È
cd # "
## −
Therefore, y , and by symmetry x 0.œœ œ # œ
M
M83
ln 2 5ln 4x
15
16
3
ˆ‰
ˆ‰


(b) x 0, y 1.09œ¸
87. (a) y cosh x tan sinh x sinh xœÊœœ œ
Hw Hww w
wH dxwHH H
dy
ˆ‰ ˆ‰ ‘ˆ‰ˆ‰9
(b) The tension at P is given by T cos H T H sec H 1 tan H 1 sinh x999œÊœ œ # œ #È É ˆ‰
#
#
w
H
H cosh x w cosh x wyœœ œ ˆ‰ ˆ‰ ˆ‰
wHw
HwH
88. s sinh ax sinh ax as ax sinh as x sinh as; y cosh ax cosh axœÊœÊœÊœ œœ
"""" " " #
aaaa È
sinh ax 1 a s 1 sœ #œ #œ #
"" " ####
aa aÈÈ É

481. Chapter 7 Practice Exercises 477
89. (a) Since the cable is 32 ft long, s 16 and x 15. From Exercise 88, x sinh as 15a sinh 16aœœ œ Êœ
" " "
a
sinh 15a 16a.ʜ
(b) The intersection is near (0.042 0.672).ß

(c) Newton's method indicates that at a 0.0417525 the curves y 16a and y sinh 15a intersect.¸œœ
(d) T wy (2 lb) 47.90 lbœ¸ ¸ ˆ‰
"
0.0417525
(e) The sag is cosh a 4.85 ft.
""
aa
ab"&¸

CHAPTER 7 PRACTICE EXERCISES
1. y 10e (10) e 2eœÊœ œ
−#%#%#x5 x5 x5dy
dx 5
ˆ‰
"
2. y 2 e 2 2 e 2eœÊœ œ
ÈÈÈ
Š‹Š‹

2x 2x 2xdy
dx
3. y xe e x 4e e (1) 4e xe e e xeœ Êœ œ œ
"" " " ""
416 dx4 16 44
dy4x 4x 4x 4x 4x 4x 4x 4x 4x
cdabab
4. y x e x e x 2x e e (2x) (2 2x)e 2e (1 x)œœ Êœ # œ# œ #
## ##−#% % % % %#2 x 2x 2x 2x 2x 2 x
dy
dx
cdab
5. y ln sin 2 cot œÊœ œœab
#
))
dy 2(sin )(cos )
dsin sin
2 cos
)) )
)) )

6. y ln sec 2 tan œÊœ œab
#
))
dy 2(sec )(sec tan )
dsec))
)))

7. y log œœÊœ œ
2Š‹

xx2
ln
ln dx ln (ln 2)x
dy





## #
"
Š‹
Š‹
x
x
8. y log (3x 7) œ œÊœ œ
5
ln (3x 7) dy
ln 5 dx ln 5 3x 7 (ln 5)(3x 7)
33 "

ˆ‰ˆ ‰
9. y 8 8 (ln 8)( 1) 8 (ln 8) 10. y 9 9 (ln 9)(2) 9 (2 ln 9)œÊœ œ œÊœ œ
−− −tt t 2t2t2tdy dy
dt dt
11. y 5x 5(3.6)x 18xœÊœ œ
36 26 26    dy
dx
12. y 2 x 2 2 x 2xœÊœ œ
ÈÈÈ
Š‹Š ‹
− −− −− $Î $Î
2 21 21dy
dx

482. 478 Chapter 7 Transcendental Functions
13. y (x 2) ln y ln (x 2) (x 2) ln (x 2) (x 2) (1) ln (x 2)œ Ê œ œ Ê œ
x2 x2œœ y
yx
œ
ˆ‰
"
#
(x 2) ln (x 2) 1ʜ
dy
dx x2œ
cd
14. y 2(ln x) ln y ln 2(ln x) ln (2) ln (ln x) 0 (ln (ln x))œÊœ œ Êœ
x2 x2##
cd ˆ‰ ˆ‰ ˆ‰ ’“
xx y
yln x###
"
œ # ˆ‰
x
y ln (ln x) 2 (ln x) (ln x) ln (ln x)Êœ œ
w "" "
##œ ‘ œ ‘ˆ‰
ln x ln x
x2 x2##
15. y sin 1 u sin 1 u œœÊœ œ œ
" " # #
"Î#


È
ab dy
du
1 u ( 2u)
11u
uu
1u 1 1u u 1u
#
w
w
w

w ww w
ab
Ê’ “ab
ÈÈ È abkk
12
, 0u1œœ
"

u
u1 u 1 u
ÈÈ
ww
16. y sin sin v œœÊœœœœ
" " "Î#"""





Š‹
ÈÈÈ
ÉÉab
È
v
dy
dv
v
1v
2v 1 v 2v
v
2v v 1
œ
#
w

w


v1
v
œ
"
2v v 1
È
17. y ln cos x yœÊœœab
" w "

Š‹
È
ʜ
w
#œ
1x
cos x
1x cosx
Î#
wÎ#
18. y z cos z 1 z z cos z 1 z cos z 1 z ( 2z)œœÊœ
" " # " # #
"Î# "Î#

"
# È
ab ab ˆ‰ dy
dz
z
1z
È
w cos z cos zœœ
" "

zz
1z 1z
ÈÈ
ww
19. y t tan t ln t tan t t tan tœ Êœœ
" " """"""
##ˆ‰ ˆ ‰ ˆ‰ˆ‰
dy
dt 1 t t 1 t 2t
t
ww
20. y 1 t cot 2t 2t cot 2t 1 tœ Ê œ ab ab ˆ‰
#" " #
œ
dy
dt 1 4t
2
w
21. y z sec z z 1 z sec z z 1 z sec z (1) z 1 (2z)œœÊœ
" " # " ##
"Î# "Î# ""

# È
ab a b ab Š‹ dy
dz
zz1kk
È
w sec z sec z, z 1œœ
zz 1z
zz1 z1 z1kk
ÈÈ È
ww w

" "
22. y 2 x 1 sec x 2(x 1) sec xœ œ
È È ˆ‰
" "Î# " "Î#
2 (x 1) sec x (x 1) 2Êœ œ œ
dy
dx xx
x
xx1 2x1 x1
sec x sec x
–—
ˆ‰ ˆ ‰
#w
Š‹
" ""
# #
"Î# " "Î# "Î#

Š‹
ÈÈÈÈ
ÈÈ
œ
#
Î# Î#
23. y csc (sec ) 1, 0œÊœ œœ
"

#
))
dy
dtan
sec tan tan
sec sec 1
))
)) ) 1
))kk
È kk
w
24. y 1 x e y 2xe 1 x 2xe eœ Ê œ œ ab ab Š‹
#w #
œ
tan x tan x tan x tan x e
1x
Î# Î# Î# Î#
Î#
w tan x
25. y ln y ln ln (2) ln x 1 ln (cos 2x) 0œÊœ œ Êœ
2x 1 2x 1
cos 2x cos 2x
y ( 2 sin 2x)
y x 1 cos 2x
2xab ab
ÈÈ
ww œ
w
œœ # ""
##

Š‹ ab ˆ‰
y tan 2x y tan 2xÊœ œ
w
œœ
œˆ‰ˆ‰
2x 2x
x1 x1
2x 1
cos 2x
ww
w
ab
È
26. y ln y ln ln (3x 4) ln (2x 4) œÊœ œ Êœ ÉÉ cd ˆ‰
#( #(
œ3x 4 3x 4 3 2
2x 4 2x 4 10 y 10 3x 4 2x 4
yœœ" "

yyÊœ œ
w "" œ" "
10 3x 4 x 2 2x 4 10 3x 4 x 2
33x43ˆ‰ ˆ‰ˆ‰ É
#(

483. Chapter 7 Practice Exercises 479
27. y ln y 5 ln (t 1) ln (t 1) ln (t 2) ln (t 3) œ Ê œ Ê’“ Š‹Š‹ cd
(t 1)(t 1) dy
(t 2)(t 3) ydt


&
"
5 5œ Êœ ˆ‰ ˆ‰ ’“
"""" """"
# #

&
t1 t1 t t3 dt (t2)(t3) t1 t1 t t3
dy (t 1)(t 1)
28. y ln y ln 2 ln u u ln 2 ln u 1 ln 2œÊœÊ œ
2u2 2u
u1 ydu u u1
dy
u
Èw
w
œ
""""
##
#
ab Š‹Š ‹ ˆ‰
ln 2ʜ
dy
du u u 1
2u2 u
u1
u
Èw
w
œ
"
œ
ˆ‰
29. y (sin ) ln y ln (sin ) ln (sin )œÊœ Ê œ ))) )))
È
)
))
)
ÈÈ
Š‹Š ‹ ˆ‰ ""
#
"Î#
yd sin
dy cos
(sin ) cot ʜ
dy ln (sin )
d 2)
) )
)
)))
È
È
Š‹
È
30. y (ln x) ln y ln (ln x) ln (ln x)œÊœ Êœ
1lnx
ln x y ln x ln x x (ln x) x
yÎ """""" ˆ‰ ˆ‰ˆ‰ˆ‰ ˆ‰ ’“
œ
w
y (ln x)ʜ
wÎ 1lnx1ln(ln x)
x(ln x)
’“ w
31. e sin e dx sin u du, where u e and du e dx''xx x x
ab œœœ
cos u C cos e Cœ œ ab
x
32. e cos 3e 2 dt cos u du, where u 3e 2 and du 3e dt''tt t t
3
abœ œ œ
"
sin u C sin 3e 2 Cœœ
""
33
t
ab
33. e sec e 7 dx sec u du, where u e 7 and du e dx
''xx x x##
abœ œ œ
tan u C tan e 7 Cœœ ab
x
34. e csc e 1 cot e 1 dy csc u cot u du, where u e 1 and du e dy''yy y y y
ababœ œœ
csc u C csc e 1 Cœ œ ab
y
35. sec x e dx e du, where u tan x and du sec x dx''ab
##tan x u
œœœ
eCe Cœœ
u tan x
36. csc x e dx e du, where u cot x and du csc x dx''ab
# #cot x
œ œ œ
u
eC e Cœ œ
ucotx
37. dx du, where u 3x 4, du 3 dx; x 1 u 7, x 1 u 1''
ww
w17
11
"""
3x 4 3 u
œ œ œ œÊœ œÊœ
ln u ln 1 ln 7 [0 ln 7]œœœœ
"" " "
(33 3 3
ln 7
cd c dkk kk kk
38. dx u du, where u ln x, du dx; x 1 u 0, x e u 1''
10
e1
È
ln x
xx
œœœœÊœœÊœ
"Î# "
u10 œ œœœ ‘œ ‘
2222
3333
$Î# $Î# $Î#
"
!
39. tan dx dx 3 du, where u cos , du sin dx; x 0 u 1, x'' '
00 1
12
œœ
ˆ‰ ˆ‰ ˆ‰
xxx
3u333
sin
cos
œœ œœ œÊœœ
ˆ‰
ˆ‰
x
3
x
3 #
""
1 u ʜ
"
#
3 ln u 3 ln ln 1 3 ln ln 2 ln 8œ œ œ œ œcd kkkk œ ‘¸¸
"Î#
"
""
##
$

484. 480 Chapter 7 Transcendental Functions
40. 2 cot x dx 2 dx du, where u sin x, du cos x dx; x u , x'''
16 16 12
14 14 1 2###
###
1111œœ œœœÊœœ
cos x 2
sin x u 6 4
1
11
œ
""""
#
u ʜ
"
È
2
ln u ln ln ln 1 ln 2 ln 1 ln 2 ln 2œœœœœ
22 2 2ln 2
211 1 11
cdkk ’“¹¹ ¸¸ œ ‘ œ ‘
12
12#
#
œ "" " "
## #È
41. dt du, where u t 25, du 2t dt; t 0 u 25, t 4 u 9''
025
49
2t
t25 u
w

" #
œ œ œ œÊœ œÊœ
w
w
ln u ln 9 ln 25 ln 9 ln 25 ln œœœœcd kk k kkk
*
#&
9
25
42. dt du, where u 1 sin t, du cos t dt; t u 2, t u''
w#
##
œ
œ22
612
cos t
1sin t u 6 ##
" "
œ œ œ œ Ê œ œ Ê œ
11
ln u ln ln 2 ln 1 ln 2 ln 2 2 ln 2 ln 4œ œ œ œ œcd kkkk œ ‘¸¸
"Î#
#
"
#
43. dv tan u du du, where u ln v and du dv'
tan (ln v)
vcos u v
sin u
œœ œ œ' '
"
ln cos u C ln cos (ln v) Cœ œ kk k k
44. dv du, where u ln v and du dv'"" "
v ln v u v
œœœ'
ln u C ln ln v Cœœ kk k k
45. dx u du, where u ln x and du dx'
(ln x)
xx
Î&
œœœ'$ "
C (ln x) Cœœ
u
Îw
# #
" #
46. dx u du, where u ln (x 5) and du dx'
ln (x 5)
x5 x5


"
œœœ'
CC œœ
u ln (x 5)
2
w
w
#
cd
47. csc (1 ln r) dr csc u du, where u 1 ln r and du dr'" "##
r r
œ œ œ'
cot u C cot (1 ln r) Cœ œ
48. dv cos u du, where u 1 ln v and du dv'
cos (1 ln v)
v v
"
œ œ œ'
sin u C sin (1 ln v) Cœ œ
49. x3 dx 3 du, where u x and du 2x dx''xu
w
œœœ
"
#
#
3C 3 Cœœ
""
## ln 3 ln 3
ux
ab Š‹
w
50. 2 sec x dx 2 du, where u tan x and du sec x dx''tan x u##
œœœ
2C Cœœ
"
##ln ln
u 2
ab
tan x
51. dx 3 dx 3 ln x 3 ln 7 ln 1 3 ln 7''
11
77
3
xx
œœœœ
" (
"
cd abkk
52. dx dx ln x ln 32 ln 1 ln 32 ln 32 ln 2''
11
32 32
"""" " " $#
"5x 5 x 5 5 5
œœœœœœ cd abkk Š‹
È
$

485. Chapter 7 Practice Exercises 481
53. dx x dx x ln x ln 4 ln 1 ln 4''
11
44
ˆ‰ ˆ ‰ œ ‘ œ ‘kk ˆ‰ˆ‰
x 16 15
8x 4 x 8 8 8 16
œ œ œœ
"""""" " " "
## # # #
#
%
"
ln 4 ln 2œ œ
15 15
16 16È
54. dx 12x dx ln x 12x ln 8 (ln 1 12)''
11
88
ˆ‰ ˆ ‰ œ ‘cdkk ˆ‰
28 2 2 2 12
3x x 3 x 3 3 8
œ œ œ
Ê
" # "
)
"
ln 8 12 ln 8 (ln 8) 7 ln 8 7 ln 4 7œ œ œ œ œ
23 2212
333
ˆ‰ˆ‰ ˆ‰
##
#Î$
55. e dx e du, where u (x 1), du dx; x 2 u 1, x 1 u 0''
Ê
Ê21
10
u
Ð Ñx1
œ œ œ œ Ê œ œ Ê œ
eeee1œ œ œ cd ab
u!
"
!"
56. e dw e du, where u 2w, du 2 dw; w ln 2 u ln , w 0 u 0''
Ê ln 2 n 1 4
00
2w u
œœœœÊœœÊœ
" "
#l 4
eee1œœœœ
"" ""
## #
!
cd c d ˆ‰
uln140
ln 1 4

48
3
57. e 3e 1 dr u du, where u 3e 1, du 3e dr; r 0 u 4, r ln 5 u 16''
14
ln 5 16
rr r r
abœ œœœÊœœÊœ
$Î# " $Î#
3
u164œ œ œ œ œ
22 2 2
33 34346
œ ‘ ˆ ‰ˆ ‰ˆ ‰ˆ ‰ˆ ‰
"Î# "Î# "Î#
"'
%
"" " "
#
58. e e 1 d u du, where u e 1, du e d ; 0 u 0, ln 9 u 8''
00
ln 9 8
## # #
abœ œœœÊœœÊœ
"Î# "Î#
))))
u8020œœœœœ
22 2 2
33 3 33
32 2
œ‘ˆ ‰ˆ ‰
$Î# $Î# $Î# *Î#
)
!
È
59. (1 7 ln x) dx u du, where u 1 7 ln x, du dx, x 1 u 1, x e u 8''
11
e8
"" "Î$ "Î$
x7 x
7
œ œœœÊœœÊœ
u81(41)œœœœ
33 3 9
14 14 14 14
œ‘ˆ ‰ˆ‰
#Î$ #Î$ #Î$
)
"
60. dx (ln x) dx u du, where u ln x, du dx; x e u 1, x e u 2'' '
ee 1
ee 2
ÊÊ
"" " "Î# "Î# #
xln x xxÈ
œœ œœœÊœœÊœ
2u 2 2 1 2 2 2œœœœ‘ Š‹
ÈÈ
"Î#
#
"
61. dv [ln (v 1)] dv u du, where u ln (v 1), du dv;'' '
11 ln2
33 ln4
[ln (v 1)]
v1 v1 v1
œ
œœ œ
##""
Ê
œ œ œœ
v 1 u ln 2, v 3 u ln 4; œÊœ œÊœ
u (ln 4) (ln 2) (2 ln 2) (ln 2) (8 1) (ln 2)œ œ œ œ œ
"" "$$$$$ $
33 3 3 3
(ln 2) 7
cd c d c d
ln 4
ln 2
&
62. (1 ln t)(t ln t) dt (t ln t)(1 ln t) dt u du, where u t ln t, du (t) (ln t)(1) dt'''
222 ln2
444 ln4
œœ œœ ˆ‰ˆ‰
"
t
(1 ln t) dt; t 2 u 2 ln 2, t 4 œ œÊœ œ
u 4 ln 4 ʜ
u (4 ln 4) (2 ln 2) (8 ln 2) (2 ln 2) (16 1) 30 (ln 2)œœ œ œ œ
"" "
## # #
##### #
cdcdcd
4ln4
2ln2 (2 ln 2)
Ê
63. d (ln ) d u du, where u ln , du d , 1 u 0, 8 u ln 8'' '
11 0
88 ln8
log
ln 4 ln 4
4)
)) )
))) ))) )œœ œœœÊœœÊœ
""" "
ˆ‰
u (ln 8) 0œœ œœ
""
#
###
ln 4 ln 16 4 ln 2 4
(3 ln 2)9 ln 2
cd c d
ln 8
&
Ê
64. d d 8 (ln ) d 8 u du, where u ln , du d ;''''
1110
eee1
8(ln 3)(log ) 8(ln 3)(ln )
(ln 3)
3))
)) ) )
)))) ))œœ œ œœ ˆ‰
""
1 u 0, e u 1))œÊœ œÊœ
4u 41 0 4œœœcd ab
###
"
!

486. 482 Chapter 7 Transcendental Functions
65. dx 3 dx 3 du, where u 2x, du 2 dx;'' '
w#w# w#
## #34 34 32
34 34 32
62
94x 3 u 3 (2x)ÈÈ È

"
www ww
œœ œœ
x u , x uœÊœ œÊœ
3333
44 ##
3sin 3sin sin 3 3œœœœœœ ‘ œ ‘œ ‘ˆ‰ˆ‰ ˆ‰ ˆ ‰ ˆ ‰
" " "
$Î#
$Î#
""
##
u
3663
11 1
1
66. dx dx du, where u 5x, du 5 dx;'' '
w#w# w
##15 15 1
15 15 1
66 5 6
4 25x 2 u
55 2 (5x)ÈÈÈ

"
www ww
œœ œœ
x u 1, x u 1œÊœ œÊœ
11
55
sin sin sinœœœœœ
6u6 6 62
525 566535
œ ‘ œ ‘ œ ‘ˆ‰ˆ‰ ˆ‰ ˆ ‰ ˆ ‰
" " "
"
"
""
##
11 11
67. dt 3 dt 3 du, where u 3t, du 3 dt;'' '
ww w22 23
22 23
3
43t 2 u
3
23t
œœ
œ
"
w ww
w
wœœ œœ
ÈÈ ÈÈ
È
Š‹
È
œ
œ
t 2 u 2 3, t 2 u 2 3 œÊœ œÊœ
ÈÈ
3 tan tan 3 tan 3œœœœ
ÈÈÈœ ‘ œ ‘ˆ‰ ˆ ‰ ’“Š‹ Š ‹
"
## #
" " "u
233
33
3
23
23
œ
œ
w ÈÈ
È
111
68. dt dt tan tan 3 tan 1''
œœ
œ
33
33 3
3
""" " "
œ
œ
" " "
3t 34 36
3t
33 3 3
t 3
w w
wœœ œœœ
Š‹
È
ÈÈ È È
È
’“Š ‹Š‹
È ˆ‰
11 1
69. dy dy du, where u 2y and du 2 dy'' '""
y 4y 1 (2y) (2y) 1
2
uu 1
ÈÈÈ
ww w œœ œœ
sec u C sec 2y Cœœ
" "
kk k k
70. dy 24 dy 24 sec C 6 sec C''24
yy 16 yy 4 44 4
yy
ÈÈ
www

"" " "
œœœ ˆ‰ ¸¸¸¸
71. dy dy du, where u 3y, du 3 dy;'' '
œœ œ
23 23 2
23 23 2##
##
""
kk k kÈÈÈ
kky 9y 1 3y (3y) 1
3
uu1
ww w œœ œœ
y u 2, y u 2œÊœ œÊœ
È
2
33
2
È
sec u sec 2 sec 2œ œ œœcd ’“
È
" " "
2
2
œ
11 1
34 12
72. dy dy du,'' '
ww w
w#w# w2/ 5 2/ 5 2
65 65 6œœ
œœ œœœ
" "


kkÈ
È
ÈÈ È È
ÊŠ ‹ Š ‹ Ê Š ‹
y 5y 3
5
5y 5y 3 u u 3
w ww w
w
œœ
where u 5y, du 5 dy; y u 2, y u 6 œ œ œ Ê œ œ Ê œ
ÈÈ È
2
55
6
ÈÈ
È
sec sec 2 sec œ œ œ œ œ œ’“’ “ ¹¹
È ˆ‰ œ ‘
" " " " " "
#

ÈÈ È ÈÈ È È
È
33 3 33 3 123
u1 2 32
46 121 36
3
w
w
œ6
2
11 1 1 1 1
73. dx dx dx du, where u x 1 and'' ' '"" ""
ÈÈ È È ab2x x 1 u 1x2x1 1(x1)
w ww w
œœœ œ
du dxœ
sin u C sin (x 1) Cœœ
" "
74. dx dx dx du'' ' '"" " "


ÈÈ ab
ÊŠ ‹ ÊŠ ‹
ÈÈ
x4x1 3x4x4
3(x2) 3u
w w ww
ww
œœ œ
where u x 2 and du dxœ œ sin C sin Cœœ
" "
Š‹ Š ‹
ux2
33
ÈÈ

487. Chapter 7 Practice Exercises 483
75. dv 2 dv 2 dv 2 du,'' ''
ÊÊ Ê
ÊÊ Ê22 20
11 11
2
v 4v 5 1 v 4v 4 1 (v 2) 1 u
ÊÊ ÊÊ
œœ œ œœ œœ œ
"""
œœœ
ab
where u v 2, du dv; v 2 u 0, v 1 u 1 œ œ œÊ œ œÊ œ
2 tan u 2 tan 1 tan 0 2 0œœœœcd ab ˆ‰
" " "
"
! #
11
4
76. dv dv dv du'' ' '
ÊÊ Ê Ê #
#11 1 12
11 1 32
33 3 3
4v 4v 4 4 4 4
vv
vu
Ê
Ê
ÊÊ
Ê
Ê
œœ
"""
œœœ
œœ œ
œœ œ
3
44
33
Š‹
Œœ ŒœŠ‹
œ
## #
œ
œœ
where u v , du dv; v 1 u , v 1 u œ œ œÊ œ œ Ê œ
""
###
3
tan tan 3 tanœœœœœ
32 2u 2
4 36 6633 3
3333
’“’ “Š‹ Š ‹
È œ ‘ˆ‰ˆ‰
ÈÈ È
ÈÈÈÈ
" " "
$Î#
"Î#
#####
" 11 11 1
†
œ
È
3
4
1
77. dt dt dt du'' ' '"" ""
(t 1) t 2t 8 (t 1) t 2t 1 9 (t 1) (t 1) 3 uu 3
ÈÈ ÈÈ ab
Ê Ê ÊÊ ÊÊ
œœœ
where u t 1 and du dtœ œ
sec C sec Cœœ
""œ" "
33 3 3
ut1
¸¸ ¸ ¸
78. dt dt dt du'' ' '"" """
œœ (3t 1) 9t 6t (3t1) 9t 6t1 1 (3t 1) (3t 1) 1 3
uu 1
ÈÈ È È ab
Ê Ê ÊÊ Ê
œœœ
where u 3t 1 and du 3 dtœ œ
sec u C sec 3t 1 Cœœ
""" "
33
kk k k
79. 3 2 ln 3 ln 2 y(ln 3) (y 1) ln 2 (ln 3 ln 2)y ln 2 ln y ln 2 y
yy1 y y1 3ln 2
ln
œ Ê œ Ê œ Ê œÊ œÊœ
œœ
# ˆ‰
Š‹
3
#
80. 4 3 ln 4 ln 3 y ln 4 (y 2) ln 3 2 ln 3 (ln 3 ln 4)y (ln 12)y 2 ln 3
yy2 y y2
œÊ œ Êœ Êœ Ê œ yÊœ
ln 9
ln 12
81. 9e x e ln e ln 2y(ln e) ln y ln ln ln ln x ln 3
2y 2y 2y xx xxxx
99 9993
œÊœÊœÊœÊœ œœœ
# "
#
ÊÊ ÊÊ Ê
Š‹ Š‹ Š‹ É ¸¸ kk
82. 3 3 ln x ln 3 ln (3 ln x) y ln 3 ln (3 ln x) y
yy ln (3 ln x) ln 3 ln (ln x)
ln 3 ln 3
œÊœÊœÊœœ
œ
83. ln (y 1) x ln y e e e e y 1 ye y ye 1 y 1 e 1œ Ê œ œ Êœ Ê œÊ œ
ln y 1 x lny x lny x x xÐÑ Ð Ñ
ab
yʜ
"
1e
x
84. ln (10 ln y) ln 5x e e 10 ln y 5x ln y e e y eœÊ œÊ œÊœÊœÊœ
ln 10 ln y ln 5x ln y x 2 x 2 xÐÑ Î Î
#
85. The limit leads to the indeterminate form : lim lim ln 10
x0
0101
0x 1
(ln 10)10
x0Ä
x x

œœ
Ä
86. The limit leads to the indeterminate form : lim lim ln 3
031
01
(ln 3)3
))ÄÄ00
œ œ

)
œœ
87. The limit leads to the indeterminate form : lim lim ln 2
x0
021
0e1 e
2 (ln 2)(cos x)
x0Ä
sin x
xx
sin x


œœ
Ä
88. The limit leads to the indeterminate form : lim lim ln 2
02
0e1 e
2 (ln 2)( cos x)
x0Ä
Î Îsin x
xx
sin x
"


œœ
x0Ä
89. The limit leads to the indeterminate form : lim lim lim 5
x0
0 5 5 cos x 5 sin x 5 cos x
0ex1 e1 e
x0Ä


xxx œœ œ
Ä
x0Ä

488. 484 Chapter 7 Transcendental Functions
90. The limit leads to the indeterminate form : lim lim 4
x0
0 4 4e 4e
0xe exe
x0Ä
ÐÐ
œ
xx
xxx
œœ
Ä91. The limit leads to the indeterminate form : lim
0
0t
tln(12t)
2t
t
lim
tÄ! Ä!
œ œ
#

œ œœ_
Š‹1
2
12t
Ð
92. The limit leads to the indeterminate form : lim lim
x4
0
0e3x e1
sin ( x) 2 (sin x)(cos x)
x4Ä
œ
ÑÑ
1111
x4 x4

œ
Ä lim lim 2
x4 x4
œœ œ
ÄÄ
11 1 1 sin(2x) 2 cos(2x)
e1 e
x4 x4ÑÑ
œ
Ð
#
1
93. The limit leads to the indeterminate form : lim lim lim 1
0e e e
0tt t 1
tttÄ! Ä! Ä!
œœœ
Š‹ Š‹
ttt
œ œ œ
"Ð"
94. The limit leads to the indeterminate form : lim e ln y lim lim
_
_
ÐÑ
Ð
yyyÄ! Ä! Ä!
œœœ
1y ln y y
e e
œœ
yyy
Ñ#Î#Îw
Î#
lim 0œ œ
yÄ!
œ
y
e
y
Ñ#
95. Let f(x) 1 ln f(x) lim ln f(x) lim ; the limit leads to theœ Ê œ Ê œˆ‰
3
xx x
ln 1 3x ln 1 3x
x ab abœœ
Ñ#Î#
Î# Î#
xxÄ_ Ä_
indeterminate form : lim lim 3 lim 1 lim e e
033
0x x 1
xx xxÄ_ Ä_ Ä_ Ä_
Š‹
œ
w#
Ð
ʜ3x
13x
3
x
Ð œ
$
ÑœœœÊœœ ˆ‰
x
ln f x
ÐÑ
96. Let f(x) 1 ln f(x) x ln 1 lim ln f(x) lim ; the limit leads to theœ Ê œ Ê œˆ‰ ˆ‰
33
xx x
ln 1 3x
x
xxÄ! Ä!
œœ
abœ
Ñ#
Î#
indeterminate form : lim lim 0 lim 1 lim e e 1
_
_
!
xx xxÄ! Ä! Ä! Ä!
œœ œ œ
ÐÑ
Š‹
œ
w#
Ð
ʜ3x
13x
xx3 x
3x 3
ÑœœœÊœœœ ˆ‰
x
ln f x
97. (a) lim lim lim same rate
xx xÄ_ Ä_ Ä_
log x
log x ln 2 ln
ln 3 ln 32
3
œœœÊ
Š‹
Š‹
ln x
ln 2
ln x
ln 3
#
(b) lim lim lim lim 1 same rate
xxxxÄ_ Ä_ Ä_ Ä_
xx2x
x
x1 x
œ
#
Š‹
œ
x
œœœ"œÊ
œ
œ
(c) lim lim lim faster
xxxÄ_ Ä_ Ä_
ˆ‰
x
100
xe 100x 100
xe e
Ñx
xxœœœ_Ê
(d) lim faster
xÄ_
x
tan x
Î#œ_ Ê
(e) lim lim lim lim 1 same rate
xx x xÄ_ Ä_ Ä_ Ä_
csc x sin x
xx
1
Î# Î# Î#
Î# Îw
Š‹
ab
Ê Š‹
œ
œ
w#
œ
ʜ
#
œ
#
x
x
1x
x
œœ œ œÊ
#w
Ð #w
Ð
"
Ð
(f) lim lim lim same rate
xÄ_
sinh x e
ee
ee
xx
xx 2xœœœÊ
xxÄ_ Ä_
abÐ
###
"Ð "
Ñ Ñ
98. (a) lim lim 0 slower
xxÄ_ Ä_
32
23
Ñ
Ñx
x
œœÊ ˆ‰
x (b) lim lim lim same rate
xx xÄ_ Ä_ Ä_
ln 2x ln 2 ln x ln 2
ln x 2 (ln x) ln x
œœœœÊ
œ""
###
ˆ‰
(c) lim lim lim lim 0 slower
xxxxÄ_ Ä_ Ä_ Ä_
10x 2x 30x 4x 60x 4 60
eeee
& œœ
œœœ
xxxœœœœÊ
x
(d) lim lim lim lim 1 same rate
xx xxÄ_ Ä_ Ä_ Ä_
tan
tan x
xx 1
Ñ#
Î# Î#
Î# Îw
Š‹ Š ‹
Š‹
ab
œ œ
œ
w#
Ð
w#
œ
#
x
x
x
1x
x
œœœœÊ
Ð
"
œ
(e) lim lim lim lim faster
xx x xÄ_ Ä_ Ä_ Ä_
sin
sin x
x2x
x
21
Ñ#
Î# Î#
Îw Î&
Ñ
ќ
Ñ
Î#
œ
#
œ
Š‹
Š‹
ab
#w
É
œ
œ
#
x
x
œœ œœ_Ê
x
1x
x
œ
Ð
Ð
(f) lim lim lim lim 2 same rate
xx x xÄ_ Ä_ Ä_ Ä_
sech x 2 2
eeeee1e
ÑÑÑÑ Ñxxxxx 2xœœ œœÊ
Š‹
ab
2
ee
xx
Ð
œ
œœ
ˆ‰

489. Chapter 7 Practice Exercises 485
99. (a) 1 2 for x sufficiently large true
Š‹
Š‹
œœ
# Ñ
œ
#
x x
x
œ
"
œ Ÿ Ê
x
œ
(b) x 1 M for any positive integer M whenever x M false
Š‹
Š‹
œœ
# Ñ
œ
Ñ
x x
x
œ
#
œ Ê
È
(c) lim lim 1 the same growth rate false
xxÄ_ Ä_
x
xln x 1œ
"
œ
œœÊ Ê
œ
x
(d) lim lim lim 0 grows slower true
xxxÄ_ Ä_ Ä_
ln (ln x)
ln x ln x
œœœÊ Ê
–—
ˆ‰
#w
œ
x
ln x
#
x
" (e) for all x true
tan x
1
Î#
ŸÊ
1
#
(f) 1 e (1 1) 1 if x 0 true
cosh x
e
xœ Ÿœ Ê
""
##
ab
œ2x
100. (a) 1 if x 0 true
Š‹
Š‹
œ
Ñ
œœ
# Ñ
x
x x
œ
"
œ
œŸ Ê
x1
œ
(b) lim lim 0 true
xxÄ_ Ä_
Š‹
Š‹
œ
Ñ
œœ
# Ñ
x
x x
œ
"
œ
œœÊ ˆ‰
x1
œ
(c) lim lim 0 true
xxÄ_ Ä_
ln x
x1 1œ
œœÊ
Š‹
œ
x
(d) 1112 if x2 true
ln 2x ln 2
ln x ln x
œ Ÿœ   Ê
(e) if x 1 true
sec x
111
cos
Ñ#
Î#
œ
œŸœÊ
Š‹ ˆ‰
œ
x
#
1
#
(f) 1 e if x 0 true
sinh x
e
xœ Ÿ Ê
""
##
ab
œ2x
101. e 1
df df df
dx dx dx e 1 1 3
œÊœÊœœœ
x
Š‹ Š‹
Ñ#Î#
xfln2 xfln2œÐ Ñ œÐ Ñ
""""
#
Š‹
ab
df
dx
xln2
xln2œ
œ
x
102. y f(x) y 1 y 1 x f (x) ; f (f(x)) x andœÊœÊœÊœÊ œ œ œœ
"" " " " "
ÐÐ
Ð" Ð"

xx y1 x1
11Š‹ Š‹
œœ
xx
ff(x) 1 1(x1)x; x;ab ¹¹
Ð" #"Ð"Ð"
Ð

œ œ œ œ œ œ
Š‹ ’“Š‹
œ
œ
œ
x1 x
df
dx (x 1)
11
Ñ#
œ œ
fx fxÐÑ ÐÑ
f (x)
w ""
œ Ê œ
xdx f(x)
df
œ œ
Î# ¹
fxÐÑ
103. y x ln 2x x y x ln (2x) 1 ln 2x;œÊœœ
wˆ‰
2
2x
solving y 0 x ; y 0 for x and y 0 for
ww w ""
##
œÊœ
x relative minimum of at x ; fÊ œ œ
"""""
####
ˆ‰
ee
and f 0 absolute minimum is at x and theˆ‰
e
###
""
œÊ œ
absolute maximum is 0 at xœ
e
#

490. 486 Chapter 7 Transcendental Functions
104. y 10x(2 ln x) y 10(2 ln x) 10xœÊœ
w " ˆ‰
x
20 10 ln x 10 10(1 ln x); solving y 0œ œ œ
w
x e; y 0 for x e and y 0 for x eʜ
ww
relative maximum at x e of 10e; y on e andÊœ !Ð!ßÓ
#
y e 10e (2 2 ln e) 0 absolute minimum is 0ab
##
œœÊ
at x e and the absolute maximum is 10e at x eœœ
#

105. A dx 2u du u 1, whereœœœœ''
10
e1
2 ln x
x
cd
#
"
!
u ln x and du dx; x 1 u 0, x e u 1œœœÊœœÊœ
"
x
106. (a) A dx ln x ln 20 ln 10 ln ln 2, and A dx ln x ln 2 ln 1 ln 2" #
" "#! #
"! "œ œ œœœ œ œ œœ''
10 1
20 2x10x
20
cd cdkk kk
(b) A dx ln x ln kb ln ka ln ln ln b ln a, and A dx ln x" #
" "œ œ œœœœ œ œ''
ka a
kb b
kb b
ka axkaax
kb b
cd cdkk kk
ln b ln aœ
107. y ln x ; x m/secœÊœ œ Êœ œÊ œ
dy dy dy dy dy
dx x dt dx dt dt x dt e
dx
x
""""
ˆ‰È ¹
È
e
Ê
108. y 9e 3e ; ; x 9 y 9eœÊœ œÊœ œÊœ
Ê#w#x3 x3 dy (dy/dt)
dx dt (dy/dx) dt 3e
dx dx
9yŠ‹È

$
œ
4
#wx3
e e 1 5 ft/secÊœ œ ¸¸ ÈÈ
dx
dt 4
9
x9Ä
Š‹É
Š‹


" $$
œ
&
&
4
9
e
3
e
109. A xy xe e (x)( 2x) e e 1 2x . Solving 0 1 2x 0œœ Ê œ œ œÊ œ
ÊÊ ÊÊxx xx
ÊÊ ÊÊ
dA dA
dx dx
ab
##
x ; 0 for x and 0 for 0 x absolute maximum of e atÊœ Ê œ
"" " "" "Î#
ÈÈ È ÈÈ
22 2 2
dA dA
dx dx 2e
x units long by y e units high.œœœ
"" "Î#
ÈÈ
2 e
110. A xy x . Solving 0 1 ln x 0 x e;œœ œ Ê œ œ œÊ œÊœˆ‰
ln x ln x dA ln x 1 ln x dA
xxdxxxx dx
ÊÊÊÊ
"
0 for x e and 0 for x e absolute maximum of at x e units long and y units
dA dA ln e
dx dx e e e
Ê œœ œ
""
Ê
high.
111. K ln (5x) ln (3x) ln 5 ln x ln 3 ln x ln 5 ln 3 ln œ œœœ
5
3
112. (a) No, there are two intersections: one at x 2œ
and the other at x 4œ

491. Chapter 7 Practice Exercises 487
(b) Yes, because there is only one intersection

113.
log x
log x ln 4 ln x ln 4 2 ln 2
ln x ln 2 ln 2 ln 2
4
2
œœ œœœ
Š‹
Š‹
ln x
ln 4
ln x
ln
#
†
"
#
114. (a) f(x) , g(x)œœ
ln 2 ln x
ln x ln #
(b) f is negative when g is negative, positive when g is
positive, and undefined when g 0; the values of fœ
decrease as those of g increase
(c) (ln 2) (ln x)
ln 2 ln x
ln x ln
œÊ œ
#
##
(ln 2 ln x)(ln 2 ln x) 0 ln x ln 2 orÊ œÊœ
ln x ln 2 e e or e e x orœÊœ œÊœ
ln x ln 1 2 ln x ln 2 "
#
x 2. Therefore, the curves cross at the two pointsœ
1 and 2 (2 ).Š‹ ˆ‰ˆ‰
""
## # #
ß œ ß ß œ ß"
ln
ln ln
ln 2
ˆ‰
#
Ê

115. (a) y yœÊœœ
ln x ln x 2 ln x
xxx 2xx 2xÈÈ È
w "

y x (2 ln x) x x ln x 2 ;Êœ œ
ww &Î# &Î# &Î# "
#
33
44 ˆ‰
solving y 0 ln x 2 x e ; y 0 for x e and
w#w#
œÊ œÊœ
and y 0 for x e a maximum of ; y 0
w# ww
Ê œ
2
e
ln x x e ; the curve is concave down onʜʜ
8
3
)Î$
0 e and concave up on e ; so there is anˆ‰ ˆ ‰ßß_
)Î$ )Î$
inflection point at e .ˆ‰
)Î$)
$
ß
e


(b) y e y 2xe y 2e 4x eœÊœ Êœ
w ww#xxxx
ÊÊÊÊ
4x 2 e ; solving y 0 x 0; y 0 forœ œÊœab
# w w x
Ê
x 0 and y 0 for x 0 a maximum at x 0 ofÊ œ
w
e 1; there are points of inflection at x ; the
! "
œœ„
È
2
curve is concave down for x and concave
""
ÈÈ
22
up otherwise.

(c) y (1 x) e y e (1 x) e xeœ Ê œ œ
ÊÊ ÊÊxx xx w
y e xe (x 1) e ; solving y 0Êœ œ œ
ww w ÊÊ Êxx x
xe 0 x 0; y 0 for x 0 and y 0Ê œ Ê œ
Êx ww
for x 0 a maximum at x 0 of (1 0) e 1;Ê œ œ
!
there is a point of inflection at x 1 and the curve isœ
concave up for x 1 and concave down for x 1.&Î

492. 488 Chapter 7 Transcendental Functions
116. y x ln x y ln x x ln x 1; solving y 0œ Êœ œ œ
ww " ˆ‰
x
ln x 1 0 ln x 1 x e ; y 0 forʜʜʜ
" w
x e and y 0 for x e a minimum of e ln e&ÎÎw
" w " " "
at x e . This minimum is an absolute minimumœ œ
" "
e
since y is positive for all x .
ww"
œ!
x

117. Since the half life is 5700 years and A(t) A e we have A e e ln (0.5) 5700kœœÊœÊœ !!
##
"
kt 5700k 5700kA(
k . With 10% of the original carbon-14 remaining we have 0.1A A e 0.1 eÊœ œ Ê œ
ln (0.5)
5700
!!
ln 0 5 ln 0 5
5700 5700
tt
ln (0.1) t t 18,935 years (rounded to the nearest year).ʜʜ ¸
ln (0.5) (5700) ln (0.1)
5700 ln (0.5)
118. T T (T T ) e 180 40 (220 40) e , time in hours, k 4 ln 4 ln 70 40œ Ê œ Êœ œ Ê
sos
kt k 4
ww # ˆ‰ ˆ‰
79
97
(220 40) e t 1.78 hr 107 min, the total time the time it took to cool fromœ Êœ ¸ ¸ Ê
w4ln 9 7 t ln 6
4 lnˆ‰
9
7
180° F to 70° F was 107 15 92 minœ
119. cot cot , 0 x 50 )1œ Ê œ
" "
œ

œˆ‰ ˆ ‰
x5x d
60 3 30 dx
1
1
)
Š‹ Š ‹
ˆ‰
Š‹
1
60 30
x
60
50 x
30
ww
œ
w
30 ; solving 0 x 200x 3200 0 x 100 20 17, butœ œÊ œÊœ „’“
È
2d
60 x 30 (50 x) dx
ww w w

" #)
100 20 17 is not in the domain; 0 for x 20 5 17 and 0 for 20 5 17 x 50
ÈÈÈ
Š‹ Š‹
dd
dx dx
))
x 20 5 17 17.54 m maximizes Êœ ¸Š‹
È
)
120. v x ln x (ln 1 ln x) x ln x 2x ln x x x(2 ln x 1); solving 0œœœÊœœ œ
## # #" "ˆ‰ ˆ‰
xdxxdx
dv dv
2 ln x 1 0 ln x x e ; 0 for x e and 0 for x e a relativeʜʜʜ Ê
"
# w#w# w#12 12 12dv dv
dx dx
maximum at x e ; x and r 1 h e e 1.65 cmœœœÊœœ¸
w# #12 12r
h È
CHAPTER 7 ADDITIONAL AND ADVANCED EXERCISES
1. lim dx lim sin x lim sin b sin 0 lim sin b 0 lim sin b
b1 b1 b1 b1 b1ÄÄÄ Ä Ä
www w w
'
0
b
b
0
"

" " " " "
#È
1x
w
œœœœœcd abab
1
2. lim tan t dt lim form
xxÄ_ Ä_
"_ "
_xx
tan t dt'
0
x
œ
'
0
x
Î#
ˆ‰
lim œœ
xÄ_
tan x
1
Î#
1
#
3. y cos x ln y ln cos x and lim lim lim œÊœ œ œˆ‰ ˆ‰ÈÈ
1x#
œœ œ
" "
#xx
ln cos x sin x tan x
2x cos x x
xx xÄ! Ä! Ä!
ˆ‰ È ÈÈ
ÈÈ È
lim lim cos x eœ œ Ê œ œ
"" "
##
xxÄ! Ä!
œœ
#
w#
#
w

#
w

x secx
x e
È
È
ˆ‰È
1x
12
4. y x e ln y lim ln y lim lim lim 2œ Ê œ Ê œ œ œ œab
x2x# 2 ln x e 2 1 e
xxe1ee
2e 2eab abœœ
œœ
xx
xxx
xx
xx x xÄ_ Ä_ Ä_ Ä_
lim x e lim e eÊœœ
xxÄ_ Ä_
ab
xy2x# #
5. lim lim
xxÄ_ Ä_
ˆ ‰ ˆ‰ ˆ‰ ˆ‰
#w–—–— –—
"" " " " "
# #
œœ œ
n1 n n n n n
11 1
112 1n
ᜠá
Š‹ Š‹ Š‹
œœ œ
nn n

493. Chapter 7 Additional and Advanced Exercises 489
which can be interpreted as a Riemann sum with partitioning x lim ?œÊ á
""""
# #nn1nn
xÄ_
ˆ‰
dx ln (1 x) ln 2œœœ'
0
1
"
œ
"
!1x
cd
6. lim e e e lim e e e which can be interpreted as a
xxÄ_ Ä_
""""
nnnn
cd œ ‘ˆ‰ ˆ‰ ˆ‰
1n 2n 1n 21n n1n##
ᜠá
Riemann sum with partitioning x lim e e e e dx e e 1?œÊ ᜠœ œ
"" "
!nn
xÄ_
cdcd
1n 2n x x
0
1## '
7. A(t) e dx e 1 e , V(t) e dx e 1 eœœœœ œœ''
00
tt
xx t 2x 2x 2t t
0
t
0
ww w w w w
cd abœ ‘1
11
##
(a) lim A(t) lim 1 e 1
ttÄ_ Ä_
œœ ab
wt
(b) lim lim
ttÄ_ Ä_
V(t)
A(t) 1 e
1e
œœ
#
w
Î
Î
ˆ‰
#
2t
t
1
(c) lim lim lim lim 1 e
tt t tÄ! Ä! Ä! Ä!
œœ œ œ
w
V(t)
A(t) 1e 1e
1e 1e 1e
œœ œœ
##
ww
ÎÎÎ
ÎÎ
ˆ‰ abab
ab

#
2t t t
tt
1
ab
t
1
8. (a) lim log 2 lim 0;
aaÄ! Ä!
œœ
aœœ
ln 2
ln a
lim log 2 lim ;
a1 a1ÄÄ
ww
aœœ_
ln 2
ln a
lim log 2 lim ;
a1 a1ÄÄ
œœ
aœœ_
ln 2
ln 1
lim log 2 lim 0
aaÄ_ Ä_
aœœ
ln 2
ln a
(b)
9. A dx dx ; A dx dx"#
"
#œœ œœœœ'' ''
11 11
ee ee e
1
2 log x (ln x) 2 log x
x ln 2 x ln 2 ln 4 ln 4 x
2ln x 2ln x2 4
’“
w
A:A 2:1œœÊœ’“
(ln x)
2 ln ln 2
w
##
"
"#
e
1
10. y tan x tan yœ Êœ
" " w ""
œ

œ ˆ‰
x1x
1
w
#
w
Š‹
Š‹
œ
#
x
x
0 tan x tan is a constantœœÊ
"" "
œœ
" "
1x 1x x
ww
ˆ‰
and the constant is for x 0; it is for x 0 since
11
##

tan x tan is odd. Next the
" " "
# ˆ‰
x

lim tan x tan and lim tan x tan 0
x xÄ! Ä!
œ w
œ ‘ˆ‰ˆ‰ˆ‰ ˆ‰
" " " " ""
## # #
œ!œ œœ
xx
11 1 1
11. ln x x ln x and ln x x ln x x ln x; then, x ln x x ln x x x ln x x x or ln x
abxx x x x x x x
x
œ œ œ œ Ê œ!Ê œ œ!Þab a b
####
ln x x ; x x x ln x 2 ln x x 2. Therefore, x x when x 2 or x .œ!Ê œ" œ Ê œ Ê œ œ œ œ"
xxx x# ab
x
ab
12. In the interval x 2 the function sin x 011ÎÎ Î
(sin x) is not defined for all values in thatÊ
sin x
interval or its translation by 2 .1

13. f(x) e f (x) e g (x), where g (x) f (2) eœÊœ œ Êœ œ
gx gx x22
1x 116 17
ÐÑ w ÐÑ w w w !
œ œ

ˆ‰

494. 490 Chapter 7 Transcendental Functions
14. (a) e 2x
df 2 ln e
dx e
x
œœ
x
x
†
(b) f(0) dt 0œœ '
1
1
2 ln t
t
(c) 2x f(x) x C; f(0) 0 C 0 f(x) x the graph of f(x) is a parabola
df
dx
œ Ê œ œÊ œÊ œ Ê
##
15. Triangle ABD is an isosceles right triangle with its right angle at B and an angle of measure at A. We
1
4
therefore have DAB DAE CAB tan tan .
1
43
œn œn n œ
" """
#
16. (a) The figure shows that ln e e ln ln e ln e
ln e ln
e
ee
&w & w & w&
1
1
11
11 11
(b) y y ; solving y 0 ln x 1 x e; y 0 for x e andœÊœ Ê œÊ œÊœ
ln x ln x 1 ln x
xxxxx
www""
ˆ‰ˆ‰
ww
y 0 for 0 x e an absolute maximum occurs at x e
w
Ê œ
17. The area of the shaded region is sin x dx sin y dy, which is the same as the area of the region to
''
00
11
" "
œ
the left of the curve y sin x (and part of the rectangle formed by the coordinate axes and dashed lines y 1,œ œ x ) . The area of the rectangle is sin y dy sin x dx, so we haveœœ
11
##
" ''
00
12
œ#
sin x dx sin x dx sin x dx sin x dx.
11
##
" "
œÊœ'' ' '
00 0 0
122 1
œœ##
18. (a) slope of L slope of L slope of L $#"
""
ÎÎ wÎÎ
bbaa
ln b ln a
(b) area of small (shaded) rectangle area under curve area of large rectangleÎÎ
(b a) dx (b a) Ê Ê
"""""
bxabbaa
ln b ln a'
a
b
19. (a) g(x) h(x) 0 g(x) h(x); also g(x) h(x) 0 g( x) h( x) 0 g(x) h(x) 0 œÊ œ œÊœÊ œ
g(x) h(x); therefore h(x) h(x) h(x) 0 g(x) 0Êœ œÊœÊœ
(b) f (x);
f(x) f( x) f (x) f (x) f (x) f (x)f (x) f ( x) f ( x)
## #

œœœ
cd
EO O E EOEO (x) f
Eœ
f(x)
f(x) f( x) f (x) f (x) f (x) f (x)f (x) f (x) f ( x) f ( x)
## #

œœœ
cdc d
EO E O EOEO
O
(c) Part b such a decomposition is unique.Ê
20. (a) g(0 0) 1 g (0) g(0) 2g(0) g(0) g (0) 2g(0) g (0) g(0) 0œ Ê œ Ê œ Ê œ
g(0) g(0)
1 g(0) g(0)
œ

#$$
cd
g(0) g (0) 1 0 g(0) 0ʜʜcd
#
(b) g (x) lim lim lim
w


œœ œ
h0 h0 h0ÄÄ Ä
g(x h) g(x) g(x) g(h) g(x) g (x) g(h)
h h h 1 g(x) g(h)
g(x)’“
cd
g(x) g(h)
1 g(x) g(h)
w
w lim 1 1 g (x) 1 g (x) 1 [g(x)]œœœœ
h0Ä
’“’ “ cd
g(h) 1 g (x)
h 1 g(x) g(h)
œ

## #
w
†
(c) 1 y dx tan y x C tan (g(x)) x C; g(0) 0 tan 0 0 C
dy dy
dx 1 y
œ Ê œ Ê œ Ê œ œ Ê œ
#"" "
œ
w
C 0 tan (g(x)) x g(x) tan xÊœÊ œÊ œ
"
21. M dx 2 tan x and M dx ln 1 x ln 2 xœœœœœœÊœ''
00
11
y
22x
1x 1x M
M
#
" #
"
!!
"
w w cd c d ab
1 y
; y 0 by symmetryœœ œ
ln 2 ln 4
ˆ‰
œ
#
1
22. (a) V dx dx ln x ln 4 ln ln 16 ln 2 ln 2œœœœœœœ1 1 ''
14 14
44##
Š‹ cd abkk ˆ‰
"" "
#
#
%
"Î%
%
Èx 4x 4 4 44 4
11 1 11
(b) M x dx x dx x ;
y
14 14
44œœœœœœ''
##
Š‹ œ ‘ˆ ‰
"""
#
"Î# $Î#
%
"Î% #Èx
1864163
23324424
M dx dx ln x ln 16 ln 2;
x
14 14
44œœœœœ''
##
"" " " " " "
# ##
%
"Î%
Š‹Š‹ œ ‘kk
ÈÈx2x
1
8x 8 8

495. Chapter 7 Additional and Advanced Exercises 491
M dx x dx x 2 ; therefore, x andœœ œœœ œœœœ''
14 14
44##
"" "
#
"Î# "Î#
%
"Î% ## #Èx 2M24314
3632217 M
œ‘ˆ‰ˆ‰
y
y ln 2œœ œ
M
M33
2ln 2xˆ‰ˆ‰
"
#
23. A(t) A e ; A(t) 2A 2A A e e 2 rt ln 2 t tœ œ Ê œ Ê œ Ê œ Êœ ʸ œ œ!!!!
rt rt rt ln 2 .7 70 70
r r 100r (r%)
24. ks k dt ln s kt C s s e
ds ds
dt s
œÊœ Ê œÊœ !
kt
the 14th century model of free fall was exponential;Ê
note that the motion starts too slowly at first and then
becomes too fast after about 7 seconds

25. (a) L k k ; solving 0œÊœ œˆ‰ Š‹
a b cot b csc dL b csc b csc cot dL
Rr R r dd
)) )))
))

w
r b csc bR csc cot 0 (b csc ) r csc R cot 0; but b csc 0 sinceÊ œÊ œ Á
%# % % %
))) ))) ) ab
r csc R cot 0 cos cos , the critical value of ))))) )ÁÊ œÊ œ Êœ
1
#
%% " rr
RR


Š‹
(b) cos cos (0.48225) 61 °)œ¸ ¸
" "
%ˆ‰
5
6
26. Two views of the graph of y 1000 1 (.99) are shown below.œœ ‘
x"
x

(a) At about x 11 there is a minimumœ
(b) There is no maximum; however, the curve is asymptotic to y 1000. The curve is near 1000 whenœ
x 643. 

496. 492 Chapter 7 Transcendental Functions
NOTES:

497. CHAPTER 8 TECHNIQUES OF INTEGRATION
8.1 BASIC INTEGRATION FORMULAS
1. ; 2 u C 2 8x 1 C
u8x 1
du 16x dx''16x dx du
8x 1 uÈÈ
œ
œ
#
#
”•
È È
œ
œ
Äœœ
2. ; 2 u C 2 1 3 sin x C
u 1 3 sin x
du 3 cos x dx''3 cos x dx du
13 sin x uÈÈœ
”•
È È
œ
œ
Äœœ
3. 3 sin v cos v dv; 3 u du 3 u C 2(sin v) C
usin v
du cos v dv''È
”•
È
œ
œ
Äœœ † 2
3
$Î# $Î#
4. cot y csc y dy; u ( du) C C
u cot y
du csc y dy''$# $
#
Ê
”•
œ
œ
Äœœ
u
44
cot y
# #
5. ; ln u ln 10 ln 2 ln 5
u8x 2
du 16x dx
x 0 u 2, x 1 u 10''
0 2
1 10
16x dx du
8x 2 u
œ
œ
#
"!
#
Ô×
ÕØ
cdkk
œ
œ
œÊœ œÊœ
Äœ œœ
6. ; du ln u ln 3 ln 1 ln 3
utan z
du sec z dz
z u 1, z u 3''
œ
œ
œ
œ4 1
3 3
3
1
sec z dz
tan z u
43
œ
Ô×
ÖÙ
ÕØ È
cdkk
ÈÈ
œ
œ
œÊœ œÊœ
Äœœœ
# "
11
œ
œ
7. ; 2 ln u C 2 ln x 1 C
ux
du dx
2 du'' dx 2 du
xx1
x
dx
x
uÈȈ‰
È
È
œ
"
#
Ô×
ÖÙ
ÖÙ
ÕØ
È
kk ˆ‰È
œ"
œ
œ
Äœ œ
8. ; 2 ln u C 2 ln x 1 C
ux1
du dx
2 du'' 'dx dx 2 du
xx xx1
x
dx
x
uÊÊ
"
#
ÈÈÈ ˆ‰
È
È
œÄœœ
œ
œ
œ
Ô×
ÖÙ
ÖÙ
ÕØ
È
kk ¸¸È
9. cot (3 7x) dx; cot u du ln sin u C ln sin (3 7x) C
u37x
du 7 dx''Äœœ
œ
œ
”• kk k k
"" "
77 7
10. csc ( x 1) dx; csc u ln csc u cot u C
ux1
du dx''1
1
1
Ĝ
œ
œ
”• kk†
du
11
Ê"
ln csc ( x 1) cot ( x 1) Cœ
"
1
kk11
11. e csc e 1 d ; csc u du ln csc u cot u C ln csc e 1 cot e 1 C
ue 1
du e d
'')) ))ˆ‰ ¸ ¸
”• kk ˆ‰ˆ‰ Ä œ œ
œ
œ
)
)
#
#
12. dx; cot u du ln sin u C ln sin (3 ln x) C
u3ln x
du''
cot (3 ln x)
x dx
x
œ
”• kk k k
œ
œ
Äœœ

498. 494 Chapter 8 Techniques of Integration
13. sec dt; 3 sec u du 3 ln sec u tan u C 3 ln sec tan C
u
du''t3 tt
3 33
t
dt
3
–—
kk ¸¸
œ
œ
Äœœ
14. x sec x 5 dx; sec u du ln sec u tan u C
ux 5
du 2x dx''ab k k ”•
#
#
""
##
Ĝ
œ
œ ln sec x 5 tan x 5 Cœ
"
#
##
kkabab
15. csc (s ) ds; csc u du ln csc u cot u C ln csc (s ) cot (s ) C
us
du ds
'' Ä œ œ
œ
œ
111
1
”• kkk k
16. csc d ; csc u du ln csc u cot u C ln csc cot C
u
du''"" ""
"
Ê)) ))
)
)
)
œ
œ )
–—
kk ¸¸
œ
œ
Ä œ œ
d
17. 2xe dx; e du e e e 2 1 1
ux
du 2x dx
x 0 u 0, x ln 2 u ln 2''
0 0
ln 2 ln 2
uuln2 ln 2
0
œ
x
œ
Ô×
ÕØ È
cd
œ
œ
œÊœ œ Êœ
Ä œ œ œœ
#
!
18. sin (y) e dy; e du e du e 1 e
u cos y
du sin y dy
y u 0, y u 1'''
œ
œ
œ #
#2 01
10
u
cos y uu e
e
Ô×
ÕØ
cd
œ
œ
œÊœ œÊœ
Ä œ œ œœ
1
#
!
Ê"
Ê" Ê"
1
19. e sec v dv; e du e C e C
utan v
du sec v dv
''tan v u u tan v#
#
”•
œ
œ
Äœœ20. ; 2e du 2e C 2e C
ut
du''e dt
t
dt
2t
uu t
œt
È
È
È–—
Ȝ
œ
Äœœ
21. 3 dx; 3 du 3 C C
ux1
du dx''x1 u u
ln 3 ln 3
3œ "
”•
ˆ‰
œ
œ
Äœœ
x1
22. dx; 2 du C C
u ln x
du''222
xln ln dx
x
u
ln x u ln x
”•
œ
œ
Äœœ
##
23. ; 2 du C C
uw
du''2 dw 2 2
w
dw
w
ln ln
œ œwuw
#
#
##È
È –—
Ȝ
œ
Äœœ
u
24. 10 d ; 10 du C C
u2
du 2 d''2u 10 10
ln 10 ln 10
)
)
)
)
”• Š‹
œ
œ
Äœœ
""
## #
u2 œ
25. ; 3 tan x C 3 tan 3u C
x3u
dx 3 du''9 du 3 dx
19u 1xœœ
Ê" Ê"
œœ”•
œ
œ
Äœ œ
26. ; 2 tan u C 2 tan (2x 1) C
u2x1
du 2 dx''4 dx 2 du
1 (2x 1) 1 uœœ œ
Ê" Ê"
œœ”•
œ
œ
Äœ œ
27. ; sin u 0
u3x
du 3 dx
x 0 u 0, x u''
0 0
16 12
œ œ
dx du
19x 1u
6
33 3618È È
Ê Ê
""
#
"" " Ê"
"Î#
!
œ œ
Ô×
ÕØ
œ ‘ˆ‰
œ
œ
œÊœ œÊœ
Äœ œœ
11

499. Section 8.1 Basic Integration Formulas 495
28. sin sin 0'
0
1
dt t
4t
6È
Ê
Ê" Ê"
##
"
!
"
œ
œœœœ ‘ˆ‰
1
29. ; sin u C sin s C
us
du 2s ds''2s ds du
1s 1u
ÈÈ
Ê
#
Ê
" " #
# œ”•
œ
œ
Äœœ
30. ; sin u C sin (2 ln x) C
u 2 ln x
du''2 dx du
x1 4 lnx 1 u
2 dx
x
ÈÈ
ÊÊ
Ê" Ê"
œœ”•
œ
œ
Äœœ
31. 5 sec 5x C 6 sec 5x C''6 dx 6 dx 6
x 25x 1 5x x
5È
É
œ œ "
œ
Ê Ê
Ê" Ê"
œœ œ
5
† kk kk
32. sec C'dr r
rr 9
33È
œ
Ê
" Ê"
œ ¸¸
33. ; tan u C tan e C
ue
du e dx'' 'dx e dx du
ee e 1 u1
xx 2x
x
œœ œ
Ê" Ê"
œÄœœ
œ
œ
”•
x
x
x
34. ; sec u C sec e C
ue
du e dy'' '
dy e dy
e1 uu1
ee 1
du
ÈÈ
Éab
2y
y
yy
ÊÊ
Ê
Ê" Ê"
œÄœœ
œ
œ
œ œ”• kk
y
y
y
35. ; sec u du ln sec u tan u
u ln x
du
x 1 u 0, x e u'''
1
dx du
x cos (ln x) cos u
dx
x
3
e 33
3
00
œœ3 Ô×
ÕØ
cdkk
œ
œ
œÊœ œ Êœ
Äœ œ
œ
œœ
œ
œœ 1
1Î$
! ln sec tan ln sec 0 tan 0 ln 2 3 ln (1) ln 2 3œœœ¸¸ kk Š‹ Š‹
ÈÈ
11
33
36. ; ln 1 4u C ln 1 4 ln x C
ulnx
du ln x dx'' 'ln x dx ln x dx du
x 4x ln x x 1 4 ln x 1 4u 8 8 2
x
#
#
"" " #
œœœÄœœ
œ
œ
ab ”• kk a b
37. 8 ; 8 8 tan u
ux1
du dx
x 1 u 0, x 2 u'' '
11
8 dx dx du
x 2x2 1(x1) 1u
22 1
0
œœ œ
Êœ œÊ œ
Ê"
"
!
œÄœ
œ
œ
œÊœ œÊœ"
Ô×
ÕØ
cd
8 tan 1 tan 0 8 0 2œœœab ˆ‰
Ê" Ê" 1
4
1
38. 2 ; 2 2 tan u
ux3
du dx
x2 u 1, x4 u1'' '
22 1
44 1
2 dx dx du
x 6x10 (x3) 1 u 1
œœ œ
Êœ Ê œ œ
Ê"
"
Ê"
œÄœ
œ
œ
œÊœ œÊœ
Ô×
ÕØ
cd
#
2 tan 1 tan ( 1) 2œœœcd œ ‘ˆ‰
Ê" Ê" 11
44
1
39. ; sin u C sin (t 2) C
ut2
du dt'' 'dt dt du
t4t3 1(t2) 1u
ÈÈ È
Êœ Ê ÊÊ Ê
Ê" Ê"
œ œ œ
œÄœœ
œ
œ
”•
40. ; sin u C sin ( 1) C
u1
du d'' 'dd du
21u 1( 1)
))
)) )ÈÈ È
ÊÊ ÊÊ
Ê" Ê"
œ œœ
œÄœœ
œ
œ
”•
)
)
)
41. ; sec u C sec x 1 C,
ux1
du dx'' 'dx dx du
(x 1) x 2x (x 1) (x 1) 1 uu 1œœ œœÊ Ê
Ê" Ê"
ÈÈ È
œ œ œ
œÄœœ
œ
œ
”• kk k k
ux11kk k kœ

500. 496 Chapter 8 Techniques of Integration
42. ; sec u C
ux2
du dx'' 'dx dx du
(x 2) x 4x 3 (x 2) (x 2) 1 uu 1ÊÊœ ÊÊÊ Ê
Ê"
ÈÈ È
œ œ œ
œÄœ
œ
œ
”• kk
sec x 2 C, u x 2 1œ œ
Ê"
kk kkkk
43. (sec x cot x) dx sec x 2 sec x cot x cot x dx sec x dx 2 csc x dx csc x 1 dx
'' 'œ œ
## # # #
ab ab ''
tan x 2 ln csc x cot x cot x x Cœ kk
44. (csc x tan x) dx csc x 2 csc x tan x tan x dx csc x dx 2 sec x dx sec x 1 dx
'' '''œ œ
## # # #
ab ab
cot x 2 ln sec x tan x tan x x Cœ kk
45. csc x sin 3x dx (csc x)(sin 2x cos x sin x cos 2x) dx (csc x) 2 sin x cos x sin x cos 2x dx
'' ' œœ ab
#
2 cos x cos 2x dx [(1 cos 2x) cos 2x] dx (1 2 cos 2x) dx x sin 2x Cœœœ œ'' 'ab
#
46. (sin 3x cos 2x cos 3x sin 2x) dx sin (3x 2x) dx sin x dx cos x C''' œœœ
47. dx 1 dx x ln x 1 C''x
x1 x1œœ
"
œ œˆ‰ kk
48. dx 1 dx x tan x C''x
x1 x1
œ
œœ
œœ
" Ê"
œ œ ˆ‰
49. dx 2x dx x ln x 1 (9 ln 8) (2 ln 1) 7 ln 8''
œœ
œ
22
33
3
2
2x 2x
x1 x1
%
œœ
ÊÊ
##
œ œœœˆ‰ cd kk
50. dx (2x 3) dx x 3x ln 2x 3 (9 9 ln 9) (1 3 ln 1) ln 9 4''
##11
33
4x 7 2
2x 3 2x 3
œ
Ê
œœ
#
$
Ê"
œ œœœœ ‘ cd kk
51. dt (4t 1) dt 2t t 2 tan C''4t t 16t 4 t
t4 t4
!œ
œœ
ʜ
#
#"
œ œ œ ‘ˆ‰
52. d 1 d ln 2 5 C''277 5 5
25 25 3
))) ) )
))
!œ !œ
ʜ
##
#
))) ) ) )œ œ œ ‘ab kk
53. dx sin x 1 x C'''1 x dx x dx
1x 1x 1x
Ê
ÊÊÊ
Ê" #
ÈÈÈ
œœœ
œœ
È
54. dx (x 1) ln x C'''
x2x1
2x x 1 2 x 1
dx dx
x
œÊ
ÊÊ
"Î#
È
ÈÈ
œœ kk
55. dx sec x sec x tan x dx tan x sec x 1 2 ( 1) 2''
00
44
œœœœ
1sin x
cos x
œ # Î%
!
œ œ œœ!œabcd Š‹
ÈÈ
1
56. dx dx tan (2x) ln 1 4x''
00
12 12
œœ
28x 2 8x
14x 14x 14x
Ê
œœœ
" #
"Î#
!
œœœœœˆ‰ cd kk
tan 1 ln 2 tan 0 ln 1 ln 2œœabab
Ê" Ê" 1
4
57. dx dx sec x sec x tan x dx tan x sec x C'' ' 'dx
1sin x 1sinx cosx
(1 sin x) (1 sin x)
œÊ
ÊÊ #
œœœ œ
ab
œœ ab
58. 1 cos x 1 cos 2 2 cos sec dx tan Cœ œ Ê œ œ œˆ‰ ˆ‰†
x x dx dx x x
1cos x 2 cos## ###
## " '' '
œ
œˆ‰
x
59. d d ; ln u C ln 1 sin C
u1sin
du cos d'' '"
œsec tan u
du
))
)) )
)
))
œÄœœ
œ
œ
”• kk k k

501. Section 8.1 Basic Integration Formulas 497
60. d d ; ln u C ln 1 cos C
u1cos
du sin d'' '"Ê
œœcsc cot 1 cos u
sin du
)) )
)
)) )
)
))
œÄœœ
œ
œ
”• kk k k
61. dx dx 1 dx 1 dx 1 csc x dx'''''""œ
ÊÊ Ê
#
1sec x cos x1 cos x1 sinx sinx
cos x 1 cos x cos x
œ œœœ ˆ‰ˆ‰ˆ ‰
œœ
1cscxcsc x cot x dxxcot xcsc xCœ œ 'ab
#
62. dx dx 1 dx 1 dx''' '""œ
ÊÊ Ê Êœ1csc x sin x1 sin x1 (sin x1)(sin x1)
sin x sin x 1
œœœ ˆ‰ Š‹
1 dx 1 sec x dx 1 sec x sec x tan x dx x tan x sec x Cœ œ œ œ'' 'ˆ‰ˆ ‰ ab
1 sin x sin x
cos x cos x
œ ##
œœ
63. dx sin dx; sin dx 2 cos 2(cos cos 0)
sin 0
for 0 2'' '
00 0
22 2
œœ œ
É ¸¸ ˆ‰ œ ‘
”•
1cos x x x x
x
x
Ê
## ##
#
#
#
!
œÄœœ
 
ŸŸ 1
1
1
( 2)( 2) 4œ œ
64. 1 cos 2x dx 2 sin x dx; 2 sin x dx 2 cos x
sin x 0
for 0 x'' '
00 0
œœ œ
È ÈÈÈ
kk ”• ’“œ Ä œ
 
ŸŸ1
1
! 2 (cos cos 0) 2 2œ œ
ÈÈ
1
65. 1 cos 2t dt 2 cos t dt; 2 cos t dt 2 sin t
cos t 0
for t'' '
œœ œ
œœ œ
œœ œ22 2
È ÈÈÈ
kk ”• ’“œ Ä œ
Ÿ
ŸŸ
1
1
1
#
Î#
1 2 sin sin 2œ œ
ÈȈ‰1
1
#
66. 1 cos t dt 2 cos dt; 2 cos dt 2 2 sin
cos 0
for t 0'' '
## #œœ œ
00 0
È ÈÈȸ¸
”• ’“œ Ä œ
 
Î** ttt
t
###
#
!
Ê
1 1
22sin 0 sin 22œœ
ÈÈœ ‘ˆ‰
1
#
67. 1 cos d sin d ; sin d cos cos 0 cos ( )
sin 0
for 0'' '
## #œœ œ
00 0
È
kk c d”•œ Äœ œ
Ÿ
Î* * #
!
Ê
)) ) ) )) ) 1
)
1)
1
1 ( 1) 2œœ
68. 1 sin d cos d ; cos d sin sin sin 1
cos 0
for '' '
œœ œ
œœ œ
œœ œ22 2
È
kk c d”•œ Äœœœ
Ÿ
ŸŸ #
#
Î# #
)) ) ) )) ) 1
)
)1 1
1
1
1
69. tan y 1 dy sec y dy; sec y dy ln sec y tan y
sec y 0
for y'' '
#
œœ œ
œœ œ
œœ œ44 4
44 4
È kk c d”• kk#
Î%
Î%
œ Ä œ
 
Î** 11
1
1
44
ln 2 1 ln 2 1œ¹¹¹¹
ÈÈ
70. sec y 1 dy tan y dy; tan y dy ln cos y ln
tan y 0
for y 0'' '
# œœ œ44 4
00 0
È kk c d”• kk Š‹#
!
Î%
"
œ Ä œ œ
Ÿ
Î** 1 1
4
2
È
ln 2œ
È
71. (csc x cot x) dx csc x 2 csc x cot x cot x dx 2 csc x 1 2 csc x cot x dx
'' '
œœ œ
œœ œ
œœ œ
œœ œ44 4
34 34 34
œ œ
## # #
abab
2 cot x x 2 csc x 2 cot 2 csc 2 cot 2 csc œ œ cd ˆ‰ˆ‰
$Î%
Î%
1
1
11 1 11 133 3
44 4 44 4
2( 1) 2 2 2(1) 2 2 4œ œ ’“’“ Š‹ Š‹
ÈÈ
3
44
111
#

502. 498 Chapter 8 Techniques of Integration
72. (sec x 4 cos x) dx sec x 8 16 dx tan x 16x 4 sin 2x''
00
44
œœœœ
œ œ
## œ
#
Î%
! œ ‘ˆ‰ cd
1 cos 2x 1
tan 4 4 sin (tan 0 0 4 sin 0) 5 4œ œˆ‰
11
4
11
#
73. cos csc (sin ) d ; csc u du ln csc u cot u C
usin
du cos d'')))
)
))
”• kk
œ
œ
Ĝ
ln csc (sin ) cot (sin ) Cœ kk ))
74. 1 cot (x ln x) dx; cot u du ln sin u C ln sin (x ln x) C
uxln x
du 1 dx''ˆ‰
”•
ˆ‰
kk k k Ä œœ
œ
œ
"
"x
x
75. (csc x sec x)(sin x cos x) dx (1 cot x tan x 1) dx cot x dx tan x dx'''' œ œ
ln sin x ln cos x Cœkk k k
76. 3 sinh ln 5 dx sinh u du 6 cosh u C 6 cosh ln 5 C
uln5
2 du dx' ˆ‰ ˆ‰
”•
x x
2
x
2
œ œ' œ œ
œ
œ '
#
77. ; 12 tan u C 12 tan y C
uy
du dy''
6 dy
y(1 y) 1 u
2y
12 du
È
È
œœ
"
Ê" Ê"
–—
È
È
œ
œ
Äœ œ
œ
78. ; sec u C sec 2x C
u2x
du 2 dx'' 'dx 2 dx du
x4x 1 uu 1 2x (2x) 1ÈÈ È
œ œœ
ÊÊ Ê
Ê" Ê"
œÄœœ
œ
œ
”• kk k k
79. ; 7 sec C
ux1
du dx'' '7 dx 7 dx 7 du u
(x1)x2x48 uu49 (x 1) (x 1) 49 77
ÊÊÊ Ê ÊÊÊ
" Ê"
ÈÈ È
œ œœ
œÄœ
œ
œ
”•
¸¸†
sec Cœ
Ê" Ê"¸¸
x
7
80. ; sec u C
u2x1
du 2 dx'' 'dx dx du
(2x 1) 4x 4x (2x 1) (2x 1) 1 2u u 1œœ œœÊ Ê
"
#
Ê"
ÈÈ È
œ œ œ
œÄœ
œ
œ
”• kk
sec 2x 1 Cœ
"
#
Ê"
kk
81. sec t tan (tan t) dt; tan u du ln cos u C ln sec u C ln sec (tan t) C
utan t
du sec t dt
''#
#
”• kk kk k k
œ
œ
Äœœœ
82. csch C' dx x
xx
ÈÈ
Ĝ
"
$
Ê"
$
œ
œ ¹¹
83. (a) cos d (cos ) 1 sin d ; 1 u du u C sin sin C
usin
du cos d'' '$# # $ "
)) ) ) ) ) )
)
))
œ Äœœ
œ
œ
ab ab ”•
u
33
%
(b) cos d (cos ) 1 sin d 1 u du 1 2u u du u u C'' '&###%$
##
)) ) ) )œœœœababa b '
2u
35

sin sin sin Cœ)))
2
35
$& "
(c) cos d cos (cos ) d 1 sin (cos ) d'' '*) #
%
)) ) ) ) ) ) )œœab a b
84. (a) sin d 1 cos (sin ) d ; 1 u ( du) u C
u cos
du sin d'' '$# #
)) ) ) )
)
))
œ Äœ
œ
œ
ab ab ”•
u
3
%
cos cos Cœ ))
" $
3
(b) sin d 1 cos (sin ) d 1 u ( du) 1 2u u du'' ' '&# # #%
##
)) ) ) )œ œœab ab a b
cos cos cos Cœ)))
2
35
$& "
(c) sin d 1 u ( du) 1 3u 3u u du cos cos cos C'' '(# #%' $&
$
)) ) ) )œ œ œ ab a b
3cos
57

)

503. Section 8.1 Basic Integration Formulas 499
(d) sin d sin (sin ) d 1 cos (sin ) d'' '"$ "# #
'
)) ) ) ) ) ) )œœab a b
85. (a) tan d sec 1 (tan ) d sec tan d tan d tan tan d'' ' '$# # # "
#
)) ) ) ) ) )) )) ) ))œ œ œab '
tan ln cos Cœ
"
#
#
))kk
(b) tan d sec 1 tan d tan sec d tan d tan tan d'' ' ' '&#$$#$%$ "
)) ) ) ) ) )) )) ) ))œ œ œabab
4
(c) tan d sec 1 tan d tan sec d tan d tan tan d'' ' ' '(#&&#&'& "
)) ) ) ) ) )) )) ) ))œ œ œabab
6
(d) tan d sec 1 tan d tan sec d tan d ;'' ' '
2k1 2k1 2k1 2k1$###
)) ) ) ) ) )) ))œ œ abab
##
u du tan d u tan d tan tan d
utan
du sec d
”•
œ
œ
Ä œ œ
)
))
)) )) ) ))
#
""
#
'' ' '
2k 1 2k 1 2k 2k 1 2k 2k 1## # #
2k k
86. (a) cot d csc 1 (cot ) d cot csc d cot d cot cot d'' ' ' '$# # # "
#
)) ) ) ) ) )) )) ) ))œ œ œab
cot ln sin Cœ
"
#
#
))kk
(b) cot d csc 1 cot d cot csc d cot d cot cot d'' ' ' '&#$$#$ %$ "
)) ) ) ) ) )) )) ) ))œ œ œabab
4
(c) cot d csc 1 cot d cot csc d cot d cot cot d'' ' ' '(#&&#& '& "
)) ) ) ) ) )) )) ) ))œ œ œabab
6
(d) cot d csc 1 cot d cot csc d cot d ;'' ' '
2k1 2k1 2k1 2k1$###
)) ) ) ) ) )) ))œ œ abab
##
u du cot d u cot d
u cot
du csc d
”•
œ
œ
Ä œ
)
))
)) ))
#
"
#
'' '
2k 1 2k 1 2k 2k 1## #
k
cot cot dœ
"
2k 2k 2k 1
)))'
#
87. A (2 cos x sec x) dx 2 sin x ln sec x tan xœœ'
#œ
œ
œ
œ4
4
cd kk
1
1
Î%
Î%
2 ln 21 2ln 21œ ’“’ “
ÈÈ ÈÈ
Š‹ Š‹
22 ln 22 lnœ œ
ÈÈ
Š‹
#$
È
È
Š‹
È
2
21
21
21
œ"
Ê
œ
Ê
œ
22 ln3 22œ
ÈÈ
Š‹
88. A (csc x sin x) dx ln csc x cot x cos xœœ
'
œ
œ
œ
œ6
2
cdkk
1
1
Î#
Î'
ln 1 0 ln 2 3 ln 2 3œ œ kk ¹¹ Š‹
ÈÈ
ÈÈ
33
##

89. V (2 cos x) dx sec x dx 4 cos x dx sec x dxœœ
''''
#
œœ œœ
œœ œœ
œœ œœ44 44
44 44
1111
## # #
2 (1 cos 2x) dx tan x 2 x sin 2x [1 ( 1)]œ œ 1111'
#œ
œ
œ
œ4
4
cd œ ‘
1
1
1
1
Î%
Î%
"
#
Î%
Î%
22212œœœ11111œ ‘ˆ‰ˆ‰ˆ ‰
11 1
44
""
## #
#
90. V csc x dx sin x dx csc x dx (1 cos 2x) dxœœ'' ''
œœ œœ
œœ œœ
œœ œœ
œœ œœ66 66
22 22
111
###
#
1
cot x x sin 2x 0 3 0œ œ 11cd œ ‘ˆ‰ ’“’ “Š‹ Š ‹
È
1
1
1111
1
1
Î#
Î' ## ## ##
""
Î#
Î' 6
3
†
È
3œ œ 11
È
Š‹Š‹
11 1
#
2
64 86
373
ÈÈ

504. 500 Chapter 8 Techniques of Integration
91. y ln (cos x) tan x sec x 1; L 1 dxœÊœÊœœœ
dy dy dy
dx cos x dx dx
sin x
Š‹ Š‹ Ê
##
##
'
a
b
1 sec x 1 dx sec x dx ln sec x tan x ln 2 3 ln 1 0 ln 2 3œœ œ œœ''
00
33
œœœœ
È ab c d kk kk ¹¹ Š‹
ÈÈ #
Î$
!
1
92. y ln (sec x) tan x sec x 1; L 1 dxœÊœÊœœœ
dy dy dy
dx sec x dx dx
sec x tan x
Š‹ Š‹ Ê
##
##
'
a
b
sec x dx ln sec x tan x ln 2 1 ln 1 0 ln 2 1œ œ œ œ '
0
4
œœ
cd kkkk ¹¹ Š‹
ÈÈ
1Î%
!
93. M sec x (sec x) dx sec x dx
x
44
44œœ''
#
œœ
œœ
œœ
ˆ‰
""
##
#
tan x 1 ( 1) 1;œœœ
""
##
Î%
Î%
cd c d
1
1
M sec x dx ln sec x tan xœœ'
#œ
œ
œ
œ4
4
cdkk
1
1
Î%
Î%
ln 2 1 ln 2 1 lnœœ¹¹¹¹Š‹
ÈÈ
È
È
2
21
œ"
Ê
ln ln 3 2 2 ; x 0 byœœœ
#$
Š‹
È
Š‹
È
2
1
œ"
#
œ
symmetry of the region, and yœœ
M
M
ln 3 2 2x "
œŠ‹
È

94. M csc x (csc x) dx csc x dx
x
66
56 56œœ''
œœ
œœ
œœ
œœ
ˆ‰
""
##
#
cot x 3 3 3;œ œ œ
""
##
&Î'
Î'
cd ’“Š‹Š‹
ÈÈÈ
1
1
M csc x dx ln csc x cot xœœ'
œ
œ
œ
œ6
56
cdkk
&Î'
Î'
1
1
ln 2 3 ln 2 3 ln œ œ¹¹Š ‹¹¹
ÈÈ
¹¹
23
23
œ
Ê
È
È
ln 2 ln 2 3 ; x by symmetryœœœ
#$
Š‹
È
Š‹
È
23
43
œ
#
œ
1
of the region, and yœœ
M
M
3
2 ln 2 3x
È
Š‹
È
œ

95. csc x dx (csc x)(1) dx (csc x) dx dx;'' œœ œ '' ˆ‰
csc x cot x csc x csc x cot x
csc x cot x csc x cot x
œœ
œœ
œ
ln u C ln csc x cot x C
u csc x cot x
du csc x cot x csc x dx
”•
ab
kk k k
œ
œ
Äœœ
#
Ê
'du
u
96. x 1 (x 1) (x 1)(x 1) (x 1) (x 1) (x 1) (x 1) (x 1)cdcdab œ ‘
# # #Î$ %Î$ # #Î$ #Î$
#Î$ #Î$
œœœ
(x 1) (x 1) 1œ œ
# #Ê
œœ
#Î$ #Î$ˆ‰ ˆ ‰
x1 2
x1 x1
(a) x 1 (x 1) dx (x 1) 1 dx;
u
du dx''cdab ˆ‰
–—
##
#Î$
œ
#Î$
"
œ
"
œ
œ
œ
œ
2
x1
x1
(x 1)
œ
(1 2u) du (1 2u) C 1 C CÄ œ œ œ ' #Î$ "Î$
###
"Î$ "Î$
Ê3323x1
x1 x 1 ˆ‰ ˆ‰
(b) x 1 (x 1) dx (x 1) dx; u''cdab ˆ‰ ˆ‰
##
#Î$ ÊÊ
œœ
#Î$
œ œ
x1 x1
x1 x1
k
du k dx 2k dx; dx duÊœ œ œˆ‰ ˆ‰
x1 x1
x 1 (x 1) (x 1) k x 1
(x 1) (x 1) (x 1) (x 1)ʜ
#
œÊÊ Êœ
k1 k1# #cd
œ
œ
k1
k1#
$
du; then, du duœœ
(x 1)
2k x1 x1 k x1 k x1
x1 x1 x1 x1œ ÊÊ"Ê"Ê
##
#Î$
œ
ˆ‰ ˆ‰ ˆ‰ ˆ‰
1k 1k 13k##
''
du u du (3k) u C u C Cœœœœœ
"Ê " " Ê
# # # # #
"Î$
kx1 k k x1
x1 3 3 x1'ˆ‰ ˆ‰
k1 3k 1
13k 1 13k 13k

'

505. Section 8.2 Integration by Parts 501
(c) x 1 (x 1) dx (x 1) dx;'cdab ˆ‰
##
#Î$ Ê
œ
#Î$
œ ' x1
x1
du;
utanx
xtan u
dx
Ô×
ÕØ
ˆ‰ˆ‰ ˆ ‰
œ
œ
œ
Ĝ
Ê"
"Ê " Ê
œœ œ œ
#Î$ #Î$
du
cos u
(tan u1) tan u1 cosu (sin ucos u) sin ucos u
tan u 1 du sin u cos u
œ
œœ œ
''

sin u cos u sin u sin u 2 sin cos u
sin u cos u sin u sin u 2 cos sin u
–—
ˆ‰ ˆ‰
ˆ‰ ˆ‰
’“
œ œ
œ œ
Ä
111
111
#
#
"
Ê
Ê
44
44
2 cos u cos
sin u
'
œˆ‰ ˆ‰
ˆ‰
#
#
4
4
uÊ
#Î$
#
4
du
tan u sec u du tan u C Cœœœ
"
###
#Î$ # "Î$
Ê
"Î$' ˆ‰ˆ‰ ˆ‰ ’“
11 1
4 4 4 1 tan u tan
33 tan u tan
#
#
4
4
C œ
3x1
x1#
Ê
"Î$
ˆ‰
(d) u tan x tan u x tan u x dx 2 tan u du du ;œÊœÊœÊœ œœ
" # "ÈÈ ˆ‰
cos u cos u cos u
2 sin u 2d(cos u)
œ %%
x1tanu1 ; x1tanu1 ;œ œ œ œ œ œ
## ÊÊ œ "sin u cos u 1 2 cos u cos u sin u
cos u cos u cos u cos u
œœ œ œœ
œœ œœ
(x 1) (x 1) dx' œ
#Î$ %Î$ Ê " Ê '
ab
ab ab
12 cosu
cos u cos u
2d(cos u)
cos u
œ

œœ
%
††
1 2 cos u ( 2) cos u d(cos u) 1 2 cos u d 1 2 cos uœ œ ''ab abab
###
#Î$ #Î$ "
#
†† † †
12 cosu C C Cœ œ œ
333x1
x1###
#
"Î$
"Î$
Ê
"Î$
ab
–—
ˆ‰
Š‹
Š‹
12 cosu
cos u
cos uœ
#
#
$
#
(e) u tan tan u x 1 2(tan u 1) dx 2d(tan u);œÊœÊœÊœœ
Ê" ÊÊ
##ˆ‰
x1 x1 2 du
cos u
œ
(x 1) (x 1) dx (tan u) (tan u 1) 2 2 d(tan u)'' œ
#Î$ %Î$ #Î$ %Î$ #
†††
1d1 1 C1Cœ œ œ
"" " "
# # #
#Î$ "Î$ "Î$'ˆ‰ˆ‰ˆ‰ ˆ‰
tan u1 tan u1 tan u1 x1
332
C œ
3x1
x1#
Ê
"Î$
ˆ‰
(f)
u cos x
x cos u
dx sin u du
Ô×
ÕØ
œ
œ
œ
Ä œ
Ê"
ʜ
'' sin u du sin u du
cos u 1 (cos u 1) sin u 2 cos
%
œœ

œ
#)%
Éab ab ˆ‰
u œ œ œ''' du du du
(sin u) 2 cos 2 sin cos
cos
sin cos
"'! œ'!
œœœ
#)% ")% &)%
œ
œ
œ
œˆ‰ ˆ‰ˆ‰
ˆ‰
uuu
u
u u
"
#
"Î$
Š‹
tan d tan tan C tan C Cœ œ œ œ ' "Î$ #Î$ #
##### # #
"Î$
Ê
"Î$ˆ‰ ˆ ‰ ˆ ‰ ˆ ‰
uu3u3 u 3cos u1
cos u 1
C œ
3x
x1#
Ê"
"Î$
ˆ‰
(g) x 1 (x 1) dx;
ucoshx
x cosh u
dx sinh u'cdab
Ô×
ÕØ
#
#Î$
Ê"
ʜ
Î# $
œ
œ
œ ' sinh u du
cosh u 1 (cosh u 1)
%
œœ
œ
Éab
œœœ '' 'sinh u du du du
sinh u (cosh u 1)
(sinh u) 4 cosh sinh cosh
% %% #&
œœœ
Èab
ÉÉ ˆ ‰ ˆ‰ ˆ‰œ
"
# uuu
tanh d tanh tanh C C Cœ œ œ œ 'ˆ‰ˆ‰ˆ‰ ˆ ‰ ˆ‰
u u 3 u 3 cosh u 1 3 x 1
coshu1 x1##### #
"Î$ #Î$
ÊÊ
"Î$ "Î$
8.2 INTEGRATION BY PARTS
1. u x, du dx; dv sin dx, v 2 cos ;œœ œ œ
xx
##
x sin dx 2x cos 2 cos dx 2x cos 4 sin C'' xxx xx ### ##
œ œ ˆ ‰ ˆ‰ ˆ‰
2. u , du d ; dv cos d , v sin ;œœ œ œ)) 1)) 1)
"
1
cos d sin sin d sin cos C'') 1) ) 1) 1) ) 1) 1)œ œ
))
11 11
""
œ

506. 502 Chapter 8 Techniques of Integration
3. cos t
tsin t
#
ïïïïî
w#8
2t cos t((((' Î
wÎ8
2 sin t((((' Î
w#8
0 t cos t dt t sin t 2t cos t 2 sin t C
'##
œ
4. sin x
x cos x
#
((((' Î
w#8
2x in x s((((' Î
wÎ8
2 cos xïïïïî
w#8
0 x sin x dx x cos x 2x sin x 2 cos x C
'##
œ
5. u ln x, du ; dv x dx, v ;œœœœ
dx x
x
œ
#
x ln x dx ln x 2 ln 2 2 ln 2 ln 4''
11
22
œ œœœ’“ ’“
xxdxx 33
x4 44
œœ œ
##
##
""
6. u ln x, du ; dv x dx, v ;œœœ œ
dx x
x4
$
#
x ln x dx ln x ''
11
ee ee
11
$ œ
œœœ’“ ’“
x x dx e x 3e 1
4 4 x 4 16 16
#####
7. u tan y, du ; dv dy, v y;œœœœ
Ê"
œ
dy
1y
œ
tan y dy y tan y y tan y ln 1 y C y tan y ln 1 y C' " " " # "
#
" #
œœœ'
y dy
1yab
œ ab È
8. u sin y, du ; dv dy, v y;œœ œœ
Ê"
Ê
dy
1yÈ
œ
sin y dy y sin y y sin y 1 y C'Ê" Ê" Ê"
Ê
#
œœ'
y dy
1yÈ
œ
È
9. u x, du dx; dv sec x dx, v tan x;œœ œ œ
#
x sec x dx x tan x tan x dx x tan x ln cos x C'' #
œ œ kk
10. 4x sec 2x dx; [y 2x] y sec y dy y tan y tan y dy y tan y ln sec y C
' ##
œÄ œ œ '' kk
2x tan 2x ln sec 2x Cœ kk
11. e
x
x e
$
ïïïïî
w#8
x
3x e
#
ïïïïî
wÎ8
x
6x eïïïïî
w#8
x
6 eïïïïî
wÎ8
x
0 x e dx x e 3x e 6xe 6e C x 3x 6x 6 e C'$$# $#xxxxx x
œ œ ab

507. Section 8.2 Integration by Parts 503
12. e
#p
p e

%
((((' Î
w#8
#p
4p

e
$
ïïïïî
wÎ8
#p
12p e
#
((((' Î
w#8
#p
24p eïïïïî
wÎ8
#p
24 e((((' Î
w#8
#p
0 p e dp p e 4p e 12p e 24pe 24e C'%%$#######pppppp
œ
p 4p 12p 24p 24 e C œ ab
%$ # #p
13. e
x
x 5x e
#
Î (((('
w#8
x
2x 5 eÎ (((('
wÎ8
x
2 e ïïïïî
w#8
x
0 x 5x e dx x 5x e (2x 5)e 2e C x e 7xe 7e C'ab ab
## #
œœ
xxxxxxx
x7x7eC œ ab
# x
14. e
r
r r 1 e
#
# # (((('
w#8
r
2r 1 e# (((('
wÎ8
r
2 e ïïïïî
w#8
r
0 r r 1 e dr r r 1 e (2r 1) e 2e C'abab
##
œ
rrrr
r r 1 (2r 1) 2 e C r r 2 e Cœ œ cdabab
## rr
15. e
x
x e
&
ïïïïî
w#8
x
5x e
%
ïïïïî
wÎ8
x
20x e
$
ïïïïî
w#8
x
60x e
#
ïïïïî
wÎ8
x
120x eïïïïî
w#8
x
120 e e ïïïïî
wÎ8
x
0 x e dx x e 5x e 20x e 60x e 120xe 120e C'& &%$#x xxxx xx
œ
x 5x 20x 60x 120x 120 e Cœ ab
&%$# x

508. 504 Chapter 8 Techniques of Integration
16. e
4t
t e
# "
ïïïïî
w#8
4
4t
2t eïïïïî
wÎ8
"
164t
2 eïïïïî
w#8
"
644t
0 t e dt e e e C e e e C'# "
#4t 4t 4t 4t 4t 4t 4t
œœ
t2t2 tt
4 1664 4 83
œœ
eC Š‹
tt
483
œ
Î# #
"
# 4t
17. sin 2 )
cos 2))
# "
((((' Î
w#8
2
2 sin 2))((((' Î
wÎ8
"
4
2 cos 2ïïïïî
w#8
"
8
)
0 sin 2 d cos 2 sin 2 cos 2'
0
2
œœ
))) ) ) )
#
##
"
Î#
!
œ ’“
))
1
œ
4
(1) 0 (1) 0 0 1 œ œ œ’“ œ ‘
11 11
œ œœ
844 488
4
††† †
"""Ê
#
18. cos 2x
x sin 2x
$ "
ïïïïî
w#8
2
3x cos 2x
# "
((((' Î
wÎ8
4
6x sin 2x((((' Î
w#8
"
8
6 os 2x cïïïïî
wÎ8
"
16
0 x cos 2x dx sin 2x cos 2x sin 2x cos 2x'
0
2
œœ
$
#
Î#
!
œ’“
x3x 3x3
448
% œ
1
0 ( 1) 0 ( 1) 0 0 0 1 œ œ œ’“ œ ‘
11 1 1 1
!œ œ œ
16 16 8 8 8 16 4 16
333 333 34
†† †† †
abÊ
19. u sec t, du ; dv t dt, v ;œœœœ
Ê"
Ê
#
dt t
tt 1
È
œ
œ
t sec t dt sec t 2 '''
23 23 23
222œœœ
œœœ
Ê" Ê"
##
#
#Î $ ÊÊ
œ œ’“ Š‹ ˆ‰
t t dt 2 t dt
tt 1 2t 1
336
œœ
œœ
È
ÈÈ
††
11
t1 3 1 3œœœœœ
55455
99393939
3353311 11 1
’“ Š ‹ Š‹
È ÈÈÉ
"" "
## #
#
#
#Î $
Ê
È
ÈÈÈ
20. u sin x , du ; dv 2x dx, v x ;œœœœ
" # #
Ê
ab
2x dx
1x
È
#
2x sin x dx x sin x x'''
0 00
12 12 12
œ œœ
œ œœ
" # # " # #
"Î #
!
ÊÊ
"
#
Ê
ab c d ab ˆ‰ˆ‰œœ
È
ÈÈ
ˆ‰
†
2x dx
1x 21x
6
d1 x
##
#
1
1x 1œ œ œ
11 1
1141
3 6312
###
%
"Î #
!
œÊ
’“
È
É
È
È
21. I e sin d ; u sin , du cos d ; dv e d , v e I e sin e cos d ;œœœœœÊœ''
#####
)) ) )) ) ) ))cd
u cos , du sin d ; dv e d , v e I e sin e cos e sin dcd Š‹œœ œœÊœ)))) ))))
## # # # ' e sin e cos I C 2I e sin e cos C I e sin e cos C, where C isœÊœÊœ œ
## ## ##
)) )) ))
ww "
##
abab
C

another arbitrary constant

509. Section 8.2 Integration by Parts 505
22. I e cos y dy; u cos y, du sin y dy; dv e dy, v eœœœœœ'# ##y yy
cd
I e cos y e ( sin y) dy e cos y e sin y dy; u sin y, du cos y dy;Êœ œ œ œ
## ##yy yy''ab c
dv e dy, v e I e cos y e sin y e cos y dy e cos y e sin y I Cdab Š‹œ œ Ê œ œ
## # # ##yy y y y yy ' w
2I e (sin y cos y) C I e sin y e cos y C, where C is another arbitrary constantÊœ Êœ œ
###yyy w "
##
ab
C

23. I e cos 3x dx; u cos 3x; du 3 sin 3x dx, dv e dx; v eœœœœœ'2x 2x 2x œ ‘
"
#
I e cos 3x e sin 3x dx; u sin 3x, du 3 cos 3x, dv e dx; v eÊœ œ œ œ œ
" "
## #2x 2x 2x 2x 3' œ ‘
I e cos 3x e sin 3x e cos 3x dx e cos 3x e sin 3x I CÊœ œ
"" "
#### #
w2x 2x 2x 2x 2x33 39
44
Š‹ '
I e cos 3x e sin 3x C (3 sin 3x 2 cos 3x) C, where C CÊœ Ê œ
13 3 e 4
4 4 13 13
"
#
ww 2x 2x
2x
24. e sin 2x dx; [y 2x] e sin y dy I; u sin y, du cos y dy; dv e dy, v e''## ##2x y yy
œÄ œœ œ œ œ
"
#
cd
I e sin y e cos y dy u cos y, du sin y; dv e dy, v eÊœ œ œ œ œ
"
#
Š‹ cd
## # #yy yy '
I e sin y e cos y e ( sin y) dy e (sin y cos y) I CÊœ œ
"" "
## #
w## # #yy y y
Š‹ ab '
2I e (sin y cos y) C I e (sin y cos y) C (sin 2x cos 2x) C, whereÊ œ Ê œ œ
""
#
w##yy
44
e
#2x
Cœ
C

#
25. e ds; e x dx xe dx; u x, du dx; dv e dx, v e ;
3s 9 x
ds x dx'''
œ3s 9 x x x x$
”• cd
œ
œ
Äœ œœœœ
#
2
3
22
33
†
xe dx xe e dx xe e C 3s 9 e e C
22 2 2 33 3 3' Š‹ Š ‹ ab È
xxx xx 3s93s9
œ œœ '
œœ$$
26. u x, du dx; dv 1 x dx, v (1 x) ;œœ œ œ
È È
2
3
$
x 1 x dx (1 x) x (1 x) dx (1 x)''
00
11
È œ ‘ œ ‘ÈÈœ œ œ22 224
33 3515
$$
"
!!
&Î#
"
27. u x, du dx; dv tan x dx, v tan x dx dx dx dxœœ œ œ œ œ œ
## "Ê ''' '' sin x cos x dx
cos x cos x cos x
œœ
œœœ
tan x x; x tan x dx x(tan x x) (tan x x) dx 3 ln cos xœ œ œ ''
00
33
œœœœ
# Î$
! #
Î$
!
cd kk Š‹’ “
È
1 11
1
33
x œ
3 ln ln 2œœ
11 1 1 1
33 183 18
3
Š‹
È
"
#
œœ È
28. u ln x x , du ; dv dx, v x; ln x x dx x ln x x x dxœ œ œ œ œ ab ab ab '
### œ
œ œ
œ"(2x 1) dx
xx x(x 1)
2x
œ
' †
x ln x x x ln x x dx x ln x x 2x ln x 1 Cœ œ œab ab ab kk
## # œœÊ"
œœ''
(2x 1) dx 2(x 1)
x1 x1
29. sin (ln x) dx; du dx (sin u) e du. From Exercise 21, (sin u) e du e C
u ln x
dx e du'
Ô×
ÕØ
ˆ‰
œ
œÄ œ
œ
" Ê
#x
sin u cos u
u
uuu
''
x cos (ln x) x sin (ln x) Cœ
"
#
cd

510. 506 Chapter 8 Techniques of Integration
30. z(ln z) dz; du dz e u e du e u du;
u ln z
dz e du''' ### "
Ô×
ÕØ
œ
œÄ œ
œ
z
uu 2u
u
†† †
e
2u
u e
# "
ïïïïî
w#8
2
2u
2u

eïïïïî
wÎ8
"
42u
2 eïïïïî
w#8
"
82u
0 u e du e e e C 2u 2u 1 C'##
##
"2u 2u 2u 2u
œ œ
uu e
44
œ 2u
cd
2(ln z) 2 ln z 1 C œ
z
4
œ
cd
#
31. (a) u x, du dx; dv sin x dx, v cos x;œœ œ œ
S x sin x dx [ x cos x] cos x dx [sin x]
" !!œœœœ''
00
œœ
11
11
(b) S x sin x dx [ x cos x] cos x dx 3 [sin x] 3
#
##œ œ œ œ''
œœ
œœ
22
”• cd
11
11
11
(c) S x sin x dx [ x cos x] cos x dx 5 [sin x] 5
$
$$
##œœœœ''
22
33œœ
œœ
11
11
11
(d) S ( 1) x sin x dx ( 1) [ x cos x] [sin x]
Ñœ"œ œ
n1 n1 n1 n1
n
n1
nn$$
'
œ
œ
œœ
œœ
cd
( 1) (n 1) ( 1) n ( 1) 0 (2n 1)œ œ
n1 n n1$$
cd 11 1
32. (a) u x, du dx; dv cos x dx, v sin x;œœ œ œ
S x cos x dx [x sin x] sin x dx [cos x] 2"
##œ œ œ œ''
œœ
œœ
œ œ
œ œ
œœ
œœ
œ œ
œ œ22
32 32
32 32
2 2
”•
ˆ‰
311
1
(b) S x cos x dx [x sin x] sin x dx [cos x] 4#
##œœœœ''
32 32
52 52
2 2
2 2œœ
œœ
œ œ
œ œ
œœ
œœ
"œ "œ

œ ‘ˆ‰
5311
1
(c) S x cos x dx [x sin x] sin x dx [cos x] 6$
##œ œ œ œ''
52 52
72 72
2 2
2 2œœ
œœ
œ œ
œ œ
œœ
œœ
!œ !œ
"œ "œ
”•
ˆ‰
7511
1
(d) S ( 1) x cos x dx ( 1) [x sin x] sin x dx
n
nnn
2n12 2n12
2n 1 2 2n 1 2
n1 2
2n 1 2œ œ ''




œ œ
œ œ
œ
œ
”• ( 1) ( 1) ( 1) [cos x] (2n 2n ) 2nœ œ œ
nn n1 n1 2
2n 1 2
’“
(2n 1) (2n 1)œÊ
## #
"11 #

œ
œ
11 11 1
33. V 2 (ln 2 x) e dx 2 ln 2 e dx 2 xe dxœœ
'''
000
ln 2 ln 2 ln 2
xxx
111
(2 ln 2) e 2 xe e dxœ11cd c dŒ Î
xxxln 2 ln 2
00
0
ln 2 '
2 ln 2 2 2 ln 2 e 2 ln 2 2 2 (1 ln 2)œ œœ11 111ˆ‰ cd
xln 2
0

34. (a) V 2 xe dx 2 xe e dxœœ
''
00
11
xxx
11
###
Œ Îcd
"
!
2e2 1œ œ11Š‹ cd ˆ‰
""" "
!eee #x
2 œ1
4
e
1

511. Section 8.2 Integration by Parts 507
(b) V 2 (1 x)e dx; u 1 x, du dx; dv e dx,œ œœœ'
0
1
xx
1
##
v e ; V 2 (1 x) e e dxœ œ
###xxx
0
1
1”•cd ab
"
!'
2[01(1)] e 21 1œ œœ11’“ cd ˆ‰
#x"
!
"
ee
21

35. (a) V 2 x cos x dx 2 [x sin x] sin x dxœœ
''
00
22
œœœœ
11 Œ Î
1Î#
! 2 [cos x] 2 0 1 ( 2)œ œœ1111Š‹ ˆ‰
11 1
##
Î#
!

(b) V 2 x cos x dx; u x, du dx; dv cos x dx, v sin x;œ œœœ œ'
0
2
œœ
1ˆ‰
11
##
V 2 x sin x 2 sin x dx 0 2 [ cos x] 2 (0 1) 2œ œ œœ11111œ ‘ˆ‰
1
1 1
#
Î#
!
Î#
! '
0
2
œœ
36. (a) V 2 x(x sin x) dx;œ '
0
œ
1
sin x
x cos x
#
((((' Î
w#8
2x in x s((((' Î
wÎ8
2 cos xïïïïî
w#8
0 V 2 x sin x dx 2 x cos x 2x sin x 2 cos x 2 4Êœ œ œ 11 11
'
0
œ
## #
!
cdab
1
(b) V 2 ( x)x sin x dx 2 x sin x dx 2 x sin x dx 2 [ x cos x sin x] 2 8œ œ œ'''
000
œœœ
11 1 1 1 1 1
### $
!
1
ab
8 œ1
37. (a) av(y) 2e cos t dtœ
"
#1'
0
2
t
œ
#
e œ
"Ê
#
#
!1
1
œ ‘ˆ‰
#tsin t cos t
(see Exercise 22) av(y) 1 eʜ
"
#1
ab
#2œ
(b)
38. (a) av(y) 4e (sin t cos t) dtœ
"
#1'
0
2
t
œ
#
e sin t dt e cos t dtœ
22
11''
00
22
tt
œœ
##
ee œ
2 sin t cos t sin t cos t
1
1
œ ‘ˆ‰ˆ‰
##ttÊÊ Ê
##
#
!
e sin t 0œ œ
2
1
1
cd
#t #
!
(b)
39. I x cos x dx; u x , du nx dx; dv cos x dx, v sin xœœœœœ
'nnn
cd
Ê"
I x sin x nx sin x dxʜ
nn ' Ê"

512. 508 Chapter 8 Techniques of Integration
40. I x sin x dx; u x , du nx dx; dv sin x dx, v cos xœœœœœ'nnn
cd
Ê"
I x cos x nx cos x dxʜ
nn ' Ê"
41. I x e dx; u x , du nx dx; dv e dx, v eœœœœœ'nax n n ax ax
a œ ‘
Ê" "
I e x e dx, aÊœ Á!
xe n
aa
ax n ax
nax
'Ê"
42. I ln x dx; u ln x , du dx; dv dx, v xœœœœ"œ'ab ab’“
nnnln x
x
ab
n!"
I x ln x n ln x dxʜ ab ab
nn'
Ê"
43. sin x dx x sin x sin y dy x sin x cos y C x sin x cos sin x C''Ê" Ê" Ê" Ê" Ê"
œ œœ ab
44. tan x dx x tan x tan y dy x tan x ln cos y C x tan x ln cos tan x C
''Ê" Ê" Ê" Ê" Ê"
œ œœ kk k k ab
45. sec x dx x sec x sec y dy x sec x ln sec y tan y C
''Ê" Ê" Ê"
œ œ kk
x sec x ln sec sec x tan sec x C x sec x ln x x 1 Cœ œ
Ê" Ê" Ê" Ê" #
kkabab ¹¹
È
46. log x dx x log x 2 dy x log x C x log x C''
22 2 2
y œœœ
2x
ln ln
y
##
47. Yes, cos x is the angle whose cosine is x which implies sin cos x 1 x .
Ê" Ê" #
ab
È
œ
48. Yes, tan x is the angle whose tangent is x which implies sec tan x 1 x .
Ê" Ê" #
ab
È
œ
49. (a) sinh x dx x sinh x sinh y dy x sinh x cosh y C x sinh x cosh sinh x C;
''Ê" Ê" Ê" Ê" Ê"
œ œœ ab
check: d x sinh x cosh sinh x C sinh x sinh sinh x dxcdab ab ’“
Ê" Ê" Ê" Ê"
œœ
"
œ
x
1x 1x
ÈÈ
œœ
sinh x dxœ
Ê"
(b) sinh x dx x sinh x x dx x sinh x 1 x 2x dx'' '" " " # ""
œ #
"Î#
œ œ Š‹ ab
È
1x œ
x sinh x 1 x Cœ
" #
"Î#
ab
check: d x sinh x 1 x C sinh x dx sinh x dx’“’“ ab
" # " "
"Î#
œœ
œ œ
xx
1x 1x
ÈÈ
œœ
50. (a) tanh x dx x tanh x tanh y dy x tanh x ln cosh y C''Ê" Ê" Ê"
œ œ kk
x tanh x ln cosh tanh x C;œ
Ê" Ê"
kkab
check: d x tanh x ln cosh tanh x C tanh x dxcd kkab ’“
Ê" Ê" Ê"
ÊÊ
"
œ
x
1x coshtanhx1x
sinh tanh x
œ
!"
ab
ab
tanh x dx tanh x dxœœœ ‘
Ê" Ê"
ÊÊ
xx
1x 1x
œœ
(b) tanh x dx x tanh x dx x tanh x dx x tanh x ln 1 x C'''" " " " #
##
""
œœ œ
x2x
1x 1x
œœ kk
check: d x tanh x ln 1 x C tanh x dx tanh x dxœ ‘œ ‘kk
" # " ""
#
œ œ
xx
1x 1x
œœ
8.3 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
1. 5x 13 A(x 2) B(x 3) (A B)x (2A 3B)
5x 13 A B
(x 3)(x 2) x 3 x 2
Ê
ÊÊ Ê Ê
œ Êœœ
B (10 13) B 3 A 2; thus,
AB5
2A 3B 13
Êʜʜʜ œ
œ
œ
Ê
5x 13 2 3
(x 3)(x 2) x 3 x
Ê
#

513. Section 8.3 Integration of Rational Functions by Partial Fractions 509
2. 5x 7 A(x 1) B(x 2) (A B)x (A 2B)
5x 7 5x 7 A B
x 3x 2 (x 2)(x 1) x 2 x 1
ÊÊ
Êœ Ê Ê Ê Ê
œ œœÊœœ
B 2 A 3; thus,
AB5
A2B7
Êʜʜ œ
œ
œ
Ê
5x 7 3 2
x 3x 2 x 2 x 1
Ê
Êœ Ê Ê
œ
3. x 4 A(x 1) B Ax (A B) A 1 and B 3;
A1
AB4
x 4 A B
(x 1) x 1 (x 1)
œ
œœœ
œœœ ÊœœÊ Êœ œ
œ
œ
Ê
thus,
x 4 1 3
(x 1) x 1 (x 1)
œ
œœœ
œœœ
4. 2x 2 A(x 1) B Ax ( A B)
A2
AB2
2x 2 2x 2 A B
x 2x 1 (x 1) x 1 (x 1)
œœ
Êœ Ê Ê Ê
œœ œ œœ ÊœœÊ
œ
œ
Ê
A 2 and B 4; thus, Êœ œ œ
2x 2 2 4
x 2x 1 x 1 (x 1)
œ
Êœ Ê Ê
œœ
5. z 1 Az(z 1) B(z 1) Cz z 1 (A C)z ( A B)z B
z 1 A B C
z (z 1) z z z 1
œ
ÊÊ
##
œœ œ Êœ Êœ
B 1 A 2 C 2; thus,
AC0
AB1
B1
Êʜʜʜœ
œ
œ
œ
Þ
ß
à
z 2 2
z (z 1) z z z 1
œ" Ê Ê"
ÊÊ
œœ
6. 1 A(z 2) B(z 3) (A B)z (2A 3B)
zAB
z z 6z z z 6 (z 3)(z 2) z 3 z
!œ œ
#
""
œœ œÊœœ
5B 1 B A ; thus,
AB0
2A 3B 1
Êʜʜʜ œ
œ
œ
Ê
""
ÊÊ Ê œ
Ê
5 5 z z 6z z 3 z 2
z
!œ
""
55
7. 1 (after long division);
t 8 5t 2 5t 2 5t 2 A B
t 5t 6 t 5t 6 t 5t 6 (t 3)(t 2) t 3 t 2
œ
œœ œ
œœ œœ
Êœ Êœ Êœ Ê Ê Ê Ê
œ œ œ
5t 2 A(t 2) B(t 3) (A B)t ( 2A 3B) B (10 2) 12
AB5
2A 3B 2
Êœœ Ê Êœœ
œ
œ
Ê B 12 A 17; thus, 1ʜʜ œ
t 8 17 12
t 5t 6 t 3 t 2
œ
œ
œÊ
Êœ Ê Ê
8. 1 1 (after long division);
t 9 9t 9 9t 9 9t 9 A B Ct D
t 9t t 9t t t 9 t t 9 t t t 9

œœ œœ œ œ
œÊœÊœ Êœ œ
œœ œœœ
œ œ œ
ab ab
9t 9 At t 9 B t 9 (Ct D)t (A C)t (B D)t 9At 9BÊœ œ
### #$#
abab
A 0 C 0; B 1 D 10; thus, 1
AC0
BD 9
9A 0
9B 9
ÊʜʜœÊœœ
œ
œ
œ
œ
Þ
á
á
ß
á
á
à
t 9 10
t 9t tt 9
#
# œ œœ
œ"Ê
œ œ
9. 1 A(1 x) B(1 x); x 1 A ; x 1 B ;
" ""
# #1 x 1 x 1 x
AB
œœ ÊœœÊœœÊœ
ln 1 x ln 1 x C'''dx dx dx
1 x 1 x 1 x###
"" "
œœœ cdkkkk
10. 1 A(x 2) Bx; x 0 A ; x 2 B ;
" ""
# #x 2x x x 2
AB
œ œ Êœ œÊ œ œÊ œ
ln x ln x 2 C'''dx dx dx
x 2x x x 2
œ
# ##
"" "
œ œ cdkk k k
11. x 4 A(x 1) B(x 6); x 1 B ; x 6 A ;
x 4 A B 5 2 2
x 5x 6 x 6 x 1 7 7 7
œ Ê
œÊ œ Ê Ê
œ œ ÊœœÊœœÊœœ
dx ln x 6 ln x 1 C ln (x 6) (x 1) C'''x 4 2 dx 5 dx 2 5
x 5x 6 7 x 6 7 x 1 7 7 7
œ "
œÊ œ Ê
#&
œ œœœ kk kk k k
12. 2x 1 A(x 3) B(x 4); x 3 B 7 ; x 4 A 9;
2x 1 A B 79
x 7x 12 x 4 x 3 1 1
œ
Êœ Ê Ê Ê
œ œ ÊœœÊœœœÊœœ
dx 9 7 9 ln x 4 7 ln x 3 C ln C'''2x 1 dx dx
x 7x 12 x 4 x 3
(x 4)
(x 3)
œ
Êœ Ê Ê
Ê
Ê
œ

œœœ kk kk ¹¹

514. 510 Chapter 8 Techniques of Integration
13. y A(y 1) B(y 3); y 1 B ; y 3 A ;
y
y 2y 3 y 3 y 1 4 4 4
AB 1 3
œ
ÊÊ Ê œ Ê
Ê"
œ ÊœœÊœœœÊœ
ln y 3 ln y 1 ln 5 ln 9 ln 1 ln 5'''
444
888
y dy dy dy
y 2y 3 4 y 3 4 y 1 4 4 4 4 4 4
33 33
œ
ÊÊ Ê œ
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)
%
œœœ œ ‘ˆ ‰ˆ ‰kk kk
ln 5 ln 3œœ
""
## #
ln 15
14. y 4 A(y 1) By; y 0 A 4; y 1 B 3 ;
y 4
y y y y 1 1
AB 3œ
œœ Ê
œœ Ê œ œ Ê œ œ Ê œ œ
dy 4 3 4 ln y 3 ln y 1 (4 ln 1 3 ln 2) 4 ln 3 ln '''
12 12 12
111œœœ
y 4 dy dy
y y y y 1
3œ
œœ ##
"
"Î#
"
œ œ œœ cdkk k k ˆ‰
ln ln ln ln 16 ln œ œ œ
"" "
816 8 88 4
27 27 27
ˆ‰††
15. 1 A(t 2)(t 1) Bt(t 1) Ct(t 2); t 0 A ; t 2
" "
œÊ œ Ê #t t 2t t t2 t1
AB C
%œ œ Êœ œÊ œ œ
B; t1 C; Êœ œÊœ œ
1dtdtdtdt
6 3 tt2t t 6 t2 3 t1
""""
# ''''
!œ
ln t ln t 2 ln t 1 Cœ
"" "
#
kk kk kk
63
16. (x 3) A(x 2)(x 2) Bx(x 2) Cx(x 2); x 0 A ; x 2
x 3 A B C 3
2x 8x x x 2 x 8
œ"
## Ê
% œ ÊœœÊœ œ
B ; x 2 C ; dxÊœ œÊœ œ
"œ"
ÊœÊ16 16 2x 8x 8 x 16 x 2 16 x 2
5 x 3 3 dx dx 5 dx''''
%
ln x ln x 2 ln x 2 C ln Cœœ
35
816 16 16x
(x 2) (x 2)
kk kk kk ¹¹
"" ʜ

w
17. (x 2) (after long division); 3x 2 A(x 1) B
x 3x 2 3x 2 A B
x 2x 1 (x 1) (x 1) x 1 (x 1)
%
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œœ œ œ œ œ
œœ
œ œ Ê œ
Ax (A B) A 3, A B 2 A 3, B 1; œÊœœÊœœ '
0
1
x dx
x 2x 1
%
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œœ
(x 2) dx 3 2x 3 ln x 1œ œ '''
000
111
dx dx x
x 1 (x 1) x 1#
"
"
!
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2 3 ln 2 (1) 3 ln 2 2œ œ ˆ‰
""
##
18. (x 2) (after long division); 3x 2 A(x 1) B
x 3x 2 3x 2 A B
x 2x 1 (x 1) (x 1) x 1 (x 1)
%
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Êœ Ê Ê Ê Ê
ÊÊ
œ œ Ê œ
Ax ( A B) A 3, A B 2 A 3, B 1; œ Ê œ œÊ œ œ '
#1
0
x dx
x 2x 1
%
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ʜ
(x 2) dx 3 2x 3 ln x 1œ œ'''
###111
000
dx dx x
x 1 (x 1) x 1#
"
!
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0 0 3 ln 1 2 3 ln 2 2 3 ln 2œœŠ‹Š‹
"" "
# #(1) ( )
19. 1 A(x 1)(x 1) B(x 1)(x 1) C(x 1) D(x 1) ;
"
Ê œÊ œ Ê
####
abx 1
AB C D
x 1 x 1 (x 1) (x 1)
œ
œ œœ
œ Êœ
x 1 C ; x 1 D ; coefficient of x A B A B 0; constant A B C DœÊ œ œ Ê œ œ Ê œ œ
"" $
44
A B C D 1 A B ; thus, A B ; Ê œ Ê œ œ Ê œ
"" "
# Ê44
dx
x 1 '
ab
œ
œ
ln Cœ œ
""" " "œ
œÊ œ Ê ÊÊ4 x 1 4 x 1 4 (x 1) 4 (x 1) 4 x 1 2 x 1
dx dx dx dx x 1 x''' '
œœ œ
¸¸
ab
20. x A(x 1) B(x 1)(x 1) C(x 1); x 1
xABC
(x 1) x 2x 1 x 1 x 1 (x 1)
œ
œœ
Êœœ Ê œ œ
##
ab
œ Êœœ
C ; x 1 A ; coefficient of x A B A B 1 B ; Êœ œÊœ œÊœÊœ
"" #
Êœœ2 4 4 (x 1) x 2x 1
3x dx '
œ
œ
ab
ln x 1 ln x 1 Cœ œ
""" "
#4 x 1 4 x 1 2 (x 1) 4 4 (x 1)
dx 3 dx dx 3'''
œ kk kk
Cœ
ln (x 1)(x 1)
42(x 1)
kkʜ "
œ
%
21. 1 A x 1 (Bx C)(x 1); x 1 A ; coefficient of x
1ABx C
(x 1) x 1 x 1 x 1œœ œ œ #
œ "# #
ab
œœœ Êœ œÊœ ab
A B A B 0 B ; constant A C A C 1 C ; œ Ê œÊ œ œ Ê œÊ œ
1 dx
2 (x 1) x 1
"
# '
0
1 ab
œ

515. Section 8.3 Integration of Rational Functions by Partial Fractions 511
dx ln x 1 ln x 1 tan xœ œ
"" " " "
# # #
ʜ #"
"
!2x 1 x 1 4
dx (x 1)''
00
11
œ
œ ‘kk a b
ln 2 ln 2 tan 1 ln 1 ln 1 tan 0 ln 2œ œœˆ‰ˆ‰ˆ‰
""" """ ""
#### #
Ê" Ê" œ
44448
( 2 ln 2)1 1
22. 3t t 4 A t 1 (Bt C)t; t 0 A 4; coefficient of t
3t t 4 A Bt C
t t t t 1
œ
!œ
œœ œ
œœ
## #
œ Ê œ œÊ œ ab
A B A B 3 B 1; coefficient of t C C 1; dtœ Ê œÊ œ œ Ê œ '
1
3
œ
3t t 4
t 1
œ
%
œœ
œ
4 dt 4 ln t ln t 1 tan tœ œ''
11
33
œœ
dt
tt 1 2
(t 1)ʜ
œ
" #"
$
"
œ
œ ‘kk a b
È
4 ln 3 ln 4 tan 3 4 ln 1 ln 2 tan 1 2 ln 3 ln 2 ln 2œ œŠ‹
ÈÈ ˆ‰
"" "
## #
Ê" Ê" 11
34
2 ln 3 ln 2 lnœœ
"
## #
11
11
9
2
Š‹
È
23. y 2y 1 (Ay B) y 1 Cy D
y 2y 1 Ay B Cy D
y 1 y 1 y 1
œ
œœ
œœ œ
œœ œ œ
œœ œ
##
ab ab
œ Êœ ab
Ay By (A C)y (B D) A 0, B 1; A C 2 C 2; B D 1 D 0;œ Ê œ œ œÊ œ œÊ œ
$#
dy dy 2 dy tan y C'''
y 2y 1 y
y 1 y 1 y 1 y 1
œ
œœ
œœ œœ
œœ
œœ
""
œœ
Ê"
ab ab
œ œ
24. 8x 8x 2 (Ax B) 4x 1 Cx D
8x 8x 2 Ax B Cx D
4x 1 4x 14x 1
œ
œœ
œœ œ
œœ œ œ
œœ œ
##
ab ab
œ Êœ ab
4Ax 4Bx (A C)x (B D); A 0, B 2; A C 8 C 8; B D 2 D 0;œ œœœÊœœÊœ
$#
dx 2 8 tan 2x C'''8x 8x 2 dx x dx
4x 1 4x 1 4x 1 4x 1
œ
œœ
œœ œœ
œœ "
œœ œœ
Ê"
ab ab
œ œ
25. 2s 2
2s 2 As B C D E
s 1 (s 1) s 1 s 1 (s 1) (s 1)
œœ
œÊ œ Ê Ê Êab
œ !œ œ ! œ Ê
(As B)(s 1) C s 1 (s 1) D s 1 (s 1) E s 1œ
$# ## #
ab ab ab
As ( 3A B)s (3A 3B)s ( A 3B)s B C s 2s 2s 2s 1 D s s s 1œ cdabab
% $ # %$# $#
Es 1##ab
#
(AC)s (3AB2CD)s (3A3B2CDE)s (A3B2CD)s(BCDE)œ
%$ #
summing all eq
A C 0
3A B 2C D 0
3A 3B 2C D E 0
A3B2CD 2
B C D E 2
Ê
œ
œ
œ
œ
œ
Þ
á
á
á
á
ß
á
á
á
á
à
uations 2E 4 E 2;ʜʜ
summing eqs (2) and (3) 2B 2 0 B 1; summing eqs (3) and (4) 2A 2 2 A 0; C 0Ê œ Ê œ Ê œ Ê œ œ
from eq (1); then 1 0 D 2 2 from eq (5) D 1; œ Ê œ
ds 2 (s1) (s1) tansC''''2s 2 ds ds ds
s 1 (s 1) s 1 (s 1) (s 1)
œ
œÊ œ Ê Ê
# " "
ab
œ !œ œ ! œ œ
26. s 81 A s 9 (Bs C)s s 9 (Ds E)s
s 81 A Bs C Ds E
ss 9 s 9ss 9
#
œœ
œœ œ
œœœ
œœ œ
%# #
#
ab ab
œ Ê œ ab ab
A s 18s 81 Bs Cs 9Bs 9Cs Ds Esœaba b
%# %$ # #
(A B)s Cs (18A 9B D)s (9C E)s 81A 81A 81 or A 1; A B 1 B 0;œ Ê œ œ œÊœ
%$ #
C 0; 9C E 0 E 0; 18A 9B D 0 D 18; ds 18œœÊœ œÊœ œ '''s81 ds s ds
ss 9 s 9 s
#
œœ
œœ
œ
œœab ab ln s Cœ kk
9
s 9ab
œ
œ
27. 2 5 8 4 (A B) 2 2 C D
2 5 8 4 A B C D
2 2 2 2 2
))) ) )
)) )) ))
% œ
œœ
œœ œ
œœœ œ œ
œœ œœ #
$# #
ab ab
œ Ê œ ))) ) )) ) ab
A (2A B) (2A 2B C) (2B D) A 2; 2A B 5 B 1; 2A 2B C 8 C 2;œ Êœ œÊœ œÊœ)) )
$#
2B D 4 D 2; d d dœÊ œ œ '''2 5 8 4 2 1 2 2
2 2 2 2 2 2
))) ) )
)) )) ))
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œ œ
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œœœ œ œ
œœ œœ œœab ab ab
)) ) dœ œ'''''2 2 d d
2 2 2 2 ( 1) 1 2
d 2 2 d 2 2
2 2
)) )
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# # #
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abab
ab

516. 512 Chapter 8 Techniques of Integration
ln 2 2 tan ( 1) Cœ
Ê"
#
#"
))
œ
2
ab)) )
28. 4 2 3 1
)))) ) ) )
))) )
# !œ
œœœ
!œ! œ
4 2 3 1 A B C D E F
1 1 1 1
ʜʜ œ œ œ
œœœ œ
%$#
ab ab ab
œ Ê ))))
(A B) 1 (C D) 1 E F (A B) 2 1 C D C D E Fœ œ )) )) ) ))) ))) )ab ab a ba b
## %#$#
#
A B 2A 2B A B C D C D E Fœabab)) ) )) ))) )
&% $ # $#
A B (2A C) (2B D) (A C E) (B D F) A 0; B 1; 2A C 4œ Êœ œ œ)))) )
&%$#
C 4; 2B D 2 D 0; A C E 3 E 1; B D F 1 F 0;Êœ œÊœ œÊœ œÊœ
d4 tan211C'''')))) ) )) ))
))) )
# !œ
œœœ
!œ! œ
4 2 3 1 d d d
1 1 1 1 4
ʜʜ "
œœœ œ
" # #
" #
ab ab ab
))))œ œ ab ab
29. 2x 2x ; 1 A(x 1) Bx; x 0 A 1;
2x 2x 1 A B
x x x x x(x 1) x(x 1) x x 1
% œ
œœ
ʜ " " "
ÊÊÊÊÊ
œ œ œ Êœ œÊœ
x 1 B 1; 2x dx x ln x ln x 1 C x ln CœÊ œ œ œ œ ''''2x 2x 1 dx dx x 1
x x x x 1 x
!œ
œ
Êœ Ê
ÊÊ
##
kk k k ¸¸
30. x 1 x 1 ; 1 A(x 1) B(x 1);
x AB
x 1 x 1 (x 1)(x 1) (x 1)(x 1) x 1 x 1
#
œœ
Ê Ê œÊ œÊ œ Ê
## """
œ œ œ Êœ ab ab
x 1 A ; x 1 B ; dx x 1 dxœ Ê œ œ Ê œ œ
"" ""
## ##
# '' ''xdxdx
x 1 x 1 x 1
#
œ
ab x x ln x 1 ln x 1 C x ln Cœœ
"" " "Ê$
## #33x 1
xx 1
kk kk ¸¸
%
31. 9 (after long division);
9x 3x 1 9x 3x 9x 3x 1 A B C
x x x (x 1) x (x 1) x x x 1
% œœ
!œ œ œ œ
ʜ ʜ" ʜ
ÊÊ Ê Ê
œ œ
9x 3x 1 Ax(x 1) B(x 1) Cx ; x 1 C 7; x 0 B 1; A C 9 A 2;Ê œ œÊ œ œÊ œ œÊ œ
##
dx 9 dx 2 7 9x 2 ln x 7 ln x 1 C'''''9x 3x 1 dx dx dx
x x x x x 1 x
%
% œœ
ʜ "
ÊÊ
œ œ kk k k
32. (4x 4) ; 12x 4 A(2x 1) B
16x 12x 4 12x 4 A B
4x 4x 1 4x 4x 1 (2x 1) 2x 1 (2x 1)
%
œœœœ
Êœ Êœ Ê Ê Ê
ÊÊ
œ œ Ê œ
A 6; A B 4 B 2; dx 4 (x 1) dx 6 2ÊœœÊœ œ ''''16x dx dx
4x 4x 1 2x 1 (2x 1)
%
œ œ
Êœ Ê Ê 2(x 1) 3 ln 2x 1 C 2x 4x 3 ln 2x 1 (2x 1) C, where C 2 Cœœ œ
##" "
Ê
" "
kk kk
2x 1
33. y ; 1 A y 1 (By C)y (A B)y Cy A
y y 1 By C
y y y y 1 y y 1 y y 1
A
# œ
!œœœ
œÊ œ
œœœœ
"" ##
œ œ Ê œ œ
abab
ab
A 1; A B 0 B 1; C 0; dy y dyÊœ œÊœ œ œ ''''
y y 1 dy y dy
y y y y 1
# œ
!œ
œÊ
œœ ln y ln 1 y Cœ
y
œ
##
" #
kk a b
34. 2y 2 ;
2y By C
y y y 1 y y y 1 y y y 1 y 1 (y 1) y 1 y 1
22 2A
#
!œ !œ !œ œ œ
ÊœÊ ÊœÊ ÊœÊ œ Ê Ê œ
œ
œ œ œ
ab
2 A y 1 (By C)(y 1) Ay A By Cy By C (A B)y ( B C)y (A C)Êœ œ œ ab a ba b
####
A B 0, B C 0 or C B, A C A B 2 A 1, B 1, C 1;Êœœ œ œœÊœ œ œ
dy 2 (y 1) dy dy'''''
2y dy y dy
y y y 1 y 1 y 1 y 1
#
!œ œœ
ÊœÊ Ê œ œ
œ (y 1) ln y 1 ln y 1 tan y C y 2y ln y 1 ln y 1 tan y C,œ œ
# # " # # " ""
##
"
kk a b kk a b
where C C 1œ
"
35. e y ln C ln C''''e dt e 1
e 3e 2 y 3y 2 y 1 y 2 y 2 e
dy dy dy y 1
t t
2t t t
#
œ œ
œœ œœ œ cd ¹¹ Š ‹
t
œ
36. dt e dt; dy y dy
ye
dy e dt'' '' 'e 2e e e 2e 1
e 1 e 1 y 1 y 1 2 y 1
y 2y 1 y 1 y y
4t 2t t 3t t
2t 2t
32
œÊ œÊ
œœ œ œ œ
œÊ Ê
œÄœœ
œ
œ
t
t
t
”• Š‹ œœœ
dyÎ'
dy
y 1
œ
œ
ln y 1 tan y C e ln e 1 tan e Cœ œ
y
22
2t 2t t
2
"""
##
#" "
ab a b ab

517. Section 8.3 Integration of Rational Functions by Partial Fractions 513
37. ; [sin y t, cos y dy dt] dt ln C'''
cos y dy dy
sin y sin y 6 t t 6 5 t 2 t 3 5 t 3
t 2
œ œ
œÊ œÊ Ê œ œ
"" " "Ê
œœÄ œœ ˆ‰¸¸
ln Cœ
" Ê
œ5sin y 3
sin y 2
¹¹
38. ; cos y ln C ln ''''sin d cos 2
cos cos 2 y y 2 3 y 2 3 y 1 3 y 1 3 cos 1
dy dy dy y 2)) )
)) )
œœ
œÊ œÊ œ Ê Ê Ê
"" " "œ œ
cd ¹¹ ¸¸)œÄ œœ œ C
ln C ln Cœœ
"œ " Ê
ʜ31 cos 3cos 2
2 cos cos 1
¸¸ ¸¸
))
))
39. dx dx 3 dx'''
(x 2) tan (2x) 12x 3x tan (2x)
4x 1 (x 2) 4x 1 (x 2)
xÊÊÊ
œÊ œ Ê
œ !" % !"
œœ œ œ
ab
œ
tan (2x) d tan (2x) 3 6 3 ln x 2 Cœœ
"
#
Ê" Ê"''' ab kk
dx dx 6
x 2 (x 2) 4 x 2
tan 2x
œ
!"
œ
ab
40. dx dx dx'''
(x 1) tan (3x) 9x x tan (3x)
9x 1 (x 1) 9x 1 (x 1)
xœœœ
œœ œ œ
œ !" % !"
œœ œ œ
ab
œ
tan (3x) d tan (3x) ln x 1 Cœœ
" Ê" Ê"
œœ œ3 x 1 (x 1) 6 x 1
dx dx 1 tan 3x''' ab kk
œ
!"
œ
ab
41. t 3t 2 1; x ln C; Ce ; t 3 and x 0ab ¸¸
#
Êœ Ê Ê Ê Ê
ÊÊ
œœ œœ œœ œ
dx dt dt dt t 2 t 2
dt t 3t 2 t 2 t 1 t 1 t 1
x '''
œ
C e x ln 2 ln t 2 ln t 1 ln 2ÊœÊ œ Êœ œ
"Ê" Ê
##
t 2 t 2
t 1 t 1
x
¸¸ˆ‰ kk kk
42. 3t 4t 1 2 3; x 2 3 3 3ab
ÈÈ È È
%#
œœ œ œ
œ œ œ
dx dt dt dt
dt t 13t 4t 1 t '''
# œ œ " œ
3 3 tan 3t 3 tan t C; t 1 and x C CœœœÊœÊœ
Ê" Ê" Ê
Š‹
ÈÈ
11
ÈÈ È
33 3
444
11 1
x 3 tan 3t 3 tan tʜ
Ê" Ê"
Š‹
ÈÈ
1
43. t 2t 2x 2; ln x 1 ln x 1 ln C;a b kk kk '' ¸¸
# """"
# # # # #
œ œ Ê œ Êœ
dx dx dt dt dt t
dt x 1 t 2t t t 2 t ''
œ t 1 and x 1 ln 2 ln C C ln 2 ln 3 ln 6 ln x 1 ln 6 x 1œ œÊ œÊœœ Ê œ Êœ
"
#3 t 2 t
t6t
kk ¸¸
x 1, t 0ʜ
6t
t #
44. (t 1) x 1 tan x ln t 1 C; t 0 and x tan ln 1 CœÊ œ Ê œœ œÊ œ
dx dx dt
dt x 1 t 1 4 4
#" "
œœ ''
œ kk kk
11
C tan 1 tan x ln t 1 1 x tan (ln (t 1) 1), t 1Êœ œÊ œ Êœ
Ê" Ê"1
4
kk
45. V y dx dx 3 dx 3 ln 3 ln 25œœ œœ œ11 1 1 1'' '
05 05 05
25 25 25** *
** *
#
ÊÊ Ê
""
#Þ&
!Þ&
9x
3x x x 3 x x 3
œ Œ Î
ˆ‰ œ ‘¸¸
46. V 2 xy dx 2 dx 4 dxœœ œ11 1 '' '
00 0
11 1
2x 2
(x 1)(2 x) 3 x 1 3 2 xœÊ œ Ê
"" "
ˆ‰ˆ‰ ˆ‰
ln x 1 2 ln 2 x (ln 2)œ œœ ‘abkk kk
44
33
11
"
!
47. A tan x dx x tan x dxœœ''
00
33
3
0
œœ
œ
Ê" Ê"
œ
cd
x
1 x
œ ln x 1 ln 2;œ œ
11
ÈÈ
33
33
œ ‘ab
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#
#
œ
3
0
x x tan x dxœ
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A'
0
3
œ
x tan x dxœ
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##
#"
A1 x
x
Œ Î
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œ
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3
0 0
3
'
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xtanxœ
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1
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3
0
1.10œœ œ
µ
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33 2
Š‹Š‹
111
ÈÈ

518. 514 Chapter 8 Techniques of Integration
48. A dx 3 2 3 ln x ln x 3 2 ln x 1 ln ;œœœœ''''
3333
5555
4x 13x 9 dx dx dx 125
x 2x 3x x x 3 x 1 9
œ
!œ
œÊ
œÊ œ Ê
&
$
cdkkkk kk
x dx 4x 3 2 (8 11 ln 2 3 ln 6) 3.90œœœœ
µ
"" " œÊ  
œÊ œ Ê $A x 2x 3x A x 3 x 1 A
x 4x 13x 9 dx dx'''
333
555
ab
œ
!œ
Š‹cd
49. (a) kx(N x) k dt k dt ln kt C;
dx dx dx dx x
dt x(N x) N x N N x N N x
œÊ œ Ê œ Ê œ'' '''
ÊÊÊ
"" "
¸¸
k , N 1000, t 0 and x 2 ln C ln lnœœ œ œÊ œÊ œ
"""""
Ê250 1000 998 1000 1000 x 250 1000 499
2xt
¸¸ ¸ ¸ ˆ‰
ln 4t e 499x e (1000 x) 499 e x 1000e xʜʜʜʜʜ¸¸ ab
499x 499x 1000e
1000 x 1000 x 499 e
4t 4t 4t 4t
ÊÊ œ
4t
4t
(b) x N 500 500 500 499 500e 1000e e 499 t ln 499 1.55 daysœœ Ê œ Ê œ Êœ Êœ ¸
" "
#
1000e
499 e 4
4t 4t 4t
4t
4t
†
50. k(a x)(b x) k dt
dx dx
dt (a x)(b x)
œ Ê œ
ÊÊ
(a) a b: k dt kt C; t 0 and x 0 C ktœœÊœœœÊœÊœ''dx
(a x) a x a a x aÊÊ Ê
""""
œ
ax xaÊœ Êœ Êœ œ
"œ"
Êœœœa x a akt 1 akt 1 akt 1
akt a a a kt
œ
(b) a b: k dt k dt ln kt C;ÁœÊœÊœ'''''dx dx dx b x
(a x)(b x) b a a x b a b x b a a xÊÊ Ê Ê Ê Ê Ê Ê
"" "Ê
¸¸
t 0 and x 0 ln C ln (b a)kt ln eœœÊ œÊ œÊœ
"Ê Ê
ÊÊ Ê
µÊ¸
ba a ax a ax a
bbx bbxb bakt
¸¸ ˆ‰
x ʜ
ab 1 e
abe
œ ‘Ê
Ê

bakt
bakt
51. (a) dx x 4x 5x 4x 4 dx''
00
11
x(x 1)
x1 x1 7
422
##
œ œ
Ê
œœ
'&%#
œ œˆ‰ 1
(b) 100% 0.04%
22
7
Ê1
1
† œ
µ
(c) The area is less than 0.003

52. P(x) ax bx c, P(0) c 1 and P (0) 0 b 0 P(x) ax 1. Next,œ œœ œÊœÊ œ
#w #
; for the integral to be a function, we must have A 0 and
ax 1 A B C D E
x(x1) x x x x1 (x1)
œ
!œ œ! œ
œ
ÊÊÊ
œ œ rational
D 0. Thus, ax 1 Bx(x 1) C(x 1) Ex (B E)x (C 2B)x (B 2C)x Cœœœ
###$$#
E B; x 1 a 1 E; therefore, 1 2B a 1 2E a 1 2(a 1) a
BE0
C2Ba
C
ÊÊœœÊœœÊœÊœ
œ
œ
œ"
Þ
ß
à
a3ʜ

519. Section 8.4 Trigonometric Integrals 515
8.4 TRIGONOMETRIC INTEGRALS
1. sin x dx sin x sin x dx cos x sin x dx 2cos x cos x sin x dx
'' ' '
00 0 0
11 1 1/2 /2 /2 /2
5 4
œœ"œ"ab a b a b
## #
##
sin x dx 2cos x sin x dx cos x sin x dx cos xœ œ#'' '
00 0
0
11 1 1/2 /2 /2
4 cos x cos x
35
/2
#
’“
35
œ!" œabˆ‰
#" )
$& "&
2. sin dx (using Exercise 1) sin dx 2cos sin dx cos sin dx''''
0000
1111
5 4x x xx xx
2 2 22 22
ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰œ
#
cos cos cosœ# œ!# œœ ‘ˆ ‰ˆ‰ ˆ‰ ˆ‰ ab
xxx
23 25 2
35%# %#"'
$& "&
1
0
3. cos x dx cos x cos x dx sin x cos x dx cos x dx sin x cos x dx'' ' ''
ÊÊ Ê ÊÊ
## #
11 1 11
11 1 11
/2 /2 /2 /2 /2
/2 /2 /2 /2 /2
3
œœ"œab a b
sin xœ œ""œ’“ ˆ‰ˆ ‰
sin x
3
3
1
1
Î#
Î#
""%
$$$
4. 3cos 3x dx cos 3x cos 3x 3dx sin 3x cos 3x 3dx sin 3x sin 3x cos 3x 3dx'' ' '
00 0 0
/6 /6 /6 /6
5
11 1 1
œ †œ" †œ"# †ab a b a b
## #%
##
cos 3x 3dx sin 3x cos 3x 3dx sin 3x cos 3x 3dx sin 3xœ † # † † œ # '' '
00 0
/6 /6 /6
sin 3x sin 3x
35
611 1 1
#%
Î
’“
35
0
œ" !œˆ‰ ab
2
$& "&
")
5. sin y dy sin y sin y dy cos y sin y dy sin y dy cos y sin y dy'' ' ''
00 0 0 0
/2 /2 /2 /2 /2
76 2 2
11 1 1 1
œœ"œ$ ab
$
cos y sin y dy cos y sin y dy cos y$ œ $$ œ!""œ''
00
/2 /2
46 cos y cos y cos y
35
11 1
’“ abˆ‰
3
(&($&
Î#
$" "'
0
6. 7cos t dt (using Exercise 5) 7 cos t dt sin t cos t dt sin t cos t dt sin t cos t dt'''''
00000
/2 /2 /2 /2 /2
7246
11111
œ$$”•
7sin t 7 7œ $$ œ""!œ’“ ˆ‰ ab
sin t sin t sin t
35
3
(&(&
Î#
$" "'
1
0
7. sin x dx dx cos x cos x dx dx cos x dx dx'' ' '' '
00 0 00 0
11 1 11 1
) œ) œ# "# # # œ# # # †# #
4 cos x cos 4x ˆ‰ ab
" # "œ
# #
#
#
x sin x dx cos 4x dx x sin 4xœ## # œ# œ# œ$cd œ ‘
1
11
1
0
00 0
'' 1111
"
#
8. cos 2 x dx dx cos 4 x cos 4 x dx dx cos 4 x dx dx'' ' '''
00 0 000
11 1 111
)œ) œ#"# œ#% #
4 cos 4 x cos x
1111 ˆ‰ ab
"œ "œ '
# #
#
#1 1
x sin 4 x dx cos x dx x sin xœ# ) œ# ) œ#"œ$œ ‘ œ ‘
""
""
)11
11 1
0 00
11
0
''
9. 16 sin x cos x dx 16 dx cos x dx dx dx'' '''
ÊÊ ÊÊÊ
## #" # " # "
## #11 111
11 111
/4 /4 /4 /4 /4
/4 /4 /4 /4 /4
cos x cos x cos 4x
œœ%"#œ%%ˆ‰ˆ‰ ˆ‰ ab
4x dx cos 4x dx 2x =2œ## œ œcd œ ‘ˆ‰ ˆ‰
1
1
11
11
1
1
11/4
/4
/4 /4
/4 /4
sin 4x
2
/4
/4Ê
ÊÊ Ê ##'' 11 1 1
10. 8 sin y cos y dy 8 dy dy cos y dy cos y dy cos y dy'' ''''
00 0000
11 1111
4 cos y cos y# #$" # " #
##
#
œ œ###ˆ‰ˆ‰
y sin 2y dy sin y cos y dy dy cos 4y dy cos y dyœ " # # œ #œ ‘ˆ‰ ab
" """œ
###
#
2
cos 4y1
11 111
0 00 000
'' ''' 1
sin y cos y dy y sin 4y sin 2y##œ† œœ'
0
0
1 1
11# """"
#)## ##
11’“
sin 2y
3
3

520. 516 Chapter 8 Techniques of Integration
11. 35 sin x cos x dx 35 sin x sin x cos x dx 35 sin x cos x dx 35 sin x cos x dx'' ''
00 00
11 11/2 /2 /2 /2
43 4 4 6
œ"œ ab
#
35 35 7 5œ œ!œ#’“ abab
sin x sin x
57
/2
57
1
0
12. cos 2x sin 2x dx'
0
0
1 1
2 cos 2x
3
œ œ œ!’“
"""
#''
3
13. 8cos 2 sin 2 d 8 cos 2'
0
0
0
1
)
1
1
/4
34 cos 2
4
/4
/4
))) ) œ œ œ!"œ"’“ˆ‰ cdabab
"
#
4
14. sin 2 cos 2 d sin 2 sin 2 cos 2 d sin 2 cos 2 d sin 2 cos 2 d'' ''
00 00
11 11/2 /2 /2 /2
23 2 2 2 4
)))))))))))))œ" œ ab
œ† † œ!’“
""
##
sin 2 sin 2
35
/2
35
))
1
0
15. dx sin dx sin dx cos '''
000 0
222
cos xxxx
2
111
1
É ¹¹ œ ‘
"Ê
####
œœœ#œ##œ%
16. cos 2x dx sin 2x dx sin 2x dx cos 2x'''
000
0
111 1
ÈÈÈÈÈÈÈ
’“" œ#llœ# œ# œ##œ##
17. sin t dt cos t dt cos t dt cos t dt sin t sin t''''
000
0
1111
1
11
1
È
cd cd" œ l l œ œ œ"!!"œ##
/2
/2
/2
/2
18. cos d sin d sin d cos '''
000
0
111
1
È
cd" œ l l œ œ œ""œ##
)) ) ) )) )
19. tan x dx sec x dx sec x dx ln sec x tan x ln ln'''
ÊÊÊ
#
Ê
111
111
1
1
/4 /4 /4
/4 /4 /4
/4
/4
È
cd Š‹Š‹
ÈÈ
" œ l l œ œ l l œ #" #"
ln lnœœ#"#Š‹ Š ‹
È
È
È
#"
#"
20. sec x dx tan x dx tan x dx tan x dx ln sec x ln sec x ''''
!
#
!
!
!
111
11 1
1
/4 /4 /4
/4 /4 /4
/4
È
cdcd"œllœ œllll
!#/4
ln ln ln ln ln lnœ " # # " œ# #œ #ab ab
ÈÈ È
21. cos 2 d sin d sin d cos sin '' '
00 0
0
11 1
1
/2 /2 /2
/2
)))))) ))))))
È ÈÈ È ÈÈ
cdab" œ #l l œ# œ# œ#"œ#
22. cos t dt sin t dt sin t dt sin t dt sin t dt cos t sin t dt
''''''
!
## $$$ #
$Î# $Î#
!!
1111 1
111 1
ab ab ab ¸¸" œ œ œ œ " cos t sin t dt sin t dt cos t sin t dt sin t dt cos t sin t dt cos t" œ œ '''''
!!!
###
!! !
Ê
1 11
11
1
ab ’“
cos t
3
3
cos t œ " " " " œ’“ ˆ‰ˆ‰
cos t
3
3
1
!
"" "")
$$ $$$
23. 2 sec x dx; u sec x, du sec x tan x dx, dv sec x dx, v tan x;'
Ê
!
$#
1/3
œœ œ œ
2 sec x dx 2 sec x tan x sec x tan x dx sec x sec x dx'' '
ÊÊ Ê
!! !
$ !
Î$
11 1
1
/3 /3 /3
22
œ # œ#†"†!#†#† $# "cd ab È
sec x dx sec x dx; 2 2 sec x dx ln sec x + tan xœ% $# # œ% $ # l lÈÈ cd '''
ÊÊÊ
!!!
$$ !
Î$
111
1
/3 /3 /3
2 2 sec x dx ln + ln ln '
Ê
!
$
1/3
œ% $# l" !l# l# $lœ% $# # $ÈÈÈÈ
Š‹
2 sec x dx ln '
Ê
!
$
1/3
œ# $ # $ÈÈ
Š‹

521. Section 8.4 Trigonometric Integrals 517
24. e sec e dx; u sec e , du sec e tan e e dx, dv sec e e dx, v tan e .'x x x x xx xx x$ #
ab ab ab ab ab abœœ œ œ
e sec e dx sec e tan e sec e tan e e dx
''xx xx x xx$#
ab ab ab ab abœ
sec e tan e sec e sec e e dxœ "ab ab aba b ab
xx x x x' #
sec e tan e sec e e dx sec e e dxœ ab ab ab ab
x x xx xx''$
e sec e dx sec e tan e ln sec e tan e C#œ'xx xx x x$
ab ab ab ab ab ¸¸
e sec e dx sec e tan e ln sec e tan e C'xx xx x x$ "
#
ab ab ab ab abˆ‰ ¸¸œ
25. sec d tan sec d sec d tan sec d tan '' ''
!! ! !
Î
!
11 1 1
)
1/4 /4 /4 /4
422222 tan
3
4
)) ) )) )) ) )) )œ" œ œab ’“
3
œ" !œˆ‰ ab
"%
$$
26. 3sec 3x dx tan 3x sec 3x 3dx sec 3x 3dx tan 3x sec 3x 3dx'' ''
!! !!
11 11/12 /12 / /12
422222
ab a b ab ab ab ababœ" œ
tan 3xœ œ"!œ’“ab ab ˆ‰
tan 3x
3
12
3
ab
1Î
!
"%
$$
27. csc d cot csc d csc d cot csc d cot'' ''
11 11
11 11
)
1
1
/4 /4 /4 /4
/2 /2 /2 /2
4 cot
3
2
/4
)) ) )) )) ) )) )œ" œ œab ’“
## # ##
Î
%
œ!" œabˆ‰
"%
$$
28. csc d cot csc d csc d cot csc d cot'' ''
11 11
11 11
)))))))
1
1
/2 /2 /2 /2
4
cot
3
2
$œ$" œ$ $ œ''
#######
## # ##
Î
)))) ˆ‰ ’“
%
œ
œ
œ '†!#†! '†"#†" œ)abab
29. 4 tan x dx 4 sec x tan x dx 4 sec x tan x dx 4 tan x dx ln sec x'' ' '
00 0 0
/4 /4 /4 /4
3 tan x
411 1 1 1
œ"œ œ%%llab ’“
##
#
Î
!
œ
2ln ln lnœ " % ##†!% "œ## #ab
È
30. 6 tan x dx 6 sec x tan x dx 6 sec x tan x dx 6 tan x dx
'' ' '
ÊÊ Ê Ê
##
11 1 1
11 1 1
/4 /4 /4 /4
/4 /4 /4 /4
4222
œ"œ ab
6 sec x tan x dx 6 sec x 1 dx 6 sec x dx 6 dxœœ''' ''
ÊÊ ÊÊ $
Î
Î
#
11 11
11 11 1
1
/4 /4 /4 /4
/4 /4 /4 /4
22 2 tan x
4
4
ab ’“
%
tan x xœ#" " ' ' œ%'" " œ$ )abcdcd abab ab
11
11
11ÎÎ
Î Î
$$
##
44
44
1
31. cot x dx csc x cot x dx csc x cot x dx cot x dx ln csc x'' ' '
11 1 1
11 1 1 1
1
/6 /6 /6 /6
/3 /3 /3 /3
32 2 cot x
3
6
œ"œ œllab ’“
œ
#
Î
Î
ln ln lnœ $ # œ $
"" # %
#$ $ $
ˆ‰ Š‹
È
È
32. 8 cot t dt 8 csc t cot t dt 8 csc t cot t dt 8 cot t dt'' ' '
11 1 1
11 1 1
/4 /4 /4 /4
/2 /2 /2 /2
422222
œ"œ ab
8 8 csc t dt cot t tœ " œ !" ) ) œ )!" % # œ# ’“ a b abcd cd ab
cot t
33
2
4
/4
/2
2 22
44
3
1
1
1
1
11
11
Î
Î
))"' ÎÎ
ÎÎ $$
' 111
33. sin 3x cos 2x dx sin x sin 5x dx cos x cos 5x''
ÊÊ
!!
""""""'
##&#&&&
!
Ê11 1
œœœ""œab œ ‘ˆ ‰
34. sin 2x cos 3x dx sin x sin 5x dx cos x cos 5x''
!!
ÎÎ
""""""#
##&##&&
Î
!
11
1
22
2
œœ œ!"œabab abab œ ‘ˆ‰

522. 518 Chapter 8 Techniques of Integration
35. sin 3x sin 3x dx cos cos 6x dx dx cos 6x dx x sin 6x'' ''
ÊÊ ÊÊ
"""""
####"### Ê11 11
11 11
1
1
11
œ!œ œ œ!œab œ ‘ 1
36. sin x cos x dx sin sin 2x dx sin 2x dx cos 2x'' '
!! !
ÎÎ Î
"""""
##%%#
Î#
!
11 1
1
22 2
œ ! œ œ œ "" œab cdab
37. cos 3x cos 4x dx cos x cos 7x dx sin x sin 7x 0 0''
!!
""""
#( !
11
1
œ œœœabababab œ ‘
22
38. cos 7x cos x dx cos 6x cos 8x dx sin 6x sin 8x 0''
Î Î
ÎÎ
""""
#
Î
Î11
11
1
122
22
26 8
2
2
œœœab œ ‘
39. x t t x ; y y ; t x 2 ;œ Ê œ œ Ê œ !Ÿ Ÿ#Ê!Ÿ Ÿ
#Î$ # $ #Î$
##
tx
œ %
A x dx; u du u
ux
du 9x dx
œ# " Ä "œ†"
œ
œ ''
!! #% $
*#% %
%
$
*Ð Ñ
$Î#
*Ð Ñ
!
22
x
9
99
2
œ)% œ'!
%
œ'!
1Š‹ ’ “É ”•
È ab
11
2œ "* "
#
#(
#Î$
$Î#
1
’“ˆ‰ˆ‰
40. y ln cos x ; y tan x; y tan x; tan x dx sec x dx ln sec x tan xœœœœ "œllœllab ab c d
È
ww#Ê #
!!
ÎÎ
#
!
sin x
cos x
33
/3 ''
11
1
ln 2 ln ln 2œ$"!œ$Š‹ Š‹
ÈÈ ab
41. y ln sec x ; y tan x; y tan x; tan x dx sec x dx ln sec x tan xœœœœ "œllœllab ab c d
È
ww# #
!!
ÎÎ
#
!
sec x tan x
sec x
44
/4 ''
11
1
ln ln lnœ #" !" œ #"Š‹ Š‹
ÈÈ
ab
42. M sec x dx ln sec x tan x ln ln lnœœllœ#"l#"lœ'
Î
Î
Î
#"
#"1
1
1
1
4
4
/4
4
cd Š‹
ÈÈ
È
È
y dx tan xœœœ""œ
""""
Î
Î
#
Î
Î
ln ln ln ln
œœœœ
œœœœ
#Î$ #Î$ #Î$ #Î$

'
1
1
1
14
4
sec x 4
4
œ
##
cd a b ab
x, y lnÊœ!ßab Œ Ί‹
È
È
#"
#"
Ê"
43. V sin x dx dx dx cos 2x dx x sin 2xœœ œ œœ!!!œ11 1'' ''
!! !!
# "Ê
## # # # !!
11 11
11 11 1 1 1 11cos 2x
244
cd c d a b a b
œ
44. A cos 4x dx cos 2x dx cos 2x dx cos 2x dx cos 2x dxœ" œ#llœ# # #'' '''
!! !Î%$Î%
Î% $ Î%11 111
11
È ÈÈ È È
sin 2x sin 2x sin 2xœ œ "! "" !" œ # #œ# #
ÈÈ È ÈÈÈ
## # ###
## # ###
Î% $ Î%
! Î% $Î%

cd cd cd ab a b ab
ÈÈ È
11
1
1
1
45. (a) m n m n and m n sin mx sin nx dx cos m n x cos m n x dx
##
# #
"
#
ÁÊÁ! Á!Ê œ ''
kk
kk11
cdab ab
sinm nx sinm nxœ
"" "
#
#
œ ‘ab ab
mn mn
k1
k
sin m n k sin m n k sin m n k sin m n kœ # #
"" " "" "
# #
ˆ‰ˆ‰abababababa b aba b ab ab
mn mn mn mn
11
sin m n k sin m n k sin m n k sin m n kœœ!
""""
# # # #ab ab ab abmn mn mn mn
ab ab ab abab ab ab ab
sin mx and sin nx are orthogonal.Ê
(b) Same as part since cos dx . m n m n and m n cos mx cos nx dx
"
#
# #
##''
k k
k k1 1
!œ ÁÊÁ! Á!Ê1
cosm nx cosm nxdx sinm nx sinm nxœœ
""""
##
#
#'
k
k
mn mn
k
1
1
c d ab abab ab œ ‘
k
sin m n k sin m n k sin m n k sin m n kœ##
""""
# # # #ab ab ab abmn mn mn mn
abababababa b aba b ab ab11
sin m n k sin m n k sin m n k sin m n kœœ!
""""
# # # #ab ab ab abmn mn mn mn
ab ab ab abab ab ab ab

523. Section 8.5 Trigonometric Substitutions 519
cos mx and cos nx are orthogonal.Ê
(c) Let m n sin mx cos nx sin sin m n x and sin dx and sin m n x dx 0œÊ œ ! ! œ! œ
"""
###
# #
ab ab ab abab ''
kk
kk11
sin mx and cos nx are orthogonal if m n.ʜ
Let m n.Á
sin mx cos nx dx sinm nx sinm nxdx cosm nx cosm nx''
kk
kk
mn mn
k
# #
""""
##
#
11
1
œœc d ab abab ab œ ‘
k
cos m n k cos m n k cos m n k cos m n kœ # #
""""
# # # #ab ab ab abmn mn mn mn
abababababa b aba b ab ab11
cos m n k cos m n k cos m n k cos m n kœ œ!
""" "
# # # #ab ab ab abmn mn mn mn
ab ab ab abab ab ab ab
sin mx and cos nx are orthogonal.Ê
46. f x sin mx dx sin nx sin mx dx. Since sin nx sin mx dx ,
for m n
1 for m n
" "
ÊÊ Ê
œ"
11 111 1
11 1'' 'ab !
œœœ
!Á
œ
n
N
a
n
the sum on the right has only one nonzero term, namely sin mx sin mx dx a .
a
mm
1 1
1
'
Ê
œ
8.5 TRIGONOMETRIC SUBSTITUTIONS
1. y 3 tan , , dy , 9 y 9 1 tan œœœœÊœœ)) )
11 ) )
))
)
##
## "
œ
3 d 9 cos
cos cos 3 3 9y
cos
œœ œ ab
È
kk
because cos 0 when ;ˆ‰ ))
11
##
3 ln sec tan C ln C ln 9 y y C'''
dy y
9y
cos d d
3 cos cos 3 3
9y
È
È
œ
ww œ
#
œ œ
œ
œœœœœ
)) )
))
kk ¹¹ ¸¸È))
2. ; 3y x ; x tan t, t , dx , 1 x ;''
3 dy
19y
dx dt
1x cos t cos tÈÈœ œ ##
# "
œ œ
œ
cd
È
œÄ œ œ œ
11
ln sec t tan t C ln x 1 x C ln 1 9y 3y C''dx dt
1x cos t
È
Š‹œ
##
œ œ
œœœœ
$
cos t
kk ¹¹
È ¸¸È
3. tan tan 1 tan ( 1)'
Ê# # # # # #
""" ""Ê" Ê" Ê"
#
#2
2
dx x
4x 444
œœœœœœ ‘ ˆ ‰ˆ ‰ ˆ ‰ˆ ‰
111
4. tan tan 1 tan 0 0''
00
22
dx dx x
82x 4x 416### ### # ##
""" """ "" Ê" Ê" Ê"
#
!
œœœœ œ œ œœ ‘ˆ ‰ˆ‰ˆ‰ˆ‰
11
5. sin sin sin 0 0'
0
2
3
dx x
9x
366
Î
Ê
Ê" Ê" Ê"
$Î#
!
"
#È
œ
œœœœœ ‘
11
6. ; t 2x sin t sin sin 0 0''
00
12 2 12 2
12
0
œœ
œ
œœ
œ
2 dx dt
14x 1t 2 44ÈÈ È
ÊÊ
Ê" Ê" Ê" "
œœ
cd cdœÄ œ œ œœ
11
7. t 5 sin , , dt 5 cos d , 25 t 5 cos ;œœ œ)) )) )
11
##
# È
25 t dt (5 cos )(5 cos ) d 25 cos d 25 d 25 C'' ''È ˆ‰œ œ œ œ # # œ
##
))) )) )
1cos 2 sin 2
4
)))
sin cos C sin C sin Cœ œ œ
25 25 t t 25 t
55 5 5
25 t t 25 t
## ##
Ê" Ê" ÊÊ
ab ’“ˆ‰ ˆ‰ ˆ‰Š‹)))
ÈÈ œœ
8. t sin , , dt cos d , 1 9t cos ;œœ œ
""
##
#
33
)) )) )
11 È
1 9t dt (cos )(cos ) d cos d sin cos C sin (3t) 3t 1 9t C'' 'È È
ab ’“œ œ œ œ # #""" " #"
336 6
))) )) ) ))
9. x sec , 0 , dx sec tan d , 4x 49 49 sec 49 7 tan ;œ œ œ œ
77
###
##
)) ))) ) )
1 ÈÈ
sec d ln sec tan C ln C''dx 2x
4x 49
sec tan d
7 tan 77
4x 49
È
ˆ‰ È
œ
œ
œ
Ê
"" "
## #
Ê
œœœœ
7
)))
)
')) ) )kk ¹¹

524. 520 Chapter 8 Techniques of Integration
10. x sec , 0 , dx sec tan d , 25x 9 9 sec 9 3 tan ;œ œ œ œ
33
55
)) ))) ) )
1
#
## ÈÈ
sec d ln sec tan C ln C''5 dx 5x
25x 9
5 sec tan d
3 tan 3 3
25x 9
È
ˆ‰ È
œ
œ
Ê
Ê
œœœœ
3
5
)))
)
')) ) )kk ¹¹
11. y 7 sec , 0 , dy 7 sec tan d , y 49 7 tan ;œœ œ)) ))) )
1
#
# È
dy 7 tan d 7 sec 1 d 7(tan ) C''
Èy49
y 7 sec
(7 tan )(7 sec tan ) dœ
Ê ##
œœœœ
))))
) '')) ) ) ) )ab
7secCœ’“ ˆ‰
Èy49
77
yœ
Ê Ê"
12. y 5 sec , 0 , dy 5 sec tan d , y 25 5 tan ;œœ œ)) ))) )
1
#
# È
dy tan cos d sin d (1 cos 2 ) d'' ' ' '
Èy25
y 125 sec 5 5 10
(5 tan )(5 sec tan ) dœ
%%
Ê """ ## #
œœœœ
))))
)
))) )) )) sin cos C sec C Cœ œ œ
"" Ê" ÊÊ
#10 10 5 y y 10 y
y y25 y255 sec
ab ’“’“ˆ‰Š‹Š‹)))
ÈÈ ˆ‰œœ !"
œ
y
5
13. x sec , 0 , dx sec tan d , x 1 tan ;œœ œ)) ))) )
1
#
# È
sin C C'' 'dx sec tan d d
xx 1
sec tan sec x
x1
œœ
œ
œÈ
È
Ê
Ê
œœœœ
))) )
)) )
)
14. x sec , 0 , dx sec tan d , x 1 tan ;œœ œ)) ))) )
1
#
# È
2 cos d 2 d sin cos C'' ' '2 dx 2 tan sec d 1 cos 2
xx 1
sec tan
% œ
%È
Ê
# œ
#
œœœ œ
))) )
))
)) ) ) ) )ˆ‰
tan cos C sec x x 1 C sec x Cœ œ œ )))
#" " # "
# Ê È ˆ‰
xx
x1
È
œ
œ
15. x 2 tan , , dx , x 4 ;œœ œ))
11 )
))##
#2 d 2
cos cos
œ
È
88 ;'' ' 'x dx sin d
x4
8 tan (cos ) d cos 1 ( sin ) d
cos cos cos
%%
œ
!œ
œ ##
È
ab a b
œ
ÊÊ
œœœ
))) ) ))
)
))
))
t cos 8 dt 8 dt 8 C 8 sec Ccd ˆ‰ ˆ ‰ Š‹œÄ œœœ)) ''t1 sec
tt tt3t 3
œ %
## œ!
Ê"" "" )
8Cx44x4Cœ œ Œ Î ab
È
È abx4 x4
83 3œ œ
!'œ
œ
#
œ "#
$Î#
#
†
16. x tan , , dx sec d , x 1 sec ;œœ œ)) )) )
11
##
# # È
CC'''dx sec d cos d
xx 1 tan sec sin sin x
x1
œœ
œ
œœ
œÈ
È
œ
" ʜ
œœœœ
)) ))
)) ) )
17. w 2 sin , , dw 2 cos d , 4 w 2 cos ;œœ œ)) )) )
11
##
# È
22 cot C C'' '8 dw 8 2 cos d d
w4w
4 sin 2 cos sin w
24 w
œœ
œœ
œÈ
È
Ê
ÊÊ
œœœœ
†
†
)) )
)) )
)
18. w 3 sin , , dw 3 cos d , 9 w 3 cos ;œœ œ)) )) )
11
##
# È
dw cot d d csc 1 d'' '' '
È
9w
w9 sin sin
3 cos 3 cos d 1 sinÊ ## Ê œ
œœ œ
œ
œœœœ
))) )
))
†
)) ) ) )Š‹ ab
cot C sin Cœ œ ))
È
9w
w3
wÊ Ê" œ
ˆ‰
19. x sin , 0 , dx cos d , 1 x cos ;œŸŸœ œ)) )) )
1
3
ab
#$
$Î#
4 d 4 sec 1 d'' ' '
00 0 0
32 3 3 3
œ
œœ œ œ
4x dx 4 sin cos d 1 cos
1x
cos cos
œœ œ
œ
!'œ !œ
abÊ
Ê #
œœ œ
œœ œ
))) )
))
Š‹ ab)))
4tan 4 3œœcd
È
))
1 1Î$
!
4
3

525. Section 8.5 Trigonometric Substitutions 521
20. x 2 sin , 0 , dx 2 cos d , 4 x 8 cos ;œŸŸœ œ)) )) )
1
6
ab
#$
$Î#
tan '' '
00 0
16 6
dx 2 cos d d
4x
8 cos 4 cos 4 12
3
43ab
È
È
Ê
"" " Î'
!
œ
!'œ !œ
œœœœœ
œœœœ
)) )
))
1
cd)
21. x sec , 0 , dx sec tan d , x 1 tan ;œœ œ)) ))) )
1
2
ab
#$
$Î#
CC'' 'dx sec tan d cos d x
x1
tan sin sin
x1ab
È
œ
!'œ !œ
œ
Ê
"
Ê
œœœœ
))) ))
)))
22. x sec , 0 , dx sec tan d , x 1 tan ;œœ œ)) ))) )
1
2
ab
#&
&Î#
dC C'' 'x dx sec sec tan d cos x
x1 3x 1
tan 3 sin sin
œœ %
œ œ
 'œ !'œ&% #
ab abÊ Ê
"
œœœœ
)))) )
)) )
†
)
23. x sin , , dx cos d , 1 x cos ;œ œ œ)) )) )
11
##
#$
$Î#
ab
cot csc d C C'''
ab È1x dx
xsin 55x
cos cos d cot 1xÊ %# " Ê
&
œ
!'œ
ww
%& œ
œœ œœ
))) )
)
†
))) Š‹
24. x sin , , dx cos d , 1 x cos ;œ œ œ)) )) )
11
##
#
"Î#
ab
cot csc d C C'''
ab È1x dx
xsin
cos cos d cot
33x
1xÊ ## " Ê
$
œ
"'œ
##
% œ
œœ œœ
))) )
)
†
))) Š‹
25. x tan , , dx sec d , 4x 1 sec ;œœ œ
""
####
## %
#
)) )) )
11
ab
4 cos d 2( sin cos ) C 2 tan 2x C'' '8 dx 4x
4x 1
8 sec d
sec 4x 1ab
ˆ‰
ab
œ
œ
"
œ
œ
# œ
œ
#"
œ
œœœœ
))
)
)) ) ) )
26. t tan , , dt sec d , 9t 1 sec ;œœ œ
""
##
## #
33
)) )) )
11
2 cos d sin cos C tan 3t C'' '6 dt 3t
9t 1
6 sec d
sec 9t 1ab
ˆ‰
ab
œ
œ
" œ
# œ
œ
#"
œ
œœœœ
3
))
)
)) ) ) )
27. v sin , , dv cos d , 1 v cos ;œœ œ)) )) )
11
##
#&
&Î#
ab
tan sec d C C'' 'v dv sin cos d tan v
1v
cos 3 3
1v
œœ %
œ
 'œ
œ
ab
È
Ê
## "
Ê
$
œœ œœ
))) )
)
))) Š‹
28. r sin , ;œ))
11
##
cot csc d C C'' '
ab È1r dr
rsin 77r
cos cos d cot 1rÊ '# " Ê
(
œ
 'œ
88
&* œ
œœ œœ
))) )
)
†
))) ’“
29. Let e 3 tan , t ln (3 tan ), tan tan , dt d , e 9 9 tan 9 3 sec ;
t 2t
œœ ŸŸ œ œ œ)) ) ) ))
Ê" Ê""%
$$
#ˆ‰ ˆ‰ È È
sec
tan
œ
)
)
sec d ln sec tan '' '
0tan13 tan13
ln 4 tan 4 3 tan 4 3
tan 4
e dt 3 tan sec d
e9 tan 3 sec
t
2t
È
œ
œœœ
#Î#Î
#Î #Î
#Î



)))
))
†
†
œ
)) ) )cdkk
3
tan 1 3Ê

#Î
ln ln ln 9 ln 1 10œ œˆ‰ Š‹ Š ‹
È
54
33 3 3
10
È
"
30. Let e tan , t ln (tan ), tan tan , dt d , 1 e 1 tan sec ;
t 2t
œœ ŸŸ œ œœ)) ) ) ))
" " # #ˆ‰ ˆ‰
34sec
43tan
œ
)
)
cos d sin '' '
ln 3 4 tan 3 4 tan 3 4
ln 4 3 tan 4 3 tan 4 3
tan 4

Î
e dt
1e
(tan ) d
sec
t
2t
ab
Š‹
œ
%'œ %œœœ
#Î#Î
#Î #Î
#Î
))
)
sec
tan
#
œ
œ
)) )cd
œÊ
3
tan
#Î œœ
43
55 5
"
31. ; u 2 t, du dt ; u tan , , du sec d , 1 u sec ;''
112 1 3
14 1œœ
œ
2 dt 2 du
t4tt t 1u 6 4ÈÈ Èœ
"
œ
###
’“ÈœœÄ œŸŸœ œ
œ
œ )) )) )
11
2 2''
13 6
14œœ
œ
œ
2 du 2 sec d
1u sec 4 6 6œ
Î%
Î'
œœ
œœœœœ
œ
œ
)) 1 1 1
)
1
1
cd ˆ‰)

526. 522 Chapter 8 Techniques of Integration
32. y e , 0 , dy e sec d , 1 (ln y) 1 tan sec ;œŸŸœ œœ
tan tan
4
)) 1
))) ))
# ## È È
d sec d ln sec tan ln 1 2'' '
1
e4 4
dy
y 1 (ln y) 00
esec
e sec Ȝ
Î%
!
œ
œ
œœœœ
œœœœ
tan
tan#
#
)
)
1
))))) cdkk Š‹
È
33. x sec , 0 , dx sec tan d , x 1 sec 1 tan ;œ œ œ œ)) ))) ) )
1
#
## È È
Csec xC''dx sec tan d
xx 1
sec tan È
œ
Ê
Ê"
œœœ
)))
))
)
34. x tan , dx sec d , 1 x sec ;œœ œ))) )
###
CtanxC''dx sec d
x1 sec
œœ
œ
œ
Ê"
œœœ
))
)
)
35. x sec , dx sec tan d , x 1 sec 1 tan ;œ œ œ œ)))) ) )
È È
##
sec d tan C x 1 C'' 'x dx sec sec tan d
x1
tan È
œ
Ê
# #
œœœœ
))))
)
†
)) )
È
36. x sin , dx cos d , ;œœ ))))
11
##
CsinxC''dx cos d
1x
cos È
Ê
Ê"
œ
œœœ
))
)
)
37. x x 4; dy x 4 ; y dx;
x 2 sec , 0
dx 2 sec tan d
x 4 2 tan
dy
dx x x
dx x4
œœ œ
œ
œ
œ
ÈÈ
Ô×
ÖÙ
ÕØÈ
## Ê
#
# '
Ȝ
))
)))
)
1
y 2tan d2sec 1 d2(tan )CÄœ œ œ œ '''
(2 tan )(2 sec tan ) d
2 sec
))))
)
##
)) ) ) ) )ab
2 sec C; x 2 and y 0 0 0 C C 0 y 2 sec œ œœÊœÊœÊœ’“ ’“ ˆ‰
È È
x4 x4x xœ œ
Ê Ê
## ##
Ê" Ê"
38. x 9 1, dy ; y ; y
x 3 sec , 0
dx 3 sec tan d
x 9 3 tan
È
Ô×
ÖÙ
ÕØÈ
#
ÊÊ
#
#
œ œ œ Äœ
œ
œ
œ
dy
dx 3 tan
dx dx 3 sec tan d
x9 x9
ÈÈ
œœ
''
))
)))
)
1
)))
)
sec d ln sec tan C ln C; x 5 and y ln 3 ln 3 ln 3 C C 0œœœœœÊœÊœ')) ) )kk ¹¹
x
33
x9
È
œ
Ê
y ln Êœ ¹¹
x
33
x9
È
œ
Ê
39. x 4 3, dy ; y 3 tan C; x 2 and y 0 0 tan 1 Cab
#""
## #
œœ œ œ œ œÊœ
dy
dx x 4 x 4
3 dx dx 3 x 3
œœ
'
C y tanʜ ʜ
33x3
88
11
##
Ê"
ˆ‰
40. x 1 x 1, dy ; x tan , dx sec d , x 1 sec ;ab ab
È
# ## $
# $Î#
#
œ
œœ œ œ œ
dy
dx
dx
x1ab
œ
!'œ
))) )
y cos d sin C tan cos C C C; x 0 and y 1œœ œœ œœœœ''sec d tan x
sec sec x1
œ
%
œ
)) )
))
)) ) ) )
È
œ
10C y 1ʜʜ
x
x1
È
œ
œ
41. A dx; x 3 sin , 0 , dx 3 cos d , 9 x 9 9 sin 3 cos ;œœŸŸœœœ'
0
3
È
9x
3
Ê
#
## œ
)) )) ) )
1 ÈÈ
A 3 cos d sin cos œœœœ''
00
22
œœœœ
3 cos 3 cos d 3 3
34
))) 1 1† #
#
Î#
!
)) ) ) )cd

527. Section 8.5 Trigonometric Substitutions 523
42. V dx 4 ;œœ ''
00
11
11ˆ‰
2dx
1x x1œ
#
œ
œ
œ
œ
ab
x tan , dx sec d , x 1 sec ;œœ œ))) )
## #
V4 4 cos dœœ11)) ''
00
44
œœœœ
sec d
sec
œ
#
))
)
#
2 (1 cos 2) d 2 1œœœ1))1)1'
0
4
œœ
œ ‘ˆ‰
sin 2)1
1
##
Î%
!

43. C C'' 'dx 2 dz 2 2
1sin x (1z) 1z
1
1tanÊÊÊ
Ê
Ê
œœœœ
Š‹
Š‹
ˆ‰
2 dz
1z
2z
1z
Î
#
Î
#
œ
œ x
44. ln 1 z C'' ' 'dx 2 dz dz
1sin xcos x 1z 2z1z 1z
1
œœ œœœÊ œ
œœ
œœœœ
Š‹
Š‹
2 dz
1z
2z 1 z
1z 1z
Î
#
ÎÎ
##
œ
#
œœ kk
ln tan 1 Cœ¸¸ˆ‰
x
#
45. (1 2) 1'' '
00 0
21 1
œœ
dx 2 dz 2
1sin x (1z) 1z
1
œœœ
œ
"
!
œ œ œ œ œ
Š‹
Š‹
2 dz
1z
2z
1z
Î
#
Î
#
œ
œ‘
46. 3 1'' '
œ
œ
œœ œ
œ313 13
21 1
dx dz 1
1cos x z z
1
Ê
Ê
"
"Î $
œœœœ
œœ
Š‹
Š‹
È
2 dz
1z
1z
1z
Î
#
œ
#
Î
#
œ
œ‘ È
47. tan tan '' ' '
00 0 0
21 1 1
œœ
d 2 dz 2 dz 2 z 2
2cos 22z 1z z 3
2 3333
)
)œœœÊœ
œ
Ê" Ê"
"
!
"
œœ œœ œ
Š‹
Š‹
ÈÈÈÈ
2 dz
1z
1z
1z
Î
#
œ
#
Î
#
œœ œ ’“
œœ
1 1
33
3
9È
È
48. '' ''
œ
œ
œ
œ21 1 1
23 3 3 3
cos d 1
sin cos sin 2z 2z 2z 2z
21 z dz))
)) )œÊœœ
œ
Ê Ê
œœœ
œœœ
Š‹Š‹
Ô×
ÕØ
Š‹
ab
1 z 2 dz
1z 1z
2z 1 z
1z
2z
1zœ
#
ÎÎ
##
œ
#
Î
#
##
Î
#$
#$
œ
%%
z
2z
œ
dz
ln z ln 3 0 (ln 3 2) ln 3 1œœ œœœ ’“Š ‹ Š‹
ÈÈ ˆ‰
"" """"
## ##
$
"
z3ln 3
44444
œ
È
49. ln C'' ''dt 2 dz 2 dz
sin t cos t 2z 1 z (z 1) 2 2z12
z1 2
ÊÊœœÊ
Ê
" œÊ
œœ
œœœœ
Š‹
Š‹
ÈÈ
È
2 dz
1z
2z 1 z
1z 1z
Î
#
ÎÎ
##
œ
#
œœ ¹¹
ln Cœ
"
œÊ
œœ
ÈÈ
ˆ‰ È
ˆ‰2tan 12
tan 1 2
ºº
t
t
#
#
50. '' ' 'cos t dt
1cos t 1z 1z 1z
1
21 z dz 21 z dz
1z 1z 1zÊ œœÊœ
Ê
ÊÊ
œÊœ Ê
œœ œ
Š‹Š‹
Š‹
ab ab
ababab aba b
1 z 2 dz
1z 1z
1z
1zœ
#
ÎÎ
##
œ
#
Î
#
œœ
œœœ
œ œœ œ
2 2 tan z C cot t Cœœœœœ'''''
ab
ab ab
1z dz
1zz z1z 1z z z 1 z
dz dz dz dz tÊ
#
" Ê"
œ
œœ œ œ œ œ œ
ˆ‰
51. sec d''''' ''))œœ œ œ œ
d 2 dz 2 dz dz dz
cos 1z (1z)(1z) 1z 1z
)
)
Š‹
Š‹
2 dz
1z
1z
1z
Î
#
œ
#
Î
#
ʜʜÊ
œ
ln 1 z ln 1 z C ln Cœ œ kkkk
»»
1tan
1tan
œ
Ê
Š‹
Š‹
œ
œ
#
#
52. csc d ln z C ln tan C''''))œœ œœœ
ddz
sin z
))
)
Š‹
Š‹
2 dz
1z
2z
1z
Î
#
Î
#
kk ¸¸
#

528. 524 Chapter 8 Techniques of Integration
8.6 INTEGRAL TABLES AND COMPUTER ALGEBRA SYSTEMS
1. tan C'dx 2 x 3
xx 3 3 3ÈÈ
Ê
Ê" Ê
œ É
(We used FORMULA 13(a) with a 1, b 3)œœ
2. ln C ln C'dx
xx4 x42 4x44
x4 4 x42
ÈÈ È È
ÈÈ È
Èœœœ
"" œÊ œÊ
œœ #
œœ¹¹ ¹¹
(We used FORMULA 13(b) with a 1, b 4)œœ
3. 2 x 2 dx 2 x 2 dx'' '' 'x dx dx
x2 x2 x2
(x 2) dx
ÈÈ È
ÊÊ Ê
Ê
"Ê"
œœ Š‹ Š‹
ÈÈ
2 x2 4Cœ œˆ‰ ˆ‰ È
’“
22
13 11 3
x2 x2
2(x 2)
Š‹ Š‹
ÈÈ
ÊÊ
Ê
% "
(We used FORMULA 11 with a 1, b 2, n 1 and a 1, b 2, n 1)œœœ œœœ
4. '''''x dx 3 dx dx 3 dx
(2x 3) (2x 3) (2x 3)
(2x 3) dx
2x 3 2x 3
œœœ
""
####
œ
œ œ
!'œ !'œ !'œ %œœ
È
ˆ‰È
2x 3 dx 2x 3 dx Cœœ
""
# # ## ##
Ê" ÊÄ
œœ''Š‹ Š‹ÈÈ ˆ‰ˆ‰ ˆ‰ˆ‰
3232 2x 3 2x 3
1(1)
ˆ‰ ˆ‰ÈÈ
"!"
(2x33)C Cœœ
"
#
œ
ÈÈ2x 3 2x 3
(x 3)
(We used FORMULA 11 with a 2, b 3, n 1 and a 2, b 3, n 3)œœœ œœœ
5. x 2x 3 dx (2x 3) 2x 3 dx 2x 3 dx 2x 3 dx 2x 3 dx'' '' 'ÈÈÈÈÈ
Š‹ Š‹œ œ ""
####
$"
33
C1CCœœœˆ‰ˆ‰ ˆ‰ˆ‰ œ ‘
" Ê
## ## #
ÊÊ
ÊÊœ232 2x3
2x 3 2x 3
53 5 5
(2x 3) (2x 3) (x 1)
Š‹ Š‹
ÈÈ
%
!'œ !'œ
(We used FORMULA 11 with a 2, b 3, n 3 and a 2, b 3, n 1)œœœ œœœ
6. x(7x 5) dx (7x 5)(7x 5) dx (7x 5) dx 7x 5 dx 7x 5 dx'' '''œœ
$Î# $Î# $Î#""
&$
7777
55 ˆ‰ ˆ‰ÈÈ
C2Cœœ ˆ‰ˆ‰ ˆ‰ˆ‰ ’“’ “
" œœ œœ
77 7 77 5 49 7
2527x 5 7x 5 (7x 5) 2(7x 5)
ˆ‰ ˆ‰ÈÈ

 'œ
Cœ’“ ˆ‰
(7x 5)
49 7
14x 4œ Ê
 'œ
(We used FORMULA 11 with a 7, b 5, n 5 and a 7, b 5, n 3)œœœ œœœ
7. dx C''
ÈÈ
È
94x 94x
xx
(4) dx
x9 4x
ÊÊ Ê
# Ê
œ œ
(We used FORMULA 14 with a 4, b 9)œ œ
2 ln Cœ
ÈÈÈ
ÈÈÈ
94x 94x 9
x 994x9
ÊÊÊ "
ʜ
Š‹¹ ¹
(We used FORMULA 13(b) with a 4, b 9)œ œ
ln Cœ
ÊÊ ÊÊ
ʜ
ÈÈ
È
94x 94x3
x3
2
94x3
¹¹
8. C''dx 4 dx
x 4x9 x4x9
4x 9
(9)x 18
œÈÈ
È
ÊÊ
Ê
Ê
œ
(We used FORMULA 15 with a 4, b 9)œœ
tan Cœ
È
È
4x 9
9x 9 9
22 4x9
9
Ê Ê" Ê
ˆ‰Š‹ É
(We used FORMULA 13(a) with a 4, b 9)œœ
tan Cœ
È
4x 9
9x 27 9
44x9Ê Ê" Ê
É

529. Section 8.6 Integral Tables and Computer Algebra Systems 525
9. x 4x x dx x 2 2x x dx sin C''ÈÈ ˆ‰œ œ ## œÊ Ê
##
Ê" Ê
†
(x 2)(2x 3 2) 2 2 x x
6
2x2†††
È
œ %
4 sin C 4 sin Cœœ
(x 2)(2x 6) 4x x (x 2)(x 3) 4x x
63
x2 x2œÊ Ê œÊ Ê Ê" Ê"ÊÊ
##
ÈÈ
œ œ
ˆ‰ ˆ‰
(We used FORMULA 51 with a 2)œ
10. dx dx 2 x x sin C x x sin (2x 1) C''
È É
xx
xx
2xx xÊ
Ê
"" "
## #
## Ê" Ê"
Ê œ
"
œ
œ "
œ
"
œ
œ œ œ
†
É Š‹
È
†
We used FORMULA 52 with aˆ‰ œ
"
#
11. ln C ln C''dx dx
x7 x
x7x
77
77x
xx
77x
ÈÈ È
ÊŠ ‹
È
ÈÈ
ÊŠ ‹
È È
œ
œ
""
œœ
œœ
œ œ
œ
œ
œ
œ
œœ œ
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
We used FORMULA 26 with a 7Š‹
È
œ
12. ln C ln C''dx dx
x7 x
x7x
77
77x
xx
77x
ÈÈ È
ÊŠ ‹
È
ÈÈ
ÊŠ ‹
È È
Ê
Ê
""
œÊ
œÊ
œ œ
œ
œ
œ
œ
œœ œ
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
We used FORMULA 34 with a 7Š‹
È
œ
13. dx dx 2 x 2 ln C 4 x 2 ln C''
ÈÈ È È
4x 2 x 2 2 x 2 4x
xx x x
ÊÊ œÊ œÊ ## #œœœ œœ œ
œ œ œ
ÈÈ
¹¹ ¹¹
(We used FORMULA 31 with a )œ#
14. dx dx x 2 2 sec C x 4 2 sec C''
ÈÈ
x4 x2
xx
xxœœœ
ÊÊ ## # Ê" Ê"
##
œœœ
ÈÈ ¸¸ ¸¸
(We used FORMULA 42 with a )œ#
15. 25 p dp 5 p dp 5 p sin C 25 p sin C''ÈÈÈ Èœ œ œ ##### #
## # #
Ê" Ê"ppp p 525
55
œ
(We used FORMULA 29 with a 5)œ
16. q 25 q dq q 5 q dq sin q 5 q 5 2q C''## " ## ### ## "ÈÈ È ˆ‰ abœ œ
5
858
q
#
sin q 25 q 25 2q Cœ
625
858
q" # " #
ˆ‰ È ab
(We used FORMULA 30 with a 5)œ
17. dr dr sin r 2 r C 2 sin r 4 r C''rr2r r
4r 2 r
œœœ
œœœ
ÈÈ
ÊÊ
### ##
Ê" Ê" "" ## #
œœœ ˆ‰ ˆ‰ÈÈ
(We used FORMULA 33 with a 2)œ
18. cosh C ln s s 2 C ln s s 2 C''ds ds s
s2
s2
2
ÈÈ
ÊŠ‹
È
œ
œ
œ
Ê
Ê
Ê" ##
#
œ œ œ œ
»»
Ê Š‹ ¹ ¹
È È
We used FORMULA 36 with a 2Š‹
È
œ
19. tan tan C tan tan C' d2 542
54 sin 2 54 4 3 3 4225 16
)1)1
)#
ÊÊ ""
Ê
Ê" Ê"
œœ
È
’“É ˆ‰ œ ‘ˆ‰)
(We used FORMULA 70 with b 5, c 4, a 2)œœœ
20. ln C ln C' d 1 5 4 sin 2 3 cos 2
4 5 sin 2 4 5 sin 2 6 4 5 sin 2225 16
5 4 sin 2 25 16 cos 2) ))
)) )
))
œœ œ
Ê"œœ
Ê
œœÊ
œœ
È
È
¹¹ ¸¸
(We used FORMULA 71 with a 2, b 4, c 5)œœœ

530. 526 Chapter 8 Techniques of Integration
21. e cos 3t dt (2 cos 3t 3 sin 3t) C (2 cos 3t 3 sin 3t) C'2t
œœ
ee
23 13
2t 2t
œœ
œ
(We used FORMULA 108 with a 2, b 3)œœ
22. e sin 4t dt ( 3 sin 4t 4 cos 4t) C ( sin 4t 4 cos 4t) C'#3t
œœ$
ee
(3) 4 25
!!
œœ3t 3t
ʜ
(We used FORMULA 107 with a 3, b 4)œ œ
23. x cos x dx x cos x dx cos x cos x'' ' 'Ê" " Ê" Ê" Ê"
##
""
ÊÊ
œœœ
x x dx x x dx
11 11
1x 1x
11 11œœ
ÈÈ œœ
œœ
(We used FORMULA 100 with a 1, n 1)œœ
cos x sin x x 1 x C cos x sin x x 1 x Cœ œ
xx
44
œ œ
### ## #
Ê" Ê" Ê" Ê""" " " " " ##
ˆ‰ Š‹
ÈÈ
(We used FORMULA 33 with a 1)œ
24. x tan x dx x tan x dx tan x tan x'' ' 'Ê" " Ê" Ê" Ê"
##
""
œ"
œ"œ" œab ab
x x dx x x dx
11 11 1x 1x
11 11œœ
ab
œœ
œœ
œ
(We used FORMULA 101 with a 1, n 1)œœ tan x dx (after long division)œ"
x
1x
œ
œ
##
Ê" ""
'ˆ‰
tan x dx dx tan x x tan x C x tan x x Cœ œœ"
x x
1x
œœ
œ
###### #
" " " # """" "" "
'' abab
25. ln C''ds ds s s 3
9s 3 s 23 3 s 43 s 3ab a b abÊÊ ÊÊ
"œ
œ !œ
œœ œœ œ
œœ
†† †
3
¸¸
(We used FORMULA 19 with a 3)œ
ln Cœ
ss3
18 9 s 108 s 3abÊÊ
"œ
œ
¸¸
26. ln C''dd
2
2 22 2 42
2
2
)) )
)
) )
)
)ab
”•Š‹
È Š‹Š‹
ÈÈ
”• Š‹
È
È
ÈÊ
Ê Ê
" œ
Ê
œ
œœ
œ
œ
œœ
œ
œœ
3¹¹
We used FORMULA 19 with a 2Š‹
È
œ
ln Cœ
)
)
)
)42 82 2
2
ab ÈÈ
È
Ê
" œ
Ê
œ ¹¹
27. dx''
ÈÈ
È
4x 9 4x 9
xx2
4dx
x4x 9
œœ
œ
œ œ
(We used FORMULA 14 with a 4, b 9)œœ
2 ln C ln Cœ œ
ÈÈ ÈÈ
ÈÈ È
È
È
4x9 4x9 9 4x9 4x93
xx3 94x99
2
4x93
œœÊ œœÊ "
œœ œœ
Š‹ ¹¹¹¹
(We used FORMULA 13(b) with a 4, b 9)œœ
28. dx C''
ÈÈ
È
9x 4 9x 4
xx2
9dx
x9x 4
ÊÊ
Ê
œ œ
(We used FORMULA 14 with a 9, b 4)œœ
tan C tan Cœ œ
ÈÈÈ
È
9x 4 9x 4 9x 4
x4x
92 9x4 9
4
ÊÊÊ
###
Ê" Ê" Ê
Š‹ É
(We used FORMULA 13(a) with a 9, b 4)œœ
29. dt 2 3t 4 ( 4)''
È
È
3t 4
t
dt
t3t 4
Ê
Ê
œ
È
(We used FORMULA 12 with a 3, b 4)œœ
2 3t 4 4 tan C 2 3t 4 4 tan Cœ œ
ÈÈ
Š‹ É
23t4
4 4
3t 4
È
È
Ê" Ê"Ê Ê
#
(We used FORMULA 13(a) with a 3, b 4)œœ

531. Section 8.6 Integral Tables and Computer Algebra Systems 527
30. dt 2 3t 9 9''
È
È
3t 9
t
dt
t3t 9
œ
œ
œÈ
(We used FORMULA 12 with a 3, b 9)œœ
2 3t 9 9 ln C 2 3t 9 3 ln Cœ œ ÈÈ Š‹ ¹¹¹¹
1
93t99
3t 9 9 3t 9 3
3t 9 3ÈÈ È
ÈÈ È
È
œÊ œÊ
œœ œœ
(We used FORMULA 13(b) with a 3, b 9)œœ
31. x tan x dx tan x dx tan x dx'''#" " "
œœœ œ
""
œ œ
xxxx
2 1 2 1 1 x 3 3 1 x
21 21ÎÎ%%
œœ
(We used FORMULA 101 with a 1, n 2);œœ dx x dx ln 1 x C x tan x dx''' 'xx dxx
1 x 1 x
!œ
œœ
##
" ##"
œ œÊ ab
tan x ln 1 x Cœ
xx
366
% œ
" # "
ab
32. dx x tan x dx tan x dx tan x dx'' ' 'tan x x x x x
x ( 2 1) ( 2 1) 1 x ( 1) 1 x
!" !" !"
œ œœ
œœ œ
# " " "
Êœ Êœ œ Ê œ
"
21 21
ab
(We used FORMULA 101 with a 1, n 2);œœ ln x ln 1 x C'' ''x dx dx dx x dx
1x x1 x x 1x
!"
œœ œ
#
" #
œœœ
ab
kk a b
dx tan x ln x ln 1 x Cʜ'tan x
xx
!"
œ
""" #
#
kk a b
33. sin 3x cos 2x dx C' œ
cos 5x cos x
10 #
(We used FORMULA 62(a) with a 3, b 2)œœ
34. sin 2x cos 3x dx C' œ
cos 5x cos x
10 #
(We used FORMULA 62(a) with a 2, b 3)œœ
35. 8 sin 4t sin dx sin sin C 8 C' t87t89t
79 79
sin sin
###
œœˆ‰ ˆ‰
–—
Š‹ Š‹
7t 9t
##
(We used FORMULA 62(b) with a 4, b )œœ
"
#
36. sin sin dt 3 sin sin C' tt t t
36 6
œˆ‰ ˆ‰
#
(We used FORMULA 62(b) with a , b )œœ
""
36
37. cos cos d 6 sin sin C' )) ) )
34 127 1
67
)œˆ‰ ˆ‰
#
(We used FORMULA 62(c) with a , b )œœ
""
34
38. cos cos 7 d sin sin C C' )))
2 13 2 15 13 15
13 1 15
sin sin
))œœ
"
#
ˆ‰ ˆ‰
Š‹ Š‹
13 15œœ
##
(We used FORMULA 62(c) with a , b 7)œœ
"
2
39. dx'''''x x 1 x dx dx dx
x 1 x 1 x 1 x 1 x 1
dx 1
%
œœ œ
œœ œ œœ
œ
œœ "
œœ œ #
œ
ab ab ab
ab
œ œ
ln x 1 tan x Cœ
""
##
#"
ab
x
21 xab
œ
(For the second integral we used FORMULA 17 with a 1)œ
40. dx 3 3''''' ''x 6x dx 6x dx 3 dx dx dx
x 3 x 3 x 3 x 3x 3
x 3
dx 3
x 3
œ
œœœ œ
œœœœœœ œ
œ
œ
œ
œ
œ
œœœ œ œ
œ
œ
œ
ab ab ab ab
Š‹
È
ab
”•Š‹
È
œ œ

532. 528 Chapter 8 Techniques of Integration
tan 3 tan Cœ
" "Ê" Ê"
œ
œ
ÈÈ Èab
Š‹Š‹
ÈÈ
Π#
Š‹
È33 3
x3 x x
x 3
23 3 x 23
Š‹ Š‹
ÎÑ
ÏÒ œ œœ
œ
%
For the first integral we used FORMULA 16 with a 3; for the third integral we used FORMULA 17 Š
È
œ
with a 3‹
È
œ
tan Cœ
" Ê"
œœ23 3
x3 x
x 3 2 x 3ÈÈ ab
Š‹ œœ
41. sin x dx; 2 u sin u du 2 sin u du
ux
xu
dx 2u du'''Ê" " Ê" Ê" #
œœ
"
ÊÈ
Ô×
ÕØ
È
Š‹
œ
œ
œ
Ĝ
uu
11 11
1 u
11 11œœ
È œ
u sin uœ
#"
Ê 'u du
1 u
œ
œ
È
(We used FORMULA 99 with a 1, n 1)œœ
u sin u sin u u 1 u C u sin u u 1 u Cœ œ
#" " # " "" " "
## # #
##
Š‹
ÈÈ ˆ‰
(We used FORMULA 33 with a 1)œ
x sinx xxCœ ˆ‰ È È
""
##
Ê" #
42. dx; 2u du 2 cos u du 2 u cos u 1 u C
ux
xu
dx 2u du'''
cos x
x
cos u
u1
!"
!"È
È Ô×
ÕØ
È
Š‹
È
œ
œ
œ
Äœœ
# Ê" Ê" " # †
(We used FORMULA 97 with a 1)œ
2x cos x 1xCœŠ‹ÈÈ È
Ê"
43. dx; du 2 du 2 sin u u 1 u C
ux
xu
dx 2u du'''
È
ÈÈÈ
x
1 x
u2u u
1 u 1 uÊ
#
ÊÊ
""
##
Ê" # Ô×
ÕØ
È
Š‹
È
œ
œ
œ
Äœ œ
†
œœ
œ
sin u u 1 u Cœ
Ê" # È
(We used FORMULA 33 with a 1)œ
sin x x1xCsin x xx Cœœ
Ê" Ê" #ÈÈ È È È
44. dx; 2u du 2 2 u du
ux
xu
dx 2u du'''
ÈÈ
È
2 x
x
2 u
u
Ê
#
Ê
#
# Ô×
ÕØ
È
ÊŠ‹
È
œ
œ
œ
Ĝ
œ
†
2 2 u sin C u 2 u 2 sin Cœ œ
–—
ÊŠ ‹ Š‹ Š‹
È È
uu u
2
22##
#
## Ê" Ê"
Š‹
È
ÈÈ
œ
We used FORMULA 29 with a 2Š‹
È
œ
2x x 2 sin Cœ
È È
# Ê"
#
x
45. (cot t) 1 sin t dt ;
usin t
du cos t dt'' 'È
”•œ Ä
œ
œ # Ê Ê
È È
1 sin t (cos t) dt
sin t u
1 u du
œ œ
1u ln Cœ
È
¹¹
# œÊ1 1 u
u
È
œ
(We used FORMULA 31 with a 1)œ
1 sin t ln Cœ
È
¹¹
# œÊ1 1 sin t
sin t
È
œ

533. Section 8.6 Integral Tables and Computer Algebra Systems 529
46. ; ln C
usin t
du cos t dt'' 'dt cos t dt du
(tan t) 4 sin t (sin t) 4 sin t u 4 u
2 4 u
uÈÈ È
È
ÊÊ Ê
"
#
œÊ
œœ œ
œ
œÄœ
œ
œ
”• ¹¹
(We used FORMULA 34 with a 2)œ
ln Cœ
"
#
œÊ
¹¹
2 4 sin t
sin t
È
œ
47. ; ln u 3 u C
u ln y
ye
dy e du'''
dy
y 3 (ln y)
e du du
e3 u 3 u
ÈÈÈœ œœ
#
œ œœ
Ô×
ÕØ
¹¹
È
œ
œ
œ
Äœœ
u
u
u
u
ln ln y 3 (ln y) Cœ¸¸ È
#
We used FORMULA 20 with a 3Š‹
È
œ
48. ; ln u 5 u C ln sin 5 sin C
usin
du cos d''cos d du
5 sin 5 u
))
)
ÈÈ
œœ
##
œœ”• ¹¹¹ ¹
ÈÈ
œ
œ
Ä œ œ
)
))
))
We used FORMULA 20 with a 5Š‹
È
œ
49. ; ln u u 1 C ln 3r 9r 1 C
u3r
du 3 dr''3 dr du
9r 1 u 1
ÈÈ
œœ
ÊÊ
##
”• ¹¹¹ ¹
ÈÈ
œ
œ
Ä œœ
(We used FORMULA 36 with a 1)œ
50. ; ln u 1 u C ln 3y 1 9y C
u3y
du 3 dy''
3 dy
1 9y
du
1 u
ÈÈœ œ
##
œ œ”• ¹¹
È ¸¸È
œ
œ
Ä œ œ
(We used FORMULA 20 with a 1)œ
51. cos x dx; 2 t cos t dt 2 cos t dt t cos t dt
tx
xt
dx 2t dt''''" " " # " #
##
"
ÊÊÈ
Ô×
ÕØ
È
Š‹
œ
œ
œ
Äœœ
tt t
1 t 1 t
œœ œ
œœ
ÈÈ
(We used FORMULA 100 with a 1, n 1)œœ
t cos t sin t t 1 t Cœ
#" " ""
##
# È
(We used FORMULA 33 with a 1)œ
x cos x sin x x 1 x C x cos x sin x x x Cœœ
Ê" Ê" Ê" Ê""" ""
## ##
#ÈÈÈ ÈÈ È È
52. tan y dy; 2 t tan t dt 2 tan t dt t tan t dt
ty
yt
dy 2t dt''''" " " # " #
##
"
È
Ô×
ÕØ
È
’“
œ
œ
œ
Äœœ
tt t
1 t 1 t
œœ œ
œœ
(We used FORMULA 101 with n 1, a 1)œœ
t tan t dt t tan t t tan t C y tan y tan y y Cœ œœ
# " # " " " "œ
œœ ''t 1 dt
t 1 1 t
œ
œœ
ÈÈÈ
53. sin 2x dx sin 2x dx sin 2x dx'' '&$
##
ÊÊ
œ œ
sin 2x cos 2x 5 1 sin 2x cos 2x 4 sin 2x cos 2x 3 1
55 10533
## œ
††
’“
(We used FORMULA 60 with a 2, n 5 and a 2, n 3)œœ œœ
sin 2x cos 2x cos 2x C Cœ œ
sin 2x cos 2x 2 8 sin 2x cos 2x 2 sin 2x cos 2x 4 cos 2x
10 15 15 10 15 15
#
# "
# ˆ‰
54. sin d sin d sin cos sin d'' '&$%
#####
ÊÊ)))))
)) )œ œ
sin cos sin cos
53
5 1 2 4 3 1
5553
# œ
œœ œœ
" "
œ œ
œœ œœ
† †
’“
We used FORMULA 60 with a , n 5 and a , n 3ˆ‰ œœ œœ
""
##
sin cos sin cos 2 cos C sin cos sin cos cos Cœ œ
28 8 2816
5 15 15 5 15 15
%# %#
## ## # ## ## #
)) )) ) )) )) )
ˆ‰

534. 530 Chapter 8 Techniques of Integration
55. 8 cos 2 t dt 8 cos 2 t dt'' %# Ê
11œŠ‹
cos 2 t sin 2 t 4 1
42 4
%
11
1†
(We used FORMULA 61 with a 2 , n 4)œœ1
6Cœ
cos 2 t sin 2 t t sin (2 2 t)
42
%
11
11
1
’“
#
††
†
(We used FORMULA 59 with a 2 )œ1
3t C 3t Cœœ
cos 2 t sin 2 t 3 sin 4 t cos 2 t sin 2 t 3 cos 2 t sin 2 t
42
%%
11 1 11 11
1111
56. 3 cos 3y dy 3 cos 3y dy'' &$ Ê
œŠ‹
cos 3y sin 3y
53 5
5 1
#
†
cos 3y dyœ
cos 3y sin 3y cos 3y sin 3y
5533 3
12 3 1
# œ
Š‹
†
Ê
'
(We used FORMULA 61 with a 3, n 5 and a 3, n 3)œœ œœ
cos 3y sin 3y cos 3y sin 3y sin 3y Cœ
" %#
515 15
48
57. sin 2 cos 2 d sin 2 cos 2 d''#$ #
# #
Ê
))) )))œ
sin 2 cos 2 3 1
(2 3) 3
% œ
))
(We used FORMULA 69 with a 2, m 3, n 2)œœœ
sin 2 cos 2 d sin 2 d(sin 2 ) Cœ œ œ
sin 2 cos 2 2 sin 2 cos 2 2 sin 2 cos 2 sin 2
10 5 10 5 10 15
!œ !œ !œ !
)) )) )) )''## "
#
))) ) ) ’“
58. 9 sin cos d 9 sin cos d'' $ $Î# $Î#
œœ
Ê
))) )))œ ’“
sin cos 3 1
3 3
œ  'œ
œœ
))
ˆ‰ ˆ‰
33
2 sin cos 4 cos sin dœ
# &Î# $Î#
)) )))'
We used FORMULA 68 with a 1, n 3, mˆ‰ œœ œ
3
#
2 sin cos 4 cos d(cos ) 2 sin cos 4 cos Cœ œ
# &Î# $Î# # &Î# &Î#
)) )) )) )' ˆ‰
2
5
2 cos sin Cœ ˆ‰ˆ‰
&Î# #
))
4
5
59. 2 sin t sec t dt 2 sin t cos t dt 2 cos t dt'' '#% #% %
ÊÊ
Ê
œœ Š‹
sin t cos t 2 1
2 4 2 4
!%
(We used FORMULA 68 with a 1, n 2, m 4)œœ œ
sin t cos t cos t dt sin t cos t sec t dt sin t cos t sec t dtœ œœ
$ % $ % $ #
ÊÊ
Ê'' ' Š‹
sec t tan t 4 2
4 1 4 1
œ
(We used FORMULA 92 with a 1, n 4)œœ
sin t cos t tan t C sec t tan t tan t C tan t sec t 1 Cœœ œ
$ # #
Š‹ ab
sec t tan t 2 2 2 2
33 3 3 3
œ
tan t Cœ
2
3
$
An easy way to find the integral using substitution:
2 sin t cos t dt 2 tan t sec t dt 2 tan t d(tan t) tan t C'''#%### $
œœœ
2
3
60. csc y cos y dy sin y cos y dy sin y cos y dy'' '#& #& #$
#
Ê
œœ
Š‹
$
sin y
cos y
52 5
5 1
#
sin y cos y dyœ
Š‹ Š‹
$$
sin y sin y
cos y cos y
333 23
43 1
# œ
#$ #
Ê #
'
(We used FORMULA 69 with n 2, m 5, a 1 and n 2, m 3, a 1)œ œ œ œ œ œ
cos y sin y d(sin y) Cœ œ
Š‹
$
sin y
cos y
3 3 sin y 3 3 sin y 3 sin y 3 sin y
48 8 cos y 4 cos y
#
# œ
Š‹
" ## '
61. 4 tan 2x dx 4 tan 2x dx tan 2x 4 tan 2x dx'''$#
#
œ œŠ‹
tan 2x
2
œ
†
(We used FORMULA 86 with n 3, a 2)œœ tan 2x ln sec 2x C tan 2x 2 ln sec 2x Cœ œ
##
#
4
kk kk

535. Section 8.6 Integral Tables and Computer Algebra Systems 531
62. tan dx tan dx tan tan dx'' '%#$#
#### ʈ‰ ˆ‰ ˆ‰ ˆ‰
xx2xx tan
(4 1) 3
œ œ
%
œ
"
œˆ‰
x
(We used FORMULA 86 with n 4, a )œœ
"
#
tan 2 tan x Cœ
2x x
3
$
##
(We used FORMULA 84 with a )œ
"
#
63. 8 cot t dt 8 cot t dt'' %#
œ Š‹
cot t
3
%
(We used FORMULA 87 with a 1, n 4)œœ
8 cot t cot t t Cœ ˆ‰
" $
3
(We used FORMULA 85 with a 1)œ
64. 4 cot 2t dt 4 cot 2t dt cot 2t 4 cot 2t dt'''$#
Ê
œ œ ’“
cot 2t
2(3 1)
œ
(We used FORMULA 87 with a 2, n 3)œœ cot 2t ln sin 2t C cot 2t 2 ln sin 2t Cœ œ
##
#
4
kk kk
(We used FORMULA 83 with a 2)œ
65. 2 sec x dx 2 sec x dx '' $
ÊÊ
Ê
11œ’“
sec x tan x 3 2
(3 1) 3 1
11
1
(We used FORMULA 92 with n 3, a )œœ 1
sec x tan x ln sec x tan x Cœ
""
11
11 1 1 kk
(We used FORMULA 88 with a )œ1
66. csc dx csc dx''"" Ê
## # #
$
Ê
x3 2x
(3 1) 3 1
œ Š‹
csc cot
xx
##
"
œ
We used FORMULA 93 with a , n 3ˆ‰ œœ
"
#
csc cot ln csc cot C csc cot ln csc cot Cœ œ
"""
### ## #### ##
œ ‘¸¸¸¸
xx x x xx x x
We used FORMULA 89 with aˆ‰ œ
"
#
67. 3 sec 3x dx 3 sec 3x dx'' %#
ÊÊ
Ê
œ’“
sec 3x tan 3x 4 2
3(4 1) 4 1
œ
(We used FORMULA 92 with n 4, a 3)œœ
tan 3x Cœ
sec 3x tan 3x 2
33
œ
(We used FORMULA 90 with a 3)œ
68. csc d csc d''%#
Ê
Ê
Ê
))
34 13 (4 1)
4 2
))œ
csc cot
#œœ
33
"
3
We used FORMULA 93 with n 4, aˆ‰ œœ
"
3
csc cot 3 cot C csc cot 2 cot Cœ œ
##)) ) )) )
333 3 33 3
2
†
We used FORMULA 91 with aˆ‰ œ
"
3
69. csc x dx csc x dx csc x dx'' '&$
ÊÊ ÊÊ
ÊÊ
œ œ
csc x cot x 5 2 csc x cot x 3 csc x cot x 3 2
5 1 5 1 4 4 3 1 3 1
%%
Š‹
(We used FORMULA 93 with n 5, a 1 and n 3, a 1)œœ œœ
csc x cot x csc x cot x ln csc x cot x Cœ
" $
488
33
kk
(We used FORMULA 89 with a 1)œ
70. sec x dx sec x dx sec x dx'' '&$
ÊÊ ÊÊ
ÊÊ
œ œ
sec x tan x 5 2 sec x tan x 3 sec x tan x 3 2
5 1 5 1 4 4 3 1 3 1
%%
Š‹
(We used FORMULA 92 with a 1, n 5 and a 1, n 3)œœ œœ
sec x tan x sec x tan x ln sec x tan x Cœ
" $
488
33
kk

536. 532 Chapter 8 Techniques of Integration
(We used FORMULA 88 with a 1)œ
71. 16x (ln x) dx 16 x ln x dx 16 x dx'' '$# $ $ ""
#
œ œ’“’ “ ’“
x(ln x) x(ln x) x(ln x)
44 4 44
2
#
(We used FORMULA 110 with a 1, n 3, m 2 and a 1, n 3, m 1)œœ œ œœ œ
16 C 4x (ln x) 2x ln x CœœŠ‹
x(ln x) x(ln x)
483
xx
##
##
%#%
72. (ln x) dx (ln x) dx x(ln x) 3 ln x dx x(ln x) 3x(ln x) dx'' ' '$#$ $# "
"
œ œ œ '
x(ln x) x(ln x)
11 11 1
32 x ln x
% œ
’“ Š‹
x(ln x) 3x(ln x) 6x ln x 6x Cœ
$#
(We used FORMULA 110 with n 0, a 1 and m 3, 2, 1)œœ œ
73. xe dx (3x 1) C (3x 1) C'3x ee
39
œœ
3x 3x
œ
(We used FORMULA 104 with a 3)œ
74. xe dx ( 2x 1) C (2x 1) C' #2x
œ œ
ee
(2) 4
!!
œ2x 2x
Ê
(We used FORMULA 104 with a 2)œ
75. x e dx 2x e 3 2 x e dx 2x e 6 2x e 2 2 xe dx
'' '$$ #$#x2 x2 x2 x2 x2 x2œœ œœœ œ
œ œ †† Š‹
2x e 12x e 24 4e 1 C 2x e 12x e 96e 1 Cœ œ
$# $#
##x2 x2 x2 x2 x2 x2œœ œ œœœ
† ˆ‰ ˆ‰
xx
We used FORMULA 105 with a twice and FORMULA 104 with aˆ‰ œœ
""
##
76. x e dx x e xe dx''## "111
11
xxx 2
œ
(We used FORMULA 105 with n 2, a )œœ 1
xe e ( x 1) C xe ( x 1) Cœ œ
""##
111 1 1
11 1xx x22e
†
œ %†11 ˆ‰
#x
(We used FORMULA 104 with a )œ1
77. x 2 dx x2 dx 2 dx C'' '#
####
"x
œ œ œ
x2 2 x2 2 x2 x2 2 x2 2
ln 2 ln 2 ln ln 2 ln 2 ln 2 ln ln ln (ln 2)
xx
œœ œ
œxxxxxx
Š‹’“
(We used FORMULA 106 with a 1, b 2, n 2, n 1)œœœœ
78. x 2 dx x2 dx 2 dx'' '# Ê
#
Ê"##xx
œ œ
x2 2 x2 2 x2
ln 2 ln 2 ln ln 2 ln 2 ln 2
x
œxxx
Š‹
Cœ
x2 2 x2 2
ln ln ln (ln 2)
œ
œ###xxx
###
’“
(We used FORMULA 106 with a 1, b 2, n 2, n 1)œœœœ
79. x dx dx C C''11
xx xx x
ln ln ln ln ln ln (ln )
œ œ œ
11111
11 111 1 1
xxxxx
""
ˆ‰
œ
(We used FORMULA 106 with n 1, b , a 1)œœœ 1
80. x2 dx 2 dx C''
ÈÈ
ÈÈ È
2x 2x x2 x2 2
2 ln 2 2 ln 2 2 ln 2 (ln 2)
œ œ
œœœ
2x x2 x2
"
#
œ (We used FORMULA 106 with a 2, b 2, n 1)œœœ
È
81. e sec e dt; sec x dx sec x dx
xe
dx e dt'''tt
t
t
sec x tan x 3 2
31 31
$$
ÊÊ
Ê
ab ”•" Ä œ
œ"
œ
(We used FORMULA 92 with a 1, n 3)œœ
ln sec x tan x C sec e 1 tan e 1 ln sec e 1 tan e 1 Cœ œ
sec x tan x tt t t
## #
""
kkc d ababk k ab ab

537. Section 8.6 Integral Tables and Computer Algebra Systems 533
82. d ; 2 csc t dt 2 csc t dt
t
t
d2t dt'''
csc csc t cot t 3 2
31 31
%È
È
)
)
)
)
)
)
Ô×
ÕØ
È
’“
œ
œ
œ
Ĝ
# $
ÊÊ
Ê
(We used FORMULA 93 with a 1, n 3)œœ
2 ln csc t cot t C csc cot ln csc cot Cœ œ œ ‘kk
ÈÈ È È
¹¹
csc t cot t
##
"
)) ) )
83. 2 x 1 dx; x tan t 2 sec t sec t dt 2 sec t dt 2 sec t dt''' '
000 0
144 4
È
cd ”•
œ ‘# #$
ÊÊ
Î%
!
Ê
œÄ œ œ
œœ œœœ œ
†
sec t tan t 3 2
31 31
†
1
(We used FORMULA 92 with n 3, a 1)œœ
sec t tan t ln sec t tan t 2 ln 2 1œœcd kk
ÈÈ
Š‹†
1Î%
!
84. ; y sin x sec x dx sec x dx''' '
000 0
32 3 3 3
œ
œœœ œ
dy
1y
cos x dx sec x tan x 4 2
cosx 41 41
abÊ
%#
ÊÊ
Î$
!
Ê
œ
 'œ
œ
cd ’“œÄ œ œ
œœ œ 1
(We used FORMULA 92 with a 1, n 4)œœ
tan x 3 3 2 3œ œœ’“ ˆ‰ ˆ‰ÈÈÈ
sec x tan x 2 4 2
33 3 3
œ
1Î$
!
85. dr; r sec (sec tan ) d tan d tan d''''
1000
2333
abr1
r sec 4 1
tan tan
œ
!'œ
% %
Ê %#
Ê
Î$
!
cd ’“œÄ œ œ ) ) ) ) )) ))
œœ œœœ œ
))
)
1
tan 3œœœ’“
È
tan
3333
33
%
)11
1
))
Î$
!
È
(We used FORMULA 86 with a 1, n 4 and FORMULA 84 with a 1)œœ œ
86. ; t tan cos d cos d''' '
000 0
13 6 6 6
œœœ œ
œ
dt sec d cos sin 5 1
t1 sec 55
ab
œ
Ÿ'œ


œ
&$
Î'
!
Ê
cd ’“ ˆ‰œÄ œ œ ))) ))
œœ œ
)) ) )
)
1
cos dœ ’“ ’“ ”•
ˆ‰
cos sin 4 cos sin 3 1
553 3
# œ
)) ))
11Î' Î'
!!
Ê '
0
6
œœ
))
cos sin sin œ ’“
cos sin 4 8
515 15
#
))
1
#
Î'
!
)) )
(We used FORMULA 61 with a 1, n 5 and a 1, n 3)œœ œœ
œ œœ œ
Š‹ ˆ‰
È
œ3
œœ
#
"
5 15 15 160 10 15 480 480
4 8 9 4 3 9 48 32 4 2033
ˆ‰ ˆ‰ ˆ‰ˆ‰Š‹
## #
#
"" " œœ ††
87. sinh 3x dx sinh 3x dx''"" Ê&$
88535
sinh 3x cosh 3x 5 1
œŠ‹
#
†
sinh 3x dxœ
sinh 3x cosh 3x sinh 3x cosh 3x 3 1
120 10 3 3 3
#
"Ê
Š‹
†
'
(We used FORMULA 117 with a 3, n 5 and a 3, n 3)œœ œœ
cosh 3x Cœ
sinh 3x cosh 3x sinh 3x cosh 3x 2
1 0 90 30 3
#
#
"
ˆ‰
sinh 3x cosh 3x sinh 3x cosh 3x cosh 3x Cœ
""
#
%
1 0 90 45
1
88. dx; 2 cosh u du 2 cosh u du
ux
du'' '
cosh x
x
dx
2x
cosh u sinh u 4 1
44
#
%È
È
È
–—
È
Š‹
œ
œ
Ĝ
%# Ê Cœ
cosh u sinh u 3 sinh 2u u
4
%
## #
ˆ‰
(We used FORMULA 118 with a 1, n 4 and FORMULA 116 with a 1)œœ œ
cosh x sinh x sinh 2 x x Cœ
"
#
$ ÈÈ ÈÈ
33
84
89. x cosh 3x dx sinh 3x x sinh 3x dx sinh 3x cosh 3x cosh 3x dx'' '# "
œ œ
x2 x2x
33 3333
œœ
Š‹
(We used FORMULA 122 with a 3, n 2 and FORMULA 121 with a 3, n 1)œœ œœ

538. 534 Chapter 8 Techniques of Integration
sinh 3x cosh 3x sinh 3x Cœ
x2x2
3927
œ
90. x sinh 5x dx cosh 5x sinh 5x C' œ
x 525
"
(We used FORMULA 119 with a 5)œ
91. sech x tanh x dx C' (
œ
sech x
7

(We used FORMULA 135 with a 1, n 7)œœ
92. csch 2x coth 2x dx C C' $
#
œ œ
csch 2x csch 2x
36
%%
†
(We used FORMULA 136 with a 2, n 3)œœ
93. u ax b x dx ;œÊœ Ê œ
ub du
aa
Ê
du ln u C ln ax b C'' 'x dx du b b b
(ax b) au a a u u a u a ax b
(u b)
œ œ
Ê "" " "
œœœ œœ œœœœœ ˆ‰ œ ‘ œ ‘kk k k
94. x a tan a x a sec 2x dx 2a sec tan dx a sec d ;œÊœ Êœ Êœ))))))
# # ## ## #
d (1 cos 2 ) d sin 2 C'' ''dx a sec d
ax a sec a sec 2a 2aab a b
œœ œœ
œœ
œ
!œ ! !
œ
"" ""
#
œœœœ
))
) )
))))) ˆ‰
sin cos C cos C Cœ œ œ
"" "
# œ
#
a 2a cos 2a 1 tan
sin tan
%% % œab œ ‘ˆ‰ˆ‰))) ) ) )
))
))
tan C tan Cœœ
"" Ê" Ê"
œ
œ2a a 2a a x 2a a
xx x x
a1
% œœ œ !œ
œ–—
Š‹
ab
x
a
95. x a sin a x a cos 2x dx 2a cos sin d dx a cos d ;œÊœ Êœ Êœ)) )))))
#### # a x dx a cos (a cos ) d a cos d (1 cos 2 ) d C'' ''È ˆ‰## ##
###
œ œ œ œ ))) )) )) )
aasin 2
œœ
)
cos sin C 1 sin sin C sin Cœ œ œ
aa axx
aaa
ax
œœ œ œœ
## #
# Ê" Ê
ab Š‹Š‹
È
))) ) )) ††
È
sin a x Cœ
axx
a
œ
##
Ê" ##
È
96. x a sec x a a tan 2x dx 2a tan sec d dx a sec tan d ;œÊœ Êœ Êœ))))))))
## ## # #
cos d sin C 1 cos C'' ''dx a tan sec d d
xx a
a sec a tan a sec a a a
œœ œ
œœ œ œ œ œÈ ab
Ê
""" #
œœœœœ
))) )
)) )
)) ) )
È
CCCCœœ œœˆ‰ ˆ‰ ˆ‰
"" "
Ê
ÊÊ
Ê"
a a sec a a x
1
sec 1 x a
œœ œœ
"
œ
"
œœœ
œ
œ
ÉÉ
ˆ‰
ÈÈ
ˆ‰
cos
cos
x
a
x
aœ
œ )
)
97. x sin ax dx x d(cos ax) (cos ax) x cos ax nx dx'' 'n
œ œ
nnn1
aaaˆ‰ ˆ ‰
""" Ê
†
cos ax x cos ax dxœ
xn
aa
n1
n
'Ê
We used integration by parts u dv uv v du with u x , v cos axŠ ‹ ''œ œ œ
n "
a
98. x (ln ax) dx (ln ax) d m(ln ax) dx'' 'nm m m1 xx
n1 n1 n1 x
x(ln ax)
œœ Š‹ Š‹ ˆ‰
n1 n1 n1 mÎÎ Î
œœ œ
Ê"
x (ln ax) dx, nœ Á"
x(ln ax)
n1 n1
m nm1
n1 mÎ
œœ
Ê
'
We used integration by parts u dv uv v du with u (ln ax) , vŠ ‹ ''œ œ œ
m x
n1
n1Î
œ
99. x sin ax dx sin ax d sin ax dx'' 'n Ê" Ê" Ê"
œœ œ Ê
œœ Š‹ Š‹
xx x a
n1 n1 n1 1(ax)
n1 n1 n1ÎÎ Î
œ
È
sin ax , n 1œ Á
xax dx
n1 n1
1ax
n1 n1ÎÎ
œœ
œœ
Ê"
Ê
'
È

539. Section 8.6 Integral Tables and Computer Algebra Systems 535
We used integration by parts u dv uv v du with u sin ax, vŠ ‹ ''œ œ œ
Ê"
œ
x
n1
n1#
100. x tan ax dx tan ax d tan ax dx'' 'n Ê" Ê" Ê"
œœ œœ
œœ Š‹ Š‹
xx xa
n1 n1 n1 1(ax)
n1 n1 n1ÎÎ Î
œ
tan ax , n 1œ Á
xax dx
n1 n1 1ax
n1 n1ÎÎ
œœ
œœœ
Ê"
'
We used integration by parts u dv uv v du with u tan ax, vŠ ‹ ''œ œ œ
Ê"
œ
x
n1
n1Î
101. S 2 y 1 (y ) dxœ '
0
2
œ
1È
w#
2 x 2 1 dxœ1 '
0
2
œ
È É#
œ
x
x2
œ
œ
2 2 x 1 dxœ
ÈÈ
1 '
0
2
œ
#
22 ln x x 1œ
ÈÈ
’“ ¹¹1
xx 1
È
œ
œ
##
" #
2
0
œ
(We used FORMULA 21 with a 1)œ
26ln23 23 2 ln23œœ
ÈÈ ÈÈ
’“Š‹
ÈÈÈ È
Š‹111
102. L 1 (2x) dx 2 x dx 2 x ln x xœœ œ ''
00
32 32 32
0
œœ œ
œœ œ
È ÉÉ É ’“ ˆ‰ˆ‰ Š‹ ### # """""
##4444
x
We used FORMULA 2 with aˆ‰ œ
"
#
14x lnx 14x 14 ln 14 ln œ œ ’“Š‹
ÈÈ
Š‹ ÉÉ ˆ‰ ˆ‰
x 33
444444
33
## ###
##"" " " ""
œ
32
0œ
ÈÈ
(2) ln 1 ln 2 ln 3 2œ œ
ÈÈ È
33 3
44 4 4
"""
##
Š‹ Š ‹
È
103. A 2 x 1 2; x œœœœ''
00
33
dx x dx
x1 x 1 AÈÈœœ
$
!
"
’“È
x 1 dxœ
""
œAA
dx
x1''
00
33
È
È
(x 1) 1 ;œœ
"
#
$Î#
$
!
†
24
33
œ ‘
(We used FORMULA 11 with a 1, b 1, n 1 andœœœ
a 1, b 1, n 1)œœœ
y ln (x 1) ln 4 ln 2 ln 2œ œ œœœ
"" ""
# #
$
!Ax14 4
dx'
0
3
cd
È
104. M x dx 18 dx 54 18x 27 ln 2x 3
y
000
333œœœ'''ˆ‰ cd kk
36 2x 3 dx
2x 3 2x 3 2x 3œœœ
œ $
!
18 3 27 ln 9 ( 27 ln 3) 54 27 2 ln 3 27 ln 3 54 27 ln 3œ œ œ††
105. S 2 x 1 4x dx;œ1 '
#1
1
# #È
u1u du
u2x
du 2 dx
”•
È
œ
œ
$#
1
4'
#2
2
# #
12u 1u lnu 1uœ
1
48 8
u
’“ab
ÈÈ
Š‹
# ##"
#
#
(We used FORMULA 22 with a 1)œ
(1 2 4) 1 4 ln 2 1 4œ
1
48 8
2
’Š‹ ÈȆ
"
(1 2 4) 1 4 ln 2 1 4 “ÈÈ Š‹## ## Î##
2
88
†
"
5 ln 7.62œ ¸
1
48
9 25
25
’“
È
Š‹
#
" œ
ʜ
È
È

540. 536 Chapter 8 Techniques of Integration
106. (a) The volume of the filled part equals the length of the
tank times the area of the shaded region shown in the
accompanying figure. Consider a layer of gasoline
of thickness dy located at height y where
r y r d. The width of this layer is
2 r y . Therefore, A 2 r y dyÈÈ
## ##
œ '
#
#$r
rd
and V L A 2L r y dyœœ † '
#
#$r
rd
È##

(b) 2L r y dy 2L sin '
#
#$ #$
#r
rd rd
r
È ’“##
Ê
##
Ê"
œ
yr y r y
r
È
œœ œ
(We used FORMULA 29 with a r)œ
2L 2rd d sin 2L 2rd d sinœœ ’“’“
ÈÈ ˆ‰ ˆ‰ ˆ‰ ˆ ‰ Š‹ ˆ‰
(d r) rdrr d-r r dr
rr
Ê
##### # #
## Ê" Ê"ÊÊ
œœ œ
11
107. The integrand f(x) x x is nonnegative, so the integral is maximized by integrating over the function'sœ
È
#
entire domain, which runs from x 0 to x 1œœ
x x dx 2 x x dx 2 x x sinÊœ œ ''
00
11
È
ÉÉ ”• Š‹ ## # ""
####
ÊÊ
Ê"
"
!
††
ˆ‰ ˆ‰xx
"""
œœœ
œ
"
œ
We used FORMULA 48 with aˆ‰ œ
"
#
x x sin (2x 1)œœœ’“
È ˆ‰
ˆ‰x
8888
Ê
###
#"""Ê"
"
!
"
œ
†
111
108. The integrand is maximized by integrating g(x) x 2x x over the largest domain on which g isœ
È
#
nonnegative, namely [ 2]!ß
x 2x x dx sin (x 1)ʜ '
0
2
È
’“ # œÊ Ê "
#
Ê"
#
!
(x 1)(2x 3) 2x x
6
È
œ
(We used FORMULA 51 with a )œ"
œœ
""
## # # #
†
111
ˆ‰
CAS
EXPLORATIONS
109. Example CAS commands:
:Maple
q1 := Int( x*ln(x), x ); # (a)
q1 = value( q1 );
q2 := Int( x^2*ln(x), x ); # (b)
q2 = value( q2 );
q3 := Int( x^3*ln(x), x ); # (c)
q3 = value( q3 );
q4 := Int( x^4*ln(x), x ); # (d)
q4 = value( q4 );
q5 := Int( x^n*ln(x), x ); # (e)
q6 := value( q5 );
q7 := simplify(q6) assuming n::integer;
q5 = collect( factor(q7), ln(x) );

541. Section 8.6 Integral Tables and Computer Algebra Systems 537
110. Example CAS commands:
:Maple
q1 := Int( ln(x)/x, x ); # (a)
q1 = value( q1 );
q2 := Int( ln(x)/x^2, x ); # (b)
q2 = value( q2 );
q3 := Int( ln(x)/x^3, x ); # (c)
q3 = value( q3 );
q4 := Int( ln(x)/x^4, x ); # (d)
q4 = value( q4 );
q5 := Int( ln(x)/x^n, x ); # (e)
q6 := value( q5 );
q7 := simplify(q6) assuming n::integer;
q5 = collect( factor(q7), ln(x) );
111. Example CAS commands:
:Maple
q := Int( sin(x)^n/(sin(x)^n+cos(x)^n), x=0..Pi/2 ); # (a)
q = value( q );
q1 := eval( q, n=1 ): # (b)
q1 = value( q1 );
for N in [1,2,3,5,7] do
q1 := eval( q, n=N );
print( q1 = evalf(q1) );
end do:
qq1 := PDEtools[dchange]( x=Pi/2-u, q, [u] ); # (c)
qq2 := subs( u=x, qq1 );
qq3 := q + q = q + qq2;
qq4 := combine( qq3 );
qq5 := value( qq4 );
simplify( qq5/2 );
109-111. Example CAS commands:
: (functions may vary)Mathematica
In Mathematica, the natural log is denoted by Log rather than Ln, Log base 10 is Log[x,10]
Mathematica does not include an arbitrary constant when computing an indefinite integral,
Clear[x, f, n]
f[x_]:=Log[x] / x
n
Integrate[f[x], x]
For exercise 111, Mathematica cannot evaluate the integral with arbitrary n. It does evaluate the integral (value is /4 in1
each case) for small values of n, but for large values of n, it identifies this integral as Indeterminate
109. (e) x ln x dx x dx, n 1''nn x ln x
n1 n1
œ Á
n1Î
œœ
"
(We used FORMULA 110 with a 1, m 1)œœ
Cln x Cœœ
x ln x x x
n1 (n1) n1 n1
n1 n1 n1ÎÎ Î
œ
œœ œ œ
"
ˆ‰
110. (e) x ln x dx x dx, n 1''ÊÊ
Êœ Ê œ
"nn x ln x
n 1 (n)1
œ Á
!În1
(We used FORMULA 110 with a 1, m 1, n n)œœœ

542. 538 Chapter 8 Techniques of Integration
Cln x Cœ œ
x ln x x x
1n 1n 1n 1n 1n
1n 1n 1n!!!
ÊÊÊ Ê Ê
""
Š‹ ˆ‰
111. (a) Neither MAPLE nor MATHEMATICA can find this integral for arbitrary n.
(b) MAPLE and MATHEMATICA get stuck at about n 5.œ
(c) Let x u dx du; x 0 u , x u 0;œÊ œ œ Ê œ œ Ê œ
111
###
I œœ œœ '' ''
02 00
20 2 2
œ œœ
œ
œ œœ
œ
sin x dx cos u du cos x dx
sinx cosx cosu sinu cosx sinx
sin u du
sin u cos u
n nn
nn nn nn
n
nn
œœœ
ÊÊ
Êœ Ê
ˆ‰
ˆ‰ ˆ‰
#
##
œ
œœ I I dx dx IÊœ œ œ Êœ ''
00
22
œœœœ
ˆ‰
sin x cos x
sin x cos x 4
nn
nn
œ
#
11
8.7 NUMERICAL INTEGRATION
1. x dx
'
1
2
I. (a) For n 4, x ;œœœœÊœ?
ba 21 x
n44 8
ÊÊ" "
#
?
mf(x ) 12 T (12) ;!
iœÊœ œ
"
#8
3
f(x) x f (x) 1 f 0 M 0œÊ œÊ œÊ œ
www
E0 Êœkk
T
x f(x ) m mf(x )
x1 1 1 1
x 5/4 5/4 2 5/2
x 3/2 3/2 2 3
x 7/4 7/4 2 7/2
x2 2 1 2
ii i
!
"
#
$
%
(b) x dx 2 E x dx T 0''
11
22
T
œœœÊœ œ’“ kk
x3
2
œ
#
"
"
##
(c) 100 0%
kkE
True Value
T
‚œ
II. (a) For n 4, x ;œœœœÊœ?
ba 21 x
n44312
ÊÊ" " ?
mf(x ) 18 S (18) ;!
iœÊœ œ
"
#12
3
f (x) 0 M 0 E 0
Ð%Ñ
œÊ œÊ œ kk
S
(b) x dx E x dx S 0''
11
22
S
œ Ê œ œœ
333
###
kk
(c) 100 0%
kkE
True Value
S
‚œ
x f(x ) m mf(x )
x1 1 1 1
x 5/4 5/4 4 5
x 3/2 3/2 2 3
x 7/4 7/4 4 7
x2 2 1 2
ii i
!
"
#
$
%
2. (2x 1) dx'
1
3
Î
I. (a) For n 4, x ;œœœœœÊœ?
ba 31 2 x
n442 4
ÊÊ " "
#
?
mf(x ) 24 T (24) 6 ;!
iœÊœ œ
"
4
f(x) 2x 1 f (x) 2 f 0 M 0œÊ œÊ œÊ œ
www
E0 Êœkk
T
x f(x ) m mf(x )
x1 1 1 1
x3/2 2 2 4
x2 3 2 6
x5/2 4 2 8
x3 5 1 5
ii i
!
"
#
$
%
(b) (2x 1) dx x x (9 3) (1 1) 6 E (2x 1) dx T 6 6 0''
1 1
3 3
T
œ œœÊ œ œœcd kk
#
$
"
(c) 100 0%
kkE
True Value
T
‚œ
II. (a) For n 4, x ;œœœœœÊœ?
ba 31 2 x
n44236
ÊÊ " " ?
mf(x ) 36 S (36) 6 ;!
iœÊœ œ
"
6
f (x) 0 M 0 E 0
Ð%Ñ
œÊ œÊ œ kk
S
(b) (2x 1) dx 6 E (2x 1) dx S''
11
33
S
œÊœ kk
660 œœ
(c) 100 0%
kkE
True Value
S
‚œ
x f(x ) m mf(x )
x1 1 1 1
x3/2 2 4 8
x2 3 2 6
x5/2 4 4 16
x3 5 1 5
ii i
!
"
#
$
%
3. x 1 dx'
Ê
#1
1
ab#
I. (a) For n 4, x œœœ œœÊœà?
ba 2 x
n44 4
1(1)Ê"" ÊÊ
##
?
mf(x ) 11 T (11) 2.75!
iœÊœ œ à
"
4
f(x) x 1 f (x) 2x f (x) 2 M 2œÊ œ Ê œÊ œ
#w ww
E (2) or 0.08333ÊŸ œkk ˆ‰
T
1(1)
11
ÊÊ
## #
""
#
x f(x ) m mf(x )
x121 2
x 1/2 5/4 2 5/2
x0 1 2 2
x 1/2 5/4 2 5/2
x1 2 1 2
ii i
!
"
#
$
%Î
Î

543. Section 8.7 Numerical Integration 539
(b) x 1 dx x 1 1 E x 1 dx T''
Ê Ê
# #
"
Ê"
"
#1 1
1 1
T
ab ab ’“ ˆ‰ˆ ‰œœœÊœ œœ
x118 811
3333 341
%
E 0.08333Êœ¸kk¸¸
T
"
#1
(c) 100 100 3%
kkE
True Value
T
‚œ ‚¸Š‹
"
œ
1
8
3 II. (a) For n 4, x œœœ œœÊœà?
ba 2 x
n44 36
1(1)Ê"" ÊÊ
#
?
mf(x ) 16 S (16) 2.66667 ;!
iœÊœ œœ
"
63
8
f (x) 0 f (x) 0 M 0 E 0
Ð$Ñ Ð%Ñ
œÊ œÊ œÊ œ kk
S
(b) x 1 dx x'
Ê
#
"
Ê"1
1
ab ’“œœ
x8
33
%
Ex1 dxS 0Ê œ œœkk a b
S
1
1
'
Ê
# 88
33
(c) 100 0%
kkE
True Value
S
‚œ
x f(x ) m mf(x )
x121 2
x 1/2 5/4 4 5
x0 1 2 2
x 1/2 5/4 4 5
x1 2 1 2
ii i
!
"
#
$
%Î
Î
4. x 1 dx
'
Ê
#2
0
abÎ
I. (a) For n 4, x œœœ œœÊœ?
ba 2 x
n44 4
0(2)Ê"" ÊÊ
##
?
mf(x ) 3 T (3) ;!
iœÊœ œ
"
44
3
f(x) x 1 f (x) 2x f (x) 2œÊ œ Ê œ
#w ww
M 2 E (2) 0.08333ÊœÊ Ÿ œœ kk ˆ‰
T
0(2)
11
ÊÊ
## #
""
#
x f(x ) m mf(x )
x231 3
x 3/2 5/4 2 5/2
x102 0
x 1/2 3/4 2 3/2
x0 11 1
ii i
!
"
#
$
%Î
Î
Î
ÎÎ Î
ÎÎ
(b) x 1 dx - x 0 2 E x 1 dx T''
Ê Ê
# #
!
#
"2 2
0 0
T
ab ab ’“ ˆ‰ œ œ œ Ê œ œœ
x82 23
333 3412
%
E Êœkk
T
"
12
(c) 100 100 13%
kkE
True Value
T
‚œ ‚¸Š‹
"
12
2
3
II. (a) For n 4, xœœœ œœ?
ba 2
n44
0(2)Ê" ÊÊ
#
; mf(x ) 4 S (4) ;Êœ œÊœ œ
?x2
36 6 3
"" !
i
f (x) 0 f (x) 0 M 0 E 0
Ð$Ñ Ð%Ñ
œÊ œÊ œÊ œ kk
S
(b) x 1 dx E x 1 dx S''
ÊÊ
##22
00
S
ab kk abœÊœ
2
3
0 œœ
22
33
(c) 100 0%
kkE
True Value
S
‚œ
x f(x ) m mf(x )
x231 3
x 3/2 5/4 4 5
x102 0
x 1/2 3/4 4 3
x0 11 1
ii i
!
"
#
$
%Î
Î
Î
ÎÎ Î
ÎÎ
5. t t dt
'
0
2
ab
$
#
I. (a) For n 4, xœœœœœ?
ba 20 2
n442
ÊÊ "
; mf(t ) 25 T (25) ;Êœ œÊœ œ
?x25
444#
"" !
i
f(t) t t f (t) 3t 1 f (t) 6tœÊ œ Ê œ
$w# ww
M 12 f (2) E (12)Êœœ Ê Ÿ œ
ww Ê" "
##
#
kk ˆ‰
T
20
12
t f(t ) m mf(t )
t0 0 1 0
t 1/2 5/8 2 5/4
t1 2 2 4
t 3/2 39/8 2 39/4
t 2 10 1 10
ii i
!
"
#
$
%
(b) t t dt 0 6 E t t dt T 6 E''
0 0
2 2
T T
a b kk a b kk’“Š ‹
$ $
##
#
!
""
œœœÊ œ œœÊ œ
tt 22 25
44 444
#
(c) 100 100 4%
kk ¸¸E
True Value 6
T
‚œ ‚¸
Ê
"
4
II. (a) For n 4, x ;œ œ œ œœÊ œ?
ba 20 2 x
n44236
ÊÊ " " ?
mf(t ) 36 S (36) 6 ;!
iœÊœ œ
"
6
f (t) 6 f (t) 0 M 0 E 0
Ð$Ñ Ð%Ñ
œÊ œÊ œÊ œ kk
S
(b) t t dt 6 E t t dt S''
00
22
S
ab kk ab
$$
œÊ œ
660 œœ
(c) 100 0%
kkE
True Value
S
‚œ
t f(t ) m mf(t )
t0 0 1 0
t 1/2 5/8 4 5/2
t1 2 2 4
t 3/2 39/8 4 39/2
t 2 10 1 10
ii i
!
"
#
$
%

544. 540 Chapter 8 Techniques of Integration
6. t 1 dt'
Ê
$1
1
ab#
I. (a) For n 4, xœœœ œœ?
ba 2
n44
1(1)Ê" ÊÊ
#
; mf(t ) 8 T (8) 2 ;Êœ œÊœ œ
?x
44#
"" !
i
f(t) t 1 f (t) 3t f (t) 6tœÊ œ Ê œ
$w#ww
M 6 f (1) E (6)Êœœ Ê Ÿ œ
ww ÊÊ
##
""
#
kk ˆ‰
T
1(1)
14
t f(t ) m mf(t )
t101 0
t 1/2 7/8 2 7/4
t0 1 2 2
t 1/2 9/8 2 9/4
t1 2 1 2
ii i
!
"
#
$
%Î
Î
(b) t 1 dt t 1 ( 1) 2 E t 1 dt T 2 2 0''
Ê Ê
$ $
"
Ê"
Ê1 1
1 1
T
ab kk ab’“ Š ‹Š ‹ œ œ œ Ê œ œœ
t1
444
(1)
## #
(c) 100 0%
kkE
True Value
T
‚œ
II. (a) For n 4, xœœœ œœ?
ba 2
n44
1(1)Ê" ÊÊ
#
; mf(t ) 12 S (12) 2 ;Êœ œÊœ œ
?x
36 6
"" !
i
f (t) 6 f (t) 0 M 0 E 0
Ð$Ñ Ð%Ñ
œÊ œÊ œÊ œ kk
S
(b) t 1 dt 2 E t 1 dt S''
ÊÊ
$$11
11
S
ab kk abœÊœ
220 œœ
(c) 100 0%
kkE
True Value
S
‚œ
t f(t ) m mf(t )
t101 0
t 1/2 7/8 4 7/2
t0 1 2 2
t 1/2 9/8 4 9/2
t1 2 1 2
ii i
!
"
#
$
%Î
Î
7. ds'
1
2
"
s
œ
I. (a) For n 4, x ;œœœœÊœ?
ba 21 x
n44 8
ÊÊ" "
#
?
mf(s ) T! Š‹
iœÊœ œ
179,573 179,573 179,573
44,100 8 44,100 352,800
"
0.50899; f(s) f (s)¸œÊœ
" w
ss
2
œ %
f(s) M 6 f(1)ʜʜœ
ww ww6
s
#
E (6) 0.03125ÊŸ œœkk ˆ‰
T
2
14 3
Ê" " "
##
#
s f(s ) m mf(s )
s1 1 1 1
s 5/4 16/25 2 32/25
s 3/2 4/9 2 8/9
s 7/4 16/49 2 32/49
s 2 1/4 1 1/4
ii i
!
"
#
$
%
(b) ds s ds E ds T 0.50899 0.00899'' '
11 1
22 2
T
""""" #
#
" ## #ss1s
11
œ œœœœœÊœœœœ‘ˆ‰
E 0.00899Êœkk
T (c) 100 100 2%
kkE
True Value 0.5
0.00899
T
‚œ ‚¸
II. (a) For n 4, x ;œœœœÊœ?
ba 21 x
n44312
ÊÊ" " ?
mf(s ) S! Š‹
iœÊœ œ
264,821 264,821 264,821
44,100 12 44,100 529,200
"
0.50042; f (s) f (s)¸œÊœ
Ð$Ñ Ð%Ñ 24 120
ss
w
M 120 E (120)Êœ Ê Ÿ kk¸¸ˆ‰
S
2
180 4
Ê" "
%
0.00260œ¸
"
384
s f(s ) m mf(s )
s1 1 1 1
s 5/4 16/25 4 64/25
s 3/2 4/9 2 8/9
s 7/4 16/49 4 64/49
s 2 1/4 1 1/4
ii i
!
"
#
$
%
(b) ds E ds S 0.50042 0.00042 E 0.00042''
11
22
S S
"""
##ss
1
œœœÊ œ œ œ Ê œ kk
(c) 100 100 0.08%
kkE
True Value 0.5
0.0004
S
‚œ ‚¸
8. ds'
2
4
"
Ê(s 1)
œ
I. (a) For n 4, x ;œœœœÊœ?
ba 42 x
n42 4
ÊÊ" "
#
?
mf(s )!
iœ
1269
450
T 0.70500;Êœ œ œ
"
4 450 1800
1269 1269
ˆ‰
f(s) (s 1) f (s)œ Ê œ
# w
Ê
2
(s 1)
%
f(s) M 6ʜ ʜ
ww
Ê
6
(s 1)
#
E (6) 0.25ÊŸ œœkk ˆ‰
T
42
14
Ê" "
##
#
s f(s ) m mf(s )
s2 1 1 1
s 5/2 4/9 2 8/9
s 3 1/4 2 1/2
s 7/2 4/25 2 8/25
s 4 1/9 1 1/9
ii i
!
"
#
$
%
(b) ds E ds T 0.705 0.03833''
2 2
4 4
T
"Ê"Ê"Ê" "
#
%
#
(s 1) (s 1) 4 1 1 3 (s 1) 3
22
œ œœœœÊœ œ¸’“ ˆ‰ˆ‰
E 0.03833ʸkk
T

545. Section 8.7 Numerical Integration 541
(c) 100 100 6%
kk
ˆ‰
E
True Value
0.03833
T
‚œ ‚¸
2
3
II. (a) For n 4, x ;œœœœÊœ?
ba 42 x
n4236
ÊÊ" " ?
mf(s )!
iœ
1813
450
S 0.67148;Êœ œ ¸
"
6 450 2700
1813 1813
ˆ‰
f (s) f (s) M 120
Ð$Ñ Ð%ÑÊ
ÊÊ
œÊ œÊœ
24 120
(s 1) (s 1)
w
E (120) 0.08333ÊŸ œ¸kk ˆ‰
S
42
180 12
Ê" "
#
%
s f(s ) m mf(s )
s2 1 1 1
s 5/2 4/9 4 16/9
s 3 1/4 2 1/2
s 7/2 4/25 4 16/25
s 4 1/9 1 1/9
ii i
!
"
#
$
%
(b) ds E ds S 0.67148 0.00481 E 0.00481''
22
44
S S
""
ÊÊ(s 1) 3 (s 1) 3
22
œœœÊ œ ¸ œ Ê ¸ kk
(c) 100 100 1%
kk
ˆ‰
E
True Value
0.00481
S
‚œ ‚¸
2
3
9. sin t dt'
0
œ
I. (a) For n 4, x ;œœœœÊœ?
ba 0 x
n44 8
ÊÊ
#
11?1
mf(t ) 2 2 2 4.8284! È
iœ ¸
T 2 2 2 1.89612;Êœ ¸
1
8
Š‹
È
f(t) sin t f (t) cos t f (t) sin tœÊœ Ê œ
www
M 1 E (1)ÊœÊ Ÿ œ kk ˆ‰
T
11 1Ê
##
#
0
14 19
%
0.16149¸
t f(t ) m mf(t )
t0 0 1 0
t/42/22 2
t/21 2 2
t 3 /4 2/2 2 2
t010
ii i
!
"
#
$
%
1
1
1
1
ÈÈ
ÈÈ
(b) sin t dt [ cos t] ( cos ) ( cos 0) 2 E sin t dt T 2 1.89612 0.10388
''
0 0
T
œ œ
œ œ œ Ê œ ¸ œ
1
!
1 kk
(c) 100 100 5%
kkE
True Value 2
0.10388
T
‚œ ‚¸
II. (a) For n 4, x ;œœœœÊœ?
ba 0 x
n44312
ÊÊ 11?1
mf(t ) 2 4 2 7.6569! È
iœ ¸
S 2 4 2 2.00456;Êœ ¸
1
12
Š‹
È
f (t) cos t f (t) sin t
Ð$Ñ Ð%Ñ
œ Ê œ
M 1 E (1) 0.00664ÊœÊ Ÿ ¸ kk ˆ‰
S
11Ê
%
0
180 4
t f(t ) m mf(t )
t0 0 1 0
t/42/2422
t/21 2 2
t3/4 2/2 4 22
t010
ii i
!
"
#
$
%
1
1
1
1
ÈÈ
ÈÈ
(b) sin t dt E sin t dt S 2 2.00456 0.00456 E 0.00456
''
00
S S
œœ
Êœ ¸ œ Ê ¸ kk
(c) 100 100 0%
kkE
True Value 2
0.00456
S
‚œ ‚¸
10. sin t dt
'
0
1
1
I. (a) For n 4, x ;œœœœÊœ?
ba 10 1 x 1
n44 8
ÊÊ
#
?
mf(t ) 2 2 2 4.828! È
iœ ¸
T 2 2 2 0.60355; f(t) sin tÊœ ¸ œ
1
8
Š‹
È
1
f (t) cos tʜ
w
11
f (t) sin t Mʜ ʜ
ww # #
11 1
E 0.05140ÊŸ ¸kk ab ˆ‰
T
10 1
14
Ê
#
#
#
1
t f(t ) m mf(t )
t0 0 1 0
t 1/4 2/2 2 2
t1/2 1 2 2
t 3/4 2/2 2 2
t1 0 1 0
ii i
!
"
#
$
%
ÈÈ
ÈÈ
(b) sin t dt [ cos t] cos cos 0 0.63662 E sin t dt T''
0 0
1 1
T
111 1œ œ œ ¸ Ê œ
""" "
!1111
ˆ‰ˆ‰ kk
2
0.60355 0.03307¸ œ
2
1
(c) 100 100 5%
kk
ˆ‰
E
True Value
0.03307
T
‚œ ‚¸
2
#

546. 542 Chapter 8 Techniques of Integration
II. (a) For n 4, x ;œœœœÊœ?
ba 10 1 x 1
n44312
ÊÊ ?
mf(t ) 2 4 2 7.65685! È
iœ ¸
S 2 4 2 0.63807;Êœ ¸
1
12
Š‹
È
f (t) cos t f (t) sin t
Ð$Ñ $ Ð%Ñ %
œ Ê œ11 11
M E 0.00211ÊœÊ Ÿ ¸11
%% Ê
%
kk ab ˆ‰
S
10 1
180 4
t f(t ) m mf(t )
t0 0 1 0
t1/4 2/24 22
t1/2 1 2 2
t3/4 2/24 22
t1 0 1 0
ii i
!
"
#
$
%
ÈÈ
ÈÈ
(b) sin t dt 0.63662 E sin t dt S 0.63807 0.00145 E 0.00145''
00
11
S S
11œ¸ Ê œ ¸ œ Ê ¸
22
11
kk
(c) 100 100 0%
kk
ˆ‰
E
True Value
0.00145
S
‚œ ‚¸
2
#
11. (a) n 8 x ;œÊ œÊ œ?
""
#816
x?
mf(x ) 1(0.0) 2(0.12402) 2(0.24206) 2(0.34763) 2(0.43301) 2(0.48789) 2(0.49608)!
iœ
2(0.42361) 1(0) 5.1086 T (5.1086) 0.31929œÊœ œ
"
16
(b) n 8 x ;œÊ œÊ œ?
""
8324
x?
mf(x ) 1(0.0) 4(0.12402) 2(0.24206) 4(0.34763) 2(0.43301) 4(0.48789) 2(0.49608)!
iœ
4(0.42361) 1(0) 7.8749 S (7.8749) 0.32812œÊœ œ
"
24
(c) Let u 1 x du 2x dx du x dx; x 0 u 1, x 1 u 0œ Ê œ Ê œ œ Ê œ œ Ê œ
# "
#
x 1 x dx u du u du u 1 0 ;'' '
01 0
10 1
È Èˆ‰ œ ‘’ “ Š‹ Š‹Š‹
È È
œœ œ œœœ# "" " " " " "
##
"Î# $Î#
"$$
!
"
!23333
u
%'œ
œ
3
E x 1 x dx T 0.31929 0.01404; E x 1 x dx S 0.32812 0.00521
T S
0 0
1 1œ¸œ œ¸œ''ÈÈ ## ""
33
12. (a) n 8 x ;œÊ œÊ œ?
3x3
816
?
#
mf( ) 1(0) 2(0.09334) 2(0.18429) 2(0.27075) 2(0.35112) 2(0.42443) 2(0.49026)! )
iœ
2(0.58466) 1(0.6) 5.3977 T (5.3977) 1.01207œÊœ œ
3
16
(b) n 8 x ;œÊ œÊ œ?
3x
838
? "
mf( ) 1(0) 4(0.09334) 2(0.18429) 4(0.27075) 2(0.35112) 4(0.42443) 2(0.49026)! )
iœ
4(0.58466) 1(0.6) 8.14406 S (8.14406) 1.01801œÊœ œ
"
8
(c) Let u 16 du 2 d du d ; 0 u 16, 3 u 16 3 25œ Ê œ Ê œ œÊœ œÊœœ)))))) )
# #"
#
d du u du 25 16 1;'' '
016 16
325 25
)
)
ÈÈ
16 u 2
u
œ
"" " "
##
"Î#
#&
"'
œ
"'œ
œ
)œ œ œ œœˆ‰ ’“Š‹
ÈÈ
1
E d T 1 1.01207 0.01207; E d S 1 1.01801 0.01801
T S
0 0
3 3œ¸œœ¸œ'' ))
))
ÈÈ
16 16œœ
œ œ
))
13. (a) n 8 x ;œÊ œ Ê œ?
1?1
816
x
#
mf(t ) 1(0.0) 2(0.99138) 2(1.26906) 2(1.05961) 2(0.75) 2(0.48821) 2(0.28946) 2(0.13429)!
iœ
1(0) 9.96402 T (9.96402) 1.95643œ Êœ ¸
1
16
(b) n 8 x ;œÊ œ Ê œ?
1?1
8324
x
mf(t ) 1(0.0) 4(0.99138) 2(1.26906) 4(1.05961) 2(0.75) 4(0.48821) 2(0.28946) 4(0.13429)!
iœ
1(0) 15.311 S (15.311) 2.00421œ ʸ ¸
1
24
(c) Let u 2 sin t du cos t dt; t u 2 sin 1, t u 2 sin 3œ Ê œ œÊœ œ œ Êœ œ
1111
####
ˆ‰
dt du 3 u du 3 3 3 2;'''
#œ
œ
œ
œ211
233
3 cos t 3 u 1
(2 sin t) u 1 3 1œÊ
#
$
"
"
œœ
!"œœ œ œœ ’“Š‹ ˆ‰ ˆ‰
E dt T 2 1.95643 0.04357; E dt S
T S
2 2
2 2œ¸œœ ''
#œ
œ œ
œ œ
œ œ
3 cos t 3 cos t
(2 sin t) (2 sin t)œœ
œ œ
2 2.00421 0.00421¸ œ

547. Section 8.7 Numerical Integration 543
14. (a) n 8 x ;œÊ œ Ê œ?
1?1
32 64
x
#
mf(y ) 1(2.0) 2(1.51606) 2(1.18237) 2(0.93998) 2(0.75402) 2(0.60145) 2(0.46364)!
iœ
2(0.31688) 1(0) 13.5488 T (13.5488) 0.66508œÊ¸ œ
1
64
(b) n 8 x ;œÊ œ Ê œ?
1?1
32 3 96
x
mf(y ) 1(2.0) 4(1.51606) 2(1.18237) 4(0.93988) 2(0.75402) 4(0.60145) 2(0.46364)!
iœ
4(0.31688) 1(0) 20.29734 S (20.29734) 0.66423œÊ¸ œ
1
96
(c) Let u cot y du csc y dy; y u 1, y u 0œÊœ œÊœœÊœ
#
#
11
4
csc y cot y dy u ( du) u du 1 0 ;'''
œ
œ
œ
œ410
201
ab È È ’“ Š‹ Š‹
È È
# "Î#
"$$
!
œœ œœœ
u2 2 2
33 3
%'œ
3
2
E csc y cot y dy T 0.66508 0.00159; E csc y cot y dy S
T S
4 4
2 2œ¸œœ ''
œ œ
œ œ
œ œ
œ œ
ab abÈÈ
# #2
3
0.66423 0.00244¸ œ
2
3
15. (a) M 0 (see Exercise 1): Then n 1 x 1 E (1) (0) 0 10œœÊœÊœœ ? kk
T
"
#
#%
1
(b) M 0 (see Exercise 1): Then n 2 (n must be even) x E (0) 0 10œœÊœÊœœ ?
"""
##
%
Ê!
kk ˆ‰
S
180
16. (a) M 0 (see Exercise 2): Then n 1 x 2 E (2) (0) 0 10œœÊœÊœœ ? kk
T
2
1#
#%
(b) M 0 (see Exercise 2): Then n 2 (n must be even) x E (1) (0) 0 10œœÊœ"Êœœ ? kk
S
2
180
!Ê!
17. (a) M 2 (see Exercise 3): Then x E (2) 10 n 10 n 10œœÊŸœÊÊ ?
2224 4 4
n12n3n 3 3
kk ab abˆ‰ É
T
#
% # % %
œ
n 115.4, so let n 116Ê œ
(b) M 0 (see Exercise 3): Then n 2 (n must be even) x E (1) (0) 0 10œœÊœ"Êœœ ? kk
S
2
180
!Ê!
18. (a) M 2 (see Exercise 4): Then x E (2) 10 n 10 n 10œœÊŸœÊÊ ?
2224 4 4
n12n3n 3 3
kk ab abˆ‰ É
T
#
% # % %
œ
n 115.4, so let n 116Ê œ
(b) M 0 (see Exercise 4): Then n 2 (n must be even) x E (1) (0) 0 10œœÊœ"Êœœ ? kk
S
2
180
!Ê!
19. (a) M 12 (see Exercise 5): Then x E (12) 10 n 8 10 n 8 10œœÊŸœÊÊ ?
2228
n12nn
kk ab abˆ‰ È
T
#
% # % %
œ
n 282.8, so let n 283Ê œ
(b) M 0 (see Exercise 5): Then n 2 (n must be even) x E (1) (0) 0 10œœÊœ"Êœœ ? kk
S
2
180
!Ê!
20. (a) M 6 (see Exercise 6): Then x E (6) 10 n 4 10 n 4 10œœÊŸœÊÊ ?
2224
n12nn
kk ab abˆ‰ È
T
#
% # % %
œ
200, so let n 201œœ
(b) M 0 (see Exercise 6): Then n 2 (n must be even) x E (1) (0) 0 10œœÊœ"Êœœ ? kk
S
2
180
!Ê!
21. (a) M 6 (see Exercise 7): Then x E (6) 10 n 10 n 10œœÊŸœÊÊ ?
1111
n12n2n
kk ab abˆ‰ É
T
#
% # % ""
##
%
œ
n 70.7, so let n 71Ê œ
(b) M 120 (see Exercise 7): Then x E (120) 10 n 10œœÊœœÊ ?
""
%
Ê! ! !
n 180 n 3
12 2
3n
kk a bˆ‰
S #
n 10 n 9.04, so let n 10 (n must be even)Ê Ê œ
% %
Éab
2
3
22. (a) M 6 (see Exercise 8): Then x E (6) 10 n 4 10 n 4 10œœÊŸœÊÊ ?
2224
n12nn
kk ab abˆ‰ È
T
#
% # % %
œ
n 200, so let n 201Ê œ
(b) M 120 (see Exercise 8): Then x E (120) 10 n 10œœÊŸœÊ ?
22264 64
n 180 n 3 3n
kk a bˆ‰
S
%
Ê! ! !
#
n 10 n 21.5, so let n 22 (n must be even)Ê Ê œ
% %
Éab
64
3

548. 544 Chapter 8 Techniques of Integration
23. (a) f(x) x 1 f (x) (x 1) f (x) (x 1) M .œÊ œ Ê œ œ Êœ œÈ
w "Î# ww $Î#"""""
#
œ
44
4x1 41
ˆ‰È
Š‹
È
%%
Then x E 10 n 10 n 10 n 75,?œÊ Ÿ œ Ê Ê Ê
3339 9 9
n 12 n 4 16n 16 16
kk ab abˆ‰ˆ‰ É
T
#
" % # % %
œ
so let n 76œ
(b) f (x) (x 1) f (x) (x 1) M . Then x
Ð$Ñ &Î# Ð%Ñ (Î#
œ
œ Ê œ œ Êœ œ œ
3151515153
816 16n
16 x 1 16 1
ˆ‰È
Š‹
È
?
E 10 n n n 10.6, so letÊ Ÿ œ Ê Ê Êkk ˆ‰ˆ ‰ É
S
33 15
180 n 16 16(180) 16(180)
3 (15)
16(180)n
3 (15) 10 3 (15) 10%
Ê! !

#
&# # &# ˆ‰ ab
n 12 (n must be even)œ
24. (a) f(x) f (x) (x 1) f (x) (x 1) M .œÊœ Êœœ Êœ œ
""
œ
w $Î# w &Î#
#
w
œ
È
ˆ‰È
Š‹
Èx1
3333
44
4x1 41

Then x E 10 n n n 129.9, so let?œÊ Ÿ œ Ê Ê Ê
33333
n 12 n 4 48n 48 48
310 310
kk ˆ‰ˆ‰ É
T
#
% # #
œ
## ## ˆ‰ ab
n 130œ
(b) f (x) (x 1) f (x) (x 1) M . Then x
Ð$Ñ (Î# Ð%Ñ *Î#
œ
œ Ê œ œ Ê œ œ œ
15 105 105 105 105 3
816 16n
16 x 1 16 1
ˆ‰È
Š‹
È
?
E 10 n n n 17.25, soÊ Ÿ œ Ê Ê Êkk ˆ‰ˆ ‰ É
S
3 3 105
180 n 16 16(180) 16(180)
3 (105)
16(180)n
3 (105) 10 3 (105) 10%
Ê! !

#
&# # &# ˆ‰ ab
let n 18 (n must be even)œ
25. (a) f(x) sin (x 1) f (x) cos (x 1) f (x) sin (x 1) M 1. Then x E (1)œ Ê œ Ê œ Êœ œÊ Ÿ
www
#
?
222
n12n
kk ˆ‰
T
10 n n n 81.6, so let n 82œ Ê Ê Ê œ
8
12n 12 1
810 810
œ
# #
% #
#
ˆ‰ ab É
(b) f (x) cos (x 1) f (x) sin (x 1) M 1. Then x E (1) 10
Ð$Ñ Ð%Ñ Ê!
%
œ Ê œ Ê œ œ Ê Ÿ œ ?
22232
n 180 n 180n
kk ˆ‰
S #
n n n 6.49, so let n 8 (n must be even)Ê Ê Ê œ
%
32 10
180 180
32 10ˆ‰ ab
# # #
É
26. (a) f(x) cos (x ) f (x) sin (x ) f (x) cos (x ) M 1. Then xœÊœÊœÊœ œ11 1?
www 2
n
E (1) 10 n n n 81.6, so let n 82Ê Ÿ œ Ê Ê Ê œkk ˆ‰ É
T
22 8
12 n 12n 12 1
810 810#
% #
#
œ
# #
ˆ‰ ab
(b) f (x) sin (x ) f (x) cos (x ) M 1. Then x E (1) 10
Ð$Ñ Ð%Ñ Ê!
%
œÊ œ Êœ œÊŸ œ11?
22232
n 180 n 180n
kk ˆ‰
S #
n n n 6.49, so let n 8 (n must be even)Ê Ê Ê œ
%
32 10
180 180
32 10ˆ‰ ab
# # #
É
27. 6.0 2 8.2 2 9.1 2 12.7 13.0 30 15,990 ft .
5
2
3
ababab ab a b ÞÞÞ œ
28. (a) Using Trapezoid Rule, x 200 100;?œÊœœ
?x 200
22
mf(x ) 13,180 Area 100 (13,180)!
iœÊ¸
1,318,000 ft . Since the average depth 20 ftœœ
#
we obtain Volume 20 (Area) 26,360,000 ft . ¸¸
$ (b) Now, Number of fish 26,360 (to the nearestœœ
Volume
1000
fish) Maximum to be caught 75% of 26,360ʜ
19,770 Number of licenses 988œÊ œœ
19,770
0#
x f(x ) m mf(x )
x 0 0 1 0
x 200 520 2 1040
x 400 800 2 1600
x 600 1000 2 2000
x 800 1140 2 2280
x 1000 1160 2 2320
x
ii i
!
"
#
$
%
'
5
1200 1110 2 2220
x 1400 860 2 1720
x 1600 0 1 0
(
)

549. Section 8.7 Numerical Integration 545
29. Use the conversion 30 mph 44 fps (ft perœ
sec) since time is measured in seconds. The
distance traveled as the car accelerates from,
say, 40 mph 58.67 fps to 50 mph 73.33 fpsœœ
in (4.5 3.2) 1.3 sec is the area of theœ
trapezoid (see figure) associated with that time
interval: (58.67 73.33)(1.3) 85.8 ft. The
"
#
œ
total distance traveled by the Ford Mustang
Cobra is the sum of all these eleven trapezoids
(using and the table below):
?t
#

s (44)(1.1) (102.67)(0.5) (132)(0.65) (161.33)(0.7) (190.67)(0.95) (220)(1.2) (249.33)(1.25)œ
(278.67)(1.65) (308)(2.3) (337.33)(2.8) (366.67)(5.45) 5166.346 ft 0.9785 miœ¸
v (mph) 0 30 40 50 60 70 80 90 100 110 120 130
v (fps) 0 44 58.67 73.33 88 102.67 117.33 132 146.67 161.33 176 190.67
t (sec) 0 2.2 3.2 4.5 5.9 7.8 10.2 12.7 16 20.6 26.2 37.1
t/2 0 1.1 0.5 0.65 0.7 0.95 1.2 1.25 1.65 2.3 2.8 5.45?
30. Using Simpson's Rule, x 4;?œœ œœ
ba 240 24
n66
ÊÊ
my 350 S (350) 466.7 in.!
iœÊœ œ¸
4 1400
33
#
x y m my
x 0 0 1 0
x 4 18.75 4 75
x 8 24 2 48
x 12 26 4 104
x 16 24 2 48
x 20 18.75 4 75
x24 0 1 0
ii i
5
!
"
#
$
%
'
31. Using Simpson's Rule, x 1 ;?œÊ œ
?x
33
"
my 33.6 Cross Section Area (33.6)!
iœÊ ¸
"
3
11.2 ft . Let x be the length of the tank. Then theœ
#
Volume V (Cross Sectional Area) x 11.2x.œœ
Now 5000 lb of gasoline at 42 lb/ft
$ V 119.05 ftÊœ œ
5000
42
$
119.05 11.2x x 10.63 ftʜʸ
x y m my
x 0 1.5 1 1.5
x 1 1.6 4 6.4
x 2 1.8 2 3.6
x 3 1.9 4 7.6
x 4 2.0 2 4.0
x 5 2.1 4 8.4
x 6 2.1 1 2.1
ii i
5
!
"
#
$
%
'
32. 4.2 L0.019 2 0.020 2 0.021 2 0.031 0.035
24
2
cd abab abÞÞÞ œ
33. (a) E x M; n 4 x ; f 1 M 1 E (1) 0.00021kk a b kk ¸¸ ˆ‰
S SŸœÊœœŸÊœÊŸ¸
ba
180 4 8 180 8
00Ê %Ð%Ñ
ÊÊ %
??
# #
œ œ
11
ˆ‰
(b) x ;?œÊ œ
1?1
834
x
#
mf(x ) 10.47208705!
iœ
S (10.47208705) 1.37079Êœ ¸
1
#4
x f(x ) m mf(x )
x0 1 1 1
x /8 0.974495358 4 3.897981432
x /4 0.900316316 2 1.800632632
x 3 /8 0.784213303 4 3.136853212
x /2 0.636619772 1 0.63
ii i 1
!
"
#
$
%
1
1
1
1 6619772
(c) 100 0.015%¸‚¸ˆ‰
0.00021
1.37079
34. (a) x 0.1 erf 1 y 4y 2y 4y 4y y?œ œ œÊ œ ÞÞÞ
ba 10 2 0.1
n10 3 3
0123 910
ÊÊ
ab a bˆ‰
È
e 4e 2e 4e 4e e 0.843
2
30
0 0.01 0.04 0.09 0.81 1
È1
abÞÞÞ¸
ÊÊÊ ÊÊ
(b) E 0.1 12 6.7 10kk a bab
SŸ¸‚
10
180
4 6Ê Ê

550. 546 Chapter 8 Techniques of Integration
35. (a) n 10 x ;œÊ œ œÊ œ?
11?1Ê
##
0x
10 10 0
mf(x ) 1(0) 2(0.09708) 2(0.36932) 2(0.76248) 2(1.19513) 2(1.57080) 2(1.79270)!
iœ
2(1.77912) 2(1.47727) 2(0.87372) 1(0) 19.83524 T (19.83524) 3.11571œÊœ œ
1
#0
(b) 3.11571 0.025881ÎÑ
(c) With M 3.11, we get E (3.11) (3.11) 0.08036œŸœ kk ˆ‰
T
11 1
1 10 1200#
#
!
36. (a) f (x) 2 cos x x sin x f (x) 3 sin x x cos x f (x) 4 cos x x sin x. From the graphs
ww Ð$Ñ Ð%Ñ
œÊœ Êœ
shown below, 4 cos x x sin x 4.8 for 0 x .kk ŸŸ 1

(b) n 10 x E (4.8) 0.00082œÊ œÊ Ÿ ¸?
111
10 180 10
kk ˆ‰
S
%
(c) mf(x ) 1(0) 4(0.09708) 2(0.36932) 4(0.76248) 2(1.19513) 4(1.57080) 2(1.79270)!
iœ
4(1.77912) 2(1.47727) 4(0.87372) 1(0) 30.0016 S (30.0016) 3.14176œÊœ œ
1
30
(c) 3.14176 0.00017kk1ÎÑ
37. T y 2y 2y 2y 2y y where x and f is continuous on [a, b]. Soœ ÞÞÞ œ
?x ba
2 n
0123 n1n
ab Ê
Ê ?
T .œ œ ÞÞÞ
ba ba
n2 n222
yyyyy y y y fx fx fx fx fx fxÊÊ œœœœœ ÁÁÁœ œ œ œ œ œa b ab ab ab ab a b ab01122 n1n1n 0 1 1 2 n1 n !! !
Š‹
Since f is continuous on each interval [x , x ], and is always between f x and f x , there is a point c k1 k k1 k k
fx fx
ÊÊ
œ
#
abab
k1 k!
ab ab in
[x , x ] with f c ; this is a consequence of the Intermediate Value Theorem. Thus our sum isk1 k k
fx fx
Ê
œ
# abœ
ababk1 k!
f c which has the form x f c with x for all k. This is a Riemann Sum for f on [a, b].!!ˆ‰ab ab
k1 k1
nn
ba ba
nn
kkkk
œœ
ÊÊ
?? œ
38. S y 4y 2y 4y 2y 4y y where n is even, x and f is continuous on [a, b]. Soœ ÞÞÞ œ
?x ba
3 n
0123 n2n1n
ab ÊÊ
Ê ?
Sœ ÞÞÞ
ba
n333 3
y4yy y4yy y4yy y 4y yÊ œœ œœ œœ œ œ
ˆ‰
012 234 456 n2n1n !!
œ ÞÞÞ
ba fx 4fx fx fx 4fx fx fx 4fx fx fx 4fx fx
666 6
Ê œœ œœ œœ œ œ
n
2
012 234 456 n2n1nŠ ‹
ab ab ab ab ab ab ab ab ab a b a b ab !!
is the average of the six values of the continuous function on the interval [x , x ], so it is b
fx 4fx fx
6
2k 2k 2
ababab2k 2k 1 2k 2œœ
œ ÎÎ
etween
the minimum and maximum of f on this interval. By the Extreme Value Theorem for continuous functions, f takes on its
maximum and minimum in this interval, so there are x and x with x x , x x and
a b 2k a b 2k 2ŸŸ œ
f x f x . By the Intermediate Value Theorem, there is c in [x , x ] withab abab k2k2k2
fx 4fx fx
6ŸŸ
ababab2k 2k 1 2k 2œœ
œ ÎÎ
f c . So our sum has the form x f c with x a Riemann sum for f on [a, b].ab ab ! k kk k
fx 4fx fx
6 n/2
k1
n/2
baœœß
ababab
ab2k 2k 1 2k 2œœ
œ
Ê ÎÎ
??
Exercises 39-42 were done using a graphing calculator with n 50œ
39. 1.08943 40. 1.37076 41. 0.82812 42. 51.05400

551. Section 8.7 Numerical Integration 547
43. (a) T 1.983523538 44. (a) S 2.00010951710 10¸¸
T 1.999835504 S 2.000000011
100 100¸¸
T 1.999998355 S 2.000000000
1000 1000¸¸
(b) n 2 T (b) n 2 SEE
10 0.016476462 1.6476462 10 10 1.09517 10
100 1.64496 10 100 1.1 10
1000 1.646 10 1000 0
kk kk TS nn
24
48
6œ œ
œ‚ ‚
‚‚
‚
ÊÊ
ÊÊ
Ê
(c) 10 (c) 10EE EEkk kk kk kkTT SS
2 410n n 10n n¸¸
Ê Ê
(d) b a , x , M 1 (d) b a , x , M 1œ œ œ œ œ œ1? 1?ab ab
2 4
n n
1 1
2 4
2 4
EEkk kk Š‹ Š‹ T S
12 n 12n 180 n 180nn n
23 45
22 44Ÿœ Ÿ œ
11 1 1 1 1
10 10EE E Ekk kk kk kkTT S S
12 10n 180 10n
2 410n n 10n n
3 5
2 4ŸŸ Ÿ Ÿ
1 1
ab ab
Ê Ê
45. (a) f x 2x cos x , f x 2x 2x sin x 2cos x 4x sin x 2cos x
www
ab ab ab a b ab ab ab abœœ†œ
222222
(b)
(c) The graph shows that 3 f x 2 so f x 3 for 1 x 1.** *Î**
ww ww
ab k k ab
(d) x 3Ekk ababT
11 2 x
12Ÿœ
ÊÊ
#
ab ab
?
?
2
(e) For 0 x 0.1, 0.005 0.01E Ÿ Ÿœ ? kk T
x 0.1
2
ab?
2
2
#
(f) n 20  œ
11
x0.1
2ÊÊab
?
46. (a) f x 4x 2x cos x 8x sin x 4x sin x 8x cos x 12x sin x
''' 2 2 2 2 3 2 2
ab ab ab ab ab abœ † œ
f x 8x 2x sin x 24x cos x 12x 2x cos x 12 sin x 16x 12 sin x 48 x cos x
ab43222 2 24222
ab ab ab ab ab a b ab abœ † † œ
(b)
(c) The graph shows that 30 f x 0 so f x 30 for 1 x 1.Î* * * Î**
ab ab44
ab ab¸¸
(d) x 30Ekk ababS
11 4 x
180 3Ÿœ
ÊÊab ab
?
?
4
(e) For 0 x 0.4, 0.00853 0.01E Ÿ Ÿ¸ ? kk S
x
33
0.4ab?
4
2
(f) n 5  œ
11
x0.4
2ÊÊab
?
47. (a) Using d , and A yields the following areas (in square inches, rounded to the nearest tenth):œœœ
CdC
24
2
11
1ˆ‰
2
2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7, 10.7, 9.3, 6.4, 3.2
(b) If C y is the circumference as a function of y, then the area of a cross section isab
A y , and the volume is C y dy.ab abŠ‹œœ1
Cy
24 4
2
Cy 1 2ab abÎ1
11
2
'
0
6

552. 548 Chapter 8 Techniques of Integration
(c) A y dy C y dy''
00
66
ab abœ
1
4
2
1
5.4 2 4.5 4.4 5.1 6.3 7.8 9.4 10.8 11.6 11.6 10.8 9.0 6.3¸ # ##### ######
160
424
2 22222 2222222
1
ˆ‰c da b
Ê
34.7 in¸
3
(d) V C y dy 5.4 4 4.5 2 4.4 4 5.1 2 6.3 4 7.8 2 9.4 10.8œ ¸ %
1160
4436
2 2222222 2
11'
0
6
ab ababababababa bˆ‰’
Ê
2 11.6 11.6 2 10.8 9.0 6.3 34.792 in%%œabababab “
22222 3
by Simpson's Rule. The Simpson's Rule estimate should be more accurate than the trapezoid estimate. The error in the
Simpson's estimate is proportional to y 0.0625 whereas the error in the trapezoid estimate is proportional toab?
4
œ
y 0.25, a larger number when y 0.5 in.ab??
2
œœ
48. (a) Displacement Volume V y 4y 2y 4y 2y 4y y , x 0, x 10 x,¸ ÞÞÞ œ œ
?x
3
0123 n2n1n0 n
ab ÊÊ ?
x 2.54, n 10 A x dx 0 4 1.07 2 3.84 4 7.82 2 12.20 4 15.18 2 16.14?œœÊ ¸ '
x
x
2.54
30
n
ab abababa ba ba b’
4 14.00 2 9.21 4 3.24 0 248.02 209.99 210 ft . œ œ¸a babab a b “
2.54
3
3
(b) The weigth of water displaced is approximately 64 120 13,440 lb.†œ
(c) The volume of a prism 2.54 16.14 409.96 410 ft . Thus, the prismatic coefficient is 0.51.œœ¸ ¸aba b
3 210 ft
410 ft
3
3
49. (a) a 1, e Length 4 1 cos t dtœœÊ œ
""
#
# '
0
2
œœ
É
4
2 4 cos t dt f(t) dt; use theœœ ''
00
22
œœœœ
È #
Trapezoid Rule with n 10 tœÊ œ œ?
ba
n10
0Ê Êˆ‰
#
œ
. 4 cos t dt mf(x ) 37.3686183œ¸œ
1
#
#
0'
0
2 10
n0
n
œœ

È ! T (37.3686183) (37.3686183)Êœ œ
?1t
40#
2.934924419 Length 2(2.934924419)œÊœ
5.870¸
(b) f (t) 1 M 1kk
ww
Ê œ
E t M 1 0.0032ÊŸ Ÿ Ÿkk a b ˆ‰
T
ba
110
0Ê
###
#
Ê #
?
ˆ‰
#
œ
1
x f(x ) m mf(x )
x 0 1.732050808 1 1.732050808
x /20 1.739100843 2 3.478201686
x /10 1.759400893 2 3.
ii i
!
"
#
1
1 518801786
x 3 /20 1.790560631 2 3.581121262
x /5 1.82906848 1 3.658136959
x /4 1.870828693 1 3.741657387
x 3 /10 1.911676881 2 3.823353762
$
%
' 1
1
1
1
5
x 7 /20 1.947791731 2 3.895583461
x 2 /5 1.975982919 2 3.951965839
x 9 /20 1.993872679 2 3.987745357
x/2 2 1 2(
)
*
"!1
1
1
1
50. x ; mf(x ) 29.184807792?œœÊœ œ
11?1Ê0x
88324 !
i
S 29.18480779 3.82028Êœ ¸
1
24
ab
x f(x ) m mf(x )
x 0 1.414213562 1 1.414213562
x /8 1.361452677 4 5.445810706
x /4 1.224744871 2 2.44948974
ii i
!
"
#
1
1 3
x 3 /8 1.070722471 4 4.282889883
x/2 1 2 2
x 5 /8 1.070722471 4 4.282889883
x 3 /4 1.224744871 2 2.449489743
x 7 /8 1.361452677 4 5.4458107
$
%
'
( 1
1
1
1
1
5
06
x 1.414213562 1 1.414213562
) 1
51. The length of the curve y sin x from 0 to 20 is: L 1 dx; cos x œœœÊˆ‰ ˆ‰ Ê Š‹ Š‹
333
0dxdx00dx
dy dy dy111
###
## '
0
20
cos x L 1 cos x dx. Using numerical integration we find L 21.07 inœÊœ ¸
93 93
400 0 400 0
11 11
œ œ
#
##
#ˆ‰ ˆ‰ É '
0
20
52. First, we'll find the length of the cosine curve: L 1 dx; sinœ œ '
#25
25
Ê Š‹ ˆ‰
dy dy
dx dx 50 50
25 x
#
11
sin L 1 sin dx. Using a numerical integrator we findÊœ Êœ Š‹ ˆ‰ ˆ‰ É
dy
dx 4 50 4 50
xx
#
# #11 11
œ œ
'
#25
25

553. Section 8.7 Improper Integrals 549
L 73.1848 ft. Surface area is: A length width (73.1848)(300) 21,955.44 ft.¸œ¸œ †
Cost 1.75A (1.75)(21,955.44) $38,422.02. Answers may vary slightly, depending on the numericalœœ œ
integration used.
53. y sin x cos x cos x S 2 (sin x) 1 cos x dx; a numerical integration givesœÊœ Ê œ Êœ
dy dy
dx dx
Š‹
È
#
# #
'
0
œ
1
S 14.4¸
54. y S 2 1 dx; a numerical integration gives S 5.28œÊœÊ œÊœ ¸
xx x xx
4dx dx 4 4 4
dy dy
œœœ œ
#
#
Š‹ Š‹ É '
0
2
1
55. y x sin 2x 1 2 cos 2x (1 2 cos 2x) ; by symmetry of the graph we have thatœ Ê œ Ê œ
dy dy
dx dx
Š‹
#
#
S 2 2 (x sin 2x) 1 (1 2 cos 2x) dx; a numerical integration gives S 54.9œ ¸'
0
23
œœ
1 È
#
56. y 36 x œÊœ œ œ
xxx
1dx1212 12 1
dy ( 2x)36 x 36 x
36 x 12 36 x 36 x
36 x x
###
# ÊÊ ""Ê
ÊÊÊ
ÊÊÈ
ÈÈ
ÈÈÈ
ab œœ
œœœ
œ œœ
†
S 36 x 1 dxœœÊœÊœ
"
#
ÊÊ Ê Ê
ÊÊ
#
#
1 dx 36 36 x 12 36 36 x
36 2x 18 x 18 x 18 x
36 x 6 36 x
dy 2xa bab ab ab
ÈÈ ab ab
œœ œ
œœ
œ
œ œ
œ
œ
Š‹
È É '
0
6
1†
36 x dx; using numerical integration we get S 41.8œ ¸'
0
6
1x18x
66Éab ˆ‰ # Ê
#
œ
57. A calculator or computer numerical integrator yields sin 0.6 0.643501109.
Ê"
¸
58. A calculator or computer numerical integrator yields 3.1415929.1¸
8.8 IMPROPER INTEGRALS
1. lim lim tan x lim tan b tan 0 0''
00
b
b
0

dx dx
x 1 x 1
œœ
##
Ê" Ê" Ê"
œœ œœœ
bb bÄ_ Ä_ Ä_
cd a b
11
2. lim lim 1000x lim 1000 1000''
11
b
0001b

#*
w
dx dx 1000
xx b
1 001 1 001 0 001œœ œœœœœ
bb bÄ_ Ä_ Ä_
cd ˆ‰
Ê
3. lim x dx lim 2x lim 2 2 b 2 0 2''
0b
11
12
b
dx
xÈ
œ œ œ œœ
bb0b0Ä! ÄÄ
$$$
œ
w
"Î#
cd Š‹
È
4. lim 4 x dx lim 2 4 b 2 4 0 4 4''
00
4b
dx
4 x
È
Ê
"Î#
œ œ œœ
b4 b4ÄÄ
##
ab ’“
È
Š‹
È
5. lim 3x lim 3x'''
## # # $110
101
b
1c
dx dx dx
xxx
œ'! œ'! œ'!œœ
b cÄ! Ä!
œ ‘ œ ‘
"Î$ "Î$
"
lim 3b 3( 1) lim 3(1) 3c (0 3) (3 0) 6œ œœ
b cÄ! Ä!
# $
œ ‘ œ ‘
"Î$ "Î$ "Î$ "Î$
6. lim x lim x'''
## # #8 $880
101
b
c
dx dx dx 3 3
xxx
")% ")% ")%œœ
b cÄ! Ä!
œ ‘ œ ‘
##
#Î$ #Î$
"
lim b ( 8) lim (1) c 0 (4) 0œ œœ
b cÄ! Ä!
# $
œ ‘ œ ‘œ ‘ˆ ‰
33 3 3 3 3 9
## # # # # #
#Î$ #Î$ #Î$ #Î$
7. lim sin x lim sin b sin 0 0'
0
1
b
0
dx
1 x
È
Ê
Ê" Ê" Ê"
##
œ
œœœœ
b1 b1ÄÄ
##
cd a b
11

554. 550 Chapter 8 Techniques of Integration
8. lim 1000r lim 1000 1000b 1000 0 1000'
0
1
0001 00011
b
dr
r
0 999œœœœœ
bbÄ! Ä!
$$

cd a b
9. lim ln x 1 lim ln x 1 lim ln '''
#'#'#'
###
## #&
222
22
bb
b
2 dx dx dx x 1
x 1 x 1 x 1 x 1
œ
ÊÊœ œ
Ê
œœ œ
bb bĜ_ Ĝ_ Ĝ_
cd cdkk kk œ ‘¸¸
lim ln ln ln 3 ln lim ln 3 ln 1 ln 3œ œ œœ
bbĜ_ Ĝ_
ˆ‰¸¸ ¸ ¸ Š‹
ÊÊ Ê"
Êœ œ
3b 1 b
1b 1 b 1
10. lim tan lim tan 1 tan'
#
2
2
b
2 dx x b 3
x 4 2 4 4
œ
# #
Ê" Ê" Ê"
œœœœ
bbĜ_ Ĝ_
œ ‘ˆ ‰ˆ‰
111
11. lim 2 ln lim 2 ln 2 ln 2 ln (1) 2 ln 0 2 ln 2 ln 4'
2
b


2 dv v b 2
v v v b
œ
Ê ##
Ê" Ê" Ê" "
œœœœœ
bbÄ_ Ä_
œ ‘ˆ ‰ ˆ‰¸¸ ¸¸ ¸¸
12. lim ln lim ln ln ln (1) ln 0 ln 3 ln 3'
2
b


2 dt t b 2
t 1 t 1 b 1 1 3
œ
#
Ê" Ê" Ê" "
œœœœœ
bbÄ_ Ä_
œ ‘ˆ ‰ ˆ‰¸¸ ¸¸ ¸¸
13. ; lim lim
ux 1
du 2x dx''' ''
#'#'
'' '
2x dx 2x dx 2x dx du du
x 1 x 1 x 1 uu u uab ab ab
œœœ
œœœ œœ
œœœ
#
""
œ Äœ
œ
œ
0 1
01
1c
b1
”•
œ‘ œ‘
b cÄ_ Ä_
lim 1 lim ( 1) ( 1 0) (0 1) 0œ œœ
b cÄ_ Ä_
ˆ‰ œ ‘
""
bc
14. ;
ux 4
du 2x dx''' ''
#'#'
'' '
x dx x dx x dx du du
x 4 x 4 x 4 2u 2uab ab ab
œœœ
%'œ !'œ !'œ !'œ !'œ
œœœ
#
œ Ä
œ
œ
0 4
04
”•
lim lim lim lim 0 0 0œ œ œ œ
bb ccÄ_ Ä_Ä_ Ä_
’“ ’“ Š ‹ Š ‹ ˆ‰ˆ‰
""""""""
####ÈÈ ÈÈuu c b


b
c
15. d ; lim lim u lim 3 b
u2
du 2( ) d'''
00b
133
3
b
)
))
du du
2 2u 2u
œ"
œ
#
ÈÈÈ
œ
)
))
))
”•
œ‘È Š‹
È È
œ
œ"
Äœ œ œ
bbbÄ! Ä! Ä!
$$$
30 3œœ
ÈÈ
16. ds ; lim
u4s
du 2s ds''' ''
000 40
222 0c
s 2s ds ds du ds
4 s 4 s 4 s 4 s u
œ" " "
ÊÊÊ Ê
##
#
ÈÈÈ ÈÈ
œœœ œ
œ Ä
œ
œ
”•
c2Ä
#
lim lim lim u lim sin œ œ
bb c2 c2Ä! Ä!ÄÄ
$$ ##
''
b0
4c
4
b
c
du ds s
2u 4 s
ÈÈ
Ê
Ê"
#!
œ
œ‘ œ ‘È
lim 2 b lim sin sin 0 (2 0) 0œ œœ
b c2Ä! Ä
$ #
Š‹
È ˆ‰ˆ‰
Ê" Ê"
###
œc4 11
17. ; lim lim 2 tan u
ux
du'''
000
b
b
0

dx 2 du 2 du
(1 x) x
dx
2x
u 1 u 1œ œœ
Ê"
È
È
–—
È
cd
œ
œ
Äœ œ
œœ
bbÄ_ Ä_
lim 2 tan b 2 tan 0 2 2(0)œœœ
bÄ_
ab ˆ‰
Ê" Ê"
#
1
1
18. lim lim ''' ' '
112 b 2
22c

$
dx dx dx dx dx
x x 1 x x 1 x x 1 x x 1 x x 1
ÈÈ È È È
œœ œ œ œ
ÊÊ Ê Ê Ê
œ œ
b1
c
Ä
Ä_
lim sec x lim sec x lim sec 2 sec b lim sec c sec 2œœ
b1 b1
cc
ÄÄ
Ä_ Ä_
$$
cd cd a b a bkk kk
Ê" Ê" Ê" Ê" Ê" Ê"
2c
b2
0œœˆ‰ˆ ‰
1111
33 ##
19. lim ln 1 tan v lim ln 1 tan b ln 1 tan 0'
0
b
0

dv
1 v 1 tan vaba bœœ
Ê" Ê" Ê"
œ !" œœ
bbÄ_ Ä_
cdcdkkkk kk
ln 1 ln (1 0) ln 1œœˆ‰ ˆ‰
11
##
20. dx lim 8 tan x lim 8 tan b 8 tan 0 8 8(0) 2'
0
b
0

16 tan x
1 x
!"
œ
œ #
" " " #
### #
œœœœ
bbÄ_ Ä_
’“’“ab abab ˆ‰
1
1

555. Section 8.7 Improper Integrals 551
21. e d lim e e 0 e e lim be e 1 lim '
#
0
0
b
bb
)) )
)))
œœœ
bbbĜ_ Ĝ_ Ĝ_
œ ‘ˆ‰ab cd†
!! Ê"b
e
!b
1 lim (l'Hopital's rule for form)^œ
bĜ_
ˆ‰
"_
Ê _e
!b
10 1œ œ
22. 2e sin d lim 2e sin d
''
00
b

ÊÊ))
)) ))œ
bÄ_
lim 2 ( sin cos ) (FORMULA 107 with a 1, b 1)œ œœ
bÄ_
’“
e
1 1
!œ
œ
))
b
0
lim 0 1œœœ
bÄ_
Êœ œ œ2(sin b cos b) 2(sin 0 cos 0) 2(0 1)
2e 2e 2
b Þ
23. e dx e dx lim e lim 1 e (1 0) 1''
#'#'
#
00
xx x b
b
##
œ œ œ œœ
bbĜ_ Ĝ_
cd a b
!
24. 2xe dx 2xe dx 2xe dx lim e lim e'''
#'#'

### # #
xxx x x
0
0
0c
b0
$$$ $ $
œœ
b cÄœ_ Ä_
cd cd
lim 1 e lim e ( 1) ( 1 0) (0 1) 0œ œœ
b cÄœ_ Ä_
cdcdab
##bc
$$
25. x ln x dx lim ln x ln 1 lim ln b lim 0'
0
1 1
b
œœœ
bbbÄ! Ä! Ä!
$$$
’“ Š‹ ˆ‰
xx bb ln b
44 44
œœ œœ
œ
## #
"" "
Š‹ 2
b
lim lim 0œ œ œ œ
""""
Ê
44444
b
bbÄ! Ä!
$$
Š‹
Š‹
$
#
b
4
b
Š‹
œ
26. ( ln x) dx lim x x ln x 1 1 ln 1 lim b b ln b 1 0 lim 1 lim '
0
1
b
œ œ œ œ
bbbbÄ! Ä! Ä! Ä!
$$$$
cdcd cd
"
Ê
ln b
Š‹ Š ‹
Š‹
$$
$
$
b
b
b
1 lim b101œ œœ
bÄ!
$
27. lim sin lim sin sin 0 0'
0
2
b
0
ds s b
4 s
È
Ê
Ê" Ê" Ê"
####
œ
œœœœ
b2 b2ÄÄ
##
œ ‘ˆ‰
11
28. lim 2 sin r lim 2 sin b 2 sin 0 2 0'
0
1
b
0
4r dr
1 r
È
Ê
" # " # "
#
#
œœœœ
b1 b1ÄÄ
##
cd c dab a b †
1
1
29. lim sec s sec 2 lim sec b 0'
1
2
b
ds
ss 1
33È
œ
Ê
Ê" Ê" Ê"
œœœœ
b1 b1ÄÄ
$$

cd
11
30. lim sec lim sec sec 0'
2
4
b
dt t 4 b
tt 4
36È
œ
Ê
"""""
## ### ## #
Ê" Ê" Ê"
œœœœ
b2 b2ÄÄ
$$

œ ‘ œ ‘ˆ‰ˆ‰ ˆ‰
11
†
31. lim lim lim 2 x lim 2 x'''
ÊÊ Ê11c
4b4
b4
1c
dx dx dx
x xxÈÈÈkk
œœ
bb ccÄ! Ä!Ä! Ä!
## $$ #
œ ‘ œ‘ÈÈ
lim 2 b 2 ( 1) 2 4 lim 2 c 0 2 2 2 0 6œ œ œ
b cÄ! Ä!
# $
Š‹
È ˆ‰È È È †
32. lim 2 1 x lim 2 x 1'''
001
212 b
c
dx dx dx
x 1 1 x x 1
ÈÈÈkkÊ ÊÊ
œœ
b1 c1Ä Ä
#
Þ
$

’“ ’“
ÈÈ
lim 21b 210 221 lim 2c1 02204œ œœ
b1 c1Ä Ä
# $
Š‹Š ‹ Š‹
È ÈÈ È
33. lim ln lim ln ln 0 ln ln 2'
#

#1
b
1
d 2 b 2 2
5 6 3 b 3 1 3
))
)) )
œ
#
œœÊ"œ"
œœœœ
bbÄ_ Ä_
œ ‘ œ ‘¸ ¸ ˆ‰¸¸ ¸¸

556. 552 Chapter 8 Techniques of Integration
34. lim ln x 1 ln x 1 tan x lim ln tan x'
0
b
0
b
0

dx x 1
(x 1) x 1 4 x 1 # # # #
"" " "œ" #" "
œab È
œ
œœœ
bbÄ_ Ä_
œ ‘kk a b ’“Š‹
lim ln tan b ln tan 0 ln 1 ln 1 0œœœ
bÄ_
’“’“Š‹
"œ"" """ " "" "
#########œ
Ê" Ê"b
b 1 1 4ÈÈ
œ
††
11
35. tan d lim ln cos lim ln cos b ln 1 lim ln cos b ,'
0
2
22 2
b
0
œ
œœ œ
œ
## #
)) )œ œ œ œ_
bb bÄÄ Ä
cd cd cdkk kk kk
the integral diverges
36. cot d lim ln sin ln 1 lim ln sin b lim ln sin b ,
'
0
2
b
œ
œ
œ
$$$
œ 
)) )œœœœ_
b0 b0 b0ÄÄÄ
cd cd cdkk kk kk
the integral diverges
37. ; x . Since 0 for all 0 x and ''' '
00 0
0
œœ œ
œ
sin d sin x dx sin x dx sin x dx
xx xx x
))
1)
ÈÈÈÈÈÈ
Ê
"
cd1) 1œ Ä œ Ÿ Ÿ ŸŸ
converges, then dx converges by the Direct Comparison Test.'
0
œ
sin x
xÈ
38. ; . Since 0
x2
d'''
#œ
œ
œœ
œ œ220
202
cos d
(2) 2x x 2x 2
x
dx
cos dx sin dx sin
))
1)
1
#
##
#
Ê
"
")% ")% ")% ")%
Ô×
ÕØ
œ
œ
œ
Äœ ŸŸ
1)
)
)
ˆ‰ ˆ‰
œ
## # #
œ
xx x
x
")% for all
0 x 2 and converges, then converges by the Direct Comparison Test.ŸŸ 1 ''
00
22
œœ
dx
2x x
sin dx
")% ")%
x
#
#
39. x e dx; y e dy lim e lim e e'''
0 1ln2
ln 2 1 ln 2
1x y y b 1ln2 b
1ln2
# "
Ê#
îœ
ϔ
œ œ ‘ cd cdc d
xy
ye dy
œÄ œ œ œ
œ!
œy
bbÄ_ Ä_
0 e e , so the integral converges.œ œ
#1ln2 1ln2
40. dx; y x 2 e dy 2 , so the integral converges.''
00
11
y
e2
x e
#œx
È
œ ‘ÈœÄ œ
#
41. . Since for 0 t , 0 and converges, then the original integral''
0 0
œ œ
dt dt
tsin t tsin t t tÈÈÈÈœœ
""
ŸŸ Ÿ Ÿ1
converges as well by the Direct Comparison Test.
42. ; let f(t) and g(t) , then lim lim lim lim '
0
1
dt t3t6t
t sin t t sin t t g(t) t sin t 1 cos t sin t
f(t)
ÊÊ ÊÊ
""
œœ œœœ
%
% œ
t0 t0 t0 t0ÄÄ Ä Ä
lim 6. Now, lim lim , which diverges œ œ œ œ œ_ Ê
t0 bbÄ Ä! Ä!
6dt dt
cos t t t b t sin t''
0 0
1 1
1
b
!œ œ
$$
œ‘ œ ‘
"" "
## #
diverges by the Limit Comparison Test.
43. and lim ln lim ln 0 , which''' '
001 0
212 1
b
0
dx dx dx dx 1 x b
1x 1x 1x 1x 1x 1b # #
"œ ""œ
œœœ œœ œ œ œ_
b1 b1ÄÄ
##
œ ‘ œ ‘¸¸ ¸¸
diverges diverges as well.Ê '
0
2
dx
1xÊ
œ
44. and lim ln (1 x) lim ln (1 b) 0 , which''' '
001 0
212 1
b
0
dx dx dx dx
1x 1x 1x 1xÊÊÊ Ê
œ œœœ_
b1 b1ÄÄ
##
cdcd
diverges diverges as well.Ê '
0
2
dx
1xÊ
45. ln x dx ln ( x) dx ln x dx; ln x dx lim x ln x x [1 0 1] lim [b ln b b]'' ''
## $$11 00
10 11
1
b
kk c dœ œ œ
bbÄ! Ä!
†
1 0 1; ln ( x) dx 1 ln x dx 2 converges.œ œ œ Ê œ
''
##11
01
kk

557. Section 8.7 Improper Integrals 553
46. x ln x dx x ln ( x) dx ( x ln x) dx lim ln x lim ln x'''
## $$110
10 1 11
bc
ab c dkk ’“’“œœ
bcÄ! Ä!
xx xx
44
œœ œœ
##
ln 1 lim ln b ln 1 - lim ln c 0 0 0 the integralœ œœÊœ ‘ œ ‘’“ ’“
"" "" ""
### # 444 444
bb cc
bcÄ! Ä!
$$
œœ œœ
converges (see Exercise 25 for the limit calculations).
47. ; 0 for 1 x and converges converges by the Direct'''
1 11

dx dx dx
1x x 1 x x 1xœœ œ
""
%%% % %ŸŸ Ÿ_ Ê
Comparison Test.
48. ; lim lim lim 1 and lim 2 x ,''
4 4
b
4

dx dx
x1 x1 x
x
1 10ÈÈ È
Š‹
Š‹
È
ÊÊ
""
Ê Ê
xxx bÄ_ Ä_ Ä_ Ä_
1
x1
x x
œ
œ œ
Î
œ œ
œœœœœ œ_ œ‘È
which diverges diverges by the Limit Comparison Test.Ê '
4

dx
x1ÈÊ
49. ; lim lim lim 1 and lim 2 v ,''
2 2
b


dv dv
v1 v1 10
v
1
vÈÈÈÈ
Š‹
Š‹
È
ÉÊÊÊ
""
Ê
vvv bÄ_ Ä_ Ä_ Ä_
1
v1
v
œ
œ
Î
œ
œœœœ œ œ_
"
v
œ‘È
which diverges diverges by the Limit Comparison Test.Ê '
2

dv
v1
È
Ê
50. ; 0 for 0 and lim e lim e 1 1''
00
b
0
b

##
dd
1e 1e e e
))
œœ
""
### #ŸŸ Ÿ_ œ œ œ)
bbÄ_ Ä_
cd a b
#
converges converges by the Direct Comparison Test.ÊÊ''
00

dd
e1e
))
##
œ
51. and lim ''''''
001011
11
b
1

dx dx dx dx dx dx
x1 x1 x1 x1 xx 2xÈÈ ÈÈ
ww ww
!! œ
œœ œœ
"
œ œ
bÄ_
œ ‘
lim converges by the Direct Comparison Test.œœÊ
bÄ_
ˆ‰
"" "
### œb
dx
x1
œ
w
'
0

È
52. ; lim lim lim 1; dx lim ln b ,''
2 2
b
2

dx x
x1 x1 1
xÈÈ
Š‹
Š‹ É
œœ
ÊÊ
""
Êxxx bÄ_ Ä_ Ä_ Ä_
$
$
Î
œ œ
$
œ
x1
x x
œœ œœœ_ cd
which diverges diverges by the Limit Comparison Test.Ê '
2

dx
x1
È
œ
Ê
53. dx; lim lim lim 1; dx'''
1 11

ÈÈÈ
Œ#
Π#
È
É
x1
x x
xx
x1
1
dx
x
œ
œ
"
œ
œ œ !'œ
xxxÄ_ Ä_ Ä_
œ
œx
x
x1
x
x
#
Ê
$
œ
œœ œœ
lim 2x lim 2 2 dx converges by the Limit Comparison Test.œ œ œÊ
bbÄ_ Ä_
œ ‘ Š‹
"Î# Ê œ
b
1 1
2
b
x1
xÈ
È '

œ
54. ; lim lim lim 1; lim ln x ,'''
2 22
b
2

x dx x dx dx
x1 x1 x
x
1
xÈÈÈ
Š‹
Š‹
È
É
###
#
ÊÊ
"
Êxxx bÄ_ Ä_ Ä_ Ä_
x
x1
x
x x
œ
œ
"
Î
"
œ
"
œœ œœœœ_ cd
which diverges diverges by the Limit Comparison Test.Ê '
2

x dx
x1
È
#
Ê
55. dx; 0 for x and lim ln x , which diverges''
œ œ
œ

2cos x 2cos x dx
xxx x
œ"œ
Ÿ   œ œ_ 1
bÄ_
cd
b
dx diverges by the Direct Comparison Test.Ê '
œ

2cos x
x
œ
56. dx; 0 for x and dx lim lim ''
œ œ œ

1 sin x sin x 2 2 2 2 2 2
xxx x x b
œ"œ
œœœ œ ŸŸ   œ œ œ 1
bbÄ_ Ä_
œ‘ˆ‰
b
11
converges dx converges by the Direct Comparison Test.ÊÊ''
œœ

2 dx 1 sin x
xx
œœ
œ

558. 554 Chapter 8 Techniques of Integration
57. ; lim 1 and lim 4t lim 2 2 converges'' '
44 4
b
4

2 dt t 2 dt 4 2 dt
t1 t1 t tb
%'œ !'œ !'œ !'œ
!'œ
ÊÊ
"Î# Êt bbÄ_ Ä_ Ä_
œ œœ œÊ œ ‘ Š‹
È
converges by the Limit Comparison Test.Ê '
4

2 dt
t1
%'œ
œ
58. ; 0 for x 2 and diverges diverges by the Direct Comparison Test.'''
222

dx dx dx
ln x x ln x x ln x
Ê
""
59. dx; 0 for x 1 and diverges diverges by the Direct Comparison Test.'''
111

e e dx e dx
xxx x x
xx x
Ê
"
60. ln (ln x) dx; x e (ln y) e dy; 0 ln y (ln y) e for y e and ln y dy lim y ln y y'' '
ee e
yy y b
e
e

cd c dœÄ   œ
bÄ_
, which diverges ln e dy diverges ln (ln x) dx diverges by the Direct Comparison Test.œ_ Ê Ê ''
ee
y

e
61. ; lim lim lim 1; e dx'''
1 11
x2


dx dx
ex ex
e
1 10
ÈÈÈÈ
Š‹
Š‹
È
É
xx
x
ÊÊ
""
Ê ÊxxxÄ_ Ä_ Ä_
$
Î
œ
œ
œ
ex
e
x
e
x
x
x
x
œœ œœœ
e
lim 2e 2e 2e e dx converges convergesœ œÊ Ê
b
lim
bÄ_
œ
Ä_
cd a b
#
w
x2 b2 12 x2
b
11 2dx
e exÈÈ''
x
Ê
by the Limit Comparison Test.
62. ; lim lim lim 1 and lim e''
1 1
xb
1

#
dx e dx
e2 e2 10 e
1
x xx
x
ÊÊÊ
""
Ê xxx bÄ_ Ä_ Ä_ Ä_
ˆ‰
ˆ‰
$
Î
œ
e2
e
2
e
xx
x
œœ œœœ
x
xx
Š‹
cd
lim e e converges converges by the Limit Comparison Test.œœÊ Ê
bÄ_
ab
#
b
11 Ê" "
Êee e2
dx dx ''
x x x
63. 2 ; and' ''''''
#'

dx dx dx dx dx dx dx
x1 x1 x1 x1 x1 x1 xÈ ÈÈÈ ÈÈ
# ### ##
œ
œ œœœ œœ
œœ
00 01 01
11
lim lim 1 1 converges by the Direct Comparison Test.''
1
b
1

#'
dx dx
xx b x1
œ
#œœœÊ
bbÄ_ Ä_
œ‘ˆ‰
""
œ
È

64. 2 ; 0 for x 0; converges 2 converges by the'' ' '
#
'' ' '
dx dx dx dx
ee ee ee e e ee
xxx x xxxx x
œœœ œ
""
### #œ Ê
000
x
Direct Comparison Test.
65. (a) ; t ln x lim t lim (ln 2)''
10
2ln2
1p 1p
ln 2
b
dx dt b
x(ln x) t p 1 p 1 1 p
pp
1pcd ’“œÄ œ œ
bbÄ! Ä!
$$
##
""
Êœ Ê Ê
#
the integral converges for p 1 and diverges for p 1Ê 
(b) ; t ln x and this integral is essentially the same as in Exercise 65(a): it converges''
2ln2

dx dt
x(ln x) t
ppcdπ
for p 1 and diverges for p 1!*
66. lim ln x 1 lim ln b 1 0 lim ln b 1 the integral dx''
0
b
0

#'
2x dx 2x
x1 x1
œ œ
œ œ
## #
œœœœ_Ê
bb bÄ_ Ä_ Ä_
cd cd abab ab
diverges. But lim lim ln x 1 lim ln b 1 ln b 1 lim ln
bb b bÄ_ Ä_ Ä_ Ä_
'
#
#
b
b
b
2x dx b
x1 b1
œ œ
œ
œ œ
### œ"
œœœcd c dab abab Š‹
lim (ln 1) 0œœ
bÄ_
67. A e dx lim e lim e eœœœ'
0
xxb0 b
0

####
bbÄ_ Ä_
cd abab
011œœ

559. Section 8.7 Improper Integrals 555
68. x xe dx lim xe e lim be e 0 e e 0 1 1;œ œ œ œœ
"
A'
0
xxxbb00 b
0

#######
bbÄ_ Ä_
cdaba b †
y e dx e dx lim e lim e e 0œœœœœœ
" " "" "" "" ""#
#######2A 44''
00
x2x 2x 2b20
b
0

## # ##
ab œ ‘ˆ‰ˆ‰
bbÄ_ Ä_
œ
69. V 2 xe dx 2 xe dx 2 lim xe e 2 lim be e 1 2œœœœœ''
00
x x xx bb b
0

# # ## ##
11 1 1 1
bbÄ_ Ä_
cd ab ’“
70. V e dx e dx lim e lim eœœœœœ''
00
x2x 2x 2b
b
0

## # #
11 1 1ab œ ‘ˆ‰
# """
####
bbÄ_ Ä_
1
71. A (sec x tan x) dx lim ln sec x tan x ln sec x lim ln 1 ln 1 0œœ œ '
0
2
22
b
0
œ
œœ
œ
##
bbÄÄ
cd kkkkkk ˆ‰¸¸
tan b
sec b
lim ln 1 sin b ln 2œœ
bÄ
œ
2
#
kk
72. (a) V sec x dx tan x dx sec x tan x dx sec x sec x 1 dxœœœ
'' ' '
00 0 0
22 2 2
œœ œ œœœ œ œ
111 1
## ## ##
abc d ab
dxœœ1 '
0
2
œœ
1
œ
#
(b) S 2 sec x 1 sec x tan x dx 2 sec x(sec x tan x) dx lim tan x
outer
00
22
2
b
0œ  œ''
œœ
œ
œœ
#
111
È
cd
## #
bÄ
lim tan b 0 lim tan b S diverges; S 2 tan x 1 sec x dxœœœ_Ê œ11 1”• cd ab
È
bbÄÄ
œœ
œ
22
outer inner
0
2
##
œ
## % '
2 tan x sec x dx lim tan x lim tan b 0 lim tan b œœœœ_'
0
2
22 2
b
0
œ
œœ œ
œ
## #
11 1 1
### #
bb bÄÄ Ä
cd cd ab”•
S divergesÊ
inner73. (a) e dx lim e lim e e 0 e e'
3
3x 3x 3b 3 3 9 9
b
3

######
œœœœ
bbÄ_ Ä_
œ ‘ˆ‰ˆ‰
"""""
33333 œ
†
0.0000411 0.000042. Since e e for x 3, then e dx 0.000042 and therefore¸ Ÿ
## #
x3x x
3
$$ '
e dx can be replaced by e dx without introducing an error greater than 0.000042.''
00
3

œœxx
$$
(b) e dx = 0.88621'
0
3
œx
$µ
74. (a) V dx lim lim (0 1)œœœœœ'
1
b
1

11 1 11ˆ‰ œ‘ˆ‰ˆ‰’“
""""
#
xxb1
bbÄ_ Ä_
(b) When you take the limit to , you are no longer modeling the real world which is finite. The comparison_
step in the modeling process discussed in Section 4.2 relating the mathematical world to the real world
fails to hold.
75. (a)

560. 556 Chapter 8 Techniques of Integration
(b) > int((sin(t))/t, t=0..infinity); answer is ˆ‰
1
#
76. (a)
(b) > f:= 2*exp( t^2)/sqrt(Pi);Î
> int(f, t=0..infinity); (answer is 1)
77. (a) f x eabœ
1
2
x/2
È
1
Ê
2

f is increasing on ( , 0]. f is decreasing on [0, ). f has a local maximum at 0, f 0_ _ œ !ß ababŠ‹
1
2
È
1
(b) Maple commands:
>f: exp( x^2/2)(sqrt(2*pi);œ
>int(f, x 1..1); 0.683œ ¸
>int(f, x 2..2); 0.954œ ¸
>int(f, x 3..3); 0.997œ ¸
(c) Part (b) suggests that as n increases, the integral approaches 1. We can take f(x) dx as close to 1 as we want by
'
#n
n
choosing n 1 large enough. Also, we can make f(x) dx and f(x) dx as small as we want by choosing n large! ''
n
n
'#
#'
enough. This is because 0 f x e for x 1. (Likewise, 0 f x e for x 1.) ab ab
Êx/2 x/2
Thus, f(x) dx e dx.''
nn


Êx/2
e dx e dx 2e 2e 2e 2e''
nn

ÊÊ Ê
Ä_ Ä_ Ä_
ÊÊÊx/2 x/2 n/2
cc c
c
x/2 c/2 n/2
c
n
œœœ œ ÎÎ#lim lim lim cdc d
As n , 2e 0, for large enough n, f(x) dx is as small as we want. Likewise for large enough n,Ä_ Ä
Ên/2 '
n

f(x) dx is as small as we want.'
#'
#
n
78. dx , since the left hand integral converges but both of the right hand'''
333

ˆ‰
""
# #xx x x
dx dx
ÎÞ Î
integrals diverge.

561. Section 8.7 Improper Integrals 557
79. (a) The statement is true since f(x) dx f(x) dx f(x) dx, f(x) dx f(x) dx f(x) dx''''''
#'#'

bab b
ab a a
œ œ
and f(x) dx exists since f(x) is integrable on every interval [a b].
'
a
b
ß
(b) f(x) dx f(x) dx f(x) dx f(x) dx f(x) dx f(x) dx
''' '''
#' #'

aabb
a aaa
œ
f(x) dx f(x) dx f(x) dx f(x) dx f(x) dxœœ
'''''
#' #'

ba b
ba b
80. (a) f(x) dx f(x) dx f(x) dx f( u) du f(x) dx''' ''
#' #' '
'' '
œœ
00
00
f( u) du f(x) dx 2 f(x) dx, where u xœ œ œ'' '
00 0
'' '
(b) f(x) dx f(x) dx f(x) dx f( u) du f(x) dx''' ''
#' #' '
'' '
œœ
00
00
f(u) du f(x) dx f(x) dx f(x) dx 0, where u xœ œ œ œ''''
00 00
'' ''81. ; diverges because lim ''''
#' #'

dx dx dx dx
x1 x1 x1 x1
ÈÈÈÈ
Š‹
Š‹
œœœœ
œœœœ
œ
1
11
xÄ_
$
$
$
Ê
x
x1
œ
lim lim 1 1 and diverges; therefore, divergesœœœ
xxÄ_ Ä_
È
È
x1
xxx
dx dx
x1œ
œ
œ
œ "
œ
É ''
1

#'82. dx converges, since dx 2 dx which was shown to converge in'''
#' #'

"""
œœœ
ÈÈÈ
1x 1x 1x
www
œ
0
Exercise 51
83. ; and lim e lim e 1 1'' '
#'#'
'' '
##
dx e dx e dx
ee e1e1 ee e e
x x 2x 2x x x x x
xx
œœœœ
""
##œœœœœ
0
xcc
0 ccÄ_ Ä_
cd a b
2 convergesʜ''
#

e dx dx
e1 ee
x
2x x x
œœ 0
#
84. ; , where u x, and since (u 1) and'''''
#' #' # #'
'#'# '
e dx e dx e dx e dx e du e
x1 x1 x1 x1 1u 1u u
####xxxxu u
œœœœ œ œ
œœœœœ œ
"
œ œ œ
11
11
diverges, the integral diverges diverges'''
11

#'
du e du e dx
u1ux1
ux
œœ
œœ Ê
#
85. e dx 2 e dx 2 lim e dx 2 lim e 2, so the integral converges.'' '
#

## # ###
xx x x
00
b
b
0
œœ œ œ
bbÄ_ Ä_
cd
86. ;'''''
#'#' #&#
'#& # '
dx dx dx dx dx
(x 1) (x 1) (x 1) (x 1) (x 1)œœœœœ
œœœœœœ
12
12
lim lim , which diverges diverges
b1 b1Ĝ Ĝ
## # #'#&

''
b
b
dx dx
(x 1) x 1 (x 1)œœ œ
"
œ œœ œ_ Ê œ‘
87. dx 2 dx 2 dx 2 lim dx''' '
#

kkk k kkk k
kk
sin x cos x sin x cos x
x 1 x1 x1 x1
sin x cos x dxœœ
œœœ œ
œ
œ œ
00 0
b
œœ
bÄ_
2 lim ln x 1 , which diverges dx divergesœœ_ Ê
bÄ_
cdkk
b
0 '
#

kkk k
kk
sin x cos x
x1
œ
œ
88. dx 0 by Exercise 80(b) because the integrand is odd and the integral'
#'

x
x1x2abab
œœ
œœ
œ
converges''
00

x dx dx
x1x2 xabab
œœ %
œœ
Ÿ
89. Example CAS commands:
:Maple
f := (x,p) -> x^p*ln(x);
domain := 0..exp(1);
fn_list := [seq( f(x,p), p=-2..2 )];

562. 558 Chapter 8 Techniques of Integration
plot( fn_list, x=domain, y=-50..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0],
legend=["p= -2","p = -1","p = 0","p = 1","p = 2"], title="#89 (Section 8.8)" );
q1 := Int( f(x,p), x=domain );
q2 := value( q1 );
q3 := simplify( q2 ) assuming p>-1;
q4 := simplify( q2 ) assuming p<-1;
q5 := value( eval( q1, p=-1 ) );
i1 := q1 = piecewise( p<-1, q4, p=-1, q5, p>-1, q3 );
90. Example CAS commands:
:Maple
f := (x,p) -> x^p*ln(x);
domain := exp(1)..infinity;
fn_list := [seq( f(x,p), p=-2..2 )];
plot( fn_list, x=exp(1)..10, y=0..100, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0],
legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#90 (Section 8.8)" );
q6 := Int( f(x,p), x=domain );
q7 := value( q6 );
q8 := simplify( q7 ) assuming p>-1;
q9 := simplify( q7 ) assuming p<-1;
q10 := value( eval( q6, p=-1 ) );
i2 := q6 = piecewise( p<-1, q9, p=-1, q10, p>-1, q8 );
91. Example CAS commands:
:Maple
f := (x,p) -> x^p*ln(x);
domain := 0..infinity;
fn_list := [seq( f(x,p), p=-2..2 )];
plot( fn_list, x=0..10, y=-50..50, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0],
legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#91 (Section 8.8)" );
q11 := Int( f(x,p), x=domain ):
q11 = lhs(i1+i2);
`` = rhs(i1+i2);
`` = piecewise( p<-1, q4+q9, p=-1, q5+q10, p>-1, q3+q8 );
`` = piecewise( p<-1, -infinity, p=-1, undefined, p>-1, infinity );
92. Example CAS commands:
:Maple
f := (x,p) -> x^p*ln(abs(x));
domain := -infinity..infinity;
fn_list := [seq( f(x,p), p=-2..2 )];
plot( fn_list, x=-4..4, y=-20..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9],
legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#92 (Section 8.8)" );
q12 := Int( f(x,p), x=domain );
q12p := Int( f(x,p), x=0..infinity );
q12n := Int( f(x,p), x=-infinity..0 );
q12 = q12p + q12n;
`` = simplify( q12p+q12n );

563. Chapter 8 Practice Exercises 559
89-92. Example CAS commands:
: (functions and domains may vary)Mathematica
Clear[x, f, p]
f[x_]:= x Log[Abs[x]]
p
int = Integrate[f[x], {x, e, 100)]
int /. p 2.5Ä
In order to plot the function, a value for p must be selected.
p = 3;
Plot[f[x], {x, 2.72, 10}]
CHAPTER 8 PRACTICE EXERCISES
1. x 4x 9 dx; u du u C 4x 9 C
u4x 9
du 8x dx''È
”•
È ab#
#
"" " $Î# #
#
$Î#
?
œ
œ
8831
2
†
2. 6x 3x 5 dx; u du u C 3x 5 C
u3x 5
du 6x dx''È
”•
È ab#
#
$Î##
$Î#
?
œ
œ
22
33
3. x(2x 1) dx; u du u du u du u u C
u2x1
du 2 dx''''?
œ
œ
"Î#$ Î#" Î#& Î#$ Î#" " "
##
”•
ˆ‰ ˆ ‰È Š‹
u 1 2 2
4453
Cœ
(2x 1) (2x 1)
10 6
œœ

4. dx; du u du u 2u C
u1x
du dx'''x 2
1 x
(1 u)
uu 3ÈÈÈ

" $Î#" Î#
”• Š‹È
œ
œ
Ä œ œ
(1 x) 2(1 x) Cœ
2
3
$Î#" Î#
5. ; 2u C C
u8x 1
du 16x dx''x dx du
8x 1 16 16 8u
8x 1
ÈÈ
È


œ
#
"" "Î# œ
”•
œ
œ
Äœ œ †
6. ; 2u C C
u94x
du 8x dx''x dx du
9 4x
88 4u
9 4x
ÈÈ
È

#
"" "Î#

”•
œ
œ
Ä œ œ †
7. ; ln u C ln 25 y C
u25y
du 2y dy''
y dy
25 y u
du
œ###
#
"" " #
”• kk a b
œ
œ
Äœœ
8. ; ln u C ln 4 y C
u4y
du 4y dy''
y dy
4y 4u 4 4
du
"
#
œ
%
$
"" " %”• kk a b
œ
œ
?
9. ; 2u C C
u94t
du 16t dt''t dt du
94t
16 16 8u
94t
"
#
#
ÈÈ
È

%
$
"" "Î# ”•
œ
œ
Ä œ œ †
10. ; tan u C tan t C
ut
du 2t dt''2t dt du
t1 u1
#
œ
#
œ
" " #”•
œ
œ
?
11. z z 1 dz; u du u C z 1 C
uz
du z dz''#Î$&Î$# Î$& Î$& Î$
#Î$ &Î$
&Î$
#Î$ #ˆ‰ ˆ‰
”•?
œ"
œ
5
3
3339
5555
†

564. 560 Chapter 8 Techniques of Integration
12. z 1 z dz; u du 2 u C 1 z C
u1z
du z dz''"Î& %Î&" Î# %Î&
"Î# "Î#
%Î&
"Î& #ˆ‰ ˆ‰
”•
È?
œ
œ
4
5
55 5
44
†
13. ; C C
u1cos 2
du sin 2 d''sin 2 d du
(1 cos 2) u 2u (1 cos 2)
))
) )##
"" "
”•
œ
œ#
Äœœ
)
))
14. ; 2u C 2 1 sin C
u1sin
du cos d''cos d du
(1 sin ) u
))
)œ
"Î#
”•
È
œ
œ
?
)
))
)
15. ; ln u C ln 3 4 cos t C
u 3 4 cos t
du 4 sin t dt''sin t dt du
34 cos t 4 u 4 4œ
"" "
”• kk k k
œ
œ
Ä œ œ
16. ; ln u C ln 1 sin 2t C
u 1 sin 2t
du 2 cos 2t dt''cos 2t dt du
1sin 2t 2 u 2 2œ
"" "
”• kk k k
œ
œ
?
17. (sin 2x) e dx; e du e C e C
u cos 2x
du 2 sin 2x dx''
cos 2x u u cos 2x
”•
œ
œ
Ä œ œ
"""
###
18. (sec x tan x) e dx; e du e C e C
u sec x
du sec x tan x dx''
sec x u u sec x
”•
œ
œ
?
19. e sin e cos e d ; u du u C cos e C
u cos e
du sin e e d''


ab ab ab ”•
ˆ‰
ˆ‰
##$$ ""
)
)
œ
œ
Ä œ œ
)
)
†
33
20. e sec e d ; sec u du tan u C tan e C
ue
du e d''


##
ab ab”•)
)
œ
œ
?
21. 2 dx C 22. 5 dx C''
x1
œ œ
2 5
ln 2 ln 52
x1 x2
x2
È "
È
Š‹

23. ; ln u C ln ln v C
u ln v
du dv''dv du
v ln v u
v
”• kk k k
œ
œ
Äœœ "
24. ; ln u C ln 2 ln v C
u2ln v
du dv''dv du
v(2 ln v) u
v
œ "
”• kk k k
œ
œ
?
25. ; ln u C ln 2 tan x C
u2tanx
du'' dx du
x12tanx u dx
x1
aba b
ï´

œœ
"
œ
" ”• kk k k
œ
œ
Äœœ
26. ; u du u C sin x C
usinx
du''sin x dx
1x
dx
1x
ï´


È
È

"

""
##
#"
# –—
ab
œ
œ
Äœœ
27. ; sin u C sin (2x) C
u2x
du 2 dx''2 dx du
14x 1u
ÈÈ

" "
”•
œ
œ
?
28. ; sin u C sin C
u
du dx'' 'dx dx du x
49 x 1 u
7 7
1
x
7
7
ÈÈ
Ɉ‰
"

"
" "

?
œ
œ
x
7
”•
ˆ‰

565. Chapter 8 Practice Exercises 561
29. ; sin u C sin C
ut
du dt'' 'dt dt du 3t
16 9t 1 u
43334
1
3
4
3
4
ÈÈ
ÊŠ‹

""""

" "

?
œ
œ
3t
4
–—
ˆ‰
30. ; sin u C sin C
ut
du dt'' 'dt dt du 2t
94t 1u
32223
1
2
3
2
3
ÈÈ
ÊŠ‹

""""

" "

?
œ
œ
2t
3
–—
ˆ‰
31. ; tan u C tan C
ut
du dt'' 'dt dt du t
9t 9 3 1u 3 3 3
1
3
3
œœ
""""
œ
"
"
" "
?
œ
œ
ˆ‰
t
3–—
ˆ‰
32. ; tan u C tan (5t) C
u5t
du 5 dt''dt du
1 25t 5 1 u 5 5œœ
"" " " "
”•
œ
œ
Äœœ
33. sec C''4 dx 4 dx 5x
5x 25x 16
25 5 4
xx
È
É



" "
œœ
16
25
¸¸
34. 3 2 sec C''6 dx dx 2x
x4x 9 xx
3È
É



"
œœ
9
4
¸¸
35. sin C''dx x 2
4x x
d(x 2)
4(x2)ÈÈ



"
#

œœ ˆ‰
36. sin (x 2) C''dx
4x x 3
d(x 2)
1(x2)ÈÈ



"

œœ
37. tan C''
dy d(y 2) y 2
y4y8 (y2)4

# #
" "
œœ ˆ‰
38. tan (t 2) C''dt
t4t5 (t2)1
d(t 2)

œœ œ œ
œ "
œœ
39. sec x 1 C'' dx
(x 1) x 2x
d(x 1)
(x 1) (x 1) 1


"
ÈÈ

œœ kk
40. sec v 1 C'' dv
(v 1) v 2v
d(v 1)
(v 1) (v 1) 1œœ
œ

"
ÈÈ

œœ kk
41. sin x dx dx C''#
##
œœ
1 cos 2x x sin 2x
4
42. cos 3x dx dx C''# "œ
##
œœ
cos 6x x sin 6x
12
43. sin d 1 cos sin d ; 2 1 u du 2u C
u cos
du sin d'' '$# #
###
#
"
##
)))
)
)
))
)
œ Ä œ
œ
œ
ˆ‰ˆ‰
–—
ab
2u
3
"
cos 2 cos Cœ
2
3
$
##
))
44. sin cos d 1 cos (sin ) cos d ; 1 u u du u u du
u cos
du sin d'' ''$# # # ## % #
))) ) ) ))
)
))
œ Ä œ
œ
œ
abab ab ab ”•
CCœœ
uu cos cos
53 5 3

))

566. 562 Chapter 8 Techniques of Integration
45. tan 2t dt (tan 2t) sec 2t 1 dt tan 2t sec 2t dt tan 2t dt;
u2t
du 2 dt'' ' '$# #
œœ
œ
œ
ab ”•
tan u sec u du tan u du tan u ln cos u C tan 2t ln cos 2t C?
""""""
## # #
## #''
44
kk k k
tan 2t ln sec 2t Cœ
""#
#4
kk
46. 6 sec t dt 6 tan t 1 sec t dt; 6 u 1 du 2u 6u C
utan t
du sec t dt
'' '% ## # $
#
?
œ
œ
abab ab ”•
2 tan t 6 tan t Cœ
$
47. csc 2x dx ln csc 2x cot 2x C'''dx dx
2 sin x cos x sin 2x
œœ œ
"
#
kk
48. ; sec u du ln sec u tan u C
u2x
du 2 dx'' ''2 dx 2 dx du
cos x sin x cos 2x cos u


?
œ
œ
”• kk
ln sec 2x tan 2x Cœkk
49. csc y 1 dy cot y dy ln sin y ln 1 ln ln 2''
œœ
œœ
œ
œ
œœ
œœ
œ
œ44
22
2
4
È cdkk
È # "
œ œ œ œ
È
2
50. cot t 1 dt csc t dt ln csc t cot t ln csc cot ln csc cot ''
œœ
œœ
œ
œ
œœ
œœ
œ
œ44
34 34
34
4
È
cdkk ¸¸¸¸#
œ œ œ
33
44 44
11 11
ln 2 1 ln 2 1 ln ln ln 3 2 2œ œ œ œ ¹¹¹¹¹¹ Š ‹
ÈÈ È
»»
È
È
Š‹Š‹
ÈÈ
2
21
221
1
œ"

œ" œ
#
51. 1 cos 2x dx sin 2x dx sin 2x dx sin 2x dx''''
0002
2
2
02
œœœœ
œ
œœ
œ
È
kk ‘ ‘œ œ œ#
##
œ
œ
œ
œ
cos 2x cos 2x
2œ œˆ‰ ‘ˆ‰
"" " "
## # #
52. 1 sin dx cos dx cos dx cos dx 2 sin 2 sin ''''
000
22 2
0
œœœœ
œ
œœ
œ
È ¸¸ ‘ ‘œ œ œ #
# #####
x xxxxx

(2 0) (0 2) 4œœ
53. 1 cos 2t dt 2 sin t dt 2 2 sin t dt 2 2 cos t 2 2 [0 ( 1)] 2 2'''

œœœ œ
œœ
œœœ œ220
222 2
0
È ÈÈ ÈÈÈ
kk ’“œ œ œ œœ
54. 1 cos 2t dt 2 cos t dt 2 cos t dt 2 cos t dt''''
œœœœ
œœœœ
22322
32
È ÈÈÈ
kkœ œ
œ
œ
2 sin t 2 sin t 2 ( 1 0) 2 [0 ( 1)] 2 2œ œ œ
ÈÈ ÈÈ È
cd cd
"œœ
œ2 #
$Î#
1
1
55. x x 2 tan C''x dx 4 dx x
x4 x4


œœ #
"
œ œ ˆ‰
56. dx x dx ln 9 x C'' 'x dx 9x x 9
9x x9 x9
xx 9 9x
"


œœ œ##
#
œœœ’“ ˆ‰ ab
ab
57. dx (2x 1) dx x x 2 ln 2x 1 C''4x 3 4
2x 1 x 1

œ
#
#
œ œ ‘ kk
58. 2 dx 2x 8 ln x 4 C''2x dx 8
x4 x4
œ œ ˆ‰ kk
59. dy ln y 4 tan C'''
2y 1 2y dy dy y
y4 y4 y4

œœœ ##
#" "
œœ ab ˆ‰

567. Chapter 8 Practice Exercises 563
60. dy 4 ln y 1 4 tan y C'''
y 4 y dy dy
y1 y1 y1
œ
œœœ#
" #"
œ œ ab
61. dt 2 4 t 2 sin C'''t 2 t dt dt t
4t 4t 4t
œ

# "
#ÈÈÈ

œ œ
È ˆ‰
62. dt 2 1 t ln t C'''
2t 1 t
t1 t 1 t
2t dt dt
t




#
È
ÈÈ
œœ
È
kk
63. dx dx''' 'tan x dx sin x dx sin x 1 cos x
tan xsec x sin x1 1sinx cosx
(sin x)(1 sin x)


œœ œ


dx tan xxCxtan xsec xCœ œ œ '''
d(cos x)
cos x cos x cos x
dx

"
64. dx dx''' 'cot x dx cos x dx cos x 1 sin x
cot xcsc x cos x1 1cosx sinx
(cos x)(1 cos x)


œœ œ


dx cot xxCxcot xcsc xCœœœ'''
d(sin x)
sin x sin x sin x
dx

"
65. sec (5 3x) dx; sec y sec y dy ln sec y tan y C
y53x
dy 3 dx'''?
œ
œ
”• Š‹ kk†
dy
33 3
""
ln sec (5 3x) tan (5 3x) Cœ
"
3
kk
66. x csc x 3 dx csc x 3 d x 3 ln csc x 3 cot x 3 C''ab abab k k abab
### ## ""
##
œ œ
67. cot dx 4 cot d 4 ln sin C''ˆ‰ ˆ‰ ˆ‰ ¸ ¸ ˆ‰
xxxx
4444
œœ
68. tan (2x 7) dx tan (2x 7) d(2x 7) ln cos (2x 7) C ln sec (2x 7) C'' œ œ œ
""" ###
kk kk
69. x 1 x dx; (1 u) u du u u du u u C
u1x
du dx'''È
”•
È ˆ‰?
œ
œ $Î#" Î#& Î#$ Î#22
53
(1 x) (1 x) C 2 Cœ œ
22
53 35
1x 1x
&Î#$ Î#

–—
Š‹Š‹
ÈÈ
"
70. 3x 2x 1 dx; 3 u du u u du u u C
u2x1
du 2 dx'''È
”•
ˆ‰ ˆ ‰ÈÄ œœ
œ
œ u1 3 3 2 3 2
44543
"
##
$Î#" Î#& Î#$ Î#
†††
(2x 1) (2x 1) C Cœœ
3
10 10
32x1 2x1
&Î#$ Î#"
##
œœˆ‰ˆ‰ÈÈ

71. z 1 dz; tan 1 sec d sec d
ztan
dz sec d'''ÈÈ
”•##
#
#$
?
œ
œ
)
))
)))))†
sec d (FORMULA 92)œ
sec tan 3 2
31 31
))

'))
ln sec tan C ln z 1 z Cœ œ
sin
2 cos
zz 1)
)


""
###
œ #
kk ¹¹
È
))
È
72. 16 z dz; cos d sin C C
z 4 tan
dz 4 sec d'''ab ”•?
œ
œ
#
$Î#
#
""
œ
)
))
)) )
4 sec d z
64 sec d 16 16 16 16 z

"

))
)) È
Cœ
z
16 16 zabœ

568. 564 Chapter 8 Techniques of Integration
73. , u ; sec d
utan
du sec d''' ''
dy dy y
25 y 55
1
du sec d
1u 1tan
ÈÈ È
Ɉ‰œ
"
œ
œœ
#


œœ œ Äœ
œ
œ
y
5
‘
”•
)
))
))
))
)
ln sec tan C ln 1 u u C ln 1 C ln Cœ œ œ œ kk ¹¹ ¹¹
È
ºº
É ˆ‰)) "" ""
#
# œœyy
55 5
25 y yÈ

ln y 25 y Cœ¸¸È
#
74. ln 1 u u C from Exercise 73'' '
dy dy
25 9y 533
1
du
1u
ÈÈ
ÊŠ‹
œ
"""
œ
œ
#
"

œœœ
3y
5
¹¹
È
ln 25 9y 3y C?
" #
3
¸¸È
75. ; csc d cot C C
xsin
dx cos d'''dx cos d
x1x
sin cos x
1x


È
È

# ”•
œ
œ
?
)
))
)) )
))
))
76. ; sin d 1 cos (sin ) d ;
xsin
dx cos d''''x dx sin cos d
1x
cos
""

È

$# ”• ab
œ
œ
?
)
))
)) ) ) )
)))
)
u cos 1 u du u C cos cos 1 x 1 x Ccdab ab ' È
œ Ä œ œ œ )))
#$# "" #
$Î#u
33 3
"
: Ans 1 x C by another methodNote ´
#x1x
33
2
È
È
77. ; sin d d sin 2 C
xsin
dx cos d''''x dx sin cos d 1 cos 2
1x
cos 4


È

# ""
## ”•
œ
œ
?
)
))
)) ) ) )
))) )
) sin cos Cœ œ
""
## # #

)))
sin x x1 x
ï´ È
78. 4 x dx; 2 cos 2 cos d 2 (1 cos 2 ) d 2 sin 2 C
x 2 sin
dx 2 cos d'''È
”•
ˆ‰?
œ
œ # "
#
)
))
))) ))) )†
2 2 sin cos C 2 sin x 1 C 2 sin Cœ œ œ )))
" "
## #
# ˆ‰ ˆ‰ ˆ‰É
xx x x4 x
2
È

79. ; sec d
x 3 sec
dx 3 sec tan d' '''dx 3 sec tan d 3 sec tan d
x9 9 sec 9
3 tan ÈÈ


”•
œ
œ
Äœœ
)
)))
))
))) )))
)
)
ln sec tan C ln 1 C ln C ln x x 9 Cœ œ œ œ kk ºº
Ɉ‰ ¹¹ ¹ ¹
È
)) """
# #xx
33 3
xx9
È

80. ; ;
xsec usin
dx sec tan d du cos d''''12 dx 12 sec tan d 12 cos d 12 du
x1
tan sin u
ab



”• ”•
œœ
œœ
Äœ Ä
))
))) ))
))) ))
)) CC Cœ œ œ
12 12 12 x
usin
x1
) È


81. dw; sec tan d tan d sec 1 d
w sec
dw sec tan d''''
È
w1
wsec
tan
##
”•
ˆ‰ ab
œ
œ
?
)
)))
))) )) ) )
)
)
†
tan C w 1 sec w Cœœ ))
È
# "
82. dz; 4 tan d 4(tan ) C
z 4 sec
dz 4 sec tan d'''
È
z16
z 4sec
4 tan 4 sec tan d
#
”•
œ
œ
?
)
)))
)) ) )
))))
)
†
z164 sec Cœ
È ˆ‰
# "z
4
83. u ln (x 1), du ; dv dx, v x;œ œ œ œ
dx
x1œ
ln (x 1) dx x ln (x 1) dx x ln (x 1) dx x ln (x 1) x ln (x 1) C''''œ œ œ
xdx
x1 x1œœ
"
(x 1) ln (x 1) x C (x 1) ln (x 1) (x 1) C, where C C 1œ œ œ ""

569. Chapter 8 Practice Exercises 565
84. u ln x, du ; dv x dx, v x ;œœœ œ
dx
x3
#$ "
x ln x dx x ln x x dx ln x C''#$$ """
œ œ
33x39
xx
ˆ‰
""
85. u tan 3x, du ; dv dx, v x;œœœœ
"
œ
3 dx
19x

tan 3x dx x tan 3x ; x tan 3x
y19x
dy 18x dx'' '" " "
œ
#
"
œ Ä
œ
œ
3x dx
19x 6 y
dy
”•
x tan (3x) ln 1 9x Cœ
" # "
6
ab
86. u cos , du ; dv dx, v x;œœœœ
"
#

ˆ‰
xdx
4x
È

cos dx x cos ; x cos
y4x
dy 2x dx'' '" " "
## ##

#
"ˆ‰ ˆ‰ ˆ‰
”•
x x x dx x
4x
dy
y
œ Ä
œ
œ
ÈÈ x cos 4 x C x cos 2 1 Cœœ
" "
###
#
#ˆ‰ ˆ‰ ˆ‰È É
xxx
87. e
x
(x 1) e ïïïïî
ÐÑ
# x
2(x 1) e ïïïïî
ÐÑ
x
2 e ïïïïî
ÐÑ
x
0 (x 1) e dx (x 1) 2(x 1) 2 e CÊ œ ' ##xx
cd
88. sin (1 x)
x cos (1 x)
#
ïïïïî
ÐÑ
2x sin (1 x)ïïïïî
ÐÑ
2 cos(1x)ïïïïî
ÐÑ
0 x sin (1 x) dx x cos (1 x) 2x sin (1 x) 2 cos (1 x) Cʜ
'##
89. u cos 2x, du 2 sin 2x dx; dv e dx, v e ;œœ œœ
xx
I e cos 2x dx e cos 2x 2 e sin 2x dx;œœ''xx x
u sin 2x, du 2 cos 2x dx; dv e dx, v e ;œœ œœ
xx
I e cos 2x 2 e sin 2x 2 e cos 2x dx e cos 2x 2e sin 2x 4I I Cœ œ Êœ
xx x xx
’“ ' e cos 2x 2e sin 2x
55
xx
90. u sin 3x, du 3 cos 3x dx; dv e dx, v e ;œœ œœ
2x 2x "
#
I e sin 3x dx e sin 3x e cos 3x dx;œœ''2x 2x 2x "
#
3
2
u cos 3x, du 3 sin 3x dx; dv e dx, v e ;œœ œœ
2x 2x "
#
I e sin 3x e cos 3x e sin 3x dx e sin 3x e cos 3x Iœ œ
"" "
### # # 2x 2x 2x 2x 2x33 39
44
’“ '
I e sin 3x e cos 3x C e sin 3x e cos 3x CÊœ œ
43 23
13 4 13 13
ˆ‰
"
# 2x 2x 2x 2x
91. 2 ln x 2 ln x 1 C'''x dx 2 dx dx
x 3x2 x2 x1


œ œ kkkk
92. ln x 3 ln x 1 C'''x dx 3 dx dx 3
x 4x3 x3 x1

œœ # œ # œ # #
""
œ œ kk kk

570. 566 Chapter 8 Techniques of Integration
93. dx ln x ln x 1 C''dx 1 1
x(x1) x x1 (x1) x1œœœ œ
" "
œ œŠ‹ kk k k
94. dx dx 2 ln C 2 ln x 2 ln x 1 C''x1 2 2 x
x(x1) x1xx x x x
""""

œ œ œ ˆ‰¸¸ kk k k
95. ; cos y ln C''''sin d
cos cos 2 y y2 3 y1 3 y 3 y1
dy dy dy y 2))
))

#
""" œ
cd ¹¹)œÄ œ œ
ln C ln Cœœ
" "
3cos 1 3cos 2
cos 2 cos 1
¸¸ ¸¸
))
))
96. ; sin x ln C''''cos d dx dx dx sin 2
sin sin 6 x x6 5 x2 5 x3 5 sin 3
)) )
)) )


"" "
cd ¸¸)œÄ œœ
97. dx dx dx 4 ln x ln x 1 4 tan x C'''3x 4x 4 4 x 4
xx x x1


"
œœ#
#"
œ œ kk a b
98. 2 tan C''4x dx 4 dx x
x4x x4
"
œœ #
"
œœ ˆ‰
99. dv ln v ln v 2 ln v 2 C''
(v 3) dv
2v 8v 4v 8(v 2) 8(v ) 8 16 16
35 3 5œ
# #
"" "
" œ œŠ‹ kk kk kk
ln Cœ
"
16 v
(v 2) (v 2)
¹¹

î
100. ln C''''
(3v 7) dv ( 2) dv (v 2)(v 3)
(v 1)(v 2)(v 3) v 1 v 2 v 3 (v 1)
dv dv

œœ ¹¹

101. tan t tan C tan t tan C'''dt dt dt t t
t4t3 t1 t3 6 33 3
3

œœ
"" " " "
#œ # œ # #
" " " "
#
œ œ œ
ÈÈ È
È
Š‹
102. ln t 2 ln t 1 C'''t dt t dt t dt
tt2 3t2 3t1 6 6


"" " "

##
œœ kk ab
103. dx x dx x dx'' '''x x 2x 2 dx 4 dx
x x2 x x2 3 x1 3 x2
"

œ

œ œ ˆ‰
ln x 2 ln x 1 Cœ
x4 2
33

#
kk kk
104. dx 1 dx 1 dx dx x ln x 1 ln x C'' ' '''x1 x dx dx xx xx x(x1) x1 x
"
""
œœ" "

œ œ œ œ ˆ‰ ’“ kkkk
105. dx x dx x dx'' '''x4x 3x 3 dx 9 dx
x 4x3 x 4x3 x1 x3
"

œ
œœ œœ # œ # œ
œ œ ˆ‰
ln x 3 ln x 1 Cœ
x9 3

## #
kk kk
106. dx (2x 3) dx (2x 3) dx'' '''2x x 21x 24 x dx 2 dx
x2x8 x2x8 3 x 3 x4
"

"
#
œ œ ‘
x 3x ln x 4 ln x 2 Cœ
# "2
33
kk kk
107. ; ln u 1 ln u 1 C
ux1
du
dx 2u du''''dx 2 u du du du
x3 x 1
dx
2x 1 3u1u3u13u13 3ˆ‰È È abœ œ
"" " "
Ô×
ÖÙ
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È
kk kk
œ
œ
œ
Ä œ œ

ln Cœ
" "
œœ3
x1
x11
¹¹
È
È

571. Chapter 8 Practice Exercises 567
108. ; 3 3 ln C 3 ln C
ux
du
dx 3u du'''dx 3u du du u
x1 x 1x
dx
3x u (1 u) u(1 u) u 1
x
ˆ‰È È
È
œ œ
#
œœœ
" "
"


"
"
Ô×
ÖÙ
ÕØ
È
¸¸ ¹¹
œ
œ
œ
Äœ œ œ
109. ; ln C ln C ln 1 e C
ue 1
du e ds
ds''''ds du du du u e
e 1 u(u 1) u 1 u u 1 e
s
s
du
u 1
s s
s

œ
"
Ô×
ÕØ
¸¸ ¸ ¸ kk
œ
œ
œ
?
s
110. ; 2 ln
ue1
du
ds'''''ds 2u du du du du u
e 1
s
e ds
2e1
2u du
u 1
u u 1 (u 1)(u 1) u 1 u 1 u È È
ab
s
s
œ œ


"
Ô×
ÖÙ
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È
¸¸
œ
œ
œ
Ä œ œœ
s


1
C
ln Cœ¹¹
È
È
e 1
e 1 1s
s
"
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111. (a) 16 y C''
y dy
16 y 16 y
d 16 y
ÈÈ
ab

"
#

#

œ œ È
(b) ; y 4 sin x 4 4 cos x C C 16 y C''
y dy
16 y
sin x cos x dx
cos x 4
4 16 y
È
È


#


cd Èœ Ä œ œ œ
112. (a) 4 x C''x dx
4 x 4 x
d4 x
ÈÈ
ab
œœ
"
#
œ
#


œœ
È
(b) ; x 2 tan y 2 sec y tan y dy 2 sec y C 4 x C''x dx
4 x
2 tan y 2 sec y dy
2 sec yÈ
œ
#


cd ' È
œÄ œ œœ
†
113. (a) ln 4 x C''x dx
4 x 4 x
d4 x
# #
"" #

œ œ
ab
kk
(b) ; x 2 sin tan d ln cos C ln C''x dx 2 sin 2 cos d
4 x 4 cos 2
4 x



cd kk ' Š‹œÄ œ œ œ ))))
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)
†
È
ln 4 x Cœ
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#
#
kk
114. (a) 4t 1 C''t dt
4t 1 4t 1
84
d4t 1
ÈÈ
ab



""
#
œœ
È
(b) ; t sec sec d C C'''t dt tan
4t 1
sec tan sec d
tan 4 4 4
4t 1
È
È

´´



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#
#
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)) ))
)
)†
115. ; ln u C ln C ln C
u9x
du 2x dx''x dx du
9 x u u 9 x
##
#
"" " "


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œ
œ
Ä œ œ œ ÈÈ
116. ln x ln 3 x ln 3 x C''''dx dx dx dx
x9 x 9 x 183 x 183 x 9 18 18ab
"" " " " "
œ œ kk kk kk
ln x ln 9 x Cœ
"" #
918
kk k k
117. ln 3 x ln 3 x C ln C'''dx dx dx x 3
9 x 6 3 x 6 3 x 6 6 6 x 3
"" " " "œ
œ œ œ kk kk ¸¸
118. ; d d C sin C
x 3 sin
dx 3 cos d''dx 3 cos x
9 x
3 cos 3È

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)
))
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)
119. sin x cos x dx cos x 1 cos x sin x dx cos x sin x dx cos x sin x dx C'' ''34 4 2 4 6 cos x cos x
57
œœœab
57
120. cos x sin x dx sin x cos x cos x dx sin x 1 sin x cos x dx'' '55 5 4 5 2
2
œœ ab
sin x cos x dx 2 sin x cos x dx sin x cos x dx Cœ œ'''579 sin x 2sin x sin x
6810
6810

572. 568 Chapter 8 Techniques of Integration
121. tan x sec x dx C' 42 tan x
5
œ
5
122. tan x sec x dx sec x 1 sec x sec x tan x dx sec x sec x tan x dx sec x sec x tan x dx'' ' '33 2 2 4 2
œ †† œ †† ††ab
Cœ
sec x sec x
53
53
123. sin 5 cos 6 d sin sin 11 d sin d sin 11 d cos cos 11 C'' '') )) ) ) ) )) )) ) )œ œ œ
"""""
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a b ab a b abab a b
cos cos 11 Cœ
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124. cos 3 cos 3 d cos 0 cos 6 d d cos 6 d sin 6 C'' ''))) )) ) ))) )œ œ œ
""""
####
ab
1
12
125. 1 cos dt 2 dt 4 2 C cos sin ''É ˆ‰ ¸ ¸ ¸ ¸ÈÈ
œ œ
ttt
244
126. e tan e 1 dt e dt ln Csec e sec e tan e''tt 2t tttÈ
kk k kœ œ
127. E ( x) M where x ; f(x) x f (x) x f (x) 2x f'''(x) 6xkk
sŸ˜ ˜œœ œœÊœÊ œ Ê œ
3 3 2
180 n n x
" " "% " w # ww $ %
f (x) 24x which is decreasing on [1 3] maximum of f (x) on [1 3] is f (1) 24 M 24. ThenÊœ ßÊ ß œÊœ
Ð%Ñ & Ð%Ñ Ð%Ñ
E 0.0001 (24) 0.0001 0.0001 (0.0001) n 10,000kk ˆ ‰ˆ‰ ˆ‰ˆ‰ ˆ‰ ˆ‰
sŸÊ ŸÊ ŸÊŸ Ê 
3 2 768 180 768
180 n 180 768 180 nn
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%
%
##
n 14.37 n 16 (n must be even)Ê  Ê 
128. E ( x) M where x ; 0 f (x) 8 M 8. Then E 10 (8) 10kk kk ˆ‰
T TŸ˜ ˜œœŸŸÊœ Ÿ Ê Ÿ
1 0 1 0
12 n n 12 n
" ""# ww $ $
#
10 1000 n n 25.82 n 26ÊŸ Ê  Ê  Ê  Ê 
2 3n 2000
3n 3


$ #
#
129. x ;˜œ œ œ Ê œ
b a 0 x
n66 1
˜
##
11 1
mf(x ) 12 T (12) ;! ˆ‰
i0
6
i#
œÊœ œ
1
12
1
x f(x ) m mf(x )
x0 0 1 0
x/61/221
x/33/223
x/2224
x 2 /3 3/2 2 3
x 5 /6 1/2 2 1
x010
ii i
!
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1
1
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1
1
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mf(x ) 18 and !
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S (18) .œœˆ‰
1
18
1
x f(x ) m mf(x )
x0 0 1 0
x/61/242
x/33/223
x/2248
x 2 /3 3/2 2 3
x 5 /6 1/2 4 2
x010
ii i
!
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1
1
1
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1
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130. f (x) 3 M 3; x . Hence E 10 (3) 10 10 n¸¸ ˆ‰ˆ‰ kk
Ð%Ñ & & & %" " " " "
%
ŸÊœ˜œœ ŸÊ ŸÊŸÊ 
22 10
n n 180 n 60 60n
s #

n 6.38 n 8 (n must be even)Ê  Ê 
131. y 37 sin (x 101) 25 dx 37 cos (x 101) 25x
av
0
365œœ
""

$'&
!365 0 365 365 2 365
2 365 2' ‘ ‘ˆ‰ ˆ ‰ ˆ‰
11
1
37 cos (365 101) 25(365) 37 cos (0 101) 25(0)œ
"
365 2 365 2 365
365 2 365 2
‘ˆ‰ˆ‰ˆ‰ ‘ˆ‰ ‘
11
11
cos (264) 25 cos ( 101) cos (264) cos ( 101) 25œ œ
37 2 37 2 37 2 2
2 365 2 365 2 365 365111
1111
ˆ‰ ˆ ‰ˆ ‰ ˆ‰ˆ ‰

573. Chapter 8 Practice Exercises 569
(0.16705 0.16705) 25 25° F¸ œ
37
#1
132. av(C ) 8.27 10 26T 1.87T dT 8.27T T T
v
20
675œœ
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& # # $
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#!675 20 655 10 10
3 0.62333'cd ab ‘

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8.27 10 26T 1.87T 5.434 1.87T 26T 283,600 0 TœÊœÊ¸
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ab
26 676 4(1.87)(283,600)
(1.87)
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396.45 °C¸
133. (a) Each interval is 5 min hour.œ
1
12
2.42 gal2.5 2 2.4 2 2.3 2 2.4 2.3
129
24 12
cd ab ab abÞÞÞ œ¸
(b) 60 mph hours/gal 24.83 mi/galab ˆ‰
12
29
¸
134. Using the trapezoid rule, x 15 7.5;˜œ Ê œ
˜x
2
mf(x ) 794.8 Area 794.8 7.5 5961 ft ;! abab
iœÊ¸ œ
2
The cost is Area $2.10/ft 5961 ft $2.10/ft†¸ababab
222
$12,518.10 the job cannot be done for $11,000.œÊ
x f(x ) m mf(x )
x 0 0 1 0
x 15 36 2 72
x 30 54 2 108
x 45 51 2 102
x 60 49.5 2 99
x 75 54 2 108
x 90 64.4 2 128.8
x1
ii i
!
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#
$
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(
5
05 67.5 2 135
x 120 42 1 42
)
135. lim lim sin lim sin sin 0''
00
3b
b
0
dx dx x b 0
9x 9x
333ÈÈ

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##

œœ œœœ
b3 b3 b3ÄÄ Ä

‘ˆ‰ˆ‰ˆ‰
11
136. ln x dx lim x ln x x (1 ln 1 1) lim b ln b b 1 lim 1 lim '
0
1
1
b
œœœœ
bbbbÄ! Ä! Ä! Ä!
´´´´
cd cd †
ln b
Š‹ Š ‹
Š‹


œ
b
b
b

10 1œ œ
137. 2 2 3 lim y 6 1 lim b 6''' '
´´110 0
101 1
1
b
dy dy dy dy
yyy y
œœ œ œ œ †
bbÄ! Ä!
‘ Š‹
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138. converges if each integral converges, but''''

2212
12
dddd
(1) (1) (1) (1)
))))
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lim 1 and diverges diverges
)Ä_
)) )
)) )
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(1) ( 1)
dd
œœ
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22
ïï

139. lim ln lim ln ln 0 ln ln 3'''
333
b
3
ïïï
2 du du du u 2 b 2 3 2
u2u u2 u u b 3 3


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bbÄ_ Ä_
‘ ‘¸¸ ˆ‰¸¸ ¸¸
140. dv dv lim ln v ln (4v 1)''
11
b
1
ïï
3v 1 4
4v v v v 4v 1 v
"" "

" œ œ ˆ‰ ‘
bÄ_
lim ln (ln 1 1 ln 3) ln 1 ln 3 1 ln œœœ
bÄ_
‘ˆ‰
b 3
4b 1 b 4 4
""
141. x e dx lim x e 2xe 2e lim b e 2be 2e ( 2) 0 2 2'
0
xxxxbbb
b
0
ï

## #
œ œ œœ
bbÄ_ Ä_
cdab
142. xe dx lim e e lim e e 0'
ï
0
3x 3x 3x 3b 3b
0
b
œ œ œ œ
bbÄ_ Ä_
‘ˆ‰
xb
39 9 39 9 9
"" """
143. 2 lim tan lim tan tan (0)'''
ï
ïïï
dx dx dx 2 2x 2 2b
4x9 4x9 33 333 x

œœ## #
"" " "
œ
" " "
œœœ œ
00
b
0
9
4
bbÄ_ Ä_
‘ ‘ˆ‰ ˆ‰

574. 570 Chapter 8 Techniques of Integration
0œœ
"
##
ˆ‰
2
36
†
11
144. 2 2 lim tan 2 lim tan tan (0) 2 0''
ï
ïï
4 dx 4 dx x b
x16 x16 4 4

œœ #
" " "
œœ œ œœ
0
b
0 bbÄ_ Ä_
‘ ‘ˆ‰ˆ‰ ˆ‰Š‹
1
1
145. lim 1 and diverges diverges
)Ä_
)) )
)) )ÈÈ

œœ11
dd
œÊ''
66
ïï
146. I e cos u du lim e cos u e sin u du 1 lim e sin u e cos u duœœœ '' '
00 0
uuu uu bb
00
ïï ï

bbÄ_ Ä_
cd cdab
I 1 0 I 2I 1 I convergesÊœÊ œ Êœ
"
#
147. dz dz dz lim 0 lim '''
11e
e eb
1e
ïï
ln z ln z ln z 1
zz z2 2 2
(ln z) (ln z) (ln b)
œ œ œ ’“ ’“Š ‹ ’ “

bbÄ_ Ä_
##
"
divergesœ_ Ê
148. 0 e for t 1 and e dt converges dt convergesŸ   Ê
ee
tt
t t
ÈÈ

ïïtt
11 ''
149. 2 converges converges''' '

ïïï ï
2 dx 2 dx 4 dx 2 dx
ee ee e ee
xxx xxx x
œœ œ
œ Ê
00
150. ;'''''


dx dx dx dx dx
x1 e x1 e x1 e x1 e x1 e

ab ab ab ab abœœœœœ
xxxxxœ
101
10 1
lim lim lim 1 e 2 and diverges diverges
x0 x0 x0ÄÄÄ
Š‹
’“
ab
ab
´



x
x

œ

x1e
œœœ Ê
x1 e
x xx1e
dx dxœ
œ
x
x
ab
x
00
11 ''
divergesÊ '
ï
ï
dx
x1 e

abœ
x
151. ; 2u 2u 2 du u u 2u 2 ln 1 u C
ux
du'''x dx u 2u du 2 2
1x
dx
x
1u 1u 3œ
#
œœ
#$#
È
È
–—
È
ˆ‰ kk
œ
œ
?

†
x2x2 ln1 x Cœ
2x
3

ÈÈ ˆ‰
152. dx x dx x dx ln x 2 ln x 2 C'' '''x 2 4x 2 3 dx 5 dx x 3 5
4x x 4 x2 x2
"

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#####
œ œ œ ˆ‰ kk kk
153. ; d(sin )
xtan
dx sec d''''dx sec d cos d 1 sin
xx 1 tan sec sin sin ab



#
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#

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œ
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Äœœ
)
))
)
)) )) )
)) ))
ln sin sin C ln Cœœ kk ¹¹Š‹))
""
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#
œœ
#
xx
x1 x1
ÈÈ

154. dx; 2 cos u du 2 sin u C 2 sin x C
ux
du'''
cos x
x
dx
x
cos u 2u du
u
È
È
È
–—
È
È
œ
œ
?
#
†
155. sin (x 1) C''dx
2x x
d(x 1)
1(x1)ÈÈ

œ

"

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156. ' u C t 2t C
ut 2t
du (2t 2) dt 2(t 1) dt''
(t 1) dt
t2t
du
u


#
"
#
#
ÈÈ
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È È
œ
œ œ
?
157. ; u tan ln sec tan C ln 1 u u C''du sec d
1u sec È
œ
#


cd k k ¹¹
È
œÄ œ œ )))
))
)

575. Chapter 8 Practice Exercises 571
158. e cos e dt sin e C'tt t
œ
159. dx 2 csc x dx csc x dx 2 cot x ln csc x cot x C''''2 cos x sin x cos x dx
sin x sin x sin x
"#
œœ kk
2 cot x csc x ln csc x cot x Cœ kk
160. d d sec d d tan C'' ' 'sin 1 cos
cos cos


))
))
)))))))œœœ
#
161. ln tan C''''9 dv dv dv dv 3 v v
81 v v9 12 3v 1 3v 12 3v 6 3
""""œ"
# #
"
# œœ ¸¸
162. tan (sin x) C''cos x dx
1sinx 1sinx
d(sin x)
œœ
"
œœ
163. cos (2 1) )
sin (2 1)))ïïïïî
ÐÑ
"
#
1 cos(2 1)ïïïïî
ÐÑ
"
4
)
0 cos(21) d sin(21) cos(21)Cʜ ')) ) ) )
)
#
"
4
164. lim lim ( 1) 0 1 1'
2
b
2
ï
dx
(x1) 1x 1b
""
œ œ œœ
bbÄ_ Ä_
‘ ‘
165. x 2 dx (x 2) dx 3'' ' ''x dx 3x 2 dx dx
x 2x1 x 2x1 x1 (x1)
"



œ œ ˆ‰
2x 3 ln x 1 Cœ
x
x1

#
"
kk
166. ; 2 x dx 2 x 4x C
x1
dx
d 2(x 1) dx''''d dx 4
1
d
2
2(x 1) dx
xx 3
)
)
)
)É È
È ÈÈ
œ
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Ô×
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È
È
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141C4 1Cœ œ
4
33
1
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)
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167. ; 2 sin 2y dy cos (2y) C cos 2 x C
yx
dy'''
2 sin x dx
x sec x
dx
2x
2 sin y 2y dy
y sec y
È
ÈÈ
È
–—
È
ˆ‰È
œ
œ
Äœ œœ
†
168. x dx dx ln C'' 'x dx 16x x 2x 2x x x 4
x16 x16 x4 x4 x4

##
# #

œ œ œ ˆ‰ ˆ ‰ ¹¹
169. 2 csc (2y) dy ln csc (2y) cot (2y) C'''
dy 2 dy
sin y cos y sin 2y
œœ œ kk
170. tan C''dd
24 (1)3 3
3
3
)) )
)) )


" "
œœ
È
È
Š‹
171. dx tan x sec x dx tan x d(tan x) tan x C''tan x
cos x
œœœ
## "
# ' †
172. sec r 1 C'' dr
(r 1) r 2r
d(r 1)
(r 1) (r 1) 1œœ
œ

"
ÈÈ

œœ kk

576. 572 Chapter 8 Techniques of Integration
173. ; u C 4 (r 2) C
u(r2)
du 2(r 2) dr'' '
(r 2) dr (r 2) dr
r4r 4(r2)
du
2u
""

#
#
ÈÈ È

?
œ%
œ
”•
È È
174. tan C''
y dy y
4y
dy
4y 4"
""
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"
#



œœ
ab
ab
Š‹
175. C sec C''sin 2 d
(1 cos 2 ) (1 cos 2 ) (1 cos 2 ) 4
d(1 cos 2 )))
)))
)
"#"#"
""" " #
œ œ œ )
176. ln C (FORMULA 19)''dx dx x x
x1 1x 21 x 4 x 1ab ab ab



"""
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177. 1 cos 4x dx 2 cos 2x dx sin 2x''
""
"" "
"
""
"" "
"44
22 2
4
È È
’“œ œ œ
ÈÈ
22
##
178. (15) dx (15) d(2x 1) C'
2x 1 2x 1
œœ
""
##' Š‹
15
ln 15
2x 1"
179. ; y 4y C (2 x) 4(2 x) C
y2x
dy dx''x dx 2 2
2x
(2 y) dy
y3 3ÈÈ

$Î#" Î#$ Î#" Î#
”•
œ
œ
Ä œ œ
222xCœ
–—
È
Š‹
È
2x
3

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180. dv; v sin csc d d cot C'''''
È ab1v
vsinsin
cos cos d 1sin d #


cdœÄ œ œ œ))))))
)))
))
))†
sin v Cœ
"
È
1v
v

181. tan (y 1) C''
dy d(y 1)
y2y2 (y1)1


"
œœ
182. ln x 1 dx; ln y 2y dy; u ln y, du ; dv 2y dy, v y
yx1
dy''È
–—
È
?
œ
œ
dx
2x 1
dy
y
È

# †
2y ln y dy y ln y y dy y ln y y C (x 1) ln x 1 (x 1) CÊœœœ'' ### ""
##
" È
(x1) ln x1 x C x ln x1 xln x1 Cœ œ
"""
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"
cdc dkk kk kk ˆ‰
183. tan d tan d ln sec C''))) )) )
#$ $$ $ ""
ab abab k kœœ
33
184. sin C''x dx x 1
82x x
dx 1
9x1
3È
ab
É ab

"""
##
"

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#




œœ Š‹
185. dz dz ln z ln z 4 tan C''z1 z1 z
zz 4 4 z z z 4 4 4z 8 8
"""""""" "
"" #
#"

ab
œ œ ˆ‰ kk a b
186. x e dx x e d x x e e C C''$### ""
## #
xx xx


œœœ ab ˆ‰
abx1e
x
187. 9 4t C''t dt
94t 94t
84
d9 4t
ÈÈ
ab

""
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œ œ
È
188. 1 cos 5 d 2 cos d sin sin 0''
00
10 10
""""
È È ˆ‰ ‘ˆ‰ˆ‰œ œ œ œ)) ) 55 2 22 22
5545
)) 1
1
##
Î"!
!
ÈÈ

577. Chapter 8 Practice Exercises 573
189. ;
xsin
dx cos d'' '''cot d cos d dx dx x dx
1 sin (sin ) 1 sin x 1 x x x 1
)) ))
)))œœ œ œ
?
œ
œ
ab ab ”•
)
))
ln sin ln 1 sin Cœkk a b))
"
#
#
190. u tan x, du ; dv , v ;œœœœ
"
œ
"dx dx
1x x x

tan x tan x''''tan x dx dx dx x dx
xx x1xx x1x
ï´

œ œ
""" "
œœab tan x ln x ln 1 x C ln x ln 1 x Cœ œ
""" #
#
#
xx
tan x
kk a b kk
È
ï´
191. ; y x ln sec x C ln sec y C''
tan y dy
2y 2x
tan x 2x dx
È
È
‘¸¸ÈÈ kkœÄ œ œ
†
192. ; e x ln x 1 ln x 2 C''''edt dx dx dx
e 3e 2 (x 1)(x 2) x 1 x
t
2t t

œœ œ œ œ œ#
c d kkkk
t
œÄ œ œ ln C ln Cœœ ¸¸ ˆ‰
xe
xe
œ" œ"
œ# œ#
t
t
193. 1 d d ln 2 ln 2 C'' ''')) ) )
)) ))


d4 dd
44 # #
œ œ œ ˆ‰ kkkk)) )) )
ln Cœ ) ¸¸
)
)
œ

2
2
194. dx tan x dx sec x 1 dx tan x x C'''1 cos 2x 1 cos 2x

œ
##
œœœ ab
195. ; cos u du sin u C sin sin x C x C
usinx
du''
cos sin x dx
1x
dx
1x
ab
È
È
ï´



"

" –—
ab
œ
œ
Ä œ œ œ
196. 2 csc 2x dx'' ' ''cos x dx cos x dx cos x dx 2 dx
sin x sin x (sin x) 1 sin x (sin x) cos x sin 2x
"

œ œ œ œ
ab ab
ln csc (2x) cot (2x) Cœkk
197. sin cos dx sin dx sin x dx cos x C'' 'xx xx
## # ## # #
"""
œœ œˆ‰
198. dx tan x 2 C'''xx2 dx x dx x
x2 x2 x2 22



" "
œœ œ#
" #
"
ab ab ÈÈ
œ œ Š‹ ab tan Cœ
"" "
#œÈÈ ab22
x
x2
Š‹
199. ln 1 e C'e dt
1e
t
t
œ
œab
t
200. tan t dt (tan t) sec t 1 dt tan t dt ln sec t C'' '$#
##
œ œ œ ab kk
tan t tan t

201. ; dx xe dx lim e e
x ln y
dx
dy e dx'''
100
x
2x 2x 2x
b
0
ïïï

ln y dy
ye 4
dy
y
xe x
"
Ô×
ÖÙ
ÕØ
‘
œ
œ
œ
?
†
x
3x
bÄ_
#
"
lim 0œœ
bÄ_
ˆ‰ˆ‰
" ""b
2e 4e 4 4
2b 2b
202. dx 3 cot x dx cos x dx 3 ln sin x ln tan x sin x C''''3secxsin x secx dx
tan x tan x
œœ

œœ kk kk
203. ; ln u C ln ln (sin v) C
u ln (sin v)
du'' 'cot v dv cos v dv du
ln (sin v) (sin v) ln (sin v) u cos v dv
sin v
?
œ
œ
”• kk k k

578. 574 Chapter 8 Techniques of Integration
204. ;
u2x1
du 2 dx'' ' 'dx 2 dx 2 dx du
(2x 1) x x (2x 1) 4x 4x u u 1 (2x 1) (2x 1) 1 ÈÈ È È

œœ Ä
œ
œ
”•
sec u C sec 2x 1 Cœœ
" "
kk k k205. e dx x dx x C''ln xÈ
œœÈ
2
3
$Î#
206. e 3 4e d ; 3 u du (3 u) C 3 4e C
u4e
du 4e d'') ))
)
)È
”•
È ab?
œ
œ
)
) "" " $Î#
$Î#
443 6
2
†
207. ; tan u C tan (cos 5t) C
u cos 5t
du 5 sin 5t dt''sin 5t dt du
1(cos 5t) 5 1u 5 5""
"" " " "
”•
œ
œ
Ä œ œ
208. ; sec x C sec e C
xe
dx e dv''dv dx
e1 xx 1
ÈÈ
2v

" "
”• ab
œ
œ
?
v
v
v

209. (27) d (27) d(3 1) (27) C C' 31 31 3)) ) "
))œœœ
"""
3 3 ln 27 3 ln 27
27' Š‹
31"
210. sin x
x cos x
&
ïïïïî
ÐÑ
5x n x si
%
ïïïïî
ÐÑ
20x cos x
$
ïïïïî
ÐÑ
60x sin x
#
ïïïïî
ÐÑ
120x cos xïïïïî
ÐÑ
120 sin xïïïïî
ÐÑ
0 x sin x dx x cos x 5x sin x 20x cos x 60x sin x 120x cos xʜ
'&& % $#
120 sin x C
211. ; 2 du 2u 2 ln 1 u C 2 r 2 ln 1 r C
ur
du'''dr 2u du 2
1r
dr
2r
1u 1u" ""È
È
–—
È
ˆ‰ ˆ‰ kk ÈÈ
œ
œ
?" ""
212. dx ln x 10x 9 C''4x 20x
x 10x 9 x 10x 9
d x 10x 9
"





% #
œœ
ˆ‰
kk
213. ln C' ''''
8 dy dy 2 dy 4 dy dy y
y(y2) y y y (y2) y2 y y
22
"
"""
œ œ ¹¹
214. ; 3u C t 2t C
ut 2t
du 2(t 1) dt''
(t 1) dt
t2t
du 3
u
"
"
#
""
## #
"Î$#
"Î$
ab

”• ab
œ
œ
? †
215. 4 sec C''8 dm 8 dm 7m
m 49m 4
7
mm
È
Ɉ‰





"
#
œœ
2
7
¸¸
216. ;
u ln t
du''' dt du du
t(1 ln t) (ln t)(2 ln t) (1 u) u(2 u)
dt
t
(u 1) (u 1) 1"" "" ÈÈÈ
”•
œ
œ
Ĝ
sec u 1 C sec ln t 1 Cœœ
" "
kk k k

579. Chapter 8 Practice Exercises 575
217. If u 1 (t 1) dt and dv 3(x 1) dx, then du 1 (x 1) dx, and v (x 1) so integrationœ œ œ œ'
0
x
ÈÈ %% #$
by parts 3(x 1) 1 (t 1) dt dx (x 1) 1 (t 1) dtÊ œ '' '
00 0
1x x
#$ %%
"
!
’“’ “ÈÈ
(x1) 1(x1) dx 1(x1) œ œ'
0
1
$ %% "
$Î#
"
!
"È ’“ab
66
8
È
218. 4v v 1
4v v1 A B C DvE
v(v 1) v 1 v v v 1 v 1
"



$
ab
œ Ê
Av(v1)v1B(v1)v1Cvv1(DvE)v(v1)œ ab ab ab ab
#### #
v0: 1 B B1;œœÊœ
v1: 42C C2;œœÊœ
coefficient of v : 0 A C D A D 2;
%
œ Ê œ
coefficient of v : 4 A B E D
$
œ
coefficient of v : 0 A B C E C D 4 D 2 (summing with previous equation);
#
œ Ê œ Ê œ
coefficient of v: 1 A B A 0;œ Ê œ
in summary: A 0, B 1, C 2, D 2 and E 1œœœœ œ
dv lim v dvʜ''
22
b
ï
4v v 1 2 2v
v(v 1) v 1 v 1 1 v 1 v
"

"

#
ab
bÄ_
ˆ‰
lim ln (v 1) tan v ln 1 vœ
bÄ_
‘ab
#" #"
v
b
2
lim ln tan b ln 1 tan 2 ln 5 0 0 0 tan 2 ln 5œœ
bÄ_
’“Š‹ ˆ‰ˆ‰ˆ‰
(b 1)
1b b

œ# ##
"" " " " "


1
ln (5) tan 2œ
1
##
" "
219. u f(x), du f (x) dx; dv dx, v x;œœ œœ
w
f(x) dx x f(x) xf (x) dx f f cos x dx'' '
œœ œ
œœ œ
œ
œ
œœ œ
œœ œ
œ
œ22 2
32 32 32
32
2
œ œcd ‘ˆ‰ ˆ‰
w
## ##
3311 11
sin x 3b a ( 1) 1 3b a 2œ œ œ ˆ‰ cd abc dab
3b a11 1 1
## # # 32
2œ
œ
œ
œ
220. tan x tan a; lim tan x lim tan b tan a tan a;''
0a
a
ab
0a
dx dx
1x 1xœœ #
" " " " " "
œœ œ œ œcd cd a b
ï
bbÄ_ Ä_
1
therefore, tan a tan a tan a a 1 since a 0.
" " "
#
œ Ê œ Êœ
11
4
CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES
1. u sin x , du ; dv dx, v x;œœœœab
"
#

2 sin x dx
1x
ï´

È sin x dx x sin x ;''ab ab
" "
##

œ
2x sin x dx
1x
ï´

È
u sin x, du ; dv , v 2 1 x ;œœ œœ
"

#dx 2x dx
1x 1x
ÈÈ

È
2 sin x 1 x 2 dx 2 sin x 1 x 2x C; thereforeœ œ ''2x sin x dx
1x
ï´

È

" " ##
ab ab
ÈÈ
sin x dx x sin x 2 sin x 1 x 2x C'ab abab
È
" " "
##
#
œ
2. ,
""
xx
œ
,
"""
œœx(x1) x x1
œ
,
""""
œœ œ #œx(x 1)(x 2) 2x x 1 (x 2)
œ
,
" """"
œ œ œ #œ #œ œx(x 1)(x 2)(x 3) 6x (x 1) (x 2) 6(x 3)
œ
the following pattern:
" """" "
œœœœ # œ œ# œ œx(x 1)(x 2)(x 3)(x 4) 4x 6(x 1) 4(x ) 6(x 3) 24(x 4)
œ Ê
; therefore
"
œœ â âœ
"
x(x 1)(x 2) (x m) (k!)(m k)!(x k) x(x 1)(x 2) (x m)
() dx
œ!
m
k0
#
k
'

580. 576 Chapter 8 Techniques of Integration
ln x k Cœ!’“ kk
m
k0
#
()
(k!)(m k)!
"

k
3. u sin x, du ; dv x dx, v ;œœ œœ
"

#
dx x
1x
È


x sin x dx sin x ; x sin x dx sin x
xsin
dx cos d'' ''" " " "
# #

œ Ä œ
œ
œ
x x dx x sin cos d
21 x
2 cos


È ”•
)
))
)))
)
sin x sin d sin x C sin x Cœ œœ
x x sin 2 x sin cos
44

## ### #
" # " """
')) ˆ‰
)) )))
sin x Cœ
x x1 x sin x
4
ï´
#
"
È
4. sin y dy; 2z sin z dz; from Exercise 3, z sin z dz
zy
dz'''" " "
È
–—
Ȝ
œ
Ä
dy
2yÈ
C sin y dy y sin y Cœ Ê œ
z sin z z1 z sin z
4
y1ysin y
ï´ ï´
ï´
# #
" "
ÈÈÈ È
' ÈÈ
y sin y Cœ
"
##
È
ÈÈyy sin y
ï´
5. d d (sec 2 1) d C'' ' 'd cos 1 cos 2
1 tan cos sin 2 cos 4
ln sec 2 tan 2 2)) )
))) )
)))
##
œ" œœ

œœœœ ))))
kk
6. u ln x 1 x , du ; dv dx, v x;œœ œ œœŠ‹Š‹Š‹È È
dx dx
x 1x 21x 2x1x xÈÈÈÈÈ Èœœ œ œ
""
#
ln x 1 x dx x ln x 1 x ; ;'''Š‹Š‹ÈÈÈÈ œ
""
## œ

x dx x dx
x1x
x
ÈÈ
Ɉ‰
´´


4
sec sec d
x sec
dx sec tan d
–—
ab
œ
œ
?
""
##
"
#
""
#
#
)
)))
)))
4
(sec 1) sec tan d
tan
''
))))
)
†
ˆ‰
´
CCœœ
tan ln sec tan
2x xln 2x12x x
)))
##

kk
ÈÈ
¹¹

ln x 1 x dx x ln x 1 x CÊœ 'Š‹Š‹ÈÈÈÈ
2x xln 2x12x x
4
ÈÈ
¹¹


7. ; ;
tsin
dt cos d
utan
du sec d
d''''dt cos d d du
t1t
sin cos tan 1 (u 1) u 1
du
u1


#
œ
È ab


”•
Ô×
ÕØ
œ
œ
Äœ Ä
œ
œ
œ
)
))
)
))
)
)) )
)) ) ln tan u C ln Cœ œ œ
"" " "" """
# # # # # # # œ
"'''du du u du u 1 tan
u1 u1 u1 sec u1

¹¹ ¸¸
È
)
)
)
ln t 1 t sin t Cœ
""
##
# "
Š‹
È
8. ; ;
ue
du e dx'''
ab
ÈÈÈ
É
2e e dx
3e 6e 1
(2u 1) du (2u 1) du
3u 6u 1 3 (u 1)
2x x
2x x




"

”•
œ
œ
Ĝ
x
x

4
3
sec 1 (sec ) d sec d sec d
u 1 sec
du sec tan d
Ô×
ÕØ
Š‹
œ
œ
?
2
3
2
3
33 3
44
3
È
È
ÈÈ È
)
)))
))) )) ))
"" #'''
tan ln sec tan C (u 1) 1 ln (u 1) (u 1) 1 Cœ œ
443 3
334 4 33
3
)))
""
" "
##
#ÈÈ
È
kk ÉÉ ¹¹†
3u 6u 1 ln u 1 (u 1) C ln œ
24
33 33
3È
¹¹Š‹É ## ""
"
#ÈÈ
È
2e 2e ln e 1 e 2e Cœ
"" "
È
3 33
’“ÉÉ ¹¹
2x x 2x x x
9. dx dx dx'' '"" "
œ x4 x2 4x x2x2x2x2
#
œœ
ab abab
dxœ
"
# 16 x 2x 2 (x 1) 1 x 2x (x 1) 1
2x222x22'’“

581. Chapter 8 Additional and Advanced Exercises 577
ln tan (x 1) tan (x 1) Cœ
"œœ"
#
" "
16 x 2x 8
x2x2
¹¹ cd


10. dx dx''""""
x1 6 x1x1xx1xx1
x2 x2
œˆ‰
ln dxœ
" " "
#

6x11 xx1 xx1
x1 2x 3 2x1 3
xx
¸¸
”• '

´´


ˆ‰ ˆ‰
33
44
ln ln 2 3 tan 2 3 tan Cœ
""" "
œ# œœ
" "
6x11 xx1
xxx 2x1 2x1
33
¸¸ ’“¹ ¹ Š‹ Š‹
ÈÈ


ÈÈ
11. lim sin t dt lim cos t lim cos x cos ( x) lim cos x cos x lim 0 0
xxx x xÄ_ Ä_ Ä_ Ä_ Ä_
'

x
x
x
x
œœœœ œcd c d a b
12. lim dt; lim lim 1 lim dt diverges since diverges
xtt xÄ! Ä! Ä! Ä!
´´´ ´
'''
x x0
1 11
cos t cos t dt
tcos ttt

Š‹
Š‹

œ
œ
t
cos t
t
œœÊ
"
; thus
lim x dt is an indeterminate 0 form and we apply l'Hopital's rule:^
xÄ!
´
'
x
1
cos t
t
†_
lim x dt lim lim lim cos x 1
xxxxÄ! Ä! Ä! Ä!
´´´´
'
x
1
cos t
t
dt
œœœœ

'
1
x
cos t
t
x
cos x
x
x
œ

œ

œ
Š‹
Š‹
13. lim ln 1 lim ln 1 k ln (1 x) dx;
u 1 x, du dx
x0 u1, x1 u2
nnÄ_ Ä_
!!É ˆ‰ˆ‰ˆ‰
”•
nn
k1 k1
0
1
##
n
œ œ
œ œ
œÊœ œÊœ
k
nnn
"" '
ln u du u ln u u (2 ln 2 2) (ln 1 1) 2 ln 2 1 ln 4 1?'
1
2
cd
#
"
14. lim lim lim
nn nÄ_ Ä_ Ä_
!! ! Š‹ ˆ‰ ˆ‰
ÎÑ
ÏÒ
n1 n1 n1
k0 k0 k0
## #
""""


ÈÈ
Ê’ “Š‹
nk nk
n
nn
1k

œœ

n
dx sin xœœœ'
0
1
"

"
"
! #È
1x

cd
1
15. cos 2x 1 1 cos 2x 2 cos x; L 1 cos 2t dt 2 cos t dt
dy dy
dx dx
œÊœœ œ œ
ÈÈ
Š‹ Š ‹ Ê
È È
##
# #
''
00
44
œœœœ
2sin t 1œœ
È
cd
1Î%
!
16. 1 ; L 1 dx
dy dy dy
dx 1 x dx 1 x dx
2x 12x x 1x 1x 4x
1x 1x
œÊ œ œ œ œ


###







Š‹ Š ‹ Š‹ Ê
ab
ab ab '
0
12
œ
dx 1 dx 1 dx x ln œœœœ'' '
00 0
12 12 12
œœ œ
Š‹ ˆ‰ˆ ‰ ‘¸¸
"œ " " œ

"Î#
!
x2 1x
1x 1x 1x 1x 1x


ln 3 (0 ln 1) ln 3œ œ ˆ‰
""
##
17. V 2 dx 2 xy dxœœ ''
a0
b1
11ˆ‰ Š‹
shell shell
radius height
6 x 1 x dx;
u1x
du dx
x(1u)
œ
œ
œ
œ
1 '
0
1
#
##È
Ô×
ÕØ
6 (1 u) u duÄ 1 '
1
0
#È
6 u2uu duœ 1
'
1
0
ˆ‰
"Î#$ Î#& Î#
6uuu 6œ œ 11 ‘ˆ ‰
242 242
357 357
$Î#& Î# (Î#
!
"
66œœœ11ˆ‰ˆ‰
70 84 30 16 32
105 105 35
1

582. 578 Chapter 8 Techniques of Integration
18. V y dx œœ ''
a1
b4
11
#

25 dx
x(5 x)

œ1 '
1
4
ˆ‰
dx 5 dx dx
xx 5x


ln ln 4 ln 5œœ111 ‘ˆ ‰ˆ ‰¸¸
x5 5
5x x 4 4
%
"
"
2 ln 4œ
15
4
1
1

19. V 2 dx 2 xe dxœœ
''
a0
b1
x
11ˆ‰ Š‹
shell shell
radius height
2xe e 2œœ11cd
xx "
!

20. V 2 (ln 2 x) e 1 dxœ
'
0
ln 2
x
1 ab
2 (ln 2) e ln 2 xe x dxœ1
'
0
ln 2
xx
cd
2 (ln 2) e (ln 2)x xe eœ1’“
xxx
ln 2
0 x

#
2 2 ln 2 (ln 2) 2 ln 2 2 2 (ln 2 1)œ11’“
#
#
(ln 2)

2ln 21œ 1’“
(ln 2)

#

21. (a) V 1 (ln x) dxœ
'
1
e
1cd
#
x x(ln x) 2 ln x dx œ 11cd
#
e
1
1
e'
(FORMULA 110)
x x(ln x) 2(x ln x x)œ 1cd
#
e
1
x x(ln x) 2x ln xœ 1cd
#
e
1
ee2e(1)œœ11cd

(b) V (1 ln x) dx 1 2 ln x (ln x) dxœ œ
''
11
ee
11
##
cd
x 2(x ln x x) x(ln x) 2 ln x dxœ 11cd
#
e
1
1
e'
x 2(x ln x x) x(ln x) 2(x ln x x)œ 1cd
#
e
1
5x 4x ln x x(ln x)œ 1cd
#
e
1
(5e4ee)(5) (2e5)œœ11cd

22. (a) V e 1 dy e 1 dy y 1œœœœœ1111 ''
00
11
y2y
‘ ‘ˆ‰ab a b ’“
#
####
"
!
"ee e3
2y 1ab
(b) V e 1 dy e 2e 1 dy 2e y 2e 1 2œœ œœ11 1 1 ''
00
11
y2yy y
ab a b ‘ˆ‰’“Š‹
#
###
"
!
"ee
2y
2e œœ1Š‹
e5 e4e5

## #
1ab
23. (a) lim x ln x 0 lim f(x) 0 f(0) f is continuous
xxÄ! Ä!
´´
œÊ œœ Ê

583. Chapter 8 Additional and Advanced Exercises 579
(b) V x (ln x) dx; lim (ln x) (2 ln x)
u (ln x)
du (2 ln x)
dv x dx
v
?
œ
œ
œ
œ ''
0 0
2 2
b
11
## #
#
#
#
Ô×
ÖÙ
ÖÙ
ÖÙ
ÕØ
Œ œ’“Š‹
dx
x
x
3
xxdx
33x
"
""
bÄ!
´
(ln 2) lim ln xœ œ11”•
ˆ‰ ˆ‰ ’“’ “
82xx 16
3339 3927
8(ln 2) 16(ln 2)#
bÄ!
´
""
2
b
24. V ( ln x) dxœ '
0
1
1
#
lim x(ln x) 2 ln x dxœ1Œ œcd
bÄ!
´
#
1
b
0
1'
2 lim x ln x x 2œ œ11
bÄ!
´
cd
1
b

25. M ln x dx x ln x x (e e) (0 1) 1;œœœœ
'
1
e
e
1
cd
M (ln x) dx (ln x) dx
x
11
eeœœ'' ˆ‰
ln x
##
" #
x(ln x) 2 ln x dx (e 2);œœ
""
##
#
Š‹cd
e
1
1
e'
M x ln x dx x dx
y
11
ee e
1œœ'' ’“
x ln x

##
"
x ln x e e 1 ;œœœ
""""
#####
###
’“’ “ Š‹ ab
xe
4

e
1
therefore, x and yœœ œœ
M
M4 M
e1 e2 My x


#

26. M 2 sin x ;œœ œ'
0
1
2 dx
1x
È

"
"
!

cd 1
M21x2;
y
0
1œœœ'2x dx
1x
È

#
"
!

’“
È
therefore, x and y 0 by symmetryœœ œ
M
M
2y
1

27. L 1 dx dx; L
xtan
dx sec d
?
œ
œ '' '
11 4
ee tane
É ”•
" œ
#xx tan
x1 sec sec d


È )
))
œœ

)))
)
†
d (tan sec csc ) d sec ln csc cot œœœ''
œœ
œ
œœ
œ44
tan e tan e
tan e
4


(sec ) tan 1
tan
))
)
ab

œ
)))))))) cd kk
1e ln 2ln1 2 1e ln 2ln1 2œ œ Š‹’“Š‹Š‹
ÈÈÈÈÈÈ
¹¹ Š‹
# #œ œ" "
È È
1e 1e
ee ee

28. y ln x 1 1 x S 2 x 1 x dy S 2 e 1 e dy;
ue
du e dy
œÊ œÊœ Êœ
œ
œ
Š‹
È È
”•
dx
dy
#
# #
11''
c0
d1
y 2y
y
y
S 2 1 u du; 2 sec sec d
utan
du sec d
Äœ Ä
œ
œ
11)))
)
)) ''
14
etane
È
”• #
#
#
œœ

†
2 sec tan ln sec tan 1 e e ln 1 e e 2 1 ln 2 1œœ1)) ))1 1ˆ‰cd kk ’“’“Š‹¹ ¹ Š‹
ÈÈ ÈÈ
"
#
## tan e

œœî
†
e1 e ln 2œ 1’“
ÈÈ
Š‹
# œœ
œ
È
È
1e e
21

29. L 4 1 dx; x y 1 y 1 x 1 x xœ œÊœÊœ'
0
1
Ê Š‹ ˆ‰ ˆ‰ˆ‰ˆ‰
dy dy
dx dx 3
32
#
#Î$# Î$# Î$# Î$" Î$
$Î#" Î#
#

584. 580 Chapter 8 Techniques of Integration
L 4 1 dx 4 6 x 6ʜʜ œ œ œŠ‹ Š ‹ Ê
‘
dy
dx
1x 1x dx
xxx
#
#Î$
"
!
œ"


''
00
11
30. S 2 f(x) 1 f (x) dx; f(x) 1 x f (x) 1 S 2 1 xœ œÊœÊœ1 1 ''
1 1
1 1
É cd cd ˆ‰ ˆ‰w
# #Î$ w #Î$
$Î#$ Î## "
x
dx
x
œ"
†
È
4 1 x dx; 4 (1 u) du 6 (1 u) d(1 u)
ux
du
œ Ä œ
œ
œ
111'''
0 00
1 11
ˆ‰ˆ‰
–—
#Î$$ Î#$ Î#
$Î#
"
#Î$
#x 2dx
3x
3
´œ"
´œ" †
6(1u)œ œ1†
212
55
‘
&Î#
"
!
1
31. y x or y x, 0 x 4Š‹ ÈÈ
dy dy
dx 4x dx x
#
" „"
#
œÊœ Êœ œ ŸŸ
È
32. The integral 1 x dx is the area enclosed by the x-axis and the semicircle y 1 x . This area is half'
1
1
ÈÈ
œ # #
the circle's area, or and multiplying by 2 gives . The length of the circular arc y 1 x from x 1 to
1
#
#
1 œ œ
È
x 1 is L 1 dx 1 dx (2 ) since L is half theœœ œ œ œ œ'' '
11 1
11 1
Ê Š‹ Š ‹ Ê
dy
dx
xdx
1x 1x
##
"

#ÈÈ

11
circle's circumference. In conclusion, 2 1 x dx .''
11
11
È
œ #

dx
1x
È

33. (b) e dx e e dx''

ïï

xe e x
xx
œ
lim e e dx lim e e dx;œ
a bÄ_ Äœ_
''
a0
0b
ex ex

xx

ue
du e dx
”•
œ
œ
Ä
x
x
lim e du lim e du
a bÄ_ Äœ_
''
e
1
uu
a


1
e
b
lim e lim eœ
a bÄ_ Ä_
cd cd
uu1e
e
a
b 1
lim e lim eœ
a bÄ_ Äœ_
‘ ‘
""
ee e
a
e
b
e0 1œ œˆ‰ˆ‰
"" !
ee

34. u , du ; dv ny dy, v y ;œœ œ œ
"
œœ1y (1y)
dy

n1 n
lim dy lim dy lim dy. Now, 0 y
nn nÄ_ Ä_ Ä_
'''
000
111
ny y y y y
1y 1y 1y 1y 1y
n
n1 n n n nï

œœœ#œœ
"
!
"
œœ ŸŸŒ œ’“
0 lim dy lim y dy lim lim 0 lim dyÊŸ Ÿ œ œ œÊ
nnnnnÄ_ Ä_ Ä_ Ä_ Ä_
'' '
00 0
11 1
n
yy ny
1y n1 n1 1y
n n1
œœœœ
"
!
"

’“
n1
0œœ
""
##
35. u x a du 2x dx;œ Ê œ
##
x x a dx u du u du C, n 2'''Š‹ Š‹
È ˆ‰È
## """
###œ
œ œ œ Á
n
n
n2
œ u
1
n2 1
nœÐ

CC Cœœ œ
u
nn n
u
xa


n2 2
n2
n2
œ# œ# œ#
ˆ‰È Š‹
È

36. sin sin
1
6
xdx dx dx du
4x 4x x 42x 4u 2
œœ œ œ
" "" "
##
"
! ‘ '' ' '
00 0 0
11 1 2ÈÈ ÈÈÈ
#
sin sin œœœœ
""" " "
##ÈÈÈ
È ÈÈ
222
u 22
48
‘ˆ‰
2
0
1 1

585. Chapter 8 Additional and Advanced Exercises 581
37. dx lim dx lim ln x 1 ln x lim ln ''
11
b
b
1
b
1
ï
ˆ‰ ˆ‰ ‘ab ’“
ax ax a
x1 x x1 x x
x1


œ# œ# # # #
"" "" # œ
œ œ œ
bb bÄ_ Ä_ Ä_
ab
a
lim ln ln 2 ; lim lim lim b if a the improperœ œœ_Ê
bbbbÄ_ Ä_ Ä_ Ä_
" "
# #
œœ
’“
ab abb1 b1
bbb
b

aa
2a
a2a
1
2
integral diverges if a ; for a : lim lim 1 1 lim ln ln 2œ œ œÊ
"" " "
## #
œ œ "Î#
bb bÄ_ Ä_ Ä_
È abb1
bb b
b1



É ”•
ln 1 ln 2 ; if a : 0 lim lim lim (b 1) 0œœ Ÿ œ œ
"" "
## # œ
œ œ
ˆ‰
ln 2
4bb1
b1 (b 1)
bbbÄ_ Ä_ Ä_
ab

a
2a
2a 1
lim ln the improper integral diverges if a ; in summary, the improper integralÊœ_Ê
bÄ_
abb1
b

œ "
#
a
dx converges only when a and has the value '
1
ï
ˆ‰
ax ln 2
x1 x 4

œ# #
""
œ
38. G(x) lim e dt lim e lim if x 0 xG(x) xœœœœœÊœ
bb bÄ_ Ä_ Ä_
'
0
b
xt xt
b
0
‘ˆ‰ Š‹
"" "
xxxx x
1e 10
xb
1 if x 0œ
39. A converges if p 1 and diverges if p 1. Thus, p 1 for infinite area. The volume of the solid of revolutionœŸŸ'
1
ï
dx
x
p
about the x-axis is V dx which converges if 2p 1 and diverges if 2p 1. Thus we wantœœ Ÿ ''
11
ïï
11ˆ‰
"
#
xx
dx
p2p
p for finite volume. In conclusion, the curve y x gives infinite area and finite volume for values of p satisfyingœ
"
# p
p1.
" #
Ÿ
40. The area is given by the integral A ;œ '
0
1
dx
x
p
p 1: A lim ln x lim ln b , diverges;œ œ œ œ_
bbÄ! Ä!
´´
cd
1
b
p 1: A lim x 1 lim b , diverges;œ œ œ_
bbÄ! Ä!
´´

cd
1p 1p1
b
p 1: A lim x lim b 1 0, converges; thus, p 1 for infinite area.œ œ" œ  
bbÄ! Ä!
´´

cd
1p 1p1
b
The volume of the solid of revolution about the x-axis is V which converges if 2p 1 or
x
0
1œ1'dx
x
2p
p , and diverges if p . Thus, V is infinite whenever the area is infinite (p 1).   
""
##
x
The volume of the solid of revolution about the y-axis is V R(y) dy which
y
11œœ11''
ïï
cd
# dy
y
2p#
converges if 1 p 2 (see Exercise 39). In conclusion, the curve y x gives infinite area and finite
2
p
Í œ
p
volume for values of p satisfying 1 p 2, as described above.Ÿ
41. e cos 3x
2x
ÐÑ
2e sin 3x
2 1
3x
ÐÑ
4e cos 3x

2 1
9x
ÐÑ
I sin 3x cos 3x I I (3 sin 3x 2 cos 3x) I (3 sin 3x 2 cos 3x) Cœ Êœ Êœ
e2e 413e e
39999 13
2x 2x 2x 2x
42. e sin 4x
3x
ÐÑ
3e cos 4x
3
4x
ÐÑ
"
9e sin 4x

3
16x
ÐÑ
"
I cos 4x sin 4x I I (3 sin 4x 4 cos 4x) I (3 sin 4x 4 cos 4x) Cœ Ê œ Ê œ
e3e925e e
416161616 25
3x 3x 3x 3x

586. 582 Chapter 8 Techniques of Integration
43. sin 3x sin xÐÑ
3 cos 3x cos xÐÑ
9 sin 3x sin x

ÐÑ
I sin 3x cos x 3 cos 3x sin x 9I 8I sin 3x cos x 3 cos 3x sin xœ Ê œ
ICʜ
sin 3x cos x 3 cos 3x sin x
8

44. cos 5x sin 4xÐÑ
sin 5x cos 4x ÐÑ
"
4
25cos 5x sin 4

ÐÑ
"
16
I cos 5x cos 4x sin 5x sin 4x I I cos 5x cos 4x sin 5x sin 4xœ Ê œ
" "
4161616416
5259 5
I (4 cos 5x cos 4x 5 sin 5x sin 4x) Cʜ
"
9
45. e sin bx
ax
ÐÑ
ae cos bx
ax
ÐÑ
" b
a e sin bx

# "ax
ÐÑ
b

I cos bx sin bx I I (a sin bx b cos bx)œ Ê œ
eaeaabe
bbbbb
ax ax ax

Š‹
œ
I (a sin bx b cos bx) Cʜ
e
ab
ax

œ
46. e cos bx
ax
ÐÑ
ae sin bx
ax
ÐÑ
" b
a e cos bx

# "ax
ÐÑ
b

I sin bx cos bx I I (a cos bx b sin bx)œ Ê œ
eae aabe
bb bbb
ax ax ax

Š‹
œ
I (a cos bx b sin bx) Cʜ
e
ab
ax

œ
47. ln (ax) 1ÐÑ
x

"
x
ÐÑ
I x ln (ax) x dx x ln (ax) x Cœ œ 'ˆ‰
"
x
48. ln (ax) xÐÑ
#
x

"" $
x3
ÐÑ
I x ln (ax) dx x ln (ax) x Cœ œ
""""$$$
3x339
x 'ˆ‰Š‹
"
49. (a) (1) e dt lim e dt lim e lim ( 1) 0 1 1>œ œ œ œ œœ''
00
ttt
b
b
0
ï

bbbÄ_ Ä_ Ä_
cd ‘
"
e
b
(b) u t , du xt dt; dv e dt, v e ; x fixed positive realœœ œ œœ
xx1 t t
(x 1) t e dt lim t e x t e dt lim 0 e x (x) x (x)Êœ œ œ œ> >> ''
00
xt xt x1t b
0
ïï

bbÄ_ Ä_
cd ˆ‰
b
e
x
x
b
!

587. Chapter 8 Additional and Advanced Exercises 583
(c) (n 1) n (n) n!:>>œ œ
n 0: (0 1) (1) 0!;œœœ>>
n k: Assume (k 1) k! for some k 0;œœ >
n k 1: (k 1 1) (k 1) (k 1) from part (b)œ œ >>
(k 1)k! induction hypothesisœ
(k 1)! definition of factorialœ
Thus, (n 1) n (n) n! for every positive integer n.>>œ œ
50. (a) (x) and n (n) n! n! n 2n>> 1¸œÊ¸œˆ‰ ˆ‰ ˆ‰ÉÉ È
x2 n2 n
ex ene
xnn
11
(b) n 2n calculator
10 3598695.619 3628800
20 2.4227868 10 2.432902 10
ˆ‰È
n
e
n
1
‚‚
") ")
$# $#
%( %(
'% '%
)" )"
30 2.6451710 10 2.652528 10
40 8.1421726 10 8.1591528 10
50 3.0363446 10 3.0414093 10
60 8.3094383 10 8.3209871 10
‚‚
‚‚
‚‚
‚‚
(c) n 2n 2n e calculator
10 3598695.619 3628810.051 3628800
ˆ‰ ˆ‰ÈÈ
nn
ee
nn
1 12n
11
œ

588. 584 Chapter 8 Techniques of Integration
NOTES:

589. CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION
9.1 SLOPE FIELDS AND SEPARABLE DIFFERENTIAL EQUATIONS
1. (a) y e y e 2y 3y 2 e 3e eœÊœÊœ œ
œœ œœœxx xxxww
ab
(b) y e e y e e 2y 3y 2 e e 3 e e eœ Êœ Ê œ œ
œ œÊ œ œÊ œ œÊ œ œÊ œx 3x2 x 3x2 x 3x2 x 3x2 xww
##
33 ˆ‰ ab
(c) y e Ce y e Ce 2y 3y 2 e Ce 3 e Ce eœ Êœ Ê œ œ
œ œÊ œ œÊ œ œÊ œ œÊ œx 3x2 x 3x2 x 3x2 x 3x2 xww
##
33 ˆ‰ ab
2. (a) y y yœ Ê œ œ œ
""" w#
#
xxx
œ
ˆ‰
(b) y y yœ Ê œ œ œ
"""
œœœ
w#
#
x 3 (x 3) (x 3)
œ’“
(c) y y yœÊœ œ œ
"""
œœœ
w#
#
xC (xC) xC
œ
œ ‘
3. y dt y dt x y dt e x dt e xy eœ Ê œ Ê œ œ œ
""" " w#w
xt xt xx t xt11 11
eeee e'' ''
xx xx
xxx
ttxt t
œ
ˆ‰ˆ ‰ Š‹
x y xy eʜ
#w x
4. y 1 t dt y 1 t dt 1 xœÊœ
"" "
œ œ
%%% w
#
œ
È È
Š‹
È1x 1x11
4x
1x
% %
Ð
%
Ð
''
xx
ÈÈÈ
–—
Š‹
y 1 t dt 1 y y 1 y y 1Êœ Êœ Ê œ
wwwÐ" Ð
œœœ œ
%
Š‹Š ‹ Š‹
È
2x 2x 2x
1x 1x 1x1x 1
Ð ÐÐ
% %%
%
È
'
x
†
5. y e tan 2e y e tan 2e e 2e e tan 2eœÊœ œ
œœœœxx xxx xxxÐ" w Ð" Ð" "
œ œ
ab ab ab ab ’“
12e
2
14eab
x 2xœ
y y y y ; y( ln 2) e tan 2e 2 tan 1 2Êœ Êœ œ œ œ œ
w w Ð" Ð"
œœ #
22
14e 14e 4
2x 2x
œ ln 2 ln 2
ab ˆ‰
11
6. y (x 2) e y e 2xe (x 2) y e 2xy; y(2) (2 2) e 0œ Ê œ Ê œ œ œ
œœœ œ œxxx x 2
œœœ œ œ
ww ˆ‰
7. y y y y xy sin x yœ Ê œ Ê œ Ê œ Ê œ
cos x x sin x cos x sin x cos x sin x
xx xxxxx
yww wwÐÐ "
œ
ˆ‰
xy y sin x; y 0Êœ œ œ
w
# ˆ‰
1 1
1
cos ( /2)
(/2)
8. y y y x y x y xy y ; y(e) e.œÊœ Êœ Ê œ Ê œ œœ
x xx e
ln x (ln x) ln x (ln x) ln x (ln x) ln e
ln x x
w w #w #w #
Ð
""
Š‹
œ
x
œœœ
œœ
9. 2 xy 1 2x y dy dx 2y dy x dx 2y dy x dx 2 yÈ ˆ‰
dy
dx 3
2
œÊ œ Ê œ Ê œ Ê
"Ñ# "Ñ# "Ñ# Ð"Ñ# "Ñ# Ð"Ñ# ÎÑ# ''
2x C y x C, where C CœÊ œ œ
"Î# $Î# "Î#
""
"
#
2
3
10. x y dy x y dx y dy x dx y dy x dx 2y C
dy
dx 3
x
œÊœ Ê œÊ œ Êœ
# # "Ñ# Ð"Ñ# # Ð"Ñ# # "Ñ#
È ''
Ð
2y x Cʜ
"Î# $"
3
11. e dy e e dx e dy e dx e dy e dx e e C e e C
dy
dx
œÊœ Ê œ Ê œ ʜʜ
xy x y y x y x y x y xœœ ''
12. 3 e dy 3 e dx e dy 3 dx e dy 3 dx e C e C
dy dx
œÊœ ÊœÊ œ ʜʜxx x xxx
$$ $ $ÐÐyyyyyy 33 ''

590. 586 Chapter 9 Further Applications of Integration
13. y cos y dy y cos y dx dy dx dy dx. In the integral on the left-hand
dy
dx y y
sec y sec y
œÊœ ʜʜÈÈ ÈÈ ˆ‰
##
œœ
ÈÈ
ÈÈ
''
side, substitute u y du dy 2 du dy, and we have sec u du dx 2 tan u x CœÊœ Ê œ œ Ê œÈ
11
2y yÈÈ ''#
x 2 tan y CÊ œ È
14. 2xy 1 dy dx 2 ydy dx 2 y dy x dx 2 y dy x dxÈ ÈÈ ÈÈ
dy
dx
11
2xy x
1/2 1/2 1/2 1/2
œÊ œ Ê œ Ê œ Ê œ
ÈÈ
ÐÐ
''
2 dy C 2 y 3 x C 2 y 3 x C, where C CÊœÊœÊ œ œ
ÈÈÈ ÈÈ ˆ‰È
y x3 3
11 1
3/2
22
3
3/2
3
2
1/2
Î
œ
15. x e dy dx e dy dx e dy dx In the integral on theÈ
dy dy
dx dx
ee ee e e
xx x x
œÊœÊœ Êœ Ê œ Þ
yx%È
yyœœ œ œxx x x
ÈÈ È È
œœyy ''
right-hand side, substitute u x du dx 2 du dx, and we have e dy 2 e duœÊœ Ê œ œÈ
""
#ÈÈxx ''œyu
e2eC e2eC, where CCÊ œ Ê œ œ
œœyu y x
11
œ
16. sec x e e cos x dy e e cos x dx e dy e cos x dxab a b
dy dy
dx dx
œÊœ Êœ Êœ
y sin x y sin x y sin x y sin xÊÊ œ
e dy e cos x dx e e C e e C, where C CÊœ ʜʜ œ''œœœyyy sin x sin x sin x
11
17. 2x 1 y dy 2x 1 y dx 2x dx 2x dx sin y x C since y
dy dy dy
dx
22
1y 1y
œÊœÊ œÊ œ Ê œ "ÈÈ kk
ÈÈÐÐ
Ð" #22
''
ysinx Cʜ ab
2
18. dy dx dy dx dx e dy e dx e dy e dx e C
dy
dx e e e e e
ee eee e 2y x 2y x x
1
œÊœ Êœ œ Ê œ Ê œ Êœ
2x y 2x y 2x y x 2y
xy xy xy 2y$$ $
!!
''
#
e 2e C where C 2CÊ œ œ
2y x
1
19. y x y slope of 0 for the line y x.
w
œÊ œ
For x, y 0, y x y slope 0 in Quadrant I. œÊ
w
For x, y 0, y x y slope 0 in Quadrant III. œÊ
w
For y x , y 0, x 0, y x y slope 0 inkk kk œÊ
w
Quadrant II above y x.œ
For y x , y 0, x 0, y x y slope 0 inkk kk œÊ
w
Quadrant II below y x.œ
For y x , x 0, y 0, y x y slope 0 inkk kk œÊ
w
Quadrant IV above y x.œ
For y x , x 0, y 0, y x y slope 0 inkk kk œÊ
w
Quadrant IV below y x.œ
All of the conditions are seen in slope field (d).

20. y y 1 slope is constant for a given value of y, slope
w
œÊ
is 0 for y 1, slope is positive for y 1 and negative forœ
y 1. These characteristics are evident in slope field (c).ßÐ

591. Section 9.1 Slope Fields and Separable Differential Equations 587
21. y slope 1 on y x and 1 on y x.
w
œ Ê œ œ œ
x
y
y slope 0 on the y-axis, excluding 0, 0 ,
w
œ Ê œ
x y
ab
and is undefined on the x-axis. Slopes are positive for
x 0, y 0 and x 0, y 0 (Quadrants II and IV),Ñ! !Ñ
otherwise negative. Field (a) is consistent with these
conditions.

22. y y x slope is 0 for y x and for y x.
w
œÊ œ œ
22
For y x slope is positive and for y x slope iskk kk kk kkÑ!
negative. Field (b) has these characteristics.

23. 24.
25-36. Example CAS commands:
:Maple
ode := diff( y(x), x ) = y(x);
icA := [0, 1];
icB := [0, 2];
icC := [0,-1];
DEplot( ode, y(x), x=0..2, [icA,icB,icC], arrows=slim, linecolor=blue, title="#25 (Section 9.1)" );
:Mathematica
To plot vector fields, you must begin by loading a graphics package.
<<Graphics`PlotField`
To control lengths and appearance of vectors, select the Help browser, type PlotVectorField and select Go.
Clear[x, y, f]
yprime = y (2 y);
pv = PlotVectorField[{1, yprime}, {x, 5, 5}, {y, 4, 6}, Axes True, AxesLabel {x, y}]; Ä Ä
To draw solution curves with Mathematica, you must first solve the differential equation. This will be done with
the DSolve command. The y[x] and x at the end of the command specify the dependent and independent variables.
The command will not work unless the y in the differential equation is referenced as y[x].
equation = y'[x] == y[x] (2 y[x]) ;
initcond = y[a] == b;
sols = DSolve[{equation, initcond}, y[x], x]
vals = {{0, 1/2}, {0, 3/2}, {0, 2}, {0, 3}}
f[{a_, b_}] = sols[[1, 1, 2]];

592. 588 Chapter 9 Further Applications of Integration
solnset = Map[f, vals]
ps = Plot[Evaluate[solnset, {x, 5, 5}];
Show[pv, ps, PlotRange { 4, 6}];Ä
The code for problems such as 33 & 34 is similar for the direction field, but the analytical solutions involve
complicated inverse functions, so the numerical solver NDSolve is used. Note that a domain interval is
specified.
equation = y'[x] == Cos[2x y[x]] ;
initcond = y[0] == 2;
sol = NDSolve[{equation, initcond}, y[x], {x, 0, 5}]
ps = Plot[Evaluate[y[x]/.sol, {x, 0, 5}];
N[y[x] /. sol/.x 2]Ä
Show[pv, ps, PlotRange {0, 5}];Ä
Solutions for 35 can be found one at a time and plots named and shown together. No direction fields here.
For 36, the direction field code is similar, but the solution is found implicitly using integrations. The plot
requires loading another special graphics package.
<<Graphics`ImplicitPlot`
Clear[x,y]
solution[c_] = Integrate[2 (y 1), y] == Integrate[3x 4x 2, x] c
2
values = { 6, 4, 2, 0, 2, 4, 6};
solns = Map[solution, values];
ps = ImplicitPlot[solns, {x, 3, 3}, {y, 3, 3}]
Show[pv, ps]
25. 26.
27. 28.

593. Section 9.2 First-Order Linear Differential Equations 589
29. 30.
9.2 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
1. x y e y , P(x) , Q(x)
dy dy
dx dx x x x x
x ee
œ Ê œ œ œˆ‰
""
xx
P(x) dx dx ln x ln x, x 0 v(x) e e x'' œœœÊœ œœ
"
x
kk
'Px dxÐÑ ln x
y v(x) Q(x) dx x dx e C , x 0œœœœ
"""œ
v(x) x x x x
eeC' 'ˆ‰ ab
x
x
x
2. e 2e y 1 2y e , P(x) 2, Q(x) e
xdy dy
dx dx
xxx
œÊœ œ œ
ÐÐ
P(x) dx 2 dx 2x v(x) e e'' œœÊœ œ
'Px dxÐÑ 2x
y e e dx e dx e C e Ceœ œ œ œ
"""
eee
2x 2x 2x
'
2x x x x x 2x
†
œœœ ' ab
3. xy 3y , x 0 y , P(x) , Q(x)
w
œ Ê œ œ œ
sin x 3 sin x 3 sin x
xdxxxxx
dy
œ ÐÐ
ˆ‰
dx 3 ln x ln x , x 0 v(x) e x'3
x
œœÊœœkk
$$ ln x
%
y x dx sin x dx ( cos x C) , x 0œœœœ
""" Ð$
xx x x x
sin x C cos x
ÐÐ Ð Ð Ð
'ˆ‰ '
4. y (tan x) y cos x, x (tan x) y cos x, P(x) tan x, Q(x) cos x
w# # #
##
œÊœ œ œ
11 dy
dx
tan x dx dx ln cos x ln (cos x) , x v(x) e (cos x)'' œœœ Êœœ
sin x
cos x
kk
Ð" Ð"
##
11 ln cos xÐÑ
ÐÑ
y (cos x) cos x dx (cos x) cos x dx (cos x)(sin x C) sin x cos x C cos xœœœœ
" Ð" #
(cos x)
$Î
'' †
5. x 2y 1 , x 0 y , P(x) , Q(x)
dy dy
dx x dx x x x x x x
22
œ Ê œ œ œ
"""""
ˆ‰
œœ
dx 2 ln x ln x , x 0 v(x) e x'2
x
œœÊœœkk
## ln x
œ
y x dx (x 1) dx x C , x 0œœœœ
""" " " ""#
##xxx x x xx
xC
œœœ œ œ
œ
''ˆ‰ Š‹
6. (1 x) y y x y , P(x) , Q(x)œÊ œ œ œ
w ""
œœ œ œÈ ˆ‰
dy
dx 1x 1 x 1x 1x
xxÈÈ
dx ln (1 x), since x 0 v(x) e 1'"
œ1x
œ Êœ œ
ln 1 xÐßÑ
y (1 x) dx x dx x Cœ œ œ œ
"""
œœœ œ œœ
$Î#
1x 1x 1x 1x 3 3(1x) 1x
x 22xC'' Š‹ È ˆ‰ˆ ‰
È Ð"œ
7. y e P(x) , Q(x) e P(x) dx x v(x) e
dy
dx
œ Ê œ œ Ê œ Ê œ
"" " " "
## # # #x2 x2 x2ÊÊ œÊ '
y e e dx e dx e x C xe CeÊœ œ œ œ
"" " " "
####e
ÐÎx2
''œ%% % % % %x2 x2 x2 x2 x2 x2ˆ‰ ˆ ‰
8. 2y 2xe P(x) 2, Q(x) 2xe P(x) dx 2 dx 2x v(x) e
dy
dx
œ Ê œ œ Ê œ œ Ê œ
œœ2x 2x 2x ''
y e 2xe dx 2x dx e x C x e CeÊœ œ œ œ
"" ##
ee
2x 2x
''2x 2x 2x 2x 2x
ab a b
œœœœ

594. 590 Chapter 9 Further Applications of Integration
9. y 2 ln x P(x) , Q(x) 2 ln x P(x) dx dx ln x, x 0
dy
dx x x x
œ Êœ œ Ê œ œ ˆ‰
"" " ''
v(x) e y x (2 ln x) dx x (ln x) C x (ln x) CxÊœœÊœ œ œ
œln x"" ##
xx 'ˆ‰ cd
10. y , x 0 P(x) , Q(x) P(x) dx dx 2 ln x ln x , x 0
dy
dx x x x x x
2cos x 2 cos x 2
œ Êœ œÊ œ œ œ ˆ‰ kk
œœ
'' #
v(x) e x y x dx cos x dx (sin x C)ÊœœÊœ œ œ œ
ln x
œ
## """ œ
xx x x x
cos x sin x C
œœ œ œ œ
''ˆ‰
11. s P(t) , Q(t) P(t) dt dt 4 ln t 1 ln (t 1)
ds 4 t 1 4 t 1 4
dt t 1 (t 1) t 1 (t 1) t 1
œÊœ œÊ œ œœˆ‰ kk
ÐÐ Ð Ð Ð
œœ %
ÐÐ
''
v(t) e (t 1) s (t 1) dt t 1 dtÊœ œÊœ œ
ln t 1Ð Ñ
!
%%# "œ""
ÐÐ Ð(t 1) (t 1)
t
(t 1)
%% Ð
'' ’“ ab
tCœœ
"
Ð ÐÐÐ(t1) 3(t1) (t1) (t1)
tttC
3
% %%%
ÐЊ‹
12. (t 1) 2s 3(t 1) s 3 P(t) , Q(t) 3 (t 1)œÊ œÊœ œ
ds ds 2 2
dt (t 1) dt t 1 (t 1) t 1
""
œœœ œ
ÐÎ

ˆ‰
P(t) dt dt 2 ln t 1 ln (t 1) v(t) e (t 1)ÊœœœÊœœ'' 2
t1œ
##
kk
ln t 1Ð Ñ
œ
s (t 1) 3 (t 1) dt 3(t 1) (t 1) dtÊœ œ
""
œœ
#ÐÎ #Ð"
(t 1) (t 1)
œœ
'' cd c d
(t1) ln t1 C (t1)(t1) ln(t1) , t 1œœ
"
œ œ
ÎÐ#
(t 1) (t 1)
C
œ œcd kk
13. (cot ) r sec P( ) cot , Q( ) sec P( ) d cot d ln sin v( ) e
dr
d)
œÊœ œÊ œ œ Êœ)))))) )))) )) '' kk
ln sin %%œ
sin because 0 r (sin )(sec ) d tan d ln sec CœÊœ œœ)) ))))) )
1
)))#
"""
sin sin sin '' abkk
(csc ) ln sec Cœ))abkk
14. tan r sin (cot ) r sin cos P( ) cot , Q( ) sin cos )) ))))))))
dr dr r sin dr
d d tan tan d)))))
)
œ Ê œ Ê œ Ê œ œ
#
œ
P( ) d cot d ln sin ln (sin ) since 0 v( ) e sin Êœœœ Êœœ'')) )) ) ) ) ) ) kk
1
# ln sinÐÑ)
r (sin ) (sin cos ) d sin cos d CÊœ œ œ œ
""" #
sin sin sin 3 3 sin
sin sin C
))))
))'')))) ))) ˆ‰Š‹
Ð œ
15. 2y 3 P(t) 2, Q(t) 3 P(t) dt 2 dt 2t v(t) e y 3e dt
dy
dt e
œÊ œ œÊ œ œÊ œ Êœ '' '
2t 2t "
2t
e C ; y(0) 1 C 1 C y eœ œÊœÊœÊœ
"""
#####e
33 3
2t
ˆ‰
2t 2t œ
16. t P(t) , Q(t) t P(t) dt 2 ln t v(t) e t y t t dt
dy 2y
dt t t t
2
œÊ œ œÊ œ Ê œ œÊœ
## ### " '' kk abab
ln t
œ
œ
t dt C ; y(2) 1 1 C yœœœœÊœÊœÊœ
""%
tt55t 54 555t
ttC 8C 12t12
œœ œ œ
Ð
' Š‹
17. y P( ) , Q( ) P( ) d ln v( ) e
dy
d
sin sin
)) ) ) )
))
œÊœ œÊ œ Êœœˆ‰ kk kk
""
)) )))) ) ' ln%%œ
y d d for 0 y sin d ( cos C)Êœ œ ÁÊœ œ
"" ""
kk)))) ) )
))'' 'kkˆ‰ ˆ‰))))) )) )
sin sin
cos ; y 1 C y cos œ œ Ê œ Ê œ
""
##)) ) )
11 1
))
C
2
ˆ‰
18. y sec tan P( ) , Q( ) sec tan P( ) d 2 ln v( ) e
dy
d
22
)) )
œ Êœ œ Ê œ Êœˆ‰ kk))) ) )))) )) ) )
## ' œ2ln%%œ
y sec tan d sec tan d (sec C) sec C ;œÊœ œ œ œ ) )))))) ))))) )))
Ð# Ð# # # # # #"
)
ޜ
''aba b
y 2 2 (2) C C 2 y sec 2ˆ‰ ˆ ‰ Š‹ Š‹
111
11399
18 18
œÊœ Ê œ Êœ
œœ
œœ
)) )
##

595. Section 9.2 First-Order Linear Differential Equations 591
19. (x 1) 2 x x y 2 y 2xy P(x) 2x,œÊ œ Êœ Êœ
dy dy x(x 1) dy
dx x 1 dx x 1 (x 1) dx (x 1)
eee
ab ’“
#
œœœ œ
œ
xx x
œœ
œœ
œ
Q(x) P(x) dx 2x dx x v(x) e y e dxœÊ œœÊœÊœ
e e
(x 1) (x 1)
x
e
x x
œ œ
œ œ
œ
œ œ
#Ð "
'' '
Ð
œ
œx
œx
’“
e dx e C Ce ; y(0) 5 C 5 C 5œ œ œ œÊœÊ"œ
xx x
(x 1) 1 x 1 0 1
(x 1) e1
œœ œ
œ
$Î œ
'"

œ
’“
x
C 6 y 6eʜʜ
x e
x1
œ
œ x
œ
20. xy x P(x) x, Q(x) x P(x) dx x dx v(x) e y e x dx
dy
dx
x x2 x2
e
œÊ œ œÊ œ œ Ê œ Êœ '' '
œ œœ
œ
Ð
#
ÎÎ " x2
†
e C 1 ; y(0) 6 1 C 6 C 7 y 1œ œ œ Ê œ Ê œ Ê œ
" Î
ee e
x2 C 7
x2 x2 x2
œœ œ
ÐÐ Ð
œŠ‹
21. ky 0 P(t) k, Q(t) 0 P(t) dt k dt kt v(t) e
dy
dt
kt
œÊ œ œÊ œ œÊ œ '' Ð
y e (0) dt e (0 C) Ce ; y(0) y C y y y eÊœ œ œ œ Ê œ Êœ
" Ð
!!!
e
kt kt kt kt
$kt
'ˆ‰
22. (a) v 0 P(t) , Q(t) 0 P(t) dt dt t v(t) e
dv k k k k kt
dt m m m m m
œÊ œ œÊ œ œœÊ œ '' kt m%
y e 0 dt ; v(0) v v C v v v eÊœ œ œ Ê œ Ê œ Êœ
" Î Ð Î Ñ
!!!!
eee
kt m kmtCC
kt m kt m k 0 mÐÐÞ&Ð
' †
(b) v dt ln v t C v e v e e . Let e C
dv k dv k k
dt m v m m
kmt C kmt C C
1
œ Ê œ Ê œ Ê œ Ê œ † œ Þ
Î ÐÎÑab
Then v C and v 0 v C C . So v v eœ† œœ †œ œ
11
ee
10110
kmt
ÞÐ & ÞÐ &kmt km 0 ab
œÊ
Ð Î Ñ
23. x dx x ln x C x ln x Cx (b) is correct'"
x
œœÊabkkkk
24. cos x dx (sin x C) tan x (b) is correct
""
cos x cos x cos x
C' œœÊ
25. Let y(t) the amount of salt in the container and V(t) the total volume of liquid in the tank at time t.œœ
Then, the departure rate is (the outflow rate).
y(t)
V(t)
(a) Rate entering 10 lb/minœœ
2 lb
gal min
5 gal
†
(b) Volume V(t) 100 gal (5t gal 4t gal) (100 t) galœœ œ
(c) The volume at time t is (100 t) gal. The amount of salt in the tank at time t is y lbs. So theÎ
concentration at any time t is lbs/gal. Then, the rate leaving (lbs/gal) 4 (gal/min)
yy
100 t 100 tœœ
œ †
lbs/minœ
4y
100 tœ
(d) 10 y 10 P(t) , Q(t) 10 P(t) dt dt
dy 4y dy
dt 100 t dt 100 t 100 t 100 t
44 4
œ Ê œ Ê œ œ Ê œ
œœ œ œ
ˆ‰ ''
4 ln (100 t) v(t) e (100 t) y (100 t) (10 dt)œÊœœÊœ
4ln 100 tÐßÑ %% "
œ(100 t)
%
'
C 2(100 t) ; y(0) 50 2(100 0) 50œœœÊœ
10 C C
(100 t) (100 t) (100 0)
(100 t)
5œœœ
œ
%%%
ÑŠ‹
C (150)(100) y 2(100 t) y 2(100 t)ʜ ʜ ʜ
%
œ
œ
(150)(100)
(100 t)
150
1
%
%%
ˆ‰
t
100
(e) y(25) 2(100 25) 188.56 lbs concentration 1.5 lb/galœ ¸ Ê œ¸¸
(150)(100) y(25)
(100 25) volume 125
188.6
%
%
œ
26. (a) 5 3 2 V 100 2t
dV
dt
œœÊœ ab
The tank is full when V 200 100 2t t 50 minœœÊœ
(b) Let y t be the amount of concentrate in the tank at time t.ab
53 y
dy gal y gal dy y dy
dt gal min 100 2t gal min dt 2 2 50 t dt 2 t 50 2
lb lb 5 3 3 5
œ ʜʜŠ‹Š‹Š ‹Š‹ ˆ‰
"
#œ œœ ab
Q(t) ; P(t) P(t) dt dt ln t 50 since t 50 0œœ Ê œ œ
531 313
2250t 2t502
ˆ‰ ab
œœ
''
vt e e t 50ab a bœœ œ
'P(t) dt ln t 50 3/2
3
2
abœ

596. 592 Chapter 9 Further Applications of Integration
yt t50 dt t50 yt t50 t50 Cab ab ab ab œ ‘abœœ Êœ ÎÎ
15 C
t50 t502
3/2 3/2 5/2
ab abœ œ
Ð
3/2 3/2
'
Apply the initial condition (i.e., distilled water in the tank at t 0):œ
y 0 0 50 C 50 y t t 50 . When the tank is full at t 50,ab abœœ Ê œ Ê œ œ
C50
50
5/2
t50
3/2 3/2
5/2
abœ
y 50 100 83.22 pounds of concentrate.abœ ¸
50
100
5/2
3/2
27. Let y be the amount of fertilizer in the tank at time t. Then rate entering 1 1 1 and theœœ
lb lb
gal min min
gal
†
volume in the tank at time t is V(t) 100 (gal) [1 (gal/min) 3 (gal/min)]t min (100 2t) gal. Henceœ œ
rate out 3 lbs/min 1 lbs/min y 1œœ Êœ Êœˆ‰ ˆ ‰ ˆ‰
y3y dy 3y dy
100 2t 100 2t dt 100 2t dt 100 2t
3
ÐÐ Ð Ð
P(t) , Q(t) 1 P(t) dt dt v(t) eÊœ œÊ œ œ Êœ
33
100 2t 100 2t
3 ln (100 2t)
ÐÐÐ#
Ð '' 3 ln 100 2t 2
(100 2t) y (100 2t) dt (100 2t) Cœ Êœ œ
ÐÎÑ# ÐÎÑ# ÐÎÑ# "
Ð
ÐÐ
Ð#(100 2t)
2(100 2t)
$

' ’“
(100 2t) C(100 2t) ; y(0) 0 [100 2(0)] C[100 2(0)] C(100) 100œ œÊ Ê œ
$Î# $Î# $Î#
C (100) y (100 2t) . Let 0 2Êœ œ Êœ œÊ œ
Ð"Ñ# " Ð ÐÐ
10 10 dt dt 10
(100 2t) dy dy (100 2t) ( 2)
Ð"œ
œ
ˆ‰
3
2 0 20 3 100 2t 400 9(100 2t) 400 900 18t 500 18tœ œ Ê œ Ê œ Ê œ Ê œ
3 100 2t
10
È
Ð È
t 27.8 min, the time to reach the maximum. The maximum amount is thenʸ
y(27.8) [100 2(27.8)] 14.8 lbœ ¸
[100 2(27.8)]
10
Ð
Ð"œ
28. Let y y(t) be the amount of carbon monoxide (CO) in the room at time t. The amount of CO entering theœ
room is ft /min, and the amount of CO leaving the room is ft /min.ˆ‰ ˆ‰ˆ‰
43 12 3
100 10 1000 4500 10 15,000
yy
‚œ œ
$ $
Thus, y P(t) , Q(t) v(t) e
dy y dy
dt 1000 15,000 dt 15,000 1000 15,000 1000
12 12 12
œ Ê œ Ê œ œ Êœ
"" t 15 000Îß
y e dt y e e C e 180e C ;Êœ Êœ œ
"
e 1000 1000
12 12 15,000
t 15 000$Ð
' t 15 000 t 15,000 t 15 000 t 15 000Îß Î Îß Îß œ%Ðt 15 000ˆ‰ ab
†
y(0) 0 0 1(180 C) C 180 y 180 180e . When the concentration of CO is 0.01%œÊœ Ê œ Êœ
œ%Ðt 15 000
in the room, the amount of CO satisfies y 0.45 ft . When the room contains this amount we
y
4500 100
.01
œÊœ
$
have 0.45 180 180e e t 15,000 ln 37.55 min.œ Ê œ Êœ ¸
œÊ "t15000 179.55 179.55
180 180 Î ßt 15 000 ˆ‰
29. Steady State and we want i 1 e 1 e eœœÊœÊœÊœ
VVVV
RRRR
"" " "
## # #
ˆ‰ ˆ‰ ab
œÞ œÞ œÞRt L Rt L Rt L
ln ln t t ln 2 secÊœÊ œÊœ
""
##
Rt L L
LR R
30. (a) i 0 di dt ln i C i e e Ce ; i(0) I I C
di R R Rt dt L i L L
œÊ œ Ê œ Êœ œ œÊœ
"
" CRtLœœ% œÞRt L
iIe ampʜ
œÞRt L
(b) I I e e ln ln 2 t ln 2 sec
"""
###
œÊœÊœœÊœ
œÞ œÞRt L Rt L Rt L
LR
(c) t i I e I e ampœÊœ œ
L
R Ð Î ÑÐ Î ÑRt L L R œt
31. (a) t i 1 e 1 e 0.9502 amp, or about 95% of the steady state valueœÊœ œ¸
3L V V V
RR R R
abab
Ð ÑÐ Î ÑR/L 3L R œ3
(b) t i 1 e 1 e 0.8647 amp, or about 86% of the steady state valueœÊœ œ¸
2L V V V
RR R R
abab
Ð ÑÐ Î ÑR/L 2L R œ2
32. (a) i P(t) , Q(t) P(t) dt dt v(t) e
di R V R V R Rt
dt L L L L L L
œÊ œ œÊ œ œÊ œ '' Rt L%
i e dt e C CeÊœ œ œ
""
eLeRLR
VLVV
Rt L Rt L%%
'Rt L Rt L R L t%% ˆ‰ œ ‘ˆ‰
(b) i(0) 0 C 0 C i eœÊ œÊ œ Êœ
VVVV
RRRR œ%Rt L
(c) i 0 i 0 i is a solution of Eq. (11); i CeœÊœÊœ œÊœ œ
Vdi diR RVV V
Rdt dtL LRL R
ˆ‰ˆ‰
RLt
33. y y y ; we have n 2, so let u y y . Then y u and 1y y
wÐÐÐÐ
œ œ œ œ œ œ Ê œ
2121122 du du
dx dx dx dx
dy dy
u u u u 1. With e e as the integrating factor, we haveÊ œ Ê œ œ
ÐÐÐ212 dxxdu du
dx dx '

597. Section 9.3 Euler's Method 593
e u e u e . Integrating, we get e u e C u 1 y
xxx xxdu d C1 1 e
dx dx ey eC 1ˆ‰ abœ œ œÊœœÊœ œ
xCx
e
x
x
œ œ
34. y y xy ; we have n 2, so let u y . Then y u and y y u .
wÐÐÐÐ
œ œ œ œ œ Ê œ œ
211222 du du du
dx dx dx dx dx
dy dy
Substituting: u u xu u x. Using e e as an integrating factor:œÊœ œ
ÐÐÐ212 dxxdu du
dx dx '
e u eu x e eu e 1 x C u y u
xxxxx 1du d e
dx dx e e xe C
e1 x Cˆ‰ ab a b œ œ Ê œ Êœ Êœ œ
x
xxx
x
ab Ð

35. xy y y y y y . Let u y y y u and y u .
wÐw Ð ÐÐ ÐÐ
œ Ê œ œ œ Êœ œ
2 2 1 2 3 1/3 2 2/311
xx ˆ‰ ˆ‰
ab
3y y y u . Thus we have
du 1 du 1 du dx dx dx 3 dx 3 dx
222/3dy dy
œÊœœ œ
wÐÐ ˆ‰ˆ ‰ ˆ‰ˆ ‰ˆ ‰ab
u u u u 1. The integrating factor, v x , isˆ‰ˆ ‰ˆ ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ab
1du 1 1 du 3 3
3dx x x dx x x
2/3 1/3 2/3ÐÐ
œ Êœ
e e e x . Thus x u x 3x x u x C u 1 y
'
3
x
3
3
dx 3ln x ln x 3 3 3 2 3 3 3 d3 C
dx x x
œœœ œ œÊœÊœœ ab ˆ‰
y1Êœ ˆ‰
C
x
1/3
3
36. x y 2xy y y y y . P x , Q x , n 3. Let u y y .
23 3 132 21 2 1
xx x x
ww ÐÐ
œÊ œ œ œ œ œ œˆ‰ ˆ ‰ ˆ‰ ˆ ‰ab ab
22
Substituting gives 2 u 2 u . Let the integrating factor, v x , be
du 2 1 du 4 2
dx x x dx x x
œ Ê œab abˆ‰ ˆ ‰ ˆ ‰
22
ÐÐ
e e x . Thus x u 2x x u x C u Cx y
'ab
$ $4
x
4
dx ln x 4 4 6 4 5 4 2 d22
dx 5 5x
œœ œÊœÊœœ
ÐÐÐÐÐ Ð
ab
yCxÊœ ˆ‰
2
5x
4
1/2Ð
9.3 EULER'S METHOD
1. y y 1 dx 1 1 (.5) 0.25,"!
Ð
#œ œ œŠ‹ ˆ‰
y
x
1

y y 1 dx 0.25 1 (.5) 0.3,#"
Ðœ œ œŠ‹ ˆ‰
y
x2.5
0.25Î
Î
y y 1 dx 0.3 1 (.5) 0.75;$#œ œ œŠ‹ ˆ‰
y
x3
0.3œ
œ
y 1 P(x) , Q(x) 1 P(x) dx dx ln x ln x, x 0 v(x) e x
dy
dx x x x
œÊœ œÊ œ œœ Êœœˆ‰ kk
"" " ' ' ln x
y x 1 dx C ; x 2, y 1 1 1 C 4 yÊœ œ œ œÊœ Ê œÊœ
""
# #xx 2 x
xCx4'† Š‹
œ
y(3.5) 0.6071Ê œœ ¸
3.5 4 4.25
3.5 7#
2. y y x (1 y ) dx 0 1(1 0)(.2) .2,"!! !œ œ œ
y y x (1 y ) dx .2 1.2(1 .2)(.2) .392,
#"" "œ œ œ
y y x (1 y ) dx .392 1.4(1 .392)(.2) .5622;
$## #œ œ œ
x dx ln 1 y C; x 1, y 0 ln 1 C C ln 1 y
dy
1y
x x
Ð# ####
"" "
œÊœœœÊœÊœÊœkk kk
œ œ
y 1 e y(1.6) .5416Êœ Ê ¸
ab1x 2œÞ
œ
3. y y (2x y 2y ) dx 3 [2(0)(3) 2(3)](.2) 4.2,"! !! !œ œ œ
y y (2x y 2y ) dx 4.2 [2(.2)(4.2) 2(4.2)](.2) 6.216,
#" "" "œ œ œ
y y (2x y 2y ) dx 6.216 [2(.4)(6.216) 2(6.216)](.2) 9.6969;
$# ## #œ œ œ
2y(x 1) 2(x 1) dx ln y (x 1) C; x 0, y 3 ln 3 1 C C ln 3 1
dy dy
dx y
œÊœ Ê œœœÊœÊœ kk
#
ln y (x 1) ln 3 1 y e e e 3e y(.6) 14.2765ʜʜ œ œ Ê ¸
# ÐÑ ÐÑx1 ln31 ln3x 2x xx2
œœ
4. y y y (1 2x ) dx 1 1 [1 2( 1)](.5) .5,"! !
##
!œ œ œ
y y y (1 2x ) dx .5 (.5) [1 2( .5)](.5) .5,
#" "
##
"œ œ œ
y y y (1 2x ) dx .5 (.5) [1 2(0)](.5) .625;
$# #
##
#œ œ œ
(1 2x) dx x x C; x 1, y 1 1 1 ( 1) C C 1 1 x x
dy
yy y
œœ Ê œ œ œ Ê œ Ê œ Ê œ
" "###
y y(.5) 4Êœ Ê œ œ
""
ÐÐ Ð Ð1xx 1.5(.5)
œœ

598. 594 Chapter 9 Further Applications of Integration
5. y y 2x e dx 2 2(0)(.1) 2,"! !œ œ œ
x
œ
Ñ
y y 2x e dx 2 2(.1) e (.1) 2.0202,#" "œ œ œ
x1
œœ
Î
Þ
y y 2x e dx 2.0202 2(.2) e (.1) 2.0618,$# #œ œ œ
x 2
œ
œ
œ
Þ
dy 2xe dx y e C; y(0) 2 2 1 C C 1 y e 1 y(.3) e 1 2.0942œÊœœÊœÊœÊœÊœ¸
xx x3
œœ œœ
Þ
6. y y y e 2 dx 2 2 e 2 (.5) 2.5,"! !
!œ œ œabab
xÑ
y y y e 2 dx 2.5 2.5 e 2 (.5) 3.5744,#" "
Þœ œ œab a b
xÎ 5
y y y e 2 dx 3.5744 3.5744 e 2 (.5) 5.7207;$# #
"œ œ œab a b
xœ
y e 2 P(x) 1, Q(x) e 2 P(x) dx x v(x) e y e e 2 dx
dy
dx e
xx x
œ Ê œ œ Ê œÊ œ Ê œ '' Ð "
$x
œxx
ab
e x 2e C ; y(0) 2 2 2 C C 0 y xe 2 y(1.5) 1.5e 2 8.7225œ œÊœÊœÊœÊ œ ¸
x x
ab
œ Ñx 15
7. y 1 1(.2) 1.2,"œ œ
y 1.2 (1.2)(.2) 1.44,
#œ œ
y 1.44 (1.44)(.2) 1.728,
$œ œ
y 1.728 (1.728)(.2) 2.0736,
%œ œ
y 2.0736 (2.0736)(.2) 2.48832;
&œ œ
dx ln y x C y Ce ; y(0) 1 1 Ce C 1 y e y(1) e 2.7183
dy
y
xx
œ Ê œ Êœ œÊœ Ê œÊœ Ê œ¸"
!
8. y 2 (.2) 2.4,"œ œˆ‰
2
1
y 2.4 (.2) 2.8,#œ œˆ‰
2.4
1.2
y 2.8 (.2) 3.2,$œ œˆ‰
2.8
1.4
y 3.2 (.2) 3.6,%œ œˆ‰
3.2
1.6
y 3.6 (.2) 4;&œ œˆ‰
3.6
1.8
ln y ln x C y kx; y(1) 2 2 k y 2x y(2) 4
dy
yx
dx
œ Ê œ Êœ œÊœÊœ Ê œ
9. y 1 (.5) .5,"
Ðœ œ’“
(1)
1
œ
È
y .5 (.5) .39794,#
Ðœ œ’“
(.5)
1.5
œ
È
y .39794 (.5) .34195,$
Ðœ œ ’“
( .39794)
2
œ
È
y .34195 (.5) .30497,%
Ðœ œ ’“
( .34195)
2.5
œ
È
y .27812, y .25745, y .24088, y .2272;&'()œ œ œ œ
2 x C; y(1) 1 1 2 C C 1 y y(5) .2880
dy
yy
dx
x 1x 15
œœ Êœ œÊ œ Ê œÊ œ Ê œ ¸
È ÈÈ
" ""
Ð# Ð#
È
10. y 1 1 e 1,"
!"œ œab ˆ‰
3
y 1 1 e 0.68408,#
#Î$"œ œˆ‰ˆ‰
3
y 0.68408 0.68408 e 0.35245,$
%Î$"œ œ ˆ‰ˆ‰
3
y 0.35245 0.35245 e 2.93295,%
'Î$"œ œˆ‰ˆ‰
3
y 2.93295 2.93295 e 8.70790,&
)Î$"œ œˆ‰ˆ‰
3
y 8.7079 8.7079 e 20.95441;'
"!Î$"œ œˆ‰ˆ‰
3
y y e P(x) 1, Q(x) e P(x) dx x v(x) e y e e dx
w ÐÐ "
œ Ê œ œ Ê œ Ê œ Ê œ
2x 2x x x 2x
e ''
$x ab
e e C ; y(0) 1 1 1 C C 2 y e 2e y(2) e 2e 39.8200œ œÊœÊ œÊœ Ê œ ¸
xx 2x x
ab
%#
11. Let z y 2y x 1 dx and y y y x 1 z x 1 dx with x 0, y 3, and dx 0.2.n n1 n1 n1 n n1 n1 n1 n n 0 0œ œ œ œ œÐÐÐ ÐÐÐ ab a b abab
The exact solution is y 3e . Using a programmable calculator or a spreadsheet (I used a spreadsheet) gives theœ
xx 2abœ
values in the following table.

599. Section 9.3 Euler's Method 595
x z y-approx y-exact Error
0 --- 3 3 0
0.2 4.2 4.608 4.658122 0.050122
0.4 6.81984 7.623475 7.835089 0.211614
0.6 11.89262 13.56369 14.27646 0.712777
12. Let z y x 1 y dx and y y dx with x 1, y 0, and dx 0.2.nn1n1 n1 nn1 0 0
x1y x1z
2œ œ œ œ œÐÐ Ð Ð
ab Š‹
n1 n1 n nÎÎabab
The exact solution is y 1 e . Using a programmable calculator or a spreadsheet (I used a spreadsheet) gives theœ
ˆ‰1x 2œÞ
2
values in the following table.
x z y-approx y-exact Error
1 --- 0 0 0
1.2 0.2 1.196 0.197481 0.001481
1.4 0.38896 0.378026 0.381217 0.003191
1.6 0.552178 0.536753 0.541594 0.004841
13. 2xe , y 0 2 y y 2x e dx y 2x e 0.1 y 0.2x e
dy
dx
n1nn nn n n
œœÊœœ œ
xxxx
22 2
nn n
ab a b œ
2
On a TI-92 Plus calculator home screen, type the following commands:
2 STO > y: 0 STO > x: y (enter)
y 0.2*x*e^(x^2) STO > y: x 0.1 STO > x: y (enter, 10 times)ÑÑ
The last value displayed gives y 1 3.45835
Eulerab¸
The exact solution: dy 2xe dx y e C; y 0 2 e C C 1 y 1 eœ Êœ œœ Ê œÊœ
xx x
22 2
ab
0
y 1 1 e 3.71828Êœ¸ exactab
14. y e 2, y 0 2 y y y e 2 dx y 0.5 y e 2
dy
dx
xxx
n1 n n n n
œ œÊ œ œ ab ab ab œ
nn
On a TI-92 Plus calculator home screen, type the following commands:
2 STO > y: 0 STO > x: y (enter)
y 0.5*(y e 2) STO > y: x 0.5 STO > x: y (enter, 4 times)ÑÑÐ Ñ
x
The last value displayed gives y 2 9.82187Eulerab¸
The exact solution: y e 2 P x 1, Q x e 2 P(x) dx x v x e
dy
dx
xx x
œ Ê œ œ Ê œÊ œab ab ab ' Ð
y e e 2 dx e x 2e C ;y 0 2 2 2 C C 0Êœ œ œÊœÊ œ
1
e
xx x x
Îx
'ÐÐ
ab a bab
y xe 2 y 2 2e 2 16.7781Êœ Ê œ ¸
x2
exact
ab
15. , y 0 y 0 y y dx y 0.1 y 0.
dy
dx y y y y
x
n1 n n n
xx x
œ ß œ"Ê œ œ œ"
ÈÈÈÈ
ab a b œ
nn n
nn n
On a TI-92 Plus calculator home screen, type the following commands:
1 STO > y: 0 STO > x: y (enter)
y 0.1*( x /y) STO > y: x 0.1 STO > x: y (enter, 10 times)ÑÑ È
The last value displayed gives y 1 1.5000
Eulerab¸ The exact solution: dy dx y dy x dx x C; 0 C CœÊœ Êœ œœœ Êœ
È ababx
y2322232
y 2112 13/2 y0 3/2
È ab
2 2
2
x y x 1 y 1 1 1.5275Êœ Êœ Ê "œ ¸
y
23 2 3 3
21 4 43/2 3/2
exact
3/2
2
É ab abÉ
16. 1 y , y 0 0 y y 1 y dx y 1 y 0.1 y 0.1 1 y
dy
dx
2222
n1 n n n nn n
œ œÊ œ œ œ ab ab abab ab œ
On a TI-92 Plus calculator home screen, type the following commands:
0 STO > y: 0 STO > x: y (enter)
y 0.1*(1 y ) STO > y: x 0.1 STO > x: y (enter, 10 times)ÑÑ Ñ
2
The last value displayed gives y 1 1.3964Eulerab¸
The exact solution: dy 1 y dx dx tan y x C; tan y 0 tan 0 0 0 C C 0œ Ê œ Ê œ œ œœÊœab ab
2111 dy
1yœ
ÐÐÐ
2

600. 596 Chapter 9 Further Applications of Integration
tan y x y tan x y tan 1 1.5574ʜʜÊ"œ¸
1
exact
ab
17. (a) 2y x 1 2 x 1 dx y dy 2x 2 dx y x 2x C
dy dy
dx y
222
œÊœÊ œÊœab ab a b
2
''"
Initial value: y 2 2 2 2 2 C C 2ab abœ Ê œ Ê œ
"
#
2
Solution: y x 2x 2 or yœ œ
"

2 1
x2x2
2
y 3 0.2abœ œ œ
11
3232 5
2
ab
(b) To find the approximation, set y 2y x 1 and use EULERT with initial values x 2 and y and step size1
2œ œ œab
"
#
0.2 for 5 Points. This gives y 3 0.1851; error 0.0149.ab¸ ¸
(c) Use step size 0.1 for 10 points. This gives y 3 0.1929; error 0.0071.ab¸ ¸
(d) Use step size 0.05 for 20 points. This gives y 3 0.1965; error 0.0035.ab¸ ¸
18. (a) y 1 dx ln y 1 x C y 1 e y 1 e e y Ae 1
dy dy
dx y 1
xC Cx x
œÊ œ Ê œÊœ Êœ„ Êœ '

œ
' kk kk
Initial value: y 0 3 3 Ae 1 A 2abœÊœ Ê œ
0
Solution: y 2e 1œ
x
y 1 2e 1 6.4366abœ¸
(b) To find the approximation, set y y 1 and use a graphing calculator or CAS with initial values x 0 and y 3
1œ œ œ
and step size 0.2 for 5 Points. This gives y 1 5.9766; error 0.4599ab¸¸
(c) Use step size 0.1 for 10 points. This gives y 1 6.1875; error 0.2491.ab¸¸
(d) Use step size 0.05 for 20 points. This gives y 1 6.3066; error 0.1300.ab¸¸
19. The exact solution is y , so y 3 0.2. To find the approximation, let z y 2y x 1 dx andœœ œ



1
x2x2
nn1 n1
2
n1
2 ab a b
y y y x 1 z x 1 dx with initial values x 2 and y . Use a spreadsheet, graphingnn1 n1 0 0
222
n1 nnœ œ œ
"
#ababab
calculator, or CAS as indicated in parts (a) through (d).
(a) Use dx 0.2 with 5 steps to obtain y 3 0.2024 error 0.0024.œ¸Ê¸ ab
(b) Use dx 0.1 with 10 steps to obtain y 3 0.2005 error 0.0005.œ¸Ê¸ ab
(c) Use dx 0.05 with 20 steps to obtain y 3 0.2001 error 0.0001.œ¸Ê¸ ab
(d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step
size.
20. The exact solution is y 2e 1, so y 1 2e 1 6.4366. To find the approximation, let z y y 1 dxœ œ¸ œ
x
nn1 n1
ab a b
and y y dx with initial value y 3. Use a spreadsheet, graphing calculator, or CAS as indicated innn1 n
yz2
2œ œ

ˆ‰
n1 nÞ
parts (a) through (d).
(a) Use dx 0.2 with 5 steps to obtain y 1 6.4054 error 0.0311.œ¸Ê¸ ab
(b) Use dx 0.1 with 10 steps to obtain y 1 6.4282 error 0.0084œ¸Ê¸ ab
(c) Use dx 0.05 with 20 steps to obtain y 1 6.4344 error 0.0022œ¸Ê¸ ab
(d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step
size.
13-16. Example CAS commands:
:Maple
ode := diff( y(x), x ) = 2*x*exp(x^2);ic := y(0)=2;
xstar := 1;
dx := 0.1;
approx := dsolve( {ode,ic}, y(x), numeric, method=classical[foreuler], stepsize=dx ):
approx(xstar);
exact := dsolve( {ode,ic}, y(x) );
eval( exact, x=xstar );

601. Section 9.3 Euler's Method 597
evalf( % );
17. Example CAS commands:
:Maple
ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2;
xstar := 3;
exact := dsolve( {ode,ic}, y(x) ); # (a)
eval( exact, x=xstar );
evalf( % );
approx1 := dsolve( {ode,ic}, y(x), # (b)
numeric, method=classical[foreuler], stepsize=0.2 ):
approx1(xstar);
approx2 := dsolve( {ode,ic}, y(x), # (c)
numeric, method=classical[foreuler], stepsize=0.1 ):
approx2(xstar);
approx3 := dsolve( {ode,ic}, y(x), # (d)
numeric, method=classical[foreuler], stepsize=0.05 ):
approx3(xstar);
19. Example CAS commands:
:Maple
ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2;
xstar := 3;
approx1 := dsolve( {ode,ic}, y(x), # (a)
numeric, method=classical[heunform], stepsize=0.2 ):
approx1(xstar);
approx2 := dsolve( {ode,ic}, y(x), # (b)
numeric, method=classical[heunform], stepsize=0.1 ):
approx2(xstar);
approx3 := dsolve( {ode,ic}, y(x), # (c)
numeric, method=classical[heunform], stepsize=0.05 ):
approx3(xstar);
21. Example CAS commands:
:Maple
ode := diff( y(x), x ) = x + y(x);ic := y(0)=-7/10;
x0 := -4;x1 := 4;y0 := -4; y1 := 4;
b := 1;
P1 := DEplot( ode, y(x), x=x0..x1, y=y0..y1, arrows=thin, title="#21(a) (Section 9.3)" ):
P1;
Ygen := unapply( rhs(dsolve( ode, y(x) )), x,_C1 ); # (b)
P2 := seq( plot( Ygen(x,c), x=x0..x1, y=y0..y1, color=blue ), c=-2..2 ): # (c)
display( [P1,P2], title="#21(c) (Section 9.3)" );
CC := solve( Ygen(0,C)=rhs(ic), C ); # (d)
Ypart := Ygen(x,CC);
P3 := plot( Ypart, x=0..b, title="#21(d) (Section 9.3)" ):
P3;
euler4 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/4 ): # (e)
P4 := odeplot( euler4, [x,y(x)], x=0..b, numpoints=4, color=blue ):

602. 598 Chapter 9 Further Applications of Integration
display( [P3,P4], title="#21(e) (Section 9.3)" );
euler8 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/8 ): # (f)
P5 := odeplot( euler8, [x,y(x)], x=0..b, numpoints=8, color=green ):
euler16 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/16 ):
P6 := odeplot( euler16, [x,y(x)], x=0..b, numpoints=16, color=pink ):
euler32 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/32 ):
P7 := odeplot( euler32, [x,y(x)], x=0..b, numpoints=32, color=cyan ):
display( [P3,P4,P5,P6,P7], title="#21(f) (Section 9.3)" );
< < N | h | `percent error` >, # (g)
< 4 | (x1-x0)/ 4 | evalf[5]( abs(1-eval(y(x),euler4(b))/eval(Ypart,x=b))*100 ) >,
< 8 | (x1-x0)/ 8 | evalf[5]( abs(1-eval(y(x),euler8(b))/eval(Ypart,x=b))*100 ) >,
< 16 | (x1-x0)/16 | evalf[5]( abs(1-eval(y(x),euler16(b))/eval(Ypart,x=b))*100 ) >,
< 32 | (x1-x0)/32 | evalf[5]( abs(1-eval(y(x),euler32(b))/eval(Ypart,x=b))*100 ) > >;
13-24. Example CAS commands:
: (assigned functions, step sizes, and values for initial conditions may vary)Mathematica
For exercises 13 - 20, find the exact solution as follows. Set up two error lists.
Clear[x, y, f]
f[x_,y_]:= 2 y (x 1)
2

a = 2; b = 1/2;
xstar = 3;
desol=DSolve[{y'[x] == f[x, y[x]], y[a] == b}, y[x], x] //Simplify
actual[x_] = desol[[1, 1, 2]];
{xstar, actual[xstar]}
errorlisteuler = { };
errorlisteulerimp = { };
pa = Plot[actual[x], {x, a, xstar}]
Euler's method with error at x*. The command is used with a sequence of commands that are repeated n times.Do
a = 2; b = -1/2;
dx = 0.2;
xstar = 3; n = (xstar a) /dx;
solnslist = {{a,b}};
Do[ {new = b + f[a,b] dx, a = a + dx, b = new, AppendTo[solnslist, {a,b}]},{n}]
solnslist
error= actual[xstar] solnslist[[n, 2]]
relativeerror= error / actual[xstar]
AppendTo[errorlisteuler, error]
pe = ListPlot[solnslist, PlotStyle {Hue[.4], PointSize[0.02]}]Ä
Show[pa, pe]
Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the
error as the step size decreases by entering the input command: errorlisteuler
Improved Euler's method. with error at x*
a = 2; b = 1/2;
dx = 0.2;
xstar = 3; n = (xstar a) /dx;
solnslist = {{a,b}};
Do[{new1 = b f[a,b] dx, new2 = b + (f[a, b] f[a+dx, new1])/2 dx, a = a dx, b = new2,
AppendTo[solnslist, {a,b}]},{n}]
solnslist
error= actual[xstar] solnslist[[n, 2]

603. Section 9.4 Graphical Solutions of Autonomous Differential Equations 599
relativeerror= error / actual[xstar]
AppendTo[errorlisteulerimp, error]
peimp = ListPlot[solnslist, PlotStyle {Hue[.8], PointSize[0.02]}]Ä
Show[pa, peimp]
Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the
error as the step size decreases by entering the input command: errorlisteulerimp
You can also type Show[pa, pe, peimp]. This would be appropriate for a fixed value of dx with each method.
You can also make a list of relative errors.
Problems 21 - 24 involve use of code from section 9.1 together with the above code for Euler's method.
9.4 GRAPHICAL SOUTIONS OF AUTONOMOUS DIFFERENTIAL EQUATIONS
1. y y 2 y 3
w
œ abab
(a) y 2 is a stable equilibrium value and y 3 is an unstable equilibrium.œ œ
(b) y 2y 1 y 2 y 2 y y 3
ww w
œ œ a b ab ab ˆ‰
1
2

(c)
2. y y 2 y 2
w
œ abab
(a) y 2 is a stable equilibrium value and y 2 is an unstable equilibrium.œ œ
(b) y 2yy 2y 2yy 2
ww w
œœ abab

(c)

604. 600 Chapter 9 Further Applications of Integration
3. y y y y 1 y y 1
w
œœ
3
abab
(a) y 1 and y 1 is an unstable equilibrium and y 0 is a stable equilibrium value.œ œ œ
(b) y 3y 1 y 3 y 1 y y y y 1
ww w
œœ a b ab ab Š‹Š‹
2 11
33
ÈÈ

(c)
4. y y y 2
w
œab
(a) y 0 is a stable equilibrium value and y 2 is an unstable equilibrium.œœ
(b) y 2y 2 y 2y y 1 y 2
ww w
œ œ a b abab

(c)
5. y y, y 0
w
œÈ
(a) There are no equilibrium values.
(b) y y y
ww w "
#
œœ œ
11
2y 2yÈÈ
È

605. Section 9.4 Graphical Solutions of Autonomous Differential Equations 601
(c)
6. y y y, y 0
w
œ È
(a) y 1 is an unstable equilibrium.œ
(b) y 1 y 1 y y y y 1
ww w "
#
œ œ œ Š‹Š‹ ˆ‰ˆ‰ˆ‰ÈÈ È
11
2y 2yÈÈ

(c)
7. y y 1 y 2 y 3
w
œababab
(a) y 1 and y 3 is an unstable equilibrium and y 2 is a stable equilibrium value.œœ œ
(b) y 3y 12y11y1y2y3 3y1y y2y y3
ww
œœa babababab ab ab Š‹Š‹
2 63 63
33
ÈÈ

(c)

606. 602 Chapter 9 Further Applications of Integration
8. y y y y y 1
w
œœ
32 2
ab
(a) y 0 and y 1 is an unstable equilibrium.œœ
(b) y 3y 2y y y y 3y 2 y 1
ww
œ œ abababab
2323

(c)
9. 1 2P has a stable equilibrium at P . 2 2 1 2P
dP d P dP
dt dt dt
œ œ œ œ
"
#
2
2
ab


10. P 1 2P has an unstable equilibrium at P 0 and a stable equilibrium at P .
dP
dt
œ œ œab
"
#
14P P14P12P
dP dP
dt dt
2
2
œ œ ababab

607. Section 9.4 Graphical Solutions of Autonomous Differential Equations 603

11. 2P P 3 has a stable equilibrium at P 0 and an unstable equilibrium at P 3.
dP
dt
œ œ œab
2 2P 3 4P 2P 3 P 3
dP dP
dt dt
2
2
œœ ab abab

12. 3P 1 P P has a stable equilibria at P 0 and P 1 an unstable equilibrium at P .
dP
dt
œ œ œ œab ˆ‰
" "
# #
6P 6P+1 P P P P P 1
dP 3 dP 3
dt dt 6 6
2 33 33
2
2
œ œ
## #
"
ab ab Š‹Š‹ ˆ‰
ÈÈ

608. 604 Chapter 9 Further Applications of Integration
13.
Before the catastrophe, the population exhibits logistic growth and P t M , the stable equilibrium. After theabÄ
0
catastrophe, the population declines logistically and P t M , the new stable equilibrium.abÄ 1
14. rP M P P m , r, M, m 0
dP
dt
œ abab

The model has 3 equilibrium oints. The rest oint P 0, P M are asymptotically stable while P m is unstable. Forœœ œ
initial populations greater than m, the model predicts P approaches M for large t. For initial populations less than m, the
model predicts extinction. Points of inflection occur at P a and P b where a andMm M mMmœœ œ k‘ ‘ k
"
3
22
‘ È
b. Mm M mMmœ kk ‘ k
"
3
22
‘ È
(a) The model is reasonable in the sense that if P m, then P 0 as t ; if m P M, then P M as t ; if¹Èȸ¹¹Èȸ
P M, then P M as t .ÉÈȸ
(b) It is different if the population falls below m, for then P 0 as t (extinction). If is probably a more realisticÄÄ_
model for that reason because we know some populations have become extinct after the population level became too
low.
(c) For P M we see that rP M P P m is negative. Thus the curve is everywhere decreasing. Moreover,œ
dP
dt
abab
P M is a solution to the differential equation. Since the equation satisfies the existence and uniqueness conditions,´
solution trajectories cannot cross. Thus, P M as t .ÄÄ_
(d) See the initial discussion above.
(e) See the initial discussion above.
15. g v , g, k, m 0 and v t 0
dv k
dt m
2
œ   ab
Equilibrium: g v 0 v
dv k
dt m k
2 mg
œ œÊœ É
Concavity: 2 v 2 v g v
dv k dv k k
dt m dt m m
2
2
2
œ œ ˆ‰ ˆ‰ˆ ‰
(a)
(b)
(c) v 178.9 122 mphterminal
160 ft
0.005 sœœ œÉ

609. Section 9.4 Graphical Solutions of Autonomous Differential Equations 605
16. F F Fœ pr
ma mg k vœ È
g v, v 0 v
dv k
dt m
0
œ œÈ ab
Thus, 0 implies v , the terminal velocity. If v , the object will fall faster and faster, approaching the
dv
dt k k
mg mg22
0
œœ ˆ‰ ˆ‰
terminal velocity; if v , the object will slow down to the terminal velocity.0
mg
k
2Ɉ‰
17. F F Fœ
pr
ma 50 5 vœ kk
50 5 v
dv 1
dt m
œab kk
The maximum velocity occurs when 0 or v 10 .
dv ft
dt sec
œœ
18. (a) The model seems reasonable because the rate of spread of a piece of information, an innovation, or a cultural fad is
proportional to the product of the number of individuals who have it (X) and those who do not (N X). When X is‘
small, there are only a few individuals to spread the item so the rate of spread is slow. On the other hand, when
(N X) is small the rate of spread will be slow because there are only a few indiciduals who can receive it during the‘
interval of time. The rate of spread will be fastest when both X and (N X) are large because then there are a lot of‘
individuals to spread the item and a lot of individuals to receive it.
(b) There is a stable equilibrium at X N and an unstable equilibrium at X 0.œœ
k N X kX k X N X N 2X inflection points at X 0, X , and X N.
d X dX dX N
dt dt dt 2
2
2
2
œœÊ œœ œab aba b

(c)
19. L Ri V i i , V, L, R 0
di di V R R V
dt dt L L L R
œÊœœ ˆ‰
Equilibrium: i 0 i
di R V V
dt L R R
œœÊœˆ‰
Concavity: i
di R di R V
dt L dt L R
2
2
2
œ œ ˆ‰ ˆ‰ˆ ‰
Phase Line:

610. 606 Chapter 9 Further Applications of Integration
If the switch is closed at t 0, then i 0 0, and the graph of the solution looks like this:œœ ab

As t , it i . (In the steady state condition, the self-inductance acts like a simple wire connector and, asÄ_ Ä œ steady state
V
R
a result, the current throught the resistor can be calculated using the familiar version of Ohm's Law.)
20. (a) Free body diagram of he pearl:
(b) Use Newton's Second Law, summing forces in the direction of the acceleration:
mg Pg kv ma g v.œ Êœ
dv m P k
dt m m
ˆ‰
œ
(c) Equilibrium: v 0
dv k
dt m k
mPg
œœŠ‹
abœ
v ʜ terminal
mPg
k
abœ
Concavity: v
dv kdv k
dt m dt m k
2mPg
2
2
œ œ ˆ‰Š‹
abœ

(d)
(e) The terminal velocity of the pearl is .
abmPg
k
œ
9.5 APPLICATIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS
1. Note that the total mass is 66 7 73 kg, therefore, v v e v 9eœ œ Êœ
0
kmt 3.9t73œÄ œ Äab
(a) s t 9e dt e Cabœœ 'œÄ œÄ3.9t 73 3.9t 732190
13
Since s 0 0 we have C and s t 1 e 168.5ab ab ˆ‰œœ œœ¸
2190 2190 2190
13 13 13
tt
3.9t 73
lim lim
Ä_ Ä_
π
The cyclist will coast about 168.5 meters.
(b) 1 9e ln 9 t 41.13 secœÊœÊœ¸
π3.9t 73 3.9t 73 ln 9
73 3.9
It will take about 41.13 seconds.

611. Section 9.5 Applications of First-Order Differential Equations 607
2. v v e v 9e v 9eœÊœ Êœ0
k m t 59,000 51,000,000 t 59t 51,000œÄ œ Ä œ Äab a b
(a) s t 9e dt e Cabœœ 'œÄ œÄ59t 51,000 59t 51,000459,0000
59
Since s 0 0 we have C and s t 1 e 7780 mab ab ˆ‰œœ œ œ¸
459,0000 459,0000 459,0000
59 59 59
tt
59t 51,000
lim lim
Ä_ Ä_
π
The ship will coast about 7780 m, or 7.78 km.
(b) 1 9e ln 9 t 1899.3 secœÊœÊœ¸
π59t 51,000 59t
51,000 59
51,000 ln 9
It will take about 31.65 minutes.
3. The total distance traveled 4.91 k 22.36. Therefore, the distance traveled is given by theœÊ œ Êœ
vm
kk
2.75 39.920 aba b
function s t 4.91 1 e . The graph shows s t and the data points.ab abˆ‰œ
œab22.36/39.92 t

4. coasting distance 1.32 k
vm
kk33
0.80 49.90 9980
œÊœÊœ
aba b
We know that 1.32 and .
vm
k m 33 49.9 33
k 998 200
œœœ
ab
Using Equation 3, we have: s t 1 e 1.32 1 e 1.32 1 eab a bˆ‰ˆ‰œ œ ¸
vm
k
k/m t 20t/33 0.606t0 œœœab
5. (a) 0.0015P 150 P P 150 P P M P
dP 0.255 k
dt 150 M
œœœab abab
Thus, k 0.255 and M 150, and Pœœœœ
M 150
1Ae 1Ae‘ ‘
œ œkt 0.255t
Initial condition: P 0 6 6 1 A 25 A 24abœÊœ Ê œ Ê œ
150
1Ae‘
0
Formula: P œ
150
1 24e‘
œ0.255t
(b) 100 1 24e 24e e 0.255t ln 48œÊœÊœÊœÊœ
150 3
1 24e 2 48
0.255t 0.255t 0.255t
‘#
œœœ ""
œ0.255t
t 17.21 weeksÊœ ¸
ln 48
0.255
125 1 24e 24e e 0.255t ln 120œÊœÊœÊœÊœ
150 6
1 24e 5 5 120
0.255t 0.255t 0.255t
‘
œœœ ""
œ0.255t
t 21.28Êœ ¸
ln 120
0.255
It will take about 17.21 weeks to reach 100 guppies, and about 21.28 weeks to reach 125 guppies.
6. (a) 0.0004P 250 P P 150 P P M P
dP 0.1 k
dt 250 M
œœœababab
Thus, k 0.1 and M 250, and Pœœ œœ
M 250
1Ae 1Ae‘ ‘
œ œkt 0.1t
Initial condition: P 0 28, where t 0 represents the year 1970abœœ
28 28 1 A 250 A 1 7.9286œÊœÊœœ¸
250 250 111
1Ae 28 14‘
0 ab
Formula: P or approximately Pœœ
250 250
1e 1 7.9286e‘ ‘
111 0.1t
14
0.1tœ œ
(b) The population P t will round to 250 when P t 249.5 249.5 249.5 1 e 250ab ab ˆ‰ Êœ Ê œ
250 111
1e 14
0.1t
‘
œ
111
14
0.1tœ
0.5 e 0.1t ln t 10 ln 55,389 ln 14 82.8.ʜʜʜʜ¸
ab ˆ‰249.5 111e
14 55,389 55,389
0.1t 14 14
œ0.1t
œ
ab
It will take about 83 years.
7. (a) Using the general solution form Example 2, part (c),
0.08875 10 8 10 y y y t
dy
dt 1 Ae 1 Ae
77 M810810
1Ae
œ‚‚Êœœ œababab
œ
‘‘
‚‚
‘
œœ œ‘È‘ÉÉk¹ ÉrMt 0.71t
77
t

612. 608 Chapter 9 Further Applications of Integration
Apply the initial condition:
y 0 1.6 10 1 4 y 1 2.69671 10 kg.ab abœ‚ œ ÊœÊ œ ¸ ‚
7 7810 8 810
1 A 1.6 1 4e
‚‚
‘‘
77
0.71
œ
(b) y t 4 10 4e 1 t 1.95253 years.abœ‚ œ Ê œÊœ ¸
7 0.71t810
14e 0.71
ln‚
‘
œ
7
0.71t
1
4
œ
ˆ‰
8. (a) If a part of the population leaves or is removed from the environment (e.g., a preserve or a region) each year, then c
would represent the rate of reduction of the population due to this removal and/or migration. When grizzly bears
become a nuisance (e.g., feeding on livestock) or threaten human safety, they are often relocated to other areas or even
eliminated, but only after relocation efforts fail. In addition, bears are killed, sometimes accidently and sometimes
maliciously. For an environment that has a capacity of about 100 bears, a realistic value for c would probably be
between 0 and 4.
(b)
Equilibrium solutions: 0 0.001 100 P P 1 P 100P 1000 0 P 11.27 (unstable) and
dP
dt
2
eq
œœ ʜʸab
P 88.73 (stable)
eq¸
(c)
For 0 P 0 11, the bear population will eventually disappear, for 12 P 0 88, the population will grow toŸ ŸŸab ab
about 89, the population will remain at about 89, and for P 0 89, the population will decrease to about 89 bears.abÄ
9. (a) 1 y dy 1 y dt dt ln 1 y t C e e 1 y e e
dy dy
dt 1 y
1
ln 1 y t C t C
œÊ œ Ê œ Ê œ Ê œ Ê œab kk kk
‘
‘‘kk
11
1 y C e y Ce 1, where C e and C C . Apply the initial condition: y 0 1 Ce 1œ„ Êœ œ œ„ œœ 222
tt C 0
1
ab
C2 y2e 1.ʜʜ
t
(b) 0.5 400 y y dy 0.5 400 y y dt 0.5 dt. Using the partial fraction decomposition in
dy dy
dt 400 y y
œÊœÊ œab ab
abœ
Example 2, part (c), we obtain dy 0.5 dt dy 200 dt
11 1 1 1
400 y 400 y y 400 y
Š‹ Š‹œÊœ
œœ
dy 200 dt ln y ln y 400 200t C ln 200t CÊ œ ÊœÊ œ''Š‹ ¹¹ kk k k
11
y 400 y y 400
11
y
œ œ
e e e e C e (where C e ) C eÊœœÊœ œÊœ„
ln
200t C 200t C 200t C 200tyy
y 400 y 400
22 2
¹¹
y
y 400 11 1
œ ‘
œœ
¹¹
Ce (where C C ) y Ce y 400 Ce 1 Ce y 400 CeÊœ œ„Êœ Ê œ
y
y 400
200t 200t 200t 200t 200t
2
œ
ab
y y , where A . Apply the initial condition:Êœ Êœ œ œ
400 Ce 400 400 1
Ce 1 1 Ae C 1e
200t
200t 1 200t
C
200t
œ
œ œ y0 2 A 199 ytab abœœ Ê œ Ê œ
400 400
1 Ae 1 199e‘‘
0 200t œ

613. Section 9.5 Applications of First-Order Differential Equations 609
10. r M P P dP r M P P dt r dt. Using the partial fraction decomposition in Example 6, part (c),
dP dP
dt M P P
œÊœ Ê œab ab
abœ
we obtain dP r dt dP rM dt dP rM dt
11 1 1 1 1 1
MP M P P M P P P M
ˆ‰ ˆ‰ ˆ‰œÊœÊœ
œœ œ
''
ln P ln P M rM t C ln rM t C e e e eÊœ Ê œ Ê œ œkk k kab ab ¸¸ 11
P
PM
ln rM t C rM t C
œ
‘¸¸ ab ab
P
PM
11
œ
C e (where C e ) C e Ce (where C C )Ê œ œ Ê œ„ Ê œ œ„¸¸
PPP
P M PM PM
22 2 2
rM t C rM t rM t
œœœ
ab ab ab
1
P Ce P MCe 1 Ce P MCe P PÊœ Ê œ Êœ Êœ
ab ab ab abrM t rM t rM t rM t MCe
Ce 1
M
1e ˆ‰

rM t
rM t 1
C
rM t
œ œ
œ
P , where A .Êœ œ
M1
1Ae Cœ
œrM t
11. (a) kP P dP k dt P kt C P
dP
dt kt C
22
œÊ œ ʜʜ''œœ" œ"
‘
Initial condition: P 0 P P CabœÊœÊœ 00
11
CP
œ
0
Solution: P œ œ
1
kt 1/P 1 kP t
P
œœab
00
0
(b) There is a vertical asymptote at tœ
1
kp
O
12. (a) r M P P m r 1200 P P 100 r 1100 r
dP dP 1 dP 1100 dP
dt dt 1200 P P 100 dt 1200 P P 100 dt
œ Êœ Ê œÊ œabab a ba b
abab ababœœ œœ
1100 r 1100 r dP 1100 r dtʜʜʜ
aba b
abab
P 100 1200 P
1200 P P 100 dt 1200 P P 100 dt 1200 P P 100
dP 1 1 dP 1 1
œœ œœ œœ
ˆ‰ˆ‰
dP 1100 r dt ln 1200 P ln P 100 1100 r t CÊœ Êœ''ˆ‰ abab
11
1200 P P 100
1
œœ
ln 1100 r t C ln 1100 r t C e e Ceʜʜʜ„Êœ¸¸ ¸¸
P 100 P 100 P 100 P 100
1200 P 1200 P 1200 P 1200 P
11
C 1100 r t 1100 r tœœœœ
œœœœ
1
where C e P 100 1200Ce CPe P 1 Ce 1200Ce 100œ„ʜʜ
C 1100 r t 1100 r t 1100 r t 1100 r t1
ab
P P where A .Êœ œ Êœ œ
1200Ce 100 1200 100Ae 1
Ce 1 1 Ae C
1200 e
1e
1100 r t 1100 r t
1100 r t 1 1100 r t
100
C
1100 r t
C
1100 r t
‘‘
‘‘
‘
‘
œ
œ
œ
œ
(b) Apply the initial condition: 300 300 300A 1200 100A A Pœ ʜʜʜ Þ
1200 100A 9 2400 900Ae
1A 2 29e
‘ ‘
‘ ‘
œ
œ1100 r t
1100 r t
(Note that P 1200 as t .)ÄÄ_
(c) r M P P m r r M m r M m
dP 1 dP M m dP dP
dt M P P m dt M P P m dt M P P m dt
Pm MP
œ Ê œÊ œÊ œabab ab ab
abab abab abab
aba b
œœ œœ œœ
œ
rM m dP rM mdtÊ œÊ œˆ‰ ˆ‰ ab ab
11dP 11
MP Pmdt MP Pmœœ œœ ''
ln M P ln P m M m r t C ln M m r t C e eÊœÊ œÊœ„ababab ab ¸¸ 11
Pm Pm
MP MP
CMm rtœœ
œœ
œ
1ab
Ce where C e P m MCe CPeÊœ œ„Êœ
Pm
MP
M m rt C M m rt M m rtœ
œ
œœœab ab ab
1
P1 Ce MCe m P P PÊ œ Êœ Êœ Êœˆ‰
ab abMm rt Mm rt MCe m M mAe
Ce 1 1 Ae
Me
1e
œœ ‘‘
‘‘
‘
‘



Mm rt Mm rt
Mm rt Mm rt
m
C
Mm rt
1
C
Mm rtœ œœ
œ œœ
œœ
œœ
A. œ
1
C
Apply the initial condition P 0 Pabœ 0
PPPAMmAAP000
MmA
1A P m
MP MP m mM P e
Pm MPeœÊœÊœÊœ
‘

œ
0
0
00
Mm rt
00
Mm rt aba b
aba b
œœ
œœ


(Note that P M as t provided P m.)Èȸ É 0
13. y mx m 0 y . So forœÊœÊ œÊœ
yxyy y
xx x
_
œ w
2
orthogonals: y dy x dx C
dy y
dx y 2 2
xx
œ Ê œ Ê œ
2 2
xyCʜ
22
1

614. 610 Chapter 9 Further Applications of Integration
14. y cx c 0 x y 2xyœÊœÊ œÊ œ
22 y x y 2xy
xx
24
2

w
y . So for the orthogonals: ʜ ?
w2y dy
xdx2y
x
2ydy xdx y C y C,ʜʜʜ„
2 xx
22
2 2
É
C0#

15. kx y 1 1 y kx k
22 2 2 1y
x
??? ? ?

2
2
0 2yx y 1 y 2xʜ??
x2yy 1 y2x
x
22
22
ab ˆ‰

w

w
abab
y . So for the orthogonals:Êœ œ
w


ˆ‰ ˆ‰ab1y 2x 1y
2xy xy
22
2
dy x dx ln y C
dy xy y
dx 1 y y 2 2
1y x
œ Ê œ Ê œ


2
2 2 2
ˆ‰

16. 2x y c 4x 2yy 0 y . For
222 4x 2x
2y y
?? ?ʜ?
ww
orthogonals: ln y ln x C
dy y dy
dx 2x y 2x
dx
?ʜ? ? w
"
#
ln y ln x ln C y C xʜ w ʜ
1/2
11
1/2
kk

17. y ce cœÊœÊ œ!
x y
e
„x
ey ye 1
e

xx
x
2
abab
ab
e y ye y y. So for the orthogonals:ʜʜ
w wxx
ydy dx x C
dy y
dx y 2
1
?? ? ? ?
2
y 2xC y 2xCʜw ʜ" w
2
11 È

18. y e ln y kx k 0œÊ œÊœÊ œ
kx ln y
x
xyln y
x
Š‹
1
y
2w

y ln y 0 y . So for the orthogonals:??ʜŠ‹
x
yx
yln yww
y ln y dy x dx
dy
dx y ln y
x
?? ?

yln y y x C??
""
##
222 1
4
ab ˆ‰
y ln y x C??
22 y
2
1
2

615. Section 9.5 Applications of First-Order Differential Equations 611
19. 2x 3y 5 and y x intersect at 1, 1 . Also, 2x 3y 5 4x 6y y 0 y y 1, 1
22 23 22 4x 2
6y 3
w? ? w??w ?ʜ? ? ab ab
ww w
y x 2y y 3x y y 1, 1 . Since y y 1, the curves are orthogonal.
1
23 2
1111 1
3x 3 2 3
2y 2 3 2
?? ? ʜ ? ? _? ?
www ww
2
1
ab ˆ‰ˆ‰
20. (a) x dx y dy 0 C is the general equation???
x
22
y
2 2
of the family with slope y . For the orthogonals:
w
?
x
y
y ln y ln x C or y C x
w
?? ? ? ? w ?
ydy
xyx
dx
1
(where C e is the general equation of the1
Cτ
orthogonals.

(b) x dy 2y dx 0 2y dx x dy?ʜʜ
dy
2y x
dx
ln y ln x C y C x isʜʜwʜ
""
##
Š‹
dy
yx
dx
1
2
the equation for the solution family.
ln y ln x C 0 y
""
##
w
???ʜ
y2y
yx x
1

slope of orthogonals is ʜ
dy
dx 2y
x
2y dy x dx y C is the generalʜʜ
2 x
2
2
equation of the orthogonals.

2 . y 4a 4ax and y 4b 4bx (at intersection) 4a 4ax 4b 4bx a b x a b"œ œÊ œ Êœ
22 22 2 2 22
ab
a b a b a b x x a b. Now, y 4a 4a a b 4a 4a 4ab 4ab y 2 ab.Ê œ Êœ œ œ œ Êœ„ababab ab
È
22 22
Thus the intersections are at a b, 2 ab . So, y 4a 4ax y which are equal to andŠ‹
È
„ œ Êœ
22
1
4a 4a
2y
22 ab
w
Š‹
È
and at the intersections. Also, y 4b 4bx y which are equal to and? ?ʜ
4a a a 4b 4b
22ab 22 ab
bb 2y
22
2
Š‹ Š‹
È È

w
ÈÈ
and at the intersections. y y . Thus the curves are orthogonal.
4b b b
22ab
aa 12
Š‹
È

ww
? _ ?ÉÉ abab

616. 612 Chapter 9 Further Applications of Integration
CHAPTER 9 PRACTICE EXERCISES
. y cos y dx 2tan y x C y tan? ? ?? ?ʜ
dy dy
dx y cos y 2
2 1xC
2
ÈÈ È ˆ‰ˆ‰
ÈÈ
2
!
2. y dy 3 x 1 dx y ln y x 1 C
w

?ʜ??
3yx1 y1 2 3
y1 y
ab ab
2
ab ab
3. yy sec y sec x sec x dx tan x C sin y 2tan x C
w
?ʜʜʜab ab
22 2 2 y dy
sec y 2
sin y
1
ab
ˆ‰
2
2
4. y cos x dy sin x dx 0 y dy dx C y C
2 sin x 1 2
cos x 2 cos x cos x
y
1
ab É?ʜʜʜ
2
2
ab ab ab

5. y xe x 2 e dy x x 2 dx e C e C
w
œÊœÊœ Êœ
yy y y 2x 2 3x 4 2x 2 3x 4
15 15ÈÈ
aba b aba b
3/2 3/2
ylnCylnCÊ œ Ê œ ’“’“
2x 2 3x 4 2x 2 3x 4
15 15
aba b aba b
3/2 3/2
6. y xye e x dx ln y e C
w "
#
?ʜ ʜ
xx xdy
y
22 2
7. sec x dy x cos y dx 0 tan y cos x x sin x C?ʜʜ
2 dy
cos y sec x
xdx
2
8. 2x dx 3 y csc x dy 0 3 y dy dx 2y 2 2 x cos x 4x sin x C
23/22 2x
csc x
?ʜʜÈÈ ab
2
y 2 x cos x 2x sin x Cʜ
3/2 2
1
ab
9. y ye dy y 1 e ln x C
w
?? ??w ? w
edx
xy x
yy
y
ab kk
10. y xe csc y y csc y dy x e dx sin y cos y x 1 e C
w w
?ʜʜ??
xy x x xe e e
ecsc y 2
xy y
y
abab
11. x x 1 dy y dx 0 x x 1 dy y dx ln y ln x 1 ln x Cab ab abab???ʜ ʜ w
dy
yxx1
dx
ab
ln y lnx 1 lnx ln C ln y ln yÊœ Êœ Êœabab Š‹ 1
Cx 1 Cx 1
xx
11ab ab
12. y y 1 x ln x C ln 2ln x ln C C x
w


? ? ?? ? w? ? w ? ?abab Š‹
21 2dy y1 y1
y1 x 2 y1 y1
dx
11
2
lnŠ‹
y1
y1
„
13. 2y y xe y y e .
ww "
#
œ Ê œ
x/2 x/2 x
2
px , vx e e .ab ab? ? ?
"
#
'ˆ‰
"
#
dx x/2
e y e y e e e y e y C y e C
w "
#
x/2 x/2 x/2 x/2 x/2 x/2 x/2 xxd x x x
22dx 2 4 4
? ?? ?? ? wʜ wˆ‰ˆ‰ˆ‰ ˆ ‰ Š‹
22
14. y e sin x y 2y 2e sin x.
y
2
xx

w? ? w ?
w
px 2, vx e e .ab abœœœ
'2dx 2x
e y 2e y 2e e sin x 2e sin x e y 2e sin x e y e sin x cos x C
2x 2x 2x x x 2x x 2x x d
dx
w
? ?? ?ʜ ab a b
y e sin x cos x Ceʜ
x2x
ab
15. xy 2y 1 x y y .
ww
? ? ?
1 211
xxx ˆ‰
2
vx e e e x.abœœœœ
22ln xln x2'
dx
x
2
xy 2xy x 1 xy x 1 xy x C y
222 dx1C
dx 2 x x
w "
#
œÊ œÊ œ Êœab
2
2

617. Chapter 9 Practice Exercises 613
16. xy y 2x ln x y y 2 ln x.
ww
œ Ê œ ˆ‰
1
x
v x e e . y y ln xab ˆ‰ ˆ‰œœœ œÊ
w'
dx
x
ln x11 1 2
xx x x
2
y ln x y C y x Cx ln x ln x
d1 2 1
dx x x x
22
ˆ‰ cd cd_? ?_? wʜ w
17. 1 e dy ye e dx 0 1 e y e y e y y .aba b ab ???ʜ ?
xxx xxx ee
1e 1e
ww
!!

xx
xx „
ab
vx e e e 1.ab?? ?
'
edx
x
1e
x
x
# ln e 1 xab!
e1y e1 y e1 e e1ye C e1yabab abc d ab ˆ‰ ab
xx x xxx ee d
1e 1e dx
x
www ? w? ??w?w w
w
!!

xx
xx „
ab
yÊœ œ
eC eC
e1 1e
„„xx
xx
!!
!!
18. x 4ye 0 x x 4ye . Let v y e e . Then e x xe 4ye xe 4ye
dx d
dy dy
y y dy y y y 2y y 2y
w ?? w? ? ? w ? ? ?
ww
ab a b
'
xe 2y 1 e C x 2y 1 e Ceʜ ʜ
y2y yy
ab ab

19. x 3y dy y dx 0 x dy y dx 3y dy xy 3y dy xy y Cab ab???ʜʜ
2223 d
dx
20. y dx 3x y cos y dx 0 x x y cos y. Let v y e e e y . ?? ? ? ? ? ?ab ab Š‹
w2 3 3ln y ln y 3 3
y
'
3dy
y
3
Then y x 3y x cos y and y x cos y dy sin y C. So x y sin y C
32 3 3w
w? ? ?w ? w ' ab
21. e e dy e dx e e C. We have y 0 2, so e e C C 2e and
dy
dx
xy2 y x2 y x2 2 2 2
?ʜ ʜ ??ʜ
ab ab
ab
e e 2e y ln e 2e
yx22 x22
œ Ê œ
ab ab ˆ‰
22. ln ln y tan x C y e . We have y 0 e e e
dy y ln y dy
dx 1 x y ln y 1 x
dx 1e 22e
?ʜ? ? wʜ ?ʜ
!!

22
tan x C tan 0 C
11 a b ab ab
„„
##
e 2tan0Cln 20Cln 2Cln 2yeʜ????ʜʜ
tan 0 C 1 e
„ #
„1 tan x ln 2
1
ab
ab

23. x 1 2y x y y . Let v x e e e x 1 .ab ab ab ˆ‰?? ? ? ? ? ?
dy
dx x1 x1
2x dx 2lnx 1 lnx 1 2w
!! '
2
x1
2
#
ab ab
So y x 1 x 1 y x 1 xx 1 yx 1 xx 1dx yx 1
w
!!
ab ab ab abab ab ‘abww w ? w? ? w? w? w w
22 2 22xd
x1 x1 dx
2
ab ab '
y x 1 C y x 1 C . We have y 0 1 1 C. So?w?wwʜw ww ?ʜab ab ab Š‹
22xx xx
32 32
32 32
yx1 1?w wwab Š‹
2xx
32
32
24. x 2y x 1 y y x . Let v x e e x . So x y 2xy x x
dy
dx x x
2ln x223 21 dx
w?w?w ?w ? ? ? w ?w
w wˆ‰ ab
'ˆ‰
2
x
2
x y x x x y C y . We have y 1 1 1 C C .? ?? ?ʜ ?ʜʜ
dxxxC 11
dx 4 2 4 x 4 4
23 2
ab ab
42 2
2
""
##
So y?w w?
x 1 x2x1
44x 4x
242
22
„!!
#
25. 3x y x . Let v x e e . So e y 3x e y x e e y x e e y e C.
dy
dx dx 3
2 2 3x dx x x 2 x 2 x x 2 x x x d1
w? ? ? w ? ? ? ʜwab Š‹
'233 3 3 3 33 3
w
We have y 0 1 e 1 e C 1 C C and e y e y eab a bœÊ œ ÊœÊ œ œ Êœ
00 xx x 1141414
3333333
33 33 3

26. xdy y cos x dx 0 xy y cos x 0 y y . Let v x e e x. œÊ œÊ œ œ œ œab ab ˆ‰
ww 1cos x
xx
dx ln x '
1
x
So xy x y cos x xy cos x xy cos x dx xy sin x C. We have y 0 0 1 C
w
w ? ? ? ʜ ʜ w ?? ?wˆ‰ ˆ‰ ˆ‰ ab
1d
xdx 22 '
11
C 1. So xy 1 sin x yʜ ? ʜ
1sin x
x
27. x dy y y dx 0 2ln y 1 ln x C. We have y 1 1 2 ln 1 1 ln 1 Cw ?? ?? w? w ?? w ? wˆ‰ ˆ‰ÈÈ ab Š‹
È
dy
yy
dx
xˆ‰È!
2 ln 2 C ln 2 ln 4. So 2 ln y 1 ln x ln 4 ln 4x ln y 1 ln 4x ln 4x? ?? ? w?w? ? w? ?
2 1/2 ˆ‰ ˆ‰ÈÈ ab ab ab
"
#

618. 614 Chapter 9 Further Applications of Integration
e e y12x y 2x1ʜ??ʜ
ln y 1 ln 4x
2ˆ‰È ab!
1/2
È ÈÈ ˆ‰
28. y dx e e C. We have y 0 1 e e C C .

!
!2xx00dx e e 1 1
dy e 1 e y 3 3 3
dy y 1
?? ?ʜ ?ʜʜ
x2x
2x x 2
3 3
„ ab
ab
So ee y3ee 1y3ee 1
y
33
xx 3 xx xx1 1/3
3
œ Êœ Êœ

ab c d ab
29. xy x 2 y 3x e y y 3x e . Let v x e e . So
ww
? ? ? ? ? ?ab ab ˆ‰
3x 2x dx x2ln xx2 e
xx 'ˆ‰
x2
x
x
2„
y y 3 y 3 y 3x C. We have y 1 0 0 3 1 C C 3
eex2 de e
xxx dxx x
xx x x
22 2 2
w
????? ?ʜʜˆ‰ ˆ ‰ ab ab y 3x3 yxe3x3ʆ œ Êœ
e
x
2x
x
2

ab
30. y dx 3x xy 2 dy 0 0 x 1 x . œÊ œÊœÊ œab Š‹
dx dx 3x 2 dx 3 2
dy y dy y y dy y y
3x xy 2
Py 1 Pydy 3ln y y vy e yeab ab abœÊ œ Ê œ œ
3
y
3ln y y 3 y'
ye x ye 1 x 2ye ye x 2ye dy 2e y 2y 2 C
3y 3y 2y 3y 2y y23
y
w
?ʜ ?Š‹ ab '
y . We have y 2 1 1 C 4e and? ? ? ? ? ? ?
3
2y 2y 2 Ce
x2
2122 Ceˆ‰ ab
2y 1
!!!
ab
„
yʜ
3
2y 2y 2 4e
x
ˆ‰
2y1

#
31. To find the approximate values let y y y cos x 0.1 with x 0, y 0, and 20 steps. Use ann1 n1 n1 0 0?w w ? ? abab
spreadsheet, graphing calculator, or CAS to obtain the values in the following table.
x y
0 0
0.1 0.1000
0.2 0.2095
0.3 0.3285
0.4 0.4568
0.5 0.5946
0.6 0.7418
0.7 0.8986
0.8 1.0649
0.9 1.2411
1.0 1.4273
x y
1.1 1.6241
1.2 1.8319
1.3 2.0513
1.4 2.2832
1.5 2.5285
1.6 2.7884
1.7 3.0643
1.8 3.3579
1.9 3.6709
2.0 4.0057
32. To find the approximate values let z y 2 y 2 x 3 0.1 and
nn1 n1 n1?w wabababa b
y y 0.1 with initial values x 3, y 1, and 20 steps. Use ann1 00
2y 2 x 3 2z2 x 3
2? ??
Š‹ ab
aba bababn1 n1 n n„„
spreadsheet, graphing calculator, or CAS to obtain the values in the following table.
x y
3 1
2.9 0.6680
2.8 0.2599
2.7 0.2294
2.6 0.8011
2.5 1.4509
2.4 2.1687
2.3 2.9374
2.2 3.7333
2.1 4.










5268
2.0 5.2840
x y
1.9 5.9686
1.8 6.5456
1.7 6.9831
1.6 7.2562
1.5 7.3488
1.4 7.2553
1.3 6.9813
1.2 6.5430











1.1 5.9655
1.0 5.2805

619. Chapter 9 Practice Exercises 615
33. To estimate y 3 , let z y 0.05 and y y 0.05 with initial valuesab ab ab Š‹ Š ‹ nn1 nn1
x2y x2y
x1 x1 x1
x2z?w ?w w

#
"
n1 n1 n1 n1
n1 n1 n
nn„„ „„
„„
x 0, y 1, and 60 steps. Use a spreadsheet, graphing calculator, or CAS to obtain y 3 0.9063.00œœ ¸ ab
34. To estimate y 4 , let z y 0.05 with initial values x 1, y 1, and 60 steps. Use aab a b Š‹ nn1 0 0
x2y1
x? ? ?

2
n1 n1
n1
„ „
„
spreadsheet, graphing calculator, or CAS to obtain y 4 4.4974.ab¸
35. Let y y dx with starting values x 0 and y 2, and steps of 0.1 and 0.1. Use a spreadsheet,nn1 0 0
1
eœ œ œ
ˆ‰ ab
xy2
n1 n1
„„
##
programmable calculator, or CAS to generate the following graphs.
(a)
(b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot
handle the calculations for x 1. (This occurs because the analytic solution is y 2 ln 2 e , which has anŸ œ ab
x
asymptote at x ln 2 0.69. Obviously, the Euler approximations are misleading for x 0.7.)œ ¸ Ÿ

36. Let z y dx and y y dx with starting values x 0 and ynn1 nn1 0
xy xy
ex ex ex
xz? ?w w ?
!!
#
" !Š‹ Š ‹ab ab
22
n1 n1 nn1 n1
yy
n1 n1
n1 n1 n
2
n
zn
„„ „„
„„
„„
0œ0,
and steps of 0.1 and 0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs.
(a) (b)
37. x 1 1.2 1.4 1.6 1.8 2.0
y 1 0.8 0.56 0.28 0.04 0.4
x dy x dx y C; x 1 and y 1
dy
dx 2
x
?? ? ʜ w ? ?
2
1 C C y exactÊ œ Ê œ Ê œ
"
#
3x3
222
ab
2
y 2 is the exact value.ʜ?ab
23
22
2
"
#

620. 616 Chapter 9 Further Applications of Integration
38. x 1 1.2 1.4 1.6 1.8 2.0
y 1 0.8 0.6333 0.4904 0.3654 0.2544
dy dx y ln x C; x 1 and y 1
dy
dx x x
11
?? ? ʜ w ? ? kk
1 ln 1 C C 1 y exact ln x 1Ê œ Ê œ Ê œ abkk
y 2 ln 2 1 0.3069 is the exact value.Êœ¸ab

39. x 1 1.2 1.4 1.6 1.8 2.0
y 1 1.2 0.488 1.9046 2.5141 3.4192
xy x dx ln y C
dy dy
dx y 2
x
?ʜ ? ? kk
2
ye e e Ce; x1 and y 1ʜ ? _ ? ? ?
xx x
22 2
22 2
!CC
1
1 C e C e y exact e e? ? ? ? ? _11
1/2 1/2 1/2 ab
x
2
2
e y 2 e 4.4817 is theœ Ê œ ¸
ˆ‰x1/2 3/2
2

ab
exact value.

40. x 1 1.2 1.4 1.6 1.8 2.0
y 1 1.2 1.3667 1.5130 1.6452 1.7688
y dy dx x C; x 1 and y 1
dy y
dx y 2
1
?? ? ? ? ? ?
2
1 C C y 2x 1
""
##
œ Ê œÊ œ
2
y exact 2x 1 y 2 3 1.7321 is theÊ œ Ê œ ¸ab ab
È È
exact value.

41. y 1 y y 1 y 1 . We have y 0 y 1 0, y 1 0 y 1, 1.
dy
dx
2
œÊœ œÊœ œÊœ
ww
abab abab
(a) Equilibrium points are 1 (stable) and 1 (unstable)
(b) y y 1 y 2yy y 2y y 1 2y y 1 y 1 . So y 0 y 0, y 1, y 1.
w ww w ww ww
œÊ œ Ê œ œ œÊœ œ œ
2 2
a babab

(c)
42. y y y y 1 y . We have y 0 y 1 y 0 y 0, 1 y 0 y 0, 1.
dy
dx
2
œ Ê œ œÊ œÊœ œÊ œ
ww
ab ab
(a) The equilibrium points are 0 and 1. So, 0 is unstable and 1 is stable.
(b) Let increasing, decreasing.ïî œ íï œ
y
yyy
01
qqíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqp
! ! !
www
y y y y y 2yy y y y 2y y y y y 2y 2y y 2y 3y y
wwwwwww ww
œ Ê œ Ê œ œ Ê œ
2 2 2 223 32
abab
y 2y 3y 1 y y 2y 1 y 1 . So, y 0 y 0, 2y 1 0, y 1 0 y 0, y ,œ Ê œ œÊœ œ œÊœ œababab
2 ww ww "
#

621. Chapter 9 Additional and Advanced Exercises 617
y1. œ
Let concave up, concave down.ïî œ íï œ
y
yyyy
01 1/2
qíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqñqqïïïïïîqp
! ! ! !
ww ww ww ww (c)
43. (a) Force Mass times Acceleration (Newton's Second Law) or F ma. Let a v . Thenœœœœ†œ
dv dv ds dv
dt ds dt ds
ma mgR s a gR s v gR s v dv gR s ds v dv gR s dsœ Ê œ Ê œ Ê œ Ê œ
22 22 22 22 22 dv
ds
''
C v 2C C. When t 0, v v and s R v Cʜʜ? ?? ?ʜ
v
2s s s R
gR 2gR 2gR 2gR
11 0
2 2
0
2 222 2
C v 2gR v v 2gRÊœ Ê œ
222
00
2gR
s
2
(b) If v 2gR, then v v , since v 0 if v 2gR. Then s ds 2gR dt0 0
2 2gR 2gR
ss dt
ds 2gR
s
2œœÊœ   œÊœÈÈ É È È
2 2 2 È
È
s ds 2gR dt s 2gR t C s 2gR t C; t 0 and s Rʜ ʜ wʜ w??''1/2 3/2 3/2 22 2 23
32
1 ÈÈ È ˆ‰
R 2gR 0 C C R s 2gR t R R 2g t Rʜ wʜʜ w? w
3/2 3/2 3/2 3/2 3/2333
222
22ˆ‰ ˆ‰ ˆ‰ÈÈ ab È
RRt1RsR R2gt1 t1 1 t???ʜ w ww
3/2 3/2 3/23
2
1/2
3 2gR
2R
3v 3v
2R 2R
2/3 ‘ ‘ ‘ˆ‰ È ’“Š‹ ˆ‰ ˆ‰

È 00
44. coasting distance 0.97 k 27.343. s t 1 e s t 0.97 1 e
vm vm
kk k
0.86 30.84 k/m t 27.343/30.84 t0 0
?ʜ? ?ʜ
aba b ab a b
ab abˆ‰ ˆ ‰

s t 0.97 1 e . A graph of the model is shown superimposed on a graph of the data.ʜ ab a b
0.8866t

CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES
1. (a) k c y dy k y c dt k dt k dt ln y c k t C
dy dy dy
dt V V y c V y c V V
AA A A A
1
œ Ê œ Ê œ Ê œ Ê œ ab ab kk

''
y c e e . Apply the initial condition, y 0 y y c C C y cÊœ„ œ Ê œ Ê œ
Ckt
00 01
A
V
ab
yc y ce .ʜ ab 0
kt
A
V
(b) Steady state solution: y y t c y c 0 ccyce _
Ä_ Ä_
?? ?? lim lim
tt
0
kt
0
ab a ba b ‘ab
A
V
2. Measure the amounts of oxygen involved in mL. Then the inflow of oxygen is 1000 mL/min (Assumed: it will take 5
minutes to deliver the 5L 5000mL); the amount of oxygen at t 0 is 210 mL; letting A the amount of oxygen in theœœœ
flask, the concentration at time t is A mL/L; the outflow rate of oxygen is A mL/L (lb/sec). The rate of change in A, ,
dA
dt
equals the rate of gain (1000 mL/min) minus rate of loss (A mL/min). Thus:
1000 A dt ln A 1000 t C A 1000 Ce . At t 0, A 210, so C 790 and
dA dA
dt 1000 A
t
?? ?? ??? ?? ?


ab
A 1000 790e . Thus, A 5 1000 790e 994.7 mL. The concentration is 99.47%.? ? ?
t5 994.7 mL
1000 mL
ab

622. 618 Chapter 9 Further Applications of Integration
3. The amount of CO in the room at time t is A t . The rate of change in the amount of CO , is the rate of internal2 2
dA
dt ab
production (R ) plus the inflow rate (R ) minus the outflow rate (R ).
12 3 R 20 30 students ft 0.04 1.39 1
breaths/min 100
student 1728 ft min
3 ft CO ft COœ¸ˆ‰ ˆ‰ ab Š‹
33
22
3
R 1000 0.0004 0.4 2
ft
min min min
ft CO ft COœœŠ‹Š ‹
3 33
22
R 1000 0.1A 3
A
10,000 min
ft COœœŠ‹
3
2
1.39 0.4 0.1A 1.79 0.1A A 0.1A 1.79. Let v t e . We have Ae 1.79e
dA d
dt dt
0.1dt 0.1dt 0.1dt
???? ? ?
w
ab Š‹
'''
Ae 1.79e dt 17.9e C. At t 0, A 10,000 0.0004 4 ft CO C 13.9ʜ ? w ?? ? ʜ
0.1t 0.1t 0.1t 3
2' abab
A 17.9 13.9e . So A 60 17.9 13.9e 17.87 ft of CO in the 10,000 ft room. The percent ofÊœ œ ¸
0.1t 0.1 60 3 3
2
ab
ab
CO is 100 0.18%2
17.87
10,000‚œ
4. F v u F v u F m v v u F m u .
dmv dmv
dt dt dt dt dt dt dt dt dt dt
dm dm dv dm dm dm dv dmab ab
?ʜʜʜab ab
b m b t C. At t 0, m m , so C m and m m b t.
dm
dt
00 0
œ Ê œ œ œ œ œ kk kk
Thus, F m b t u b m b t g g v gt u ln Cœ œ Êœ Êœ abkkabkkkk kk Š‹ 00 1
dv dv
dt dt m b t m
ub m btkk kk
kk
00
0

v 0 at t 0 C 0. So v gt u ln y dt and u c, y 0 at gt u ln??ʜ ? ?ʜ ?? 1
mbt
mdt
dy mbt
m Š‹ Š‹
0
0
0
0 kk kk '’“
t0 y gt c tln?ʜ w w
"
#

2
mbt mbt
bm
’“Š‹Š‹
00
0kk kk
kk
5. (a) Let y be any function such that v x y v x Q x dx C, v x e . Thenab ab ab ab?? '
'Px dxab vx y vx y y v x vxQx. We have vx e v x e Px vxPx.
d
dx
Px dx Px dx
a b ab ab ab ab ab ab ab ababab_? _w_ ? ? ? ?? ?
ww w ''ab ab
Thus v x y y v x P x v x Q x y y P x Q x the given y is a solution.ab ab ab ab ab ab ab ? ? ? ?
ww
(b) If v and Q are continuous on and x a, b , then v t Q t dt v x Q xa, bc d a b ab ab a b a b’“−œ
d
dx x
x'
0
v t Q t dt v x Q x dx. So C y v x v x Q x dx. From part (a), v x y v x Q x dx C.ʜ ? ?'
x
x
00
0
ab ab ab ab a b ab ab ab ab ab'' '
Substituting for C: v x y v x Q x dx y v x v x Q x dx v x y y v x when x x .ab ab ab a b ab ab ab a b?ʜ ? ''
00 00 0
6. (a) y P x y 0, y x 0. Use v x e as an integrating factor. Then v x y 0 v x y C
w
? ? ? ?ʜab a b ab a b ab ab 0
Px dx d
dx
'ab
y Ce and y C e , y C e , y x y x 0, y y C C eÊœ œ œ œ œ œ

#''' 'Px dx Px dx Px dx Px dx
11 2 10 20 12 12
ab ab ab ab
ab ab a b
C e and y y 0 0 0. So y y is a solution to y P x y 0 with y x 0.œ œœ œ œ
312 12 0
Px dx w
'ab
ab a b (b) v x e C C C .yx yx eCC
dd dd
dx dx dx dx
12
Px dx Px dx
12 12 3
ab abababc dab ab Š‹ ‘
ab ????
' 'ab ab
v x dx v x dx Cyx yx yx yx''d
dx
12 12
abababc d abc dab ab ab ab œœ!œ
(c) y C e , y C e , y y y . So y x 0 C e C e
11 2 12 0 1
Px dx Px dx Px dx Px dx??????

# #'' ''ab ab ab ab
ab
CC0CC yxyx for axb.??ʜ? ?
12 1 2 1 2 ab ab

623. CHAPTER 10 CONIC SECTIONS AND POLAR COORDINATES
10.1 CONIC SECTIONS AND QUADRATIC EQUATIONS
1. x 4p 8 p 2; focus is (2 0), directrix is x 2œÊ œÊœ ß œ
y
8
œ
2. x 4p 4 p 1; focus is ( 1 0), directrix is x 1œ Ê œ Ê œ ß œ
y
4
œ
3. y 4p 6 p ; focus is , directrix is yœ Ê œ Ê œ !ß œ
x333
6
œ
## #
ˆ‰
4. y 4p 2 p ; focus is , directrix is yœÊ œÊœ !ß œ
x111
2
œ
## #
ˆ‰
5. 1 c 4 9 13 foci are 13 ; vertices are 2 0 ; asymptotes are y x
x 3
49
y
œ œ
œÊœ œ Ê „ ß! „ß œ„È ÈÈ
Š‹ ab
#
6. 1 c 9 4 5 foci are 0 5 ; vertices are 0 3
x
49
y
œ œ
œÊœ œ Ê ß„ ß„
È ÈÈ
Š‹ ab
7. y 1 c 2 1 1 foci are 1 0 ; vertices are 2
x
2
œ
œÊœ œÊ „ß „ ß!
# ÈÈ
ab Š‹
8. x 1 c 4 1 5 foci are 0 5 ; vertices are 2 ; asymptotes are y 2x
y
4
œ
œÊœ œ Ê ß„ !ß„ œ„
# È ÈÈ
Š‹ ab
9. y 12x x 4p 12 p 3; 10. x 6y y 4p 6 p ;
# #
# #
œ Êœ Ê œ Êœ œ Êœ Ê œÊœ
y
1 6
x3
œ œ
focus is ( ), directrix is x 3 focus is , directrix is y$ß ! œ !ß œ ˆ‰
33
##

11. x 8y y 4p 8 p 2; 12. y 2x x 4p 2 p ;
# #
œ œ# #
"
œ Êœ Ê œÊœ œ Êœ Ê œÊœ
x
8
y
œ œ
focus is ( 2), directrix is y 2 focus is , directrix is x!ß œ ß ! œ ˆ‰
""
##

624. 620 Chapter 10 Conic Sections and Polar Coordinates
13. y 4x y 4p p ; 14. y 8x y 4p p ;œÊœÊœÊœ œÊœÊœÊœ
# #"" ""x x
416 832
œ œ
Ê Ê
ˆ‰ ˆ‰
4 8
focus is , directrix is y focus is , directrix is yˆ‰ ˆ ‰!ß œ !ß œ
"" ""
#16 16 32 3

15. x 3y x 4p p ; 16. x 2y x 4p p ;œ Êœ Ê œÊœ œ Êœ Ê œÊœ
# #"" ""
##
y y
31 8
œ œ
Ê Ê
œ
ˆ‰ ˆ‰
3
focus is , directrix is x focus is , directrix is xˆ‰ ˆ‰ß! œ ß! œ
"" " "
##11 8 8

17. 16x 25y 400 1 18. 7x 16y 112 1
## ##
#
œÊœ œÊœ
x x
516 16 7
y y
œ œœ œ
cab 25163 cab 1673Êœ œ œ Êœ œ œ
ÈÈ ÈÈ
## ##

19. 2x y 2 x 1 20. 2x y 4 1
## # ##
##
œÊ œ œÊ œ
y yx
4
œ œœ
cab 211 cab 422Êœ œ œ Êœ œ œ
ÈÈ ÈÈÈ
## ##

625. Section 10.1 Conic Sections and Quadratic Equations 621
21. 3x 2y 6 1 22. 9x 10y 90 1
## # #
#
œÊœ œÊœ
x xy y
3 10 9
œ œœ œ
cab 321 cab 1091Êœ œ œ Êœ œ œ
ÈÈ ÈÈ
## ##

23. 6x 9y 54 1 24. 169x 25y 4225 1
## ##
œÊœ œ Êœ
x x
96 25 169
y y
œ œœ œ
c a b 9 6 3 c a b 169 25 12Êœ œ œ Êœ œ œ
ÈÈ ÈÈ È
## ##

25. Foci: 2 , Vertices: 2 0 a 2, c 2 b a c 4 2 2 1Š‹ Š‹
ÈÈÈ
ab„ ß! „ß Êœ œ Ê œœ œÊ œ
###
#
#
x
4
y
œ œ
26. Foci: 4 , Vertices: 0 5 a 5, c 4 b 25 16 9 1ab ab!ß„ ß„ Êœ œÊ œœÊ œ
#
#
x
95
y
œ œ
27. x y 1 c a b 1 1 2 ; 28. 9x 16y 144 1
## ####
œÊœ œ œ œ Ê œ
ÈÈ È
x
16 9
y
œ œ
asymptotes are y x c a b 16 9 5;œ„ Ê œ œ œ
È È
##
asymptotes are y x œ„
3
4

626. 622 Chapter 10 Conic Sections and Polar Coordinates
29. y x 8 1 c a b 30. y x 4 1 c a b
## ## ## ##
œÊ œÊœ œÊ œÊœ
y y
88 44
x x
œ œœ œ
ÈÈ
8 8 4; asymptotes are y x 4 4 2 2; asymptotes are y xœœ œ„ œœ œ„ÈÈ È

31. 8x 2y 16 1 c a b 32. y 3x 3 x 1 c a b
## ## #
#
## ##
œÊœÊœ œÊœÊœ
x yy
83
œ œ œ
ÈÈ
2 8 10 ; asymptotes are y 2x 3 1 2; asymptotes are y 3xœœ œ„ œœ œ„ÈÈ È È

33. 8y 2x 16 1 c a b 34. 64x 36y 2304 1 c a b
## # #
#
## ##
œ Ê œÊœ œ Ê œÊœ
y yx x
8 36 64
œ œœ œ
ÈÈ
2 8 10 ; asymptotes are y 36 64 10; asymptotes are yœœ œ„ œ œ œ„ÈÈ È
x 4
3#

35. Foci: 2 , Asymptotes: y x c 2 and 1 a b c a b 2a 2 2aŠ‹
ÈÈ
!ß „ œ „ Ê œ œ Ê œ Ê œ œ Ê œ
a
b
### # #
a 1 b 1 y x 1ÊœÊœÊ œ
##
36. Foci: 2 , Asymptotes: y x c 2 and b c a b aab„ß! œ„ Ê œ œ Ê œ Ê œ œ œ
"" ####
ÈÈÈ
333
ba a4a
a33
œœ
4 a 3 a 3 b 1 y 1Êœ Ê œÊœ ÊœÊ œ
4a x
33
œ œ
## È
37. Vertices: 3 0 , Asymptotes: y x a 3 and b (3) 4 1ab„ß œ„ Ê œ œ Ê œ œ Ê œ
4b44x
3a33916
y
œ œ
38. Vertices: 2 , Asymptotes: y x a 2 and b 2(2) 4 1ab!ß „ œ „ Ê œ œ Ê œ œ Ê œ
1a1 x
2b2 416
y
œ œ

627. Section 10.1 Conic Sections and Quadratic Equations 623
39. (a) y 8x 4p 8 p 2 directrix is x 2,
#
œ Ê œÊœÊ œ
focus is ( ), and vertex is ( 0); therefore the new#ß ! !ß
directrix is x 1, the new focus is (3 2), and theœ ß
new vertex is (1 2)ß

40. (a) x 4y 4p 4 p 1 directrix is y 1,
#
œ Ê œ Ê œ Ê œ
focus is ( 1), and vertex is ( 0); therefore the new!ß !ß
directrix is y 4, the new focus is ( 1 2), and theϧ
new vertex is ( 1 3)ß

41. (a) 1 center is ( 0), vertices are ( 4 0)
x
16 9
y
œ œ
œÊ !ß ß
and ( ); c a b 7 foci are 7 0%ß ! œ œ Ê ß
È ÈÈ
Š‹
##
and 7 ; therefore the new center is ( ), theŠ‹
È
ß! %ß$
new vertices are ( 3) and (8 3), and the new foci are!ß ß
47 Š‹
È
„ß$

42. (a) 1 center is ( 0), vertices are (0 5)
x
925
y
œ œ
œÊ !ß ß
and (0 5); c a b 16 4 foci areß œ œ œ Ê
È È
##
( 4) and ( 4) ; therefore the new center is ( 3 2),!ß !ß ß
the new vertices are ( 3 3) and ( 3 7), and the newß ß
foci are ( 3 2) and ( 3 6)ß ß

43. (a) 1 center is ( 0), vertices are ( 4 0)
x
16 9
y
œ œ
œÊ !ß ß
and (4 0), and the asymptotes are orßœ„
x
43
y
y ; c a b 25 5 foci areœ„ œ œ œ Ê
3x
4 È È##
( 5 0) and (5 0) ; therefore the new center is (2 0), theß ß ß
new vertices are ( 2 0) and (6 0), the new fociß ß
are ( 3 0) and (7 0), and the new asymptotes areß ß
y œ„
3(x 2)
4
œ

628. 624 Chapter 10 Conic Sections and Polar Coordinates
44. (a) 1 center is ( 0), vertices are (0 2)
y
45
x
œ œ
œÊ !ß ß
and (0 2), and the asymptotes are orßœ„
y
2
x
5
È
y ; c a b 9 3 foci areœ„ œ œ œ Ê
2x
5
È È È##
(0 3) and (0 3) ; therefore the new center is (0 2),ßß ß
the new vertices are (0 4) and (0 0), the new fociß ß
are (0 1) and (0 5), and the new asymptotes areßß
y2 œ„
2x
5
È

45. y 4x 4p 4 p 1 focus is ( 0), directrix is x 1, and vertex is (0 0); therefore the new
#
œ Ê œÊœÊ "ß œ ß
vertex is ( 2 3), the new focus is ( 1 3), and the new directrix is x 3; the new equation isß ß œ
(y 3) 4(x 2)œ
#
46. y 12x 4p 12 p 3 focus is ( 3 0), directrix is x 3, and vertex is (0 0); therefore the new
#
œ Ê œ Ê œ Ê ß œ ß
vertex is (4 3), the new focus is (1 3), and the new directrix is x 7; the new equation is (y 3) 12(x 4)ßß œ œ
#
47. x 8y 4p 8 p 2 focus is (0 2), directrix is y 2, and vertex is (0 0); therefore the new
#
œ Ê œÊœÊ ß œ ß
vertex is (1 7), the new focus is (1 5), and the new directrix is y 9; the new equation isß ß œ
(x 1) 8(y 7)œ
#
48. x 6y 4p 6 p focus is , directrix is y , and vertex is (0 0); therefore the new
#
## #
œ Ê œ Ê œ Ê !ß œ ß
33 3
ˆ‰
vertex is ( 3 2), the new focus is 3 , and the new directrix is y ; the new equation isß ß œ ˆ‰
"
##
7
(x 3) 6(y 2)œ
#
49. 1 center is ( 0), vertices are (0 3) and ( 3); c a b 9 6 3 foci are 3
x
69
y
œ œ
œ Ê !ß ß !ß œ œ œ Ê !ß
È ÈÈ È
Š‹
##
and 3 ; therefore the new center is ( 1), the new vertices are ( 2 2) and ( 4), and the new fociŠ‹
È
!ß #ß ß #ß
are 1 3 ; the new equation is 1Š‹
È
#ß „ œ
(x 2) (y 1)
69

œœ
50. y 1 center is ( 0), vertices are 2 and 2 ; c a b 2 1 1 foci are
x
2
œ
œÊ !ß ß! ß! œ œ œÊ
# ##
Š‹Š ‹
ÈÈ È È
( 1 0) and ( ); therefore the new center is (3 4), the new vertices are 3 2 4 , and the new fociß "ß! ß „ ß Š‹
È
are (2 4) and (4 4); the new equation is (y 4) 1ßß œ
(x 3)œ
#
#
œ
51. 1 center is ( 0), vertices are 3 and 3 ; c a b 3 2 1 foci are
x
3
y
œ œ
œÊ !ß ß! ß! œ œ œÊ
#
##
Š‹Š ‹
ÈÈ È È
( 1 0) and ( ); therefore the new center is (2 3), the new vertices are 2 3 3 , and the new fociß "ß! ß „ ß Š‹
È
are (1 3) and (3 3); the new equation is 1ßß œ
(x 2) (y 3)
3
œœ
#
œœ
52. 1 center is ( 0), vertices are ( ) and ( 5); c a b 25 16 3 foci are
x
16 5
y
œ œ
œ Ê !ß !ß & !ß œ œ œ Ê
#
##
È È
(0 3) and (0 3); therefore the new center is ( 4 5), the new vertices are ( 4 0) and ( 4 10), and the newß ß ß ß ß
foci are ( 4 2) and ( 4 8); the new equation is 1ß ß œ
(x 4) (y 5)
16 5

#
œœ
53. 1 center is ( 0), vertices are (2 0) and ( 2 0); c a b 4 5 3 foci are ( ) and
x
45
y
œ œ
œ Ê !ß ß ß œ œ œ Ê $ß !
È È
##
( 3 0); the asymptotes are y ; therefore the new center is (2 2), the new vertices areß „ œ Ê œ„ ß
x y
5
5x
##È
È
(4 2) and (0 2), and the new foci are (5 2) and ( 1 2); the new asymptotes are y 2 ; the newßß ßß œ„
È
5(x 2)œ
#

629. Section 10.1 Conic Sections and Quadratic Equations 625
equation is 1
(x 2) (y 2)
45
œœ
œœ
œ
54. 1 center is ( 0), vertices are (4 0) and ( 4 0); c a b 16 9 5 foci are ( 5 )
x
16 9
y
œ œ
œ Ê !ß ß ß œ œ œ Ê ß!
È È
##
and (5 0); the asymptotes are y ; therefore the new center is ( 5 1), the new vertices areß „ œ Ê œ„ ß
x3x
43 4
y
( 1 1) and ( 9 1), and the new foci are ( 10 1) and (0 1); the new asymptotes are y 1 ;ß ß ß ß œ„
3(x 5)
4

the new equation is 1
(x 5) (y 1)
16 9

œœ
œ
55. y x 1 center is ( 0), vertices are (0 1) and (0 1); c a b 1 1 2 foci are
## ##
œÊ !ß ß ß œ œ œ Ê
ÈÈ È
2 ; the asymptotes are y x; therefore the new center is ( 1 1), the new vertices are ( 1 0) andŠ‹
È
!ß „ œ „ ß ß
( 1 2), and the new foci are 1 1 2 ; the new asymptotes are y 1 (x 1); the new equation isß ß „ œ„ Š‹
È
(y 1) (x 1) 1œ
##
56. x 1 center is ( 0), vertices are 0 3 and 3 ; c a b 3 1 2 foci are ( )
y
3
œ
œÊ !ß ß !ß œ œ œÊ !ß#
# ##
Š‹Š ‹
ÈÈ È È
and ( 2); the asymptotes are x y 3x; therefore the new center is (1 3), the new vertices!ß „ œ Ê œ „ ß
y
3
È È
are 3 , and the new foci are ( ) and (1 1); the new asymptotes are y 3 3 (x 1); the newŠ‹
ÈÈ
"ß $ „ "ß & ß œ „
equation is (x 1) 1
(y 3)
3
œ #
œ
œ
57. x 4x y 12 x 4x 4 y 12 4 (x 2) y 16; this is a circle: center at
## # # ##
œ Ê œÊ œ
C( 2 0), a 4ß œ
58. 2x 2y 28x 12y 114 0 x 14x 49 y 6y 9 57 49 9 (x 7) (y 3) 1;
## # # # #
œÊœÊœ
this is a circle: center at C(7 3), a 1ß œ
59. x 2x 4y 3 0 x 2x 1 4y 3 1 (x 1) 4(y 1); this is a parabola:
## #
œÊ œÊ œ
V( 11), F( 10)ß ß
60. y 4y 8x 12 0 y 4y 4 8x 12 4 (y 2) 8(x 2); this is a parabola:
## #
œÊ œÊ œ
V( 2), F( )#ß !ß #
61. x 5y 4x 1 x 4x 4 5y 5 (x 2) 5y 5 y 1; this is an ellipse: the
## # # ## #
œÊœÊœÊ œ
(x 2)
5
œ
center is ( 2 0), the vertices are 2 5 0 ; c a b 5 1 2 the foci are ( 4 0) and ( 0)ß „ ß œ œ œ Ê ß !ߊ‹
È È È
##
62. 9x 6y 36y 0 9x 6 y 6y 9 54 9x 6(y 3) 54 1; this is an ellipse:
## # # # #
œÊ œÊ œÊ œ ab
x
69
(y 3)
œ œ
the center is (0 3), the vertices are ( 0) and ( 6); c a b 9 6 3 the foci areß !ß !ß œ œ œ Ê
È ÈÈ
##
03 3Š‹
È
ß „
63. x 2y 2x 4y 1 x 2x 1 2 y 2y 1 2 (x 1) 2(y 1) 2
## # # # #
œÊ œÊœ ab
(y 1) 1; this is an ellipse: the center is (1 1), the vertices are 2 ;Ê œ ß "„ß"
(x 1)
2
œ #
œ
Š‹
È
c a b 2 1 1 the foci are (2 1) and (0 1)œœœÊ ß ß
È È
##
64. 4x y 8x 2y 1 4 x 2x 1 y 2y 1 4 4(x 1) (y 1) 4
## # # # #
œÊ œÊ œab
(x 1) 1; this is an ellipse: the center is ( 1 1), the vertices are ( 1 3) andÊ œ ß ß
# œ(y 1)
4
œ

630. 626 Chapter 10 Conic Sections and Polar Coordinates
( 1 1); c a b 4 1 3 the foci are 1 3ß œ œ œ Ê ß"„
È È ÈÈ
Š‹
##
65. x y 2x 4y 4 x 2x 1 y 4y 4 1 (x 1) (y 2) 1; this is a hyperbola:
## # # # #
œÊ œÊ œ ab
the center is (1 2), the vertices are (2 2) and ( 2); c a b 1 1 2 the foci are 1 2 ;ß ß !ß œ œ œ Ê „ ß #
ÈÈ È È Š‹
##
the asymptotes are y 2 (x 1)œ„
66. x y 4x 6y 6 x 4x 4 y 6y 9 1 (x 2) (y 3) 1; this is a hyperbola:
## # # # #
œÊ œÊ œ ab
the center is ( 2 3), the vertices are ( 1 3) and ( 3 3); c a b 1 1 2 the foci areß ß ß œ œ œ Ê
ÈÈ È
##
2 2 3 ; the asymptotes are y 3 (x 2)Š‹
È
„ ß œ„
67. 2x y 6y 3 2x y 6y 9 6 1; this is a hyperbola: the center is ( ),
## # # œ
œ Ê œ Ê œ !ß $ab
(y 3)
63
x
œ œ
the vertices are 3 6 ; c a b 6 3 3 the foci are (0 6) and ( 0); the asymptotes areŠ‹
È È È!ß „ œ œ œ Ê ß !ß
##
y2x3
y3
63
xœ
ÈÈ
œ„ Ê œ„
È
68. y 4x 16x 24 y 4 x 4x 4 8 1; this is a hyperbola: the center is (2 0),
## # # œ
œÊ œÊ œ ß ab
y(x2)
82
œœ
the vertices are 2 8 ; c a b 8 2 10 the foci are 2 10 ; the asymptotes areŠ‹ Š ‹
ÈÈÈ È Èß„ œ œ œ Ê ß„
##
y 2(x 2)
y
8
x2
2
ÈÈ
œ„ Ê œ„
œ
69. 70.
71. 72.

631. Section 10.1 Conic Sections and Quadratic Equations 627
73. 74. x y 1 1 x y 1 1 x y and kk
## ## ##
ŸÊŸŸÊŸ
x y 1 1 y x and x y 1
## ## ##
ŸÊ  Ÿ
75. Volume of the Parabolic Solid: V 2 x h x dx 2 h x dx 2 h"
#
!œœ œ''
00
b2 b2 b2
œœ œ
11 1ˆ‰ Š‹ ’“
4h 4x x x
bb2b
œœœ

; Volume of the Cone: V h h ; therefore V Vœœœœœ
1 1hb b b hb 3
833412
œ œœ
#"#
""
##
# 11ˆ‰ Š‹
76. y x dx C C; y 0 when x 0 0 C C 0; therefore y is theœœœœœÊœÊœ œ'wwx wx wx
H H 2H 2H 2H
w(0)
Š‹
œœ œœ
#
equation of the cable's curve
77. A general equation of the circle is x y ax by c 0, so we will substitute the three given points into
##
œ this equation and solve the resulting system: c and a b ; therefore
a c 1
b c 1
2a 2b c 8
Þ
ß
à
œ
œ
œ
Êœ œœ
47
33
3x 3y 7x 7y 4 0 represents the circle
##
œ
78. A general equation of the circle is x y ax by c 0, so we will substitute each of the three given points
##
œ
into this equation and solve the resulting system: a 2, b 2, and c 23;
2a 3b c 13
3a 2b c 13
4a 3b c 25
Þ
ß
à
œ
œ
œ
Êœ œ œ
therefore x y 2x 2y 23 0 represents the circle
##
œ
79. r ( 2 1) (1 3) 13 (x 2) (y 1) 13 is an equation of the circle; the distance from the
### ##
œ œ Ê œ
center to (1.1 2.8) is ( 1.1) (1 2.8) 12.85 13 , the radius the point is inside the circleß#œ ÊÈ ÈÈ
##
80. (x 2) (y 1) 5 2(x 2) 2(y 1) 0 ; y 0 (x 2) (0 1) 5 œÊ œÊ œ œÊ œ
## ##œ
œ
dy dy
dx dx y 1
x2
(x 2) 4 x 4 or x 0 the circle crosses the x-axis at (4 0) and ( 0); x 0Ê œ Ê œ œ Ê ß !ß œ
#
(0 2) (y 1) 5 (y 1) 1 y 2 or y 0 the circle crosses the y-axis at ( 2) and ( ).Ê œ Ê œ Ê œ œ Ê !ß !ß !
## #
At (4 0): 2 the tangent line is y 2(x 4) or y 2x 8ßœœÊ œ œ
dy
dx 0 1
42œ
œ
At ( ): 2 the tangent line is y 2x!ß ! œ œ Ê œ
dy
dx 0 1
02œ
œ
At ( ): 2 the tangent line is y 2 2x or y 2x 2!ß # œ œ Ê œ œ
dy
dx 2 1
02œ
œ

632. 628 Chapter 10 Conic Sections and Polar Coordinates
81. (a) y kx x ; the volume of the solid formed by
#
œÊœ
y
k
œ
revolving R about the y-axis is V dy""
# œ'
0
kx
œ
1Š‹
y
k
œ
y dy ; the volume of the rightœœ
1 1
k5
xkx
œ
œ
'
0
kx
œ
%
È
circular cylinder formed by revolving PQ about the
y-axis is V x kx the volume of the solid#
#œÊ1
È
formed by revolving R about the y-axis is
#
V V V . Therefore we can see the$#"œœ
4x kx
5
1
œÈ
ratio of V to V is 4:1.$"

(b) The volume of the solid formed by revolving R about the x-axis is V kt dt k t dt#"
# œœ''
00
xx
11Š‹
È
. The volume of the right circular cylinder formed by revolving PS about the x-axis isœ
1kx
œ
#
V kx x kx the volume of the solid formed by revolving R about the x-axis is# "
#
#œœÊ11Š‹
È
V V V kx . Therefore the ratio of V to V is 1:1.$#" $"
#
##œœ œ 1
11kx kx
œœ
82. Let P ( p y ) be any point on x p, and let P(x y) be a point where a tangent intersects y 4px. Now""
#ß œ ß œ
y 4px 2y 4p ; then the slope of a tangent line from P is
#
"
œ
œœ
œÊ œÊœ œœ
dy dy 2p y y dy 2p
dx dx y x ( p) dx y !
y yy 2px 2p . Since x , we have y yy 2p 2p y yy y 2pÊ œ œ œ Ê œ
## # ####
"""
"
#
yy
4p 4p
œœ
Š‹
y yy 2p 0 y y y 4p . Therefore the slopes of the twoʜʜ œ„
"
##
##
""
„
##
2y 4y 16p !
œ œ
È
œ È
"
tangents from P are m and m m m 1"" # "#
œœÊœœ
2p 2p 4p
y y 4p y y 4p yy4p
!!
œœ œœ
œ
œœ œ
ÈÈ ab
œœ œœ
the lines are perpendicularÊ
83. Let y 1 on the interval 0 x 2. The area of the inscribed rectangle is given byœ ŸŸÉ
x
4
œ
A(x) 2x 2 1 4x 1 (since the length is 2x and the height is 2y)œœŠ‹ÉÉ
xx
44
œœ
A (x) 4 1 . Thus A (x) 0 4 1 0 4 1 x 0 x 2Ê œ œÊ œÊ œÊ œ
ww ##
œœ
ÉÉ Š‹
xx xx x
444
11
œ œœ œœ
œ œ
ÉÉ
x x
4 4
x 2 (only the positive square root lies in the interval). Since A(0) A(2) 0 we have that A 2 4Êœ œ œ œ
È È
Š‹
is the maximum area when the length is 2 2 and the height is 2.
ÈÈ
84. (a) Around the x-axis: 9x 4y 36 y 9 x y 9 x and we use the positive root
## # # #
œÊœ Êœ„
99
44
É
V 2 9 x dx 2 9 x dx 2 9x x 24Êœ œ œ œ''
00
22
1111Š‹É ˆ‰ œ ‘
993
444
#
#
#$
#
!
(b) Around the y-axis: 9x 4y 36 x 4 y x 4 y and we use the positive root
## # # #
œÊœ Êœ„
44
99
É
V 2 4 y dy 2 4 y dy 2 4y y 16Êœ œ œ œ''
00
33
1111Š‹É ˆ‰ œ ‘
444
9927
#
#
#$
$
!
85. 9x 4y 36 y y x 4 on the interval 2 x 4 V x 4 dx
## # œ
##
##
#
œÊœ Êœ„ ŸŸÊœ
9x 36 3 3
4
œ
ÈÈ
Š‹ '
2
4
1
x 4 dx 4x 16 8 8 (56 24) 24œœœœœœ
9 9 x 9 64 8 9 56 3
443433434
111 11'
2
4
ab ’“ œ ‘ˆ ‰ˆ‰ˆ‰
#
%
#
"
1
86. x y 1 x 1 y on the interval 3 y 3 V 1 y dy 2 1 y dy
## ###
##
œÊœ„ ŸŸÊ œ œ ÈÈÈ ˆ‰ ˆ‰ ''
30
33
11
2 1 y dy 2 y 24œœœ111'
0
3
ab ’“
#
$
!
y
3
"

633. Section 10.1 Conic Sections and Quadratic Equations 629
87. Let y 16 x on the interval 3 x 3. Since the plate is symmetric about the y-axis, x 0. For aœ ŸŸ œÉ
16
9
#
vertical strip: x y x , length 16 x , width dx area dA 16 x dxab

ÉÉ
µµ
ßœß œ œÊ œœ
É16 x
16 16
99
œ
#
##
16
9œ
mass dm dA 16 x dx. Moment of the strip about the x-axis:Êœœœ $$ É
16
9
#
y dm 16 x dx 8 x dx so the moment of the plate about the x-axis is
µ
œœ
É16 x
16 8
99
œ
#
# #
16
9œ
Š‹É ˆ‰$$
M y dm 8 x dx 8x x 32 ; also the mass of the plate is
x
3
3œœ œ œ
µ' '

$$ $ˆ‰ œ ‘
88
927
#$
$
œ!
M 16 x dx 4 1 x dx 4 3 1 u du where u 3 du dx; x 3œœœ œÊœœ'' '
33 1
33 1
$$$É É ˆ‰ È
16 x
93 3
## "
#
u 1 and x 3 u 1. Hence, 4 3 1 u du 12 1 u duÊœ œÊœ œ $$ ''
11
11
ÈÈ ##
12 u 1 u sin u 6 y . Therefore the center of mass is .œœÊœœœ !ß$1$’“Š‹
È ˆ‰
" # œÊ
"
œÊ
2M633
M 32 16 16 x $
1$ 1 1
88. y x 1 x 1 (2x) 1 1œÊœ œ Ê œ Ê œ
È
ab Š‹ Š‹ Ê É
# "
#
#
œÊ$#

##
dy dy dy
dx dx x 1 dx x 1
xx x
x1
È
œ
œ
œœ
œ
S 2 y 1 dx 2 x 1 dx 2 2x 1 dx ;œÊœ œ œ ÉÉ Ê Š‹
ÈÈ
2x 1 2x 1
x1 dx x1
dy
œ œ
œ œ


#
## '' '
00 0
22 2
œœ œ
11 1
u1 du uu1lnu u1 25ln2 5
u2x
du 2 dx
–—
È
È
ÈÈÈ
’“’“Š‹Š‹ Š‹
ÈÈ
œ
œ
Ä œ œ
22
22 2 2
11 1
ÈÈ È'
0
2
### "
#
!
89. (r r ) 0 r r C, a constant the points P(t) lie on a hyperbola with foci at A
dr dr
dt dt dt
d
AB
œÊ œÊœ Ê
AB AB
and B
90. (a) tan m tan f (x ) where f(x) 4px ;""œÊ œ œ
L
w
! È
f (x) (4px) (4p) f (x )
% œÊ$# %"
#
!
œœÊœ
2p 2p
4px 4pxÈÈ
% tan .œÊ œ
2p 2p
yy
%%
"
(b) tan m9œœ œ
FP
y0 y
xp xp%%
%%œ
œœ
(c) tan !œœ
tan tan
1tan tan
9"
9"
œ

Š‹
Š‹Š‹
y
xp y
2p
y
xp y
2p
œ
œœ
"
œ
œœ
"
œ
"1
œœ œœ
y 2p(x p)
y (x p 2p) y (x p) y (x p) y
4px 2px 2p 2p(x p) 2p
œ
% %
%% %% %% %
%% %
œ
œœ



91. PF will always equal PB because the string has constant length AB FP PA AP PB.œœ
92. (a) In the labeling of the accompanying figure we have
tan t so the coordinates of A are (1 tan t). The
y
1
ϧ
coordinates of P are therefore (1 r tan t). Since„%
1 y (OA) , we have 1 tan t (1 r)
## # # # #
œ œ
1 r 1 tan t sec t r sec t 1.Êœ œ Êœ
È
#
The coordinates of P are therefore (x y) (sec t tan t)ßœ ß
xysecttant1Êœ œ
## # #

634. 630 Chapter 10 Conic Sections and Polar Coordinates
(b) In the labeling of the accompany figure the coordinates
of A are (cos t sin t), the coordinates of C are (1 tan t),ßß
and the coordinates of P are (1 d tan t). By similar„%
triangles,
dOC d
AB OA 1 cos t 1
1tant
œÊ œ
œ

È
œ
d (1 cos t)(sec t) sec t 1. The coordinatesÊœ œ
of P are therefore (sec t tan t) and P moves on theß
hyperbola x y 1 as in part (a).
##
œ
93. x 4py and y p x 4p x 2p. Therefore the line y p cuts the parabola at points ( 2p p) and
###
œœÊœÊœ„ œ ß
(2p p), and these points are [2p ( 2p)] (p p) 4p units apart.ßœ È
##
94. lim x x a lim x x a lim
xxxÄ_ Ä_ Ä_
Š‹Š‹
ÈÈ
–—
bb b b
aa a a
xxaxxa
xxa
œ œ
## ##


Š‹Š‹
ÈÈ
È
œœ œœ
œœ
lim lim 0œœœ
bba
aa
xxa
xxa xxa
xxÄ_ Ä_
’“ ’“
œœœ
œœ œœ
œ
œœ

ab
ÈÈ
10.2 CLASSIFYING CONIC SECTIONS BY ECCENTRICITY
1. 16x 25y 400 1 c a b
##
#
##
œÊœÊœ
x
516
y
œ œ
È
25 16 3 e ; F 3 0 ;œœÊœœ „ß
È
ab
c3
a5
directrices are x 0œ „ œ„ œ„
a525
e3 ˆ‰
3
5

2. 7x 16y 112 1 c a b
## ##
œÊœÊœ
x
16 7
y
œ œ
È
16 7 3 e ; F 3 0 ;œœÊœœ „ß
È
ab
c3
a4
directrices are x 0œ „ œ„ œ„
a416
e3 ˆ‰
3
4

3. 2x y 2 x 1 c a b
## # ##
œÊ œÊœ
y
2
œ
È
2 1 1 e ; F 0 1 ;œœÊœœ ß„
È
ab
c1
a 2
È
directrices are y 0 2œ „ œ„ œ„
a
e
2
È
Š‹
1
2
œ

635. Section 10.2 Classifying Conic Sections by Eccentricity 631
4. 2x y 4 1 c a b
##
#
##
œÊ œÊœ
x y
4
œ œ
È
4 2 2 e ; F 0 2 ;œœ Êœœ ß„
ÈÈ È
Š‹
c
a2
2
È
directrices are y 0 2 2œ „ œ„ œ„
a2
e
Š‹
œ
2
2
È

5. 3x 2y 6 1 c a b
##
#
##
œÊœÊœ
x y
3
œ œ
È
3 2 1 e ; F 0 1 ;œœÊœœ ß„
È
ab
c1
a 3
È
directrices are y 0 3œ „ œ„ œ„
a
e
3
È
Š‹
1
3
œ

6. 9x 10y 90 1 c a b
## ##
œÊœÊœ
x
10 9
y
œ œ
È
10 9 1 e ; F 1 0 ;œœÊœœ „ß
È
ab
c1
a 10
È
directrices are x 0 10œ „ œ„ œ„
a
e
10
È
Š‹
1
10
œ

7. 6x 9y 54 1 c a b
## ##
œÊœÊœ
x
96
y
œ œ
È
9 6 3 e ; F 3 0 ;œœ Êœœ „ß
ÈÈ È
Š‹
c
a3
3
È
directrices are x 0 3 3œ „ œ„ œ„
a3
e
Š‹
œ3
3
È

636. 632 Chapter 10 Conic Sections and Polar Coordinates
8. 169x 25y 4225 1 c a b
## ##
œ ʜʜ
x
25 169
y
œ œ
È
169 25 12 e ; F 0 12 ;œœÊœœ ß„
È
ab
c12
a13
directrices are y 0œ „ œ„ œ„
a 13 169
e12 ˆ‰
12
13

9. Foci: 0 3 , e 0.5 c 3 and a 6 b 36 9 27 1abß„ œ Ê œ œ œ œ Ê œ œ Ê œ
c3 x
e0.5 736
y#
#
œ œ
10. Foci: 8 0 , e 0.2 c 8 and a 40 b 1600 64 1536 1ab„ß œ Ê œ œ œ œ Ê œ œ Ê œ
c8 x
e 0. 1600 1536
y
#
#
œ œ
11. Vertices: 0 70 , e 0.1 a 70 and c ae 70(0.1) 7 b 4900 49 4851 1abß„ œ Ê œ œ œ œ Ê œ œ Ê œ
# x
4851 4900
y
œ œ
12. Vertices: 10 0 , e 0.24 a 10 and c ae 10(0.24) 2.4 b 100 5.76 94.24ab„ß œ Êœ œ œ œ Ê œ œ
#
1Ê œ
x
100 94.24
y
œ œ
13. Focus: 5 , Directrix: x c ae 5 and eŠ‹
ÈÈ
ß! œÊœœ œÊœÊœÊœ
9a9ae995
5555 ee e 9
5
ÈÈÈÈ
È
œœ
#
e . Then PF PD x 5 (y 0) x x 5 y xÊœ œ ʜʜ
ÈÈ È
ÈÈ
55 5
33 3 9
959
55
ÊŠ‹ ¹¹Š‹ Š‹
ÈÈ
###
# #
x25x5y x x xy4 1Ê œ Ê œÊœ
### ##È
Š‹ 518814 x
959 94 5
y
È
œ œ
14. Focus: ( 0), Directrix: x c ae 4 and e e . Then%ß œÊœœ œÊœÊœÊœÊœ
16 a16 ae16 4 16 3
3e3e3e34
3
œœ
#
#
È
PF PD (x 4) (y 0) x (x 4) y x x 8x 16 yœ Ê œ Êœ Ê
ÈÈ
33 16 3 16
343##
## ## # #
#
È ¸¸ ˆ‰
xx xy 1œÊœÊœ
3 32 256 16 x
4394 3
y
ˆ‰
### "
œ œ
ˆ‰ ˆ‰
64 16
33
15. Focus: ( 0), Directrix: x 16 c ae 4 and 16 16 16 e e . Then%ß œÊœœ œÊœÊœÊœÊœ
aae 4 1
ee e 4
œœ
# "
#
PF PD (x 4) (y 0) x 16 (x 4) y (x 16) x 8x 16 yœ Ê œ Êœ Ê
111
4##
## ## # # #È kk
x 32x 256 x y 48 1œÊœÊœ
13x
4 4 64 48
y
ab
###
œ œ
16. Focus: 2 , Directrix: x 2 2 c ae 2 and 2 2 2 2 2 2 eŠ‹
ÈÈÈÈÈÈ
ß! œÊœœ œÊœÊœÊœ
aae
ee e
2
œœ
È
# "
#
e . Then PF PD x 2 (y 0) x 2 2 x 2 yÊœ œ Êœ Ê
11 1
22 2
ÈÈ È
ÊŠ‹ ¹ ¹Š‹
ÈÈÈ
##
# #
x22 x22x2y x42x8 xy2 1œ Ê œ ʜʜ
"""
####
#
### ##
Š‹ Š ‹
ÈÈ È
x
4
y
œ œ

637. Section 10.2 Classifying Conic Sections by Eccentricity 633
17. e take c 4 and a 5; c a bœÊ œ œ œ
4
5
###
16 25 b b 9 b 3; thereforeʜʜʜ
##
1
x
59
y
œ œ
#
œ

18. The eccentricity e for Pluto is 0.25 e 0.25Êœœ œ
c
a4
"
take c 1 and a 4; c a b 1 16 bÊœ œœÊœ
### #
b 15 b 15 ; therefore, 1 is aʜʜ œ
# È x
16 15
y
œ œ
model of Pluto's orbit.

19. One axis is from A( ) to B( 7) and is 6 units long; the"ß " "ß
other axis is from C( ) to D( 1 4) and is 4 units long.$ß % ß
Therefore a 3, b 2 and the major axis is vertical. Theœœ
center is the point C( 4) and the ellipse is given by"ß
1; cab325
(x 1) (y 4)
49
œœ #####
œœ
œ œœœ
c 5 ; therefore the foci are F 1 4 5 , theʜ ߄
ÈÈ
Š‹
eccentricity is e , and the directrices areœœ
c
a3
5
È
y4 4 4 .œ„œ„ œ„
a3
e5
95
Š‹
È
œ5
3

20. Using PF e PD, we have (x 4) y x 9 (x 4) y (x 9) x 8x 16 yœ œ Ê œ Ê † È kk
## ## # # #24
39
x 18x 81 x y 20 5x 9y 180 or 1.œÊœÊœ œ
45 x
9 9 36 20
y
ab
#####
œ œ
21. The ellipse must pass through ( 0) c 0; the point ( 1 2) lies on the ellipse a 2b 8. The ellipse!ß Ê œ ß Ê œ
is tangent to the x-axis its center is on the y-axis, so a 0 and b 4 the equation is 4x y 4y 0.ÊœœÊœ
##
Next, 4x y 4y 4 4 4x (y 24) 4 x 1 a 2 and b 1 (now using the
## # # # œ
œÊ œÊ œÊœ œ
(y 2)
4
œ
standard symbols) c a b 4 1 3 c 3 e .Ê œ œœÊœ Êœœ
###
# È c
a
3
È
22. We first prove a result which we will use: let m , and"
m be two nonparallel, nonperpendicular lines. Let be# !
the acute angle between the lines. Then tan .!œ
mm
1mmʜ
Êœœ
Ê
To see this result, let be the angle of inclination of the) "
line with slope m , and be the angle of inclination of the"#)
line with slope m . Assume m m . Then and we
#"#"# ))
have . Then tan tan ( )!) ) ! ) )œ œ
"# "#
, since m tan and andœœ œ
tan tan m m
1tan tan 1mm
))
))ʜ ʜ
Êœ Êœœœ
ÊÊ
""
)
mtan .
##œ )

638. 634 Chapter 10 Conic Sections and Polar Coordinates
Now we prove the reflective property of ellipses (see the
accompanying figure): If 1, then
x
ab
y
œ
œœ
œ
‹Œ
b x a y a b and y a x y .
## ## ## w ## œ
œ
œ œ Êœ
bbx
a
aa xÈ
Èœœ
Let P(x y ) be any point on the ellipse!!ß y (x ) . Let F (c 0) and F ( c 0)Êœ œ ß ß
w
!"#
œœ
œ
bx b x
aa x
ay ww
ŒŠœ
w
œ
œ
É
be the foci. Then m and m . Let andPF PF
yy
xc xcʜ
ww
wwœœ
œÊ
!
be the angles between the tangent line and PF and PF ,"
"#
respectively. Then

tan !œœ œ œ
œÊ
Š‹
œœ
œ
bx y
ay xc
bxy
ay(x c)
œ
ÊÊ
œ
Ê Ê
w
œ
ÊÊ
œ
ÊÊ
w
1
œÊ œ œ Ê
œœ œÊœ
œbx bxc ay bxc bx ay
ayx ayc bxy ayc a b xy
bxc
œœ œ œœ œ œœ œœ
ww wwwww
œœœ œœœ
ww w ww w ww
œ
ab
ab
ab
ayc cxy cy
b
œœ
œœ
www w
œ
œÊ
œ.
Similarly, tan . Since tan tan , and and are both less than 90°, we have ."!"!" !"œœ œ
b
cy
œ
w
23. x y 1 c a b 1 1 2 e
## ##
œÊœ œ œ Êœ
ÈÈ È
c
a
2 ; asymptotes are y x; F 2 ;œœ œ„ „ß!
È
2
1
ÈÈ
Š‹
directrices are x 0œ„œ„
a
e 2
"
È

24. 9x 16y 144 1 c a b
## ##
œÊœÊœ
x
16 9
y
œ œ
È
16 9 5 e ; asymptotes areœœÊœœÈ
c5
a4
y x; F 5 ; directrices are x 0œ„ „ ß! œ „
3a
4e
ab
œ„
"6
5

25. y x 8 1 c a b
## ##
œÊ œÊœ
y
88
x
œ œ
È
8 8 4 e 2 ; asymptotes areœœÊœœœÈ È
c4
a 8
È
y x; F 0 4 ; directrices are y 0œ„ ß„ œ „ab
a
e
2œ„ œ„
È
È
8
2

639. Section 10.2 Classifying Conic Sections by Eccentricity 635
26. y x 4 1 c a b
## ##
œÊ œÊœ
y
44
x
œ œ
È
4 4 2 2 e 2 ; asymptotesœœ Êœœ œÈ ÈÈ
c
a2
22
È
are y x; F 0 2 2 ; directrices are y 0œ„ ß„ œ „Š‹
È
a
e
2œ„ œ„
2
2
È È

27. 8x 2y 16 1 c a b
## ##
œÊœÊœ
x
28
y
œ œ
È
2 8 10 e 5 ; asymptotesœœ Êœœ œÈ È È
c
a
10
2
È
È
are y 2x; F 10 ; directrices are x 0œ„ „ ß! œ „Š‹
È
a
e
œ„ œ„
È
ÈÈ
2
510
2

28. y 3x 3 x 1 c a b
## # ##
œÊœÊœ
y
3
œ
È
3 1 2 e ; asymptotes areœœÊœœÈ
c2
a 3
È
y 3 x; F 0 2 ; directrices are y 0œ„ ß„ œ „
È
ab
a
e
œ„ œ„
È
Š‹
3 3
2
3
œ
#

29. 8y 2x 16 1 c a b
## ##
œÊœÊœ
y
28
x
œ œ
È
2 8 10 e 5 ; asymptotesœœ Êœœ œÈ È È
c
a
10
2
È
È
are y ; F 0 10 ; directrices are y 0œ„ ß„ œ „
xa
e#
Š‹
È
œ„ œ„
È
ÈÈ
2
510
2

30. 64x 36y 2304 1 c a b
## ##
œ ʜʜ
x
36 64
y
œ œ
È
36 64 10 e ; asymptotes areœœÊœœœÈ
c105
a63
y x; F 10 ; directrices are x 0œ„ „ ß! œ „
4a
3e
ab
œ„ œ„
618
5ˆ‰
5
3

31. Vertices 1 and e 3 a 1 and e 3 c 3a 3 b c a 9 1 8 y 1ab!ß„ œÊœ œœÊœ œÊ œœœÊ œ
c x
a 8
### #
œ

640. 636 Chapter 10 Conic Sections and Polar Coordinates
32. Vertices 2 and e 2 a 2 and e 2 c 2a 4 b c a 16 4 12 1ab„ß! œÊœ œœÊœ œÊ œœ œ Ê œ
c x
a 41
y###
#
œ œ
33. Foci 3 and e 3 c 3 and e 3 c 3a a 1 b c a 9 1 8 x 1ab„ß! œÊœ œœÊœ ÊœÊ œœœÊ œ
c
a 8
y### #
œ
34. Foci 5 and e 1.25 c 5 and e 1.25 c a 5 a a 4 b c aab!ß„ œ Êœ œœ œÊœÊœÊœÊœ
c55 5
a44 4
###
25 16 9 1œœÊ œ
y
16 9
x
œ œ
35. Focus (4 0) and Directrix x 2 c ae 4 and 2 2 2 e e 2 . ThenߜʜœœÊœÊœÊœ#Êœ
aae 4
ee e
œœ
# È
PF 2 PD (x 4) (y 0) 2 x 2 (x 4) y 2(x 2) x 8x 16 yœ Ê œ Êœ Ê
ÈÈ È kk
## ## # # #
2 x 4x 4 x y 8 1œÊœÊœab
### x
88
y
œ œ
36. Focus 10 and Directrix x 2 c ae 10 and 2 2 2 e 5Š‹
ÈÈ ÈÈÈÈ È
ß! œÊœœ œÊœÊ œÊœ
aae
ee e
10
œœ
È
#
e 5. Then PF 5PD x10(y0) 5 x2 x10yÊœ œ ÊœÊ
%% %
##
# #ÈÈ È
ÊŠ‹ ¹¹Š‹
ÈÈ È
5x 2 x 210x10y 5x 22x2 1 5x y 2510œÊ œ Ê œ
ÈÈÈÈ
Š‹ Š ‹Š‹
ÈÈ È
#
### ##
1 1ʜʜ
Š‹
È
ÈÈ ÈÈ
15x
510 2510 25 1025
yy x
œ
#œ œ œ
œ
œœ œ
37. Focus ( 2 0) and Directrix x c ae 2 and e 4 e 2. Thenß œÊœ œ œ Ê œ Ê œ Ê œÊœ
""""
####
#aae 2
ee e
œœ
PF 2PD (x 2) (y 0) 2 x (x 2) y 4 x x 4x 4 yœ Ê œ Êœ ÊÈ ¸¸ ˆ‰
## ""
##
## # #
#
4 x x 3x y 3 x 1œÊœÊœˆ‰
#### "
43
y
œ
38. Focus ( 6 0) and Directrix x c ae 6 and e 3 e 3. Then ß œ# Ê œ œ œ# Ê œ# Ê œ# Ê œ Ê œ
aae 6
ee e
œœ
# È
PF 3 PD (x 6) (y 0) 3 x 2 (x 6) y 3(x 2) x 12x 36 yœ Ê œ Êœ Ê
ÈÈ È kk
## ## # # #
3 x 4x 4 2x y 24 1œÊœÊœab
###
#
x
124
y
œ œ
39. (x 1) (y 3) y 2 x 2x 1 y 6y 9 y 4y 4 4x 5y 8x 60y 4 0È kk a b œ Ê œ Ê œ
##
#
## # ##39
4
4 x 2x 1 5 y 12y 36 4 4 180 1ÊœÊ œaba b
## œœ(y 6) (x 1)
36 45
œœ
40. c a b b c a ; e c ea c e a b e a a a e 1 ; thus,
### ### ### ######
œÊœ œÊœÊœ Êœ œ
c
a
ab
1; the asymptotes of this hyperbola are y e 1 x as e increases, the
xx
ab aae1
yy
œœ
œœ œœœ
œœ
œ"Ê œ œ„ Ê
abœ
#
ab
absolute values of the slopes of the asymptotes increase and the hyperbola approaches a straight line.
41. To prove the reflective property for hyperbolas:
1 a y b x a b and .
xxb
a b dx ya
ydy
œ œ
œœ œ
œ
œÊ œ œ
## ## ##
Let P(x y ) be a point of tangency (see the accompanying!!ß
figure). The slope from P to F( c 0) is and fromw$
y
xcw
w
Ê
P to F (c 0) it is . Let the tangent through P meet#
ϧ
y
xcw
w
the x-axis in point A, and define the angles F PAnœ " !
and F PA . We will show that tan tan . Fromnœ œ
# "!"
the preliminary result in Exercise 22,

641. Section 10.3 Quadratic Equations and Rotations 637
tan . In a similar ma!œœ œœ
œÊ
Š‹Š ‹
xb y
ya xc
xb y
ya xc
ÊÊ
œ
Ê
œ
Ê
!
ÊÊ
œ
Ê
œ
Ê
!
œ
Ê1
xb xbc y a
xya yac xyb xyc yac yc
ab xbc b
œœ œ œ œ
ww w w
ww w ww ww w w
œœ œ œœ
œœ œ œ
ʜ
ÊÊ Ê
Ê
nner,
tan . Since tan tan , and and are acute angles, we have ."!"!"!"œœ œ œ
Œ œ
Š‹Š‹
yxb
xc ya
yxb
xc ya
ww
w
!
œ
w
œ
ww
w
!
œ
w
œ
œ
w
œ
Ê1
b
yc
42. From the accompanying figure, a ray of light emanating from
the focus A that met the parabola at P would be reflected
from the hyperbola as if it came directly from B
(Exercise 41). The same light ray would be reflected off the
ellipse to pass through B. Thus BPC is a straight line.
Let be the angle of incidence of the light ray on the"
hyperbola. Let be the angle of incidence of the light ray!
on the ellipse. Note that is the angle between the!"!
tangent lines to the ellipse and hyperbola at P. Since BPC is
a straight line, 2 2 180°. Thus 90 °.!" !"‹Œ ‹Œ

10.3 QUADRATIC EQUATIONS AND ROTATIONS
1. x 3xy y x 0 B 4AC ( 3) 4(1)(1) 5 0 Hyperbola
## # #
œÊ œ œÊ
2. 3x 18xy 27y 5x 7y 4 B 4AC ( 18) 4(3)(27) 0 Parabola
## # #
œÊœ œÊ
3. 3x 7xy 17y 1 B 4AC ( 7) 4(3) 17 0.477 0 Ellipse
####
œÊ œ ¸ Ê
ÈÈ
4. 2x 15 xy 2y x y 0 B 4AC 15 4(2)(2) 1 0 Ellipse
## #
#
œÊ œ œÊ
ÈÈ
Š‹
5. x 2xy y 2x y 2 0 B 4AC 2 4(1)(1) 0 Parabola
## # #
œÊ œ œÊ
6. 2x y 4xy 2x 3y 6 B 4AC 4 4(2)( 1) 24 0 Hyperbola
## # #
œÊ œ œ Ê
7. x 4xy 4y 3x 6 B 4AC 4 4(1)(4) 0 Parabola
## ##
œÊ œ œÊ
8. x y 3x 2y 10 B 4AC 0 4(1)(1) 4 0 Ellipse (circle)
## # #
œ Ê œ œÊ
9. xy y 3x 5 B 4AC 1 4(0)(1) 1 0 Hyperbola œÊ œ œÊ
###
10. 3x 6xy 3y 4x 5y 12 B 4AC 6 4(3)(3) 0 Parabola
## ##
œÊ œ œÊ
11. 3x 5xy 2y 7x 14y 1 B 4AC ( 5) 4(3)(2) 1 0 Hyperbola
## # #
œÊ œ œÊ
12. 2x 4.9xy 3y 4x 7 B 4AC ( 4.9) 4(2)(3) 0.01 0 Hyperbola
## # #
œÊœ œÊ
13. x 3xy 3y 6y 7 B 4AC ( 3) 4(1)(3) 3 0 Ellipse
## # #
œÊ œ œÊ
14. 25x 21xy 4y 350x 0 B 4AC 21 4(25)(4) 41 0 Hyperbola
## ##
œÊœ œÊ
15. 6x 3xy 2y 17y 2 0 B 4AC 3 4(6)(2) 39 0 Ellipse
## ##
œÊ œ œÊ

642. 638 Chapter 10 Conic Sections and Polar Coordinates
16. 3x 12xy 12y 435x 9y 72 0 B 4AC 12 4(3)(12) 0 Parabola
## ##
œÊœ œÊ
17. cot 2 0 2 ; therefore x x cos y sin ,!!! !!œœœÊœÊœ œ
AC 0
B1 4
œ
#
ww11
yx sin y cos xx y , yx yœ Êœ œ
w w ww ww
## ##
!!
ÈÈ ÈÈ
22 22
x y x y 2 x y 2 x y 4 HyperbolaÊ œÊœÊœÊŠ‹Š‹
ÈÈ ÈÈ
22 22
## ## ##
ww ww ww ww ""## ##
18. cot 2 0 2 ; therefore x x cos y sin ,!!! !!œœœÊœÊœ œ
AC 11
B1 4
œœ
#
ww11
yx sin y cos xx y , yx yœ Êœ œ
w w ww ww
## ##
!!
ÈÈ ÈÈ
22 22
xy xy xy xy1Ê œŠ‹Š‹Š‹Š‹
ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22
## ## ## ##
ww ww ww ww
# #
xxyyxyxxyy1 xy1 3xy2 E llipseÊ œÊ œÊœÊ
" """" " "
# #### # ##
wwwwwwwwww ww ww# #### # ## ## 3
19. cot 2 2 ; therefore x x cos y sin ,!!!!!œœœÊœÊœ œ
AC 31
B36 23 3
œœ" ww
ÈÈ
11
y x sin y cos x x y , y x yœ Êœœ
ww wwww
## ##
"
!!
ÈÈ
33 1
3xy23 xyxy xy8xyÊ Š‹Š‹Š‹Š‹Š‹
È
ÈÈÈÈÈ
333331111
## #### ## ##
ww ww w w w w ww
# #
"
8 3 x y 0 4x 16y 0 ParabolaœÊœÊ
È
Š‹
"
##
ww ww #È
3
20. cot 2 2 ; therefore x x cos y sin ,!!!!!œœœÊœÊœ œ
AC 12
B36 33
œœ"
œ
ww
ÈÈ
11
y x sin y cos x x y , y x yœ Êœœ
ww wwww
## ##
"
!!
ÈÈ
33 1
xy 3 xy x y2x y 1 x y1Ê œÊœŠ‹Š‹Š‹Š‹
È
ÈÈÈÈ
33331111 5
## #### ## ##
ww ww w w w w w w
# #
"##
x 5y 2 EllipseÊ œÊ
ww##
21. cot 2 0 2 ; therefore x x cos y sin ,!!! !!œœœÊœÊœ œ
AC 11
B2 2 4
œœ
œ
ww11
y x sin y cos x x y , y x yœ Êœ œ
w w ww ww
## ##
!!
ÈÈ ÈÈ
22 22
xy2xy xy xy2 y1Ê œÊœŠ‹Š‹Š‹Š‹
ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22
## ## ## ##
ww ww ww ww w
# #
#
Parallel horizontal linesÊ
22. cot 2 2 ; therefore x x cos y sin ,!!!!!œœ œÊœÊœ œ
AC 31 2
B33 23 3
œœ "
œ
ww
ÈÈ
11
y x sin y cos x x y , y x yœ Êœ œ
ww ww ww
## ##
!!
11 33
ÈÈ
3 x y 2 3 x y x y x y 1 4y 1Ê œÊœŠ‹Š‹Š‹Š‹
È
1111 3333
## ## ## ##
ww ww ww ww w
# #
#ÈÈÈÈ
Parallel horizontal linesÊ
23. cot 2 0 2 ; therefore x x cos y sin ,!!!!!œœ œÊœÊœ œ
AC
B24
22
22
œ œ ww
ÈÈ
È
11
y x sin y cos x x y , y x yœ Êœ œ
w w ww ww
## ##
!!
ÈÈ ÈÈ
22 22
2x y 22x y x y 2x yÊ
ÈÈ È
Š‹Š‹Š‹Š‹
ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22
## ## ## ##
ww ww ww ww
# #
8 x y 8 x y 0 2 2x 8 2 y 0 ParabolaœÊœÊŠ‹Š‹
ÈÈ
ÈÈ ÈÈ
22 22
## ##
ww ww w w #
24. cot 2 0 2 ; therefore x x cos y sin ,!!! !!œœœÊœÊœ œ
AC 00
B1 2 4
œœ ww11
y x sin y cos x x y , y x yœ Êœ œ
w w ww ww
## ##
!!
ÈÈ ÈÈ
22 22

643. Section 10.3 Quadratic Equations and Rotations 639
xy xy xy xy10 xy22x2Ê œÊŠ‹Š‹Š‹Š‹
È
ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22
## ## ## ##
ww ww ww ww ww w ##
0 HyperbolaœÊ
25. cot 2 0 2 ; therefore x x cos y sin ,!!! !!œœœÊœÊœ œ
AC 33
B2 2 4
œœ ww11
y x sin y cos x x y , y x yœ Êœ œ
w w ww ww
## ##
!!
ÈÈ ÈÈ
22 22
3 x y 2 x y x + y 3 x y 19 4x 2y 19Ê œÊœŠ‹Š‹Š‹Š‹
ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22
## ## ## ##
ww ww ww ww ww
# #
##
EllipseÊ
26. cot 2 2 ; therefore x x cos y sin ,!!!!!œœ œÊœÊœ œ
AC
B36
3(1)
43 3
œ" œœ ww
ÈÈ
11
y x sin y cos x x y , y x yœ Êœœ
ww wwww
## ##
!!
ÈÈ
33 11
3 x y 4 3 x y x y x y 7 5x 3y 7Ê œÊœŠ‹Š‹Š‹Š‹
È
ÈÈÈÈ
33331111
## #### ##
ww ww w w w w w w
# #
##
HyperbolaÊ
27. cot 2 cos 2 (if we choose 2 in Quadrant I); thus sin !!! !œœÊ œ œ œ œ
14 2 3 3 1 cos 2
16 4 5 2
1
5
œ œ" œ
#
É É
!
ˆ‰
È
3
5
and cos (or sin and cos )!!!œœœ œ œÉ É
1cos 2 2 2 1
55 5
ʜ"
##
Ê!
ˆ‰
ÈÈ È
3
5
28. cot 2 cos 2 (if we choose 2 in Quadrant II); thus sin !!!!œœœÊ œ œ
AC 4 3 3 1cos 2
B44 5 2
œœ" œ
œ
É
!
and cos (or sin and cos )œœ œ œœ œ œÉÉ É
11 21cos 2 112
5555
œœ Êœ
###
ʜ
ˆ‰ ˆ‰
ÈÈÈÈ
3 3
5 5
!!!
!
29. tan 2 2 26.57 ° 13.28 ° sin 0.23, cos 0.97; then A 0.9, B 0.0,!!!!!œœÊ¸ ʸ Ê ¸ ¸ ¸ ¸
œ" "
œ#
ww
13
C 3.1, D 0.7, E 1.2, and F 3 0.9 x 3.1 y 0.7x 1.2y 3 0, an ellipse
www w wwww ##
¸ ¸ ¸ œ Ê œ
30. tan 2 2 11.31 ° 5.65 ° sin 0.10, cos 1.00; then A 2.1, B 0.0,!!!!!œœÊ¸ ʸʸ ¸ ¸¸
""
œœ
ww
2(3) 5
C 3.1, D 3.0, E 0.3, and F 7 2.1 x 3.1 y 3.0x 0.3y 7 0, a hyperbola
www w wwww ##
¸ ¸ ¸ œÊ œ
31. tan 2 2 53.13 ° 26.5 ° sin 0.45, cos 0.89; then A 0.0, B 0.0,!!!!!œœÊ¸ ʸ(Ê ¸ ¸ ¸ ¸
œ
œ
ww44
14 3
C 5.0, D 0, E 0, and F 5 5.0 y 5 0 or y 1.0, parallel lines
www w w w #
¸¸¸ œÊœœ„
32. tan 2 2 36.87 ° 18.43 ° sin 0.32, cos 0.95; then A 0.0, B 0.0,!!!!!œœÊ¸ ʸ Ê ¸ ¸ ¸ ¸
œ
œ
ww12 3
218 4
C 20.1, D 0, E 0, and F 49 20.1 y 49 0, parallel lines
www w w #
¸¸¸ œÊœ
33. tan 2 5 2 78.69 ° 39.35 ° sin 0.63, cos 0.77; then A 5.0, B 0.0,!!! !!œœÊ¸ ʸ Ê ¸ ¸ ¸ ¸
5
32œ
ww
C 0.05, D 5.0, E 6.2, and F 1 5.0 x 0.05 y 5.0x 6.2y 1 0, a hyperbola
www w wwww ##
¸ ¸ ¸ œ Ê œ
34. tan 2 1 2 45.00 ° 22.5 ° sin 0.38, cos 0.92; then A 0.5, B 0.0,!!!!!œ œ Ê ¸ Ê ¸ Ê ¸ ¸ ¸ ¸
7
29œ
ww
C 10.4, D 18.4, E 7.6, and F 86 0.5 x 10.4 y 18.4x 7.6y 86 0, an ellipse
www w wwww ##
¸¸¸ œÊ œ ab
35. 90° x x cos 90°y sin 90° y and y x sin 90°y cos 90°x!œÊœ œ œ œ
ww www w
(a) 1 (b) 1 (c) x y a
xx
ba ab
yy
" "œ œ
œœ œ œ
""œœ
œ œ œ
ww###
(d) y mx y mx 0 D m and E 1; 90 ° D 1 and E m my x 0 y xœÊœÊœ œ œ Êœ œÊœÊœ !
ww wwww "
m
(e) y mx b y mx b 0 D m and E 1; 90 ° D 1, E m and F bœÊœÊœ œ œ Êœ œ œ !
ww w
my x b 0 y xʜʜ
ww w w "
mm
b

644. 640 Chapter 10 Conic Sections and Polar Coordinates
36. 180° x x cos 180°y sin 180° x and y x sin 180°y cos 180° y!œ Ê œ œ œ œ
ww www w
(a) 1 (b) 1 (c) x y a
xx
ab ab
yy
" "œ œ
œœ œœ
" "œ œ
œ œ œ
ww###
(d) y mx y mx 0 D m and E 1; 180 ° D m and E 1 y mx 0 œÊœÊœ œ œ Êœ œÊœÊ !
ww ww
ymx
ww
œ
(e) y mx b y mx b 0 D m and E 1; 180 ° D m, E 1 and F bœÊœÊœ œ œ Êœ œ œ !
ww w
ymxb0 ymxbÊ œ Ê œ
ww w w
37. (a) A cos 45° sin 45° , B 0, C cos 45 ° sin 45° , F 1
wwww
## # #
""
œ œ œ œ œ œ œŠ‹Š‹
ÈÈ
22
x y 1 x y 2ʜʜ
""
##
ww ww## ##
(b) A , C (see part (a) above), D E B 0, F a x y a x y 2a
ww wwww ww ww"" ""
## ##
## ##
œœ œœœœÊœÊœ
38. xy 2 x y 4 1 (see Exercise 37(b)) a 2 and b 2 c 4 4 2 2œÊ œÊ œ Êœ œÊœ œ
ww## x
44
y
"œ "œ
È È
e2Êœœ œ
c
a
22
È
# È
39. Yes, the graph is a hyperbola: with AC 0 we have 4AC 0 and B 4AC 0.
#
40. The one curve that meets all three of the stated criteria is the ellipse x 4xy 5y 1 0. The reasoning:
##
œ
The symmetry about the origin means that ( x y) lies on the graph whenever (x y) does. Addingw$w $
Ax Bxy Cy Dx Ey F 0 and A( x) B( x)( y) C( y) D( x) E( y) F 0 and dividing
## # #
œ œ
the result by 2 produces the equivalent equation Ax Bxy Cy F 0. Substituting x 1, y 0 (because
##
œ œœ
the point (1 0) lies on the curve) shows further that A F. Then Fx Bxy Cy F 0. By implicitßœœ
##
differentiation, 2Fx By Bxy 2Cyy 0, so substituting x 2, y 1, and y 0 (from Property 3) œ œœ œ
ww w
gives 4F B 0 B 4F the conic is Fx 4Fxy Cy F 0. Now substituting x 2 and y 1œ Ê œ Ê œ œ œ
##
again gives 4F 8F C F 0 C 5F the equation is now Fx 4Fxy 5Fy F 0. Finally, œÊ œ Ê œ
##
dividing through by F gives the equation x 4xy 5y 1 0.œ
##
41. Let be any angle. Then A cos sin 1, B 0, C sin cos 1, D E 0 and F a!!!!!
w# # w w# # ww w#
œœœœœœœ œ
xya.ʜ
ww###
42. If A C, then B B cos 2 (C A) sin 2 B cos 2 . Then 2 B B cos 0 so theœœœ œÊœÊœœ
w w
##
!!!!!
11 1
4
xy-term is eliminated.
43. (a) B 4AC 4 4(1)(4) 0, so the discriminant indicates this conic is a parabola
##
œ œ
(b) The left-hand side of x 4xy 4y 6x 12y 9 0 factors as a perfect square: (x 2y 3) 0
## #
œ œ
x 2y 3 0 2y x 3; thus the curve is a degenerate parabola (i.e., a straight line).Ê œÊ œ
44. (a) B 4AC 6 4(9)(1) 0, so the discriminant indicates this conic is a parabola
##
œ œ
(b) The left-hand side of 9x 6xy y 12x 4y 4 0 factors as a perfect square: (3x y 2) 0
## #
œ œ
3x y 2 0 y 3x 2; thus the curve is a degenerate parabola (i.e., a straight line).ʜʜ

645. Section 10.3 Quadratic Equations and Rotations 641
45. (a) B 4AC 1 4(0)(0) 1 hyperbola
#
œ œÊ
(b) xy 2x y 0 y(x 1) 2x yœÊ œÊœ
œ
œ
2x
x1
(c) y and we want 2,œÊœ œ
œœ
œœ
2x 2 1
x 1 dx (x 1)
dy
œ
Š‹
dy
dx
the slope of y 2x 2œ Ê œ
(x 1)œ
#
œ
(x 1) 4 x 3 or x 1; x 3Ê œÊœ œ œ
#
y 3 (3 3) is a point on the hyperbolaÊœÊß
where the line with slope m 2 is normalœ
the line is y 3 2(x 3) or y 2x 3;Êœœ
x 1 y 1 ( 1 1) is a point on theœ Ê œ Ê ß
hyperbola where the line with slope m 2 isœ
normal the line is y 1 2(x 1) orʜ
y2x3œ

46. (a) False: let A C 1, B 2 B 4AC 0 parabolaœœ œÊ œÊ
#
(b) False: see part (a) above
(c) True: AC 0 4AC 0 B 4AC 0 hyperbolaÊ Ê Ê
#
47. Assume the ellipse has been rotated to eliminate the xy-term the new equation is A x C y 1 theÊœÊ
ww ww##
semi-axes are and the area is . Since B 4ACÉÉ ÉÉ Š‹Š‹
"" "" #
AC AC AC 4AC
2
"" ""
"" "" Êœœ 1
11
ÈÈ
B 4A C 4A C (because B 0) we find that the area is as claimed.œ œ œ
www ww w#
œ
2
4AC B
1
È
œ
48. (a) A C A cos B cos sin C sin A sin B cos sin C sin
ww####
œ abab!!! ! !!! !
A cos sin C sin cos A Cœœabab
## # #
!! ! !
(b) D E (D cos E sin ) ( D sin E cos ) D cos 2DE cos sin E sin
ww # ### ####
œ œ !! ! ! ! !! !
D sin 2DE sin cos E cos D cos sin E sin cos D E œ œ
## # # # # # # # # # #
!!!! !! !! abab
49. B 4A C
www#
w
B cos 2 (C A) sin 2 4 A cos B cos sin C sin A sin B cos sin C cosœ ababab! ! !!! !!!! !
# ### #
B cos 2 2B(C A) sin 2 cos 2 (C A) sin 2 4A cos sin 4AB cos sin œ
## ## ## # $
!!!!!!!!
4AC cos 4AB cos sin 4B cos sin 4BC cos sin 4AC sin 4BC cos sinww!ww!
%$###$%$
!!!!! !! !!!
4C cos sinw
## #
!!
B cos 2 2BC sin 2 cos 2 2AB sin 2 cos 2 C sin 2 2AC sin 2 A sin 2œ
## ## # ##
!!!!!! !!
4A cos sin 4AB cos sin 4AC cos 4AB cos sin B sin 2 4BC cos sin w!ww!w
###$%$##$
! ! !! ! ! ! ! !!
4AC sin 4BC cos sin 4C cos sinw! w
%$###
!!!!!
B 2BC(2 sin cos ) cos sin 2AB(2 sin cos ) cos sin C 4 sin cosœ
### #####
!! ! ! !! ! ! ! !ab abab
2AC 4 sin cos A 4 sin cos 4A cos sin 4AB cos sin 4AC cosw!w!wabab
## ### ### $ %
!! !! !! !! !
4AB cos sin 4BC cos sin 4AC sin 4BC cos sin 4C cos sinwww!w !! !! ! !! !!
$$ % $###
B 8AC sin cos 4AC cos 4AC sinœ
### % %
!! ! !
B 4AC cos 2 sin cos sinœ
#%##%
ab!!!!
B4ACcos sinœ
###
#
ab!!
B4ACœ
#

646. 642 Chapter 10 Conic Sections and Polar Coordinates
10.4 CONICS AND PARAMETRIC EQUATAIONS; THE CYCLOID
1. x cos t, y sin t, 0 t 2. x sin (2 (1 t)), y cos (2 (1 t)), 0 t 1œ œ ŸŸ œ œ ŸŸ 111
cos t sin t 1 x y 1 sin (2 (1 t)) cos (2 (1 t)) 1ʜʜ Ê œ
## ## # #
11
xy1 ʜ
##

3. x 4 cos t, y 5 sin t, 0 t 4. x 4 sin t, y 5 cos t, 0 t 2œ œ ŸŸ œ œ ŸŸ 11
1 1 1 1Ê œÊ œ Ê œÊ œ
16 cos t 25 sin t x 16 sin t 25 cos t x
16 25 16 25 16 25 16 5
y y
œœ œ œ œ œ œ œ
#

5. x t, y t, t 0 y x 6. x sec t 1, y tan t, tœœ  Êœ œ œ È È
#
##
11
sect1tant xy ʜʜ
## #

7. x sec t, y tan t, t 8. x csc t, y cot t, 0 tœ œ œ œ
11
##
1
sec t tan t 1 x y 1 1 cot t csc t 1 y x x y 1Ê œÊ œ Ê œ Êœ Ê œ
## ## # # ####

647. Section 10.4 Conics and Parametric Equations; The Cycloid 643
9. x t, y 4 t , 0 t 2 10. x t , y t 1, t 0œœ ŸŸ œ œ  
È È
# %#
y 4 x y x 1, x 0Êœ Êœ  
È È
# #

11. x cosh t, y sinh t, 1 12. x 2 sinh t, y 2 cosh t, tœ œ _ _ œ œ _ _
cosh t sinh t 1 x y 1 4 cosh t 4 sinh t 4 y x 4Ê œÊ œ Ê œÊ œ
# # ## # # ##

13. Arc PF Arc AF since each is the distance rolled andœ
FCP Arc PF b( FCP);
Arc PF Arc AF
ba
œn Ê œ n œ )
Arc AF a a b( FCP) FCP ;ʜʜnÊnœ )) )
a
b
OCG ; OCG OCP PCEnœnœnn
1
#
)
OCP . Now OCP FCPœn n œ nˆ‰
1
#
!1
. Thus OCG œ n œ Ê 1) 1) ! )
aa
bb
11
##
.œ Ê œ œ1) !!1))1 )
aaab
bbb
1
#
œ
ˆ‰

Then x OG BG OG PE (a b) cos b cos (a b) cos b cosœœœ œ )! ) 1) ˆ‰
ab
b
œ
(a b) cos b cos . Also y EG CG CE (a b) sin b sin œ œ œ œ )) )!ˆ‰
ab
b
œ
(a b) sin b sin (a b) sin b sin . Thereforeœ œ )1) ) )ˆ‰ ˆ‰
ab ab
bb
œœ
x (a b) cos b cos and y (a b) sin b sin .œ œ )) ))ˆ‰ ˆ‰
ab ab
bb
œœ
If b , then x a cos cosœœ
aaa
444
a
ˆ‰ Š‹))
œˆ‰
ˆ‰
a
4
a
4
cos cos 3 cos (cos cos 2 sin sin 2 )œ œ
3a a 3a a
44 44
)))))))
cos (cos ) cos sin (sin )(2 sin cos )œ
3a a
44
)))))))ab ab
##
cos cos cos sin sin cos œ
3a a a 2a
44 4 4
)))) ))
$##
cos cos (cos ) 1 cos a cos ;œ œ
3a a 3a
44 4
)) ) ) )
$#$
ab
y a sin sin sin sin 3 sin (sin cos 2 cos sin 2 )œ œ œ ˆ‰ Š‹
aa 3aa 3aa
4 4 44 44
a
)))))))))
œˆ‰
ˆ‰
a
4
a
4
sin (sin ) cos sin (cos )(2 sin cos )œ
3a a
44
)))))))ab ab
##
sin sin cos sin cos sin œ
3a a a 2a
44 4 4
)))) ))
#$ #
sin sin cos sinœ
3a 3a a
44 4
))))
#$
sin (sin ) 1 sin sin a sin .œ œ
3a 3a a
44 4
)) ) )) ab
#$$

648. 644 Chapter 10 Conic Sections and Polar Coordinates
14. P traces a hypocycloid where the larger radius is 2a and the smaller is a x (2a a) cos a cosÊœ )) ˆ‰
2a a
a
Ê
2a cos , 0 2 , and y (2a a) sin a sin a sin a sin 0. Therefore P traces the# %% #w w #w#))1 ) ) ) ) ˆ‰
2a a
a
Ê
diameter of the circle back and forth as goes from 0 to 2 .)1
15. Draw line AM in the figure and note that AMO is a rightn
angle since it is an inscribed angle which spans the diameter
of a circle. Then AN MN AM . Now, OA a,
###
#" #
tan t, and sin t. Next MN OP
AN AM
aa
œœ œ
OP AN AM a tan t a sin tÊœœ
#######
OP a tan t a sin tʜ
È
## ##
(a sin t) sec t 1 . In triangle BPO,#w#
È
# a sin t
cos t
Ê
x OP sin t a sin t tan t andœœœ
a sin t
cos t
œ
#
y OP cos t a sin t x a sin t tan t and y a sin t.œœÊœ œ
## #

16. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid
(see the accompanying figure).

Let denote the angle through which the wheel turns. Then h a and k a. Next introduce x y -axes)) œœ
ww
parallel to the xy-axes and having their origin at the center C of the wheel. Then x b cos and
w
œ !
y b sin , where . It follows that x b cos b sin and y b sin
www
## #
##w #w#w#w!! ) ) ) )
33 311 1
ˆ‰ ˆ‰
b cos x h x a b sin and y k y a b cos are parametric equations of the trochoid.œ Êœœ œœ))) )
ww
17. D (x 2) y D (x 2) y (t 2) t D t 4tœ Ê œ œ Ê œÉ ˆ‰ ˆ‰ ˆ ‰
# """
###
###
## ## #% 17
4
4t 4 0 t 1. The second derivative is always positive for t 0 t 1 gives a localÊœœÊœ ÁÊœ
dD
dt
ab
Ê
$
minimum for D (and hence D) which is an absolute minimum since it is the only extremum the closest
#
Ê
point on the parabola is (1 1).ß
18. D 2 cos t (sin t 0) D 2 cos t sin t œÊœÊɈ‰ ˆ‰
33
44dt
dD##
# ## ab
Ê
2 2 cos t ( 2 sin t) 2 sin t cos t ( 2 sin t) 3 cos t 0 2 sin t 0 or 3 cos t 0œ œ œÊœ œˆ‰ ˆ‰
333
4 ##
t 0, or t , . Now 6 cos t 3 cos t 6 sin t so that (0) 3 relativeÊœ œ œ œÊ1
11
3 3 dt dt
5 dD dD
ÊÊ ÊÊ
Ê Ê
ab ab ##
maximum, ( ) 9 relative maximum, relative minimum, and
dD dD
dt dt 3 2
9
ÊÊ ÊÊ
Ê Ê
ab ab
1œ Ê œ Ê ˆ‰
1
relative minimum. Therefore both t and t give points on the ellipse closest to
dD
dt 3 3 3
59 5
ÊÊ
Ê
ab
ˆ‰
111
œÊ œ œ
#
the point 1 and 1 are the desired points.ˆ‰ Š‹Š ‹
3
4
33
ß! Ê ß ß
ÈÈ
##

649. Section 10.4 Conics and Parametric Equations; The Cycloid 645
19. (a) (b) (c)
20. (a) (b) (c)
21.
22. (a) (b) (c)
23. (a) (b)

650. 646 Chapter 10 Conic Sections and Polar Coordinates
24. (a) (b)
25. (a) (b) (c)
26. (a) (b)
(c) (d)

651. Section 10.5 Polar Coordinates 647
10.5 POLAR COORDINATES
1. a, e; b, g; c, h; d, f 2. a, f; b, h; c, g; d, e
3. (a) 2 2n and 2 (2n 1) , n an integerˆ‰ˆ ‰ß ß
11
##
11
(b) ( 2n ) and ( (2n 1) ), n an integer#ß #ß 11
(c) 2 2n and 2 (2n 1) , n an integerˆ‰ˆ ‰ß ß
3311
##
11
(d) ( (2n 1) ) and ( 2n ), n an integer#ß #ß11

4. (a) 3 2n and 3 2n , n an integerˆ‰ˆ ‰ß ß
11
44
5
11
(b) 3 2n and 3 2n , n an integerˆ‰ˆ‰ß ß
11
44
5
11
(c) 3 2n and 3 2n , n an integerˆ‰ˆ ‰ß ß
11
44
3
11
(d) 3 2n and 3 2n , n an integerˆ‰ˆ‰ß ß
11
44
3
11

5. (a) x r cos 3 cos 0 3, y r sin 3 sin 0 0 Cartesian coordinates are ( 0)œœ œœœœÊ $ß))
(b) x r cos 3 cos 0 3, y r sin 3 sin 0 0 Cartesian coordinates are ( 0)œœœœœœÊ $ß))
(c) x r cos 2 cos 1, y r sin 2 sin 3 Cartesian coordinates are 1 3œœ œœœ œÊ ß))
22
33
11 ÈÈ
Š‹
(d) x r cos 2 cos 1, y r sin 2 sin 3 Cartesian coordinates are 1 3œœ œœœ œÊ ß))
77
33
11 ÈÈ
Š‹
(e) x r cos 3 cos 3, y r sin 3 sin 0 Cartesian coordinates are (3 0)œœœœœœÊ ß)1 )1
(f) x r cos 2 cos 1, y r sin 2 sin 3 Cartesian coordinates are 1 3œœ œœœ œÊ ß))
11
33 ÈÈ
Š‹
(g) x r cos 3 cos 2 3, y r sin 3 sin 2 0 Cartesian coordinates are ( 3 0)œœ œœœ œÊ ß)1 )1
(h) x r cos 2 cos 1, y r sin 2 sin 3 Cartesian coordinates are 1 3œœœœœœÊ ß)) ˆ‰ ˆ‰ ÈÈ
Š‹
11
33
6. (a) x 2 cos 1, y 2 sin 1 Cartesian coordinates are (1 1)œœœœÊ ß
ÈÈ
11
44
(b) x 1 cos 0 1, y 1 sin 0 0 Cartesian coordinates are (1 0)œœœœÊ ß
(c) x 0 cos 0, y 0 sin 0 Cartesian coordinates are ( 0)œœœœÊ !ß
11
##
(d) x 2 cos 1, y 2 sin 1 Cartesian coordinates are ( 1 1)œ œ œ œ Ê ß
ÈÈ ˆ‰ ˆ‰
11
44
(e) x 3 cos , y 3 sin Cartesian coordinates are œ œ œ œ Ê ß
553 3
62 6
33 3311
È È
###
Š‹
(f) x 5 cos tan 3, y 5 sin tan 4 Cartesian coordinates are ( 4)œœœœÊ $߈‰ ˆ‰
" "44
33
(g) x 1 cos 7 1, y 1 sin 7 0 Cartesian coordinates are (1 0)œ œ œ œ Ê ß11
(h) x 2 3 cos 3, y 2 3 sin 3 Cartesian coordinates are 3 3œœœœÊ ß
ÈÈÈ È
Š‹
22
33
11

652. 648 Chapter 10 Conic Sections and Polar Coordinates
7. 8. 9.
10. 11. 12.
13. 14. 15.
16. 17. 18.
19. 20. 21.

653. Section 10.5 Polar Coordinates 649
22.
23. r cos 2 x 2, vertical line through ( 0) 24. r sin 1 y 1, horizontal line through (0 1)))œÊœ #ß œÊœ ß
25. r sin 0 y 0, the x-axis 26. r cos 0 x 0, the y-axis))œÊœ œÊœ
27. r 4 csc r r sin 4 y 4, a horizontal line through (0 4)œÊœÊœÊœ ß))
4
sin )
28. r 3 sec r r cos 3 x 3, a vertical line through ( 3 0)œ Ê œ Ê œ Ê œ ß))
ß3
cos )
29. r cos r sin 1 x y 1, line with slope m 1 and intercept b 1))œÊœ œ œ
30. r sin r cos y x, line with slope m 1 and intercept b 0))œÊœ œ œ
31. r 1 x y 1, circle with center C ( 0) and radius 1
###
œÊ œ œ!ß
32. r 4r sin x y 4y x y 4y 4 4 x (y 2) 4, circle with center C (0 2) and radius 2
##### ##
œÊœÊœÊœ œß)
33. r r sin 2r cos 5 y 2x 5, line with slope m 2 and intercept b 5œÊœÊœ œ œ
5
sin 2 cos ))ß
))
34. r sin 2 2 2r sin cos 2 (r sin )(r cos ) 1 xy 1, hyperbola with focal axis y x
##
)))))œÊ œÊ œÊ œ œ
35. r cot csc r sin cos r sin r cos y x, parabola with vertex (0 0)œœ ʜʜʜ ß)) ) ) ) )ˆ‰ˆ‰
cos
sin sin
)
))
" ### #
which opens to the right
36. r 4 tan sec r 4 r cos 4 sin r cos 4r sin x 4y, parabola withœÊœÊœÊœÊœ)) ) ) ) ) ˆ‰
sin
cos
)
)
ß
### #
vertex ( 0) which opens upwardœ!ß
37. r (csc ) e r sin e y e , graph of the natural exponential functionœÊœÊœ))
r cos r cos xßß
38. r sin ln r ln cos ln (r cos ) y ln x, graph of the natural logarithm function)))œ œ Êœ
39. r 2r cos sin 1 x y 2xy 1 x 2xy y 1 (x y) 1 x y 1, two parallel
## ## # # #
œÊ œÊ œÊœÊœ„))
straight lines of slope 1 and y-intercepts b 1œ„
40. cos sin r cos r sin x y x y x y, two perpendicular
########
)) ) )œÊ œ ʜʜʄœ kk kk
lines through the origin with slopes 1 and 1, respectively.w
41. r 4r cos x y 4x x 4x y 0 x 4x 4 y 4 (x 2) y 4, a circle with
#########
œ ʜʜʜʜ)
center C( 2 0) and radius 2ß

654. 650 Chapter 10 Conic Sections and Polar Coordinates
42. r 6r sin x y 6y x y 6y 0 x y 6y 9 9 x (y 3) 9, a circle with
####### ##
œ ʜʜʜʜ)
center C(0 3) and radius 3ß
43. r 8 sin r 8r sin x y 8y x y 8y 0 x y 8y 16 16œ Êœ Êœ ʜʜ))
#######
x (y 4) 16, a circle with center C(0 4) and radius 4Êœ ß
##
44. r 3 cos r 3r cos x y 3x x y 3x 0 x 3x yœ Êœ Êœ ʜʜ))
####### 99
44
x y , a circle with center C and radius Ê œ ß!ˆ‰ ˆ‰
39 3 3
4###
#
#
45. r 2 cos 2 sin r 2r cos 2r sin x y 2x 2y x 2x y 2y 0œÊœ ʜʜ)) ) )
#####
(x 1) (y 1) 2, a circle with center C(1 1) and radius 2Ê œ ß
## È
46. r 2 cos sin r 2r cos r sin x y 2x y x 2x y y 0œÊœ ʜʜ)) ) )
#####
(x 1) y , a circle with center C 1 and radius Ê œ ß
# ""
###
#ˆ‰ ˆ ‰
5
4
5
È
47. r sin 2 r sin cos cos sin 2 r sin r cos 2 y x 2ˆ‰ ˆ ‰))) ))œÊ œÊ œÊ œ
111
666
33
ÈÈ
## ##
""
3 y x 4, line with slope m and intercept bÊœ œ œ
È
"
ÈÈ
33
4
48. r sin 5 r sin cos cos sin 5 r cos r sin 5 x y 5ˆ‰ ˆ ‰
222
333
33111
œÊ œÊ œÊ œ)))))
ÈÈ
## ##
""
3 x y 10, line with slope m 3 and intercept b 10Êœ œ œ
ÈÈ
49. x 7 r cos 7 50. y 1 r sin 1œÊ œ œÊ œ))
51. x y r cos r sin 52. x y 3 r cos r sin 3œÊ œ Êœ œÊ œ))) ))
1
4
53. x y 4 r 4 r 2 or r 2
## #
œÊ œÊœ œ
54. x y 1 r cos r sin 1 r cos sin 1 r cos 2 1
## ## ## # # # #
œÊ œÊ œÊ œ)) )) ) ab
55. 1 4x 9y 36 4r cos 9r sin 36
x
94
y
ß ß
œÊ œ Ê œ
## ## ##
))
56. xy 2 (r cos )(r sin ) 2 r cos sin 2 2r cos sin 4 r sin 2 4œÊ œÊ œÊ œÊ œ) ) )) )) )
###
57. y 4x r sin 4r cos r sin 4 cos
### #
œÊ œ Ê œ )) ))
58. x xy y 1 x y xy 1 r r sin cos 1 r (1 sin cos ) 1
## ## ## #
œÊ œÊ œÊ œ )) ))
59. x (y 2) 4 x y 4y 4 4 x y 4y r 4r sin r 4 sin
# # ## ## #
œÊ œÊ œ Ê œ Êœ ))
60. (x 5) y 25 x 10x 25 y 25 x y 10x r 10r cos r 10 cos œ Ê œ Êœ Êœ Êœ
## # # ## #
))
61. (x 3) (y 1) 4 x 6x 9 y 2y 1 4 x y 6x 2y 6 r 6r cos 2r sin 6 œÊ œÊ œÊœ
## # # ## #
))
62. (x 2) (y 5) 16 x 4x 4 y 10y 25 16 x y 4x 10y 13 rœ Ê œ Êœ Ê
## # # ## #
4r cos 10r sin 13œ ))
63. ( ) where is any angle!ß))

655. Section 10.6 Graphing in Polar Coordinates 651
64. (a) x a r cos a r r a sec œÊ œÊœ Êœ))
a
cos )
(b) y b r sin b r r b csc œÊ œÊœ Êœ))
b
sin )
10.6 GRAPHING IN POLAR COORDINATES
1. 1 cos ( ) 1 cos r symmetric about theœ œÊ))
x-axis; 1 cos ( ) r and 1 cos ( )Á )1)
1 cos r not symmetric about the y-axis;œ ÁÊ)
therefore not symmetric about the origin

2. 2 2 cos ( ) 2 2 cos r symmetric about theœ œÊ))
x-axis; 2 cos ( ) r and 2 2 cos ( )# Á )1)
2 2 cos r not symmetric about the y-axis;œ Á Ê )
therefore not symmetric about the origin

3. 1 sin ( ) 1 sin r and 1 sin ( )œ Á )) 1)
1 sin r not symmetric about the x-axis;œ ÁÊ)
1 sin ( ) 1 sin r symmetric aboutœœÊ1) )
the y-axis; therefore not symmetric about the origin

4. 1 sin ( ) 1 sin r and 1 sin ( )œ Á )) 1)
1 sin r not symmetric about the x-axis;œ ÁÊ)
1 sin ( ) 1 sin r symmetric about theœœÊ1) )
y-axis; therefore not symmetric about the origin

5. 2 sin ( ) 2 sin r and 2 sin ( )œ Á )) 1)
2 sin r not symmetric about the x-axis;œ ÁÊ)
2 sin ( ) 2 sin r symmetric about theœœÊ1) )
y-axis; therefore not symmetric about the origin

656. 652 Chapter 10 Conic Sections and Polar Coordinates
6. 1 2 sin ( ) 1 2 sin r and 1 2 sin ( )œ Á )) 1)
1 2 sin r not symmetric about the x-axis;œ ÁÊ)
1 2 sin ( ) 1 2 sin r symmetric about theœœÊ1) )
y-axis; therefore not symmetric about the origin

7. sin sin r symmetric about the y-axis;ˆ‰ ˆ‰œ œÊ
))
##
sin sin , so the graph symmetric about theˆ‰ ˆ‰
2
2
1) )ß
#
œ is
x-axis, and hence the origin.

8. cos cos r symmetric about the x-axis;ˆ‰ ˆ‰œ œÊ
))
##
cos cos , so the graph symmetric about theˆ‰ ˆ‰
2
2
1) )ß
#
œ is
y-axis, and hence the origin.

9. cos ( ) cos r (r ) and ( r ) are on theœ œ Êß ß)) ) )
#
graph when (r ) is on the graph symmetric about theßÊ)
x-axis and the y-axis; therefore symmetric about the origin

10. sin ( ) sin r (r ) and ( r ) are on1) ) 1) 1)œ œ Êß ß
#
the graph when (r ) is on the graph symmetric aboutßÊ)
the y-axis and the x-axis; therefore symmetric about the
origin

657. Section 10.6 Graphing in Polar Coordinates 653
11. sin ( ) sin r (r ) and ( r )œœÊß ß1) ) 1) 1)
#
are on the graph when (r ) is on the graph symmetricßÊ)
about the y-axis and the x-axis; therefore symmetric about
the origin

12. cos ( ) cos r (r ) and ( r ) are onœ œÊß ß)) ) )
#
the graph when (r ) is on the graph symmetric aboutßÊ)
the x-axis and the y-axis; therefore symmetric about the
origin

13. Since r are on the graph when (r ) is on the graphab„ß ß))
r 4 cos 2( ) r 4 cos 2 , the graph isˆ‰ab„œ Êœ
# #
))
symmetric about the x-axis and the y-axis the graph isÊ
symmetric about the origin

14. Since (r ) on the graph ( r ) is on the graphßÊß))
r 4 sin 2 r 4 sin 2 , the graph isˆ‰ab„œ Êœ
# #
))
symmetric about the origin. But 4 sin 2( ) 4 sin 2œ))
r and 4 sin 2( ) 4 sin (2 2 ) 4 sin ( 2 )Áœœ
#
1) 1 ) )
4 sin 2 r the graph is not symmetric aboutœ Á Ê)
#
the x-axis; therefore the graph is not symmetric about
the y-axis

15. Since (r ) on the graph ( r ) is on the graphßÊß))
r sin 2 r sin 2 , the graph isˆ‰ab„œ Êœ
# #
))
symmetric about the origin. But sin 2( ) ( sin 2 )œ))
sin 2 r and sin 2( ) sin (2 2 ))1)1)Á œ
#
sin ( 2 ) ( sin 2 ) sin 2 r the graphœ œ œ Á Ê)))
#
is not symmetric about the x-axis; therefore the graph is
not symmetric about the y-axis

16. Since r are on the graph when (r ) is on theab„ß ß))
graph r cos 2( ) r cos 2 , theˆ‰ab„œ Êœ
# #
))
graph is symmetric about the x-axis and the y-axis theÊ
graph is symmetric about the origin.

658. 654 Chapter 10 Conic Sections and Polar Coordinates
17. r 1 1 , and r 1))œ Ê œ Ê ß œ Ê œ
111
###
ˆ‰
1 ; r sin ; SlopeÊß œ œ œˆ‰
1))
)))#ß
w #dr r sin r cos
d r cos r sin
)
"
"
Slope at 1 isœÊß
ß#
ßß #
sin r cos
sin cos r sin
ß
)) 1
)) )
ˆ‰
1; Slope at 1 is
ß#ß
ßßß #
sin ( 1) cos
sin cos ( 1) sin
߈‰
ßß
ßß ß
ßß
ßß ß
œ ß ˆ‰
1 1
ßß#ß ß
ßß ßßß ß
sin ( 1) cos
sin cos ( 1) sin
߈‰ ˆ‰
ˆ‰ ˆ‰ ˆ‰
ßß
ßß ß
ßß
ßß ß
œ

18. 0 r 1 ( 0), and r 1))1œ Ê œ Ê "ß œ Ê œ
( ); r cos ;Ê"ß œ œ1)
w dr
d)
Slopeœœ
r sin r cos cos sin r cos
r cos r sin cos cos r sin
"
"
)) )))
)) )))
##
ßß Slope at ( 0) is œÊ"ß
cos sin r cos
cos r sin cos 0 ( 1) sin 0
cos 0 sin 0 ( 1) cos 0)) )
))
#
ßßß
#ß
ßß
1; Slope at ( ) is 1œ "ß œ 1
cos sin ( 1) cos
cos ( 1) sin
11 1
11
#ß
ßß
ß

19. r 1 ; r 1))œ Ê œ Ê "ß œ Ê œ
111
444
ˆ‰
1; r1 ;Êß œ Ê œÊ"߈‰ ˆ‰
11 1
44 4
33
)
r1 1 ;)œ Ê œ Ê ß
33
44
11
ˆ‰
r2 cos 2;
w
œœ
dr
d)
)
Slopeœœ
r sin r cos 2 cos 2 sin r cos
r cos r sin 2 cos 2 cos r sin
"
"
)) )))
)) )))
##
ßß Slope at 1 is 1;Êß œ ˆ‰
1
4
2 cos sin (1) cos
2 cos cos (1) sin
ˆ‰ ˆ‰ ˆ‰
ˆ‰ ˆ‰ ˆ‰
ßß ß
ßß ß
ß
ß
44
44
#
ß

Slope at 1 is 1;ˆ‰ß œ
1
4
2 cos sin ( 1) cos
2 cos cos ( 1) sin
ˆ‰ˆ‰ ˆ‰
ˆ‰ ˆ‰ ˆ‰
ßß#ßß
ßßßßß
ßß ß
ßß ß
ß
ß
44
44
Slope at 1 is 1;ˆ‰ß œ
3
4
2 cos sin ( 1) cos
2 cos cos ( 1) sin
1
Š‹ Š‹ Š‹
Š‹ Š‹ Š‹
33 3
44
33 3
44ßß ß
ßß ß
ß
ß
#ß
ßß
Slope at 1 is 1ˆ‰ß œ
3
4
2 cos sin (1) cos
2 cos cos (1) sin
1
Š‹Š‹ Š‹
Š‹Š‹ Š‹
ßß#ß
ßßßß
33 3
44
33 3
44ßß ß
ßß ß
ß
ß20. 0 r 1 (1 0); r 1 1 ;))œÊœÊß œ ÊœÊß
11
22
ˆ‰
r1 ; r1))1œ Ê œ Ê "ß œ Ê œ
11
#
ˆ‰
2
(1 ); r 2 sin 2 ;Êß œ œ1)
w dr
d)
Slopeœœ
r sin r cos 2 sin 2 sin r cos
r cos r sin 2 sin 2 cos r sin
"
"
)) )))
)) )))
#ß #
ßß ß
Slope at (1 0) is , which is undefined;Êß
ß#
ßß
2 sin 0 sin 0 cos 0
2 sin 0 cos 0 sin 0
Slope at 1 is 0;ˆ‰ß œ
1
2
2 sin 2 sin ( 1) cos
2 sin 2 cos ( 1) sin
ß#ß
ßßß
ˆ‰ ˆ‰ ˆ‰
ˆ‰ ˆ‰ ˆ‰
ßß ß
ßß ß
22 2
22 2

Slope at 1 is 0;ˆ‰ß œ
1
2
2 sin 2 sin ( 1) cos
2 sin 2 cos ( 1) sin
ßß ß#ßß
ßß ßßßß
ˆ‰ˆ‰ ˆ‰
ˆ‰ ˆ‰ ˆ‰
ßß ß
ßß ß
ßß ß
ßß ß
Slope at ( ) is , which is undefined"ß1
ß#
ßß
2 sin 2 sin cos
2 sin 2 cos sin
11 1
11 1

659. Section 10.6 Graphing in Polar Coordinates 655
21. (a) (b)
22. (a) (b)
23. (a) (b)
24. (a) (b)
25.

660. 656 Chapter 10 Conic Sections and Polar Coordinates
26. r 2 sec r r cos 2 x 2œÊœÊœÊœ))
2
cos )

27. 28.
29. is the same point as 2 ; r 2 sin 2 2 sin 2 2 is on the graphˆ ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ ‰#ß ß œ œ œÊß
3
444 4
11111
#
is on the graphÊ#߈‰
3
4
1
30. is the same point as ; r sin sin is on the graph ˆ‰ ˆ‰ ˆ‰ ˆ‰ Š‹
"" """
## #####
ßßœœœÊßÊß
3 3
2236
11111
ˆ‰


is on the graph
31. 1 cos 1 cos cos 0 , œ Ê œÊœ))))
11
##
3
r 1; points of intersection are and .Ê œ "ß "ß ˆ‰ˆ‰
11
##
3
The point of intersection ( 0) is found by graphing.!ß

32. 1 sin 1 sin sin 0 0, r 1;œ Ê œÊœÊœ))))1
points of intersection are (1 0) and (1 ). The point ofßß 1
intersection ( 0) is found by graphing.!ß

661. Section 10.6 Graphing in Polar Coordinates 657
33. 2 sin 2 sin 2 sin sin 2 sin )))))œÊœÊ
2 sin cos sin 2 sin cos 0œÊœ)) ) ))
(sin )(1 2 cos ) 0 sin 0 or cos ʜʜ œ)) ))
"
#
0, , , or ; 0 or r 0,Êœ œ Êœ)1 ) 1
11
33
r 3 , and r 3 ; points of))œÊœ œ Êœ
11
33 ÈÈ
intersection are ( 0), 3 , and 3!ß ß ß Š‹Š ‹
ÈÈ
11
33

34. cos 1 cos 2 cos 1 cos ))) )œ Ê œ Ê œ
"
#
, r ; points of intersection areʜ ʜ)
11
33
"
#
and , . The point (0 0) is found byˆ‰ˆ ‰
""
##
ß ß
11
33
graphing.

35. 2 4 sin sin , ; pointsŠ‹
È
#
"
#
œÊœÊœ)))
11
66
5
of intersection are 2 and 2 . TheŠ‹Š ‹
ÈÈ
ßß
11
66
5
points 2 and 2 are found byŠ‹Š ‹
ÈÈ
ß ß
11
66
5
graphing.

36. 2 sin 2 cos sin cos , ;
ÈÈ
)))))œÊœÊœ
11
44
5
r 1 r 1 and r 1))œÊœÊœ„ œ Êœ
11
44
5##
no solution for r; points of intersection are 1 .Ê„ß ˆ‰
1
4
The points ( 0) and 1 are found by graphing.!ß „ ߈‰
3
4
1

37. 1 2 sin 2 sin 2 2 , , , œÊœÊœ)) )
"
#
1111
66 6 6
51317
, , , ; points of intersection areʜ)
1111
12 12 12 12
51317
1 , 1 , 1 , and 1 . No otherˆ‰ˆ‰ˆ‰ˆ‰ßßß ß
11 1 1
12 12 1 1
513 17
##
points are found by graphing.

662. 658 Chapter 10 Conic Sections and Polar Coordinates
38. 2 cos 2 2 sin 2 cos 2 sin 2
ÈÈ
))))œÊœ
2 , , , , , , ;ʜ ʜ))
111 1 111 1
44 4 4 88 8 8
5913 5913
, r 1 r 1; , ))œÊœÊœ„œ
11 1 1
88 8 8
9513 #
r 1 no solution for r; points of intersection areÊœÊ
#
1 and 1 . The point of intersection ( 0) is foundˆ‰ˆ‰ßß !ß
11
88
9
by graphing.

39. r sin 2 and r cos 2 are generated completely for
##
œœ))
0 . Then sin 2 cos 2 2 is the onlyŸŸ œ Ê œ))))
11
# 4
solution on that interval r sin 2Êœ Ê œ œ)
11
88 2
# "
ˆ‰
È
r ; points of intersection are .Êœ„ „ ß
""
ÈÈ
22 8
Š‹
1
The point of intersection ( 0) is found by graphing.!ß

40. 1 sin 1 cos sin cos , œ Êœ Êœ
)))))11
#####
37
44
, ; r 1 cos 1 ;Ê œ œ Ê œ œ))
37 3 3
4
211 1 1
## # #
È
r 1 cos 1 ; points of)œ Ê œ œ
77
4
211
##
È
intersection are and 1 . TheŠ‹Š‹" ß ß
ÈÈ
2237
## ##
11
three points of intersection (0 0) and 1 are߄ߊ‹
È
2
##
1
found by graphing and symmetry.

41. 1 2 sin 2 sin 2 2 , , , œÊœÊœ)) )
"
#
1111
66 6 6
51317
, , , ; points of intersection areʜ)
1111
11 1 1
51317
## # #
, , 1 , and . The pointsˆ‰ˆ‰ˆ‰ˆ‰"ß "ß ß "ß
11 1 1
11 1 12
513 17
## #
of intersection 1 , , andˆ‰ˆ‰ˆ‰ß"ß "ß
71119
11 1
111
## #
are found by graphing and symmetry.ˆ‰"ß
23
1
1
#

42. r 2 sin 2 is completely generated on 0 so
#
#
œŸŸ ))
1
that 1 2 sin 2 sin 2 2 , ,œÊœÊœÊœ))) )
"
#
11 1
66 12
5
; points of intersection are 1 and . The
55
111
111
###
ˆ‰ˆ‰ß"ß
points of intersection and 1 are foundˆ‰ˆ‰"ß ß
11
11
5
##
by graphing.

43. Note that (r ) and ( r ) describe the same point in the plane. Then r 1 cos 1 cos ( )ßß œÍ))1 ) )1
1 (cos cos sin sin ) 1 cos (1 cos ) r; therefore (r ) is on the graph ofœ œ œ œ ß)1 )1 ) ) )
r 1 cos ( r ) is on the graph of r 1 cos the answer is (a).œ Íß œ Ê))1 )

663. Section 10.6 Graphing in Polar Coordinates 659

44. Note that (r ) and ( r ) describe the same point in the plane. Then r cos 2 sin 2( ))ßß œÍ ))1 ) )1 ˆ‰
1
#
sin 2 sin (2 ) cos cos (2 ) sin cos 2 r; therefore (r ) is on the graph ofœ œ œ œ ߈ ‰ ˆ‰ ˆ‰)) ) ) )
555111
###
r sin 2 the answer is (a).œ ʈ‰)
1
#

45. 46.
47. (a) (b) (c) (d)

664. 660 Chapter 10 Conic Sections and Polar Coordinates
48. (a) (b) (c)
(d) (e)
49. (a) r 4 cos cos ; r 1 cos r 1 0 r 4r 4 (r 2) 0
# ##
œ Ê œ œ Ê œ Ê œ Ê œ)) )
rr
44
œœ
Š‹
r 2; therefore cos 1 (2 ) is a point of intersectionÊ œ œ œ Ê œ Ê ß))11
2
4
œ
(b) r 0 0 4 cos cos 0 , or is on the graph; r 0 0 1 cos œÊ œ Ê œÊœ Ê!ß !ß œÊœ
#
## # #
)) ) )
11 1 133
ˆ‰ˆ ‰
cos 1 0 (0 0) is on the graph. Since ( 0) for polar coordinates, the graphsÊ œ Ê œ Ê ß !ß œ !ß)) ˆ‰
1
#
intersect at the origin.
50. (a) Let r f( ) be symmetric about the x-axis and the y-axis. Then (r ) on the graph (r ) is on theœßÊß)))
graph because of symmetry about the x-axis. Then ( r ( )) ( r ) is on the graph because ofß œ ß))
symmetry about the y-axis. Therefore r f( ) is symmetric about the origin.œ)
(b) Let r f( ) be symmetric about the x-axis and the origin. Then (r ) on the graph (r ) is on theœßÊß)))
graph because of symmetry about the x-axis. Then ( r ) is on the graph because of symmetry aboutß)
the origin. Therefore r f( ) is symmetric about the y-axis.œ)
(c) Let r f( ) be symmetric about the y-axis and the origin. Then (r ) on the graph ( r ) is on theœßÊß)))
graph because of symmetry about the y-axis. Then ( ( r) ) (r ) is on the graph because of ß œ ß))
symmetry about the origin. Therefore r f( ) is symmetric about the x-axis.œ)
51. The maximum width of the petal of the rose which lies along the x-axis is twice the largest y value of the curve
on the interval 0 . So we wish to maximize 2y 2r sin 2 cos 2 sin on 0 . LetŸŸ œ œ ŸŸ)))))
1 1
4 4
f( ) 2 cos 2 sin 2 1 2 sin (sin ) 2 sin 4 sin f ( ) 2 cos 12 sin cos . Then))) ))))))))œœ œÊœ ab
#$w#
f ( ) 0 2 cos 12 sin cos 0 (cos ) 1 6 sin 0 cos 0 or 1 6 sin 0 or
w# # #
#
))))))) ))œÊ œÊ œÊ œ œÊœ ab
1
sin . Since we want 0 , we choose sin f( ) 2 sin 4 sin))))))œŸŸœÊœ
„" œ" $1
66 4ÈÈ
1
Š‹
2 4 . We can see from the graph of r cos 2 that a maximum does occur in theœœ œŠ‹
""
ÈÈ
È
666
26
9
† )
interval 0 . Therefore the maximum width occurs at sin , and the maximum widthŸŸ œ))
1
4 6
œ" "
Š‹
È
is .
26
9
È
52. We wish to maximize y r sin 2(1 cos )(sin ) 2 sin 2 sin cos . Thenœœ œ))))))
2 cos 2(sin )( sin ) 2 cos cos 2 cos 2 sin 2 cos 2 cos 4 cos 2; thus
dy
d)
œ œ œ ))))))))))
## #
0 4 cos 2 cos 2 = 0 2 cos cos 1 0 (2 cos 1)(cos 1) 0 cos
dy
d)
œÊ Ê œÊ œÊ œ
## "
#
)) )) ) ) ) or cos 1 , , . From the graph, we can see that the maximum occurs in the first quadrant so))1œ Ê œ
11
33
5
we choose . Then y 2 sin 2 sin cos . The x-coordinate of this point is x r cos )œœ œ œ
1111 1
3333 3
33
È
#
2 1 cos cos . Thus the maximum height is h occurring at x .œ œ œ œˆ‰ˆ‰
11
33
3 333
###
È

665. Section 10.7 Area and Lengths in Polar Coordinates 661
10.7 AREA AND LENGTHS IN POLAR COORDINATES
1. A (4 2 cos ) d 16 16 cos 4 cos d 8 8 cos 2 dœ œ œ'' '
00 0
22 2
œœ œ
""
## #
##
)) ) )) ) )ab œ ‘ˆ‰
1cos 2)
(9 8 cos cos 2 ) d 9 8 sin sin 2 18œ œ œ'
0
2
œ
))))) ) 1 œ ‘
"
#
!2
1
2. A [a(1 cos )] d a 1 2 cos cos d a 1 2 cos dœ œ œ '' '
00 0
22 2
œœ œ
"" "
## # #
## ##
)) ) )) ) )ab ˆ‰
1cos 2)
a 2 cos cos 2 d a 2 sin sin 2 aœ œœ
""""
## # ## #
###
#
!'
0
2
œ
ˆ‰ œ ‘
333
4
))) )))1
1
3. A 2 cos 2 d dœœœœ''
00
44
œœœœ
""
###
#
Î%
!
)) ) )
1cos 4 sin 4
48
))1
1
œ ‘
4. A 2 2a cos 2 d 2a cos 2 d 2a 2aœœœœ''

œœ
œœ
œœ44
44
"
#
## # #
Î%
Î%
ab œ‘)) ))
sin 2
2
)
1
1
5. A (4 sin 2 ) d 2 sin 2 d cos 2 2œœœœ''
00
22
œœœœ
"
#
Î#
!
)) )) ) cd
1
6. A (6)(2) (2 sin 3 ) d 12 sin 3 d 12 4œœœœ ''
00
66
œœœœ
"
#
Î'
!
)) )) œ ‘
cos 3
3
)
1
7. r 2 cos and r 2 sin 2 cos 2 sin œœÊœ))))
cos sin ; thereforeʜʜ)))
1
4
A 2 (2 sin ) d 4 sin dœœ ''
00
44
œœœœ
"
#
##
)) ))
4 d (2 2 cos 2 ) dœœ''
00
44
œœœœ
ˆ‰
1cos 2œ
#
)
)))
2sin 2 1œ œcd))
1 1Î%
! #

8. r 1 and r 2 sin 2 sin 1 sin œœ Ê œÊœ )) )
"
#
or ; thereforeʜ)
11
66
5
A (1) (2 sin ) 1 dœ 1))
### "
#'
œ
œ
œ
œ6
56
cd
2 sin dœ 1)) '
œ
œ
œ
œ6
56
ˆ‰
# "
#
1 cos 2 dœ 1)) '
œ
œ
œ
œ6
56
ˆ‰
"
#
cos 2 dœ œ 1))1)'
œ
œ
œ
œ6
56
ˆ‰ œ ‘
""
##
&Î'
Î'2
sin 2)
1
1
sin sin œ œ1ˆ‰ˆ‰
55
131236
4331111 1
## #
"" œ
È

666. 662 Chapter 10 Conic Sections and Polar Coordinates
9. r 2 and r 2(1 cos ) 2 2(1 cos )œœÊœ ))
cos 0 ; thereforeʜʜ„))
1
#
A 2 [2(1 cos )] d area of the circleœ '
0
2
œœ
""
##
#
))
4 1 2 cos cos d (2)œ '
0
2
œœ
ab ˆ‰)))1
## "
#
4 1 2 cos d 2œ '
0
2
œœ
ˆ‰ ))1
1cos 2
#
)
(4 8 cos 2 2 cos 2 ) d 2œ '
0
2
œœ
)))1
6 8 sin sin 2 2 5 8œ œcd)))11
1Î#
!

10. r 2(1 cos ) and r 2(1 cos ) 1 cos œ œ Ê)))
1 cos cos 0 or ; the graph alsoœ Ê œ Ê œ)) )
11
##
3
gives the point of intersection (0 0); thereforeß
A 2 [2(1 cos )] d 2 [2(1 cos )] dœ''
02
2
œœ
œ
œ
œ
""
##
##
)) ))
4 1 2 cos cos dœ
'
0
2
œœ
ab )))
#
4 1 2 cos cos dÎÎÎ '
œ
œ
œ2
ab )))
#

4 1 2 cos d 4 1 2 cos dœ ''
02
2
œœ
œ
œ
œ
ˆ‰ˆ‰ )) ))
1cos 2 1cos 2
##
))
(6 8 cos 2 cos 2 ) d (6 8 cos 2 cos 2 ) dœ ''
02
2
œœ
œ
œ
œ
))) )))
6 8 sin sin 2 6 8 sin sin 2 6 16œ œcdcd))) )))1
1 1
1
Î#
! Î#
11. r 3 and r 6 cos 2 3 6 cos 2 cos 2œœÊœÊœ
È
# "
#
)))
(in the 1st quadrant); we use symmetry of theʜ)
1
6
graph to find the area, so
A4 (6 cos 2) 3 dœ '
0
6
œœ
”• Š‹
È
""
##
#
))
2 (6 cos 2 3) d 2 3 sin 2 3œœ
'
0
6
œœ
)) )) cd
1Î'
!
33œ
È
1

12. r 3a cos and r a(1 cos ) 3a cos a(1 cos )œœÊœ))))
3 cos 1 cos cos or ;ʜʜʜ))))
"
#
11
33
the graph also gives the point of intersection (0 0); thereforeß
A 2 (3a cos ) a (1 cos ) dœ '
0
3
œœ
"
#
## #
cd )))
9a cos a 2a cos a cos dœ
'
0
3
œœ
ab
## # # ##
))))
8a cos 2a cos a dœ
'
0
3
œœ
ab
## # #
)))
4a (1 cos 2 ) 2a cos a dœ
'
0
3
œœ
cd
###
)))
3a 4a cos 2 2a cos dœ
'
0
3
œœ
ab
## #
)))

3a 2a sin 2 2a sin a 2a 2a a 1 3œ œ œcd ˆ‰ Š‹ Š ‹
È
## # ## # #
Î$
!
"
##
)))1 1
1 È
3

667. Section 10.7 Area and Lengths in Polar Coordinates 663
13. r 1 and r 2 cos 1 2 cos cos œœÊœÊœ )))
"
#
in quadrant II; thereforeʜ)
2
3
1
A 2 ( 2 cos ) 1 d 4 cos 1 dœœ ''
23 23œœ
œœ
œœ
"
#
## #
cd ab)) ))
[2(1 cos 2 ) 1] d (1 2 cos 2 ) dœœ
''
23 23œœ
œœ
œœ
)) )) sin 2œ œcd))
1
1
1
#Î$ #3
3
È

14. (a) A 2 (2 cos 1) d 4 cos 4 cos 1 d [2(1 cos 2 ) 4 cos 1] dœœœ'' '
00 0
23 23 23
œœ œœœ œ
"
#
##
)) ) )) ) ))ab
(3 2 cos 2 4 cos ) d 3 sin 2 4 sin 2 2œ œ œœ'
0
23
œœ
)))))) 1 1 cd
#Î$
! ## #
1 ÈÈ È
343 33
(b) A 2 3 3 (from 14(a) above and Example 2 in the text)œ œŠ‹Š‹
È
111
33 33
ÈÈ
##
15. r 6 and r 3 csc 6 sin 3 sin œœ Ê œÊœ )) )
"
#
or ; therefore A 6 9 csc dÊœ œ )))
11
66
5 '
œ
œ
œ
œ6
56
"
#
##
ab
18 csc d 18 cot œ œ'
œ
œ
œ
œ6
56
ˆ‰ œ ‘
99
##
#
&Î'
Î'
)) ) )
1
1
15 3 3 3 12 9 3œ œŠ‹Š‹
ÈÈÈ
111
99
##

16. r 6 cos 2 and r sec sec 6 cos 2 cos cos 2 cos 2 cos 1
#####
#
œœÊœÊœÊœ )))) )) ))
39 9 3
4248
aba b 2 cos cos 2 cos cos 0 16 cos 8 cos 3 0Êœ Ê œÊ œ
33
88
%# %# % #
)) )) ) )
4 cos 1 4 cos 3 0 cos or cos cos (the second equation has no realÊ œÊœ œÊœ„abab
## ## "
#
)) )) )
3
44
3
È
roots) (in the first quadrant); thus A 2 6 cos 2 sec d 6 cos 2 sec dÊœ œ œ )))))))
1
644
99 ''
00
66
œœœœ
"
#
##
ˆ‰ˆ‰
3 sin 2 tan 3œ œ œœœ ‘ Š‹))
99
444
3333333
43
1Î'
! ##
ÈÈÈÈ
È
17. (a) r tan and r csc tan csc œœ Êœ)))) Š‹ Š‹
ÈÈ
22
##
sin cos 1 cos cos ʜ ʜ
##
##
))))Š‹ Š‹
ÈÈ
22
cos cos 1 0 cos 2 orÊ œÊœ
#
#
)) )Š‹
È
È
2
(use the quadratic formula) (the solution
È
2
4#
ʜ)
1
in the first quadrant); therefore the area of R is"

A tan d sec 1 d tan tan ;"
"" " " "
## # # #
## Î%
!œœœœœ''
00
44
œœœœ
)) ) ) ) )ab cd ˆ‰
1 11 1
44 8
AO csc and OB csc 1 AB 1œœœœÊœœŠ‹ Š‹ Š‹ Ê
ÈÈ È ÈÈ
22 2 22
4### # # #
#
#
11
the area of R is A ; therefore the area of the region shaded in the text isÊœœ ##
""
## # Š‹Š‹
ÈÈ
22
4
2 . Note: The area must be found this way since no common interval generates the region. Forˆ‰
""
##
œ
11
84 4
3
example, the interval 0 generates the arc OB of r tan but does not generate the segment AB of the lineŸŸ œ))
1
4
r csc . Instead the interval generates the half-line from B to on the line r csc .œ_œ
È È
2 2
# #
) )
(b) lim tan and the line x 1 is r sec in polar coordinates; then lim (tan sec )
)1 )1ÄÎ ÄÎ2 2
œ
)) ))œ_ œ œ

= lim lim lim 0 r tan approaches
)1 )1 )1ÄÎ ÄÎ ÄÎ222

ˆ‰ ˆ‰ ˆ‰
sin sin 1 cos
cos cos cos sin
)))
)) ) )
œ œ œÊœ
"œ
œ
)
r sec as r sec (or x 1) is a vertical asymptote of r tan . Similarly, r sec œÄÊœœ œ œ)) ) ) )
1

#

668. 664 Chapter 10 Conic Sections and Polar Coordinates
(or x 1) is a vertical asymptote of r tan .œ œ )
18. It is not because the circle is generated twice from 0 to 2 . The area of the cardioid is)1œ
A 2 (cos 1) d cos 2 cos 1 d 2 cos 1 dœœœ'' '
00 0
œœ œ
"
# #
##
)) ) )) ))ab ˆ‰
1cos 2)
2 sin . The area of the circle is A the area requested is actuallyœ œ œ œÊœ ‘ˆ‰
3sin 2 3
24 4
)) 1 1
1
)1
! ##
"
#

35
44
11 1
#
œ
19. r , 0 5 2 ; therefore Length (2 ) d 4 dœŸŸÊœ œ œ )) ) ) )) )))
# ## %#
# È Éab
È dr
d) ''
00
55
œœ
4 d (since 0) 4 d ; u 4 du d ; 0 u 4,œ œ  œÊœœÊœ''
00
55
œœ
kk
ÈÈ œ)) ) ) )) ) ) )))
## # "
#
5 u 9 u du u “
È È œ ‘)œÊœÄ œ œ '
4
9
""
##
$Î#
*
%
219
33
20. r , 0 ; therefore Length d 2 dœŸŸÊœ œ œ
edre ee e
22 22 d
œœ œœ œ
ÈÈ ÈÈ
)1 ) )
)
''
00
œœ
ÊŠ‹ Š‹ Š‹ Ê
##
#
2
e d e e 1œœœ '
0
œ
))1
1
)œ‘
!
21. r 1 cos sin ; therefore Length (1 cos ) ( sin ) dœ Ê œ œ )) )))
dr
d) '
0
2
œ
È ##
2 2 2 cos d 2 d 4 d 4 cos d 4 2 sin 8œ œ œ œ œ œ''''
0000
œœœœ
È É É ˆ‰ œ ‘)) ) ) )
4(1 cos ) 1cos
2

###

!
) )))
1
22. r a sin , 0 , a 0 a sin cos ; therefore Length a sin a sin cos dœŸŸÊœ œ
#
### ###
#
##
))) )))
)
)1 )
dr
d '
0
œ
Ɉ ‰ ˆ ‰
a sin a sin cos d a sin sin cos d (since 0 ) a sin dœ œ œŸŸ''
00
0
œœ œ
ÉÉ ¸¸ ˆ‰ '
#% ## # # #
### ### #
))) ))) )
)))1)
2a cos 2aœ œœ ‘
)
1
2!
23. r , 0 ; therefore Length dœŸŸÊœ œ
6 dr 6 sin 6 6 sin
1cos d (1cos ) 1cos (1cos )#
#
#
))) ))
1) )
) )
œ œ
'
0
2
œœ
ʈ‰ Š‹
d6 1 dœœ ''
00
22
œœœœ
É ¸¸ É
36 36 sin sin
(1 cos ) 1 cos (1 cos )1cos
"
)))
))
)
œ œ
œœ
!
ab
))
since 0 on 0 6 dœŸŸˆ‰ˆ‰ É
""
#1cos 1cos (1cos )
1 2 cos cos sin
)))
1)))
)) '
0
2
œœ
œœ
œ
6 d 6 2 6 2 3 sec dœœœœ''''
0000
2222
œœœœœœœœ
ˆ‰ ¸¸É
ÈÈ
"
#
$
1cos (1cos )
22 cos d d
(1 cos )
2 cos
))
))))
)
œ
œ
œ))
ˆ‰
œ
3 sec d 6 sec u du (use tables) 6 sec u duœœ œ '' '
00 0
24 4
œœ œœœ œ
$$
# #
Î%
!
")
1
) Œ
œ ‘
sec u tan u
2
6 ln sec u tan u 3 2 ln 1 2œ œŠ‹’“œ ‘kk
ÈÈ
Š‹
""
Î%
!È
2 2
1
24. r , ; therefore Length dœŸŸÊœ œ
2 dr 2 sin 2 2 sin
1cos d (1cos ) 1cos (1cos )œ# œ œ œ
œœ
#
#
))) ))
1) )
)1 )
œ œ
'
œ
œ
œ2
ʈ‰ Š‹
1 d dœœ''
œœ
œœ
œœ22
Ê Š‹ ¸¸ É
4sin 2
(1 cos ) 1 cos (1 cos )1cos
(1 cos ) sin
œœœ œ

)))
)
)
))
œ œ
œ
œ
œœ
ab
))
since 1 cos 0 on 2 dœ ŸŸˆ‰ˆ‰ É))1 )
1)))
))#œœ
" '
œ
œ
œ21cos (1cos )
1 2 cos cos sin
œœ
œ
2 d 2 2 2 2 csc dœœœœ''''
œœœœ
œœœœ
œœœœ2222
ˆ‰ ¸¸É
ÈÈ
"œ
œœ #œ
$
1cos (1cos )
22 cos d d
(1 cos )
2 sin
))
))))
)
œ
œ
œ))
ˆ‰
œ
csc d since csc 0 on 2 csc u du (use tables)œœ ŸŸ œ''
œ œ
œ œ
œ œ
œ2 4
2
$$
###ˆ‰ ˆ ‰
))1
))1

669. Section 10.7 Area and Lengths in Polar Coordinates 665
2 csc u du 2 ln csc u cot u 2 ln 2 1Œ
œ ‘ œ ‘Š‹’“ kk Š‹
È
œœ
csc u cot u
22 22
1
1 1
1Î#
Î% Î%
""" ""
# #
Î# '
œ
œ
œ
œ4
2 ÈÈ
2ln1 2œ
ÈÈ
Š‹
25. r cos sin cos ; therefore Length cos sin cos dœÊœ œ
$# $#
##
))) )))
)3d 3 3 3 3 3
dr '
0
4
œœ
Ɉ ‰ ˆ ‰ )
cos sin cos d cos cos sin d cos dœ œ œ'''
000
444
œœœœœœ
ÉɈ‰ ˆ‰ ˆ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰
'#% ## ##))) ) )) )
333 3 33 3
)))
d sin œœœ'
0
4
œœ
1cos 32 3
23 88

##
"
Î%
!
ˆ‰
2
3œ
))œ ‘
)1
1
26. r 1 sin 2 , 0 2 (1 sin 2 ) (2 cos 2 ) (cos 2 )(1 sin 2 ) ; thereforeœ ŸŸ Êœ œ È È
))1 ) ) ) )
dr
d)
"
#
"Î# "Î#
Length (1 sin 2 ) d dœ œ ''
00
22
œœ
œœ
É É)) )
cos2 12 sin 2sin2cos2
(1 sin 2 ) 1 sin 2
œœœ
))))
))

d 2 d 2 2œœœœ''
00
22
œœ
œœ
É
ÈÈ
’“
22 sin 2
1sin 2


#
!
)
)
1
)))1
È
27. r 1 cos 2 (1 cos 2 ) ( 2 sin 2 ); therefore Length (1 cos 2 ) dœ Êœ œ È É))) ))
dr sin 2
d (1 cos 2 )) )
)"
#
"Î# '
0
2
œ
œ
œ
d d 2 d 2 2œœœœœ'''
000
222
œœœ
œœœ
É É
ÈÈ
’“
12 cos 2cos2sin2 22 cos 2
1cos 2 1cos 2


#
!
))) )
))
1
œœ
))))1
È
28. (a) r a 0; Length a 0 d a d a 2 aœÊ œ œ œ œ œ
dr
d)
1 ''
00
22
œœ
È
kk c d##
#
!
)))1
(b) r a cos a sin ; Length (a cos ) ( a sin ) d a cos sin dœÊœ œ œ )) ))))))
dr
d) ''
00
œœ
È Èab ## ###
a d a aœœœ '
0
œ
kk c d)) 1
1
!
(c) r a sin a cos ; Length (a cos ) (a sin ) d a cos sin dœÊœ œ œ )) ))) )))
dr
d) ''
00
œœ
È Èab ## ###
a d a aœœœ '
0
œ
kk c d)) 1
1
!
29. r cos 2 , 0 (cos 2 ) ( sin 2 )(2) ; therefore Surface AreaœŸŸÊœ œ
È
)) ) )
1)
) )4d
dr sin 2
cos 2
"œ
#
"Î#
È
(2 r cos ) cos 2 d 2 cos 2 (cos ) cos 2 dœœ ''
0 0
4 4
œ œœ œ
1) ) ) 1 ) ) ) )ÊŠ‹Š‹ Š ‹
ÈÈ É
##
œsin 2 sin 2
cos 2 cos 2
))
) )È
œ
2 cos 2 (cos ) d 2 cos d 2 sin 2œœœœ''
00
44
œœœœ
Š‹
È É cd
È
1)) ) 1))1)1
" Î%
!cos 2)
1
30. r 2e , 0 2 e e ; therefore Surface AreaœŸŸÊœ œ
ÈÈ ˆ‰
œœœ222
)
1
)###
"dr
d
2
È
2 2e (sin ) 2e e d 2 2e (sin ) 2e e dœœ''
0 0
2 2
2 222
œ œ

œ œ
œ œœœ
Š‹Š‹Š‹ Š‹
ÈÈ È
Ê É1) )1) )
##
##
"
È
2
2 2 e (sin ) e d 2 2 e (sin ) e d 2 5 e sin dœœ œ'' '
00 0
22 2
222
œœ œ

œœ œ
œœœ
Š‹ Š‹Š‹
ÈÈ É È
1))1))1 ))
5 5
2#
È
È
2 5 (sin cos ) 5 e 1 where we integrated by partsœœ1))1ÈÈœ ‘ ab
e
2
œ 1Î#
!
œœ2
31. r cos 2 r cos 2 ; use r cos 2 on 0 (cos 2 ) ( sin 2 )(2) ;
# "Î#"œ
#
œÊœ„ œ ßÊœ œ))) ))
ÈÈ œ‘
1)
) )4d
dr sin 2
cos 2
È
therefore Surface Area 2 2 cos 2 (sin ) cos 2 d 4 cos 2 (sin ) dœœ ''
0 0
4 4
œ œœ œ
Š‹
ÈÈ É É1)) ) )1 )) )
sin 2
cos 2 cos 2
œ
)
))
"
4 sin d 4 cos 4 ( 1) 2 2 2œœœœ1))1)1 1'
0
4
œœ
cd ’“Š‹
È
1Î%
! #
È
2

670. 666 Chapter 10 Conic Sections and Polar Coordinates
32. r 2a cos 2a sin ; therefore Surface Area 2 (2a cos )(cos ) (2a cos ) ( 2a sin ) dœÊœ œ )) 1)))))
dr
d) '
0
œ
È ##
4a cos 4a cos sin d 8a cos a d 8a cos dœœœ1) )))1))1))'''
000
œœœ
ab a b ab kkÈ
#### ## #
8a d 4a (1 cos 2 ) d 4a sin 2 4aœœœœ
## ### "
# !
1)1))1))1''
00
œœ
ˆ‰ œ ‘
1cos 2
2
)
1
33. Let r f( ). Then x f( ) cos f ( ) cos f( ) sin f ( ) cos f( ) sin œœÊœÊœ) ) ) ) ))) ) )))
dx dx
dd))
ww
# #
ˆ‰ cd
f ( ) cos 2f ( ) f( ) sin cos [f( )] sin ; y f( ) sin f ( ) sin f( ) cos œ œÊœcd
w#w ## w#
) ) )))) ) ) )) ))) )
dy
d)
f ( ) sin f( ) cos f ( ) sin 2f ( )f( ) sin cos [f( )] cos . ThereforeÊœ œ Š‹ cdcd
dy
d)
#
ww#w## ##
))) ) ) ) )))) ) )
f( ) cos sin [f( )] cos sin f( ) [f( )] r .ˆ‰ ˆ‰Š‹ cdabab cd
dx dr
dd d
dy
)) )
# #
#
w## ###w ####
œ œ œ))))))))
Thus, L d r d .œœ ''
!!
ÎÎ
ʈ‰ É ˆ‰Š‹
dx dr
dd d
dy
)) )
##
#
#
))
34. (a) r a(1 cos ) d sin a
av
0
2œœœ
"
œ
#
!20 2
a
11
1'
œ
)) ) )cd
(b) r a d a a
av
0
2œœœ
""
œ#
#
!2011
1'
œ
))cd
(c) r a cos d a sin
av
2
2œœœ
""
œœ
Î#
Î#ˆ‰ ˆ ‰

œœ
'
œ
œ
œ
œ
)) )
11
1
1
cd
2a
35. r 2f( ), 2f ( ) r [2f( )] 2f ( ) Length 4[f( )] 4 f ( ) dœŸŸÊœÊœ Êœ )! ) " ) ) ) ) ) )
dr dr
dd))
w# #w
# #
#w
#
ˆ‰ cd cd É '
!
Î
2 [f( )] f ( ) d which is twice the length of the curve r f( ) for .œ œŸŸ'
!
Î
É cd))) )!)"#w
#
36. Again r 2f( ) r [2f( )] 2f ( ) Surface Area 2 [2f( ) sin ] 4[f( )] 4 f ( ) dœÊœ Ê œ ))) 1)))))
#w
# #
#w
#ˆ‰ cd cd É
dr
d
2
) '
!
Î
4 2 [f( ) sin ] [f( )] f ( ) d which is four times the area of the surface generated by revolvingœ '
!
Î
1) ) ) ) )É cd
#w
#
r f( ) about the x-axis for .œŸŸ)!)"
37. xœœ œ
22 2
33 3'' '
''
00 0
22 2
00
22


r cos d [a(1 cos )] (cos ) d a 1 3 cos 3 cos cos (cos ) d
r d [a(1 cos )] d

œœ
)) ) ) ) ) ) ) ) )
)))


ab
a 1 2 cos cos d
œœ'
0
2

ab )))
(After considerable algebra usingœœ
2 1 cos 2 1 cos 2
3
1 cos 2
a cos 3 3 1 sin (cos ) d
12 cos d'
'
0
2
0
2


’“ ˆ‰ ˆ‰ ab
œ ‘ˆ‰
))))
))


''
œœ
œ
œ
'
œ
œœ
œ
the identity cos A ‰
#
#


œ
1 cos 2A
a cos cos 2 2 cos sin cos 4 d
2 cos cos 2 d
'
'
0
2
0
2


ˆ‰
ˆ‰
15 8 4
12 3 3 12
3
)))) ))
)))
œ &
œœ
&
a;œœœ
a sin sin 2 sin sin 4
2 sin sin 2
a
36
5
œ ‘
œ ‘
ˆ‰
15 8 2 2
12 3 3 3 48
3
4
15
6
)) ) ) )
)) )
1
1


Î &
œ
*
œ
&
œ
*


y ; u a(1 cos ) du sin d ; 0 u 2a;œœ œÊœœÊœ
22
33
r sin d [a(1 cos )] (sin ) d
r d
3a''
'
00
22
0
2


ÎÎ
œ
)) ) ) )
)
1

"
œ ))))
2 u 2a 0. Therefore the centroid is y a 0d ab ˆ‰)1œÊœ Ä œœ Bßœß
2
3a
u du
33 6
05'
2a
2a
œ
&Î
11
38. r d a d a a ; x 0;''
00
œœ
####
!
)))1œœœœ œ œ œœcd
1) )) ))
)
111
22
33
2
3
r cos d a cos d
r d
aaa
asin 0''
'
00
0


ÎÎ
œ
œœœ
Î
*
cd

y . Therefore the centroid is x y 0 .œœœœœ ßœß
22
33
24
33
r sin d a sin d
r d
aaa3 3
acos a 4a 4a''
'
00
0


ÎÎ
œ
œœœ
ÎÎ
*
)) ))
)
1111 1
)cd ˆ‰œ

ab ˆ‰

671. Section 10.8 Conic Sections in Polar Coordinates 667
10.8 CONIC SECTIONS IN POLAR COORDINATES
1. r cos 5 r cos cos sin sin 5 r cos r sin 5 x y 5 3 x yˆ‰ ˆ ‰ È
))) )) œÊ œÊ œÊ œÊ
111
666
33
ÈÈ
## ##
""
10 y 3 x 10œÊœ
È
2. r cos 2 r cos cos sin sin 2 r cos r sin 2ˆ‰ ˆ ‰))) ))œÊ œÊ œ
333
444
22111
ÈÈ
##
x y 2 2 x 2 y 4 y x 2 2Ê œ Ê œ Ê œ
ÈÈ
22
##
ÈÈ È
3. r cos 3 r cos cos sin sin 3 r cos r sin 3ˆ‰ ˆ ‰))) ))œÊ œÊ œ
4441
333
3111
##
È
x y 3 x 3 y 6 y x 2 3Ê œ Ê œ Ê œ
1 33
3##
ÈÈ ÈÈ
4. r cos 4 r cos 4 r cos cos sin sin 4ˆ‰ ˆ‰ ˆ ‰ˆ‰)))) œ Ê œ Ê œ
11 11
44 44
r cos r sin 4 x y 4 2 x 2 y 8 y x 4 2Ê œÊ œÊ œÊœ
ÈÈ ÈÈ
22 22
## ##
))
ÈÈ È
5. r cos 2 r cos cos sin sin ˆ‰ ˆ ‰È
)))œ Ê
111
444
2 r cos r sin 2 x yœÊ œÊ
ÈÈ
"" ""
ÈÈ ÈÈ
22 22
))
2 x y 2 y 2 xœÊœÊœ
È

6. r cos 1 r cos cos sin sin 1ˆ‰ ˆ ‰)))œÊ œ
333
444
111
r cos r sin 1 x y 2Ê œ Ê œ
ÈÈ
22
22
))
È
yx 2ʜ
È

7. r cos 3 r cos cos sin sin 3ˆ‰ ˆ ‰)))œÊ œ
222
333
111
r cos r sin 3 x y 3Ê œ Ê œ
1
22
33
))
ÈÈ
"
##
x3y6 y x23Ê œ Ê œ
ÈÈ
È
3
3

672. 668 Chapter 10 Conic Sections and Polar Coordinates
8. r cos 2 r cos cos sin sin 2ˆ‰ ˆ ‰)))œÊ œ
111
333
r cos r sin 2 x y 2Ê œÊœ
1
22
33
))
ÈÈ
"
##
x3y4 y xÊ œÊœ
È
ÈÈ
343
33

9. 2 x 2 y 6 2 r cos 2 r sin 6 r cos sin 3 r cos cos sin sin
ÈÈ È È
Š‹ ˆ‰œÊ œÊ œÊ )) )) ))
ÈÈ
22
44##
11
3 r cos 3œÊ œ ˆ‰)
1
4
10. 3 x y 1 3 r cos r sin 1 r cos sin r cos cos sin sin
ÈÈ
Š‹ ˆ‰œ Ê œ Ê œ Ê )) ) ) ) )
È
3 1
66## #
" 11
r cosœÊ œ
""
##
ˆ‰)
1
6
11. y 5 r sin 5 r sin 5 r sin ( ) 5 r cos ( ) 5 r cos 5œÊ œÊ œÊ œÊ œÊ œ))) )) ˆ‰ ˆ‰
11
##
12. x 4 r cos 4 r cos 4 r cos ( ) 4œ Ê œ Ê œ Ê œ)))1
13. r 2(4) cos 8 cos 14. r 2(1) sin 2 sin œ œ œ œ)) ) )
15. r 2 2 sin 16. r 2 cos cos œœœ
È ˆ‰)))
"
#
17. 18.
19. 20.

673. Section 10.8 Conic Sections in Polar Coordinates 669
21. (x 6) y 36 C (6 0), a 6 22. (x 2) y 4 C ( 2 0), a 2œ Êœß œ œÊœß œ
## ##
r 12 cos is the polar equation r 4 cos is the polar equationʜ ʜ ))

23. x (y 5) 25 C ( 5), a 5 24. x (y 7) 49 C ( 7), a 7
## ##
œ Ê œ!ß œ œ Ê œ!ß œ
r 10 sin is the polar equation r 14 sin is the polar equationʜ ʜ ))

25. x 2x y 0 (x 1) y 1 26. x 16x y 0 (x 8) y 64
# # ## # # ##
œÊ œ œÊ œ
C ( 1 0), a 1 r 2 cos is C (8 0), a 8 r 16 cos is theÊœß œÊœ Êœß œÊœ ))
the polar equation polar equation

27. x y y 0 x y 28. x y y 0 x y
## # ## # ""
#
# #
œÊ œ œÊ œˆ‰ ˆ‰
4339
424
C , a r sin is the C 0 , a r sin is theÊ œ !ß œ Ê œ Ê œ ß œ Ê œˆ‰ ˆ‰
""
##
))
22 4
33 3
polar equation polar equation

674. 670 Chapter 10 Conic Sections and Polar Coordinates
29. e 1, x 2 k 2 rœœÊœÊœ œ
2(1)
1 (1) cos 1 cos
2
))
30. e 1, y 2 k 2 rœœÊœÊœ œ
2(1)
1 (1) sin 1 sin
2
))
31. e 5, y 6 k 6 rœœÊœÊœ œ
6(5)
1 5 sin 1 5 sin
30
œœ ))
32. e 2, x 4 k 4 rœœÊœÊœ œ
4(2)
1 2 cos 1 2 cos
8
))
33. e , x 1 k 1 rœœÊœÊœ œ
"
#
ˆ‰
ˆ‰
&
œ
&
œ
(1)
1 cos
1
2cos ) )
34. e , x 2 k 2 rœœÊœÊœ œ
"
œ œ44cos
(2)
1 cos
2
ˆ‰
ˆ‰
&
&
4
4
) )
35. e , x 10 k 10 rœœÊœÊœ œ
"
œ œ55sin
(10)
1 sin
10
ˆ‰
ˆ‰
&
&
5
5
) )
36. e , y 6 k 6 rœœÊœÊœ œ
"
33sin
(6)
1 sin
6
ˆ‰
ˆ‰
&
&
3
3
) )
37. r e 1, k 1 x 1œÊœœÊœ
"
1cos )

38. r e , k 6 x 6;œœ ÊœœÊœ
63
2cos 1 cos #
"
) )ˆ‰
&
œ
a 1 e ke a 1 3 a 3ab ’“ˆ‰œÊ œÊ œ
# "
#
#
3
4
a 4 ea 2ÊœÊ œ

39. r rœÊœ œ
25
10 5 cos 1 cos 1 cos œ œœ) ))
ˆ‰ ˆ‰
ˆ‰ ˆ‰
25 5
10
5
10
œ
&
œ
e , k 5 x 5; a 1 e keÊœ œÊœ œ
"
#
#
ab
a 1 a a eaÊ œÊ œÊœ Ê œ’“ˆ‰
"
## #
#
535 10 5
433

40. r r e 1, k 2 x 2œÊœÊœœÊœ
42
22 cos 1cos œœ ))

675. Section 10.8 Conic Sections in Polar Coordinates 671
41. r r rœÊœ Êœ
400 25
16 8 sin 1 sin 1 sin ) ) )
ˆ‰
ˆ‰ ˆ‰
400
16
8
16

"
e , k 50 y 50; a 1 e keœœÊœ œ
"
#
#
ab
a 1 25 a 25 aÊ œÊ œÊœ’“ˆ‰
"
#
#
3 100
43
eaʜ
50
3

42. r r e 1, 43. r r e 1,œÊœÊœ œÊœÊœ
12 4 8 4
33 sin 1sin 22 sin 1sin )) ))
k4 y4 k4 y 4œÊœ œÊœ

44. r r e , k 4œÊœ Êœœ
42
2sin 1 sin #
"
) )ˆ‰

"
y 4; a 1 e ke a 1 2Êœ œÊ œab ’“ˆ‰
# "
#
#
a2 a eaʜʜʜ
384
433

45. 46.
47. 48.

676. 672 Chapter 10 Conic Sections and Polar Coordinates
49. 50.
51. 52.
53. 54.
55. 56.
57. (a) Perihelion a ae a(1 e), Aphelion ea a a(1 e)œœ œœ
(b) Planet Perihelion Aphelion
Mercury 0.3075 AU 0.4667 AU
Venus 0.7184 AU 0.7282 AU
Earth 0.9833 AU 1.0167 AU
Mars 1.3817 AU 1.6663 AU
Jupiter 4.9512 AU 5.4548 AU
Saturn 9.0210 AU 10.0570 AU
Uranus 18.2977 AU 20.0623 AU
Neptune 29.8135 AU 30.3065 AU
Pluto 29.6549 AU 49.2251 AU

677. Section 10.8 Conic Sections in Polar Coordinates 673
58. Mercury: rœœ
(0.3871) 1 0.2056
1 0.2056 cos 1 0.2056 cos
0.3707ab

"
))
Venus: rœœ
(0.7233) 1 0.0068
1 0.0068 cos 1 0.0068 cos
0.7233ab

"
))
Earth: rœœ
1 1 0.0167
1 0.0167 cos 1 0.0617 cos
0.9997ab

"
))
Mars: rœœ
(1.524) 1 0.0934
1 0.0934 cos 1 0.0934 cos
1.511ab

"
))
Jupiter: rœœ
(5.203) 1 0.0484
1 0.0484 cos 1 0.0484 cos
5.191ab

"
))
Saturn: rœœ
(9.539) 1 0.0543
1 0.0543 cos 1 0.0543 cos
9.511ab

"
))
Uranus: rœœ
(19.18) 1 0.0460
1 0.0460 cos 1 0.0460 cos
19.14ab

"
))
Neptune: rœœ
(30.06) 1 0.0082
1 0.0082 cos 1 0.0082 cos
30.06ab

"
))
59. (a) r 4 sin r 4r sin x y 4y;œÊœ Ê "œ))
###
r 3 sec r r cos 3œÊœÊœ
ÈÈ
))
È
3
cos )
x3; x3 3y4yÊœ œ Ê "œ
ÈÈ È
Š‹
#
#
y 4y 3 0 (y 3)(y 1) 0 y 3Ê œÊ œÊœ
#
or y 1. Therefore in Cartesian coordinates, the pointsœ
of intersection are 3 3 and 3 1 . In polarŠ‹Š‹
ÈÈ
ßß
coordinates, 4 sin 3 sec 4 sin cos 3))))œÊ œ
ÈÈ
2 sin cos sin 2 2 orʜʜʜ)) ) )
ÈÈ
33
3##
1
or ; r 2, and
2
3636 3
1111 1
Êœ œ Êœ œ)) )
r 2 3 2 and 2 3 are the pointsÊœ Ê ß ß
ÈÈ ˆ‰ Š‹
11
63
of intersection in polar coordinates.
(b)
60. (a) r 8 cos r 8r cos x y 8xœÊœ Ê "œ))
###
x 8x y 0 (x 4) y 16;Ê œÊ œ
## ##
r 2 sec r r cos 2œÊœÊœ))
2
cos )
x2; x2 2 8(2)y 0Êœ œÊ œ
##
y 12 y 2 3. Therefore 2 2 3ʜʜ„ ß„
# ÈÈ
Š‹
are the points of intersection in Cartesian coordinates.
In polar coordinates, 8 cos 2 sec 8 cos 2)) )œÊ œ
#
cos cos , , , orʜʜ„Êœ
# ""
#
)))
4333
24111
; and r 4, and and
55 24
333 33
111 11
))œÊœœ
r 4 4 and 4 are the points ofÊœ Êß ßˆ‰ˆ‰
11
33
5
(b)
intersection in polar coordinates. The points 4 and 4 are the same points.ˆ‰ˆ‰ß ß
24
33
11
61. r cos 4 x 4 k 4: parabola e 1 r)œÊœÊœ ʜʜ
4
1cos )
62. r cos 2 r cos cos sin sin 2 r sin 2 y 2 k 2: parabola e 1ˆ‰ ˆ ‰))))œÊ " œÊ œÊœÊœ Êœ
111
###
rʜ
2
1sin )

678. 674 Chapter 10 Conic Sections and Polar Coordinates
63. (a) Let the ellipse be the orbit, with the Sun at one focus.
Then r a c and r a c max min
rr
rrœ" œÊ
max min
max min

e œœœœ
(a c) (a c)
(a c) (a c) 2a a
2c c

(b) Let F , F be the foci. Then PF PF 10 where"# " # " œ
P is any point on the ellipse. If P is a vertex, then
PF a c and PF a c
"#œ" œ
(a c) (a c) 10Ê œ
2a 10 a 5. Since e we have 0.2ʜʜ œ œ
cc
a5
c 1.0 the pins should be 2 inches apart.Êœ Ê

64. e 0.97, Major axis 36.18 AU a 18.09, Minor axis 9.12 AU b 4.56 (1 AU 1.49 10 km)œœÊœ œÊœ¸‚
)
(a) r AUœœœ œ
ke 1.07
1e cos 1e cos 10.97 cos 10.97 cos
a1 e (18.09) 1 (0.97)


)) ) )
ab cd
" "
(b) 0 r 0.5431 AU 8.09 10 km)œÊœ ¸ ¸ ‚
1.07
10.97
(
(c) r 35.7 AU 5.32 10 km)1œÊœ ¸ ¸ ‚
1.07
10.97
*
65. x y 2ay 0 (r cos ) (r sin ) 2ar sin 0
## # #
œÊ œ)) )
r cos r sin 2ar sin 0 r 2ar sin Ê œÊ œ
## ## #
)) ) )
r 2a sin ʜ )

66. y 4ax 4a (r sin ) 4ar cos 4a r sin
## # ###
œ " Êœ " Ê)) )
4ar cos 4a r 1 cos 4ar cos 4aœ " Ê œ ")))
## # #
ab
r r cos 4ar cos 4a rÊ œ " Ê
### # #
))
r cos 4ar cos 4a r (r cos 2a)œ "" Êœ "
## # # #
)) )
r (r cos 2a) r r cos 2a orÊœ„ " Ê œ))
r r cos 2a r or r ;" œÊœ œ)
2a 2a
1cos 1cos

))
the equations have the same graph, which is a parabola
opening to the right

67. x cos y sin p r cos cos r sin sin p!! )!)!" œÊ " œ
r(cos cos sin sin ) p r cos ( ) pÊ " œÊ œ)! )! )!

68. x y 2ax x y a y 0ab ab
## ## ##
#
œ
r 2a(r cos ) r a (r sin ) 0Ê œab ab
####
#
))
r 2ar cos a r sin 0Ê œ
%$ ###
))
r r 2ar cos a 1 cos 0 (assume r 0)Ê œÁ
## # #
cd ab))
r 2ar cos a a cos 0Ê œ
####
))
r 2ar cos a cos a 0Ê œab
####
))
(r a cos ) a r a cos aÊ" œÊ " œ„))
##
r a(1 cos ) or r a(1 cos );Êœ œ))

679. Chapter 10 Practice Exercises 675
the equations have the same graph, which is a cardioid
69 - 70. Example CAS commands:
:Maple
with( plots );#69
f := (r,k,e) -> k*e/(1+e*cos(theta));
elist := [3/4,1,5/4]; # (a)
P1 := seq( plot( f(r,-2,e), theta=-Pi..Pi, coords=polar ), e=elist ):
display( [P1], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)k=-2" );
P2 := seq( plot( f(r,2,e), theta=-Pi..Pi, coords=polar ), e=elist ):
display( [P2], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)k=2" );
elist2 := [7/6,5/4,4/3,3/2,2,3,5,10,20]; # (b)
P3 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist2 ):
display( [P3], insequence=true, view=[-20..20,-20..20], title="#69(b) (Section 10.8)k=-1, e>1" );
elist3 := [1/2,1/3,1/4,1/10,1/20];
P4 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist3 ):
display( [P4], insequence=true, title="#69(b) (Section 10.8)k=-1, e<1" );
klist := -5..-1; # (c)
P5 := seq( plot( f(r,k,1/2), theta=-Pi..Pi, coords=polar ), k=klist ):
display( [P5], insequence=true, title="#69(c) (Section 10.8)e=1/2, k<0" );
P6 := seq( plot( f(r,k,1), theta=-Pi..Pi, coords=polar ), k=klist ):
display( [P6], insequence=true, view=[-4..50,-50..50], title="#69(c) (Section 10.8)e=1, k<0" );
P7 := seq( plot( f(r,k,2), theta=-Pi..Pi, coords=polar ), k=klist ):
display( [P5], insequence=true, title="#69(c) (Section 10.8)e=2, k<0" );
: (assigned function and values for parameters and bounds may vary):Mathematica
To do polar plots in Mathematica, it is necessary to first load a graphics package
In the command, it is assumed that the variable r is given as a function of the variable .PolarPlot )
<<Graphics`Graphics`
f[ _, k_,ec_]:= (k ec) / (1 ec Cos[ ]))) "
PolarPlot[{ f[ , 2, 3/4], f[ , 2, 1], f[ , 2, 5/4]}, { , 0, 2 }, PlotRange { 20, 20},))))1 Ä
PlotStyle {RGBColor[1, 0, 0], RGBColor[0, 1, 0], RGBColor[0, 0, 1]}];Ä
PolarPlot[{ f[ , 1, 1], f[ , 2, 1], f[ , 3, 1], f[ , 4, 1], f[ , 5, 1]}, { , 0, 2 }, PlotRange { 20, 20},))))) )1 Ä
PlotStyle Ä
{RGBColor[1, 0, 0], RGBColor[0, 1, 0], RGBColor[0, 0, 1, RGBColor[.5, .5, 0], RGBColor[0, .5, .5]}];ß
The limitation on the range is primarily needed when plotting hyperbolas.
Problem 70 can be done in a similar fashion.
CHAPTER PRACTICE EXERCISES"!
1. x 4y y 4p 4 p 1; 2. x 2y y 4p 2 p ;
# #
##
"
œÊœ Ê œÊœ œ Ê œÊ œÊœ
xx
4
" "
therefore Focus is (0 1), Directrix is y 1 therefore Focus is ; Directrix is yß œ!ßœ ˆ‰
""
##

680. 676 Chapter 10 Conic Sections and Polar Coordinates
3. y 3x x 4p 3 p ; 4. y x x 4p p ;
# #
œÊœÊœÊœ œ Êœ ʜʜ
y y
34 3 33
38 82
" "
ˆ‰
8
3
therefore Focus is 0 , Directrix is x therefore Focus is , Directrix is xˆ‰ ˆ ‰
33 22
44 33
ßœ ß! œ

5. 16x 7y 112 1 6. x 2y 4 1 c 4 2 2
## ## #
#
" œÊ "œ " œÊ "œÊ œ œ
xx
716 4
yy
" "" "
c 16 7 9 c 3; e c 2 ; eÊœ œÊœ œœ Êœ œœ
#
#
c3 c
a4 a
2 È
È

7. 3x y 3 x 1 c 1 3 4 8. 5y 4x 20 1 c 4 5 9
## # # # # #
œÊ œÊ œ "œ œÊ œÊ œ "œ
y y
3 45
x
" " "
c 2; e 2; the asymptotes are c 3, e ; the asymptotes are y xÊœ œœœ Êœ œœ œ„
c2 c3 2
a1 a 5# È
y3xœ„
È

9. x 12y y 4p 12 p 3 focus is ( 3), directrix is y 3, vertex is (0 0); therefore new
#
#
œ ÊœÊ œ ÊœÊ !ß œß
x
1
"
vertex is (2 3), new focus is (2 0), new directrix is y 6, and the new equation is (x 2) 12(y 3)ßß œ œ
#
10. y 10x x 4p 10 p focus is 0 , directrix is x , vertex is (0 0); therefore new
#
## #
œÊœÊœÊœÊ ß œ ß
y
10
55 5
"
ˆ‰
vertex is 1 , new focus is (2 1), new directrix is x 3, and the new equation is (y 1) 10 xˆ‰ ˆ‰ß ß œ œ "
" "
# #
#
11. 1 a 5 and b 3 c 25 9 4 foci are 4 , vertices are 5 , center is
x
95
y
" "
"œÊœ œÊœ œÊ !ß„ !ß„
#
È
ab ab
(0 0); therefore the new center is ( 5), new foci are ( 3 1) and ( 3 9), new vertices are ( 10) andß $ß ß ß $ß
( 0), and the new equation is 1$ß " œ
(x 3) (y 5)
95

#
""
12. 1 a 13 and b 12 c 169 144 5 foci are 5 0 , vertices are 13 0 , center
x
169 144
y
" "
" œÊœ œ Êœ œÊ „ß „ß
È
ab a b
is (0 0); therefore the new center is (5 12), new foci are (10 12) and (0 12), new vertices are (18 12) andßßßßß

681. Chapter 10 Practice Exercises 677
( 8 12), and the new equation is 1ß " œ
(x 5) (y 12)
169 144

""
13. 1 a 2 2 and b 2 c 8 2 10 foci are 0 10 , vertices are
y
82
x
" "
œÊœ œ Êœ "œÊ ß„
ÈÈ È ÈÈ
Š‹
0 2 2 , center is (0 0), and the asymptotes are y 2x; therefore the new center is 2 2 2 , new foci areŠ‹ Š‹
È È
ß„ ß œ„ ß
2 2 2 10 , new vertices are 2 4 2 and ( 0), the new asymptotes are y 2x 4 2 2 andŠ‹ Š‹
ÈÈ ÈÈ
ß„ ß #ß œ
y 2x 4 2 2; the new equation is 1œ œ
È
Š‹
È
y22
8
(x 2)


#
"
"
14. 1 a 6 and b 8 c 36 64 10 foci are 10 0 , vertices are 6 0 , the center
x
36 64
y
" "
œÊœ œÊœ "œÊ „ß „ßÈ ab ab
is (0 0) and the asymptotes are or y x; therefore the new center is ( 10 3), the new foci areßœ„œ„ ß
y
86 3
x4
( 20 3) and (0 3), the new vertices are ( 16 3) and ( 4 3), the new asymptotes are y x andß ß ß ß œ "
431
33
y x ; the new equation is 1œ œ
449
3 3 36 64
(x 10) (y 3)
""
15. x 4x 4y 0 x 4x 4 4y 4 (x 2) 4y 4 y 1, a hyperbola; a 2 and
### # ## #
œÊ œÊ œÊ œœ
(x 2)
4
"
b 1 c 1 4 5 ; the center is (2 0), the vertices are ( 0) and (4 0); the foci are 2 5 0 andœÊœ "œ ß !ß ß „ ßÈ ÈÈ
Š‹
the asymptotes are yœ„
x2
#
16. 4x y 4y 8 4x y 4y 4 4 4x (y 2) 4 x 1, a hyperbola; a 1 and
## ## # # #
œÊ œÊ œÊ œœ
(y 2)
4
"
b 2 c 1 4 5 ; the center is ( 2), the vertices are (1 2) and ( 2), the foci are 5 2 andœÊœ "œ!ß ß "ß „ ßÈ ÈÈ
Š‹
the asymptotes are y 2x 2œ„ "
17. y 2y 16x 49 y 2y 1 16x 48 (y 1) 16(x 3), a parabola; the vertex is ( 1);
## #
œÊ œ Ê œ $ß
4p 16 p 4 the focus is ( 7 1) and the directrix is x 1œÊœÊ ßœ
18. x 2x 8y 17 x 2x 1 8y 16 (x 1) 8(y 2), a parabola; the vertex is (1 2);
## #
œÊ œ Ê œ ß
4p 8 p 2 the focus is (1 4) and the directrix is y 0œÊœÊ ß œ
19. 9x 16y 54x 64y 1 9 x 6x 16 y 4y 1 9 x 6x 9 16 y 4y 4 144
## # # # #
œÊ œÊ œabab a ba b
9(x 3) 16(y 2) 144 1, an ellipse; the center is ( 3 2); a 4 and b 3Ê œÊ " œ ßœ œ
## (x 3) (y 2)
16 9
""
c 16 9 7 ; the foci are 7 2 ; the vertices are (1 2) and ( 7 2)Êœ œ $„ ß ß ß
È ÈÈ
Š‹
20. 25x 9y 100x 54y 44 25 x 4x 9 y 6y 44 25 x 4x 4 9 y 6y 9 225
## # # # #
œÊ œÊ œabab a ba b
1, an ellipse; the center is (2 3); a 5 and b 3 c 25 9 4; the foci areÊ " œß œœÊœ œ
(x 2) (y 3)
925

""
È
(2 1) and (2 7); the vertices are (2 2) and (2 8)ßß ßß
21. x y 2x 2y 0 x 2x 1 y 2y 1 2 (x 1) (y 1) 2, a circle with center (1 1) and
## # # # #
œÊ œÊ œß
radius 2œ
È
22. x y 4x 2y 1 x 4x 4 y 2y 1 6 (x 2) (y 1) 6, a circle with center ( 2 1)
## # # # #
""" œÊ """"" œÊ """ œ ß
and radius 6œ
È
23. B 4AC 1 4(1)(1) 3 0 ellipse 24. B 4AC 4 4(1)(4) 0 parabola
# ##
œ œ Ê œ œÊ
25. B 4AC 3 4(1)(2) 1 0 hyperbola 26. B 4AC 2 4(1)( 2) 12 0 hyperbola
## ##
œ œÊ œ œÊ

682. 678 Chapter 10 Conic Sections and Polar Coordinates
27. x 2xy y 0 (x y) 0 x y 0 or y x, a straight line
## #
œÊ œÊ œœ
28. B 4AC ( 3) 4(1)(4) 7 0 ellipse
##
œ œ Ê
29. B 4AC 1 4(2)(2) 15 0 ellipse; cot 2 0 2 ; x x y and
## ww
###
œ œ ÊœœÊœÊœœ !!!
AC
B4
2211
ÈÈ
y x y 2 x y x y x y 2 x y 15 0œ " Ê œ
ÈÈ ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22 22
## ## ## ## ##
ww ww ww ww ww
# #
Š‹Š‹Š‹Š‹
5x 3y 30Ê " œ
ww##
30. B 4AC 2 4(3)(3) 32 0 ellipse; cot 2 0 2 ; x x y and
## ww
###
œ œ ÊœœÊœÊœœ !!!
AC
B4
2211
ÈÈ
yxy 3xy2 xy xy3xy19œ " Ê œ
ÈÈ ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22 22
## ## ## ## ##
ww ww ww ww ww
# #
Š‹Š‹Š‹Š‹
4x 2y 19Ê " œ
ww##
31. B 4AC 2 3 4(1)( 1) 16 hyperbola; cot 2 2 ; x x y
# ww
#
"
##
œ œÊ œ œ Ê œÊœ œ Š‹
È
!!!
AC 1
B36 3
3
È
È
11
and yx y xy 23 xy x y x y 4œ " Ê œ
11111 33 3 3 3
## ## #### ##
ww ww wwww ww
# #
ÈÈ È È È
Š‹Š‹Š‹Š‹
È
2x 2y 4 y x 2Ê œÊ œ
ww ww## ##
32. B 4AC ( 3) 4(1)(1) 5 0 hyperbola; cot 2 0 2 ; x x y
## ww
###
œ œÊœœÊœÊœœ !!!
AC
B4
2211
ÈÈ
and yxy xy3 xy xy xy5œ " Ê œ
ÈÈ ÈÈ ÈÈ ÈÈ ÈÈ
22 22 22 22 22
## ## ## ## ##
ww ww ww ww ww
# #
Š‹Š‹Š‹Š‹
y x 5 or 5y x 10Ê œ œ
5
##
ww ww## ##"
33. x tan t and y sec t x tan t 34. x 2 cos t and y 2 sin t x 4 cos t andœœÊœ œ œÊœ
"" "
##
## ##
4
and y sec t 4x tan t and y 4 sin t x y 4
#### ####"
œÊœ œÊ "œ
4
4y sec t 4x 1 4y 4y 4x 1
## # # ##
œÊ "œÊ œ

35. x cos t and y cos t y ( x) x 35. x 4 cos t and y 9 sin t x 16 cos t andœ œÊœ œœœÊœ
### ##
y 81 sin t 1
##
œÊ "œ
x
16 81
y
" "

683. Chapter 10 Practice Exercises 679
37. 38.

39. d 40. e 41. l 42. f
43. k 44. h 45. i 46. j
47. r sin and r 1 sin sin 1 sin 0 1œœ " Êœ " Êœ))))
so no solutions exist. There are no points of intersection
found by solving the system. The point of intersection
(0 0) is found by graphing.ß

48. r cos and r 1 cos cos 1 cos œœ Êœ ))))
cos , ; r ; ʜʜ œÊœ œ )) ) )
""
##
11 1 1
33 3 3
r . The points of intersection are andÊœ ß
""
##
ˆ‰
1
3
. The point of intersection (0 0) is foundˆ‰
"
#
ß ß
1
3
by graphing.

49. r 1 cos and r 1 cos 1 cos 1 cos œ" œ Ê" œ))))
2 cos 0 cos 0 , ; or ʜʜʜ œ))))
11 1 1
## # #
33
r 1. The points of intersection are 1 and 1 .Êœ ß ß ˆ‰ˆ‰
11
##
3
The point of intersection (0 0) is found by graphing.ß

684. 680 Chapter 10 Conic Sections and Polar Coordinates
50. r 1 sin and r 1 sin 1 sin 1 sin œ œ Ê œ))))
2 sin 0 sin 0 0, ; 0 or ʜʜʜœ)))1)1
r 1. The points of intersection are (1 0) and (1 ).Êœ ß ß 1
The point of intersection ( 0) is found by graphing.!ß

51. r 1 sin and r 1 sin intersect at all points ofœ œ))
r 1 sin because the graphs coincide. This can beœ )
seen by graphing them.

52. r 1 cos and r 1 cos intersect at all points ofœ œ))
r 1 cos because the graphs coincide. This can beœ )
seen by graphing them.

53. r sec and r 2 sin sec 2 sin œœÊœ))))
1 2 sin cos 1 sin 2 2 Êœ Êœ Ê œ Êœ)) ) ) )
11
# 4
r 2 sin 2 the point of intersection isÊœ œ Ê
1
4È
2 . No other points of intersection exist.Š‹
È
ß
1
4

54. r 2 csc and r 4 cos 2 csc 4 cos œ œ Ê œ))))
1 2 sin cos 1 sin 2 2 , Êœ Êœ Ê œ)) ) )
11
##
5
, ; r 4 cos 2 2 ;Ê œ œ Ê œ œ))
11 1 1
44 4 4
5 È
r 4 cos 2 2 . The point of)œÊœ œ
55
44
11 È
intersection is 2 2 and the point 2 2 is theŠ‹ Š ‹
ÈÈ
$!$
5
44
11
same point.

685. Chapter 10 Practice Exercises 681
55. r cos 2 3 r cos cos sin sin ˆ‰ ˆ ‰ È
)))œ Ê
111
333
2 3 r cos r sin 2 3œÊ œ
ÈÈ
"
##
))
È
3
r cos 3 r sin 4 3 x 3 y 4 3Ê œÊœ))
ÈÈÈÈ
yx4ʜ
È
3
3

56. r cos r cos cos sin sin ˆ‰ ˆ ‰)))œ Ê
333
444
2111
È
#
r cos r sin x y 1œÊ œÊœ
ÈÈ È È
22 2 2
## # #
))
yx1ʜ

57. r 2 sec r r cos 2 x 2œÊœÊœÊœ))
2
cos )

58. r 2 sec r cos 2 x 2œ Ê œ Ê œ
ÈÈÈ
))

59. r csc r sin yœ Ê œ Ê œ
333
###
))

60. r 3 3 csc r sin 3 3 y 3 3œÊœÊœ
ÈÈÈ
))

686. 682 Chapter 10 Conic Sections and Polar Coordinates
61. r 4 sin r 4r sin x y 4y 0œ Ê œ Ê œ))
###
x (y 2) 4; circle with center ( 2) andÊ œ !ß
##
radius 2.

62. r 3 3 sin r 3 3 r sin œÊœ
ÈÈ
))
#
xy33y0 x y ;Ê œÊ œ
## #
#
# È
Š‹
33 27
4
È
circle with center and radius Š‹!ß
33 33
ÈÈ
##

63. r 2 2 cos r 2 2 r cos œÊœ
ÈÈ
))
#
xy22x0 x 2 y2;Ê œÊ œ
## #
# ÈÈ
Š‹
circle with center 2 0 and radius 2Š‹
ÈÈ
ß

64. r 6 cos r 6r cos x y 6x 0œ Ê œ Ê œ))
###
(x 3) y 9; circle with center ( 3 0) andÊœ ß
##
radius 3

65. x y 5y 0 x y C
## #
##
#
œÊ œ Ê œ!߈‰ ˆ ‰
525 5
4
and a ; r 5r sin 0 r 5 sin œ œÊœ
5
#
#
))

66. x y 2y 0 x (y 1) 1 C ( 1) and
## # #
œÊ œÊ œ!ß
a 1; r 2r sin 0 r 2 sin œ œÊœ
#
))

687. Chapter 10 Practice Exercises 683
67. x y 3x 0 x y C
## #
##
#
œÊ œÊ œß!ˆ‰ ˆ‰
39 3
4
and a ; r 3r cos 0 r 3 cos œ œÊœ
3
#
#
))

68. x y 4x 0 (x 2) y 4 C ( 2 0)
## ##
œÊ œÊ œß
and a 2; r 4r cos 0 r 4 cos œ œÊœ
#
))

69. r e 1 parabola with vertex at (1 0)œÊœÊ ß
2
1cos œ )

70. r r e ellipse;œÊœ ÊœÊ
84
2cos 1 cos œ# œ
"
) )ˆ‰
œ

ke 4 k 4 k 8; k ea 8 aœÊ œÊœ œÊœ
""
##
aa
e ˆ‰
œ

a ea ; therefore the center isÊœ Ê œ œ
16 16 8
333
ˆ‰ˆ ‰
"
#
; vertices are ( ) and 0ˆ‰ ˆ‰
88
33
ß)ßß11

71. r e 2 hyperbola; ke 6 2k 6œ ÊœÊ œÊ œ
6
12 cos )
k 3 vertices are (2 ) and (6 )ÊœÊ ß ß 11

72. r r e ; ke 4œÊœ Êœœ
12 4
3sin 3 1 sin œ œ
"
) )ˆ‰
œ
3
k 4 k 12; a 1 e 4 a 1ʜʜ œÊ
"" #
#
33
ab ’“ˆ‰
4 a ea ; therefore theœÊœÊ œ œ
993
3###
"
ˆ‰ˆ‰
center is ; vertices are 3 and 6ˆ ‰ ˆ‰ˆ‰
33 3
## # #
ßßß
111

73. e 2 and r cos 2 x 2 is directrix k 2; the conic is a hyperbola; r rœ œÊœ Êœ œ Êœ)
ke
1 e cos 1 cos
(2)(2)
œœ#))
rʜ
4
1 cos œ# )

688. 684 Chapter 10 Conic Sections and Polar Coordinates
74. e 1 and r cos 4 x 4 is directrix k 4; the conic is a parabola; r rœœÊœÊœ œÊœ)
ke
1e cos 1cos
(4)(1)
))
rʜ
4
1cos )
75. e and r sin 2 y 2 is directrix k 2; the conic is an ellipse; r rœ œÊœ Êœ œ Êœ
"
# œ œ
)
ke
1e sin
(2)
1 sin ) )
ˆ‰
ˆ‰
œ

œ

rʜ
2
2sin œ)
76. e and r sin 6 y 6 is directrix k 6; the conic is an ellipse; r rœœÊœÊœ œÊœ
"
3 1e sin
ke (6)
1 sin
)
) )
ˆ‰
ˆ‰
œ
œ
3
3
rʜ
6
3sin )
77. A 2 r d (2 cos ) d 4 4 cos cos d 4 4 cos dœœœœ'' ' '
00 0 0
œœ œ œ
" œ
# #
## #
))) ))) )) ab ˆ‰
1cos 2)
4 cos d 4 sin œ œ œ'
0
œ
ˆ‰ œ ‘
9 cos 2 9 sin 2 9
24## # !
)))) 1
))
1
78. A sin 3 d d sin 6œœœœ''
00
33
œœœœ
"""
##
#
Î$
!
ab ˆ‰ œ ‘)) ) ) )
1cos 6
46 12
)1
1
79. r 1 cos 2 and r 1 1 1 cos 2 0 cos 2 2 ; thereforeœ œ Ê œ Ê œ Ê œ Ê œ)))))
11
# 4
A 4 (1 cos 2 ) 1 d 2 1 2 cos 2 cos 2 1 dœœ''
00
44
œœœœ
"
#
## #
cd ab)) )))
2 2 cos 2 d 2 sin 2 2 1 0 2œœœœ'
0
4
œœ
ˆ‰ œ ‘ˆ ‰))))
""
##
Î%
!
cos 4 sin 4
28 8 4
))11
1
80. The circle lies interior to the cardioid (see the graphs in Exercises 61 and 63). Thus,
A 2 [2(1 sin )] d (the integral is the area of the cardioid minus the area of the circle)œ '

œ
œ
œ2
2
"
#
#
))1
4 1 2 sin sin d (6 8 sin 2 cos 2 ) d 6 8 cos sin 2œ œ œ
''

œœ
œœ
œœ22
22
ab cd) ))1 ) ))1 ) ) ) 1
# Î#
Î#
1
1
3(3) 5œœcd1 111
81. r 1 cos sin ; Length ( 1 cos ) ( sin ) d 2 2 cos dœ Ê œ œ œ )) ))) ))
dr
d) ''
00
22
œœ
È È ##
d 2 sin d 4 cos ( 4)( 1) ( 4)(1) 8œœœœœ''
00
22
œœ
É œ ‘
4(1 cos )
2

##
#
!
) ))
1
))
82. r 2 sin 2 cos , 0 2 cos 2 sin ; r (2 sin 2 cos ) (2 cos 2 sin )œ ŸŸÊœ œ ))) )) )) ))
1
))#
###
#
dr dr
dd
ˆ‰
8sin cos 8 L 8 d 2 2 2 2 2œœÊœ œ œ œab
È
’“
ÈÈÈ ˆ‰
##
Î#
!
#
)) ) ) 1 '
0
2
œœ 1
1
83. r 8 sin , 0 8 sin cos ; r 8 sin 8 sin cos œŸŸÊœ œ
$##$#
## #ˆ‰ ˆ‰ ˆ‰ ˆ ‰ œ ‘œ ‘ˆ‰ ˆ‰ ˆ‰
)1 )) ) ))
))34d33d 3 33
dr dr
)
64 sin L 64 sin d 8 sin d 8 dœÊœ œ œ
%# %

#ˆ‰ ˆ‰ ˆ‰ É ’“
)))
333
1cos '''
000
444
œœœœœœ
)) )
ˆ‰
2
3œ
4 4 cos d 4 6 sin 4 6 sin 0 3œ œ œ œ'
0
4
œœ
œ ‘ œ ‘ˆ‰ˆ‰ˆ‰ ˆ‰
22
3346
))11
1
)) 1
Î%
!
84. r 1 cos 2 (1 cos 2 ) ( 2 sin 2 ) œ Êœ œ Ê œÈ ˆ‰)))
dr sin 2 dr sin 2
dd1cos 2 1cos 2)))
))
)
"
# œ
"Î#
œ
#
È

r1cos 2Ê œ œ œ
#
#
œœ œ
œœ œœœˆ‰
dr sin 2 1 2 cos 2 cos 2 sin 2
d 1 cos 2 1 cos 2 1 cos 2
(1 cos 2 ) sin 2
))))
)))) ))
)

2 L 2 d 2 2œœÊœ œœ
22 cos 2
1cos 2
œ
œ##
)11
) '

œ
œ
œ2
2
ÈÈ È œ ‘ˆ‰)1

689. Chapter 10 Practice Exercises 685
85. r cos 2 ; Surface Area 2 (r sin ) r dœÊœ œ
È Éˆ‰)1))
dr sin 2 dr
dd cos 2))
)
)
#
#
È '
0
4
œœ
2 cos 2 (sin ) cos 2 d 2 cos 2 (sin ) d 2 sin dœœ œ'''
0 00
4 44
œ œœœ œœ
1)) ) ) 1)) ) 1))
ÈÈ É É
sin 2
cos 2 cos 2

)
))
"
2( cos ) 2 1 2 2œ œ œcd Š‹Š ‹
È
1) 1 1
1Î%
! #
È
2
86. r sin 2 2r 2 cos 2 r cos 2 ; Surface Area 2 2 (r cos ) r d
# #
#
œÊœ Êœ œ ))) 1))
dr dr dr
dd d)) ) '
0
2
œœ
Ɉ‰
2 2 (cos ) r r d 2 2 (cos ) (sin 2 ) (cos 2 ) d 2 2 cos dœœ œ'' '
00 0
22 2
œœ œœœ œ
1) ) 1) ) )) 1))Ɉ‰ È
%##
#
dr
dt
2 2 sin 4œœcd1) 1
1Î#
!
87. (a) Around the x-axis: 9x 4y 36 y 9 x y 9 x and we use the positive root:
## # # #
œÊœ Êœ„
99
44
É
V 2 9 x dx 2 9 x dx 2 9x x 24œœœœ ''
00
22
1111Š‹É ˆ‰ œ ‘
993
444
#
#
#$
#
!
(b) Around the y-axis: 9x 4y 36 x 4 y x 4 y and we use the positive root:
## # # #
œÊœ Êœ„
44
99
É
V 2 4 y dy 2 4 y dy 2 4y y 16œœœœ ''
00
33
1111Š‹É ˆ‰ œ ‘
444
9927
#
#
#$
$
!
88. 9x 4y 36, x 4 y y x 4 ; V x 4 dx x 4 dx
## # #
##
##
#
œ œÊœ Êœ œ œ
9x 36 3 3 9
4 4

ÈÈ
Š‹ ab ''
22
44
1
1
4x 16 8 (32) 24œœ œœ œ
9x 9 64 8 956 24 3
43 4 3 3 43 3 4
11 11
’“ œ ‘ˆ ‰ˆ‰ˆ‰
!
%
#
1
89. Each portion of the wave front reflects to the other focus, and since the wave front travels at a constant speed
as it expands, the different portions of the wave arrive at the second focus simultaneously, from all directions,
causing a spurt at the second focus.
90. The velocity of the signals is v 980 ft/ms. Let t be theœ
"
time it takes for the signal to go from A to S. Then
d 980t and d 980(t 1400)
""# "œœ
d d 980(1400) 1.372 10 ft or 259.8 miles.Êœ œ ‚
#"
'
The ship is 259.8 miles closer to A than to B.
The difference of the distances is always constant (259.8
miles) so the ship is traveling along a branch of a hyperbola
with foci at the two towers. The branch is the one having
tower A as its focus.

91. The time for the bullet to hit the target remains constant, say t t . Let the time it takes for sound toœ
!
travel from the target to the listener be t . Since the listener hears the sounds simultaneously, t t t#"!# œ
where t is the time for the sound to travel from the rifle to the listener. If v is the velocity of sound, then
"
vt vt vt or vt vt vt . Now vt is the distance from the rifle to the listener and vt is the distance"!#"#! " #œ œ
from the target to the listener. Therefore the difference of the distances is constant since vt is constant so
!
the listener is on a branch of a hyperbola with foci at the rifle and the target. The branch is the one with the
target as focus.
92. Let (r ) be a point on the graph where r a . Let (r ) be on the graph where r a and
"" " " ## # #ߜߜ))))
2 . Then r and r lie on the same ray on consecutive turns of the spiral and the distance between)) 1
#" " #œ
the two points is r r a a a( ) 2 a, which is constant.
#" # " # "œœ œ)) )) 1

690. 686 Chapter 10 Conic Sections and Polar Coordinates
93. (a) r r er cos k x y ex k x y k ex x yœ Ê œÊœÊœÊ
k
1e cos œ
## ## ##
)
) ÈÈ
k 2kex e x x e x y 2kex k 0 1 e x y 2kex k 0œÊœÊ œ
######## ####
ab
(b) e 0 x y k 0 x y k circle;œÊ œÊ œ Ê
### ## #
0 e 1 e 1 e 1 0 B 4AC 0 4 1 e (1) 4 e 1 0 ellipse;Ê Ê Ê œ œ Ê
## # # # #
ab ab
e 1 B 4AC 0 4(0)(1) 0 parabola;œÊ œ œÊ
##
e 1 e 1 B 4AC 0 4 1 e (1) 4e 4 0 hyperbolaÊ Ê œ œ Ê
## ###
ab
94. (a) The length of the major axis is 300 miles 8000 miles 1000 miles 2a a 4650 miles. If theœÊœ
center of the earth is one focus and the distance from the center of the earth to the satellite's low point is
4300 miles (half the diameter plus the distance above the North Pole), then the distance from the center
of the ellipse to the focus (center of the earth) is 4650 miles 4300 miles 350 miles c. Thereforeœœ
e. œœ œ
c 350 miles 7
a 4650 miles 93
(b) r r mileœÊœ œ
a1 e
1 e cos 93 7 cos
4650 1
1 cos
430,000ab
’“ˆ‰
ˆ‰

œœ

œ


)) )
7
93
7
93
CHAPTER 10 ADDITIONAL AND ADVANCED EXERCISES
1. Directrix x 3 and focus (4 0) vertex is œßÊß! ˆ‰
7
#
p the equation is xÊœÊ œ
"
###
7 y


2. x 6x 12y 9 0 x 6x 9 12y y vertex is (3 0) and p 3 focus is (3 3) and the
##
œ Ê œ Ê œ Ê ß œ Ê ß
(x 3)
12

directrix is y 3œ
3. x 4y vertex is ( 0) and p 1 focus is ( 1); thus the distance from P(x y) to the vertex is x y
# ##
œ Ê !ß œ Ê !ß ß È
and the distance from P to the focus is x (y 1) x y 2 x (y 1)ÈÈÈ
######
Ê œ
x y 4 x (y 1) x y 4x 4y 8y 4 3x 3y 8y 4 0, which is a circleÊœ ÊœÊ œ
## # # ## # # # #
cd
4. Let the segment a b intersect the y-axis in point A and
intersect the x-axis in point B so that PB b and PA aœœ
(see figure). Draw the horizontal line through P and let it
intersect the y-axis in point C. Let PBOnœ )
APC . Then sin and cos Ên œ œ œ)) )
y
ba
x
cos sin 1.Êœ œ
x
ab
y



##
))

5. Vertices are 2 a 2; e 0.5 c 1 foci are 0 1ab ab!ß „ Ê œ œ Ê œ Ê œ Ê ß „
cc
a #
6. Let the center of the ellipse be (x 0); directrix x 2, focus (4 0), and e c 2 2 cߜߜʜʜ
2a a
3e e
a (2 c). Also c ae a a 2 a a a a a ; x 2Êœ œ œ Êœ Êœ Ê œ Êœ œ
2222445412a
333339935e
ˆ‰
x 2 x the center is 0 ; x 4 c c 4 so that c a bÊœ œ Êœ Ê ß œÊœ œ œˆ‰ˆ‰ ˆ ‰
12 3 18 28 28 28 8
55 5 5 55#
###
; therefore the equation is 1 or 1œœ œ œˆ‰ ˆ‰
12 8 80
5 5 25 144 16
x25x y5y## ˆ‰ ˆ‰
ˆ‰ ˆ‰
28 28
55
144 80
25 25

691. Chapter 10 Additional and Advanced Exercises 687
7. Let the center of the hyperbola be (0 y).ß
(a) Directrix y 1, focus (0 7) and e 2 c 6 c 6 a 2c 12. Also c ae 2aœ ß œÊœÊœÊœ œœ
aa
ee
a 2(2a) 12 a 4 c 8; y ( 1) 2 y 1 the center is (0 1); c a bÊœ ʜʜ œœœÊœÊ ß œ
a4
e #
###
b c a 64 16 48; therefore the equation is 1Ê œœœ œ
### (y 1)
16 48
x

(b) e 5 c 6 c 6 a 5c 30. Also, c ae 5a a 5(5a) 30 24a 30 aœÊœÊœÊœ œœÊœ Ê œÊœ
aa 5
ee 4
c ; y ( 1) y the center is ; c a b b c aÊœ œœ œ ÊœÊ !ß œ Ê œ
25 a 3 3
4e544 4
ˆ‰
5
4
" ### ###
ˆ‰
; therefore the equation is 1 or 1œœ œ œ
625 25 75 x 2x
16 16 25 75
y16y
#
œœˆ‰ ˆ‰
ˆ‰ ˆ‰
33
44
25
16
75



8. The center is (0 0) and c 2 4 a b b 4 a . The equation is 1 1ߜʜʜ œÊœ
## # # y
ab a b
x 49 144



1 49 4 a 144a a 4 a 196 49a 144a 4a a a 197a 196Ê œÊ œ Ê œ Ê
49 144
a4a

ab
# ### ###%%#
ab ab
0 a 196 a 1 0 a 14 or a 1; a 14 b 4 (14) 0 which is impossible; a 1œÊ œÊœ œ œ Ê œ œabab
## # # b 4 1 3; therefore the equation is y 1Ê œœ œ
## x
3

9. (a) b x a y a b ; at (x y ) the tangent line is y y (x x )
## ## ##
"" " "
œ Êœ ß œ
dy
dx ay ay
bx bx



œ
œ
Š‹
ayy bxx bx ay ab bxx ayy ab 0ÊœœÊœ
# # ## ## ## # # ##
"" ""
!!
(b) b x a y a b ; at (x y ) the tangent line is y y (x x )
## ## ##
"" " "
œ Êœ ß œ
dy
dx ay ay
bx bx



œ
œ
Š‹
b xx a yy b x a y a b b xx a yy a b 0ÊœœÊœ
# # ## ## ## # # ##
"" ""
!!
10. Ax Bxy Cy Dx Ey F 0 has the derivative ; at (x y ) the tangent line is
##
œœ
""
œ œ ß
dy 2Ax By D
dx Bx 2Cy E
yy (xx) Byx2CyyEyByx2CyEyœ Ê """""""

œœ
#Š‹
2AxByD
Bx 2Cy Eœœ
œœ
!
2Axx Bxy Dx 2Ax Bx y Dx 2Axx B(yx xy ) 2Cyy Dx Dx Ey Eyœ Ê "" """ " "" " " "
# !
2Ax 2Bx y 2Cy . Now add 2Dx 2Ey to both sides of this last equation, divide the result byœ
##
"" " "
!!
2, and represent the constant value on the right by F to get:!
AxxB CyyD E F""
œœ
###
œœˆ ‰ ˆ‰ˆ‰
yx xy y y xxœœ œ œ
11. 12.
13. 14.

692. 688 Chapter 10 Conic Sections and Polar Coordinates
15. 9x 4y 36 4x 9y 16 0abab
## ##
Ÿ
9x 4y 36 0 and 4x 9y 16 0ÊŸ  
## ##
or 9x 4y 36 0 and 4x 9y 16 0
## ##
  Ÿ

16. 9x 4y 36 4x 9y 16 0, which is theabab
## ##

complement of the set in Exercise 15

17. x y 9 0 x y 9 0 or
%# ##
#
œÊœab ab
x y 9 0 y x 9 or x y 9
# # ## ##
œÊ œ œab

18. x xy y 3 tan 2 which is undefined
## "

Ê œ !
11
2 90° 45° AÊœ Êœ Ê!!
w
cos 45° cos 45° sin 45° sin 45° , B 0,œ œœ
##w
#
3
C sin 45° sin 45° cos 45° cos 45°
w# # "
#
œ œ
x y 3 which is the interior of aÊ
3
##
ww##"
rotated ellipse

19. Arc PF Arc AF since each is the distance rolled;œ
PCF Arc PF b( PCF); nœ Ê œn œ
Arc PF Arc AF
ba
)
Arc AF a a b( PCF) PCF ;ʜʜnÊnœ )) ) ˆ‰
a
b
OCB and OCB PCF PCEnœ nœnn
1
#
)
PCF œn œ Ê ˆ‰ˆ‰ˆ‰
111
###
!)! )
a
b
œÊœˆ‰ ˆ ‰ ˆ‰
aa
bb
)! ))!
11 1
## #
.Êœ Êœ!1) ) !1 )ˆ‰ ˆ ‰
aab
bb
œ

Now x OB BD OB EP (a b) cos b cos (a b) cos b cosœœœ œ )! ) 1 ) ˆ‰ˆ‰
ab
b
œ
(a b) cos b cos cos b sin sin (a b) cos b cos andœ œ )1 ) 1 ) ) )ˆ‰ ˆ‰ ˆ‰ˆ‰ ˆ‰ ˆ‰
ab ab ab
bb b
œœ œ
y PD CB CE (a b) sin b sin (a b) sin b sinœœœ œ )! ) ) ˆ‰ˆ‰
ab
b
œ
(a b) sin b sin cos b cos sin (a b) sin b sin ;œ œ )1 ) 1 ) ) )ˆ‰ ˆ‰ ˆ‰ˆ‰ ˆ‰ ˆ‰
ab ab ab
bb b
œœ œ
therefore x (a b) cos b cos and y (a b) sin b sinœ œ )) ))ˆ‰ ˆ‰ˆ‰ ˆ‰
ab ab
bb
œœ

693. Chapter 10 Additional and Advanced Exercises 689
20. (a) x a(t sin t) a(1 cos t) and let 1 dm dA y dx y dtœ Ê œ œÊ œ œ œ
dx dx
dt dt
$ ˆ‰
a(1 cos t) a (1 cos t) dt a (1 cos t) dt; then A a (1 cos t) dtœ œ œ
## ## '
0
21 a 1 2 cos t cos t dt a 1 2 cos t cos 2t dt a t 2 sin tœ œ œ
### # ""
##
#
!''ab ˆ‰ œ ‘
00
22 3sin 2t
24
11 1
3 a ; x = x a(t sin t) and y = y a(1 cos t) M y dm y dAœœ œÊœœ
µµ µµ
1$
# ""
##
x ''
a(1 cos t) a (1 cos t) dt a (1 cos t) dt 1 3 cos t 3 cos t cos t dtœ œ œ''' ab
000
222 a111""
###
##$ $ #$
!
1 3 cos t 1 sin t (cos t) dt t 3 sin t sin tœ œ
a 3 3 cos 2t a 5 3 sin 2t sin t
0
2
243
! !!
### #
#
#
!
'œ ‘ab ’“
1
1
. Therefore y a. Also, M x dm x dAœœœœœœ
µµ
5a 5 M
M3a 6
y
1
1
!
!


#
x
5a
Š‹

'' $
a(t sin t) a (1 cos t) dt a t 2t cos t t cos t sin t 2 sin t cos t sin t cos t dtœ œ '' ab
00
2211
##$ # # a 2 cos t 2t sin t t cos 2t sin 2t cos t sin t 3 a . Thusœ œ
$# ##$ ""
#
!
’“
ttcost
2484 3
!
1
1 x a a a is the center of mass.œœ œ Ê ß
M
M3a 6
3a 5y 1
1
!

11 ˆ‰ (b) x t t and y 2t t ; let 1 dm dA y dx y dtœÊœ œÊœ œÊœœœ
2dx dx
3dt dt dt
dy$Î# "Î# "Î# "Î#
$ ˆ‰
2t t dt 2t dt; x = x t and y = t M y dm t (2t dt)ÊœœœÊœœ
µµ µ
ˆ‰ˆ‰ ''
"Î# "Î# $Î# "Î# "Î#
#
2
3
y
x
0
3
È
2t dt t 3. Also, M x dm x dA t (2t dt)œœœ œœœ
µµ'''' œ ‘ È
0 0
3 3412 2
55 3
y
È ÈÈ
$Î# &Î# $Î#
$
!
4
t dt t 27.œœœ' œ ‘ È
0
348 8
321 7
È È
&Î# (Î#
$
!
4
21. (a) x e cos t and y e sin t x y e cos t e sin t e . Also tan tœœÊœœœœ
2t 2t 4t 4t 4t ## # # y
xe cos t
e sin t
2t
2t
t tan x y e is the Cartesian equation. Since r x y andÊœ Ê œ œ
" # # # # #ˆ‰
y
x %Îtan y x

ab
tan , the polar equation is r e or r e for r 0)œœœ
" #ˆ‰
y
x 42
(b) ds r d dr ; r e dr 2e d
### #
œœÊœ))
22))
ds r d 2e d e d 4e dÊœ œ
### # #
##
)) ))ˆ‰ˆ‰
22 4)) )
5e d ds 5 e d L 5 e dœÊœ Êœ
42 2)) )
)) )
# ÈÈ '
0
2
œ
e1 œœ’“ ab
ÈÈ
5e 5
2
4 2œ#
!
#
1
1

22. r 2 sin dr 2 sin cos d ds r d dr 2 sin d 2 sin cos dœÊœ Êœœ
$# ####$##
##ˆ‰ ˆ‰ ˆ‰ œ ‘ œ ‘ˆ‰ ˆ‰ ˆ‰
))) )))
333 333
)) ) )
4 sin d 4 sin cos d 4 sin sin cos d 4 sin dœ œ œ
'# % ## % # # # %#ˆ‰ ˆ‰ ˆ‰ œ ‘œ ‘ˆ‰ˆ‰ ˆ‰ ˆ‰
))) ))) )
333 333 3
)) ))
ds 2 sin d . Then L 2 sin d 1 cos d sin 3Êœ œ œ œ œ
##
$
!ˆ‰ ˆ‰ œ ‘ œ ‘ˆ‰ ˆ‰
))))
1
33323
232
))))1 ''
00
33
œœ
23. r 1 cos and S 2 ds, where y r sin ; ds r d drœ œ œœ œ )133))' È
## #
(1 cos ) d sin d 1 2 cos cos sin d 2 2 cos d 4 cos dœ œ œÈ È È É ˆ‰) ) )) ) ) )) )) )
## # # # # #
#
)
2 cos d since 0 . Then S 2 (r sin ) 2 cos d 4 (1 cos ) sin cos dœŸŸœ œˆ‰ ˆ‰ ˆ‰
)1 ) )
## # #
)) 1) )1))) ''
00
22
œœœœ
††
4 2 cos 2 sin cos cos d 16 cos sin dœœœ''
0 0
2 2
œ œœ œ
1)1)œ ‘œ ‘ˆ‰ˆ‰ˆ‰ ˆ‰ ˆ‰ ˆ‰ ’“
#%
## # ##

Î#
!
) ))) )) 1
1
25
32 cos
Î
ˆ‰
œ
œœ
(32)
555
32 32 4 2


1
1 11
Š‹ È
œ
2

Î
ˆ‰

694. 690 Chapter 10 Conic Sections and Polar Coordinates
24. The region in question is the figure eight in the middle.
The arc of r 2a sin in the first quadrant givesœ
#
#ˆ‰
)
of that region. Therefore the area is A 4 r d
""
#
#
4
œ'
0
2
œœ
)
4 2a sin d 8a sin dœœ ''
00
22
œœœœ
"
## #
##%
#
œ ‘ˆ‰ˆ‰
))
))
8a sin 1 cos dœ
## #
##'
0
2
œœ
ˆ‰œ ‘ˆ‰
))
)
8a sin sin cos dœ
## ##
###'
0
2
œœ
œ ‘ˆ‰ ˆ‰ ˆ‰
)))
)
8a dœ
#
#'
0
2
œœ
Š‹
1cos sin
4
))

)

2a 2 2 cos d a (3 4 cos cos 2 ) d a 3 4 sin sin 2œ œœ
### "
#
Î#
!''
00
22
œœœœ
ˆ‰ œ ‘)) ))))))
1cos 2
2
)
1
a4œ
#
#ˆ‰
31
25. e 2 and r cos 2 x 2 is the directrix k 2; the conic is a hyperbola with rœ œÊœ Êœ œ)
ke
1e cos œ )
rÊœ œ
(2)(2)
12 cos 12 cos
4
œœ ))
26. e 1 and r cos 4 x 4 is the directrix k 4; the conic is a parabola with rœœÊœ Êœ œ)
ke
1e cos )
rÊœ œ
(4)(1)
1cos 1cos
4
))
27. e and r sin 2 y 2 is the directrix k 2; the conic is an ellipse with rœ œÊœ Êœ œ
"
# œ
)
ke
1e sin )
rÊœ œ
2
1 sin
2
2sin
ˆ‰
ˆ‰
œ

œ

œ œ) )
28. e and r sin 6 y 6 is the directrix k 6; the conic is an ellipse with rœœÊœ Êœ œ
"
3 1e sin
ke
)
)
rÊœ œ
6
1 sin
6
3sin
ˆ‰
ˆ‰
œ
œ
3
3
) )
29. The length of the rope is L 2x 2c y 8c.œ 
(a) The angle A ( BED) occurs when the distancen
CF is maximized. Now x c yœj jœ
È
##
xcL2x2cÊjœ
È
##
x c (2x) 2 2.Êœ œ
dx
dx
xc
j"
#
##
"Î#

ab
È
Thus 0 2 0 x 2 x c
dx
dx
xc
j

##
œÊ œÊœ
È
È
x 4x 4c 3x 4c Êœ Ê œ Êœ
### ## c3
x4


. Since sin we have sin Êœ œ œ
ccAA
xx
33
ÈÈ
####
60 ° A 120 °Êœ Êœ
A
#

(b) If the ring is fixed at E (i.e., y is held constant) and E is moved to the right, for example, the rope will slip
around the pegs so that BE lengthens and DE becomes shorter BE ED is always 2x L y 2c,Ê œ
which is constant the point E lies on an ellipse with the pegs as foci.Ê
(c) Minimal potential energy occurs when the weight is at its lowest point E is at the intersection of theÊ
ellipse and its minor axis.

695. Chapter 10 Additional and Advanced Exercises 691
30. d d 30;
dd dd
cc c cc c
30 30œ
œ Ê œ œ "#
d d 30. Therefore P and Q lie on an ellipse withʜ $%
F and F as foci. Now 2a d d 30 a 15 and"# "# œœ Êœ
the focal distance is 10 b 15 10 125Êœœ
###
an equation of the ellipse is 1. Nextʜ
x
225 125
y

xxvtxv x10.#"! "! "œ œ œ Š‹
10
v
%

If the plane is flying level, then P and Q must be symmetric to the y-axis x x x x 10ʜʜ
"###
x 5 1 y y since y must be positive. Therefore the position ofÊœÊ œÊœ Êœ###
#5 1000
225 125 9 3
y 10 10

œ
Î
È
the plane is where the origin (0 0) is located midway between the two stations.Š‹&ß ß
10 10
3
È
31. If the vertex is ( 0), then the focus is (p 0). Let P(x y) be the present position of the comet. Then!ß ß ß
(x p) y 4 10 . Since y 4px we have (x p) 4px 4 10 (x p) 4px 16 10 .ÈÈœ‚ œ œ‚ Ê œ‚
## # (# ( # "%
Also, x p 4 10 cos 60 °210 xp210. Therefore 210 4pp210 1610œ‚ œ‚ Ê œ‚ ‚ ‚ œ ‚
(( ( ( ("%
#
aba b
4 10 4p 8p 10 16 10 4p 8p 10 12 10 0 p 2p 10 3 10 0Ê‚ ‚ œ‚ Ê ‚ ‚ œÊ ‚ ‚ œ
"% # ( "% # ( "% # ( "%
p 3 10 p 10 0 p 3 10 or p 10 . Since p is positive we obtain p 10 miles.Ê‚ œÊœ‚ œ œabab
(( ( ( (
32. x 2t and y t y ; let D (x 0) 3 x x 9 x 9œœÊœ œœœ
# ####
#
##
"xxx3x
4 4 16 16
$$
É ˆ‰ ÉÉ
x 8x 144 be the distance from any point on the parabola to (0 3). We want to minimize D. Thenœ ß
" %#
4È
x 8x 144 4x 16x 0 x 2x 0 x 4x 0 x 0 or
dD
dx 8
x2x
x8x144
œ œ œÊœÊœÊœ
" "%# $ $ $
"Î#
#
abab
ˆ‰
È
œ

!

x 2. Now x 0 y 0 and x 2 y 1. The distance from (0 0) to (0 3) is D 3. The distanceœ„ œ Ê œ œ„ Ê œ ß ß œ
from 2 1 to ( 3) is D 2 (1 3) 2 2 which is less than 3. Therefore the points closest toab ab É È
„ ß !ß œ „ œ
#
#
(0 3) are 2 1 .ß„ßab
33. cot 2 0 45 ° is the angle of rotation A cos 45°cos 45° sin 45°sin 45°, B 0,!!œœÊœ Êœ œ œ
AC 3
B
w# # w
#
and C sin 45°sin 45° cos 45°cos 45° x y 1 b and a 2 c a b
w# # w w ### ""
## #
##
œ œÊœÊœ œÊœ
32
3
É È
2 c . Therefore the eccentricity is e 0.82.œœ Ê œ œœ œ ¸
24 2 c 2
33 a 3 3 2
ÈÈ
Š‹
2
3
œ
É
34. The angle of rotation is A sin cos , B 0, and C sin cos 1!œÊ œ œ œ œ œÊ œ
111 11
444 44
x ywww ""
####
' '
a 2 and b 2 c a b 4 c 2. Therefore the eccentricity is e 2 .Êœ œ Ê œ œÊœ œœ œ
ÈÈ È
### c2
a 2
È
35. x y 1 x 2 xy y 1 2 xy 1 (x y) 4xy 1 2(x y) (x y)È ÈÈ È œÊ œÊ œ Ê œ
#
4xy x 2xy y 2x 2y 1 x 2xy y 2x 2y 1 0 B 4AC ( 2) 4(1)(1) 0Ê œ Ê œÊ œ œ
## ## # #
the curve is part of a parabolaÊ
36. A 2 sin cos 1, B 0, C 2 sin cos 1, D 2 sin 1, E 2 cos !œÊœ œœœ œœ œœ
111 11 1 1
444 44 4 4
www ww ÈÈ
1, F2 xyxy20 xx yy 2 xx yyœ œÊ œÊ œÊ
w wwww ww ww ww ww## # # # # ""
Š‹Š‹ Š ‹Š ‹
44
2 1. The center is (x y ) x cos sin andœ Ê œ ß œ ß Ê œ œ
ˆ‰ˆ‰ Èyx
44
2
''œœ



## ## # # #
ww "" " "
ˆ‰ ˆ‰
11
y sin cos 0 or the center is (x y) 0 . Next a 2 the vertices areœ œ ßœß œÊ
""
## #
11
44
2
Š‹
È
È
(x y ) 2 and 2 x cos 2 sin 1 and
ww """ " " "
### # # # #
ßœß ß Êœ œ Š‹Š ‹ Š‹
ÈÈ È
11
44
2
È
y sin 2 cos 1 or (x y) 1 1 is one vertex, and x cos 2 sin œ œßœß œ
"" ""
## # # #
11 11
44 44
2
Š‹ Š ‹ Š ‹
È È
È

696. 692 Chapter 10 Conic Sections and Polar Coordinates
1 and y sin 2 sin 1 or (x y) 1 1 is the other vertex. Alsoœ œ œ ßœ ß
È È
2 2
44## # #
"" 11
Š‹ Š ‹
È
c 2 2 4 c 2 the foci are (x y ) and x cos sin and
#ww "" "
## # # # # #
œœ Ê œ Ê ß œ ß ß Ê œ œ ˆ‰ ˆ ‰
35 3
44
211
È
y sin cos 2 or (x y) 2 is one focus, and x cos sin andœ œ ßœß œ œ
" "
## # # # #
11 11
44 44
3 5232 ÈÈ
Š‹
ÈÈ
y sin cos 2 or (x y) 2 is the other focus. The asymptotes areœ œ ßœß
"
## #
11
44
5 32 ÈÈ
Š‹
È
y x in the rotated system. Since x x y and y x y x y x
w w ww ww w" " "" ""
##
œ„ œ œ Êœˆ‰
ÈÈ ÈÈ È
22 22 2
2
x y x and x y y x y y ; the asymptotes areÊœ œ Êœ
ÈÈ ÈÈ
È
22 22 2
2## ##
ww w
x y x y the asymptotes are 2 x 1 0 or x and 2 y 0 orœ„ Ê œ œ œ
ÈÈ ÈÈ
È
22 22
2### ###
"" "
Š‹
ÈÈ
y 0. Finally, the x -axis is the line through 0 with a slope of 1 recall that y x .œßœÊœ
w
# #
Š‹ ˆ‰
È È
2 2
4
!
1
The y -axis is the line through 0 with a slope of 1 y x .
w
##
Š‹
ÈÈ
22
ßÊœ
37. (a) The equation of a parabola with focus ( 0) and vertex (a 0) is r and rotating this parabola!ß ß œ
2a
1cos œ )
through 45 ° gives r .!œœ
2a
1cosˆ‰)

4
(b) Foci at ( 0) and (2 0) the center is (1 0) a 3 and c 1 since one vertex is at (4 0). Then e!ß ßÊ ßÊœ œ ß œ
c
a
. For ellipses with one focus at the origin and major axis along the x-axis we have rœ œ
"
3 1e cos
a1 eab

)
. œœ
31
1 cos
8
3cos
ˆ‰
ˆ‰


œ
œ
9
3
) )
(c) Center at and focus at ( 0) c 2; center at 2 and vertex at a 1. Then eˆ‰ ˆ‰ ˆ‰#ß !ß Ê œ ß "ß Ê œ œ
111
###
c
a
2. Also k ae (1)(2) . Therefore r .œœ œ œ œ œ œ œ
2a3ke3
1 e 1e sin 12 sin 12 sin
(2)"
## œ œ œ )))
ˆ‰
3

38. Let (d ) and (d ) be the polar coordinates of P and P , respectively. Then , and we have"" ## " # # "ßß œ)) ))1
d and d . Therefore "#
œœœ
"" œ œœœœ œ
33
2cos 2cos( ) d d 3 3
2cos 2cos( )
))1
) )1
œœœ
œ œ
.œœ
4 cos cos cos sin sin
33
4 ))1)1œœ œ
39. Arc PT Arc TO since each is the same distance rolled. Now Arc PT a( TAP) and Arc TO a( TBO)œœnœn
TAP TBO. Since AP a BO we have that ADP is congruent to BCO CO DP OP isÊn œn œ œ Ê œ Ê ??
parallel to AB TBO TAP . Then OPDC is a square r CD AB AD CB AB 2CBÊn œn œ Êœœœ )
r 2a 2a cos 2a(1 cos ), which is the polar equation of a cardioid.Êœ œ ))
40. Note first that the point P traces out a circular arc as the door
closes until the second door panel PQ is tangent to the circle.
This happens when P is located at , since OPQ isŠ‹
""
ÈÈ
22
ßn
90° at that time. Thus the curve is the circle x y 1 for
##
œ
0 x . When x , the second door panel isŸŸ  
""
ÈÈ
22
tangent to the curve at P. Now let t represent POQ so thatn
as t runs from to 0, the door closes. The coordinates of P
1
#
are given by (cos t sin t), and the coordinates of Q byß

(2 cos t 0) (since triangle POQ is isosceles). Therefore at a fixed instant of time t, the slope of the lineß
formed by the second panel PQ is m tan t the tangent line PQ isœœ œ Ê
?
?
y
xcos t2 cos t
sin t 0

y 0 ( tan t)(x 2 cos t) y ( tan t) x 2 sin t. Now, to find an equation of the curve forœ Ê œ
x 1, we want to find, for x, the largest value of y as t ranges over the interval 0 t . We
"
È
2 4
ŸŸ ŸŸfixed
1
solve 0 sec t x 2 cos t 0 sec t x 2 cos t x 2 cos t. (Note that
dy
dt
œÊ œÊ œ Êœab ab
## $
2 sec t tan t x 2 sin t 0 on 0 t , so a maximum occurs for y.) Now x 2 cos t the
dy
dt


œ Ÿ Ÿ œ Êab
# $
#
1

697. Chapter 10 Additional and Advanced Exercises 693
corresponding y value is y ( tan t) 2 cos t 2 sin t 2 sin t cos t 2 sin t (2 sin t) cos t 1œ œ œ ab a b
$# #
2 sin t. Therefore parametric equations for the path of the curve are given by x 2 cos t and y 2 sin tœ œœ
$ $$
for 0 t . In Cartesian coordinates, we have the curve x y 2 cos t 2 sin tŸŸ œ
1
4
#Î$ #Î$ $ $
#Î$ #Î$
abab
2 cos t sin t 2 the curve traced out by the door is given byœœÊ
#Î$ # # #Î$
ab

x y 1 for 0 x
xy2 for x1
Þ
ß
à
## "
#Î$ #Î$ #Î$ "
œ ŸŸ
œ ŸŸ
È
È
2
2
41. tan tan ( ) ;"< < " < <œ Ê œ œ#" #"

œ
tan tan
1tan tan
<<
<<
œ

the curves will be orthogonal when tan is undefined, or"
when tan < #
" "œÊœ
tan g ( )
r
<)œ
'
'
’“
r
f( )
œ
rf()g()ʜ
#ww
))

42. r sin sin cos tan tanœÊœ Êœ œ
%$ˆ‰ ˆ‰ ˆ‰ ˆ‰
))) )
)4d 4 4 4
dr sin
sin cos
<
$
!ˆ‰
ˆ‰ ˆ‰
œ
œœ
4
44
43. r 2a sin 3 6a cos 3 tan tan 3 ; when , tan tan œÊœ Êœœœ œœ))< ))<
dr r 2a sin 3
d 6a cos 3 3 6 3))
)11
ˆ‰
dr
d
œ
""
#
ʜ<
1
#
44. (a) (b) r 1 r tan )) )<œÊœ Ê œ Ê
" # dr
d)
k
$1
lim tan œœÊ œ_
)
)
Ÿœ


)<
)Ä_
from the right as the spiral winds inÊÄ<
1
#
around the origin.
45. tan cot is at ; tan tan is 3 at ; since the product of<))<)) "#

"œœœ œœ œ
È
ÈÈ
3 cos
3 sin 3 3cos 3
sin )
)
1) 1
)
È
these slopes is 1, the tangents are perpendicular!
46. a(1 cos ) 3a cos 1 2 cos cos orœ Êœ Êœ)) ))
"
#
; tan is 3 at ;)< )œœ œ
11 )
)3a sin 3
1
a(1 cos )œ
È
tan is at . Then<) 2
3a cos
3a sin 3 3œœ
)1
)
"
È
tan ""œœœÊœ


#
"
œœ
œ
œœ
œ
33
3
Š‹
ÈÈ
Š‹Š‹
È È
33
13 3 6
1

698. 694 Chapter 10 Conic Sections and Polar Coordinates
47. r ; r ; "#
" "
œÊœ œÊœ œ
1cos d (1cos ) 1cos d (1cos ) 1cos 1cos
dr dr sin 3 3 sin 3
)) ) )) ) ) )
))
œ

1 cos 3 3 cos 4 cos 2 cos r r 2 the curves intersect at theÊ œ Ê œÊ œÊœ„Ê œœÊ)))))
"
#
"#
1
3
points ; tan is at ; tan ˆ‰#ß „ œ œ œ œ œ
1)1
)3sin 3
1cos 1cos
3
<)<"#
"
ˆ‰
’“ ’“
ˆ‰
œ


!
Î Î
!
1cos
sin 3 sin
(1 cos ) (1 cos )
3
1cos œ
œ œ
œ œ
œ
È
)
)sin
is
3 at ; therefore tan is undefined at since 1 tan tan 1 3 0 ;
È È
Š‹Š‹)")<< "œœœœÊœ
11 1
33 3
"#
"
#È
tan and tan 3 tan is also undefinedkk
È
<< " "#


"
œ Î$œ$
œ œ œ œ Ê
1 cos 1 cos
sin sin 3
ˆ‰ ˆ‰
ˆ‰ ˆ‰ È


33
33
)1
at )"œ Ê œ
11
3 #
48. (a) We need , so that tan tan ( )<)1 < 1)œ œ
tan . Now tan œ œ œ)<
r a(1 cos )
a sin ˆ‰
dr
d
œ
œ

)
)
tan cos cos sinœ œ Ê œ))))
sin
cos
)
)
##
cos cos 1 cosʜ)) )
##
2 cos cos 1 0ʜ
#
))
cos or cos 1; cos Êœ œ œÊœ„))))
""
##
1
3
r ; cos 1 r 0.Êœ œÊ œÊœ
3a
#
))1
Therefore the points where the tangent line
is horizontal are and ( ).ˆ‰
3a
3#
ß„ !ß
1
1

(b) We need so that tan tan cot . Thus tan cot <) < ) ) < )œ œ œ œ œ œ
11 )
)##
œ
ˆ‰
r a(1 cos )
a sin ˆ‰
dr
d
œ sin sin cos sin cos cos or sin 0; cos œÊ œ Ê œ œ œÊœ„
cos 2
sin 3
) 1
)
))) )) ) ) ) )
""
##
r ; sin 0 0 (not , see part (a)) r 2a. Therefore the points where the tangent lineÊœ œ Ê œ Êœ
a
#
))1
is vertical are and (2a 0).ˆ‰
a2
3#
ß„ ß
1
49. r and r ; then"#
œÊœ œÊœ
a a sin b b sin
1cos d (1cos ) 1cos d (1cos )
dr dr
)) ) )) )
)) œ

tan and tan 1 tan tan << << "# "#

œœ œœÊ
ˆ‰
’“ ’“
ˆ‰
a
1cos
a sin b sin
(1 cos ) (1 cos )
b
1cos
!

ÎÎ


œ
œœ
œœ
œ 1cos 1cos
sin sin
))
))
1 1 0 is undefined the parabolas are orthogonal at eachœ œ œ Ê Êˆ‰ˆ‰
1cos 1cos 1cos
sin sin sin


)) )
)) )


"
point of intersection
50. tan is 1 at <)<œœ œÊœ
r a(1 cos )
a sin 4ˆ‰
dr
d
œ

#
)
)
11
51. r 3 sec r ; 4 4 cos 3 4 cos 4 cos (2 cos 3)(2 cos 1) 0œÊœ œÊœ Ê œ) ) ))))
33
cos cos ))
#
cos or cos or (the second equation has no solutions); tan Êœ œÊœ œ))) <
"
##
#
œ35
3 3 4 sin
4(1 cos )11 )
)
is 3 at and tan cot is at . Then tan is undefined sinceœ œ œ
1 cos 3 sec
sin 3 3 sec tan 3 3
œ"
"
)1 ) 1
))) È
<) "
È
1 tan tan 1 3 0 . Also, tan 3 and tan œœÊœ œ œ<< " < < "# # "
" "
# Š‹Š ‹
ÈÈ
kk
È È
3 3
1 53 53œœœœ
1 tan tan 1 3 0 tan is also undefined .Ê œ œÊ Ê œ<< " " "#
"
# Š‹Š‹
È
È
3
1
52. tan 1 at ; m tan ( ) tan 1<)<)<œœœÊœœœœ
a tan
sec 44
3
ˆ‰
ˆ‰
œ
œ



a
11 1
#
tan
53. 1 cos 1 sin cos sin ; tan ;
""

"
1cos 1sin 4sin
1cos
)) )
1)
œÊœÊœÊœ œ œ )))))<
ˆ‰
’“
œ



Î
1cos
sin
(1 cos )œ
œ
œ
tan . Thus at , tan 1 2 and<)< #"

œœ œœœ
ˆ‰
’“
œ


Î
1sin
cos
(1 sin )œ
œ
œ 1sin
cos 4
1cos
sin
)1
)
ˆ‰
ˆ‰


4
4 È

699. Chapter 10 Additional and Advanced Exercises 695
tan 2 1. Then tan 1 <" " #




œœ œ œœÊœ
1sin
cos
21 1 2
12112
22 2
22 2 4
ˆ‰ È
ˆ‰
Š‹Š‹
ÈÈ
Š‹Š‹
ÈÈ È


4
4È 1
54. (a) (b) r 2 csc 2
#
œœœ )
22
sin 2 2 sin cos )))
r sin cos 1 xy 1, a hyperbolaʜʜ
#
))
(c) At , x y 1 1)œ œ œ Ê œ œ
1
4dxx
dy "

m œÊœÊœœœ
tan9<9)
33
444
1111
#
55. (a) tan ln r C (by integration) r Be for some constant B;!œÊœ Êœ Êœ
rdrd
rtan tan ˆ‰
dr
d
œ
))
!! !œtan
A Be d Be Beœœœ
"
#
#Î # #'
)
)œ

2 2 tan 2 tan abtan B(tan )e
44
tan ! !
)
)
!
)’“ cd

œ 2tanœ


œœ
r r since r B e and r B e ; constant of proportionality Kœ œ œ œ
tan tan
4 4
! !
ab
## ## ##
Î !Î !
œœ
2tan 2tan œ !
(b) tan r r r!œÊœ Ê œ Ê œœ
rdrr dr r dr r tan1
d tan d tan d tan tanˆ‰
dr
d
œ
)! ) ! ) ! !
!
ˆ‰ ˆ‰ Š‹
##
### œ

r Length r d Be d B (sec ) eœÊœ œ œ
#
Š‹ ˆ‰ cd
sec sec sec
tan tan tan



œ
!!!
!!!
)
) ''
))
))œœ

))!
œœ tan tan
†
(sec ) Be Be K (r r ) where K sec is the constant of proportionalityœœœ!!cd
œ !œ tan tan
#"
56. r sin 2 2a r sin cos a xy a and
### ##
)))œÊ œÊœ . If P(x y ) is a point on the curve, the tangent
dy
dx x
a
œ ß


""
line is y y (x x ), so the tangent line crossesœ ""
a
x


!
the x-axis when y 0 y (x x )œÊœ ""
a
x


!
x x x x x x 2xÊ œ Êœ œœ
xy x y
aa

œœ

!!
"""""
since 1. Let Q be (2x 0). Then
xy
aœœ

œß " PQ (2x x ) (0 y ) x y andœœÈ È "" "
#### !!

OP r (x 0) (y 0) x y OP PQ and the triangle is isosceles.œœ œ Ê œÈ È ""
#### !!

700. 696 Chapter 10 Conic Sections and Polar Coordinates
NOTES:

701. CHAPTER 11 INFINITE SEQUENCES AND SERIES
11.1 SEQUENCES
1. a 0, a , a , a"#$%
œ"œ"œ œ
#œœ œœ œœ œœ
11 2 13 2 14 3
1 4 39 416
œœœœ
2. a 1, a , a , a"#$%
""
#œœ œœ œœ œœ
11111
1! ! 2 3! 6 4! 24
3. a 1, a , a , a"# $ %
œœ" œ œ
#œ œ œ œ
"" "œœœœœœ œœ
(1) ( ) (1) (1)
1 413 615 817

4. a 2 ( 1) 1, a 2 ( 1) 3, a 2 ( 1) 1, a 2 ( 1) 3"#$%
"#$%œ œ œ œ œ œ œ œ
5. a , a , a , a"#$%
#####
""""
#œœ œœ œœ œœ
2222
22


'
6. a , a , a , a"#$%
œ""œœœ"
##œœœœœœœœ
2213217215
24 28 16 2

'
7. a 1, a 1 , a , a , a , a ,"# $ % & '
"" " "
## ## # #œ œœ œ œ œ œ œ œ œ
3 3 7 7 15 15 31 63
4 4 8 8 16 32
'
a, a, a, a()*"!œœœ œ
127 255 511 1023
64 128 256 512
8. a 1, a , a , a , a , a , a , a ,"# $ % & ' ( )
"" " """"
####œœœœœœœœœœ œ
ˆ‰ ˆ‰ ˆ ‰
)) )
œœ
3 6 4 4 5 1 0 7 0 5040 40,320
64
a, a*"!
""œœ
362,880 3,628,800
9. a 2, a 1, a , a , a ,"# $ % &
œœ
####
""" œœ œœœœ œœ œ œ œ œ œ
(1)(2) (1)(1)
248
(1) (1)
' )
œ
( ) ˆ‰ ˆ‰
4
a , a , a , a , a'( ) * "!
"""""
##œœœœ œ
16 3 64 1 8 256
10. a 2, a 1, a , a , a , a ,"#$%&'
œœ
##
œœ ""œ œ œ œ œ œ œ œ œ œ
1( 2) 2( 1)
33 4 5 5 3
22 34†† ††ˆ‰ ˆ‰
2
3 )
œ
a, a, a, a()*"!
""œ œ œ œ
22
749 5
11. a 1, a 1, a 1 1 2, a 2 1 3, a 3 2 5, a 8, a 13, a 21, a 34, a 55"#$%&' ()*"!œ œ œœ œœ œœ œ œ œ œ œ
12. a 2, a 1, a , a , a 1, a 2, a 2, a 1, a , a"# $ % & ' () * "!
"" ""
#œ# ##
œ
œœ œ œ œ œ œ œ œ œ œ œ œ
ˆ‰ ˆ‰
ˆ‰
))
œœ
)
œ
1
13. a ( 1) , n 1, 2, 14. a ( 1) , n 1, 2, n n
n1 nœ œ á œ œ á
_
15. a ( 1) n , n 1, 2, 16. a , n 1, 2, n n
n1 ()
nœ œ á œ œ á
_# œ"
n1*
œ
17. a n 1, n 1, 2, 18. a n 4 , n 1, 2, n nœœá œœá
#
19. a 4n 3, n 1, 2, 20. a 4n 2 , n 1, 2, n nœ œ á œ œ á
21. a , n 1, 2, 22. a , n 1, 2, n n
1(1) n(1) nœœá œ œÚÛœá
_œ
# ##
œ_œ
n1 n* ))
œœ ˆ‰
23. lim 2 (0.1) 2 converges (Theorem 5, #4)
nÄ_
œÊ
n

702. 698 Chapter 11 Infinite Sequences and Series
24. lim lim 1 1 converges
nnÄ_ Ä_
n( ) (1)
nn
_œ" œ
nn
œœÊ
25. lim lim lim 1 converges
nn nÄ_ Ä_ Ä_
"œ œ
_# #
œ
_
2n 2
1n
2
2
œœœÊ
ˆ‰
ˆ‰
Ú
Ú
n
n
26. lim lim diverges
nnÄ_ Ä_
2n
13n
2n
3
_"
œ
_
œ
È
È Š‹
Š‹
œœ_Ê
œ
œ
œ
œ
n
n
27. lim lim 5 converges
nnÄ_ Ä_
"œ
_
œ
_
5n
n8n
5
1
#
'&
Ú
#
œœÊ
Š‹
ˆ‰
n
8
n
28. lim lim lim 0 converges
nn nÄ_ Ä_ Ä_
n3 n3
n 5n 6 (n 3)(n 2) n
__"
__ _ _ _#
œ œœœÊ
29. lim lim lim (n 1) diverges
nn nÄ_ Ä_ Ä_
n2n1
n1 n1
(n 1)(n 1)
œ
œ_
œœ
œœ
œœœ_Ê
30 lim lim diverges
nnÄ_ Ä_
"œ
œ
œ
œ
n
70 4n
n
4
&
œ
œœ_Ê
Š‹
Š‹
œ
&
&
n
70
n
31. lim 1 ( 1) does not exist diverges 32. lim ( 1) 1 does not exist diverges
nnÄ_ Ä_
ab ˆ‰'& ! & & !
n n "
n
33. lim 1 lim 1 converges
nnÄ_ Ä_
ˆ‰ˆ‰ ˆ ‰ˆ‰
n
nn nn
_" " " " " "
####
œ œÊ
34. lim 2 3 6 converges 35. lim 0 converges
nnÄ_ Ä_
ˆ‰ˆ‰ œÊ œÊ
""
## #œ
œ"
nn
n1
()
n1
Û
36. lim lim 0 converges
nnÄ_ Ä_
ˆ‰œ œÊ
"
##
œ"
n ()
n
n
37. lim lim lim 2 converges
nnnÄ_ Ä_ Ä_
É É Ê Š‹
È
2n 2n
n1 n1__
œœ œÊ
2
1œ
œ
n
38. lim lim diverges
nnÄ_ Ä_
""
(0.9) 9
0
nœœ_ʈ‰
n
39. lim sin sin lim sin 1 converges
nnÄ_ Ä_
ˆ‰ ˆ‰ Š‹
111
###
""
œ œ œÊ
nn
40. lim n cos (n ) lim (n )( 1) does not exist diverges
nnÄ_ Ä_
11 1 œ Ê
n
41. lim 0 because converges by the Sandwich Theorem for sequences
nÄ_
sin n sin n
nnnn
œŸŸÊ
""
42. lim 0 because 0 converges by the Sandwich Theorem for sequences
nÄ_
sin n sin n
œœ
###
"
nnnœŸŸÊ
43. lim lim 0 converges (using l'Hopital's rule)^
nnÄ_ Ä_
n
ln 2##
"
nnœœÊ
44. lim lim lim lim diverges (using l'Hopital's rule) ^
nn n nÄ_ Ä_ Ä_ Ä_
33 ln 3
n3n 6n 6
3(ln 3) 3(ln 3)
nn nn

œœœœ_Ê
45. lim lim lim lim 0 converges
nnnnÄ_ Ä_ Ä_ Ä_
ln (n )
n
2n
n1
1
_"
_
_
È
È Š‹
Š‹
œœœ œÊ
ˆ‰
Š‹
œ
#
œ
&
œ
n1
n
2
n
n
œ
œ

703. Section 11.1 Sequences 699
46. lim lim 1 converges
nnÄ_ Ä_
ln n
ln 2n
œœÊ
ˆ‰
ˆ‰
œ
n
2
2n
47. lim 8 1 converges (Theorem 5, #3)
nÄ_
1nœ
œÊ
48. lim (0.03) 1 converges (Theorem 5, #3)

nÄ_
1nœ
œÊ
49. lim 1 e converges (Theorem 5, #5)
nÄ_
ˆ‰œÊ
7
n
n
(
50. lim 1 lim 1 e converges (Theorem 5, #5)
nnÄ_ Ä_
ˆ‰ ’“œ œ Ê
" œ" œ"
nn
()
n
n
51. lim 10n lim 10 n 1 1 1 converges (Theorem 5, #3 and #2)
nnÄ_ Ä_
È
n
œœœÊ
1n 1nÎÎ
††
52. lim n lim n 1 1 converges (Theorem 5, #2)
nnÄ_ Ä_
È ˆ‰È
n
n
#
#
#
œœœÊ
53. lim 1 converges (Theorem 5, #3 and #2)
nÄ_
ˆ‰
3
n1
1n lim 3
lim n
Î
"
œœœÊ
n
n&'
&'
1n
1nÊ
Ê
54. lim (n 4) lim x 1 converges; (let x n 4, then use Theorem 5, #2)
n xÄ_ Ä_
œ œÊ œ
1n4 1xÎÐ Ñ Î
55. lim diverges (Theorem 5, #2)
nÄ_
ln n
n lim n
lim ln n
1
1n 1nÊÊœœœ_Ê
n
n&'
&' _
56. lim ln n ln (n 1) lim ln ln lim ln 1 0 converges
nnnÄ_ Ä_ Ä_
cd ˆ‰ Š‹œ œ œœÊ
nn
n1 n1__
57. lim 4 n lim 4 n 4 1 4 converges (Theorem 5, #2)
nnÄ_ Ä_
È È
n nn
œœœÊ †
58. lim 3 lim 3 lim 3 3 9 1 9 converges (Theorem 5, #3)
nn nÄ_ Ä_ Ä_
È
n
2n 1 21n 1n_ _Ÿ # Ÿ
œœœœÊ
ab
††
59. lim lim lim 0 and 0 lim 0 converges
nn n nÄ_ Ä_ Ä_ Ä_
n! n! n!
n nnn nn n n n
2 3 (n 1)(n)
n nnœŸœ ÊœÊ
"ߜ
â
"††
†† †
ˆ‰
60. lim 0 converges (Theorem 5, #6)
nÄ_
(4)
n!
œ
n
œÊ
61. lim lim diverges (Theorem 5, #6)
nnÄ_ Ä_
n!
10
6n
(10 )
n
n!œœ_Ê
"
Š‹
_
62. lim lim diverges (Theorem 5, #6)
nnÄ_ Ä_
n!
23
nn 6
n
n!œœ_Ê
"
ˆ‰
63. lim lim exp ln lim exp e converges
nn nÄ_ Ä_ Ä_
ˆ‰ ˆ ‰ ˆ ‰ ˆ‰
"""œ
ÎÐ Ñ
œ"
nln nnln n
1lnn
ln 1 ln n
œœœÊ
64. lim ln 1 ln lim 1 ln e 1 converges (Theorem 5, #5)
nnÄ_ Ä_
ˆ‰ ˆ‰ Š‹œ œœÊ
""
nn
nn
65. lim lim exp n ln lim exp
nn nÄ_ Ä_ Ä_
ˆ‰ ˆ ‰ ˆ‰ Š‹
3n 3n
3n 1 3n 1
n ln (3n 1) ln (3n 1)_" _"
œœ
_œ œ
œœ
Ú
n

704. 700 Chapter 11 Infinite Sequences and Series
lim exp lim exp exp e convergesœœœœÊ
nnÄ_ Ä_
Š‹ ˆ‰
33
3n 1 3n 1
n
#"
œ
&
&
&Š‹
6n 6
(3n 1)(3n 1) 9
œ
_œ
#Î$66. lim lim exp n ln lim exp lim exp
nn n nÄ_ Ä_ Ä_ Ä_
ˆ‰ ˆ ‰ ˆ‰ Š‹

nn
n1 n1
n ln nln(n1)
ÝÝ
œ_
œœ œ
ˆ‰
Š‹
œ
œœ
#
œ
&
n
nn1
n
&
&
lim exp e convergesœœÊ
nÄ_
Š‹
n
n(n 1)
œ
Ý
œ"
67. lim lim x x lim exp ln x lim exp
nn n nÄ_ Ä_ Ä_ Ä_
ˆ‰ ˆ‰ ˆ ‰ ˆ‰ Š‹
x
2n 1 n 1 n n 1 n
1n 1n ln (2n 1)
n
Ý#Ý #Ý
Î
"""
Î œ_
œœ œ
x lim exp xe x, x 0 convergesœœœÊ
nÄ_
ˆ‰
œ
Ý
!2
2n 1
68. lim 1 lim exp n ln 1 lim exp lim exp
nn n nÄ_ Ä_ Ä_ Ä_
ˆ‰ ˆ ‰ ˆ‰
–—
œ œ œ
""
nn
n
œœ
ln 1 1Š‹ ‚
ˆ‰
Š‹Š ‹
Š‹
& &
&
œ
&
œ
)&
œ
œ
&
n
n
2
nn
n
lim exp e 1 convergesœœœÊ
nÄ_
ˆ‰
œ
œ
!2n
n1
œ
69. lim lim 0 converges (Theorem 5, #6)
nnÄ_ Ä_
36 36
2n! n!
nn n
n
†
†
œœÊ
70. lim lim lim 0 converg
nn nÄ_ Ä_ Ä_
ˆ‰ ˆ‰ˆ‰ ˆ ‰
ˆ‰ ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ‰ ˆ ‰
10 12 10 120
11 11 11 121
nnnn
9 11 12 9 12 11 108
10 12 11 10 11 12 110
n n nn nn n
ÝÝ Ý
œœœÊ
1
es
(Theorem 5, #4)
71. lim tanh n lim lim lim lim 1 converges
nn nnnÄ_ Ä_ Ä_ Ä_ Ä_
œœœœ"œÊ
ee e 2e
ee e 1 2e
n n 2n 2n
n n 2n 2n
œœ"
ÝÝ


72. lim sinh (ln n) lim lim diverges
nnnÄ_ Ä_ Ä_
œœœ_Ê
ee
2
n
ln n ln n
n
œ œ
#

Ú ˆ‰
73. lim lim lim lim converges
nn n nÄ_ Ä_ Ä_ Ä_
n sin
2n 1
œ Úˆ‰
n
œ #
"
œœ œœÊ
sin cos
cosˆ‰ ˆ‰
Š‹ Š ‹
ˆ‰ˆ‰Š‹
ˆ‰
œ œ
œ
&&)
œœ
&
n n
222
nnnn
n n
2
n

&
&
#
74. lim n 1 cos lim lim lim sin 0 converges
nnnnÄ_ Ä_ Ä_ Ä_
ˆ‰ ˆ‰œ œ œ œÊ
" "
n n
ˆ‰
ˆ‰
œ ‘ˆ‰Š‹
Š‹
(&cos
sin
œ
œ
œœ
&
œ
&
n
n
n n
n75. lim tan n converges 76. lim tan n 0 0 converges
nnÄ_ Ä_
œ" œ"
# #
"
œÊ œ œÊ
1 1
Èn
†
77. lim lim 0 converges (Theorem 5, #4)
nnÄ_ Ä_
ˆ‰ ˆ‰ Š‹ Š‹
"" " "
33
nn
22
n
œ œÊ
ÈÈn
78. lim n n lim exp lim exp e 1 converges
nn nÄ_ Ä_ Ä_
È
’“ ˆ‰
n#
Ý Ý
Ý
!
œ œ œ œ Ê
ln n n
nnn
2n 1ab
œ
œ
79. lim lim lim lim 0 converges
nn n nÄ_ Ä_ Ä_ Ä_
(ln n) 200 (ln n) 200 199 (ln n)
nn n n
200!
œ Ú Ú
œœ œáœœÊ
†
80. lim lim lim lim lim 0 converges
nn n n nÄ_ Ä_ Ä_ Ä_ Ä_
(ln n) 10(ln n) 80(ln n)
nnnn
3840
('&
ÈÈÈÈ
œ œœœáœœÊ
–—
Š‹
Š‹
5(ln n)
n
n
Û
œ
&œ

705. Section 11.1 Sequences 701
81. lim n n n lim n n n lim lim
nn nnÄ_ Ä_ Ä_ Ä_
Š‹Š‹Š‹
ÈÈ
œ œ œ
## _œ
_œ _œ
"
_œ
nnn
nnn nnn
n
11
È
ÈÈ
É
œ
œœ
œ
n
convergesœÊ
"
#
82. lim lim lim
nn nÄ_ Ä_ Ä_
""
œœ _ œœ _ œ_ _
œ_ _ œ_ _
œœÈÈÈ ÈÈÈ
ÈÈ ÈÈ
n1 nn n1 nn n1 nn
n1 nn n1 nn
1n
œœ œœ œ œ
œœ œ œ
œœ Š‹Š‹
lim 2 convergesœœÊ
nÄ_
É É
ˆ‰
11
1
œ_ _
œœ
ÚÚ
œ
Ú
n n
n
83. lim dx lim lim 0 converges (Theorem 5, #1)
nnnÄ_ Ä_ Ä_
"" "
nx n n
ln n'
1
n
œœœÊ
84. lim dx lim lim 1 if p 1 converges
nn nÄ_ Ä_ Ä_
'
1
n n
1
""""""
œœœx1px1pnp1
pp1p1œœœÊ’“ ˆ‰

85. 1, 1, 2, 4, 8, 16, 32, 1, 2 , 2 , 2 , 2 , 2 , 2 , x 1 and x 2 for n 2ᜠá Ê œ œ  
!"#$%&
"
n
n2
á
86. (a) 1 2(1) 1, 3 2(2) 1; let f(a b) (a 2b) 2(a b) a 4ab 4b 2a 4ab 2b
## ## # ## ## #
œ œ ßœ œ
2b a ; a 2b 1 f(a b) 2b a 1; a 2b 1 f(a b) 2b a 1œ œÊßœœ œÊßœœ
### # ## # # ##
(b) r 2 2 r 2
# ___œœœ „" "
___
# œœ
#
n n
œ œ œ œ Ê œ „ˆ‰ Ê Š‹
a 2b a 4ab 4b 2a 4ab 2b
a b (a b) (a b) y y
a2b
œœœœ
œœœ
œœ ab
n n
In the first and second fractions, y n. Let represent the (n 1)th fraction where 1 and b n 1
nÐ&ÐÐ&
aa
bb
for n a positive integer 3. Now the nth fraction is and a b 2b 2n 2 n y n. Thus,Ð'ÐÐ&Ð!Ð
a2b
ab
_
_
n
lim r 2.
nÄ_
nœ
È
87. (a) f(x) x 2; the sequence converges to 1.414213562 2œ ¸
# È
(b) f(x) tan (x) 1; the sequence converges to 0.7853981635œ ¸
1
4
(c) f(x) e ; the sequence 1, 0, 1, 2, 3, 4, 5, divergesœ á
x
88. (a) lim nf lim lim f (0), where x
n
xx
Ä_
Ä! Ä!
ˆ‰
" "_œ w
nx x n
f( x) f(0 x) f(0)
œœ œ œ
??
ÚÚ
??
??
?
(b) lim n tan f (0) 1, f(x) tan x
nÄ_
œ" ¸ œ"""
_ˆ‰
n10
œœ œ œ
œ
(c) lim n e 1 f (0) e 1, f(x) e 1
nÄ_
ab
1n xœ
œ œœ œ
w!
(d) lim n ln 1 f (0) 2, f(x) ln (1 2x)
nÄ_
ˆ‰œ œ œ œ
22
n 1 2(0)
w
_
89. (a) If a 2n 1, then b 2n 2n 2n 2n, c 2n 2nœ œÚ ÛœÚ ÛœÚ Ûœ œÜ ÝœÜ Ý
a4n4n1 a
œœ œ
## # # #
__ " " ## #
2n 2n 1 and a b (2n 1) 2n 2n 4n 4n 1 4n 8n 4nœ œ œ
# ## # # # %$#
#
ab
4n 8n 8n 4n 1 2n 2n 1 c .œœ œ
%$# # #
#
ab
(b) lim lim 1 or lim lim sin lim sin 1
aa aa
2
Ä_ Ä_ Ä_ Ä_
ÄÎ
ÚÛ ÚÛ
ÜÝ ÜÝ
aa
aa
&&
&&
&&
&&
œœœœœ
2n 2n
2n 2n 1
œ
œ
_
__
))
)1
90. (a) lim (2n ) lim exp lim exp lim exp e 1;
nn n nÄ_ Ä_ Ä_ Ä_
1
12n ln 2n
2n n
Î !
##
"ab
Š‹
œœ œœœ ˆ‰ ˆ‰

1
2
2nœ
œ
n! 2n , Stirlings approximation n! (2n ) for large values of n¸Ê¸¸ˆ‰ ˆ‰ÈÈ
nnn
eee
12nnn
11
Îab
(b) n n!
40 15.76852702 14.71517765
50 19.48325423 18.393
È
n n
e
97206
60 23.19189561 22.07276647

706. 702 Chapter 11 Infinite Sequences and Series
91. (a) lim lim lim 0
nn nÄ_ Ä_ Ä_
ln n
ncn cn
cc1cœœœ
ˆ‰
œ
n

"
(b) For all 0, there exists an N e such that n e ln n ln n ln%œÊÊ
Ð ÑÎ Ð ÑÎ "ln c ln c c ln
c
%% %
% ˆ‰
n 0 lim 0
n
ÊÊÊ Ê œ
Ä_
c
nn n
"" " "
%
cc c%%¸¸
92. Let {a } and {b } be sequences both converging to L. Define {c } by c b and c a , where
nn n2nn2n1n œœ
(
n 1, 2, 3, . For all 0 there exists N such that when n N then a L and there exists Nχ %% "" # kk
n
such that when n N then b L . If n 1 2max{N N }, then c L , so {c } converges to L.Ñ&†Ñ'â &† #"#kk kk
n nn %%
93. lim n lim exp ln n lim exp e 1
nn nÄ_ Ä_ Ä_
1n
nn
Î! ""
œœœœ ˆ‰ ˆ‰
94. lim x lim exp ln x e 1, because x remains fixed while n gets large
nnÄ_ Ä_
1n
n
Î! "
œœœ ˆ‰
95. Assume the hypotheses of the theorem and let be a positive number. For all there exists a N such that%%
"
when n N then a L a L L a , and there exists a N such that whenÑ&†!&†&†!&†" # kknn n %% % %
n N then c L c L c L . If n max{N N }, thenÑ&†!&†&†!†'Ñâ
# "#kknnn %% % %
L abcL bL lim bL. Ÿ Ÿ Ê Ê œ%%%
nnn n n kk
nÄ_
96. Let . We have f continuous at L there exists so that x L f(x) f(L) . Also, a L there%$$%ÑÄ ! & † ! & † Þ ! kk k k n
exists N so that for n N a L . Thus for n N, f(a ) f(L) f(a ) f(L).Ñ&† Ñ &†! Þkk k k nnn $%
97. a a 3n 3n 4n 4 3n 6n n 2n1 n
3(n 1) 1
(n 1) 1 n 1 n n 1
3n 1 3n 4 3n 1
_
__
__ _ _# _
___ ##Ð! Ñ!Ñ!'''Ñ'''
4 2; the steps are reversible so the sequence is nondecreasing; 3 3n 1 3n 3!Ñ †! '† '
3n
n1
_"
_
1 3; the steps are reversible so the sequence is bounded above by 3!†
98. a a n1 n
(2(n 1) 3)! (2n 3)! (2n 5)! (2n 3)! (2n 5)! (n 2)!
((n 1) 1)! (n 1)! (n 2)! (n 1)! (2n 3)! (n 1)!
_
_______
__ _ _ _ _ _Ð! Ñ!Ñ!Ñ
(2n 5)(2n 4) n 2; the steps are reversible so the sequence is nondecreasing; the sequence is not!' 'Ñ'
bounded since (2n 3)(2n 2) (n 2) can become as large as we please
(2n 3)!
(n 1)!
_
_
œ â
99. a a 2 3 n 1 which is true for n 5; the steps aren1 n
23 23 23
(n 1)! n! 2 3 n!
(n 1)!
_
_
_Î! Î ! Î ! Î' Ð
n1n1 nn n1n1
nn** **
†
reversible so the sequence is decreasing after a , but it is not nondecreasing for all its terms; a 6, a 18,
&"# œœ
a 36, a 54, a 64.8 the sequence is bounded from above by 64.8$%&œœœœÊ
324
5
100. a a 2 2 ; the steps aren1 n
2222 2
n1 n n n1 n(n 1)
_
_# # _ # # _ #
"" "" "Ð!& & Ð&&!& Ð &! Ð&
n1 n n1 n n1***
reversible so the sequence is nondecreasing; 2 2 the sequence is bounded from above&& Î !
2
n
"
#
n
101. a 1 converges because 0 by Example 1; also it is a nondecreasing sequence bounded above by 1
nœ Ä
"" nn
102. a n diverges because n and 0 by Example 1, so the sequence is unbounded
nœ Ä_ Ä
"" nn
103. a 1 and 0 ; since 0 (by Example 1) 0, the sequence converges; also it isn
21
2nnœœ Ä ÊÄ
n
nn n n
œ" """ "
## #
a nondecreasing sequence bounded above by 1
104. a ; the sequence converges to by Theorem 5, #4n
21 2
333
nœœ !
n
nn
œ"
ˆ‰

707. Section 11.1 Sequences 703
105. a ( 1) 1 diverges because a 0 for n odd, while for n even a 2 1 converges to 2; itnnn
n n1
n nœ œ œ ab ˆ‰ ˆ ‰
_ "
diverges by definition of divergence
106. x max {cos 1 cos 2 cos 3 cos n} and x max {cos 1 cos 2 cos 3 cos (n 1)} x with x 1
nn1 nnœ ßßßáß œ ßßßáß   Ÿ _
so the sequence is nondecreasing and bounded above by 1 the sequence converges.Ê
107. If {a } is nonincreasing with lower bound M, then { a } is a nondecreasing sequence with upper bound M.
nn &&
By Theorem 1, { a } converges and hence {a } converges. If {a } has no lower bound, then { a } has no&&
nnn n
upper bound and therefore diverges. Hence, {a } also diverges.n
108. a a n 2n 1 n 2n 1 0 and 1; thus the sequence isnn1
n1 n1
nn1 n
(n 1)ÐÓÐ Ó''Ð'ÓÐ Ð_
_ ___"
_
##
nonincreasing and bounded below by 1 it convergesÊ
109. a a n 1 2n 2n n 2n 2n n 1 nnn1
12n
n
2(n 1)
n1ÐÓ Ð Ó'' 'Ð' 'Ó'Ð_
_ "_ _
_
##
ÈÈ
ÈÈ
ÈÈ ÈÈ ÈÈ
and 2 ; thus the sequence is nonincreasing and bounded below by 2 it converges
12n
n
_
È
È
 Ê
ÈÈ
110. a a 2 2 4 2 2 4 2 2 2 4 2 4nn1
14 14
2
n1 n1n n nn1 n1 n n1n nn1ÐÓÐÓ&Ð&Ó&Ð&_
œœ
#
__ _ _ _ _
nn1
nn1 *
*
2 1 2 4 4 1 4 (2 4) 1 ( 2) 4 ; thus the sequence is nonincreasing. However,Ó&Ð & ÓÐ & ÓÐ&††
nn1 n n_
a 2 which is not bounded below so the sequence divergesn
4
2
nœœ
""
##
nn n
n
111. 4 so a a 4 4 1 and
43 3 3 3 3 3 3
44 4444 4
nnnnn1
nn1
n1 n
n*
_
_
_" _
œ   Í   Í   Í  ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
4 4; thus the sequence is nonincreasing and bounded below by 4 it converges'Ð !ˆ‰
3
4
n
112. a 1, a 2 3, a 2(2 3) 3 2 2 3, a 2 2 2 3 3 2 2 1 3,"# $ %
##$$œ œ œ œ " œ " œ ab a b ab ab
22
††
a 2 2 2 1 3 3 2 2 1 3, , a 2 2 1 3 2 3 2 3
&
$ $ % % œ" œ" œ" œœœáœ œ cd ab a bab n
nn n n1 †
2(13)3 23; aa 23 2 3 2 2 12œ œ   Í  Í  Í Ÿ
n1 n n n1 n n1
nn1
œ __
_
so the sequence is nonincreasing but not bounded below and therefore diverges
113. Let 0 M 1 and let N be an integer greater than . Then n N n n nM M†† Ñ!Ñ !& Ñ
MM
1M 1Mœœ
n M nM n M(n 1) M.!Ñ' !Ñ ' ! Ñ
n
n1_
114. Since M is a least upper bound and M is an upper bound, M M . Since M is a least upper bound and M"#"##" Ÿ
is an upper bound, M M . We conclude that M M so the least upper bound is unique.
#" "#Ÿœ
115. The sequence a 1 is the sequence , , , , . This sequence is bounded above by ,n
() 33 3œ á
œ"
# #### #
""
n
but it clearly does not converge, by definition of convergence.
116. Let L be the limit of the convergent sequence {a }. Then by definition of convergence, for theren
%
#
corresponds an N such that for all m and n, m N a L and n N a L . NowÑ! &† Ñ! &†kk kk mn
%%
##
a a a L L a a L L a whenever m N and n N.kkk kkkkkmn m n m n œ Ÿ œ
%%
##
%
117. Given an 0, by definition of convergence there corresponds an N such that for all n N,%
La and La . Now LL LaaL La aL 2.kk kk k kk kkkkk
"# #"#"#" œ Ÿ œnn nnnn%% %%%
L L 2 says that the difference between two fixed values is smaller than any positive number 2 .kk
#"&† % %
The only nonnegative number smaller than every positive number is 0, so L L 0 or L L .kk
"# " #œ œ

708. 704 Chapter 11 Infinite Sequences and Series
118. Let k(n) and i(n) be two order-preserving functions whose domains are the set of positive integers and whose
ranges are a subset of the positive integers. Consider the two subsequences a and a , where a L ,
kn in knÐÑ ÐÑ ÐÑ " Ä
a L and L L . Thus a a L L 0. So there does not exist N such that for all m, n N
in kn inÐÑ ÐÑ ÐÑ#"# "#Ä Á Ä ¸¸ kk
a a . So by Exercise 116, the sequence a is not convergent and hence diverges.Ê Ö×kk
mn n %
119. a L given an 0 there corresponds an N such that 2k N a L . Similarly,
2k 2kÞÓ Ñ Ñ ! &† %% "" cd kk
a L 2k 1 N a L . Let N max{N N }. Then n N a L whether
2k 1 2k 1 n_#_ "#ÄÍ Ê œ ß Ê cd kkkk %%
n is even or odd, and hence a L.
nÄ
120. Assume a 0. This implies that given an 0 there corresponds an N such that n N a 0
n nÞÑ Ñ!&† %% kk
a a a 0 a 0. On the other hand, assume a 0. This implies that!†! †! &†! Þ Þkk k k k k kk kkkk kk
nn n n n%% %
given an 0 there corresponds an N such that for n N, a 0 a a%%%%ÑÑ&†!†!† kk kk kkkk kk
nnn
a 0 a 0.!&†!Þkk
nn %
121. 0.5 1 10 1 n n 692.8¹¹
È ˆ‰ ˆ‰ ˆ‰
n
999
1000
&† !& † &† ! †† !Ñ !Ñ
œ$ "" " "
##
Î
1000 1000 1000 1000
1n
999 1001
nn ln
ln
ˆ‰
ˆ‰
Ú
œ
N 692; a and lim a 1Êœ œ œ nn
1n
ˆ‰
"
#
Î
nÄ_
122. n 1 10 n 1 n n 9123 N 9123;¸ ¸ ˆ‰ ˆ‰È
n
Ê Ê Ê Ê œ
œ$ Ÿ ""
1000 1000 1000 1000
1n 999 1001
nn
a n n and lim a 1nn
1nœœ œÈ
n Î
nÄ_
123. (0.9) 10 n ln (0.9) 3 ln 10 n 65.54 N 65; a and lim a 0
n 3 ln 10 9
ln (0.9) 10
nn
n
Ê Ê ¸ Êœ œ œ
œ$ œ ˆ‰
nÄ_
124. 10 n! 2 10 and by calculator experimentation, n 14 N 14; a and lim a 0
2 2
n! n!
n
nn
n n
Ê Êœ œ œ
œá á
nÄ_
125. (a) f(x) x a f (x) 2x x x xœÊ œ Ê œ Ê œ œ œ
#w
__
œœœ_
##
_
n1 n n1
xa 2x xa xa
x2x2x
x
œœœœ
nnnnn
nnn
n
a
x ab
ˆ‰
(b) x 2, x 1.75, x 1.732142857, x 1.73205081, x 1.732050808; we are finding the positive"# $ % &œœ œ œ œ
number where x 3 0; that is, where x 3, x 0, or where x 3 .
##
œ œ œ
È
126. x 1.5, x 1.416666667, x 1.414215686, x 1.414213562, x 1.414213562; we are finding the
"#$%&œœœœœ
positive number x 2 0; that is, where x 2, x 0, or where x 2 .
##
œ œ œ
È
127. x 1, x 1 cos (1) 1.540302306, x 1.540302306 cos (1 cos (1)) 1.570791601,
"# $œœ œ œ œ
x 1.570791601 cos (1.570791601) 1.570796327 to 9 decimal places. After a few steps, the%
#œ œœ
1
arc x and line segment cos x are nearly the same as the quarter circle.ab ab
n1 n1áá
128. (a) S 6.815, S 6.4061, S 6.021734, S 5.66042996, S 5.320804162, S 5.001555913,"# $ % & 'œœ œ œ œ œ
S 4.701462558, S 4.419374804, S 4.154212316, S 3.904959577, S 3.670662003,
()*"!""œœœœœ
S 3.450422282 so it will take Ford about 12 years to catch up
"#œ
(b) x 11.8¸
129-140. Example CAS Commands:
:Maple
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;

709. Section 11.1 Sequences 705
plot( f(x), x=a..b, title="#23(a) (Section 5.1)" );
N := [ 100, 200, 1000 ]; # (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do # (c)
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot); # (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
: (sequence functions may vary):Mathematica
Clear[a, n]
a[n_]; = n
1 / n
first25= Table[N[a[n]],{n, 1, 25}]
Limit[a[n], n 8]Ä
The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging your table
to more than the first 25 values.
If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01 of the
limit, do the following.
Clear[minN, lim]
lim= 1
Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}]&
minN
For sequences that are given recursively, the following code is suggested. The portion of the command a[n_]:=a[n] stores
the elements of the sequence and helps to streamline computation.
Clear[a, n]
a[1]= 1;
a[n_]; = a[n]= a[n 1] (1/5)&'
(n 1)œ
first25= Table[N[a[n]], {n, 1, 25}]
The limit command does not work in this case, but the limit can be observed as 1.25.
Clear[minN, lim]
lim= 1.25
Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}]&
minN
141. Example CAS Commands:
:Maple
with( Student[Calculus1] );
A := n->(1+r/m)*A(n-1) + b;
A(0) := A0;
A(0) := 1000; r := 0.02015; m := 12; b := 50; # (a)
pts1 := [seq( [n,A(n)], n=0..99 )]:
plot( pts1, style=point, title="#141(a) (Section 11.1)");

710. 706 Chapter 11 Infinite Sequences and Series
A(60);
The sequence { A[n] } is not unbounded;
limit( A[n], n=infinity ) = infinity.
A(0) := 5000; r := 0.0589; m := 12; b := -50; # (b)
pts1 := [seq( [n,A(n)], n=0..99 )]:
plot( pts1, style=point, title="#141(b) (Section 11.1)");
A(60);
pts1 := [seq( [n,A(n)], n=0..199 )]:
plot( pts1, style=point, title="#141(b) (Section 11.1)");
# This sequence is not bounded, and diverges to -infinity:
limit( A[n], n=infinity ) = -infinity.
A(0) := 5000; r := 0.045; m := 4; b := 0; # (c)
for n from 1 while A(n)<20000 do end do; n;
It takes 31 years (124 quarters) for the investment to grow to $20,000 when the interest rate is 4.5%, compounded
quarterly.
r := 0.0625;
for n from 1 while A(n)<20000 do end do; n;
When the interest rate increases to 6.25% (compounded quarterly), it takes only 22.5 years for the balance to reach
$20,000.
B := k -> (1+r/m)^k * (A(0)+m*b/r) - m*b/r; # (d)
A(0) := 1000.; r := 0.02015; m := 12; b := 50;
for k from 0 to 49 do
printf( "%5d %9.2f %9.2f %9.2f", k, A(k), B(k), B(k)-A(k) );
end do;
A(0) := 'A(0)'; r := 'r'; m := 'm'; b := 'b'; n := 'n';
eval( AA(n+1) - ((1+r/m)*AA(n) + b), AA=B );
simplify( % );
142. Example CAS Commands:
:Maple
r := 3/4.; # (a)
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 99 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title="#142(a) (Section 11.1)" );
R1 := [1.1, 1.2, 1.5, 2.5, 2.8, 2.9]; # (b)
for r in R1 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 99 do
A := r*A*(1-A);
L := L, [n,A];

711. Section 11.1 Sequences 707
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(b) (Section 11.1)r = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R1)], insequence=true );
R2 := [3.05, 3.1, 3.2, 3.3, 3.35, 3.4]; # (c)
for r in R2 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 99 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(c) (Section 11.1)r = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R2)], insequence=true );
R3 := [3.46, 3.47, 3.48, 3.49, 3.5, 3.51, 3.52, 3.53, 3.542, 3.544, 3.546, 3.548]; # (d)
for r in R3 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 199 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(d) (Section 11.1)r = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R3)], insequence=true );
R4 := [3.5695]; # (e)
for r in R4 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(e) (Section 11.1)r = %f", r);

712. 708 Chapter 11 Infinite Sequences and Series
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R4)], insequence=true );
R5 := [3.65]; # (f)
for r in R5 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(f) (Section 11.1)r = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R5)], insequence=true );
R6 := [3.65, 3.75]; # (g)
for r in R6 do
for a in [0.300, 0.301, 0.600, 0.601 ] do
A := a;
L := [0,a];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,a] := [L];
end do:
t := sprintf("#142(g) (Section 11.1)r = %f", r);
P[r] := plot( [seq( pt[r,a], a=[0.300, 0.301, 0.600, 0.601] )], style=point, title=t );
end do:
display( [seq(P[r], r=R6)], insequence=true );
11.2 INFINITE SERIES
1. s lim s 3nn
a1 r
(1 r)
21
1 1
2œœ Ê œœ
ab ˆ‰ˆ‰
ˆ‰ ˆ‰
œ
œ
œ
œœ
n
3
n
33 Ú
ÚÚ
nÄ_
2. s lim snn
a1 r
(1 r) 11
1
1 1 œœ Ê œœ
ab ˆ‰ˆ ‰ ˆ‰ˆ‰
ˆ‰ ˆ‰
œ
œ
œ
œœ
"
n 99
100 100 100
n
100 100 Ú
ÚÚ
nÄ_
3. s lim snn
a1 r
(1 r) 3
1
1
2œœ Ê œœ
ab ˆ‰
ˆ‰ ˆ‰
œ
œ
œœ
œœ
"
n
n
3 Ú
œ
Ú
œ œ
nÄ_
4. s , a geometric series where r 1 divergencen
1 ( 2)
1 ( 2)œÊ
œœ
œœ
n
kk
5. s lim s
""" """" """" "
_ _# _ _# # _ _# # _# #(n 1)(n ) n 1 n 3 3 4 n 1 n n
n n
œ Êœá œ Ê œˆ‰ˆ‰ ˆ ‰
nÄ_

713. Section 11.2 Infinite Series 709
6. s 5 5
5 5 5 5 55 55 5 5 5 5 5
n(n 1) nn 1 2 23 34 n 1n nn 1 n 1
n
__ œ _ _
œ Êœá œˆ‰ˆ‰ˆ‰ˆ‰ˆ‰
lim s 5ʜ
nÄ_
n
7. 1 , the sum of this geometric series is á œ œ
"" " " "
œœ _41664 51 1
4
ˆ‰ ˆ‰
ÚÚ
44
8. , the sum of this geometric series is
"" " "
œ #16 64 256 1 1
á œ
ˆ‰
ˆ‰
Ú
Ú
16
4 9. , the sum of this geometric series is
77 7 7
41664 3 1
á œ
ˆ‰
ˆ‰
7
4
4
œ
Ú10. 5 , the sum of this geometric series is 4á œ
55 5 5
41664 1 œœˆ‰
Ú
4
11. (5 1) , is the sum of two geometric series; the sum is''''''' '(ˆ‰ˆ‰ˆ ‰
555
34987##
"" "
10
5323
1 1 œœ
"
##ˆ‰ ˆ‰
ÚÚ
œœœ
3
12. (5 1) , is the difference of two geometric series; the sum is&'&'&'& '(ˆ‰ˆ‰ˆ ‰
555
34987##
"" "
10
5317
1 1 œœ
"
##ˆ‰ ˆ‰
ÚÚ
œœœ
3
13. (1 1) , is the sum of two geometric series; the sum is''&'' '& '(ˆ‰ˆ ‰ˆ ‰
11 1
5425815##
"" "
2
1517
1 1 66œ_
"
ˆ‰ ˆ‰
ÚÚ
œœœ
5
14. 2 2 1 ; the sum of this geometric series is 2 áœ á œ
48 16 24 8 10
525125 525125 31
ˆ‰ Š‹
"
œˆ‰
2
5
15. s 1
4
(4n 3)(4n 1) 4n 3 4n 1 5 5 9 9 13 4n 7 4n 3
n
œ_ œ _ œœ
" " """"" " "
œ Êœ á ˆ‰ˆ‰ˆ ‰ ˆ ‰
1 lim s lim 1 1œÊ œ œˆ‰ ˆ‰
"" " "
œ_ _ _4n 3 4n 1 4n 1 4n 1
n
nnÄ_ Ä_
16. A(2n 1) B(2n 1) 6
6AB
(2n 1)(2n 1) 2n 1 2n 1 (2n 1)(2n 1)
A(2n 1) B(2n 1)
œ_ œ _ œ_
__ œ
œœ Êœ
(2A 2B)n (A B) 6 2A 6 A 3 and B 3. Hence,
2A 2B 0 A B 0
A B6 AB6
Ê œÊ Ê Ê œÊœ œ
œ œ
œ œ
œœ
3 3!! ˆ‰ Š‹
kk
n1 n1œœ
6
(2n 1)(2n 1) n 1 n 1 133557 (k 1) 12k 1 k 1œ _ #œ #_ # œ _ œ #_
" " """""" " " "
œ œ á
3 1 the sum is lim 3 1 3œ Ê œˆ‰ ˆ‰
""
#_ #_k 1 k 1
kÄ_
17.
40n A B C D
(2n 1) (2n 1) (2n 1) (2n 1) (2n 1) (2n 1)œ_ œœ __
œœ œ œœ
œ
A(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1) D(2n 1)
(2n 1) (2n 1)
œ _____ œ_œ
œ_
œœ œ œ
œœ
A(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1) D(2n 1) 40nÊ œ
## ##
A8n4n2n1B4n4n1C8n4n2n1D4n4n140nÊ œ œabababab
$##$# #
(8A 8C)n (4A 4B 4C 4D)n ( 2A 4B 2C 4D)n ( A B C D) 40nÊ œ
$#

8A 8C 0 8A 8C 0
4A 4B 4C 4D 0 A B C D 0
2A 4B 2C 4D 40 A 2
A B C D 0
ÊÊ
œ œ
œ œ
œ
œ
ÚÚ
ÝÝ
ÝÝ
ÛÛ
ÝÝ
ÝÝ
ÜÜ
BC2D20 2B2D20
A BC D 0
4B 20 B 5
B D 0
œ œ
œ
Êʜʜ
œ
œ
and D 5 C 0 and A 0. Hence,
A C 0
A5C50
œ Ê Ê œ œ
œ
œ
œ
!’“
k
n1œ
40n
( n 1) (2n 1)#œ _
œœ

714. 710 Chapter 11 Infinite Sequences and Series
5 5œ œ á !’“Š ‹
k
n1œ
" " """ " " " " "
#œ #_ # # œ _ #œ #_( n 1) ( n 1) 1 9 9 5 5 (2(k 1) 1) ( k 1) ( k 1)
œœ œœœ
5 1 the sum is lim 5 1 5œ Ê œŠ‹ Š‹
""
__(2k 1) (2k 1)
œœ
nÄ_
18. s 1
2n 1
n (n 1) n (n 1) 4 4 9 9 16 (n 1) n n (n 1)
n
_ "" """"" "" ""
__ œ_
œœœ œ œœ œ œœ Êœ á ˆ‰ˆ‰ˆ ‰ ’“’“
lim s lim 1 1Êœœ
nnÄ_ Ä_
n
(n 1) ’“
"
_
œ
19. s 1 1 n
22 4 n 1 33 nn n 1 n 1œá œŠ‹Š‹Š‹Š ‹Š ‹
""""" """" "
œ __ÈÈÈÈÈ ÈÈÈÈ È
lim s lim 1 1Êœœ
nnÄ_ Ä_
n
n 1 Š‹
"
_È
20. s nœá œˆ‰ˆ‰ˆ‰ˆ ‰ˆ ‰
"" "" "" " " " " " "
# ###### # ### #
Ú 1n1 1n 1n 1n1 1n1
lim sÊ œœ
nÄ_
n
1
"" "
##
21. sn
ln 3 ln ln 4 ln 3 ln 5 ln 4 ln (n 1) ln n ln (n 2) ln (n 1)œá ˆ‰ˆ‰ˆ‰ Š‹Š ‹
"" "" "" " " " "
#___
lim sœ Ê œ
"" "
#_ #ln ln (n 2) ln
n
nÄ_
22. s tan (1) tan (2) tan (2) tan (3) tan (n 1) tan (n)
nχcdcdc d
œ" œ" œ" œ" œ" œ"
tan (n) tan (n 1) tan (1) tan (n 1) lim s tan (1) œ Ê œ œœcd
œ" œ" œ" œ" œ"
##
nÄ_
n
44
111 1
23. convergent geometric series with sum 2 2
"
œ œ1
2
21Š‹
È
È
œ
œ
2
œœ
È
24. divergent geometric series with r 2 1 25. convergent geometric series with sum 1kk
È
œ œ
Š‹
Š‹
3
&
œ
&
1 œ&
26. lim ( 1) n 0 diverges 27. lim cos (n ) lim ( 1) 0 diverges
nnnÄ_ Ä_ Ä_
ÁÊ œ ÁÊ
n1 n_
1
28. cos (n ) ( 1) convergent geometric series with sum 1œ Ê œ
n
1
5
6
"
œŠ‹
&
œ
5
29. convergent geometric series with sum
"
œ
œ
1
e
e1
Š‹
œ
&
e
œ
œ
œ
30. lim ln 0 diverges
nÄ_
"
n
œ_Á Ê
31. convergent geometric series with sum 2
220182
1
999
œŠ‹
œ
10
œœ
32. convergent geometric series with sum
"
œ
œ
1
x
x 1
Š‹
œ
x
œ
33. difference of two geometric series with sum 3
""
œœ
##
1 1
33
Š‹ Š‹
2
33
œœ
œ
34. lim 1 lim 1 e 0 diverges
nnÄ_ Ä_
ˆ‰ ˆ ‰œ œÁÊ
"œ" œ"
nn
nn

715. Section 11.2 Infinite Series 711
35. lim 0 diverges 36. lim lim lim n diverges
nnnnÄ_ Ä_ Ä_ Ä_
n! nnnn
1000 n! 1 n
n
nœ_Á Ê œ œ_ Ê
†
†
â
#â
37. ln ln (n) ln (n 1) s ln (1) ln (2) ln (2) ln (3) ln (3) ln (4)!!ˆ‰ c dcdcdcd
__
œœn1 n1
n
n 1_
œ Êœá
n
ln (n 1) ln (n) ln (n) ln (n 1) ln (1) ln (n 1) ln (n 1) lim s , diverges œ œ Ê œ_ Êcdcd
nÄ_
n
38. lim a lim ln ln 0 diverges
nnÄ_ Ä_
n
n
2n 1œœÁÊ ˆ‰ˆ‰
_#
"
39. convergent geometric series with sum
"
œ œ1 eˆ‰
e
œ
œ
1
1
40. divergent geometric series with r 1kkœ¸
e 23.141
22.459
œ
1
e
41. ( 1) x ( x) ; a 1, r x; converges to for x 1!! kk
__
œœn0 n0
œ œœ œ
nn n ""
œœ _1 ( x) 1 x
42. ( 1) x x ; a 1, r x ; converges to for x 1!! ab kk
__
œœn0 n0
œ œœ
n2n
n
1 x
## "
_
œ
43. a 3, r ; converges to for 1 1 or 1 x 3œœ œ
x 1 3 6 x
1
3 x
œœ"
#œ#
œŠ‹
x áœ
&
44. ; a , r ; converges to !! ˆ‰ ˆ‰
__
œœn0 n0
(1)
3 sin x 3 sin x 3 sin x
nn
1
œ
#_ #_ # _
""œ""œ"
œ
n
œœœ
ˆ‰
Š‹
Ú
œ
áœ
'
3 sin x
for all x since for all xœœ ŸŸ
3 sin x 3 sin x
2(4 sin x) 8 2 sin x 4 3 sin x
__ """
__ _#
ˆ‰
45. a 1, r 2x; converges to for 2x 1 or xœœ
""
œ#1 2x
kk kk
46. a 1, r ; converges to for 1 or x 1.œœ œ
""
œ
_xx 1x
1
x1
œœœ
œ
Š‹
áœ
&
x
¸¸ kk
47. a 1, r (x 1) ; converges to for x 1 1 or 2 x 0œ œ œ
n
1 (x 1) x
""
__ #_
kk
48. a 1, r ; converges to for 1 or 1 x 5œœ œ
3 x 2 3 x
1
x 1
œ"œ
#œ#
œŠ‹
3 xá
&
¸¸
49. a 1, r sin x; converges to for x (2k 1) , k an integerœœ Á
"
œ#1 sin x
1
50. a 1, r ln x; converges to for ln x 1 or e x eœœ
"
œ
œ"
1 ln x
kk
51. 0.23 52. 0.234 œœœ œ œœ!!ˆ‰ ˆ‰
_ _
œ œn0 n0
23 23 234 234
100 10 99 1000 10 999
n n
1 1
" "
œ œ
œ &
Š‹ Š‹
ˆ‰
Š‹
23 234
100 1000
1000
œ œ
100
53. 0.7 54. 0.d œœœ œœœ!!ˆ‰ ˆ‰
_ _
œ œn0 n0
77 dd
10 10 9 10 10 9
n n
1 1
" "
œ œ
Š‹ Š‹
Š‹ Š‹
7 d
10 10
10 10
œ œ
55. 0.06 œœœœ!ˆ‰ˆ‰ˆ‰
_
œn0
16 6
10 10 10 90 15
n
1
""
œ
Š‹
Š‹
6
100
10
œ

716. 712 Chapter 11 Infinite Sequences and Series
56. 1.414 1 1 1œ œ œ œ! ˆ‰
_
œn0
414 414 413
1000 10 999 999
n
1
""
œ
&
Š‹
Š‹
414
1000
1000
œ
57. 1.24123 œ œ œ œ œ œ
124 123 124 124 123 124 123
100 10 10 100 100 10 10 100 99,900 99,900 33,300
n
1
123,999 41,333! ˆ‰
_
œn0

"
œ
œ
Š‹
Š‹
123
10
10
!
œ
)
58. 3.142857 3 3 3œ œ œ œ œ! ˆ‰
_
œn0
142,857 142,857 3,142,854 116,402
10 10 10 1 999,999 37,037
n
1
__ _
"
œ
œ
Š‹
Š‹
142,857
10
10
_
œ
_
59. (a) (b) (c) !!!
___
œœ œœn 2 n0 n5
"" "
__ __ œœ#(n 4)(n 5) (n 2)(n 3) (n 3)(n )
60. (a) (b) (c) !!!
___
œœ œœn 1 n3 n20
55 5
(n 2)(n 3) (n 2)(n 1) (n 19)(n 18)__ œœ œ œ
61. (a) one example is 1
""" "
#
œ
ᜠœ
4816
1
Š‹
Š‹
œ
&
œ
&
(b) one example is 3 ᜠœ
333 3
4816
1
#
œ
œ
Š‹
Š‹
3
&
œ
&
(c) one example is 1 ; the series k where k is any positive or á ᜠœ
""" "
##
œ
4816 48
kkk
1
Š‹
Š‹
k
&
œ
&
negative number.
62. The series k is a geometric series whose sum is k where k can be any positive or negative number.!ˆ‰
_
œn0
1
2
n1
1
_
œ
Š‹
Š‹
k
&
œ
&
œ63. Let a b . Then a b 1, while (1) diverges.nn n n
nn
a
bœœ œ œ œ œˆ‰ ˆ‰ !!! ! ! Š‹
""
##
__ _ _ _
œœ œ œ œn1 n1 n1 n1 n1
n
n
64. Let a b . Then a b 1, while a b AB.nn n n nn
nnn
43œœ œ œ œ œ œÁˆ‰ ˆ‰ ˆ‰!!! ! ! ab
""""
##
__ _ _ _
œœ œ œ œn1 n1 n1 n1 n1
65. Let a and b . Then A a , B b 1 and 1 .nn n n
43bB
nn n
a Aœœ œœœœ œœÁˆ‰ ˆ‰ ˆ‰ !!!! Š‹
"" " "
# #
____
œœœœn1 n1 n1 n1
n
n
66. Yes: diverges. The reasoning: a converges a 0 diverges by the!! !Š‹ Š‹
" ""
a aa
nn
n nn
Ê Ä Ê Ä_Ê
nth-Term Test.
67. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series
that diverges does not change the divergence of the series.
68. Let A a a a and lim A A. Assume a b converges to S. Let
nnnnnœá œ "#
nÄ_
!ab
S(ab)(ab) (ab) S(aa a)(bb b)
nnnnnnœá Êœáá"" ## "# "#
b b b S A lim b b b S A b converges. ThisÊáœ Ê á œÊ"# "# nnn n n
nÄ_
ab !
contradicts the assumption that b diverges; therefore, a b diverges.!! ab
nnn '

717. Section 11.3 The Integral Test 713
69. (a) 5 1 r r ; 2 2 2
22 333
1r 5 5 5 5œ
#
œÊœÊœ á ˆ‰ ˆ‰
(b) 5 1 r r ;
Š‹
13
2
1 r 10 10 2 10 10 10
13 3 13 13 3 13 3 13 3
œ###
#$
œÊœÊœá ˆ‰ ˆ‰ ˆ‰
70. 1 e e 9 1 e e b ln ᜠœÊœ Ê œÊœ
b2b b b
1e 999
88""
œ
b
ˆ‰
71. s 1 2r r 2r r 2r r 2r , n 0, 1,
n
2n 2n 1œ á œ á
#$%&
s 1rr r 2r2r2r 2r lim sÊ œ á á Ê œ n n
2n 2n 1
1r 1r
2raba b
#% $ & "
œœ
nÄ_
œœ
, if r 1 or r 1œ
12r
1r
_
œ
#
œkk kk
72. L sœ œn
aar
1r 1r 1r
a1 r
œœ œ
œab
n n
73. distance 4 2 (4) (4) 4 2 28 mœ ᜠœ’“ˆ‰ ˆ‰

33 3
44
1
#
œŠ‹
3
4
74. time 2 2 2 2œáœáÉÉ ÉÉ ˆ ‰ˆ‰ ˆ ‰ˆ‰ ˆ ‰ˆ‰ ˆ‰É ÉÉ
”•É
443 43 43 4433
4.9 4.9 4 4.9 4 4.9 4 4.9 4.9 4 4
#$ #
12.58 secœ œ œ œ ¸
24 24
4.9 4.9 4.9 4.9 2 3
1
3423
423 43
4.9 2 3 4.9 2 3
ÈÈ ÈÈ È
ÈÈ Š‹
ÈÈ
ÈÈ ÈÈ
Š‹ Š‹
Š‹ Š‹Š ‹
–—
É
É
3
4
3
4
œ
œ
œ_
œœ
_75. area 2 2 (1) 4 2 1 8 mœ á œ á œ œ
## #
##
""
# œ
Š‹ Š‹
È
È
2
4
1
œ
&
76. area 2 4 8œáœáœ œ
–—–—–—
ˆ‰
111
1
Š‹ Š‹ Š‹ Š‹
Š‹
œœœ œ
&
œ
&
œœœ
### #
""
œ
48 4
11
4816
1
177. (a) L 3, L 3 , L 3 , , L 3 lim L lim 3"# $
#œœ œ áœ Ê œ œ_ˆ‰ ˆ‰ ˆ‰ ˆ‰
44 4 4
33 3 3
nn
n1 n1
((
nnÄ_ Ä_
(b) Using the fact that the area of an equilateral triangle of side length s is s , we see that A ,
ÈÈ
33
44
2
"
œ
AA3 , AA34 ,#" $#
""
##œ œ œ œ Š‹ Š‹ˆ‰ ˆ ‰ ab
È ÈÈ È ÈÈÈ
3 33 3 333
43 4 1 43 4 12 7
22
2
A A 3 4 , A A 3 4 , . . . ,%$
""œ œab abŠ‹ Š‹ˆ‰ ˆ‰
2333
43 43
22
54
ÈÈ
34
A 34 334 33
n
n2 n2 n2œ œ œ
ÈÈÈ È
333 3
4434 949
n2 n n1 n1
4
!! !ab abŠ‹ˆ‰ ˆ‰ ÈÈ
Π'
'' '
** *
œœ$ ""
œœ
2 n1
n
&

33 33œ œ œ
ÈÈÈ
3323
44205 1
1
ÈÈ
Π'
ˆ‰
1
36
4
9
œ
78. Each term of the series represents the area of one of the squares shown in the figure, and all of the!
_
œn1
"
n
œ squares lie inside the rectangle of width 1 and length 2. Since the squares do not fill the!ˆ‰
'
*
n0
""
# œ
n
1
œœ
œ
&
rectangle completely, and the area of the rectangle is 2, we have 2.!
_
œn1
"
n
Ϡ
11.3 THE INTEGRAL TEST
1. converges; a geometric series with r 1 2. converges; a geometric series with r 1œ œ
" "
10 e

718. 714 Chapter 11 Infinite Sequences and Series
3. diverges; by the nth-Term Test for Divergence, lim 1 0
nÄ_
n
n1_
œÁ
4. diverges by the Integral Test; dx 5 ln (n 1) 5 ln 2 dx ''
11
n
55
x1 x1__
œÊ Ä_
'
5. diverges; 3 , which is a divergent p-series (p )!!
__
œœn1 n1
3
nnÈÈ
œœ
""
#
6. converges; 2 , which is a convergent p-series (p )!!
__
œœn1 n1
œ"
#
23
nn nÈ
œ œ
&ʜ
7. converges; a geometric series with r 1œ
"
8
8. diverges; 8 and since diverges, 8 diverges!! ! !
__ _ _
œœ œ œn1 n1 n1 n1
œ"81 1
nn n n
œ
9. diverges by the Integral Test: dx ln n ln 2 dx ''
22
n
ln x ln x
xx
œÊ Ä_
"
#
#
ab
'
10. diverges by the Integral Test: dx; te dt lim 2te 4e
t ln x
dt
dx e dt ''
2ln2
''
ln x
x
dx
x
t
t2 t2 t2
È
Ô×
ÕØ
( ‘
œ
œ
œ
Ĝ
ÎÎÎ
bÄ_
b
ln 2
lim 2e (b 2) 2e (ln 2 2)œœ_
bÄ_
( ‘
b2 ln2 2ÎÐÑÎ
11. converges; a geometric series with r 1œ
2
3
12. diverges; lim lim lim 0
nn nÄ_ Ä_ Ä_
55 ln 5ln 55
43 4 ln 4 ln 4 4
n
nn
nn
_
œœ œ_Á ˆ‰ˆ‰
13. diverges; 2 , which diverges by the Integral Test!!
__
œœn0 n0
œ"
__
2
n1 n 1
œ
14. diverges by the Integral Test: ln (2n 1) as n '
1
n
dx
2x 1œ#
"
œ Ä_ Ä_
15. diverges; lim a lim lim 0
nn nÄ_ Ä_ Ä_
n
22 ln 2
n1 1œœ œ_Á
nn
_
16. diverges by the Integral Test: ; ln n 1 ln 2
ux
du ''
12
nn1
dx du
xx1
dx
x
uÈȈ‰
È
_
–—
È
ˆ‰È
œ"
œ
Ĝ
œ)
as n Ä_ Ä_
17. diverges; lim lim lim 0
nn nÄ_ Ä_ Ä_
ÈÈ Š‹
Š‹
nn
ln n
œœœ_Á
œ
œ
2n
nœ
#
18. diverges; lim a lim 1 e 0
nnÄ_ Ä_
n
n
nœœÁˆ‰
"
19. diverges; a geometric series with r 1.44 1œ¸
"
#ln
20. converges; a geometric series with r 0.91 1œ¸
"
ln 3

719. Section 11.3 The Integral Test 715
21. converges by the Integral Test: dx; du
u ln x
du dx ''
3ln3
'' Š‹
ÈÈ
œ
x
(ln x) (ln x) 1
x
uu 1
œ œ
œ
"
"
œ
”•
œ
œ
Ä
lim sec u lim sec b sec (ln 3) lim cos sec (ln 3)œœœ
bb bÄ_ Ä_ Ä_
cd c dkk ( ‘ˆ‰
œ" œ" œ" œ" œ" "
b
ln 3
b
cos (0) sec (ln 3) sec (ln 3) 1.1439œ œ ¸
œ" œ" œ"
#
1
22. converges by the Integral Test: dx dx; du
u ln x
du dx '' '
11 0
'' '
""
__ _ "x 1 ln x 1 (ln x) 1 u
x
ab
Š‹
œœ œœÄ
œ
œ
œ
x
”•
lim tan u lim tan b tan 0 0œœœœ
bbÄ_ Ä_
cd ab
œ" œ" œ"
##
b
0 11
23. diverges by the nth-Term Test for divergence; lim n sin lim lim 1 0
nn x0Ä_ Ä_ Ä
ˆ‰
"
nx
sin x
œœœÁ
sinˆ‰
ˆ‰
œ
œ
n
n
24. diverges by the nth-Term Test for divergence; lim n tan lim lim
nnnÄ_ Ä_ Ä_
ˆ‰
"
n
œœ
tan
secˆ‰
ˆ‰
Š‹ ˆ‰
Š‹
œ
œ
œ
&
&œ
œ
&
n
n
n n
n
œ
œ
lim sec sec 0 1 0œœœÁ
nÄ_
##"ˆ‰
n
25. converges by the Integral Test: dx; du lim tan u
ue
du e dx ''
1e
x
x
''
e
1e 1u
x
2x
__
" œ" ”• cd
œ
œ
Äœ œ
nÄ_
b
e
lim tan b tan e tan e 0.35œœ¸
bÄ_
ab
œ" œ" œ"
#
1
26. converges by the Integral Test: dx; du du
ue
du e dx
dx du '''
1ee
'''
2222
1e u(1u) u u1
x
x
u
___
"
x
Ô×
ÕØ
ˆ‰
œ
œ
œ
Ĝ
lim 2 ln lim 2 ln 2 ln 2 ln 1 2 ln 2 ln 0.63œ œ œœ¸
bbÄ_ Ä_
( ‘ˆ‰ˆ‰ ˆ‰ˆ‰
ube ee
u1 b1 e1 e1 e1___ __
b
e
27. converges by the Integral Test: dx; 8u du 4u 4
utanx
du ''
14
2
2
4
' œ
œ
œ
œ
8 tan x 3
1x 4 16 4 dx
1x
) œœ œ
œ
œ
_
œ"
_
# ”• cd Š‹
œ
œ
Äœœœ
œ
œ
œ
œ
11 1
28. diverges by the Integral Test: dx; lim ln u
ux 1
du 2x dx ''
12
''
xdu
x1 4
œ
_##
#
"" ”•
( ‘
œ
œ
Ĝ
bÄ_
b
2
lim (ln b ln 2)œœ_
bÄ_
"
#
29. converges by the Integral Test: sech x dx 2 lim dx 2 lim tan e''
11
b
b
1
'
œœ
bbÄ_ Ä_
e
1e
x
x
x
_
œ"
ab
œ cd
2 lim tan e tan e 2 tan e 0.71œœ¸
bÄ_
ab
œ" œ" œ"b
1
30. converges by the Integral Test: sech x dx lim sech x dx lim tanh x lim (tanh b tanh 1)
''
11
b
b
1
'
##
œœœ
bbbÄ_ Ä_ Ä_
cd
1 tanh 1 0.76œ ¸
31. dx lim a ln x 2 ln x 4 lim ln ln ;'
1
'
ˆ‰ ˆ‰ cdkkkk
a 3
x2 x4 b4 5
(b 2)
__ _
" _
œ œ
bbÄ_ Ä_
b
1
a a
lim a lim (b 2) the series converges to ln if a 1 and diverges to if
, a 1
1, a 1
bbÄ_ Ä_
(b 2)
b4 3
a1 5_
_
œ
a
œœ Ê œ _
_
œ
œ
ˆ‰
a 1. If a 1, the terms of the series eventually become negative and the Integral Test does not apply. Fromц
that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges.

720. 716 Chapter 11 Infinite Sequences and Series
32. dx lim ln lim ln ln ; lim '
3
b
3
'
ˆ‰ ˆ‰ ’“¹¹
"œœœ"
œ_ _ _ _x1 x1 (x1) (b1) 4 (b1)
2a x1 b1 2 b
œ œ
bb bÄ_ Ä_ Ä_
2a 2a 2a 2a
lim the series converges to ln ln 2 if a and diverges to if
1, a
, a
œœÊ œœ_
œ
_
bÄ_
" "
#_ # #
"
#
"
#
a(b 1)
4
2a 1 )
ˆ‰
if a . If a , the terms of the series eventually become negative and the Integral Test does not apply.†Ñ
""
##
From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it
diverges.
33. (a)
(b) There are (13)(365)(24)(60)(60) 10 seconds in 13 billion years; by part (a) s 1 ln n whereab
*
nÎ'
n (13)(365)(24)(60)(60) 10 s 1 ln (13)(365)(24)(60)(60) 10œÊŸ ab a b ab
**
n
1 ln (13) ln (365) ln (24) 2 ln (60) 9 ln (10) 41.55œ ¸
34. No, because and diverges!!!
___
œœœn1 n1 n1
"" " "
nx x n n
œ
35. Yes. If a is a divergent series of positive numbers, then a also diverges and a .!!! ˆ‰ ˆ ‰
_ _
œ œn1 n1
n nn
aa"
## #
_
n1*
œ
nn
There is no “smallest" divergent series of positive numbers: for any divergent series a of positive!
_
œn1
n
numbers has smaller terms and still diverges.!ˆ‰
_
œn1
an
#
36. No, if is a convergent series of positive numbers, then 2 a 2a also converges, and 2a a .!!!
_ __
œ œœn1 n1 n1
a n nn nnœ 
There is no “largest" convergent series of positive numbers.
37. Let A and B 2 a , where {a } is a nonincreasing sequence of positive terms converging to
nkn k
k
2œœ!!
nn
k1 k1œœ
a
ab
k
0. Note that {A } and {B } are nondecreasing sequences of positive terms. Now,nn
B2a4a8a 2a 2a 2a2a 2a2a2a2an
n
2œá œ á#%) # %% )))) ab
n aba b
2a 2a 2a 2a 2a 2a 2a 2a 2a 2a 2a' ' '(' Î'' ' ' ''' '(
ðóóóóóóóóóóóóóóñóóóóóóóóóóóóóóò
ˆ‰ aba b
ab ab ab22 2
nn n
"# $% &'()
2 terms
n1œ
2a 2a 2a 2A 2 a . Therefore if a converges, ᜟˆ‰ !!
ab a b ab ab221 22 kk
n1 n1 nn
_
_
œk1

then {B } is bounded above 2 a converges. Conversely,n Ê!
k
2
&'
k
Aa aa aaaa aa2a4a 2a aBa 2a.nnn
nk
2 2œ á á œ " #$ %&'( " # % " "aba b !
ab ab
n k
_
œk1

721. Section 11.4 Comparison Tests 717
Therefore, if 2 a converges, then {A } is bounded above and hence converges.!
_
œk1
k
2 nab
k
38. (a) a 2 a 2 , which diverges
ab ab ab22 2 ln 2 2 n(ln 2) n(ln 2) ln n
nn
nn
nn n nœœ Ê œ œ
"" """
##†† !! !
__ _
œœ œn2 n2 n2
diverges.Ê!
_
œn2
"
n ln n
(b) a 2 a 2 , a geometric series that
ab ab
ab
22
nn
2
n
nn
np np p 1 p 1 nœÊ œ œ œ
""""
###!!!! ˆ‰
____
œœœœn1 n1 n1 n1
†
converges if 1 or p 1, but diverges if p 1.
"
#
p1 †Ñ Î
39. (a) ; u du lim lim b (ln 2)
u ln x
du''
2ln2
p
b
ln 2
''
(
dx u
x(ln x) p 1 1 pdx
x
p1 p1
p
p1”• ’“ Š‹ cd
œ
œ
Äœ œ
bbÄ_ Ä_
*
œ_ œ
" œ_ œ_
the improper integral converges if p 1 and diverges
(ln 2) , p 1
, p
œÊ

_"
œ
"
œp1 ()p1
if p 1. For p 1: lim ln (ln x) lim ln (ln b) ln (ln 2) , so the improperœ œ œ œ_ '
2
b
2
'
dx
x ln x
bbÄ_ Ä_
cd c d
integral diverges if p 1.œ
(b) Since the series and the integral converge or diverge together, converges if and only if p 1.!
_
œn2
"
n(ln n)
p
40. (a) p 1 the series divergesœÊ
(b) p 1.01 the series convergesœÊ
(c) ; p 1 the series diverges!!
__
œœn2 n2
"" "
n ln n 3 n(ln n)ab
&œœÊ
(d) p 3 the series convergesœÊ
41. (a) From Fig. 11.8 in the text with f(x) and a , we have dx 1œ œ Ÿá
"" " """
#xk x 3n
k '
1
n1
)
1 f(x) dx ln (n 1) 1 1 ln n 0 ln (n 1) ln nÎ' ! ' Î'''('Î' ! Î ' &'
1
n
"" "
# 3n
1 ln n 1. Therefore the sequence 1 ln n is bounded aboveÎ '''(' & Î '''(' &ˆ‰ ˜ ™ ˆ‰
"" " "" "
# #3n 3n
by 1 and below by 0.
(b) From the graph in Fig. 11.8(a) with f(x) , dx ln (n 1) ln nœ œ
"" "
_xn1 x'
n
n1
)
0 ln (n 1) ln n 1 ln (n 1) 1 ln n .Ê œ á á
"""""""
_#_#n1 3 n1 3 n
cd ˆ‰ˆ‰
If we define a 1 ln n, then 0 a a a a {a } is a decreasing sequence ofnn1nn1nn
3nœœ Ê Ê
"""
#
__
nonnegative terms.
42. e e for x 1, and e dx lim e lim e e e e dx converges by
œœ œ œœxx x b1
œ œ
Ÿ  œœœÊ ''
1 1
x 1x
' '
( ((
bbÄ_ Ä_
cd ˆ‰
b
"
the Comparison Test for improper integrals e 1 e converges by the Integral Test.ʜ!!
__
œœn0 n1
œœnn
œœ
11.4 COMPARISON TESTS
1. diverges by the Limit Comparison Test (part 1) when compared with , a divergent p-series:!
_
œn1
"
Èn
lim lim lim
nnnÄ_ Ä_ Ä_
Π&
Š‹
œ
&'
)
œ
œœ
œ
nn
n
œœ œ
È
ÈÈ
n
2n n n_
""
#_ #
& Î!
ˆ‰
16

722. 718 Chapter 11 Infinite Sequences and Series
2. diverges by the Direct Comparison Test since n n n n n 0 , which is the nth''Ñ' ' ! ÑÈ
3
nn n_
"
È
term of the divergent series or use Limit Comparison Test with b!
'
*
n1
""
nn
n
œ
3. converges by the Direct Comparison Test; , which is the nth term of a convergent geometric series
sin n
2
œ
nnŸ
"
#
4. converges by the Direct Comparison Test; and the p-series converges
1cos n 2
nn n
_"
œœ œŸ !
5. diverges since lim 0
nÄ_
2n 2
3n 1 3œ
œÁ
6. converges by the Limit Comparison Test (part 1) with , the nth term of a convergent p-series:
"
n
&ʜ
lim lim 1
nnÄ_ Ä_
Š‹
Š‹
n
nn
n
&
œ
)Î&
œ
œœ ˆ‰
n
n
_" 7. converges by the Direct Comparison Test; , the nth term of a convergent geometricˆ‰ˆ‰ˆ‰
nn
3n13n3
nn n
_
"
œ
series
8. converges by the Limit Comparison Test (part 1) with , the nth term of a convergent p-series:
"
n
&ʜ
lim lim lim 1 1
nnnÄ_ Ä_ Ä_
Š‹
Š‹
œ
)Î&
œ
)
'
n
n2
œ
œœœÉ É
n2 2
nn
&
&&
_
9. diverges by the Direct Comparison Test; n ln n ln n ln ln n and Ñ!Ñ !††
"" " "
n ln n ln (ln n) n !
_
œn3
diverges
10. diverges by the Limit Comparison Test (part 3) when compared with , a divergent p-series:!
_
œn2
"
n
lim lim lim lim lim lim n
nnn nnnÄ_ Ä_ Ä_ Ä_ Ä_ Ä_
Š‹
ˆ‰
œ
&
œ
œœ
(ln n)
n
nn
œœ œœ œœ_
nn
(ln n) ln n
(ln n)
œ
"" " ""
#
## #
Š‹ Š‹
11. converges by the Limit Comparison Test (part 2) when compared with , a convergent p-series:!
_
œn1
"
n
œ
lim lim lim 2 lim 0
nnn nÄ_ Ä_ Ä_ Ä_
’“
Š‹
(ln n)
n
n
n
&
)
œ
&
œ
œœ œœ
(ln n)
n1 n
2(ln n)
ln n
œ Š‹
12. converges by the Limit Comparison Test (part 2) when compared with , a convergent p-series:!
_
œn1
"
n
œ
lim lim lim 3 lim 3 lim 6 lim
nnn n n nÄ_ Ä_ Ä_ Ä_ Ä_ Ä_
’“
Š‹
(ln n)
n
n
nn
)
)
œ
&
œœ
œœ œ œ œ
(ln n) (ln n)
n1 n 1 n
3(ln n) 2(ln n)
ln n
& œ
œ Š‹ Š‹
60 0œœ†
13. diverges by the Limit Comparison Test (part 3) with , the nth term of the divergent harmonic series:
"
n
lim lim lim lim
nnnnÄ_ Ä_ Ä_ Ä_
’“ Š‹
ˆ‰ ˆ‰
1
n ln n 2 n
nn
œœ
œœ
œ
œœ œœ_
ÈÈnn
ln n 2

723. Section 11.4 Comparison Tests 719
14. converges by the Limit Comparison Test (part 2) with , the nth term of a convergent p-series:
"
n
(!'
lim lim lim 8 lim 8 lim 32 l
nnn nnÄ_ Ä_ Ä_ Ä_ Ä_
’“
Š‹ Š‹ Š‹
ˆ‰ ˆ‰
(ln n)
n
n4n 4n
2 ln n
nn
&
)Î&
œœ œ
ÊŸÛ ÚŸÛ ÚŸÛ
œ
œœ œœ œ
(ln n)
nn
ln n
œ
)!' )!'
im 32 0 0
nÄ_
"
n
Ú!'œœ†
15. diverges by the Limit Comparison Test (part 3) with , the nth term of the divergent harmonic series:
"
n
lim lim lim lim n
nnnnÄ_ Ä_ Ä_ Ä_
ˆ‰
ˆ‰
œ
'
œ
œ
1ln n
n
n
œœœœ_
n
1ln n_
"
Š‹
16. diverges by the Limit Comparison Test (part 3) with , the nth term of the divergent harmonic series:
"
n
lim lim lim lim lim lim
nnnnnnÄ_ Ä_ Ä_ Ä_ Ä_ Ä_
Š‹
ˆ‰
’“
œ
'
&
œ
'
(1 ln n)
n
2(1 ln n)
n
2
n
œœ œ œœœ_
nnn
(1 ln n) (1 ln n)_#_#
"
œ
"
Š‹
17. diverges by the Integral Test: dx u du lim u lim b ln 3''
2ln3
''
ln (x 1)
x1 2
_
_#
""###
œœ œ œ_
bbÄ_ Ä_
á‘ ab
b
ln 3
18. diverges by the Limit Comparison Test (part 3) with , the nth term of the divergent harmonic series:
"
n
lim lim lim lim lim lim
nnnnnnÄ_ Ä_ Ä_ Ä_ Ä_ Ä_
Š‹
ˆ‰ ˆ‰
œ
'
&
œ
1lnn
n
2 ln n
n
2
n
œœœœœœ_
nnn
1lnn ln n_##
"
œ
"
Š‹
19. converges by the Direct Comparison Test with , the nth term of a convergent p-series: n 1 n for
" #
n
&Ñ
n 2 n n 1 n n n 1 n or use Limit Comparison Test with .Ð! &Ñ ! &Ñ ! Ñ
# # $ $Î# # ""
œ
ab
È
n nn 1
1
n
&ʜ
œ
œ È
20. converges by the Direct Comparison Test with , the nth term of a convergent p-series: n 1 n
" ##
n
'Ñ
n 1 nn n or use Limit Comparison Test with .!'Ñ ! Ñ ! †
# $Î# $Î# _" "
_ È
n1
n
n
n1 nn
œ
œ
È
È
21. converges because which is the sum of two convergent series:!!!
___
œœœn1 n1 n1
"œ " œ"
#
n
n2 n2
nnnœ
converges by the Direct Comparison Test since , and is a convergent geometric!!
_ _
œ œn1 n1
"""œ"
##n2 n 2
n nn n †
series
22. converges by the Direct Comparison Test: and , the sum of!! ˆ‰
__
œœn1 n1
n2
n2 n2 n n2 n n
_""""""
#
n
nn nn
œœœœœ Ÿ
the nth terms of a convergent geometric series and a convergent p-series
23. converges by the Direct Comparison Test: , which is the nth term of a convergent geometric
""
_313
n1 n1 †
series
24. diverges; lim lim 0
nnÄ_ Ä_
Š‹ ˆ‰
3
3333
n1
nn
_" " " "
œœÁ
25. diverges by the Limit Comparison Test (part 1) with , the nth term of the divergent harmonic series:
"
n
lim lim 1
n x0Ä_ Ä
ˆ‰
ˆ‰
sin
œ
œ
n
n
œœ
sin x
x

724. 720 Chapter 11 Infinite Sequences and Series
26. diverges by the Limit Comparison Test (part 1) with , the nth term of the divergent harmonic series:
"
n
lim lim lim 1 1 1
nn x0Ä_ Ä_ Ä
ˆ‰ ˆ‰
ˆ‰ ˆ‰
tan sin
cos
œœ
œœ œ
nn
nn n
œœœœŠ‹ ˆ‰ˆ‰
""
cos x x
sin x
†27. converges by the Limit Comparison Test (part 1) with , the nth term of a convergent p-series:
"
n
œ
lim lim lim lim 10
nnnnÄ_ Ä_ Ä_ Ä_
Š‹
Š‹
10n
n(n 1)(n 2)
n
''
œ
&
œœœœ
10n n 20n 1 20
n3n2 2n3 2
œ
œ
__
__ _
28. converges by the Limit Comparison Test (part 1) with , the nth term of a convergent p-series:
"
n
œ
lim lim lim lim 5
nnnnÄ_ Ä_ Ä_ Ä_
'(
Š‹
5n 3n
n(n 2) n 5
n
)
(
&&
('
œ
&
&'
œœœœ
5n 3n 15n 3 30n
n2n5n10 3n4n5 6n4
& œ

œœ
œ_œ œ_ œ
29. converges by the Direct Comparison Test: and is the product of a
tan n
nn n n
)
ww w w
11 11 11 11œ
œœ
&&
!!
__
œœn1 n1
1
#
"
convergent p-series and a nonzero constant
30. converges by the Direct Comparison Test: sec n and is the
œ"
##
"
Ê œ
11 sec n
nn n n
)
œœ
13 13 13 13œœ œ œ
ˆ‰ ˆ‰
œœ
!!
__
œœn1 n1
product of a convergent p-series and a nonzero constant
31. converges by the Limit Comparison Test (part 1) with : lim lim coth n lim
"_
œnee
ee
œ

nnnÄ_ Ä_ Ä_
Š‹
Š‹
coth n
n
n
&
œ
&
œœ
nn
nn
lim 1œœ
nÄ_
"_
œ
e
1e

2n
2n
32. converges by the Limit Comparison Test (part 1) with : lim lim tanh n lim
"œ
_nee
ee
œ

nnnÄ_ Ä_ Ä_
Š‹
Š‹
tanh n
n
n
&
œ
&
œœ
nn
nn
lim 1œœ
nÄ_
"œ
_
e
1e

2n
2n
33. diverges by the Limit Comparison Test (part 1) with : lim lim 1.
11
n n
nnÄ_ Ä_
Š‹
ˆ‰
1
nn
n
1
n
œ
œœ
Èn34. converges by the Limit Comparison Test (part 1) with : lim lim n 1
"
n
œ
œ
)
œ
nnÄ_ Ä_
Š‹
Š‹
n
n
n
n
n
œ
œœ È35. . The series converges by the Limit Comparison Test (part 1) with :
" "
_ _ _Ü_ _123 n n(n1) n
2
œœ
"
ˆ‰
n(n 1)'
& œ
lim lim lim lim 2.
nnnnÄ_ Ä_ Ä_ Ä_
Š‹
Š‹
2
nn 1
n
()'
œ
&
œœœœ
2n 4n 4
nn 2n1 2
œ
œ
__
36. the series converges by the Direct
"
_ _ _Ü_ _ _12 3 n n(n1)(2n1) n
66
œœ œ & œœ ŸÊ
"
n(n1)(2n1)
6''
Comparison Test
37. (a) If lim 0, then there exists an integer N such that for all n N, 0 1 1 1
nÄ_
a aa
b bbn nn
n nn
œÊ ¹¹
a b . Thus, if b converges, then a converges by the Direct Comparison Test.!†
nn n n
!!

725. Section 11.4 Comparison Tests 721
(b) If lim , then there exists an integer N such that for all n N, 1 a b . Thus, if
nÄ_
a a
b b
nnn n
n n
œ_ Ê
b diverges, then a diverges by the Direct Comparison Test.!!
nn
38. Yes, converges by the Direct Comparison Test because a!
_
œn1
aa
nn
nn n
†
39. lim there exists an integer N such that for all n N, 1 a b . If a converges,
nÄ_
a a
b b
nn nn n
n n
œ_ Ê Ê !
then b converges by the Direct Comparison Test!
n
40. a converges lim a 0 there exists an integer N such that for all n N, 0 a 1 a a! nn nn n ÊœÊ ŸÊ
nÄ_
#
a converges by the Direct Comparison TestÊ!
#
n
41. Example CAS commands:
:Maple
a := n -> 1./n^3/sin(n)^2;
s := k -> sum( a(n), n=1..k ); # (a)]
limit( s(k), k=infinity );
pts := [seq( [k,s(k)], k=1..100 )]: # (b)
plot( pts, style=point, title="#41(b) (Section 11.4)" );
pts := [seq( [k,s(k)], k=1..200 )]: # (c)
plot( pts, style=point, title="#41(c) (Section 11.4)" );
pts := [seq( [k,s(k)], k=1..400 )]: # (d)
plot( pts, style=point, title="#41(d) (Section 11.4)" );
evalf( 355/113 );
:Mathematica
Clear[a, n, s, k, p]
a[n_]:= 1 / ( n Sin[n] )
32
s[k_]= Sum[ a[n], {n, 1, k}]
points[p_]:= Table[{k, N[s[k]]}, {k, 1, p}]
points[100]
ListPlot[points[100]]
points[200]
ListPlot[points[200]
points[400]
ListPlot[points[400], PlotRange All]Ä
To investigate what is happening around k = 355, you could do the following.
N[355/113]
N[ 355/113]1&
Sin[355]//N
a[355]//N
N[s[354]]
N[s[355]]
N[s[356]]

726. 722 Chapter 11 Infinite Sequences and Series
11.5 THE RATIO AND ROOT TESTS
1. converges by the Ratio Test: lim lim lim
nn nÄ_ Ä_ Ä_
a
a
(n 1) 2
nn1
n
2
n1
n
2*
*
œœ
”•
”•
(n 1)
2
2
n1
n
2
n'
'
&
œ
œ
_
#
œ
œ
†
lim 1 1œœ
nÄ_
ˆ‰ˆ‰
"""
##n
2
È
2. converges by the Ratio Test: lim lim lim 1 1
nn n
lim
nÄ_ Ä_ Ä_
œ
Ä_
a
aennee
(n 1) en1
n
2
n1 2
n*
*
œœ œ
Š‹
Š‹
(n 1)
2
e
n1
n
e
n'
'
&
_ "" "
#
† ˆ‰ˆ‰ 3. diverges by the Ratio Test: lim lim lim lim
nn n nÄ_ Ä_ Ä_ Ä_
a
aen!e
(n )! enn1
n
n1
n*
*
œœ œœ_
Š‹
ˆ‰
(n 1)!
e
n1
n!
e
n'
'
_" _"
†
4. diverges by the Ratio Test: lim lim lim lim
nn n nÄ_ Ä_ Ä_ Ä_
a
a10n!10
(n )! 10 nn1
n
n1
n*
*
œœ œœ_
Š‹
ˆ‰
(n 1)!
10
n1
n!
10
n'
'
_"
†
5. converges by the Ratio Test: lim lim lim lim 1
nn n nÄ_ Ä_ Ä_ Ä_
a
a10nn
(n ) 10n1
n
n1
n*
)Ð
*)Ð
œœ œ
Š‹
Š‹
(n 1)
10
n1
n
10
n'
œ 
'
œ 
_" "
"!
† ˆ‰ˆ
‰
"
10
1œ
"
10
6. diverges; lim a lim lim 1 e 0
nn nÄ_ Ä_ Ä_
n
n2 2
nn
nnœœœÁˆ‰ ˆ ‰
œœ œ#
7. converges by the Direct Comparison Test: 2 ( 1) (3) which is the n term of a convergent
2(1)
(1.25) 5 5
44
nn
nth_œ
n
n
œŸˆ‰ ˆ‰cd
geometric series
8. converges; a geometric series with r 1kk¸¸œ
2
3
9. diverges; lim a lim 1 lim 1 e 0.05 0
nn nÄ_ Ä_ Ä_
n
33
nn
nnœœœ¸Áˆ‰ ˆ ‰
œ œ$
10. diverges; lim a lim 1 lim 1 e 0.72 0
nn nÄ_ Ä_ Ä_
n
3n n
n
nœœ œ¸Áˆ‰

"
œ
œ"Ÿ$
Š‹
œ
3
11. converges by the Direct Comparison Test: for n 2, the n term of a convergent p-series.
ln n n
nnn
th
&& œœ  
"
12. converges by the nth-Root Test: lim a lim lim lim
nn n nÄ_ Ä_ Ä_ Ä_
È É
n
n
n
n
n1n
n
1n
n
(ln n)
nn
(ln n)
n
ln nœœ œ
ab
ab
!
!
lim 0 1œœ
nÄ_
Š‹
œ
n
1
13. diverges by the Direct Comparison Test: for n 2 or by the Limit Comparison Test (part 1)
"" œ ""
#nn n n
n1
œ
œœ
ˆ‰
with .
"
n
14. converges by the nth-Root Test: lim a lim lim
nn nÄ_ Ä_ Ä_
È Éˆ‰ ˆ ‰ ˆ‰
n
n
n
nn nn
nn 1nœœ
"" ""
Î
œœ
lim 0 1œœ
nÄ_
ˆ‰
""
nn
œ

727. Section 11.5 The Ratio and Root Tests 723
15. diverges by the Direct Comparison Test: for n 3
ln n
nn
 
"
16. converges by the Ratio Test: lim lim 1
nnÄ_ Ä_
a
an ln(n)
(n 1) ln (n 1) 2n1
n
n1
nÛ
Û
œœ
__
##
"
†
17. converges by the Ratio Test: lim lim 0 1
nnÄ_ Ä_
a
a (n 1)! (n 1)(n 2)
(n 2)(n 3) n!n1
nÛ
œœ
__
___
†
18. converges by the Ratio Test: lim lim 1
nnÄ_ Ä_
a
aene
(n 1) en1
n
n1
nÛ
&
*&
œœ
_ "
†
19. converges by the Ratio Test: lim lim lim 1
nn nÄ_ Ä_ Ä_
a
a 3!(n1)!3 (n3)! 3(n1) 3
(n 4)! 3! n! 3 n 4n1
n
n1
nÛ
Û
œœœ
_
__ _
_"
†
20. converges by the Ratio Test: lim lim
nnÄ_ Ä_
a
a 3 (n 1)! n2 (n 1)!
(n 1)2 (n 2)! 3n!n1
n
n1
n1 n
nÛ
Û
Û
œ
__
__
†
lim 1œœ
nÄ_
ˆ‰ˆ‰ˆ‰
n12n2 2
n3n1 3
__
_
21. converges by the Ratio Test: lim lim lim 0 1
nn nÄ_ Ä_ Ä_
a
a (2n 3)! n! (2n 3)(2n 2)
(n 1)! (2n 1)! nn1
nÛ
œœœ
__
___
_"
†
22. converges by the Ratio Test: lim lim lim lim
nn n nÄ_ Ä_ Ä_ Ä_
a
a(n1)n!n1
(n 1)! nn
nn1
n
n1
nÛ
Û
œœœ
_
__
"
† ˆ‰
ˆ‰
n
n
n'œ
lim 1œœ
nÄ_
""
ˆ‰1'
œ
n
n
e
23. converges by the Root Test: lim a lim lim lim 0 1
nn nnÄ_ Ä_ Ä_ Ä_
È É
n n
n
n
n
n
(ln n) ln n ln n
nœ œœœ
È
"
24. converges by the Root Test: lim a lim lim 0 1
nn nÄ_ Ä_ Ä_
È É
n n
n2
n n
n
n
(ln n)
n
ln n lim ln n
lim nœœœœ
Ê
È
ÈÈ
È
n
n&'
&'
lim n 1Š‹ È
nÄ_
n
œ
25. converges by the Direct Comparison Test:
n! ln n ln n n
n(n 2)! n(n 1)(n 2) n(n 1)(n 2) (n 1)(n ) n_______#
""
œœ
œ
which is the nth-term of a convergent p-series
26. diverges by the Ratio Test: lim lim lim 1
nn nÄ_ Ä_ Ä_
a
a (n1)2 3 (n1)
3n2 n33n1
n
n1 n 3
n1 n 3Û
*&
&*
œœœ
__##
† ˆ‰
27. converges by the Ratio Test: lim lim 0 1
nnÄ_ Ä_
a
aa
an1
nn
n
nÛ
œœ
ˆ‰
1sin n'
28. converges by the Ratio Test: lim lim lim 0 since the numerator
nn nÄ_ Ä_ Ä_
a
aa n
a
tan nn1
nn
nÛ
ŸÚ
œœœ
Š‹
1tan n
n'
áœ
"_
approaches 1 while the denominator tends to _
1
#
29. diverges by the Ratio Test: lim lim lim 1
nn nÄ_ Ä_ Ä_
a
aa2n1
a
3n 1 3n 1 2n 1
nn
nÛÛ
œœœ
ˆ‰
3n 1á
œ
_#
30. diverges; a a a a a an1 n n1 n1 n1 n2
nnn1 nn1n2
n1 n1 n n1 n n1
__ œ_ œ
__ _œ
œœœœÊœ Êœ ˆ‰ˆ ‰ ˆ‰ˆ‰ˆ ‰
a a a a , which is a constant times theÊœ â ʜʜn1 n1 n1
nn1n2 3
n1nn1 n1 n1
a
_"__
_œ# _ _
œœ "
ˆ‰ˆ‰ˆ‰ˆ‰
Ú
general term of the diverging harmonic series
31. converges by the Ratio Test: lim lim lim 0 1
nn nÄ_ Ä_ Ä_
a
aan
a
2n1
nn
nÛ
œœœ
Š‹
2
n

728. 724 Chapter 11 Infinite Sequences and Series
32. converges by the Ratio Test: lim lim lim 1
nn nÄ_ Ä_ Ä_
a
aa n
an1
nn
n*
œœœ
Œœ
œn
n
n
& Èn
"
#
33. converges by the Ratio Test: lim lim lim lim 0 1
nn n nÄ_ Ä_ Ä_ Ä_
a
aa nn
a
ln nn1
nn
n*
œœœœ
Š‹
1ln n
n'
"_ "
34. 0 and a a 0; ln n 10 for n e n ln n n 10 1
nln n nln n
n10 n10
n
_" _
_# _
"
"!
œÊÊÊ
a a a ; thus a a lim a 0, so the series diverges by the nth-Term TestÊœ  Ê Án1 n n n1 n n
nln n
n10
__
_"
_#
nÄ_
35. diverges by the nth-Term Test: a , a , a , a , ,"# $ %
" " "" ""œœ œ œ œ œá
3 3 33 33
ÉÉÉ ÉÉÊÊ Ë
22 2 33 6!
'
'
a lim a 1 because is a subsequence of whose limit is 1 by Table 8.1
nnœÊ œÉÉÉ š› š›
n! n! n"""
333
nÄ_
36. converges by the Direct Comparison Test: a , a , a , a ,"# $ %
"" "" ""
## ## ##
##''#%
$%œœœœœœáˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰Š‹ Š‹
a which is the nth-term of a convergent geometric seriesʜ n
n! n
ˆ‰ ˆ‰
""
##
37. converges by the Ratio Test: lim lim lim
nn nÄ_ Ä_ Ä_
a
a (2n 2)! 2 n! n! (2n
2 (n 1)! (n 1)! (2n)! 2(n 1)(n 1)n1
n
n1
n*
*
œœ
__ __
_
†
)(2n 1)_# _
lim 1œœ
nÄ_
n
2n 1
_" "
_#
38. diverges by the Ratio Test: lim lim
nnÄ_ Ä_
a
a (n 1)! (n 2)! (n 3)! (3n)!
(3n 3)! n! (n 1)! (n 2)!n1
n*
œ
___
___
† lim lim 3 3 3 3 27 1œœœœ
nnÄ_ Ä_
(3n 3)(3 2)(3n 1)
(n 1)(n 2)(n 3) n n 3
3n 2 3n 1__ _
___ _# _
__
ˆ‰ˆ‰ ††
39. diverges by the Root Test: lim a lim lim 1
nn nÄ_ Ä_ Ä_
È É
n n
n
n
n
(n!)
n
n!
n´œœ_
ab
œ œ
40. converges by the Root Test: lim lim lim lim
nnnnÄ_ Ä_ Ä_ Ä_
É É ˆ‰ˆ‰ˆ‰ ˆ ‰ˆ‰
n
nn
n
n
n
n n(n!) (n!)
n n
n! 2 3 n 1 n
n nnn n n
œœœœ â
ab
"œ
lim 0 1Ÿœ
nÄ_
"
n
41. converges by the Root Test: lim a lim lim lim 0 1
nn nnÄ_ Ä_ Ä_ Ä_
È É
n n
n
n
nn
n
nn
2 ln 2œœœœ
œ
##
"
42. diverges by the Root Test: lim a lim lim 1
nn nÄ_ Ä_ Ä_
È É
n n
n
n
n
nn
4œœœ_
ab#
œ
43. converges by the Ratio Test: lim lim
nnÄ_ Ä_
a
a 4 2 (n 1)! 1 3 (2n 1)
1 3 (2n 1)(2n 1) 42n!n1
n
n1n1
nn*
**
œ
†† †
†† †
ߜ _
_ߜ
†
lim 1œœ
nÄ_
2n
(4 )(n 1) 4
_" "
#_†
44. converges by the Ratio Test: an
13 (2n 1) 1234 (2n 1)(2n) (2n)!
(2 4 n) 3 1 (2 4 2n) 3 1 2n! 3 1œœ œ
† †††
††
ߜ ߜ
ß# _ ß _ _ab ab aba b
nn nn œ œ
lim lim ʜ
nnÄ_ Ä_
(2n 2)!
2 (n 1)! 3 1
2 n! 3 1 (2n )(2n 2) 3 1
(2n)! 2 (n 1) 3 1
_
__
__"__
__cd ab
abab ab
ab
n1 n1
nn n
n1**
œ
œ
œœÛ
†
lim 1 1œœœ
nÄ_
Š‹
4n 6n 2
4n 8n 4 3 3 3 3
1 3
œ
œ

__ " "
__ _
_ab
ab
n
n
†
45. Ratio: lim lim lim 1 1 no conclusion
nn nÄ_ Ä_ Ä_
a
a (n 1) 1 n 1
nn
pn1
n
p
p*
œœœœÊ
"
__
† ˆ‰
p
Root: lim a lim lim 1 no conclusion
nnnÄ_ Ä_ Ä_
È É
n
n
pp n
p
n
n(1) nœœ œœÊ
"""
ˆ‰È

729. Section 11.6 Alternating Series, Absolute and Conditional Convergence 725
46. Ratio: lim lim lim lim lim
nn n n nÄ_ Ä_ Ä_ Ä_ Ä_
a
a (ln (n 1)) 1 ln (n 1) n
(ln n) ln n n
pp
pn1
n
p
pÛ
œœœœ
" _"
__
† ’“ Š‹ ”•
ˆ‰
ˆ‰
œ
œ
'
n
n1
(1) 1 no conclusionœœÊ
p
Root: lim a lim ; let f(n) (ln n) , then ln f(n)
nnÄ_ Ä_
È É
n n
p p
n
(ln n) n

1n ln (ln n)œœ œ œ
"" Î
Š‹
lim (ln n)
nÄ_
1n
lim ln f(n) lim lim lim 0 lim (ln n)ÊœœœœÊ
nnnn nÄ_ Ä_ Ä_ Ä_ Ä_
ln (ln n)
n1n ln n
1n
ˆ‰
œ
n ln n " Î
lim e e 1; therefore lim a 1 no conclusionœœœ œ œœÊ
nnÄ_ Ä_
ln f n
n

(1)
ÐÑ ! ""
È
n
p p
Š‹ lim (ln n)
nÄ_
1n
47. a for every n and the series converges by the Ratio Test since lim 1n
nn 2
2 2n
(n )Ÿ œ
nn n1
n
!
'
Û
n1
# #
_" " nÄ_
Û†
a converges by the Direct Comparison TestÊ!
'
Û
n1
n
11.6 ALTERNATING SERIES, ABSOLUTE AND CONDITIONAL CONVERGENCE
1. converges absolutely converges by the Absolute Convergence Test since a which is aʜ !!kk
__
œœn1 n1
n
n
" œ
convergent p-series
2. converges absolutely converges by the Absolute Convergence Test since a which is aʜ !!kk
__
œœn1 n1
n
n
" &ʜ
convergent p-series
3. diverges by the nth-Term Test since for n 10 1 lim 0 ( 1) diverges Ê Ê ÁÊ
nn n
10 10 10
nn
n1
nÄ_
ˆ‰ ˆ‰ !
_
œn1
_
4. diverges by the nth-Term Test since lim lim (after 10 applications of L'Hopital's^
nnÄ_ Ä_
" "0
n 10!
0 (ln 10)
n n
Ú
Ú œœ_
rule)
5. converges by the Alternating Series Test because f(x) ln x is an increasing function of x is decreasingœÊ
"
ln x
u u for n 1; also u 0 for n 1 and lim 0Ê        œnn1 n
ln n
_
"
nÄ_
6. converges by the Alternating Series Test since f(x) f (x) 0 when x e f(x) isœÊ œ Ê
ln x 1 ln x
xx
w œ
œ
decreasing u u ; also u 0 for n 1 and lim u lim lim 0Ê      œ œ œ nn1 n n
ln n
n1
_
nn nÄ_ Ä_ Ä_
Š‹
œ
n
7. diverges by the nth-Term Test since lim lim lim 0
nn nÄ_ Ä_ Ä_
ln n ln n
ln n 2 ln n
œœœœÁ
""
##
8. converges by the Alternating Series Test since f(x) ln 1 x f (x) 0 for x 0 f(x) isœ Ê œ Êab
œ" ¸ œ"
_x(x 1)
decreasing u u ; also u 0 for n 1 and lim u lim ln 1 ln lim 1 ln 1 0Ê      œ œ œ œnn1 n n
nn
_
""
nn nÄ_ Ä_ Ä_
ˆ‰ ˆ‰ Š‹
9. converges by the Alternating Series Test since f(x) f (x) 0 f(x) is decreasingœÊœ Ê
ÈÈ
È
x 1x2x
x1 2x(x 1)
_" œ œ
_
w
_
œ
u u ; also u 0 for n 1 and lim u lim 0Ê      œ œ
nn1 nÚ
nnÄ_ Ä_
n
n
n1
È_"
_

730. 726 Chapter 11 Infinite Sequences and Series
10. diverges by the nth-Term Test since lim lim 3 0
nnÄ_ Ä_
3n 1
n1
31
1
È
È
É
Š‹
_
_
_
_
œœÁ
œ
œ
n
n
œ
11. converges absolutely since a a convergent geometric series!!kk ˆ‰
''
ÛÛ
n1 n1
n
10
nœ
"
12. converges absolutely by the Direct Comparison Test since which is the nth¹¹ ˆ‰
( 1) (0.1)
n (10) n 10
nœ ""
n1 n
nÛ
œ
term of a convergent geometric series
13. converges conditionally since 0 and lim 0 convergence; but a
"" " "
_ÈÈ Ènn n1
n
n
œÊ œ
nÄ_
!!kk
__
œœn1 n1
ÚÊœ
is a divergent p-series14. converges conditionally since 0 and lim 0 convergence; but
"" "
__ __1n 1n 1n1ÈÈ È
œÊ
nÄ_
a is a divergent series since and is a divergent p-series!! !kk
__ _
œœ œn1 n1 n1
n
1n 1n n nœ 
""""
__#ÈÈÈ
ÚÊœ
15. converges absolutely since a and which is the nth-term of a converging p-series!!kk
__
œœn1 n1
n
nn
n1 n1 nœ
&& œ
__
"
16. diverges by the nth-Term Test since lim
nÄ_
n!
#
nœ_
17. converges conditionally since 0 and lim 0 convergence; but a
"" "
___ _n3 (n1)3 n3
n
œÊ
nÄ_
!kk
_
œn1
diverges because and is a divergent seriesœ !!
__
œœn1 n1
""""
__n3 n3 4n n
18. converges absolutely because the series converges by the Direct Comparison Test since !¸¸ ¸¸
_
œn1
sin n sin n
nnn
œ œœ Ÿ
"
19. diverges by the nth-Term Test since lim 1 0
nÄ_
3n
5n
_
_
œÁ
20. converges conditionally since f(x) ln x is an increasing function of x is decreasingœÊœ
""
3 ln x ln xab
&
0 for n 2 and lim 0 convergence; but a Ê   œÊ œ
"" " "
_3 ln n 3 ln (n 1) 3 ln n ln n
n
nÄ_
!!kk
__
œœn2 n2
ab
& diverges because and divergesœ!!
__
œœn2 n2
""""
3 ln n 3 ln n 3n n
21. converges conditionally since f(x) f (x) 0 f(x) is decreasing and henceœÊ œ Ê
"" " w
xx xx
2
œ
ˆ‰
u u 0 for n 1 and lim 0 convergence; but a
nn1 n   œÊ œ
Ú
nÄ_
ˆ‰ !!kk
"" _
nn n
1n
œ œ
__
œœn1 n1
is the sum of a convergent and divergent series, and hence divergesœ!!
__
œœn1 n1
""
nn
œ
22. converges absolutely by the Direct Comparison Test since 2 which is the nth term¹¹ ˆ‰
(2)
n5 n5 5
22
nœ
__
n1
nn
n1Û Û
œ
of a convergent geometric series

731. Section 11.6 Alternating Series, Absolute and Conditional Convergence 727
23. converges absolutely by the Ratio Test: lim lim 1
nnÄ_ Ä_
Š‹ ”•
u
u3
2n1
n*
œœ
(n )
n
'&
&
'
&ˆ‰
ˆ‰
2
3
n1
2
3
n
24. diverges by the nth-Term Test since lim a lim 10 1 0
nnÄ_ Ä_
n
1nœœÁ
Î25. converges absolutely by the Integral Test since tan x dx lim '
1
b
1
'
ab ˆ‰ ’“
œ" "
_#1x
tan x
œ
Î)
œ œ
bÄ_
ab
lim tan b tan 1œœœ
bÄ_
’“’“abab ˆ‰ ˆ‰
œ" œ"
## "
##
##
11 1
432
3
œ
26. converges conditionally since f(x) f (x) 0 f(x) is decreasingœÊœ Ê
" w _
x ln x (x ln x)
ln (x) 1cd
œ
u u 0 for n 2 and lim 0 convergence; but by the Integral Test,Ê   œÊ
nn1 )
nÄ_
"
n ln n
lim dx lim ln (ln x) lim ln (ln b) ln (ln 2)''
22
b
b
2
'
dx
x ln x ln x
œœœœ_
bbbÄ_ Ä_ Ä_

cd c d
Š‹
œ
x
a divergesʜ!!kk
__
œœn1 n1
n
n ln n
"
27. diverges by the nth-Term Test since lim 1 0
nÄ_
n
n1_
œÁ
28. converges conditionally since f(x) f (x)œÊœ
ln x
x ln x (x ln x)
(x ln x) (ln x) 1
œœ
w
œœ œŠ‹ Š ‹
œœ
xx
œ
0 u u 0 when n e and lim œœÊ 
1ln x
(x ln x) (x ln x) n ln n
1 ln x ln n
nn1
œœ_
œœ œ
œ
_
Š‹ Š‹
ln x ln x
xx
œœ
nÄ_
lim 0 convergence; but n ln n n so thatœœÊ ÊÊ
nÄ_
Š‹
Š‹
œ
œ
n
n
1
nln n n nln n n
ln n
œ
"" "
œœ
a diverges by the Direct Comparison Test!!kk
_
œn1
n
ln n
nln nœ
'
*
n1
œ
29. converges absolutely by the Ratio Test: lim lim lim 0 1
nn nÄ_ Ä_ Ä_
Š‹
u
u (n 1)! (100) n 1
( 00) n! 00n1
n
n1
n*
*
œœœ
"
__
"
†
30. converges absolutely since a is a convergent geometric series!!kk ˆ‰
__
œœn1 n1
n
5
nœ
"
31. converges absolutely by the Direct Comparison Test since a and!!kk
__
œœn1 n1
n
n2n1œ
"
__
œ
which is the nth-term of a convergent p-series
""
__n2n1 n
œœ †
32. converges absolutely since a is a convergent!! ! !kk ˆ‰ ˆ ‰ ˆ‰
__ _ _
œœ œ œn1 n1 n1 n1
n
ln n ln n
ln n 2 ln nœœ œ
œ
nn n
"
#
geometric series
33. converges absolutely since a is a convergent p-series!! !kk ¹¹
__ _
œœ œn1 n1 n1
n
(1)
nn nœœ
œ "
n
È
&ʜ
34. converges conditionally since is the convergent alternating harmonic series, but!!
__
œœn1 n1
cos n
nn
(1)1
œ
œ
n
a diverges!!kk
__
œœn1 n1
n
nœ
"

732. 728 Chapter 11 Infinite Sequences and Series
35. converges absolutely by the Root Test: lim a lim lim 1
nn nÄ_ Ä_ Ä_
Èkk Š‹
n
n
n
nœœœ
(n 1)
(2n) n
1n
n_
Î
_" "
##
36. converges absolutely by the Ratio Test: lim lim lim
nn nÄ_ Ä_ Ä_
¹¹
a
a ((2n 2)!) (n!) (2n 2)(2n 1)
(n 1)! (2n)! (n 1)n1
nÛ
œ
œ
œ
œœœ
ab_
___
_
†
"
4
†1
37. diverges by the nth-Term Test since lim a lim lim
nn nÄ_ Ä_ Ä_
kkn
(2n)! (n )(n 2) (2n)
2n!n 2nœœ
n
_" _ ß
n
lim lim 0œœ_Á
nnÄ_ Ä_
(n 1)(n 2) (n (n 1)) n1
n1_ _ß_ œ
##
_
œ
n1
ˆ‰
38. converges absolutely by the Ratio Test: lim lim
nnÄ_ Ä_
¹¹
a
a (2n 3)! n! n! 3
(n 1)! (n 1)! 3 (2n 1)!n1
n
n1
nÛ
Û
œ
__ _
_
†
lim 1œœ
nÄ_
(n 1) 3
(2n 2)(2n 3) 4
3_
__
œ
39. converges conditionally since and is a
ÈÈ ÈÈ
ÈÈ È ÈÈ È
n1 n n1 n
1 n1 n n1 n n1 n
_œ __
__ __ __
""
† œ š›
decreasing sequence of positive terms which converges to 0 converges; butÊ!
_
œn1
()
n1 n
œ"
__
n
È È
a diverges by the Limit Comparison Test (part 1) with ; a divergent p-series:!!kk
__
œœn1 n1
n
n1 n nœ
" "
__È ÈÈ
lim lim lim
nnnÄ_ Ä_ Ä_
œ
''
œ
œ œ
œ
n1 n
n
1
n
œœœ
È
È È
É
n
n1 n
1
11
__
__
"
#
40. diverges by the nth-Term Test since lim n n n lim n n n
nnÄ_ Ä_
Š‹Š‹Š‹
ÈÈ
## __
__
œ †
È
È
nnn
nnnœ
œ lim lim 0œœœÁ
nnÄ_ Ä_
n
nnn 11
È
É
œ Ú
__
""
__
#
n
41. diverges by the nth-Term Test since lim n n n lim n n n
nnÄ_ Ä_
Š‹Š‹ÉÉÈÈ ÈÈ
–—
œ
É ÈÈ
É ÈÈ
nnn
nnn
__
__
lim lim 0œœœÁ
nnÄ_ Ä_
È
ÉÉÈÈ
n
nnn 11__
""
__
#
œ
œ
n
42. converges conditionally since is a decreasing sequence of positive terms converging to 0š›
"
__È Ènn1
converges; but lim lim lim Êœœœ!
_
œn1
()
nn1 nn1
n
11
œ"
__ __
""
__
#
n
n
ÈÈÈÈ
È
É nnnÄ_ Ä_ Ä_
Š‹
Š‹
œ
''
œ
œ œ
œ
nn1
n
Ú
so that diverges by the Limit Comparison Test with which is a divergent p-series!!
_ _
œ œn1 n1
""
__È ÈÈnn1 n
43. converges absolutely by the Direct Comparison Test since sech (n) which is theœœœ
22e2e2
ee e 1 e e
n n 2n 2n n
nn
__

nth term of a convergent geometric series
44. converges absolutely by the Limit Comparison Test (part 1): a !!kk
__
œœn1 n1
n
2
eeœ
nn
œ
Apply the Limit Comparison Test with , the n-th term of a convergent geometric series:
1
e
n
lim lim lim 2
nnnÄ_ Ä_ Ä_
Π'
2
ee
nn
1
e
n
á
á
œœœ
2e 2
ee 1e
n
nn 2n
œœ

45. error ( 1) 0.2 46. error ( 1) 0.00001kk kk¸¸ ¸ ¸ˆ‰ ˆ ‰ œ œ
' '" "
5 10
á

733. Section 11.6 Alternating Series, Absolute and Conditional Convergence 729
47. error ( 1) 2 10 48. error ( 1) t t 1kk kkk k¹¹ œ‚ œ
' "" % % %(0.01)
5
á
49. (2n)! 200,000 n 5 1 0.54030
" """"
#(2n)! 10 5 !4!6!8!
510
Ê œ ʠʸ
_
_
50. n! n 9 1 1 0.367881944
" """""""
#n! 10 5 ! 3! 4! 5! 6! 7! 8!
510
† ! † !Ð!&'&'&'&'w
_
_
51. (a) a a fails since nn1
3І_
""
#
(b) Since a is the sum of two absolutely convergent!! ! !kk á ‘ˆ‰ ˆ‰ˆ‰ ˆ‰
__ _ _
œœ œ œn1 n1 n1 n1
n
33
nn n nœœ
"" " "
##
series, we can rearrange the terms of the original series to find its sum: 1 ˆ‰ˆ‰
"" " """ " "
### œœ3927 48 11
á ᜠœœ
ˆ‰ ˆ‰
ˆ‰ ˆ‰
ÚÚ
ÚÚ
œ
œ
3
3
52. s 1 0.6687714032 s 0.692580927#! #!
""" " " ""
###œá ¸ Ê ¸
3 4 19 20 1
†
53. The unused terms are ( 1) a ( 1) a a ( 1) a a! ab ab
'
ÛÚ
ÚÚ Ú
ÚÚ ÚÚ
jn1
j1 n1 n3
j n1n2 n3n4
œ á
( 1) a a a a . Each grouped term is positive, so the remainderœ á
n1
n1 n2 n3 n4Ú
ÚÚ ÚÚ
cdabab
has the same sign as ( 1) , which is the sign of the first unused term.&
n1Ú
54. s n
1 2 3 3 4 n(n 1) k(k 1) k k 1œá œ œ
""" " " ""
#___††† !! ˆ‰
nn
k1 k1œœ
1 which are the first 2n termsœáˆ‰ˆ‰ˆ‰ˆ‰ ˆ ‰
""""""" ""
## _ 33445 nn1
of the first series, hence the two series are the same. Yes, for
s 1 1n
kk1 3 34 45 n1n nn1 n1œ œá œ!ˆ ‰ˆ‰ˆ‰ˆ‰ˆ‰ ˆ ‰ˆ ‰
n
k1œ
"" """"""" """" "
_## œ _ _ lim s lim 1 1 both series converge to 1. The sum of the first 2n 1 terms of the firstÊœœÊ
nnÄ_ Ä_
n
n1
ˆ‰
"
_
series is 1 1. Their sum is lim s lim 1 1.ˆ‰ ˆ‰œ œ œ
"" "
__ _n1 n1 n1
n
nnÄ_ Ä_
55. Theorem 16 states that a converges a converges. But this is equivalent to a diverges a!! !!kk kk
__ __
œœ œœn1 n1 n1 n1
nn nn ÊÊ
diverges.
56. a a a a a a for all n; then a converges a converges and thesek k kk kk kk kk !!
"# " #''(' Î ' '(' !nnnn
__
œœn1 n1
imply that a aºº
!! kk
__
œœn1 n1
nnŸ
57. (a) a b converges by the Direct Comparison Test since a b a b and hence!kk kkkkkk
_
œn1
nn nn n n''Î'
a b converges absolutely!ab
_
œn1
nn'
(b) b converges b converges absolutely; since a converges absolutely and!! !kk
__ _
œœ œn1 n1 n1
nn n !&
b converges absolutely, we have a ( b ) a b converges absolutely by part (a)!!! cd ab
___
œœœn1 n1 n1
œnnnnn
(c) a converges k a ka converges ka converges absolutely!!!!kk kk kk k k
____
œœœœn1 n1 n1 n1
nnnn Êœ Ê

734. 730 Chapter 11 Infinite Sequences and Series
58. If a b ( 1) , then ( 1) converges, but a b divergesnn nn
nn
nn nœœ œ
"" "
ÈÈ !!!
___
œœœn1 n1 n1
59. s , s 1 ,"#
"""
###œ œ œ
s 1 0.5099,$
" """"""""""
####œ¸
468101 141618 02
s s 0.1766,%$
"œ¸
3
s s 0.512,&%
"""""""""""
### #œ¸
4 6 8303 343638404244
s s 0.312,'&
"œ¸
5
s s 0.51106('
"""""""""""œ¸
46 48 50 52 54 56 58 60 62 64 66

60. (a) Since a converges, say to M, for 0 there is an integer N such that a M!!kk kk ºº n n %Ñ&† "
#
N1
n1œœ
œ
%
a a a a a . Also, aÓ& ' †Ó&†Ó†
»»»»
!!! ! ! !kk kk kk kk kk

N1 N1
n1 n1 nN nN nNœœ
œœœœœ
œœœ œ œ
___
nnn n n n
%%%
###
converges to L for 0 there is an integer N (which we can choose greater than or equal to N ) suchÍ% #"
that s L . Therefore, a and s L .kk kk kk ! NNœ œ
&† † &†
%%%
###
_
œnN
œ
n
(b) The series a converges absolutely, say to M. Thus, there exists N such that a M!!kk kk ºº
_
œ œn1 n1
k
n n" &† %
whenever k N . Now all of the terms in the sequence b appear in a . Sum together all of the
" ef efkk kknn
terms in b , in order, until you include all of the terms a , and let N be the largest index in theef efkk kk nn
N
n1œ
œ #
sum b so obtained. Then b M as well b converges to M.!! !kk kk kk ºº
NN
n1 n1 n1&&
œœ œ
_
nn n &† !%
61. (a) If a converges, then a converges and a a !!!!!kk kk
_____
œœœœœn1 n1 n1 n1 n1
nn nn
aa""
## #
_ œ
nnkk
converges where b .
a , if a 0
0, if a 0 n
aa nn
nœœ
 
†
nn_
#
kk
œ
(b) If a converges, then a converges and a a !!!!!kk kk
_____
œœœœœn1 n1 n1 n1 n1
nn nn
aa""
## #
œ œ
nnkk
converges where c .
0, if a 0
a , if a 0 n
aa n
nnœœ
 
†
nnœ
#
kk
œ
62. The terms in this conditionally convergent series were not added in the order given.

735. Section 11.7 Power Series 731
63. Here is an example figure when N 5. Notice that u u u and u u u , but u u forœ  $#" $&% nn1
n5. 
11.7 POWER SERIES
1. lim 1 lim 1 x 1 1 x 1; when x 1 we have ( 1) , a divergent
nnÄ_ Ä_
¹¹ ¹¹ kk !
u
ux
xn1
n
n1
n*
*
Ê Ê Ê œ
_
œn1
n
series; when x 1 we have 1, a divergent seriesœ !
_
œn1
(a) the radius is 1; the interval of convergence is 1 x 1&† †
(b) the interval of absolute convergence is 1 x 1&† †
(c) there are no values for which the series converges conditionally
2. lim 1 lim 1 x 5 1 6 x 4; when x 6 we have
nnÄ_ Ä_
¹¹ ¹ ¹ kk
u
u(x5)
(x 5)n1
n
n1
n*
*
Ê Ê Ê œ
_
_
( 1) , a divergent series; when x 4 we have 1, a divergent series!!
_ _
œ œn1 n1
œ
n
(a) the radius is 1; the interval of convergence is 6 x 4&† †&
(b) the interval of absolute convergence is 6 x 4&† †&
(c) there are no values for which the series converges conditionally
3. lim 1 lim 1 4x 1 1 1 4x 1 1 x 0; when x we
nnÄ_ Ä_
¹¹ ¹ ¹ kk
u
u (4x 1)
(4x 1)n1
n
n1
n*
*
Ê Ê Ê Ê œ
_
_ ##
""
have ( 1) ( 1) ( 1) 1 , a divergent series; when x 0 we have ( 1) (1)!!! !
___ _
œœœ œn1 n1 n1 n1
œ œ œ
nn 2n n nn
( 1) , a divergent seriesœ!
_
œn1
n
(a) the radius is ; the interval of convergence is x 0
""
#4
&††
(b) the interval of absolute convergence is x 0&††
"
#
(c) there are no values for which the series converges conditionally
4. lim 1 lim 1 3x 2 lim 1 3x 2 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kk ˆ‰
u
u n 1 (3x 2) n 1
(3x 2) nnn1
n
n1
n*
*
†! †! & †! &†
œ
_œ _
†
1 3x 2 1 x 1; when x we have which is the alternating harmonic series and isÊ Ê œ
"" œ"
33n
() !
_
œn1
n
conditionally convergent; when x 1 we have , the divergent harmonic seriesœ !
_
œn1
"
n (a) the radius is ; the interval of convergence is x 1
"" 33
Ά
(b) the interval of absolute convergence is x 1
" 3
††

736. 732 Chapter 11 Infinite Sequences and Series
(c) the series converges conditionally at xœ
"
3
5. lim 1 lim 1 1 x 2 10 10 x 2 10
nnÄ_ Ä_
¹¹ ¹ ¹ kk
u
u10(x2)10
(x 2) 10 x2n1
n
n1
n1 n
n*
*
*
†! †! †! &† !&†&†
œ
œ
œ
†
kk
8 x 12; when x 8 we have ( ) , a divergent series; when x 12 we have 1, a divergentÊ œ " œ !!
_ _
œ œn1 n1
n
series
(a) the radius is 0; the interval of convergence is 8 x 12_&††
(b) the interval of absolute convergence is 8 x 12&† †
(c) there are no values for which the series converges conditionally
6. lim 1 lim 1 lim 2x 1 2x 1 x ; when x we have
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kk
u
u (2x)
(2x)n1
n
n1
n*
*
Ê Ê Ê Ê œ
"" "
## #
( ) , a divergent series; when x we have 1, a divergent series!!
__
œœn1 n1
" œ
n "
# (a) the radius is ; the interval of convergence is x
""" ###
&††
(b) the interval of absolute convergence is x&††
""
##
(c) there are no values for which the series converges conditionally
7. lim 1 lim 1 x lim 1 x 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kk
u
u (n3) nx (n3)(n)
(n 1)x (n 2) (n 1)(n 2)n1
n
n1
n*
*
†! †! †! †
__ __
__
†
1 x 1; when x 1 we have ( ) , a divergent series by the nth-term Test; when x weÊ œ " œ" !
_
œn1
nn
n_#
have , a divergent series!
_
œn1
n
n_#
(a) the radius is ; the interval of convergence is x_&_††_
(b) the interval of absolute convergence is x&_ † † _
(c) there are no values for which the series converges conditionally
8. lim 1 lim 1 x 2 lim 1 x 2 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kk ˆ‰
u
un1(x2) n1
(x 2) nnn1
n
n1
n*
*
†! †! ' †! '†
_
__ _
†
1 x 2 1 3 x 1; when x 3 we have , a divergent series; when x we haveÊ Ê œ œ" !
_
œn1
"
n
, a convergent series!
_
œn1
(1)
n
œ
n
(a) the radius is ; the interval of convergence is 3 x_&†Î&_
(b) the interval of absolute convergence is 3 x&† †&_
(c) the series converges conditionally at x 1œ
9. lim 1 lim 1 lim lim 1
nn nnÄ_ Ä_ Ä_ Ä_
¹¹ ¹ ¹ Š ‹Š ‹ É
u
ux3n1n1
xnn
(n 1) n 1 3
nn3 xn1
n
n1
n1
n
n*
*
*
†! †! †
__ __È
È kk
†
(1)(1) 1 x 3 3 x 3; when x 3 we have , an absolutely convergent series;ÊÊÊ œ
kkx
3
()
n
kk !
_
œn1
œ"
n
&Êœ when x 3 we have , a convergent p-seriesœ !
_
œn1
1
n
&ʜ
(a) the radius is 3; the interval of convergence is 3 x 3&Î Î
(b) the interval of absolute convergence is 3 x 3&Î Î
(c) there are no values for which the series converges conditionally
10. lim 1 lim 1 x 1 lim 1 x 1 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kkÉ
u
u(x1)n1
(x 1)
n1
n nn1
n
n1
n*
*
†! †! & †! &†
œ
_ œ_È
È
†

737. Section 11.7 Power Series 733
1 x 1 1 0 x 2; when x 0 we have , a conditionally convergent series; when x 2Ê Ê œ œ !
_
œn1
()
n
œ"
n
ÚÊœ
we have , a divergent series!
_
œn1
1
n
ÚÊœ
(a) the radius is 1; the interval of convergence is 0 x 2Ά
(b) the interval of absolute convergence is 0 x 2††
(c) the series converges conditionally at x 0œ
11. lim 1 lim 1 x lim 1 for all x
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk ˆ‰
u
u(n1)!x n1
xn!n1
n
n1
nÛ
Û
†! †! †
__
"
†
(a) the radius is ; the series converges for all x_
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
12. lim 1 lim 1 3 x lim 1 for all x
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk ˆ‰
u
u (n1)! 3x n1
3x n!n1
n
n1 n1
nnÛ
ÛÛ
†! †! †
__
"
†
(a) the radius is ; the series converges for all x_
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
13. lim 1 lim 1 x lim 1 for all x
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ˆ‰
u
u(n1)!x n1
xn!n1
n
2n 3
2n 1Û
Û
Û
†! †! †
__
# "
†
(a) the radius is ; the series converges for all x_
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
14. lim 1 lim 1 (2x 3) lim 1 for all x
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ˆ‰
u
u (n 1)! (2x 3) n 1
(2x 3) n!n1
n
2n 3
2n 1Û
Û
Û
†! †! ' †
_
__ _
# "
†
(a) the radius is ; the series converges for all x_
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
15. lim 1 lim 1 x lim x 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kkÉ
u
uxn2n4
xn3
(n 1) 3
n3n1
n
n1
nÛ
Û
œ
œ œ
œ
†! †! †_! †
È
È
__
_ _
__
†
1 x 1; when x 1 we have , a conditionally convergent series; when x 1 we haveÊ œ œ !
_
œn1
()
n3
œ"
_
n
Ȝ , a divergent series!
_
œn1
"
_
È
n3
œ
(a) the radius is 1; the interval of convergence is 1 x 1&Î †
(b) the interval of absolute convergence is 1 x 1&† †
(c) the series converges conditionally at x 1œ
16. lim 1 lim 1 x lim x 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kkÉ
u
uxn2n4
xn3
(n 1) 3
n3n1
n
n1
nÛ
Û
œ
œ œ
œ
†! †! †_! †
È
È
__
_ _
__
†
1 x 1; when x 1 we have , a divergent series; when x 1 we have ,Ê œ œ !!
_ _
œ œn1 n1
"
__
œ"
ÈÈ
n3 n3
()
œ œ
n
a conditionally convergent series
(a) the radius is 1; the interval of convergence is 1 x 1&† Î
(b) the interval of absolute convergence is 1 x 1&† †
(c) the series converges conditionally at x 1œ

738. 734 Chapter 11 Infinite Sequences and Series
17. lim 1 lim 1 lim 1 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ˆ‰
u
u5n(x3)5n5
(n 1)(x 3) 5n x3 x3n1
n
n1
n1 n
n*
*
*
†! †! †! †
__
_
__ _"
†
kk kk
x 3 5 5 x 3 5 8 x 2; when x 8 we have ( 1) n, a divergentÊ Ê Ê œ œ kk !!
__
œœn1 n1
n( 5)
5
œ
n
n
n
series; when x 2 we have n, a divergent seriesœœ !!
__
œœn1 n1
n5
5
n
n
(a) the radius is 5; the interval of convergence is 8 x 2&† †
(b) the interval of absolute convergence is 8 x 2&† †
(c) there are no values for which the series converges conditionally18. lim 1 lim 1 lim 1 x 4
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ¹ ¹ kk
u
u 4 n2n2 nx 4 nn2n2
(n 1)x 4n 1 (n 1)n 1 xn1
n
n1
n1 n
n*
*
* œ œ
œœ
†! †! †! †
_
__ __
___
ab ab
ab ab kk
†
4 x 4; when x 4 we have , a conditionally convergent series; when x 4 we haveÊ œ œ !
_
œn1
n( 1)
n1
œ
_
n
œ , a divergent series!
_
œn1
n
n1
œ
_
(a) the radius is 4; the interval of convergence is 4 x 4&Î †
(b) the interval of absolute convergence is 4 x 4&† †
(c) the series converges conditionally at x 4œ
19. lim 1 lim 1 lim 1 1 x 3
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ É ˆ‰ kk
u
u33n3
n1x x x 3n1
nxn1
n
n1
n1
n
n*
*
*
†! †! †! †! †
È
È
kk kk_ _
† 3 x 3; when x 3 we have ( 1) n , a divergent series; when x 3 we haveÊ œ œ ! È
_
œn1
n
n, a divergent series!È
_
œn1
(a) the radius is 3; the interval of convergence is 3 x 3&† †
(b) the interval of absolute convergence is 3 x 3&† †
(c) there are no values for which the series converges conditionally
20. lim 1 lim 1 2x 5 lim 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ Š ‹ kk
u
u
n 1 (2x 5) n 1
n (2x 5) nn1
n
n1 n1n1
n nn*
* **
†! †! ' †
ÈÈ
ÈÈ
__ _
_
2x 5 1 2x 5 1 1 2x 5 1 3 x 2; when x 3 we haveÊ Ê Ê Ê œkk kkŒ '
lim t
lim n
t
n&'
&'
È
È
t
n
( 1) n, a divergent series since lim n 1; when x 2 we have n, a divergent series!! ÈÈÈ
_ _
œ œn1 n1
œœ
n
nÄ_
n n
(a) the radius is ; the interval of convergence is 3 x 2
"
#
&† †&
(b) the interval of absolute convergence is 3 x 2&† †&
(c) there are no values for which the series converges conditionally
21. lim 1 lim 1 x 1 x 1 x 1
nnÄ_ Ä_
¹¹
»»
kk kk kk ˆ‰
u
u e
1 x lim 1
1 x lim 1
en1
nn
n1 t
n1
nn*
*
*
†! †! †! †! †
Š‹ Š‹
Š‹ Š‹
__
__
œœ
'
œœ
n1 t
nn t
n&'
&'
1 x 1; when x 1 we have ( 1) 1 , a divergent series by the nth-Term Test sinceÊ œ ! ˆ‰
_
œn1
n
n
n
"
lim 1 e 0; when x 1 we have 1 , a divergent series
nÄ_
ˆ‰ ˆ‰ !œÁ œ
""
nn
nn
_
œn1
(a) the radius is ; the interval of convergence is 1 x 1_&††
(b) the interval of absolute convergence is 1 x 1&† †
(c) there are no values for which the series converges conditionally

739. Section 11.7 Power Series 735
22. lim 1 lim 1 x lim 1 x lim 1 x 1
nn n nÄ_ Ä_ Ä_ Ä_
¹¹ ¹ ¹ kk kk kkºº
ˆ‰
u
ux ln n n1
ln (n 1)x nn1
n
n1
n*
*
†! †! †! †! †
_
_
ˆ‰
ˆ‰
œ
'
œ
n1
n
1 x 1; when x 1 we have ( 1) ln n, a divergent series by the nth-Term Test sinceÊ œ !
_
œn1
n
lim ln n 0; when x 1 we have ln n, a divergent series
nÄ_
Áœ !
_
œn1
(a) the radius is 1; the interval of convergence is 1 x 1&† †
(b) the interval of absolute convergence is 1 x 1&† †
(c) there are no values for which the series converges conditionally
23. lim 1 lim 1 x lim 1 lim (n 1) 1
nn nnÄ_ Ä_ Ä_ Ä_
¹¹ ¹ ¹ Š ‹Š ‹ kk ˆ‰
u
unx n
(n 1) x nn1
n
n1n1
nn*
**
†! †! ' ' †
_ "
e x lim (n 1) 1 only x 0 satisfies this inequalityÊÊœkk
nÄ_
(a) the radius is 0; the series converges only for x 0œ
(b) the series converges absolutely only for x 0œ
(c) there are no values for which the series converges conditionally
24. lim 1 lim 1 x 4 lim (n 1) 1 only x 4 satisfies this
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk
u
un!(x4)
(n 1)! (x 4)n1
n
n1
n*
*
Ê Ê Ê œ
_œ
œ
inequality
(a) the radius is 0; the series converges only for x 4œ
(b) the series converges absolutely only for x 4œ
(c) there are no values for which the series converges conditionally
25. lim 1 lim 1 lim 1 1 x 2 2
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ˆ‰ kk
u
u (n1)2 (x2) n1
(x 2) n2 n x2 x2n1
n
n1
n1 n
n*
*
*
†! †! †! †! '†
_
__ # _ #
__
†
kk kk
2 x 2 2 4 x 0; when x 4 we have , a divergent series; when x 0 we haveÊ Ê œ œ !
_
œn1
œ"
n
, the alternating harmonic series which converges conditionally!
_
œn1
(1)
n
œ
n1*
(a) the radius is 2; the interval of convergence is 4 x 0&† Î
(b) the interval of absolute convergence is 4 x 0&† †
(c) the series converges conditionally at x 0œ
26. lim 1 lim 1 2 x 1 lim 1 2 x 1 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kk ˆ‰
u
u ( 2) (n 1)(x 1) n 1
( 2) (n 2)(x 1) n2n1
n
n1 n1
nn*
**
†! †! & †! &†
œ_œ
œ_œ _
_
x 1 x 1 x ; when x we have (n 1) , a divergent series; when xÊ Ê Ê œ œkk !
"" "" "
## ## # # #
3 3
_
œn1
we have ( 1) (n 1), a divergent series!
_
œn1
n
&' (a) the radius is ; the interval of convergence is x
""
###
††
3
(b) the interval of absolute convergence is x
" ##
††
3
(c) there are no values for which the series converges conditionally
27. lim 1 lim 1 x lim lim 1
nn nnÄ_ Ä_ Ä_ Ä_
¹¹ ¹ ¹ Š ‹Š ‹ kk
u
uxn1ln(n1)
xnln n
(n 1) ln (n 1)
n(ln n)n1
n
n1
n*
*
œ
œ
†! †! †
__ __
#
ab
†
x (1) lim 1 x lim 1 x 1 1 x 1; when x 1 we haveÊÊÊÊœkk kk kkŒ ' Š‹
nnÄ_ Ä_
ˆ‰
ˆ‰
œ
œ
'
n
n1
#
_
#
n1
n
which converges absolutely; when x 1 we have which converges!!
_ _
œ œn1 n1
(1)
n(ln n) n(ln n)
œ "
n
œ œ œ
(a) the radius is ; the interval of convergence is 1 x 1_&ÎÎ

740. 736 Chapter 11 Infinite Sequences and Series
(b) the interval of absolute convergence is 1 x 1&Î Î
(c) there are no values for which the series converges conditionally
28. lim 1 lim 1 x lim lim 1
nn nnÄ_ Ä_ Ä_ Ä_
¹¹ ¹ ¹ Š ‹Š ‹ kk
u
u (n 1) ln (n 1) x n 1 ln (n 1)
xn n ln (n) ln (n)n1
n
n1
nÛ
Û
†! †! †
__ _ _
†
x (1)(1) 1 x 1 1 x 1; when x 1 we have , a convergent alternating series;ÊÊÊ œkk kk !
_
œn2
(1)
n ln n
œ
n
when x 1 we have which diverges by Exercise 38, Section 11.3œ !
_
œn2
"
n ln n
(a) the radius is ; the interval of convergence is 1 x 1_&Ά
(b) the interval of absolute convergence is 1 x 1&† †
(c) the series converges conditionally at x 1œ
29. lim 1 lim 1 (4x 5) lim 1 (4x 5) 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ Š ‹
u
u (4x 5) n 1
(4x 5)
(n 1)
nnn1
n
2n 3
2n 1Û
Û


Û
†! †! & †! & †
œ
_ œ_
##
$Î#
†
4x 5 1 1 4x 5 1 1 x ; when x 1 we have which isÊÊÊ œ œkk !!
3 (1)
nn#
œ œ"
__
œœn1 n1
2n 1Û

absolutely convergent; when x we have , a convergent p-seriesœ
3 ()
n#
" !
_
œn1
2n 1Û
(a) the radius is ; the interval of convergence is 1 x
"
#4
3
ŸŸ
(b) the interval of absolute convergence is 1 xŸŸ
3
#
(c) there are no values for which the series converges conditionally
30. lim 1 lim 1 3x 1 lim 1 3x 1 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kk ˆ‰
u
u 2n4 (3x1) 2n4
(3x 1) 2n 2 2n 2n1
n
n2
n1Û
Û
Û
†! †! ' †! '†
_
__ _
__
†
1 3x 1 1 x 0; when x we have , a conditionally convergent series;Ê Ê œ
22
332n1
(1) !
_
œn1
œ
_
n1Û
when x 0 we have , a divergent seriesœœ !!
__
œœn1 n1
()
2n 1 n 1
"
_#_
"
n1Û
(a) the radius is ; the interval of convergence is x 0
"
33
2
&Ά
(b) the interval of absolute convergence is x 0&††
2
3
(c) the series converges conditionally at xœ
2
3
31. lim 1 lim 1 x lim 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ¹ ¹ kk É
u
u(x) n1
(x )
n1
n nn1
n
n1
nÛ
Û
†! †! ' †
_
_ __
1
1È
È
† 1
x lim 1 x 1 1 x 1 1 x 1 ;! ' †! '†!&†'†!&&††&kk kkÉ ˆ‰11111
nÄ_
n
n1_
when x 1 we have , a conditionally convergent series; when x 1 we haveœ œ œ 1 1!!
__
œœn1 n1
(1) ( )
n n
œœ"
nn
È
ÚÊœ
, a divergent p-series!!
__
œœn1 n1
""
n
Èn n
œ
ÚÊœ
(a) the radius is ; the interval of convergence is ( 1 ) x (1 )_&&Ά& 11
(b) the interval of absolute convergence is 1 x 1&& † † &11
(c) the series converges conditionally at x 1œ 1
32. lim 1 lim 1 lim 1 1
nn nÄ_ Ä_ Ä_
¹¹
»»
kk
u
u2
x2 x2
2
x2n1
n
2n 3
n 1 2n 1
nÛ
Û œ
ÛÛ
†! †! †
Š‹ Š‹
ÈÈ
Š‹
È
œœ
œ
#
†
1 x 2 2 x 2 2 2 x 2 2 0 x 2 2 ; when!†!&†!&†!&†&†!††
Š‹
È
x2œ
#
#
œ
Š‹ ¹¹
ÈÈÈÈÈÈÈ
x 0 we have 2 which diverges since lim a 0; when x 2 2œœœ Áœ!!! ÈÈ
___
œœœn1 n1 n1
Š‹
È
œ
#
2
2
2
n
2n 1
nn
n12Û
ÛÊ
nÄ_

741. Section 11.7 Power Series 737
we have 2, a divergent series!!! È
___
œœœn1 n1 n1
Š‹
È
2
22
2
2n 1
nn
n12*
*!
œœ
(a) the radius is 2; the interval of convergence is 0 x 2 2
ÈÈ
††
(b) the interval of absolute convergence is 0 x 2 2††
È
(c) there are no values for which the series converges conditionally
33. lim 1 lim 1 lim 1 1 (x 1) 4 x 1 2
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk k k
u
u4(x1)4
(x 1) (x 1)4n1
n
2n 2
n1 2n
n*
* œ
*
†! †! †! & †! &†
œœ
œ
#
†
2 x 1 2 1 x 3; at x 1 we have 1, which diverges; at x 3Ê Ê œ œ œ œ !!!
___
œœœn0 n0 n0
(2)
44
4œ
2n
nn
n
we have 1, a divergent series; the interval of convergence is 1 x 3; the series!!!
___
œœœn0 n0 n0
24
44
2n n
nn
œœ
is a convergent geometric series when 1 x 3 and the sum is!! Š‹ˆ‰
__
œœn0 n0
(x )
4
x1
n
œ" œ
#
#
2n
n
œ
""
œ
œ_ œ _ œ
1
44
4 x 2x 1 3 2x x
Š‹ ’“
x 4(x )
4(œ
&
ááœ
&
œ œœœœ œ
34. lim 1 lim 1 lim 1 1 (x 1) 9 x 1 3
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk k k
u
u9(x1)9
(x 1) (x 1)9n1
n
2n 2
n1 2n
n*
* œ
*
†! †! †! ' †! '†
__
_
#
†
3 x 1 3 4 x 2; when x 4 we have 1 which diverges; at x 2 we haveÊ Ê œ œ œ !!
__
œœn0 n0
(3)
9
œ
2n
n
which also diverges; the interval of convergence is 4 x 2; the series!!
__
œœn0 n0
3
9
2n
n
œ"
is a convergent geometric series when 4 x 2 and the sum is!! Š‹ˆ‰
__
œœn0 n0
(x 1)
93
x1
n
_ _
#
2n
n
œ
""
œ
œœ œ œ œ
1
99
9 x 2x 1 8 2x x
Š‹ ’“
x1
3
9(x1)
9' ('
&
œ œœœœ œ
35. lim 1 lim 1 x 2 2 2 x 2 2 0 x 4
nnÄ_ Ä_
¹¹ ºº
¸¸ÈÈÈ
u
u2
x2 2
x2n1
n
n1
n1
n
n*
*
*
†! †! &†!&† &†!† †
ˆ‰È
ˆ‰È
œ
œ
†
0 x 16; when x 0 we have ( 1) , a divergent series; when x 16 we have (1) , a divergentÊ œ œ !!
_ _
œ œn0 n0
nn
series; the interval of convergence is 0 x 16; the series is a convergent geometric series when†† !Š‹
_
œn0
Èx2œ
#
n 0 x 16 and its sum is œ œ
""
œ
œ
1
2
4x
Œ œ Œ œ
È
œœx2 2 x2(( '
&&
36. lim 1 lim 1 ln x 1 1 ln x 1 e x e; when x e or e we
nnÄ_ Ä_
¹¹ ¹ ¹ kk
u
u(ln x)
(ln x)n1
n
n1
n*
*
Ê Ê Ê Ê œ
œ" œ"
obtain the series 1 and ( 1) which both diverge; the interval of convergence is e x e;!!
__
œœn0 n0
nn
&††
œ"
(ln x) when e x e!
_
œn0
n
œ
"
œ
œ"
1ln x
37. lim 1 lim 1 lim 1 1 1 x 2
nn nÄ_ Ä_ Ä_
¹¹ Š ‹ ºº
ˆ‰ kk
u
u3x13 3
x1 3 x
n1
n x1n1
n*
œ œ
œ
œ
†! †! †! †! †
_ _"
_
_
_ #
†
ab
x 2 2 x 2 ; at x 2 we have (1) which diverges; the interval of convergence isÊ Ê œ„kk
ÈÈ È È !
_
œn0
n
2 x 2 ; the series is a convergent geometric series when 2 x 2 and its sum is&†† &††
ÈÈ ÈÈ !Š‹
_
œn0
x1
3
n
œ
_

""
œ
#œ
1
3
x
Š‹Š ‹
x1 3x1
33
&&
'((œœ
œ

742. 738 Chapter 11 Infinite Sequences and Series
38. lim 1 lim 1 x 1 2 3 x 3 ; when x 3 we
nnÄ_ Ä_
¹¹ ¹ ¹ kk
ÈÈ È
u
u2
x1 2
x1n1
n
n1
n1
n
n*
œ
*
* œ
Ê Ê Ê œ„
ab
ab
œ
_
#
†
have 1 , a divergent series; the interval of convergence is 3 x 3 ; the series is a! !ÈÈ
Š‹
_ _
œ œn0 n0
n
&††
x1
2
n
œ
œ
convergent geometric series when 3 x 3 and its sum is œ œ
ÈÈ
""
œ
œ
1
2
3x
Š‹
&'
x1
2
2x1
&
( ((
&
&
&' œ
39. lim 1 x 3 2 1 x 5; when x 1 we have (1) which diverges;
nÄ_
¹¹ kk !
(x 3) 2
(x 3)
nœ
#œ
n1
n1 n
n*
*
† Ê Ê œ
'
*
n1
when x 5 we have ( 1) which also diverges; the interval of convergence is 1 x 5; the sum of thisœ !
_
œn1
n
convergent geometric series is . If f(x) 1 (x 3) (x 3) (x 3)
""""
_
œ# #
#
1
2
x1 4
n
n
Š‹
x3(
&
œ œá á ˆ‰
then f (x) (x 3) n(x 3) is convergent when 1 x 5, and divergesœœáá
2
x1
n
n1
œ## #
¸œ "" "
ˆ‰
when x 1 or 5. The sum for f (x) is , the derivative of .œ
w œ
œœ
22
(x 1) x 1
œ
40. If f(x) 1 (x 3) (x 3) (x 3) then f(x) dxœ á áœ
"" "
##œ
#
4x1
n
n 2
ˆ‰ '
x . At x 1 the series diverges; at x 5œá á œ œ
(x 3) (x 3) (x 3)
412 n1 n1
n
2œœ œ "œ
#_ _
œ &*
ˆ‰ !
n1
_
œn1
the series converges. Therefore the interval of convergence is 1 x 5 and the sum is!
_
œn1
(1)2
n1
œ
_
n
†Î 2 ln x 1 (3 ln 4), since dx 2 ln x 1 C, where C 3 ln 4 when x 3.kk kk œ œ œ '2
x1œ
41. (a) Differentiate the series for sin x to get cos x 1œ á
3x 5x 7x 9x 11x
3! 5! 7! 9! 11!
œ
1 . The series converges for all values of x sinceœ á
xxxxx
!4!6!8!10!

#
lim x lim 0 1 for all x.
nnÄ_ Ä_
¹¹Š‹
x
(2n 2)! x 2n 1 2n 2
n! 2
2n 2*
φ
___
# "
†
ab
abab
œœ
(b) sin 2x 2x 2xœ ᜠá
2 x 2 x 2 x 2 x 2 x 8x 32x 128x 512x 2048x
3! 5! 7! 9! 11! 3! 5! 7! 9! 11!
&& (( œœ ÑÑ )))) & ( œ Ñ ))
(c) 2 sin x cos x 2 (0 1) (0 0 1 1)x 0 1 0 0 1 x 0 0 1 0 0 1 xœ ( ˆ‰ˆ ‰† †† † †† ††††
œ" " "
##
#$
3!
0100001x00100 001x''&&' ''''''ˆ‰ˆ ‰††††† ††††††
""" """ "
##
%&
4! 3! 4! 3! 5!
010000001x 2x áœá ‘ˆ‰ ’“†††††††
"""""
#
'
6! 4! 3! 5! 3! 5!
4x 16x
& (
2x œ á
2x 2x 2x 2x 2 x
3! 5! 7! 9! 11!
&& (( œœ ÑÑ ))))
42. (a) e 1 1 x e ; thus the derivative of e is e itself
d 2x3x4x5x xxx x 2!3!4!5! !3!4!
x xxx
abœáœ áœ

#
(b) e dx e C x C, which is the general antiderivative of e'xx xxxx
3! 4! 5!
œœá
œ &'(
#
x
(c) e 1 x ; e e 1 1 (1 1 1 1)x 1 1 1 1 x
œ #
# ##
""x xxxx
!3!4!5! ! !
œá œ

(xx
†††† ††† ˆ‰
11 11x11 11x' &'& ' &' &'ˆ‰ˆ ‰†† †† †† † ††
"""" """"""
## ##
$%
3! ! ! 3! 4! 3! ! ! 3! 4!
1 1 1 1x 100000 áœáˆ‰††††††
" " "" "" " "
##
&
5! 4! ! 3! 3! ! 4! 5!
43. (a) ln sec x C tan x dx x dxkk Š‹œ œ á '' x 2x 17x 62x
3 15 315 2835
& (œ Ñ
C; x 0 C 0 ln sec x ,œ á œÊ œÊ œ á
x x x 17x 31x x x x 17x 31x
1 45 2520 14,175 12 45 2520 14,175

## #
kk
converges when x&††
11
##
(b) sec x x 1 x , converges
# #
œ œ áœá
d(tan x)
dx dx 3 15 315 2835 3 45 315
d x 2x 17x 62x 2x 17x 62x
Š‹
&
when x &††
11
##

743. Section 11.8 Taylor and Maclaurin Series 739
(c) sec x (sec x)(sec x) 1 1
#
##
œœáኋŠ‹
x5x61x x5x61x
24 720 24 720
œ
1x x xœ ሉˆ ‰ˆ ‰
"" "
##
#% '5 5 615561
24 4 24 720 48 48 720
1 x , xχ
#
##
2x 17x 62x
3 45 315

11
44. (a) ln sec x tan x C sec x dx 1 dxkk Š‹œ œá '' x5x61x
224720
œ
x C; x 0 C 0 ln sec x tan xœ á œÊ œÊ
x x 61x 277x
6 24 5040 72,576
&( œ Ñ
kk
x , converges when xœ á
x x 61x 277x
6 24 5040 72,576
&( œ Ñ
11
##
(b) sec x tan x 1 x , convergesœœáœá
d(sec x)
dx dx 24 720 6 120 1008
d x 5x 61x 5x 61x 277x
Š‹
œ
#
when x &††
11
##
(c) (sec x)(tan x) 1 xœ á ኋŠ‹
x5x61x x2x17x
24 720 3 15 315
œ
#
xx x xx ,œ ᜠሉˆ ‰ˆ ‰
"" " "
#
$& (
3 15 6 24 315 15 72 720 6 120 1008
2 5 17 5 61 5x 61x 277x
&( œ
x &††
11
##
45. (a) If f(x) a x , then f (x) n(n 1)(n 2) (n (k 1)) a x and f (0) k!aœ œ â œ!!
__
œœn0 nk
nnk
nk nkkÐÑ ÐÑ
a ; likewise if f(x) b x , then b a b for every nonnegative integer kÊœ œ œ Êœ knkkk
f(0) f(0)
k! k!
n
ß„ ß„k k
!
_
œn0
(b) If f(x) a x 0 for all x, then f (x) 0 for all x from part (a) that a 0 for everyœœ œ Ê œ!
_
œn0
n k
nk ÐÑ
nonnegative integer k
46. 1 x x x x x x 1 2x 3x 4x
""
œœ œ
#$% # $
1 x (1 x) (1 x)
x
œáÊ œ áÊ ’“ ab
œ œ
x2x 3x 4x x x14x9x 16x œáÊ œ áÊ
#$% # $ __
œœ
’“ ab
1x xx
(1 x) (1 x)
& &
œ
x 4x 9x 16x 6œ á Ê œ á Ê œ
#$ % "
#
ˆ‰
œœ
&
'
4
ˆ‰
Ú
œ
8
n
4916 n
4816 2 !
'
Û
n1
47. The series converges conditionally at the left-hand endpoint of its interval of convergence [ 1 1 ; the!
_
œn1
x
n
n
ß Ñ
series converges absolutely at the left-hand endpoint of its interval of convergence [ 1 1]!
_
œn1
x
n
n
ab
œ &â
48. Answers will vary. For instance:
(a) (b) (x 1) (c) !!!ˆ‰ ˆ‰
___
œœœn1 n1 n1
x x3
3
n
n
n
'
œ
#
11.8 TAYLOR AND MACLAURIN SERIES
1. f(x) ln x, f (x) , f (x) , f (x) ; f(1) ln 1 0, f (1) 1, f (1) 1, f (1) 2 P (x) 0,œ œ œ œ œ œ œ œ œ Ê œ
w ww www w ww www""
!
xxx
2
œ &
P (x) (x 1), P (x) (x 1) (x 1) , P (x) (x 1) (x 1) (x 1)"# $
"""
##
##$œ œ œ
3
2. f(x) ln (1 x), f (x) (1 x) , f (x) (1 x) , f (x) 2(1 x) ; f(0) ln 1 0,œ œ œ œ œ œœ
¸ œ"¸¸ œ#¸¸¸ œ$"
_1x
f (0) 1, f (0) (1) 1, f (0) 2(1) 2 P (x) 0, P (x) x, P (x) x , P (x)
¸ ¸¸ œ# ¸¸¸ œ$
!"# $
#
œ œ œ œ œ œ Ê œ œ œ
1 x
1
œ
xœ
xx
3

#

744. 740 Chapter 11 Infinite Sequences and Series
3. f(x) x , f (x) x , f (x) 2x , f (x) 6x ; f(2) , f (2) , f (2) , f (x)œ œ œ œ œ œ œ œ œ
" """œ" ¸ œ# ¸¸ œ$ ¸¸¸ œ% ¸ ¸¸ ¸¸¸
#x 44 8
3
P (x) , P (x) (x 2), P (x) (x 2) (x 2) ,Ê œ œ œ!" #
""" "" "
## #
#
448
P (x) (x 2) (x 2) (x 2)$
"" " "
#
#$œ
48 16
4. f(x) (x 2) , f (x) (x 2) , f (x) 2(x 2) , f (x) 6(x 2) ; f(0) (2) , f (0) (2)œ œ œ œ œ œ œ
œ" ¸ œ# ¸¸ œ$ ¸¸¸ œ% œ" ¸ œ# "
#
, f (0) 2(2) , f (0) 6(2) P (x) , P (x) , P (x) ,œ œ œ œ œ Ê œ œ œ
"" """¸¸ œ$ ¸¸¸ œ%
!" #
## #44 8 448
3xxx
œ
P(x)$
"
#œ
xx x
4816

5. f(x) sin x, f (x) cos x, f (x) sin x, f (x) cos x; f sin , f cos ,œ œ œ œ œ œ œ œ
w ww www w
## ˆ‰ ˆ‰
11 1 1
44 4 4
22
ÈÈ
f sin , f cos P , P (x) x ,
ww www
#####
!"ˆ‰ ˆ‰ ˆ ‰
11 1 1 1
44 4 4 4
22222
œ œ œ œ Ê œ œ
ÈÈÈÈÈ
P (x) x x , P (x) x x x#$
## ## #
##$œ œ
ÈÈ È ÈÈ È È
22 2 22 2 2
44 4 44 4 1 4
ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
11 11 1
6. f(x) cos x, f (x) sin x, f (x) cos x, f (x) sin x; f cos ,œœœ œœœ
w ww www " ˆ‰
11
44 2
È
f sin , f cos , f sin P (x) ,
wwwwww """"
!ˆ‰ ˆ‰ ˆ‰
11 1 1 11
44 4 4 44 2222
œ œ œ œ œ œ Ê œ
ÈÈÈÈ
P (x) x , P (x) x x ,"#
"" "" "
#
#œ œ
ÈÈ ÈÈ È
22 22 2 444
ˆ‰ ˆ‰ ˆ‰
111
P (x) x x x$
"" " "
#
#$œ
ÈÈ È È
22 2 62 44 4
ˆ‰ ˆ‰ ˆ‰
11 1
7. f(x) x x , f (x) x , f (x) x , f (x) x ; f(4) 4 2,œœ œ œ œ œœÈ ˆ‰ ˆ ‰ ˆ‰ È
"Î# w "Î# ww $Î# www &Î#""
# 48
3
f (4) 4 , f (4) 4 ,f (4) 4 P (x) 2, P (x) 2 (x 4),
w "Î# ww $Î# www &Î#"" " " "
#
!"
œœœœœœÊœœˆ‰ ˆ ‰ ˆ‰
4 4 32 8 256 4
33
P(x)2 (x4) (x4), P(x)2 (x4) (x4) (x4)#$
"" "" " ##$
#œ œ
4 64 4 64 51
8. f(x) (x 4) , f (x) (x 4) , f (x) (x 4) , f (x) (x 4) ; f(0) (4) 2,œ œ œ œ œ œ
"Î# w "Î# ww $Î# www &Î# "Î#""
# ˆ‰ ˆ ‰ ˆ‰
48
3
f (0) (4) , f (0) (4) , f (0) (4) P (x) 2,
w "Î# ww $Î# www &Î#"" " "
#
!
œœœœœœÊœˆ‰ ˆ ‰ ˆ‰
4 4 32 8 256
33
P (x) 2 x, P (x) 2 x x , P (x) 2 x x x"# $
""" """ ##$œ œ œ
4 4 64 4 64 512
9. e e 1 x
xx x xxx
n! n! ! 3! 4!
(x)
œ Ê œ œá!!
__
œœn0 n0
n n
œ œ
#
œ &'
10. e e 1
x xxxxx
n! n! 4 2! 2 3! 24!
œ Êœ œá!!
__
œœn0 n0
n
n
x2œ
ˆ‰
x
2
#
œ &'
& '
†† †
11. f(x) (1 x) f (x) (1 x) , f (x) 2(1 x) , f (x) 3!(1 x) f (x)œ Ê œ œ œ Êá
œ" ¸ œ# ¸¸ œ$ ¸¸¸ œ% Ê_k
( 1) k!(1 x) ; f(0) 1, f (0) 1, f (0) 2, f (0) 3!, f (0) ( 1) k!œ œ œ œ œ áß œ
kk1 œ œ ¸ ¸¸ ¸¸¸ Ê_kk
1 x x x (x) (1)xÊ œáœ œ
"
_
#$
1x
nn !!
__
œœn0 n0
n
12. f(x) (1 x) f (x) (1 x) , f (x) 2(1 x) , f (x) 3!(1 x) f (x)œ Ê œ œ œ Êá
œ" ¸ œ# ¸¸ œ$ ¸¸¸ œ% Ê_k
k!(1 x) ; f(0) 1, f (0) 1, f (0) 2, f (0) 3!, f (0) k!œ œ œ œ œáß œ
(( Ê_k1 k wwwwww
1xx x xÊ œáœ
"
œ
#$
1x
n !
_
œn0
13. sin x sin 3x 3xœ Ê œ œ œá!!!
___
œœœn0 n0 n0
()x ()(3x) ()3 x
(n 1)! (n 1)! (n 1)! 3! 5!
3x 3xœ" œ" œ"
#_ #_ #_
n2n1 n 2n1 n2n12n1**** && ((

745. Section 11.8 Taylor and Maclaurin Series 741
14. sin x sin œ Ê œ œ œá!!!
___
œœœn0 n0 n0
()x ()x
( n 1)! ( n 1)! (2n 1)! 2 3! 2 5!
xxxx ()œ" œ"
#_ # #_ # _ #
œ"
n2n1 n2n1 nx
2n 1
2n 1* *
œ
*
*&(
&( ˆ‰
††
15. 7 cos ( x) 7 cos x 7 7 , since the cosine is an even functionœ œ œá !
_
œn0
()x
(2n)! ! 4! 6!
7x 7x 7xœ"
#
n2n œ
16. cos x 5 cos x 5 5œ Ê œ œá!!
__
œœn0 n0
(1)x (1)(x)
(2n)! ( n)! 2! 4! 6!
5x 5x 5xœœ
#
n2n n 2n
1
1 111
œœ
17. cosh x 1 x 1 x 1œ œ á á œá
ee xxx xxx xxx
!3!4! !3!4! !4!6!
xx
_"
## # # #
#

’“Š‹Š‹
œ!
_
œn0
x
(2n)!
2n
18. sinh x 1 x 1 x xœ œ á á œá
ee xxx xxx xxx
! 3! 4! ! 3! 4! 3! 5! 6!
xx
œ"
## # #

’“Š‹Š‹
œ!
_
œn0
x
(2n 1)!
2n 1*
_
19. f(x) x 2x 5x 4 f (x) 4x 6x 5, f (x) 12x 12x, f (x) 24x 12, f (x) 24œ Ê œ œ œ œ
% $ w $ # ww # www Ð Ñ 4
f (x) 0 if n 5; f(0) 4, f (0) 5, f (0) 0, f (0) 12, f (0) 24, f (0) 0 if n 5Êœ œœœ œ œ œ 
ÐÑ w ww www ÐÑ ÐÑn 4n
x 2x 5x 4 4 5x x x x 2x 5x 4 itselfÊ œ œ
%$ $ %%$ 12 24
3! 4!
20. f(x) (x 1) f (x) 2(x 1); f (x) 2 f (x) 0 if n 3; f(0) 1, f (0) 2, f (0) 2, f (0) 0 ifœ Ê œ œÊ œ   œ œ œ œ
#w ww ÐÑ wwwÐÑ nn
n3 (x1) 12x x 12xx Ê œ œ
###
#
2
!
21. f(x) x 2x 4 f (x) 3x 2, f (x) 6x, f (x) 6 f (x) 0 if n 4; f(2) 8, f (2) 10,œÊ œ œ œÊ œ   œ œ
$ w # ww www Ð Ñ w n
f (2) 12, f (2) 6, f (2) 0 if n 4 x 2x 4 8 10(x 2) (x 2) (x 2)
ww www Ð Ñ $ # $
œ œ œ   Ê œ
n 12 6
2! 3!
8 10(x 2) 6(x 2) (x 2)œ
#$
22. f(x) 2x x 3x 8 f (x) 6x 2x 3, f (x) 12x 2, f (x) 12 f (x) 0 if n 4; f(1) 2,œ Ê œ œ œ Ê œ   œ
$ # w # ww www Ð Ñ n
f (1) 11, f (1) 14, f (1) 12, f (1) 0 if n 4 2x x 3x 8
wwwwwwÐÑ $#
œœ œ œ Ê
n
211(x1) (x1) (x1) 211(x1)7(x1) 2(x1)œ œ
14 12
2! 3!
#$ #$
23. f(x) x x 1 f (x) 4x 2x, f (x) 12x 2, f (x) 24x, f (x) 24, f (x) 0 if n 5;œÊ œ œ œ œ œ  
% # w $ ww # www Ð Ñ Ð Ñ 4n
f( 2) 21, f ( 2) 36, f ( 2) 50, f ( 2) 48, f ( 2) 24, f ( 2) 0 if n 5 x x 1œ œ œ œ œ œ   Ê
w ww www ÐÑ ÐÑ % # 4n
21 36(x 2) (x 2) (x 2) (x 2) 21 36(x 2) 25(x 2) 8(x 2) (x 2)œ œ
50 48 24
2! 3! 4!
#$% #$%
24. f(x) 3x x 2x x 2 f (x) 15x 4x 6x 2x, f (x) 60x 12x 12x 2,œÊ œ œ
&% $# w % $ # ww $ #
f (x) 180x 24x 12, f (x) 360x 24, f (x) 360, f (x) 0 if n 6; f( 1) 7,
www # Ð Ñ Ð Ñ Ð Ñ
œœœ œ œ
45n
f ( 1) 23, f ( 1) 82, f ( 1) 216, f ( 1) 384, f ( 1) 360, f ( 1) 0 if n 6
w ww www ÐÑ ÐÑ ÐÑ
œ œ œ œ œ œ  
45n
3x x 2x x 2 723(x1) (x1) (x1) (x1) (x1)Ê œ
&%$# #$%& 82 216 384 360
2! 3! 4! 5!
7 23(x 1) 41(x 1) 36(x 1) 16(x 1) 3(x 1)œ
#$%&
25. f(x) x f (x) 2x , f (x) 3! x , f (x) 4! x f (x) ( 1) (n 1)! x ;œÊ œ œ œ Ê œ
# w $ ww % www & Ð Ñ nnn2
f(1) 1, f (1) 2, f (1) 3!, f (1) 4!, f (1) ( 1) (n 1)! œœœ œ œÊ
w ww www Ð Ñ "nn
x
œ
1 2(x 1) 3(x 1) 4(x 1) ( 1) (n 1)(x 1)œáœ
#$ !
_
œn0
nn

746. 742 Chapter 11 Infinite Sequences and Series
26. f(x) f (x) (1 x) , f (x) 2(1 x) , f (x) 3! (1 x) f (x) n! (1 x) ;œÊ œ œ œÊ œ
x
1x
nn1
œ
w #ww $www % ÐÑ
f(0) 0, f (0) 1, f (0) 2, f (0) 3! x x x xœ œ œ œ Ê œáœ
¸¸¸¸¸¸ #$ _
œ
x
1x
n1 !
_
œn0
27. f(x) e f (x) e , f (x) e f (x) e ; f(2) e , f (2) e , f (2) eœÊ œ œÊ œ œ œá œ
xxxnx nwww ÐÑ #w#ÐÑ#
e e e(x2) (x2) (x2) (x2)Êœáœ
x nee e
3! n!
## # $
#
œ
!
_
œn0
28. f(x) 2 f (x) 2 ln 2, f (x) 2 (ln 2) , f (x) 2 (ln 2) f (x) 2 (ln 2) ; f(1) 2, f (1) 2 ln 2,œÊ œ œ œ Ê œ œ œ
xx x xnxnw ww # www $ Ð Ñ w
f (1) 2(ln 2) , f (1) 2(ln 2) , , f (1) 2(ln 2)
ww # www $ Ð Ñ
œœáœ
nn
2 2 (2 ln 2)(x 1) (x 1) (x 1) Êœ áœ
x 2(ln 2) 2(ln 2) 2(ln 2) (x 1)
3! n!
œ &
#
#$ œ
!
_
œn0
nn
29. If e (x a) and f(x) e , we have f (a) e f or all n 0, 1, 2, 3,
x
œ œ œ œá!
_
œn0
f(a)
n!
nxna
ß„n
ÐÑ
e e e 1 (x a) at x aÊœ áœáœ
xa a (x a) (x a) (x a) (x a)
0! 1! 2! 2!
’“’“
œœœ œ
 Úœ œ
30. f(x) e f (x) e for all n f (1) e for all n 0, 1, 2, œÊ œ Ê œ œ á
xn x nÐÑ ÐÑ
e ee(x1) (x1) (x1) e1(x1)ʜᜠá
x ee
!3! 2!3!
(x 1) (x 1)
#
#$ œœ
’“

31. f(x) f(a) f (a)(x a) (x a) (x a) f (x)œ áÊ
w#$w
#
f (a) f (a)
3!
†† †††
f (a) f (a)(x a) 3(x a) f (x) f (a) f (a)(x a) 4 3(x a)œ áÊ œ á
w ww # ww ww www # f (a) f (a)
3! 4!
††† ß„ 4
†
f (x) f (a) f (a)(x a) (x a)Êœ á
ÐÑ ÐÑ ÐÑ #
#
nnn1 f(a)
ßÛ„n2
f(a) f(a) 0, f (a) f (a) 0, , f (a) f (a) 0Êœ œá œ
w w ÐÑ ÐÑ nn
32. E(x) f(x) b b (x a) b (x a) b (x a) b (x a)χ !" # $
#$
n
n
0 E(a) f(a) b b f(a); from condition (b),Êœ œ Ê œ !!
lim 0
xaÄ
f(x) f(a) b (x a) b (x a) b (x a) b (x a)
(x a)
œ œ œœ œœ œœÜœ œ
œ Ú
n
n
n
œ
lim 0ʜ
xaÄ
f(x) b 2b(x a) 3b(x a) nb(x a)
n(x a)



œ œ œ œ œ œÜœ œ
œ n
n1
n1
b f (a) lim 0Êœ Ê œ"
w œ œ œ œÜœ œ" œ
œœ
xaÄ
f (x) 2b 3! b (x a) n(n )b (x a)
n(n 1)(x a)
†† Î


n
n2
n2 b f (a) lim 0Êœ Ê œ#
"
#œœ#œ
ww œ œÜœ œ œ œ
xaÄ
f (x) 3! b n(n 1)(n 2)b (x a)
n(n 1)(n )(x a)
††† Î
&

n
n3
n3
b f (a) lim 0 b f (a); therefore,œœ Ê œÊ œ$
""www Ð Ñ œ
3! n! n!
f (x) n! b
n
n
xaÄ
ß„n
n
g(x) f(a) f (a)(x a) (x a) (x a) P (x)œ á œ
w# f (a) f (a)
2! n!
n
n
†† â „ n
33. f(x) ln (cos x) f (x) tan x and f (x) sec x; f(0) 0, f (0) 0, f (0) 1œ Ê œ œ œ œ œ
www#www
L(x) 0 and Q(x)Êœ œ
x
2
œ
34. f(x) e f (x) (cos x)e and f (x) ( sin x)e (cos x) e ; f(0) 1, f (0) 1,œÊœ œ œ œ
sin x sin x sin x sin xwww #w
f (0) 1 L(x) 1 x and Q(x) 1 x
ww
#
œ Ê œ œ
x
œ
35. f(x) 1 x f (x) x 1 x and f (x) 1 x 3x 1 x ; f(0) 1,œ Ê œ œ œab ab ab ab
#w #ww ###
"Î# $Î# $Î# &Î#
f (0) 0, f (0) 1 L(x) 1 and Q(x) 1
www
#
œœÊœ œ
x
œ
36. f(x) cosh x f (x) sinh x and f (x) cosh x; f(0) 1, f (0) 0, f (0) 1 L(x) 1 and Q(x) 1œÊœ œ œœ œÊœ œ
www www
#
x
œ

747. Section 11.9 Convergence of Taylor Series; Error Estimates 743
37. f(x) sin x f (x) cos x and f (x) sin x; f(0) 0, f (0) 1, f (0) 0 L(x) x and Q(x) xœÊœ œ œ œ œÊœ œ
www www
38. f(x) tan x f (x) sec x and f (x) 2 sec x tan x; f(0) 0, f (0) 1, f 0 L(x) x and Q(x) xœÊœ œ œ œœÊœ œ
w # ww # w ww
11.9 CONVERGENCE OF TAYLOR SERIES; ERROR ESTIMATES
1. e 1 x e 1 ( 5x) 1 5x
x5x x x 5x 5x
!n! ! !3! n!
( 5x) ( 1) 5 x
œ áœ Ê œ ᜠáœ
œ œœ &&œ
###
œ œœ
!!
_ _
œ œn0 n0
n nnn
2. e 1 x e 1 1
xx2 xx x xxx
!n! ! 2!23!
œ áœ Ê œ ᜠá
œ œ &
œ
œ

#####
œŸ œ œ
! ˆ‰
_
œn0
n
x ˆ‰
œ!
_
œn0
(1)x
2n!
œ
nn
n
3. sin x x 5 sin ( x) 5 ( x)œáœ Ê œ á
xx
3! 5! ( n 1)! 3! 5!
(1)x (x) (x)
& á *&(
! ’“
_
œn0
œœœ
#_
n2n1
œ!
_
œn0
5( 1) x
(n 1)!
œ
#_
n12n1ÛÛ
4. sin x x sin œáœ Ê œá
xx x x
3! 5! ( n 1)! 3! 5! 7!
(1)x
&( Û
œœœ
&(œ
!
_
œn0
œ
#_ # #
n2n1 xxx
11
ˆ‰ ˆ‰ ˆ‰
œœœ
œ!
_
œn0
(1) x
2(n1)!
œ
#_
n 2n12n1
2n 1
1
ÛÛ
Û
5. cos x cos x 1 1œÊœ œ œá!!! È
___
œœœn0 n0 n0
(1)x
(2n)! (2n)! (2n)! ! 4! 6!
(1) x1
(1)x1 x1 x1 x1œ
œ_
œ_ _ _ _
#
n2n
n
2n
n n ’“ab
ab ab ab
ÚÊœ

6. cos x cos cos œÊœ œ œ!!! Š‹ Š‹ Œ '
__
œœn0 n0
(1)x (1)x
(2n)! ( n)! 2 (2n)!
xx
2
(1)
n0
œ œ
##
"Î# œ _
œ
n2n n3n
n x
2n
n &
&
œ
ÚÊœ
È
Œ œŠ‹ 1œá
xx x
22! 2 4! 2 6!


†††
7. e xe x x x
xx x x x xxx
n! n! n! ! 3! 4!
n0
œ Ê œ œ œá!!!
Π'
__
œœn0 n0
nnn1
_
œ
#
#
Û &'(
8. sin x x sin x x xœ Ê œ œ œá!!!
Π'
___
œœœn0 n0 n0
(1)x (1)x (1)x
(2n 1)! ( n 1)! (2n 1)! 3! 5! 7!
xxxœœœ
_#__
## $
n2n1 n2n1 n2n3ÛÛÛ áÝ
9. cos x 1 cos x 1 1 1œ Ê œ œ á!!
__
œœn0 n0
(1)x (1)x
(2n)! ( n)! 2 4! 6! 8! 10!
x x x xxxxxœœ
####
n2n n2n œœ œœ
œ áœ
xxxx
4! 6! 8! 10! ( n)!
(1)x

!
_
œn2
œ
#
n2n
10. sin x sin x x xœÊœ !!
Π'
__
œœn0 n0
(1)x (1)x
(2n 1)! 3! ( n 1)! 3!
xxœœ
_#_
n2n1 n2n1ÛÛ &&
xx œ ᜠᜊ‹ !
xxxxx x xxxx
3! 5! 7! 9! 11! 3! 5! 7! 9! 11! (2n 1)!
(1)x
&(œÑ)) & (œÑ)) Û
_
œn2
œ
_
n2n1
11. cos x x cos x x xœÊœ œ œá!!!
___
œœœn0 n0 n0
(1)x (1)(x) ( ) x
(2n)! ( n)! ( n)! 2! 4! 6!
xxxœœœ"
##
n2n n 2n n 2n2n1
1
11 111
Û

748. 744 Chapter 11 Infinite Sequences and Series
12. cos x x cos x x xœ Ê œ œ œá!!! ab
___
œœœn0 n0 n0
(1)x ( )x
(2n)! ( n)! ( n)! 2! 4! 6!
(1) x xx xœœ" ### # œ
##
n2n n4n2n
2n
ab
œ *
13. cos x 1
# """ ""
# # ## ##
œ
œ œ œ á
cos 2x ( 1) (2x) (2x) (2x) (2x) (2x)
(2n)! 2!4!6!8!! ’“
_
œn0
n2n œ
11 1 œáœ œ
(2x) (2x) (2x) (2x) ( 1) (2x) ( 1) 2 x
2 2! 2 4! 2 6! 2 8! 2 (2n)! (2n)!

†††† †
!!
__
œœn1 n1
œœ
n 2n n 2n 1 2n
14. sin x cos 2x 1
# œ"" ""
### ###
œ œ œ áœáˆ‰ Š‹
1 cos 2x (2x) (2x) (2x) (2x) (2x) (2x)
!4!6! 22!24!26!
œ
†††
œœ!!
__
œœn1 n1
(1) (2x) (1) 2 x
(2n)! (2n)!
œœ
#
n1 2n n 2n1 2n*
†
15. x x (2x) 2 x x 2x 2 x 2 x
x
12x 12x
nnn2
œ
œœ
# # # $#%$&"
œ œ œ œáˆ‰ !!
__
œœn0 n0
16. x ln (1 2x) x 2xœ œ œá!!
__
œœn1 n1
(1) (2x) (1) 2x
nn 45
2x 2x 2xœœ #
#
n1 n n1nn1 *
17. x 1 x x x 1 2x 3x nx
"""
œœœ
#$ # œ
1x dx 1x (1x)
n n1d
œ œáÊ œ œ áœ!! ˆ‰
_ _
œ œn0 n1
œ
(n 1)xœ!
_
œn0
n
18. 1 2x 3x 2 6x 12x n(n 1)x
2d d d
1x dx 1 x dx (1 x) dx
n2
abœ
""
œœ
##œ
&
œ
œœœœ œáœáœˆ‰ Š‹ ab !
_
œn2
(n 2)(n 1)xœ!
_
œn0
n
19. By the Alternating Series Estimation Theorem, the error is less than x 5! 5 10
kkx
5!
(
Ê ‚kk a ba b
& œ%
x 600 10 x 6 10 0.56968Ê‚ Ê‚ ¸kk kk
È
& œ% œ#
(
20. If cos x 1 and x 0.5, then the error is less than 0.0026, by Alternating Series Estimation Theorem;œ œ
x (.5)
24
œ '
#
kk ¹¹
since the next term in the series is positive, the approximation 1 is too small, by the Alternating Series Estimation&
x
œ
#
Theorem
21. If sin x x and x 10 , then the error is less than 1.67 10 , by Alternating Series Estimation Theorem;œ ¸‚kk
œ$ œ ab10
3!
10
&
&
The Alternating Series Estimation Theorem says R (x) has the same sign as . Moreover, x sin x# &†
x
3!
&
0 sin x x R (x) x 0 10 x 0.Ê œ ÊÊ #
œ$
22. 1 x 1 . By the Alternating Series Estimation Theorem the errorÈ kk ¹¹œ á
xx x x
816 88
(0.01)
#
œ
œ & œ œ
1.25 10œ‚
&
23. R (x) 1.87 10 , where c is between 0 and xkk ¹¹ #
œœ ‚
ex
3! 3!
3 (0.1) 4
c 01& âw„ &
24. R (x) 1.67 10 , where c is between 0 and xkk ¹¹ #
œ%œœ‚
ex
3! 3!
(0.1)
c& &

749. Section 11.9 Convergence of Taylor Series; Error Estimates 745
25. R (x) x (1.13) 0.000294kk ¸¸ ¹¹ %
& _
##
_œ œ ¸
cosh c e e x
5! 5! 5! 5!
1.65 (0.5) (0.5)
cc
1.65 (
) ((
†
26. If we approximate e with 1 h and 0 h 0.01, then error h
h eh e hh e(0.0)
'ÎÎ †ÎÎ kk ¹¹ Š ‹
c001 001œ¸ w
## #
"†
0.00505h 0.006h (0.6%)h, where c is between 0 and h.œœ
27. R x .01 x (1%) x .01 0 x .02k k kk kk kk kk¹¹ ¸¸ ¸¸"
"
_# # # #œœœÊÊ
(1 c) !
xxx x
œ
œœ
28. tan x x tan 1 1 ; error .01
œ" œ" """ "
#_
œáÊ œ œ &'&'( † †
xxx
357 4 357 n1
&(œ
1
kk
2n 1 100 n 49!'Ñ !Ñ
29. (a) sin x x 1 , s 1 and s 1 ; if L is the sum of theœáÊ œá œ œ
xxx sin x xxx x
3! 5! 7! x 3! 5! 7! 6
&
"#
series representing , then by the Alternating Series Estimation Theorem, L s 1 0 and
sin x sin x
x x
œ "
L s 1 0. Therefore 1 1œ #
sin x x x sin x
x6 6x Š‹
œœ
(b) The graph of y , x 0, is bounded below by theœÁ
sin x
x
graph of y 1 and above by the graph of y 1 asœ œ
x
6
œ
derived in part (a).

30. (a) cos x 1 1 cos x ;œáÊ œáÊ œá
xxx xxxx 1cos x xxx
!4!6! !4!6!8! x 4!6!8!
œ
œ
## #
œ"
if L is the sum of the series representing , then by the Alternating Series Estimation Theorem
1cos x
x
œ
œ
L s - 0 and 0. Therefore .œ "
œ" œ " " œ"
## ###
1 cos x 1 cos x x x 1 cos x
xx4! 4xœœ œ
œœ Š‹
(b) The graph of y is bounded below byœ
1cos x
x
œ
œ
the graph of y and above by the graph ofœ
"
#
x
24
œ
y as indicated in part (a).œ
"
#

31. sin x when x 0.1; the sum is sin (0.1) 0.099833417œ¸
32. cos x when x ; the sum is cos 0.707106781œœ¸
11
44 2
ˆ‰
"
È
33. tan x when x ; the sum is tan 0.808448
œ" œ"
œ¸
11
33
ˆ‰
34. ln (1 x) when x ; the sum is ln (1 ) 1.421080œ ¸ 11

750. 746 Chapter 11 Infinite Sequences and Series
35. e sin x 0 x x x x x x
x
3! ! 3! 3! 5! ! 3! 4! 5! 3! 3! 5!
œ á
#$%& ' "" "" """" """"
##ˆ‰ˆ‰ˆ ‰ˆ ‰
xxxxxχ
#$&'"" "
33090
36. e cos x 1 x x x x x
x
! ! ! 3! 4! ! 2! 4! 4! 2! 3! 5!
œ á
#$% &"" "" """" """"
## # #ˆ‰ˆ‰ˆ ‰ˆ ‰
1x x x xœ á
"" "$% &
3630
37. sin x cos 2x 1
# œ"" ""
### ## #
œ œ œ áœáˆ‰ Š‹
1 cos 2x 2x 2x 2x(2x) (2x) (2x)
2! 4! 6! ! 4! 6!
œ
sin x 2x 2 sin x cos xÊ œ ᜠáÊ
d d 2x 2x 2x
dx dx 2! 4! 6! 3! 5! 7!
(2x) (2x) (2x)
ab Š‹
#
&(œ
2x sin 2x, which checksœáœ
(2x) (2x) (2x)
3! 5! 7!
&(œ
38. cos x cos 2x sin x 1
##
##
œ œá &'&'(Š‹Š‹
(2x) (2x) (2x) (2x)
!4!6!8! !4!6!8!
2x 2x 2x 2x
œ
11xxxxœáœ á
2x 2x 2x 2
! 4! 6! 3 45 315

#
#% ' )""
39. A special case of Taylor's Theorem is f(b) f(a) f (c)(b a), where c is between a and bœ Ê
w
f(b) f(a) f (c)(b a), the Mean Value Theorem.œ
w
40. If f(x) is twice differentiable and at x a there is a point of inflection, then f (a) 0. Therefore,œœ
ww
L(x) Q(x) f(a) f (a)(x a).œœ
w
41. (a) f 0, f (a) 0 and x a interior to the interval I f(x) f(a) (x a) 0 throughout I
ww w #
#
Ÿœ œ ÊœŸ
f(c)
††
œ
f(x) f(a) throughout I f has a local maximum at x aÊŸ Ê œ
(b) similar reasoning gives f(x) f(a) (x a) 0 throughout I f(x) f(a) throughout I f has aœ   Ê   Ê
f(c)
††
œ
#
#
local minimum at x aœ
42. (a) f(x) (1 x) f (x) (1 x) f (x) 2(1 x) f (x) 6(1 x)œ Ê œ Ê œ Ê œ
"w #ww $ÐÑ % 3
f (x) 24(1 x) ; therefore 1 x x xÊœ ¸
ÐÑ & # $ "
œ
4
1x
(b) x 0.1 x the errorkk ¹¹ ¹¹ˆ‰ ˆ‰†!††!†!† !
10 10 10 x 10
11 1x 9 (1x) 9 (1x) 9
""
œœ œ
&&
%
áá
'
e (0.1) 0.00016935 0.00017, since .$
%
&
"
œŸœ œ¹¹ ¹¹¹¹ ˆ‰
max f (x) x f (x)
4! 9 4! (1 x)
10
â„ ' ß„
á4 4
43. (a) f(x) (1 x) f (x) k(1 x) f (x) k(k 1)(1 x) ; f(0) 1, f (0) k, and f (0) k(k 1)œ Ê œ Ê œ œ œ œ
kk1 k2¸ œ ¸¸ œ ¸ ¸¸
Q(x) 1 kx xʜ
k(k )œ"
#
#
(b) R (x) x x 0 x or 0 x .21544kk kk¸¸ #
"" " "$$œÊÊ
32
3! 100 100 100
††
Ú!&
44. Let P x x P .5 10 since P approximates accurate to n decimals. Then,œ Ê œ ‚11 1kk k k
œn
P sin P ( x) sin ( x) ( x) sin x (x sin x) (P sin P) œ œ œ Ê 111 1 1 kk
sin x x 10 .5 10 P sin P gives an approximation to correct to 3nœŸ‚‚ Êkk
kkx
3! 3!
0.125 3n 3n
&
œœ
1
decimals.
45. If f(x) a x , then f (x) n(n 1)(n 2) (n k 1)a x and f (0) k! aœ œ â œ!!
__
œœn0 nk
nnk
nk nkkÐÑ ÐÑ
a for k a nonnegative integer. Therefore, the coefficients of f(x) are identical with theʜ k
f(0)
k!
ß„k
corresponding coefficients in the Maclaurin series of f(x) and the statement follows.

751. Section 11.9 Convergence of Taylor Series; Error Estimates 747
46. : f even f( x) f(x) f ( x) f (x) f ( x) f (x) f odd;Note Êœ Êœ Ê œ Ê
www ww
f odd f( x) f(x) f ( x) f (x) f ( x) f (x) f even;Êœ Êœ Ê œ Ê
wwwww
also, f odd f( 0) f(0) 2f(0) 0 f(0) 0Êœ Ê œÊ œ
(a) If f(x) is even, then any odd-order derivative is odd and equal to 0 at x 0. Therefore,œ
a a a 0; that is, the Maclaurin series for f contains only even powers.
"$&œœœáœ
(b) If f(x) is odd, then any even-order derivative is odd and equal to 0 at x 0. Therefore,œ
a a a 0; that is, the Maclaurin series for f contains only odd powers.
!#%œœœáœ
47. (a) Suppose f(x) is a continuous periodic function with period p. Let x be an arbitrary real number. Then f
!
assumes a minimum m and a maximum m in the interval [x x p]; i.e., m f(x) m for all x in"#!!"# â' Î Î
[x x p]. Since f is periodic it has exactly the same values on all other intervals [x p x 2p],
!! !!â' 'â'
[x 2p x 3p], , and [x p x ], [x 2p x p], , and so forth. That is, for all real numbers
!! !!!!'â' ( &â &â& (
x we have m f(x) m . Now choose M max m , m . Then_ _ Ÿ Ÿ œ
"# "# efkkkk
M m m f(x) m m M f(x) M for all x.&Î& Î Î Î Î Î ! Îkk kk k k
"" ##
(b) The dominate term in the nth order Taylor polynomial generated by cos x about x a is (x a) orœ
sin (a)
n!
n
(x a) . In both cases, as x increases the absolute value of these dominate terms tends to ,
cos (a)
n!
n
& _kk
causing the graph of P (x) to move away from cos x.
n
48. (b) tan x x
œ" œ
œáÊ
xx xtanx
35 x
& (Î)
&
; from the Alternating Seriesœ á
"
35
x
œ
Estimation Theorem, 0
xtanx
x3
œ"
Î)
&
&†
0 !&&Ñ!†
xtanx x xtanx
x35 3x
œ" "œ
ŸÚ œ ŸÚ
&&
Š‹
; therefore, the lim œ
"œ"
35 x 3
xxtanx
œŸÚ
&
x0Ä

49. (a) e cos ( ) i sin ( ) 1 i(0) 1
œi1
œ œœ11
(b) e cos i sin (1 i)
i4
44 22 2
i1 11Î ""
œ œœ ˆ‰ ˆ‰ Š‹
ÈÈ È
(c) e cos i sin 0 i( 1) i
œŸ
##
i21 11
œ œœˆ‰ ˆ‰
50. e cos i sin e e cos ( ) i sin ( ) cos i sin ;
iii())))
œ Êœœœ)) ) )))
œœ
e e cos i sin cos i sin 2 cos cos ;
ii ee))
œ œ Ê œ
œ _
#
)))) ) )
ii&&
e e cos i sin (cos i sin ) 2i sin sin
ii ee
i
))
œ œ Ê œ
œ œ
#
)))) ) )
ii&&
51. e 1 x e 1 i and
xi xxx
!3!4! 2! 3! 4!
(i ) (i ) (i )
œáÊ œ ''''(
œ &'
#
) )))
)
e1i 1i
œ œœœ
#
i ( i) ( i) ( i) (i) (i) (i)
2! 3! 4! ! 3! 4!
) ))) )))
œ áœá))
œ
ʜ
ee
ii&&
_
##
Š‹Š‹1i 1iáá))
(i ) (i ) (i ) (i ) (i ) (i )
!3!4! !3!4!œœœ œœœ
& )* &)*
&&
1 cos ;œáœ
)))
œ
#!4!6!
)

ee
ii
ii&&
œ
##

œ
Š‹Š‹1i 1iáá))
(i ) (i ) (i ) (i ) (i ) (i )
!3!4! !3!4!œœœ œœœ
& )* &)*
&& sin œáœ))
)))
& (œ
3! 5! 7!

752. 748 Chapter 11 Infinite Sequences and Series
52. e cos i sin e e cos ( ) i sin ( ) cos i sin
iii)))
œ Êœœœ)) ) )))
œ œ
(a) e e (cos i sin ) (cos i sin ) 2 cos cos cosh i
ii ee))
œ œ Ê œ œ
œ _
#
)) )) ) ) )
ii&&
(b) e e (cos i sin ) (cos i sin ) 2i sin i sin sinh i
ii ee
2
))
œ œ Ê œ œ
œ œ
)) )) ) ) )
ii&&
53. e sin x 1 x x
x xxx xxx
!3!4! 3!5!7!
œ á ኋŠ‹
œ &' &(œ
#
(1)x (1)x x x x x x x x ;œ ᜠá
#$% &#$&"" "" " " " " "
####ˆ‰ˆ‰ˆ ‰
6661014 330
e e e e (cos x i sin x) e cos x i e sin x e sin x is the series of the imaginary part
xix 1ix x x x x
†œœ œ Ê
 _
ab
of e which we calculate next; e 1 (x ix)
 _  _ _ ___
#
1ix 1ix (x ix) (x ix) (x ix) (x ix)
n! !3!4!
œ œá!
_
œn0
n œ &'
1 x ix 2ix 2ix 2x 4x 4x 4ix 8ix the imaginary partœ áÊ
"" " " "
#
#$$%&& '
!3! 4! 5! 6!
ab a b a b a b a b
of e is xxxxx xxxxx in agreement with our
ÐÑ #$&' #$ & '
#
"" "1ix 2248
!3!5!6! 33090
áœá
product calculation. The series for e sin x converges for all values of x.
x
54. e e (cos bx i sin bx) ae (cos bx i sin bx) e ( b sin bx bi cos bx)
dd
dx dx
aib ax ax ax
ˆ‰ cd
Ð Ñ
œœ
ae (cos bx i sin bx) bie (cos bx i sin bx) ae ibe (a ib)eœ œœ
ax ax aibx aibx aibx Ð Ñ Ð Ñ Ð Ñ
55. (a) e e (cos i sin )(cos i sin ) (cos cos sin sin ) i(sin cos sin cos )
ii))Ú œ
œ œ ) ) ) ) )) )) )) ))" " # # "# "# "# #"
cos( ) i sin( ) eœ œ)) )) "# "#
 _i))
ڜ
(b) e cos( ) i sin( ) cos i sin (cos i sin )
œ _""
__
i cos i sin
cos i sin cos i sin e
) ))
)) ))
œ œ œ œ œ) ))))) ˆ‰
i&
56. e C iC e (cos bx i sin bx) C iC
abi abi
ab ab
abix axœœ
__
Ð Ñ
"# "#
œœ œœ œ ˆ‰
(a cos bx ia sin bx ib cos bx b sin bx) C iCœ
e
ab
ax
œœ
_
"#
[(a cos bx b sin bx) (a sin bx b cos bx)i] C iCœ
e
ab
ax
œœ
_
"#
CiC;œ
e (a cos bx b sin bx) ie (a sin bx b cos bx)
ab ab
ax ax
_œ
__
"#
œœ œœ
e e e e (cos bx i sin bx) e cos bx ie sin bx, so that given
Ð Ñabix axibx ax ax ax
œœ œ
e dx e C iC we conclude that e cos bx dx C''Ð Ñ Ð Ñ œ
_ _
"# "
_abix abix axabi
ab ab
e (a cos bx b sin bx)
œ œ
œœ œœ
ax
and e sin bx dx C'ax e (a sin bx b cos bx)
ab
œ
ax
œ
_
#
œœ
57-62. Example CAS commands:
:Maple f := x -> 1/sqrt(1+x);
x0 := -3/4;
x1 := 3/4;
# Step 1:
plot( f(x), x=x0..x1, title="Step 1: #57 (Section 11.9)" );
# Step 2:
P1 := unapply( TaylorApproximation(f(x), x = 0, order=1), x );
P2 := unapply( TaylorApproximation(f(x), x = 0, order=2), x );
P3 := unapply( TaylorApproximation(f(x), x = 0, order=3), x );
# Step 3:
D2f := D(D(f));
D3f := D(D(D(f)));
D4f := D(D(D(D(f))));
plot( [D2f(x),D3f(x),D4f(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 3: #57 (Section 11.9)" );
c1 := x0;
M1 := abs( D2f(c1) );

753. Section 11.9 Convergence of Taylor Series; Error Estimates 749
c2 := x0;
M2 := abs( D3f(c2) );
c3 := x0;
M3 := abs( D4f(c3) );
# Step 4:
R1 := unapply( abs(M1/2!*(x-0)^2), x );
R2 := unapply( abs(M2/3!*(x-0)^3), x );
R3 := unapply( abs(M3/4!*(x-0)^4), x );
plot( [R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 4: #57 (Section 11.9)" );
# Step 5:
E1 := unapply( abs(f(x)-P1(x)), x );
E2 := unapply( abs(f(x)-P2(x)), x );
E3 := unapply( abs(f(x)-P3(x)), x );
plot( [E1(x),E2(x),E3(x),R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green],
linestyle=[1,1,1,3,3,3], title="Step 5: #57 (Section 11.9)" );
# Step 6:
TaylorApproximation( f(x), view=[x0..x1,DEFAULT], x=0, output=animation, order=1..3 );
L1 := fsolve( abs(f(x)-P1(x))=0.01, x=x0/2 ); # (a)
R1 := fsolve( abs(f(x)-P1(x))=0.01, x=x1/2 );
L2 := fsolve( abs(f(x)-P2(x))=0.01, x=x0/2 );
R2 := fsolve( abs(f(x)-P2(x))=0.01, x=x1/2 );
L3 := fsolve( abs(f(x)-P3(x))=0.01, x=x0/2 );
R3 := fsolve( abs(f(x)-P3(x))=0.01, x=x1/2 );
plot( [E1(x),E2(x),E3(x),0.01], x=min(L1,L2,L3)..max(R1,R2,R3), thickness=[0,2,4,0], linestyle=[0,0,0,2],
color=[red,blue,green,black], view=[DEFAULT,0..0.01], title="#57(a) (Section 11.9)" );
abs(`f(x)`-`P`[1](x) ) <= evalf( E1(x0) ); # (b)
abs(`f(x)`-`P`[2](x) ) <= evalf( E2(x0) );
abs(`f(x)`-`P`[3](x) ) <= evalf( E3(x0) );
: (assigned function and values for a, b, c, and n may vary)Mathematica
Clear[x, f, c]
f[x_]= (1 x)
3/2
{a, b}= { 1/2, 2};
pf=Plot[ f[x], {x, a, b}];
poly1[x_]=Series[f[x], {x,0,1}]//Normal
poly2[x_]=Series[f[x], {x,0,2}]//Normal
poly3[x_]=Series[f[x], {x,0,3}]//Normal
Plot[{f[x], poly1[x], poly2[x], poly3[x]}, {x, a, b},
PlotStyle {RGBColor[1,0,0], RGBColor[0,1,0], RGBColor[0,0,1], RGBColor[0,.5,.5]}];Ä
The above defines the approximations. The following analyzes the derivatives to determine their maximum values.
f''[c]
Plot[f''[x], {x, a, b}];
f'''[c]
Plot[f'''[x], {x, a, b}];
f''''[c]
Plot[f''''[x], {x, a, b}];
Noting the upper bound for each of the above derivatives occurs at x = a, the upper bounds m1, m2, and m3 can be defined
and bounds for remainders viewed as functions of x.
m1=f''[a]
m2=-f'''[a]

754. 750 Chapter 11 Infinite Sequences and Series
m3=f''''[a]
r1[x_]=m1 x /2!
2
Plot[r1[x], {x, a, b}];
r2[x_]=m2 x /3!
3
Plot[r2[x], {x, a, b}];
r3[x_]=m3 x /4!
4
Plot[r3[x], {x, a, b}];
A three dimensional look at the error functions, allowing both c and x to vary can also be viewed. Recall that c must be a
value between 0 and x, so some points on the surfaces where c is not in that interval are meaningless.
Plot3D[f''[c] x /2!, {x, a, b}, {c, a, b}, PlotRange All]
2
Ä
Plot3D[f'''[c] x /3!, {x, a, b}, {c, a, b}, PlotRange All]
3
Ä
Plot3D[f''''[c] x /4!, {x, a, b}, {c, a, b}, PlotRange All]
4
Ä
11.10 APPLICATIONS OF POWER SERIES
1. (1 x) 1 x 1 x x x œ ᜠá
"Î# #$""""
## #
ˆ‰ˆ ‰ ˆ‰ˆ ‰ˆ ‰
œœ œœ
&& &&&
&)
œœœxx
!3! 816
3
2. (1 x) 1 x 1 x x x œ ᜠá
"Î$ #$"""
#3 ! 3! 3 9 81
5
ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰
œœ &)
33 333
225
œœœxx
3. (1 x) 1 ( x) 1 x x x œ á œ á
œ"Ÿ# #$" "
## #
ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰œœœ œœ œœ
œœ
&& &&&
&) 335
(x) (x)
!3! 816
35
4. (1 2x) 1 ( 2x) 1 x x x œ áœá
"Î# #$"
##
œœœ
ˆ‰ˆ ‰
œœ
&&
&
œœ
&& &
œœ( 2x)
!3! 22
( 2x)
11
Š‹Š ‹Š ‹
3 &
5. 1 1 1 x x xˆ‰ ˆ‰ œ # á œ
xx 3
!3! 4### #
œ#
#$"
(2)(3) (2)(3)(4)œœ œœœˆ‰ ˆ‰
xx
&&
& )
6. 1 1 1 x x xˆ‰ ˆ‰ œ# áœá
xx 3
!3! 4### #
œ#
#$"
(2)(3) (2)(3)(4)œœ œ œœœ œˆ‰ ˆ‰
xx
&&
& )
7. 1 x 1 x 1 x x xab œ ᜠá
$$ $' *
œ"Ÿ# " "
## #
ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰ab abœœ œœœ
œœ
&& &&&
))
&) 335
xx
!3! 816
35
8. 1 x 1 x 1 x x xab œ ᜠá
## #% '
œ"Ÿ$ " "
#3 ! 3! 3 9 81
214
ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰ab abœœ œœ œ
œœ &&
&)
33 333
447
xx
9. 1 1 1ˆ‰ ˆ‰œ áœ
11 1
x x ! 3! x 8x 16x
"Î#
"""
## #
ˆ‰ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰ˆ‰
œœ œœ
&& &&&
&)
œœœ
131
xx
œ &
10. 1 1 1ˆ‰ ˆ‰ œ ᜠá
22 24 40
x 3 x ! 3! 3x 9x 81x
"Î$
"
#
ˆ‰ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰ˆ‰
œœ
&)
33x 333x
22 252
œœ œœœ
œ &
11. (1 x) 1 4x 1 4x 6x 4x xœ œ
% #$%
#
(4)(3)x (4)(3)(2)x (4)(3)(2)x
!3! 4!

12. 1 x 1 3x 1 3x 3x xab œ œ
## #%'
$
#
(3)(2) x (3)(2)(1) x
!3!
ab ab
œœ
œ &
13. (1 2x) 1 3( 2x) 1 6x 12x 8xœ œ
$ #$œœ
#
(3)(2)( 2x) (3)(2)(1)( 2x)
!3!

14. 1 1 4 1 2x x x xˆ‰ ˆ‰œ œ
xx 3
!3! 4! 216### #
%
#$ %""
(4)(3) (4)(3)(2) (4)(3)(2)(1)ˆ‰ ˆ‰ ˆ‰œœ œ
xx x
&& &
& )*

755. Section 11.10 Applications of Power Series 751
15. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na xÊœ á á
dy
dx
n
n1
"#
œ
y (a a ) (2a a )x (3a a )x (na a )xÊœ á áœ!
dy
dx
nn1
n1
"Ê #" $# œ
#œ
a a 0, 2a a 0, 3a a 0 and in general na a 0. Since y 1 when x 0 we haveÊœœœ œ œ œ"Ê #" $# œ nn1
a 1. Therefore a 1, a , a , , a!"#$
œœ ""
##
œ œœœœœœœáœœ
aa
21 3 3 n n!
n
a (1)Ú œ
††
n1
n
y1x x x x eÊœ á ᜠœ
""
#
#$ œ œœ"
3! n! n!
(1) ( )xnx
nnn
!
_
œn0
16. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n a 2a x na xÊœ á á
dy
dx
n
n1
"#
œ
2y (a 2a ) (2a 2a )x (3a 2a )x (na 2a )xÊœ á áœ!
dy
dx
nn1
n1
"Ê #" $# œ
#œ
a 2a 0, 2a 2a 0, 3a 2a 0 and in general na 2a 0. Since y 1 when x 0 we haveÊœœœ œ œ œ"Ê #" $# œ nn1
a 1. Therefore a 2a 2(1) 2, a a (2) , a a , ,!"!#"$#
## # # #œœœœœœœœœœá
22 2 2222
33 3
œœ &
Š‹
†
a a a y 1 2x x x xnn1 n2
222 2 22 2
n n n 1 n! 3! n!
nœ œ œÊœáሉ ˆ‰ Š‹ œœ
œ#
#$
n1 n n
1 (2x) eœáᜠœ
(2x) (2x) (2x) (2x)
2! 3! n! n!
2x
nn
!
_
œn0
17. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n a 2a x na xÊœ á á
dy
dx
n
n1
"#
œ
y (a a ) (2a a )x (3a a )x (na a )x 1Êœ á áœ
dy
dx
nn1
n1
"Ê #" $# œ
#œ
a a 1, 2a a 0, 3a a 0 and in general na a 0. Since y 0 when x 0 we haveÊœœœ œ œ œ"Ê #" $# œ nn1
a 0. Therefore a 1, a , a , a , , a!"#$%
## # #
"" " "œ œ œœ œœ œœ á œ œ
aa a
33 443 n n!
n
aÚ
†††
n1
y 0 1x x x x xÊœ á á
"" " "
## #
#$ %
343 n!
n
†††
11x x x x x 1 1e 1œ á ᜠœ ˆ‰ !
"" " "
##
#$ %
3 4 3 2 n! n!
nx x
†††
_
œn0
n
18. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na xÊœ á á
dy
dx
n
n1
"#
œ
y (a a ) (2a a )x (3a a )x (na a )x 1Êœ á áœ
dy
dx
nn1
n1
"Ê #" $# œ
#œ
a a 1, 2a a 0, 3a a 0 and in general na a 0. Since y 2 when x 0 we haveÊœœœ œ œ œ"Ê #" $# œ nn1
a 2. Therefore a 1 a 1, a , a , , a!"!#$
œœ
## #
"" œ œ"œœœœœœœáœœ
aa
133 nn!
n
a ()Ú œ
††
n1
n
y2xx x x 11xx x xÊ œ á ᜠá á
"" ""
## ##
#$ #$ œ" œ"
3n! 3n!
() ()nn
††
n n
Š‹
1 1eœ œ!
_
œn0
(1)x
n!
xœ œ
nn
19. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n a 2a x na xÊœ á á
dy
dx
n
n1
"#
œ
y (a a ) (2a a )x (3a a )x (na a )x xÊœ á áœ
dy
dx
nn1
n1
"Ê #" $# œ
#œ
a a 0, 2a a 1, 3a a 0 and in general na a 0. Since y 0 when x 0 we haveÊœœœ œ œ œ"Ê #" $# œ nn1
a 0. Therefore a 0, a , a , a , , a!"#$%
_
## # #
"" " "œœœœœœœœáœœ
1a a a
33 443 n n!
n
a Ú
†††
n1
y 0 0x x x x xÊœ á á
"" " "
## #
#$ %
343 n!
n
†††
11x x x x x 1x 1xe x1œ á ᜠœ ˆ‰ !
"" " "
## #
#$ %
343 n! n!
nx x
†††
_
œn0
n
20. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na xÊœ á á
dy
dx
n
n1
"#
œ

756. 752 Chapter 11 Infinite Sequences and Series
y (a a ) (2a a )x (3a a )x (na a )x 2xÊœ á áœ
dy
dx
nn1
n1
"Ê #" $# œ
#œ
a a 0, 2a a 2, 3a a 0 and in general na a 0. Since y 1 when x 0 we haveÊœœœ œ œ œ"Ê #" $# œ nn1
a 1. Therefore a 1, a , a , , a!"#$
œœ
## #
"" œ œ"œ œ œ œ œ œ á œ œ
2a a
33 nn!
n
a ()Ú œ
†
n1
n
y 1 1x x x xÊœ á á
""
##
#$ œ"
3n!
() n
†
n
11x x x x 22x 22xe 2x2œ á ᜠœ Š‹ !
""
##
#$ œœ" œ
3n! n!
() (1)xnx
†
nnn
_
œn0
21. y xy a (2a a )x (3a a )x (na a )x 0 a 0, 2a a 0, 3a a 0,
w œ
"#Ê$" œ "#Ê$"
œá áœÊœ œ œ nn2
n1
4a a 0 and in general na a 0. Since y 1 when x 0, we have a 1. Therefore a ,%# œ Ê #
##
"œ œ œ œ œ œœ nn2
a

a 0, a , a 0, , a and a 0$% &
""
##✜ œœ œœá œ œ
aa a
3445 462n
2n 2n 1
†††
y 1 x x x x eÊœ á ᜠœ œ
"" " "
## # #â
#% ' Î
††† ††4 4 6 4 6 2n 2 n! n!
2n x 2 x !!
__
œœn0 n0
2n
n
x
n Š‹
œ
œ œ
22. y x y a 2a x (3a a )x (4a a )x (na a )x 0 a 0, a 0,
¸# # $ œ
" # $Ê %" œ " #
œá áœÊœœ nn3
n1
3a a 0, 4a a 0 and in general na a 0. Since y 1 when x 0, we have a 1. Therefore$Ê %" œ Êœ œ œ œ œ œ nn3
a , a 0, a 0, a , , a , a 0 and a 0$ %&'
"""
✜ œœ œœ œœ á œ œ œ
a aaa
3 3 4 5 6 36 369 3n
3n 3n 1 3n 2
†††
y 1 x x x x eÊœ á ᜠœ œ
"" " "$' * Î
â3 36 369 369 3n 3 n! n!
3n x 3 x
††† †† !!
__
œœn0 n0
3n
n
x
3
n Š‹
&
&
23. (1 x)y y (a a ) (2a a a )x (3a 2a a )x (4a 3a a )xœ á
w #$
"! #"" $ ## % $$
(na (n 1)a a )x 0 a a 0, 2a 2a 0, 3a 3a 0 and in áœÊœ œ œnn1n1
n1
œ œ "Ê #" $#
œ
general (na na ) 0. Since y 2 when x 0, we have a 2. Thereforenn1œ œ œ œœÊ
a 2, a 2, , a 2 y 2 2x 2x 2x"#
#
œœœáœÊœáœ œn
n 2
1x
!
_
œn0
24. 1 x y 2xy a (2a 2a )x (3a 2a a )x (4a 2a 2a )x (na na )xabœ á á
#¸ # $ œ
" #Ê $"" %## œ nn2
n1
0 a 0, 2a 2a 0, 3a 3a 0, 4a 4a 0 and in general na na 0. Since y 3 whenœÊœœœœ œ œ"#Ê$"%# œ nn2
x 0, we have a 3. Therefore a 3, a 0, a 3, , a 0, a ( 1) 3œœ œœœáœœ Ê#$%_ 2n 1 2n
n y 3 3x 3x 3( 1) x 3 xʜᜠœ œ
#% #
_ !! ab
__
œœn0 n0
n2n
n 3
1x
œ
25. y a a x a x a x y 2a 3 2a x n(n 1)a x y yœ á á Ê œ á á Ê !" # # $
#¸¸ œ¸¸
nn
nn2 †
(2a a ) (3 2a a )x (4 3a a )x (n(n 1)a a )x 0 2a a 0,œá áœÊœ
#Ê $" %# œ #Ê
#œ †† nn2
n2
3 2a a 0, 4 3a a 0 and in general n(n 1)a a 0. Since y 1 and y 0 when x 0,†† $" %# œ
wœ œ œ œ œ œ nn2
we have a 0 and a 1. Therefore a 0, a , a 0, a , , a and! " #$ %&
"" "
###_œ œ œœ œœ á œ
3 543 ( n 1)!
2n 1
† †††
a 0 y x x x sinh x2n
3! 5! (2n 1)!
xœÊœáœ œ
""$&
_ !
_
œn0
2n 1Û
26. y a a x a x a x y 2a 3 2a x n(n 1)a x y yœ á á Ê œ á á Ê !" # # $
#¸¸ œ¸¸
nn
nn2 †
(2a a ) (3 2a a )x (4 3a a )x (n(n 1)a a )x 0 2a a 0,œá áœÊœ
#Ê $" %# œ #Ê
#œ †† nn2
n2
3 2a a 0, 4 3a a 0 and in general n(n 1)a a 0. Since y 0 and y 1 when x 0,†† $" %# œ
wœ œ œ œ œ œ nn2 we have a 1 and a 0. Therefore a , a 0, a , a 0, , a 0 and a!" #$%&
""
## #
œ"œœ œœœœáœ œ
43 ( n)!
2n 1 2n
()
††
n
y1 x x cos xÊœ ᜠœ
""#% œ
2 4! (2n)!
(1)x !
_
œn0
n2n
27. y a a x a x a x y 2a 3 2a x n(n 1)a x y yœ á á Ê œ á á Ê !" # # $
#¸¸ œ¸¸
nn
nn2 †
(2a a ) (3 2a a )x (4 3a a )x (n(n 1)a a )x x 2a a 0,œá áœÊœ
#Ê $" %# œ #Ê
#œ †† nn2
n2
3 2a a 1, 4 3a a 0 and in general n(n 1)a a 0. Since y 1 and y 2 when x 0,†† $" %# œ
wœ œ œ œ œ œ nn2

757. Section 11.10 Applications of Power Series 753
we have a 2 and a 1. Therefore a , a 0, a , a 0, , a 2 and!" #$%&
" œ
#œœ œ"œœœáœ
43 ( n)!
2n
(1)
†
†
n1*
a 0 y2xx 2 2x2 xcos 2x2n 1
x
4! (2n)!
(1) x
_
# œœ Ê œ ᜠœ †
' *
!
_
œn1
n12n
28. y a a x a x a x y 2a 3 2a x n(n 1)a x y yœ á á Ê œ á á Ê !" # # $
#¸¸ œ¸¸
nn
nn2 †
(2a a ) (3 2a a )x (4 3a a )x (n(n 1)a a )x x 2a a 0,œá áœÊœ
#Ê $" %# œ #Ê
#œ †† nn2
n2
3 2a a 1, 4 3a a 0 and in general n(n 1)a a 0. Since y 2 and y 1 when x 0,†† $" %# œ
wœ œ œ œ œ œ nn2
we have a 1 and a 2. Therefore a , a , a , a , , a!" #$%&
œ" " œ" " œ
## # #œ œ œ œ œ œ œ á œ
234 54 5! ( n)!
31
2n
†† ††
and a y 1 2x x x 1 2x 2n 1
33x3x
(2n 1)! 3! (2n)! (2n 1)!
_
_# _
"#$œÊœáœ !!
__
œœn1 n1
2n 2n 1 *
29. y a a x 2 a x 2 a x 2œ á á!" #
# abab ab n
n
y 2a 32ax2 n(n1)ax2 y yÊœ á áÊ
ww ww
#$
œ
†ab ab n
n2
(2a a ) (3 2a a )(x 2) (4 3a a )(x 2) (n(n 1)a a )(x 2) xœ á á œ#Ê $" %# œ
#œ †† nn2
n2
x 2 2 2a a 2, 3 2a a 1, and n(n 1)a a 0 for n 3. Since y 0 when x 2,œ Ê œ œ œ œ œab #Ê $" œ † nn2
we have a 0, and since y 2 when x 2, we have a 2. Therefore a 1, a , a 1 ,!"#$%
w "œ
# † †††œ œ œ œ œ œ œ œ
12
43 4321
ab
a , . . . , a , and a . Since a 2, we have a x 2 2 x 2 and&""
† # ††††
"œ œ œœœ œ œ œ œ
13 2 3
54 54321 2n ! (2n 1)!
2n 2n 1
ˆ‰ ababab
ab
2x2 31x2 3x2 1x2 x23x2.ababa bababababab ab œ œ œ
yx23x2 x2 x2 x2 x2 . . .ʜ abab ab ab ab
2323
2! 3! 4! 5!
2345
y x 2 x 2 x 2 . . . 3 x 2 x x 2 x 2 . . .ʜ
22 33
2! 4! 3! 5!
24 35
ab ab ab ab ab
y x2 3 ʜ !!
__
œœn0 n0
(x 2) (x 2)
(2n)! (2n 1)!
œœ
_
2n 2n 1 *
30. y x y 2a 6a x (4 3a a )x (n(n 1)a a )x 0 2a 0, 6a 0,
¸¸ # # œ
#$ %Ê œ # $
œ á áœÊ œ œ† nn4
n2
4 3a a 0, 5 4a a 0, and in general n(n 1)a a 0. Since y b and y a when x 0,†† %! &"
wœ œ œ œ œ œ nn4
we have a a, a b, a 0, a 0, a , a , a 0, a 0, a , a!"#$% & '() *œœœœœ œ œœœ œ
ab a b
34 45 3478 4589† † ††† †††
y a bx x x x xÊœá
ab a b
34 45 3478 4589† † ††† †††
%&)*
31. y x y 2a 6a x (4 3a a )x (n(n 1)a a )x x 2a 0, 6a 1,
¸¸ # # œ
#$ %Ê œ # $
œ á áœÊ œ œ† nn4
n2
4 3a a 0, 5 4a a 0, and in general n(n 1)a a 0. Since y b and y a when x 0,†† %! &"
wœ œ œ œ œ œ nn4
we have a a and a b. Therefore a 0, a , a , a , a 0, a!" #$%&'(
"œ"
##œ œ œ œ œ œ œ œ
††† †††33445 367
ab
y a bx x x x xÊœ á
1ab 1 axbx
23 34 45 2367 3478 4589† † † ††† ††† †††
$%& (

32. y 2y y (2a 2a a (2 3a 4a a )x (3 4a 2 3a a )x
ww w #
#"! $#" % $#
œ Ñ á †††
((n 1)na 2(n 1)a a )x 0 2a 2a a 0, 2 3a 4a a 0, áœÊ œ œ
nnn2
n2
œ" œ # " Ê $ # "
œ †
3 4a 2 3a a 0 and in general (n 1)na 2(n 1)a a 0. Since y 1 and y 0 when††
%$# œ"œ
wœ œ œ œ nnn2
when x 0, we have a 0 and a 1. Therefore a 1, a , a , a and aœ œ œ œœœœ œ ! " #$%&
""
#œ
11
624 (n1)!
n
y x x x x x x xeÊœ ᜠœ œ œ
#$% &"" "
## œ 64 (n1)! n! n!
xx x x !!!
__ _
œœ œn1 n0 n0
nn1 n *
33. sin x dx x dx 0.00267 with error''
00
02 02

##
!Þ# !Þ#
!!
œáœá¸¸Š‹’“’“
xx x x x
3! 5! 3 7 3! 3
_ )Ð & œ &
†
E 0.0000003kkŸ¸
(.2)
73!
œ
†
34. dx 1 x 1 dx 1 dx'' '
00 0
02 02 02

e xxx xxx
x x !3!4! 6 24
x
œ" "
##
œ á œ ኋŠ‹
x 0.19044 with error E 0.00002œ á ¸ Ÿ ¸’“ kk
xx
418 96
(0.2)
œ & '
!Þ#
!

758. 754 Chapter 11 Infinite Sequences and Series
35. dx 1 dx x [x] 0.1 with error''
00
01 01

"
_
!Þ"
!
!Þ"
!È
1 x
x3x x
28 10
'
' _(
œáœá¸¸Š‹’“
E 0.000001kkŸœ
(0.1)
10
(
36. 1 x dx 1 dx x x 0.25174 with error''
00
025 025

$ #
!Þ#& !Þ#&
!!È
Š‹’“’“ œ á œ á ¸ ¸xx xx x
39 945 9
œ '&( &
E 0.0000217kkŸ¸
(0.25)
45
(
37. dx 1 dx x x''
00
01 01

sin x xxx xxx xx
x 3! 5! 7! 3 3! 5 5! 7 7! 3 3! 5 5!
œ á œ &'&'( w&'Š‹’ “’“
œ
††† ††
!Þ" !Þ"
!!
0.0999444611, E 2.8 10¸Ÿ¸‚ kk
(0.1)
77!
12
7
†
œ
38. exp x dx 1 x dx x x''
00
01 01

ab Š‹’“’“ œ á œ á ¸
##
!Þ" !Þ"
!!
xxx xx x xx x
2! 3! 4! 3 10 42 3 10 42
'
0.0996676643, E 4.6 10¸Ÿ¸‚ kk
(0.1)
216
12
9
œ
39. 1 x (1) (1) x (1) x (1) xa b ab ab abœ
% "Î# "Î# % $Î# % &Î# %
"Î# # $
#
Š‹
œ
&
œœ œœ
&& &&&
1! 3!
ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰œœœ
3
(1) x 1 ᜠá
ˆ‰ˆ‰ˆ‰ˆ‰
œœ
&&&&
œœœ
35
4! 8 16 128
xxx 5xœáŸ# %
%
#
ab
' "Úœ Ú_
1 dx x 0.100001, E 1.39 10Ê á ¸ ¸ Ÿ ¸ ‚'
0
01

Š‹’“ kk
xxx 5x x
8 16 128 10 72
(0.1) 11
' "Úœ Ú_ á
#
!Þ"
!
œ
9
40. dx dx''
00
11
ˆ‰ Š‹’“
1cos x xxx x xx x x x
x 4! 6! 8! 10! 3 4! 5 6! 7 8! 9 10!
œ"
##
"
!
œ
œ œ á ¸ &'&'
††††
0.4863853764, E 1.9 10¸Ÿ¸‚ kk
1
11 12!
10
†
œ
41. cos t dt 1 dt t error .00011''
00
11
#
#
"
!
"
œ á œ á Ê ¸Š‹’ “ kk
ttt t t t
4! 6! 10 9 4! 13 6! 13 6!
'
†† †
42. cos t dt 1 dt t''
00
11
È Š‹’ “œ á œ '&'&(
tttt t t t t
4! 6! 8! 4 3 4! 4 6! 5 8!#
"
!
œ
†††
error 0.000004960!†wkk
"
58!†
43. F(x) t dt œ á œ á ¸ '
0
x x
Š‹’ “
#
!
tt t t t t t x x x
3! 5! 7! 3 7 3! 11 5! 15 7! 3 7 3! 11 5!
_ )Ð )' & œ )) )( & œ ))
††† ††
error 0.000013!†wkk
"
15 7!†
44. F(x) t t dtœ á œ á'
0
x x
Š‹’ “
#%
!
ttt t tt t t t t
2! 3! 4! 5! 3 5 7 2! 9 3! 11 4! 13 5!
_
††††
error 0.00064w&'&' ! † w
xx x x x
3 5 7 2! 9 3! 11 4! 13 5!
& (œ Ñ ))
†† † †
kk
"
45. (a) F(x) t dt error .00052œ á œ á ¸ Ê ¸'
0
x x
Š‹’“ kk
ttt t t t xx
357 21 30 1 30
(0.5)
& _
###
!
(b) error .00089 when F(x) ( 1) kk† w w&'&'('&
"
#
"&
33 34 3 4 5 6 7 8 31 32
xxxx x
† ††† †
œ
46. (a) F(x) 1 dt t xœ á œá¸ &'&' '
0
x x
Š‹’ “
ttt tttt xxxx
2 3 4 22334455 3 4 5
œ
œœœœ
††††
!
#
error .00043!†wkk
(0.5)
6
_
œ
(b) error .00097 when F(x) x ( 1) kk† w w&'&'('&
"
#
$"
32 3 4 31
xxx x
œ œœœ œ
œ &' &)

759. Section 11.10 Applications of Power Series 755
47. e (1 x) 1 x 1 x lim
"" "
##
œ_
x x 3! 3! 4! x
x xx xx e(1x)
œœ œ
ab Š‹Š‹ œ á œáÊ
x0Ä
x
lim œáœ
x0Ä
Š‹
""
##
xx
3! 4!
œ
48. e e 1 x 1 x 2x
"" "œ
##x x !3!4! !3!4! x 3! 5! 7!
xx xxx xxx 2x2x2x
ab ’“Š‹Š‹Š‹ œ á á œ ''''(
œ
2 lim lim 2 2œáÊ œ áœ
2x 2x 2x e e 2x 2x 2x
3! 5! 7! x 3! 5! 7!

x0 xÄ Ä_
xx
œ
Š‹
49. 1 cos t 1 1 lim
"" "
###
"œ œ
tt t
ttttt tt
4! 6! 4! 6! 8!
cos t
'' '
Š‹’ “ Š‹ œ á œáÊ
t0Ä
Š‹
t
&
&
lim œ áœ
t0Ä
Š‹
""
4! 6! 8! 24
tt
œ '
50. sin lim
"" "
œ_
)) )
) ) )) ))
))
(( (
Š‹Š ‹ œ áœáÊ))))
6 6 3! 5! 5! 7! 9!
sin
)Ä0
Š‹
œ
)
6
lim œ áœ
)Ä0
Š‹
""
#5! 7! 9! 1 0
))
œ '
51. y tan y y y lim lim
"" " " œ" œ
yy35357 y357
yy yy ytany yy
&& &
ab ’“ Š‹Š‹ œ á œáÊ œ á
y0 y0ÄÄ
œ
"
3
52.
tan y sin y
y cos y y cos y y cos y cos y
yy
)
&& &
œ
œ_œÜœœ_œÜœ_œÜœ_œÜ
œœœ
Œ œŒ œ Œ œ Œ œ
y y y y y 23y 23y
3 5 3! 5! 6 5! 6 5!
) !)! )! &
œ
lim lim Êœ œ
y0 y0ÄÄ
tan y sin y
y cos y cos y 6
)
&
œ
œ_ œÜ
"
Œ œ
œ
&
65!
23y
53. x 1 e x 1 1 1 lim x e 1
#œŸ# #œŸ"" " " "
# #
Š‹ Š‹ ˆ‰ œ áœáÊ
1x 1x
x6x xx6x
œ œ
œ _œ''
xÄ_
lim 1 1œ áœ
xÄ_
ˆ‰
""
#x 6x
œ '
54. (x 1) sin (x 1) 1 œ áœáˆ‰ Š‹
" """ ""
____ _ _x 1 x 1 3!(x 1) 5!(x 1) 3!(x 1) 5!(x 1)
'
lim (x 1) sin lim 1 1Ê œ áœ
xxÄ_ Ä_
ˆ‰ Š‹
"""
__ _x 1 3!(x 1) 5!(x 1)
œ '
55. lim lim
ln 1 x ln 1 x
1cos x 1cos x
x1 1
11
ab ab
Œ œ Œ œ Œ œ
Š‹ Š‹ Š‹
_ _
œ œ
œ_œÜ œ_œÜ œ_œÜ
œœ_œÜ œ_Ü
œ œ
œ
œœÊœ
xx xx xx
33 3
xx
!4! !4!
x

&& &
&*
&
œœ
&&
&
x0 x0ÄÄ
!4!
x
œ_Ü
& œœ2! 2
56. lim
x4 x2 x4
ln (x 1) ln (x 1)
(x 2)(x 2)
(x 2) 1
œ œ
œ_œ
œ œ
œ_
œœ_œÜ œ_œÜ
œœÊ
’“’“
(x 2) (x 2) (x 2)
33
x2(( (
&) &
&&
(
x2Ä
lim 4œœ
x2Ä
x2
1
_
œ_ œÜ’“
x2 (x 2)
3(
&
(
&
57. ln ln (1 x) ln (1 x) x x 2 xˆ‰ Š‹Š ‹Š‹
1x xxx xxx xx
1x 34 34 35
_
œ##
œ œ áᜠá
œ
58. ln (1 x) x error when x 0.1;œá áÊ œ œ œ
xxx
3 4 n n n10
(1) x ( ) x

#
œœ" "
n1n n1n
n
kk ¹¹
n10 10 when n 8 7 terms
"" )
n10 10
n
n†! Ñ Ð!

760. 756 Chapter 11 Infinite Sequences and Series
59. tan x x error when x 1;
œ" œ" œ
œœ#œ
"
œá áÊ œ œ œ
xxxx
3579 2n1 2n1 n1
()x (1)x
&(τ n12n1 n12n1
kk ¹¹
n 500.5 the first term not used is the 501 we must use 500 terms
""
#œ #n1 10
1001
Ê œ Ê Ê
&
st
60. tan x x and lim x lim x
œ" ##œ
œ_#_
œœ
œá á œ œ
xxxx x 2n1 2n1
3579 2n1 2n1x n1
(1) x
& áÝ Û
n12n1 2n 1
2n 1
nnÄ_ Ä_
¹¹ ¸¸†
tan x converges for x 1; when x 1 we have which is a convergent series; when x 1Êœ œ
œ" œ
œ
kk !
_
œn1
(1)
2n 1
n we have which is a convergent series the series representing tan x diverges for x 1! kk
_
œn1
(1)
2n 1
œ
œ
œ"
n1
Ê
61. tan x x and when the series representing 48 tan has an
œ" œ"œ
œ
"
œá á
xxxx
3579 2n1 18
(1) x
&(τ n12n1
ˆ‰
error less than 10 , then the series representing the sum
" '
3
†
48 tan 32 tan 20 tan also has an error of magnitude less than 10 ; thus
" " " '"" "
#ˆ‰ ˆ‰ ˆ ‰
18 57 39
'&
error 48 n 4 using a calculator 4 termskkœÊ  Ê
Š‹
œ
18
2n 1
_
# †
"
n1 310
62. ln (sec x) tan t dt t dtœ œ á ¸á''
00
xx
Š‹
t2t xx x
315 1245
&
#
63. (a) 1 x 1 sin x x ; Using the Ratio Test:ab&w'''! w'''
#œ"
œ"Ÿ#
#
x3x5x x3x5x
8 16 6 40 112

lim 1 x lim
nnÄ_ Ä_
¹¹¹¹
1 3 5 (2n 1)(2n 1)x 2 4 6 (2n)(2n )
2 4 6 (2n)(2n 2)(2n 3) 1 3 5 (2n 1)x
†† ††
†† ††
ßœ _ ß _"
ß__ ßœ
#
2n 3
2n 1Û
Û
† †!
(2n 1)(2n 1)
(2n 2)
''
'(2n 3)'
†1
x 1 the radius of convergence is 1. See Exercise 69.!†!kk
(b) cos x 1 x cos x sin x x x
d x 3x 5x x 3x 5x
dx 6 40 112 6 40 112
abab Š‹
œ" # œ" œ"
œ"Ÿ#
## #
œ Ê œ ¸ ¸
11 1
& (œ &(œ
64. (a) 1 t (1) (1) tab ab ˆ‰'w'& ' '
# œ"Ÿ# œ$Ÿ# #
œ"Ÿ# œ œ œ œ œ "
##
ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰ ab ab
ÚÚ
œœ œœœ
ŸáÊœ œ ŸÝÊœ œ
335
(1) t (1) t
!3!
1 sinh x 1 dt xœ Ê ¸ œ
t 3t 3 5t t 3t 5t x 3x 5x
2 2! 2 3! 8 16 6 40 112


##
œ"
††
†
'
0
x
Š‹
(b) sinh 0.24746908; the error is less than the absolute value of the first unused
œ""""ˆ‰
4 4 384 40,960
3
¸ œ
term, , evaluated at t since the series is alternating error 2.725 10
5x
112 4 112
5
Ý
Ú
Ý
œÊ¸‚
" '
kk
ˆ‰
4
65. 1 x x x 1 x x x
œ" " œ "
_œœ __
#$ #$
1x 1(x) dx 1x 1x dx
d1 d
œ œáÊ œ œ á ˆ‰ ab
œ
12x3x 4xχ
#$
66. 1 x x x 1 x x x 2x 4x 6x
""
œœ
#%' #%' $ &
œ1x dx 1x dx
d2xd
1x
œœ
œ
œœáÊ œ œ ᜠሉ ab
ab
67. Wallis' formula gives the approximation 4 to produce the table1¸’“
244668 (2n 2)(2n)
335577 (2n 1)(2n 1)
††††† †
††††† †
ߜ
ßœ œ
n µ1
10 3.221088998
20 3.181104886
30 3.167880758
80 3.151425420
90 3.150331383
93 3.150049112
94 3.149959030
95 3.149870848
100 3.149456425

761. Section 11.11 Fourier Series 757
At n 1929 we obtain the first approximation accurate to 3 decimals: 3.141999845. At n 30,000 we still doœ œ
not obtain accuracy to 4 decimals: 3.141617732, so the convergence to is very slow. Here is a CAS1 Maple
procedure to produce these approximations:
pie :=
proc(n)
local i,j;
a(2) := evalf(8/9);
for i from 3 to n do a(i) := evalf(2*(2*i 2)*i/(2*i 1)^2*a(i 1)) od;&&&
[[j,4*a(j)] $ (j = n 5 .. n)]&
end
68. ln 1 0; ln 2 ln 2 0.69314; ln 3 ln 2 ln ln 2 ln œ œ ¸ ¸ œ œ
1 1
1 1
3357
3
_ _
œ œ
"
#
Š‹ Š‹ Š‹ Š‹ Š‹
Š‹ Š‹
œ œœœ œ
œ œ
3 333 5
3 5

ˆ‰
& (œ
ln 2 2 1.09861; ln 4 2 ln 2 1.38628; ln 5 ln 4 ln ln 4 ln ¸¸œ¸œœ

ˆ‰
"
_
œ
5357 4
5
1
1
Š‹ Š‹ Š‹ Š‹
Š‹
œœœ œ
œ
555 9
9
& (œ
1.60943; ln 6 ln 2 ln 3 1.79175; ln 7 ln 6 ln ln 6 ln 1.94591; ln 8 3 ln 2¸ œ¸ œ œ ¸ œ ˆ‰
7
6
1
1
_
œ
Š‹
Š‹
œ
œ
13
13
2.07944; ln 9 2 ln 3 2.19722; ln 10 ln 2 ln 5 2.30258¸ œ ¸ œ¸
69. 1 x 1 x (1) (1) xab a b ab ab ˆ‰œ œ
# # œ"Ÿ# œ$Ÿ# #
œ"Ÿ# œ"Ÿ# "
##
ˆ‰ˆ‰ abœœ œ
œ
&&
(!Î& &
& 3
(1) x
!
11 ᜠáœ
ˆ‰ˆ‰ˆ‰ abœœœ œ
œ
&&&
(ÑÎ& &
) 35
(1) x
3! 2 ! 2 3! n!
x 13x 135x 1 3 5 (2n 1)x
œ

## #
ßœ†††
†† †
††
!
_
œn1
2n
n
sin x 1 t dt 1 dt x ,Êœ œ œ
œ" #
œ"Ÿ# ßœ ßœ
##ß_''
00
xx
ab Π'
!!
__
œœn1 n1
1 3 5 (2n 1)x 1 3 5 (2n 1)x
n! 4 (2n)(2n 1)
†† ††
††
2n 2n 1
n *
where x 1kk†
70. tan t tan x dt 1 dtcd
–—
ˆ‰
œ" œ"
_
#_
""""
x
xx x
œ œ œ œ á
1 '' '
'' '
dt
1t t t t t
œœœ _'
Š‹
Š‹
1
t
t
&
œ
&
1'
dt lim œ á œ ᜠ&'&'('
x
b
x
'
ˆ‰ ( ‘
"""" " " " " " " " "
t t t t 3t 5t x 3x 5xt7t7x
œ ' œœ
bÄ_
tan x , x 1; tan t tan x Ê œá œ œ
œ" œ" œ"
# #_
"" "1 1
x3x 5x 1t
dt
& ( œ cd
x
x
('
('
'
lim tan x ,œ áœáÊ œá
bĜ_
( ‘
"""" "" " " "" " œ"
#t 3t 5t x 3x 5x x 3x 5x7t 7x
& (&( &(œœ
x
b
1
x1†&
71. (a) tan tan (n 1) tan (n 1)ab
œ" œ" _œ œ
__ œ__œ
_œœ
œ œ œ
tan tan (n 1) tan tan (n 1)
1tantan(n1) tantan(n1) 1(n1)(n1) n
(n 1) (n 1) 2abab
abab
)Î)
ŸÚ ŸÚ œ
(b) tan tan (n 1) tan (n 1) tan 2 tan 0 tan 3 tan 1!! ˆ‰ cd abab
NN
n1 n1œœ
œ" œ" œ" œ" œ" œ" œ"2
n
œœ œ
tan 4 tan 2 tan (N 1) tan (N 1) tan (N 1) tan N á œ aba b
œ" œ" œ" œ" œ" œ" 1
4
(c) tan lim tan (N 1) tan N
n
! ˆ‰ ( ‘
_
œn1
œ" œ" œ"
##
2 3
n444
œœ œœ
Ä_
1 111 1
11.11 FOURIER SERIES
1. a 1 dx 1, a cos kx dx 0, b sin kx dx 0.0k k
111 11
2
sin kx cos kx
kk
22
00œœœ œ œœ œœ
111 11
11
'' '
00 0
22 2
œœ œ
( ‘ ( ‘
Thus, the Fourier series for f x is 1.ab

762. 758 Chapter 11 Infinite Sequences and Series

2. a 1 dx 1 dx 0, a cos kx dx cos kx dx 0,0k
111
2
sin kx sin kx
kk
0
2œœœ œ œ &
11 1
11
1”•” •”• ¹¹
'' ' '
00
22
œœ œ œ
œœ
b sin kx dx sin kx dx cos k 1 cos 2 k cos kk
111 cos kx cos kx
kk
0
2
kœœ œ ' &'' &
111
11
1”•”• ¹¹ cdaba b
''
0
2
œœ
œ
111
22 cos k .
, k odd
0, k even
œ œ
1
k
4
k
1
1
ab œ1
Thus, the Fourier series for f x is .sin x . . . ab ( ‘
4 sin 3x sin 5x
351
'''

3. a x dx x 2 dx 4 2 0. Note,0
11
22
2222œœœ
11”• ab a b ( ‘
''
0
2
œœ
œ
11111
""
##
x 2 cos kx dx u cos ku du (Let u 2 x). So a x cos kx dx x 2 cos kx dx 0.'' ''
œ œ
œœ œœ
2 2
00
ab ab”•œ œœ œ11 1 k
1
1
Note, x 2 sin kx dx u sin ku du (Let u 2 x). So b x sin kx dx x 2 sin kx dx'' ''
œ œ
œœ œœ
2 2
00
ab ab ”•œ œœ 11 1 k
1
1
x sin kx dx cos k 1 . cos kx sin kxœœ œœ &'
22 22 x1
kk 0 kk
k1
11
1'
0
œ
( ‘ ab
2 1
_
Thus, the Fourier series for f x is 1 .ab a b!
k1
k12 sin kx
k
œ
_
_
&

4. a f x dx x dx , a f x cos kx dx x cos kx dx0k
1111 1
226
22 2œœœœ œ
11 1 1
'' ' '
00 0 0
22
œœ œ œ
ab ab 1
sin kx x cos kx cos k 1 , b f x sin kx dx x sin kx dxœ œ œ œ œ œ
1x 2 11
kk k k k
0
k
k
2
1 11
1
’“Š‹ ab abˆ‰
2
32 2 2
## #
1 ''
00
2
œœ
cos kx x sin kx 1 1 1 1œ œœ
12x 12 1
kk k kk k k k
0
kkk
111
1
11
’“’“’“Š‹ Š‹ Š ‹ ab ab ab32 33 3
22
###

763. Section 11.11 Fourier Series 759
.
, k odd
, k even
œ
&'
&
œ
4
kk
k
1
1
1
3
Thus, the Fourier series for f x is 2 cos x sin x cos 2x sin 2x cos 3x sin 3x . . .ab Š‹ Š ‹
14 294
62927
2
1&' '& &' '
111
11
2 2
œ" œ
#

5. a e dx e 1 , a e cos kx dx , cos kx k sin kx0k
11 1 1 e1
22 1k
x2 x e
1k 0
2œœœ œ œ '
11 1 1 1
1
1
''
00
22
œœ
ab ( ‘ab
x
2
2
2
_
œ
_
œ
ab
b e sin kx dx . sin kx k cos kxk
11 x e
1k 0
2 k1 e
1kœœ œ &
11 1
1
'
0
2
œ
( ‘ab
x
2
2
2
_
œ
_
ˆ‰
ab
œ
Thus, the Fourier series for f x is e 1 .ab a b !ˆ‰
1 e 1 cos kx k sin kx
21k1k
2
k1
11
1
&' &
2
22œ
œ
œ
_
__

6. a f x dx e dx , a f x cos kx dx e cos kx dx cos kx k sin kx0k
11e11 1 1
222
xx e
1k 0œœœœ œ œ '
1111 1 1
1
'' ' '
00 0 0
22
œœ œ œ
ab ab ( ‘ab
œ
œ
_
x
2
. b f x sin kx dx e sin kx dxe1 1
, k odd
, k even
œœ œœ &&
111
1k
k
1e
1k
e1
1k
k
x
111
1
1
1
ab
ab
ab
ab
_
œ_
_
œ
_
2
2
2
( ‘ab )
ab
œ
œ
''
00
2
œœ
.sin kx k cos kx e1 1
, k odd
, k even
œœœ & &&
1ke
1k 0 1k
k
k1 e
1k
1e
1k
11
1
1
1
1
( ‘ ( ‘ab ab )
x
2 2
2
2
_
œ
_
_
_
œ
_
ab
ab
ab
ab
œ
œ
Thus, the Fourier series for f x isab
cos x sin x cos 2x sin 2x cos 3x sin 3x . . .
e1 e1
22 2 5 5 10 10
1e 1e 21e 1e 31e
œœ œœ œ œ œ
œœ __ œ _ _
11 1 1 1 1 1
&''' & ' '
ab ab ab ab ab

7. a f x dx cos x dx 0, a cos x cos kx dx
,k1
0k
11 1
22
1sink 1x sink 1x
2k 1 2k 1
0œœ œœ œ
' Á
11 1
1
1
'' '
00 0
22 2
œœ œ
ab
Ú
Û
Ü
’“
ab ab
ab ab
œ_
œ_
x sin 2x , k 1
1
4 01
1
( ‘
""
#
œ

764. 760 Chapter 11 Infinite Sequences and Series
.
,k 1
,k 1
œ
!Á
œ
œ
"
#
b cos x sin kx dx
,k1
cos 2x , k 1
k
1
1cos k 1 x cos k 1 x
2k 1 2k 1
0
1
4
0œœ œ
Á '
œ
1
1
1
1
1
'
0
2
œ
Ú
Ý
Û
Ý
Ü
’“
¹
)
ab ab
ab ab
œ_
œ_
, k odd
, k even
.
!
2k
k11ab
2
œ
Thus, the Fourier series for f x is cos x sin kx.ab !
"
#œ
'
k even
2k
k11ab
2

8. a f x dx 2 dx x dx 1 , a f x cos kx dx0 k
11 31
22 4œœœœ
11 1
''' '
00 0
22 2
œœœ œ
œ
ab ab ”• 1
2 cos kx dx x cos kx dx .
, k odd
0, k even
œœ œœ '
&
11 cos kx x sin kx
kk
2 11
k
2
k
111
1
1
1
”• œ
( ‘
''
0
2
œœ
œ
2
k
2
2
œ_œab
b f x sin kx dx 2 sin kx dx x sin kx dx cos kxk
11 1 2 x cos kx sin kx
kkk
0
2œœœ '&
11 1
11
1
'''
00
22
œœœ
œ
ab ”•”• ¹¹ˆ‰
2
.
3 , k odd
, k even
œ
'
)
ˆ‰
14
k
1
k
1
Thus, the Fourier series for f x is 1 cos x 3 sin x sin 2x cos 3x 3 sin 3x . . . .ab ˆ‰ ˆ‰&& '' ' & ' ' '
32 4 2 14
493
1
11 1 1
"
#
9. cos px dx sin px 0 if p 0.'
0
2
œ
œœÁ
1
p
2
0
¹
1
10. sin px dx cos px 0 if p 0.11'
0
2
œ
œ œ œ Á &
11
pp
2
0
¹ cd
1
11. cos px cos qx dx dx 0 if p q.cos pqxcospqx sin pqx sinpqx ''
00
22
œœ
œœ œÁ '' & '' &
""
## _œ
cdab ab ( ‘ab ab
11
pq pq
2
0
1
If p q then cos px cos qx dx cos px dx 1 cos 2px dx x sin 2px .œœœœœ '''
000
222
œœœ
2 1
2p
2
0
""
##
ab Š‹ ¹
1
1
12. sin px sin qx dx dx 0 if p q.cos pqxcospqx sin pqx sinpqx ''
00
22
œœ
œœ œÁ && ' && '
""
## œ_
cdab ab ( ‘ab ab
11
pq pq
2
0
1
If p q then sin px sin qx dx sin px dx 1 cos 2px dx x sin 2px .œœœœœ '''
000
222
œœœ
2 1
2p
2
0
""
##
ab Š‹ ¹
1
1

765. Chapter 11 Practice Exercises 761
13. sin px cos qx dx dx sin pqxsinpqx cos pqx cospqx ''
00
22
œœ
œœ '' & '' &
""
## _œ
cdab ab ( ‘ab ab
11
pq pq
2
0
1
0. If p q then sin px cos qx dx sin px cos px dx sin 2px dx11 11œ œ œ œ œ&'&
" "
# #_œ
( ‘ab ab
11
pq pq '''
000
222
œœœ
cos 2px 1 1 0.œ œ œ
11
44
2
0
11
1
¹ ab
14. Yes. Note that if f is continuous at c, then the expression f c since f c f x f c and
fc fc
2
xc
ab ab
*
*
_ _
Ä
œœœab a b ab ablim
f c f x f c . Now since the sum of two piecewise continuous functions on 0, 2 is also continuous on 0, 2 ,ab ab ab
œ
Ä
œœ ÒÓ ÒÓlim
xc

11
the function f g satisfies the hypothesis of Theorem 24, and so its Fourier series converges to '
ababababfgc fgc
2
___
*
for 0 c 2 . Let s x denote the Fourier series for f x . Then for any c in the interval 0, 2†† 11 fab ab a b
sc
fgx fgx fx gx fx gxfg
fgc fgc
2 xc xc xc xc xc xc
_
___ ""
##Ä Ä ÄÄ ÄÄab ’“’ “
abab abab ab ab ab ab
œœ œ
''' '''
abababab
*
**
lim lim lim lim lim lim
+
s c s c , since f and g satisfy the hypothesis of Theorem 24.fc gc fc gcœœ '''
"
#
__ œœ
cd ab abababab ab ab ab fg
15. (a) f x is piecewise continuous on 0, 2 and f x 1 for all x f x is piecewise continuous on 0, 2 . Thenab ab ab ÒÓœ ÁÊ ÒÓ11 1
ww
by Theorem 24, the Fourier series for f x converges to f x for all x and converges to f fab ab a b ab abÁ111
"
#
_œ
0 at x .œ œ œ
"
#
ab11 1
(b) The Fourier series for f x is 1 . If we differentiate this series term by term we get the seriesab a b!
k1
k12 sin kx
k
œ
_
_
&
1 2 cos kx, which diverges by the n term test for divergence for any x since 1 2 cos kx 0.!ab ab
k1
k1 k1th
k
œ
_
_ _
Ä_
& Álim
16. Since the Fourier series in discontinuous at x , by Theorem 24, the Fourier series will converge to . Thus,œ1
fc fc
2
ab ab
*
_
at x we have 2 cos x sin x cos 2x sin 2x cos 3x sin 3x . . . .œ œ 11
ff
26 2 9 27
14 2942ab ab11 111
11
*
_ œ" œ
#
Š‹ Š ‹
2 2
2 cos sin cos 2 sin 2 cos 3 sin 3 . . .ʜ
01 4 2 94
26 2 9 27
2_œ" œ
#
11 1 1
11
22 2
11 11 11 1Š‹ Š ‹
2 . . . 2 1 . . . 2 2Ê œ œ œ Ê œ
01 2 1 11 1 1 1
26 9 6 49 6 n26 n
22 2
n1 n1
_"
#
œœ
__
1 11
2 22
22
11 1 ˆ‰ !!
22 .
1111
2222
222
26 n3 n6 n
n1 n1 n1
111
œ Êœ Êœ!!!
œœœ
___
CHAPTER 11 PRACTICE EXERCISES
1. converges to 1, since lim a lim 1 1
nnÄ_ Ä_
n
(1)
nœœ Š‹
œ
n
2. converges to 0, since 0 a , lim 0 0, lim 0 using the Sandwich Theorem for SequencesŸŸ œ œ n
22
nnÈÈ
nnÄ_ Ä_
3. converges to 1, since lim a lim lim 1 1œœœ
nn nÄ_ Ä_ Ä_
n
12
2
ˆ‰ ˆ ‰
œ"
#
n
nn
4. converges to 1, since lim a lim 1 (0.9) 1 0 1
nnÄ_ Ä_
nœ œœcd
n
5. diverges, since sin 0 1 0 1 0 1˜™ ef
n1
#
œ ßßßßßßá
6. converges to 0, since {sin n } {0 0 0 }1œ ßßßá
7. converges to 0, since lim a lim 2 lim 0
nn nÄ_ Ä_ Ä_
n
ln n
n1œœ œ
œ Š‹
œ
n

766. 762 Chapter 11 Infinite Sequences and Series
8. converges to 0, since lim a lim lim 0
nn nÄ_ Ä_ Ä_
n
ln (2n )
n1œœœ
_"
Š‹
2
2n 1
'
9. converges to 1, since lim a lim lim 1
nn nÄ_ Ä_ Ä_
n
nln n
n1
1œœœ ˆ‰
_
_Š‹
œ
n
10. converges to 0, since lim a lim lim lim lim 0
nn n n nÄ_ Ä_ Ä_ Ä_ Ä_
n
ln 2n 1
n16nn
12n 2œœœœœ
ab
Š‹
&
œ
_
6n
2n 1
&
Ú
'
11. converges to e , since lim a lim lim 1 e by Theorem 5
á á5 5
n
n
n
nn nÄ_ Ä_ Ä_
œœœˆ‰ Š‹
n5
nn
(5)œ œ
12. converges to , since lim a lim 1 lim by Theorem 5
""""
_ene
n
1
nn nÄ_ Ä_ Ä_
œœ œˆ‰
án
ˆ‰
Ú
n
n
13. converges to 3, since lim a lim lim 3 by Theorem 5
nn nÄ_ Ä_ Ä_
n
333
n1
1n
nœœœœˆ‰
n
1nÎ
Ê
14. converges to 1, since lim a lim lim 1 by Theorem 5
nn nÄ_ Ä_ Ä_
n
331
n1
1n
nœœœœˆ‰
Î 1n
1nÊ
Ê
15. converges to ln 2, since lim a lim n 2 1 lim lim lim 2 ln 2
nn n n nÄ_ Ä_ Ä_ Ä_ Ä_
n
21œœ œ œab
1n 1nœ œ
1nÊ
œ
Š‹
œ
á

&
áœ
&
n
2 ln 2
1n
n
n
–—
Š‹

2 ln 2 ln 2œœ
!
†
16. converges to 1, since lim a lim 2n 1 lim exp lim exp e 1
nn n nÄ_ Ä_ Ä_ Ä_
n
ln (2n 1)
n1œœ œ œœÈ Š‹ Œ '
n _ !
2
2n 1
'
17. diverges, since lim a lim lim (n 1)
nn nÄ_ Ä_ Ä_
n
(n 1)!
n!œœœ_
_
18. converges to 0, since lim a lim 0 by Theorem 5
nnÄ_ Ä_
n
(4)
n!œœ
œ
n
19. s
"
œœ #œ œ #œ œ œ(2n3)(2n1) n3 2n1 3 5 5 7 n3 2n1 3 2n1
n
œÊœá œ
Š‹ Š‹ Š‹ Š‹ Š‹ Š‹ Š‹ Š‹ Š‹ Š‹
œ œ œœ œœ œ œ œ œ
& & && && & & & &
–—–—– —
lim s lim Êœœ
nnÄ_ Ä_
n
62n1 6 –—
""
œ
Š‹
œ
&
20. s lim s
œœ œ œ œ
__ # _#_
222 22 22 22 22
n(n1) n n1 3 3 4 n n1 n1
n n
œ Êœ á œ ʈ‰ˆ‰ ˆ ‰
nÄ_
lim 1 1œœ
nÄ_
ˆ‰
2
n1_
21. s
9 3 3 333333 3 3
(3n1)(3n2) 3n1 3n2 5 5 8 8 11 3n1 3n2
n
œ_ œ _ # œ _
œ Êœ á ˆ‰ˆ‰ˆ ‰ ˆ ‰
lim s lim œ Ê œ œ
33 3 3 3
3n 3n 2
n
#_# # _ #
nnÄ_ Ä_
ˆ‰
22. s
œœ œœœ œ
œ_ œ _ œ_
8 22 222222 22
(4n 3)(4n 1) 4n 3 4n 1 9 13 13 17 17 21 4n 3 4n 1
œ Êœ á
n
ˆ‰ˆ‰ˆ‰ ˆ ‰
lim s lim œ Ê œ œ
22 22 2
94n1 94n1 9__
nnÄ_ Ä_
n
ˆ‰
23. e , a convergent geometric series with r and a 1 the sum is !!
__
œœn0 n0
œ """
œ
œ
n
eee1
1
e
œœœÊœ
n
Š‹
œ
e

767. Chapter 11 Practice Exercises 763
24. ( 1) a convergent geometric series with r and a the sum is!! ˆ‰ˆ‰
__
œœn1 n0
œ œ œÊ
n
n33 3
444 44
n
œ" " œ

ˆ‰
ˆ‰
œ
œ
3
4
4
1
3
5
ڜ
25. diverges, a p-series with pœ
"
#
26. 5 , diverges since it is a nonzero multiple of the divergent harmonic series!!
__
œœn1 n1
œ"5
nn
œ
27. Since f(x) f (x) 0 f(x) is decreasing a a , and lim a lim 0, the
n
œÊ œÊ Ê œ œ
Ä_
"" w
#
_
xx
n1 n n
1
n
Ú
nÄ_ È
series converges by the Alternating Series Test. Since diverges, the given series converges!!
_ _
œ œn1 n1
()
n n
œ" "
n
È È
conditionally.
28. converges absolutely by the Direct Comparison Test since for n 1, which is the nth term of a
""
#nn
&&†Ð
convergent p-series
29. The given series does not converge absolutely by the Direct Comparison Test since , which is
""
__ln (n 1) n 1

the nth term of a divergent series. Since f(x) f (x) 0 f(x) isœÊœ Ê
""
___
w
ln (x 1) (ln (x 1)) (x 1)
œ
decreasing a a , and lim a lim 0, the given series converges conditionally by theÊ œ œ n1 n
ln (n 1)
_
"
_
nnÄ_ Ä_
n
Alternating Series Test.
30. dx lim dx lim (ln x) lim the series''
22
b
b
2
'
"" """ œ"
#x(ln x) x(ln x) ln b ln 2 ln
œœœœœœÊ
bb bÄ_ Ä_ Ä_
cd ˆ‰
converges absolutely by the Integral Test
31. converges absolutely by the Direct Comparison Test since , the nth term of a convergent p-series
ln n n
nnn
&& œœ
"
32. diverges by the Direct Comparison Test for e n ln e ln n n ln n ln n ln (ln n)
nn n n
nn
Ñ! Ñ!Ñ! ш‰
n ln n ln (ln n) , the nth term of the divergent harmonic seriesÊ Ê
ln n
ln (ln n) n
"
33. lim lim 1 1 converges absolutely by the Limit Comparison Test
nnÄ_ Ä_
Š‹
Š‹
œ
&
'
œ
&
nn 1
n
œ
œœœÊÉ
È
n
n1
œ
œ
_
34. Since f(x) f (x) 0 when x 2 a a for n 2 and lim 0, theœÊœ  Ê   œ
3x 3n
x1 n1
3x 2 x
x1
n1 n
œ œ
& &
&
&
œ
_ _
w œ
_
_
ab
ab nÄ_
series converges by the Alternating Series Test. The series does not converge absolutely: By the Limit
Comparison Test, lim lim 3. Therefore the convergence is conditional.
nnÄ_ Ä_
Š‹
ˆ‰
3n
n 1
n
&
Ú
'
œ
œœ
3n
n1
&
&
_
35. converges absolutely by the Ratio Test since lim lim 0 1
nnÄ_ Ä_
’“
n2 n! n 2
(n 1)! n 1 (n 1)
__
__ _
† œœ
œ
36. diverges since lim a lim does not exist
nnÄ_ Ä_
n
()n 1
2n n 1œ
œ" _
_œ
n
ab
œ
œ
37. converges absolutely by the Ratio Test since lim lim 0 1
nnÄ_ Ä_
’“
3n! 3
(n 1)! 3 n 1
n1
nÛ
__
†œœ

768. 764 Chapter 11 Infinite Sequences and Series
38. converges absolutely by the Root Test since lim a lim lim 0 1
nn nÄ_ Ä_ Ä_
È É
n
n
nn
n
nœœœ
23 6
nn
39. converges absolutely by the Limit Comparison Test since lim lim 1
nnÄ_ Ä_
Š‹
Š‹
œ
)Î&
œ
''
n
n(n 1)(n 2)
œ
œœÉ
n(n 1)(n 2)
n
__
&
40. converges absolutely by the Limit Comparison Test since lim lim 1
nnÄ_ Ä_
Š‹
Š‹
œ
&
œ
&
(
n
nn 1
œ
œœÉ
nn 1
n
œœ
'
abœ
41. lim 1 lim 1 lim 1 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ˆ‰
u
u (n 1)3 (x 4) 3 n 1 3
(x 4) n3 n x4 x4n1
n
n1
n1 n
n*
*
*
†! †! †! †
_
__ _
__
†
kk kk
x 4 3 3 x 4 3 7 x 1; at x 7 we have , theÊ Ê Ê œ œkk !!
__
œœn1 n1
(1)3 ( )
n3 n
œœ"
nn n
n
alternating harmonic series, which converges conditionally; at x 1 we have , the divergentœ œ !!
__
œœn1 n1
3
n3 n
n
n
"
harmonic series
(a) the radius is 3; the interval of convergence is 7 x 1&Î †&
(b) the interval of absolute convergence is 7 x 1&† †&
(c) the series converges conditionally at x 7œ42. lim 1 lim 1 (x 1) lim 0 1, which holds for
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹
u
u (2n 1)! (x 1) ( n)(2n 1)
(x 1) (2n 1)!n1
n*
Ê Ê œ
œœ
_œ #_
# "
2n
2n 2
†
œ
all x
(a) the radius is ; the series converges for all x_
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
43. lim 1 lim 1 3x 1 lim 1 3x 1 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kk
u
u (n 1) (3x 1) (n 1)
(3x 1) nnn1
n
n1
n*
*
œ œ
œœ
†! †! & †! &†
œ
_œ _
†
1 3x 1 1 0 3x 2 0 x ; at x 0 we have Ê Ê Ê œ œ
2
3nn
(1) (1) ( ) !!
__
œœn1 n1
œœ œ"
n1 n 2n1
œœ
, a nonzero constant multiple of a convergent p-series, which is absolutely convergent; at x weœ œ!
_
œn1
"
n 3
2
œ
have , which converges absolutely!!
__
œœn1 n1
(1) (1) ( )
nn
œœ"
n1 n n1
œœ
œ
(a) the radius is ; the interval of convergence is 0 x
"
33
2
ŸŸ
(b) the interval of absolute convergence is 0 xŸŸ
2
3
(c) there are no values for which the series converges conditionally
44. lim 1 lim 1 lim 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ¸¸
u
u 2n3 2 n1 (2x1) 2 2n3 n1
n2 2n1 2 n2 2n(2x 1) 2x 1n1
n
n1
n1 n
n*
*
*
†! †! †
__ __"
___ __
_ _
††† †
kk
(1) 1 2x 1 2 2 2x 1 2 3 2x 1 x ; at x we haveÊ Ê Ê Ê Ê œ
kk2x 1 33_
# ## #
"
kk
which diverges by the nth-Term Test for Divergence since!!
__
œœn1 n1
n1
2n 1 2n 1
(2) ( )(n1)_
_# _
œœ"_
†
nn
n
œ lim 0; at x we have , which diverges by the nth-
nÄ_
ˆ‰ !!
n1 n1 2 n
2n 1 2n 1 2n 1
_" " _ _"
_# # _# _
œÁ œ œ
__
œœn1 n1
†
n
n
Term Test
(a) the radius is 1; the interval of convergence is x&††
3
##
"
(b) the interval of absolute convergence is x&††
3
##
"
(c) there are no values for which the series converges conditionally

769. Chapter 11 Practice Exercises 765
45. lim 1 lim 1 x lim 1 lim 1
nn n nÄ_ Ä_ Ä_ Ä_
¹¹ ¹ ¹ kk ¸¸ ˆ‰ˆ‰ˆ‰
u
u (n1) x n1 n1 e n1
xn n
n xn1
n
n1 n
n1 n*
*
*
†! †! †! †
____
""
†
kk
0 1, which holds for all x!†
kkx
e
†
(a) the radius is ; the series converges for all x_
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
46. lim 1 lim 1 x lim 1 x 1; when x 1 we have
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk kkÉ
u
uxn1
xn
n1
nn1
n
n1
n*
*
Ê Ê Ê œ
È
È
_ _
†
, which converges by the Alternating Series Test; when x 1 we have , a divergent!!
_
œn1
(1)
n n
œ "
n
È È
œ
'
*
n1
p-series
(a) the radius is 1; the interval of convergence is 1 x 1&Î †
(b) the interval of absolute convergence is 1 x 1&† †
(c) the series converges conditionally at x 1œ
47. lim 1 lim 1 lim 1 3 x 3;
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ˆ‰ ÈÈ
u
u3(n1)x3n1
(n 2)x 3xn2n1
n
2n 1
n 1 2n 1
n*
*
*
œ
†! †! †!& ††
_
__
_
†
the series and , obtained with x 3, both diverge!! È
__
œœn1 n1
œ„
n1 n1
33
__
ÈÈ
(a) the radius is 3; the interval of convergence is 3 x 3
ÈÈÈ
&††
(b) the interval of absolute convergence is 3 x 3&††
ÈÈ
(c) there are no values for which the series converges conditionally
48. lim 1 lim 1 (x 1) lim 1 (x 1) (1) 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ˆ‰
u
u 2n 3 (x 1) 2n 3
(x 1)x 2n 1 2n 1n1
n
2n 3
2n 1*
*
*
†! †! & †! & †
œ
_œ _
__ ##
†
(x 1) 1 x 1 1 1 x 1 1 0 x 2; at x 0 we have Ê Ê Ê Ê œ
# œœ
#_
kk !
_
œn1
(1)(1)
n1
n2n1 * which converges conditionally by the Alternating Series Test and the factœœ!!
__
œœn1 n1
(1) (1)
2n 1 2n 1
œœ
__
3n 1* n1œ
that diverges; at x 2 we have , which also converges!!!
___
œœœn1 n1 n1
"
___
œœ
2n 1 2n 1 2n 1
(1)(1) (1)
œœ
n2n1 n*
conditionally
(a) the radius is 1; the interval of convergence is 0 x 2ŸŸ
(b) the interval of absolute convergence is 0 x 2††
(c) the series converges conditionally at x 0 and x 2œœ
49. lim 1 lim 1 x lim 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ kk
»»
u
ucsch(n)x
csch (n 1)xn1
n
n1
n*
*
†! †! †
_
Š‹
ˆ‰
2
ee
n1 n1
2
ee
nn
' ((
(
(
(
x lim 1 1 e x e; the series e csch n, obtained with x e,ÊÊÊ„ œ„kk a b¹¹ !
nÄ_
ee
1e e
x n
)ÎÎ

œ
œ
2n 1
2n 2 kk
_
œn1
both diverge since lim e) csch n 0
nÄ_
a„Á
n
(a) the radius is e; the interval of convergence is e x e&† †
(b) the interval of absolute convergence is e x e&† †
(c) there are no values for which the series converges conditionally
50. lim 1 lim 1 x lim 1 x 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ¹ ¹ kk kk
u
ux coth(n) 1e1e
x coth(n1) 1e 1en1
n
n1
n2n22n
2n 2 2n*
*

†! †! †! †
_ _œ
œ_
†
1 x 1; the series 1 coth n, obtained with x 1, both diverge since lim 1 coth n 0Ê „ œ„ „ Á !ab ab
_
œn1
n n
nÄ_
(a) the radius is 1; the interval of convergence is 1 x 1&† †
(b) the interval of absolute convergence is 1 x 1&† †

770. 766 Chapter 11 Infinite Sequences and Series
(c) there are no values for which the series converges conditionally
51. The given series has the form 1 x x x ( x) , where x ; the sum is áᜠœ œ
#$ "" "
_ _
n
1x 4 5 1
4
ˆ‰
Ú
4
52. The given series has the form x ( 1) ln (1 x), where x ; the sum isá ᜠœ
xx x 2
3n 3
n1

#
œ
n
ln 0.510825624ˆ‰
5
3
¸
53. The given series has the form x ( 1) sin x, where x ; the sum isá ᜠœ
xx x
3! 5! (2n 1)!
n
& áÛ 2n 1
_
1
sin 01œ
54. The given series has the form 1 ( 1) cos x, where x ; the sum is cos á ᜠœ œ
xx x
2! 4! (2n)! 3 3
n
œ ' 2n
11 "
#
55. The given series has the form 1 x e , where x ln 2; the sum is e 2áᜠœ œ
xx x
2! 3! n!
xln2
œœ n
ÐÑ
56. The given series has the form x ( 1) tan x, where x ; the sum isá ᜠœ
xx x
35 (2n1)
n
3
&( Î 2n 1
œ
œ" "
È
tan
œ" "
Š‹
È
3 6
œ
1
57. Consider as the sum of a convergent geometric series with a 1 and r 2x
" "
œ œ12x 12x
œœÊ
1 (2x) (2x) (2x) (2x) 2 x where 2x 1 xœáœ œ Ê
#$ "
# !! kk kk
__
œœn0 n0
nnn
58. Consider as the sum of a convergent geometric series with a 1 and r x
" ""
_ _œœ
$
1x 1x 1 x
& && œœÊœ
ab
1 x x x ( 1) x where x 1 x 1 x 1œáœ Ê Êababab kk kk kk !
$$ $ $ $
#$
_
œn0
n3n
59. sin x sin x œÊœ œ!!!
___
œœœn0 n0 n0
(1)x (1)(x) (1) x
(2n 1)! (2n 1)! ( n 1)!
œœœ
__#_
n2n1 n 2n1 n 2n12n1ÛÛÛÛ
1
11
60. sin x sin œÊœ œ!!!
___
œœœn0 n0 n0
(1)x (1)2 x
(2n 1)! 3 (2n 1)! 3 ( n 1)!
2x
(1)
œœ
__#_
œ
n2n1 n2n12n1
n
2n 1
2n 1Û ÛÛ
Û
Û Š‹
2x
3
61. cos x cos x œÊœ œ!!! ˆ‰
___
œœœn0 n0 n0
(1)x (1)x
(2n)! (2n)! ( n)!
(1) xœœ &Î#
œ
#
n2n n5nn
2n ˆ‰
áÊœ
62. cos x cos 5x cos (5x) œÊœ œ œ!!! È ˆ‰
___
œœœn0 n0 n0
(1)x (1)5x
(2n)! (2n)! ( n)!
( 1) (5x)œ œ"Î#
œ
#
n2n nnnn
2n ˆ‰
ÚÊœ
63. e e
xx2 xx
n! n! n!
n0
œÊœ œ!!!
__
œœn0 n0
nnn
x
n
n
ÐÎÑ
_
œ
#
1 1
ˆ‰
œ
œ
64. e e
xx x
n! n! n!
x (1)x
œÊœ œ!!!
___
œœœn0 n0 n0
n
n
n2n
œ œ œ
œ œ ab
65. f(x) 3 x 3 x f (x) x 3 x f (x) x 3 x 3 xœœ Ê œ Ê œ
È
ab ab ab ab
# #w # ww## #
"Ÿ# œ"Ÿ# œ$Ÿ# œ"Ÿ#
f (x) 3x 3 x 3x 3 x ; f( 1) 2, f ( 1) , f ( 1) ,Êœ œœœœ
www $ # # w ww
&Î# $Î# """
##
ab ab
88
3
f ( 1) 3 x 2
www # __ _
œœÊœá
33 9
32 8 32 2 1! 2 2! 2 3!
(x 1) 3(x 1) 9(x 1)È
†† †
œ &
&(

771. Chapter 11 Practice Exercises 767
66. f(x) (1 x) f (x) (1 x) f (x) 2(1 x) f (x) 6(1 x) ; f(2) 1, f (2) 1,œœÊ œÊ œÊ œ œ œ
"
œ
œ" ¸ œ# ¸¸ œ$ ¸¸¸ œ% ¸
1x
f (2) 2, f (2) 6 1 (x 2) (x 2) (x 2)
ww www # $ "
œ
œ œÊ œ á
1x
67. f(x) (x 1) f (x) (x 1) f (x) 2(x 1) f (x) 6(x 1) ; f(3) ,œœÊ œÊ œÊ œ œ
" "
_
œ" ¸ œ# ¸¸ œ$ ¸¸¸ œ%
x1 4
f (3) , f (3) , f (2) (x 3) (x 3) (x 3)
w ww www # $"œ"""""
_
œ œ œÊœá
44 x1444
26
44
' '
68. f(x) x f (x) x f (x) 2x f (x) 6x ; f(a) , f (a) , f (a) ,œœ Ê œ Ê œ Ê œ œ œ œ
" ""œ" ¸ œ# ¸¸ œ$ ¸¸¸ œ% ¸ ¸¸
x aaa
2

f (a) (x a) (x a) (x a)
www # $œ""" " "
œ Êœ á
6
aa xaa a
' '
69. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na x yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a a 2a a x (3a a )x (na a )x 0 a a 0, 2a a 0,œ á áœÊœ œaba b "Ê #" $# œ "Ê #"
#œ
nn1
n1
3a a 0 and in general na a 0. Since y 1 when x 0 we have a 1. Therefore a 1,$# œ Ê "œ œ œ œ œ œ nn1
a , a , a , , a #$%
œœœ "" " œ
##œ
œ œœœœ œœ œœ ᜠœ œ
aaa
21 3 32 4 43 n n (n 1)! n!
n
a 1(1) (1)Ú
Û
††††
n1
nn1
y1xx x x eÊœ á ᜠœ
""
##
#$ œ œœ
3n! n!
(1) (1)xnx
†
n1 nnÛ
!
_
œn0
70. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na x yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a a 2a a x (3a a )x (na a )x 0 a a 0, 2a a 0,œ á áœÊœ œaba b "Ê #" $# œ "Ê #"
#œ
nn1
n1
3a a 0 and in general na a 0. Since y 3 when x 0 we have a 3. Therefore a 3,$# œ Ê "œ œ œ œ œ œ nn1
a , a , a y 3 3x x x x#$
œœ œ œ
# #
#$œœ œœ œ œ Êœ á á
aa
2332nn! 213n!
33 3 333
n
a nÚ œ
†††
n1
31 x 3 3eœ á á œ œŠ‹ !
xx x x
!3! n! n!
x

#
nn
_
œn0
71. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n a 2a x na x 2yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a 2a 2a 2a x (3a 2a )x (na 2a )x 0. Since y 3 when x 0 weœ á ᜠœ œaba b "Ê #" $# œ
#œ
nn1
n1
have a 3. Therefore a 2a 2(3) 3(2), a a ( 2 3) 3 , a a!"! #" $#
## #œœœœœœœœ
22 2 2
3
† Š‹
œ
3 3 , , a a 3 3œ œ á œ œ œ
22 2 2 2
33 nn(n1)!n!
nn1
(1) 2 (1)2
’“Š‹ Š ‹Š‹Š‹ Š ‹ ˆ‰ ˆ‰
œ &
## œ
œ
œœ
†
n1n1 nn
y 3 3(2x) 3 x 3 x 3 xÊœá á
(2) (2) ( 1) 2
3n!
n
œ &
##
#$ œ
†
nn
3 1 (2x) 3 3eœá ᜠœ’“ !
(2x) (2x) ( 1) (2x) ( 1) (2x)
2! 3! n! n!
2x
œ &
œœ œ
nn nn
_
œn0
72. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n a 2a x na x yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a a 2a a x (3a a )x (na a )x 1 a a 1, 2a a 0,œ á áœÊœ œaba b "Ê #" $# œ "Ê #"
#œ
nn1
n1
3a a 0 and in general na a 0 for n 1. Since y 0 when x 0 we have a 0. Therefore$# œ Êœ œ œ œ œ nn1
a 1 a 1, a , a , a , , a"!# $ %
œœœ "" "
##œœ œœ œœ œœ á
aaa
21 3 32 4 43
nÚ
††††
y0x x x xœœ œ Êœá á
œ œ""œœ œ
œ##
#$a
n n (n 1)! n! 3 n!
1(1) (1) (1) nn1
nn1 n1
Û Ûˆ‰
†
11 x x x x 1 1 1 eœ á ᜠœ’“ !
""
##
#$ œœœ
3n! n!
(1) (1)xnx
†
nnn
_
œn0
73. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na x yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a a 2a a x (3a a )x (na a )x 3x a a 0, 2a a 3,œ á áœÊœ œaba b "Ê #" $# œ "Ê #"
#œ
nn1
n1

772. 768 Chapter 11 Infinite Sequences and Series
3a a 0 and in general na a 0 for n 2. Since y 1 when x 0 we have a 1. Therefore$# œ Êœ œ œ œ œ nn1
a 1, a , a , a , , a"# $ %
_
##œ œ œ œœ œœ á œ œ
3a a a
2332443 nn!
22 2 2
n
a Ú
†††
n1
y1x x x x xÊœ á ሉ
232 2
343 n!
n
###
#$ %
†††
21x x x x x 33x2 33x2e 3x3œá᜜ˆ‰ !
"" " "
## #
#$ %
343 n! n!
nx x
†††
_
œn0
n
74. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na x yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a a 2a a x (3a a )x (na a )x x a a 0, 2a a 1,œ á áœÊœ œaba b "Ê #" $# œ "Ê #"
#œ
nn1
n1
3a a 0 and in general na a 0 for n 2. Since y 0 when x 0 we have a 0. Therefore$# œ Êœ œ œ œ œ nn1
a 0, a , a , , a"# $
"œ œ ""
#
œ œœœ œ œœáœ œ
aa
2332nn!
n
a (1)Ú œ
†
n1
n
y 0 0x x x x 1 x x x x 1 xÊ œ á á œ á á
"" ""
## ##
#$ #$ œœ
3n! 3n!
(1) (1)nn
††
n n
Š‹
1xe x1œœ!
_
œn0
(1)x
n!
xœ œ
nn
75. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n a 2a x na x yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a a 2a a x (3a a )x (na a )x x a a 0, 2a a 1,œ á áœÊœ œaba b "Ê #" $# œ "Ê #"
#œ
nn1
n1
3a a 0 and in general na a 0 for n 2. Since y 1 when x 0 we have a 1. Therefore$# œ Êœ œ œ œ œ nn1
a 1, a , a , a , , a"# $ %
_
##œœ œ œœ œœ ᜜
1a a a
2332443 nn!
22 2 2
n
a Ú
†††
n1
y1x x x x xÊœ á ሉ
222 2
342 n!
n
###
#$ %
†††
21x x x x x 1x2 1x2e x1œá᜜ˆ‰ !
"" " "
## #
#$ %
343 n! n!
nx x
†††
_
œn0
n
76. Assume the solution has the form y a a x a x a x a xœá á Ê" # œ
#œ
n1 n
n1 n
a 2a x na x yÊœ á áÊ
dy dy
dx dx
n
n1
"#
œ
a a 2a a x (3a a )x (na a )x x a a 0, 2a a 1,œ á áœÊ œ œaba b "Ê #" $# œ "Ê #"
#œ
nn1
n1
3a a 0 and in general na a 0 for n 2. Since y 2 when x 0 we have a 2. Therefore$# œ Êœ œ œ œ œ nn1
a 2, a , a , a , , a"# $ %
œ_
##œœ œ œœ œœ ᜜
1a a a
2332443nn!
11 1 1
n
a Ú
†††
n1
y 2 2x x x x xÊœ á á
11 1 1
343 n!
n
## #
#$ %
†††
1x x x x x 1x 1xe x1œ á ᜠœ ˆ‰ !
"" " "
## #
#$ %
343 n! n!
nx x
†††
_
œn0
n
77. exp x dx 1 x dx x''
00
12 12
œœ
ab Š‹’ “ œ á œ á
$$
"Î#
!
xxx x x x x
2! 3! 4! 4 7 2! 10 3! 13 4!
_
†††
0.484917143w&'&'&w
""""""
# ##
'œ
††† †† †† ††472! 2 10 3! 2 13 4! 2 16 5!
78. x sin x dx x x dx x dx'' '
00 0
11 1
ab Š‹Š ‹
$$ %
œ á œ á
xxxx xxxx
3! 5! 7! 9! 3! 5! 7! 9!
ÚáœÚœÝ Ú Ú_œœœ"
0.185330149œá¸’“
xxxxx
5 113! 175! 237! 299!

††††
"
!
79. dx 1 dx x''
11
12 12
œœ
tanx xxxxx xx x x x
x 357911 9254981121

œ á œ '&'& '(Š‹’ “
"Î#
!
w&'&'&'&'&'
"""""""""""
## ####9 2 5 9 2 11 2 13 2 15 2 19 21717†† †††† †† ††
& œÝ œÚÝ
0.4872223583¸

773. Chapter 11 Practice Exercises 769
80. dx x dx xxxx dx'' '
00 0
164 164 164
œœ œ
tanx xxx
xx 357 3 5 7
) &(œ
ÈÈ
œ á œ á
" """ "Î# &Î# *Î# "$Î#
Š‹ ˆ‰
x x x x 0.0013020379œ á œ á ¸( ‘ˆ ‰
22 2 2 2222
3 1 55 105 3 8 55 8 105 8 21 8
$Î# (Î# ""Î# "&Î#
#
"Î'%
! ††† †
& )) )(œ
81. lim lim lim
x0 x0 x0ÄÄ Ä
7 sin x 7
e1
2x
œ #
œœœ
7x 71
2x 2
Š‹Š‹
Š‹Š‹
á á
&&&' &&&'
xx xx
3! 5! 3! 5!
2x 2x 2x 2x
2! 3! ! 3!
) !&*
&& )) & )&
&
82. lim lim lim
)) )ÄÄ Ä00 0
ee 2
sin
&&
œœ
œ

)
))
œœ
Š‹Š‹Š‹
Š‹ Š‹
1122á á á
á
)))
))
œœ œœ œœ
œœ œœ
& )&) )!
&&
)! )!
! 3! ! 3! 3! 5!
3! 5! 3! 5!
&'
lim 2œœ
)Ä0
2Š‹
Š‹
œ
&
œ
&
3! 5!
3! 5!
&&'
á
œ
œ
83. lim lim lim lim
t0 t0 t0 t0ÄÄÄ Ä
ˆ‰
"" œ_
#œ œ
œ_
2 cos t t 2t (1 cos t)
t22 cos t
t22 2
2t
œ œ œ
œœ
œ
œ
œ
Œ œ Œ œ
Š‹1
11
á
á
tt t
4! 4!
tt
4!
&**
&
&*
&
á
á
t
6!
2t
4!
_
*
_
Š‹t
lim œœ
t0Ä
2
1
Š‹
Š‹
œ
&
&
4! 6!
t
2t
4!
á
á1
"
#
84. lim lim
h0 h0ÄÄ
Š‹ Œ œŒ œ
sin h
h
hh hh
3! 5! ! 4!
œ
œ
cos h
hh
œœ œ
11á á
& *&*
&
lim lim œ œ áœ
h0 h0ÄÄ
Œ œ
hhhhhh
!3!5!4!6!7!
&& ÛÛ__
&
á
h ! 3! 5! 4! 6! 7! 3
hhhh
œ
œœ áá Š‹
"" "
#
85. lim lim lim
z0 z0 z0ÄÄ Ä
"œ
œ_
œœ_
œ
cos z
ln (1 z) sin z
11z z
zz
œ
œœ
œœ
Š‹ Š‹
Š‹Š‹Š ‹
zz
33
zz zz z2zz
33!5! 34
**
&) )! & )*
&&
á á
áá á
lim 2œœ
z0Ä
Š‹
Š‹1á
á
z
3
2z z
34
&
œ
&
&
86. lim lim lim
y0 y0 y0ÄÄ Ä
yy y
cos y cosh y
11
œœ œ
œ
œœ_
œœ
Œ œŒ œ Œ œ
yyy yyy 2y2y
4!6! !4!6! 6!
&
&& &
á á á
lim 1œœ
y0Ä
"
œœ œÜŒ œ1
2y
6!
*
87. lim s lim s lim s 0
x0 x0 x0ÄÄ Ä
ˆ‰
–—
Š‹
sin 3x r r 3 9 81x r
xx x x x 40 x
3x
&
&(
œœ œ á œ
Š‹œ_œÜ
#
(3x) (3x)
6 120
0 and s 0 r 3 and sÊœ œÊœ œ
r3 9 9
xx
œœ
##
88. (a) csc x csc x sin xw' ! w ! w
"_
_x6 6x 6x
x6x 6x
œ
œ
(b) The approximation sin x is better than¸
6x
6x_
œ
sin x x.¸

774. 770 Chapter 11 Infinite Sequences and Series
89. (a) sin sin sin sin sin sin sin sin sin sin !ˆ ‰ˆ ‰ˆ‰ˆ‰ˆ ‰
_
œn1
" " " " "" "" " "
#_ # ##_2nn1 34567 nn1
œ á
( 1) sin ; f(x) sin f (x) 0 if x 2 sin sin , andá œ œ Ê œ   Ê !
_
œn2
n
nx x n1n
cos
"" "" w
œ
_
Š‹
œ
x
œ
lim sin 0 ( 1) sin converges by the Alternating Series Test
nÄ_
""
nn
œÊ !
_
œn2
n (b) error sin 0.02381 and the sum is an underestimate because the remainder is positivekk ¸¸†w
"
42
90. (a) tan tan ( 1) tan (see Exercise 89); f(x) tan f (x) 0!!ˆ‰
__
œœn1 n2
"" " "
#_
w
2n n 1 n x x
œ œÊœ
n
œsec
&œˆ‰
x
œ tan tan , and lim tan 0 ( 1) tan converges by the Alternating SeriesÊ œÊ
"" " "
_n1 n n n
n
nÄ_
!
á
*
n2
Test
(b) error tan 0.02382 and the sum is an underestimate because the remainder is positivekk ¸¸†w
"
42
91. lim 1 x lim 1 x
nnÄ_ Ä_
¹¹ kk kk¸¸
2 5 8 (3n 1)(3n 2)x 2 4 6 (2n)
4 6 (2n)(2n 2) 5 8 (3n 1)x 2n 2 3
3n 2 2†† ††
†† ††
ßœ _ ß
#ß _ #ßœ _
_
n1
n*
† †! †! †
the radius of convergence is Ê
2
3
92. lim 1 x lim 1 x
nnÄ_ Ä_
¹¹ kk kk¸¸
3 5 7 (2n 1)(2n 3)(x 1) 4 9 14 (5n 1)
4 9 14 (5n 1)(5n 4) 3 5 7 (2n 1)x 5n 4 2
2n 3 5†† ††
†† ††
ß_ _ œ ßœ
ßœ _ ß_ _
_
n1
n*
† †! †! †
the radius of convergence is Ê
5
2
93. ln 1 ln 1 ln 1 ln (k 1) ln k ln (k 1) ln k!! !ˆ‰ ( ‘ˆ‰ˆ‰ cd
nn n
k2 k2 k2œœ œ
œ œ
"""
kkk
œ
ln 3 ln 2 ln 1 ln 2 ln 4 ln 3 ln 2 ln 3 ln 5 ln 4 ln 3 ln 4 ln 6 ln 5 ln 4 ln 5œ cdcdcdcd
ln (n 1) ln n ln (n 1) ln n ln 1 ln 2 ln (n 1) ln n after cancellationáœcdcdcd
ln 1 ln ln 1 lim ln ln is the sumÊœ Ê œ œ!!ˆ‰ˆ‰ ˆ‰ ˆ‰
n
k2 k2œœ
"_ " _"
#k2n k 2n
n1 n 1
œœ
á
nÄ_
94. - !! ˆ‰ (ˆ‰ˆ‰ˆ‰ˆ‰ ˆ ‰
nn
k2 k2œœ
" " " " " " "" "" "" " "
œ# œ_ # # œ#k 1 k1 k1 1 3 4 3 5 4 6 n n
1
œœ œ á
œ œ œ œ‘ˆ ‰ˆ ‰ˆ‰ ’“
" " """" " " " " " œœ
œ_ # # _ ## _ # _ _
_œ _œ
n1 n1 1 n n1 n n1 2n(n1) 4n(n1)
33nn2 3n(n 1) 2(n 1) 2n
œ
lim Êœ œ! ˆ‰
á
k2œ
""
œ# _k1 2nn1 4
31 1 3
œ
nÄ_
95. (a) lim 1 x lim
nnÄ_ Ä_
¹¹ kk
1 4 7 (3n 2)(3n 1)x (3n)! (3n )
(3n 3)! 1 4 7 (3n 2)x (3n 1)(3n 2)(3n 3)
††
††
ߜ _ _"
_ ߜ ___
$
3n 3
3n*
† †!
x 0 1 the radius of convergence is œÊ _kk
$
†
(b) y 1 x xœ Ê œ!!
__
œœn1 n1
147 (3n 2) dy 147 (3n 2)
(3n)! dx (3n 1)!
†† ††ßœ ßœ
œ 3n 3n 1 (
xx xÊœ œ
d y 147 (3n 2) 147 (3n 5)
dx (3n 2)! (3n 3)!
œ
œ
!!
__
œœn1 n2
†† ††ßœ ßœ
œœ 3n 2 3n 2((
x 1 x xy 0 a 1 and b 0œ œÊœ œŒ á
!
_
œn1
147 (3n 2)
(3n)!
††ßœ 3n
96. (a) x x ( x) x ( x) x ( x) xxxx (1)x which
xx
1x 1(x)
nn
œœ
_œœ
## # ## $ #$%&
œ œáœ ᜠ!
_
œn2
converges absolutely for x 1kk†
(b) x 1 ( 1) x ( 1) which divergesœÊ œ !!
__
œœn2 n2
nn n

775. Chapter 11 Practice Exercises 771
97. Yes, the series a b converges as we now show. Since a converges it follows that a 0 a 1!!
__
œœn1 n1
nn n nnÞ!†
for n some index N a b b for n N a b converges by the Direct Comparison Test with bÑ!†Ñ!
nn n nn n
!!
_ _
œ œn1 n1
98. No, the series a b might diverge (as it would if a and b both equaled n) or it might converge (as it!
_
œn1
nn n n
would if a and b both equaled ).nn
n
"
99. (x x ) lim (x x ) lim (x x ) lim (x ) x both the series and!!
__
œœn1 1
n1 n k1 k n1 n1___"_"œ œ œ Ê
nnnÄ_ Ä_ Ä_
k
sequence must either converge or diverge.
100. It converges by the Limit Comparison Test since lim lim 1 because a converges
nnÄ_ Ä_
Š‹
an
1an
n
'
a
œœ
"
_1a
n
n
!
_
œn1
and so a 0.nÄ
101. Newton's method gives x x x , and if the sequence {x } has the limit L, thenn1 n n n
x1
40 x 1
39
40 40
_
œ
œ
"œ œ
ab
abn
n
'
&Ñ
L L L 1 and x converges since 1œÊœ Ö× œ
39 39
40 40 f x 40
n
fxf x"
ÒÓ
¹¹
ab ab
ab
††
†
2
102. a a a a a! ˆ‰ ˆ ‰ ˆ ‰
_
œn1
a
n 3 4 34 5678
aaan
œá  ""#%)
##
" "" """"
œ &'
a (aaaa ) which is a divergent series' ' ' '( ' '( Ð ' ' ' '(ˆ‰
"" " " "
"' #%)"'
#91011 16
103. a for n 2 a a a , and n
ln n ln ln 4 ln 8 ln ln 2 3 ln 2œ  Ê    á ᜠá
" """ " " "
#$%
###
1 which diverges so that 1 diverges by the Integral Test.œ á
""" "
##ln 3 n ln n
ˆ‰ !
_
œn2
104. (a) T 0 2 e e e e 0.885660616œ œ¸

84
ˆ‰
Ú
œ
##
"""
#
"Î# "Î#
Š‹ˆ‰
(b) x e x 1 x x x x x dx 0.68333
# # #$ #$
## #
"
!
x xx xxxx41
3 4 10 60
œ áœáÊ œ œ œŠ‹ Š‹’“
œ ''&'(
'
0
1
(c) If the second derivative is positive, the curve is concave upward and the polygonal line segments used in
the trapezoidal rule lie above the curve. The trapezoidal approximation is therefore greater than
the actual area under the graph.
(d) All terms in the Maclaurin series are positive. If we truncate the series, we are omitting positive terms
and hence the estimate is too small.
(e) x e dx x e 2xe 2e e 2e 2e 2 e 2 0.7182818285
'
0
1
##
"
!
xxxx
œ œ œ¸cd
105. a f x dx 1 dx a f x cos kx dx cos kx dx 0.0k
11 1 1
22œœœßœ œ œ
11 1 1
'' ' '
00
22 2 2
œœ œ œ
œœ
ab ab
"
#
b f x sin kx dx sin kx dx 1 1 .
, k odd
0, k evenk
11cos kx1
kk
2
k
2
kœ œ œ œ œ
&
1111
1
1
1
''
0
22
œœ
œ
ab a b ¹ Š‹ œ
Thus, the Fourier series of f x is sin kxab !
"
#
&
k odd
2
k1

776. 772 Chapter 11 Infinite Sequences and Series

106. a x dx 1 dx , a x cos kx dx cos kx dx0k
111
24
cos kx x sin kx
kk 0œœœ œ '
111
1”•” •
( ‘
'' ' '
00
2 2
œœ œ œ
œ œ
""
#
1
2
1 1 .
, k odd
0, k even
œœ
&
1
k
k
2
k
1
1
2
2Š‹ab œ
b x sin kx dx sin kx dx 1 1k
11cos kx1 sin kx x cos kx
kk 0 kkk
2
1kœœ œ &
1111
1
1
1”•
( ‘ ¹ Š‹ab
''
0
2
œœ
œ
2
k1
abœ
*
.
1 , k odd
, k even
œ
&
&
)
ˆ‰
12
k
1
k
1
Thus, the Fourier series of f x is cos x 1 sin x sin 2x cos 3x 1 sin 3x . . .ab ˆ‰ ˆ‰
"" "
##
'& '& & & ' & '
493
22 212
1
11 1 1

107. a x dx x 2 dx x dx u du 0 where we used the0
11
22œœœ
11”•ab a b ab ab ’“
'' ''
000
2
œœ œœ
œ
1111
substitution u x in the second integral. We havea x cos kx dx x 2 cos kx dx . Usingœ œ 111 k
1
1 ”•ab a b
''
0
2
œœ
œ
the substitution u x in the second integral gives x 2 cos kx dx u cos ku k duœ œ 1111 ''
œ
œœ
2
0
ab aba b
.
u cos ku du, k odd
u cos ku du, k even
œ
&
&&
Ú
Û
Ü
ab
ab
'
'
0
0
œ
œ
1
1
Thus, a .
x cos kx dx, k odd
0, k evenkœ
&
)
ab
#
1'
0
œ
1
Now, since k is odd, letting v x x cos kx dx v cos kv dv , k odd. (Seeœ Ê œ œ œ11
###
1111''
00
œœ
ab ˆ‰
24
kk
22
Exercise 106). So, a .
, k odd
0, k even k
4
kœœ
1
2
Using similar techniques we see that b
u sin ku du, k odd
0, k even
, k odd
0, k
2
kœœ
&
)
ab
œ
#
1'
0
œ
1
k even
.
Thus, the Fourier series of f x is cos kx sin kx .ab!ˆ‰
k odd
42
kk1
2 '

777. Chapter 11 Additional and Advanced Exercises 773

108. a sin x dx sin x dx . We have a sin x cos kx dx0k
1121
2œœœœ
1111
'' '
00 0
2 2
œœ œ
kk kk
sin x cos kx dx sin x cos kx dx . Using techniques similar to those used in Exercise 107, we findœ
1
1
’“''
0
œœ
œ

a .
0, k odd
sin x cos kx dx, k even
0, k odd
, k evenk 2 4
k1œœ
) œ
1 1
'
0
œ
œ
œab
2
b sin x sin kx dx sin x sin kx dx sin x sin kx dx
0, k odd
sin x sin kx k
11
2œœœ
11
1
'''
'
00
2
0
œœœ
œ
œ
kk ’“
)

dx, k even
0œ
for all k.
Thus, the Fourier series of f x is cos kx .ab !Š‹
24
k even
k111
'
œ
œab
2

CHAPTER 11 ADDITIONAL AND ADVANCED EXERCISES
1. converges since and converges by the Limit Comparison Test:
"" "
œ# œ œ(3n ) (3n 2) (3n 2)
2n 1 2† !
_
œn1
lim lim 3
nnÄ_ Ä_
Š‹
Š‹
œ
)Î&
œ
(
)Î&
n
(3n 2)
œœ ˆ‰
3n 2
n
œ
$Î#
$Î#
2. converges by the Integral Test: tan x lim lim '
1
b
'

ab ’“ ’ “
œ"
#
_
dx
x 1 3 3 192
tan x tan b
œ
Î) Î)
&&
&œœ
bbÄ_ Ä_
ab ab 1
œ œŠ‹
11 1
&& &
24 192 192
7
3. diverges by the nth-Term Test since lim a lim ( 1) tanh n lim ( 1) lim ( 1)
nn n bÄ_ Ä_ Ä_Ä_
n
nn n 1e
1eœ œ œ Š‹
œ
_

2n
2n
does not exist 4. converges by the Direct Comparison Test: n! n ln (n!) n ln (n) n†! † ! †
n ln (n!)
ln (n)
log (n!) n , which is the nth-term of a convergent p-series!†!† n
log (n!)
nn
n

"
5. converges by the Direct Comparison Test: a 1 , a , a"# $œœ œ œ œ
12 12 12 23 12
(1)(3)(2) 3 4 (2)(4)(3) 4 5 3 4
œœ
†††
†††
ˆ‰ˆ‰
, a , 1 represents theœœ œáÊ
12 34 23 12 2 12
(3)(5)(4) 5 6 4 5 3 4 (4)(6)(5) (n 1)(n 3)(n 2)
œœœ%
"
___
ˆ‰ˆ‰ˆ‰ !
†††
†††
_
œn1

778. 774 Chapter 11 Infinite Sequences and Series
given series and , which is the nth-term of a convergent p-series
12 12
(n 1)(n 3)(n 2) n___
œ '†
6. converges by the Ratio Test: lim lim 0 1
nnÄ_ Ä_
a
a (n 1)(n 1)
nn1
n*
œœ
œ_
7. diverges by the nth-Term Test since if a L as n , then L L L 1 0 Ln
1L
15ÄÄ_œÊœÊœ
"
_#
# œ„
È
0Á
8. Split the given series into and ; the first subseries is a convergent geometric series and the!!
__
œœn1 n1
"
33
2n
2n 1 2n*
second converges by the Root Test: lim lim 1
nnÄ_ Ä_
É
n
2n
nn2n 1
3999
2 n
œœœ
ÈÈ
""†
9. f(x) cos x with a f 0.5, f , f 0.5, f , f 0.5;œ œ Ê œ œ œ œ œ
11 1 1 1 1
33 3 3 3 3
33
ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
w www
##
ww
ÈÈ Ê_4
cos x x x xœ á
""
## #
#$ÈÈ
33
34 3 1 3
ˆ‰ˆ‰ ˆ‰
11 1
10. f(x) sin x with a 2 f(2 ) 0, f (2 ) 1, f (2 ) 0, f (2 ) 1, f (2 ) 0, f (2 ) 1,œœÊœœœœœœ 11 1 1 1 1 1
w www ww Ê_ Ê_45
f (2 ) 0, f (2 ) 1; sin x (x 2 )
Ê_ Ê_67
11 1œ œœá
(x 2 ) (x 2 ) (x 2 )
3! 5! 7!
œœœ111
& (œ
11. e 1 x with a 0
x
œá œ
xx
!3!

#
12. f(x) ln x with a 1 f(1) 0, f (1) 1, f (1) 1,f (1) 2, f (1) 6;œœÊœœœœœ
w www ww Ê_4
ln x (x 1)χ
(x 1) (x 1) (x 1)
34
œœœ
#
œ &'
13. f(x) cos x with a 22 f(22 ) 1, f (22 ) 0, f (22 ) 1, f (22 ) 0, f (22 ) ,œœÊœœœœœ" 11 1 1 1 1
w www ww Ê_4
f (22 ) 0, f (22 ) 1; cos x 1 (x 22 ) (x 22 ) (x 22 )
Ê_ Ê_56
11 1 1 1œ œ œ á
"" "
#
#%'
4! 6!
14. f(x) tan x with a 1 f(1) , f (1) , f (1) , f (1) ;œœÊœœœœ
œ" ¸ ¸¸¸ """
###
ww1
4
tan x
œ" œœ œ
χ
1
42 4 12
(x 1) (x 1) (x 1)
œ &
15. Yes, the sequence converges: c a b c b 1 lim c ln b
nœ Êœ Ê œ ab ˆ‰ˆ‰
nn 1n
nn
a
bn
n 1n
n
ln 1Î Î
Ä_
_
nÄ_
lim
ˆ‰ˆ‰
a
b
n
ln b ln b ln b since 0 a b. Thus, lim c e b.œ œ œ œœlim
n
ln 0 ln
1 01
n
ln b
Ä_ _
†
_
ˆ‰ ˆ‰ ˆ‰
ˆ‰
aa a
bb b
n
a
b
n
nÄ_
16. 1 1 áœ
237237 2 3 7
10 10 10 10 10 10 10 1010
œ '
!!!
__ _
œœ œn1 n1 n1
3n 2 3n 1 3n 1 1œ œ !! !
__ _
œœ œn0 n0 n0
237
10 10 10
111
3n 1 3n 2 3n 3
2
10
10 10 10
37
10 10***
)))
&&&

ˆ‰
ˆ‰ ˆ‰ ˆ‰
Š‹ Š‹
œœœ
1œœ œ
200 30 7 999 237 412
999 999 999 999 333
_
17. s s snn n
dx dx dx dx dx
1 x 1x 1x 1x 1xœÊœáÊœ!
n1
k0
œ
œ
'''''
k01n10
k1 1 2 n n
)
(
_____
œœœœœ lim s lim tan n tan 0Êœ œ
nnÄ_ Ä_
n ab
œ" œ"
#
1
18. lim lim lim 1
nn nÄ_ Ä_ Ä_
¹¹ ¹ ¹ ¹ ¹ ¸¸
u
u (n2)(2x1) nx 2x1 n(n2) 2x1
(n1)x (n1)(2x1) (n1) xxn1
n
n1 n
n1 n*
* œ
*
œœœ
___ _
__ _ _ _
†† x 2x 1 ; if x 0, x 2x 1 x 2x 1 x 1; if x 0, x 2x 1!†' Ñ †'!†'!Ñ&&†† †'kk k k kk k k kk k k
"
#
x 2x 1 3x 1 x ; if x , x 2x 1 x 2x 1 x 1. Therefore,!& † ' ! Ñ& ! Ñ& †& † ' !& †& & ! †&
""
#3
kk k k

779. Chapter 11 Additional and Advanced Exercises 775
the series converges absolutely for x 1 and x .†& Ñ&
"
3
19. (a) Each A fits into the corresponding upper triangular region, whose vertices are:
n1Ú
(n f(n) f(n 1)), (n 1 f(n 1)) and (n f(n)) along the line whose slope is f(n 1) f(n).â&' 'â' â '&
All the A 's fit into the first upper triangular region whose area is A
n
n1
f(1) f(2) f(1) f(2)
2
n
œœ
#
!†!
á
Û
(b) If A f(x) dx, thenk
f(k 1) f(k)œ
__
# '
k
k1
Ú
A f(x) dx f(x) dx f(x) dx!
n1
k1
œ
œ
k
f(1) f(2) f(2) f(3) f(3) f(n 1) f(n)œ á
_____Ü_œ_
# '' '
12 n1
23 n á
f(k) f(x) dx A f(k) f(x) dx , fromœ Êœ
f(1) f(n) f(1) f(n) f(1) f(2)
k
_ _œ
# ##
!!!
n1 n1 n
k2 k1 k1
œœ
œœœ
''
1 1
n n
part (a). The sequence A is bounded above and increasing, so it converges and the limit inœ Û
!
n1
k1
œ
œ
k
question must exist.
(c) Let L lim f(k) f(x) dx f 1 f n , which exists by part (b). Since f is positive andœ
nÄ_
”•
! abab ab
n
k1œ
'
1
n
"
#
decreasing f n M 0 exists. Thus lim f(k) f(x) dx L f 1 M .lim
nÄ_
"
#
ab a b ”•
! abœ  œ
nÄ_
n
k1œ
'
1
n
20. The number of triangles removed at stage n is 3 ; the side length at stage n is ; the area of a triangle
n1 b
2
œ
n1
at stage n is .
È
3
42
b
ˆ‰
n1
#
(a) b 3 3 3 b b
ÈÈÈÈ È È
3333 3 3
4424 42 4244
bbb 3 3
2
n
##$ ##
ᜠœŠ‹ Š‹ Š‹ !! ˆ‰
œœœ
œ _á
__
œœn0 n0
n
2n (b) a geometric series with sum 3b
Š‹
ˆ‰
œ3
4
3
4
b
1
œ
œ
#
œ
È
(c) No; for instance,the three vertices of the original triangle are not removed. However the total area removed
is 3b which equals the area of the original triangle. Thus the set of points not removed has area 0.
È
#
21. (a) No, the limit does not appear to depend on the value of the constant a
(b) Yes, the limit depends on the value of b
(c) s 1 ln s lim ln sœ Ê œ Ê œŠ‹
cos
n
n ln 1
ˆ‰
Œ œ
ˆ‰
&' Œ œ
Š‹
a
n
n œ
œ
cos
a
n
n
1
cos
a
n
n
aa a
nn n
sin cos
n
n


Ú
nÄ_
œ
á
('
&
œ
&
lim 1 lim s e 0.3678794412; similarly,œœœÊœ¸
nnÄ_ Ä_
aa a
nn n
cos
a
n
n
sin cos
1
ˆ‰ ˆ‰œ
œ

01
10
œ
œ
œ"
lim 1 e
nÄ_
Š‹œ
cos
bn
n
1b
ˆ‰
a
n
œŸ
22. a converges lim a 0; lim lim ! ’“ˆ‰ ˆ‰
_
œn1
nn
1sin a 1sin a
n
1n 1 sin lim a
1sin 0 Êœ œ œ œ
nn nÄ_ Ä_ Ä_
__
####
Î _
_ n n
n
Š‹
n&'
the series converges by the nth-Root TestœÊ
"
#
23. lim 1 lim 1 bx 1 x 5 b
nnÄ_ Ä_
¹¹ ¹ ¹ kk
u
uln(n1)bx bb5
bx ln nn1
n
n1n1
nnÛ
ÛÛ
Ê Ê ÊœÊœ„
_
"" "
†
24. A polynomial has only a finite number of nonzero terms in its Taylor series, but the functions sin x, ln x and
e have infinitely many nonzero terms in their Taylor expansions.
x

780. 776 Chapter 11 Infinite Sequences and Series
25. lim lim
x0 x0ÄÄ
sin (ax) sin x x
xx
ax x x
œœ
œ _Üœœ_Üœ
&& œ
Š‹Š‹
ax x
3! 3!
)) )
lim x is finite if a 2 0 a 2;œá œÊœ
x0Ä
’“ Š‹
a2 a a
x 3!3! 5!5!
œ" " #
œ
&(
lim
x0Ä
sin 2x sin x x 2 7
x3!3!6
œœ "
&
& œ œ
26. lim 1 lim 1 lim 1
x0 x0 x0ÄÄ Ä
cos ax b ba ax
xx2x448
1b
œ "œ
##
œ_œÜœ
œœœ
œœœœ Ê œ Ê á œ
Œ œ
ax ax
4!
&& **
&
Š‹
b 1 and a 2Êœ œ„
27. (a) 1 C 2 1 and converges
u
un nn n
(n 1) 2n
n1
*
œ
œœ œ
œœÊœ
_ ""
!
_
œn1
(b) 1 C 1 1 and diverges
u
un nn n
n1 1 0n
n1
*
œ
œœÊœŸ
_" !
_
œn1
28. 1 1 after long division
u
u (2n 1) 4n 4n 1 n 4n 4n 1 n n
2n(2n 1) 4n 2n 5n
n1
*
œœ œ œ
œ
œœ œ œ
_
œœ_ œ_
_
Š‹ Š‹
–—
63
4
5n
4n 4n 1
&
&
&
(''(
C 1 and f(n) 5 u converges by Raabe's TestÊœ œ œ ŸÊ
35n5
4n 4n 1
4
n
#œ_
œ_
kk !
œ
œ
Š‹
4
nn œ
&
_
œn1
29. (a) a L a a a a L a converges by the Direct Comparison Test!!!
___
œœœn1 n1 n1
nnnn nnœÊ Ÿ œ Ê
##
(b) converges by the Limit Comparison Test: lim lim 1 since a converges and
nnÄ_ Ä_
Š‹
an
1an
(
a1a
nnn
œœ
"
œ !
_
œn1
therefore lim a 0
xÄ_
nœ30. If 0 a 1 then ln (1 a ) ln (1 a ) a a a a , œ œ á áœnnnn n
aa
31a nn
a kk
œ &
nn n
n
#œ
#$
a positive term of a convergent series, by the Limit Comparison Test and Exercise 29b
31. (1 x) 1 x where x 1 (1 x) nx and when x we haveœ Ê œ œ œ
œ" œ" œ ""
œ# !! kk
__
œœn1 n1
nn1
(1 x) dx
d
œ
412 3 4 nœá ሉ ˆ‰ ˆ‰ ˆ‰
"" " "
## # #
#$ œ n1
32. (a) x (n 1)x n(n 1)x n(n 1)x!! ! !
__ _ _
œœ œ œn1 n1 n1 n1
n1nn1nx2xx 2 2x
1 x (1 x) (1 x) (1 x)
_œ
œœ œ œ
œ
œÊ œ Ê œ Ê œ
œœ
œ &&
, x 1Êœœ ! kk
_
œn1
n(n 1)
x(x1)
1
2x_
œ
œ
n
2
x
x
Š‹
œ
&
œ
&
(b) x x x 3x x 1 0 x 1 1 1œ Êœ Ê œÊœ ! Š‹Š‹
_
œn1
n(n )
x(x1) 9 9
2x 57 57_"
œ
$#
"Î$ "Î$
n
œ
& ÈÈ
2.769292, using a CAS or calculator¸
33. The sequence {x } converges to from below so x 0 for each n. By the Alternating Seriesnnn
11
## %œ
Estimation Theorem ( ) with error ( ) , and since the remainder is negative this is an%% % n1 n n
3! 5!
_
"" $&w† kk
overestimate 0 ( ) .!† †%% n1 n
6
_
" $
34. Yes, the series ln (1 a ) converges by the Direct Comparison Test: 1 a 1 a!
_
œn1
' '†''''(nnn
aa
!3!
œ &
nn
#
1 a e ln (1 a ) a!'† ! ' † nnn
a
n

781. Chapter 11 Additional and Advanced Exercises 777
35. (a) 1 x x x 1 2x 3x 4x nx
""
œœ
#$ # $ œ
(1 x) dx 1 x dx
dd n1
œœ œ ᜠᜈ‰ ab !
_
œn1
(b) from part (a) we have n 6!ˆ‰ ˆ‰ ˆ‰ ’“
_
œn1
5
666
n1
1
2
œ
"""
œ
œœ
ˆ‰
5
6
(c) from part (a) we have np q!
_
œn1
n1 qq
(1 p) q q
œ
œ
"
œœœ
œœ
36. (a) p 2 1 and E(x) kp k2 k2 2!! ! ! ! ˆ‰
__ _ _ _
œœ œ œ œk1 k1 k1 k1 k1
kk
kk1k
1 1œœœ œœ œ œ œ
œœœ
œ
"""
##
œ
ˆ‰
ˆ‰ á ‘ˆ‰
Ú
œ
Ú œ
œ
Ú
œ
by Exercise 35(a)
(b) p 1 and E(x) kp k k!! ! ! ! ! ˆ‰ ˆ‰ ˆ‰’“
__ _ _ _ _
œœ œ œ œ œk1 k1 k1 k1 k1 k1
k k
55 5 5
6 656 5 6 6
k k1
1œœ œ œ œœ œ
k1 k1
k k
5
6
5
6
"" "
œ
œˆ‰
ˆ‰
6 œœˆ‰
""
œ
6
1á ‘ˆ‰
5
6
œ
(c) p lim 1 1 and E(x) kp k
k
!! ! ! ! ˆ‰ ˆ‰ Š‹
__ _ __
œœ œ œœk1 k1 k1 k1 k1
k k
k(k1) k k1 k1 k(k1)œ œ œ œ œ œ
Ä_
""" " "
__ _ _
, a divergent series so that E(x) does not existœ!
_
œk1
"
_k1
37. (a) R C e C e C e R lim R
n n n
kt 2kt nkt
Ce 1 e
1e 1 e e 1
Ce Cœ á œ Êœ œœ
Ä_!! !
œœ œ
œ
œœœ
    








kt nkt
kt kt kt
ktˆ‰
(b) R R e 0.36787944 and R 0.58195028;n
e1e e1e
1e 1eœÊœ¸ œ¸
ŸÚ ŸÚ 
ŸÚ ŸÚ1n
ab a bœœ
œœ
""!
œ"
R 0.58197671; R R 0.00002643 0.0001œ¸ ¸ Ê
"
œ
"!
œ
e1 R
RR Ú 
(c) R , 4.7541659; R nn
e1e
1e e1 e1 e1
RR1eœœ¸Ê
Ÿ¸ Ÿ¸
Ÿ¸ ¸ ¸ ¸
Ÿ¸11n
11 11
1nˆ‰œ
œ##œ # œ#œ
"" œ " "
ˆ‰ ˆ‰ˆ‰
1 e e ln ln n 6.93 n 7Ê Ê Ê Ê Ê Êœ
œŸ œŸ "" " "
## # #
n10 n10 nn
10 10 ˆ‰ ˆ‰
38. (a) R Re R C C e t lnœÊœœÊœÊœ
C
e1
kt kt
H
CC
CkC


kt
HH
LL
œ
!!
"
Š‹
(b) t ln e 20 hrs!
"œœ
0.05
(c) Give an initial dose that produces a concentration of 2 mg/ml followed every t ln 69.31 hrs!
"
#œ¸
0.0 0.5
2
ˆ‰
by a dose that raises the concentration by 1.5 mg/ml
(d) t ln 5 ln 6 hrs!
"œœ¸
0.2 0.03 3
0.1 10
ˆ‰ ˆ‰
39. The convergence of a implies that lim a 0. Let N 0 be such that a 1 a
n
!kk kk kk kk
_
œn1
nn nn
Ä_
œ Ê
""
##
2 a for all n N. Now ln 1 a a aÊ œ ០''''(
kk
kk
a
1a 34 3 4
nnnn
aaa a a an
n
nnn n n n
œ##
kk k k kk ab ¹ ¹ ¹¹ ¹¹ ¹¹
œ
a a a a 2 a . Therefore ln (1 a ) converges by the Directᜠkk kk kk kk kk ! nnnn n n
a
1a
#$%
œ
kk
kk
n
n
_
œn1
Comparison Test since a converges.!kk
_
œn1
n
40. converges if p 1 and diverges otherwise by the Integral Test: when p 1 we have!
_
œn3
"
n ln n(ln (ln n))
p œ lim lim ln (ln (ln x)) ; when p 1 we have lim
bb bÄ_ Ä_ Ä_
''
3 3
b b
b
dx dx
x ln x(ln (ln x)) x ln x(ln (ln x))
œœ_Ácd
$
p
lim
,if p1
, if p 1
œœ

_
bÄ_
’“
Ú
(ln (ln x))
1p
(ln (ln 3))
1p
Û
ŸÛp1
p1
œ
œ
b

782. 778 Chapter 11 Infinite Sequences and Series
41. (a) s2n 1
ccc ttttt
1 3 2n 1 1 2 3 2n 1
ctt
_
#_ _
œœ œœ á œ á
Ú ÛÛ 2n 1 2n 1 2n
t1 t t .œá œ "#
""" ""
## #__ __
ˆ‰ˆ‰ ˆ ‰ !
3 2n n1 2n1 k(k1) 2n1
2n
tt t
2n 1 2n 1 kÛÛ
2n
k1œ
(b) c ( 1) convergesef e f ! n
n (1)
nœ Ê
_
œn1
œ
n
(c) c 1 1 1 1 1 1 1 1 1 the series 1 convergesef e f n
34567œ ßßßßßßßßßá Ê á
""""""
#
42. (a) 1 t t t ( 1) t (1 t) 1 t t t ( 1) t t t t t ( 1) tabá œá á
#$ #$ #$% _ nn nn nn 1
1(1)t 1tt t (1)tœ Ê á œ
nn 1 nn (1)t
1t 1t
_#$ œ
__
"
nn 1Û
(b) dt 1 t t ( 1) t dt ln 1 t''
00
xx
x
"
__
# œ
!1t 1t
nn (1) t
œá Ê’“ cdkk
n1n1ÛÛ
t dt ln 1 xœá Ê ’“ kk
tt
2 3 n1 n1
(1)t (1) t
ÛÛÛ
œœ
__
!
nn 1 n 1n 1
x
0
x
'
x R , where R dtœá œ
xx
3n1 n1
(1)x (1) t
n1 n1
œ & ÛÛÛ
#_ _
œœ
__
nn1 n1n1
'
0
x
(c) x 0 and R ( 1) dt R dt t dtœ ÊœŸœ n1 n1
n1 n1 ttx
1t 1t n
__
__
___#
'''
000
xxx
n1 n2Û Û
kk
n1œ
(d) 1 x 0 and R ( 1) dt R dt dt œ Ê œ Ÿ
n1
n1
000
xxxÚ
Ú
'''ttt
1t 1t 1t
n1
n1 n1 n1ÛÛÛ
___
_
kk ¹¹¹¹
dx since 1 t 1 xŸœ  '
0
x
kk k k
kk abkk
tx
1x 1x(n2)
n1 n2Û Û
œœ_
kk kk
(e) From part (d) we have R the given series converges sincekk n1
x
1x(n2)
_
œ_ŸÊ
kk
abkk
n2Û
lim 0 R 0 when x 1. If x 1, by part (c) R 0.
nÄ_
kk kk
abkk
x x
1x(n2) n2 n2
n1 n1
1
n2 n2Û Û
œ_ __
__
œÊ Ä œ Ÿ œ Äkk kk kk
Thus the given series converges to ln 1 x for 1 x 1.ab'&†Î

783. CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
1. The line through the point ( ) parallel to the z-axis#ß $ß !
2. The line through the point ( 1 0 ) parallel to the y-axis? ?!
3. The x-axis
4. The line through the point (1 ) parallel to the z-axisß!ß!
5. The circle x y 4 in the xy-plane
##
œ
6. The circle x y 4 in the plane z = 2
##
œ
7. The circle x z 4 in the xz-plane
##
œ
8. The circle y z 1 in the yz-plane
##
œ
9. The circle y z 1 in the yz-plane
##
œ
10. The circle x z 9 in the plane y 4
##
œ œ
11. The circle x y 16 in the xy-plane
##
œ
12. The circle x z 3 in the xz-plane
##
œ
13. (a) The first quadrant of the xy-plane (b) The fourth quadrant of the xy-plane
14. (a) The slab bounded by the planes x 0 and x 1œœ
(b) The square column bounded by the planes x 0, x 1, y 0, y 1œœœœ
(c) The unit cube in the first octant having one vertex at the origin
15. (a) The solid ball of radius 1 centered at the origin
(b) The exterior of the sphere of radius 1 centered at the origin
16. (a) The circumference and interior of the circle x y 1 in the xy-plane
##
œ
(b) The circumference and interior of the circle x y 1 in the plane z 3
##
œ œ
(c) A solid cylindrical column of radius 1 whose axis is the z-axis
17. (a) The closed upper hemisphere of radius 1 centered at the origin
(b) The solid upper hemisphere of radius 1 centered at the origin
18. (a) The line y x in the xy-planeœ
(b) The plane y x consisting of all points of the form (x x z)œßß

784. 780 Chapter 12 Vectors and the Geometry of Space
19. (a) x 3 (b) y 1 (c) z 2¢¢ ¢ 
20. (a) x 3 (b) y 1 (c) z 2¢¢ ¢
21. (a) z 1 (b) x 3 (c) y 1¢¢¢ 
22. (a) x y 4, z 0 (b) y z 4, x 0 (c) x z 4, y 0
## ## ##
¡¢ ¢ ¡¢ ¢ ¡¢ ¢
23. (a) x (y 2) 4, z 0 (b) (y 2) z 4, x 0 (c) x z 4, y 2
# # ## ##
¡  ¢ ¢   ¡¢ ¢ ¡¢ ¢
24. (a) (x 3) (y 4) 1, z 1 (b) (y 4) (z 1) 1, x 3 ¡¡ ¢ ¢  ¡ ¢ ¢ 
## ##
(c) (x 3) (z 1) 1, y 4¡¡ ¢ ¢
##
25. (a) y 3, z 1 (b) x 1, z 1 (c) x 1, y 3¢¢  ¢¢  ¢¢
26. x y z x (y 2) z x y z x (y 2) z y y 4y 4 y 1ÈÈ
### # ## #### ## ##
¡¡¢ ¡  ¡ £ ¡¡¢¡  ¡ £ ¢ ¡£¢
27. x y z 25, z 3 x y 16 in the plane z 3
### ##
¡¡¢ ¢£¡¢ ¢
28. x y (z 1) 4 and x y (z 1) 4 x y (z 1) x y (z 1) z 0, x y 3
## # ## # ## # ## # ##
¡¡  ¢ ¡¡¡ ¢£ ¡¡  ¢¡¡¡ £¢ ¡¢
29. 0 z 1 30. 0 x 2, 0 y 2, 0 z 2ŸŸ ŸŸ ŸŸ ŸŸ
31. z 0 32. z 1 x y¸¢   È
##
33. (a) (x 1) (y 1) (z 1) 1 (b) (x 1) (y 1) (z 1) 1 ¡ ¡ ¹  ¡ ¡ Ê
### ###
34. 1 x y z 4¸¡¡¸
###
35. P P 3 1 3 1 0 1 9 3kk abababÉ È "#
###¢   ¡  ¡  ¢ ¢
36. P P 2 1 5 1 0 5 50 5 2kk abababÉ È È "#
###¢ ¡ ¡  ¡  ¢ ¢
37. P P 4 1 2 4 7 5 49 7kk aba babÉ È "#
###¢ ¡  ¡ ¢ ¢
38. P P 2 3 3 4 4 5 3kk abababÉ È "#
###¢   ¡  ¡  ¢
39. P P 2 0 2 0 2 0 3 4 2 3kkabababÉ ÈÈ "#
###œœ†œ
40. P P 0 5 0 3 0 2 38kk abababÉ È "#
###¢   ¡  ¡¡ ¢
41. center ( 2 0 2), radius 2 2 42. center , radius   ? ?   ?  ? 
È ˆ‰
"""
### #
È
21
43. center 2 2 2 , radius 2 44. center , radius Š‹
ÈÈ È È ˆ‰?? !??
""
33 3
29
È

785. Section 12.2 Vectors 781
45. (x 1) (y 2) ( 3) 14 46. x (y 1) (z 5) 4 ¡ ¡‘ ¢ ¡¡¡ ¢
### ###
47. (x 2) y z 3 48. x (y 7) z 49¡ ¡¡¢ ¡¡ ¡¢
### # ##
49. x y z 4x 4z 0 x 4x 4 y z 4z 4 4 4
### # # #
¡¡¡ ¢£ ¡¡¡¡  ¡¢¡ abab
(x 2) (y 0) (z 2) 8 the center is at ( 2 0 2) and the radius is 8£¡¡ ¡  ¢ £  ??
###
#
Š‹
ÈÈ
50. x y z 6y 8z 0 x y 6y 9 z 8z 16 9 16 (x 0) (y 3) (z 4) 5
### # # # # # # #
¡¡ ¡¢£ ¡  ¡¡ ¡¡ ¢¡ £   ¡  ¡¡ ¢aba b
the center is at (0 3 4) and the radius is 5$??51. 2x 2y 2z x y z 9 x x y y z z
### # # # """
####
¡ ¡ ¡¡¡¢ £ ¡ ¡ ¡ ¡ ¡ ¢
9
xx yy zz x y z£ ¡¡¡¡¡¡¡¡¢¡£¡¡¡¡¡¢ˆ‰ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰ Š‹
###"" "" "" " " "
####
###
#
16 16 16 16 4 4 4 4
93 53
È
the center is at and the radius is £   ?  ?  ˆ‰
"""
444 4
53
È
52. 3x 3y 3z 2y 2z 9 x y y z z 3 x y y z z 3
### ## # ## # ""
¡ ¡ ¡ ¢£¡¡¡ ¢£¡ ¡¡¡ ¡¢¡
22 2 2 2
33 39 399
ˆ‰ˆ‰
(x 0) y z the center is at 0 and the radius is £ ¡¡ ¡  ¢ £ ? ?
# "" ""
##
#ˆ‰ˆ‰ ˆ ‰ Š‹
333 33 3
29 29
ÈÈ
53. (a) the distance between (x y z) and (x 0 0) is y z?? ?? ¡ È
##
(b) the distance between (x y z) and (0 y 0) is x z?? ?? ¡
È
##
(c) the distance between (x y z) and (0 0 z) is x y?? ?? ¡ È
##
54. (a) the distance between (x y z) and (x y 0) is zßß ßß
(b) the distance between (x y z) and (0 y z) is xßß ßß
(c) the distance between (x y z) and (x 0 z) is yßß ßß
55. AB 1 1 1 2 3 1 4 9 4 17kk a b a b a bÉ ab È È
¢  ¡  ¡ ¢¡¡¢
###
BC 31 4 1 53 4254 33kk aba babÉ ab È È
¢ ¡  ¡ ¢¡¡¢
###
CA 13 24 15 16416 366kk a bababÉ È È
¢  ¡ ¡ ¢ ¡¡¢ ¢
###
Thus the perimeter of triangle ABC is 17 33 6.
ÈÈ
¡¡
56. PA 2 3 1 1 3 2 1 4 1 6kk aba babÉ È È
¢ ¡  ¡ ¢¡¡¢
###
PB 43 31 12 141 6kk abababÉ È È
¢   ¡  ¡  ¢ ¡¡¢
###
Thus P is equidistant from A and B.
12.2 VECTORS
1. (a) 3 3 , 3 2 9, 6 2. (a) 2 2 , 2 5 4, 10 ¡ ¡   ¡ ¡abab ab ab ¢      ¢ 
(b) 9 6 117 3 13 (b) 4 10 116 2 29ÉÉ ab a b
ÈÈ ÈÈ
2 2
2 2
¡  ¢ ¢ ¡  ¢ ¢
3. (a) 3 2 , 2 5 1, 3 4. (a) 3 2 , 2 5 5, 7 ¡ ¡  ¡ ¡ab ab¡   ¡ ¢       ¢  
(b) 1 3 10 (b) 5 7 74
È ÈÈ É ab
22 2
2
¡¢ ¡  ¢

786. 782 Chapter 12 Vectors and the Geometry of Space
5. (a) 2 2 3 , 2 2 6, 4 6. (a) 2 2 3 , 2 2 6, 4uu¢ ¢   ¢   ¢   ¡ ¡   ¡ ¡ab a b ab a b
3 3 2 , 3 5 6, 15 5 5 2 , 5 5 10, 25vv¢  ¢  ¢  ¢  ¡ ¡  ¡ ¡abab abab
2 3 6 4 , 4 15 12, 19 2 5 6 10 , 4 25 16, 29uv uv ¢     ¢    ¡¢ ¡  ¡ ¢   ¡ ¡   ¡ ¡ab a b
(b) 12 19 505 (b) 16 29 1097ÉÉ ab ab
ÈÈ
2 2
22
¡  ¢   ¡ ¢
7. (a) 3 , 2 , 8. (a) 3 , 2 ,
3 3 3 9 6 5 5 5 15 10
555 55 13 1313 1313
uu¢  ¢    ¢    ¢ ¢£¢£ ¢ £¢£ab a b ab a b
2, 5 , 4 2, 5 ,
4 4 4 8 12 12 12 24 60
5 5 5 5 13 13 13 13 13
vv¢   ¢  ¢   ¢ ¢£¢£ ¢ £¢£abab ab ab
, 4 , , 3,
34 9 86 114 5 12 15 241060 70
55 5 55 55 1313 13 131313 13
uv u v¡¢¡  ¡¢   ¡ ¢ ¡  ¡ ¢ ¢£¢£ ¢ £¢£ˆ‰ ˆ ‰
(b) (b) 3ÉɈ‰ ˆ‰ ˆ‰ ab
114 70
55 5 13 13
22 297 64212
¡¢  ¡¢
È È
"
9. 2 1, 1 3 1, 4 10. 0, 0 1, 1  ¡ ¡  ¡ ¢£    ¢       ¢ 
24 3 ¡
##
¡" ab
11. 0 2, 0 3 2, 3 ¡ ¡  ¢  
12. AB 2 1, 0 1 1, 1 , CD 2 1 , 2 3 1, 1 , AB CD 0, 0
ÄÄ ÄÄ
¢     ¢ ¢     ¢   ¡ ¢  ¡ ¡   ¡  ¡  ¡ab ab
13. cos , sin , 14. cos , sin , ¢£¢£ ¢ £¢£ ˆ‰ ˆ‰
22 33
33 2 4 4
3
22
11 11
¢      ¢   
" ""
#
È
ÈÈ
15. This is the unit vector which makes an angle of 120 90 210 with the positive x-axis;
‰‰ ‰
¡¢
cos 210 , sin 210 ,  ¡ ¢£
‰‰ "
#
¢   
È
3
2
16. cos 135 , sin 135 ,  ¡ ¢£
‰‰ ""
¢ 
ÈÈ
22
17. P P 2 5 9 7 2 1 3 2"#
Ä
¢   ¡   ¡    ¢  ¡  ababa b abij kijk
18. P P 3 1 0 2 5 0 4 2 5
"#
Ä
¢   ¡   ¡   ¢  ¡a bababijkijk  
19. AB 10 7 8 8 1 1 3 16
Ä
¢     ¡    ¡   ¢  ¡abababab abijkij
20. AB 1 1 4 0 5 3 2 4 2
Ä
¢   ¡   ¡   ¢  ¡ ¡a bababijkijk
21. 5 5 1, 1, 1 2, 0, 3 5, 5, 5 2, 0, 3 5 2, 5 0, 5 3 3, 5, 8 3 5 8uv ijk ¢    ¢   ¢    ¢ ¢¡  ¡ ¡ ¡ ¡  ¡ ¡
22. 2 3 2 1, 0, 2 3 1, 1, 1 2, 0, 4 3, 3, 3 5, 3, 1 5 3 ¡ ¢   ¡ ¢  ¡ ¢  ¢¡ uv ijk  ¡ ¡ ¡ ¡ ¡

787. Section 12.2 Vectors 783
23. The vector is horizontal and 1 in. long. The vectors and are in. long. is vertical and makes a 45 angle withvuwwu
" ‰1
16
the horizontal. All vectors must be drawn to scale.
(a) (b)
(c) (d)
24. The angle between the vectors is 120 and vector is horizontal. They are all 1 in. long. Draw to scale.
‰
u
(a) (b)
(c) (d)
25. length 2 2 2 1 ( 2) 3, the direction is 2 2 3¢¡  ¢ ¡¡ ¢ ¡  £¡ ¢ ¡ kk È ˆ‰ij k i j k ij k i j k
## # ""22 22
333 333
26. length 9 2 6 81 4 36 11, the direction is 9 2 6¢ ¡ ¢ ¡¡¢   ¡ £ ¡kk Èijk i j k ijk
926
11 11 11
11¢ ¡ˆ‰
926
11 11 11
ijk
27. length 5 25 5, the direction is 5 5( )œœ œ Êœkk
È
kkkk
28. length 1, the direction is 1¢¡ ¢ ¡¢ ¡ £¡¢ ¡¸¸ ˆ‰ É
34 916 34 34 34
55 2525 55 55 55
ik ik ik ik
29. length 3 , the direction is ¢   ¢ ¢   ¹¹Š‹ Ê É
11 11
666 6 333
ÈÈÈ È ÈÈÈ
ijk ijk
""" "
#
#
£  ¢   
11 11
666 333
ÈÈÈ ÈÈÈ
ijk ijk
"" "
#
ÉŠ‹

788. 784 Chapter 12 Vectors and the Geometry of Space
30. length 3 1, the direction is ¢¡¡ ¢ ¢ ¡¡¹¹Š‹ Ê
11 11
333 3 333
ÈÈÈ È ÈÈÈ
ijk ijk
"" "
#
1£¡¡¢ ¡¡
11 11
333 333
ÈÈÈ ÈÈÈ
ijk ijk
""
Š‹
31. (a) 2 (b) 3 (c) (d) 6 2 3ikjkijk  ¡ ¡
È
32
10 5
32. (a) 7 (b) (c) (d)      ¡ jikijkijk
32 42
55 43
11 a a a
2 36
ÈÈ
ÈÈÈ
33. 12 5 169 13; (12 5 ) the desired vector is (12 5 )kk
È È
v v ik ik¢¡¢ ¢ ¢¢  £  
## ""v
vkk 13 13 13
7
34. ; the desired vector is 3kkÉ Š‹vijkijk¢¡¡¢ ¢   £     
""" """
#444
3 111
333 333
È
kk ÈÈÈ ÈÈÈ
v
v
333¢  ¡ ¡
ÈÈÈ
ijk
35. (a) 3 4 5 5 2 the direction is ijk i j k i j k¡  ¢ ¡   £ ¡  
È
Š‹
34 34
52 52 2 52 52 2
ÈÈÈ ÈÈÈ
""
(b) the midpoint is 3ˆ‰
"
##
ßß
5
36. (a) 3 6 2 7 the direction is ijk i j k i j k ¡ ¢   ¡ £   ¡ˆ‰
362 362
777 777
(b) the midpoint is 1 6ˆ‰
5
#
ßß
37. (a) 3 the direction is    ¢     £    ijk i j k i j k
È
Š‹
111 111
333 333
ÈÈÈ ÈÈÈ
(b) the midpoint is ˆ‰
579
###
ßß

38. (a) 2 2 2 2 3 the direction is ijk i j k i j k   ¢     £    
È
Š‹
111 111
333 333
ÈÈÈ ÈÈÈ
(b) the midpoint is ( 1)"ß "ß
39. AB (5 a) (1 b) (3 c) 4 2 5 a 1, 1 b 4, and 3 c 2 a 4, b 3, and
Ä
¢  ¡  ¡  ¢¡  £ ¢  ¢  ¢ £¢ ¢ ijkijk
c 5 A is the point (4 3 5)‰$ ??
40. AB (a 2) (b 3) (c 6) 7 3 8 a 2 7, b 3 3, and c 6 8 a 9, b 0,
Ä
¢¡ ¡¡ ¡  ¢ ¡¡ £¡¢  ¡¢  ¢£¢  ¢ijkijk
and c 14 B is the point ( 9 0 14)¢£  ??
41. 2 a( ) b( ) (a b) (a b) a b 2 and a b 1 2a 3 a andij ij ij i j¡¢ ¡¡   ¢ ¡ ¡   £ ¡¢  ¢ £ ¢ £ ¢
3
#
baœ"œ
"
#
42. 2 a(2 3 ) b( ) (2a b) (3a b) 2a b 1 and 3a b 2 a 3 andij ij ij i j  ¢ ¡ ¡ ¡ ¢ ¡ ¡ ¡ £ ¡¢ ¡¢ £ ¢ 
b 1 a 7 a(2 3 ) 6 9 and b( ) 7¢ #¢ £ ¢ ¡ ¢   ¢ ¡ ¢uijijuiji7j
"# ¡
43. If x is the magnitude of the x-component, then cos 30° x F cos 30° (10) 5 3 lbkk kk kk Š‹
È
œÊœœœ
kk
kk
Èx
F
3
#
53;ÊœFi x
È
if y is the magnitude of the y-component, then sin 30° y F sin 30 °(10) 5 lb 5 .kk kk kk ˆ‰œÊœ œ œ Êœ
kk
kk
y
F
1
y
#
Fj

789. Section 12.2 Vectors 785
44. If x is the magnitude of the x-component, then cos 45° x F cos 45 °(12) 6 2 lbkk kk kk Š‹
È
œÊœœœ
kk
kk
Èx
F
2
#
6 2 (he negative sign is indicated by the diagram)£¢ Fi x
È
if y is the magnitude of the y-component, then sin 45° y F sin 45 °(12) 6 2 lbkk kk kk Š‹
È
œÊœ œ œ
kk
kk
Èy
F
2
#
6 2 (the negative sign is indicated by the diagram)£¢ Fj y
È
45. 25 west of north is 90 25 115 north of east. 800 cos 155 , sin 115 338.095, 725.046
‰‰‰‰ ‰‰
¡¢ !   ¡  ¡
46. 10 east of south is 270 10 280 ”north” of east. 600 cos 280 , sin 280 104.189, 590.885
‰‰‰‰ ‰‰
¡¢ !    ¡  ¡
47. (a) The tree is located at the tip of the vector OP (5 cos 60°) (5 sin 60°) P
Ä
¢¡¢¡£¢?ijij
55 53 53
## ##
ÈÈ
Š‹
(b) The telephone pole is located at the point Q, which is the tip of the vector OP PQ
ÄÄ
¡
(10 cos 315 °) (10 sin 315°)¢¡ ¡ ¡ ¢¡ ¡  Š‹ Š‹Š ‹
55
2
53 5310 2 10 2
ij i j i j
ÈÈÈÈ
#####
Q Êœ ߊ‹
5102 53 102 
##
¡
ÈÈ È
48. Let t and s . Choose T on OP so that TQ isœœ
qp
pq pq
1
  
parallel to OP , so that TP Q is similar to OP P . Then21 12 ˜˜
t OT t OP so that T t x , t y , t z .
kk
kk
OT
OP
1111
1
œÊ œ œ
ÄÄ
ab
Also, s TQ s OP s x , y , z .
kk
kk
TQ
OP
2222
2
œÊ œ œ
ÄÄ
 ¡
Letting Q x, y, z , we have thatœab
TQ x tx, y ty, z tz sx, y, z
Ä
¢      ¢ ¡ ¡
111222
Thus x t x s x , y t y s y , z t z s z .¢¡ ¢¡ ¢¡ 12 12 12
(Note that if Q is the midpoint, then 1 and t s
p
q
œœœ
"
#

so that x x x , y , z so that this result agress with the midpoint formula.)¢¡¢ ¢ ¢
""
##
    
12
xx zz
222
yy 12 12 12
49. (a) the midpoint of AB is M 0 and CM 1 1 ( 3) 3ˆ ‰ ˆ‰ˆ‰
55 5 5 3 3
## # # # #
?? ¢  ¡  ¡!  ¢¡ 
Ä
ijkijk
(b) the desired vector is CM 3 2ˆ‰ ˆ ‰
2233
33
Ä
¢ ¡   ¢¡ 
##
ijkijk
(c) the vector whose sum is the vector from the origin to C and the result of part (b) will terminate
at the center of mass the terminal point of ( 3 ) ( 2 ) 2 2 is the point£ ¡¡ ¡¡  ¢¡¡ ij k ij k i jk
(2 2 1), which is the location of the center of massßß
50. The midpoint of AB is M 0 and CM 1 (0 2) 1 2???? † D? ? ˆ‰ ˆ‰
35 2 23 5 25 7
33 3## # # # #
?? ¢ ¡ ¡   ¡ ¡ ¢   ¡
Ä
ij kijk
. The terminal point of OC ( 2 )¢ ¡  ¡ ¡ ¢  ¡ ¡ ¡ 
Ä
547 547 547
333 333 333
ijk ijk ijk ijk ˆ‰ˆ‰
is the point which is the location of the intersection of the medians.¢¡¡ ??
2 2 4 224
3 3 3 333
ijk ˆ‰
51. Without loss of generality we identify the vertices of the quadrilateral such that A(0 0 0), B(x 0 0),ßß ßß
b
C(x y 0) and D(x y z ) the midpoint of AB is M 0 0 , the midpoint of BC iscc ddd AB
xßß ßß Ê ßß ˆ‰
b
#
M 0 , the midpoint of CD is M and the midpoint of AD isBC CD
xx xx zyyy
ˆ‰ ˆ ‰
b ddccc c d  
## # ##
 
ßß ß ß
M the midpoint of M M is , which is the same as the midpointAD AB CD
xz zyyy
44
ˆ‰
Œ
dd dd dc
### #
 
ßß Ê ß
xxx
bdc
  
¡
 
of M M , .AD BC
yy
44
z œßŒ
xx x
bdc¡
  
 
#
 
c d d

790. 786 Chapter 12 Vectors and the Geometry of Space
52. Let V , V , V , , V be the vertices of a regular n-sided polygon and denote the vector from the center to"#$ á ni v
V for i 1, 2, 3, , n. If and the polygon is rotated through an angle of whereii
i(2 )
n œá œ Sv!
n
i1œ
1
i 1, 2, 3, , n, then would remain the same. Since the vector does not change with these rotations we concludeχ SS
that .S0œ
53. Without loss of generality we can coordinatize the vertices of the triangle such that A(0 0), B(b 0) andßß C(x y ) a is located at , b is at and c is at 0 . Therefore, Aa ,cc
bx x xyy y bb?£ ? ? ? ¢¡ ¡
Ä
ˆ‰ˆ‰ ˆ‰ ˆ‰ˆ‰
 
## ## # ## #cc ccc c
ij
Bb b , and Cc x ( y ) Aa Bb Cc .
ÄÄ
¢ ¡ ¢  ¡  £ ¡¡¢
ÄÄÄ
ˆ‰ˆ‰ ˆ‰
x y b
ccc c
## #
ij i j 0
54. Let be any unit vector in the plane. If is positioned so that its initial point is at the origin and terminal point is auu t x, y ,ab
then makes an angle with , measured in the counter-clockwise direction. Since 1, we have that x cos andui u ) )kkœœ
y sin . Thus cos sin . Since was assumed to be any unit vector in the plane, this holds for unit¢¢¡)))uiju every
vector in the plane.
12.3 THE DOT PRODUCT
NOTE: In Exercises 1-8 below we calculate proj as the vector , so the scalar multiplier of is
vuv v Š‹
kk
kk
u
v
cos )
the number in column 5 divided by the number in column 2.
cos cos proj vu v u u u† kk kk kk ))
v
1. 25 5 5 1 5 2 4 5     ¡  ij k
È
2. 3 1 13 3 3
334
13 5 5
ˆ‰ik¡
3. 25 15 5 (10 11 2 )
""
339
5
ijk¡ 
4. 13 15 3 (2 10 11 )
13 13 13
45 15 225
ijk¡ 
5. 2 34 3 (5 3 )
ÈÈ
22
334 34 17ÈÈ È
"
jk 
6. 3 2 2 3 ( )
È ÈÈ
   ¡
ÈÈÈÈÈÈ
ÈÈ
32 32 32
32 2
¡¡¡
#
ij
7. 10 17 26 21 ( 5 )¡  ¡
ÈÈ È
10 17 10 17 10 17
546 26 26
   
ÈÈÈ
ÈÈÈ
ij
8. ,
""""""
#6665 5
30 30
30 3
ÈÈ
ÈÈÈ
¢£
9. cos cos cos cos 0.75 rad)œœ œ œ¸
¡" ¡" ¡" ¡"   ¡
     ¡
Š‹ Š ‹ Š ‹ Š‹
uv
uv
†
kkkk ÈÈÈÈ È
(2)(1) (1)(2) (0)( 1)
210 12(1)
44
5 6 30
        
10. cos cos cos cos 0.84 rad)œœ œ œ¸
¡" ¡" ¡" ¡"  ¡  
 ¡      
Š‹ Š ‹ Š ‹ ˆ‰
uv
uv
†
kkkk È ÈÈÈ 3
(2)(3) ( 2)(0) (1)(4)
2(2)1 304
10 2
9 25
        
11. cos cos cos)œœ œ
¡" ¡" ¡"
 ¡   ¡
 ¡      ¡
¡
Š‹ Š ‹
ÎÑ
ÏÒ
uv
uv
†
kkkk
Š‹Š‹
ÈÈ
ÊŠ ‹ ÊŠ ‹
ÈÈ
ÈÈ
3 3 (7)(1) (0)(2)
3(7)0 3(1)(2)
37
52 8
  
      

791. Section 12.3 The Dot Product 787
cos 1.77 radœ¸
¡"¡
Š‹
1
26
È
12. cos cos cos)œœ œ
¡" ¡" ¡"
¡   ¡
  ¡ ¡  
¡
Š‹ Š ‹
ÎÑ
ÏÒ
uv
uv
†
kkkk
Š‹ Š ‹
ÈÈ
ÊŠ‹Š‹
ÈÈ È
ÈÈ
(1)( 1) 2 (1) 2 (1)
(1) 2 2 ( 1) (1) (1)
1
5 3
    
  
cos 1.83 radœ¸
¡"¡
Š‹
1
15
È
13. AB 3, 1 , BC 1, 3 , and AC 2, 2 . BA 3, 1 , CB 1, 3 , CA 2, 2 .
ÄÄ Ä Ä ÄÄ
¢ ¢    ¢   ¢    ¢ ¢  ¡  ¡  ¡  ¡  ¡  ¡
AB BA 10, BC CB 10, AC CA 2 2,¹¹¹¹ ¹¹¹¹ ¹¹¹¹
ÈÈ È
ÄÄ ÄÄ ÄÄ
œœ œœ œœ
Angle at A cos cos cos 63.435œœ œ¸
¡" ¡" ¡" ¢
ÄÄ
ÄÄ
 ¡
Œ ‚ 
Š‹
AB AC
AB AC
32 1 2
10 2 2
1
5
†
¹¹¹¹ È
ab a b
Š‹Š‹
È È
Angle at B cos cos cos 53.130 , andœœ œ¸
¡" ¡" ¡" ¢
ÄÄ
ÄÄ
¡¡ ¡¡
Œ ‚ 
ˆ‰
BC BA
BC BA
13 31
10 10
3
5
†
¹¹¹¹ È È
abababab
Š‹Š‹
Angle at C cos cos cos 63.435œœ œ¸
¡" ¡" ¡" ¢
ÄÄ
ÄÄ
¡ 
Œ ‚ 
Š‹
CB CA
CB CA
12 32
10 2 2
1
5
†
¹¹¹¹ È
ab ab
Š‹Š‹
È È
14. AC 2, 4 and BD 4, 2 . AC BD 2 4 4 2 0, so the angle measures are all 90 .
ÄÄ
œœ†œœ
ÄÄ
 ¡   ¡ ab a b
‰
15. (a) cos , cos , cos and!"#œœ œœ œ œ
iv kv
iv v jv v kv v
jv†† †
kkkk kk kkkk kk kkkk kk
ab c
cos cos cos 1
###
###
  
!"#¡¡¢¡¡¢ ¢¢ Š‹ Š‹ Š‹
abcabc


kk kk kk kkkk kkkk
kkkk
v v v vv vv
vv
   
(b) 1 cos a, cos b and cos c are the direction cosines of kkv vœÊ œœ œœ œœ!" #
ab c
kk kk kkvv v
16. 10 2 is parallel to the pipe in the north direction and 10 is parallel to the pipe in the eastuik vjk¢¡ ¢¡
direction. The angle between the two pipes is cos cos 1.55 rad 88.88 °.)œœ ¸¸
¡" ¡"
Š‹ Š ‹
uv
uv
†
kkkk ÈÈ
2
104 101
17. ( ) (3 4 ) ( ) 4 , whereu v u v ij j k ij i j i j k¢ ¡  ¢¡¡¡ ¡¢¡¡ ¡¡??? ? † D? ?? ?
vu vu
vv vv
††
††
3 3 33 33
# # ## ##
3 and 2vu vv††œœ
18. ( ) ( ) ( ) ,uvuvvuvijjkijij ijk¢ ¡  ¢   ¡¡¡ ¡¢¡¡ ¡¡??? ? † D? ?? ?
vu vu
vv vv
††
††
""
# # # # ## ##
¡£ˆ‰
1 1 11 11
where 1 and 2vu vv††œœ
19. ( 2 ) (8 4 12 )uvuvijkijkijk¢¡ ¢¡ ¡¡  ¡ ??? ? † D ˆ‰
vu vu
vv vv
††
††
14 14 28 14
3333
 
, where 28 and 6¢¡  ¡   ¢ ¢ˆ‰ˆ‰
14 28 14 10 16 22
333 333
i j k i j k vu vv ††
20. ( ) ( ) ( ( ), where 1 and 1; yesuvuvAijkAi)jk vu vv¢¡ ¢¡¡ ¢¡¡ ¢ ¢??? ? † D ˆ‰
vu vu
vv vv
††
††
""
11
¡¢¢
21. The sum of two vectors of equal length is orthogonal to their difference, as we can see from the equationalways
()() 0v v v v vv vv vv vv v v
" # " # "" #" "# ## " #
##¡  ¢¡  ¢ ¢† †††† kk kk
22. CA CB ( ( )) ( ) 0 because since both equal
ÄÄ
¢ ¡  ¡¢ ¡ ¢ ¢ ¢††††††v u v u vv vu uv uu v u u v kk kk kk kk
##
the radius of the circle. Therefore, CA and CB are orthogonal.
ÄÄ

792. 788 Chapter 12 Vectors and the Geometry of Space
23. Let and be the sides of a rhombus the diagonals are and uv duvd uv £¢¡¢ ¡ "#
( ) ( ) 0 because , since a rhombus£ ¢¡ ¡¢ ¡ ¡¢ ¢ ¢dd uv uv uuuvvuvv v u u v"#
##†† †††† kk kk kk kk
has equal sides.
24. Suppose the diagonals of a rectangle are perpendicular, and let and be the sides of a rectangle the diagonals areuv Ê
and . Since the diagonals are perpendicular we have 0duv d uv dd
"# "#¢¡ ¢ ¡ ¢ †
( ) ( ) 0 0 0Ï¡ ¡¢ ¡ ¡¢Ï ¢Ï¡  ¢uv uv uuuvvuvv v u v u v u††††† kkkk abab kk kk kk kk
##
0 which is not possible, or 0 which is equivalent to the rectangle is a square.Ï¡¢  ¢ ¢£ab ab kkkkkk kk kk kkvu vu vu
25. Clearly the diagonals of a rectangle are equal in length. What is not as obvious is the statement that equal
diagonals happen only in a rectangle. We show this is true by letting the adjacent sides of a parallelogram be
the vectors (v v ) and (u u ). The equal diagonals of the parallelogram are
"# "#ij ij¡¡
(v v ) (u u ) and (v v ) (u u ). Hence (v v ) (u u )d ij ij d ij ij d d ij ij
" "# "# # "# "# " # "# "#¢¡¡¡ ¢¡ ¡ ¢¢ ¡¡¡ kk kk k k
(v v ) (u u ) (v u ) (v u ) (v u ) (v u )¢ ¡   ¡ £ ¡ ¡¡ ¢   ¡ k kkkkk
" # " # "" ## "" ##ij ij i j i j
(v u) (v u) (v u) (v u) v 2vu u v 2vu u£ ¡ ¡¡ ¢   ¡  £¡ ¡¡¡ ¡ÈÈ "" ## "" ## "" ##
## ## #### 11   
v 2vu u v 2vu u 2(vu vu) 2(vu vu) vu vu 0¢  ¡¡  ¡ £ ¡ ¢  ¡ £ ¡ ¢
####
"" ## "" ## "" ## "" ##
11   
(v v ) (u u ) 0 the vectors (v v ) and (u u ) are perpendicular and the parallelogram£¡ ¡¢£ ¡ ¡"# "# "# "#ij ij ij ij†
must be a rectangle.
26. If and is the indicated diagonal, then ( )kk kk kk kku v uv uvuuuvu u vuuv v¢ ¡ ¡ ¢¡¢¡¢¡ ††† ††
##
( ) the angle cos between the diagonal and and the angle¢¡¢¡ £uv vv u v v u†† †
¡"  
 
Š‹
()

uvu
uvu
†
kkkk
cos between the diagonal and are equal because the inverse cosine function is one-to-one.
¡"  
 
Š‹
()

uvv
uvv
†
kkkk
v
Therefore, the diagonal bisects the angle between and .uv
27. horizontal component: 1200 cos 8 1188 ft/s; vertical component: 1200 sin 8 167 ft/sab ab
‰‰
¸¸
28. cos 33 15 2.5 lb, so . Then cos 33 , sin 33 2.205, 1.432kk a b kk  ¡ ¡www
‰‰ ‰ ‰
 ¢ ¢ ¢ !
2.5 lb 2.5 lb
cos 18 cos 18
¡¡
29. (a) Since cos 1, we have cos (1) .k k k k kkk k kkkk kkkk))ŸŸœ uv uv uv uv†œkk
(b) We have equality precisely when cos 1 or when one or both of and is . In the case of nonzerokk)œ uv0
vectors, we have equality when 0 or , i.e., when the vectors are parallel.)1œ
30. (x y ) x y cos 0 when . Thisijv ijv¡¢¡ ¸ ¸¸†kkkk ))1
1
#
means (x y) has to be a point whose position vector makesß
an angle with that is a right angle or bigger.v

31. (a b ) a b a b( ) a(1) b(0) avu uuuuuuu u uu†††† †
" " # " "" #" " #"
# #¢¡ ¢ ¡ ¢ ¡ ¢ ¡¢ kk
32. No, need not equal . For example, 2 but ( ) 1 0 1 andv v iji j iij iiij
"# ¡Ò¡ ¡ ¢ ¡ ¢¡¢†††
(2) 2 1201.ii j ii ij††††¡¢ ¡ ¢¡ ¢

793. Section 12.3 The Dot Product 789
33. P(x y ) P x , x and Q(x y ) Q x x are any two points P and Q on the line with b 0"" " " ## # #?¢   ?¢ ?  Òˆ‰ ˆ‰
ca ca
bb bb
PQ (x x ) (x x ) PQ (x x ) (x x ) (a b ) a(x x ) b (x x )£¢ ¡  £ ¢ ¡   ¡¢ ¡  
ÄÄ #" "# #" "# #" "#ijvijij
aa a
bb b
†††D ??
0 is perpendicular to PQ for b 0. If b 0, then a is perpendicular to the vertical line ax c.œÊ Á œ œ œ
Ä
vvi
Alternatively, the slope of is and the slope of the line ax by c is , so the slopes are negativev
ba
ab
¡¢  
reciprocals the vector and the line are perpendicular.Ê v
34. The slope of is and the slope of bx ay c is , provided that a 0. If a 0, then b is parallel tovvj
bb
aa
 ¢ Ò ¢ ¢
the vertical line bx c. In either case, the vector is parallel to the line ax by c.¢ ¢ v
35. 2 is perpendicular to the line x 2y c;vi j¢¡ ¡ ¢
P(2 1) on the line 2 2 c x 2y 4?$‰$‰

36. 2 is perpendicular to the line 2x y c;vij¢        ¢
P( 1 2) on the line ( 2)( 1) 2 c ? £       ¢
2x y 0£    ¢

37. 2 is perpendicular to the line 2x y c;vij¢  ¡   ¡ ¢
P( 2 7) on the line ( 2)( 2) 7 c ?  £       ¢
2x y 3£  ¡ ¢ 

38. 2 3 is perpendicular to the line 2x 3y c;vij¢    ¢
P(11 10) on the line (2)(11) (3)(10) c?$‰
2x 3y 8£ ¢ 

794. 790 Chapter 12 Vectors and the Geometry of Space
39. is parallel to the line x y c;vij¢      ¢
P( 2 1) on the line 2 1 c x y 1 ? £     ¢ £    ¢ab
or x y 1.¡¢ 

40. 2 3 is parallel to the line 3x 2y c;vij¢¡   ¢
P(0 2) on the line 0 2( 2) c 3x 2y 4?  £     ¢ £   ¢

41. 2 is parallel to the line 2x y c;vij¢      ¡ ¢
P(1 2) on the line 2(1) 2 c 2x y 0?$‰$‰
or 2x y 0. ¢

42. 3 2 is parallel to the line 2x 3y c;vij¢     ¢
P(1 3) on the line ( 2)(1) (3)(3) c?$‰
2x 3y 11 or 2x 3y 11£    ¢  ¡ ¢

43. P(0 0), Q(1 1) and 5 PQ and PQ (5 ) ( ) 5 N m 5 J?? ¢£¢¡ ¢ ¢ ¡¢ ¢
ÄÄ
Fj ijWF jij †† †
44. (distance) cos (602,148 N)(605 km)(cos 0) 364,299,540 N km (364,299,540)(1000) N mWFœœ œ œkk ) ††
3.6429954 10 Jœ‚
""
45. PQ cos (200)(20)(cos 30 °) 2000 3 3464.10 N m 3464.10 JWFœœ œœœ
Ä
kk¹¹
È
) †

795. Section 12.3 The Dot Product 791
46. PQ cos (1000)(5280)(cos 60 °) 2,640,000 ft lbWFœœ œ
Ä
kk¹¹ ) †
In Exercises 47-52 we use the fact that a b is normal to the line ax by c.nij¢¡ ¡ ¢
47. 3 and 2 cos cos cosnijnij"#
¡" ¡" ¡" ¡"¢¡ ¢ £¢ ¢ ¢ ¢ ) Š‹ Š ‹ Š‹
nn
nn¢ 
¢ †
kkkk ÈÈ È 4
61
10 5 2
1
48. 3 and 3 cos cos cosnijnij"#
¡" ¡" ¡" ¡  "¢  ¡ ¢ ¡ £ ¢ ¢ ¢   ¢
ÈÈ
Š‹ Š‹ ˆ‰)
nn
nn¢ 
¢ †
kkkk ÈÈ 23
31 2
4 4
1
49. 3 and 3 cos cos cosnijnij"#
¡" ¡" ¡"  ¢  ¢  £¢ ¢ ¢ ¢
ÈÈ
Š‹ Š ‹ Š‹)
nn
nn¢ 
¢ †
kkkk
ÈÈ È
ÈÈ 26
33 3
4 4
1
50. 3 and 1 3 1 3 cosni j n i j"#
¡"¢¡ ¢   ¡ ¡ £ ¢
ÈÈÈ
Š‹Š‹ Š‹ )
nn
nn¢ 
¢ †
kkkk
cos cos cosœœœœ
¡" ¡" ¡" ¡  
  ¡      
"
‚ 
Š‹ Š‹
1333
13 12331233
4
28 2 4
ÈÈ
È É ÈÈ
ÈÈ
1
51. 3 4 and cos cos cos 0.14 radnijnij"#
¡" ¡" ¡"  ¢  ¢ £¢ ¢ ¢ ! ) Š‹ Š ‹ Š‹
nn
nn¢ 
¢ †
kkkk ÈÈÈ
34 7
25 2 5 2
52. 12 5 and 2 2 cos cos cos 1.18 radnijnij"#
¡" ¡" ¡" ¡¢¡ ¢ £¢ ¢ ¢ ! ) Š‹ Š ‹ Š‹
nn
nn¢ 
¢ †
kkkk ÈÈ È
24 10 14
169 8 26 2
53. The angle between the corresponding normals is equal to the angle between the corresponding tangents. The points of intersection are and . At the tangent line for f(x) x isŠ‹Š‹Š‹ ? ?  ? ¢
ÈÈ È
33 333 3
44 4## #
#
y f x y 3 x y 3x , and the tangent line for ¢      £ ¢  ¡ ¡ £ ¢   
3 33
4 44
33 3w
## #
Š‹Š ‹ Š ‹Š‹
ÈÈ
ÈÈ È
f(x) x is y f x y 3 x 3x . The corresponding¢    ¢      £ ¢ ¡ ¡¢ ¡ˆ‰ Š‹Š ‹ Š ‹Š‹
ÈÈ
33 39
4 44
33 3
####
#w
ÈÈ È
normals are 3 and 3 . The angle at is cosnijn ij "#
#
¡"¢¡ ¢ ¡  ? ¢
ÈÈ
Š‹ Š‹
È
kkkk
33
4
)
nn
nn¢ 
¢ †
cos cos , the angle is and . At the tangent line for f(x) x is¢¢ ¢ ? ¢
¡" ¡" #¡  "
##
Š‹ Š‹ ˆ‰
31 2 2 3
4 4 3334
3
ÈÈ
È
111
y 3 x 3x and the tangent line for f(x) x is y 3 x¢¡¡¢¡ ¢ ¢ ¡¡
ÈÈ È
Š‹ Š‹
È È
3 339 3 3
44 4###
#
3x . The corresponding normals are 3 and 3 . The angle at is¢    ¢  ¡ ¢ ¡ ?
ÈÈÈ
Š‹
3 3
4 4
3
nijnij"#
#
È
cos cos cos , the angle is and .)¢¢¢ ¢
¡" ¡" ¡" ¡  "
#
Š‹ Š‹ ˆ‰
nn
nn¢ 
¢ †
kkkk ÈÈ 333
31 2 2
4 4
111
54. The points of intersection are and . The curve x y has derivative theŠ‹Š ‹!? !?   ¢   ¢   £
ÈÈ
33 3
4dxy
dy
## #
# "
tangent line at is y (x 0) is normal to the curve at that point. TheŠ‹!?   ¢     £ ¢ ¡
ÈÈ
ÈÈ
33
33##
""
"
nij
curve x y has derivative the tangent line at is y (x 0)¢   ¢ £ !?   ¢  
# ""
###
3
4dxy
dy 33
3
Š‹
ÈÈ
È
is normal to the curve. The angle between the curves is cos£¢  ¡ ¢nij #
" ¡"
È kkkk3 ) Š‹
nn
nn¢ 
¢ †
cos cos cos and . Because of symmetry the angles betweenœœœœ
¡" ¡" ¡"
¡ 
  
"
#
‚ 
Š‹ ˆ‰
¢
¢¢
33
33
2
4
3
1
1 1
33
2
ÉÉ
ˆ‰
ˆ‰
11
the curves at the two points of intersection are the same.
55. The curves intersect when y x y y y 0 or y 1. The points of intersection are ( ) andœ œ œ Ê œ œ !ß !
$# '
$
ab
( ). Note that y 0 since y y . At ( 0) the tangent line for y x is y 0 and the tangent line for"ß "   œ !ß œ œ
'$

796. 792 Chapter 12 Vectors and the Geometry of Space
y x is x 0. Therefore, the angle of intersection at (0 0) is . At (1 1) the tangent line for y x isœœ ß ß œÈ
1
#
$
y 3x 2 and the tangent line for y x is y x . The corresponding normal vectors are¢  ¢ ¢ ¡ È
""
##
3 and cos cos , the angle is and .nijn ij"#
""
#
¡" ¡"¢  ¡ ¢  ¡ £ ¢ ¢ ¢ ) Š‹ Š‹
nn
nn¢ 
¢ †
kkkk È 444 2
3111
56. The points of intersection for the curves y x and y x are ( 0) and ( 1 1). At ( 0) the tangent¢  ¢ !?   ?  !?
# $ È
line for y x is y 0 and the tangent line for y x is x 0. Therefore, the angle of intersection at ( 0)¢  ¢ ¢ ¢ !?
# $ È
is . At ( 1 1) the tangent line for y x is y 2x 1 and the tangent line for y x is y x .
1
#
# $ "
 ?  ¢  ¢ ¡ ¢ ¢   È
33
2
The corresponding normal vectors are 2 and cosnijn ij "#
" ¡"¢  ¢  £¢
3
) Š‹
nn
nn¢ 
¢ †
kkkk
cos cos cos , the angle is and .œœœœ
¡" ¡" ¡"
 "
 
"
‚  Œ Š‹
25
33
9
5 10
3
ÈÉ
ˆ‰
È
5 1 2 444
3 ¢
  
111
12.4 THE CROSS PRODUCT
1. 3 length 3 and the direction is ;22
0
uv i j k i j k
ij k
‚œ œ Ê œ "
""
ââ
ââ
ââ
ââ
ââ
ââ
ˆ‰
22 22
333 333
""
( 3 length 3 and the direction is vu uv) i j k i j k‚œ‚ œ Ê œ ˆ‰
22 22
333 333
""
2. 5( ) length 5 and the direction is 230
10
uv k k
ijk
‚œ œ Ê œ
"
ââ
ââ
ââ
ââ
ââ
ââ
( 5( ) length 5 and the direction is vu uv) k k‚œ‚ œ Ê œ
3. length 0 and has no direction224
12
uv 0
ijk
‚œ œ Ê œ 
"
ââ
ââ
ââ
ââ
ââ
ââ
( length 0 and has no directionvu uv)0‚œ‚ œ Ê œ
4. length 0 and has no direction11 1
00 0
uv 0
ij k
‚œ œ Ê œ  
ââ
ââ
ââ
ââ
ââ
ââ
( length 0 and has no directionvu uv)0‚œ‚ œ Ê œ
5. 6( ) length 6 and the direction is 200
030
uv k k
ijk
‚œ œ Ê œ
 
ââ
ââ
ââ
ââ
ââ
ââ
( 6( ) length 6 and the direction is vu uv) k k‚œ‚ œ Ê œ
6. ( ) ( ) length 1 and the direction is 001
100
uv ij jk ki j j
ijk
‚œ‚‚‚ œ‚œ œÊ œ
ââ
ââ
ââ
ââ
ââ
ââ
( length 1 and the direction is vu uv) j j‚œ‚ œÊ œ
7. 6 12 length 6 5 and the direction is 824
221
uv i k i k
ijk
‚œ œ Ê œ    
ââ
ââ
ââ
ââ
ââ
ââ
È
"
ÈÈ
55
2
( (6 12 length 6 5 and the direction is vu uv) i k) i k‚œ‚ œ Ê œ
È
"
ÈÈ
55
2

797. Section 12.4 The Cross Product 793
8. 1 2 2 2 length 2 3 and the direction is
112
uv i j k i j k
ijk
‚œ œ Ê œ
ââ
ââ
ââ
ââ
ââ
ââ
È
3 1
333##
" ""
ÈÈÈ
( ( 2 2 2 length 2 3 and the direction is vu uv) i j k) i j k‚œ‚ œ Ê œ
È
""
ÈÈÈ
333
1
9. 10. 100 10 1
010 01 0
uv k uv ik
ijk ij k
‚œ œ ‚œ œ  
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â

11. 12. 510 1 2 10
01 1 1 2 0
uv ijk uv k
ij k i j k
‚œ œ ‚œ œ  
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ

13. 2 14. 2110 012
110 100
uv k uv jk
ijk ijk
‚œ œ ‚œ œ
 
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ

15. (a) PQ PR 8 4 4 Area PQ PR 64 16 16 2 611 3
13 1
ÄÄ
‚ œ œ Ê œ ‚ œ œ
ÄÄ
 
  
ââ
ââ
ââ
ââ
ââ
ââ
¹¹ È È
ijk
ijk
""
##
(b) (2 )uijk¢â ¢â ¡ ¡
PQ PR
PQ PR 6
Ä
‚
Ä
Ä
‚
Ä
"
¹¹
È

798. 794 Chapter 12 Vectors and the Geometry of Space
16. (a) PQ PR 4 4 2 Area PQ PR 16 16 4 3102
220
ÄÄ
‚ œ œ Ê œ ‚ œ œ
ÄÄ
 
ââ
ââ
ââ
ââ
ââ
ââ
¹¹ È
ijk
ijk
""
##
(b) (2 2 )uijk¢â ¢â ¡  
PQ PR
PQ PR
3
Ä
‚
Ä
Ä
‚
Ä
"
¹¹
17. (a) PQ PR Area PQ PR 1 1111
110
ÄÄ
‚œ œÊ œ ‚ œ œ
ÄÄ
ââ
ââ
ââ
ââ
ââ
ââ
¹¹ È
ijk
ij
""
###
È
2
(b) ( ) ( )uijij¢â ¢â  ¡ ¢â  
PQ PR
PQ PR 22
Ä
‚
Ä
Ä
‚
Ä
""
¹¹
ÈÈ
18. (a) PQ PR 2 3 Area PQ PR 4 9 1211
10 2
ÄÄ
‚ œ œÊ œ ‚ œ œ
ÄÄ
  
 
ââ
ââ
ââ
ââ
ââ
ââ
¹¹ È
ij k
ijk
""
## #
È
14
(b) (2 3 )uijk¢â ¢â ¡ ¡
PQ PR
PQ PR 14
Ä
‚
Ä
Ä
‚
Ä
"
¹¹
È
19. If a a a , b b b , and c c c , then ( ) ,
aaa
bbb
ccc
uijkv ijkwijk uvwœ œ œ ‚œ"#$ "# $ "#$
"#$
"#$
"#$ †
ââ
ââ
ââ
ââ
ââ
ââ
( ) and ( ) which all have the same value, since the
bbb ccc
ccc aaa
aaa bbb
vw u wu v††‚œ ‚œ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
"#$ "#$
"#$ "#$
"#$ "#$
interchanging of two pair of rows in a determinant does not change its value the volume isÊ
() abs 8
200
020
002
kk
ââ
ââ
ââ
ââ
ââ
ââ
uvw‚œ œ†
20. ( ) abs 4 (for details about verification, see Exercise 19)
111
21 2
12 1
kk
ââ
ââ
ââ
ââ
ââ
ââ
uvw‚œ œ
 
 
  
†
21. ( ) abs 7 7 (for details about verification, see Exercise 19)
210
211
102
kk kk
ââ
ââ
ââ
ââ
ââ
ââ
uvw‚œ œœ  †
22. ( ) abs 8 (for details about verification, see Exercise 19)
11 2
10 1
24 2
kk
ââ
ââ
ââ
ââ
ââ
ââ
uvw‚œ œ
 
  
 
†
23. (a) 6, 81, 18 noneuv uw vw†† †¢  ¢  ¢ £
(b) , , 511 5 11 015
01 5 153 3 153 3
uv 0uw 0vw 0
ij k i j k i jk
‚œ Á ‚ œ œ ‚ œ Á    
     
ââ â â â â
ââ â â â â
ââ â â â â
ââ â â â â
ââ â â â â
ââ â â â â
and are parallelÊuw
24. (a) 0, 0, 3 , 0, 0, 0 , , , uv u w ur vw vr wr u vu wv wv r†††††œ‚œ œ œ œ œÊ¼¼¼¼ 1
and wr¼

799. Section 12.4 The Cross Product 795
(b) , , 12 1 12 1
11 1 10 1
12 1uv 0uw 0ur 0
ijk ijk ijk
‚œ Á ‚ œ Á ‚œ œ  
 
 
  
ââ ââ â â
ââ ââ â â
ââ ââ â â
ââ ââ â â
ââ ââ â â
ââ ââ â â
11
##
1
, , 111
101
111 1 01vw 0vr 0wr 0
ijk ijk ijk
‚œ Á ‚œ Á ‚œ Á   
     
ââââ ââ
ââââ ââ
ââââ ââ
ââââ ââ
ââââ ââ
ââââ ââ
11 11
## ##
11
and are parallelÊur
25. PQ PQ sin (60 °) 30 ft lb 10 3 ft lb¹¹¹¹ kk
È
ÄÄ
‚œ œ œFF
2
3
3
†† † †
È
#
26. PQ PQ sin (135 °) 30 ft lb 10 2 ft lb¹¹¹¹ kk
È
ÄÄ
‚œ œ œFF
2
3
2
†† † †
È
#
27. (a) true, a a akkÈ Èuuu¢¡¡¢
###
13  
†
(b) not always true, uu u†œkk
#
(c) true, 0 0 0 and 0 0 0aaa 000
000 aaa
u0 ijk0 0u ijk0
ijk ijk
‚œ œ œ ‚œ œ œ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ "#$
"#$
(d) true, ( ) ( aa aa) ( aa aa) ( aa aa)aaa
aaa
uu i j k0
ijk
‚ œ œ œ
   
ââ
ââ
ââ
ââ
ââ
ââ "#$
"#$
#$ #$ "$ "$ "# "#
(e) not always true, for exampleijk kji‚œ Áœ‚
(f) true, distributive property of the cross product
(g) true, ( ) ( ) 0uvvuvv u0‚œ‚œœ†† †
(h) true, the volume of a parallelpiped with , , and along the three edges is ( ) ( ) ( ),uv w uvw vwu uvw ‚œ‚œ‚†††
since the dot product is commutative.
28. (a) true, a b a b a b b a b a b auv vu††¢¡¡¢¡¡¢
"" ## $$ "" ## $$
(b) true, ( ) aaa bbb
bbb aaa
uv vu
ijk ijk
‚ œ œ œ ‚
ââââ
ââââ
ââââ
ââââ
ââââ
ââââ "#$ "#$
"#$ "#$
(c) true, ( ) ( ) aaa aaa
bbb bbb
‚œ œ œ‚   uv uv
ijk ijk
ââââ
ââââ
ââââ
ââââ
ââââ
ââââ "#$ "#$
"#$ "#$
(d) true, (c ) (ca )b (ca )b (ca )b a (cb ) a (cb ) a (cb ) (c ) c(a b a b a b )uv u v††¢¡¡¢¡¡¢ ¢¡¡ "" ## $$ "" ## $$ "" ## $$
c( )œuv†
(e) true, c( ) c (c ) (c )a a a ca ca ca a a a
b b b b b b cb cb cb
uv u v u v
ijk i j k i j k
‚œ œ œ ‚œ œ‚
âââ â â â
âââ â â â
âââ â â â
âââ â â â
âââ â â â
âââ â â â
"#$ " # $ " # $
"#$ " # $ " # $
(f) true, aaa aaauu u†¢¡¡¢ ¡¡ ¢
### ###
#
#
13 13   
ˆ‰È kk
(g) true, ( ) 0uuu0u‚œœ††
(h) true, and ( ) ( ) 0uvu uvv uvuvuv‚¼ ‚¼ Ê ‚ œ ‚ œ ††
29. (a) proj (b) ( ) (c) ( ) (d) ( )
vu v uv uv w uvwœ„‚ „‚‚‚Š‹ ab kk
uv
vv
†
kkkk
†
30. (a) ( ) ( )uv uw‚‚‚
(b) ()()() ()uv uv uv u uv vuu vu uv vv‚ œ ‚ ‚œ‚‚‚‚
2( ), or simply œ‚‚œ ‚ ‚0vuuv0 vu uv
(c) (d) kk k kuuw
v
vkk
‚

800. 796 Chapter 12 Vectors and the Geometry of Space
31. (a) yes, and are both vectors (b) no, is a vector but is a scalaruv w u vw‚ †
(c) yes, and are both vectors (d) no, is a vector but is a scalaruuw u vw‚ †
32. ( ) is perpendicular to , and is perpendicular to both and ( ) isuv w uv uv u v uv w‚‚ ‚ ‚ Ê ‚‚
parallel to a vector in the plane of and which means it lies in the plane determined by and .uv uv
The situation is degenerate if and are parallel so and the vectors do not determine a plane.uv uv0 ‚œ
Similar reasoning shows that ( ) lies in the plane of and provided and are nonparallel.uvw v w v w‚‚
33. No, need not equal . For example, , but ( ) andv w ij ij i ij iiij0kk Á ‚ œ‚‚œ œ
() .i ij iiij0kk‚ œ‚‚œ œ
34. Yes. If and , then ( ) and ( ) 0. Suppose now that .uvuw uvuw u vw 0 uvw vw‚œ‚ œ ‚ œ œ Á†† †
Then ( ) implies that k for some real number k 0. This in turn implies thatuvw0 vw u‚ œ œ Á
( ) (k ) k 0, which implies that . Since , it cannot be true that , souvw u u u u 0 u 0 v w†† ¢ ¢ ¢ ¢ Ò Ò kk
#
.vwœ
35. AB and AD AB AD 2 area AB AD 2 110
110
Ä Ä ÄÄ ÄÄ
œ œ Ê ‚ œ œ Ê œ ‚ œ  
  
ij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
36. AB 7 3 and AD 2 5 AB AD 29 area AB AD 29730
250
Ä Ä ÄÄ ÄÄ
œ œ Ê ‚ œ œ Ê œ ‚ œij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
37. AB 3 2 and AD 5 AB AD 13 area AB AD 13320
510
Ä Ä ÄÄ ÄÄ
œ œÊ ‚ œ œ Ê œ ‚ œ  ij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
38. AB 7 4 and AD 2 5 AB AD 43 area AB AD 43740
250
Ä Ä ÄÄ ÄÄ
œ œ Ê ‚ œ œ Ê œ ‚ œ  ij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
39. AB 2 3 and AC 3 AB AC 11 area AB AC 230
310
Ä Ä ÄÄ ÄÄ
œ œ Ê ‚ œ œ Ê œ ‚ œ  ij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
"
##
11
40. AB 4 4 and AC 3 2 AB AC 4 area AB AC 2 440
320
Ä Ä ÄÄ ÄÄ
œ œ Ê ‚ œ œ Ê œ ‚ œij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
"
#
41. AB 6 5 and AC 11 5 AB AC 25 area AB AC 650
11 5 0
Ä Ä ÄÄ ÄÄ
œ œ Ê ‚ œ œ Ê œ ‚ œ  
 
ij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
"
##
25
42. AB 16 5 and AC 4 4 AB AC 84 area AB AC 4216 5 0
440
Ä Ä ÄÄ ÄÄ
œ œÊ ‚ œ œ Ê œ ‚ œ  ij ij k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
"
#

801. Section 12.5 Lines and Planes in Space 797
43. If a a and b b , then and the triangle's area isaa0
bb0
aa
bb
AijBij AB k
ijk
œ œ ‚œ œ"# "# "#
"#
"#
"#
ââ
ââ
ââ
ââ
ââ
ââ
ºº
. The applicable sign is ( ) if the acute angle from to runs counterclockwise
aa
bb
""
##
"#
"#
kk ººAB AB‚œ„
in the xy-plane, and ( ) if it runs clockwise, because the area must be a nonnegative number. 
44. If a a , b b , and c c , then the area of the triangle is AB AC . Now,AijBijCijœ œ œ ‚
ÄÄ "# "# "#
"
# ¹¹
AB AC AB ACbaba0
caca0
ba ba
ca ca
ÄÄ ÄÄ
‚œ œ Ê ‚  
  
  
  
ââ
ââ
ââ
ââ
ââ
ââ
ºº ¹¹
ijk
k ""##
""##
"" ##
"" ##
"
#
(b a )(c a ) (c a )(b a ) a (b c ) a (c b ) (b c c b )¢         ¢  ¡  ¡  
""
##
""## ""## "## #"" "#"#
kkk k
. The applicable sign ensures the area formula gives a nonnegative number.
aa1
bb1
cc1
œ„
"
#
"#
"#
"#
ââ
ââ
ââ
ââ
ââ
ââ
12.5 LINES AND PLANES IN SPACE
1. The direction and P(3 4 1) x 3 t, y 4 t, z 1 tijk¡ ¡ ?  ?  £ ¢ ¡ ¢  ¡ ¢  ¡
2. The direction PQ 2 2 2 and P(1 2 1) x 1 2t, y 2 2t, z 1 2t
Ä
¢   ¡ ??  £ ¢  ¢  ¢ ¡ijk
3. The direction PQ 5 5 5 and P( 2 0 3) x 2 5t, y 5t, z 3 5t
Ä
¢¡   ?? £¢ ¡ ¢ ¢ ijk
4. The direction PQ and P(1 2 0) x 1, y 2 t, z t
Ä
¢    ? ? £ ¢ ¢   ¢ jk
5. The direction 2 and P( ) x 0, y 2t, z tjk !?!?! $ ‰ ‰ ‰
6. The direction 2 3 and P(3 2 1) x 3 2t, y 2 t, z 1 3tij k ¡ ? ? £ ¢ ¡ ¢   ¢ ¡
7. The direction and P(1 1 1) x 1, y 1, z 1 tk ?? £ ¢ ¢ ¢ ¡
8. The direction 3 7 5 and P(2 4 5) x 2 3t, y 4 7t, z 5 5tijk¡  ?? £¢¡ ¢¡ ¢ 
9. The direction 2 2 and P(0 7 0) x t, y 7 2t, z 2tijk¡¡ ? ? £¢ ¢ ¡ ¢
10. The direction is 2 4 2 and P( 0) x 2 2t, y 3 4t, z 2t123
345
AB i j k
ijk
¢ ¢   ¡   #? $? £ ¢   ¢ ¡ ¢  
ââ
ââ
ââ
ââ
ââ
ââ
11. The direction and P(0 0 0) x t, y 0, z 0i ßß Ê œ œ œ
12. The direction and P(0 0 0) x 0, y 0, z tk ßß Ê œ œ œ

802. 798 Chapter 12 Vectors and the Geometry of Space
13. The direction PQ and P(0 0 0) x t,
Ä
œ ßß Ê œij k
3
#
y t, z t, where 0 t 1œœ ŸŸ
3
#

14. The direction PQ and P(0 0 0) x t, y 0, z 0,
Ä
œßßÊœœœi
where 0 t 1ŸŸ

15. The direction PQ and P(1 1 0) x 1, y 1 t,
Ä
œßßÊœœj
z 0, where 1 t 0œŸŸ

16. The direction PQ and P(1 1 0) x 1, y 1, z t,
Ä
œßßÊœœœk
where 0 t 1ŸŸ

17. The direction PQ 2 and P(0 1 1) x 0,
Ä
œ ß ß Ê œj
y 1 2t, z 1, where 0 t 1œ œ ŸŸ

18. The direction PQ 2 and P(0 2 0) x 3t,
Ä
œ$ ß ß Ê œij
y 2 2t, z 0, where 0 t 1œ œ ŸŸ

803. Section 12.5 Lines and Planes in Space 799
19. The direction PQ 2 2 2 and P(2 0 2)
Ä
¢  ¡   ? ?ijk
x 2 2t, y 2t, z 2 2t, where 0 t 1£¢ ¢¢  ¸¸

20. The direction PQ 3 and P(1 0 1)
Ä
¢  ¡ ¡ ? ? ijk
x 1 t, y 3t, z 1 t, where 0 t 1£¢  ¢ ¢ ¡ ¸¸

21. 3(x 0) ( 2)(y 2) ( 1)(z 1) 0 3x 2y z 3 ¡   ¡  ¡¢£    ¢ 
22. 3(x 1) (1)(y 1) (1)(z 3) 0 3x y z 5 ¡ ¡¡  ¢£ ¡¡¢
23. PQ 3 , PS 3 2 PQ PS 7 5 4 is normal to the plane113
132
ÄÄ ÄÄ
œ œ Ê ‚ œ œ  
  
ij k i j k i j k
ijk
ââ
ââ
ââ
ââ
ââ
ââ
7(x 2) ( 5)(y 0) ( 4)(z 2) 0 7x 5y 4z 6£  ¡   ¡   ¢£   ¢
24. PQ 2 , PS 3 2 3 PQ PS 3 is normal to the plane 112
323
ÄÄ ÄÄ
œ œ Ê ‚ œ œ  
 
ij k i j k i jk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
( 1)(x 1) ( 3)(y 5) (1)(z 7) 0 x 3y z 9£    ¡    ¡   ¢ £ ¡  ¢
25. 3 4 , P(2 4 5) (1)(x 2) (3)(y 4) (4)(z 5) 0 x 3y 4z 34ni j k¢¡¡ ??¢  ¡  ¡  ¢£¡ ¡ ¢
26. 2 , P(1 2 1) (1)(x 1) ( 2)(y 2) (1)(z 1) 0 x 2y z 6ni jk¢  ¡ ? ? ¢   ¡  ¡ ¡   ¢ £   ¡ ¢
27. t 0 and s 1; then z 4t 3 4s 1
x 2t 1 s 2 2t s 1 4t 2s 2
y 3t 2 2s 4 3t 2s 2 3t 2s 2
œœœ
¢¡¢¡  ¢  ¢
¢¡¢¡  ¢  ¢
£££¢¢ ¢¡¢  
4(0) 3 ( 4)( 1) 1 is satisfied the lines do intersect when t 0 and s 1 the point of£¡¢    £ ¢ ¢ £
intersection is x 1, y 2, and z 3 or P(1 2 3). A vector normal to the plane determined by these lines isœœ œ ßß
20 12 , where and are directions of the lines the plane23 4
12 4
nn i jk n n
ij k
"# " #‚œ œ Ê
 
ââ
ââ
ââ
ââ
ââ
ââ
containing the lines is represented by( 20)(x 1) (12)(y 2) (1)(z 3) 0 20x 12y z 7.  ¡  ¡ ¢£ ¡¡¢
28. s 1 and t 0; then z t 1 5s 6 0 1 5( 1) 6
x t 2s 2 t 2s 2
yt2 s3 t s1
œœ
¢¢¡  ¢
¢  ¡ ¢ ¡     ¢
£ £ ¢  ¢ ¢¡¢ ¡ £ ¡¢ ¡
is satisfied the lines do intersect when s 1 and t 0 the point of intersection is x 0, y 2 and z 1£¢ ¢£¢¢¢
or P(0 2 1). A vector normal to the plane determined by these lines is 11
215
ßß ‚ œ "nn
ijk
"#
ââ
ââ
ââ
ââ
ââ
ââ

804. 800 Chapter 12 Vectors and the Geometry of Space
6 3 3 , where and are directions of the lines the plane containing the lines is represented by¢    ¡ £ijk n n "#
( 6)(x 0) ( 3)(y 2) (3)(z 1) 0 6x 3y 3z 3.   ¡   ¡  ¢£ ¡ ¢
29. The cross product of and 4 2 2 has the same direction as the normal to the planeijk i j k¡    ¡  
6 6 . Select a point on either line, such as P( 1 2 1). Since the lines are given11
42 2
£¢ ¢¡  ?? 
  
njk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
to intersect, the desired plane is 0(x 1) 6(y 2) 6(z 1) 0 6y 6z 18 y z 3.¡¡  ¡  ¢£ ¡ ¢ £¡¢
30. The cross product of 3 and has the same direction as the normal to the planeijk ijk   ¡¡
2 2 4 . Select a point on either line, such as P(0 3 2). Since the lines are131
11 1
nijk
ij k
¢¢  ¡ ??   
ââ
ââ
ââ
ââ
ââ
ââ
given to intersect, the desired plane is ( 2)(x 0) ( 2)(y 3) (4)(z 2) 0 2x 2y 4z 14   ¡   ¡ ¡¢£  ¡¢ 
xy2z7.£¡  ¢
31. 3 3 3 is a vector in the direction of the line of intersection of the planes21 1
12 1
nn ijk
ij k
"#‚ œ œ  
ââ
ââ
ââ
ââ
ââ
ââ
3(x 2) ( 3)(y 1) 3(z 1) 0 3x 3y 3z 0 x y z 0 is the desired plane containing£  ¡   ¡ ¡¢£  ¡¢£ ¡¢
P(21 1)
!??
32. A vector normal to the desired plane is P P 2 12 2 ; choosing P (1 2 3) as a20 2
412
"# "
Ä
¢ ¢     ?? 
 
nijk
ij k
ââ
ââ
ââ
ââ
ââ
ââ
point on the plane ( 2)(x 1) ( 12)(y 2) ( 2)(z 3) 0 2x 12y 2z 32 x 6y z 16£    ¡    ¡    ¢ £      ¢  £ ¡ ¡¢
is the desired plane
33. S(0 0 12), P(0 0 0) and 4 2 2 PS 24 48 24( 2 )0012
422
?? ?? ¢   ¡ £ ¢ ¢ ¡ ¢ ¡
Ä
 
vijk v i j ij
ijk
ââ
ââ
ââ
ââ
ââ
ââ
d 5 24 2 30 is the distance from S to the lineÊœ œ œ œ œ
¹¹
kk
ÈÈ
ÈÈ
PS
24 1 4
1644
24 5
24
Ä
‚
 
  
v
v
ÈÈ
†
34. S(0 0 0), P(5 5 3) and 3 4 5 PS 13 16 5 553
34 5
?? ??  ¢ ¡   £ ¢ ¢    
Ä
  
 
vijk v i jk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
d 9 3 is the distance from S to the lineÊœ œ œ œ œ
¹¹
kk
ÈÈ
ÈÈ
PS
169 256 25
91625
450
50
Ä
‚
  
  
v
v
È
35. S(2 1 3), P(2 1 3) and 2 6 PS d 0 is the distance from S to the line?? ?? ¢ ¡ £ ¢ £ ¢ ¢ ¢
Ä
vij v0
¹¹
kk È
PS
0
40
Ä
‚v
v
(i.e., the point S lies on the line)
36. S(2 1 1), P(0 1 0) and 2 2 2 PS 2 6 4 20 1
22 2
??  ?? ¢¡¡ £ ¢ ¢ ¡
Ä
 vijk v ijk
ij k
ââ
ââ
ââ
ââ
ââ
ââ
d is the distance from S to the lineÊœ œ œ œ
¹¹
kk
ÈÈ
ÈÈ
PS
43616
444
56
12
14
3
Ä
‚
  
  
v
v
É

805. Section 12.5 Lines and Planes in Space 801
37. S(3 1 4), P(4 3 5) and 2 3 PS 30 6 6 149
123
ß ß ß ß œ Ê ‚ œ œ
Ä


v ijk v ijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
d is the distance from S to the lineÊœ œ œœœ œ
¹¹
kk
È ÈÈÈ È
ÈÈÈÈ
PS
900 36 36
149
972 486 81 6
14 77
942
7
Ä
‚
ßß
ßß
v
v
†
38. S( 1 4 3), P(10 3 0) and 4 4 PS 28 56 28 28( 2 )11 7 3
404
ß ß ßß œ Ê ‚ œ œ œ
Ä
vik v i j k ijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
d 7 3 is the distance from S to the lineÊœ œ œ
¹¹
kk
È
È
PS
28141
41 1
Ä
‚
ßß
ß
v
v
È
39. S(2 3 4), x 2y 2z 13 and P(13 0 0) is on the plane PS 11 3 4 and 2 2ßß œ ßß Ê œ œ
Ä
ijk nijk
dPS 3Êœ œ œ œ
Ä
¹¹¹ ¹¹¹†
n
nkk ÈÈ

ßß
1168 9
144 9
40. S(0 0 0), 3x 2y 6z 6 and P(2 0 0) is on the plane PS 2 and 3 2 6ßß œ ßß Ê œ œ
Ä
inijk
dPSÊœ œ œ œ
Ä
¹¹¹ ¹†
n
nkk ÈÈ

ßß
666
9436 49 7
41. S(0 1 1), 4y 3z 12 and P(0 3 0) is on the plane PS 4 and 4 3ßß œ ßß Ê œ œ
Ä
jk n j k
dPSÊœ œ œ
Ä
¹¹¹¹†
n
nkk È
16 3 19
16 9 5
ß
ß
42. S(2 2 3), 2x y 2z 4 and P(2 0 0) is on the plane PS 2 3 and 2 2ßß œ ßß Ê œ œ
Ä
jk nijk
dPSÊœ œ œ
Ä
¹¹¹ ¹†
n
nkk È
26 8
414 3
ß
ßß
43. S(0 1 0), 2x y 2z 4 and P(2 0 0) is on the plane PS 2 and 2 2ß ß œ ß ß Ê œ œ
Ä
ij n ij k
dPSÊœ œ œ
Ä
¹¹¹ ¹†
n
nkk È

ßß
410 5
414 3
44. S(1 0 1), 4x y z 4 and P( 1 0 0) is on the plane PS 2 and 4ßß œ ßß Ê œ œ
Ä
ik n ijk
dPSÊœ œ œ œ
Ä
¹¹¹ ¹†
n
nkk ÈÈ
È

ßß #
81 9
1611 18
32
45. The point P(1 0 0) is on the first plane and S(10 0 0) is a point on the second plane PS 9 , andßß ßß Ê œ
Ä
i
2 6 is normal to the first plane the distance from S to the first plane is d PSni j k#" " ! #
Ä
¹¹†
n
nkk
, which is also the distance between the planes.œœ¹¹
99
1436 41
ÈÈßß
46. The line is parallel to the plane since ( 2 6 ) 1 2 3 0. Also the pointvn i j k i j k††œ œœˆ‰
"
#
S(1 0 0) when t 1 lies on the line, and the point P(10 0 0) lies on the plane PS 9 . The distanceßß œ ßß Ê œ
Ä
i
from S to the plane is d PS , which is also the distance from the line to theœœ œ
Ä
¹¹¹ ¹†
n
nkk ÈÈ

ßß
99
1436 41
plane.
47. and 2 2 cos cos cosnij n ijk"#
" " " "œ œ Ê œ œ œ œ ) Š‹ Š‹ Š‹
nn
nn
†
kkkk ÈÈÈ 4
21
2 9 2
1

806. 802 Chapter 12 Vectors and the Geometry of Space
48. 5 and 2 3 cos cos cos (0)nijknijk"#
" " "
#œ œ Êœ œ œ œ ) Š‹ Š ‹
nn
nn
†
kkkk ÈÈ
523
27 14
1
49. 2 2 2 and 2 2 cos cos cos 1.76 radnijknijk"#
" " " "œ œÊœ œ œ ¸ ) Š‹ Š ‹ Š‹
nn
nn
†
kkkk ÈÈÈ
442
12 9 33
50. and cos cos 0.96 radnijk nk"#
" "œ œ Ê œ œ ¸ ) Š‹ Š‹
nn
nn
†
kkkk ÈÈ
1
3 1
51. 2 2 and 2 cos cos cos 0.82 radn i jk n i jk"#
" " " œ œÊœ œ œ ¸ ) Š‹ Š ‹ Š‹
nn
nn
†
kkkk ÈÈ È
241 5
9 6 3 6
52. 4 3 and 3 2 6 cos cos cos 0.73 radnjknijk"#
" " " ßœ œ Êœ œ œ ¸ ) Š‹ Š ‹ ˆ‰
nn
nn
†
kkkk ÈÈ 35
818 26
25 49
53. 2x y 3z 6 2(1 t) (3t) 3(1 t) 6 2t 5 6 t x , y and z œÊ œÊœÊœÊœ œ œ
""
####
33
is the pointÊß߈‰
33
###
"
54. 6x 3y 4z 12 6(2) 3(3 2t) 4( 2 2t) 12 14t 29 12 t x 2, y 3 , œÊ œ Ê œÊœ Êœ œ
41 41
14 7
and z 2 2 is the pointœ Ê ß ß
41 20 27
777
ˆ‰
55. x y z 2 (1 2t) (1 5t) (3t) 2 10t 2 2 t 0 x 1, y 1 and z 0""#! " "" " #! "#!#!# # #
(1 1 0) is the pointÊßß
56. 2x 3z 7 2( 1 3t) 3(5t) 7 9t 2 7 t 1 x 1 3, y 2 and z 5 œ Ê œ Êœ ÊœÊ œ œ œ
( 4 2 5) is the pointÊ ß ß
57. and , the direction of the desired line; (1 1 1)111
110
nijk nij nn ij
ijk
"#"#œ œ Ê ‚ œ œ ßß
ââ
ââ
ââ
ââ
ââ
ââ
is on both planes the desired line is x 1 t, y 1 t, z 1Êœœœ
58. 3 6 2 and 2 14 2 15 , the direction of the 362
21 2
nijknij2knn ijk
ij k
"#"#œ œ Ê ‚ œ œ

ß
ââ
ââ
ââ
ââ
ââ
ââ
desired line; (1 0 0) is on both planes the desired line is x 1 14t, y 2t, z 15tßß Ê œ œ œ
59. 2 4 and 2 6 3 , the direction of the 124
11 2
nijk nijk nn jk
ij k
"#"#œ œ Ê ‚ œ œ

ââ
ââ
ââ
ââ
ââ
ââ
desired line; (4 3 1) is on both planes the desired line is x 4, y 3 6t, z 1 3tßß Ê œ œ œ
60. 5 2 and 4 5 10 25 20 , the direction of the520
04 5
nijnjknn ijk
ij k
"# "#œ œÊ‚œ œ

ââ
ââ
ââ
ââ
ââ
ââ
desired line; (1 3 1) is on both planes the desired line is x 1 10t, y 3 25t, z 1 20tß ß Ê œ œ œ
61. L1 & L2: x 3 2t 1 4s and y 1 4t 1 2s
2t 4s 2 2t 4s 2
4t 2s 2 2t s 1
œ œ œ œ Ê Ê
œ œ
œ œ
œœ
3s 3 s 1 and t 1 on L1, z 1 and on L2, z 1 L1 and L2 intersect at (5 3 1).Ê œ Ê œ œ Ê œ œ Ê ß ß

807. Section 12.5 Lines and Planes in Space 803
L2 & L3: The direction of L2 is (4 2 4 ) (2 2 ) which is the same as the direction
""
63
ijk ijk"" # ""
(2 2 ) of L3; hence L2 and L3 are parallel.
"
3
ij k""
L1 & L3: x 3 2t 3 2r and y 1 4t 2 r 3t 3
2t 2r 0 t r 0
4t r 3 4t r 3
œ œ œ œÊ Ê Ê œ
œ œ
œ œ
œœ
t 1 and r 1 on L1, z 2 while on L3, z 0 L1 and L2 do not intersect. The direction of L1Êœ œ Ê œ œ Ê
is (2 4 ) while the direction of L3 is (2 2 ) and neither is a multiple of the other; hence
""
È
21 3
ijk ijk
L1 and L3 are skew.
62. L1 & L2: x 1 2t 2 s and y 1 t 3s 5s 3 s and t on L1,
2t s 1
t3s1
œ œ œœ Ê Ê œ Ê œ œ Ê
"#
œ
œ
34
55
z while on L2, z 1 L1 and L2 do not intersect. The direction of L1 is (2 3 )œœœÊ
12 3 2
555 14
"
È
ij k
while the direction of L2 is ( 3 ) and neither is a multiple of the other; hence, L1 and L2 are
"
È
11
ijk
skew.
L2 & L3: x 2 s 5 2r and y 3s 1 r 5s 5 s 1 and r 2 on L2,
s2r3
3s r 1
œœ œ œÊ Ê œ Ê œ œÊ
œ
"#
œ
z 2 and on L3, z 2 L2 and L3 intersect at (1 3 2).œœÊ ßß
L1 & L3: L1 and L3 have the same direction (2 3 ); hence L1 and L3 are parallel.
"
È
14
ij k
63. x 2 2t, y 4 t, z 7 3t; x 2 t, y 2 t, z 1 tœ œ œ œ œ œ
"
##
3
64. 1(x 4) 2(y 1) 1(z 5) 0 x 4 2y 2 z 5 0 x 2y z 7; œÊ œÊ œ
2 (x 3) 2 2 (y 2) 2 (z 0) 0 2x 2 2y 2z 7 2 œÊ œ
ÈÈÈ ÈÈÈÈ
65. x 0 t , y , z ; y 0 t 1, x 1, z 3 ( 1 0 3); z 0œ Ê œ œ œ Ê !ß ß œ Ê œ œ œ Ê ß ß œ
"" "
### ##
33
ˆ‰
t 0, x 1, y 1 (1 1 0)ÊœœœÊßß
66. The line contains (0 0 3) and 3 1 3 because the projection of the line onto the xy-plane contains the originßß ßߊ‹
È
and intersects the positive x-axis at a 30° angle. The direction of the line is 3 0 the line in question
È
ij k"" !
is x 3t, y t, z 3.œœœ
È
67. With substitution of the line into the plane we have 2(1 2t) (2 5t) ( 3t) 8 2 4t 2 5t 3t 8œÊœ
4t 4 8 t 1 the point ( 1 7 3) is contained in both the line and plane, so they are not parallel.Ê œÊœÊ ßß
68. The planes are parallel when either vector A B C or A B C is a multiple of the other or
""" ###ijk ijk"" ""
when (A B C ) (A B C 0. The planes are perpendicular when their normals arekk
""" ###ijk ijk ‚ œ
perpendicular, or(A B C ) (A B C ) 0.
""" ###ijk ijk"" "" # †
69. There are many possible answers. One is found as follows: eliminate t to get t x 1 2 yœœœ
z3
#
x 1 2 y and 2 y x y 3 and 2y z 7 are two such planes.Ê œ œ Ê œ œ
z3
#
70. Since the plane passes through the origin, its general equation is of the form Ax By Cz 0. Since it meets""#
the plane M at a right angle, their normal vectors are perpendicular 2A 3B C 0. One choice satisfying!""#
this equation is A 1, B 1 and C 1 x y z 0. Any plane Ax By Cz 0 with 2A 3B C 0œ œ œÊœ œ œ
will pass through the origin and be perpendicular to M.

808. 804 Chapter 12 Vectors and the Geometry of Space
71. The points (a 0 0), (0 b 0) and (0 0 c) are the x, y, and z intercepts of the plane. Since a, b, and c are allßß ßß ßß
nonzero, the plane must intersect all three coordinate axes and cannot pass through the origin. Thus,
1 describes all planes except those through the origin or parallel to a coordinate axis.
xz
abc
y
""#
72. Yes. If and are nonzero vectors parallel to the lines, then is perpendicular to the lines.vv vv0
"# "# ‚Á
73. (a) EP cEP x y z c (x x ) y z x c(x x ), y cy and z cz ,
ÄÄ
œÊœÊœœ œ
"! "!"" !"! " "ijk i j k cd
where c is a positive real number
(b) At x 0 c 1 y y and z z ; at x x x 0, y 0, z 0; lim c lim """"!!

œÊœÊœ œ œ Ê œ œ œ œ
xxßßÄ_ Ä_
x
xx#
ߜ
lim 1 c 1 so that y y and z zœœÊÄ ÄÄ
xßÄ_
"

""
1
74. The plane which contains the triangular plane is x y z 2. The line containing the endpoints of the line""#
segment is x 1 t, y 2t, z 2t. The plane and the line intersect at . The visible section of the lineœ œ œ ßß ˆ‰
222
333
segment is 1 unit in length. The length of the line segment is 1 2 2 3 ofɈ‰ ˆ‰ ˆ‰ È
"
# ##
###
333 3
22 2
""# ""#!
the line segment is hidden from view.
12.6 CYLINDERS AND QUADRIC SURFACES
1. d, ellipsoid 2. i, hyperboloid 3. a, cylinder
4. g, cone 5. l, hyperbolic paraboloid 6. e, paraboloid
7. b, cylinder 8. j, hyperboloid 9. k, hyperbolic paraboloid
10. f, paraboloid 11. h, cone 12. c, ellipsoid
13. x y 4 14. x z 4 15. z y 1
## ## #
œ œ œ

16. x y 17. x 4z 16 18. 4x y 36#"#"#
#####

809. Section 12.6 Cylinders and Quadric Surfaces 805
19. z y 1 20. yz 1 21. 9x y z 9
## ###
œ œ œ

22. 4x 4y z 16 23. 4x 9y 4z 36 24. 9x 4y 36z 36
### ### ## #
""# ""# "" #

25. x 4y z 26. z x 9y 27. z 8 x y
## ## ##
œ œ œ

28. z 18 x 9y 29. x 4 4y z 30. y 1 x zœ œ œ
# # ## ##

810. 806 Chapter 12 Vectors and the Geometry of Space
31. x y z 32. y z x 33. 4x 9z 9y
### ## # # # #
"# "# " #

34. 9x 4y 36z 35. x y z 1 36. y z x 1
# # # ### ###
œ œ œ

37. 1 38. 1 39. z x y 1
yy
494 449
zx x z

œ œ œ
###

40. z 1 41. x y 1 42. y 1
y
44 4 4 4
xzxz

œ œ œ
### #

811. Section 12.6 Cylinders and Quadric Surfaces 807
43. y x z 44. x y z 45. x y z 4
## # # ###
œ œœ œ

46. 4x 4y z 47. z 1 y x 48. y z 4
### ## ##
œ œ œ

49. y x z 50. z 4x 4y 4 51. 16x 4y 1œ œ œab
## ### ##

52. z x y 1 53. x y z 4 54. x 4 yœ œ œ
## ### #

55. x z y 56. z y 1 57. x z 1
## # # ##
œ œ œ
x
4

812. 808 Chapter 12 Vectors and the Geometry of Space
58. 4x 4y z 4 59. 16y 9z 4x 60. z x y 1
### ## # ##
œ œ œ

61. 9x 4y z 36 62. 4x 9z y 63. x y 16z 16
### ### ## #
œ œ œ

64. z 4y 9 65. z x y 66. y x z 1
# # ## ###
œ œ œ ab

67. x 4y 1 68. z 4x y 4 69. 4y z 4x 4
## ## ###
œ œ œ

70. z 1 x 71. x y z 72. y z 1œ œ œ
### ## x
4

813. Section 12.5 Lines and Planes in Space 809
73. yz 1 74. 36x 9y 4z 36 75. 9x 16y 4zœ œ œ
### # ##

76. 4z x y 4
###
!!$

77. (a) If x 1 and z c, then x 1 A ab
##
œ œ œ Ê œÊ œ
yyy
49 4 9
z9cx

$
$

Š‹ ’“
9c
9
49 c
9 1
œœ1Š‹Š ‹
ÈÈ ab9c 29c
33 9
29c 1
(b) From part (a), each slice has the area , where 3 z 3. Thus V 2 9 z dz
29z
99
21 1ab #

Ÿ Ÿ œ '
0
3
ab
9 z dz 9z (27 9) 8$!$!$!$
44z4
9939
111'
0
3
ab ’“
#
$
!
!
1
(c) 1 1 A
xz x
abc cc
yy ac z bc z



œÊ œÊ œ
–—–—
ÈÈ
ac z bc z
cc

$$


1Š‹Š‹

V 2 c z dz c z c . Note that if r a b c,"$ ! $ ! $ $ $$$'
0
c c
1111ab 2ab z 2ab 2 4abc
cc3c33

!ab ’“ ˆ‰
## # $
!
then V , which is the volume of a sphere.œ
4r
3
1
!
78. The ellipsoid has the form 1. To determine c we note that the point (0 r h) lies on the surface
xz
RRc
y



œ ßß
#
of the barrel. Thus, 1 c . We calculate the volume by the disk method:
rh hR
Rc Rr


œÊ œ
#
V y dz. Now, 1 y R 1 R 1 R zœœÊœœœ1 '
h
h
###### y
Rc c hR h
zz Rr zR r



Š‹’ “ Š‹
ab
V R z dz R z z 2 R h R r h 2Êœ œ œ œ 1111 '

h
h h
h
’“’ “ Š‹Š‹ Š‹ ‘ab
### $###" "Rr Rr 2Rh rh
h3h 333


R h r h, the volume of the barrel. If r R, then V 2 R h which is the volume of a cylinder ofœ œ œ
42
33
11 1
## #
radius R and height 2h. If r 0 and h R, then V R which is the volume of a sphere.œœ œ
4
3
1
$
79. We calculate the volume by the slicing method, taking slices parallel to the xy-plane. For fixed z,
xz
ab c
y



œ
gives the ellipse 1. The area of this ellipse is a b (see Exercise 77a). Hence
xzzabzy
ccc

Š‹ Š‹
za zb
cc
œ œ 1ˆ‰ˆ‰ÈÈ
1
the volume is given by V dz . Now the area of the elliptic base when z h isœœœ œ '
0
h h
111abz abz abh
c2cc
’“

!
A , as determined previously. Thus, V h (base)(altitude), as claimed.œœœœ
111abh abh abh
ccc

""
##
ˆ‰

814. 810 Chapter 12 Vectors and the Geometry of Space
80. (a) For each fixed value of z, the hyperboloid 1 results in a cross-sectional ellipse
xz
abc
y



œ
1. The area of the cross-sectional ellipse (see Exercise 77a) is
x y

–—–—
ac z bc z
cc

!!


œ
A(z) c z c z c z . The volume of the solid by the method of slices isœ œ1Š‹Š‹
ÈÈ
ab
ab ab
cc c
## ## ##1

V A(z) dz c z dz c z z c h h 3c hœœœœœ''
00
hh
h
11 1 1ab ab ab abh
cc3c33c
ab a b ‘ˆ ‰
## # $ # $ ## ""
!
(b) A A(0) ab and A A(h) c h , from part (a) V 3c h!
## ##œœ œœ Êœ 1
h
11ab abh
c3c
ab a b
21 2 2ab c h (2A A)œœœ œ
11 1abh h abh c h h ab h
3c3c3c 3
Š‹Š‹ ‘ab


Ÿ ##
!
1
h (c) A A c 4c h (A 4A A )
m mhœœ œ ʈ‰ Š‹ ab
hab h ab h
c44c 6#
###
!
11


ab 4c h c h c 4c h c h 6c 2hœ œ œ
hab ab abh abh
6c c 6c 6c
‘ababa ba b1
111 1

## ## # #### # #
3c h V from part (a)œœ
1abh
3c
ab
##
81. y y , a parabola in the plane y y vertex when 0 or c 0 x 0$"$! $" $ $!$"$""
z x dz dz 2x
cb a dx dx a
y



1
Vertex 0 y ; writing the parabola as x z we see that 4p pÊßß œ œÊœŠ‹ "
#cy a y
bcbc4c
aaa



1 1
Focus 0 y"ww! Š‹ "
cy
b4c
a



1
82. The curve has the general form Ax By Dxy Gx Hy K 0 which is the same form as Eq. (1) in
##
œ
Section 10.3 for a conic section (including the degenerate cases) in the xy-plane.
83. No, it is not mere coincidence. A plane parallel to one of the coordinate planes will set one of the variables
x, y, or z equal to a constant in the general equation Ax By Cz Dxy Eyz Fxz Gx Hy Jz K
###

0 for a quadric surface. The resulting equation then has the general form for a conic in that parallel plane.œ
For example, setting y y results in the equation Ax Cz D x E z Fxz Gx Jz K 0 whereœœ
"
##ww w
D Dy , E Ey , and K K By Hy , which is the general form of a conic section in the plane y y
ww w #
"" " "
œœ œ œ
1
by Section 10.3.
84. The trace will be a conic section. To see why, solve the plane's equation Ax By Cz 0 for one of theœ
variables in terms of the other two and substitute into the equation Ax By Cz K 0. The result
###
áœ
will be a second degree equation in the remaining two variables. By Section 10.3, this equation will represent a
conic section. (See also the discussion in Exercises 82 and 83.)
85. z y 86. z 1 y$$!
# #

815. Section 12.5 Lines and Planes in Space 811
87. z x yœ
##

88. z x 2yœ
##
(a) (b)
(c) (d)
89-94. Example CAS commands:
:Maple
with( plots );
eq := x^2/9 + y^2/36 = 1 - z^2/25;
implicitplot3d( eq, x=-3..3, y=-6..6, z=-5..5, scaling=constrained,
shading=zhue, axes=boxed, title="#89 (Section 12.6)" );
: (functions and domains may vary):Mathematica
In the following chapter, you will consider contours or level curves for surfaces in three dimensions. For the purposes of
plotting the functions of two variables expressed implicitly in this section, we will call upon the function .ContourPlot3D

816. 812 Chapter 12 Vectors and the Geometry of Space
To insert the stated function, write all terms on the same side of the equal sign and the default contour equating that
expression to zero will be plotted.
This built-in function requires the loading of a special graphics package.
<<Graphics`ContourPlot3D`
Clear[x, y, z]
ContourPlot3D[x /9 y /16 z /2 1, {x, 9, 9}, {y, 12, 12}, {z, 5, 5},
22 2
!!!! ! !
Axes True, AxesLabel {x, y, z}, Boxed False,ÄÄÄ
PlotLabel "Elliptic Hyperboloid of Two Sheets"]Ä
Your identification of the plot may or may not be able to be done without considering the graph.
CHAPTER 12 PRACTICE EXERCISES
1. (a) 3 3, 4 4 2, 5 9 8, 12 20 17, 32  ¡  ¡  ¡  ¡ œ œ
(b) 17 32 1313
È È
22
œ
2. (a) 3 2, 4 5 1, 1 3. (a) 2 3 , 2 4 6, 8  ¡  ¡   ¡  ¡ab ab œ œ
(b) 1 1 2 (b) 6 8 10ÉÉab ab ab
È
œ œ
22 2
2
4. (a) 5 2 , 5 5 10, 25  ¡  ¡ab a b!$ !
(b) 10 25 725 5 29É ab
ÈÈ
2
2
œ œ
5. radians below the negative x-axis: , [assuming counterclockwise].
1
6
3
¢£!!
È
##
"
6. , ¢£
È
3
##
"
7. 2 4 8. 5 3 4Š‹ Š ‹ab a b
!"
ˆ‰
182 134
41 17 17 55ÈÈÈ
Ɉ ‰ ˆ ‰
22
34
55
22
Ÿ
Ÿ
ij i j i j i jœ œ
9. length 2 2 2 2 2, 2 2 2 the direction is œœœ œ Ê ¹¹ Š‹
ÈÈ ÈÈ Èij ij ij ij
"" ""
ÈÈ ÈÈ
22 22
10. length 1 1 2, 2 the direction is œ œ œ œ Ê kk È ÈÈ
Š‹ij ij i j i j
"" ""
ÈÈ ÈÈ
22 22
11. t ln 2 e cos ln 2 e sin ln 2 e sin ln 2 e cos ln 2œÊœ vijˆ‰ˆ‰ab ab ab ab
ln 2 ln 2 ln 2 ln 2
2 cos ln 2 2 sin ln 2 2 sin ln 2 2 cos ln 2 2 cos ln 2 sin ln 2 sin ln 2 cos ln 2œ œ abab cdab ab ab ab ababab ab ab abij ij
length 2 2 cos ln 2 sin ln 2 cos ln 2 sin ln 2cos ln 2 sin ln 2 sin ln 2 cos ln 2œœ kk ababcd ab ab ab abababab ab ab ab Éij
22
2 2cos ln 2 2sin ln 2 2 2;œœÈ ab ab
È
22
222cos ln 2 sin ln 2 sin ln 2 cos ln 2cdababab ab ab ab
È
Š‹ œij
ababab ab ab ab
È
cos ln 2 sin ln 2 sin ln 2 cos ln 2
2
ij
directionʜ
ababab ab ab ab
ÈÈ
cos ln 2 sin ln 2 sin ln 2 cos ln 2
22

ij
12. t ( 2 sin ) 2 cos 2 ; length 2 4 0 2; 2 2 the direction isœÊœ œ œ œ œœÊ
111
222
viji i ii i ˆ‰ kk abÈ
13. length 2 3 6 4 9 36 7, 2 3 6 7 the direction is œ œ œ œ Ê kk È ˆ‰ijk ijk i j k i j k
236 236
777 777

817. Chapter 12 Practice Exercises 813
14. length 2 1 4 1 6, 2 6 the direction isœ œ œ œ Êkk È ÈÈ
Š‹ijk ijk i j k
121
666
ÈÈÈ

121
666
ÈÈÈ
ijk
15. 2 2 2
v
v
ij k ij k
kk ÈÈÈÈÈ
œ œ œ††
44 44
4(1)4 33 33 33 33
828


ijk
16. 5 5 5 3 4!$! $! $!!
v
v
ik ik
kk
ˆ‰ ˆ‰ ˆ‰ ˆ‰
Ɉ ‰ ˆ ‰ É
††
34 34
55 55
34
55
916
25 25
ŸŸ
Ÿ Ÿ

ik
17. 1 1 2, 4 1 4 3, 3, 3, 2 2 , 11 0
21 2
kk kkÈÈ È
ââ
ââ
ââ
ââ
ââ
ââ
vu vuuvvu ijk
ij k
œœ œœœœ‚œ œ
!
††
( ) 2 2 , 4 4 1 3, cos cos ,uv vu i jkvu‚œ‚œ ‚œ œ œ œ œ kk È Š‹ Š‹)
" " "vu
vu
†
kkkk È 4 2
1
cos , proj ( )kkuuvij)œœ
33
2 2È kkkk
vœŠ‹
vu
vv
†
18. 1 1 2 6, ( 1) ( 1) 2, (1)( 1) (1)(0) (2)( 1) 3,kk kk
ÈÈ È Èvu vuœœ œœ œ œ
### # #
†
3, , ( ) , 112
10 1
uv v u i j ku v v u i j k
ijk
†œ ‚ œ œ ‚ œ ‚ œ
!!
ââ
ââ
ââ
ââ
ââ
ââ
( 1) ( 1) 1 3, cos cos coskk È È
Š‹ Š ‹ Š‹vu‚œ œ œ œ œ
### " " "
)
vu
vu
†
kkkk ÈÈÈ
33
6 2 12
cos , cos 2 , proj ( 2 ) ( )œ œ œ † œ œ œ
"
# #
"
Š‹ Š‹ kk
È
ÈÈÈ
kkkk
336 5 3
622 6
1
uuvijkijk)
vœŠ‹
vu
vv
†
19. (2 ) ( 5 ) (2 ) (2 ) (5 11 ),u v u v ijk ij k ijk ijk ij kœŠ‹
vu vu
vv vv
††
kkkk kkkk
œ œ ’“Š‹ ‘
444
3333
"
where 8 and 6vu vv††œœ
20. ( 2 ) ( ) ( 2 ) ( 2 ) ,uvuvijijkijijijkœŠ‹
vu vu
vv vv
††
kkkk kkkk
œ œ ’“Š‹ ‘ˆ‰ˆ‰
11145
33333

where 1 and 3vu vv††$! $
21. 100
110
AB k
ijk
‚œ œ
ââ
ââ
ââ
ââ
ââ
ââ

22. 2 110
110
AB k
ijk
‚œ œ !
ââ
ââ
ââ
ââ
ââ
ââ

818. 814 Chapter 12 Vectors and the Geometry of Space
23. Let v v v and w w w . Then 2 (v v v ) 2(w w w )vijkwijk vw ijk ijkœ œ œ "#$ " # $ "#$ " # $
## kkk k
(v 2w ) (v 2w ) (v 2w ) (v 2w ) (v 2w ) (v 2w )œ œ kk ˆ‰È "" ## $$ "" ## $$
#
###
# ijk
v v v 4(vw vw vw) 4 w w w 4 4 œ œ ab a b kk k k
### # # #
"" ## $$
##
$! ! 313 vvww†
4 cos 4 4 4(2)(3) cos 36 40 24 40 12 28 2 28œ œ œ œœ Ê œkk kkk k k k k k ˆ‰ ˆ‰ È
vvw w vw
## "
#
)
1
3
27œ
È
24. and are parallel when (4a 40) (20 2a) (0)24 5
48a
uv uv0 0 i j k0
ijk
‚œ Ê œ Ê œ!
!!
ââ
ââ
ââ
ââ
ââ
ââ
4a 40 0 and 20 2a 0 a 10"!$ !$"$
25. (a) area abs 2 3 4 9 1 1411 1
21 1
œ‚œ œ œ œ !kk k k
ââ
ââ
ââ
ââ
ââ
ââ
È È
uv i jk
ij k
(b) volume ( ) 1(3 2) 1( 1 6) 1( 4 1) 1
11 1
211
123
œ‚œ œœ
!
!!
uvw†
ââ
ââ
ââ
ââ
ââ
ââ
26. (a) area abs 1 110
010
œ‚œ œ œkk kk
ââ
ââ
ââ
ââ
ââ
ââ
uv k
ijk
(b) volume ( ) 1(1 0) 1(0 0) 0 1
110
010
111
œ ‚ œ œ œuvw†
ââ
ââ
ââ
ââ
ââ
ââ
27. The desired vector is or since is perpendicular to both and and, therefore, also parallel tonvvn nv n v‚‚ ‚
the plane.
28. If a 0 and b 0, then the line by c and are parallel. If a 0 and b 0, then the line ax c and areœÁ œ Áœ œ ij
parallel. If a and b are both 0, then ax by c contains the points and 0 the vectorÁœ ß!ßÊ ˆ‰ ˆ‰
cc
ab
ab c(b a ) and the line are parallel. Therefore, the vector b a is parallel to the lineˆ‰
cc
ab
i j ij ij!$! !
ax by c in every case.œ
29. The line L passes through the point P(0 0 1) parallel to . With PS 2 2 andßß œ œ
Ä
vijk ijk
PS (2 1) ( 1 2) (2 2) 3 4 , we find the distance221
111
Ä
‚œ œ œ
!
vijkijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
d.œœ œœ
¹¹
kk
ÈÈÈ
ÈÈ
PS
1916
111
26 78
3 3
Ä
‚
ŸŸ
ŸŸ
v
v
30. The line L passes through the point P(2 2 0) parallel to . With PS 2 2 andßß œ œ
Ä
vijk i jk
PS (2 1) (1 2) ( 2 2) 3 4 , we find the distance221
111
Ä
‚œ œ œ!vijkijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
d.œœ œœ
¹¹
kk
ÈÈÈ
ÈÈ
PS
1916
111
26 78
3 3
Ä
‚
ŸŸ
ŸŸ
v
v
31. Parametric equations for the line are x 1 3t, y 2, z 3 7t.œ œ œ

819. Chapter 12 Practice Exercises 815
32. The line is parallel to PQ 0 and contains the point P(1 2 0) parametric equations are
Ä
œ ßß Êijk
x 1, y 2 t, z t for 0 t 1.œœœ ŸŸ
33. The point P(4 0 0) lies on the plane x y 4, and PS (6 4) 0 ( 6 0) 2 6 with ßß œ œ œ œ
Ä
ij kik nij
d2.Êœ œ œ œ
¹¹
kk ÈÈ
n
n
†PS
200 2
110 2
Ä
ŸŸ
ŸŸ
¹¹
È
34. The point P(0 0 2) lies on the plane 2x 3y z 2, and PS (3 0) (0 0) (10 2) 3 8 withßß œ œ œ
Ä
ij kik
2 3 d 14.nijkœ Êœ œ œ œ
¹¹
kk ÈÈ
n
n
†PS
608 14
491 14
Ä
ŸŸ
ŸŸ
¹¹
È
35. P(3 2 1) and 2 (2)(x 3) (1)(y ( 2)) (1)(z 1) 0 2x y z 5ßß œ Ê œ Ê œnijk
36. P( 1 6 0) and 2 3 (1)(x ( 1)) ( 2)(y 6) (3)(z 0) 0 x 2y 3z 13ßß œ Ê œÊœni j k
37. P(1 1 2), Q(2 1 3) and R( 1 2 1) PQ 2 , PR 2 3 3 and PQ PRß ß ß ß ß ß Ê œ œ ‚
ÄÄ ÄÄ
ijk ijk
9 7 is normal to the plane ( 9)(x 1) (1)(y 1) (7)(z 2) 0121
23 3
œœ Êœ
!!
ââ
ââ
ââ
ââ
ââ
ââ
ijk
ij k
9xy7z4Ê œ
38. P(1 0 0), Q(0 1 0) and R(0 0 1) PQ , PR and PQ PRßß ßß ßß Ê œ œ ‚
ÄÄ ÄÄ
ij ik
is normal to the plane (1)(x 1) (1)(y 0) (1)(z 0) 0110
101
œ œ Ê œ!
!
ââ
ââ
ââ
ââ
ââ
ââ
ijk
ijk
xyz1Ê œ
39. 0 , since t , y and z when x 0; ( 0 3), since t 1, x 1 and z 3ˆ‰w! w! $! $! $! $ !w w! $! $! $!
"""
## # # #
33
when y 0; (1 1 0), since t 0, x 1 and y 1 when z 0$w!w $$ $! $
40. x 2t, y t, z t represents a line containing the origin and perpendicular to the plane 2x y z 4; this$ $! $! !!$
line intersects the plane 3x 5y 2z 6 when t is the solution of 3(2t) 5( t) 2( t) 6 œ œ
t is the point of intersection"$ " w!w!
2 422
3 333
ˆ‰
41. and 2 the desired angle is cos cosni nij k"#
" " "
#œ œ Ê œ œ
È
Š‹ ˆ‰
nn
nn
†
kkkk 3
1
42. and the desired angle is cos cosnij njk"#
" " "
#œ œ Ê œ œ Š‹ ˆ‰
nn
nn
†
kkkk 3
1
43. The direction of the line is 5 3 . Since the point ( 5 3 0) is on121
112
nn ijk
ijk "#†$ $!! !ww
!
ââ
ââ
ââ
ââ
ââ
ââ
both planes, the desired line is x 5 5t, y 3 t, z 3t.œ œ œ
44. The direction of the intersection is 6 9 12 3(2 3 4 ) and is the12 2
521
nn ij k ijk
ij k
"#‚ œ œ œ !
!!
ââ
ââ
ââ
ââ
ââ
ââ
same as the direction of the given line.

820. 816 Chapter 12 Vectors and the Geometry of Space
45. (a) The corresponding normals are 3 and 2 2 and since niknijk nn "# "#œ' œ †
(3)(2) (0)(2) (6)( 1) 6 0 6 0, we have that the planes are orthogonalœ œœ
(b) The line of intersection is parallel to 12 15 6 . Now to find a point in30 6
22 1
nn i jk
ij k
"#‚œ œ
!
ââ
ââ
ââ
ââ
ââ
ââ
the intersection, solve 15x 12y 19 x 0 and y
3x 6z 1 3x 6z 1
2x 2y z 3 12x 12y 6z 18
œœ
œ œ
œ œ
Êʜʜœ
19
12
is a point on the line we seek. Therefore, the line is x 12t, y 15t and z 6t.Ê!ßß œ œ œˆ‰
19 19
16 12 6#
" "
46. A vector in the direction of the plane's normal is 7 3 5 and P( 2 3) on231
112
nuv i j k
ijk
œ‚œ œ "ßß
!
ââ
ââ
ââ
ââ
ââ
ââ
the plane 7(x 1) 3(y 2) 5(z 3) 0 7x 3y 5z 14." !! !! !$" !!$!
47. Yes; (2 4 ) (2 0 ) 2 2 4 1 1 0 0 the vector is orthogonal to the plane's normalvn i j k i j k† † †††œ œ œÊ
is parallel to the planeÊv
48. PP 0 represents the half-space of points lying on one side of the plane in the direction which the normal n n†
!
Ä

points
49. A normal to the plane is AB AC 2 2 the distance is d20 1
210
nijk
ij k
$†$ $!!!" $
ÄÄ
!
!
ââ
ââ
ââ
ââ
ââ
ââ
¹¹
AP
Ä
†n
n
3œœœ¹¹ ¸¸
(4)( 2 )
144
180
3
ij ijk#
ŸŸ
†
È
50. P(0 0 0) lies on the plane 2x 3y 5z 0, and PS 2 2 3 with 2 3 5 ßß œ œ œ Ê
Ä
ijk nijk
dœœ œ¹¹¹ ¹
n
n
†PS 4615 25
4925 38
Ä
ŸŸ
ŸŸkk ÈÈ
51. 2 is normal to the plane 0 3 3 3 3 is orthogonal211
11 1
nijk nv ijk jk
ij k
œ Ê‚œ œ œ !!
ââ
ââ
ââ
ââ
ââ
ââ
to and parallel to the planev
52. The vector is normal to the plane of and ( ) is orthogonal to and parallel to the plane of BC B CABC A B‚Ê‚‚
and :C
5 3 and ( ) 2 312 1 2 1 1
11 2 5 3 1
BC i jk A BC i jk
ij k i j k
‚œ œ ‚ ‚ œ œ
!!!
!
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
( ) 4 9 1 14 and ( 2 3 ) is the desired unit vector.Ê‚‚œ œ œ kk È È
ABC u i jk
"
È
14
53. A vector parallel to the line of intersection is 5 3121
112
vn n ij k
ijk
$†$ $!!
! "#
ââ
ââ
ââ
ââ
ââ
ââ
25 1 9 35 2 (5 3 ) is the desired vector.Êœ œ Ê œ kkÈ È
Š‹vijk
v
vkk È
2
35
54. The line containing (0 0 0) normal to the plane is represented by x 2t, y t, and z t. This lineww $ $! $!
intersects the plane 3x 5y 2z 6 when 3(2t) 5( t) 2( t) 6 t the point is . œ œ Êœ Ê ßß
2 422
3 333
ˆ‰

821. Chapter 12 Practice Exercises 817
55. The line is represented by x 3 2t, y 2 t, and z 1 2t. It meets the plane 2x y 2z 2 whenœ œ œ œ
2(3 2t) (2 t) 2( 2t) 2 t the point is . " œÊœ Ê ßß
811267
9999
ˆ‰
56. The direction of the intersection is 3 5 cos21 1
11 2
vn n i jk
ij k
œ‚œ œÊœ ! "#
"
ââ
ââ
ââ
ââ
ââ
ââ
Š‹)
vi
vi
†
kkkk
cos 59.5°œ¸
"
Š‹
3
35
È
57. The intersection occurs when (3 2t) 3(2t) t 4 t 1 the point is (1 2 1). The required line œÊœÊ ßß
must be perpendicular to both the given line and to the normal, and hence is parallel to 22 1
13 1
ââ
ââ
ââ
ââ
ââ
ââ
ij k
!
5 3 4 the line is represented by x 1 5t, y 2 3t, and z 1 4t.œ Ê œ œ œ ijk
58. If P(a b c) is a point on the line of intersection, then P lies in both planes a 2b c 3 0 andßß Ê œ
2a b c 1 0 (a 2b c 3) k(2a b c 1) 0 for all k.œ Ê œ
59. The vector AB CD (2 7 2 ) is normal to the plane and A( 2 0 3) lies on the324
0
ÄÄ
‚œ œ ßß!
!
ââ
ââ
ââ
ââ
ââ
ââ
ij k
ijk
26 26
55
26
5
plane 2(x 2) 7(y 0) 2(z ( 3)) 0 2x 7y 2z 10 0 is an equation of the plane.ʜʜ
60. Yes; the line's direction vector is 2 3 5 which is parallel to the line and also parallel to the normalijk
4 6 10 to the plane the line is orthogonal to the plane. Êij k
61. The vector PQ PR 11 3 is normal to the plane.213
301
Ä
†$ $!!!
Ä
!
!
ââ
ââ
ââ
ââ
ââ
ââ
ijk
ijk
(a) No, the plane is not orthogonal to PQ PR .
Ä
‚
Ä
(b) No, these equations represent a line, not a plane.
(c) No, the plane (x 2) 11(y 1) 3z 0 has normal 11 3 which is not parallel to PQ PR . œ ‚
Ä Ä
ijk
(d) No, this vector equation is equivalent to the equations 3y 3z 3, 3x 2z 6, and 3x 2y 4œ œ œ
x t, y t, z 1 t, which represents a line, not a plane."$!! $ $!
42
33
(e) Yes, this is a plane containing the point R( 2 1 0) with normal PQ PR .!w w †
Ä Ä
62. (a) The line through A and B is x 1 t, y t, z 1 5t; the line through C and D must be parallel andœ œ œ
is L : x 1 t, y 2 t, z 3 5t. The line through B and C is x 1, y 2 2s, z 3 4s; the line
"œ œ œ œ œ œ
through A and D must be parallel and is L : x 2, y 1 2s, z 4 4s. The lines L and L intersect
#"#œœ œ
at D(2 1 8) where t 1 and s 1.ßß œ œ
(b) cos )œœ
(2 4 ) ( 5 )
20 27 15
3jkijk†
ÈÈ È
(c) BC BC ( 2 ) where BA 5 and BC 2 4Š‹
BA BC 18 9
BC BC
20 5
ÄÄ
ÄÄ
†
†
ÄÄ Ä Ä
œ œ œ œ jk ijk jk
(d) area (2 4 ) ( 5 ) 14 4 2 6 6œ‚ œœkkkk
È
jk ijk ijk
(e) From part (d), 14 4 2 is normal to the plane 14(x 1) 4(y 0) 2(z 1) 0nijkœ Ê œ
7x 2y z 8.ʜ
(f) From part (d), 14 4 2 the area of the projection on the yz-plane is 14; the area of thenijk niœÊ œ kk†
projection on the xy-plane is 4; and the area of the projection on the xy-plane is 2.kk k knj nk††œœ

822. 818 Chapter 12 Vectors and the Geometry of Space
63. AB 2 , CD 4 and AC 2 5 9 the distance is 21 1
14 1
ÄÄ Ä
œ œ œ Ê œ Ê !
!
ijk i jk, ij n ij k
ijk
œ
ââ
ââ
ââ
ââ
ââ
ââ
dœœ¹¹
(2 ) ( 5 9 )
25181
11
107
ij ij k
ŸŸ
†
ÈÈ
64. AB 2 4 , CD 2 , and AC 3 3 7 3 2 the distance 24 1
112
ÄÄ Ä
œ œ œ Êœ œ Ê !!
!
ijk ijk ij n ijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
is dœœ¹¹
(3 3)(7 3 2)
4994
12
62

ŸŸ
ijijk†
ÈÈ
65. x y z 4 66. x (y 1) z 1 67. 4x 4y z 4
### # ## # ##
œ œ œ

68. 36x 9y 4z 36 69. z x y 70. y x z
### ## ##
œ œ œ ab ab

71. x y z 72. x z y 73. x y z 4
### ## # ###
œ œ œ

823. Chapter 12 Additional and Advanced Exercises 819
74. 4y z 4x 4 75. y x z 1 76. z x y 1
## # ### ###
œ œ œ

CHAPTER 12 ADDITIONAL AND ADVANCED EXERCISES
1. Information from ship A indicates the submarine is now on the line L : x 4 2t, y 3t, z t;"
"œœœ
3
information from ship B indicates the submarine is now on the line L : x 18s, y 5 6s, z s. The#$$!$!
current position of the sub is 6 3 and occurs when the lines intersect at t 1 and s . The straightˆ‰ww! $ $
""
33
line path of the submarine contains both points P 2 1 and Q 6 3 ; the line representing this pathˆ‰ˆ‰w! w! w w!
""
33
is L: x 2 4t, y 1 4t, z . The submarine traveled the distance between P and Q in 4 minutes œ œ œ Ê
"
3
a speed of 2 thousand ft/min. In 20 minutes the submarine will move 20 2 thousand ft from
¹¹ ÈPQ
44
32
Ä
œœ
ÈÈ
Q along the line L 20 2 (2 4t 6) ( 1 4t 3) 0 800 16(t 1) 16(t 1) 32(t 1)Ê œ Ê œ œ
È È
### ## #
(t 1) 25 t 6 the submarine will be located at 26 23 in 20 minutes."! $ $ "$" ww!
# "800
32 3 ˆ‰
2. H stops its flight when 6 110t 446 t 4 hours. After 6 hours, H is at P(246 57 9) while H is at
# "# œÊœ ßß
(446 13 0). The distance between P and Q is (246 446) (57 13) (9 0) 204.98 miles. At 150ßß ¸ È
###
mph, it would take about 1.37 hours for H to reach H ."#
3. Torque PQ 15 ft-lb PQ sin ft 20 lbœ‚Ê œ œ Êœ
ÄÄ
¹¹ ¹¹ kk kk kkFFFF
1
#
3
4
†
4. Let be the vector from O to A and 3 2 be the vector from O to B. The vector orthogonal to aijk bi j k v aœ œ
and is parallel to (since the rotation is closkwise). Now 2 ; proj 2 2 2bv ba baij k b a i j kÊ ‚ œ œ œ ‚ a
ab
aa
ˆ‰
†
†
2, 2, 2 is the center of the circular path 1, 3, 2 takes radius 1 1 0 2 arc length perÊÊœœÊab ab ab É È
22
2
second covered by the point is 2 units/sec (velocity is constant). A unit vector in the direction of is
3
# ‚
‚È
kkœvv ba
bakk
23œ Êœ œ œ
"" ‚ ""
‚#ÈÈÈ ÈÈÈ kk
ÈÈ
666 666
232 33
22
ijkvv ijk ij k kkŠ‹ Š ‹
È È
ba
ba
5. (a) If P(x y z) is a point in the plane determined by the three points P (x y z ), P (x y z ) andßß ßß ßß "" "" ## ##
P (x y z ), then the vectors PP , PP and PP all lie in the plane. Thus PP (PP PP ) 0$$ $$ " # $ " # $ßß ‚ œ
ÄÄ Ä Ä Ä Ä
†
0 by the determinant formula for the triple scalar product in Section 10.4.
xxyyzz
xxyyzz
xxyyzz
ʜ
!!!
!!!
!!!
ââ
ââ
ââ
ââ
ââ
ââ
"""
###
$$$
(b) Subtract row 1 from rows 2, 3, and 4 and evaluate the resulting determinant (which has the same value
as the given determinant) by cofactor expansion about column 4. This expansion is exactly the
determinant in part (a) so we have all points P(x y z) in the plane determined by P (x y z ),ßß ß ß
"" ""
P(xyz), and P(xyz).#### $$$$ßß ßß

824. 820 Chapter 12 Vectors and the Geometry of Space
6. Let L : x a s b , y a s b , z a s b and L : x c t d , y c t d , z c t d . If L L ," " " # # $ $ # "" ## $$ "#œ œ œ œ œ œ ²
then for some k, a kc , i 1, 2, 3 and the determinant
acb d kccb d
acb d kccb d
acb d
iiœœ œ
!!
!!
!
âââ â
âââ â
âââ â
âââ â
âââ â
âââ â
""" " """ "
### # ### #
$$$ $
kccb d
0,$$$ $ !
œ
since the first column is a multiple of the second column. The lines L and L intersect if and only if the
"#
system has a nontrivial solution the determinant of the coefficients i
as ct (b d) 0
as ct (b d) 0
as ct (b d) 0
Ú
Û
Ü
"" ""
## ##
$$ $$ œ
œ
œ
Í s zero.
7. (a) BD AD AB
Ä
$!
ÄÄ
(b) AP AB BD AB AD
ÄÄ ÄÄ
œ œ
Ä
""
##
Š‹
(c) AC AB AD, so by part (b), AP AC
ÄÄÄ Ä Ä
œ œ
"
#
8. Extend CD to CG so that CD DG. Then CG t CF CB BG and t CF 3 CE CA, since ACBG is a
ÄÄ ÄÄ Ä ÄÄÄ Ä ÄÄ
œœœœ
parallelogram. If t CF 3 CE CA , then t 3 1 0 t 4, since F, E, and A are collinear.
ÄÄÄ
!!$ !!$"$ 0
Therefore, CG 4 CF CD 2 CF F is the midpoint of CD.
ÄÄ ÄÄ
œÊœÊ
9. If Q(x y) is a point on the line ax by c, then P Q (x x ) (y y ) , and a b is normal to theßœœœ
Ä
""" ijnij
line. The distance is proj P Q¹¹¹ ¹
n"

ŸŸ

Ä
œœ
[(xx) (yy)](a b)
ab ab
a(x x ) b(y y )""

""ijij †
ÈÈ
kk
, since c ax by.œœ
kk
È
ax by c
ab""

Ÿ
10. (a) Let Q(x y z) be any point on Ax By Cz D 0. Let QP (x x ) (y y ) (z z ) , andßß œ œ
Ä "" " " ijk
. The distance is proj QP ((x x ) (y y ) (z z ) )nijkœœ
Ä
ABC ABC
ABC ABC
ijk ijkŸŸ ŸŸ
ŸŸ ŸŸ
""""È È

¹¹¹ ¹ Š‹n †
. œœ
kkkk
ÈÈ
Ax By Cz (Ax By Cz) Ax By Cz D
ABC ABC""" """

ŸŸ ŸŸ
(b) Since both tangent planes are parallel, one-half of the distance between them is equal to the radius of the
sphere, i.e., r 3 (see also Exercise 17a). Clearly, the points (1 2 3) and ( 1 2 3)$ $ w w !w!w!
"
#

ŸŸ
kk
È
39
111 È
are on the line containing the sphere's center. Hence, the line containing the center is x 1 2t,œ
y 2 4t, z 3 6t. The distance from the plane x y z 3 0 to the center is 3œ œ œ
È
3 from part (a) t 0 the center is at (1 2 3). ThereforeʜʜÊßß
kk
È
(1 2t) (2 4t) (3 6t) 3
111

ŸŸ È
an equation of the sphere is (x 1) (y 2) (z 3) 3.œ
###
11. (a) If (x y z ) is on the plane Ax By Cz D , then the distance d between the planes is""" "ßß œ
d , since Ax By Cz D , by Exercise 10(a).œœ œ
kkkk
È kk
Ax By Cz D D D
ABC ABC

ŸŸ ŸŸ
""""
ijk (b) dœœ
kk
ÈÈ
12 6
491
6
14

ŸŸ
(c) D 8 or 4 the desired plane is
kkk k
ÈÈ
2(3) ( 1)(2) 2( 1) 4 2(3) ( 1)(2) 2( 1) D
14 14

$"$!"
2xy2x8 œ
(d) Choose the point (2 0 1) on the plane. Then 5 D 3 5 6 the desired planes areßß œ Ê œ „ Ê
kk
È
3D
6
È
x2yz356 and x2yz356.œ œ
ÈÈ
12. Let AB BC and D(x y z) be any point in the plane determined by A, B and C. Then the point D lies innœ‚ ßß
ÄÄ
this plane if and only if AD 0 AD (AB BC) 0.
ÄÄÄÄ
œÍ ‚ œ††n

825. Chapter 12 Additional and Advanced Exercises 821
13. 2 6 is normal to the plane x 2y 6z 6; 4 5 is parallel to the111
126
ni j k vn i jk
ijk
œ œ ‚œ œ
ââ
ââ
ââ
ââ
ââ
ââ
plane and perpendicular to the plane of and ( ) 32 23 13 is a126
451
vnwnvn i j k
ijk
Êœ‚‚œ œ
!
ââ
ââ
ââ
ââ
ââ
ââ
vector parallel to the plane x 2y 6z 6 in the direction of the projection vector proj . Therefore,œ
Pv
proj proj
Pv vv w w wwijkœ œ œ œ œ œ œ
w Š‹ Š‹ ˆ‰†
ww vw
ww wkk kk kk
†

32 23 13 42 32 23 13
32 23 13 1722 41 41 41 41
"
ŸŸ
14. proj proj and proj proj ( proj ) proj ( proj ) proj
zz z z zz zzwvwwvvwwwwvvwœ œ Ê œ œ
2 proj 2 $! $!vvv z
z Š‹
vz
z
†
kk

15. (a) 2 2 4 ( ) ; ( ) ( ) 0 0 ; 4 ( ) ;uv i j k uv C0uwv vwu v u0vw i u vw 0†$ † $ " † † $ ! $ ! $ † $ "† † $††
()()00uwv uvw v w 0††!$!$
(b) 43 ( ) 1026;111 143
21 2 12 1
uv ijk uvw ijk
ij k ijk
‚œ œ Ê ‚ ‚ œ œ !
!!!
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
( ) ( ) 4(2 2 ) 2( ) 10 2 6 ;uwv vwu i j k i j k i j k†† œœ
vw i j k u vw i j k
ijk ijk
‚œ œ Ê‚‚ œ œ!!
!!
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
21 2 1 11
12 1 3 4 5
345 ( ) 927;
( ) ( ) 4(2 2 ) ( 1)( 2 ) 9 2 7uwv uvw i j k i j k i j k†† œœ
(c) 2 4 ( ) 4 6 2 ;210 124
211 102
uv ijk uvw ijk
ijk ij k
‚œ œ Ê ‚ ‚ œ œ
!
!!
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
( ) ( ) 2(2 ) 4(2 ) 4 6 2 ;uwv vwu i j k i j i j k††œœ
vw i jk u vw i j k
ijk i jk
‚œ œÊ‚‚ œ œ!
!!
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
211 210
102 2 31
23 ( ) 24;
( ) ( ) 2(2 ) 3( 2 ) 2 4uwv uvw i j k i k i j k†† œ œ
(d) 3 ( ) 10 10 ;11 2 131
10 1 2 4 2
uv i jk uv w i k
ijk ijk
‚ œ œ Ê ‚ ‚ œ œ !!
!! !
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
( ) ( ) 10( ) 0( 2 ) 10 10 ;uwv vwu ik ij k i k††œœ
vw i j k u vw i j k
ijk ij k
†$ $!! "†† $ $! !!!! !
!!!
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
ââ ââ
10 1 1 1 2
242 444
4 4 4 ( ) 12 4 8 ;
( ) ( ) 10( ) 1(2 4 2 ) 12 4 8uwv uvw i k i j k i j k††œœ
16. (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )uvwvwuwuv uwvuvwvuwvwuwvuwuv0‚‚‚‚‚‚œœ †††† ††
(b) [ ( )] [( ( )] [( ( )] [( ) ] [( ) ] [( ) ]uvii uvjj uvkk uvii uvjj uvkkuv††† †††‚ ‚ ‚ œ‚ ‚ ‚ œ‚
(c) ( ) ( ) [ ( )] [( ) ( ) ] ( )( ) ( )( )uv wr uv wr u vrw vwr uwvr urvw††$††$ ! $ !†† †††††††
œºº
uw vw
ur vr
††
††
17. The formula is always true; [ ( )] [( ) ( ) ]uuuvwu uvuuuvw††† $† ! ††††
[( ) ( ) ] $†!†$!†$!†uvuu uuuvw uuvw uuvw††† †† kk kk
##

826. 822 Chapter 12 Vectors and the Geometry of Space
18. If (cos ) (sin ) and (cos ) (sin ) , where , then sin ( )AijBij ABAB kœ œ ‚œ !! "" "! "! cdkkkk
(cos sin sin cos ) sin ( ) cos sin sin cos , sincecos sin 0
cos sin 0
$$!"!$!
ââ
ââ
ââ
ââ
ââ
ââ
ijk
k!!
""
!" ! " "! !" ! "
1 and 1.kk kkABœœ
19. If a b and c d , then cos ac bd a b c d cos Aij Bij ABABœ œ œ Ê œ † kkkk
ÈÈ
))
####
(ac bd) a b c d cos (ac bd) a b c d , since cos 1.Ê œ Ê Ÿ Ÿ
###### ##### #
abab abab ))
20. proj and CA BDACA B$ $ $!$!
B Š‹ Š‹
AB AB
BB BB
††
kk kkkk kk
21. ( ) ( ) 2 2 k k kk kkkk kk ab kkkk kkAB AB AB AA ABBB A AB B A B ABœ œ Ÿ œ Ê
####
††††
Ÿkk kkAB
22. Let denote the angle between and , and the angle between and . Let a and b . Then!" CA CB A B œœkk kk
cos ,!œœœœœœ
CA BA
CA CA CA CA C C
BAA BAAA BAAA BA† †††† †† †
kkkk kkkk kkkk kkkk kk kk
ab
a
(a b ) (a b ) (a b ) aba baŸŸŸ Ÿ Ÿ

and likewise, cos . Since the angle between A and B is always and cos cos , we have"!"œŸœ
AB
C
†Ÿ
#
ba
kk
1
that bisects the angle between and .!"œÊ CAB
23. (a b ) (b a ) a b b b a a b a b a b a b aB A A B BA AA BB AB AB AA BB AB œœ† ††††† †††
##
b a a b 0, where a and b$!$ $ $
## ##
kk kkAB
24. If a b c , then a b c 0 and 0 iff a b c 0.A i j k AA AAœ œ  œ œœœ ††
###
25. (a) The vector from (0 d) to (kd 0) is kd d . Thew w $!"$ "$rij
k
1
dk 1 dk 1
k
kk kk ab abrr
r ij
kk
k
!!


"
ŸŸ

total force on the mass (0 d) due to the masses Q for k n, n 1, , n 1, n isßœá k
() Fjœ á
GMm GMm GMm GMm GMm
d2d 5d n1d 2d 2n12
2n
5

Š‹ Š ‹ Š ‹ Š ‹
ij ij ij ij
Ÿ Ÿ
ÈÈ È È ab
á
GMm GMm
5d n 1 d
2n
5 n1

Š‹ Š ‹

Ÿ Ÿ
ij ij
ÈÈ ab
The components cancel, givingi 1 the magnitude of the force isFj$ !!!!Ä! "
GMm 2 2 2
d 22 55 n1n1


Š‹
ÈÈ
ababŸŸ
1 .kk Œ Ÿ
!Fœ
GMm 2
d
i1



n
i1œ
abŸ
(b) Yes, it is finite: lim 1 is finite since converges.
nÄ_
kk Œ Ÿ
!Fœ
GMm 2 2
d
i1 i1



!
_
œ
_
œi1 i1
ab abŸŸ
26. (a) If x y 0, then x (x y) (x y)x (x x)y (x x)y. This means thattt t t t ttt ttt ttt$ ††$!$!††††
x y x y (x x) y x 1 y. Since x and y aretttt tttt t t tŠœ œ
""
Ÿ
t
t
c
1
x
ccc x


††
É1
xx
c
œ

ab
!"
kk
É kk
orthogonal, then x y x 1 y . A calculation will show thatk k kk kk
!"
tt t tŠœ
## # t
t
#
kk
É kk
x
ccc x


x 1 c c . Since y c, then y c sokk kk kk
!"
ttt œ
# #t
t
#
## #kk
É kk
x
ccc x


1 y 1 c . This means that
!"!"
kk t
kk kk
ÉÉ kk kk
xx
ccc x ccc x
tt
tt
#
# #

827. Chapter 12 Additional and Advanced Exercises 823
xy x 1 y x 1 c c.k k kk kk kk
!"!"
tt t t tŠœ œ
## ## tt
tt
##
##kk kk
ÉÉ kk kk
xx
ccc x ccc x


We now have x y c , so x y c.kk kktt ttŠ Š
# #
(b) If x and y are parallel, then x (x y) 0. This gives x y .tt ttt tt ‚‚œ Šœ t
xy
1
ttŸ
Ÿ
xy
c
$

(i) If x and y have the same direction, then x y and x y .t t tt tt Šœ Š œ
xy
11
xyttŸ
ŸŸ
ttŸ
xy xy
cc cc
††
kk
kk kk
Since y c, x c, we have y 1 c 1 y c xkk kk kk kk kk Š‹Š‹tt t t t Ê
kk kk kkkkxx y x
cc c
tt tt
x y c c 1 c. This means that x y c.Ê œ Ê Štt ttkk kk k k Š‹
kkkk kk kk kk kkx y x y x y
ccc
1
tt t t t t Ÿ
Ÿ
†
xy
cc
† (ii) If x and y have opposite directions, then x y x y and x y .tt tt tt tt œ Š œ† kkkk
xy
1
ttŸ

x y
c
$ Assume x y , then x y . Since x c, we have x 1 c 1kk kk k k kk kk Š‹Š‹tt tt t t Šœ
kk kk kk kkxy y y
1
cc
tt t t

x y
c
$ x c y x y c c 1 c.Ê Ê œ Ê ttttkk kk kk kk Š‹
kkkk kkkk kkkk kk kkx y x y x y x y
ccc
1
tt tt tt t t

x y
c
$
This means that x y c. A similar argument holds if x y .k k kk kktt t tŠ
(c) lim x y x y.
cÄ_
ttttŠœ

828. 824 Chapter 12 Vectors and the Geometry of Space
NOTES:

829. CHAPTER 13 VECTOR-VALUED FUNCTIONS
AND MOTION IN SPACE
13.1 VECTOR FUNCTIONS
1. x t 1 and y t 1 y (x 1) 1 x 2x; 2t 2 2 and 2œ œÊœ œ œ œ Êœ œ Êœ œ
###
vij a jvijaj
dd
dt dt
rv
at t 1œ
2. x t 1 and y 2t 1 x x (y 1) 1; 2t 2 2œ œÊœ "Êœ œ œ Êœ œ
# #œ
#
#
" ˆ‰
y1
4dt dt
dd
vijai
rv
2 and 2 at tÊœ œ œvi j a i
"
#
3. x e and y e y x ; e e e e 3 4 and 3 8 at t ln 3œ œ Êœ œ œ Êœ Êœ œ œ
t2t t2tt2t22d4 8
99dt9 9
#
vijaijvijaij
r
4. x cos 2t and y 3 sin 2t x y 1; ( 2 sin 2t) (6 cos 2t) œœÊœœœÊœ
##"
9dt dt
dd
vija
rv
( 4 cos 2t) ( 12 sin 2t) 6 and 4 at t 0œ Ê œ œ œijvjai
5. (cos t) (sin t) and (sin t) (cos t)vija ijœœ œœ
dd
dt dt
rv
for t , andÊœ œ
11
44
22
vijˆ‰
ÈÈ
##
; for t , andaijvjˆ‰ ˆ‰
111
4
22
œ œ œ
ÈÈ
## ##
aiˆ‰
1
#
œ

6. 2 sin 2 cos and vijaœœ œ
dttd
dt dt
rv
ˆ‰ˆ‰
##
cos sin for t , ( ) 2 andœ Ê œ œˆ‰ˆ‰
tt
##
ij vi 11
() ; for t , 2 2 andaj v ij1œ œ œ
3311
##
ˆ‰ ÈÈ
aijˆ‰
3 221
###
œ
ÈÈ

7. (1 cos t) (sin t) and vijaœœ œ
dd
dt dt
rv
(sin t) (cos t) for t , ( ) 2 and ( ) ;œ Êœ œ œij viaj 11 1
for t , and œœ œ
33 311 1
## #
vija iˆ‰ ˆ‰

8. 2t and 2 for t 1,vija jœœ œœ Ê œ
dd
dt dt
rv
( 1) 2 and ( 1) 2 ; for t 0, (0) andvija j viœ œ œ œ
(0) 2 ; for t 1, (1) 2 and (1) 2aj vijajœœœ œ

830. 826 Chapter 13 Vector-Valued Functions and Motion in Space
9. (t 1) t 1 2t 2t 2 2 ; Speed: (1) 1 (2(1)) 2 3;ri jkvijka j vœ Ê œ œ Ê œ œ œ œab kk È
# ###dd
dt dt
rr
œ
œ
Direction: (1) 3
vijk
v
(1) 2(1) 2
(1) 3 3 3 3 3 3 3
22 22
kk
œ œ Ê œ
œœ ""
ijkv ijk ˆ‰
10. (1 t) t 2t ; Speed: (1)rijkvijka jk vœ Êœ œ Êœ œ
tt d 2t d2
2223dt dt
œ " œ
œ
ÈÈÈ
rr #
kk
1 (1 ) 2; Direction: (1)œ œ œ œÊÊ Š‹
###
# œœ
## #
"" "2(1) (1)
2 2(1)
(1 )
È Èkk
v
v
ijk
2(1)
2
œ
œ
ijkv
2œŠ‹
"" "
##
ijk
È
2
11. (2 cos t) (3 sin t) 4t ( 2 sin t) (3 cos t) 4 ( 2 cos t) (3 sin t) ;rijkv ijka ijœÊœœÊœœ
dd
dt dt
rr
œ
œ
Speed: 2 sin 3 cos 4 2 5; Direction: ¸¸ ˆ ‰ˆ ‰ˆ‰ É Èv
111
##
##
#
œ œ
2
v
v
ˆ‰
¸¸ˆ‰
œ
œ
œ
œ
sin cos 2 5œ œ Ê œ Š‹Š‹ Š ‹ ˆ‰ È
234 2 2
55555 55### ## #
""
ÈÈÈÈÈ ÈÈ
11 1
ijkikv ik
12. (sec t) (tan t) t (sec t tan t) sec t rijkv i jkaœÊœœ Êœ
4d 4d
3dt 3dt
rr
ab
#
œ
œ
sec t tan t sec t 2 sec t tan t ; Speed: sec tan sec 2;œ œ œabab ¸¸ ˆ ‰ˆ ‰ˆ‰ˆ‰ É
#$ #
##
#
#
ijv
1111
66663
4
Direction: 2
vijk
v
ˆ‰ ˆ ‰
¸¸ˆ‰
œœœœ
œ
66663
6
4
œ œ Ê œ
sec tan secœœ
#
""ˆ‰
œ
333 6 333
22 22
ijkv ijk ˆ‰ ˆ ‰
1
13. (2 ln (t 1)) t 2t t 2 ;rijkvijka ijkœÊœœ Êœœ
#
œœ
Êtd2 d2
2dtt1 dt(t1)
œ œ
œœ
rr
ˆ‰ ’“
Speed: (1) (2(1)) 1 6; Direction: kk Ɉ‰ È
vœœ œ
2
11 (1)
(1)
2(1) (1)
6œ
#
##
œœ
v
v
ijk
kk
Š‹
È
2
1
"$
(1) 6œ Ê œ
"""
ÈÈÈ ÈÈÈ
666 666
21 2
ijkv ijk
È
Š‹
14. e (2 cos 3t) (2 sin 3t) e (6 sin 3t) (6 cos 3t) ri j kv i j kaœ Êœœ Êœab a b
ÊÊtt dd
dt dt
rr
œ
œ
e (18 cos 3t) (18 sin 3t) ; Speed: (0) e [ 6 sin 3(0)] [6 cos 3(0)] 37;œ œ œab k k a b É È
Ê !##
#t
ijkv
Direction: (0) 37
v
v
ij k(0)
(0)
e 6 sin 3(0) 6 cos 3(0)
37 37 37 37 37
66
kk
ab
ÈÈÈ ÈÈ
œ œ Êœ
ÊÊ œ ""
$
ikv ik
È
Š‹
15. 3 3 2t and 2 (0) 3 3 and (0) 2 (0) 3 3 0 12 andvi j kak v i ja k vœ œ Ê œ œ Ê œ œ
ÈÈ È
kk Ê Š‹
È
##
#
(0) 2 2; (0) (0) 0 cos 0 kk
È
avaœ œ œÊ œÊœ
#
#
† ))
1
16. 32t and 32 (0) and (0) 32 (0)vi ja jv ija jvœ œÊ œ œÊ œ
ÈÈ ÈÈ È È
22 22 2 2
## ## # #
##
Š ‹ Š‹ Š‹ kk Ê
1 and (0) ( 32) 32; (0) (0) ( 32) 16 2 cos œ œ œ œ œ Ê œ œ Ê œkk È Š‹
È
ava
#
##
Ê
†
ÈÈÈ
21622
1(32) 4
3
))
1
17. t t 1 and (0) andvij ka i j kvjœ œ Êœˆ‰ˆ‰ ab ’“’“’ “
2t 1 2t 2 2t
t1 t1 t1 t1 t1
œœ
œ
œœ
œœ
œ

œœ
#
"Î# "
œœ œab ab ab
(0) 2 (0) 1 and (0) 2 1 5; (0) (0) 0 cos 0 aikv a vaœÊ œ œ œ œÊ œÊœkk kk
È È
##
† ))
1
2
18. (1 t) (1 t) and (1 t) (1 t) (0) andvijka i jvijkœ œ Ê œ
22 22
33 33 3 333
"Î# "Î# "Î# "Î# "" " "
(0) (0) 1 and (0) ; (0) (0)aijv a vaœ Ê œ œ œ œ œ
"" " " "
## ###
33 3 3 3 3 3 3 99
22 222
kk kkÉɈ‰ ˆ ‰ ˆ‰ ˆ‰ ˆ‰
È
†
0 cos 0 œÊ œÊœ ))
1
#

831. Section 13.1 Vector Functions 827
19. (1 cos t) (sin t) and (sin t) (cos t) (sin t)(1 cos t) (sin t)(cos t) sin t. Thus,vijaijvaœ œ Ê œ œ †
0 sin t 0 t 0, , or 2va†œÊ œÊœ 11
20. (cos t) (sin t) and ( sin t) (cos t) sin t cos t sin t cos t 0 for all t 0vijka ikvaœ œ Êœ œ   †
21. t 7 (t 1) dt [7t] t 7'
0
1
cd ’“ ’ “
$"
""
!!
!
"
#
ij k i j k ij k œ œ
tt 3
424
"œ
22. (6 6t) 3 t dt 6t 3t 2t 4t 3 4 2 2 2'
1
2
œ ‘ œ‘È ˆ‰ cd cd Š‹
È
œ œ ijk i j ki jk
4
t
œ
# $Î# "
#
""
#
"
#
23. (sin t) (1 cos t) sec t dt cos t t sin t tan t'

"$
œ
œ
cdcdcdcd abijk i jk œ
# Î% Î% Î%
Ê ß Ê ß Ê ß
111
111
2œŠ‹
1œ
#
22
È
jk
24. (sec t tan t) (tan t) (2 sin t cos t) dt [(sec t tan t) (tan t) (sin 2t) ] dt
''
0 0
3 3
œ œ" "
cd ij k ij k œ sec t ln (cos t) cos 2t (ln 2)œ œcd c d œ ‘
11 1Î$ Î$
!!
"
#
Î$
!
ijkijk
3
4
25. dt ln t ln (5 t) ln t (ln 4) (ln 4) (ln 2)'
1
4
ˆ‰ œ ‘cd c d
"" " "
Ê# #
%%
""
%
"t5t t
ijk i j k i j k œœ œ
26. dt 2 sin t 3 tan t'
0
1
Š‹ ’“ cd
È
2
1t
33
1t 4È
ÈÈ
Ê
œ
" "
"
!
"
!
œ
œ
ik i kikœ œ 1
1
27. ( t t t ) dt ; (0) 0 0 0 2 3 2 3r ijk i j kCr ijkCijk Cijkœ œ œ Ê œ' ttt
œœ œ
###
œ
123Ê œ rijkŠ‹Š‹Š‹
ttt
œœ œ
###
28. (180t) 180t 16t dt 90t 90t t ; (0) 90(0) 90(0) (0)r i j i jCr i jCœœœ'cd ab ˆ‰ œ ‘
###$ ##$ 16 16
33
100 100 90t 90t t 100œÊœ ÊœjC j r i j
##$ˆ‰
16
3
29. (t 1) e dt (t 1) e ln (t 1) ;rijkijkCœ œ'œ ‘ˆ‰ˆ‰
3
t1#œ
"Î# $Î# " œœtt
(0) (0 1) e ln (0 1) rijkCkCijkœ œÊ œ
$Î# œÊ
(t 1) 1 1 e [1 ln (t 1)]Êœ rijkœ ‘ab
$Î# œt
30. t 4t t 2t dt 2t ; (0) 2(0)rijk ijkCr ijkCœ œ œ 'cdab Š‹ ’ “
$## # tt2t 0 0
423 4 23
2(0)
"
2t 1 1œ Ê œ Ê œ ij Cij r i j k Š‹Š‹
tt2t
43

#
#
31. ( 32 ) dt 32t ; (0) 8 8 32(0) 8 8 8 8
dd
dt dt
rr
œ œ œÊ œÊ œ' kkC ij kCijCij"""
8 8 32t ; (8 8 32t ) dt 8t 8t 16t ; (0) 100Ê œ œ œ œ
d
dt
r
ij kr ij k i j kCr k'
#
#
8(0) 8(0) 16(0) 100 100 8t 8t 100 16tÊ œÊœÊœi j kC kC krij k
# #
##
ab
32. ( ) dt (t t t ) ; (0) (0 0 0 )
dd
dt dt
rr
œ œ œÊ œÊ œ'ijk i j k C 0 i j k C 0 C 0"""
(t t t ) ; (t t t ) dt ; (0) 10 10 10Ê œ œ œ œ
dttt
dt
r
ijkr ijk i j k Cr i j k' Š‹
œœ œ
###
#
10 10 10 10 10 10Ê œ Ê œŠ‹
000
œœ œ
###
##
ijkC ijkC ijk

832. 828 Chapter 13 Vector-Valued Functions and Motion in Space
10 10 10Ê œ rijkŠ‹Š‹Š‹
ttt
œœ œ
###
33. (t) (sin t) t cos t e (t) (cos t) (2t sin t) e ; t 0 (t ) andr i jk v i jk v ikœÊœ œÊœab
#
! tt
0
(t ) P (0 1 1) x 0 t t, y 1, and z 1 t are parametric equations of the tangent liner 0œœßßÊœœ œ œ!
34. (t) (2 sin t) 2 cos t 5t (t) (2 cos t) (2 sin t) 5 ; t 4 (t ) 2 5 andrijkv ijk vikœ Êœ œÊœab !1 0
(t ) P (0 2 20 ) x 0 2t 2t, y 2, and z 20 5t are parametric equations of the tangent liner 0œœßß Êœœ œ œ ! 11
35. (t) (a sin t) a cos t bt (t) (a cos t) (a sin t) b ; t 2 (t ) a b andrijkv ijk vikœ Êœ œÊœab
!1 0
(t ) P (0 a 2b ) x 0 at at, y a, and z 2 b bt are parametric equations of the tangent liner 0œœßß Êœœ œ œ ! 11
36. (t) (cos t) sin t (sin 2t) (t) ( sin t) (cos t) (2 cos 2t) ; t (t ) 2 andrijkv ij k vikœ Êœ œÊœab !
#
1
0
(t ) P (0 1 0) x 0 t t, y 1, and z 0 2t 2t are parametric equations of the tangent liner 0œœßßÊœœœ œœ!
37. (a) (t) (sin t) (cos t) (t) (cos t) (sin t) ;vija ijœ Ê œ
(i) (t) ( sin t) (cos t) 1 constant speed;kk Èvœ œÊ
##
(ii) (sin t)(cos t) (cos t)(sin t) 0 yes, orthogonal;va†œœÊ
(iii) counterclockwise movement;
(iv) yes, (0) 0rijœ
(b) (t) (2 sin 2t) (2 cos 2t) (t) (4 cos 2t) (4 sin 2t) ;vija ijœ Ê œ
(i) (t) 4 sin 2t 4 cos 2t 2 constant speed;kk
È
vœœÊ
##
(ii) 8 sin 2t cos 2t 8 cos 2t sin 2t 0 yes, orthogonal;va†œœÊ
(iii) counterclockwise movement;
(iv) yes, (0) 0rijœ
(c) (t) sin t cos t (t) cos t sin t ;vija ijœ Ê œ ˆ‰ ˆ‰ ˆ‰ ˆ‰
11 11
## ##
(i) (t) sin t cos t 1 constant speed;kk É ˆ‰ ˆ‰vœœÊ
##
##
11
(ii) sin t cos t cos t sin t 0 yes, orthogonal;va†œ œÊˆ‰ˆ‰ ˆ‰ˆ‰
11 11
## ##
(iii) counterclockwise movement;
(iv) no, (0) 0 instead of 0rij ijœ
(d) (t) (sin t) (cos t) (t) (cos t) (sin t) ;vija ijœ Ê œ
(i) (t) ( sin t) ( cos t) 1 constant speed;kk Èvœ œÊ
##
(ii) (sin t)(cos t) (cos t)(sin t) 0 yes, orthogonal;va†œœÊ
(iii) clockwise movement;
(iv) yes, (0) 0rijœ
(e) (t) (2t sin t) (2t cos t) (t) (2 sin t 2t cos t) (2 cos t 2t sin t) ;vija i jœ Ê œ
(i) (t) (2t cos t) 4t sin t cos t 2 t 2t, t 02t sin tkk c d a b kkÉab Èvœœœœ $
#
####
variable speed;Ê
(ii) 4 t sin t t sin t cos t 4 t cos t t cos t sin t 4t 0 in generalva†œ œÁaba b
## ##
not orthogonal in general;Ê
(iii) counterclockwise movement;
(iv) yes, (0) 0rijœ
38. Let 2 2 denote the position vector of the point 2, 2, 1 and let, and .pijk u i jv i j kœ œ œ ab
"" """
ÈÈ ÈÈÈ
22 333
Then (t) (cos t) (sin t) . Note that (2 2 1) is a point on the plane and 2 is normal torp u v nijkœ ßß œ
the plane. Moreover, and are orthogonal unit vectors with 0 and are parallel to theuv unvn uv ††œœÊ
plane. Therefore, (t) identifies a point that lies in the plane for each t. Also, for each t, (cos t) (sin t)ruv "

833. Section 13.1 Vector Functions 829
is a unit vector. Starting at the point 2 , 2 , 1 the vector t traces out a circle of radius 1 andŠ‹ ab"$
11
22
ÈÈ
r
center (2 2 1) in the plane x y 2z 2.ßß œ
39. 3 (t) 3t t t ; the particle travels in the direction of the vector
d
dt
v
œœ Ê œ aijk v ijkC "
(4 1) (1 2) (4 3) 3 (since it travels in a straight line), and at time t 0 it has speed œ œijkijk
2 (0) (3 ) (t) 3t t tÊœ œÊœœ vijkCv ijk
2d622
911 dt 11 11 11
ÈÈÈÈœœ
"
r
Š ‹Š‹Š‹
(t) t t t t t t ; (0) 2 3Êœ œœrijkCrijkCŠ‹Š‹Š‹
36 2 2
11 11 11###
### ""
##ÈÈÈ
(t) t t1 t t2 t t3ÊœrijkŠ‹Š‹Š‹
36 2 2
11 11 11###
### ""
ÈÈÈ
tt(3 )(23)œ Š‹
"
#
# 2
11
È
ijk i j k
40. 2 (t) 2t t t ; the particle travels in the direction of the vector
d
dt
v
œœ Ê œ aijk v ijkC "
(3 1) (0 ( 1)) (3 2) 2 (since it travels in a straight line), and at time t 0 it has speed 2 œ œijkijk
(0) (2 ) (t) 2t t tÊœ œÊœœ vijkCvijk
2d422
411 dt 666
ÈÈÈÈœœ
"
r
Š ‹Š‹Š‹
(t) t t t t t t ; (0) 2Ê œ œ œrijkCrijkCŠ‹Š‹Š‹
### ""
##
##
422
666
ÈÈÈ
(t) t t1 t t1 t t2 t t(2 )( 2)Êœ œ rijkijkijkŠ‹Š‹Š‹Š‹
### # "" "
## #
422 2
666 6
ÈÈÈ È
41. The velocity vector is tangent to the graph of y 2x at the point ( ), has length 5, and a positive
#
œ#ß# i
component. Now, y 2x 2y 2 the tangent vector lies in the direction of the
#
Ð#ß#Ñ
##
"
œÊ œÊ œœÊ
dy dy
dx dx 2
2
¹
†
vector the velocity vector is 2 5 5ij v ij ij i jÊ œ œ œ
"""
###
œ
55
1É
Œœ
"
œ
4
5
ˆ‰ ˆ‰ ÈÈ
œ
42. (a)
(b) (1 cos t) (sin t) and (sin t) (cos t) ; (1 cos t) sin t 2 2 cos t is at a maxvijaijv vœ œ œ œ Ê kk kk
## ##
when cos t 1 t , 3 , 5 , etc., and at these values of t, 4 max 4 2; is at a minœ Ê œ œ Ê œ œ111 kk kk kk
È
vvv
##
when cos t 1 t 0, 2 , 4 , etc., and at these values of t, 0 min 0; sin t cos t 1œÊœ œÊ œ œ œ11 kk kk kkvva
## ##
for every t max min 1 1Êœœœkk kk
È
aa
43. ( 3 sin t) (2 cos t) and ( 3 cos t) (2 sin t) ; 9 sin t 4 cos t vjka jkv vœ œ œ Ê kk kk ˆ‰
## ## d
dt
18 sin t cos t 8 cos t sin t 10 sin t cos t; 0 10 sin t cos t 0 sin t 0 or cos t 0œœ œÊ œÊœœ
d
dt
ˆ‰kkv
#
t 0, or t , . When t 0, , 4 4 2; when t , , 9 3.Êœ œ œ œ Ê œ œ œ œ œ11
11 11
## ##
#3 3
kk kk kk
È È
vv v
Therefore max is 3 when t , , and min 2 when t 0, . Next, 9 cos t 4 sin tkk kk kkvva œœœœ
11
##
# ##3
1
18 cos t sin t 8 sin t cos t 10 sin t cos t; 0 10 sin t cos t 0 sin t 0 orÊœ œ œÊ œÊœ
d d
dt dt
ˆ‰ ˆ‰kk kkaa
# #
cos t 0 t 0, or t , . When t 0, , 9 3; when t , , 4 2.œÊœ œ œ œÊœ œ œÊœ11
11 11
## ##
##33
kk kk kk kkaa aa
Therefore, max 3 when t 0, , and min 2 when t , .kk kkaaœœ œœ 1
11
##
3

834. 830 Chapter 13 Vector-Valued Functions and Motion in Space
44. (a) (t) (r cos ) (r sin ) , and the distance traveled along the circle in time t is t (rate times time)rijœ !!)) v
which equals the circular arc length r (t) r cos r sin !!!))Êœ Ê œ
vvvttt
rrr
$$$
rijŠ‹Š‹
(b) (t) sin cos (t) cos sin vija ijœœ Ê œœ
dtt d t t
dt r r dt r r r r
rv
Š‹Š‹ Š ‹Š ‹vv
vv vvvv
$$ $$$$
œœ
r cos r sin (t)œ œ
vv vv
œœ
œœ
$$
$$
rr
tt
rr
’“Š‹Š‹!! ijr
(c) m m Fa rœÊ œ Ê œÊœŠ‹ Š‹
GmM GmM m GM
rrrrrr

$$$$
$ $
œœ
r vv
v
#
(d) T is the time for the satellite to complete one full orbit T circumference of circle T 2 rʜ ʜvv 1 !
(e) Substitute into T T is proportional to r since is avvœœÊœÊœÊ
2r
T r T r GM GM
GM GM 44r 4r1 11 1$$$
$$

œ
œ ###$
!
constant
45. ( ) 2 2 0 0 is a constant is constant
dddd
dt dt dt dt
vv v v v vv v vv†† † †† † †œœ œœÊ Êœ
vv v
kkÈ
46. (a) ( ) ( ) ( ) ( )
dd dd dd
dt dt dt dt dt dt
uvw vw u vw vw u wv††† ††‚œ ‚ ‚œ ‚ ‚‚
uuvw
ˆ‰
() œ‚‚‚
ddd
dt dt dt
uvw
†††vw u wuv
(b) Each of the determinants is equivalent to each expression in Eq. 7 in part (a) because of the formual in Section 12.4
expressing the triple scalar product as a determinant.
47. , since ( ) 0
d dd d dd d d dd dd
dt dt dt dt dt dt dt dt dt dt dt dt
’ “Š‹Š ‹Š‹Š‹Š‹rrrrAAB††††† †
rr rrr rr rr rr
‚ œ ‚ ‚ ‚œ ‚ ‚œ
œœœœ ""

and ( ) 0 for any vectors and ABB A B†‚œ
48. a b c with a, b, c real constants 0 0 0uCijk i j kijk0œœ Ê œ œ œ
ddadbdc
dt dt dt dt
u
49. (a) f(t) g(t) h(t) c cf(t) cg(t) ch(t) (c ) c c c uijku i j k u i j kœ Êœ Ê œ
ddf dh
dt dt dt dt
dg
cc œœŠ‹
df dh d
dt dt dt dt
dg
ijk
u
(b) f(t) g(t) h(t) ( ) f(t) g(t) h(t)ff f f f f f fuijk u i j kœÊœ
ddf dh
dt dt dt dt dt dt dt
dddgd
’“’“’“
ff f
[f(t) g(t) h(t) ]œœ
ddgd
dt dt dt dt dt dt
df dh dff
ijk ijk u ff’“
u
50. Let f (t) f (t) f (t) and g (t) g (t) g (t) . Thenuijkv ijkœ œ "#$ " # $
uv i j kœ [f (t) g (t)] [f (t) g (t)] [f (t) g (t)]"" ## $$
( ) [f (t) g (t)] [f (t) g (t)] [f (t) g (t)]Ê œ
d
dt
uv i j k
ww ww ww
"" ## $$
[f (t) f (t) f (t) ] [g (t) g (t) g (t) ] ;œ œ
www w w w
"$"#$
ijk i j k
!
dd
dt dt
uv
[f (t) g (t)] [f (t) g (t)] [f (t) g (t)]uv i j kœ "" ## $$
( ) [f (t) g (t)] [f (t) g (t)] [f (t) g (t)]Ê œ
d
dt
uv i j k
ww ww ww
"" ## $$
[f (t) f (t) f (t) ] [g (t) g (t) g (t) ]œ œ
www w w w
"#$ " # $
ijk i j k
dd
dt dt
uv
51. Suppose is continuous at t t . Then lim (t) (t ) lim [f(t) g(t) h(t) ]rrrijk œœÍ !!
tt ttÄÄÊÊ
f(t ) g(t ) h(t ) lim f(t) f(t ), lim g(t) g(t ), and lim h(t) h(t ) f, g, and h areœ œ œ œÍ!!! ! ! !ijk Í
tt tt ttÄÄ ÄÊÊ Ê
continuous at t t .œ !
52. lim [ (t) (t)] lim f(t) f(t) f(t)
g(t) g(t) g(t)
lim f (t) lim f (t)
tt tt
tt tt
ÄÄ
ÄÄÊÊ
ÊÊ
rr
ijk
ijk"# "#$
"#$
"#‚œ œ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
lim f (t)
lim g (t) lim g (t) lim g (t) tt
tt tt tt
Ä
ÄÄÄÊ
ÊÊÊ
$
"#$
lim (t) lim (t)œ‚œ‚
tt ttÄÄÊÊ
rrAB"#

835. Section 13.1 Vector Functions 831
53. (t ) exists f (t ) g (t ) h (t ) exists f (t ), g (t ), h (t ) all exist f, g, and h are continuous atrijk
w w w w www
! ! ! ! !!!
Ê Ê Ê
t t (t) is continuous at t tœÊ œ
!!r
54. (a) k (t) dt [kf(t) kg(t) kh(t) ] dt [kf(t)] dt [kg(t)] dt [kh(t)] dt
'' ' ' '
aa a a a
bb b b b
rijk ijkœœ
k f(t) dt g(t) dt h(t) dt k (t) dtœœŒ "
''' '
aa a a
bb b b
ij kr
(b) [ (t) (t)] dt f (t) g (t) h (t) f (t) g (t) h (t) dt
''
aa
bb
rr i j k i j k"# " " " # # #„œ „ abcdcd
f (t) f (t) [g (t) g (t ] [h (t) h (t)] ) dtœ „ „ „
'
a
b
abcd"# "# "#ijk
f (t) f (t) dt g (t) g (t) dt h (t) h (t) dtœ„„ „
'' '
aa a
bb b
cdcdcd"# "# "# ij k
f (t) dt f (t) dt g (t) dt g (t) dt h (t) dt h (t) dtœ„„ „”•” •” •
'' '' ''
aa aa a a
bb bb b b
"# "# " #ii jj kk
(t) dt (t) dtœ„
''
aa
bb
rr"#
(c) Let c c c . Then (t) dt c f(t) c g(t) c h(t) dtCijk Crœ œ "#$ " # $
''
aa
bb
† cd
c f(t) dt c g(t) dt c h(t) dt = (t) dt;œ
"# $
'' ' '
aa a a
bb b b
Cr†
(t) dt c h(t) c g(t) c f(t) c h(t) c g(t) c f(t) dt
''
aa
bb
Cr i j k‚œ c dcdcd #$ $" "#
c h(t) dt c g(t) dt c f(t) dt c h(t) dt c g(t) dt c f(t) dtœ ”•”•”• #$ $" "#
'' '' ''
aa aa aa
bb bb bb
ijk
(t) dtœ‚Cr
'
a
b
55. (a) Let u and be continuous on [a b]. Then lim u(t) (t) lim [u(t)f(t) u(t)g(t) u(t)h(t) ]rrijk ߜ
tt ttÄÄÊÊ
u(t )f(t ) u(t )g(t ) u(t )h(t ) u(t ) (t ) u is continuous for every t in [a b].œ œ Ê ß!! !! !! !! !ijkrr
(b) Let u and be differentiable. Then (u ) [u(t)f(t) u(t)g(t) u(t)h(t) ]rrijk
dd
dt dt
œ
f(t) u(t) g(t) u(t) h(t) u(t)œ ˆ‰ ˆ ‰ Š‹
du df du du dh
dt dt dt dt dt dt
dg
ijk
[f(t) g(t) h(t) ] u(t) uœ œijk ijkr
du df dh du d
dt dt dt dt dt dt
dg
Š‹
r
56. (a) If (t) and (t) have identical derivatives on I, then RR ijkijk"#
ddfdhdfdh
dt dt dt dt dt dt dt
dg dgR
     œ œ  œ
œ œ
, , f (t) f (t) c , g (t) g (t) c , h (t) h (t) cœÊœ œ œÊ œ œ œ
ddfdf dhdh
dt dt dt dt dt dt dt
dg dgRœ œ  œ  œ
"#"" ##" #$
f (t) g (t) h (t) [f (t) c ] [g (t) c ] [h (t) c ] (t) (t) , whereÊ œÊ œ""" #"###$ "#ijk i j kRRC
ccc.Cijkœ
"#$
(b) Let (t) be an antiderivative of (t) on I. Then (t) (t). If (t) is an antiderivative of (t) on I, thenRrRrUr
w
œ
(t) (t). Thus (t) (t) on I (t) (t) .Ur UR URC
www
œœÊœ
57. ( ) d [f( ) g( ) h( ) ] d f( ) d g( ) d h( ) d
dd dd d
dt dt dt dt dt'' '''
aa a aa
tt t tt
rijkijk77 7 7 7 7 77 77 77œœ
f(t) g(t) h(t) (t). Since ( ) d (t), we have that ( ) d is an antiderivative ofœ œ œijkr r r r
d
dt''
aa
tt
77 77
. If is any antiderivative of , then (t) ( ) d by Exercise 56(b). Then (a) ( ) drR r R r C R r C œ œ
''
a a
t a
77 77
(a) ( ) d (t) (t) (a) ( ) d (b) (a).œÊ Ê œœ Ê œ 0C C R r R CR R r R Rœ
''
aa
tb
77 77
58-61. Example CAS commands:
:Maple
> with( plots );
r := t -> [sin(t)-t*cos(t),cos(t)+t*sin(t),t^2];

836. 832 Chapter 13 Vector-Valued Functions and Motion in Space
t0 := 3*Pi/2;
lo := 0;
hi := 6*Pi;
P1 := spacecurve( r(t), t=lo..hi, axes=boxed, thickness=3 ):
display( P1, title="#58(a) (Section 13.1)" );
Dr := unapply( diff(r(t),t), t ); # (b)
Dr(t0); # (c)
q1 := expand( r(t0) + Dr(t0)*(t-t0) );
T := unapply( q1, t );
P2 := spacecurve( T(t), t=lo..hi, axes=boxed, thickness=3, color=black ):
display( [P1,P2], title="#58(d) (Section 13.1)" );
62-63. Example CAS commands:
:Maple
a := 'a'; b := 'b';
r := (a,b,t) -> [cos(a*t),sin(a*t),b*t];
Dr := unapply( diff(r(a,b,t),t), (a,b,t) );
t0 := 3*Pi/2;
q1 := expand( r(a,b,t0) + Dr(a,b,t0)*(t-t0) );
T := unapply( q1, (a,b,t) );
lo := 0;
hi := 4*Pi;
P := NULL:
for a in [ 1, 2, 4, 6 ] do
P1 := spacecurve( r(a,1,t), t=lo..hi, thickness=3 ):
P2 := spacecurve( T(a,1,t), t=lo..hi, thickness=3, color=black ):
P := P, display( [P1,P2], axes=boxed, title=sprintf("#62 (Section 13.1) a=%a",a) );
end do:
display( [P], insequence=true );
58-63. Example CAS commands:
: (assigned functions, parameters, and intervals will vary)Mathematica
The x-y-z components for the curve are entered as a list of functions of t. The unit vectors , , are not inserted.ijk
If a graph is too small, highlight it and drag out a corner or side to make it larger.
Only the components of r[t] and values for t0, tmin, and tmax require alteration for each problem.
Clear[r, v, t, x, y, z]
r[t_]={ Sin[t] t Cos[t], Cos[t] t Sin[t], t2}
t0= 3 / 2; tmin= 0; tmax= 6 ;11
ParametricPlot3D[Evaluate[r[t]], {t, tmin, tmax}, AxesLabel {x, y, z}];Ä
v[t_]= r'[t]
tanline[t_]= v[t0] t r[t0]
ParametricPlot3D[Evaluate[{r[t], tanline[t]}], {t, tmin, tmax}, AxesLabel {x, y, z}];Ä
For 62 and 63, the curve can be defined as a function of t, a, and b. Leave a space between a and t and b and t.
Clear[r, v, t, x, y, z, a, b]
r[t_,a_,b_]:={Cos[a t], Sin[a t], b t}
t0= 3 / 2; tmin= 0; tmax= 4 ;11
v[t_,a_,b_]= D[r[t, a, b], t]
tanline[t_,a_,b_]=v[t0, a, b] t r[t0, a, b]
pa1=ParametricPlot3D[Evaluate[{r[t, 1, 1], tanline[t, 1, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}];Ä

837. Section 13.2 Modeling Projectile Motion 833
pa2=ParametricPlot3D[Evaluate[{r[t, 2, 1], tanline[t, 2, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}];Ä
pa4=ParametricPlot3D[Evaluate[{r[t, 4, 1], tanline[t, 4, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}];Ä
pa6=ParametricPlot3D[Evaluate[{r[t, 6, 1], tanline[t, 6, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}];Ä
Show[GraphicsArray[{pa1, pa2, pa4, pa6}]]
13.2 MODELING PROJECTILE MOTION
1. x (v cos )t (21 km) (840 m/s)(cos 60°)t t 50 secondsœÊ œ Êœ œ!! ˆ‰
1000 m
1 km (840 m/s)(cos 60 °)
21,000 m
2. R sin 2 and maximum R occurs when 45 ° 24.5 km (sin 90 °)œœÊœ
v v
g 9.8 m/s
œ œ
$ $
œ
!! Š‹ v (9.8)(24,500) m /s 490 m/sÊœ œ !
##
È
3. (a) t 72.2 seconds; R sin 2 (sin 90 °) 25,510.2 mœœ ¸ œ œ ¸
2v sin
g 9.8 m/s g 9.8 m/s
2(500 m/s)(sin 45°) (500 m/s) v$ $
œ œ
œ œ!
!
(b) x (v cos )t 5000 m (500 m/s)(cos 45 °)t t 14.14 s; thus,œÊœ Êœ ¸ !!
5000 m
(500 m/s)(cos 45°)
y (v sin )t gt y (500 m/s)(sin 45 °)(14.14 s) 9.8 m/s (14.14 s) 4020 mœÊ¸ ¸ !
""
##
###! ab
(c) y 6378 m
maxœœ ¸
(v sin ) ((500 m/s)(sin 45°))
2g 2 9.8 m/s$
œœ
œ!
ab
4. y y (v sin )t gt y 32 ft (32 ft/sec)(sin 30 °)t 32 ft/sec t y 32 16t 16t ;œ Êœ Êœ !!
""
##
#### ! ab
the ball hits the ground when y 0 0 32 16t 16t t 1 or t 2 t 2 sec since t 0; thus,œÊœ Êœ œÊœ
#
x (v cos ) t x (32 ft/sec)(cos 30°)t 32 (2) 55.4 ftœÊœ œ ¸!
#! Š‹
È
3
5. x x (v cos )t 0 (44 cos 45 °)t 22 2t and y y (v sin )t gt 6.5 (44 sin 45 °)t 16tœ œ œ œ œ !! !!
"
#
## !!
È
6.5 22 2t 16t ; the shot lands when y 0 t 2.135 sec since t 0; thusœ œÊœ ¸
È
# „œ
#
22 2 968 416
3
È È
x 22 2t 22 2 (2.135) 66.43 ftœ¸ ¸
ÈÈ
Š‹
6. x 0 (44 cos 40°)t 33.706t and y 6.5 (44 sin 40°)t 16t 6.5 28.283t 16t ; y 0œ ¸ œ ¸ œ
##
t 1.9735 sec since t 0; thus x (33.706)(1.9735) 66.52 ft theʸ ¸ ¸ ¸ Ê
28.283 (28.283) 416
3
œœ
#
È
œ
difference in distances is about 66.52 66.43 0.09 ft or about 1 inchœ
7. R sin 2 10 m (sin 90 °) v 98 m s v 9.9 m/s;œÊœ ʜʸ
vv
g 9.8 m/s
œœ
$$
œ
! Š‹
###
! ! 6m (sin 2 ) sin 2 0.59999 2 36.87 ° or 143.12° 18.4 ° or 71.6°¸Ê¸Ê¸ ʸ
(9.9 m/s)
9.8 m/s
œ
œ
!! ! !
8. v 5 10 m/s and x 40 cm 0.4 m; thus x (v cos )t 0.4m 5 10 m/s (cos 0 °)t
!!
' 'œ‚ œ œ œ Ê œ ‚ ! ab
t 0.08 10 s 8 10 s; also, y y (v sin )t gtÊœ ‚ œ‚ œ
ÊÄ Ê¸ #
!!
"
#
!
y 5 10 m/s (sin 0 °) 8 10 s 9.8 m/s 8 10 s 3.136 10 m orÊœ‚ ‚ ‚ œ ‚abababab
' ) # ) "% "
#
#
3.136 10 cm. Therefore, it drops 3.136 10 cm.$Ñ Ñ
"# "#
9. R sin 2 3(248.8) ft (sin 18 °) v 77,292.84 ft /sec v 278.02 ft/sec 190 mphœÊ œ ʸ ʸ ¸
vv
g 32 ft/sec
œœ
$$
œ
! Š‹
###
! !
10. v ft/sec and R 200 ft 200 (sin 2 ) sin 2 0.9 2 64.2 ° 32.1 °; or!œœÊœÊœÊ¸Ê¸
80 10
332
È Š‹
80 10
3œ
œ
!! ! !
2 115.8 ° 57.9 °; If 32.1 , y 31.4 ft. If 57.9 , y 79.7 ft 75 ft. In!!! !¸Ê¸ ¸ œ ¸ ¸ ¸
‰‰
max max
’“Š‹
80 10
3œ
(sin 32.1°)
2(32)
œ
order to reach the cushion, the angle of elevation will need to be about 32.1°. At this angle, the circus performer will go

838. 834 Chapter 13 Vector-Valued Functions and Motion in Space
31.4 ft into the air at maximum height and will not strike the 75 ft high ceiling.
11. x (v cos )t 135 ft (90 ft/sec)(cos 30°)t t 1.732 sec; y (v sin )t gtœÊœ ʸ œ! !
"
#
#!!
y (90 ft/sec)(sin 30°)(1.732 sec) 32 ft/sec (1.732 sec) y 29.94 ft the golf ball will clipʸ ʸ Ê
"
#
##
ab
the leaves at the top
12. v 116 ft/sec, 45°, and x (v cos )t
!!œœœ !!
369 (116 cos 45 °)t t 4.50 sec;Êœ ʸ
also y (v sin )t gtœ !
"
#
#!
y (116 sin 45 °)(4.50) (32)(4.50)Êœ
"
#
#
45.11 ft. It will take the ball 4.50 sec to travel¸
369 ft. At that time the ball will be 45.11 ft in
the air and will hit the green past the pin.

13. We do part b first.
(b) x (v cos )t 315 ft (v cos 20 °)t v ; also y (v sin )t gtœÊœ Êœ œ !!!!
"
#
#!!
315
t cos 20°
34 ft (t sin 20 °) (32)t 34 315 tan 20 °16t t 5.04 sec t 2.25 secÊœ Êœ ʸ ʸˆ‰
315
t cos 20°
"
#
####
(a) v 149 ft/sec!œ¸
315
(2.25)(cos 20°)
14. R sin 2 (2 sin cos ) [2 cos (90 °) sin (90°)] [sin 2(90°)]œœ œ œ
vv v v
gg g g
œœ œ œ
$$ $ $
!!! !! !
15. R sin 2 16,000 m sin 2 sin 2 0.98 2 78.5 ° or 2 101.5° 39.3 °œÊœ ʜʸ ¸Ê¸
v
g 9.8 m/s
(400 m/s)
œ
$
œ
œ
!!!!!!
or 50.7°
16. (a) R sin 2 sin 2 4 sin or 4 times the original range.œœœ
(2v )
gg g
4v v$
œ œœ
$$
!! ! Š‹
(b) Now, let the initial range be R sin 2 . Then we want the factor p so that pv will double the rangeœ
v
g
œ
$
! !
sin 2 2 sin 2 p 2 p 2 or about 141%. The same percentage will approximatelyʜ ʜʜ
(pv )
gg
v$
œ œ
$
!!Š‹
È
#
double the height: p 2 p 2.
ababpv sin 2 v sin
2g 2g00
22!!
œÊœÊœ
# È
17. x x (v cos )t 0 (v cos 40 °)t 0.766 v t and y y (v sin )t gt 6.5 (v sin 40 °)t 16tœ œ ¸ œ œ !! ! ! !! !
"
#
## !!
6.5 0.643 v t 16t ; now the shot went 73.833 ft 73.833 0.766 v t t sec; the shot lands¸ Ê œ ʸ ! !
# 96.383
v
$
when y 0 0 6.5 (0.643)(96.383) 16 0 68.474 vœÊœ ʸ Ê ¸ Š‹ É
96.383
v 68.474
148,635 148,635
v
$
œ
$
#
!
46.6 ft/sec, the shot's initial speed¸
18. y y and y (v sin )t gt (v sin )t gtmax max
(v sin ) 3(v sin ) 3(v sin )
2g 4 8g 8g
3œÊœ œ Ê œ
$$ $
œœ œ!! !
!!
""
##
##
!!
3(v sin ) (8gv sin )t 4g t 4g t (8gv sin )t 3(v sin ) 0 2gt 3v sin 0 orʜʜʜ
!! !! !
##### #!! !! !
2gt v sin 0 t or t . Since the time it takes to reach y is t ,œÊœ œ œ!!
3v sin v sin v sin
2g 2g g
max max$$ $!! !
then the time it takes the projectile to reach of y is the shorter time t or half the time it takes
3
42g
max
v sin
œ
$!
to reach the maximum height.
19. ( g ) dt gt and (0) (v cos ) (v sin ) g(0) (v cos ) (v sin )
dd
dt dt
rr
œ œ œ Ê œ 'jjC i j jC i j"!! "!! !! !!
(v cos ) (v sin ) (v cos ) (v sin gt) ; [(v cos ) (v sin gt) ] dtÊœ Êœ œ Cij i jr i j"! ! ! ! ! !!! !! !!
d
dt
r '
(v t cos ) v t sin gt and (0) x y [v (0) cos ] v (0) sin g(0)œœÊ !! #!!! ! #
" "
# #
# # !! ! !ijCriji jCˆ‰ œ ‘
x y x y (x v t cos ) y v t sin gt x x v t cos andœ Ê œ Êœ Êœ!! #!! !! !! !!
"
#
#ijC ijr i j !! !ˆ‰

839. Section 13.2 Modeling Projectile Motion 835
y y v t sin gtœ !!
"
#
# !
20. From Example 3(b) in the text, v sin (68)(64) v sin 56.5°65.97 v 79 ft/sec !!!!œÊ ¸Ê¸È
21. The horizontal distance from Rebollo to the center of the cauldron is 90 ft the horizontal distance to theÊ
nearest rim is x 90 (12) 84 84 x (v cos )t 0 t 84 tœ œ Ê œ ¸ Ê œ
"
#
!!
! Š‹
90g (90)(32)
v sin (68)(64)
$! È t 1.92 sec. The vertical distance at this time is y y (v sin )t gtÊœ œ !!
"
#
# !
6 (68)(64) (1.92) 16(1.92) 73.7 ft the arrow clears the rim by 3.7 ft¸ ¸ ÊÈ
#
22. The projectile rises straight up and then falls straight down, returning to the firing point.
23. Flight time 1 sec and the measure of the angle of elevation is about 64° (using a protractor) so thatœ
t 1 v 17.80 ft/sec. Then y 4.00 ft andœÊœ ʸ œ ¸
2v sin 2v sin 64 °
g 32 2(32)
max
(17.80 sin 64°)$$
œ!
!
R sin 2 R sin 128 °7.80 ft the engine traveled about 7.80 ft in 1 sec the engineœÊœ ¸Ê Ê
v
g32
(17.80)
œ
$
œ
!
velocity was about 7.80 ft/sec
24. When marble A is located R units downrange, we have x (v cos )t R (v cos )t t . AtœÊœÊœ !!!!
R
v cos
$!
that time the height of marble A is y y (v sin )t gt (v sin ) gœ œ !! !
""
##
#
# !! Š‹Š‹
RR
v cos v cos
$$!!
y R tan g . The height of marble B at the same time t seconds isÊœ œ !
"
#
Š‹
RR
v cos v cos
œ
œ
$
œ
$
! !
h R tan gt R tan g . Since the heights are the same, the marbles collide regardlessœœ!!
""
##
#
Š‹
R
v cos
œ
œ
$
œ
!
of the initial velocity v .!
25. (a) At the time t when the projectile hits the line OR we have tan ; x [v cos ( )]t and"!"œœ
y
x
!
y [v sin ( )]t gt 0 since R isœ !
"
#
#!"
below level ground. Therefore let
y gt [v sin ( )]t 0kkœ
"
#
#
!
!"
so that tan "œœ
" ‘ " ‘
ÎÎ
œœ
œ
$$
$$
gt (v sin ( ))t gt v sin ( )
[v cos ( )]t v cos ( )
!" !"
!" !"
ÊÊÊ
ÊÊ v cos ( ) tan gt v sin ( )Êœ !!
"
# !" " !"
t , which is the timeʜ
2v sin ( ) 2v cos ( ) tan
g$$!" !" "Êœ Ê
when the projectile hits the downhill slope. Therefore,

x [v cos ( )] cos ( ) tan sin ( ) cos ( ) . If x isœ œ !
Êœ Ê # !" !" " !" !"’“ cd
2v sin ( ) 2v cos ( ) tan
gg
2v$$
œ
$!" !" "
maximized, then OR is maximized: [ sin 2( ) tan cos 2( )] 0
dx
dg
2v
!
œ œ
œ
$
!" " !"
sin 2( ) tan cos 2( ) 0 tan cot 2( ) 2( ) 90 °Ê œÊ œ Ê œ !" " !" " !" !" " (90 °) (90 °) of AOR.Êœ Êœ œ n!" " ! "
"""
###
(b) At the time t when the projectile hits OR we have
tan ; x [v cos ( )]t and"!"œœ
y
x
!
y [v sin ( )]t gtœ !
"
#
#!"
tan Êœ œ"
cd " ‘v sin( )t gt v sin( ) gt
[v cos ( )]t v cos ( )$$
ÎÎ
œœ
œ
$$!" !"
!" !"
œÊ œÊ
œœ
v cos( ) tan v sin( ) gtʜ !!
"
# !" " !"
t , which is the timeʜ
2v sin ( ) 2v cos ( ) tan
g$$!" !" "œÊ œ
when the projectile hits the uphill slope. Therefore,

840. 836 Chapter 13 Vector-Valued Functions and Motion in Space
x [v cos ( )] sin ( ) cos ( ) cos ( ) tan . If x isœ œ !
œÊ œ # !" !" !" !" "’“ cd
2v sin ( ) 2v cos ( ) tan
gg
2v$$
œ
$!" !" "
maximized, then OR is maximized: [cos 2( ) sin 2( ) tan ] 0
dx
dg
2v
!
œœ
œ
$
!" !" "
cos 2( ) sin 2( ) tan 0 cot 2( ) tan cot 2( ) tan Ê œ Ê œ! Ê œ!" !" " !" " !" "
tan ( ) 2( ) 90 °() 90 ° (90 °) of AOR. Therefore v would bisectœÊ œœÊœ œ n"!" " "! "
""
##
!
AOR for maximum range uphill.n
26. (a) t x t y t ; where x t 145 cos 23 14 t and y t 2.5 145 sin 23 t 16t .rijab a b a b ab a b ab a bab abœ œ œ
‰‰ 2
(b) y 2.5 2.5 52.655 feet, which is reached at t 1.771 seconds.max
v sin 145sin 23
2g 64 g 32
vsin 145sin 23œœ ¸ œœ ¸
ab a b0
22
0! !
% %
(c) For the time, solve y 2.5 145 sin 23 t 16t 0 for t, using the quadratic formulaœ œab
‰ 2
t 3.585 sec. Then the range at t 3.585 is about x 145 cos 23 14 3.585œ¸ ¸œ
145 sin 23 145 sin 23 160
32
% %
œœ
‰
Éab
2
abab
428.311 feet.¸
(d) For the time, solve y 2.5 145 sin 23 t 16t 20 for t, using the quadratic formulaœ œab
‰ 2
t 0.342 and 3.199 seconds. At those times the ball is aboutœ¸
145 sin 23 145 sin 23 1120
32
% %
œÊÉab
2
x 0.342 145 cos 23 14 0.342 40.860 feet from home plate and x 3.199 145 cos 23 14 3.199aba bab aba babœ¸ œ
‰ ‰
382.195 feet from home plate.¸
(e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate.
27. (a) (Assuming that "x" is zero at the point of impact:)
t x t y t ; where x t 35 cos 27 t and y t 4 35 sin 27 t 16t .rijab a b a b ab a b ab a bab abœ œ œ
‰‰ 2
(b) y 4 4 7.945 feet, which is reached at t 0.497 seconds.max
v sin 35sin 27
2g 64 g 32
vsin 35sin 27œœ ¸ œœ¸
ab a b0
22
0! !
% %
(c) For the time, solve y 4 35 sin 27 t 16t 0 for t, using the quadratic formulaœ œab
‰ 2
t 1.201 sec. Then the range is about x 1.201 35 cos 27 1.201œ¸ œ
35 sin 27 35 sin 27 256
32
% %
œÊ œ
‰
Éab
2
aba bab
37.453 feet.¸
(d) For the time, solve y 4 35 sin 27 t 16t 7 for t, using the quadratic formulaœ œab
‰ 2
t 0.254 and 0.740 seconds. At those times the ball is aboutœ¸
35 sin 27 35 sin 27 192
32
% %
œÊ ÊÉab
2
x 0.254 35 cos 27 0.254 7.921 feet and x 0.740 35 cos 27 0.740 23.077 feet the impact point,aba bab aba babœ¸ œ¸
‰‰
or about 37.453 7.921 29.532 feet and 37.453 23.077 14.376 feet from the landing spot.$) $ )
(e) Yes. It changes things because the ball won't clear the net (y 7.945).
max¸
28. The maximum height is y and this occurs for x sin 2 . These equations describeœœœ
(v sin )
ggg
v v sin cos $
œ œœ
$$! !!
##
!
parametrically the points on a curve in the xy-plane associated with the maximum heights on the parabolic trajectories in
terms of the parameter (launch angle) . Eliminating the parameter , we have x!!
#
Ê
œœ
v sin cos
gg
v sin 1 sin
"œ œ
$
œœ
"œ œ
$
!! !!ˆ‰ ab
(2y) (2y) x 4y y 0 x 4 y yœœÊ œÊœ
v sin v sin v 2v v v v
g g g g 2g 16g 4g
"œ "" œ œ œ " "
$$ $ $ $$$
œœ œœ
!! ### # #
Š‹ ’ “ Š‹
x 4 y , where x 0.Ê œ  
#
#
Š‹
vv
4g 4g
œ !
$$
œ
29. k g P t k and t g P t dt kt v t e e v t t dt
dd d1
dt dt dt v t
P t dt kt
2
2
rr r
œÊ œ œÊ œ Ê œ œ Ê œjQj Qab ab ab ab ab ab ''
'ab
ab
ge e dt ge e , where g ; apply the initial condition:œ œ œ œ "
ÊÊ Êkt kt kt kt e
k
1
g
k
1'jjCCC jCœ ‘
kt
vcos vsin vcos vsin
d
dt k k
t0
00 0 0
ggr
¹ abab ab ˆ‰
œ
œœÊœ!! ! !ijjCC i j
v e cos e v sin , dt v e cos e v sin Êœ œ "$ " "
d
dt k k
00
kt kt gg
00
kt kt gg
kk
r
ˆ‰ˆ ‰ ˆ‰ ˆ‰ˆ ‰ ˆ‰
ÊÊ ÊÊ
!! !!ijr ij 'cd
; apply the initial condition:e cos v sin œ$"$$"ˆ‰ ˆ‰ Š‹
v
kkkk
kt gt ge
0 20
ktÊ
!!ij C
Ð

841. Section 13.3 Arc Length and the Unit Tangent Vector 837 T
0 cos cos r0 CC ij ijab ˆ‰ˆ ‰ ˆ‰ˆ ‰œœ Ê œ$"$$ ""
v v sin v v sin
kkk kkk
gg
2200 00
2 2
!!
!!
t 1e cos 1e sin 1kteÊœ $"$"$$r ijabˆ‰ˆ ‰ˆ‰ ˆ‰ ˆ ‰
vv
kkk
kt kt kt g00
2 ÊÊ Ê
!!
30. (a) t x t y t ; where x t 1 e cos 20 andrijababab ab a ba bab ab ˆ‰œ œ
152
0.12
0.12t‰
y t 3 1 e sin 20 1 0.12t eab a ba b a bˆ‰ ˆ ‰œ
152 32
0.12 0.12
0.12t 0.12t‰
2
(b) Solve graphically using a calculator or CAS: At t 1.484 seconds the ball reaches a maximum height of about 40.435¸
feet.
(c) Use a graphing calculator or CAS to find that y 0 when the ball has traveled for 3.126 seconds. The range isœ¸
about x 3.126 1 e cos 20 372.311 feet.ab a bˆ‰ˆ ‰œ ¸
152
0.12
0.12 3.126‰ab
(d) Use a graphing calculator or CAS to find that y 30 for t 0.689 and 2.305 seconds, at which times the ball is aboutœ¸
x 0.689 94.454 feet and x 2.305 287.621 feet from home plate.ab ab¸¸
(e) Yes, the batter has hit a home run since a graph of the trajectory shows that the ball is more than 14 feet above the
ground when it passes over the fence.
31. (a) t x t y t ; where x t 1 e 152 cos 20 17.6 andrijababab ab a ba bab ab ˆ‰œ œ
1
0.08
0.08t‰
y t 3 1 e sin 20 1 0.08t eab a ba b a bˆ‰ ˆ ‰œ
152 32
0.08 0.08
0.08t 0.08t‰
2
(b) Solve graphically using a calculator or CAS: At t 1.527 seconds the ball reaches a maximum height of about 41.893¸
feet.
(c) Use a graphing calculator or CAS to find that y 0 when the ball has traveled for 3.181 seconds. The range isœ¸
about x 3.181 1 e 152 cos 20 17.6 351.734 feet.ab a bˆ‰ˆ ‰œ ¸
1
0.08
0.08 3.181‰ab
(d) Use a graphing calculator or CAS to find that y 35 for t 0.877 and 2.190 seconds, at which times the ball is aboutœ¸
x 0.877 106.028 feet and x 2.190 251.530 feet from home plate.ab ab¸¸
(e) No; the range is less than 380 feet. To find the wind needed for a home run, first use the method of part (d) to find that
y 20 at t 0.376 and 2.716 seconds. Then define x w 1 e 152 cos 20 w , and solveœ¸ œ ab a bˆ‰ˆ ‰
1
0.08
0.08 2.716‰ab
x w 380 to find w 12.846 ft/sec.abœ¸
13.3 ARC LENGTH AND THE UNIT TANGENT VECTOR T
1. (2 cos t) (2 sin t) 5t ( 2 sin t) (2 cos t) 5rijkv ijkœÊœ
ÈÈ
( 2 sin t) (2 cos t) 5 4 sin t 4 cos t 5 3; Êœ œ œ œkkÊ Š‹
È È
vT
## ##
#
v
vkk
sin t cos t and Length dt 3 dt 3t 3œ œ œ œ œˆ‰ˆ‰ kk c d
22
33 3
5
ijk v
È
''
00
œœ
1
!
1
2. (6 sin 2t) (6 cos 2t) 5t (12 cos 2t) ( 12 sin 2t) 5rijkv i jkœ Êœ
(12 cos 2t) ( 12 sin 2t) 5 144 cos 2t 144 sin 2t 25 13; Êœ œ œ œkkÈ È
v T
##### v
vkk
cos 2t sin 2t and Length dt 13 dt 13t 13œ œœœœˆ‰ˆ‰ kk c d
12 12 5
13 13 13
ijk v ''
00
œœ
1
!
1
3. t t t 1 t 1 t ; ri k vi k v T i kœ Ê œ Ê œ œ œ œ
2
3 1t 1t
t$Î# "Î# # "Î#
# "
œœ
kk a bÉ È
v
vkk ÈÈ
È
and Length 1 t dt (1 t)œœœ '
0
8
È œ ‘ 252
33
$Î#
)
!
4. (2 t) (t 1) t 1 ( 1) 1 3 ; rijkvijkv T ijkœ Êœ Ê œ œ œ œ kkÈ È
### ""v
vkk ÈÈÈ
333
1
and Length 3 dt 3t 3 3œœœ '
0
3
ÈÈ È
’“
$
!

842. 838 Chapter 13 Vector-Valued Functions and Motion in Space
5. cos t sin t 3 cos t sin t 3 sin t cos t rjkv j kvœ Êœ Êabab a ba b kk
$$ # #
3 cos t sin t 3 sin t cos t 9 cos t sin t cos t sin t 3 cos t sin t ;œ œ œÉababababkk È
## ####
##
( cos t) (sin t) , if 0 t , andTjkjkœœ œ ŸŸ
v
vkkkkkk
Ê
#
3 cos t sin t 3 sin t cos t
3 cos t sin t 3 cos t sin t
œœ
1
Length 3 cos t sin t dt 3 cos t sin t dt sin 2t dt cos 2tœœœœœ '''
000
222
œœœ"""
kk œ ‘
333
4##
Î#
!
1
6. 6t 2t 3t 18t 6t 9t 18t 6t 9t 441t 21t ;rijkv ijkvœ Êœ Êœ œ œ
$$$ ### #### %
###
kk abababÉ È
and Length 21t dt 7t 49Tijkijkœœ œ œ œ œ
v
vkk
" #$
#
"
8t 6t 9t 6 2 3
21t 21t 21t 7 7 7
œœ œ
œœ œ
'
1
2
cd
7. (t cos t) (t sin t) t (cos t t sin t) (sin t t cos t) 2 trijkv i j kœ Êœ
22
3
È
$Î# "Î#
Š‹
È
(cos t t sin t) (sin t t cos t) 2 t 1 t 2t (t 1) t 1 t 1, if t 0;Ê œ œ œ œœ  kk kkÊ Š‹
ÈÈ Èv
## ##
#
and Length (t 1) dt tTijkœœ œœœ
v
vkk
È
ˆ‰ˆ‰ Š‹ ’“
cos t t sin t sin t t cos t t
t1 t1 t1 2 2
2tʜ
œœœ
!
 Êœ œœ
'
0
œ 1
1
1
8. (t sin t cos t) (t cos t sin t) (sin t t cos t sin t) (cos t t sin t cos t)rijv i jœÊœ (t cos t) (t sin t) (t cos t) ( t sin t) t t t if 2 t 2; œ Êœ œœœŸŸœijv T kk kkÈ È È
### v
vkk
(cos t) (sin t) and Length t dt 1œœ œœœˆ‰ˆ‰ ’“
t cos t t sin t t
tt 2
ijij '
œ
2
2
œ
#
#
È
9. Let P(t ) denote the point. Then (5 cos t) (5 sin t) 12 and 26 25 cos t 25 sin t 144 dt!
## vijkœ œ 1 '
0
t
œ
È
13 dt 13t t 2 , and the point is P(2 ) (5 sin 2 5 cos 2 24 ) (0 5 24 )œœÊœ œßßœßß
'
0
t
œ
!! 111111
10. Let P(t ) denote the point. Then (12 cos t) (12 sin t) 5 and
! vijkœ
13 144 cos t 144 sin t 25 dt 13 dt 13t t , and the point isœ œ œ Êœ11 ''
00
tt
œ œ
È ##
!!
P( ) (12 sin ( ) 12 cos ( ) 5 ) (0 12 5 ) œ ß ß œ ß ß11111
11. (4 cos t) (4 sin t) 3t ( 4 sin t) (4 cos t) 3 ( 4 sin t) (4 cos t) 3rijkv ijkvœÊœÊœ kkÈ
###
25 5 s(t) 5 d 5t Length sœœÊœ œÊ œœ
È ˆ‰ '
0
t
7
11
##
5
12. (cos t t sin t) (sin t t cos t) ( sin t sin t t cos t) (cos t cos t t sin t)rijv i jœ Êœ
(t cos t) (t sin t) (t cos t) (t cos t) t t, since t s(t) dœ Êœ œœœ ŸŸÊœœijv kkÈ È
###
##
1
177 '
0
t
t
œ
Length s( ) sÊœœœ 1 ˆ‰
11 1
###
œœ
œ
œ ˆ‰
œ
3
8
13. e cos t e sin t e e cos t e sin t e sin t e cos t er i jk v i jkœÊœabab abab
ttt ttttt
e cos t e sin t e sin t e cos t e 3e 3 e s(t) 3 e dÊœ œœ œ Ê œkkabababÉ È ÈÈ
v
tt tt t 2t t### '
0
t
7
7
3 e 3 Length s(0) s( ln 4) 0 3 e 3œÊ œœ œ
ÈÈ È È
Š‹
tln 4 33
4
Ê
È
14. (1 2t) (1 3t) (6 6t) 2 3 6 2 3 ( 6) 7 s(t) 7 d 7trij kvijkvœ Êœ Ê œ œÊ œ œ kkÈ
## # '
0
t
7
Length s(0) s( 1) 0 ( 7) 7Êœœœ

843. Section 13.3 Arc Length and the Unit Tangent Vector 839 T
15. 2t 2t 1 t 2 2 2t 2 2 ( 2t) 4 4trij kvijkvœ ʜʜ œŠ‹Š‹ Š‹Š‹
ÈÈ ÈÈ ÈÈ È
ab kk Ê
#
##
##
2 1 t Length 2 1 t dt 2 1 t ln t 1 t 2 ln 1 2œÊ œ œ œ
ÈÈÈÈÈÈ
’“Š‹Š‹ Š‹
#### "
#
"
! '
0
1
t
2
16. Let the helix make one complete turn from t 0 to t 2 .œœ 1
Note that the radius of the cylinder is 1 theÊ
circumference of the base is 2 . When t 2 , the point P is11 œ
(cos 2 sin 2 2 ) (1 0 2 ) the cylinder is 2 units111 1 1ßßœßßÊ
high. Cut the cylinder along PQ and flatten. The resulting
rectangle has a width equal to the circumference of the
cylinder 2 and a height equal to 2 , the height of theœ11
cylinder. Therefore, the rectangle is a square and the portion
of the helix from t 0 to t 2 is its diagonal.œœ 1

17. (a) (cos t) (sin t) ( cos t) , 0 t 2 x cos t, y sin t, z 1 cos t x yrij kœ" ŸŸÊœœœÊ 1
##
cos t sin t 1, a right circular cylinder with the z-axis as the axis and radius 1. Thereforeœœ œ
##
P(cos t sin t 1 cos t) lies on the cylinder x y 1; t 0 P(1 0 0) is on the curve; t Q( 1 1)ßß œœÊßß œÊ!ßß
##
#
1
is on the curve; t R( 1 0 2) is on the curve. Then PQ and PR 2 2œ Ê ß ß œ œ
Ä Ä
1 ijk i k
ik
ijk
Ê PQ PR 2 2 is a vector normal to the plane of P, Q, and R. Then the1
202
Ä
‚œ œ
Ä
$!!
$
Ô×
ÕØ
plane containing P, Q, and R has an equation 2x 2z 2(1) 2(0) or x z 1. Any point on the curveœ œ
will satisfy this equation since x z cos t (1 cos t) 1. Therefore, any point on the curve lies on theœ œ
intersection of the cylinder x y 1 and the plane x z 1 the curve is an ellipse.
##
œ œÊ
(b) ( sin t) (cos t) (sin t) sin t cos t sin t 1 sin t vijkv Tœ Ê œ œ Ê œ kk
ÈÈ
### # v
vkk
(0) , , ( ) , œÊœœœœ
( sin t) (cos t) (sin t)
1sint 2 2
3Êœ œ
œ ##
Êœ Êijk ik ik
ÈÈÈ
œ
TjT T jTˆ‰ ˆ ‰
11
1
(c) ( cos t) (sin t) (cos t) ; isaijknikœ œ
normal to the plane x z 1 cos t cos tœÊ œ na†
0 is orthogonal to is parallel to theœÊ Êana
plane; (0) , , ,aika jaikœ œ œ ˆ‰ ab
1
#
1
ajˆ‰
31
#
œ

(d) 1 sin t (See part (b) L 1 sin t dtkk
ÈÈ
vœ Êœ
## '
0
2
œ
(e) L 7.64 (by )¸ Mathematica
18. (a) (cos 4t) (sin 4t) 4t ( 4 sin 4t) (4 cos 4t) 4 ( 4 sin 4t) (4 cos 4t) 4rijkv i jkvœÊœ Êœ kkÈ
###
32 4 2 Length 4 2 dt 4 2 t 2 2œœ Ê œ œ œ
È ÈÈÈÈ
’“ '
0
2
œ" 1Î#
!
1
(b) cos sin sin cos r i jkv i jkœÊœ ˆ‰ˆ‰ ˆ ‰ˆ ‰
ttt t t
### #####
"""
sin cos Length dt t 2 2Êœ œ œ Ê œ œ œkkɈ‰ˆ‰ˆ‰ É ’“
È
v
"" """
## # # # # #
###
%
!
tt
44 2
222
ÈÈÈ '
0
4
œ 1
1
(c) (cos t) (sin t) t ( sin t) (cos t) ( sin t) ( cos t) ( 1) 1 1rijkv ijkvœÊœÊœœ kkÈ È
###
2 Length 2 dt 2 t 2 2œÊ œ œ œ
ÈÈÈÈ
’“ '
œ2
0œ
!
Ê#1
1

844. 840 Chapter 13 Vector-Valued Functions and Motion in Space
19. PQB QOB t and PQ arc (AQ) t sincenœnœœœ
PQ length of the unwound string length of arc (AQ);œœ
thus x OB BC OB DP cos t t sin t, andœœœ
y PC QB QD sin t t cos tœœ œ

20. cos t t sin t sin t t cos t sin t t cos t sin t cos t t sin t cos trijv i jœ Êœababa ba b abab
t cos t t sin t t cos t t sin t t t t, t 0œÊœ œœœ Êœœababkkabab kk É È
ijv T ij
22
2 t cos t t sin t
tt
v
vkk
cos t sin tœ ij
13.4 CURVATURE AND THE UNIT NORMAL VECTOR N
1. t ln (cos t) (tan t) 1 ( tan t) sec t sec t sec t, sinceri j vi ji j vœ Ê œ œ Ê œ œ œ œˆ‰ kk k kÈ È
Ê ###sin t
cos t
t (cos t) (sin t) ; ( sin t) (cos t) Ê œ œ œ œ
11
##
"
Tijij ij
vT
vkk
ˆ‰ ˆ‰
sec t sec t dt
tan t d
( sin t) ( cos t) 1 ( sin t) (cos t) ;Êœ œÊœœ ¸¸ È
d
dt
T ##
Nij
ˆ‰
¸¸
d
dt
d
dt
T
T
1 cos t.,œ† œ †œ
1d
dt sec tkkv
T
¸¸
"
2. ln (sec t) t (tan t) ( tan t) 1 sec t sec t sec t,r i j v ij ij vœÊœ œÊœ œœœ ˆ‰ kk k kÈ È
sec t tan t
sec t
## #
since t (sin t) (cos t) ; (cos t) (sin t) Ê œ œ œ œ
11
##
Tijijij
vT
vkk
ˆ‰ ˆ‰
tan t 1 d
sec t sec t dt
(cos t) ( sin t) 1 (cos t) (sin t) ;Êœ œÊœœ ¸¸ È
d
dt
T ##
Nij
ˆ‰
¸¸
d
dt
d
dt
T
T
1 cos t.,œ† œ †œ
1d
dt sec tkkv
T
¸¸
"
3. (2t 3) 5 t 2 2t 2 ( 2t) 2 1 t rijvijv T ijœ Êœ Ê œ œ Êœœ ab kk È È
# ## #
œœ
Êv
vkk ÈÈ
22t
21 t 21 t
œœ
; œ œ Êœ
"" "
œœ
œœ œ œ
##
ÈÈ
Š‹Š‹ Š‹ Š‹
ÈÈ È È1t 1t
td t d t
dt dt
1t 1t 1t 1t
œœ
œœ œ œ
"" " "
ij i j
TT
¸¸
Í
Í
Í
Ì

;œœÊœœÉ
"" "
œ œ œœab
ˆ‰
¸¸ ÈÈ1t 1t
t
1t 1t
œ
œ œ
œœ
Nij
d
dt
d
dt
T
T
,œ† œ † œ
1d
dt 1 t 1t 1tkk È
abv
T
¸¸
"" "
#œ œ #œ
œ
œ
œ
3/2
4. (cos t t sin t) (sin t t cos t) (t cos t) (t sin t) ( t cos t) (t sin t) t trijvijvœ Êœ Êœ œœ kk kkÈ È
###
t, since t 0 (cos t) (sin t) ; ( sin t) (cos t)œÊœœ œ œ Tijij
vT
v
ij
kk
(t cos t) (t sin t)
tdt
dœ
( sin t) (cos t) 1 ( sin t) (cos t) ; 1Êœ œÊœœ œ†œ†œ¸¸ ¸¸È
d 1d
dt dt t t
T T
v
## ""
Nij
ˆ‰
¸¸ kk
d
dt
d
dt
T
T
,
5. (a) x . Now, f x x 1 fx,ab ab k k c d¹¹ abÉ abœ† œ Ê œ Êœ
1
xdt
dT x 2
kkab
ab
v v
v
vi j v T
w w
kk
1 f x 1 . Thus xfx fxœŠ‹ Š‹ œcd ab cd abab ab
ww
ÎÎ
w Ê
œœ
22
12 12
d
dt
fxfx fx
11fx fx
œœ
ij ij
T


ÊÊ
ab ab ab
cd cdab abŠ‹Š‹
22
32 32
Êœ œ œ¹¹
Í
Í
Í
Ì
”• Ë
dx fx
dt
fxf x
1fx
2
1fx
2
fx fx1
1fx
fx
1fx
Tab ab
ab ab
cdab
cdab
cd cdab abŠ‹
cdab
kkab
¹¹cdab
Ê
œ
œ
œ
œ œ


Ê


Ê




Š‹
Š‹ Š‹
2
32
2
32
22
2
3
2

845. Section 13.4 Curvature and the Unit Normal Vector 841 N
Thus x ,abœ†œ
1
1fx
fx fx
1fx
1fx
ab
Š‹
œn !
œn !
œ

ÊÊ



ab
kk kkab ab
kk ab
cdab
2
12 32 2
2
(b) y ln (cos x) ( sin x) tan x sec x œÊœœÊœÊœ œ
dy d y
dx cos x dx sec x
sec x
1 ( tan x)
sec x
ˆ‰
" # Ê
œÊ
œ
œ "
œ
œ

œ
,
kk
cd
kk
cos x, since xœœ
"
##sec x
11
(c) Note that f (x) 0 at an inflection point.
ww
œ
6. (a) f(t) g(t) x y x y x yrijijvijv T i jœ œÊœÊœ Êœœ
ÞÞ Þ Þ
Þ Þ
kkÈ
## v
vkk x
xy xy
y
ÈÈ
ÞÞ ÞÞ
œœ
œœ œœ

dd
dt dt
2
TT
œÊœ
yyx xy xxy yx
xy xy
yyx xy
xy
xxy yx
xy
Þ ÞÞÞ ÞÞÞ Þ ÞÞÞ ÞÞÞ
""
ÞÞ ÞÞ
œœ
ÞÞÞÞ ÞÞÞ
"
ÞÞ
œ
ÞÞÞÞ ÞÞÞ
"
ÞÞ
œ
abab
ab ab
ab
ab
ab
ab
œœ œœ œœ œ
3/2 3/2 3/2ij ¸¸ Ê’“’“
œ
œœ
œœ
3/2
2
3
2
œÊ
aba b
ab
yxyxxy
xy
ÞÞÞÞÞÞÞÞ

ÞÞ
œ
; .œœ†œ†œ
kk kkkk
kk kk È
ab
yx xy yx xy yx xy
xy xy
1
xy xy
ÞÞÞ ÞÞÞ ÞÞÞ ÞÞÞ ÞÞÞ ÞÞÞ
"""
ÞÞ ÞÞ
œœ
ÞÞ
œ
ÞÞ
œ
œœ œœ œœ œœ ,
1d
dtkkv
T
¸¸
3/2
(b) (t) t ln (sin t) , 0 t x t and y ln (sin t) x 1, x 0; y cot t, y csc tri jœ Êœ œ Êœ œ œ œ œ
ÞÞÞÞ ÞÞ
1
cos t
sin t
#
sin tÊœ œ œ,
kk
ab
ÊÊ
œ
csc t 0
1cott)
csc t
csc t
œ
œ

œ
"
(c) (t) tan (sinh t) ln (cosh t) x tan (sinh t) and y ln (cosh t) xrijœÊœ œÊœœ
Þ
" "
œ
"cosh t
1sinht cosh t
œ
sech t, x sech t tanh t; y tanh t, y sech t sech tœœ œœœÊœ œ
ÞÞ Þ ÞÞ
sinh t
cosh t sech t tanh t
sech t sech t tanh t# œ
œ
,
kk
ab

œœ
kk
sech tœ
7. (a) (t) f(t) g(t) f (t) g (t) is tangent to the curve at the point (f(t) g(t));rijvijœ Êœ ß
ww
g (t) f (t) f (t) g (t) g (t)f (t) f (t)g (t) 0; ( ) 0; thus,nv i j i j nv nv†† ††œ œ œ œ œcdcd
ww w w wwww
and are both normal to the curve at the pointnn$
(b) (t) t e 2e 2e points toward the concave side of the curve; andrijvi jn ij Nœ Ê œ Ê œ œ
2t 2t 2t n
nkk
4e 1 kk
È
nNijœÊœ
4t 2e
14e 14e
"
œœ
2t
4t 4t
ÈÈ (c) (t) 4 t t points toward the concave side of the curve;rijvijnijœÊœ Êœ
È
# Ê
ÊÊ
tt
4t 4t
ÈÈ
œœ
and 1 4 t tNn N ijœœœÊœ
n
nkk È
kkÉ Š‹
È
t2
4t
4t
œ
œ
œ
Ê#
Ê
" #
8. (a) (t) t t t t points toward the concave side of the curve when t 0 andrijvijnijœ Ê œ Ê œ
"$##
3
t points toward the concave side when t 0 t for t 0 andœ Ê œ nij N ij
# #"
œ
È
1t
!
ab
t for t0Nijœ
"
œ
#
È
1t
!
ab
(b) From part (a), 1 t kk
È ¸¸ ÉvTij ijœÊœ Êœ Ê œ
% "
œœ œœ œ
ÈÈ
ab ab ab1t 1t
t d 2t 2t d 4t 4t
dt dt
1t 1t 1t
!!

!!
!
"
TT
62
; ; t 0œœœ œ Á
2t
1t
1 t 2t 2t t t
2t
1t 1t t1t t1t
kk ˆ‰
¸¸ kk
ab ab kk kk
ÈÈœ
œÊ Ê
œœ œœ
!
!" "
!!
!!Nijij
d
dt
d
dt
T
T
Š‹ does not exist at t 0, where the curve has a point of inflection; 0 so the curvature N œœœ ¸¸¸
dd
dt ds
TT
t0œ
,
0 at t 0 is undefined. Since x t and y t y x , the curve is theœœœÊœ œœÊœ¸¸
ddt d
dt ds ds 3 3
TT
† N
""" $$
,
cubic power curve which is concave down for x t 0 and concave up for x t 0.œ œ
9. (3 sin t) (3 cos t) 4t (3 cos t) ( 3 sin t) 4 (3 cos t) ( 3 sin t) 4rijkv i jkvœ Êœ Êœ kkÈ
###
25 5 cos t sin t sin t cos tœœÊœœ Êœ
È ˆ‰ˆ‰ ˆ ‰ˆ‰Tijk ij
vT
vkk
334d33
555dt55
sin t cos t ( sin t) (cos t) ; Ê œ œ Ê œ œ œ†œ¸¸ ˆ ‰ ˆ ‰É
d3 33 133
dt 5 5 5 5 5 25
T
##
Nij
ˆ‰
¸¸
d
dt
d
dt
T
T
,
10. (cos t t sin t) (sin t t cos t) 3 (t cos t) (t sin t) (t cos t) (t sin t) trijkvijvœ Êœ Êœ œ kkÈ È
###
t t, if t 0 (cos t) (sin t) , t 0 ( sin t) (cos t)œœ Êœ œ Ê œ kk Tij ij
vT
vkk
d
dt
( sin t) (cos t) 1 ( sin t) (cos t) ; 1Êœ œÊœœ œ†œ¸¸ È
d
dt tt
T ## ""
Nij
ˆ‰
¸¸
d
dt
d
dt
T
T
,

846. 842 Chapter 13 Vector-Valued Functions and Motion in Space
11. e cos t e sin t 2 e cos t e sin t e sin t e cos t rijkv i jœÊœÊabababab
tt tttt
e cos t e sin t e sin t e cos t 2e e 2 ;kkababÉ È È
vœœœ
tt tt 2t t##
Tij ijœœ Ê œ
vT
vkk ÈÈ È È
Š‹Š‹ Š ‹Š‹
cos t sin t sin t cos t d sin t cos t cos t sin t
22 2 2 dt
Êœ ÊÊ Ê
1 ;Êœ œÊœœ ¸¸ ÊŠ‹Š‹ Š‹Š‹
d sin t cos t cos t sin t cos t sin t sin t cos t
dt 22 2 2
T ÊÊ Ê Ê Ê Êœ
##
ÈÈ È È
ˆ‰
¸¸
Nij
d
dt
d
dt
T
T
1,œ† œ †œ
1d 1 1
dt 22kk ÈÈv
T
¸¸
ee
tt
12. (6 sin 2t) (6 cos 2t) 5t (12 cos 2t) (12 sin 2t) 5rijkv i jkœ Êœ
(12 cos 2t) ( 12 sin 2t) 5 169 13 Êœ œ œÊœkkÈ È
vT
### v
vkk
cos 2t sin 2t sin 2t cos 2tœÊœˆ‰ˆ‰ ˆ ‰ˆ‰
12 12 5 d 24 24
13 13 13 dt 13 13
ijk i j
T
sin 2t cos 2t ( sin 2t) (cos 2t) ;Êœ œÊœœ ¸¸ ˆ ‰ ˆ ‰É
d24 24 24
dt 13 13 13
T
##
Nij
ˆ‰
¸¸
d
dt
d
dt
T
T
.,œ† œ†œ
1d 124 24
dt 13 13 169kkv
T
¸¸
13. , t 0 t t t t t t 1, since t 0 rij vijv Tœ ʜʜœ ÊœŠ‹ Š‹ kk
È È
tt
3

#
# %# # v
vkk
œ Êœ Êœ
t1d1 t d t
tt t1 dt dt
t1 t1 t1 t1
ÈÈ
ab ab ab ab
œœ œœ œ œ

œœ œœ œ œ
"
##
ij i j
TT
¸¸ ÊŠ‹Š‹
; .œœÊœœ œ†œ†œÉ ¸¸
1t 1 t 1 d 1 1
t1 t1 dt t1 t1 t1 tt1 tt 1
" "
œ œ œœœ œ œ
œ
œ
" œ œ
œœ œ œ

ab
ˆ‰
¸¸ ÈÈ È kk
ab
Nij
d
dt
d
dt
T
T
,
v
T
14. cos t sin t , 0 t 3 cos t sin t 3 sin t cos trij v i jœ Êœ abab a ba b
$$ # #
#
1
3 cos t sin t 3 sin t cos t 9 cos t sin t 9 sin t cos t 3 cos t sin t, since 0 tÊœ œ œ kk a b a bÉ È
v
## %#%#
##
#
1
( cos t) (sin t) (sin t) (cos t) sin t cos t 1 Êœœ Ê œ Ê œ œÊœTijij N
vTT
vkk
ˆ‰
¸¸
dd
dt dt
¸¸ È ##
d
dt
d
dt
T
T
(sin t) (cos t) ; 1 .œ œ†œ †œij ,
1d 1 1
dt 3 cos t sin t 3 cos t sin tkkv
T
¸¸
15. t a cosh , a 0 sinh 1 sinh cosh cosh ri j vi j vœ Ê œ Ê œ œ œˆ ‰ ˆ ‰ ˆ‰ ˆ‰ kkÉÉ
ttttt
aaaaa
##
sech tanh sech tanh sech Êœœ Ê œ Tij ij
vT
vkk
ˆ‰ˆ‰ ˆ ‰ˆ ‰
ttd tt t
aadtaaaaa
"" #
sech tanh sech sech tanh sech ;Êœ œ Êœœ ¸ ¸ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ ‰ ˆ ‰É
dtttt tt
dt a a a a a a a a a
T """ ## %
œœ Nij
ˆ‰
¸¸
d
dt
d
dt
T
T
sech sech .,œ† œ † œ
1d 1 t t
dt a a a acosh kkv
T
¸ ¸ ˆ‰ ˆ‰
t
a
"" #
16. (cosh t) (sinh t) t (sinh t) (cosh t) sinh t ( cosh t) 1 2 cosh trijkvijkvœÊœÊœ œ kkÈ È
##
tanh t sech t sech t sech t tanh tÊœœ Ê œ Tijk i k
vT
vkk ÈÈÈ È È
Š‹ Š‹ Š ‹Š ‹
""" " " #
222 2 2
d
dt
sech t sech t tanh t sech t (sech t) (tanh t) ;Êœ œ Êœœ ¸¸ É
d
dt 2
T "" "
##
%##
È
ˆ‰
¸¸
Nik
d
dt
d
dt
T
T
sech t sech t.,œ† œ † œ
1d 1
dt 2 cosh t 2kk ÈÈv
T
¸¸
""
#
#
17. y ax y 2ax y 2a; from Exercise 5(a), (x) 2a 1 4a xœÊœÊœ œ œ
#w ww ##
œ
$Î#
,
kk
ab
2a
14ax
œœ

kka b
(x) 2a 1 4a x 8a x ; thus, (x) 0 x 0. Now, (x) 0 for x 0 and (x) 0 forÊœ œÊœ ,,,,
w###w w w
#
ÊŸ #3
kka b a b
x 0 so that (x) has an absolute maximum at x 0 which is the vertex of the parabola. Since x 0 is theœ œ ,
only critical point for (x), the curvature has no minimum value.,

847. Section 13.4 Curvature and the Unit Normal Vector 843 N
18. (a cos t) (b sin t) ( a sin t) (b cos t) ( a cos t) (b sin t) r i jv i ja i jvaœ Êœ Êœ Ê‚
ab ab ab, since a b 0; (t)a sin t b cos t 0
a cos t b sin t 0
œœÊœœœ_
__
ââ
ââ
ââ
ââ
ââ
ââ
kkkk
ijk
kva ‚ ,
kk
kk
va
v
‚
"
ab a sin t b cos t ; (t) (ab) a sin t b cos t 2a sin t cos t 2b sin t cos tœ œ ab aba b
## # # w ## # # # #
$Î# &Î#
#
,
3
(ab) a b (sin 2t) a sin t b cos t ; thus, (t) 0 sin 2t 0 t 0, identifyingœ œÊ œÊœ
3
#
## ## ## w
ÊŸ #
ab a b ,1
points on the major axis, or t , identifying points on the minor axis. Furthermore, (t) 0 forœ
11
##
w3
,
0 t and for t ; (t) 0 for t and t 2 . Therefore, the points associatedœœ œœ ‰ œœ œœ
1111
####
w
1, 1 1
33
with t 0 and t on the major axis give absolute maximum curvature and the points associated with tœœ œ1
1
#
and t on the minor axis give absolute minimum curvature.œ
31
#
19. ; 0 a b 0 a b a b since a, b 0. Now, 0 if,œÊœ œÊœÊœ„Êœ  
adabd d
a b da da daab
ÄÄ
ÄÄ
ÄÄ
Ä
Ä
ʜ
Ä
##,, ,
ab
a b and 0 if a b is at a maximum for a b and (b) is the maximum value of .Ê œ œœ
d b
da bb 2b
,
,, ,
ÄÄ
Ä
"
20. (a) From Example 5, the curvature of the helix (t) (a cos t) (a sin t) bt , a, b 0 is ; alsorijkœ œ ,
a
ab
ÄÄ
Ä
a b . For the helix (t) (3 cos t) (3 sin t) t , 0 t 4 , a 3 and b 1 kk
È
vrijkœ œ ŸŸ œ œÊœ œ
##
Ä
1,
33
31 10
ÄÄ
and 10 K 10 dt tkk
ÈÈ
’“vœÊœ œ œ '
0
4
Ä
3312
10 10 10
ÈÈ
%
!
1
1
(b) y x x t and y t , t (t) t t 2t 1 4t ;œÊœ œ__Ê œÊœ Ê œ
## # #
rijvijv kk
È
; ; . ThusTij i jœ œ œ œ
1 2t d 4t 2 d 16t 4 2
14t 14t dt dt 1 4t14t 14t 14tÈÈ
ab ab abÄÄ
ʜ
ÄÄ Ä Ä
ÄÄ ÄÄ Ä
Ä
TT
3/2 3/2
2
3
¸¸ É
. Then K 1 4t dt dt,œ†œ œ œ
12 2 2 2
14t 14t 14t
14t 14t
È
Š‹
È
Š‹
ÈÄ Ä Ä
Ä Ä
#
Ä
Ä Ä
Ä Ä
"
3
''
Ä
""
Š‹
È
lim dt lim dt lim tan 2t lim tan 2tœœ
aa bbÄ_ Ä_ Ä_ Ä_
''
a0
0b
a0
b
22
14t 1 4tÄÄ
" "
!
ÄÄ cd cd
lim tan 2a lim tan 2bœ œœ
a bÄ_ Ä_
abab
" "
##
11
1
21. t (sin t) (cos t) 1 (cos t) 1 cos t 1 cos 1; ri j vi j v v Tœ Ê œ Ê œ œ Ê œ œ œ kkÈ È ¸¸ ˆ‰ˆ‰ É
## # #
##
11 v
vkk
; œÊœ Êœ œ œœ
ij TTTÄ
Ä ÄÄ
Ê
Ä œ Ä
cos t
1cost
d sin t cos t sin t d d 1
dt dt 1 cos t dt 11cost 1cost
sin t
t
sin
1cosÈ
abab
kk ¸¸
ˆ‰
2 22
3/2 3/2 2
2
2
2
2
ij ¸¸ ¸¸ Ä
Ä
Ä
1. Thus 1 1,ˆ‰
1
21
1
œ†œ
1 and the center is 0 x y 1Êœœ ß Ê œ3
"
##
#
#
1
ˆ‰ ˆ ‰
11
22. (2 ln t) t 1 1 ;ri jvi jv T ijœÊœÊœœÊœœˆ‰ ˆ‰ˆ‰ ˆ‰ kkÉ
""
ÄÄt t t t t t t1 t1
241t12tt1
2
Ä 222 22
22
v
vkk
. Thus 1
d4td 21dt22t2
dt dt t 1 dt t 1 t 1 2
2t 1
t1 t1 t1 t1
4 t 1 16tTT T
v
œ Êœ œ œ†œ†œ Êœ
ÊÊ
ÄÄ Ä Ä
ʜ
ÄÄÄ
ˆ‰
ab ab ab ab
ab
kk
2
22 2 2
22 42 22 2 2
22
2
22
ij ¸¸ ¸¸
Ê ab,,
2. The circle of curvature is tangent to the curve at P(0 2) circle has same tangent as the curveœÊœœ ßÊ
""
#
3
,
(1) 2 is tangent to the circle the center lies on the y-axis. If t 1 (t 0), then (t 1) 0Êœ Ê Á vi
#
t 2t 1 0 t 1 2t 2 since t 0 t 2 t 2 y 2 on bothÊ Ê Ê Ê Ê Ê
## ""t1
ttt
Ä
ˆ‰
sides of (0 2) the curve is concave down center of circle of curvature is (0 4) x (y 4) 4ß Ê Ê ß Ê œ
##
is an equation of the circle of curvature

848. 844 Chapter 13 Vector-Valued Functions and Motion in Space
23. y x f (x) 2x and f (x) 2œÊ œ œ
#w ww
Êœ œ,
kk
abab
2
1(2x) 14x
2
œœ
œœ


24. y f (x) x and f (x) 3xœÊ œ œ
x
4
!
w$ww #
Êœ œ,
kk
Š‹ab
ab
3x
1x
3x
1x
œ
"
œ

œ

œ
œ

25. y sin x f (x) cos x and f (x) sin xœÊœ œ
www
Êœ œ,
kk kk
abab
Ê
œœ
sin x sin x
1cosx 1cosx
œœ


26. y e f (x) e and f (x) eœÊ œ œ
xx x www
Êœ œ,
kk
Š‹
ˆ‰
e
1
e
1
x x
œ
œabe
e
x
2x
œ
"

27-34. Example CAS commands:
:Maple
with( plots );
r := t -> [3*cos(t),5*sin(t)];
lo := 0;
hi := 2*Pi;
t0 := Pi/4;
P1 := plot( [r(t)[], t=lo..hi] ):
display( P1, scaling=constrained, title="#27(a) (Section 13.4)" );
CURVATURE := (x,y,t) ->simplify(abs(diff(x,t)*diff(y,t,t)-diff(y,t)*diff(x,t,t))/(diff(x,t)^2+diff(y,t)^2)^(3/2));
kappa := eval(CURVATURE(r(t)[],t),t=t0);
UnitNormal := (x,y,t) ->expand( [-diff(y,t),diff(x,t)]/sqrt(diff(x,t)^2+diff(y,t)^2) );
N := eval( UnitNormal(r(t)[],t), t=t0 );
C := expand( r(t0) + N/kappa );
OscCircle := (x-C[1])^2+(y-C[2])^2 = 1/kappa^2;
evalf( OscCircle );
P2 := implicitplot( (x-C[1])^2+(y-C[2])^2 = 1/kappa^2, x=-7..4, y=-4..6, color=blue ):

849. Section 13.5 Torsion and the Unit Binormal Vector 845 B
display( [P1,P2], scaling=constrained, title="#27(e) (Section 13.4)" );
: (assigned functions and parameters may vary)Mathematica
In Mathematica, the dot product can be applied either with a period "." or with the word, "Dot".
Similarly, the cross product can be applied either with a very small "x" (in the palette next to the arrow) or with the word,
"Cross". However, the Cross command assumes the vectors are in three dimensions
For the purposes of applying the cross product command, we will define the position vector r as a three dimensional vector
with zero for its z-component. For graphing, we will use only the first two components.
Clear[r, t, x, y]
r[t_]={3 Cos[t], 5 Sin[t] }
t0= /4; tmin= 0; tmax= 2 ;11
r2[t_]= {r[t][[1]], r[t][[2]]}
pp=ParametricPlot[r2[t], {t, tmin, tmax}];
mag[v_]=Sqrt[v.v]
vel[t_]= r'[t]
speed[t_]=mag[vel[t]]
acc[t_]= vel'[t]
curv[t_]= mag[Cross[vel[t],acc[t]]]/speed[t] //Simplify
3
unittan[t_]= vel[t]/speed[t]//Simplify
unitnorm[t_]= unittan'[t] / mag[unittan'[t]]
ctr= r[t0] + (1 / curv[t0]) unitnorm[t0] //Simplify
{a,b}= {ctr[[1]], ctr[[2]]}
To plot the osculating circle, load a graphics package and then plot it, and show it together with the original curve.
<<Graphics`ImplicitPlot`
pc=ImplicitPlot[(x a)2 + (y b)2 == 1/curv[t0] , {x, 8, 8},{y, 8, 8}]$$ $$
2
radius=Graphics[Line[{{a, b}, r2[t0]}]]
Show[pp, pc, radius, AspectRatio 1]Ä
13.5 TORSION AND THE UNIT BINORMAL VECTOR B
1. By Exercise 9 in Section 13.4, cos t sin t and ( sin t) (cos t) so that Ti jkNijBTNœ œ œ‚ˆ‰ˆ ‰
334
555
cos t sin t cos t sin t . Also (3 cos t) ( 3 sin t) 4
sin t cos t 0
œ œ œ
$$
ââ
ââ
ââ
ââ
ââ
ââ
ˆ‰ˆ‰
ijk
ijkvi jk
3344 4 3
5555 5 5
( 3 sin t) ( 3 cos t) ( 3 cos t) (3 sin t) and 3 cos t 3 sin t 4
3 sin t 3 cos t 0
Êœ Ê œ ‚œ $
$$
aij ijva
ijk
d
dt
a
ââ
ââ
ââ
ââ
ââ
ââ
(12 cos t) (12 sin t) 9 12 cos t 12 sin t 9 225. Thusœ Ê ‚ œ œijkva kka ba bab
2 ###
7œœœœ
ââ
ââ
ââ
ââ
ââ
ââ ab
3 cos t 3 sin t 4
3 sin t 3 sin t 0
3 cos t 3 sin t 0
225 225 225 25
49 sint9 cost 36 4
Ê
ÊÊ
Ê ÞÊ Ê Ê
œœ
2. By Exercise 10 in Section 13.4, (cos t) (sin t) and ( sin t) (cos t) ; thus TijN ijBTNœ œ œ‚
cos t sin t . Also (t cos t) (t sin t)cos t sin t 0
sin t cos t 0
œœœœ
$
ââ
ââ
ââ
ââ
ââ
ââ
ab
ijk
kk v i j
##
t sin t cos t t cos t sin t t cos t sin t sin t t sin t cos t cos tÊœ Ê œ aij i jababa ba bab
d
dt
a
t cos t 2 sin t 2 cos t t sin t . Thus t cos t t sin t 0
t sin t cos t t cos t sin t 0
œ ‚ œ
$" "
abab
ââ
ââ
ââ
ââ
ââ
ââabab
ijva
ijk
[(t cos t)(t cos t sin t) (t sin t)( t sin t cos t)] t t t . ThusœœÊ‚œœ kk va
##
#
kkab
2 4

850. 846 Chapter 13 Vector-Valued Functions and Motion in Space
07œœœ
ââ
ââ
ââ
ââ
ââ
ââ
t cos t t sin t 0
cos t t sin t sin t t cos t 0
2 sin t t cos t 2 cos t t sin t 0
tt
0
ʜ
ÊÊ Ê
44
3. By Exercise 11 in Section 13.4, and ; ThusTijN i jœ œ Š‹Š‹ Š ‹Š ‹
cos t sin t sin t cos t cos t sin t sin t cos t
22 2 2
Êœ ÊÊÊœ
ÈÈ È È

0
0
BTN
ijk
œ‚œ œ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
’“Š‹Š‹
cos t sin t sin t cos t
22
cos t sin t sin t cos t
22
cos t 2 cos t sin t sin t sin t 2 sin t
2
ʜ
ÊÊ Êœ
Êœ œ
ÈÈ
ÈÈ
œœœ
cos t cos t
2
œ
œ
k
. Also, e cos t e sin t e sin t e cos tœ œœ’“Š‹Š‹ abab
1sin 2t 1sin 2t
22
tt ttÊœab ab
kk v i j
= 2e sin t 2e cos te sin t cos t e cos t sin t e cos t sin t e sin t cos tʜ $$ " $ $ " "aijijcdcdababa bababab
tt tt tt
2e cos t sin t 2e sin t cos t . Thus 2e e cos t sin t e sin t cos t 0
2e sin t 2e cos t 0
Êœ ‚œ œ $"
$
d
dt
tt 2ttt
tt
a
aba b
ââ
ââ
ââ
ââ
ââ
ââ
ababijva k
ijk
2e 4e . Thus Ê‚ œ œ œkkabva
2 2t 4t
cos t sin t sin t cos t 0
2 sin t 2 cos t 0
2 cos t sin t 2 sin t cos #
ʜ
Ê
ʜ ʜ
7
ââ
ââ
ââ
ââ
ââ
ââ
abab
aba b
ee
ee
ee
tt
tt
tt
t0
4
e
4t œ0
4. By Exercise 12 in Section 13.4, cos 2t sin 2t and ( sin 2t) (cos 2t) soTijkNijœœˆ‰ˆ‰
12 12 5
13 13 13
cos 2t sin 2t cos 2t sin 2t . Also,
sin 2t cos 2t 0
BTN i j k
ijk
œ‚œ œ
$$
ââ
ââ
ââ
ââ
ââ
ââ
ˆ‰ˆ ‰ ˆ‰ˆ‰
aba b
12 12 5 5 5 12
13 13 13 13 13 13
(12 cos 2t) (12 sin 2t) 5 ( 24 sin 2t) (24 cos 2t) and ( 48 cos 2t) (48 sin 2t)vijka ij ijœÊœ œ
d
dt
a
(120 cos 2t) (120 sin 2t) 288 12 cos 2t 12 sin 2t 5
24 sin 2t 24 cos 2t 0
va i j k va
ijk
‚œ œ Ê ‚ $
$$
ââ
ââ
ââ
ââ
ââ
ââ
kk
2
(120 cos 2t) ( 120 sin 2t) ( 288) 120 cos 2t sin 2t 288 97344. Thusœœœ
#######
ab
7œœœ
ââ
ââ
ââ
ââ
ââ
ââ ab
12 cos 2t 12 sin 2t 5
24 sin 2t 24 cos 2t 0
48 cos 2t 48 sin 2t 0
97344 97344 169
52448 10
Ê
ÊÊ
Ê ÞÊ Þ
5. By Exercise 13 in Section 13.4, and so that T i j N i j BTNœ œ œ
t1 1t
t1 t1 t1 t1ab ab
ÈÈ
œœ œœ
œœ œœ
1/2 1/2 ‚
. Also, t t 2t 2 so that 0 0
0
0
tt0
2t 0
200
œœœÊœÊœœÊœ "
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ijk
kvijaij i
t1
t1 t1
t1 t1
t
d
dt
ÈÈ
ÈÈ
œœ
œœ
œœ
"
œœ
#
#
a 7
6. By Exercise 14 in Section 13.4, ( cos t) (sin t) and (sin t) (cos t) so that T i j N i j BTNœ œ œ ‚
. Also, 3 cos t sin t 3 sin t cos tcos t sin t 0
sin t cos t 0
œœœ$
ââ
ââ
ââ
ââ
ââ
ââ
abab
ijk
kv i j
##
3 cos t sin t 3 sin t cos t 3 cos t sin t 3 sin t cos tÊÊaij i jœ œ
ddddd dd
dt dt dt dt dt dt dt
abab abab ˆ‰ˆ‰
## # # a

3 cos t sin t 3 sin t cos t 0
3 cos t sin t 3 sin t cos t 0
3 cos t sin t 3 sin t cos
Ê
ââ
ââ
ââ
ââ
ââ
ââ
ââ
abab
ˆ‰ˆ‰abab
$
$
$
##
##
##
dd
dt dt
dd dd
dt dt dt dt
t0
0 0œÊ œ7
7. By Exercise 15 in Section 13.4, sech tanh and tanh sech so that T i j N i j BTNœœ œ œ
v
vkk
ˆ‰ˆ‰ ˆ ‰ˆ‰
tt tt
aa aa
‚
. Also, sinh cosh sinh so thatsech tanh 0
tanh sech 0
œœœÊœÊœ
$
ââ
ââ
ââ
ââ
ââ
ââ
ˆ‰ ˆ‰
ˆ‰ ˆ‰
ˆ‰ ˆ ‰ ˆ‰
ijk
kvi ja j j
tt
aa
tt
aa
ttdt
aaadtaa
"" a
œ

851. Section 13.5 Torsion and the Unit Binormal Vector 847 B
0 0
1sinh 0
0 cosh 0
0 sinh 0
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ˆ‰
ˆ‰
ˆ‰
t
a
aa
t
aa
t
"
"
œ
œÊ œ7
8. By Exercise 16 in Section 13.4, tanh t sech t and (sech t) (tanh t) so thatTijkNikœ œŠ‹ Š‹
"""
ÈÈÈ
222
tanh t sech t . Also, (sinh t) (cosh t) tanh t sech t
sech t 0 tanh t
BTN i j k v i jk
ijk
œœ œ œ
$
‚
ââ
ââ
ââ
ââ
ââ
ââ
ââ
Š‹ Š‹
""" """
ÈÈÈ ÈÈÈ222 222
(cosh t) (sinh t) (sinh t) (cosh t) and sinh t cosh t 1
cosh t sinh t 0
aij ijva
ijk
œ Êœ ‚œ $
$
d
dt
a
ââ
ââ
ââ
ââ
ââ
ââ
(sinh t) (cosh t) cosh t sinh t (sinh t) (cosh t) sinh t cosh t 1. Thusœ œ Ê‚œij kijkvaab kk
22 # ##
.7œœœ
ââ
ââ
ââ
ââ
ââ
ââ
sinh t cosh t 1
cosh t sinh t 0
sinh t cosh t 0
sinh t cosh t 1 sinh t cosh t 1 cosh t
Ê
Ê
Ê
œœ œ œ#
" "
œœ œœ œ
9. (a cos t) (a sin t) bt ( a sin t) (a cos t) b ( a sin t) (a cos t) brijkv ijkvœÊœÊœ kkÈ
###
a b a 0; ( a cos t) ( a sin t) ( a cos t) ( a sin t) a aœÊœ œœ Êœ œœ
È
kk kk kk È È
## ###
T
d
dt
va i ja
aa0a (0)aaÊœ œ œœÊœ œ
N T
ÉÉkk kk kk kk kk kkaaaaTNN
##
##
10. (1 3t) (t 2) 3t 3 3 3 1 ( 3) 19 a 0; rijkvijkv va0œ Êœ Ê œ œ Ê œ œ œ kk kkÈ È
## #
T
d
dt
a a 0 (0) (0)Êœ œÊœ œ
N T
ÉkkaaTN0
#
#
11. (t 1) 2t t 2 2t 1 2 (2t) 5 4t a 5 4t (8t)rijkvijkvœ Êœ Ê œ œ Êœ
# ### # # "
#
"Î#
kk a bÈ È
T
4t 5 4t a (1) ; 2 (1) 2 (1) 2 a a 2œ Ê œœœÊ œÊ œÊœ œab kk kk É É ˆ‰
#
"Î# #
##
#
T N T
44 4
9 3 3È
ak a k a a
(1)œœÊœÉ
20 4
93 3 3
25 25
ÈÈ
aTN
12. (t cos t) (t sin t) t (cos t t sin t) (sin t t cos t) 2trijkv i jkœÊœ
#
(cos t t sin t) (sin t t cos t) (2t) 5t 1 a 5t 1 (10t)Êœ œ Êœ kk a bÈ È
v
#### "
#
#
"Î#
T
a (0) 0; ( 2 sin t t cos t) (2 cos t t sin t) 2 (0) 2 2 (0)œÊœœÊœÊ
5t
5t 1
È
œ
œ
T aijkajka kk
2 2 22 a a 22 0 22 (0) (0) 22 22œœÊœ œ œÊœ œ
ÈÈ È È ÈÈ Ékk ÊŠ‹
## # #
#
#
N T aaTNN
13. t t t t t 2t 1 t 1 t (2t) 1 t 1 tri j kvi j k vœ Êœ Ê œ
#$ $ ## "" ## #
##ˆ‰ˆ‰ abab kk abab É
33
2 t 2t 1 2 1 t a 2t 2 a (0) 0; 2 2t 2t (0) 2 (0) 2œœÊœÊœœÊœÊœÈabab kk
ÈÈ
%# #
TT ai j k a i a
a a 2 0 2 (0) (0) 2 2Êœ œ œÊ œ œ
N T
Ékk
È
aaTNN
#
###
14. e cos t e sin t 2e e cos t e sin t e sin t e cos t 2er i jkv i jkœÊœabab abab
ÈÈ
tt t tttt t
e cos t e sin t e sin t e cos t 2e 4e 2e a 2e a (0) 2;Êœ œ œ Êœ Ê œkkababÊ Š‹
È È
v
tt tt t 2t tt
TT
##
#
e cos t e sin t e sin t e cos t e sin t e cos t e cos t e sin t 2eaijkœ abab
È
t ttt tt t t t
2e sin t 2e cos t 2e (0) 2 2 (0) 2 2 6œ Ê œ Ê œ œabab kk
ÈÈ È
Ê Š‹
È
tt t
ijkajka
#
#

852. 848 Chapter 13 Vector-Valued Functions and Motion in Space
a a 6 2 2 (0) 2 2Êœ œ œ Ê œN T
Ékk ÊŠ‹
È ÈÈ
aaTN
# #
#
#
15. (cos t) (sin t) ( sin t) (cos t) ( sin t) (cos t) 1 rijkv ijv TœÊœ Êœ œÊœ kkÈ
## v
vkk
( sin t) (cos t) ; ( cos t) (sin t) ( cos t) ( sin t)œ Ê œ œ Ê œ ijT ij ij ˆ‰ ¸ ¸ È
1
4dt dt
22 dd
ÈÈ
##
##TT
1 ( cos t) (sin t) ; sin t cos t 0
cos t sin t 0
œÊ œ œ Ê œ œ‚œ œ $
$$
NijNijBTN k
ijk
ˆ‰
¸¸
ÈÈ
d
dt
d
dt
T
T
ˆ‰
ââ
ââ
ââ
ââ
ââ
ââ
1
4
22
##
, the normal to the osculating plane; P 1 lies on theÊ œ œÊœßßBk r ijkˆ‰ ˆ‰ Š‹
11
44
22 22
ÈÈ ÈÈ
## ##
osculating plane 0 x 0 y (z ( 1)) 0 z 1 is the osculating plane; is normalʜʜŠ‹Š‹
ÈÈ
22
##
T
to the normal plane x y 0(z ( 1)) 0 x y 0Ê œÊœŠ‹Š‹Š‹Š‹
ÈÈÈÈ ÈÈ
2222 22
#### ##
x y 0 is the normal plane; is normal to the rectifying planeÊ œ N
x y 0(z ( 1)) 0 x y 1 x y 2 is theÊ œ Ê œÊœŠ‹Š ‹Š‹Š ‹
È
ÈÈ ÈÈ ÈÈ
22 22 22
## ## ##
rectifying plane
16. (cos t) (sin t) t ( sin t) (cos t) sin t cos t 1 2 rijkv ijkv TœÊœÊœ œÊœ kk
ÈÈ
## v
vkk
sin t cos t cos t sin t œ Ê œ ÊŠ‹Š‹ Š ‹Š‹ ¸¸
""" " "
ÈÈÈ È È
222 2 2
dd
dt dt
ijk i j
TT
cos t sin t ( cos t) (sin t) ; thus (0) and (0)œœÊœœ œ œÉ
"" " ""
##
##
ÈÈÈ
ˆ‰
¸¸222
NijTjkNi
d
dt
d
dt
T
T
(0) , the normal to the osculating plane; (0) P(1 0 0) lies on0
10 0
Êœ œ œÊßß
$
Bjk ri
ijk
ââ
ââ
ââ
ââ
ââ
ââ
ââ
"" ""
ÈÈ ÈÈ22 22
the osculating plane 0(x 1) (y 0) (z 0) 0 y z 0 is the osculating plane; is normalÊ œÊœ
""
ÈÈ
22
T
to the normal plane 0(x 1) (y 0) (z 0) 0 y z 0 is the normal plane; is normal toÊ œÊœ
""
ÈÈ
22
N
the rectifying plane 1(x 1) 0(y 0) 0(z 0) 0 x 1 is the rectifying planeÊ œÊœ
17. Yes. If the car is moving along a curved path, then 0 and a 0 a a .,,ÁœÁÊœÁ
NTNkkvaTN0
#
18. constant a 0 a is orthogonal to the acceleration is normal to the pathkk kkvvaNT Êœ œÊœ Ê
TN
d
dt
19. a 0 0 is constantav aT v v¼Ê¼ Ê œÊ œÊ
T
d
dt
kk kk
20. (t) a a , where a (10) 0 and a 100 0 100 . Now, fromaTN v v aT Nœ œ œ œ œ œ Êœ
TN T N
dd
dt dt
kk kk ,, ,
#
Exercise 5(a) Section 13.4, we find for y f(x) x that ; also,œœ œ œ œ
#
œ œœ
,
kk
" ‘ab cd ab
f(x)
1f(x)
22
1(2x) 14x

œœœ
(t) t t is the position vector of the moving mass 2t 1 4trij vijvœ Ê œ Ê œ
# #
kk
È
( 2t ). At (0 0): (0) , (0) and (0) 2 m m(100 ) 200m ;Êœ ß œ œ œÊœ œ œTij TiNj Fa Nj
"
œ
È
14t
œ
,,
At 2 2 : 2 2 2 , 2 , and 2 mŠ ‹ Š‹ Š ‹ Š‹ Š‹
ÈÈ È È È
ßœœ œ œÊœTijijN ij Fa
"" "
333 33 27
22 22 2
ÈÈ
,
m(100 ) m m mœœœ,Nijijˆ‰ Š‹
200 200
27 3 3 81 81
2 2 400 2
ÈÈ
"
21. a a , where a (constant) 0 and a m m m aTN v v Fa vN F vœ œ œ œ œ Êœœ Êœ
TN T N
dd
dt dt
kk kk kk kk kk ,,,
###
m , a constant multiple of the curvature of the trajectoryœˆ‰kkv
#
,,

853. Section 13.5 Torsion and the Unit Binormal Vector 849 B
22. a 0 0 0 (since the particle is moving, we cannot have zero speed) the curvature is zero
NœÊ œÊœ Ê,,kkv
#
so the particle is moving along a straight line
23. From Example 1, t and a t so that a , t 0 tkk kkvvœœ œÊœœœÁÊœœ
NN ,, 3
# ""a t
tt
N
kkv
œ œ
,
24. (x At) (y Bt) (z Ct) A B C 0. Since the curverijkvijka0va0œ Êœ ʜʂœÊœ!!! ,
is a plane curve, 0.7œ
25. If a plane curve is sufficiently differentiable the torsion is zero as the following argument shows:
f(t) g(t) f (t) g (t) f (t) g (t) f (t) g (t)rijv i ja i j i jœ Êœ Êœ Êœ
ww w w w w ww w w ww
d
dt
a
0Êœ œ7
ââ
ââ
ââ
ââ
ââ
ââ
kk
f (t) g (t) 0
f (t) g (t) 0
f (t) g (t) 0



œ
va‚
26. From Example 2, (b) ; (b) 0 0 a b 0 b a77 7œÊœ œÊ œÊœÊœ„
bab ab
ab ab ab
œœ
œœ œœ
œœ œœ
œœ
œ
ww ## ÊÊ
œœab ab
b a since a, b 0. Also b a 0 and b a 0 so occurs when b a Êœ Ê Ê œÊ œ 777 7
ww
œ
max max
a
aa
œœ œ
"
2a
27. (t) f(t) g(t) h(t) f (t) g (t) h (t) ; 0 h (t) 0 h(t) Crijkvijkvkœ Êœ œÊ œÊ œ
www w
†
(t) f(t) g(t) C and (a) f(a) g(a) C f(a) 0, g(a) 0 and C 0 h(t) 0.Êœ œ œÊœ œ œÊœr i jkr i jk0
28. From Example 2, (a sin t) (a cos t) b a b vijkv Tœ Ê œ Ê œ kk
È
## v
vkk
(a sin t) (a cos t) b ; (a cos t) (a sin t) œ œ Êœ
""
œœ
ÈÈ
ˆ‰
¸¸ab ab
d
dt
œœ œœ
cdcd ijk ijN
T
d
dt
d
dt
T
T
(cos t) (sin t) ;
cos t sin t 0
œ œ ‚ œ $
$$
ijBTN
ijk
ââ
ââ
ââ
ââ
ââ
ââ
ââ
a sin t a cos t b
ab ab ab
ÈÈÈ
œœ œœ œœ
œœœ
(b cos t) (b sin t) œ Êœ Êœ
b sin t b cos t a d d b
ab ab ab ab ab dt dtÈÈÈ È È
œœ œœ œœ œœ œœ
œœœ œ œ
"
ijk ijN
BB
cd †
, which is consistent with the result inÊœ œ œ7
""
œœ œkk ÈÈv
B
ˆ‰ Š‹Š‹
dbb
dt a b ab ab
†N
œœ œœ
œœ
Example 2.
29-32. Example CAS commands:
:Maple
with( LinearAlgebra );
r := < t*cos(t) | t*sin(t) | t >;
t0 := sqrt(3);
rr := eval( r, t=t0 );
v := map( diff, r, t );
vv := eval( v, t=t0 );
a := map( diff, v, t );
aa := eval( a, t=t0 );
s := simplify(Norm( v, 2 )) assuming t::real;
ss := eval( s, t=t0 );
T := v/s;
TT := vv/ss ;
q1 := map( diff, simplify(T), t ):
NN := simplify(eval( q1/Norm(q1,2), t=t0 ));

854. 850 Chapter 13 Vector-Valued Functions and Motion in Space
BB := CrossProduct( TT, NN );
kappa := Norm(CrossProduct(vv,aa),2)/ss^3;
tau := simplify( Determinant(< vv, aa, eval(map(diff,a,t),t=t0) >)/Norm(CrossProduct(vv,aa),2)^3 );
a_t := eval( diff( s, t ), t=t0 );
a_n := evalf[4]( kappa*ss^2 );
: (assigned functions and value for t0 will vary)Mathematica
Clear[t, v, a, t]
mag[vector_]:=Sqrt[vector.vector]
Print["The position vector is ", r[t_]={t Cos[t], t Sin[t], t}]
Print["The velocity vector is ", v[t_]= r'[t]]
Print["The acceleration vector is ", a[t_]= v'[t]]
Print["The speed is ", speed[t_]= mag[v[t]]//Simplify]
Print["The unit tangent vector is ", utan[t_]= v[t]/speed[t] //Simplify]
Print["The curvature is ", curv[t_]= mag[Cross[v[t],a[t]]] / speed[t] //Simplify]
3
Print["The torsion is ", torsion[t_]= Det[{v[t], a[t], a'[t]}] / mag[Cross[v[t],a[t]]] //Simplify]
2
Print["The unit normal vector is ", unorm[t_]= utan'[t] / mag[utan'[t]] //Simplify]
Print["The unit binormal vector is ", ubinorm[t_]= Cross[utan[t],unorm[t]] //Simplify]
Print["The tangential component of the acceleration is ", at[t_]=a[t].utan[t] //Simplify]
Print["The normal component of the acceleration is ", an[t_]=a[t].unorm[t] //Simplify]
You can evaluate any of these functions at a specified value of t.
t0= Sqrt[3]
{utan[t0], unorm[t0], ubinorm[t0]}
N[{utan[t0], unorm[t0], ubinorm[t0]}]
{curv[t0], torsion[t0]}
N[{curv[t0], torsion[t0]}]
{at[t0], an[t0]}
N[{at[t0], an[t0]}]
To verify that the tangential and normal components of the acceleration agree with the formulas in the book:
at[t]== speed'[t] //Simplify
an[t]==curv [t] speed[t] //Simplify
2
13.6 PLANETARY MOTION AND SATELLITES
1. T a T (6,808,000 m)
T4 4 4
aGM GM 6.6726 10 Nm kg 5.975 10 kg
œœ œ œ
" Á   œ Áœ œ"
œÊœ Êœ
11 1 #$# $
‚‚abab 3.125 10 sec T 3125 10 sec 55.90 10 sec 93.2 min¸‚ ʸ ‚ ¸‚ ¸
(# # %# È
2. e 0.0167 and perihelion distance 149,577,000 km and e 1œœœ
rv
GM$
œ
$
0.0167 1 v 9.03 10 m /secÊœ ʸ‚
(149,577,000,000 m)v
6.6726 10 Nm kg 1.99 10 kg
œ
$

abab‚‚
#)##
! v 9.03 10 m /sec 3.00 10 m/secʸ ‚ ¸ ‚!
)# # %
È
3. 92.25 min 5535 sec and a TœœÊœ
T4 GM
aGM 4
œœ

1
1
$#
a (5535 sec) 3.094 10 m a 3.094 10 mÊœ œ ‚ ʸ ‚
$##!$
‚‚
#! $
ab ˆ‰6.6726 10 Nm kg 5.975 10 kg
4
Р  œ Áœ œ"
œ
"
1
È
6.764 10 m 6764 km. Note that 6764 km 12,757 km 183 km 589 km .œ‚ ¸ ¸
' "
#
ab
4. T 1639 min 98,340 sec and mass of Mars 6.418 10 kg a Tœœ œ‚Êœ
#$ $ # GM
41
œ
1.049 10 m a 1.049 10 mœ¸‚ʸ‚
abab6.6726 10 Nm kg 6.418 10 kg (98,340 sec)
4
‚‚ ## $ ## $

œ
"
1
È
2.19 10 m 21,900 kmœ‚ œ
(

855. Section 13.6 Planetary Motion and Satellites 851
5. 2a diameter of Mars perigee height apogee height D 1499 km 35,800 kmœœ
2(21,900) km D 37,299 km D 6501 kmʜʜ
6. a 22,030 km 2.203 10 m and T aœœ‚ œ
(#$ 4
GM
1
Ä
T (2.203 10 m) 9.856 10 secÊœ ‚ ¸ ‚
#($*#
‚‚
4
6.6720 10 Nm kg 6.418 10 kg
1
Ä

abab
T 9.856 10 sec 9.928 10 sec 1655 minʸ ‚ ¸ ‚ ¸
È
)# %
7. (a) Period of the satellite rotational period of the Earth period of the satellite 1436.1 minœÊœ
86,166 sec; a aœœÊœ
$$
‚‚GMT
44
6.6726 10 Nm kg 5.975 10 kg (86,166 sec)
Ä
ÄÄ
Á   œ Áœ œ" œ
11
ab ˆ‰
7.4980 10 m a 74.980 10 m 4.2168 10 m 42,168 km¸‚ ʸ ‚ ¸‚œ
## $ ( #" $
"
È
(b) The radius of the Earth is approximately 6379 km the height of the orbit is 42,168 6379 35,789 kmʜ
(c) Symcom 3, GOES 4, and Intelsat 5
8. T 1477.4 min 88,644 sec aœœÊœ
$ GMT
4
Ä
Ä
1 8.524 10 m a 8.524 10 mœœ‚ʸ‚
abab6.6726 10 Nm kg 6.418 10 kg (88,644 sec)
4
‚‚ #" $ #" $
Ð
Ä
"
1
È
2.043 10 m 20,430 km¸‚ œ
(
9. Period of the Moon 2.36055 10 sec aœ‚Êœ
'$ GMT
4
Ä
Ä
1 5.627 10 m a 5.627 10 mœ¸‚ʸ‚
ab ˆ‰6.6726 10 Nm kg 5.975 10 kg (2.36055 10 sec)
4
‚‚‚
#& $ #& $
Р  œ Áœ œ" ‚ œ
Ä
"
1
È
3.832 10 m 383,200 km from the center of the Earth.¸‚ œ
)
10. r v v 1.9967 10 r m/secœÊœÊœ œ ¸ ‚
GM GM GM
vr r r
6.6726 10 Nm kg 5.975 10 kg
Ä
Á   œ Áœ œ"
# ( "Î#‚‚
kkÉ É
abab
11. Solar System: 2.97 10 sec /m ;
T4
a 6.6726 10 Nm kg 1.99 10 kg
ÄÄ

œ¸‚
1
abab‚‚
"* # $
Earth: 9.902 10 sec /m ;
T4
a 6.6726 10 Nm kg 5.975 10 kg
ÄÄ
" Á   œ Áœ œ"
œ¸‚
1
abab‚‚
"% # $
Moon: 8.045 10 sec /m ;
T4
a 6.6726 10 Nm kg 7.354 10 kg
ÄÄ

œ¸‚
1
abab‚‚
"# # $
12. e 1 v v ;œ Êœ Êœ
rv
GM r r
GM(e 1) GM(e 1)_
Ä
_
__ #
!
ÄÄ
!
É
Circle: e 0 vœÊ œ !É
GM
r
_
Ellipse: 0 e 1 v Ê ÉÉ
GM 2GM
rr
__
!
Parabola: e 1 vœÊ œ !É
2GM
r
_
Hyperbola: e 1 vÊ !É
2GM
r
_
13. r v v which is constant since G, M, and r (the radius of orbit) are constantœÊœÊœ
GM GM GM
vr r
Ä
#
É
14. A (t t) (t) (t) (t)??œ‚Êœ ‚œ ‚
"""
###
œœÊœ
kk ¹¹¹ ¹rr r r
?
?? ?
??A
tt t
(t t) (t t) (t) (t)rrrr
(t) (t) (t) (t) lim (t)œ‚‚œ‚Êœ ‚
""" "
###
œÊ œÊ œÊ
¹ ¹¹¹ ¹¹
rr rr rr(t t) (t) (t t) (t) (t t) (t)
tt t dt t
dA???
?? ? ?
rrr r r
?t0Ä
(t) (t) œ‚œ ‚œ‚
"""
###
¸¸¸¸ kk
dd
dt dt
rr
rr rr
Þ

856. 852 Chapter 13 Vector-Valued Functions and Motion in Space
15. T 1 e T 1 e 1 1 (from Equation 32)œÊœœŠ‹ Š‹ Š‹ Š ‹
È
ab ”•
2a 4a 4a
rv GMrv rv
rv111
œœ "œ"
$$
œœ œœ
$$ $$
$
œ
$
# ##
#
2œ œ œŠ‹’ “Š‹’ “ Š‹
4a 4a
rv rv
rv rv 2GMrv rv
GM GM GM rGM
4a 2GM rv11 1
œ"œ"
œœ œœ
$$ $$
œ" œ œ œ"
$$ $ $ $$
œœ œœ œœ
$$
œ" œ
$$
$ Ê Êˆ‰ ab
4 a 4 a (from Equation 35) T œœ ʜʜab abŠ‹ ˆ‰ ˆ‰ˆ‰11
#% #% # Ê "2GM r v
2r GM GM 2a GM GM a GM
22 4aT4 $
œ
$
$

" 11
16. Let (t) denote the vector from planet A to planet B at time t. Then (t) (t) (t)rrrr
AB AB B A œ
[3 cos ( t) 2 cos (2 t)] [3 sin ( t) 2 sin (2 t)]œ 11 11 ij
3 cos ( t) 2 cos ( t) sin ( t) [3 sin ( t) 4 sin ( t) cos ( t)]œ cd ab111 111
##
ij
3 cos ( t) 4 cos ( t) 2 [(3 4 cos ( t)) sin ( t)] parametric equations for the path areœ Êcd11 11
#
ij
x(t) 2 [3 4 cos ( t)] cos ( t) and y(t) [3 4 cos ( t)] sin ( t)œ œ 11 11
17. The graph of the path of planet B is the limacon¸
at the right.

18. (i) Perihelion is the time t such that (t) is a minimum.kkr
(ii) Aphelion is the time t such that (t) is a maximum.kkr
(iii) Equinox is the time t such that (t) 0 .rw†œ
(iv) Summer solstice is the time t such that the angle between (t) and is a maximum.rw
(v) Winter solstice is the time t such that the angle between (t) and is a minimum.rw
CHAPTER 13 PRACTICE EXERCISES
1. (t) (4 cos t) 2 sin t x 4 cos tri jœ Êœ Š‹
È
and y 2 sin t 1;œÊœ
È
x
16
y
œ œ
#
vi jœ ( 4 sin t) 2 cos t andŠ‹
È
( 4 cos t) 2 sin t ; (0) 4 , (0) 2 ,ai jrivjœ œ œ Š‹
ÈÈ
(0) 4; 22 , 22 ,air ijv ijœ œ œ ˆ‰ ˆ‰ÈÈ
11
44
2 2 ; 16 sin t 2 cos taijvˆ‰ ÈÈ
kk
1
4
œ œ
##

a ; at t 0: a 0, a 0 4, 2;Êœ œ œ œ œ œ œ œœTN
d 14 sin t cos t 4
dt 216 sin t 2 cos t
a
kk kk Éva
È kk œœ
œ
œ
#
T ,
N
v
at t : a , a 9 , œœœœœ œœ
1
439327
77 49
81
42 42 a
T N È
ÈÈ
kkœ
É ,
N
v
œ

857. Chapter 13 Practice Exercises 853
2. (t) 3 sec t 3 tan t x 3 sec t and y 3 tan t sec t tan t 1;rijœ Êœ œÊœœŠ‹Š‹
ÈÈ È È
x
33
y
œ œ
##
x y 3; 3 sec t tan t 3 sec tÊœ œ
## #
vijŠ‹Š‹
ÈÈ
and
3 sec t tan t 3 sec t 2 3 sec t tan t ;aijœŠ‹Š‹
ÈÈ È
#$ #
(0) 3 , (0) 3 , (0) 3 ;rivjaiœœœ
ÈÈÈ
3 sec t tan t 3 sec tkkvœ
È
## %
a ;Êœ œ T
d 6 sec t tan t 18 sec t tan t
dt 2 3 sec t tan t 3 sec t
kkv
œ "!
œœ "
œ
œ
È at t 0: a 0, a 0 3,œœœ œ T N
Ékk
È
a
#
,œœœ
a 3
3 3N
kk
È
Èv
œ
"

3. t 1 t 1 trijv i jœ Êœ
"
œœ
##
$Î# $Î#
ÈÈ
1t 1t
t
œœ
ab ab
t 1 t 1 t . We want to maximize : andÊœ œ œkk ab ab kkÊ’“’“vv
##
$Î# $Î#
##
"
œ œ1t dt
d 2t
1t
œ
œ
œ
kk
ab
v
0 0 t 0. For t 0, 0; for t 0, 0 occurs when
d
dt
2t 2t 2t
1t 1t 1t
kk
ab ab ab
v
œÊ œÊœ Ê
ÊÊÊ
œœœ
œœœ
œœœ
kkv
max
t0 1œÊ œkkv
max
4. e cos t e sin t e cos t e sin t e sin t e cos trijv i jœ Êœabababab
tt tttt
e cos t e sin t e sin t e cos t e sin t e cos t e cos t e sin tʜ aijabab
t ttt tt t t
2e sin t 2e cos t . Let be the angle between and . Then cosœ œabab Š‹
tt

ij ra ))
" ra
ra
†
kkkk
cos cos cos 0 for all tœœœœ
" " "
#

Š‹


2e sin t cos t 2e sin t cos t 0
e cos t e sin t 2e sin t 2e cos t
2e
2t 2t
tt t t
2t
ÉÉababa ba b
œœ œ œ
1
5. 3 4 and 5 15 25 25; 3 4 5 340
550
vijai j va k va v
ijk
œ œ Ê‚œ œ Ê œ œ œ
"
ââ
ââ
ââ
ââ
ââ
ââ
kk kk
È
‚
##
Êœ œœ,
kk
kk
va
v
‚ "
" "
25
55
6. e 1 e e 1 e e 1 e 2e,œœÊœ
kk
" ‘ab
y
1y
x2x x2x x 2x 2x d3
dx

œ

œ
$Î# $Î# &Î#
#
ab ab abab ’“
,
e 1 e 3e 1 e e 1 e 1 e 3e e 1 e 1 2e ;œ œ œ
x 2x 3x 2x x 2x 2x 2x x 2x 2x
ab ab abc daba b ab
$Î# &Î# &Î# &Î#
0 1 2e 0 e 2x ln 2 x ln 2 ln 2 y ; therefore is at a
d
dx
2x 2x
2
,
œ Ê œ Ê œ Ê œ Ê œ œ Ê œab
È
"""
## È
,
maximum at the point ln 2Š‹
È
$%
"
È
2
7. x y and y y. Since the particle moves around the unit circlerij v i jviœ Êœ œÊ œ
dx dx
dt dt dt
dy
†
x y 1, 2x 2y 0 (y) x. Since y and x, we have
##
œ œ Ê œ Ê œ œ œ œ
dx x dx x dx
dt dt dt y dt dt y dt dt
dy dy dy dy
y x at (1 0), and the motion is clockwise.vij v jœ Ê ß œ
8. 9y x 9 3x x . If x y , where x and y are differentiable functions of t,œÊ œ Êœ œ
$# # "dy dy
dt dt dt 3 dt
dx dx
rij
then . Hence 4 4 and x (3) (4) 12 at (3 3). Also,vij vi vjœ œÊœ œœ œ œ ß
dx dx dx
dt dt dt dt 3 dt 3
dy dy
††
""##
and x x . Hence 2 2 andaij aiœ œ œÊœ
dx 2 dx dx dx
dt dt dt 3 dt 3 dt dt
dy dy
œ œœ
œœ œ œ œ
œœ
ˆ‰ˆ‰ ˆ ‰
#
"#
†
(3)(4) (3) ( 2) 26 at the point (x y) (3 3).aj†œœ œ ßœß
dy
dt 3 3
2
œ
œ
##"

858. 854 Chapter 13 Vector-Valued Functions and Motion in Space
9. orthogonal to 0 ( ) K, a constant. If x y , where
ddddd
dt dt dt dt dt
rrrr
r r r r rr rr r i jÊœ œ œ Ê œ œ††† ††
"""
###
x and y are differentiable functions of t, then x y x y K, which is the equation of a circlerr†œ Ê œ
## ##
centered at the origin.
10. (b) ( cos t) ( sin t)vijœ 11 1 1 1
sin t cos t ;ʜ aijaba b11 1 1
##
(0) and (0) ;v0a jœœ 1
#
(1) 2 and (1) ;via jœœ11
#
(2) and (2) ;v0a jœœ 1
#
(3) 2 and (3)via jœœ11
#

(c) Forward speed at the topmost point is (1) (3) 2 ft/sec; since the circle makes revolution perkkkkvvœœ 1
"
#
second, the center moves ft parallel to the x-axis each second the forward speed of C is ft/sec.11 Ê
11. y y (v sin )t gt y 6.5 (44 ft/sec)(sin 45 °)(3 sec) 32 ft/sec (3 sec) 6.5 66 2 144œ Êœ œ !!
""
##
# ## ! ab
È
44.16 ft the shot put is on the ground. Now, y 0 6.5 22 2t 16t 0 t 2.13 sec (the¸ Ê œ Ê œ Ê ¸
È
#
positive root) x (44 ft/sec)(cos 45°)(2.13 sec) 66.27 ft or about 66 ft, 3 in. from the stopboardʸ ¸
12. y y 7 ft 57 ftmaxœ œ ¸!
#
(v sin ) [(80 ft/sec)(sin 45 °)]
g (2) 32 ft/sec
$
œœ
œ!
ab
13. x (v cos )t and y (v sin )t gt tan œœÊœœœ!!
"
#
#
ÊÊ!!9
y
x (v cos )t v cos
(v sin )t gt (v sin ) gt$$
ÎÎ
œœ
œ
$$!!
!!
v cos tan v sin gt t , which is the time when the golf ballʜʜ !!
"
#
Ê!9 !
2v sin 2v cos tan
g$$!!9
hits the upward slope. At this time
x(v cos )œ !
Ê!Š‹
2v sin 2v cos tan
g$$!!9
v sin cos v cos tan . NowœŠ‹ab
2
g
###
!!
!! !9
OR ORœÊœ
x2
cos g cos
v sin cos v cos tan
99
!! !9
Š‹Š ‹
œœœ
$$
Ê
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2v cos
gcos cos
sin cos tan
œ
$
! !
99
!9
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2v cos
gcos
sin cos cos sin
œ
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! !9 !9
9
Ê
[sin ( )]. The distance OR is maximizedœŠ‹
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g cos
œ
$
œ
!
9
!9

when x is maximized: (cos 2 sin 2 tan ) 0 (cos 2 sin 2 tan ) 0 cot 2 tan 0
dx
dg
2v
!
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œ
$
!!9 !!9 !9
cot 2 tan ( ) 2 ʜʜʜ!9!9!
11 9
## 4
14. R sin 2 v ; for 4325 yards: 4325 yards 12,975 ft vœÊœ œÊœ
v
gsin 2 (sin 90°)
Rg (12,975 ft) 32 ft/sec
œ
$
œ
! ! !É É
!
ab
644 ft/sec; for 4752 yards: 4752 yards 14,256 ft v 675 ft/sec¸œÊœ¸ !É
(14,256 ft) 32 ft/sec
(sin 90°)
ab
œ
15. (a) R sin 2 109.5 ft (sin 90 °) v 3504 ft /sec v 3504 ft /secœÊœ Êœ Êœ
vv
g 32 ft/sec
œœ
$$
œ
! Š‹
È
###
!
##
Ê
59.19 ft/sec¸ (b) x (v cos )t and y 4 (v sin )t gt ; when the cork hits the ground, x 177.75 ft and y 0œœ œœ !!
"
#
#!!
177.75 v t and 0 4 v t 16t 16t 4 177.75 tÊ œ œ Ê œ ÊœŠ‹ Š‹ !!
"" ##
ÈÈ
È
22
181.75
4
v 74.58 ft/secÊœ œ ¸ !
(177.75) 2 4(177.75) 2
t 181.75
ÈÈ
È

859. Chapter 13 Practice Exercises 855
16. (a) x v (cos 40°)t and y 6.5 v (sin 40°)t gt 6.5 v (sin 40°)t 16t ; x 262 ft and y 0 ftœœœœœ !! !
"
#
## 5
12
262 v (cos 40 °)t or v and 0 6.5 (sin 40 °)t 16t t 14.1684Êœ œ œ Êœ
5 262.4167 262.4167
12 (cos 40 °)t (cos 40 °)t
!!
##
’“
t 3.764 sec. Therefore, 262.4167 v (cos 40 °)(3.764 sec) v v 91 ft/secʸ ¸ Ê ¸ Ê ¸ !!!
262.4167
(cos 40°)(3.764 sec)
(b) y y 6.5 60 ftmaxœ ¸ ¸!
(v sin )
2g (2)(32)
(91)(sin 40°)
$
œ! ab
2
17. x v cos t and y gt v sin t x y gt v t
# ### # ### # # ##
!! !
""
##
##
œœÊœab ab ˆ‰ ˆ‰!!
18. s x y x y s x y
ÞÞ ÞÞ
œœ Êœ
Þ Þ ÞÞ ÞÞ ÞÞ ÞÞ
d
dtÈ## ## ## # xx yy
xy
xx yy
xy
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"
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"
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ab
œœ
œ
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ababa b a bx y x y xx 2xxyyyy xyyx
xy xy xy
x y y x 2x x y y
ÞÞ ÞÞ Þ Þ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ

ÞÞ ÞÞ ÞÞ
"""
Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ

œœœœ œœ œœ
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xys Êœ Ê œ œœ
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È
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xy yx x y
xy
xy
xys xy yx
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,
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19. (t) cos d sin d (t) cos sin 1;rijvijvœ ʜʜ’ “ ’ “ Š‹ Š‹ˆ‰ ˆ‰ kk ''
00
tt
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##
1) ) 1) )
11tt
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(t) t sin t cos
cos sin 0
t sin t cos 0
aijva
ijk
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11
11
Š‹ Š‹
ââ
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11
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tt
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t t; (t) 1 s t C; (0) s(0) 0 C 0 sœÊœ œ œœÊœ œÊ œÊœÊœ1, 1 ,1kv r0
kk
kk
va
v
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ds
dt
20. s a since a 0œÊœÊœÊœÊœœ )) 9 ,
ss
aadsa aa
d1 9
#
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¸¸
21. (2 cos t) (2 sin t) t ( 2 sin t) (2 cos t) 2t ( 2 sin t) (2 cos t) (2t)rijkv ijkvœÊœÊœ
# ###
kkÈ
2 1 t Length 2 1 t dt t 1 t ln t 1 t 1 ln 1œÊ œ œ œ
ÈÈÈÈ
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####
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! '
0
4
œ" 1
11 1 1
4164 16
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22. (3 cos t) (3 sin t) 2t ( 3 sin t) (3 cos t) 3tr i jkv i jkœÊœ
$Î# "Î#
( 3 sin t) (3 cos t) 3t 9 9t 3 1 t Length 3 1 t dt 2(1 t)Êœ œ œ Ê œ œ kk a bÉ ÈÈ È œ ‘v
# # "Î#
#
$Î#
$
! '
0
3
14œ
23. (1 t) (1 t) t (1 t) (1 t)rijkvijkœ Êœ
44 22
993333
$Î# $Î# "Î# "Î# ""
(1 t) (1 t) 1 (1 t) (1 t)Êœ œÊœ kkÉœ ‘œ ‘ˆ‰vTijk
22 22
333333
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##
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#
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(0) ; (1 t) (1 t) (0) (0)Ê œ œ Ê œ Ê œTijk i j ij
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2"" " "" "Î# "Î#TTT
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ijk
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2 2 32 32 32
22
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4
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3 3 33 333
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22
333
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999 3 1 3
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kk
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0
0
7
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66 318
2
2
3 Î
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va
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t 0 0 is the point on the curveœÊ ß߈‰
44
99

860. 856 Chapter 13 Vector-Valued Functions and Motion in Space
24. e sin 2t e cos 2t 2e e sin 2t 2e cos 2t e cos 2t 2e sin 2t 2er i jkv i jkœ Êœ abababab
tt ttt tt t
e sin 2t 2e cos 2t e cos 2t 2e sin 2t 2e 3e Êœ œ ÊkkabababÉvT
tt t t t t###
œ
v
vkk
sin 2t cos 2t cos 2t sin 2t (0) ;œ Ê œˆ‰ˆ‰
"" "
33 3 3 3 333
22222
ijkTijk
cos 2t sin 2t sin 2t cos 2t (0) (0) 5
d2 4 2 4 d 24 d 2
dt 3 3 3 3 dt 3 3 dt 3
T TT
œ ʜʜˆ‰ˆ ‰ ¸¸ È
ijij
(0) ; (0) (0) (0) ;

0
Êœ œ œ‚œ œ
$
NijBTN ijk
ijk
ˆ‰
Š‹
ÈÈ È È È
ÈÈ
24
33
25
3
ijÊ "
"
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5 5 35 35 35
2425
212
333
55
2
ââ
ââ
ââ
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ââ
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4e cos 2t 3e sin 2t 3e cos 2t 4e sin 2t 2e (0) 4 3 2 and (0) 2 2a i jkaijkvijkœ Ê œ œaba b
tt tt t (0) (0) 8 4 10 64 16 100 6 5 and (0) 322
432
Ê ‚ œ œ Ê ‚œ œ œ"
$
va ijkva v
ijk
ââ
ââ
ââ
ââ
ââ
ââ
kk kkÈ È
(0) ;Êœœ,
65 25
39
ÈÈ
"
4e cos 2t 8e sin 2t 3e sin 2t 6e cos 2t 3e cos 2t 6e sin 2t 4e sin 2t 8e cos 2t 2eai jk
Þ
œ aba b
t ttt t ttt t
2e cos 2t 11e sin 2t 11 e cos 2t 2e sin 2t 2e (0) 2 11 2œ Ê œ
Þ
aba b
tt tt t
ijkaijk
(0) ; t 0 ( 2) is on the curveÊœ œœœÊ!ß"ß7
ââ
ââ
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kk
212
432
2112
80 4
180 9
Ê
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‚
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va
œ
25. t e e 1 e (ln 2) ;ri j vi j v T i j T i jœ Ê œ Ê œ Ê œ Ê œ
"""
#
œœ
2t 2t 4t
11 17 17
4
kk
È
ÈÈ ÈÈ
ee
e
4t 4t
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(ln 2) (ln 2) ;
d2 2 d 328 4
dt dt
11 17 17 17 17 17 17
TT
œ ʜʜ
"
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ee
4t 2t
4t 4t
ˆ‰ ˆ‰
ÈÈ ÈÈ
" ij ijN ij
(ln 2) (ln 2) (ln 2) ; 2e (ln 2) 8 and (ln 2) 4
0
0
BTN kaja jvij
ijk
œ‚œ œœÊ œ œ
$
ââ
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ââ
ââ
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"
"
ÈÈ
ÈÈ
17 17
4
4
17 17
2t
(ln 2) (ln 2) 8 8 and (ln 2) 17 (ln 2) ; 4e40
080
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Þ
va kva v aj
ijk
ââ
ââ
ââ
ââ
ââ
ââ
kk k k
È
,
8
17 17
2t
È
(ln 2) 16 (ln 2) 0; t ln 2 (ln 2 2 0) is on the curveʜʜ œœÊßß
Þ
aj 7
ââ
ââ
ââ
ââ
ââ
ââ
kk
140
080
0160
va‚
œ
26. (3 cosh 2t) (3 sinh 2t) 6t (6 sinh 2t) (6 cosh 2t) 6rijkvijkœÊœ
36 sinh 2t 36 cosh 2t 36 6 2 cosh 2t tanh 2t sech 2tÊœ œ Êœœ kk
ÈÈ
Š‹ Š‹vTijk
## """v
vkk ÈÈÈ
222
(ln 2) ; sech 2t sech 2t tanh 2t (ln 2)Êœ œ ÊTijk i k
15 8 d 2 2 d
17 2 2 17 2 2 2 dt dtÈÈ È È È
" #TT
Š‹Š ‹
(ln 2)œ œÊœœŠ‹ Š‹ Š ‹ Š ‹ˆ‰ ˆ‰ˆ‰ ¸ ¸ Ê
2 8 2 8 15 128 240 d 128 240
2 2 289 2 289 2 289 2 289 217 17 17 dt 17
82
ÈÈ ÈÈ ÈÈ
È#
##
ikik
T
(ln 2) ; (ln 2) (ln 2) (ln 2) ;
0
Êœ œ‚œ œ
$
NikBTN ijk
ijk
815 15 8
17 17
15 8
17 2 2 17 2
815
17 17
17 2 2 17 2
ââ
ââ
ââ
ââ
ââ
ââ
ââ
ÈÈ È
ÈÈ È
"
"
(12 cosh 2t) (12 sinh 2t) (ln 2) 12 12 andaija ijijœ Êœœ ˆ‰ ˆ‰
17 15 51 45
88 ##
vijkijkva
ijk
(ln 2) 6 6 6 6 (ln 2) (ln 2) 6
0
œœÊ‚œˆ‰ ˆ‰
ââ
ââ
ââ
ââ
ââ
ââ
15 17 45 51
88 44
45 51
44
51 45
2 #
135 153 72 153 2 and (ln 2) 2 (ln 2) ;œ Ê ‚ œ œ Ê œ œijkva v kk k k
ÈÈ
51 32
4 867
153 2
2
,
È
Š‹
È
51
4
"

861. Chapter 13 Practice Exercises 857
(24 sinh 2t) (24 cosh 2t) (ln 2) 45 51 (ln 2) ;aijaij
ÞÞ
œ ʜʜœ 7
ââ
ââ
ââ
ââ
ââ
ââ
kk45 51
44
545
22
6
0
45 51 0 32
867
Î
œ
va‚
t ln 2 6 ln 2 is on the curveœÊßß ˆ‰
51 45
88
27. 2 3t 3t 4t 4t (6 cos t) (3 6t) (4 8t) (6 sin t)rijkvijkœ Êœ abab
##
(3 6t) (4 8t) (6 sin t) 25 100t 100t 36 sin tÊœ œ kkÈ È
v
### ##
25 100t 100t 36 sin t (100 200t 72 sin t cos t) a (0) (0) 10;Êœ Ê œ œ
d d
dt dt
kk kkv v"
#
##
"Î#
ab
T
6 8 (6 cos t) 6 8 (6 cos t) 100 36 cos t (0) 136aij k a aœ Ê œ œ Ê œ kk k kÈ È È
## # #
a a 136 10 36 6 (0) 10 6Êœ œ œ œÊ œ
N T
Ékk
È È
aaTN
#
##
28. (2 t) t 2t 1 t (1 4t) 2t 1 (1 4t) (2t)ri j kvi jkvœ Êœ Ê œ abab kk È
## ###
2 8t 20t 2 8t 20t (8 40t) a (0) 2 2; 4 2œ Ê œ Êœ œ œ
ÈÈ
ab
# "
#
#
"Î#dd
dt dt
kk kkvv
T ajk
4 2 20 a a 20 2 2 12 2 3 (0) 2 2 2 3Êœ œ Êœ œ œ œ Ê œ kk kk
ÈÈÈÈ ÈÈÈ É Ê Š‹aa aTN
## #
#
#
N T
29. (sin t) 2 cos t (sin t) (cos t) 2 sin t (cos t)ri jkv i jkœ Êœ Š‹ Š‹
ÈÈ
(cos t) 2 sin t (cos t) 2 cos t (sin t) cos t ;Êœ œ Êœœ kkÊ Š ‹ Š‹ Š‹
ÈÈ
vTijk
##
#
""v
vkk ÈÈ
22
sin t (cos t) sin t sin t ( cos t) sin t 1
dd
dt dt 22 2 2
TT
œ Ê œ œŠ‹ Š‹ Š‹ Š‹ ¸¸ Ê
"" " "
##
#
ÈÈ È È
ij k
sin t (cos t) sin t ;
cos t sin t cos t
sin t cos t sin t
Êœ œ œ‚œ
$
$$$
NijkBTN
ijk
ˆ‰
¸¸ ÈÈ
ÈÈ
ÈÈ
d
dt
d
dt
T
T
Š‹ Š‹
ââ
ââ
ââ
ââ
ââ
ââ
ââ
""
""
""
22
22
22
; ( sin t) 2 cos t (sin t) cos t 2 sin t cos t
sin t 2 cos t sin t
œ œ Ê œ $
$$ $
""
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22
ika i j kva
ijk
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È
ââ
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ââ
ââ
ââ
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È
‚
2 2 4 2 ; ( cos t) 2 sin t (cos t)œ Ê‚œœÊœœ œ œ
Þ
ÈÈ È È
kk Š‹ikva a i j k ,
kk
Š‹
È
È
va
v
‚ "
kk
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2
2
2
0Êœ œ œ7
ââ
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È
È
È
kk
Š‹Š ‹ Š ‹
ÈÈ È
cos t 2 sin t cos t
sin t 2 cos t sin t
cos t 2 sin t cos t
(cos t) 2 2 sin t (0) (cos t) 2
4
Ê
ÊÊ Ê
ÊÊ
‚
ÊœÊ
va
œ
30. (5 cos t) (3 sin t) ( 5 sin t) (3 cos t) ( 5 cos t) (3 sin t)ri jkv j ka jkœ Ê œ Ê œ
25 sin t cos t 9 sin t cos t 16 sin t cos t; 0 16 sin t cos t 0 sin t 0 or cos t 0Ê œ œ œÊ œÊ œ œva va††
t 0, or ʜ
1
#
1
31. 2 4 sin 3 0 ( ) 2(1) 4 sin ( 1) 0 2 4 sin sin ri j k rijœ Êœ œ Êœ Ê œÊœˆ‰ˆ‰ ˆ‰
tt t ttt
6######
"
1
1
†
t (for the first time)ʜ
1
3
32. (t) t t t 2t 3t 1 4t 9t (1) 14rijkvijkv vœ Ê œ Ê œ Ê œ
#$ # #%
kk k k
È È
(1) , which is normal to the normal planeÊœTijk
"
ÈÈÈ
14 14 14
23
(x 1) (y 1) (z 1) 0 or x 2y 3z 6 is an equation of the normal plane. Next weÊ œ œ
"
ÈÈÈ
14 14 14
23
calculate (1) which is normal to the rectifying plane. Now, 2 6t (1) 2 6 (1) (1)Najkajkva œ Ê œ Ê ‚

862. 858 Chapter 13 Vector-Valued Functions and Motion in Space
6 6 2 (1) (1) 76 (1) ; (t) 23
026
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kk kk
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¹
ijk
ijk v a v ,
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È
76 19
14
714
ds d s
dt dt
"
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œ
t1$
1 4t 9t 8t 36t , so 2 6œ œ œ Ê ¹abab ˆ‰
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#
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t1$
22 d s ds
14 dt dtÈ
aT Njk
œ
œ
,
14 (x 1) (y 1) (z 1)œ ÊœÊ
22 11 8 9 11 8 9
14 14 7 14
23 14 19
219 777 7 7 7ÈÈ È È
ÈÈ
Š‹ Š‹
È ˆ‰
ijkœœ
#
NN ijk
0 or 11x 8y 9z 10 is an equation of the rectifying plane. Finally, (1) (1) (1)œ œ œ ‚ BTN
(3 3 ) 3(x 1) 3(y 1) (z 1) 0 or 3x 3y z23
11 8 9
œœÊœ "
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14
219 19 14 7
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ijk
ijk
1 is an equation of the osculating plane.œ
33. e (sin t) ln (1 t) e (cos t) (0) ; (0) (1 0 0) is on the lineri j kvi j kv ijkr iœ Ê œ Ê œ œ Ê ßß
tt
1t ˆ‰
"
Ê
x 1 t, y t, and z t are parametric equations of the lineÊœ œ œ
34. 2 cos t 2 sin t t 2 sin t 2 cos t rijkv ijkvœÊœÊŠ‹Š‹ Š ‹Š‹
ÈÈ ÈÈ ˆ‰
1
4
2 sin 2 cos is a vector tangent to the helix when t the tangent lineœ œ œ ÊŠ‹Š‹
ÈÈ
11 1
44 4
ijkijk
is parallel to ; also 2 cos 2 sin the point 1 1 is on the linevr i jkˆ‰ ˆ‰ ˆ ‰ Š‹Š‹
ÈÈ
11 1 11 1
44 4 44 4
œÊßß
x 1 t, y 1 t, and z t are parametric equations of the lineÊœ œ œ
1
4
35. (a) SOT TOD ?? ¸ÊœÊœ
DO OT 6380
OT SO 6380 6380 437
y $
œ
y y 5971 km;ʜ ʸ !!
6380
6817
œ
(b) VA 2 x 1 dyœ '
5971
6380
1Ê Š‹
dx
dy
#
2 6380 y dyœ1 '
5971
6817
È Š‹ ##
Ê
6380
6380 yÈ
œœ
2 6380 dy 2 6380yœœ11 '
5971
6817
cd
')"(
&*("
16,395,469 km 1.639 10 km ;œ¸‚
#(#
(c) percentage visible 3.21%¸¸
16,395,469 km
4 (6380 km)
œ
œ
1
CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES
1. (a) The velocity of the boat at (x y) relative to land is the sum of the velocity due to the rower and theß
velocity of the river, or (y 50) 10 20 . Now, 20 y 20t c; y(0) 100vijœ œ Ê œ œœ ‘
" #
250 dt
dy
c 100 y 20t 100 ( 20t 50) 10 20 t 8t 20Êœ Êœ Êœ œ vijijœ ‘ˆ‰
" ##
250 5
8
(t) t 4t 20t ; (0) 0 100 100 (t)Êœ œ Ê œÊrijCrijjCrˆ‰
8
15
$#
""
t 4t (100 20t)œ ˆ‰
8
15
$#
ij
(b) The boat reaches the shore when y 0 0 20t 100 from part (a) t 5œÊœ Êœ
(5) 125 4 25 (100 20 5) 100 ; the distance downstream isÊœ œ œrijiiˆ‰ ˆ‰
8 200 100
15 3 3
†† †
therefore m
100
3
2. (a) Let a b be the velocity of the boat. The velocity of the boat relative to an observer on the bank of theij"
river is a b . The distance x of the boat as it crosses the river is related to time byvi jœ ’“
3x(20 x)
100
Ê
xat a b a b (t)at bt ;œÊœ œ Ê œ vi ji j r i jC’“Š‹ Š ‹
3at(20 at)
100 100 100 100
3a t 60at a t 30atÊ Ê
œœ œ

863. Chapter 13 Additional and Advanced Exercises 859
(0) 0 0 0 (t) at bt . The boat reaches the shore when x 20rijC ri jœ Ê œÊ œ œ Š‹
a t 30at
100

Ê
20 at t and y 0 0 bʜʜ œÊœ œ
20 20 20b
a a 100 a 100a
a 30a (20) 30(20)
ˆ‰
œ

ˆ‰ ˆ‰
20 20
aa
Ê Ê
b 2; the speed of the boat is 20 a b a 4 a 16œ Êœ œœœÊœ
2000b 8000 12,000
100a
œÊ ## # # È
kk
ÈÈ
v
a 4; thus, 4 2 is the velocity of the boatÊœ œ vij
(b) (t) at bt 4t 2t by part (a), where 0 t 5ri ji jœ œ ŸŸŠ‹Š ‹
a t 30at 16t 120t
100 100 100
œ
Ê
(c) x 4t and y 2tœœ
16t 120t
100 100

t t 2t t 2t 15t 25œœ
46 2
25 5 25
$# #
ab
t(2t 5)(t 5), which is the graph ofœ
2
25
the cubic displayed here

3. (a) ( ) (a cos ) (a sin ) b [( a sin ) (a cos ) b ] ; 2gzrijk ijkv)))) ) )œÊœ œœ
ddd
dt dt dt
r r)
kkÈ ¸¸
ab 2œ Êœ œ Ê œ œ
È É ÉÉÉ ¸
##
œœ œ œ œ#
dd d
dt dt a b a b dt a b a b
2gz 2gb 4 gb gb)) ) )11
)1
œœ œœ œœ œœ
(b) dt 2 t C; t 0 0 C 0
dd
dt a b a b a b
2gb 2gb 2gb)) )
)
œ Ê œ Ê œ œÊœÊ œÉÉ É
œœ œœ œœ
œœ œ
"Î#
È
))
2 t ; z b zÊœ Êœ œÊœ)))
"Î#
œœ œ
É
2gb gbt gb t
a b 2 a b 2 a b
œœ œœ œœ
œœœ
ab ab
(c) (t) [( a sin ) (a cos ) b ] [( a sin ) (a cos ) b ] , from part (b)vijkijkœœ œ
dd
dt dt a b
gbtr
)) ))
)
Š‹œœ
œ
(t) ;Êœ œvT’“Š‹
( a sin ) (a cos ) b gbt gbt
a b a b a b
Êœ œ
œœœ
))i j k
ÈÈÈ
œœ œœ œœ
[( a cos ) (a sin ) ] [( a sin ) (a cos ) b ]
dd d
dt dt dt
œ œ
œ œ
r
œ )) ) )ij i jk ˆ‰
))
#
[( a cos ) (a sin ) ] [( a sin ) (a cos ) b ]œŠ‹ Š‹
gbt gb
a b a b
œœ œœ
œ œ
#
)) ) )ij i jk
a [( cos ) (sin ) ]œ’“Š‹Š‹
( a sin ) (a cos ) b gb gbt
a b a b a b
Êœ œ
œœ œ
#
))i j k
ÈÈ
œœ œœ
œœ
))ij
a (there is no component in the direction of ).œ
gb gbt
a b a bÈ
œœ
œœ
œ œ
#
TN BŠ‹
4. (a) ( ) (a cos ) (a sin ) b [(a cos a sin ) (a sin a cos ) b ] ;rijk i jk) ) ) )) ) ))) )) )œÊœ
d d
dt dt
r )
2gz a a b kk a bÈ ¸¸ ˆ‰vœœœ Êœ
ddd
dt dt dt
2gb
a a b
r ####
"Î#
œœ
)
)) )
)
È
È
œœœ œ
(b) s dt a a b dt a a b d a a u b duœœ œ œ'' ' '
00 0 0
tt t
kkabababv
#### #### ####
"Î# "Î# "Î#
)))
d
dt
)
"
a u du a c u du, where cœœ œ''
00
""
É È
a b
aa
a b
œœ
œ
œœ
œ ### œ
È
kk
s a c u ln u c u c c ln c c ln cÊœ œ ’“Š ‹
ÈÈÈÈ
¹¹ ¹¹
uc a
## #
## ## ## ##
!
##
œ
)
)) ) )
5. r ; 0 0 (1 e)r (e sin œÊœ œÊ œÊ
(1 e)r (1 e)r (e sin ) (1 e)r (e sin )
1 e cos d (1 e cos ) d (1 e cos )
dr drœœ œ
œœ œ
! $$ $
œœ
)) )) )
))
))0œ
sin 0 0 or . Note that 0 when sin 0 and 0 when sin 0. Since sin 0 onʜʜ ))1 ) ) )
dr dr
dd))
0 and sin 0 on 0 , r is a minimum when 0 and r(0) r œ œ œ1) ) )1 )
(1 e)r
1 e cos 0
œ
œ
! $

864. 860 Chapter 13 Vector-Valued Functions and Motion in Space
6. (a) f(x) x 1 sin x 0 f(0) 1 and f(2) 2 1 sin 2 since sin 2 1; since f is continuousœ œ Ê œ œ   Ÿ
"""
###
kk
on [0 2], the Intermediate Value Theorem implies there is a root between 0 and 2ß
(b) Root 1.4987011335179¸
7. (a) and r [(cos ) (sin ) ] r [( sin ) (cos ) ] andvijvu u i j i jviœ œ œ Ê œ
dx dr d dr d dx
dt dt dt dt dt dt dt
dy
r
))
)
ˆ‰ ˆ ‰)) ) ) †
cos r sin cos r sin ; and sin r cos vi vj vj†††œ Êœ œ œ
dr d dx dr d dr d
dt dt dt dt dt dt dt dt
dy
)) )) ) )
)) )
sin r cos ʜ
dy
dt dt dt
dr d
))
)
(b) (cos ) (sin ) cos sin uijvu
rrœ Êœ )) ) ) †
dx
dt dt
dy
cos r sin (cos ) sin r cos (sin ) by part (a),œ ˆ‰ˆ‰
dr d dr d
dt dt dt dt
))) ) ))
))
; therefore, cos sin ;Êœ œ vu†
r
dr dr dx
dt dt dt dt
dy
))
uijvu))œ Ê œ (sin ) (cos ) sin cos )) ) ) †
dx
dt dt
dy
cos r sin ( sin ) sin r cos (cos ) by part (a) r ;œ Êœˆ‰ˆ‰
dr d dr d d
dt dt dt dt dt
))) ) ))
)) )
)
vu†
therefore, r sin cos
ddx
dt dt dt
dy)
œ ))
8. f( ) f ( ) f ( ) f ( ) ; r rvuuœÊœ Êœ œ)) ))
dr d d r d d dr d
dt dt dt dt dt dt dt
wwww
#
))))
)
œœ
œœ
ˆ‰
r
cos r sin sin r cos r f f ;œ Êœ œˆ‰ˆ‰ˆ‰ˆ‰ ˆ‰ kk a b’“’“)) ) )
dr d dr d dr d d
dt dt dt dt dt dt dt
)) ) )
ijv
##
#w#
"Î# "Î#
#
x y y x , where x r cos and y r sin . Then ( r sin ) (cos ) kkk kva‚œ œ œ œ
ÞÞÞ ÞÞÞ
)) ))
dx d dr
dt dt dt
)
( 2 sin ) (r cos ) (r sin ) (cos ) ; (r cos ) (sin ) Êœ œ
dx d dr d d dr d dr
dt dt dt dt dt dt dt dt dt
dy
œ œœ
œ œœ
)))) ))
))) )
ˆ‰
#
(2 cos ) (r sin ) (r cos ) (sin ) . Then Êœ ‚
dy
dt dt dt dt dt dt
ddr d d dr
œ
œ œœ
œœ
)) ))
)))
ˆ‰ kk
#
va
(after algebra) r r r 2 f f f 2 fœœ†much
# ww w
$#$ˆ‰ ˆ‰ ˆ‰ Š‹ ab
d d dr d d r d dr d
dt dt dt dt dt dt dt dt
2 2)))) )
œœ
œœ
Êœ œ,
kk
kk
ab
" ‘ab
va
v
‚ ÊÞ œ
œ
f ff 2f
f f
2 2
#œœ

9. (a) Let r 2 t and 3t 1 and 3 0. The halfway point is (1 3) t 1;œ œ Ê œ œ Ê œ œ ß Êœ)
dr d d r d
dt dt dt dt
))
œœ
œœ
r (1) 3 ; r r 2 (1) 9 6vu uv uua u ua uuœ Êœ œ Êœ
dr d d r d d dr d
dt dt dt dt dt dt dt
rr r r
))))
)) ) )
’“’ “ˆ‰
œœ
œœ #
(b) It takes the beetle 2 min to crawl to the origin the rod has revolved 6 radiansÊ L [f( )] f ( ) d 2 d 4 dÊœ œ œ '' '
00 0
66 6
É cd Ɉ ‰ ˆ ‰ É))) ) )#w
# #
""
#
)))
33 399
4
œ
d ( 6) 1 d ( 6) 1 ln 6 ( 6) 1œ œ œ ''
00
66
É ÈÈ È ’“ ¸¸
37 12
93 3
(6) " " " ## # Ê
##
'
!
)) )
œ
))) ) ))
37 ln 37 6 6.5 in.œ ¸
ÈÈ
Š‹
"
6
10. (t) (t) m (t) m m ( m ) ( m ) m ; m Lr v vr vvraraFa rœ‚ Êœ‚ ‚ Êœ‚ ‚ œ‚ œ Ê
dd d d c
dt dt dt dt
Lr r L
r
ˆ‰ Š‹
œ
œ "
kk m m ( ) constant vectorœ Ê œ‚ œ‚ œ ‚ œ Ê œararr rr0L
dcc
dt
L
rr
Š‹
kk kk
""
11. (a) a right-handed frame of unit vectorscos sin 0
sin cos 0
uu k
ijk
r‚œ œÊ
$)
ââ
ââ
ââ
ââ
ââ
ââ
))
))
(b) ( sin ) (cos ) and ( cos ) (sin )
d
dd
du u
r
))
)
œ œ œ œ)) ))iju iju
œ
r
(c) From Eq. (7), r r z r r r r r zvu u k av u u u u u kœ Êœœ
ÞÞÞÞÞÞÞÞÞÞ
ÞÞÞ ÞÞ Þ
rrr)))))))) ab ˆ‰
rr r 2r zœ
ÞÞ ÞÞÞÞÞ
ÞÞ
Š‹ ˆ‰)))
#
uuk
r )
12. (a) x r cos dx cos dr r sin d ; y r sin dy sin dr r cos d ; thusœÊœ œÊœ) ) )) ) ) ))
dx cos dr 2r sin cos dr d r sin d and
### ###
œ ))))))

865. Chapter 13 Additional and Advanced Exercises 861
dy sin dr 2r sin cos dr d r cos d dx dy dz dr r d dz
### ### #######
œ Ê œ )))))) )
(c) r e dr e d œÊœ
))
)
L dr r d dzʜ '
0
ln 8
È### #
)
eee dœ '
0
ln 8
È###)))
)
3e d 3 eœœ '
0
ln 8 ln 8
0
ÈÈ
’“))
)
83 3 73œœ
ÈÈ È
(b)

866. 862 Chapter 13 Vector-Valued Functions and Motion in Space
NOTES:

867. CHAPTER 14 PARTIAL DERIVATIVES
14.1 FUNCTIONS OF SEVERAL VARIABLES
1. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
(c) level curves are straight lines y x c parallel to the line y xœ œ
(d) no boundary points
(e) both open and closed
(f) unbounded
2. (a) Domain: set of all (x y) so that y x 0 y xß Ê 
(b) Range: z 0 
(c) level curves are straight lines of the form y x c where c 0œ  
(d) boundary is y x 0 y x, a straight lineÈœ Ê œ
(e) closed
(f) unbounded
3. (a) Domain: all points in the xy-plane
(b) Range: z 0 
(c) level curves: for f(x y) 0, the origin; for f(x y) c 0, ellipses with center ( 0) and major and minorßœ ßœ !ß
axes along the x- and y-axes, respectively
(d) no boundary points
(e) both open and closed
(f) unbounded
4. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
(c) level curves: for f(x y) 0, the union of the lines y x; for f(x y) c 0, hyperbolas centered atßœ œ„ ßœÁ
(0 0) with foci on the x-axis if c 0 and on the y-axis if c 0Ð$
(d) no boundary points
(e) both open and closed
(f) unbounded
5. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
(c) level curves are hyperbolas with the x- and y-axes as asymptotes when f(x y) 0, and the x- and y-axesßÁ
when f(x y) 0ߜ
(d) no boundary points
(e) both open and closed
(f) unbounded
6. (a) Domain: all (x y) (0 y)ßÁß
(b) Range: all real numbers
(c) level curves: for f(x y) 0, the x-axis minus the origin; for f(x y) c 0, the parabolas y cx minus theßœ ßœÁ œ
#
origin
(d) boundary is the line x 0œ

868. 864 Chapter 14 Partial Derivatives
(e) open
(f) unbounded
7. (a) Domain: all (x y) satisfying x y 16ß
##
(b) Range: z 
"
4
(c) level curves are circles centered at the origin with radii r 4
(d) boundary is the circle x y 16
##
œ
(e) open
(f) bounded
8. (a) Domain: all (x y) satisfying x y 9ߟ
##
(b) Range: 0 z 3ŸŸ
(c) level curves are circles centered at the origin with radii r 3Ÿ
(d) boundary is the circle x y 9
##
œ
(e) closed
(f) bounded
9. (a) Domain: (x y) (0 0)ßÁß
(b) Range: all real numbers
(c) level curves are circles with center ( 0) and radii r 0%Ð $
(d) boundary is the single point (0 0)ß
(e) open
(f) unbounded
10. (a) Domain: all points in the xy-plane
(b) Range: 0 z 1Ÿ
(c) level curves are the origin itself and the circles with center (0 0) and radii r 0Ð$
(d) no boundary points
(e) both open and closed
(f) unbounded
11. (a) Domain: all (x y) satisfying 1 y x 1ß ŸŸ
(b) Range: zŸŸ
11
##
(c) level curves are straight lines of the form y x c where 1 c 1œ ŸŸ
(d) boundary is the two straight lines y 1 x and y 1 xœ œ
(e) closed
(f) unbounded
12. (a) Domain: all (x y), 0ßBÁ
(b) Range: z
11
##
(c) level curves are the straight lines of the form y cx, c any real number and x 0œÁ
(d) boundary is the line x 0œ
(e) open
(f) unbounded
13. f 14. e 15. a
16. c 17. d 18. b

869. Section 14.1 Functions of Several Variables 865
19. (a) (b)
20. (a) (b)
21. (a) (b)
22. (a) (b)

870. 866 Chapter 14 Partial Derivatives
23. (a) (b)
24. (a) (b)
25. (a) (b)

871. Section 14.1 Functions of Several Variables 867
26. (a) (b)
27. (a) (b)
28. (a) (b)
29. f(x y) 16 x y and 2 2 2 z 16 2 2 2 6 6 16 x y x y 10ßœ ß Êœ œÊœ Ê œ
## ## ##
##
Š‹ Š‹Š‹
ÈÈ È È
30. f(x y) x 1 and (1 0) 1 1 0 x 1 0 x 1 or x 1ßœ ß ÊDœ œÊ œÊœ œ
ÈÈ
## #
31. f(x y) dt at 2 2 z tan y tan x; at 2 2 z tan 2 tan 2ßœ ß Êœ ß Êœ '
x
y
1
1t
œ" œ" œ" œ"
Š‹ Š‹ Š‹
ÈÈ ÈÈ È È
2 tan 2 tan y tan x 2 tan 2œÊœ
œ" œ" œ" œ"ÈÈ
32. f(x y) at (1 2) z ; at (1 2) z 2 2 2y 2x yßœ ßÊœ œ ßÊœ œÊœ Ê œ!Š‹

œ
n0
n
x2
yyx1yx
1
yy"
œ
œ#œœ
Š‹
x
y
y2xʜ

872. 868 Chapter 14 Partial Derivatives
33. 34.
35. 36.
37. 38.
39. 40.
41. f(x y z) x y ln z at (3 1 1) w x y ln z; at (3 1 1) w 3 ( 1) ln 1 2ßßœ ßßÊœ ßßÊœœÈÈ È
xyln z2ÊœÈ

873. Section 14.1 Functions of Several Variables 869
42. f(x y z) ln x y z at ( ) w ln x y z ; at ( ) w ln (1 2 1) ln 4ß ß œ "ß #ß " Ê œ "ß #ß " Ê œ œab ab
## ##
ln 4 ln x y z x y z 4ʜ ʜab
####
43. g(x y z) at (ln 2 ln 4 3) w e ; at (ln 2 ln 4 3) w eßß œ ß ß Ê œ œ ß ß Ê œ!!
__
œœn0 n0
(x y) (x y)
n! z n! z

nn
nn
Ð ÑÎ Ð ÑÎxy z ln2ln4 3 e e 2 2 e ln 2œœœÊœ Êœ
ÐÑÎ ÐÑÎ ln 8 3 ln 2 x y z xy
z
44. g(x y z) at 0 2 w sin sec tßß œ ßß Ê œ ''
x2
yz
yz
x2
ddt
1tt1
)
)
ÈÈ
œœ
"
#
œ" œ"


ˆ‰ cdcd)
sin y sin x sec z sec 2 w sin y sin x sec z ; at 0 2œ Êœ ßß
œ" œ" œ" œ" œ" œ" œ" "
#
Š‹
È ˆ‰
1
4
w sin sin 0 sec 2 sin y sin x sec zÊœ œÊœ
œ" œ" œ" œ" œ" œ""
##
11 1
44
45. f(x y z) xyz and x 20 t, y t, z 20 w (20 t)(t)(20) along the line w 400t 20tßßœ œœœÊœ Êœ
#
400 40t; 0 400 40t 0 t 10 and 40 for all t yes, maximum at t 10Êœ œÊ œÊœ œ Ê œ
dw dw d w
dt dt dt


x 20 10 10, y 10, z 20 maximum of f along the line is f(10 10 20) (10)(10)(20) 2000Êœœ œ œ Ê ßß œ œ
46. f(x y z) xy z and x t 1, y t 2, z t 7 w (t 1)(t 2) (t 7) t 4t 5 along the lineßß œ œ œ œ Ê œ œ
#
2t 4; 0 2t 4 0 t 2 and 2 for all t yes, minimum at t 2 x 2 1 1,Ê œ œ Ê œ Êœ œ Ê œ Ê œœ
dw dw d w
dt dt dt


y 2 2 0, and z 2 7 9 minimum of f along the line is f(1 0 9) (1)(0) 9 9œœ œœ Ê ßß œ œ
47. w 4 4 124.86 km must be (124.86) 63 km south of Nantucketœœ ¸ Ê ¸ˆ‰ ’“
Th
d5 K/km
(290 K)(16.8 km)"Î#
"Î#
"
#
48. The graph of f(xxxx) is a set in a five-dimensional space. It is the set of points"#$%ßßß
(xxxxf(xxxx)) for (xxxx) in the domain of f. The graph of f(xxx x) is a set
"#$% "#$% "#$% "#$ßßßß ßßß ßßß ßßßáß
n
in an (n 1)-dimensional space. It is the set of points (x x x x f(x x x x )) for ßßßáßß ßßßáß "#$ "#$ nn
(x x x x ) in the domain of f."#$ßßßáß
n
49-52. Example CAS commands:
:Maple
with( plots );
f := (x,y) -> x*sin(y/2) + y*sin(2*x);
xdomain := x=0..5*Pi;
ydomain := y=0..5*Pi;
x0,y0 := 3*Pi,3*Pi;
plot3d( f(x,y), xdomain, ydomain, axes=boxed, style=patch, shading=zhue, title="#49(a) (Section 14.1)" );
plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, orientation=[-90,0],
title="#49(b) (Section 14.1)" ); # (b)
L := evalf( f(x0,y0) ); # (c)
plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, contours=[L],
orientation=[-90,0], title="#49(c) (Section 14.1)" );
53-56. Example CAS commands:
:Maple
eq := 4*ln(x^2+y^2+z^2)=1;
implicitplot3d( eq, x=-2..2, y=-2..2, z=-2..2, grid=[30,30,30], axes=boxed, title="#53 (Section 14.1)" );

874. 870 Chapter 14 Partial Derivatives
57-60. Example CAS commands:
:Maple
x := (u,v) -> u*cos(v);
y := (u,v) ->u*sin(v);
z := (u,v) -> u;
plot3d( [x(u,v),y(u,v),z(u,v)], u=0..2, v=0..2*Pi, axes=boxed, style=patchcontour, contours=[($0..4)/2], shading=zhue,
title="#57 (Section 14.1)" );
49-60. Example CAS commands:
: (assigned functions and bounds will vary)Mathematica
For 49 - 52, the command draws 2-dimensional contours that are z-level curves of surfaces z = f(x,y).ContourPlot
Clear[x, y, f]
f[x_, y_]:= x Sin[y/2] y Sin[2x]
xmin= 0; xmax= 5 ; ymin= 0; ymax= 5 ; {x0, y0}={3 , 3 };1111
cp= ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading False];Ä
cp0= ContourPlot[[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, Contours {f[x0,y0]}, ContourShading False,ÄÄ
PlotStyle {RGBColor[1,0,0]}];Ä
Show[cp, cp0]
For 53 - 56, the command will be used and requires loading a package. Write the function f[x, y, z] so thatContourPlot3D
when it is equated to zero, it represents the level surface given.
For 53, the problem associated with Log[0] can be avoided by rewriting the function as x2 + y2 +z2 - e1/4
<<Graphics`ContourPlot3D`
Clear[x, y, z, f]
f[x_, y_, z_]:= x y z Exp[1/4]
222

ContourPlot3D[f[x, y, z], {x, 5, 5}, {y, 5, 5}, {z, 5, 5}, PlotPoints {7, 7}]; Ä
For 57 - 60, the command ParametricPlot3D will be used and requires loading a package. To get the z-level curves here,
we solve x and y in terms of z and either u or v (v here), create a table of level curves, then plot that table.
<<Graphics`ParametricPlot3D`
Clear[x, y, z, u, v]
ParametricPlot3D[{u Cos[v], u Sin[v], u}, {u, 0, 2}, {v, 0, 2p}];
zlevel= Table[{z Cos[v], z sin[v]}, {z, 0, 2, .1}];
ParametricPlot[Evaluate[zlevel],{v, 0, 2 }];1
14.2 LIMITS AND CONTINUITY
1. lim
Ðß ÑÄÐß Ñxy 00
3x y 5 3(0) 0 5
xy2 002
5



#
œœ
2. lim 0
Ðß ÑÄÐß Ñxy 04
x0
y 4
ÈÈ
œœ
3. lim x y 1 3 4 1 24 2 6
Ðß ÑÄÐß Ñxy 34
È ÈÈ È## ##
œ œ œ
4. lim
Ðß ÑÄÐßÑxy 2 3
Š‹ ‘ˆ‰ˆ‰
"" " " " "
#
#œ
##
xy 3 6 36
œ œ œ
5. lim sec x tan y (sec 0) tan (1)(1) 1
Ðß ÑÄ ßxy 0ˆ‰

4
œœœ ˆ‰
1
4

875. Section 14.2 Limits and Continuity 871
6. lim cos cos cos 0 1
Ðß ÑÄÐß Ñxy 00
Š‹Š‹
xy
xy1 001
00




œœœ
7. lim e e e
Ðß ÑÄÐß Ñxy 0ln2
xy 0 ln2 ln "
#
œœœ
ˆ‰
1
2
8. lim ln 1 x y ln 1 (1) (1) ln 2
Ðß ÑÄÐß Ñxy 11
kkk k œ œ
## # #
9. lim lim e e lim 1 1 1
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Äxy 00 xy 00 x0
e sin x sin x sin x
xxx
y
y
œœœœ abˆ‰ ˆ‰
!
††
10. lim cos xy 1 cos (1)(1) 1 cos 0 1
Ðß ÑÄÐß Ñxy 11
ˆ‰ˆ ‰Èkk È
$$
œ œœ
11. lim 0
Ðß ÑÄÐß Ñxy 10
x sin y
x1 11 2
1sin 0 0


œœœ
†
12. lim 2
Ðß ÑÄ ßxy 0ˆ‰

2
cos y 1 (cos 0)
ysin x 10sin
11"


œœœ
ˆ‰


13. lim lim lim (x y) ( 1) 0
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
Á
xy 11 xy 11 xy 11
xy
x 2xy y (x y)
xy xy



œœ œ"œ
14. lim lim lim (x y) (1 1) 2
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
Á
xy 11 xy 11 xy 11
xy
x y (x y)(x y)
xy xy



œœ œœ
15. lim lim lim (y 2) (1 2) 1
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
ÁÁ
xy 11 xy 11 xy 11
x1 x1
xyy2x2 (x1)(y2)
x1 x1


œœ œœ
16. lim lim lim
Ðß ÑÄÐßÑÐßÑÄÐß ÑÐßÑÄÐß Ñ
Á ÁÁ ÁÁ
xy 2 4 xy 2 4 xy 2 4
y 4, xx y 4, xx xx

y4 y4
x y xy 4x 4x x(x 1)(y 4) x(x 1)
1
#
"
œœœ
(2 1)#
"
œ
17. lim lim lim x y 2
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
ÁÁ
xy 00 xy 00 xy 00
xy xy
xy2x2y x y x y2
xy xy


ÈÈÈÈÈÈ
ÈÈÈÈ
ˆ‰ˆ ‰
œœ ˆ‰È È
0022œ œŠ‹
ÈÈ
Note: (x y) must approach (0 0) through the first quadrant only with x y.ßß Á
18. lim lim lim x y 2
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
Á Á Á
xy 22 xy 22 xy 22
xy 4 xy 4 xy 4
xy4
xy2 xy2
xy2 xy2


ÈÈ
ˆ‰ˆ‰ÈÈ
œœ ˆ‰È
222 224œ œœŠ‹È
19. lim lim lim
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
Á Á
xy 20 xy 20 xy 20
2x y 4 2x y 4
ÈÈ
ˆ‰ˆ‰ÈÈ È
2xy2 2xy2
2xy4 2xy2 2xy2 2xy

#
"
œœ
œœœ
"""
# È(2)(2) 0 22 4
20. lim lim lim
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
Á Á
xy 43 xy 43 xy 43
xy 1 xy 1
ÈÈÈÈ
ˆ‰ˆ‰ÈÈ ÈÈÈ È
xy1 xy1
xy1 xy1xy1 xy1


"
œœ

876. 872 Chapter 14 Partial Derivatives
œœœ
"""
È È431 22 4
21. lim
TÄÐßßÑ134
Š‹
""" """
xyz 134 12 12
1243 19
œ œœ
22. lim
TÄÐßßÑ111
2xy yz 2(1)( 1) ( 1)( 1)
xz 1(1) 11
2
#
" "
œœœ
23. lim sin x cos y sec z sin 3 cos 3 sec 0 1 1 2
TÄÐßßÑ330
abab
### ## # #
œ œœ24. lim tan (xyz) tan 2 tan
TÄß߈‰
1
42
2
" " " "
#
œ œ ˆ‰ˆ‰
44
††
11
25. lim ze cos 2x 3e cos 2 (3)(1)(1) 3
TÄÐßßÑ103
2y 2 0
œœœ 1
26. lim ln x y z ln 0 ( 2) 0 ln 4 ln 2
TÄÐßßÑ020
ÈÈ È### # ##
œ œœ
27. (a) All (x y)ß
(b) All (x y) except (0 0)ßß
28. (a) All (x y) so that x yßÁ
(b) All (x y)ß
29. (a) All (x y) except where x 0 or y 0ßœœ
(b) All (x y)ß
30. (a) All (x y) so that x 3x 2 0 (x 2)( 1) 0 x 2 and x 1ß ÁÊ BÁÊÁ Á
#
(b) All (x y) so that y xßÁ
#
31. (a) All (x y z)ßß
(b) All (x y z) except the interior of the cylinder x y 1ßß œ
##
32. (a) All (x y z) so that xyz 0ßß "
(b) All (x y z)ßß
33. (a) All (x y z) with z 0ßß Á
(b) All (x y z) with x z 1ßß Á
##
34. (a) All (x y z) except (x 0 0)ßß ßß
(b) All (x y z) except ( y 0) or (x 0 0)ß ß !ß ß ß ß
35. lim lim lim lim lim ;
Ðß ÑÄÐß Ñ
œ
"
ÄÄÄÄxy 00
along y x
x0

x0 x0 x0 x0
œ œ œ œ œ
xxxx
xy xx 2 x 2x 2 2
ÈÈÈÈÈÈ
kk


"" !!!!
lim lim lim lim
Ðß ÑÄÐß Ñ
œ
#
ÄÄ Äxy 00
along y x
x0

x0 x0 x0
œ œ œœ
xxx
xy 2 x 2( x) 2 2
ÈÈÈÈÈ
kk


""

877. Section 14.2 Limits and Continuity 873
36. lim lim 1; lim lim lim
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
œ
ÄÄÄ
œ
xy 00 xy 00
along y 0

x0 x0 x0
along y x
xx x xx
xy x0 xy 2x xx
"" " ""



œœ œ œœ

ab
"
#
37. lim lim lim different limits for d
Ðß ÑÄÐß Ñ
œ
ÄÄxy 00
along y kx

x0 x0

xy
xy xkx
xkx
xkx
xkx 1k
1k














œœœÊ
ab
ab
ifferent values of k
38. lim lim lim lim ; if k 0, the limit is 1; but if k
Ðß ÑÄÐß Ñ
œ
Á
ÄÄÄxy 00
along y kx
k0

x0 x0 x0
xy x(kx)
xy x(kx) kx k
kx k
kk k k k k kk
œœœ "


0, the limit is 1
39. lim lim different limits for different values of k, k 1
Ðß ÑÄÐß Ñ
œ
Á
Äxy 00
along y kx
k1

x0
xy
xy xkx 1k
xkx 1k


œœÊ Á
40. lim lim different limits for different values of k, k 1
Ðß ÑÄÐß Ñ
œ
Á
Äxy 00
along y kx
k1

x0
xy
xy xkx 1k
xkx 1k


œœÊ Á
41. lim lim different limits for different values of k, k 0
Ðß ÑÄÐß Ñ
œ
Á
Äxy 00
along y kx
k0

x0

xy
ykxk
xkx 1k



œœÊ Á
42. lim lim different limits for different values of k, k 1
Ðß ÑÄÐß Ñ
œ
Á
Äxy 00
along y kx
k1

x0

xx1
xy xkx 1k



œœÊ Á
43. No, the limit depends only on the values f(x y) has when (x y) (x y )ßßÁß
!!
44. If f is continuous at (x y ), then lim f(x y) must equal f(x y ) 3. If f is not continuous at!! !!ßßßœ
Ðß ÑÄÐ ß Ñxy x y $$
(x y ), the limit could have any value different from 3, and need not even exist.!!ß
45. lim 1 1 and lim 1 1 lim 1, by the Sandwich Theorem
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñxy 00 xy 00 xy 00
Š‹ œœÊœ
xy tan xy
3xy
#!
46. If xy 0, lim lim lim 2 2 and" œœ œ
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñxy 00 xy 00 xy 00
2 xy 2xy
xy xy 6
xy
kkŠ‹ Š‹
kk

xy xy
66

ˆ‰
lim lim 2 2; if xy 0, lim lim
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñxy 00 xy 00 xy 00 xy 00
2 xy
xy xy xy
2 xy 2xy
kk
kk kk
kkŠ‹ Š‹
œœ # œ


xy xy
66

lim 2 2 and lim 2 lim 2, by the Sandwichœ œœÊ œ
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñxy 00 xy 00 xy 00
ˆ‰
xy
6xy xy
2 xy 44 cos xykk
kk kk
Èkk
Theorem
47. The limit is 0 since sin 1 1 sin 1 y y sin y for y 0, and y y sin y for¸ ¸ ˆ‰ ˆ‰ ˆ‰ˆ‰
"" " "
xx x x
ŸÊ ŸŸÊ ŸŸ     
y 0. Thus as (x y) ( ), both y and y approach 0 y sin 0, by the Sandwich Theorem.ŸßÄ!ß! ÊÄ ˆ‰
"
x
48. The limit is 0 since cos 1 1 cos 1 x x cos x for x 0, and x x cos x¹ ¹ Š‹ Š‹ Š‹Š‹
"" " "
yy y y
ŸÊ ŸŸÊ ŸŸ     
for x 0. Thus as (x y) ( ), both x and x approach 0 x cos 0, by the Sandwich Theorem.ŸßÄ!ß! ÊÄ Š‹
"
y

878. 874 Chapter 14 Partial Derivatives
49. (a) f(x y) sin 2 . The value of f(x y) sin 2 varies with , which is the line'skßœœ œ ßœ
ymxœ
2m 2 tan
1m 1tan

)
)
)))
angle of inclination.
(b) Since f(x y) sin 2 and since 1 sin 2 1 for every , lim f(x y) varies from 1 to 1kßœ ŸŸ ß
ymxœ
)))
Ðß ÑÄÐß Ñxy 00
along y mx.œ
50. xy x y xy x y x y x y x y x y x y x y x yk kkkkkkkkkkk kk kkab
ÈÈÈÈ
## ## ## ## ## ## ####
œ Ÿ œ Ÿ
xy xy xy xyœ ÊŸœ Ê ŸŸ ab ab ab ¹¹
## ## ## ##
#

xyxy xy xyxy
xy xy xy
abab ab



lim xy 0 by the Sandwich Theorem, since lim x y 0; thus, defineÊœ „ œ
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñxy 00 xy 00
xy
xy




##
ab
f(0 0) 0ߜ
51. lim lim lim 0
Ðß ÑÄÐß Ñ ÄÄxy 00 r0 r0
xxy
xy r cos r sin 1
r cos (r cos ) r sin r cos cos sin






œœœ
))) )))
))
ab a b
52. lim cos lim cos lim cos cos 0 1
Ðß ÑÄÐß Ñ ÄÄxy 00 r0 r0
Š‹ Š ‹ ’ “
xy
xy r cos r sin 1
r cos r sin rcos sin






œœœœ
))
))
))ab
53. lim lim lim sin sin ; the limit does not exist since sin is between
Ðß ÑÄÐß Ñ ÄÄxy 00 r0 r0
y
xy r
r sin




## #
œœ œ
)
ab)) )
0 and 1 depending on )
54. lim lim lim ; the limit does not exist for co
Ðß ÑÄÐß Ñ ÄÄxy 00 r0 r0
2x 2r cos 2 cos 2 cos
x x y r r cos r cos cos


œœœ
)))
)))
s 0)œ
55. lim tan lim tan lim tan ;
Ðß ÑÄÐß Ñ ÄÄxy 00 r0 r0
" " "

’“ ’ “ ’ “
kk kk k k k k kka b kkkkx y r cos r sin r cos sin
xy r r
œœ
)) ))
if r 0 , then lim tan lim tan ; if r 0 , thenÄœœÄ
" "
#
rrÄ! Ä!
!!
’“ ’“
kka b k k k kkkkkr cos sin cos sin
rr
)) )) 1

lim tan lim tan the limit is
rrÄ! Ä!

" "
# #
’“ Š‹
kka b k k k kkkkkr cos sin cos sin
rr
)) )) 11
œœÊ
56. lim lim lim cos sin lim (cos 2 ) which ranges between
Ðß ÑÄÐß Ñ ÄÄ Äxy 00 r0 r0 r0
xy
xy r
r cos r sin





##
œœ œ
))
ab)) )
1 and 1 depending on the limit does not exist Ê)
57. lim ln lim ln
Ðß ÑÄÐß Ñ Äxy 00 r0
Š‹Š ‹
3x x y 3y
xy r
3r cos r cos sin 3r sin






œ
))) )
lim ln 3 r cos sin ln 3 define f(0 0) ln 3œ œÊ ßœ
r0Ä
ab
## #
))
58. lim lim lim 2r cos sin 0 define f(0 0) 0
Ðß ÑÄÐß Ñ ÄÄxy 00 r0 r0
2xy
xy r
(2r cos ) r sin




#
œœœÊßœ
))ab
))
59. Let 0.1. Then x y x y 0.1 x y 0.01 x y 0 0.01 f(x y) f( )$$œ Ê Ê Ê Êß !ß !ÈÈ kk k k
## ## ## ##
0.01 .# œ%
60. Let 0.05. Then x and y f(x y) f(0 0) 0 y 0.05 .$$$ %œ ## Êß ßœ œŸ # œkk kk k k kk ¸¸¸¸
yy
x1 x1


61. Let 0.005. Then x and y f(x y) f(0 0) 0 x y x y$$$œ ## Êß ßœ œŸ kk kk k k k k kk kk ¸¸¸¸
xy xy
x1 x1



0.005 0.005 0.01 . œœ %

879. Section 14.3 Partial Derivatives 875
62. Let 0.01. Since 1 cos x 1 1 2 cos x 3 1 x y$œ ŸŸÊŸ ŸÊŸ ŸÊ Ÿ Ÿ
""
#

3 cos x 3 2 cos x
xy xykk
¸¸ kk
x y . Then x and y f(x y) f(0 0) 0 x y 0.01 0.01Ÿ Êß ßœ œŸ kk kk kk kk k k kk kk ¸¸¸¸$$
xy xy
2cos x 2cos x


0.02 .œœ %
63. Let 0.015. Then x y z f(x y z) f( 0 0) x y z 0 x y z$$œ Êßß !ß ß œ œ
È È kkkkkk
### ### ###
x t x 0.015 0.015 .œ œœŠ‹Š‹
È È
## #
##
%
64. Let 0.2. Then x , y , and z f(x y z) f( 0 0) xyz 0 xyz x y z (0.2)$$$$œ ## # Êßß !ß ß œ œœ #kk kk kk k k k k k k kkkkkk
$
0.008 .œœ %
65. Let 0.005. Then x , y , and z f(x y z) f( 0 0) 0$$$$œ ## # Êßß !ß ß œ kk kk kk k k ¹¹
xyz
xyz1



x y z x y z 0.005 0.005 0.005 0.015 .œŸ Ÿ œœ¹¹ k k kk kk kk
xyz
xyz1


%
66. Let tan (0.1). Then x , y , and z f(x y z) f( 0 0) tan x tan y tan z$$$$œ ## # Êßß !ß ß œ
" ###
kk kk kk k k k k
tan x tan y tan z tan x tan y tan z tan tan tan 0.01 0.01 0.01 0.03Ÿ œ œ œkkkkkk
#########
$$$
.œ%
67. lim f(x y z) lim (x y z) x y z f(x y z ) f is continuous at
Ðß ß ÑÄÐ ß ß Ñ Ðß ß ÑÄÐ ß ß Ñxyz x y z xyz x y z$$$ $$$
ßß œ œ œßßÊ !!! !!!
every (x y z )!!!ßß
68. lim f(x y z) lim x y z x y z f(x y z ) f is continuous at
Ðß ß ÑÄÐ ß ß Ñ Ðß ß ÑÄÐ ß ß Ñxyz x y z xyz x y z$$$ $$$
ßß œ œ œßßÊab
### ###
!!! !!!
every point (x y z )!!!ßß
14.3 PARTIAL DERIVATIVES
1. 4x, 3 2. 2x y, x 2y
`` ` `
`` ` `
ff f f
xy x y
œœ œ œ
3. 2x(y 2), x 1 4. 5y 14x 3, 5x 2y 6
`` ` `
`` ` `
#ff f f
xy x y
œ œ œ œ
5. 2y(xy 1), 2x(xy 1) 6. 6(2x 3y) , 9(2x 3y)
`` ` `
`` ` `
##ff f f
xy x y
œ œ œ œ
7. , 8. ,
`` ` `"
`` ` `

fxf f2xf
xy x yxy xy
y
x3x
œœ œ œ
ÈÈ
ÉÉ ˆ‰ ˆ‰


yy
9. (x y) , (x y)
`"` "`"` "
`` ``
ff
x(xy)x (xy)y(xy)y (xy)
œ œ œ œ
††
10. ,
``
``


ff
xy
x y (1) x(2x) x y (0) x(2y)
xy xy xy xy
y x 2xy
œœœœ
ab ab
ab ab ab ab




11. ,
``"
``
"ffx
x (xy 1) (xy 1) y (xy 1) (xy 1)
(xy 1)(1) (x y)(y) y 1 (xy )(1) (x y)(x)
œœ œ œ œ


12. ,
`"` `"`
`` ``

ff1x
xxx xyyyx xy
11
yy y y
x1 x1
œœ œ œœœ
ˆ‰ ˆ‰
’“ ’“ˆ‰ ˆ‰
y y
x x
y y
x x

††ˆ‰ ˆ‰
13. e (x y 1) e , e (x y 1) e
`` ``
`` ``
ÐÑ ÐÑ ÐÑ ÐÑff
xx yy
xy1 xy1 xy1 xy1
œ œœ œ††

880. 876 Chapter 14 Partial Derivatives
14. e sin (x y) e cos (x y), e cos (x y)
``
``
ff
xy
xx x
œ œ
15. (x y) , (x y)
`"` "`"` "
`` ``
ff
xxyx xyyxyy xy
œ œœ œ††
16. e (xy) ln y ye ln y, e (xy) ln y e xe ln y
`` `` "
`` ``
ff e
xx yy y y
xy xy xy xy xy
œœœ œ †† †† †
xy
17. 2 sin (x 3y) sin (x 3y) 2 sin (x 3y) cos (x 3y) (x 3y) 2 sin (x 3y) cos (x 3y),
`` `
`` `
f
xx x
œ œ œ ††
2 sin (x 3y) sin (x 3y) 2 sin (x 3y) cos (x 3y) (x 3y) 6 sin (x 3y) cos (x 3y)
`` `
`` `
f
yy y
œ œ œ ††
18. 2 cos 3x y cos 3x y 2 cos 3x y sin 3x y 3x y
`` `
`` `
## ###f
xx x
œ œ ab ab ababab††
6 cos3xy sin3xy,œ abab
##
2 cos 3x y cos 3x y 2 cos 3x y sin 3x y 3x y
`` `
`` `
## ###f
yy y
œ œ ab ab ababab††
4y cos 3x y sin 3x yœ abab
##
19. yx , x ln x 20. f(x y) and
`` `" `
`` ``
ff ln x f f ln x
xy ln y x x ln y y y(ln y)
œœ ߜʜ œ
y1 y

21. g(x), g(y)
``
``
ff
xy
œ œ
22. f(x y) (xy) , xy 1 f(x y) (1 xy) andßœ #Êßœ Ê œ œ !
_
œn0
n
kk
"` "`
` ` 1 xy x (1 xy) x (1 xy)
f y
†
(1 xy)
`"`
``
fx
y (1 xy) y (1 xy)
œ œ
†
23. f 1 y , f 2xy, f 4z 24. f y z, f x z, f y x
x yz xyzœ œœ œ œ œ
#
25. f 1, f , fxy z
y
yz yz
zœœ œ
ÈÈ

26. f x x y z , f y x y z , f z x y zxyzœ œ œab ab ab
### ### ###
$Î# $Î# $Î#
27. f , f , fxyz
yz xy
1 xyz 1 xyz 1 xyz
xzœœœ
ÈÈÈ
28. f , f , fxy z
xyz (xyz)1 xyz (xyz) 1 xyz (xyz) 1
z yœœ œ
"
kk kk kkÈÈ È

29. f , f , f
xyzœœœ
"
x2y3z x2y3z x2y3z
23
30. f yz (xy) , f z ln (xy) yz ln (xy) z ln (xy) (xy) z ln (xy) z,xy
xy x xy x y xy y
(yz)(y) yz yzœœœœ œ œ †† † †
"` ` `
```
f y ln (xy) yz ln (xy) y ln (xy)z
zœ œ†
`
`
31. f 2xe , f 2ye , f 2zexyz
xyz xyz xyzœ œ œ
ab ab ab

32. f yze , f xze , f xyexyz
xyz xyz xyzœ œ œ

33. f sech (x 2y 3z), f 2 sech (x 2y 3z), f 3 sech (x 2y 3z)xy zœ œ œ
###
34. f y cosh xy z , f x cosh xy z , f 2z cosh xy zxyzœ œ œab ab ab
## #

881. Section 14.3 Partial Derivatives 877
35. 2 sin (2 t ), sin (2 t )
``
``
ff
t
œ œ 11! 1!
!
36. v e 2ve , 2ve v e 2ve 2ue
``
`` ` `
# Ð ÎÑ Ð ÎÑ Ð ÎÑ # Ð ÎÑ Ð ÎÑ Ð ÎÑ``gg
uuv v vv
2u v 2u v 2u v 2u v 2u v 2u v2u 2u
œœœ œ ††ˆ‰ ˆ‰
37. sin cos , cos cos , sin sin
`` `
`` `
hh h
39 )
œœ œ 9) 39) 39)
38. 1 cos , r sin , 1
```
```
ggg
rz
œ œœ ))
)
39. W V, W P , W , W , Wpv v g
v Vv 2Vv Vv Vv
2g 2g 2g g gœœ œœœœ
$$$$
$


#
40. m, , , c,
```` `
``#`` ` #
AAAmAkAkmh
chkqmqqq
q
œœœœ œ

41. 1 y, 1 x, 0, 0, 1
``````
` ` ``````
ff ffff
x y xyyxxy
œ œ œœ œœ


42. y cos xy, x cos xy, y sin xy, x sin xy, cos xy xy sin xy
``` ` ``
````````
##ff f f ff
xyxyyxxy
œœ œ œ œœ


43. 2xy y cos x, x sin y sin x, 2y y sin x, cos y, 2x cos x
`` ````
` ` ` ` `` ``
#gg gggg
x y x y yx xy
œ œ œ œ œœ


44. e , xe 1, 0, xe , e
`` `` ``
` ` ` ` `` ``
hh hh hh
x y x y yx xy
œœ œœ œœ
yy y y


45. , , , ,
`"`"` "` "` ` "
``` ` ````
rr r r rr
x xy y xy x (xy) y (xy) yx xy (xy)
œœ œ œ œœ


46. , ,
`"` " `"` "
`` ``

ss1x
xxxxxyyyxxxy
1111
yy y y
œœ œœ œ œ”• ”• ”• ”•
ˆ‰ ˆ ‰ ˆ‰ ˆ‰
ˆ‰ ˆ‰ ˆ‰ ˆ‰
yyyy
xxxx ††
, ,
``
``





ss
xy
y(2x) 2xy x(2y) 2xy
xy xy xy xy
œœ œœ
abab ab ab
``
`` ``







ss
yx xy
x y ( 1) y(2y)
xy xy
yx
œœ œ
ab
ab ab
47. , , , and
` ` ` `
```` ``
w2w3 w 6 w 6
x 2x 3y y 2x 3y y x (2x 3y) x y (2x 3y)
œœ œ œ


48. e ln y , ln x, , and
```""`""
` ` `` ``
wwxw w
x x y y yx y x xy y x
y
œ œ œœ œ
x

49. y 2xy 3x y , 2xy 3x y 4x y , 2y 6xy 12x y , and
`` `
`` ``
# $ #% ## $$ # #$ww w
xy yx
œ œ œ

2y 6xy 12x y
`
``
##$

w
xy
œ
50. sin y y cos x y, x cos y sin x x, cos y cos x 1, and
`` `
````
ww w
xyyx
œ œ œ

cos ycos x1
`
``

w
xy
œ
51. (a) x first (b) y first (c) x first (d) x first (e) y first (f) y first
52. (a) y first three times (b) y first three times (c) y first twice (d) x first twice

882. 878 Chapter 14 Partial Derivatives
53. f (1 2) lim lim lim x
f(1 h 2) f(1 2)
hh h
1(1h)26(1h) (26) h612hh 6ßœ œ œ
h0 h0 h0ÄÄ Ä
ß ß cd ab

lim lim ( 13 6h) 13,œœ œ
h0 h0ÄÄ
13h 6h
h

f (1 2) lim lim lim y
f(1 2 h) f(1 2) (2 6 2h) (2 6)
hh h
11(2h)3(2h) (26)ßœ œ œ
h0 h0 h0ÄÄ Ä
ß ß cd
lim ( 2) 2œ œ
h0Ä
54. f ( 2 1) lim lim x
f( 2 h 1) f( 2 1)
hh
42(2h)3(2h) (32)ßœ œ
h0 h0ÄÄ
ß ß cd
lim lim 1 1,œœœ
h0 h0ÄÄ
(2h1h)1
h

f ( 2 1) lim lim y
f( 2 1 h) f( 2 1)
hh
443(1h)2(1h) (32)ßœ œ
h0 h0ÄÄ
ß ß cd

lim lim lim (1 2h) 1œœœ œ
h0 h0 h0ÄÄÄ
ab 33h24h2h 1
hh
h2h

55. f (x y z ) lim ;z
f(xyz h) f(x,yz)
h
!!!
ßß ßßß œ
h0Ä
$$$ $$$
f (1 2 3) lim lim lim lim (12 2h) 12z
f(1 2 3 h) f(1, 2 3) 2(3 h) 2(9)
hhh
2h 2hßß œ œ œ œ œ
h0 h0 h0 h0ÄÄÄÄ
ßß ß "

56. f (x y z ) lim ;y
f(x y h z ) f(x , y z )
h
!!!
ßß ßßß œ
h0Ä
$$ $ $$$
f ( 1 0 3) lim lim lim (2h 9) 9y
f( 1h3) f( 1,03)
hh
2h 9h 0ßß œ œ œ œ
h0 h0 h0ÄÄÄ
ß ß ß ab

57. y 3z x z 2y 0 3xz 2y y z at (1 1 1) we have (3 2) 1 1 or œÊ œ Êßß œˆ‰ ab
#$ # $`` ` `
`` ` `
zz z z
xx x x
2
`
`
z
x
œ
58. z x 2x 0 z 2x x at (1 1 3) we have ( 3 1 2) 1 orˆ‰ ˆ‰ ˆ ‰
``` ` `
``` ` `
xxx x x
zxzz xz z
yy
œÊ œÊß ß œ

`"
`
x
z6
œ
59. a b c 2bc cos A 2a (2bc sin A) ; also 0 2b 2c cos A (2bc sin A)
### `` `
`` `
œ Êœ Ê œ œ
AAa A
a a bc sin A b
2c cos A 2b (2bc sin A) Ê œÊœ
``
``
A A c cos A b
b b bc sin A
60. 0 (sin A) a cos A 0 ; also
ab a aa cos A
sin A sin B sin A x A sin A
(sin A) a cos A
œÊ œÊ œÊ œ
`
`
a
A
``
`` b( csc B cot B) b csc B cot B sin Aˆ‰
"` `
``sin A B B
aa
œ Êœ
61. Differentiating each equation implicitly gives 1 v ln u u and 0 u ln v v orœ œ xxxx
vu
uv
ˆ‰ ˆ‰
v
(ln u) v u 1
v (ln v) u 0
Ÿ
ˆ‰
ˆ‰
xx
v
u
u
v
xx
0ln v
ln u
ln v
ln v
(ln u)(ln v) 1 œ
œ
Êœ œ
x
ºº
ºº
"

v
u
v
u
u
v
62. Differentiating each equation implicitly gives 1 (2x)x (2y)y and 0 (2x)x y orœ œ uu uu
x and
(2x)x (2y)y 1
(2x)x y 0

uu
uu
2y
01
2x 2y
2x 1
11
2x 4xy 2x 4xy œ
œ
Êœ œ œ
u
ºº
ºº
"





y ; next s x y 2x 2y
uœœœœ œ Êœ
ºº
2x
2x 0
x 4xy 2x 4xy 2x 4xy 1 2y u u u
2x 2x 1 s x y
"
# ` ` `
`` ## `
2x 2yœ œ œŠ‹Š‹
"""
#

2x 4xy 1 2y 1 y 1 2y 1 2y
2y 1 2y
63. 2x, 2y, 4z 2, 2, 4 2 2 ( 4) 0
``` ``` ```
``` ``` ```
fff fff fff
xyz xyz xyz
œœœ Êœ œ œ Ê œ œ

883. Section 14.3 Partial Derivatives 879
64. 6xz, 6yz, 6z 3 x y , 6z, 6z, 12z
``` ``` ```
``` ``` ```
###fff fff fff
xyz xyz xyz
œ œ œ œ œ œÊ ab


6z 6z 12z 0œ œ
65. 2e sin 2x, 2e cos 2x, 4e cos 2x, 4e cos 2x
`` ` ` ``
`` ` ` ``
ffff ff
xy x y xy
œ œ œ œÊ
2y 2y 2y 2y


4e cos 2x 4e cos 2x 0œ œ
2y 2y
66. , , , 0
``` ` ``
``` ` ``


fxf f f ff
xxyyxyx y x y
yyxxy yxxy
xy xy xy xy
œœ œ œÊ œ œ




ab ab abab
67. x y z (2x) x x y z , x y z (2y)
`" `"
`# `#
### ### ###
$Î# $Î# $Î#f f
x y
œ œ œab ab ab
yxyz , xyz (2z) zxyz ;œ œ œab ab ab
### ### ###
$Î# $Î# $Î#`"
`#
f
z
xyz 3xxyz , xyz 3yxyz ,
``
``
### #### ### ####
$Î# &Î# $Î# &Î#


ff
xy
œ œ abab abab
xyz 3zxyz
` ```
` ```
### ####
$Î# &Î#


f fff
z xyz
œ Ê abab
xyz 3xxyz xyz 3yxyzœ ’“’“abababab
### #### ### ####
$Î# &Î# $Î# &Î#
xyz 3zxyz 3xyz 3x3y3zxyz œ ’“abababa bab
### #### ### # # ####
$Î# &Î# $Î# &Î#
0œ68. 3e cos 5z, 4e cos 5z, 5e sin 5z; 9e cos 5z, 16e cos 5z,
``` ` `
``` ` `
fff f f
xyz x y
3x 4y 3x 4y 3x 4y 3x 4y 3x 4y
œœœ œœ


25e cos 5z 9e cos 5z 16e cos 5z 25e cos 5z 0
````
` ```



ffff
z xyz
3x 4y 3x 4y 3x 4y 3x 4y
œ Ê œ œ
69. cos (x ct), c cos (x ct); sin (x ct), c sin (x ct) c [ sin (x ct)]
`` ` ` `
`` ` ` `
##ww w w w
xt x t t
œ œ œ œ Êœ


c œ
#`
`


w
x
70. 2 sin (2x 2ct), 2c sin (2x 2ct); 4 cos (2x 2ct), 4c cos (2x 2ct)
`` ` `
`` ` `
#ww w w
xt x t
œ œ œ œ


c [ 4 cos (2x 2ct)] cÊœ œ
``
``
##


ww
tx
71. cos (x ct) 2 sin (2x 2ct), c cos (x ct) 2c sin (2x 2ct);
``
``
ww
xt
œ œ
sin (x ct) 4 cos (2x 2ct), c sin (x ct) 4c cos (2x 2ct)
``
``
##


ww
xt
œ œ c [ sin (x ct) 4 cos (2x 2ct)] c Êœ œ
``
``
##


ww
tx
72. , ; , c c
`"` ` ` ` " `
``` ` ` `
##wwcw1wcw w
x xct t xct x (xct) t (xct) t (xct) x
œœ œ œÊœ œ


’“
73. 2 sec (2x 2ct), 2c sec (2x 2ct); 8 sec (2x 2ct) tan (2x 2ct),
`` `
`` `
###ww w
xt x
œ œ œ


8c sec (2x 2ct) tan (2x 2ct) c [8 sec (2x 2ct) tan (2x 2ct)] c
`` `
`` `
## # # #


ww w
tt x
œ Êœ œ
74. 15 sin (3x 3ct) e , 15c sin (3x 3ct) ce ; 45 cos (3x 3ct) e ,
`` `
`` `
ww w
xt x
œ œ œ
xct xct xct!!!


45c cos (3x 3ct) c e c 45 cos (3x 3ct) e c
`` `
`` `
### #


ww w
tt x
œ Êœ œ
xct xct!!
cd
75. (ac) (ac) (ac) a c ; a a a
```` ` ` ````` ` `
```` ` ` `` ``` ` `
##wfuf w f fwfuf w f
tutu t u uxuxu x u
œœ Êœ œ œœ Êœ


Š‹ Š ‹ ††
a ac c a c œÊœ œ œ
######`` ` ` `
`` ` ` `


fw f f w
ut u u x
Š‹

884. 880 Chapter 14 Partial Derivatives
76. If the first partial derivatives are continuous throughout an open region R, then by Theorem 3 in this section of the
text, f(x y) f(x y ) f (x y ) x f (x y ) y x y, where , 0 as x, y 0. Then asßœ ß ß ß ÄÄ
!! !! !! " # "#xy ? ? %? %? % % ? ?
(x y) (x y ), x 0 and y 0 lim f(x y) f(x y ) f is continuous at every pointßÄ ß Ä ÄÊ ßœ ß Ê
!! !!??
Ðß ÑÄÐ ß Ñxy x y $$
(x y ) in R.!!ß
77. Yes, since f , f , f , and f are all continuous on R, use the same reasoning as in Exercise 76 with
xx yy xy yx
f(xy) f(xy) f(xy) x f(xy) y x y andx x xx xyßœ ß ß ß !! !! !! " # ? ? %? %?
f (x y) f (x y ) f (x y ) x f (x y ) y x y. Then lim f (x y) f (x y )
y y yx yy x xßœ ß ß ß ßœ ßss !! !! !! " # !! ? ? %? %?
Ðß ÑÄÐ ß Ñxy x y $$
and lim f (x y) f (x y ).
Ðß ÑÄÐ ß Ñxy x y $$
yyßœ ß !!
14.4 THE CHAIN RULE
1. (a) 2x, 2y, sin t, cos t 2x sin t 2y cos t 2 cos t sin t 2 sin t cos t
``
``
w w dx dw
x y dt dt dt
dy
œœœ œÊœ œ
0; w x y cos t sin t 1 0œœ œ œÊ œ
## # # dw
dt
(b) ( ) 0
dw
dt
1œ
2. (a) 2x, 2y, sin t cos t, sin t cos t
``
``
w w dx dw
x y dt dt dt
dy
œœœ œ Ê
(2x)( sin t cos t) (2y)( sin t cos t)œ
2(cos t sin t)(cos t sin t) 2(cos t sin t)(sin t cos t) 2 cos t 2 sin t 2 cos t 2 sin tœ œ abab
## ##
0; w x y (cos t sin t) (cos t sin t) 2 cos t 2 sin t 2 0œœ œ œ œÊ œ
## # # # # dw
dt
(b) (0) 0
dw
dt
œ
3. (a) , , , 2 cos t sin t, 2 sin t cos t,
`"`"` "
```
w w w dx dz
x z y z z z dt dt dt t
(x y) dy
œœœ œ œœ

cos t sin t sin t cos t 1; w t 1Êœ œœœ œ œÊ œ
dw 2 2 cos t sin t x cos t sin t dw
dt z z z t z z dt
xy y
t




Š ‹ Š‹ Š‹ab


t tt
(b) (3) 1
dw
dt
œ
4. (a) , , , sin t, cos t, 2t
```
```
"Î#w2xw w2zdx dz
x xyz y xyz z xyz dt dt dt
2y dy
œœœœ œœ

Êœ œ
dw 2x sin t 4zt
dt xyz xyz xyz costsint16t
2y cos t 2 cos t sin t 2 sin t cos t 4 4t t





ˆ‰
; w ln x y z ln cos t sin t 16t ln (1 16t) œœ œ œ Êœ
16 dw 16
1 16t dt 1 16t
### # #
aba b
(b) (3)
dw 16
dt 49
œ
5. (a) 2ye , 2e , , , , e
```" "
```
wwwdx2t dzdw 2ee
x y z z dt t 1 dt t 1 dt dt t 1 t 1 z
xx t dy 4yte
œœœ œœœÊœ

x xt
4t tan t 1; w 2ye ln z 2 tan t t 1 tœ œ œ œ
(4t) tan t t 1 2 t 1
t1 t1 e
e xa bab ab




" " #
t
t
abab t 1 2 tan t (2t) 1 4t tan t 1Êœ œ
dw 2
dt t 1
ˆ‰aba b


#" "
(b) (1) (4)(1) 1 1
dw
dt 4
œ œˆ‰
1
1
6. (a) y cos xy, x cos xy, 1, 1, , e y cos xy e
``` "
```
w w w dx dz dw
x y z dt dt t dt dt t
dy x cos xy
œ œ œœœ œÊœ
t1 t1
(ln t)[cos (t ln t)] e (ln t)[cos (t ln t)] cos (t ln t) e ; w z sin xyœ œ œ
t cos (t ln t)
t t1 t1
e sin (t ln t) e [cos (t ln t)] ln t t e (1 ln t) cos (t ln t)œ Êœ œ
t1 t1 t1 dw
dt t
‘ˆ‰
"
(b) (1) 1 (1 0)(1) 0
dw
dt
œ œ

885. Section 14.4 The Chain Rule 881
7. (a) 4e ln y (sin v)
````
`````
`z z x z cos v 4e 4e sin v
uxuyu u cos v y u y
y4e ln yx
œ œ œ ab ˆ‰ Š‹
xx x
(4 cos v) ln (u sin v) 4 cos v;œ œ
4(u cos v) ln (u sin v) 4(u cos v)(sin v)
uu sin v
4e ln y (u cos v) 4e ln y (tan v)
````
`````
`z z x z u sin v 4e 4e u cos v
vxvyv u cos v y y
y xx
œ œ œab abˆ‰ Š‹
xx
[ 4(u cos v) ln (u sin v)](tan v) ( 4u sin v) ln (u sin v) ;œ œ
4(u cos v)(u cos v)
u sin v sin v
4u cos v

z 4e ln y 4(u cos v) ln (u sin v) (4 cos v) ln (u sin v) 4(u cos v)œœ Êœ
x zsin v
uu sin v
`
` ˆ‰
(4 cos v) ln (u sin v) 4 cos v; also ( 4u sin v) ln (u sin v) 4(u cos v)œ œ
`
`
zu cos v
vu sin v
ˆ‰
( 4u sin v) ln (u sin v)œ
4u cos v
sin v

(b) At 2 : 4 cos ln 2 sin 4 cos 2 2 ln 2 2 2 2 (ln 2 2);ˆ‰ ˆ ‰ ÈÈ ÈÈ
ßœ œ œ
1111
4u 4 4 4
z`
`
( 4)(2) sin ln 2 sin 4 2 ln 2 4 2 2 2 ln 2 4 2
`
`
z
v44
(4)(2) cos
sin
œ œ œ
11
ˆ‰ ÈÈ È È È
ˆ‰
ˆ‰


4
4
8. (a) cos v sin v 0;
`
`

zx sin v
uxyxyu
11
y cos v (u sin v)(cos v) (u cos v)(sin v)
œ œ œœ
–— –—
Š‹
Š‹ Š‹
Š‹
"

y
xx
yy
x
y

( u sin v) u cos v
`
`

z xu cos v
vxyxyu
11
yu sin v (u sin v)(u sin v) (u cos v)(u cos v)
œ œ œ
–— –—
Š‹
Š‹ Š‹
Š‹
"

y
xx
yy
x
y

sin v cos v 1; z tan tan (cot v) 0 and csc vœ œ œœ Êœœ
# # " " # ``"
``
Š‹ ˆ‰ ab
xzz
yuv1cotv

1 œœ
"
sin v cos v

(b) At 1.3 : 0 and 1ˆ‰ßœ œ
1
6u v
zz``
``
9. (a) (y z)(1) (x z)(1) (y x)(v) x y 2z v(y x)
```` ``
```````
`wwxw wz
uxuyuzu
y
œ œ œ
(u v) (u v) 2uv v(2u) 2u 4uv; œ œ œ
```` ``
```````
`wwxw wz
vxvyvzv
y
(y z)(1) (x z)( 1) (y x)(u) y x (y x)u 2v (2u)u 2v 2u ;œ œ œ œ
#
w xy yz xz u v u v uv u v uv u v 2u v 2u 4uv andœ œ œ Êœ ababab
######### `
`
w
u
2v 2u
`
`
#w
v
œ
(b) At 1 : 2 4 (1) 3 and 2(1) 2ˆ ‰ ˆ‰ ˆ‰ ˆ‰
"` " " ` "
#` # # ` # #
#
ßœ œœ œ
ww3
uv
10. (a) e sin u ue cos u e cos u ue sin u e
`
`
w2x 2z
u xyz xyz xyz
vv v v v 2y
œ Š‹ Š‹ Š‹ababab

e sin u ue cos uœ ˆ‰ ab
2ue sin u
ue sin u ue cos u ue
vv
v
2v 2v 2v


e cos u ue sin uˆ‰ ab
2ue cos u
ue sin u ue cos u ue
vv
v
2v 2v 2v


e; œˆ‰ ab
2ue 2
ue sin u ue cos u ue u
v
v
2v 2v 2v


ue sin u ue cos u ue
`
`
w2x 2z
v xyz xyz xyz
vv v 2y
œ Š‹ Š‹ Š‹ab a b ab

ue sin uœˆ‰ ab
2ue sin u
ue sin u ue cos u ue
v
v
2v 2v 2v


ue cos uˆ‰ ab
2ue cos u
ue sin u ue cos u ue
v
v
2v 2v 2v


ue 2; w ln u e sin u u e cos u u e ln 2u e œœ œˆ‰ ab a b a b
2ue
ue sin u ue cos u ue
v2v2v2v2v
v
2v 2v 2v


### ## #
ln 2 2 ln u 2v and 2œ Êœ œ
``
``
w2 w
uu v
(b) At ( ): 1 and 2#ß ! œ œ œ
``
`# `
w2 w
uv
11. (a) 0;
`````" ```````
` ` uuuur
x p x q x r x qr (qr) (qr) (qr)
p q rp pq qrrppq
œ œ œœ


`````"
```````
` ` uuuur
y p y q y r y qr (qr) (qr) (qr) (qr)
p q rp pq qrrppq 2p2r
œ œ œœ

; œœœ
(2x2y2z)(2x2y2z) p q
(2z 2y) (z y) z p z q z r z
zuuuur ` `
```````
`````

;œ œœœœ
"


q r (q r) (q r) (q r) (q r) (2z 2y) (z y)
rp pq qrrppq 2q2p 4y y

886. 882 Chapter 14 Partial Derivatives
u 0, , and œœ œ Êœ œ œ œ
p q 2y y (z y) y( 1) (z y)(0) y(1)
q r 2z 2y z y x y (z y) (z y) z (z y)
uu z u
` ` `
`` `

œ
y
(z y)

(b) At 3 2 1 : 0, 1, and 2Š‹
È
ßß œ œ œ œ œ
``" `
`` `
uu u2
xy(12) z(12)

12. (a) (cos x) re sin p (0) qe sin p (0) y if x ;
` ` ##
" "

u e e cos x e cos x
x 1p 1p
qr qr z
1sinx
œ œœ œ
qr qr z ln y
ÈÈÈ
aba b
11
(0) re sin p qe sin p (0) xzy ;
`
`
" " ue z
yyyy1p
qr qr z 1 z re sin p zyx
œ œœœ
qr qr
z
z
È
ˆ‰
#! !
aba b Š‹
(0) re sin p (2z ln y) qe sin p 2zre sin p (ln y)
` "
`
" " "ue
z zz1p
qr qr qr qe sin p
œ œ
qr qr
È
#!
ab a ba b ˆ‰
(2z) y x ln y xy ln y; u e sin (sin x) xy if x y ,œ œœ œ ŸŸ Ê œˆ‰ab
" `"
##`zz x
zzzlnyz z z ln y y x uabab

z
11
xzy , and xy ln y from direct calculations
``
``
uu
yz
z1 z
œœœ
(b) At : 2, , lnˆ ‰ ˆ‰ ˆ‰ˆ ‰ˆ‰ ˆ‰ˆ‰ ˆ‰È
111 1
4x y4 4z4
uu u 2
ßß œœœ œ œ
""` " ` "" ` " "
##` # ` ## ` # #
"Î# Ð"Î#Ñ" "Î# È
œ
1
È
2 ln 2
4
13. 14.
dz z dx z dz z du z dv x dw
dt x dt y dt dt u dt v dt w dt
dy
œ œ
`` ```
`` ```

15.
```` `` ```` ``
` `` `` `` ` `` `` ``
` `wwxw wz wwxw wz
u xu yu zu v xv yv zv
y y
œ œ

16.
``````` ```````
``````` ```````
wwrwswt wwrwswt
xrxsxtx yrysyty
œ œ

887. Section 14.4 The Chain Rule 883
17.
```` ````
` `` `` ` `` ``
` `wwxw wwxw
u xu yu v xv yv
y y
œ œ

18.
` `` `` ` `` ``
` `` `` ` `` ``
w wu wv w wu wv
x ux vx y uy vy
œ œ

19.
```` ````
` `` `` ` `` ``
` `zzxz zzxz
t xt yt s xs ys
y y
œ œ

20. 21.
`
`` ````
`````ydy
rdur s dus t dut
u w dw u w dw u
œœœ

22.
```` ````
` ````````
`wwxw wzwv
p xpypzpvp
y
œ

888. 884 Chapter 14 Partial Derivatives
23. since 0 since 0
`` ` ` `` ` `
```` ````
w w dx w w dx w w dx w w dx
r x dr y dr x dr dr s x ds y ds y ds ds
dy dy dy dy
œ œœ œ œœ

24.
````
`````
`wwxw
sxsys
y
œ

25. Let F(x y) x 2y xy 0 (x y) 3x yßœ œÊJßœ
$# #
x
and F (x y) 4y x y
dy 3x y
dx F ( 4y x)
Fßœ Êœ œ
x
y



(1 1)Êßœ
dy
dx 3
4
26. Let F(x y) xy y 3x 3 0 F (x y) y 3 and F (x y) x 2y ßœ œÊ ßœ ßœ Êœ œ
#

xy
dy y 3
dx F x 2y
F x
y
( 1 1) 2Ê ßœ
dy
dx
27. Let F(x y) x xy y 7 0 F (x y) 2x y and F (x y) x 2y ßœ œÊ ßœ ßœ Êœ œ
##

xy
dy 2x y
dx F x 2y
F x
y
(1 2)Êßœ
dy
dx 5
4
28. Let F(x y) xe sin xy y ln 2 0 F (x y) e y cos xy and F (x y) xe x sin xy 1ßœ œÊ ßœ ßœ
yyy
xy
( ln 2) (2 ln 2)Êœ œ Ê!ß œ
dy e y cos xy dy
dx F xe x sin xy 1 dx
F
x
y
y
y

29. Let F(x y z) z xy yz y 2 0 F (x y z) y, F (x y z) x z 3y , F (x y z) 3z yßß œ œÊ ßßœ ßß œ ßß œ
$$ ##
xy z (111) ; Êœ œ œÊßßœœ œ œ
``"`
```
zzz
x F 3z y 3z y x 4 y F 3z y 3z y
F y y xz3y xz3y F x
z z
y


(111)Êßßœ
`
`
z3
y4
30. Let F(x y z) 1 0 F (x y z) , F (x y z) , F (x y z)ßß œ œÊ ßßœ ßß œ ßß œ
""" " " "
xyz x y z
xyz

(236) 9; (236) 4Êœ œ œÊßßœ œœ œÊßßœ
````
````


zzzzzz
xF xx yF yy
F F x
z z
x y
z z
y
Š‹ Š‹
Š‹ Š‹
! !

! !



31. Let F(x y z) sin (x y) sin (y z) sin (x z) 0 F (x y z) cos (x y) cos (x z),ßß œ œÊ ßßœ x
F (x y z) cos (x y) cos (y z), F (x y z) cos (y z) cos (x z) yz
z
xF
Fßß œ ßß œ Êœ
`
` x
z
()1; ()1œ Êßßœ œœ Êßßœ
cos (x y) cos (x z) cos (x y) cos (y z)
cos (y z) cos (x z) x y F cos (y z) cos (x z) y
zz z F
` ` `
`` `
111 111
y
z
32. Let F(x y z) xe ye 2 ln x 2 3 ln 2 0 F (x y z) e , F (x y z) xe e , F (x y z) yeßß œ œÊ ßßœ ßß œ ßß œ
yz y yz z
xy z
2
x
(1 ln 2 ln 3) ; (1 ln 2 ln 3)Êœ œ Êßßœ œœ Êßßœ
````
````
zz4zxeez5
x F ye x 3 ln 2 y F ye y 3 ln 2
F e F xx
z z
y2
z z
y
yz
ˆ‰
33. 2(x y z)(1) 2(x y z)[ sin (r s)] 2(x y z)[cos (r s)]
```` ``
```````
`wwxw wz
rxryrzr
y
œ œ
2(x y z)[1 sin (r s) cos (r s)] 2[r s cos (r s) sin (r s)][1 sin (r s) cos (r s)]œ œ

889. Section 14.4 The Chain Rule 885
2(3)(2) 12Êœœ¸
`
`
w
r
r1s 1œßœ
34. y x(1) (0) (u v) (1)
```` `` " `
``````` `
`w w x w w z 2v 2v v w 4 4
vxvyvzv u z u u v 1 1
y
œ œ œ Êœ ˆ‰ ˆ‰ ˆ‰ ¸ ˆ ‰ ˆ ‰

u1v2œßœ
8œ
35. 2x ( 2) (1) 2(u 2v 1) ( 2)
```` " "
`````
`w w x w 2uv2
v x v y v x x (u2v1) u2v1
yy
œ œ œ ˆ‰ ˆ‰ ’“

7Êœ ¸
`
`
w
v
u0v0œßœ
36. (y cos xy sin y)(2u) (x cos xy x cos y)(v)
````
`````
`zzxz
uxuyu
y
œ œ
uv cos u v uv sin uv (2u) u v cos u v uv u v cos uv (v)œ cdc dab ababab
$$ ## $$##
0(cos 0cos 0)(1)2Êœ œ¸
`
`
z
u
u0v1œßœ
37. e e (2) 2;
`` `
`` `
zdzx 5 5 z 5
udxu 1x u 1(2)
uu
1eln v
œœ œ Ê œ œˆ‰ ¸ ’“ ’“

ab
u uln2v1œßœ
(1) 1
`` " "`
`` `
zdzx 5 5 z 5
vdxv 1xv v v 1(2) 1eln v
œœ œ Ê œ œˆ ‰ˆ‰ ˆ‰ ¸’“ ’“

ab
u uln2v1œßœ
38.
`" " "
``
`

zdz
udqu q 1u 1u tanu1u
q v3 v3
v3 tanu
œœ œ œŠ‹Š‹Š ‹Š‹
ÈÈ
È abab
#! ; Êœ œœœ¸ Š‹Š ‹
`"`"
```
`

z2zdztanu
utan111vdqvq
q
2v 3
u1v 2œßœ abab È

#!
1
œœÊœŠ‹Š‹ ¸
""`"
# ` #ÈÈv3 tanu 2v3
tan u z
(v 3) v
#!
#! u1v 2œßœ
39. V IR R and I; R I 0.01 volts/secœÊ œ œ œ œ Ê
`` ``
`` ``
V V dV V dI V dR dI dR
I R dt I dt R dt dt dt
(600 ohms) (0.04 amps)(0.5 ohms/sec) 0.00005 amps/secœ Êœ
dI dI
dt dt
40. V abc (bc) (ac) (ab)œÊœ œ
dV V da V db V dc da db dc
dt a dt b dt c dt dt dt dt
```
`` `
(2 m)(3 m)(1 m/sec) (1 m)(3 m)(1 m/sec) (1 m)(2 m)( 3 m/sec) 3 m /secÊœ œ¸
dV
dt
a1b2c3œßœßœ
$
and the volume is increasing; S 2ab 2ac 2bc œ Êœ
dS S da S db S dc
dt a dt b dt c dt
```
`` `
2(b c) 2(a c) 2(a b) œ Ê
da db dc dS
dt dt dt dt
¸
a1b2c3œßœßœ
2(5 m)(1 m/sec) 2(4 m)(1 m/sec) 2(3 m)( 3 m/sec) 0 m /sec and the surface area is not changing;œ œ
#
D a b c a b c œ Êœ œ Ê
È ˆ‰¸
### ``` "
`` `
dD D da D db D dc da db dc dD
dt a dt b dt c dt dt dt dt dt abc
È
a1b2c3œßœßœ
[(1 m)(1 m/sec) (2 m)(1 m/sec) (3 m)( 3 m/sec)] m/sec 0 the diagonals areœ œ ÊŠ‹
"
È È
14 m 14
6
decreasing in length
41. (1) (0) ( 1) ,
``````` ` ` ` ``
` `````` ` ` ` ``
ffufvfwf f f ff
x uxvxwx u v w uw
œ œ œ
( 1) (1) (0) , and
``````` ` ` ` ``
` `````` ` ` ` ``
ffufvfwf f f ff
y uyvywy u v w uv
œ œ œ
(0) ( 1) (1) 0
``````` ` ` ` `` ```
` `````` ` ` ` `` ```
ffufvfwf f f ff fff
z uzvzwz u v w vw xyz
œ œ œ Ê œ
42. (a) f f f cos f sin and f ( r sin ) f (r cos ) f sin f cos
`` ` "`
``` ` `
`wx w w
rrr r
xy x y x y x y
y
œ œ œ Êœ )) ) ) ) )
))
(b) sin f sin cos f sin and f sin cos f cos
``
``
##wcos w
rr
xy xy
))) ) )) )œ œˆ‰
)
)
f (sin ) ; then f cos (sin ) (sin ) f cos Êœ œ Ê yx x
wcos w w wcos w
rr r rr)))))
``` ` `
``` ` `
ˆ‰ ‘ˆ‰
))
))
sin 1 sin f (cos ) œ œ Êœ
`` ` ` ` ``
`` ` ` ` ``
##w w sin cos w w sin cos w w sin w
rrr rr rr
x
ab a b ˆ‰ ˆ‰ ˆ‰)) )
)) )) )
)))
(c) f cos andab a b ˆ‰ ˆ ‰ˆ ‰ ˆ‰ Š‹ x
w 2 sin cos w w sin w
rrrr
# # ````
````
##œ )
)) )
))


fsin ffab a b ab abˆ‰ ˆ ‰ˆ ‰ ˆ‰ ˆ‰ ˆ‰ Š‹ y xy
w 2 sin cos w w cos w w w
rrr r rr
# ### ```` `"`
```` ``
####œ Ê œ )
)) )
)) )

890. 886 Chapter 14 Partial Derivatives
43. w x y w x y xxx
wwuwv w w w w w
x ux vx u v u xu xvœœ œ Êœ
` `` `` ` ` ` `` ``
` `` `` ` ` ` `` ``
ˆ‰ ˆ‰
x y x x y y x yœ œ
` ```` ````` `` ``
`````` `````` ``` ```
w wuwv wuwvw ww ww
uuxvux uvxvxu uvu uvv
Š‹Š‹Š‹Š‹


x 2xy y ; w y x œ œœ œ
`` ` ` ````` ``
` ` `` ` ` `` `` ` `
##ww w w wwuwv ww
u u vu v y uy vy u v
y


wy x ʜ yy
wwuwv wuwv
uuyvuy uvyvy
````` ````
`````` `````Š‹Š‹


y y x x y x y 2xy x ; thusœ œ
``` ``````
` ` `` `` ` ` ` `` `
##www wwwwww
u u vu uv v u u vu v
Š‹Š‹


w w xy xy xy(w w)0, since w w 0xx yy uu vv uu vv
ww
uv œ œ œ œababab
## ## ##``
``


44. f (u)(1) g (v)(1) f (u) g (v) w f (u)(1) g (v)(1) f (u) g (v);
`
`
w w ww ww w wwww ww
x
œ œ Êœ œ
xx
f (u)(i) g (v)( i) w f (u) i g (v) i f (u) g (v) w w 0
`
`
w w ww # ww # ww www
y
œ Êœ œ Ê œ
yy xx yy ab ab
45. f (x y z) cos t, f (x y z) sin t, and f (x y z) t t 2 xy z
df f dx f f dz
dt x dt y dt z dt
dyßß œ ßß œ ßß œ Êœ
# ```
```
(cos t)( sin t) (sin t)(cos t) t t 2 (1) t t 2; 0 t t 2 0 t 2œ œ œÊ œÊœ ab
## # df
dt
or t 1; t 2 x cos ( 2), y sin ( 2), z 2 for the point (cos ( 2) sin ( 2) 2); t 1 x cos 1,œœ Êœ œ œ ßß œÊœ
y sin 1, z 1 for the point (cos 1 sin 1 1)œœ ßß
46. 2xe cos 3z ( sin t) 2x e cos 3z 3x e sin 3z (1)
dw wdx w wdz
dt x dt y dt z dt t
dy 2y 2y 2y
œ œ
``` "
``` #
##
aba ba b ˆ‰
2xe cos 3z sin t 3x e sin 3z; at the point on the curve z 0 t z 0œ œÊœœ
2y 2y 2x e cos 3z
t
2y
#
#
004Êœ œ¸
dw
dt
2(1) (4)(1)
Ðß ßÑ1ln20

#
47. (a) 8x 4y and 8y 4x (8x 4y)( sin t) (8y 4x)(cos t)
`` ``
`` ``
TTdTTdxT
x y dt x dt y dt
dy
œ œ Êœ œ
(8 cos t 4 sin t)( sin t) (8 sin t 4 cos t)(cos t) 4 sin t 4 cos t 16 sin t cos t;œ œ Êœ
## dT
dt


0 4 sin t 4 cos t 0 sin t cos t sin t cos t or sin t cos t t , , , on
dT 537
dt 4444
œÊ œÊ œ Ê œ œ Êœ
## ## 1111
the interval 0 t 2 ;ŸŸ 1
16 sin cos 0 T has a minimum at (x y) ;¹Š‹
dT
dt 4 4
22


tœ

4
œ "Êßœß
11
ÈÈ
##
16 sin cos 0 T has a maximum at (x y) ;¹Š‹
dT 3 3
dt 4 4
22


tœ
3
4
œ #Êßœ ß
11
ÈÈ
##
16 sin cos 0 T has a minimum at (x y) ;¹Š‹
dT 5 5
dt 4 4
22


tœ
5
4
œ "Êßœ ß
11
ÈÈ
##
16 sin cos 0 T has a maximum at (x y)¹Š‹
dT 7 7
dt 4 4
22


tœ
7
4
œ #Êßœß
11
ÈÈ
##
(b) T 4x 4xy 4y 8x 4y, and 8y 4x so the extreme values occur at the four pointsœ Êœ œ
## ``
``
TT
xy
found in part (a): T T 4 4 4 6, the maximum andŠ‹Š‹ ˆ‰ ˆ ‰ ˆ‰ßœ ß œ œ
ÈÈ È È
22 2 2
## # # # # #
"""
T T 4 4 4 2, the minimumŠ‹Š ‹ ˆ‰ ˆ‰ ˆ‰
ÈÈ È È
22 2 2
## ## ###
"""
ßœ ß œ œ
48. (a) y and x y 2 2 sin t x 2 cos t
`` ``
`` ``
TTdTTdxT
x y dt x dt y dt
dy
œœÊœ œŠ‹Š‹
ÈÈ
2 sin t 2 2 sin t 2 2 cos t 2 cos t 4 sin t 4 cos t 4 sin t 4 1 sin tœ œ œŠ‹Š ‹Š ‹Š ‹
ÈÈ ÈÈ
ab
## # #
4 8 sin t 16 sin t cost t; 0 4 8 sin t 0 sin t sin t t ,œ Êœ œÊ œÊ œÊ œ„ Êœ
### ""
#
dT dT
dt dt 42



È
1
, , on the interval 0 t 2 ;
357
444
111
ŸŸ 1
8 sin 2 8 T has a maximum at (x y) (2 1);¹ ˆ‰
dT
dt 4


tœ

4
œ œÊßœß
1
8 sin 2 8 T has a minimum at (x y) ( 2 1);¹ ˆ‰
dT 3
dt 4


tœ
3
4
œ œÊ ßœ ß
1

891. Section 14.5 Directional Derivatives and Gradient Vectors 887
8 sin 2 8 T has a maximum at (x y) ( 2 1);¹ ˆ‰
dT 5
dt 4


tœ
5
4
œ œÊßœ ß
1
8 sin 2 8 T has a minimum at (x y) (2 1)¹ ˆ‰
dT 7
dt 4


tœ
7
4
œ œÊ ßœß
1
(b) T xy 2 y and x so the extreme values occur at the four points found in part (a):œÊœ œ
``
``
TT
xy
T(2 1) T( 2 1) 0, the maximum and T( 2 1) T(2 1) 4, the minimumßœ ßœ ßœ ß œ
49. G(u x) g(t x) dt where u f(x) g(u x)f (x) g (t x) dt; thusßœ ß œ Ê œ œß ß ''
a a
u u
x
dG G du G dx
dx u dx x dx
``
``
w
F(x) t x dt F (x) x x (2x) t x dt 2x x x dtœ Êœ œ '''
0 00
x xx

ÈÈ Éab
È%$ # $ %$ )$ w % `
` x
3x
2t x


È
50. Using the result in Exercise 49, F(x) t x dt t x dt F (x)œ œ Ê ''
x1
1x


ÈÈ$# $# w
xx x dtxxx tx dtœœ ’“Éab
È È ## $#
$ # `
`
# '#
' '
1
x
x
1


x
x
tx
È

14.5 DIRECTIONAL DERIVATIVES AND GRADIENT VECTORS
1. 1, 1 f ; f(2 1) 1
``
``
ff
xy
œ œÊ œ ßœ™ ij
1 y x is the level curveʜ

2. ( ) 1;
```
` ` `
f2x f f
xxy x yxy
2y
œ Ê "ß " œ œ

( ) 1 f ; f(1 1) ln 2 ln 2Ê"ß"œÊ œ ßœ Ê
`
`
f
y
™ ij
ln x y 2 x y is the level curveœ Êœ ab
## ##

3. 2x ( 1 0) 2; 1
`` `
`` `
gg g
xx y
œÊ ßœ œ
g2 ; g(10) 1Êœ ßœ™ ij
1 y x is the level curveʜ
#

4. x 2 2; y
`` `
`` `
gg g
xx y
œÊ ß"œ œ Š‹
ÈÈ
21 g2;Êß"œ Êœ
`
`
g
y
Š‹
ÈÈ
™ ij
g 2 or 1 x y is the levelŠ‹
È
ß" œ Ê œ œ
""
####
##x y

curve

5. 2x (1 1 1) 3; 2y ( ) 2; 4z ln x ( ) 4;
````` `
````` `
fzf f f f f
xxxyyz z
œÊ ß ß œ œ Ê "ß "ß " œ œ Ê "ß"ß" œ
thus f 3 2 4™œijk

892. 888 Chapter 14 Partial Derivatives
6. 6xz (1 ) ; 6yz ( ) 6; 6z 3 x y
`````
``#`` `
###fzf11f f f x
xxz1x y y z xz1
œ Êß"ß"œ œ Ê "ß"ß" œ œ
ab
( ) ; thus f 6Ê "ß"ß" œ œ
`" "
`# ##
f11
z
™ ij k
7. ( 1 2 2) ; ( 1 2 2) ;
`"` `"`
````

fx f 26f f 23
xxx27yyy54
xyz xyz
y
œ Ê ßßœ œ Ê ßßœ
ab ab


( 1 2 2) ; thus f
`"`
``

fz f 23 262323
zzz54275454
xyz
œ Ê ßßœ œ
ab


™ ijk
8. e cos z 1; e cos z sin x 0 0 ;
````
``#``#
"

ffff
xx6yy6
xy xy y1
1x
33
œ Ê!ß!ßœ œ Êßßœ
È
ÈÈ
ˆ‰ ˆ‰
11
e sin z ; thus f
``" "
``####
ff
zz6
xy 32 3
œ Ê!ß!ßœ œ ˆ‰ Š‹
1
™
ÈÈ
ijk
9. ; f (x y) 2y f (5 5) 10; f (x y) 2x 6y f (5 5) 20uijœœ œ ßœ Ê ßœ ßœ Êßœ
v
v
ij
kk È
43
43
43
55
xxy y



f10 20 (Df) f 10 20 4Êœ Êœœ œ™™† ij u uP
43
55$
ˆ‰ ˆ‰
10. ; f (x y) 4x f ( 1 1) 4; f (x y) 2y f ( 1 1) 2uijœœ œ ßœ Ê ßœ ßœ Ê ßœ
v
v
ij
kk È
34
3(4)
34
55
xxyy



f42 (Df) f 4Êœ Êœœ œ™™† ij u uP
12 8
55$
11. ; g (x y) 1 g (1 1) 3; g (x y)uijœœ œ ßœ Êßœ ß
v
v
ij
kk ÈÈ
È
12 5 y
12 5
12 5
13 13 x
xxy
2y 3
2xy 4x y 1





g (1 1) 1 g 3 (D g) gœ Êßœ Êœ Êœœ œ
2y
x 13 13 13
2x 3
2xy 4x y 1
yP
36 5 31
È
È
$

™™† ij u u
12. ; h (x y) h (1 1) ;uijœœ œ ßœ Êßœ
v
v
ij
kk ÈÈÈ
Š‹
ˆ‰
ˆ‰È
Ê Š‹
32
3(2)
32
13 13
xx
1
3
1



"
#

#




y
x
y
x
y
xy
4
h (x y) h (1 1) h (D h) hyy P
1
3
1
33 36
213 213ßœ ÊßœÊ œ Êœœ
ˆ‰
ˆ‰
ˆ‰È
Ê Š‹
ÈÈ
!



$
x
y
x
x
xy
4


###
"
™™† ij u u
œ
3
213
È
13. ; f (x y z) y z f (1 1 2) 1; f (x y z) x zuijkœœ œ ßß œ Êß ßœ ßßœ
v
v
ijk
kk È
36
36(2)
362
777
xxy
#


f (1 1 2) 3; f (x y z) y x f (1 1 2) 0 f 3 (D f) f 3Êß ßœ ßßœ Êß ßœÊ œ Êœœ œ yz z P
318
77 ™™†ij u u $
14. ; f (x y z) 2x f (1 1 1) 2; f (x y z) 4yuijkœœ œ ßß œ Ê ßß œ ßß œ
v
v
ijk
kk ÈÈÈÈ

111
111
333
xxy

f(111) 4; f(xyz) 6z f(111) 6 f 2 4 6 (Df) fÊßßœ ßßœ Êßßœ Êœ Êœ yz z P ™™†ijk u u $
2460œ œŠ‹ Š‹ Š‹
"""
ÈÈÈ
333
15. ; g (x y z) 3e cos yz g (0 0 0) 3; g (x y z) 3ze sin yzuijkœœ œ ßß œ Ê ßß œ ßß œ
v
v
ij k
kk È
22
21(2)
212
333
xxy
xx


g (0 0 0) 0; g (x y z) 3ye sin yz g (0 0 0) 0 g 3 (D g) g 2Êßßœ ßßœ ÊßßœÊ œÊ œ œyz z P
x ™™† iu u $
16. ; h (x y z) y sin xy h 1 0 1;uijkœœ œ ßß œ Êßßœ
v
v
ijk
kk È


""
#
22
122
122
333 x
xx

ˆ‰
h (x y z) x sin xy ze h ; h (x y z) ye h 2 h 2yyzz
yz yz
zßß œ Ê"ß!ßœ ßßœ Ê"ß!ßœÊ œ ˆ‰ ˆ‰
"" " " "
## # #
™ ijk
(D h) h 2Êœœ œ uP
333
4 $
™†u
""
17. f (2x y) (x 2y) f( 1 1) ; f increases™™œ Ê ßœ Êœ œ œ ij iju ij
™
™
f
f (1) 1 22kk ÈÈÈ


""ij

most rapidly in the direction and decreases most rapidly in the direction ;uij uijœ œ
"" ""
ÈÈ ÈÈ
22 22
(D f) f f 2 and (D f) 2uuPP$ $
œœœ œ ™† ™ukk
ÈÈ

893. Section 14.5 Directional Derivatives and Gradient Vectors 889
18. f 2xy ye sin y x xe sin y e cos y f(1 0) 2 ; f increases most™™œ Êߜʜœaba b
xy xy xy f
f
ijjuj
# ™
™kk
rapidly in the direction and decreases most rapidly in the direction ; (D f) f fuj u j uœ œ œœ uP $™† ™ kk
2 and (D f) 2œœ
uP $19. f z y f(4 ) 5 ™™œ Êß"ß"œ Êœ œ
"
yy f
x f 5
1(5)(1)
ijk ijkuŠ‹

™
™kk È
ijk
; f increases most rapidly in the direction of and decreasesœ œ
"" ""
33 33 33 33 33 33
5 5
ÈÈÈ ÈÈÈ
ijk uijk
most rapidly in the direction ; (D f) f f 3 3 andœ œœœuijk u
""
33 33 33
5
PÈÈÈ u
$
™† ™ kk
È
(D f) 3 3uP œ
È
20. g e xe 2z g 1 ln 2 2 2 ;™™œ Êßßœ Êœ œ œ
yy g
g333
22
221
22
ijk ijku ijk ˆ‰
" "
#


™
™kk È
ijk

g increases most rapidly in the direction and decreases most rapidly in the directionuijkœ
22
333
"
; (D g) g g 3 and (D g) 3œ œœœ œ uijk u
22
333
PP
"
uu
$$
™† ™ kk
21. f f( ) 2 2 2 ;™™œ Ê "ß"ß" œ Êœ œ ˆ‰ ˆ‰ Š‹
"" "" "" " " "
xx yy zz f
f
333
ijk ijku ijk
™
™kk ÈÈÈ
f increases most rapidly in the direction and decreases most rapidly in the directionuijkœ
"""
ÈÈÈ
333
; (D f) f f 2 3 and (D f) 2 3œ œœœ œ uijk u
"""
ÈÈÈ
333
PP
™† ™ uu$ $ kk
ÈÈ
22. h 1 6 h( 0) 2 3 6 ™™œ Ê"ß"ßœ Êœ œŠ‹Š ‹
2x
xy1 xy1 h
2y 236h
236





ijk ijku
™
™kk È
ijk
; h increases most rapidly in the direction and decreases most rapidly in theœ œ
236 236
777 777
ijk uijk
direction ; (D h) h h 7 and (D h) 7œ œœœ œ uijk u
236
777
PPuu
$$
™† ™ kk
23. f 2x 2y f 2 2 2 2 2 2™™œ Êßœ ij i j Š‹
ÈÈ È È
Tangent line: 2 2 x 2 2 2 y 2 0Ê œ
ÈÈÈÈ
Š‹ Š‹
2x 2y 4Ê œ
ÈÈ

24. f 2x f 2 1 2 2™™œ Êßœ ij i j Š‹
ÈÈ
Tangent line: 2 2 x 2 (y 1) 0Ê œ
ÈÈ
Š‹
y 2 2x 3ʜ
È

25. f y x f(2 2) 2 2™™œÊß œij ij
Tangent line: 2(x 2) 2(y 2) 0Ê œ
yx4ʜ

894. 890 Chapter 14 Partial Derivatives
26. f (2x y) (2y x) f( 1 2) 4 5™™œ Ê ßœij ij
Tangent line: 4(x 1) 5(y 2) 0Ê œ
4x 5y 14 0Ê œ

27. f y (x 2y) f(3 2) 2 7 ; a vector orthogonal to f is 2 ™™ ™œœ Êßœ Êœ œi j ij v7ij u
v
v
ij
kk È
72
7(2)



and are the directions where the derivative is zeroœ œ
72 72
53 53 53 53
ÈÈ ÈÈ
iju ij
28. f f( ) ; a vector orthogonal to f is ™™ ™œ Ê"ß"œ œ
4xy 4x y
xy xy



ab ab
ij ij vij and are the directions where the derivative is zeroÊœ œ œ œuijuij
v
v
ij
kk ÈÈÈ ÈÈ

11 22 22
11 11

29. f (2x 3y) ( 3x 8y) f(1 2) 4 13 f(1 2) ( 4) (13) 185 ; no, the™™™œ Êßœ Êßœ œij ij kk È È
##
maximum rate of change is 185 14
È
#
30. T 2y (2x z) y T(1 1 1) 2 T(1 1 1) ( 2) 1 1 6 ; no, the™™ ™œ Êß ßœ Êß ßœ œijk ijk kk È È
###
minimum rate of change is 6 3
È
31. f f ( ) f ( ) and (D f)(1 2) f (1 2) f (1 2)™œ"ß# "ß # œ œ Êßœß ß
xy xyiju ij "


"" " "ij
ÈÈÈ È È
11 2 2 2 2
!
u Š‹ Š‹
2 2 f (1 2) f (1 2) 4; (D f)(1 2) f (1 2)(0) f (1 2)( 1) 3 f (1 2) 3œÊß ßœ œ Êßœß ßœÊßœ
È
xy x y y uj# u
f (1 2) 3; then f (1 2) 3 4 f (1 2) 1; thus f(1 2) 3 and Êßœ ß œÊ ßœ ßœ œœ
yx x ™ ij u
v
v
ij
kk È


2
(1) (2)

(D f) fœ Êœ œ œ
12 6 7
55 55 5
ÈÈ ÈÈ È
ij u
uP ™†
"
32. (a) (D f) 2 3 f 2 3; ; thus
uPœÊ œ œœ œ œ
ÈÈ
kk™ uijku
v
v
ijk
kk k kÈÈÈÈ


"
11(1)
11
333
f
f

™
™
ff f23 222Êœ Êœ œ™™ ™kk
È
Š‹uijkijk
"""
ÈÈÈ
333
(b) (D f) f 2 2 2(0) 2 2vij u i j uœÊœ œ œ Êœ œ œ
v
v
ij
kk ÈÈÈ ÈÈ


"" " "
11 2 2 2 2

uP ™† Š‹ Š‹
È
33. The directional derivative is the scalar component. With f evaluated at P , the scalar component of f in™™
!
the direction of is f (D f) .uu™† œ
uP $
34. D f f (f f f ) f ; similarly, D f f f and D f f f
i jkœœ œœœœœ™† † ™† ™†iijki j k
xy z x y z
35. If (x y) is a point on the line, then (x y) (x x ) (y y ) is a vector parallel to the line 0ßßœ ÊœTij TN !! †
A(x x ) B(y y ) 0, as claimed.Ê œ
!!
36. (a) (kf) k k k k k f™ ™œ œ œ œ
```
`` ` ` ` ` ```
``` ```(kf) (kf) (kf)
xy z x y z xyz
fff fff
ijk i j k ijk ˆ‰ ˆ‰Š‹ Š ‹
(b) (f g)™œ œ
` ` ` ` ` `
`` ` ``````
```(f g) (f g) (f g) g g g
xy z xxyyzz
fff
ijk i j k Š‹Š‹Š‹
fgœ œ œ
`` ` ```
````` ` ``` ```
`` ` ```ff f fff
xxyyz z xyz xyz
gg g ggg
iijjkk ijk ijk Š‹Š‹ ™™
(c) (f g) f g (Substitute g for g in part (b) above)™™™œ

895. Section 14.6 Tangent Planes and Differentials 891
(d) (fg) g f g f g f™ œ œ
``` ` ` `
`` ` `` `` ``
```(fg) (fg) (fg) g g g
xy z xx yy zz
fff
ijk i j k Š‹Š‹Š‹
gf g fg fœ ˆ‰ ˆ‰Š‹Š‹Š‹ Š‹
`` `
`` ` `` `
```ff f
xx y yz z
ggg
iijjkk
fgfggfœ œ Š‹Š‹
```
`` ` ```
```ggg
xy z xyz
fff
ijk ijk ™™
(e) ™Š‹ Š ‹ Š ‹ Œ "
f
gx y z g g g
gf gf gf
œ œ
```
`` `
Š‹ Š‹ Š‹
ff f
gg g
ijk i j k
``
`` ``
``

`
``
`ff
xx zz
gg f
yy
g
œ œ Œ "Œ "
ggg fff
gggg
gf
```
``` ```

`` ` ```
``` `` `
`` `fff
xyz xyz
gg g fff
xyz xy z
gg g
ijk ijk ijk ijk Š‹Š‹
œ œ
gf fg gffg
gg g
™™ ™™


14.6 TANTGENT PLANES AND DIFFERENTIALS
1. (a) f 2x 2y 2z f(1 1 1) 2 2 2 Tangent plane: 2(x 1) 2(y 1) 2(z 1) 0™™œ Êßßœ Ê œijk ijk
xyz3;Ê œ
(b) Normal line: x 1 2t, y 1 2t, z 1 2tœ œ œ
2. (a) f 2x 2y 2z f(3 5 4) 6 10 8 Tangent plane: 6(x 3) 10(y 5) 8(z 4) 0™™œ Êßß œ Ê œijk ijk
3x 5y 4z 18;Ê œ
(b) Normal line: x 3 6t, y 5 10t, z 4 8tœ œ œ
3. (a) f 2x 2 f(2 0 2) 4 2 Tangent plane: 4(x 2) 2(z 2) 0™™œ Êßßœ Ê œik ik
4x 2z 4 0 2x z 2 0;Ê œÊ œ
(b) Normal line: x 2 4t, y 0, z 2 2tœ œœ
4. (a) f (2x 2y) (2x 2y) 2z f(1 1 3) 4 6 Tangent plane: 4(y 1) 6(z 3) 0™™œ Êß ßœ Ê œijk jk
2y 3z 7;Ê œ
(b) Normal line: x 1, y 1 4t, z 3 6tœœ œ
5. (a) f sin x 2xy ze x z xe y f(0 1 2) 2 2 Tangent plane:™™œ Êßßœ Êababab11
xz xz
ijk ijk
#
2(x 0) 2(y 1) 1(z 2) 0 2x 2y z 4 0; œÊ œ
(b) Normal line: x 2t, y 1 2t, z 2 tœœ œ
6. (a) f (2x y) (x 2y) f(1 1 1) 3 Tangent plane:™™œ Êßß œ Êijk ijk
1(x 1) 3(y 1) 1(z 1) 0 x 3y z 1; œÊ œ
(b) Normal line: x 1 t, y 1 3t, z 1 tœ œ œ
7. (a) f for all points f(0 1 0) Tangent plane: 1(x 0) 1(y 1) 1(z 0) 0™™œ Êßßœ Ê œijk ijk
xyz10;Ê œ
(b) Normal line: x t, y 1 t, z tœœ œ
8. (a) f (2x 2y 1) (2y 2x 3) f(2 3 18) 9 7 Tangent plane:™™œ Êß ßœ Êijk ijk
9(x 2) 7(y 3) 1(z 18) 0 9x 7y z 21; œÊ œ
(b) Normal line: x 2 9t, y 3 7t, z 18 tœ œ œ
9. z f(x y) ln x y f (x y) and f (x y) f (1 0) 2 and f (1 0) 0 fromœßœ Ê ßœ ßœ Ê ßœ ßœÊab
##

xy xy
2x
xy xy
2y

Eq. (4) the tangent plane at (1 0 0) is 2(x 1) z 0 or 2x z 2 0ßß œ œ

896. 892 Chapter 14 Partial Derivatives
10. z f(x y) e f (x y) 2xe and f (x y) 2ye f (0 0) 0 and f ( ) 0œßœ Ê ßœ ßœ Êßœ !ß!œ
ab ab abxy xy xy
xy xy

from Eq. (4) the tangent plane at (0 0 1) is z 1 0 or z 1Êßß œœ
11. z f( y) y x f (x y) (y x) and f (x y) (y x) f (1 2) and f ( )œBß œ Êßœ ßœ Êßœ "ß # œÈ xyxy
"" ""
## ##
"Î# "Î#
from Eq. (4) the tangent plane at (1 2 1) is (x 1) (y 2) (z 1) 0 x y 2z 1 0Êßß œÊ œ
""
##
12. z f( y) 4x y f (x y) 8x and f (x y) y f (1 1) 8 and f ( 1) from Eq. (4) theœBß œ Êßœ ßœ#Êßœ "ßœ#Ê
##
xy xy
tangent plane at (1 1 5) is 8(x 1) 2(y 1) (z 5) 0 or 8x 2y z 5 0ßß œ œ
13. f 2y 2 f(1 1 1) 2 2 and g for all points; f g™™ ™ ™™œ Êßßœ œœ‚ijk ijk i v
2 2 Tangent line: x 1, y 1 2t, z 1 2t22
00
Êœ œ Êœœ œ"
"
vjk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
14. f yz xz xy f(1 1 1) ; g 2x 4y 6z g(1 1 1) 2 4 6 ;™™™ ™œ Êßßœ œ Êßßœ ijk ijk ijk ijk
f g 2 4 2 Tangent line: x 1 2t, y 1 4t, z 1 2t11
246
Êœ ‚ Ê œ Êœ œ œ"vijk
ijk
™™
ââ
ââ
ââ
ââ
ââ
ââ
15. f 2x 2 2 f 1 1 2 2 2 and g for all points; f g™™ ™ ™™œ Êßßœ œœ‚ijk ijk j v ˆ‰
"
#
2 2 Tangent line: x 1 2t, y 1, z 2t222
010
Êœ œ Êœ œœ vik
ijk
ââ
ââ
ââ
ââ
ââ
ââ
"
#
16. f 2y f 1 2 and g for all points; f g™™ ™ ™™œ Êßßœ œœ‚ijk ijk j v ˆ‰
""
##
Tangent line: x t, y 1, z t121
010
Êœ œ Êœ œœ vik
ijk
ââ
ââ
ââ
ââ
ââ
ââ
""
##
17. f 3x 6xy 4y 6x y 3y 4x 2z f(1 1 3) 13 13 6 ; g 2x 2y 2z™™™œ Êßßœ œ abab
## ##
ijk ijkijk
g( ) 2 2 6 ; f g 90 90 Tangent line: 313 6
22 6
Ê "ß"ß$ œ œ‚ Êœ œ Ê" ™™™ ijkv v i j
ijk
ââ
ââ
ââ
ââ
ââ
ââ
x 1 90t, y 1 90t, z 3œ œ œ
18. f 2x 2y f 2 2 4 2 2 2 2 ; g 2x 2y g 2 2 4™™ ™ ™œ Êßßœ œ Êßßij i j ijkŠ‹ Š‹
ÈÈ È È ÈÈ
22 22 ; f g 22 22 Tangent line: 2222 0
2222 1
œ œ‚ Êœ œ Ê

ÈÈ È È
ââ
ââ
ââ
ââ
ââ
ââ
ÈÈ
ÈÈ
ijkv v i j
ijk
™™
x222t, y222t, z4œ œ œ
ÈÈ ÈÈ
19. f f(3 4 12) ;™™œ Êßßœ Š‹Š‹Š‹
xz3412
xyz xyz xyz 169 169 169
y


ijk ijk
f and df ( f ) ds (0.1) 0.0008uijkuuœœ œ Êœ œ œ ¸
v
v
ijk
kk È
362
36(2)
362 9 9
7 7 7 1183 1183



™† ™† ˆ‰
20. f e cos yz ze sin yz ye sin yz f(0 0 0) ; ™™œ Êßßœœœaba ba b
xxx 222
22(2)
ijk iu
v
v
ijk
kk È


f and df ( f ) ds (0.1) 0.0577œ Êœ œ œ¸
111 1 1
333 3 3
ÈÈÈ È È
ijk u u ™† ™†

897. Section 14.6 Tangent Planes and Differentials 893
21. g (1 cos z) (1 sin z) ( x sin z y cos z) g(2 1 0) 2 ; P P 2 2 2™™œ Êß ßœ œœ
Ä
ij k ijkA ijk !"
g 0 and dg ( g ) ds (0)(0.2) 0Êœ œ œ Êœœ œœuijkuu
v
v
ijk
kk ÈÈÈÈ


222
(2) 2 2
111
333

™† ™†
22. h y sin ( xy) z x sin ( xy) 2xz h( 1 1 1) ( sin 1) ( sin ) 2™™œ Ê ßßœ cdcd11 11 11 11
#
ijk ijk
2 ; P P where P ( ) œ œœ œ !ß!ß! Ê œ œ œ
Ä
ikv ijk u i j k!" "


v
v
ijk
kk ÈÈÈÈ
111
111
333

h 3 and dh ( h ) ds 3(0.1) 0.1732Êœœ œ œ¸™† ™†uu
3
3
ÈÈÈ
23. (a) The unit tangent vector at in the direction of motion is ;Š‹
""
## # #
ߜ
ÈÈ
33
uij
T (sin 2y) (2x cos 2y) T sin 3 cos 3 D T T™™ ™†œ Êßœ Êßœij ij u Š‹Š ‹Š ‹ Š‹
ÈÈ
""
## ##
ÈÈ
33
u
sin 3 cos 3 0.935° C/ftœ ¸
È
3
##
"
ÈÈ
(b) (t) (sin 2t) (cos 2t) (t) (2 cos 2t) (2 sin 2t) and 2; rijv ijvœ Êœ œœ kk
dT T dx T
dt x dt y dt
dy``
``
T T (D T) , where ; at we have from part (a)œœ œ œ ß œ ™† ™†vvvu uijŠ‹ Š‹ kk kk
vv
vvkk kk
ÈÈ
u
""
## # #
33
sin 3 cos 3 2 3 sin 3 cos 3 1.87 ° C/secÊœ œ ¸
dT
dt
3
Š‹
ÈÈÈÈÈ
È
##
"
†
24. (a) T (4x yz) xz xy T(8 6 4) 56 32 48 ; (t) 2t 3t t the particle is™™œ Êßß œ œ Êijk ijkr ijk
##
at the point P( 6 4) when t 2; (t) 4t 3 2t (2) 8 3 4 )ß ß œœ Êœ Êœvijkvijku
v
vkk
D T(8 6 4) T [56 8 32 3 48 ( 4)] ° C/mœ Êßß œœ œ
834 736
89 89 89 89 89
ÈÈÈ È È
ijk u
u ™† †††
"
(b) T ( T ) at t 2, D T (2) 89 736 °C/sec
dT T dx T dT 736
dt x dt y dt dt
dy
89
œ œœ Êœœ œ œ
``
``
™† ™†vuv v kk ¸ Š‹
È
u
t2œ È
25. (a) f( 0) 1, f (x y) 2x f (0 0) 0, f (x y) 2y f (0 0) 0 L(x y) 1 0(x 0) 0(y 0) 1!ßœ ßœ Ê ßœ ßœ Ê ßœÊ ßœ œ xxyy
(b) f(1 1) 3, f (1 1) 2, f (1 1) 2 L(x y) 3 2(x 1) 2(y 1) 2x 2y 1ßœ ßœ ßœÊ ßœ œ xy
26. (a) f( 0) 4, f (x y) 2(x y 2) f (0 0) 4, f (x y) 2(x y 2) f (0 0) 4!ß œ ß œ Êßœ ßœ Êßœ xxyy
L(x y) 4 4(x 0) 4(y 0) 4x 4y 4Êßœ œ
(b) f(1 2) 25, f (1 2) 10, f (1 2) 10 L(x y) 25 10(x 1) 10(y 2) 10x 10y 5ßœ ßœ ßœ Ê ßœ œ
xy
27. (a) f(0 0) 5, f (x y) 3 for all (x y), f (x y) 4 for all (x y) L(x y) 5 3(x 0) 4(y 0)ßœ ßœ ß ßœ ßÊ ßœ xy
3x 4y 5œ
(b) f(1 1) 4, f (1 1) 3, f (1 1) 4 L(x y) 4 3(x 1) 4(y 1) 3x 4y 5ßœ ßœ ßœ Êßœ œ
xy
28. (a) f(1 1) 1, f (x y) 3x y f (1 1) 3, f (x y) 4x y f (1 1) 4ßœ ßœ Ê ßœ ßœ Ê ßœxxyy
#% $$
L(x y) 1 3(x 1) 4(y 1) 3x 4y 6Êßœ œ
(b) f(0 0) 0, f ( 0) 0, f (0 0) 0 L(x y) 0ßœ !ßœ ßœÊ ßœ
xy
29. (a) f(0 0) 1, f (x y) e cos y f (0 0) 1, f (x y) e sin y f (0 0) 0ßœ ßœ Ê ßœ ßœ Êßœ xxy y
xx
L(x y) 1 1(x 0) 0(y 0) x 1Êßœ œ
(b) f 0 0, f 0 0, f 0 1 L(x y) 0 0(x 0) 1 y yˆ‰ ˆ‰ ˆ‰ ˆ ‰ßœ ßœ ßœ Êßœ œ
111 11
### ##
xy
30. (a) f(0 0) 1, f (x y) e f ( ) 1, f (x y) 2e f (0 0) 2ßœ ßœ Ê!ß!œ ßœ Ê ßœ xxyy
2y x 2y x
L(xy)11(x0)2(y0) x2y1Êßœ œ
(b) f(12) e, f(12) e, f(12) 2e L(xy) e e(x 1) 2e(y 2)ßœ ßœ ßœ Ê ßœ
$$$ $$$
xy
e x 2e y 2eœ
$$$

898. 894 Chapter 14 Partial Derivatives
31. f(2 1) 3, f (x y) 2x 3y f (2 1) 1, f (x y) 3x f (2 1) 6 L(x y) 3 1(x 2) 6(y 1)ßœ ßœ Êßœ ßœ Êßœ Êßœ xxyy
7 x 6y; f (x y) 2, f (x y) 0, f (x y) 3 M 3; thus E(x y) (3) x 2 y 1œ ßœ ßœ ßœ Êœ ߟ xx yy xy kk a bˆ‰ kkkk
"
#
#
(0.1 0.1) 0.06Ÿ œˆ‰
3
#
#
32. f(2 2) 11, f (x y) x y 3 f (2 2) 7, f (x y) x 3 f (2 2) 0ßœ ßœ Êßœ ßœ Êßœ xxy y
y
#
L(x y) 11 7(x 2) 0(y 2) 7x 3; f (x y) 1, f (x y) , f (x y) 1Êßœ œ ßœ ßœ ßœ xx yy xy
"
#
M 1; thus E(x y) (1) x 2 y 2 (0.1 0.1) 0.02Êœ ߟ Ÿ œkk a bˆ‰ ˆ‰kkkk
"
##
# #1
33. f(0 0) 1, f (x y) cos y f (0 0) 1, f (x y) 1 x sin y f (0 0) 1ßœ ßœ Ê ßœ ßœ Êßœ xxy y
L(x y) 1 1(x 0) 1(y 0) x y 1; f (x y) 0, f (x y) x cos y, f (x y) sin y 1;Êßœ œ ßœ ßœ ßœ ÊQœ xx yy xy
thus E(x y) (1) x y (0.2 0.2) 0.08kk a bˆ‰ ˆ‰kk kkߟ Ÿ œ
"
##
# #1
34. f( ) 6, f (x y) y y sin (x 1) f (1 2) 4, f (x y) 2xy cos (x 1) f (1 2) 5"ß # œ ß œ Êßœ ßœ Êßœ xxyy
#
L(x y) 6 4(x 1) 5(y 2) 4x 5y 8; f (x y) y cos (x 1), f (x y) 2x,Êßœ œ ßœ ßœ xx yy
f (x y) 2y sin (x 1); x 1 0.1 0.9 x 1.1 and y 2 0.1 1.9 y 2.1; thus the max ofxyßœ ŸÊŸŸ ŸÊŸŸkk kk
f (x y) on R is 2.1, the max of f (x y) on R is 2.2, and the max of f (x y) on R is 2(2.1) sin (0.9 1)kk kk kk
xx yy xyßß ß
4.3 M 4.3; thus E(x y) (4.3) x 1 y 2 (2.15)(0.1 0.1) 0.086ŸÊœ ߟ Ÿ œkk a bˆ‰ kkkk
"
#
# #
35. f( 0) 1, f (x y) e cos y f (0 0) 1, f (x y) e sin y f (0 0) 0!ßœ ßœ Ê ßœ ßœ Êßœ xxy y
xx
L(x y) 1 1(x 0) 0(y 0) 1 x; f (x y) e cos y, f (x y) e cos y, f (x y) e sin y;Êßœ œ ßœ ßœ ßœ xx yy xy
xxx
x 0.1 0.1 x 0.1 and y 0.1 0.1 y 0.1; thus the max of f (x y) on R is e cos (0.1)kk kk k kŸÊ ŸŸ Ÿ Ê ŸŸ ß xx
01Þ
1.11, the max of f (x y) on R is e cos (0.1) 1.11, and the max of f (x y) on R is e sin (0.1)Ÿß Ÿ ß kk kk yy xy
01 01ÞÞ
0.12 M 1.11; thus E(x y) (1.11) x y (0.555)(0.1 0.1) 0.0222ŸÊœ ߟ Ÿ œkk a bˆ‰ kk kk
"
#
# #
36. f(1 1) 0, f (x y) f (1 1) 1, f (x y) f (1 1) 1 L(x y) 0 1(x 1) 1(y 1)ßœ ßœÊ ßœ ßœÊ ßœÊ ßœ xxyy
xy
""
x y 2; f (x y) , f (x y) , f (x y) 0; x 1 0.2 0.98 x 1.2 so the max ofœ ßœ ßœ ßœ ŸÊ ŸŸ xx yy xy
xy
"" kk
f (x y) on R is 1.04; y 1 0.2 0.98 y 1.2 so the max of f (x y) on R iskk kk kkxx yy
(0.98)ߟ ŸÊ ŸŸ ß
"

1.04 M 1.04; thus E(x y) (1.04) x 1 y 1 (0.52)(0.2 0.2) 0.0832
""
#
# #
(0.98)
ŸÊœ ߟ Ÿ œkk a bˆ‰ kkkk
37. (a) f( ) 3, f (1 1 1) y z 2, f (1 1 1) x z 2, f (1 1 1) y x 2"ß "ß " œ ß ß œ œßßœ œßßœ œkkk
xyz 111 111111ÐßßÑ ÐßßÑ ÐßßÑ
L(x y z) 3 2(x 1) 2(y 1) 2(z 1) 2x 2y 2z 3Êßßœ œ
(b) f(100) 0, f(100) 0, f(100) 1, f(100) 1 L(xyz) 0 0(x 1) (y 0) (z 0)ßß œ ßß œ ßß œ ßß œ Ê ßß œ
xyz
yz œ
(c) f(000) 0, f(000) 0, f(000) 0, f(000) 0 L(xyz) 0ßß œ ßß œ ßß œ ßß œ Ê ßß œ
xyz
38. (a) f(1 1 1) 3, f (1 1 1) 2x 2, f (1 1 1) 2y 2, f (1 1 1) 2z 2ßßœ ßßœ œ ßßœ œ ßßœ œkkk xyz Ð"ß"ß"Ñ Ð"ß"ß"Ñ Ð"ß"ß"Ñ
L(x y z) 3 2(x 1) 2(y 1) 2(z 1) 2x 2y 2z 3Êßßœ œ
(b) f(0 1 0) 1, f (0 1 0) 0, f ( 1 0) 2, f (0 1 0) 0 L(x y z) 1 0(x 0) 2(y 1) 0(z 0)ßß œ ßß œ !ßß œ ßß œ Ê ßß œ
xyz
2y 1œ
(c) f(1 0 0) 1, f (1 0 0) 2, f (1 0 0) 0, f (1 0 0) 0 L(x y z) 1 2(x 1) 0(y 0) 0(z 0)ßß œ ßß œ ßß œ ßß œ Ê ßß œ
xyz
2x 1œ
39. (a) f(1 0 0) 1, f (1 0 0) 1, f (1 0 0) 0,ßßœßßœ œßßœ œ xy
x
xyz xyz
y ¹¹
ÈÈ

100 100"``# "``#
f (1 0 0) 0 L(x y z) 1 1(x 1) 0(y 0) 0(z 0) xz
z
xyzßß œ œ Ê ßß œ œ¹
È

100"``#

899. Section 14.6 Tangent Planes and Differentials 895
(b) f(110) 2, f(110) , f(110) , f(110) 0ßß œ ßß œ ßß œ ßß œ
È xyz
22
""
ÈÈ
L(x y z) 2 (x 1) (y 1) 0(z 0) x yÊßßœ œ
È
"" ""
ÈÈ ÈÈ
22 22
(c) f(122) 3, f(122) , f(122) , f(122) L(xyz) 3 (x 1) (y 2) (z 2)ßß œ ßß œ ßß œ ßß œ Ê ßß œ xyz
3 3 3 333
22 22""
xyzœ
"
333
22
40. (a) f 1 1 1, f 1 1 0, f 1 1 0,ˆ‰ ˆ‰ ¸ ˆ‰ ¸
11 1
2z z
xy
y cos xy x cos xy
ßßœßßœ œßßœ œ
## ß"ß" ß"ß"ˆ‰ ˆ‰


f 1 1 1 L(x y z) 1 0 x 0(y 1) 1(z 1) 2 zz
sin xy
z
ˆ‰ ˆ‰ ¹
1 1
# #

ß"ß"
ßß œ œ Êßßœ œ


ˆ‰

(b) f(2 0 1) 0, f (2 0 1) 0, f (2 0 1) 2, f (2 0 1) 0 L(x y z) 0 0(x 2) 2(y 0) 0(z 1) 2yßß œ ßß œ ßß œ ßß œ Ê ßß œ œ xyz
41. (a) f(0 0 0) 2, f (0 0 0) e 1, f (0 0 0) sin (y z) 0,ßßœßßœ œßßœ œ xy
x kk
Ð!ß!ß!Ñ Ð!ß!ß!Ñ
f (0 0 0) sin (y z) 0 L(x y z) 2 1(x 0) 0(y 0) 0(z 0) 2 xzßß œ œÊ ßßœ œk
Ð!ß!ß!Ñ
(b) f 0 0 1, f 0 0 1, f 0 0 1, f 0 0 1 L(x y z)ˆ‰ ˆ‰ ˆ‰ ˆ‰ßß œ ßß œ ßß œ ßß œÊßß
111 1
### #
xy z
11(x0)1y 1(z0)xyz 1œ œ ˆ‰
11
2 #
(c) f 0 1, f 0 1, f 0 1, f 0 1 L(x y z)ˆ‰ ˆ‰ ˆ‰ ˆ‰ßß œ ßß œ ßß œ ßß œÊßß
11 11 11 11
44 44 44 44
xy z
11(x0)1y 1z xyz 1œ œ ˆ‰ˆ‰
11 1
44 #
42. (a) f(1 0 0) 0, f (1 0 0) 0, f (1 0 0) 0,ßßœßßœ œßßœ œ xy
yz
(xyz) 1 (xyz) 1
xz ¹¹

Ð"ß!ß!Ñ Ð"ß!ß!Ñ
f(100) 0 L(xyz) 0z
xy
(xyz) 1ßß œ œ Ê ßß œ¹


Ð"ß!ß!Ñ
(b) f(1 1 0) 0, f (1 1 0) 0, f (1 1 0) 0, f (1 1 0) 1 L(x y z) 0 0(x 1) 0(y 1) 1(z 0) zßß œ ßß œ ßß œ ßß œ Ê ßß œ œ xyz
(c) f(111) , f(111) , f(111) , f(111) L(xyz) (x 1) (y 1) (z 1)ßß œ ßß œ ßß œ ßß œ Ê ßß œ
1 1
4 4
xyz
" " " """
# # # ###
xyzœ
"""
### #
1
4
3
43. f(x y z) xz 3yz 2 at P (1 1 2) f(1 1 2) 2; f z, f 3z, f x 3y L(x y z)ßß œ ßß Ê ßß œ œœ œ Êßß ! xy z
2 2(x 1) 6(y 1) 2(z 2) 2x 6y 2z 6; f 0, f 0, f 0, f 0, f 3œ œ œœœœœ xx yy zz xy yz
M 3; thus, E(x y z) (3)(0.01 0.01 0.02) 0.0024Êœ ßߟ œkk ˆ‰
"
#
#
44. f(x y z) x xy yz z at P (1 1 2) f(1 1 2) 5; f 2x y, f x z, f y zßß œ ßß Ê ßß œ œ œ œ
## " "
!
#4
xyz
L(x y z) 5 3(x 1) 3(y 1) 2(z 2) 3x 3y 2z 5; f 2, f 0, f , f 1, f 0,Êßßœ œ œœœœœ xx yy zz xy xz
"
#
f 1 M 2; thus E(x y z) (2)(0.01 0.01 0.08) 0.01yzœÊ œ ßß Ÿ œkk ˆ‰
"
#
#
45. f(x y z) xy 2yz 3xz at P (1 1 0) f(1 1 0) 1; f y 3z, f x 2z, f 2y 3xßß œ ßß Ê ßß œ œ œ œ ! xyz
L(x y z) 1 (x 1) (y 1) (z 0) x y z 1; f 0, f 0, f 0, f 1, f 3,Êßßœ œ œœœœœ xx yy zz xy xz
f 2 M 3; thus E(x y z) (3)(0.01 0.01 0.01) 0.00135yzœÊ œ ßß Ÿ œkk ˆ‰
"
#
#
46. f(x y z) 2 cos x sin (y z) at P 0 0 f 0 0 1; f 2 sin x sin (y z),ßß œ ßß Ê ßß œ œ
ÈÈ ˆ‰ˆ‰ !
11
44
x
f 2 cos x cos (y z), f 2 cos x cos (y z) L(x y z) 1 0(x 0) (y 0) zyz
4œ œ Êßßœ
ÈÈ ˆ‰
1
y z 1; f 2 cos x sin (y z), f 2 cos x sin (y z), f 2 cos x sin (y z),œ œ œ œ
1
4
xx yy zz ÈÈÈ
f 2 sin x cos (y z), f 2 sin x cos (y z), f 2 cos x sin (y z). The absolute value ofxy xz yzœ œ œ
ÈÈÈ
each of these second partial derivatives is bounded above by 2 M 2; thus E(x y z)
ÈÈ
kkÊœ ßß
2 (0.01 0.01 0.01) 0.000636.Ÿ œˆ‰Š‹
È
"
#
#

900. 896 Chapter 14 Partial Derivatives
47. T (x y) e e and T (x y) x e e dT T (x y) dx T (x y) dyxy xy
yy yyßœ ßœ Êœ ß ß

ab
e e dx x e e dy dT 2.5 dx 3.0 dy. If dx 0.1 and dy 0.02, then theœ Êœ ŸŸabab k kk kk
yy yy
ln 2

Ð#ß Ñ
maximum possible error in the computed value of T is (2.5)(0.1) (3.0)(0.02) 0.31 in magnitude. œ
48. V 2 rh and V r dV V dr V dh dr dh; now 100 1 andrh rh
dV 2 rh dr r dh 2 dr
Vrhrh rœœÊœ Êœ œ Ÿ11
# "11
1


¸¸† 100 1 100 2 (100) (100) 2 100 100 2(1) 1 3 3%¸¸ ¸ ¸¸ ¸¸¸¸¸ ˆ‰ ˆ‰
dh dV dr dh dr dh
hVrhrh
†† ††ŸÊ Ÿ Ÿ Ÿ œÊ
49. V 2 rh and V r dV V dr V dh dV 2 rh dr r dh dV 120 dr 25 dh;
rh rh 512œœÊœ Êœ Êœ 11 11 11
##
Ðß Ñ
k
dr 0.1 cm and dh 0.1 cm dV (120 )(0.1) (25 )(0.1) 14.5 cm ; V(5 12) 300 cmkk k kŸŸÊŸ œßœ11 1 1
$$
maximum percentage error is 100 4.83%Ê„‚œ„
14.5
300
1
1
50. (a) dR dR dR dR dR dR
""" " " "
"# " #
##
RR R R R R RR
RR
œ Ê œ Êœ



Š‹ Š‹
(b) dR R dR dR dR R dR dR R will be moreœ Êœ Ê
##"" ""
"# "# ÐÑ
’“’“Š‹ Š‹ k
RR (100) (400)



100 400`
sensitive to a variation in R since "
""
(100) (400)"
(c) From part (a), dR dR dR so that R changing from 20 to 20.1 ohms dR 0.1 ohmœ ÊœŠ‹ Š‹
RR
RR

##
"#" "
and R changing from 25 to 24.9 ohms dR 0.1 ohms; R ohms##
""" Êœ œ Êœ
RR R 9
100
dR (0.1) ( 0.1) 0.011 ohms percentage change is 100Êœ ¸Ê ‚k ¸
ÐÑ Ð20 25 20 25` `#
ˆ‰ ˆ‰
100 100
99


(20) (25) R
dR
100 0.1%œ‚¸
0.011
ˆ‰
100
9
51. A xy dA x dy y dx; if x y then a 1-unit change in y gives a greater change in dA than a 1-unitœÊ œ
change in x. Thus, pay more attention to y which is the smaller of the two dimensions.
52. (a) f (x y) 2x(y 1) f (1 0) 2 and f (x y) x f (1 0) 1 df 2 dx 1 dy df is more
xxyyßœ Êßœ ßœÊߜʜ Ê
#
sensitive to changes in x
(b) df 0 2 dx dy 0 2 1 0 œÊ œÊ œÊ œ
dx dx
dy dy
"
#
53. (a) r x y 2r dr 2x dx 2y dy dr dx dy dr| 0.01 0.01
###
Ð$ß%Ñ
œÊœ Êœ Êœ„ „
x34
rr 5 5
y
ˆ‰ ˆ‰abab
0.014 100 100 0.28%; d dx dyœ„ œ„ Ê ‚ œ „ ‚ œ œ
0.07 dr 0.014
5r5
11
¸¸¸ ¸ )
Š‹ Š‹
ˆ‰ ˆ‰


y
x
yy
xx
x



dx dy d | 0.01 0.01œ Êœ„ „œ

#Ð$ß%Ñ
„ …y
y x y x 25 25 25 5
x43 0.03 0.04
) ˆ‰ ˆ‰ab ab
maximum change in d occurs when dx and dy have opposite signs (dx 0.01 and dy 0.01 or viceÊœœ )
versa) d 0.0028; tan 0.927255218 100 100Êœ ¸„ œ ¸ Ê ‚ œ ‚))
„ „
#
"0.07 4 d 0.0028
5 3 0.927255218
ˆ‰ ¸ ¸ ¸ ¸
)
)
0.30%¸
(b) the radius r is more sensitive to changes in y, and the angle is more sensitive to changes in x)
54. (a) V r h dV 2 rh dr r dh at r 1 and h 5 we have dV 10 dr dh the volume isœÊœ Êœ œ œ Ê111 11
##
about 10 times more sensitive to a change in r
(b) dV 0 0 2 rh dr r dh 2h dr r dh 10 dr dh dr dh; choose dh 1.5œÊœ œ œ Êœ œ11
# "
10
dr 0.15 h 6.5 in. and r 0.85 in. is one solution for V dV 0Êœ Êœ œ ¸ œ ?
55. f(a b c d) ad bc f d, f c, f b, f a df d da c db b dc a dd; since
ab
cd
ßßß œ œ Êœ œ œ œÊ œ ºº
ab c d
a is much greater than b , c , and d , the function f is most sensitive to a change in d.kk kk kk kk

901. Section 14.7 Extreme Values and Saddle Points 897
56. u e , u xe sin z, u y cos z du e dx xe sin z dy (y cos z) dzxy z
yy y yœœ œÊœ ab
du 3 dx 7 dy 0 dz 3 dx 7 dy magnitude of the maximum possible errorÊœ œ Êk
2ln3ßß

2
3(0.2) 7(0.6) 4.8Ÿ œ
57. Q , Q , and Q
KM hœœ œ
"" "
## #
"Î# "Î# "Î#
ˆ‰ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ ‰
2KM 2M 2KM 2K 2KM 2KM
hh hh h h

dQ dK dM dhʜ
"" "
## #
"Î# "Î# "Î#
ˆ‰ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ ‰
2KM 2M 2KM 2K 2KM 2KM
hh hh h h

dK dM dh dQœ Ê
"
#
"Î#
ˆ‰ ‘ k
2KM2M2K 2KM
hhhh

"` `Ð #22000.05
dK dM dh (0.0125)(800 dK 80 dM 32,000 dh)œ œ
"
#
"Î#
’“’ “
(2)(2)(20) (2)(20) (2)(2) (2)(2)(20)
0.05 0.05 0.05 (0.05)

Q is most sensitive to changes in hÊ
58. A ab sin C A b sin C, A a sin C, A ab cos CœÊœ œ œ
""""
####
abc
dA b sin C da a sin C db ab cos C dC; dC 2 °0.0349 radians, da 0.5 ft,Êœ œœ œˆ‰ˆ‰ˆ ‰ kk k k k k
"""
###
db 0.5 ft; at a 150 ft, b 200 ft, and C 60°, we see that the change is approximatelyœœœ œkk
dA (200)(sin 60 °) 0.5 (150)(sin 60°) 0.5 (200)(150)(cos 60°) 0.0349 338 ftœ œ„
"""
###
#
kk kk k k
59. z f(x y) g(x y z) f(x y) z 0 g (x y z) f (x y), g (x y z) f (x y) and g (x y z) 1œßÊ ßßœß œ Ê ßß œ ß ßß œ ß ßß œ
xxyyz
g(x y f(x y)) f(x y), g(x y f(x y)) f(x y) and g(x y f(x y)) 1 the tangentÊ ßßß œ ß ßßß œ ß ßßß œ Ê xxyyz!! !! !! !! !! !! !! !!
plane at the point P is f (x y )(x x ) f (x y )(y y ) [z f(x y )] 0 or!!! ! !! ! !!xy ß ß ßœ
z f (x y )(x x ) f (x y )(y y ) f(x y )œß ß ß
xy!! ! !! ! !!
60. f 2x 2y 2(cos t t sin t) 2(sin t t cos t) and (t cos t) (t sin t) ™œ œ œ Êœij i jv i ju
v
vkk
(cos t) (sin t) since t 0 (D f) fœœ Êœ
(t cos t) (t sin t)
(t cos t) (t sin t)
P
ij
u

È
$
ij u ™†
2(cos t t sin t)(cos t) 2(sin t t cos t)(sin t) 2œ œ
61. f 2x 2y 2z (2 cos t) (2 sin t) 2t and ( sin t) (cos t) ™œ œ œ Êœijk i jkv i jku
v
vkk
(D f) fœœ Êœ
( sin t) (cos t)
(sin t) (cos t) 1
sin t cos t
222
P


"ijk
uÈÈÈÈ
$ Š‹Š‹ijk u ™†
(2 cos t) (2 sin t) (2t) (D f) , (D f)(0) 0 andœ œÊ œ œŠ‹ Š‹ Š‹ ˆ‰
"sin t cos t 2t
2222 22 4ÈÈÈÈ È uu
11
(D f)u
ˆ‰
11
4 22
œ
È
62. r t t (t 3) t t ; t 1 x 1, y 1, z 1 P (1 1 1)œ Êœ œÊœ œ œ Êœßß ÈÈij kv i jk
""""
##
"Î# "Î#
!
44
and (1) ; f(x y z) x y z 3 0 f 2x 2yvijk ijkœ ßß œ œÊ œ
"""
##
##
4
™
f(1 1 1) 2 2 ; therefore ( f) the curve is normal to the surfaceÊßß œ œÊ™™ ijk v
"
4
63. r t t (2t 1) t t 2 ; t 1 x 1, y 1, z 1 P (1 1 1) andœ Êœ œÊœ œ œÊ œßßÈÈij kv i jk
""
##
"Î# "Î#
!
(1) 2 ; f(xyz) x y z 1 0 f 2x 2y f(111) 2 2 ;vijk ijk ijkœ ßß œ œÊ œ Êßßœ
""
##
##
™™
now 1 f 1 1 1 0, thus the curve is tangent to the surface when t 1vab a b†™ ßß œ œ
14.7 EXTREME VALUES AND SADDLE POINTS
1. f (x y) 2x y 3 0 and f (x y) x 2y 3 0 x 3 and y 3 critical point is ( 3 3);
xyßœ œßœ œÊœ œÊ ß
f ( 3 3) 2, f ( 3 3) 2, f ( 3 3) 1 f f f 3 0 and f 0 local minimum of
xx yy xy xx yy xx xyßœ ßœ ßœÊ œ"" Ê
#
f( 3 3) 5ߜ

902. 898 Chapter 14 Partial Derivatives
2. f (x y) 2x 3y 6 0 and f (x y) 3x 6y 3 0 x 15 and y 8 critical point is (15 8);xyßœ œßœ œÊœ œ Êß
f (15 8) 2, f (15 8) 6, f (15 8) 3 f f f 3 0 and f 0 local minimum of
xx yy xy xx yy xx xyßœß œß œÊ œ"" Ê
#
f(15 8) 63ߜ
3. f (x y) 2y 10x 4 0 and f (x y) 2x 4y 4 0 x and y critical point is ;xy
24 24
33 33ßœ œßœ œÊœ œÊ ß ˆ‰
f 10, f 4, f 2 f f f 36 0 and f 0 local maximum ofxx yy xy xx yy xx
24 24 24
33 33 33 xy
ˆ‰ ˆ‰ ˆ‰ßœ ßœ ßœÊ œ"# Ê
#
f0ˆ‰
24
33
ߜ
4. f (x y) 2y 10x 4 0 and f (x y) 2x 4y 0 x and y critical point is ;xy
42 42
99 99ßœ œßœ œÊœ œÊ ß ˆ‰
f 10, f 4, f 2 f f f 36 0 and f 0 local maximum ofxx yy xy xx yy xx
42 42 42
99 99 99 xy
ˆ‰ ˆ‰ ˆ‰ßœ ßœ ßœÊ œ"# Ê
#
fˆ‰
42 28
99 9
ߜ
5. f (x y) 2x y 3 0 and f (x y) x 2 0 x 2 and y 1 critical point is ( 2 1);
xyßœ œßœ œÊœ œÊ ß
f ( 2 1) 2, f ( 2 1) 0, f ( 2 1) 1 f f f 1 0 saddle point
xx yy xy xx yy xyßœ ßœ ßœÊ œ Ê
#
6. f (x y) y 2 0 and f (x y) 2y x 2 0 x 2 and y 2 critical point is ( 2 2);xyßœ œßœ œÊœ œÊ ß
f ( 2 2) 0, f ( 2 2) 2, f ( 2 2) 1 f f f 1 0 saddle point
xx yy xy xx yy xyßœ ßœ ßœÊ œ Ê
#
7. f (x y) 5y 14x 3 0 and f (x y) 5x 6 0 x and y critical point is ;xy
669 669
55 525ßœ œßœ œÊœ œ Ê ß
#
ˆ‰
f 14, f 0, f 5 f f f 25 0 saddle pointxx yy xy xx yy
669 669 669
525 525 525 xy
ˆ‰ ˆ‰ ˆ‰ßœ ßœ ßœÊ œ Ê
#
8. f (x y) 2y 2x 3 0 and f (x y) 2x 4y 0 x 3 and y critical point is 3 ;xy
33
2ßœ œßœ œÊœ œÊ ß
#
ˆ‰
f 3 2, f 3 4, f 3 2 f f f 4 0 and f 0 local maximum ofxx yy xy xx yy xx
333
222 xy
ˆ‰ ˆ‰ ˆ‰ßœ ßœ ßœÊ œ"# Ê
#
f3ˆ‰ßœ
317
##
9. f (x y) 2x 4y 0 and f (x y) 4x 2y 6 0 x 2 and y 1 critical point is (2 1);xyßœ œßœ œÊœ œÊ ß
f (2 1) 2, f (2 1) 2, f (2 1) 4 f f f 12 0 saddle point
xx yy xy xx yy xyßœ ßœ ßœ Ê œ Ê
#
10. f (x y) 6x 6y 2 0 and f (x y) 6x 14y 4 0 x and y critical point is ;xy
13 3 13 3
14 124ßœ œßœ œÊœ œ Êß
#
ˆ‰
f 6, f 14, f 6 f f f 48 0 and f 0 local minimum ofxx yy xy xx yy xx
13 3 13 3 13 3
14 14 14 xy
ˆ‰ ˆ‰ ˆ‰
## #
#
ßœß œß œÊ œ"" Ê
fˆ‰
13 3 31
12 4 1
ߜ
#
11. f (x y) 4x 3y 5 0 and f (x y) 3x 8y 2 0 x 2 and y 1 critical point is (2 1);xyßœ œßœ œÊœ œ Êß
f (2 1) 4, f (2 1) 8, f (2 1) 3 f f f 23 0 and f 0 local minimum of
xx yy xy xx yy xx xyßœß œß œÊ œ"" Ê
#
f(2 1) 6ߜ
12. f (x y) 8x 6y 20 0 and f (x y) 6x 10y 26 0 x 1 and y 2 critical point is (1 2);
xyßœ œßœ œÊœ œ Êß
f (1 2) 8, f (1 2) 10, f (1 2) 6 f f f 44 0 and f 0 local minimum of
xx yy xy xx yy xx xyßœß œß œÊ œ"" Ê
#
f(1 2) 36ߜ
13. f (x y) 2x 2 0 and f (x y) 2y 4 0 x 1 and y 2 critical point is (1 2); f (1 2) 2,
xy xxßœ œßœ œÊœ œÊ ß ßœ
f (1 2) 2, f (1 2) 0 f f f 4 0 saddle point
yy xy xx yy xyßœ ßœÊ œ Ê
#
14. f (x y) 2x 2y 2 0 and f (x y) 2x 4y 2 0 x 1 and y 0 critical point is (1 0);xyßœ œßœ œÊœ œÊ ß
f (1 0) 2, f (1 0) 4, f (1 0) 2 f f f 4 0 and f 0 local minimum of
xx yy xy xx yy xx xyßœ ßœ ßœ Ê œ"" Ê
#
f(1 0) 0ߜ

903. Section 14.7 Extreme Values and Saddle Points 899
15. f (x y) 2x 2y 0 and f (x y) 2x 0 x 0 and y 0 critical point is (0 0); f (0 0) 2,xy xxßœ œßœœÊœœÊ ßßœ
f (0 0) 0, f (0 0) 2 f f f 4 0 saddle point
yy xy xx yy xyßœ ßœÊ œ Ê
#
16. f (x y) 2 4x 2y 0 and f (x y) 2 2x 2y 0 x 0 and y 1 critical point is (0 1);xyßœ œßœ œÊœ œÊ ß
f (0 1) 4, f (0 1) 2, f (0 1) 2 f f f 4 0 and f 0 local maximum of f(0 1) 4
xx yy xy xx yy xx xyßœ ßœ ßœÊ œ"# Êßœ
#17. f (x y) 3x 2y 0 and f (x y) 3y 2x 0 x 0 and y 0, or x and y critical pointsxy
22
33ßœ œßœ œÊœ œ œ œÊ
##
are (0 0) and ; for (0 0): f (0 0) 6x 0, f (0 0) 6y 0, f (0 0) 2ß ßßßœœßœ œßœ ˆ‰ kk
22
33
xx yy xy 00 00ÐßÑ ÐßÑ
f f f 4 0 saddle point; for : f 4, f 4, f 2Ê œ Ê ß ßœ ßœ ßœ xx yy xx yy xyxy
22 22 22 22
33 33 33 33
#
ˆ‰ˆ‰ ˆ‰ ˆ‰
f f f 12 0 and f 0 local maximum of fÊ œ"# Ê ßœ xx yy xxxy
2 2 170
33 27
#
ˆ‰
18. f (x y) 3x 3y 0 and f (x y) 3x 3y 0 x 0 and y 0, or x 1 and y 1 critical points
xyßœ œßœ œÊœ œ œ œÊ
##
are (0 0) and ( 1 1); for ( ): f (0 0) 6x 0, f (0 0) 6y 0, f (0 0) 3 f f fß ß !ß ! ß œ œ ß œ œ ß œ Ê kk xx yy xy xx yy00 00 xyÐßÑ ÐßÑ
#
9 0 saddle point; for ( 1 1): f ( 1 1) 6, f ( 1 1) 6, f ( 1 1) 3 f f fœ Ê ß ßœ ßœ ßœÊ xx yy xy xx yy xy
#
27 0 and f 0 local maximum of f( 1 1) 1œ"# Ê ßœ xx
19. f (x y) 12x 6x 6y 0 and f (x y) 6y 6x 0 x 0 and y 0, or x 1 and y 1 criticalxyßœ œßœ œÊœœœœ Ê
#
points are (0 0) and (1 1); for ( ): f (0 0) 12 12x 12, f (0 0) 6, f (0 0) 6 f f fßß !ß ! ß œ œßœßœÊ k xx yy xy xx yy 00 xyÐßÑ
#
36 0 and f 0 local minimum of f(0 0) 0; for (1 1): f (1 1) 0, f (1 1) 6,œ"" Êßœß ßœß œ xx xx yy
f (1 1) 6 f f f 36 0 saddle pointxy xx yy xyßœÊ œ Ê
#
20. f (x y) 6x 6y 0 x y; f (x y) 6y 6y 6x 0 12y 6y 0 6y(2 y) 0 y 0 orxyßœ œÊœ ßœ œÊ œÊ œÊœ
##
y 2 (0 0) and (2 2) are the critical points; f (x y) 6, f (x y) 6 12y, f (x y) 6; for (0 0):œÊß ß ßœ ßœ ßœ ß xx yy xy
f (0 0) 6, f (0 0) 6, f (0 0) 6 f f f 72 0 saddle point; for (2 2): f (2 2) 6,xx yy xy xx yy xx xyßœ ßœ ßœÊ œ Êßßœ
#
f (2 2) 18, f (2 2) 6 f f f 72 0 and f 0 local maximum of f(2 2) 8yy xy xx yy xx xyßœ ßœÊ œ"# Êßœ
#
21. f (x y) 27x 4y 0 and f (x y) y 4x 0 x 0 and y 0, or x and y critical points arexy
44
93ßœ œßœ œÊœ œ œ œÊ
##
(0 0) and ; for ( ): f (0 0) 54x 0, f (0 0) 2y 0, f (0 0) 4 f f fß ß !ß ! ß œ œ ß œ œ ß œ Ê ˆ‰ kk
44
93
xx yy xy xx yy 00 00 xyÐßÑ ÐßÑ
#
16 0 saddle point; for : f 24, f , f 4 f f f 48 0œ Ê ß ßœ ßœ ßœ Ê œ"ˆ‰ ˆ‰ ˆ‰ ˆ‰
44 44 44 8 44
93 93 93 3 93
xx yy xy xx yy xy
#
and f 0 local minimum of fxx
44 64
93 81"Êßœ ˆ‰
22. f (x y) 24x 6y 0 y 4x ; f (x y) 3y 6x 0 3 4x 6x 0 16x 2x 0
xyßœ œÊœ ßœ œÊ œÊ œ
####%
#
ab
2x 8x 1 0 x 0 or x ( 0) and 1 are the critical points; f (x y) 48x,Ê œÊœ œ Ê!ß ß ßœab ˆ‰
$ ""
##
xx
f (x y) 6y, and f (x y) 6; for (0 0): f (0 0) 0, f (0 0) 0, f (0 0) 6 f f f 36 0yy xy xx yy xy xx yy xyßœ ßœ ß ßœ ßœ ßœÊ œ
#
saddle point; for 1 : f 1 24, f 1 6, f 1 6Ê ß ßœ ßœ ßœˆ‰ˆ‰ ˆ‰ ˆ‰
11 1 1
22 2 2
xx yy xy
f f f 108 0 and f 0 local maximum of f 1 1Ê œ "# Ê ßœ xx yy xxxy
1
2
#
ˆ‰
23. f (x y) 3x 6x 0 x 0 or x 2; f (x y) 3y 6y 0 y 0 or y 2 the critical points are
xyßœ œÊœ œ ßœ œÊœ œÊ
##
(0 0), (0 2), ( 2 0), and ( 2 2); for ( ): f (0 0) 6x 6 6, f (0 0) 6y 6 6,ßß ß ß!ß!ßœ œßœ œkk xx yy 00 00ÐßÑ ÐßÑ
f (0 0) 0 f f f 36 0 saddle point; for (0 2): f (0 2) 6, f (0 2) 6, f (0 2) 0xy xx yy xx yy xy xyßœÊ œ Ê ß ßœ ßœ ßœ
#
f f f 36 0 and f 0 local minimum of f(0 2) 12; for ( 2 0): f ( 2 0) 6,Ê œ"" Êßœ ß ßœ xx yy xx xxxy
#
f ( 2 0) 6, f ( 2 0) 0 f f f 36 0 and f 0 local maximum of f( 2 0) 4;yy xy xx yy xx xyßœ ßœÊ œ"# Ê ßœ
#
for ( 2 2): f ( 2 2) 6, f ( 2 2) 6, f ( 2 2) 0 f f f 36 0 saddle pointß ßœ ßœ ßœÊ œ Ê xx yy xy xx yy xy
#

904. 900 Chapter 14 Partial Derivatives
24. f (x y) 6x 18x 0 6x(x 3) 0 x 0 or x 3; f (x y) 6y 6y 12 0 6(y 2)(y 1) 0x yßœ œÊ œÊœ œ ßœ œÊ œ
# #
y 2 or y 1 the critical points are (0 2), (0 1), (3 2), and (3 1); f (x y) 12x 18,Êœ œÊ ß ßß ßßœ xx
f (x y) 12y 6, and f (x y) 0; for ( 2): f (0 2) 18, f (0 2) 18, f (0 2) 0yy xy xx yy xyßœ ßœ !ß ßœ ßœ ßœ
f f f 324 0 and f 0 local maximum of f(0 2) 20; for (0 1): f (0 1) 18,Ê œ "# Êß œßßœ
xx yy xx xxxy
#
f (0 1) 18, f (0 1) 0 f f f 324 0 saddle point; for (3 2): f (3 2) 18,yy xy xx yy xx xyßœ ßœ Ê œ Êß ßœ
#
f (3 2) 18, f (3 2) 0 f f f 324 0 saddle point; for (3 1): f (3 1) 18,yy xy xx yy xx xyßœ ßœÊ œ Êßßœ
#
f (3 1) 18, f (3 1) 0 f f f 324 0 and f 0 local minimum of f(3 1) 34yy xy xx yy xx xyßœ ßœÊ œ "" Êßœ
#
25. f (x y) 4y 4x 0 and f (x y) 4x 4y 0 x y x 1 x 0 x 0, 1, 1 the criticalxyßœ œßœ œÊœÊ œÊœ Ê
$$ #
ab
points are (0 0), (1 1), and ( 1 1); for ( ): f (0 0) 12x 0, f (0 0) 12y 0,ßß ß !ß ! ß œ œßœ œkk
xx yy 00 00
##
ÐßÑ ÐßÑ
f (0 0) 4 f f f 16 0 saddle point; for (1 1): f (1 1) 12, f (1 1) 12, f (1 1) 4xy xx yy xx yy xy xyßœÊ œ Êßßœ ßœ ßœ
#
f f f 128 0 and f 0 local maximum of f(1 1) 2; for ( 1 1): f ( 1 1) 12,Ê œ "# Êßœ ß ßœ xx yy xx xxxy
#
f ( 1 1) 12, f ( 1 1) 4 f f f 128 0 and f 0 local maximum of f( 1 1) 2yy xy xx yy xx xyßœ ßœÊ œ "# Ê ßœ
#
26. f (x y) 4x 4y 0 and f (x y) 4y 4x 0 x y x x 0 x 1 x 0 x 0, 1, 1xyßœ œßœ œÊœ Ê œÊ œÊœ
$$ $#
ab
the critical points are (0 0), (1 1), and ( 1 1); f (x y) 12x , f (x y) 12y , and f (x y) 4;Êßß ßßœ ßœ ßœ
xx yy xy
##
for ( 0): f (0 0) 0, f (0 0) 0, f (0 0) 4 f f f 16 0 saddle point; for (1 1):!ß ßœ ßœ ßœÊ œ Êß xx yy xy xx yy xy
#
f (1 1) 12, f (1 1) 12, f (1 1) 4 f f f 128 0 and f 0 local minimum ofxx yy xy xx yy xx xyßœß œß œÊ œ "" Ê
#
f( 1) 2; for ( 1 1): f ( 1 1) 12, f ( 1 1) 12, f ( 1 1) 4 f f f 128 0 and"ßœ ß ßœ ßœ ßœÊ œ " xx yy xy xx yy xy
#
f 0 local minimum of f( 1 1) 2xx"Ê ßœ
27. f (x y) 0 and f (x y) 0 x 0 and y 0 the critical point is ( 0);xy
2x
xy1 xy1
2yßœ œ ßœ œÊœ œÊ !ß



ab ab


f , f , f ; f ( ) 2, f (0 0) 2, f (0 0) 0xx yy xy xx yy xy
4x 2y 2 2x 4y 2 8xy
xy1 xy1 xy1œœ œ!ß!œ ßœ ßœ




ab ab ab
f f f 4 0 and f 0 local maximum of f(0 0) 1Ê œ"# Êßœ xx yy xxxy
#
28. f (x y) y 0 and f (x y) x 0 x 1 and y 1 the critical point is (1 1);xy
11
xyßœ œßœ œÊœ œÊ ß

f , f , f 1; f (1 1) 2, f (1 1) 2, f (1 1) 1 f f f 3 0 and f 2 localxx yy xy xx yy xy xx yy xx
22
xy xyœ œ œ ßœ ßœ ßœÊ œ"" Ê

#
minimum of f(1 1) 3ߜ
29. f (x y) y cos x 0 and f (x y) sin x 0 x n , n an integer, and y 0 the critical points are
xyßœ œ ßœ œÊœ œÊ 1
(n 0), n an integer (Note: cos x and sin x cannot both be 0 for the same x, so sin x must be 0 and y 0);1ß œ
f y sin x, f 0, f cos x; f (n 0) 0, f (n 0) 0, f (n 0) 1 if n is even and f (n 0) 1
xx yy xy xx yy xy xyœ œ œ ßœ ßœ ßœ ßœ 111 1
if n is odd f f f 1 0 saddle point.Ê œ Ê
xx yy xy
#
30. f (x y) 2e cos y 0 and f (x y) e sin y 0 no solution since e 0 for any x and the functionsxy
2x 2x 2xßœ œ ßœ œÊ Á
cos y and sin y cannot equal 0 for the same y no critical points no extrema and no saddle pointsÊÊ
31. (i) On OA, f(x y) f(0 y) y 4y 1 on 0 y 2;ßœ ßœ ŸŸ
#
f (0 y) 2y 4 0 y 2;
w
ßœ œÊœ
f(0 0) 1 and f( ) 3ßœ !ß#œ
(ii) On AB, f(x y) f(x 2) 2x 4x 3 on 0 x 1;ßœ ßœ ŸŸ
#
f (x 2) 4x 4 0 x 1;
w
ßœ œÊœ
f(0 2) 3 and f(1 ) 5ßœ ß# œ
(iii) On OB, f(x y) f(x 2x) 6x 12x 1 onßœ ß œ
#
0 x 1; endpoint values have been found above;ŸŸ

f (x 2x) 12x 12 0 x 1 and y 2, but ( ) is not an interior point of OB
w
ßœ œ Ê œ œ "ß #

905. Section 14.7 Extreme Values and Saddle Points 901
(iv) For interior points of the triangular region, f (x y) 4x 4 0 and f (x y) 2y 4 0xyßœ œßœ œ
x 1 and y 2, but (1 2) is not an interior point of the region. Therefore, the absolute maximum isÊœ œ ß
1 at (0 0) and the absolute minimum is 5 at ( ).ß "ß #
32. (i) On OA, D(x y) D(0 y) y 1 on 0 y 4;ßœ ßœ ŸŸ
#
D(0y)2y0 y0; D()1 and
w
ßœ œÊœ !ß!œ
D( ) 17!ß % œ
(ii) On AB, D(x y) D(x 4) x 4x 17 onߜ ߜ
#
0 x 4; D (x 4) 2x 4 0 x 2 and (2 4)ŸŸ ß œ œÊœ ß
w
is an interior point of AB; D( ) 13 and#ß % œ
D( ) D( ) 17%ß % œ !ß % œ
(iii) On OB, D(x y) D(x x) x 1 on 0 x 4;ßœ ßœ ŸŸ
#

D (x x) 2x 0 x 0 and y 0, which is not an interior point of OB; endpoint values have been found
w
ßœ œÊœ œ
above
(iv) For interior points of the triangular region, f (x y) 2x y 0 and f (x y) x 2y 0 x 0 and y 0,
xyßœ œßœ œÊœ œ
which is not an interior point of the region. Therefore, the absolute maximum is 17 at ( ) and ( ), and the!ß % %ß %
absolute minimum is 1 at (0 0).ß
33. (i) On OA, f(x y) f( y) y on 0 y 2;ß œ !ß œ Ÿ Ÿ
#
f (0 y) 2y 0 y 0 and x 0; f(0 0) 0 and
w
ßœ œÊœ œ ßœ
f(0 ) 4ß# œ
(ii) On OB, f(x y) f(x 0) x on 0 x 1;ßœ ßœ ŸŸ
#
f (x 0) 2x 0 x 0 and y 0; f(0 0) 0 and
w
ßœ œÊœ œ ßœ
f(1 0) 1ߜ
(iii) On AB, f(x y) f(x 2x 2) 5x 8x 4 onßœ ß œ
#
0x1; f(x2x2)10x80 xŸŸ ß œ œÊœ
w 4
5

and y ; f ; endpoint values have been found above.œßœ
242 4
555 5
ˆ‰
(iv) For interior points of the triangular region, f (x y) 2x 0 and f (x y) 2y 0 x 0 and y 0, but ( 0) is
xyßœ œ ßœ œÊœ œ !ß
not an interior point of the region. Therefore the absolute maximum is 4 at (0 2) and the absolute minimum is 0 atß
(0 0).ß
34. (i) On AB, T(x y) T( y) y on 3 y 3;ß œ !ß œ ŸŸ
#
T(0y)2y0 y0 and x0; T(00)0,
w
ßœ œÊœ œ ßœ
T( 3) 9, and T( 3) 9!ߜ!ߜ
(ii) On BC, T(x y) T(x 3) x 3x 9 on 0 x 5;ßœ ßœ ŸŸ
#
T (x 3) 2x 3 0 x and y 3;
w
#
ßœ œÊœ œ
3
T 3 and T(5 3) 19ˆ‰
327
4#
ߜ ߜ
(iii) On CD, T(x y) T(5 y) y 5y 5 onߜ ߜ
#
3 y 3;T (5 y) 2y 5 0 y andŸŸ ß œ œÊœ
w
#
5

x 5;T 5 , T( 3) 11 and T(5 3) 19œß œ &ßœ ßœˆ‰
545
4#
(iv) On AD, T(x y) T(x 3) x 9x 9 on 0 x 5; T (x 3) 2x 9 0 x and y 3;ßœ ß œ ŸŸ ß œœÊœ œ
#w
#
9
T 3 , T( 3) 9 and T( 3) 11ˆ‰
945
4#
ßœ !ßœ&ß œ
(v) For interior points of the rectangular region, T (x y) 2x y 6 0 and T (x y) x 2y 0 x 4
xyßœ œßœ œÊœ
and y 2 (4 2) is an interior critical point with T(4 2) 12. Therefore the absolute maximumœÊß ßœ
is 19 at (5 3) and the absolute minimum is 12 at (4 2).ß ß

906. 902 Chapter 14 Partial Derivatives
35. (i) On OC, T(x y) T(x 0) x 6x 2 onߜ ߜ
#
0x5; T(x0)2x60 x3 andŸŸ ß œ œÊœ
w
y 0; T(3 0) 7, T(0 0) 2, and T(5 0) 3œßœ ßœ ßœ
(ii) On CB, T(x y) T(5 y) y 5y 3 onߜ ߜ
#
3 y 0; T (5 y) 2y 5 0 y andŸŸ ß œ œÊœ
w
#
5
x 5; T 5 and T(5 3) 9œß œ ßœˆ‰
537
4#
(iii) On AB, T(x y) T(x 3) x 9x 11 onßœ ß œ
#
0x5; T(x3)2x90 x andŸŸ ß œœÊœ
w
#
9
y 3; T 3 and T( 3) 11œ ßœ !ßœˆ‰
937
4#
(iv) On AO, T(x y) T( y) y 2 on 3 y 0; T (0 y) 2y 0 y 0 and x 0, but (0 0) isß œ !ß œ ŸŸ ß œ œ Ê œ œ ß
#w
not an interior point of AO
(v) For interior points of the rectangular region, T (x y) 2x y 6 0 and T (x y) x 2y 0 x 4
xyßœ œßœ œÊœ
and y 2, an interior critical point with T( 2) 10. Therefore the absolute maximum is 11 atœ %ßœ
( 3) and the absolute minimum is 10 at (4 2).!ß ß
36. (i) On OA, f(x y) f( y) 24y on 0 y 1;ß œ !ß œ ŸŸ
#
f (0 y) 48y 0 y 0 and x 0, but (0 0) is
w
ßœ œÊœ œ ß
not an interior point of OA; f( 0) 0 and!ß œ
f( 1) 24!ß œ
(ii) On AB, f(x y) f(x 1) 48x 32x 24 onߜ ߜ
$
0 x 1; f (x 1) 48 96x 0 x andŸŸ ß œ œÊœ
w# "
È
2
y 1, or x and y 1, but 1 is not inœœ œ ß
""
ÈÈ
22
Š‹

the interior of AB; f 1 16 2 24 and f(1 1) 8Š‹
È
"
È
2
ߜ ߜ
(iii) On BC, f(x y) f( y) 48y 32 24y on 0 y 1; f ( y) 48 48y 0 y 1 and x 1, butß œ "ß œ Ÿ Ÿ "ß œ œÊœ œ
#w
( ) is not an interior point of BC; f( 0) 32 and f( ) 8"ß " "ß œ "ß " œ
(iv) On OC, f(x y) f(x 0) 32x on 0 x 1; f (x 0) 96x 0 x 0 and y 0, but (0 0) is not anßœ ßœ ŸŸ ß œ œÊœ œ ß
$w#
interior point of OC; f( 0) 0 and f( 0) 32!ß œ "ß œ
(v) For interior points of the rectangular region, f (x y) 48y 96x 0 and f (x y) 48x 48y 0
xyßœ œßœ œ
#
x 0 and y 0, or x and y , but (0 0) is not an interior point of the region; f 2.Êœ œ œ œ ß ß œ
"" ""
## ##
ˆ‰
Therefore the absolute maximum is 2 at and the absolute minimum is 32 at (1 0).ˆ‰
""
##
ß ß
37. (i) On AB, f(x y) f(1 y) 3 cos y on y ;ßœ ßœ ŸŸ
11
44
f (1 y) 3 sin y 0 y 0 and x 1;
w
ßœ œÊœ œ
f( 0) 3, f 1 , and f 1"ß œ ß œßœˆ‰ ˆ‰
11
44
32 32
ÈÈ
##
(ii) On CD, f(x y) f( y) 3 cos y on y ;ß œ $ß œ ŸŸ
11
44
f (3 y) 3 sin y 0 y 0 and x 3;
w
ßœ œÊœ œ
f(3 0) 3, f 3 and f 3ßœ ß œßœˆ‰ ˆ‰
11
44
32 32
ÈÈ
##
(iii) On BC, f(x y) f x 4x x onßœ ß œ ˆ‰ ab
1
4
2
È
#
#

1 x 3; f x 2(2 x) 0 x 2 and y ; f 2 2 2, f 1 , andŸŸ ß œ œÊœ œ ßœ ßœ
w
#ˆ‰ ˆ‰ ˆ‰ÈÈ
1111
4444
32
È
f3 ˆ‰ßœ
1
4
32
È
#
(iv) On AD, f(x y) f x 4x x on 1 x 3; f x 2(2 x) 0 x 2 and y ;ßœ ß œ ŸŸ ß œ œÊœ œ ˆ‰ ˆ‰ ab
È
11 1
44 4
2
È
#
#w
f 2 2 2, f 1 , and f 3ˆ‰ ˆ‰ ˆ‰È
ßœß œß œ
11 1
44 4
32 32
ÈÈ
##
(v) For interior points of the region, f (x y) (4 2x) cos y 0 and f (x y) 4x x sin y 0 x 2xyßœ œßœ œÊœab
#
and y 0, which is an interior critical point with f(2 0) 4. Therefore the absolute maximum is 4 atœßœ

907. Section 14.7 Extreme Values and Saddle Points 903
(2 0) and the absolute minimum is at 3 , 3 , 1 , and 1 .ßß ßß ß
32
44 4 4
È
#
ˆ‰ˆ‰ˆ‰ˆ‰
11 1 1
38. (i) On OA, f(x y) f( y) 2y 1 on 0 y 1;ß œ !ß œ ŸŸ
f (0 y) 2 no interior critical points; f(0 0) 1
w
ßœÊ ßœ
and f(0 1) 3ߜ
(ii) On OB, f(x y) f(x 0) 4x 1 on 0 x 1;ßœ ßœ ŸŸ
f (x 0) 4 no interior critical points; f(1 0) 5
w
ßœÊ ßœ
(iii) On AB, f(x y) f(x x 1) 8x 6x 3 onßœ ß œ
#
0x1; f(xx1)16x60 xŸŸ ß œ œÊœ
w 3
8
and y ; f , f(0 1) 3, and f( 0) 5œßœßœ "ßœ
535 15
888 8
ˆ‰

(iv) For interior points of the triangular region, f (x y) 4 8y 0 and f (x y) 8x 2 0
xyßœ œßœ œ
y and x which is an interior critical point with f 2. Therefore the absolute maximum is 5 atÊœ œ ß œ
"" ""
## 44
ˆ‰
(1 0) and the absolute minimum is 1 at ( 0).ß!ß
39. Let F(a b) 6 x x dx where a b. The boundary of the domain of F is the line a b in theßœ Ÿœ
'
a
b
ab
#
ab-plane, and F(a a) 0, so F is identically 0 on the boundary of its domain. For interior critical points weߜ
have: 6 a a 0 a 3, 2 and 6 b b 0 b 3, 2. Since a b, there is only
``
``
##FF
ab
œ œÊœ œ œÊœ Ÿab ab
one interior critical point ( 3 2) and F( 3 2) 6 x x dx gives the area under the parabolaß ßœ
'
3
2
ab
#
y 6 x x that is above the x-axis. Therefore, a 3 and b 2.œ œ œ
#
40. Let F(a b) 24 2x x dx where a b. The boundary of the domain of F is the line a b andßœ Ÿœ '
a
b
ab
#
"Î$
on this line F is identically 0. For interior critical points we have: 24 2a a 0 a 4, 6
`
`
#
"Î$F
a
œ œÊœ ab
and 24 2b b 0 b 4, 6. Since a b, there is only one critical point ( 6 4) and
`
`
#
"Î$F
b
œ œÊœ Ÿ ßab
F( 6 4) 24 2x x dx gives the area under the curve y 24 2x x that is above the x-axis.ßœ œ
'
6
4
ab ab
##
"Î$
Therefore, a 6 and b 4.œ œ
41. T (x y) 2x 1 0 and T (x y) 4y 0 x and y 0 with T 0 ; on the boundaryxy
4ßœ œßœœÊœœ ßœ
"""
##
ˆ‰
x y 1: T(x y) x x 2 for 1 x 1 T (x y) 2x 1 0 x and y ;
## # w "
##
œßœ ŸŸ Ê ß œ œÊœ œ„
È
3
T , T , T( 1 0) 2, and T( 0) 0 the hottest is 2 ° at andŠ‹Š ‹ Š‹ßœ ß œ ßœ "ßœÊ ß
"" ""
## # # ##
ÈÈ È
33 399
44 4
; the coldest is ° at 0 .Š‹ ˆ‰ß ß
"""
## #
È
3
4
42. f (x y) y 2 0 and f (x y) x 0 x and y 2; f 2 8,xy xx
2 2
xy x 2ßœ œßœ œÊœ œ ßœ œ
"" "
## ß
ˆ‰ ¸
ˆ‰
1
2
f 2 , f 2 1 f f f 1 0 and f 0 a local minimum of f 2¹ˆ‰ ˆ‰ ˆ‰yy xy xx yy xx
y4
2
xy
"" "" "
## #
ß
#ßœ œ ßœÊ œ"" Êß

ˆ‰
1
2
2 ln 2 ln 2œ œ
"
#
43. (a) f (x y) 2x 4y 0 and f (x y) 2y 4x 0 x 0 and y 0; f (0 0) 2, f (0 0) 2,x y xx yyßœ œßœ œÊœ œ ßœ ßœ
f (0 0) 4 f f f 12 0 saddle point at (0 0)
xy xx yy xyßœÊ œ Êß
#
(b) f (x y) 2x 2 0 and f (x y) 2y 4 0 x 1 and y 2; f (1 2) 2, f (1 2) 2,x y xx yyßœ œßœ œÊœ œ ßœ ßœ
f (1 2) 0 f f f 4 0 and f 0 local minimum at ( )
xy xx yy xx xyßœÊ œ"" Ê"ß#
#
(c) f (x y) 9x 9 0 and f (x y) 2y 4 0 x 1 and y 2; f (1 2) 18x 18,xy xx 12ßœ œßœ œÊœ„ œ ßœœ
#
ÐßÑ
k
f (1 2) 2, f (1 2) 0 f f f 36 0 and f 0 local minimum at ( );
yy xy xx yy xx xyßœß œÊ œ"" Ê"ß #
#
f ( 1 2) 18, f ( 2) 2, f ( 2) 0 f f f 36 0 saddle point at ( 2)xx yy xy xx yy xyßœ "ßœ "ßœÊ œ Ê "ß
#

908. 904 Chapter 14 Partial Derivatives
44. (a) Minimum at (0 0) since f(x y) 0 for all other (x y)ßß " ß
(b) Maximum of 1 at ( ) since f(x y) 1 for all other (x y)!ß ! ß # ß
(c) Neither since f(x y) 0 for x 0 and f(x y) 0 for x 0ß## ß""
(d) Neither since f(x y) 0 for x 0 and f(x y) 0 for x 0ß## ß""
(e) Neither since f(x y) 0 for x 0 and y 0, but f(x y) 0 for x 0 and y 0ß## " ß"" "
(f) Minimum at (0 0) since f(x y) 0 for all other (x y)ßß " ß
45. If k 0, then f(x y) x y f (x y) 2x 0 and f (x y) 2y 0 x 0 and y 0 ( 0) is the onlyœßœ Êßœœ ßœœÊœ œÊ!ß
##
xy
critical point. If k 0, f (x y) 2x ky 0 y x; f (x y) kx 2y 0 kx 2 x 0Áßœ œÊœ ßœ œÊ œ xy
22
kk
ˆ‰
kx 0 k x 0 x 0 or k 2 y (0) 0 or y x; in any case (0 0) is aÊ œÊ œÊœ œ„Êœ œœ„ ß
4x 4 2
kk k
ˆ‰ ˆ‰
critical point.
46. (See Exercise 45 above): f (x y) 2, f (x y) 2, and f (x y) k f f f 4 k ; f will have a
xx yy xy xx yy xyßœ ßœ ßœÊ œ
##
saddle point at (0 0) if 4 k 0 k 2 or k 2; f will have a local minimum at (0 0) if 4 k 0ß Ê ß
# #
2 k 2; the test is inconclusive if 4 k 0 k 2.Ê œÊœ„
#
47. No; for example f(x y) xy has a saddle point at (a b) (0 0) where f f 0.ßœ ßœß œœ xy
48. If f (a b) and f (a b) differ in sign, then f (a b) f (a b) 0 so f f f 0. The surface must therefore have axx yy xx yy xx yy xyßß ßß
#
saddle point at (a b) by the second derivative test.ß
49. We want the point on z 10 x y where the tangent plane is parallel to the plane x 2y 3z 0. Toœ œ
##
find a normal vector to z 10 x y let w z x y 10. Then w 2x 2y is normal toœ œ œ
## ##
™ ijk
z 10 x y at (x y). The vector w is parallel to 2 3 which is normal to the plane x 2y 3zœ ß
##
™ ijk
0 if 6x 6y 3 2 3 or x and y . Thus the point is 10 or .œ œ œœ ßß ßßijkijk
" " "" " " ""
63 633696336
355
ˆ‰ˆ‰
50. We want the point on z x y 10 where the tangent plane is parallel to the plane x 2y z 0. Letœ œ
##
w z x y 10, then w 2x 2y is normal to z x y 10 at (x y). The vector wœ œ œ ß
## ##
™™ ijk
is parallel to 2 which is normal to the plane if x and y 1. Thus the point 1 1 10ijk œœ ßß
"""
##
ˆ‰
4
or 1 is the point on the surface z x y 10 nearest the plane x 2y z 0.ˆ‰
"
#
##
ßß œ œ
45
4
51. No, because the domain x 0 and y 0 is unbounded since x and y can be as large as we please. Absolute  
extrema are guaranteed for continuous functions defined over closed domains in the plane.and
bounded
Since the domain is unbounded, the continuous function f(x y) x y need not have an absolute maximumߜ
(although, in this case, it does have an absolute minimum value of f(0 0) 0).ߜ
52. (a) (i) On x 0, f(x y) f(0 y) y y 1 for 0 y 1; f (0 y) 2y 1 0 y and x 0;œßœßœ ŸŸ ß œ œÊœ œ
#w "
#
f 0 , f(0 0) 1, and f(0 1) 1ˆ‰ß œ ßœ ßœ
"
#
3
4
(ii) On y 1, f(x y) f(x 1) x x 1 for 0 x 1; f (x 1) 2x 1 0 x and y 1,œßœßœ ŸŸ ß œ œÊœ œ
#w "
#
but 1 is outside the domain; f(0 1) 1 and f( ) 3ˆ‰ßßœ"ß"œ
"
#
(iii) On x 1, f(x y) f( y) y y 1 for 0 y 1; f (1 y) 2y 1 0 y and x 1, butœßœ"ßœ ŸŸ ß œ œÊœ œ
#w "
#
1 is outside the domain; f(1 0) 1 and f( ) 3ˆ‰ß ßœ "ß"œ
"
#
(iv) On y 0, f(x y) f(x 0) x x 1 for 0 x 1; f (x 0) 2x 1 0 x and y 0;œßœßœ ŸŸ ß œ œÊœ œ
#w "
#
f 0 ; f(0 0) 1, and f( 0) 1ˆ‰
"
#
ß œ ß œ "ß œ
3
4
(v) On the interior of the square, f (x y) 2x 2y 1 0 and f (x y) 2y 2x 1 0 2x 2y 1xyßœ œßœ œÊ œ
(xy) . Then f(xy)x y 2xyxy1(xy) (xy)1 is the absoluteÊ œßœ œ œ
"
#
## # 3
4
minimum value when 2x 2y 1.œ

909. Section 14.7 Extreme Values and Saddle Points 905
(b) The absolute maximum is f( ) 3."ß " œ
53. (a) 2 sin t 2 cos t 0 cos t sin t x y
df f dx f dx
dt x dt y dt dt dt
dy dy
œ œœ œÊ œ Êœ
``
``
(i) On the semicircle x y 4, y 0, we have t and x y 2 f 2 2 2 2. At the
##
œ   œœœÊßœ
1
4 ÈÈÈÈ
Š‹
endpoints, f( 2 0) 2 and f( ) 2. Therefore the absolute minimum is f( 2 0) 2 when t ;ßœ #ß ! œ ßœ œ1
the absolute maximum is f 2 2 2 2 when t .Š‹
ÈÈ È
ßœ œ
1
4
(ii) On the quartercircle x y 4, x 0 and y 0, the endpoints give f( 2) 2 and f( 0) 2.
##
œ    !ß œ #ß œ
Therefore the absolute minimum is f(2 0) 2 and f( 2) 2 when t 0, respectively; the absoluteßœ !ßœ œ
1
#
maximum is f 2 2 2 2 when t .Š‹
ÈÈ È
ßœ œ
1
4
(b) y x 4 sin t 4 cos t 0 cos t sin t x y.
dg g g dy dy
dt x dt y dt dt dt
dx dx
œ œ œ œÊ œ„ Êœ„
``
``
##
(i) On the semicircle x y 4, y 0, we obtain x y 2 at t and x 2, y 2 at
##
œ   œœ œ œ œ
ÈÈÈ
1
4
t . Then g 2 2 2 and g 2 2 2. At the endpoints, g( 2 0) g( 0) 0.œßœ ßœ ß œ #ß œ
3
4
1
Š‹ Š ‹
ÈÈ ÈÈ
Therefore the absolute minimum is g 2 2 2 when t ; the absolute maximum isŠ‹
ÈÈ
ßœ œ
3
4
1
g22 2 when t .Š‹
ÈÈ
ßœ œ
1
4
(ii) On the quartercircle x y 4, x 0 and y 0, the endpoints give g( 2) 0 and g( 0) 0.
##
œ    !ß œ #ß œ
Therefore the absolute minimum is g(2 0) 0 and g( 2) 0 when t 0, respectively; the absoluteßœ !ßœ œ
1
#
maximum is g 2 2 2 when t .Š‹
ÈÈ
ßœ œ
1
4
(c) 4x 2y (8 cos t)( 2 sin t) (4 sin t)(2 cos t) 8 cos t sin t 0
dh h dx h dx
dt x dt y dt dt dt
dy dy
œ œ œ œ œ
``
``
t 0, , yielding the points (2 0), (0 2) for 0 t .Êœ ß ß ŸŸ
1
#
11
(i) On the semicircle x y 4, y 0 we have h(2 0) 8, h(0 2) 4, and h( 2 0) 8. Therefore,
##
œ   ßœ ßœ ßœ
the absolute minimum is h( 2) 4 when t ; the absolute maximum is h(2 0) 8 and h( 2 0) 8!ß œ œ ß œ ßœ
1
#
when t 0, respectively.œ1
(ii) On the quartercircle x y 4, x 0 and y 0 the absolute minimum is h(0 2) 4 when t ; the
##
#
œ    ßœ œ
1
absolute maximum is h(2 0) 8 when t 0.ßœ œ
54. (a) 2 3 6 sin t 6 cos t 0 sin t cos t t for 0 t .
df f dx f dx
dt x dt y dt dt dt 4
dy dy
œ œ œ œÊ œ Êœ ŸŸ
``
``
1
1
(i) On the semi-ellipse, 1, y 0, f(x y) 2x 3y 6 cos t 6 sin t 6 6 6 2
x
94
y 22

œ   ßœ œ œ œŠ‹ Š‹
È
ÈÈ
##
at t . At the endpoints, f( 3 0) 6 and f(3 0) 6. The absolute minimum is f( 3 0) 6 whenœ ßœ ßœ ßœ
1
4
t ; the absolute maximum is f 2 6 2 when t .œßœœ1 Š‹
ÈÈ
32
4
È
#
1
(ii) On the quarter ellipse, at the endpoints f(0 2) 6 and f(3 0) 6. The absolute minimum is f(3 0) 6ߜ ߜ ߜ
and f(0 2) 6 when t 0, respectively; the absolute maximum is f 2 6 2 when t .ßœ œ ß œ œ
1 1
##
Š‹
ÈÈ
32
4
È
(b) y x (2 sin t)( 3 sin t) (3 cos t)(2 cos t) 6 cos t sin t 6 cos 2t 0
dg g g dy dy
dt x dt y dt dt dt
dx dx
œ œ œ œ œœ
``
``
##
ab
t , for 0 t .Êœ ŸŸ
11
44
3
1
(i) On the semi-ellipse, g(x y) xy 6 sin t cos t. Then g 2 3 when t , andßœ œ ß œ œ Š‹
È
32
4
È
#
1
g 2 3 when t . At the endpoints, g( 3 0) g( 0) 0. The absolute minimum isŠ‹
È
ßœ œ ß œ $ß œ
32 3
4
È
#
1
g 2 3 when t ; the absolute maximum is g 2 3 when t .Š‹ Š‹
ÈÈ
ßœ œßœœ
32 323
4 4
È È
# #
1 1
(ii) On the quarter ellipse, at the endpoints g( 2) 0 and g( 0) 0. The absolute minimum is g(3 0) 0!ߜ $ߜ ߜ
and g(0 2) 0 at t 0, respectively; the absolute maximum is g 2 3 when t .ßœ œ ß œ œ
1 1
##
Š‹
È
32
4
È

910. 906 Chapter 14 Partial Derivatives
(c) 2x 6y (6 cos t)( 3 sin t) (12 sin t)(2 cos t) 6 sin t cos t 0
dh h dx h dx
dt x dt y dt dt dt
dy dy
œ œ œ œœ
``
``
t 0, , for 0 t , yielding the points (3 0), (0 2), and ( 3 0).Êœ ŸŸ ß ß ß
1
#
11
(i) On the semi-ellipse, y 0 so that h(3 0) 9, h(0 2) 12, and h( 3 0) 9. The absolute minimum is  ßœ ßœ ßœ
h(3 0) 9 and h( 3 0) 9 when t 0, respectively; the absolute maximum is h( 2) 12 whenßœ ßœ œ !ßœ 1
t. œ
1
#
(ii) On the quarter ellipse, the absolute minimum is h(3 0) 9 when t 0; the absolute maximum isßœ œ
h( 2) 12 when t .!ß œ œ
1
#
55. y x
df f dx f dx
dt x dt y dt dt dt
dy dy
œ œ
``
``
(i) x 2t and y t 1 (t 1)(2) (2t)(1) 4t 2 0 t x 1 and y withœœ Êœ œœÊœ Êœ œ
df
dt
""
##
f 1 . The absolute minimum is f 1 when t ; there is no absoluteˆ‰ ˆ‰ßœ ßœ œ
"" "" "
## ## #
maximum.
(ii) For the endpoints: t 1 x 2 and y 0 with f( 2 0) 0; t 0 x 0 and y 1 withœÊœ œ ßœ œÊœ œ
f( 1) 0. The absolute minimum is f 1 when t ; the absolute maximum is f(0 1) 0!ß œ ßœ œ ßœˆ‰
"" "
## #
and f( 0) 0 when t 1, 0 respectively.#ß œ œ
(iii) There are no interior critical points. For the endpoints: t 0 x 0 and y 1 with f(0 1) 0;œÊœ œ ßœ
t 1 x 2 and y 2 with f(2 2) 4. The absolute minimum is f(0 1) 0 when t 0; the absoluteœÊœ œ ßœ ßœ œ
maximum is f(2 2) 4 when t 1.ßœ œ
56. (a) 2x 2y
df f dx f dx
dt x dt y dt dt dt
dy dy
œ œ
``
``
(i) x t and y 2 2t (2t)(1) 2(2 2t)( 2) 10t 8 0 t x and y withœœ Êœ œ œÊœÊœ œ
df 4 4 2
dt 5 5 5
f . The absolute minimum is f when t ; there is no absoluteˆ‰ ˆ‰
42 6 4 4 42 4 4
55 5 25 5 55 5 5
ßœ œßœœ
"
#
maximum along the line.
(ii) For the endpoints: t 0 x 0 and y 2 with f(0 2) 4; t 1 x 1 and y 0 with f(1 0) 1.œÊœ œ ßœ œÊœ œ ßœ The absolute minimum is f at the interior critical point when t ; the absolute maximum isˆ‰
42 4 4
55 5 5
ßœ œ
f(0 2) 4 at the endpoint when t 0.ßœ œ
(b)
dg g g dy 2y dy
dt x dt y dt dt dt
dx 2x dx
xy xy
œ œ
``
``


’“’“
ab ab

(i) x t and y 2 2t x y 5t 8t 4 5t 8t 4 [( 2t)(1) ( 2)(2 2t)( 2)]œœ Ê œ Êœ
## # #
#dg
dt
ab
5t 8t 4 ( 10t 8) 0 t x and y with g . The absoluteœ œÊœÊœ œ ß œ œab ˆ‰
#
# "44242 5
55555 4 ˆ‰
4
5
maximum is g when t ; there is no absolute minimum along the line since x and y can beˆ‰
42 5 4
55 4 5
ßœ œ
as large as we please.
(ii) For the endpoints: t 0 x 0 and y 2 with g(0 2) ; t 1 x 1 and y 0 with g(1 0) 1.œÊœ œ ßœ œÊœ œ ßœ
"
4
The absolute minimum is g(0 2) when t 0; the absolute maximum is g when t .ßœ œ ß œ œ
"
45545
42 5 4
ˆ‰
57. m andœœ
(2)( 1) 3( 4)
(2) 3(10) 13
20"


b1 (2)œ œ
"
31313
20 9
‘ˆ‰
yx; yÊœ œ
20 9 71
13 13 13
¸
x4œ
k x y x x y
112 1 2
20 1 0 0
33 4 9 12
2 1 10 14
kk kk k
#


D
58. m andœœ
(0)(5) 3(6)
(0) 3(8) 4
3


b5(0)œ œ
"
34 3
35
‘
yx; yÊœ œ
35 14
43 3
¸
x4œ
k x y x x y
120 4 0
20 2 0 0
32 3 4 6
058 6
kk kk k
#

D

911. Section 14.7 Extreme Values and Saddle Points 907
59. m andœœ
(3)(5) 3(8)
(3) 3(5) 2
3


b5(3)œ œ
"
32 6
31
‘
yx; yÊœ œ
31 37
26 6
¸
x4œ
k x y x x y
10 0 0 0
21 2 1 2
32 3 4 6
355 8
kk kk k
#
D
60. m andœœ
(5)(5) 3(10)
(5) 3(13) 14
5


b5(5)œ œ
"
314 14
515
‘
yx; yÊœ œœ
515 355
14 14 14
¸
x4œ #
k x y x x y
10 1 0 0
22 2 4 4
33 2 9 6
5 5 13 10
kk kk k
#
D
61. m 0.122 andœ¸
(162)(41.32) 6(1192.8)
(162) 6(5004)



b 41.32 (0.122)(162) 3.59œ ¸
"
6
cd
y 0.122x 3.59ʜ

k x y x x y
1 12 5.27 144 63.24
2 18 5.68 324 102.24
3 24 6.25 576 150
4 30 7.21 900 216.3
5 36 8.20 1296 295.2
6 42 8.7
kk kk k
#
1 1764 365.82
162 41.32 5004 1192.8D
62. m 51,545œ¸
(0.001863)(91) 4(0.065852)
(0.001863) 4(0.000001323)



and b (91 51,545(0.001863)) 1.26œ ¸
"
4
F 51,545 1.26ʜ
"
D


k F F
1 0.001 51 0.000001 0.051
2 0.0005 22 0.00000025 0.011
3 0.00024 14 0.0000000576 0.00336
40.
ˆ‰ ˆ‰ ˆ‰
"""
#
DDD

kkk
kk
000123 4 0.0000000153 0.000492
0.001863 91 0.000001323 0.065852D

912. 908 Chapter 14 Partial Derivatives
63. (b) mœ
(3201)(17,785) 10(5,710,292)
(3201) 10(1,430,389)



0.0427 and b [17,785 (0.0427)(3201)]¸œ
"
10
1764.8 y 0.0427K 1764.8¸Êœ

(c) K 364 y (0.0427)(364)œÊœ
y (0.0427)(364) 1764.8 1780Êœ ¸
k K y K K y
1 1 1761 1 1761
2 75 1771 5625 132,825
3 155 1772 24,025 274,660
4 219 1775 47,961 388,725
5 271 1777
kk kk
#
73,441 481,567
6 351 1780 123,201 624,780
7 425 1783 180,625 757,775
8 503 1786 253,009 898,358
9 575 1789 330,625 1,028,675
10 626 1791 391,876 1,121,166
D3201 17,785 1,430,389 5,710,292
64. m 1.04 andœ¸
(123)(140) 16(1431)
(123) 16(1287)



b [140 (1.04)(123)] 0.755œ ¸
"
16
y 1.04x 0.755ʜ
k x y x x y
13399
22244
3461624
42346
5542520
6532515
7 9 11 81 99
8 12 9 144 108
9 8 10 64 80
10 13 16 169 208
11 14 13 196 182
12 3 5 9 15
13 4 6
kk kk k
#
16 24
14 13 19 169 247
15 10 15 100 150
16 16 15 256 240
123 140 1287 1431D
65-70. Example CAS commands:
:Maple
f := (x,y) -> x^2+y^3-3*x*y;
x0,x1 := -5,5;
y0,y1 := -5,5;
plot3d( f(x,y), x=x0..x1, y=y0..y1, axes=boxed, shading=zhue, title="#65(a) (Section 14.7)" );
plot3d( f(x,y), x=x0..x1, y=y0..y1, grid=[40,40], axes=boxed, shading=zhue, style=patchcontour, title="#65(b)
(Section 14.7)" );
fx := D[1](f); # (c)
fy := D[2](f);
crit_pts := solve( {fx(x,y)=0,fy(x,y)=0}, {x,y} );
fxx := D[1](fx); # (d)
fxy := D[2](fx);
fyy := D[2](fy);
discr := unapply( fxx(x,y)*fyy(x,y)-fxy(x,y)^2, (x,y) );
for CP in {crit_pts} do # (e)
eval( [x,y,fxx(x,y),discr(x,y)], CP );

913. Section 14.8 Lagrange Multipliers 909
end do;
# (0,0) is a saddle point
# ( 9/4, 3/2) is a local minimum
: (assigned functions and bounds will vary)Mathematica
Clear[x,y,f]
f[x_,y_]:= x y 3x y
23

xmin= 5; xmax= 5; ymin= 5; ymax= 5;
Plot3D[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, AxesLabel {x, y, z}]Ä
ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading False, Contours 40]ÄÄ
fx= D[f[x,y], x];
fy= D[f[x,y], y];
critical=Solve[{fx==0, fy==0},{x, y}]
fxx= D[fx, x];
fxy= D[fx, y];
fyy= D[fy, y];
discriminant= fxx fyy fxy
2
{{x, y}, f[x, y], discriminant, fxx} /.critical
14.8 LAGRANGE MULTIPLIERS
1. f y x and g 2x 4y so that f g y x (2x 4y ) y 2x and x 4y™™ ™™œ œ œÊ œ Êœ œij i j ij i j -- --
x 8x or x 0.Êœ Êœ„ œ--
#
È
2
4
CASE 1: If x 0, then y 0. But (0 0) is not on the ellipse so x 0.œœß Á
CASE 2: x 0 x 2y 2y 2y 1 y .ÁÊœ„ Êœ„ Ê„ œÊœ„-
È
2
4
ÈÈ
Š‹
#
# "
#
Therefore f takes on its extreme values at and . The extreme values of f on the ellipseŠ‹Š ‹„ß „ß
ÈÈ
22
22
""
##
are .„
È
2
#
2. f y x and g 2x 2y so that f g y x (2x 2y ) y 2x and x 2y™™ ™™œ œ œÊ œ Êœ œij i j ij i j -- --
x 4x x 0 or .Êœ Êœ œ„--
# 1
2
CASE 1: If x 0, then y 0. But (0 0) is not on the circle x y 10 0 so x 0.œœß œÁ
##
CASE 2: x 0 y 2x x x x 10 0 x 5 y 5.ÁÊœ„Êœ „ œ„Ê „ œÊœ„ Êœ„-
1
2
ˆ‰ ab
ÈÈ
"
#
# #
Therefore f takes on its extreme values at 5 5 and 5 5 . The extreme values of f on theŠ‹Š ‹
ÈÈ È È
„ß „ß
circle are 5 and 5.
3. f 2x 2y and g 3 so that f g 2x 2y ( 3 ) x and y™™™™œ œ œÊ œ Êœ œij ij ijij --
--
##
3
3 10 2 x 1 and y 3 f takes on its extreme value at (1 3) on the line.Ê œÊœ Êœ œÊ ߈‰ ˆ ‰
--
##
3
-
The extreme value is f( ) 49 1 9 39."ß $ œ œ
4. f 2xy x and g so that f g 2xy x ( ) 2xy and x™™™™œ œ œÊ œÊœ œij ij ij ij
###
----
2xy x x 0 or 2y x.ʜʜ œ
#
CASE 1: If x 0, then x y 3 y 3.œ œÊœ
CASE 2: If x 0, then 2y x so that x y 3 2y y 3 y 1 x 2.Áœ œÊ œÊœÊœ
Therefore f takes on its extreme values at ( 3) and ( ). The extreme values of f are f(0 3) 0 and!ß #ß " ß œ
f( 1) 4.#ß œ

914. 910 Chapter 14 Partial Derivatives
5. We optimize f(x y) x y , the square of the distance to the origin, subject to the constraintߜ
##
g(x y) xy 54 0. Thus f 2x 2y and g y 2xy so that f g 2x 2yßœ œœ œ œÊ
##
™™ ™™ij i j ij -
y 2xy 2x y and 2y 2 xy.œ Êœ œ---ab
##
ij
CASE 1: If y 0, then x 0. But (0 0) does not satisfy the constraint xy 54 so y 0.œœß œÁ
#
CASE 2: If y 0, then 2 2 x x 2 y y . Then xy 54 54ÁœÊœÊœÊœ œÊ œ--
"" " ## #
-- - --
ˆ‰ ˆ‰ˆ ‰
22

x 3 and y 18 x 3 and y 3 2.ʜʜʜ œÊœ œ„--
$#""
27 3 È
Therefore 3 2 are the points on the curve xy 54 nearest the origin (since xy 54 has pointsŠ‹
È
$ß „ œ œ
##
increasingly far away as y gets close to 0, no points are farthest away).
6. We optimize f(x y) x y , the square of the distance to the origin subject to the constraint g(x y)ßœ ß
##
x y 2 0. Thus f 2x 2y and g 2xy x so that f g 2x 2xy and 2y xœ œœ œ œÊœ œ
# ##
™™ ™™ij ij ---
, since x 0 y 0 (but g(0 0) 0). Thus x 0 and 2x 2xy x 2yÊœ œÊœ ßÁ Á œ Ê œ-
2y 2y
x x

ˆ‰
##
2y y 2 0 y 1 (since y 0) x 2 . Therefore 2 1 are the points on the curveÊ œÊœ "Êœ„ „ ßab
ÈÈ
Š‹
#
x y 2 nearest the origin (since x y 2 has points increasingly far away as x gets close to 0, no points are
##
œœ
farthest away).
7. (a) f and g y x so that f g (y x ) 1 y and 1 x y and™™ ™™œ œ œÊ œ Êœ œ Êœij i j ij i j -- --
"
-
x 16 . Use since x 0 and y 0. Then x 4 and y 4 the minimumœÊ œ Êœ„ œ "" œœÊ
"" " "
--
--
44
value is 8 at the point (4 4). Now, xy 16, x 0, y 0 is a branch of a hyperbola in the first quadrantߜ ""
with the x-and y-axes as asymptotes. The equations x y c give a family of parallel lines with m 1.œœ
As these lines move away from the origin, the number c increases. Thus the minimum value of c occurs
where x y c is tangent to the hyperbola's branch.œ
(b) f y x and g so that f g y x ( ) y x y y 16 y 8™™ ™™œ œ œÊ œÊœœ œÊœij ij ij ij ---
x 8 f( ) 64 is the maximum value. The equations xy c (x 0 and y 0 or x 0 and y 0ÊœÊ)ß)œ œ ""##
to get a maximum value) give a family of hyperbolas in the first and third quadrants with the x- and y-
axes as asymptotes. The maximum value of c occurs where the hyperbola xy c is tangent to the lineœ
x y 16.œ
8. Let f(x y) x y be the square of the distance from the origin. Then f 2x 2y andßœ œ
##
™ ij
g (2x y) (2y x) so that f g 2x (2x y) and 2y (2y x) ™™™œ œÊœ œ Êœij -- - -
2y
2y x
2x (2x y) x(2y x) y(2x y) x y y x.Êœ Ê œ ʜʜ„Š‹
2y
2y x
##
CASE 1: y x x x(x) x 1 0 x and y x.œÊ œÊœ„ œ
## "
È
3
CASE 2: y x x x( x) ( x) 1 0 x 1 and y x. Thus fœÊ œÊœ„ œ ßœ
## ""
Š‹
ÈÈ
33
2
3
f and f(1 1) 2 f( 1 1).œß ßœœ ߊ‹
""
ÈÈ
33
Therefore the points (1 1) and ( 1 1) are the farthest away; and are the closestß ßß ßŠ‹Š ‹
"" " "
ÈÈ È È
33 3 3
points to the origin.
9. V r h 16 r h 16 r h g(r h) r h 16; S 2 rh 2 r S (2 h 4 r) 2 r andœÊœÊœÊßœ œ Êœ 111 11 111
### # #
™ ij
g 2rh r so that S g (2 rh 4 r) 2 r 2rh r 2 rh 4 r 2rh and™™™œ œÊ œ Ê œij i j ij
# #
-111- 11- ab
2 r r r 0 or . But r 0 gives no physical can, so r 0 2 h 4 r1- - - 1 1œÊœ œ œ ÁÊœÊ
# 22
rr
11
2rh 2r h 16 r (2r) r 2 h 4; thus r 2 cm and h 4 cm give the only extremeœ Ê œÊ œ ʜʜ œ œˆ‰
2
r
1 #
surface area of 24 cm . Since r 4 cm and h 1 cm V 16 cm and S 40 cm , which is a larger111
#$#
œœÊœ œ
surface area, then 24 cm must be the minimum surface area.1
#

915. Section 14.8 Lagrange Multipliers 911
10. For a cylinder of radius r and height h we want to maximize the surface area S 2 rh subject to the constraintœ1
g(r h) r a 0. Thus S 2 h 2 r and g 2r so that S g 2 h 2 r andßœ œœ œ œÊœ
##
# #
#ˆ‰
h h
™™™™11 - 1-ij ij
2 r and 2 r 4r h h 2r r a 2r a r1-1œÊœ œ Ê œÊœÊ œÊ œÊœ
-1 1hh hh 4r a
rr 4 2##
## # # ##
ˆ‰ˆ‰

È
h a 2 S 2 a 2 2 a .Êœ Êœ œ
ÈÈ
Š‹Š ‹11
a
2
È
#
11. A (2x)(2y) 4xy subject to g(x y) 1 0; A 4y 4x and g so that Aœœ ßœ œœ œ
xx
16 9 8 9
y2y

™™ ™ ij ij
g 4y 4x 4y and 4x and 4xœÊ œ Êœ œ Êœ œ-- ---™ ij ij ˆ ‰ ˆ‰ ˆ‰ ˆ‰ˆ ‰
xx
89 8 9 x 9x
2y 2y 32y 2y 32y
y x 1 x 8 x 2 2 . We use x 2 2 since x represents distance.Êœ„ Ê œÊ œÊœ„ œ
3x
4169
x


ˆ‰
3
4
# ÈÈ
Then y 2 2 , so the length is 2x 4 2 and the width is 2y 3 2.œœ œ œ
3
4
32
Š‹
ÈÈÈ
È
#
12. P 4x 4y subject to g(x y) 1 0; P 4 4 and g so that P gœ ßœ œœ œ œ
x2x
ab a b
y2y



™™ ™™ij i j -
4 and 4 and 4 y x 1 Êœ œ Êœ œ Êœ Ê œÊ ˆ‰ ˆ‰ ˆ‰ Š‹ Š‹
2x 2a 2a b x x b x
abxbxaaba
2y 2y
x
a






"---
Š‹
b
a
1 a b x a x , since x 0 y x width 2xœÊ œÊœ "Êœ œ Ê œ œab Š‹
###%

abb2a
ab ab ab a



ÈÈÈ and height 2y perimeter is P 4x 4y 4 a bœœ Ê œ œœ
2b 4a 4b
ab ab


ÈÈ

##
È
13. f 2x 2y and g (2x 2) (2y 4) so that f g 2x 2y [(2x 2) (2y 4) ]™™ ™™œ œ œœ œ ij i j ij i j -- 2x (2x 2) and 2y (2y 4) x and y , 1 y 2x x 2x (2x) 4(2x)Êœ œ Êœ œ ÁÊœ Ê -- -
--
--
##
11
2
0 x 0 and y 0, or x 2 and y 4. Therefore f(0 0) 0 is the minimum value and f(2 4) 20 is theœÊœœœœ ßœ ßœ
maximum value. (Note that 1 gives 2x 2x 2 or 2, which is impossible.)-œœ !œ
14. f 3 and g 2x 2y so that f g 3 2 x and 1 2 y and 1 2 y™™ ™™œ œ œÊœ œÊœ œij i j ----
33
2x 2x
ˆ‰
y x 4 10x 36 x x and y , or x andÊœ Ê œÊ œ Êœ„ Êœ œ œ
xx 66 26
33 10 10 10 10
##
#
ˆ‰
ÈÈ È È
y . Therefore f 6 2 10 6 12.325 is the maximum value, andœß œ œ ¸
26220
10 10 10 10
ÈÈÈÈ
Š‹
È
f 2 10 6 0.325 is the minimum value.Š‹
È
ßœ ¸
62
10 10
ÈÈ
15. T (8x 4y) ( 4x 2y) and g(x y) x y 25 0 g 2x 2y so that T g™™™™œ ßœ œÊ œ œij ij
##
-
(8x 4y) ( 4x 2y) (2x 2y ) 8x 4y 2 x and 4x 2y 2 y y , 1Ê œ Ê œ œÊœ Áijij ----


2x
1-
8x 4 2 x x 0, or 0, or 5.Ê œÊœœœˆ‰


2x
1-
---
CASE 1: x 0 y 0; but (0 0) is not on x y 25 so x 0.œÊœ ß œÁ
## CASE 2: 0 y 2x x (2x) 25 x 5 and y 2x.-œÊœ Ê œÊœ„ œ
## È
CASE 3: 5 y x 25 x 2 5 x 2 5 and y 5, or x 2 5-œÊœ œ Ê œÊœ„ Êœ œ œ

##
#
#
2x x x
4
ˆ‰ ÈÈ È È
and y 5 . œÈ
Therefore T 5 2 5 0 °T 5 2 5 is the minimum value and T 2 5 5 125 °Š‹ Š ‹ Š ‹
ÈÈ È È È Èßœœ ß ß œ
T 2 5 5 is the maximum value. (Note: 1 x 0 from the equation 4x 2y 2 y; but weœ ߜʜ œŠ‹
ÈÈ --
found x 0 in CASE 1.)Á
16. The surface area is given by S 4 r 2 rh subject to the constraint V(r h) r r h 8000. Thusœ ßœ œ11 11
#$# 4
3
S (8 r 2 h) 2 r and V 4 r 2 rh r so that S V (8 r 2 h) 2 r™™ ™™œ œ œœ 11 1 1 1 1 - 11 1ij ij ij ab
##
4r 2rh r 8r 2h 4r 2rh and 2r r r 0 or 2 r. But r 0œ Ê œ œÊœ œ Á-1 1 1 1 1 -1 1 1 -1 -cd abab
## # #
ij
so 2 r 4r h 2r rh h 0 the tank is a sphere (there is no cylindrical part) andœÊœÊ œ ÊœÊ--
22
rr
ab
#
r 8000 r 10 .
46 3
1
$
"Î$
œÊœ ˆ‰
1

916. 912 Chapter 14 Partial Derivatives
17. Let f(x y z) (x 1) (y 1) (z 1) be the square of the distance from (1 1 1). Thenßß œ ßß
###
f 2(x 1) 2(y 1) 2(z 1) and g 2 3 so that f g™™™™œ œ œijk ijk -
2(x 1) 2(y 1) 2(z 1) ( 2 3 ) 2(x 1) , 2(y 1) 2 , 2(z 1) 3Ê œ Ê œ œ œijkijk ----
2(y 1) 2[2(x 1)] and 2(z 1) 3[2(x 1)] x z 2 3 or z ; thusÊ œ œ Êœ Ê œœ
y1 y1 3y1
###
ˆ‰
2y 3 13 0 y 2 x and z . Therefore the point 2 is closest (since no
y1 3y1 35 35
## ## ##
œÊœÊœ œ ß߈‰ ˆ ‰
point on the plane is farthest from the point (1 1 1)).ßß
18. Let f(x y z) (x 1) (y 1) (z 1) be the square of the distance from (1 1 1). Thenßß œ ßß
###
f 2(x 1) 2(y 1) 2(z 1) and g 2x 2y 2z so that f g x 1 x, y 1 y™™™™œ œ œÊ œ œijk ijk ---
and z 1 z x , y , and z for 1 4œÊœ œ œÁÊ œ--
"" " """

###
1 1 1 111- - - ---
ˆ‰ˆ‰ˆ‰
x , y , z or x , y , z . The largest value of fÊœ„Êœ œ œœ œœ
"
"-
2222 222
3333 333
ÈÈÈÈ ÈÈÈ
occurs where x 0, y 0, and z 0 or at the point on the sphere. ßߊ‹
22 2
33 3
ÈÈ È
19. Let f(x y z) x y z be the square of the distance from the origin. Then f 2x 2y 2z andßß œ œ
###
™ ijk
g 2x 2y 2z so that f g 2x 2y 2z (2x 2y 2z ) 2x 2x , 2y 2y ,™™™œ œÊ œ Êœ œ ijk ijk ijk -- --
and 2z 2z x 0 or 1.œ Êœ œ--
CASE 1: 1 2y 2y y 0; 2z 2z z 0 x 1 0 x 1 and y z 0.-œÊ œ Êœ œ ÊœÊ œÊœ„ œœ
#
CASE 2: x 0 y z 1, which has no solution.œÊ œ
##
Therefore the points on the unit circle x y 1, are the points on the surface x y z 1 closest to the origin
## ###
œ œÞ
The minimum distance is 1.
20. Let f(x y z) x y z be the square of the distance to the origin. Then f 2x 2y 2z andßß œ œ
###
™ ijk
g y x so that f g 2x 2y 2z (y x ) 2x y, 2y x, and 2z™™™œ œÊ œ Êœ œ œ ijk i j k ijk -----
x 2y y 0 or 2.Êœ Ê œ Êœ œ„
--yy
##
--Š‹
CASE 1: y 0 x 0 z 1 0 z 1.œÊœÊ œÊœ
CASE 2: 2 x y and z 1 x ( 1) 1 0 x 2 0, so no solution.-œÊœ œ Ê œÊ œ
##
CASE 3: 2 x y and z 1 ( y)y 1 1 0 y 0, again.-œÊœ œÊ œÊœ
Therefore (0 0 1) is the point on the surface closest to the origin since this point gives the only extreme valueßß
and there is no maximum distance from the surface to the origin.
21. Let f(x y z) x y z be the square of the distance to the origin. Then f 2x 2y 2z andßß œ œ
###
™ ijk
g y x 2z so that f g 2x 2y 2z ( y x 2z ) 2x y , 2y x , and™™™œ œÊ œ Êœ œij k i j k ij k -- --
2z 2z 1 or z 0.œÊœ œ--
CASE 1: 1 2x y and 2y x y 0 and x 0 z 4 0 z 2 and x y 0.-œÊ œ œÊœ œÊ œÊœ„ œœ
#
CASE 2: z 0 xy 4 0 y . Then 2x , and x xœÊ œÊœ œÊœ œÊœ
44x8 8x
xx x x
-- -

##
Š‹
x 16 x 2. Thus, x 2 and y 2, or x = 2 and y 2.ʜʜ„ œ œ œ
%
Therefore we get four points: ( 2 0), ( 2 2 0), (0 0 2) and ( 0 2). But the points ( 0 2) and ( 2)#ßßßß ßß !ßß !ß ß !ß !ß
are closest to the origin since they are 2 units away and the others are 2 2 units away.
È
22. Let f(x y z) x y z be the square of the distance to the origin. Then f 2x 2y 2z andßß œ œ
###
™ ijk
g yz xz xy so that f g 2x yz, 2y xz, and 2z xy 2x xyz and 2y yxz™™™œ œÊœ œ œÊœ œijk ------
##
x y y x z x x x x 1 x 1 the points are (1 1 1), ( 1 1),Ê œ Êœ„Êœ„Ê „ „œÊœ„Ê ßß "ß ß
##
abab
( ), and ( 1 1, 1)."ß"ß " ß
23. f 2 5 and g 2x 2y 2z so that f g 2 5 (2x 2y 2z ) 1 2x ,™™ ™™œ œ œÊ œ Êœijk i j k ijk i j k -- -
2 2y , and 5 2z x , y 2x, and z 5x x ( 2x) (5x) 30 x 1.œœÊœœ œ œœ Ê œÊœ„--
""
##
###
-- -
5

917. Section 14.8 Lagrange Multipliers 913
Thus, x 1, y 2, z 5 or x 1, y 2, z 5. Therefore f(1 2 5) 30 is the maximum value andœœ œœ œœ ßßœ
f( 1 2 5) 30 is the minimum value.ßßœ
24. f 2 3 and g 2x 2y 2z so that f g 2 3 (2x 2y 2z ) 1 2x ,™™ ™™œ œ œÊ œ Êœijk i j k ijk i j k -- -
2 2y , and 3 2z x , y 2x, and z 3x x (2x) (3x) 25 x .œœÊœœœœœÊ œÊœ„--
""
##
###
-- -
35
14
È
Thus, x , y , z or x , y , z . Therefore fœœœ œ œ œ ßß
5 10 15 5 10 15 5 10 15
14 14 14 14 14 14 14 14 14
È È È È È È ÈÈÈ
Š‹
5 14 is the maximum value and f , 5 14 is the minimum value.œ ß œ
ÈÈ
Š‹
51015
14 14 14
ÈÈÈ
25. f(x y z) x y z and g(x y z) x y z 9 0 f 2x 2y 2z and g so thatßß œ ßß œ œÊ œ œ
###
™™ ijk ijk
f g 2x 2y 2z ( ) 2x , 2y , and 2z xyz xxx90™™œÊ œ Êœ œ œÊœœÊ œ----- ijkijk
x 3, y 3, and z 3.Êœ œ œ
26. f(x y z) xyz and g(x y z) x y z 16 0 f yz xz xy and g 2z so thatßß œ ßß œ œÊ œ œ
#
™™ ijk ijk
f g yz xz xy ( 2z ) yz , xz , and xy 2z yz xz z 0 or y x.™™œÊ œ Êœ œ œ ʜʜ œ----- ijkijk
But z 0 so that y x x 2z and xz . Then x 2z(xz) x 0 or x 2z . But x 0 so that" œÊ œ œ œ Êœ œ "
## #
--
x 2z y 2z 2z 2z z 16 z . We use z since z 0. Then x and yœÊœÊ œÊœ„ œ " œœ
##### 4 4 32 32
55 55ÈÈ
which yields f .Š‹
32 32 4 4096
55 5255
ßß œ
ÈÈ
27. V 6xyz and g(x y z) x y z 1 0 V 6yz 6xz 6xy and g 2x 2y 2z so thatœßßœ œÊ œ œ
###
™™ ijk ijk
V g 3yz x, 3xz y, and 3xy z 3xyz x and 3xyz y y x z x™™œÊœ œ œÊœ œÊœ„Êœ„------
##
x x x 1 x since x 0 the dimensions of the box are by by for maximumÊ œÊœ "Ê
### "
È ÈÈÈ
3 333
222
volume. (Note that there is no minimum volume since the box could be made arbitrarily thin.)
28. V xyz with x y z all positive and 1; thus V xyz and g(x y z) bcx acy abz abc 0œßß œœ ßßœ œ
xz
abc
y
V yz xz xy and g bc ac ab so that V g yz bc, xz ac, and xy abÊœ œ œÊœ œ œ™™ ™™ijk ijk ----
xyz bcx, xyz acy, and xyz abz 0. Also, bcx acy abz bx ay, cy bz, andÊœ œ œ ÊÁ œ œ Êœ œ-- -- ---
cx az y x and z x. Then 1 x x 1 1 xœÊœ œ œÊ œÊ œÊœ
bcxcxbc 3x a
aaabzabaca a 3
y ""
ˆ‰ ˆ‰
y and z V xyz is the maximum volume. (Note thatÊœ œ œ œÊ œ œ œˆ‰ˆ‰ ˆ‰ˆ‰ ˆ‰ˆ‰ˆ‰
ba b ca c abc abc
a3 3 a3 3 333 27
there is no minimum volume since the box could be made arbitrarily thin.)
29. T 16x 4z (4y 16) and g 8x 2y 8z so that T g 16x 4z (4y 16)™™™™œ œ œÊ ij k ijk ij k -
(8x 2y 8z ) 16x 8x , 4z 2y , and 4y 16 8z 2 or x 0.œ Êœ œ œÊœ œ-----ijk
CASE 1: 2 4z 2y(2) z y. Then 4z 16 16z z y . Then-œÊ œ Êœ œÊœ Êœ
44
33
4x 4 16 x .
#
##
œÊœ„ˆ‰ ˆ‰
44 4
33 3
CASE 2: x 0 4y 16 8z y 4y 4z 4(0) y y 4y 16 0œÊœ Ê œÊ œÊ œ-
2z 2z
yy
Š‹ ab
#####
y 2y 8 0 (y 4)(y 2) 0 y 4 or y 2. Now y 4 4z 4 4(4)Ê œÊ œÊœ œ œÊ œ
# ##
z 0 and y 2 4z ( 2) 4( 2) z 3.Êœ œ Êœ Êœ„
## È
The temperatures are T 642 , T(0 4 0) 600 °, T 0 2 3 600 24 3 , andˆ‰ Š‹Š ‹
ÈÈ
„ßß œßßœß ßœ
444 2
333 3
°
°
T 0 2 3 600 24 3 641.6 °. Therefore are the hottest points on the space probe.Š‹Š ‹
ÈÈ ˆ‰ßß œ ¸„ß ß
°
444
333
30. T 400yz 400xz 800xyz and g 2x 2y 2z so that T g™™™™œ œ œ
##
ijk ijk -
400yz 400xz 800xyz (2x 2y 2z ) 400yz 2x , 400xz 2y , and 800xyz 2z .Ê œ Êœ œ œ
## # #
ijkijk ----
Solving this system yields the points 1 0 , 1 0 0 , and . The correspondingabab Š‹!ß„ß „ßß „ß„ß„
""
## #
È
2

918. 914 Chapter 14 Partial Derivatives
temperatures are T 1 0 0, T 1 0 0 0, and T 50. Therefore 50 is theabab Š‹!ß „ ß œ „ ß ß œ „ ß „ ß „ œ „
""
## #
È
2
maximum temperature at and ; 50 is the minimum temperature atŠ‹Š ‹
"" " "
## # # # #
ßß„ ßß„
ÈÈ
22
and .Š‹Š‹
"" ""
## # ## #
ßß„ ßß„
ÈÈ
22
31. U (y 2) x and g 2 so that U g (y 2) x (2 ) y 2 and™™™™œ œ œÊ œ Ê#œij ij ij ij ---
x y 2 2x y 2x 2 2x (2x 2) 30 x 8 and y 14. Therefore U(8 14) $128œÊ œÊœ Ê œÊœ œ ßœ-
is the maximum value of U under the constraint.
32. M (6 z) 2y x and g 2x 2y 2z so that M g (6 z) 2y x™™™™œ œ œÊ ijk ijk ijk -
(2x 2y 2z ) 6 z 2x , 2y 2y , x 2z 1 or y 0.œ Êœ œœÊœ œ-----ijk
CASE 1: 1 6 z 2x and x 2z 6 z 2( 2z) z 2 and x 4. Then-œÊœ œÊœ Êœ œ
(4)y2360 y 4. œÊœ„
###
CASE 2: y 0, 6 z 2x , and x 2z 6 z 2x 6z z xœ œœÊœÊ œÊ œ---
xx
2z 2z
ˆ‰
##
6z z 0 z 36 z 6 or z 3. Now z 6 x 0 x 0; z 3Ê œÊœ œœ ʜʜ œab
### #
x 27 x 3 3.ʜʜ„
# È
Therefore we have the points 3 3 0 3 , (0 0 6), and 4 4 2 . Then M 3 3 0 3Š‹ Š‹
ÈÈ
ab„ßß ßß ß„ ß ß ß
27 3 60 106.8, M 3 3 0 3 60 27 3 13.2, M(0 0 6) 60, and M( 4 4 2) 12œ ¸ ßß œ ¸ßß œ ßß œ
ÈÈÈ
Š‹
M( 4 4 2). Therefore, the weakest field is at 4 4 2 .œßß ß„ ßab
33. Let g (x y z) 2x y 0 and g (x y z) y z 0 g 2 , g , and f 2x 2 2z
"# "#ßß œ œßßœ œÊ œ œ œ ™™ ™ij jk i j k
so that f g g 2x 2 2z (2 ) ( ) 2x 2 2z 2 ( )™™ ™œ Ê œ Ê œ -. - . -.-.
"# ij k ij jk ij k i jk
2x 2 , 2 , and 2z x . Then 2 2z x x 2z 2 so that 2x y 0Êœ œ œÊœ œ Êœ œ-.- . -
2( 2z 2) y 0 4z 4 y 0. This equation coupled with y z 0 implies z and y .Ê œÊ œ œœ œ
44
33
Then x so that is the point that gives the maximum value f 2œßß ßßœ
2244 244244
3333 333333
ˆ‰ ˆ‰ˆ‰ˆ‰ˆ‰
##
.œ
4
3
34. Let g (x y z) x 2y 3z 6 0 and g (x y z) x 3y 9z 9 0 g 2 3 ,"# "ßß œ œßßœ œÊ œ ™ ijk
g 3 9 , and f 2x 2y 2z so that f g g 2x 2y 2z™™ ™™™
# "#œ œ œ Ê ijk i j k i j k -.
( 2 3 ) ( 3 9 ) 2x , 2y 2 3 , and 2z 3 9 . Then 0 x 2y 3z 6œ Êœ œ œ œ- . -. -. -.ijk ijk
( ) (2 3 ) 6 7 17 6; 0 x 3y 9z 9œ Ê œœ
"
###
-. - . - . - . ˆ‰
927
( ) 3 9 34 91 18. Solving these two equations for and givesÊ Ê œ
"
####
-. - . - . - . - .ˆ‰ˆ ‰
92781
and x , y , and z . The minimum value is-.œœ Êœ œ œ œ œ œ
240 78 81 123 9
59 59 59 59 59
23 39-. -. -.
## #
f . (Note that there is no value of f subject to the constraints becauseˆ‰
81 123 9 369
59 59 59 59 59
21,771
ßß œ œ
maximum
at least one of the variables x, y, or z can be made arbitrary and assume a value as large as we please.)
35. Let f(x y z) x y z be the square of the distance from the origin. We want to minimize f(x y z) subjectßß œ ßß
###
to the constraints g (x y z) y 2z 12 0 and g (x y z) x y 6 0. Thus f 2x 2y 2z ,"#ßß œ œßßœ œœ ™ ijk
g 2 , and g so that f g g 2x , 2y , and 2z 2 . Then™™™™™
"# "#œ œ œ Êœ œ œjk ij -. .-. -
0 y 2z 12 2 12 12 5 24; 0 x y 6 6œ œ Ê œÊ œœ œˆ‰ ˆ‰
- -. ..
## # # # ##
"
--.-.
5
6 12. Solving these two equations for and gives 4 and 4 x 2,Ê œÊ #œ œ œ Ê œ œ
"
# #
-. - . - . - .
.
y 4, and z 4. The point (2 4 4) on the line of intersection is closest to the origin. (There is noœœ œœ ßß
-.
#
-
maximum distance from the origin since points on the line can be arbitrarily far away.)
36. The maximum value is f from Exercise 33 above.ˆ‰
24 4 4
33 3 3
ßßœ

919. Section 14.8 Lagrange Multipliers 915
37. Let g (x y z) z 1 0 and g (x y z) x y z 10 0 g , g 2x 2y 2z , and"# "#
###ßß œ œßßœ œÊ œ œ ™™ kijk
f 2xyz x z x y so that f g g 2xyz x z x y ( ) (2x 2y 2z )™™™™œ œ Ê œ ijk ijkk ijk
## ##
"#
-. -.
2xyz 2x , x z 2y , and x y 2z xyz x x 0 or yz y since z 1.Êœ œ œ ʜʜ œÊœ œ.. .- . ..
##
CASE 1: x 0 and z 1 y 9 0 (from g ) y 3 yielding the points 0 3 1 .œœÊ œÊœ„ ß„ß
#
#
ab
CASE 2: y x z 2y x 2y (since z 1) 2y y 1 10 0 (from g ) 3y 9 0.œÊ œ Ê œ œ Ê œÊ œ
#### ## #
#
y 3 x 2 3 x 6 yielding the points 6 3 .Êœ„ Ê œ „ Êœ„ „ ß„ ß"
ÈÈÈ ÈÈ
Š‹ Š ‹
#
#
Now f 3 1 1 and f 6 3 6 3 1 1 6 3. Therefore the maximum of f isab Š‹Š‹
ÈÈ È È
!ß „ ß œ „ ß „ ß " œ „ œ„
1 6 3 at 6 3 1 , and the minimum of f is 1 6 3 at 6 3 . „ßß „ßß"
ÈÈÈ ÈÈÈ
Š‹ Š ‹
38. (a) Let g (x y z) x y z 40 0 and g (x y z) x y z 0 g , g , and
"#"#ßß œ œßßœ œÊ œ œ™™ ijk ijk
w yz xz xy so that w g g yz xz xy ( ) ( )™™™™œ œ Ê œ ijk ijkijk ijk -. - .
"#
yz , xz , and xy yz xz z 0 or y x.Êœ œ œÊœÊœ œ-. -. -.
CASE 1: z 0 x y 40 and x y 0 no solution.œÊ œ œÊ
CASE 2: x y 2x z 40 0 and 2x z 0 z 20 x 10 and y 10 w (10)(10)(20)œÊ œ œÊœ Êœ œ Ê œ
2000 œ
(b) 2 2 is parallel to the line of intersection the line is x 2t 10, nij
ij k
œœ Êœ "" "
"" "
ââ
ââ
ââ
ââ
ââ
ââ
y 2t 10, z 20. Since z 20, we see that w xyz ( 2t 10)(2t 10)(20) 4t 100 (20)œ œœ œœ œab
#
which has its maximum when t 0 x 10, y 10, and z 20.œÊœ œ œ
39. Let g ( y z) y x 0 and g (x y z) x y z 4 0. Then f y x 2z , g , and
"# "
###Bß ß œœßßœ œœ œ™™ ij k ij
g 2x2y2z so that f g g yx2z ( ) (2x2y2z)™™™
#"#œ ™œ Ê œ ijk ijk ij ijk -. - .
y 2x , x 2y , and 2z 2z z 0 or 1.Êœ œ œÊœ œ-.-. . .
CASE 1: z 0 x y 4 0 2x 4 0 (since x y) x 2 and y 2 yielding the pointsœÊ œÊ œœÊœ„œ„
## # ÈÈ
22. Š‹
ÈÈ
„ß„ß!
CASE 2: 1 y 2x and x 2y x y 2(x y) 2x 2(2x) since x y x 0 y 0.--œÊœ œ Êœ Ê œ œÊœÊœ
z 4 0 z 2 yielding the points 2 .Ê œ Ê œ „ !ß !ß „
#
ab
Now, f 2 4 and f 2 2 2. Therefore the maximum value of f is 4 at 2 and theab ab Š‹
ÈÈ
!ß !ß „ œ „ ß „ ß ! œ !ß !ß „
minimum value of f is 2 at 2 2 .Š‹
ÈÈ
„ß„ß!
40. Let f(x y z) x y z be the square of the distance from the origin. We want to minimize f(x y z) subjectßß œ ßß
###
to the constraints g (x y z) 2y 4z 5 0 and g (x y z) 4x 4y z 0. Thus f 2x 2y 2z ,"#
###ßß œ œßßœ œœ ™ ijk
g 2 4 , and g 8x 8y 2z so that f g g 2x 2y 2z™™ ™™™
"# "#œ œ œ Ê jk i j k i j k -.
(2 4 ) (8x 8y 2z ) 2x 8x , 2y 2 8y , and 2z 4 2z x 0 or .œ Êœ œ œ Êœ œ-. .-.-. .jk i j k
"
4
CASE 1: x 0 4(0) 4y z 0 z 2y 2y 4(2y) 5 0 y , or 2y 4( 2y) 5 0œÊ œÊœ„ Ê œÊœ œ
### "
#
y yielding the points and .Êœ !ß ß " !ßß
555
663
ˆ‰ˆ ‰
"
#
CASE 2: y y 0 2z 4(0) 2z z 0 2y 4(0) 5 y and.--œÊœ ÊœÊ œ ÊœÊ œÊœ
""
#44
5
ˆ‰
(0) 4x 4 no solution.
##
#
#
œ ʈ‰
5
Then f 1 and f 25 the point 1 is closest to the origin.ˆ‰ ˆ ‰ˆ ‰ ˆ‰!ß ß œ !ß ßœ œ Ê !ß ß
""""
# #
5 5 5 125
46336936
41. f and g y x so that f g (y x ) 1 y and 1 x y x™™ ™™œ œ œÊ œ Êœ œ Êœij i j ij i j -- --
y 16 y 4 (4 4) and ( 4) are candidates for the location of extreme values. But as x ,ʜʜ„Êß %ß Ä_
#
y and f(x y) ; as x , y 0 and f(x y) . Therefore no maximum or minimum valueÄ_ ß Ä_ Ä _Ä ßÄ _
exists subject to the constraint.

920. 916 Chapter 14 Partial Derivatives
42. Let f(A B C) (Ax By C z ) C (B C 1) (A B C 1) (A C 1) . We wantßß œ œ !
4
k1œ
kk k
## # # #
to minimize f. Then f (A B C) 4A 2B 4C, f (A B C) 2A 4B 4C 4, and
ABßß œ ßß œ f (A B C) 4A 4B 8C 2. Set each partial derivative equal to 0 and solve the system to get A ,
Cßß œ œ
"
#
B , and C or the critical point of f is .œœ ßß
33
44###
"""
ˆ‰
43. (a) Maximize f(a b c) a b c subject to a b c r . Thus f 2ab c 2a bc 2a b c andßß œ œœ
### # # # # ## # # ##
™ ijk
g 2a 2b 2c so that f g 2ab c 2a , 2a bc 2b , and 2a b c 2c™™™œ œÊ œ œ œijk ----
## # # ##
2a b c 2a 2b 2c 0 or a b c .ÊœœœÊœœœ
### # # # # # #
----
CASE 1: 0 a b c 0.-œÊ œ
###
CASE 2: a b c f(abc) aaa and 3a r f(abc) is the maximum value.
# # # ### # #
$
œœ Ê ßßœ œÊ ßßœ Š‹
r
3

(b) The point a b c is on the sphere if a b c r . Moreover, by part (a), abc f a b cŠ‹ Š‹ÈÈ ÈÈÈ È
ßß œœßß
#
(abc) , as claimed.ŸÊ ŸœŠ‹
rrabc
333

$
"Î$
44. Let f(x x x ) a x a x a x a x and g(x x x ) x x x 1. Then we" # "" ## " #
## #ßßáß œ œ á ßßáß œ á
nii nn n n
!
n
i1œ

want f g a (2x ), a (2x ), , a (2x ), 0 x 1™™œÊœ œáœ ÁÊœÊ á œ--- -- ""## nn i
a
244 4
aa a
i
--- -




n
4 a 2 a f(x x x ) a x a a a isÊœ Êœ Êßßáßœ œ œ œ--
## # ##
"Î# "Î#
"#
##
"!! !!!!
Π" Π"
ˆ‰
n
i1
ii iiniii

nnnnn
i1 i1 i1 i1 i1œœœœœ
a
i
--
the maximum value.
45-50. Example CAS commands:
:Maple
f := (x,y,z) -> x*y+y*z;
g1 := (x,y,z) -> x^2+y^2-2;
g2 := (x,y,z) -> x^2+z^2-2;
h := unapply( f(x,y,z)-lambda[1]*g1(x,y,z)-lambda[2]*g2(x,y,z), (x,y,z,lambda[1],lambda[2]) ); # (a)
hx := diff( h(x,y,z,lambda[1],lambda[2]), x ); #(b)
hy := diff( h(x,y,z,lambda[1],lambda[2]), y );
hz := diff( h(x,y,z,lambda[1],lambda[2]), z );
hl1 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[1] );
hl2 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[2] );
sys := { hx=0, hy=0, hz=0, hl1=0, hl2=0 };
q1 := solve( sys, {x,y,z,lambda[1],lambda[2]} ); # (c)
q2 := map(allvalues,{q1});
for p in q2 do # (d)
eval( [x,y,z,f(x,y,z)], p );
``=evalf(eval( [x,y,z,f(x,y,z)], p ));
end do;
Mathematica
: (assigned functions will vary)
Clear[x, y, z, lambda1, lambda2]
f[x_,y_,z_]:= x y y z
g1[x_,y_,z_]:= x y 2
22

g2[x_,y_,z_]:= x z 2
22

h = f[x, y, z] lambda1 g1[x, y, z] lambda2 g2[x, y, z];
hx= D[h, x]; hy= D[h, y]; hz= D[h,z]; hL1=D[h, lambda1]; hL2= D[h, lambda2];
critical=Solve[{hx==0, hy==0, hz==0, hL1==0, hL2==0, g1[x,y,z]==0, g2[x,y,z]==0},

921. Section 14.9 Partial Derivatives with Constrained Variables 917
{x, y, z, lambda1, lambda2}]//N
{{x, y, z}, f[x, y, z]}/.critical
14.9 PARTIAL DERIVATIVES WITH CONSTRAINED VARIABLES
1. w x y z and z x y :œ œ
### ##
(a) w ; 0 and 2x 2y
y
z
xx(yz)
yy
zz
Œ"
ÎÑ
ÏÒ
Š‹ÄÄÊœ œœ
ϧ
œ
œ
```` ``` ` `
```````` ` ``
``wwxw wzz z x
yxyyyzyy y yy
yy
z
2x 2y 0 2x 2y (2x) (2y)(1) (2z)(0) 2y 2y 0œ Êœ Êœ Êœ œ œ
````
````
xxxw
yyyxyx
yy
Š‹ ˆ‰
z
(b) w ; 0 and 2x 2y
x
z
xx
yy(xz)
zz
Œ"
ÎÑ
ÏÒ
ˆ‰ÄÄÊœ œœ
œ
ϧ
œ
```` ``` ` `
```````` ` ` `
``wwxw wzx z x
zxzyzzzz z z z
yy
x
1 2y (2x)(0) (2y) (2z)(1) 1 2zÊœ Ê œ Ê œ œ
``
``# `
"` "yy
zzy z 2y
w
ˆ‰ Š‹
x
(c) w ; 0 and 2x 2y
y
z
xx(yz)
yy
zz
Œ"
ÎÑ
ÏÒ
ˆ‰ÄÄÊœ œœ
ϧ
œ
œ
```` `` ` `
```````` ` ` `
`` `wwxw wz z x
zxzyzzzz z z z
yy y
y
1 2x (2x) (2y)(0) (2z)(1) 1 2zÊœ Ê œ Ê œ œ
`` ` "
`` ` #
xx1 w
zz2x z x
ˆ‰ ˆ‰
y
2. w x y z sin t and x y t:œ œ
#
(a) w ; 0, 0, and
x
y
z
xx
yy
zz
txy
ÎÑ
ÏÒ
ÎÑ
ÐÓ
ÐÓ
ÏÒ
Š‹ÄÄÊœ œœ
œ
œ
œ
œ
``````````
```````````
`wwxwwzwtxz
yxyyyzytyyy
y
xz`
1 (2x)(0) (1)(1) ( 1)(0) (cos t)(1) 1 cos t 1 cos (x y)
``
``
tw
yy
œÊ œ œ œŠ‹
xt`
(b) w ; 0 and 0
y
z
t
xty
yy
zz
tt
ÎÑ
ÏÒ
ÎÑ
ÐÓ
ÐÓ
ÏÒ
Š‹ÄÄÊœ œœ
œ
œ
œ
œ
````````` `
`````````` `
`wwxwwzwtz t
yxyyyzytyy y
y
zt`
1 (2x)( 1) (1)(1) ( 1)(0) (cos t)(0) 1 2 t y 1 2y 2tÊœ œÊœ œ œ
`` `
``` `
`xt w
yyy y
y
Š‹ ab
zt`
(c) w ; 0 and 0
x
y
z
xx
yy
zz
txy
ÎÑ
ÏÒ
ÎÑ
ÐÓ
ÐÓ
ÏÒ
ˆ‰ÄÄÊœ œœ
œ
œ
œ
œ
`````````
`````````` `
``wwxwwzwtx
zxzyzzztzz z
yy
xy`
(2x)(0) (1)(0) ( 1)(1) (cos t)(0) 1Êœ œˆ‰
`
`
w
z
xy`
(d) w ; 0 and 0
y
z
t
xty
yy
zz
tt
ÎÑ
ÏÒ
ÎÑ
ÐÓ
ÐÓ
ÏÒ
ˆ‰ÄÄÊœ œœ
œ
œ
œ
œ
```````` `
`````````` `
``wwxwwzwt t
zxzyzzztzz z
yy
yt`
(2x)(0) (1)(0) ( 1)(1) (cos t)(0) 1Êœ œˆ‰
`
`
w
z
yt`
(e) w ; 0 and 0
x
z
t
xx
ytx
zz
tt
ÎÑ
ÏÒ
ÎÑ
ÐÓ
ÐÓ
ÏÒ
ˆ‰ÄÄÊœ œœ
œ
œ
œ
œ
` `` ` `` ``` `
`````````` `
`w wx w wz wt x z
txtytztttt t
y
xz`
(2x)(0) (1)(1) ( 1)(0) (cos t)(1) 1 cos tÊœ œˆ‰
`
`
w
t
xz`

922. 918 Chapter 14 Partial Derivatives
(f) w ; 0 and 0
y
z
t
xty
yy
zz
tt
ÎÑ
ÏÒ
ÎÑ
ÐÓ
ÐÓ
ÏÒ
ˆ‰ÄÄÊœ œœ
œ
œ
œ
œ
` `` ` `` `` `
`````````` `
``w wx w wz wt z
txtytztttt t
yy
yz`
(2x)(1) (1)(0) ( 1)(0) (cos t)(1) cos t 2x cos t 2(t y)Êœ œ œ ˆ‰
`
`
w
t
yz`
3. U f(P V T) and PV nRTœßß œ
(a) U (0)
P
V
PP
VV
T
Œ"
ÎÑ
ÏÒ
ˆ‰ ˆ‰ ˆ‰ˆ‰ÄÄÊœ œ
œ
œ
œ
PV
nR
UUPUVUTUU UV
PPPVPTPPV TnR
````````` `
````````` `
V
œ
``
``
UUV
PTnR
ˆ‰ˆ‰
(b) U (0)
V
T
P
VV
TT
Œ"
ÎÑ
ÏÒ
ˆ‰ ˆ‰ˆ‰ˆ‰ÄÄÊœ œ
œ
œ
œ
nRT
V
UUPUVUTUnRU U
TPTVTTTPVV T
```````` ` `
```````` ` `
V
œ ˆ‰ˆ‰
``
``
UnR U
PV T
4. w x y z and y sin z z sin x 0œ œ
###
(a) w ; 0 and
x
y
xx
yy
zz(xy)
Œ"
ÎÑ
ÏÒ
ˆ‰ÄÄÊœ œ
œ
œ
ϧ
```` ``
````````
``wwxw wz
xxxyxzxx
yy
y
(y cos z) (sin x) z cos x 0 . At (0 1 ),
`` ` `
`` ` `
zz zz cos x z
xx xy cos zsin x x1
œÊ œ ßß œ œ 11
1
(2x)(1) (2y)(0) (2z)( ) 2Êœ œˆ‰ k
`
`
#w
xk
y
(0 1 )
01
``
"`` #


11
(b) w (2x) (2y)(0) (2z)(1)
y
z
xx(yz)
yy
zz
Œ"
ÎÑ
ÏÒ
ˆ‰ÄÄÊœ œ
ϧ
œ
œ
```` `` `
``````` `
`wwxw wz x
zxzyzzz z
y
y
(2x) 2z. Now (sin z) y cos z sin x (z cos x) 0 and 0œ œœ
``
`` ``
``xx
zz zz
yy
y cos z sin x (z cos x) 0 . At ( ), Ê œÊ œ !ß"ß œ œ
`` `"
`` `
xx x10
z z z cos x z ( )(1)
y cos z sin x
1
11
2(0) 2 2Êœ œˆ‰ ˆ‰
`"
`C
!"ß Ñ
w
z (,
k
1 1
11
5. w x y yz z and x y z 6œ œ
## $ # # #
(a) w
x
y
xx
yy
zz(xy)
Œ"
ÎÑ
ÏÒ
Š‹ÄÄÊœ
œ
œ
ϧ
```` ``
```````
`wwxw wz
yxyyyzy
y
x
2xy (0) 2x y z (1) y 3z 2x y z y 3z . Now (2x) 2y (2z) 0 andœ œ œaba bab ab
## ## # ````
````
zzxz
yyyy
0 2y (2z) 0 . At (w x y z) (4 2 1 1), 1
``` `"` ``` ``
xzz zw
yyyz y1y
y
œÊ œÊ œ ßßß œ ßßß œœÊ ¹Š‹
x(421 1)```
(2)(2) (1) ( 1) 1 3( 1) (1) 5œ œcdcd
##
(b) w
y
z
xx(yz)
yy
zz
Œ"
ÎÑ
ÏÒ
Š‹ÄÄÊœ
ϧ
œ
œ
```` ``
```````
`wwxw wz
yxyyyzy
y
z
2xy 2x y z (1) y 3z (0) 2x y 2x y z. Now (2x) 2y (2z) 0 andœ œ œab a b a b ab
## ###````
````
xxxz
yyyy
0 (2x) 2y 0 . At (w x y z) (4 2 1 1),
`` ` `"`
`` ` ``
zx x xw
y y yx y2 y
y
(
œÊ œÊ œ ßßß œ ßßß œÊ ¹Š‹
z421 1)``` (2)(2)(1) (2)(2) (1) ( 1) 5œ œ
##"
#ˆ‰
6. y uv 1 v u ; x u v and 0 0 2u 2v 1œÊœ œ œÊœ Êœ Ê
`` ` ` ` ` `
`` ` ` ` ` `
##uv x u v v uu
yy y y y y vy
ˆ‰
v u . At (u v) 2 1 , 1œ œÊœßœßœœ
```` `"
`` `` `

uuuvuuuv u
yvyvyyvu y
12
Š‹Š‹ Š‹
È




Š‹
È

923. Section 14.10 Taylor's Formula for Two Variables 919
1Êœ Š‹
`
`
u
y
x
7. cos ; x y r 2x 2y 2r and 0 2x 2r
rxr cos
y r sin
Œ" Œ "
ˆ‰
))
)
)ÄÊœ œÊ œœÊœ
œ
œ
```
`````
### ``xrr
rxxxx
yy
)
ÊœÊ œ
``
``
rx r x
xr x xy
ˆ‰
y È
8. If x, y, and z are independent, then ˆ‰
````````
`````````
`wwxwwzwt
xxxyxzxtx
y
yz`
œ
(2x)(1) ( 2y)(0) (4)(0) (1) 2x . Thus x 2z t 25 1 0 0 1œ œ œÊ œÊ œ ˆ‰
`` ``
`` ``
tt tt
xx xx
2x 1. On the other hand, if x, y, and t are independent, then Êœ ˆ‰ ˆ‰
` `
` `
w w
x x
yz yt` `
(2x)(1) ( 2y)(0) 4 (1)(0) 2x 4 . Thus, x 2z t 25œ œ œ œ
`` ` `` ` ` ` `
`` `` `` `` ` `
`wx w wz w t z z
xx yx zx t x x x
y
12 00 2x4 2x2.Ê œÊ œ Êœ œ
``"` "
``#` #
zzw
xxx
ˆ‰ ˆ ‰
yt`
9. If x is a differentiable function of y and z, then f(x y z) 0 0 0ßß œ Ê œÊ œ
`` ` `` ` `
`` `` `` ` ``
``fx f fz f f
xx yx zx x yx
yy
. Similarly, if y is a differentiable function of x and z, and if z is aÊœ œŠ‹ Š‹
` ``
``` ```
`` `x f/ z
yf/z zf/x
f/ y y
z x
differentiable function of x and y, . Then ˆ‰ ˆ‰ Š‹Š‹
``` ` `
``` ```
`zf/x x z
xf/y yzx
y
yy
zx
œ
1.œ œŠ‹ Š‹ˆ‰
``
`` `` ``
`` ``f/ y
f/ z f/ x f/ y
f/ z f/ x
10. z z f(u) and u xy 1 1 y ; also 0 x so that x y œ œÊœ œ œ œ
`` `` ``
`` `` ``
zdfu dfzdfudf zz
xdux duyduydu xy
x1 y yx xœ œˆ‰ˆ‰
df df
du du
11. If x and y are independent, then g(x y z) 0 0 and 0 0ßß œ Ê œœÊ œ
` ``` ``
`` `` `` ` ` ``
``` `g gyg gg
xy yy zy y y zy
xzx z
, as claimed.Êœ Š‹
`
```
``z
yg/z
g/ y
x
12. Let x and y be independent. Then f(x y z w) 0, g(x y z w) 0 and 0ßßß œ ßßß œ œ
`
`
y
x
0 andÊ œ œ
`` ` `` ` ` ` `` ` `
`` `` `` ` ` ` `` ` `
`fx f fz f w f fz f w
xx yx zx w x x zx w x
y
0 imply
```` ` `` `
`` `` `` ` ` ` `` ` `
`````ggyg g gg g
xx yx zx w x x zx w x
xzwzw
œ œ



#
ˆ‰
`` ` ` `
`` ` ` `
`` `
`` ` ` `
``
`
`

fz f w f
zx w x x
gg g
zx wx x
zw
z
x


œ
œ
Êœ œ
y
»»
»»


##
##
##
##
##
##
##
##ff
xw
gg
xw
ff
zw
gg
zw
``
`` ``
``
`
`
`ff
xw xw
gg
f
z
gg g g
wzw zwwz
fff
ff
xw wx
gg
``` ````
``````
``
`` ` `
``




œ , as claimed.
Likewise, f(x y z w) 0, g(x y z w) 0 and 0 ßßß œ ßßß œ œ Ê
````````
` ````````
`xfxffzfw
y xyyyzywy
y
0 and (similarly) 0 implyœ œ œ
``` `` ` `
````` `````
`` `ffzfw z w
yzywy yzywy
gg g



#
Š‹
`` ` ` `
`` ` ` `
`` `
`` ` ` `
``
`
`

fz f w f
zy w y y
gg g
zy wy y
zw
w
y


œ
œ
Êœ œ
x
»»
»»
##
##
##
##
##
##
##
##ff
zy
gg
zy
ff
zw
gg
zw


``
`` ``
``
`
`
`ff
zy zy
gg
f
z
gg g g
wzw zwwz
fff
ff
zy yz
gg
``` ````
``````
``
`` ``
``




œ , as claimed.
14.10 TAYLOR'S FORMULA FOR TWO VARIABLES
1. f(x y) xe f e , f xe , f 0, f e , f xeßœ Ê œ œ œ œ œ
yyy yy
x y xx xy yy
f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß ß ß ß ß x y xx xy yy
"
#
## cd
0 x 1 y 0 x 0 2xy 1 y 0 x xy quadratic approximation;œ œ†† † † †
"
#
##
ab
f 0, f 0, f e , f xe
xxx xxy xyy yyy
yyœœœ œ

924. 920 Chapter 14 Partial Derivatives
f(x y) quadratic x f ( ) 3x yf (0 0) 3xy f ( ) y f (0 0)Ê߸ !ß ! ß !ß ! ß
"$# #$
6
xxx xxy xyy yyy
cd
x xy x 0 3x y 0 3xy 1 y 0 x xy xy , cubic approximationœ œ
""$# #$ #
#6
ab††††
2. f(x y) e cos y f e cos y, f e sin y, f e cos y, f e sin y, f e cos yßœ Ê œ œ œœ œ
xxxxxx
x y xx xy yy
f(x y) f(0 0) xf (0 0) yf ( 0) x f ( ) 2xyf ( ) y f (0 0)Êß¸ß ß !ß !ß ! !ß ! ß x y xx xy yy
"
#
## cd
1 x 1 y 0 x 1 2xy 0 y ( 1) 1 x x y , quadratic approximation;œ œ †† † † †
""
##
## ##
cd ab
f e cos y, f e sin y, f e cos y, f e sin y
xxx xxy xyy yyy
xxxxœœ œ œ
f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
1 x x y x 1 3x y 0 3xy ( 1) y 0œ
""
#
## $ # # $
ab cd
6
††† †
1 x x y x 3xy , cubic approximationœ
""
#
## $ #
aba b
6
3. f(x y) y sin x f y cos x, f sin x, f y sin x, f cos x, f 0ßœ Ê œ œ œ œœ x y xx xy yy
f(x y) f(0 0) xf (0 0) yf ( 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß !ß ß ß ß x y xx xy yy
"
#
## cd
0 x 0 y 0 x 0 2xy 1 y 0 xy, quadratic approximation;œ œ†† † † †
"
#
##
ab
f y cos x, f sin x, f 0, f 0
xxx xxy xyy yyyœ œ œœ f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
xy x 0 3x y 0 3xy 0 y 0 xy, cubic approximationœ œ
"$# #$
6
ab††††
4. f(x y) sin x cos y f cos x cos y, f sin x sin y, f sin x cos y, f cos x sin y,ßœ Ê œ œ œ œ
x y xx xy
f sin x cos y f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)yy x y xx xy yyœ Êß¸ß ß ß ß ß ß
"
#
##
cd
0 x 1 y 0 x 0 2xy 0 y 0 x, quadratic approximation;œ œ†† † † †
"
#
##
ab
f cos x cos y, f sin x sin y, f cos x cos y, f sin x sin y
xxx xxy xyy yyyœ œœ œ f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
x x ( 1) 3x y 0 3xy ( 1) y 0 x x 3xy , cubic approximationœ œ
""$##$ $#
66
cd ab††††
5. f(x y) e ln (1 y) f e ln (1 y), f , f e ln (1 y), f , fßœ Êœ œœ œœ
xx x
x y xx xy yy
eee
1 y 1 y (1 y)
xxx


f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß ß ß ß ß x y xx xy yy
"
#
## cd
0 x 0 y 1 x 0 2xy 1 y ( 1) y 2xy y , quadratic approximation;œ œ†† † † †
""
##
## #
cd ab
f e ln (1 y), f , f , fxxx xxy xyy yyy
x ee2e
1 y (1 y) (1 y)œ œœ œ
xxx


f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
y 2xy y x 0 3x y 1 3xy ( 1) y 2œ
"" #$ # # $
26
ab cd††† †
y 2xy y 3x y 3xy 2y , cubic approximationœ
""
#
###$
aba b
6
6. f(x y) ln (2x y 1) f , f , f , f ,ßœ Êœ œ œ œ xyxx xy
242
2x y 1 x y 1 (2x y 1) (2x y 1) #
"
f f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)yy x y xx xy yy
(2x y 1)œÊß¸ß ß ß ß ß ß
" "
#
##
cd
0 x 2 y 1 x ( 4) 2xy ( 2) y ( 1) 2x y 4x 4xy yœ œ †† † † †
""
##
## ##
cd ab
(2x y) (2x y) , quadratic approximation;œ
"
#
#
f, f, f, fxxx xxy xyy yyy
16842
(2x y 1) (2x y 1) (2x y 1) (2x y 1)œœœœ


f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
(2x y) (2x y) x 16 3x y 8 3xy 4 y 2œ
""
#
#$ # # $
6
ab††††
(2x y) (2x y) 8x 12x y 6xy yœ
""
#
#$###
3
ab
(2x y) (2x y) (2x y) , cubic approximationœ
""
#
#$
3
7. f(x y) sin x y f 2x cos x y , f 2y cos x y , f 2 cos x y 4x sin x y ,ßœ Êœ œ œ ab ab ab ab ab
## ## ## ## # ##
xyxx
f 4xy sin x y , f 2 cos x y 4y sin x yxy yyœ œ ab ab ab
## ## # ##

925. Section 14.10 Taylor's Formula for Two Variables 921
f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß ß ß ß ß x y xx xy yy
"
#
## cd
0 x 0 y 0 x 2 2xy 0 y 2 x y , quadratic approximation;œ œ†† † † †
"
#
####
ab
f 12x sin x y 8x cos x y , f 4y sin x y 8x y cos x y ,
xxx xxyœ œ ab ab ab ab
## $ ## ## # ##
f 4x sin x y 8xy cos x y , f 12y sin x y 8y cos x yxyy yyyœ œ ab ab ab ab
## # ## ## $ ##
f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
x y x 0 3x y 0 3xy 0 y 0 x y , cubic approximationœ œ
## $ # # $ ##"
6
ab††††
8. f(x y) cos x y f 2x sin x y , f 2y sin x y ,ßœ Êœ œab ab ab
## ## ##
xy
f 2 sin x y 4x cos x y , f 4xy cos x y , f 2 sin x y 4y cos x yxx xy yyœ œ œ ab ab ab ab ab
## # ## ## ## # ##
f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß ß ß ß ß x y xx xy yy
"
#
## cd
1 x 0 y 0 x 0 2xy 0 y 0 1, quadratic approximation;œ œ†† † † †
"
#
##
cd
f 12x cos x y 8x sin x y , f 4y cos x y 8x y sin x y ,
xxx xxyœ œ ab ab ab ab
## $ ## ## # ##
f 4x cos x y 8xy sin x y , f 12y cos x y 8y sin x yxyy yyyœ œab ab ab ab
## # ## ## $ ##
f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
1 x 0 3x y 0 3xy 0 y 0 1, cubic approximationœ œ
"$# #$
6
ab††††
9. f(x y) f f , f f fßœ Ê œ œ œ œ œ
""
1 x y (1 x y) (1 x y)
x y xx xy yy
2

f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß ß ß ß ß x y xx xy yy
"
#
## cd
1 x 1 y 1 x 2 2xy 2 y 2 1 (x y) x 2xy yœ œ †† † † †
"
#
## ##
abab
1 (x y) (x y) , quadratic approximation; f f f fœ œ œœœ
#

xxx xxy xyy yyy
6
(1 x y)
"
f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
1(xy)(xy) x63xy63xy6y6œ
#$ # # $"
6
ab††††
1 (x y) (x y) x 3x y 3xy y 1 (x y) (x y) (x y) , cubic approximationœ œ
#$ # #$ # $
ab
10. f(x y) f , f , f ,ßœ Ê œ œ œ
"
"

1 x y xy (1 x y xy) ( x y xy) (1 x y xy)
xyxx
1 y 2(1 y) 1 x


f, fxy yy
1
( x y xy) (1 x y xy)
2( x)œœ
"
"


f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß ß ß ß ß x y xx xy yy
"
#
## cd
1x1y1 x22xy1y2 1xyx xyy, quadratic approximation;œ œ †† † † †
"
#
## ##
ab
f, f ,xxx xxy
6(1 y) [ 4(1 x y xy) 6(1 y)(1 x)](1 y)
(1 x y xy) (1 x y xy)œœ



""
f, fxyy yyy
[ 4(1 x y xy) 6(1 x)(1 y)](1 x) 6(1 x)
(1 x y xy) (1 x y xy)œœ


""

f(x y) quadratic x f (0 0) 3x yf ( 0) 3xy f (0 0) y f (0 0)Ê߸ ß !ß ß ß
"$# #$
6
xxx xxy xyy yyy
cd
1xyx xyy x63xy23xy2y6œ
##$# #$ "
6
ab††††
1xyx xyy x xyxy y, cubic approximationœ
# #$# #$
11. f(x y) cos x cos y f sin x cos y, f cos x sin y, f cos x cos y, f sin x sin y,ßœ Ê œ œ œ œ xyxx xy
f cos x cos y f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)yy x y xx xy yyœ Êß¸ß ß ß ß ß ß
"
#
##
cd
1 x 0 y 0 x ( 1) 2xy 0 y ( 1) 1 , quadratic approximation. Since all partialœ œ†† † † †
"
###
##
cd
x y

derivatives of f are products of sines and cosines, the absolute value of these derivatives is less than or equal
to 1 E(x y) (0.1) 3(0.1) 3(0.1) 0.1) 0.00134.Êߟ Ÿ
" $$$$
6
cd
12. f(x y) e sin y f e sin y, f e cos y, f e sin y, f e cos y, f e sin yߜʜœœœœ
xxxxxx
x y xx xy yy
f(x y) f(0 0) xf (0 0) yf (0 0) x f (0 0) 2xyf (0 0) y f (0 0)Êß¸ß ß ß ß ß ß x y xx xy yy
"
#
## cd
0 x 0 y 1 x 0 2xy 1 y 0 y xy , quadratic approximation. Now, f e sin y,œ œ œ†† † † †
"
#
##
ab xxx
x
f e cos y, f e sin y, and f e cos y. Since x 0.1, e sin y e sin 0.1 0.11 andxxy xyy yyy
xx x x01œœ œ ŸŸ¸kk k k k k
Þ
e cos y e cos 0.1 1.11. Therefore,kkk k
x01
Ÿ¸
Þ

926. 922 Chapter 14 Partial Derivatives
E(x y) (0.11)(0.1) 3(1.11)(0.1) 3(0.11)(0.1) (1.11)(0.1) 0.000814.ߟ Ÿ
" $$$$
6
cd
CHAPTER 14 PRACTICE EXERCISES
1. Domain: All points in the xy-plane
Range: z 0 
Level curves are ellipses with major axis along the y-axis
and minor axis along the x-axis.

2. Domain: All points in the xy-plane
Range: 0 z##_
Level curves are the straight lines x y ln z withœ
slope 1, and z 0.

3. Domain: All (x y) such that x 0 and y 0ßÁÁ
Range: z 0Á
Level curves are hyperbolas with the x- and y-axes
as asymptotes.

4. Domain: All (x y) so that x y 0ß  
#
Range: z 0 
Level curves are the parabolas y x c, c 0.œ 
#

5. Domain: All points (x y z) in spaceßß
Range: All real numbers
Level surfaces are paraboloids of revolution with
the z-axis as axis.

927. Chapter 14 Practice Exercises 923
6. Domain: All points (x y z) in spaceßß
Range: Nonnegative real numbers
Level surfaces are ellipsoids with center (0 0 0).ßß

7. Domain: All (x y z) such that (x y z) (0 0)ßß ßß Á ß!ß
Range: Positive real numbers
Level surfaces are spheres with center (0 0 0) andßß
radius r 0.

8. Domain: All points (x y z) in spaceßß
Range: (0 1]ß
Level surfaces are spheres with center (0 0 0) andßß
radius r 0.

9. lim e cos x e cos (2)( 1) 2
Ðß ÑÄÐß Ñxy ln21
yln2
œœœ1
10. lim 2
Ðß ÑÄÐß Ñxy 00
2y
xcos y 0cos 0
20


œœ
11. lim lim lim
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ
Á„ Á„
xy 11 xy 11 xy 11
xy xy
xy xy
x y (x y)(x y) x y 1 1
1œœ
#
""
œœœœ
12. lim lim lim x y xy 1 1 1 1 1 1 3
Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñ Ðß ÑÄÐß Ñxy 11 xy 11 xy 11
xy 1
xy 1 xy 1
(xy1)xy xy1
œœ
œ
œœ
## # #
œœœœ
ab
ab ††
13. lim ln x y z ln 1 ( 1) e ln e 1
P11eÄÐ ßßÑ
kkk k œ œ œ14. lim tan (x y z) tan (1 ( 1) ( 1)) tan ( 1)
P 111ÄÐ ßßÑ
" " "
œ œ œ
1
4
15. Let y kx , k 1. Then lim lim which gives different limits forœÁ œ œ
#
œœœ
Ðß ÑÄÐß Ñ
Á
ßÄÐßÑxy 00
yx
xkx 00


y
xy xkx 1k
kx k


ab
different values of k the limit does not exist.Ê
16. Let y kx, k 0. Then lim lim which gives different limits forœÁ œ œ
Ðß ÑÄÐß Ñ ß ÑÄÐß Ñ
Á
xy 00 (xkx 00
xy 0
xy x(kx)
xy x(kx) k
1k

928. 924 Chapter 14 Partial Derivatives
different values of k the limit does not exist.Ê
17. Let y kx. Then lim which gives different limits for different valuesœœœ
Ðß ÑÄÐß Ñxy 00
xy
xy xkx 1k
xkx 1k



œ

œœ
of k the limit does not exist so f(0 0) cannot be defined in a way that makes f continuous at the origin.Êß
18. Along the x-axis, y 0 and lim lim , so the limit fails to exist
1, x 0
, x 0
œœœ

#j $
Ðß ÑÄÐß Ñ Äxy 00 x0
sin (x y)
xy x
sin xœ
kk kk
œ
f is not continuous at (0 0).Êß
19. cos sin , r sin r cos
``
``
gg
r
œ œ )) ) )
)
20. ,
`"
`#
œ

œf2x x
xxy xyxyxy
1
yxy
œ œœŠ‹

Š‹
ˆ‰
y
x
y
x

`"
`#

fx
yxy xyxyxy
2y y x y
1
œ œœŠ‹

Š‹
ˆ‰
1
x
y
x
21. , ,
`"`"`"
```
fff
RRR RRR


œ œ œ

22. h (x y z) 2 cos (2 x y 3z), h (x y z) cos (2 x y 3z), h (x y z) 3 cos (2 x y 3z)
xyzßß œ ßß œ ßß œ 11 1 1
23. , , ,
````
````
P RT P nT P nR P nRT
nVRVTVV V
œœœœ

24. f (r T w) , f (r T w) , f (r T w)
r Tßjß ß œ ßjß ß œ ßjß ß œ
"""""
j#j#j
j
2r w r w r
TT
w 2T
ÉÉ ˆ‰Š‹Š‹
11 1ÈÈ
, f (r T w) w ?Î?? ##
"" " " " "
jj #j# j
$Î#
4r T w 4r T w r 4r w w
TT T
ÉÉ É É ˆ‰ ˆ ‰
11 1 1
w
25. , 1 0, ,
`` `` ``
` ` ` ` `` ``
" "gg gg gg
xyy y x y y yx xy y
x2x
œœÊœœ œœ


26. g (x y) e y cos x, g (x y) sin x g (x y) e y sin x, g (x y) 0, g (x y) g (x y) cos x
x y xx yy xy yx
xx?" ? ` ?# ? ? ?
27. 1 y 15x , x 30x , 0, 1
ÊÊÊœÊÊÊ
````````
#

f2xff22xfff
x x 1 y x y yx xy x1
œ œ Ê œ œ œ œ




ab
28. f (x y) 3y, f (x y) 2y 3x sin y 7e f (x y) 0, f (x y) 2 cos y 7e , f (x y) f (x y)
x y xx yy xy yx
yy?# ? # # " ` ? ?# " ? ?
3œ
29. y cos (xy ), x cos (xy ), e ,
`` "
``
wwdx
x y dt dt t 1
dy
œœœœ 11
t
[y cos (xy )]e [x cos (xy )] ; t 0 x 1 and y 0Êœ œÊœ œ
dw
dt t 1
11
t ˆ‰
"

0 1 [1 ( 1)] 1Êœ œ¸ˆ‰
dw
dt 0 1
t0œ
††
"

30. e , xe sin z, y cos z sin z, t , 1 ,
`` ` "
`` `
"Î#ww w dx dz
xy z dt dttdt
dy
œœ œ œ œœ
yy
1
e t xe sin z 1 (y cos z sin z) ; t 1 x 2, y 0, and zÊœ œÊœ œ œ
dw
dt t yy"Î# "
ab ˆ‰ 11
1 1 (2 1 0)(2) (0 0) 5Êœœ¸
dw
dt
t1œ
†† 1
31. 2 cos (2x y), cos (2x y), 1, cos s, s, r
`` ``
`` ````
``ww xx
xy rsrs
yy
œœœœœœ
[2 cos (2x y)](1) [ cos (2x y)](s); r and s 0 x and y 0Êœ œ œÊœ œ
`
`
w
r
11

929. Chapter 14 Practice Exercises 925
(2 cos 2 ) (cos 2 )(0) 2; [2 cos (2x y)](cos s) [ cos (2x y)](r)Êœ œœ ¸
``
``
ww
rs
0
11
(2 cos 2 )(cos 0) (cos 2 )( ) 2Êœ œ¸
`
`
w
s
0
1111
32. 2e cos v ; u v 0 x 2 (2) ;
`` " ` "
`` `
wdwx x w 2 2
udxu 1xx1 u 55 5
œ œ œœÊœÊ œ œˆ‰ ¸ˆ‰ ab

u
00
2e sin v (0) 0
`` " ` "
`` `
wdwx x w 2
vdxv 1xx1 v 55
œœ Ê œœˆ‰ ¸ˆ‰ ab

u
00

33. y z, x z, y x, sin t, cos t, 2 sin 2t
```
```
fffdx dz
x y z dt dt dt
dy
œ œ œ œ œ œ
(y z)(sin t) (x z)(cos t) 2(y x)(sin 2t); t 1 x cos 1, y sin 1, and z cos 2Êœ œÊœ œ œ
df
dt
(sin 1 cos 2)(sin 1) (cos 1 cos 2)(cos 1) 2(sin 1 cos 1)(sin 2)Êœ ¸
df
dt
t1œ
34. (5) and (1) 5 5 5 0
`` `` ``
`` `` ``
w dw s dw w dw s dw dw w w dw dw
x ds x ds y ds y ds ds x y ds ds
œœ œœ œÊœœ
35. F(x y) 1 x y sin xy F 1 y cos xy and F 2y x cos xy ßœ Ê œ œ Ê œœ
# œœ
œœ
xy
dy 1 y cos xy
dx F 2y x cos xy
F
x
y
at (x y) ( 1) we have 1œ Ê ß œ !ß œ œ
1 y cos xy dy
2y x cos xy dx 2
1
œœ œ
"
¹
ÐßÑ01
36. F(x y) 2xy e 2 F 2y e and F 2x e ß œ Ê œ œ Ê œ œ
xy xy xy
xy
dy 2y e
dx F 2x e
F
x
y
xy
xy
"
"
at (x y) ( ln 2) we have (ln 2 1)Êßœ!ß œ œ ¹
dy
dx 0 2
2 ln 2 2
Ðß Ñ0ln2


37. f ( sin x cos y) (cos x sin y) f f ;™™™œ Ê œ Ê œ œ œij ij kkk Ɉ‰ˆ‰


44
#
"" " " "
## # # #
##
È
È
2
2
f increases most rapidly in the direction and decreases mostuij uijœœ Ê œ
™
™
f
f
22 22
kk
ÈÈ ÈÈ
## ##
rapidly in the direction ; (D f) f and (D f) ;œ œ œ œuij
ÈÈ È È
22 2 2
## # #
uuPP""kk™
`
(D f) fuiju""


""
##œœ œ Ê œ œ œ
v
v
ij
kk È
34
34
34 3 4 7
55 5 5 10

uP™† ˆ‰ˆ‰ˆ‰ˆ‰
38. f 2xe 2x e f f 2 ( 2) 2 2; ™™™œ Ê œ##Êœœ œœ
``
2y 2y
10
ij ij u ij
# ##
kkk È È
™
™
f
f
11
22kk ÈÈ
f increases most rapidly in the direction and decreases most rapidly in the directionʜ uij
11
22
ÈÈ
; (D f) f 2 2 and (D f) 2 2 ; œ œ œ œ œ œ œ uij u ij
11 11
22 11 2 2
ÈÈ ÈÈÈkk
uuPP""kk
ÈÈ
™
` "


v
v
ij

(D f) f (2) ( 2) 0Êœ œ œ
uP™†u "
"" Š‹ Š‹
ÈÈ
22
39. f f 2 3 6 ;™™œ ÊœŠ‹Š‹Š‹ k
236
2x 3y 6z 2x 3y 6z 2x 3y 6z
ijk ijk
111
f increases most rapidly in the direction andu ijk uijkœ œ œ Ê œ
™
™
f
f777 777
236
236
236 236
kk È
ijk


decreases most rapidly in the direction ; (D f) f 7, (D f) 7;œ œ œ œuijk
236
777
uuPP""kk™ œ
(D f) (D f) 7uijk"œœ Ê œ œ
v
vkk
236
777
uuPP`
40. f (2x 3y) (3x 2) (1 2z) f 2 ; f increases most™™œ Ê œ œ œ Êijk jku jk k
000
™
™
f
f
2
55kk ÈÈ
"
rapidly in the direction and decreases most rapidly in the direction ;ujk u jkœ œ
2 2
55 55
ÈÈ ÈÈ
" "
(D f) f 5 and (D f) 5 ;
uuPP""œ œ œ œœ œkk
ÈÈ
™ "


""" uijk
v
v
ijk
kk ÈÈÈÈ
111 333

(D f) f (0) (2) (1) 3Êœœœœ
uP`
™†u "
""" Š‹ Š‹ Š‹
È
ÈÈÈÈ
3333
3

930. 926 Chapter 14 Partial Derivatives
41. (cos 3t) (sin 3t) 3t (t) ( 3 sin 3t) (3 cos 3t) 3 3 3rijkv i jkvjkœÊœ Êœ ˆ‰
1
3
; f(x y z) xyz f yz xz xy ; t yields the point on the helix ( 1 0 )Ê œ ßß œ Ê œ œ ßßujk ijk
""
ÈÈ
22 3
™
1
1
f f()ʜʜœk Š‹™™††
10
11j ujjk
""
ÈÈ È
22 2
1
42. f(x y z) xyz f yz xz xy ; at (1 1 1) we get f the maximum value ofßß œ Ê œ ßß œ Ê™™ ijk ijk
Df f 3kkk
È
u111
œœ™
43. (a) Let f a b at (1 2). The direction toward (2 2) is determined by (2 1) (2 2)™œ ß ß œ œœij v i jiu
"
so that f 2 a 2. The direction toward (1 1) is determined by (1 1) (1 2)™†uvijjuœÊœ ß œ œœ #
so that f 2 b 2 b 2. Therefore f 2 2 ; f 1, 2 f 1, 2 2.™† ™uijœ Ê œ Ê œ œ œ œ xyab ab
(b) The direction toward (4 6) is determined by (4 1) (6 2) 3 4 ? vijijuij $
34
55
f. Êœ™†u
14
5
44. (a) True (b) False (c) True (d) True
45. f 2x 2z ™œ Êij k
f2,k™
011
œjk
f, k™
000
œj
f2 k™
011
œjk

46. f 2y 2z ™œ Êjk
f4,k™
220
œj
f4,k™
220
œj
f4,k™
202
œk
f4 k™
20 2
œk

47. f 2x 5 f 4 5 Tangent Plane: 4(x 2) (y 1) 5(z 1) 0™™œ Ê œ Ê œij k ij k k
211
4x y 5z 4; Normal Line: x 2 4t, y 1 t, z 1 5tÊœ œ œœ
48. f 2x 2y f 2 2 Tangent Plane: 2(x 1) 2(y 1) (z 2) 0™™œÊ œÊ œijk ijk k
112
2x 2y z 6 0; Normal Line: x 1 2t, y 1 2t, z 2 tÊœ œ œ œ
49. 0 and 2; thus the tangent plane is
`` ``
` ` ` `
z2x z z z
xxy x yxy y
2y
œÊ œ œÊ œ

¸ ¹

010
010
2(y 1) (z 0) 0 or 2y z 2 0œ œ

931. Chapter 14 Practice Exercises 927
50. 2x x y and 2y x y ; thus the tangent
``"``"
``#``#
## ##
# #zzzz
xxyy
œ Ê œ œ Ê œab ab ¸ ¹

11
11##
##
1
2
1
2
plane is (x 1) (y 1) z 0 or x y 2z 3 0œ œ
"" "
## #
ˆ‰
51. f ( cos x) f the tangent™™œ Ê œ Êij ij k
1
line is (x ) (y 1) 0 x y 1; theœÊœ11
normal line is y 1 1(x ) y x 1œ Ê œ11

52. f x y f 2 the tangent™™œ Ê œ Êij ij k
12
line is (x 1) 2(y 2) 0 y x ; the normal œÊœ
"
##
3
line is y 2 2(x 1) y 2x 4œ Êœ

53. Let f(x y z) x 2y 2z 4 and g(x y z) y 1. Then f 2x 2 2 2 2 2ßß œ ßß œ œ œ
#
k™ ijk ijk
ab11##
1
2
and g f g 2 2 the line is x 1 2t, y 1, z 2t222
00
™™™œÊ ‚ œ œ Ê œ œ œ
"
jik
ijk
ââ
ââ
ââ
ââ
ââ
ââ
"
#
54. Let f(x y z) x y z 2 and g(x y z) y 1. Then f 2y 2 andßß œ ßß œ œ œ
#
k™ ijk ijk
ab
11
22
##1
g f g the line is x t, y 1, z t121
00
™™™œÊ ‚ œ œ Ê œ œ œ
"
jik
ijk
ââ
ââ
ââ
ââ
ââ
ââ
""
##
55. f , f cos x cos y , f sin x sin yˆ‰ ˆ‰ ˆ‰ kk
11 11 11
11 1144 44 44
xy 44 44
ßœ ßœ œ ßœ œ
"" "
## # ÐÎßÎÑ ÐÎßÎÑ
L(x y) x y x y; f (x y) sin x cos y, f (x y) sin x cos y, and`?" ## #"# ?# ?#
"" " "" "
## # ## #
ˆ‰ˆ‰
11
44
xx yy
f (x y) cos x sin y. Thus an upper bound for E depends on the bound M used for f , f , and f .xy xx xy yy?# kkkk kk
With M we have E(x y) x y (0.2) 0.0142;?Ñ#"#ÑÑ
ÈÈ È
22 2
444###
"
#
#
kk Š‹ˆ‰¸¸¸¸
11
with M 1, E(x y) (1) x y (0.2) 0.02.œßŸ œ œkk ˆ‰¸¸¸¸
""
##
#
#11
44
56. f(1 1) 0, f (1 1) y 1, f (1 1) x 6y 5 L(x y) (x 1) 5(y 1) x 5y 4;ßœ ßœ œ ßœ œÊ ßœ œ kk xy 11 11ÐßÑ ÐßÑ
f (x y) 0, f (x y) 6, and f (x y) 1 maximum of f , f , and f is 6 M 6xx yy xy xx yy xyßœ ßœ ßœÊ Ê œ kkkk kk
E(x y) (6) x 1 y 1 (6)(0.1 0.2) 0.27Ê ß Ÿ œ œkk a b kkkk
""
##
# #
57. f(1 0 0) 0, f (1 0 0) y 3z 0, f (1 0 0) x 2z 1, f (1 0 0) 2y 3x 3ßß œ ßß œ œ ßß œ œ ßß œ œkk k
xyz 100 100100
L(x y z) 0(x 1) (y 0) 3(z 0) y 3z; f(1 1 0) 1, f (1 1 0) 1, f (1 1 0) 1, f ( ) 1` ?? #"## ## ?? ?? ?? j?j?!#
xyz
L(x y z) 1 (x 1) (y 1) 1(z 0) x y z 1` ??"#"## #"##
58. f 0 1, f 0 2 sin x sin (y z) 0, f 0 2 cos x cos (y z) 1,ˆ‰ ˆ‰ ˆ‰ ¹¹
ÈÈ
ß!ßœ !ßßœ œ !ßßœ œ
11 1
44 4
x y
00 00
## ##

4 4 f 0 2 cos x cos (y z) 1 L(x y z) 1 1(y 0) 1 z 1 y z ;
z
00
ˆ‰ ˆ‰ ¹
È
!ß ß œ œ Ê ß ß œ œ
1 11
4 44
##

4 f 0 , f 0 , f 0 , f 0ˆ‰ ˆ‰ ˆ‰ ˆ‰
11 11 11 11
44 44 44 44
2222
?? ??# ?? ??
ÈÈÈÈ
####
xyz
L(xyz) x y (z 0) x y z`??##"#"##""
ÈÈÈÈÈÈÈÈ
22222222
44########
ˆ‰ ˆ‰
11

932. 928 Chapter 14 Partial Derivatives
59. V r h dV 2 rh dr r dh dV 2 (1.5)(5280) dr (1.5) dh 15,840 dr 2.25 dh.kœÊœ Ê œ œ 111 1 1 11
## #
155280
You should be more careful with the diameter since it has a greater effect on dV.
60. df (2x y) dx ( x 2y) dy df 3 dy f is more sensitive to changes in y; in fact, near the pointkœ Ê œ Ê
12
(1 2) a change in x does not change f.ß61. dI dV dR dI dV dR dI 0.01 (480)(.0001) 0.038,¸¸œ Ê œ Ê œ œ
""
R R 100 100
V24

24 100 dV 1 dR 20œßœ
or increases by 0.038 amps; % change in V (100) 4.17%; % change in R (100) 20%;œ ¸ œ œˆ‰ ˆ ‰
"
24 100
20
I 0.24 estimated % change in I 100 100 15.83% more sensitive to voltage change.œœ Ê œ‚œ ‚¸ Ê
24 dI 0.038
100 I 0.24
62. A ab dA b da a db dA 16 da 10 db; da 0.1 and db 0.1kœÊœ Ê œ œ„ œ„111 11
10 16
dA 26 (0.1) 2.6 and A (10)(16) 160 100 100 1.625%Êœ„ œ„ œ œ Ê‚œ ‚¸111 1 ¸¸¸ ¸
dA 2.6
A 160
1
1
63. (a) y uv dy v du u dv; percentage change in u 2% du 0.02, and percentage change in v 3%œÊœ Ÿ Ê Ÿ Ÿ kk
dv 0.03; 100 100 100 100 100ÊŸ œ œÊ‚œ‚‚Ÿ‚‚kk ¹¹ ¸ ¸¸¸¸¸
dy dy
yuvuvy u v u v
v du u dv du dv du dv du dv
2% 3% 5%Ÿœ
(b) z u v (since u 0, v 0)œÊœ œŸ
dz du dv du dv du dv
zuvuvuvuv


100 100 100 100Ê ‚ Ÿ‚ ‚ œ‚¸¸¸ ¸ ¹¹
dz du dv
zuvy
dy
64. C C and CœÊœ œ
7
71.84w h 71.84w h 71.84w h
wh
( 0.425)(7) ( 0.725)(7)
0 425 0 725 1 425 0 725 0 425 1 725jj jj jj
œœ
dC dw dh; thus when w 70 and h 180 we haveÊœ œ œ
œœ2.975 5.075
71.84w h 71.84w h
1 425 0 725 0 425 1 725jj jj
dC (0.00000225) dw (0.00000149) dh 1 kg error in weight has more effectk
Ðß Ñ70 180
%# # `
65. f (x y) 2x y 2 0 and f (x y) x 2y 2 0 x 2 and y 2 ( 2 2) is the critical point;
xyß œ œ ß œ œ Ê œ œ Ê ß
f ( 2 2) 2, f ( 2) 2, f ( 2) 1 f f f 3 0 and f 0 local minimum value
xx yy xy xx yy xx xyßœ #ßœ #ßœÊ œ Ê
#
of f( 2) 8#ß œ
66. f (x y) 10x 4y 4 0 and f (x y) 4x 4y 4 0 x 0 and y 1 (0 1) is the critical point;
xyßœ œ ßœ œÊœ œÊß
f (0 1) 10, f (0 1) 4, f (0 1) 4 f f f 56 0 saddle point with f(0 1) 2
xx yy xy xx yy xyßœ ßœ ßœ Ê œ Ê ßœ
#
67. f (x y) 6x 3y 0 and f (x y) 3x 6y 0 y 2x and 3x 6 4x 0 x 1 8x 0xyßœ œ ßœ œÊœ œÊ œ
###%$
ab a b
x 0 and y 0, or x and y the critical points are (0 0) and . For ( ):Ê œ œ œ œ Ê ß ß !ß !
"" ""
## ##
ˆ‰
f ( ) 12x 0, f ( ) 12y 0, f ( 0) 3 f f f 9 0 saddle point withkk
xx yy xy xx yy 00 00 xy!ß ! œ œ !ß ! œ œ !ß œ Ê œ Ê
ÐßÑ ÐßÑ
#
f(0 0) 0. For : f 6, f 6, f 3 f f f 27 0 and f 0 local maximumßœ ß œœœÊ œ ʈ‰
""
##
#
xx yy xy xx yy xx xy
value of fˆ‰ß œ
"" "
## 4
68. f (x y) 3x 3y 0 and f (x y) 3y 3x 0 y x and x x 0 x x 1 0 the criticalxyßœ œ ßœ œÊœ œÊ œÊ
###%$
ab
points are (0 0) and (1 1) . For ( ): f ( ) 6x 0, f ( ) 6y 0, f ( 0) 3ß ß !ß ! !ß ! œ œ !ß ! œ œ !ß œ kk
xx yy xy 00 00ÐßÑ ÐßÑ
f f f 9 0 saddle point with f(0 0) 15. For (1 1): f (1 1) 6, f (1 1) 6, f (1 1) 3Ê œ Ê ß œ ß ß œ ß œ ß œxx yy xx yy xyxy
#
f f f 27 0 and f 0 local minimum value of f(1 1) 14Êœ Ê ßœxx yy xxxy
#
69. f (x y) 3x 6x 0 and f (x y) 3y 6y 0 x(x 2) 0 and y(y 2) 0 x 0 or x 2 andxyßœ œ ßœ œÊ œ œÊœ œ
##
y 0 or y 2 the critical points are (0 0), (0 2), ( 2 0), and ( 2 2) . For ( ): f ( ) 6x 6œ œ Ê ß ß ß ß !ß ! !ß ! œ k xx 00ÐßÑ
6, f ( ) 6y 6 6, f ( 0) 0 f f f 36 0 saddle point with f(0 0) 0. Forœ !ß ! œ œ !ß œ Ê œ Ê ß œk yy xy xx yy
00 xyÐßÑ
#
(0 2): f ( 2) 6, f (0 ) 6, f ( 2) 0 f f f 36 0 and f 0 local minimum value ofß !ßœ ß#œ !ßœÊ œ Êxx yy xy xx yy xx xy
#

933. Chapter 14 Practice Exercises 929
f( 2) 4. For ( 0): f ( 2 0) 6, f ( 0) 6, f ( 2 0) 0 f f f 36 0 and f 0!ß œ #ß ß œ #ß œ ß œ Ê œ xx yy xy xx yy xx xy
#
local maximum value of f( 2 0) 4. For ( 2 2): f ( 2 2) 6, f ( 2 2) 6, f ( 2 2) 0Ê ß œ ß ß œ ß œ ß œ xx yy xy
f f f 36 0 saddle point with f( 2 2) 0.`##$` #?xx yy xy
#
70. f (x y) 4x 16x 0 4x x 4 0 x 0, 2, 2; f (x y) 6y 6 0 y 1. Therefore the criticalx yßœ œÊ œÊœ ßœ œÊœ
$#
ab
points are (0 1), (2 1), and ( 2 1). For ( 1): f ( 1) 12x 16 16, f ( 1) 6, f ( 1) 0ß ß ß !ß !ß œ œ !ß œ !ß œ k
xx yy xy 01
#
ÐßÑ
f f f 96 0 saddle point with f(0 1) 3. For (2 1): f (2 1) 32, f (2 1) 6,Ê œÊ ßœ ß ßœ ßœxx yy xx yyxy
#
f (2 1) 0 f f f 192 0 and f 0 local minimum value of f(2 1) 19. For ( 1):xy xx yy xx xyßœÊ œ Ê ßœ #ß
#
f ( 2 1) 32, f ( 1) 6, f ( 2 1) 0 f f f 192 0 and f 0 local minimum value ofxx yy xy xx yy xx xyß œ #ß œ ß œ Ê œ Ê
#
f( 2 1) 19.ß œ
71. (i) On OA, f(x y) f(0 y) y 3y for 0 y 4? ? " ÑÑ
#
f ( y) 2y 3 0 y . But !? !?
w
##
33 ˆ‰
is not in the region.
Endpoints: f(0 0) 0 and f(0 4) 28.ߜ ߜ
(ii) On AB, f(x y) f(x x 4) x 10x 28? ?#" # "
#
for 0x4 f(xx4)2x100 ?
w
x 5, y 1. But (5 1) is not in the region.Êœ œ ß
Endpoints: f(4 0) 4 and f( 4) 28.ß œ !ß œ

(iii) On OB, f(x y) f(x 0) x 3x for 0 x 4 f (x 0) 2x 3 x and y 0 0 is aßœ ßœ ŸŸ Ê ßœ Êœ œÊ ß
#w
##
33 ˆ‰
critical point with f .ˆ‰
39
4#
ß! œ
Endpoints: f(0 0) 0 and f( 0) 4.ß œ %ß œ
(iv) For the interior of the triangular region, f (x y) 2x y 3 0 and f (x y) x 2y 3 0 x 3
xyß œ œ ß œ œ Ê œ
and y 3. But (3 3) is not in the region. Therefore the absolute maximum is 28 at (0 4) and the ? ?
absolute minimum is at .?!
93
4
ˆ‰
#
72. (i) On OA, f(x y) f(0 y) y 4y 1 for? ?#" "
#
0 y 2 f ( y) 2y 4 0 y 2 andŸŸ Ê !ß œ œ Ê œ
w
x 0. But (0 2) is not in the interior of OA.ϧ
Endpoints: f(0 0) 1 and f(0 2) 5.ߜ ߜ
(ii) On AB, f(x y) f(x 2) x 2x 5 for 0 x 4? ? # " ÑÑ
#
f (x 2) 2x 2 0 x 1 and y 2ÊßœœÊœ œ
w
(1 2) is an interior critical point of AB withÊß

f(1 2) 4. Endpoints: f(4 2) 13 and f( 2) 5.ߜ ߜ !ߜ
(iii) On BC, f(x y) f(4 y) y 4y 9 for 0 y 2 f (4 y) 2y 4 0 y and x 4. Butßœ ßœ ŸŸ Ê ßœœÊœ# œ
#w
(4 2) is not in the interior of BC. Endpoints: f(4 0) 9 and f( 2) 13.ßßœ%ßœ
(iv) On OC, f(x y) f(x 0) x 2x 1 for 0 x 4 f (x 0) 2x 2 0 x 1 and y 0 (1 0)ßœ ßœ ŸŸÊ ßœ œÊœ œÊß
#w
is an interior critical point of OC with f(1 0) 0. Endpoints: f(0 0) 1 and f(4 0) 9.ߜ ߜ ߜ
(v) For the interior of the rectangular region, f (x y) 2x 2 0 and f (x y) 2y 4 0 x 1 and
xyßœ œ ßœœÊœ
y 2. But (1 2) is not in the interior of the region. Therefore the absolute maximum is 13 at (4 2)œß ß
and the absolute minimum is 0 at (1 0).ß

934. 930 Chapter 14 Partial Derivatives
73. (i) On AB, f(x y) f( 2 y) y y 4 for?#? ##
#
2y2 f(2y)2y1 y andŸ Ÿ Ê ß œ Ê œ
w "
#
x 2 2 is an interior critical point in AB# ` # ?ˆ‰
"
#
with f 2 . Endpoints: f( 2 2) 2 andˆ‰ß œ ß œ
"
#
17
4
f(2 2) 2.?#
(ii) On BC, f(x y) f(x 2) 2 for 2 x 2? ?# #ÑÑ
f (x 2) 0 no critical points in the interior ofÊßœÊ
w
BC. Endpoints: f( 2 2) 2 and f(2 2) 2.ß œ ß œ
(iii) On CD, f(x y) f(2 y) y 5y 4 for? ? # "
#

2 y 2 f (2 y) 2y 5 0 y and x 2. But is not in the region.Ÿ Ÿ Ê ß œ œ Ê œ œ #ß
w
##
55 ˆ‰
Endpoints: f(2 2) 18 and f(2 2) 2.ß œ ß œ
(iv) On AD, f(x y) f(x 2) 4x 10 for 2 x 2 f (x 2) 4 no critical points in the interiorß œ ß œ Ÿ Ÿ Ê ß œ Ê
w
of AD. Endpoints: f( 2 2) 2 and f(2 2) 18.ß œ ß œ
(v) For the interior of the square, f (x y) y 2 0 and f (x y) 2y x 3 0 y 2 and x 1
xyßœœ ßœ œÊœ œ
(1 2) is an interior critical point of the square with f(1 2) 2. Therefore the absolute maximum`? ?#
is 18 at (2 2) and the absolute minimum is at .?# # #Ð?
17
4
ˆ‰
"
#
74. (i) On OA, f(x y) f(0 y) 2y y for 0 y 2? ? # ÑÑ
#
f ( y) 2 2y 0 y 1 and x 0 Ê!ßœœÊœ œÊ
w
( 1) is an interior critical point of OA with!ß
f(0 1) 1. Endpoints: f(0 0) 0 and f(0 2) 0.ߜ ߜ ߜ
(ii) On AB, f(x y) f(x 2) 2x x for 0 x 2? ? # ÑÑ
#
f (x 2) 2 2x 0 x 1 and y 2ÊßœœÊœ œ
w
(1 2) is an interior critical point of AB withÊß
f(1 2) 1. Endpoints: f(0 2) 0 and f(2 2) 0.ߜ ߜ ߜ
(iii) On BC, f(x y) f(2 y) 2y y for 0 y 2? ? # ÑÑ
#
f (2 y) 2 2y 0 y 1 and x 2ÊßœœÊœ œ
w

(2 1) is an interior critical point of BC with f(2 1) 1. Endpoints: f(2 0) 0 and f(2 2) 0.Êß ßœ ßœ ßœ
(iv) On OC, f(x y) f(x 0) 2x x for 0 x 2 f (x 0) 2 2x 0 x 1 and y 0 (1 0)ßœ ßœ ŸŸÊ ßœ œÊœ œÊß
#w
is an interior critical point of OC with f(1 0) 1. Endpoints: f(0 0) 0 and f(0 2) 0.ߜ ߜ ߜ
(v) For the interior of the rectangular region, f (x y) 2 2x 0 and f (x y) 2 2y 0 x 1 and
xyßœ œ ßœ œÊœ
y 1 (1 1) is an interior critical point of the square with f(1 1) 2. Therefore the absolute maximumœÊß ßœ
is 2 at (1 1) and the absolute minimum is 0 at the four corners (0 0), (0 2), (2 2), and (2 0).ß ßßß ß
75. (i) On AB, f(x y) f(x x 2) 2x 4 for? ?"#"
2 x 2 f (x x 2) 2 0 no criticalŸ Ÿ Ê ß œœ Ê
w
points in the interior of AB. Endpoints: f( 2 0) 8ß œ
and f(2 4) 0.ߜ
(ii) On BC, f(x y) f(2 y) y 4y for 0 y 4? ?#" ÑÑ
#
f (2 y) 2y 4 0 y 2 and x 2ÊßœœÊœ œ
w
(2 2) is an interior critical point of BC withÊß
f(2 2) 4. Endpoints: f(2 0) 0 and f(2 4) 0.ߜ ߜ ߜ

(iii) On AC, f(x y) f(x 0) x 2x for 2 x 2? ? # #ÑÑ
#
f (x 0) 2x 2 x 1 and y 0 (1 0) is an interior critical point of AC with f(1 0) 1.Êߜʜ œÊß ßœ
w
Endpoints: f( 2 0) 8 and f(2 0) 0.ß œ ß œ
(iv) For the interior of the triangular region, f (x y) 2x 2 0 and f (x y) 2y 4 0 x 1 and
xyßœ œ ßœœÊœ
y 2 (1 2) is an interior critical point of the region with f(1 2) 3. Therefore the absolute maximumœÊß ßœ
is 8 at ( 2 0) and the absolute minimum is 1 at (1 0).#? # ?

935. Chapter 14 Practice Exercises 931
76. (i) On AB, f(x y) f(x x) 4x 2x 16 for? ? # "
#%
2x2 f(xx)8x8x 0 x0Ÿ Ÿ Ê ß œ œ Ê œ
w$
and y 0, or x 1 and y 1, or x 1 and y 1œœ œœœ
(0 0), ( 1), ( 1 1) are all interior points of AB` ? j? # ? #
with f(0 0) 16, f(1 1) 18, and f( 1 1) 18.? ? #?#
Endpoints: f( 2 2) 0 and f(2 2) 0.ß œ ß œ
(ii) On BC, f(x y) f(2 y) 8y y for 2 y 2? ? # #ÑÑ
%
f (2 y) 8 4y 0 y 2 and x 2ÊßœœÊœ œ
w$ $ È
2 2 is an interior critical point of BC withÊߊ‹
È
$

f 2 2 6 2. Endpoints: f(2 2) 32 and f(2 2) 0.Š‹
ÈÈ
? ?## ?
$$
(iii) On AC, f(x y) f(x 2) 8x x for 2 x 2 f (x 2) 8 4x 0 x 2 and y 2ß œ ß œ Ÿ Ÿ Ê ß œ œ Ê œ œ
%w$ $ È
2 2 is an interior critical point of AC with f 2 2 6 . Endpoints:`#?# #?#Њ‹ Š‹
ÈÈ È
$$$
f( 2 2) 0 and f(2 2) 32.ß œ ß œ
(iv) For the interior of the triangular region, f (x y) 4y 4x 0 and f (x y) 4x 4y 0 x 0 and
xyßœ œ ßœ œÊœ
$$
y 0, or x 1 and y 1 or x 1 and y 1. But neither of the points (0 0) and (1 1), or ( 1 1) are interiorœœ œœœ ß ßß
to the region. Therefore the absolute maximum is 18 at (1 1) and ( 1 1), and the absolute minimum is 32 at?#?# #
(2 2).?#
77. (i) On AB, f(x y) f( 1 y) y 3y 2 for?#? # "
$#
1y1 f(1y)3y 6y0 y0ŸŸÊ ß œ œ Ê œ
w#
and x 1, or y 2 and x 1 ( 1 0) is anœ œ œ Ê ß
interior critical point of AB with f( 1 0) 2; ( 1 2)ß œ ß
is outside the boundary. Endpoints: f( 1 1) 2ß œ
and f( 1 1) 0.ß œ
(ii) On BC, f(x y) f(x 1) x 3x 2 for? ? " #
$#
1x1 f(x1)3x 6x0 x0Ÿ Ÿ Ê ß œ œ Ê œ
w#
and y 1, or x 2 and y 1 (0 1) is an#`?

interior critical point of BC with f( 1) 2; ( 2 1) is outside the boundary. Endpoints: f( 1) 0 and!ß œ ß "ß œ
f( 1) 2."ß œ
(iii) On CD, f(x y) f( y) y 3y 4 for 1 y 1 f ( y) 3y 6y 0 y 0 and x 1, orßœ"ßœ ŸŸÊ "ßœ œÊœ œ
$# w #
y 2 and x 1 ( 0) is an interior critical point of CD with f( 0) 4; (1 2) is outside the boundary.œœÊ"ß "ßœß
Endpoints: f(1 1) 2 and f( 1) 0.? j?#
(iv) On AD, f(x y) f(x 1) x 3x 4 for 1 x 1 f (x 1) 3x 6x 0 x 0 and y 1,ßœ ßœ ŸŸÊ ßœ œÊœ œ
$# w #
or x 2 and y 1 (0 1) is an interior point of AD with f(0 1) 4; ( 1) is outside theœ œ Ê ß ß œ #ß
boundary. Endpoints: f( 1 1) 2 and f( 1) 0.ß œ "ß œ
(v) For the interior of the square, f (x y) 3x 6x 0 and f (x y) 3y 6y 0 x 0 or x 2, and
xyßœ œ ßœ œÊœ œ
##
y 0 or y 2 (0 0) is an interior critical point of the square region with f( 0) 0; the points (0 2),œœÊß !ßœ ß
( 2 0), and ( 2 2) are outside the region. Therefore the absolute maximum is 4 at (1 0) and the#? #? ?
absolute minimum is 4 at (0 1).#?#

936. 932 Chapter 14 Partial Derivatives
78. (i) On AB, f(x y) f( 1 y) y 3y for 1 y 1?#? # #ÑÑ
$
f ( 1 y) 3y 3 0 y 1 and x 1Êßœ œÊœ„ œ
w#
yielding the corner points ( 1 1) and ( 1 1) with#?# #?
f( 1 1) 2 and f( 1 1) 2.ß œ ß œ
(ii) On BC, f(x y) f(x 1) x 3x 2 for? ? " "
$
1 x 1 f (x 1) 3x 3 0 noŸ Ÿ Ê ß œ œ Ê
w#
solution. Endpoints: f( 1) 2 and f( 1) 6.#j?# j?
(iii) On CD, f(x y) f( y) y 3y 2 forß œ "ß œ
$
1 y 1 f ( y) 3y 3 0 no#ÑÑ` j? "`
w#
solution. Endpoints: f(1 1) 6 and f( 1) 2.ß œ "ß œ

(iv) On AD, f(x y) f(x 1) x 3x for 1 x 1 f (x 1) 3x 3 0 x 1 and y 1ß œ ß œ Ÿ Ÿ Ê ß œ œ Ê œ „ œ
$w#
yielding the corner points ( 1 1) and (1 1) with f( 1 1) 2 and f(1 1) 2ß ß ß œ ß œ
(v) For the interior of the square, f (x y) 3x 3y 0 and f (x y) 3y 3x 0 y x and
xyßœ œ ßœ œÊœ
###
x x 0 x 0 or x 1 y 0 or y 1 ( 0) is an interior critical point of the square
%
œÊœ œÊœ œÊ!ß
region with f(0 0) 1; ( 1 1) is on the boundary. Therefore the absolute maximum is 6 at ( 1) and? #?# j?
the absolute minimum is 2 at (1 1) and ( 1 1).#?# #?
79. f 3x 2y and g 2x 2y so that f g 3x 2y (2x 2y ) 3x 2x and™™ ™™œ œ œ Ê œ Êœ
# ##
ij ij ij ij -- -
2y 2y 1 or y 0.œÊœ œ--
CASE 1: 1 3x 2x x 0 or x ; x 0 y 1 yielding the points (0 1) and ( 1); x-œÊ œ Êœ œ œÊœ„ ß !ß œ
# 2 2
3 3
y yielding the points and . ? ?
ÈÈÈ
555
33333
22
Š‹Š ‹
CASE 2: y 0 x 1 0 x 1 yielding the points (1 0) and ( 1 0).œÊ œÊœ„ ß ß
#
Evaluations give f 1 1, f , f( 0) 1, and f( 0) 1. Therefore the absoluteab Š‹!ß„œ ß„ œ "ßœ "ßœ
223
33 27
5
È
maximum is 1 at 1 and (1 0), and the absolute minimum is 1 at ( ).ab!? w ? # #j? !
80. f y x and g 2x 2y so that f g y x (2x 2y ) y 2 x and™™ ™™œ œ œ Ê œ Êœij i j ij i j -- -
xy 2 y x 2 (2 x) 4 x x 0 or 4 1.œÊœ œ Êœ œ---- -
##
CASE 1: x 0 y 0 but (0 0) does not lie on the circle, so no solution.œÊœ ß
CASE 2: 4 1 or . For , y x 1 x y 2x x yielding the--- -
# ## #"" " "
## #
œÊœ œ œ œÊœœ ÊœCœ„
È
2
points and , . For , y x 1 x y 2x x andŠ‹Š ‹
"" " " " "
#
## #
ÈÈ È È È
22 2 2 2
ß œ œÊœœ Êœ„ -
y x yielding the points and , .# # ? # Š‹Š‹
"" " "
ÈÈ È È
22 2 2
Evaluations give the absolute maximum value f f and the absolute minimumŠ‹Š ‹
"" " " "
#ÈÈ È È
22 2 2
??
value f f .Š‹Š‹ß œ ß œ
"" " " "
#ÈÈ È È
22 2 2
81. (i) f(x y) x 3y 2y on x y 1 f 2x (6y 2) and g 2x 2y so that f gßœ œÊ œ œ œ
## ##
™™™™ij ij -
2x (6y 2) (2x 2y ) 2x 2x and 6y 2 2y 1 or x 0.Êœ Êœ œ Êœ œijij ----
CASE 1: 1 6y 2 2y y and x yielding the points .-œÊ œ Êœ œ„ „ ß
""
## ##
ÈÈ
33
Š‹
CASE 2: x 0 y 1 y 1 yielding the points 1 .œÊ œÊœ„ !ß„
#
ab Evaluations give f , f(0 1) 5, and f( 1) 1. Therefore and 5 are the extremeŠ‹? ? !?
È
3
### #
"" "
values on the boundary of the disk.
(ii) For the interior of the disk, f (x y) 2x 0 and f (x y) 6y 2 0 x 0 and y
xyßœ œ ßœ œÊœ œ
"
3
is an interior critical point with f . Therefore the absolute maximum of f on theÊ !ß !ß œ ˆ‰ ˆ‰
1
333
""
disk is 5 at (0 1) and the absolute minimum of f on the disk is at .? # !? #
""
33
ˆ‰

937. Chapter 14 Practice Exercises 933
82. (i) f(x y) x y 3x xy on x y 9 f (2x 3 y) (2y x) and g 2x 2y so thatßœ œÊ œ œ
## ##
™™ ij ij
f g (2x 3 y) (2y x) (2x 2y ) 2x 3 y 2x and 2y x 2y™™œ Ê œ Ê œ œ---- ijij
2x( ) y 3 and x 2y(1 ) 0 1 and (2x) y 3 x y 3yÊ"œ œÊœ œÊœ---
xx
2y 2y
Š‹
##
x y 3y. Thus, 9 x y y 3y y 2y 3y 9 0 (2y 3)(y 3) 0Ê œ œœ Ê œÊ œ
## ### # #
y 3, . For y 3, x y 9 x 0 yielding the point (0 3). For y , x y 9Êœ œ œÊœ ß œ œ
3 3
# #
## ##
x 9 x x . Evaluations give f(0 3) 9, f 9ʜʜʜ„ ßœ ßœ
##
###
927 3
44 4
33 33 273
ÈÈÈ
Š‹
20.691, and f , 9 2.691.¸œ¸ Š‹
33 2733
4
ÈÈ
##
(ii) For the interior of the disk, f (x y) 2x 3 y 0 and f (x y) 2y x 0 x 2 and y 1
xyßœ œ ßœ œÊœ œ
(2 1) is an interior critical point of the disk with f(2 1) 3. Therefore, the absolute maximum of f on`? ?#
the disk is 9 at and the absolute minimum of f on the disk is 3 at (2 1)."#? #?
27 3 3 3
4
3
ÈÈ
Š‹
##
83. f and g 2x 2y 2z so that f g (2x 2y 2z ) 1 2x ,™™ ™™œ œ œ Ê œ Ê œijk i j k ijk i j k -- -
1 2y , 1 2z x y z . Thus x y z 1 3x 1 x yielding the pointsœ œ Êœœœ œÊ œÊœ„--
"" ### #
- È
3
, and , , . Evaluations give the absolute maximum value ofŠ‹Š ‹
""" """
ÈÈÈ ÈÈÈ
333 333
?# # #
f 3 and the absolute minimum value of f 3.Š‹ Š ‹
ÈÈ
""" "" "
ÈÈÈ È ÈÈÈ
333 3 333
3
?#? #??##
84. Let f(x y z) x y z be the square of the distance to the origin and g(x y z) z xy 4. Thenßß œ ßß œ
### #
f 2x 2y 2z and g y x 2z so that f g 2x y, 2y x, and 2z 2 z™™ ™™œ œ œ Ê œ œ œijk ijk ----
z 0 or 1.Êœ œ -
CASE 1: z 0 xy 4 x and y 2 y and 2 x y and xœ Ê œ Ê œ œ Ê œ œ Ê œ œ
444 4 88
yxy x
Š‹ ˆ‰--
--
##
y x y x. But y x x 4 leads to no solution, so y x x 4 x 2ʜʜ„ œÊœ œÊœÊœ„
## # #
yielding the points ( 2 2 0) and (2 2 0).#? ? ?#?
CASE 2: 1 2x y and 2y x 2y 4y y y 0 x 0 z 4 0 z 2-œ Ê œ œ Ê œ Ê œ Ê œ Ê œ Ê œ Ê œ „ˆ‰
y
#
#
yielding the points (0 0 2) and ( 0 2).??# !??
Evaluations give f( 2 2 0) f(2 2 0) 8 and f( 2) f( 2) 4. Thus the points ( 2) and ß ß œ ß ß œ !ß !ß œ !ß !ß œ !ß !ß
( 2) on the surface are closest to the origin.!ß !ß
85. The cost is f(x y z) 2axy 2bxz 2cyz subject to the constraint xyz V. Then f gßß œ œ œ ™™-
2ay 2bz yz, 2ax 2cz xz, and 2bx 2cy xy 2axy 2bxz xyz, 2axy 2cyz xyz, andÊœ œ œ Ê œ œ-- - - -
2bxz 2cyz xyz 2axy 2bxz 2axy 2cyz y x. Also 2axy 2bxz 2bxz 2cyz z x.œ Ê œ Êœ œ Êœ- ˆ‰ ˆ‰
ba
cc
Then x x x V x width x , Depth y , andˆ‰ˆ‰ ˆ‰ Š‹ Š‹ Š‹
b a cV cV b cV bV
c c ab ab c ab ac
œ Ê œ Ê œœ œœ œ
$
"Î$ "Î$ "Î$

Height z .œœ œˆ‰Š‹ Š‹
acV aV
cab bc

"Î$ "Î$
86. The volume of the pyramid in the first octant formed by the plane is V(a b c) ab c abc. The pointßß œ œ
"" "
#36
ˆ‰
(2 1 2) on the plane 1. We want to minimize V subject to the constraint 2bc ac 2ab abc.ßß Ê œ œ
22
abc
"
Thus, V and g (c 2b bc) (2c 2a ac) (2b a ab) so that V g™™ ™™œ œ œ
bc ac ab
666
ijk i j k -
(c 2b bc), (2c 2a ac), and (2b a ab) (ac 2ab abc),Êœ œ œ Ê œ
bc ac ab abc
66 6 6
-- - -
(2bc 2ab abc), and (2bc ac abc) ac 2 bc and 2 ab 2 bc. Now 0 since
abc abc
66
œ œÊœ œ Á-------
a 0, b 0, and c 0 ac 2bc and ab bc a 2b c. Substituting into the constraint equation givesÁÁ ÁÊœ œÊœœ
1 a 6 b 3 and c 6. Therefore the desired plane is 1 or x 2y z 6.
222 xz
aaa 636
y
œÊœÊœ œ œ œ
87. f (y z) x x , g 2x 2y , and h z x so that f g h™™™™™™œ œ œ œ ijk i j ik -.
(y z) x x (2x 2y ) (z x ) y z 2 x z, x 2 y, x x x 0Êœ Êœ œ œ Êœijk i j ik-. -.-.

938. 934 Chapter 14 Partial Derivatives
or 1..œ
CASE 1: x 0 which is impossible since xz 1.œœ
CASE 2: 1 y z 2 x z y 2 x and x 2 y y (2 )(2 y) y 0 or.-----œÊœ Êœ œ Êœ Êœ
4 1. If y 0, then x 1 x 1 so with xz 1 we obtain the points (1 0 1)-
##
œœ œÊœ„ œ ßß
and ( 1 0 1). If 4 1, then . For , y x so x y 1 xß ß œ œ„ œ œ œ Ê œ-- -
#### "" "
## #
x with xz 1 z 2, and we obtain the points 2 andÊœ„ œÊœ„ ß ß
" ""
È ÈÈ
2 22 ÈÈ
Š‹
2 . For , y x x x with xz 1 z 2,Š‹
ÈÈ
ß ß œ œÊ œÊœ„ œÊœ„
"" " " "
##
#
ÈÈ È
22 2
-
and we obtain the points , 2 and 2 .Š‹Š ‹
ÈÈ
"" " "
ÈÈ È È
22 2 2
? # ?# ?#
Evaluations give f(1 0 1) 1, f( 1 0 1) 1, f 2 , f , 2 ,ßß œ ßß œ ß ß œ ß œŠ‹Š ‹
ÈÈ
"" " "" "
##ÈÈ ÈÈ
22 22
f 2 , and f 2 . Therefore the absolute maximum is atŠ‹Š ‹
ÈÈ
"" " "
## #ÈÈ È È
22 2 2
33 3
ß ß œ ß ß œ
2 and 2 , and the absolute minimum is at 2 andŠ‹Š ‹ Š ‹
ÈÈ È
"" " " " ""
#ÈÈ È È ÈÈ
22 2 2 22
?? #?#?# #??#
2.Š‹
È
""
ÈÈ
22
? ?
88. Let f(x y z) x y z be the square of the distance to the origin. Then f 2x 2y 2z ,ßßœ œ
###
™ ijk
g , and h 4x 4y 2z so that f g h 2x 4x , 2y 4y ,™™ ™™™œ œ œ Ê œ œ ijk i j k - . -. -.
and 2z 2z 2x(1 2 ) 2y(1 2 ) 2z(1 2 ) x y or .œ Êœ œ œ Êœ œ-.- . . . .
"
#
CASE 1: x y z 4x z 2x so that x y z 1 x x 2x 1 or x x 2x 1œ Ê œ Ê œ„ œ Ê œ œ
##
(impossible) x y and z yielding the point .ʜʜ œ ßß
" " " """
##44 44
ˆ‰
CASE 2: 0 0 2z(1 1) z 0 so that 2x 2y 0 x y 0. But the origin.-œÊœÊœ Êœ œÊœœ
"
#
##
( 0 0) fails to satisfy the first constraint x y z 1.!?? ""
Therefore, the point on the curve of intersection is closest to the origin.ˆ‰
"""
#44
ßß
89. (a) y, z are independent with w x e and z x y œœÊœ
### ```` ``
```````
`yz wwxw wz
yxyyyzy
y
2xe zx e (1) yx e (0); z x y 0 2x 2y ; therefore,œ œÊœÊœab a b a b
yz yz yz` ``
` ``
## ##x xx
y yyx
y
2xe zx e 2y zx eŠ‹ ab a bˆ‰
`
`
##w
yx
y
z
yz yz yz
œœ
(b) z, x are independent with w x e and z x y œœÊœ
### ```` ``
```````
`yz wwxw wz
zxzyzzz
y
2xe (0) zx e yx e (1); z x y 1 0 2y ; therefore,œ œÊœÊœab a b a b
yz yz yz ## ##```
```#
"yyy
zzzy
zxe yxe xe yˆ‰ ab Š‹ Š ‹
`
`
###w1 z
z2y 2y
x
yz yz yz
œœ
(c) z, y are independent with w x e and z x y œœÊœ
### ```` ``
```````
`yz wwxw wz
zxzyzzz
y
2xe zx e (0) yx e (1); z x y 1 2x 0 ; therefore,œ œÊœÊœab a b a b
yz yz yz` ``"
` ``#
####x xx
z zzx
2xe yx e 1 x y eˆ‰ ˆ‰ab a b
`
`
##w1
z2x
y
yz yz yz
œœ
90. (a) T, P are independent with U f(P V T) and PV nRT œßß œ Ê œ
```````
```````
UUPUVUT
TPTVTTT
(0) (1); PV nRT P nR ; therefore,œ œÊœÊœˆ‰ ˆ‰ˆ‰ˆ‰
```` ` `
```` ` `
UUVU V VnR
PVTT T TP
ˆ‰ ˆ‰ˆ‰
`` `
`` `
UUnRU
TVPT
P
œ
(b) V, T are independent with U f(P V T) and PV nRT œßß œ Ê œ
```````
```````
UUPUVUT
VPVVVTV
(1) (0); PV nRT V P (nR) 0 ; therefore,œ œÊœœÊœˆ‰ˆ‰ˆ‰ ˆ‰ ˆ‰
`` ` ` ` ` `
`` ` ` ` ` `
UP U U P T P P
PV V T V V V V
ˆ‰ ˆ‰ˆ ‰
`` `
`` `
UUPU
VPVV
T
œ
91. Note that x r cos and y r sin r x y and tan . Thus,œœÊœœ)) ) È ˆ‰
## "y
x
(cos ) ;
````` ` ` ` `
````` ` ` ` `
œwwrw w x w wsin w
xrx x r xy r r xy
y
œœ œ
)) )
) )
ˆ‰ ˆ‰ ˆ ‰Š‹ Š‹
È
)

939. Chapter 14 Practice Exercises 935
(sin )
````` ` ` ` `
```` ` ` ` `
wwrw w w x wcos w
yry y r xy r r
y
xy
œœ œ
)) ) )
) )
ˆ‰ ˆ‰ ˆ ‰Š‹ Š‹
È
)
92. z f f af af , and z f f bf bf
xu v u v yu v u vœœ œœ
`` ``
`` ``
uv uv
xx yy
93. b and a a and b and
` ` `` `` "` "` ` ` `` `` ` `
u u wdwu dw wdwu dw wdw wdw
y x x du x du y du y du a x du b y du
œœÊœœ œœÊœ œ
b a ʜʜ
"` "` ` `
`` ``ax by x y
ww ww
94. - , ,
`"`
` `#
w2x w rs
x x y 2z (rs) (rs) 4rs 2r 2rss rs y x y 2z (rs) (rs)
2(r s) 2(r s) 2y 2(r s)
œœ œ œœœœ

ab
and (2s)
`"``````""
` ```````
`w 2 w w x w w z r s 2r 2s
z x y 2z (r s) r x r y r z r r s (r s) (r s) (r s)
y
œœÊœœ œ
’“
and (2r)œœœ œ
2wwxw wz rs 2
rs s x s y s z s rs (rs) (rs) rs
y
` `` `` ``
```` `` " " `
’“
95. e cos v x 0 e cos v e sin v 1; e sin v y 0 e sin v e cos v 0.
uuuuuu uv u v
xx x x
œ Ê œ œ Ê œa bab aba b
`` ` `
`` ` `
Solving this system yields e cos v and e sin v. Similarly, e cos v x 0
`` ``
œœuv
xx
uuu
œœ œ
e cos v e sin v 0 and e sin v y 0 e sin v e cos v 1. Solving thisÊœœÊœa bab aba b
uu u uu uv u v
yy y y
`` ` `
`` ` `
second system yields e sin v and e cos v. Therefore
` ` `` `` ` ` `` ``
œœu v uu vv
y y xy xy
uu
œœ Š‹Š‹ij ij†
e cos v e sin v e sin v e cos v 0 the vectors are orthogonal the angleœ œÊ Êcdc dababa bab
œœ œœuu uu
ij i j †
between the vectors is the constant .
1
#
96. ( r sin ) (r cos )
``
````` ` `
`` ` ` `gy fx f f f
xy x y)))
œœ ))
( r sin ) (r cos ) (r cos ) (r sin ) ʜ
`` `
` ` ` ``` ` ``` ` ` `
`` ` ` ` ` ` `



gy y fx f f f x f f
xyx x xyy y))) ))
))))Š‹ Š‹
( r sin ) (r cos ) (r cos ) (r sin )œ ))))Š‹ Š‹
``
`` ``
``xx yy
)) ))
( r sin r cos )( r sin r cos ) (r cos r sin ) ( 2)( 2) (0 2) 4 2 2 atœ œ œ œ)))) ))
(r ) 2 .ßœß)ˆ‰
1
#
97. (y z) (z x) 16 f 2(z x) 2(y z) 2(y 2z x) ; if the normal line is parallel to theœÊ œ
##
™ ij k
yz-plane, then x is constant 0 2(z x) 0 z x (y z) (z z) 16 y z 4.ʜʜʜʜʜ„
`
`
##f
x
Let x t z t y t 4. Therefore the points are (t t 4 t), t a real number.œÊ œÊ œ„ ß„ß
98. Let f(x y z) xy yz zx x z 0. If the tangent plane is to be parallel to the xy-plane, then f is??""##
#
™
perpendicular to the xy-plane f 0 and f 0. Now f (y z 1) (x z) (y x 2z)Ê œ œ œ ™† ™† ™ij ijk
so that f y z 1 0 y z 1 y 1 z, and f x z 0 x z. Then™† ™†ijœœ Ê œ Ê œ œœ Ê œ
z(1 z) ( z)z z( z) ( z) z 0 z 2z 0 z or z 0. Now z x and y " œ Ê œ Ê œ œ œ Ê œ œ
## """"
####
is one desired point; z 0 x 0 and y 1 (0 1 0) is a second desired point.Êßß œ Ê œ œ Ê ß߈‰
"""
###
99. f (x y z ) x f(x y z) x g(y z) for some function g y™œÊœÊßßœ ß Êœœ--- -ijk
`" `
`# ``
# `f f
x yy
g
g(y z) y h(z) for some function h z h (z) h(z) z C for some arbitraryÊßœ Êœœœ Ê œ
"`"
#``#
#w# `
-- -
f
zz
g
constant C g(y z) y z C f(x y z) x y z C f(0 0 a) a C`?" "`??"""`??"
" " """ "
# # ### #
## ### #
- - --- -ˆ‰
and f(0 0 a) ( a) C f(0 0 a) f(0 0 a) for any constant a, as claimed.ßß œ Ê ßß œ ßß
"
#
#
-
100. lim , s 0
s0
ˆ‰
df
ds
f(0su0su0su)f(000)
u###(000)
œ
Ä
ßß ßß
s
lim , s 0
s0
œ
Ä
Ésu su su 0



s

940. 936 Chapter 14 Partial Derivatives
lim lim 1;
s0 s0
œœœ
ÄÄ
su u uÉ



s
kku
however, f fails to exist at the origin (0 0 0)™"" ??
xz
xyz xyz xyz
y
ÈÈÈ


ijk
101. Let f(x y z) xy z 2 f y x . At (1 1 1), we have f the normal line isßß œ Ê œ ßß œ Ê™™ ijk ijk
x 1 t, y 1 t, z 1 t, so at t 1 x 0, y 0, z 0 and the normal line passes through theœ œ œ œÊ œ œ œ
origin.
102. (b) f(x y z) x y z 4??#"
###
f 2x 2y 2z at (2 3 3)Ê œ Ê ßß™ ijk
the gradient is f 4 6 6 which is™œijk
normal to the surface
(c) Tangent plane: 4x 6y 6z 8 orœ
2x 3y 3z 4œ
Normal line: x 2 4t, y 3 6t, z 3 6tœ œ œ

CHAPTER 14 ADDITIONAL AND ADVANCED EXERCISES
1. By definition, f ( 0) lim so we need to calculate the first partial derivatives in the
xy!ß œ
h0Ä
f(0h) f(00)
hxx‚œ ‚
numerator. For (x y) (0 0) we calculate f (x y) by applying the differentiation rules to the formula forßÁß ß
x
f(x y): f (x y) (xy) f (0 h) h.ß ßœ œ Êßœœ
x x
xy y xy y 4xy
xy xy h
x y (2x) x y (2x)
xy xy
h





œ
œœ



abab
ab ab
For (x y) (0 0) we apply the definition: f ( 0) lim lim 0. Then by definitionßœß !ßœ œ œ
x
h0 h0ÄÄ
f(h 0) f(0 0)
hh
00‚œ ‚ œ
f (0 0) lim 1. Similarly, f (0 0) lim , so for (x y) (0 0) we have
xy yxßœ œ ßœ ßÁß
h0 h0ÄÄ
œœ ‚œ%‚h0
hh
f(h0) f( 0) yy
f (x y) f (h 0) h; for (x y) (0 0) we obtain f (0 0) lim
y yßœ Ê ßœ œ ßœß ßœ
xxy 4xy f(0 h) f( 0)
xy h hxy
y
h




œ
œ ‚œ%‚
ab
h0Ä lim 0. Then by definition f (0 0) lim 1. Note that f (0 0) f (0 0) in this case.œœ ßœœ ßÁß
h0 h0ÄÄ
00 h0
hh
xy yx
œœ
yx
2. 1 e cos y w x e cos y g(y); e sin y g (y) 2y e sin y g (y) 2y
``
``
wwww
xy
xx x x
œ Ê œ œ œ Ê œ
g(y) y C; w ln 2 when x ln 2 and y 0 ln 2 ln 2 e cos 0 0 C 0 2 CÊœœ œ œÊœ Êœ
# #ln 2
C 2. Thus, w x e cos y g(y) x e cos y y 2.Êœ œ œ
xx #
3. Substitution of u u(x) and v v(x) in g(u v) gives g(u(x) v(x)) which is a function of the independent"? ? variable x. Then, g(u v) f(t) dt f(t) dt f(t) dt ßœ Êœœ '''
uuu
vvv
dg g g
dx u dx v dx u dx v dx
du dv du dv``
`` ` `
``
Š‹Š‹
f(t) dt f(t) dt f(u(x)) f(v(x)) f(v(x)) f(u(x)) œ œœŠ‹Š‹
``
``u dx v dx dx dx dx dx
du dv du dv dv du''
vu
uv
4. Applying the chain rules, f f . Similarly, f and
xxx yyœÊœ œ
df r d f r df r d f r df r
dr x dr x dr x dr y dr y
``` ``
``` ``
#
#
Š‹ Š‹Š‹ˆ‰


f . Moreover, ;
zzœ œÊœ œŠ‹ˆ‰
df r df r r x r r
dr z dr z x x y xyz xyz
yz y
xyz




`` ` ` `
`` ` ` `
#



ÈÈ
ˆ‰È
3
; and . Next, f f f 0Êœ œ Êœ œ
` ` `
```







rxz r z r
yzz
xyz xyz xyz
xy
xx yy zz
ˆ‰ ˆ‰ÈÈ È
3 3
`" ""Š‹Š ‹ Š‹Š ‹ ˆ‰ ˆ‰
Œ Œ
df x df df df x z
dr x y z dr dr x y z dr
yz y
xyz xyz








ˆ‰ ˆ‰ÈÈ
3 3

941. Chapter 14 Practice Exercises 937
0 0 0 œÊœÊœŠ‹Š ‹ Š ‹ ˆ‰
Œ
df z df df 2 df df 2 df
dr x y z dr dr dr dr r dr
xy
xyz xyz







ˆ‰È È
3 (f ) f , where f ln f 2 ln r ln C f Cr , orÊœ œÊœÊœÊœ
d 2 df df 2 dr
dr r dr f r
www w w#
ˆ‰
$
$
Cr f(r) b b for some constants a and b (setting a C)
df C a
dr r r
œÊœœ œ
#
5. (a) Let u tx, v ty, and w f(u v) f(u(t x) v(t y)) f(tx ty) t f(x y), where t, x, and y areœ œ œßœßßߜߜß
n
independent variables. Then nt f(x y) x y . Now,
n1`
?""
````` ` `
````` ` `
wwuwv w w
tutvt u v
(t) (0) t . Likewise,
````` ` ` ` ` "`
````` ` ` ` ` `
wwuwv w w w w w
xuxvx u v u utx
œœ œÊœ ˆ‰ ˆ‰ ˆ‰ˆ‰
(0) (t) . Therefore,
````` ` ` ` "`
````` ` ` ` `
wwuwv w w w w
yuyvy u v vty
œœ Êœ ˆ‰ ˆ‰ ˆ‰ Š‹
nt f(x y) x y . When t 1, u x, v y, and w f(x y)
n1`
ßœœ œœœ œß
`` ` `
`` ` `
wwxw w
uvtxty
y
ˆ‰ˆ ‰ ˆ‰ Š‹
and nf(x y) x y , as claimed.Êœ œÊ ßœ
`` `` ` `
`` `` ` `
wf wf f f
xx yx x y
(b) From part (a), nt f(x y) x y . Differentiating with respect to t again we obtain
n1`
? "
``
``
ww
uv
n(n 1)t f(x y) x x y y x 2xy y .#?"""""
n2` `` `` `` `` ` ` `
`` ``` ``` `` ` `` `
##


wu wv wu wv w w w
ut vut uvt vt u uv v
Also from part (a), t t t t ,
``` `` ```` ````
``` `` ````` ````
#


ww wwuwvww w
xxx xu uxvux uyyy
œœ œ œ œˆ‰ ˆ ‰ Š‹
t t t t , and t t t œ œ œ œœ œ
`` `` `` ` ` `` `` `` ``
` ` `` ` ` ` ` `` ` ` ` ` ` ` `` `
#
y v uv y v y v yx y x y u u y vu y
wwuwvww w wwuwv
ˆ‰ ˆ‰ ˆ‰


t , , and œÊ œ œ œ
#`"``"`` "``
`` ` ` ` ` `` ``


wwwww ww
vu t x u t y v t yx vu
ˆ‰ ˆ‰ ˆ‰ n(n 1)t f(x y) for t 0. When t 1, w f(x y) andÊ ßœ Á œ œß
n2`
Š‹Š ‹ Š ‹ Š‹Š ‹ˆ‰
xw w w
tx tyx ty
2xy y



```
````
we have n(n 1)f(x y) x 2xy y as claimed.#? " "
##```
````
Š‹ Š ‹ Š‹


fff
xxyy
6. (a) lim lim 1, where t 6r
r0 t0ÄÄ
sin 6r sin t
6r t
œœ œ
(b) f (0 0) lim lim lim lim
rßœ œ œ œ
h0 h0 h0 h0ÄÄÄÄ
f(0 h 0) f(0 0)
h h 6h 12h
1 sin 6h 6h 6 cos 6h 6ß ß œ œœ
ˆ‰
sin 6h
6h

lim 0 (applying l'Hopital's rule twice)œœ s
h0Ä
œ36 sin 6h
12
(c) f (r ) lim lim lim 0
œßœ œ œ œ)
h0 h0 h0ÄÄÄ
f(r h) f(r )
hhh
0ß ß œ))
ˆ‰ˆ‰
sin 6r sin 6r
6r 6r
7. (a) x y z r x y z and rrijk r i j kœ Êœ œ œ kkÈ
###

™
xz
xyz xyz xyz
y
ÈÈÈ

œ
r
r
(b) r x y z
n
n
œˆ‰È
###
r nxxyz nyxyz nzxyzÊ œ ™abababab
n
n2 1 n2 1 n2 1### ### ###
Ð

ijk
nr œ
n2`
r
(c) Let n 2 in part (b). Then r r x y z is the function.œœÊœÊœ
"""
####
## ###
™™ab a b ˆ‰rr
r

(d) d dx dy dz d x dx y dy z dz, and dr r dx r dy r dz dx dy dzrijkrrœ Ê œ œ œ †
xyz
xz
rrr
y
r dr x dx y dy z dz dÊœœ rr†
(e) a b c ax by cz ( ) a b cA i j k Ar Ar i j k Aœ Ê œ Ê œœ†™†
8. f(g(t) h(t)) c 0 , where is the tangent vectorߜʜœœ
df f dx f f f dx dx
dt x dt y dt x y dt dt dt dt
dy dy dy`` ``
`` ``
Š‹Š‹ij ij ij†
f is orthogonal to the tangent vectorÊ™
9. f(x y z) xz yz cos xy 1 f z y sin xy ( z x sin xy) (2xz y) f(0 0 1)ßß œ Ê œ Ê ßß œ
##
™™ab ijk ij
the tangent plane is x y 0; (ln t) (t ln t) t (ln t 1) ; x y 0, z 1ÊœœÊœœœœ ri jkri jk
w "ˆ‰
t
t 1 (1) . Since ( ) ( ) r (1) f 0, is parallel to the plane, andÊ œ Ê œ œ œrijk ijkij r
ww
††™
(1) 0 0 is contained in the plane.rijkrœÊ

942. 938 Chapter 14 Partial Derivatives
10. Let f(x y z) x y z xyz f 3x yz 3y xz 3z xy f(0 1 1) 3 3ßßœ Ê œ Ê ßßœ
$$$ ###
™™abababijk ijk
the tangent plane is x 3y 3z 0; 2 3 (cos (t 2))Ê œ œ rij kŠ‹ ˆ‰
t4
4t
œ
(sin (t 2)) ; x 0, y 1, z 1 t 2 (2) 3 . SinceÊ œ œ œ œÊœÊ œrij k rij
w w
Š‹ ˆ‰
3t 4
4t


(2) f 0 is parallel to the plane, and (2) is contained in the plane.rr rikr
w
†™œÊ œÊ
11. 3x 9y 0 and 3y 9x 0 y x and 3 x 9x 0 x 9x 0
`` """
``
####%
#
zz
xy 333
œ œ œ œÊœ œÊ œ ˆ‰
x x 27 0 x 0 or x 3. Now x 0 y 0 or ( 0) and x 3 y 3 or (3 3). NextÊ œÊœ œ œÊœ !ß œÊœ ßab
$ 6x, 6y, and 9. For ( 0), 81 no extremum (a saddle point),
`` ` ```
` ` `` ` ` ``
#


zz z zzz
x y xy x y xy
œœ œ!ß œÊ Š‹
and for (3 3), 243 0 and 18 0 a local minimum.?#`
`` ` `
`` `` `
#


zz z z
xy xy x
Š‹
12. f(x y) 6xye f (x y) 6y(1 2x)e 0 and f (x y) 6x(1 3y)e 0 x 0 andßœ Êßœ œ ßœ œÊœ
Ð Ñ Ð Ñ Ð Ñ2x 3y 2x 3y 2x 3y
xy
y 0, or x and y . The value f(0 0) 0 is on the boundary, and f . On the positive y-axis,œœ œ ßœ ßœ
"" """
# #33e
ˆ‰
2
f(0 y) 0, and on the positive x-axis, f(x 0) 0. As x or y we see that f(x y) 0. Thus theßœ ßœ Ä_ Ä_ ß Ä
absolute maximum of f in the closed first quadrant is at the point .
"""
#e3
2
ˆ‰ß
13. Let f(x y z) 1 f an equation of the plane tangent at the pointßß œ Ê œ Ê
x z 2x 2z
abc a b c
y2y



™ ijk
P (x y y ) is x y z 2 or x y z 1.!!!!ßß œ œ œˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰ ˆ‰
2x 2z x z
abcabc abc
2y y 2x 2y 2z`` ``

` `
"""
The intercepts of the plane are 0 0 , 0 0 and . The volume of the tetrahedron formedŠ‹Š‹Š‹
ab c
xy z

`` `
ß ß ß ß !ß !ß
by the plane and the coordinate planes is V we need to maximizeœÊˆ‰ˆ‰Š‹Š‹Š‹
""
#3 xyz
abc

```
V(x y z) (xyz) subject to the constraint f(x y z) 1. Thus,ßß œ ßß œ œ
(abc) y
6abc
xz


"
, , and . Multiply the first equation’“Š‹ ’“Š‹ ’“Š‹œœœ
(abc) (abc) 2y (abc)
6 x yz a 6 xy z b 6 xyz c
2x 2z


"" "
-- -
by a yz, the second by b xz, and the third by c xy. Then equate the first and second a y b x
## # ####
ʜ
y x, x 0; equate the first and third a z c x z x, x 0; substitute into f(x y z) 0Êœ Ê œ Êœ ßßœ
b c
a a
## ##
x y z V abc.Êœ Êœ Êœ Ê œ
abc
333
3
ÈÈÈ
È
#
14. 2(x u) , 2(y v) , 2(x u) , and 2(y v) 2 v x u v y, x u , and œ œ œ œ Ê œ œ-- . .
.
#
y v v x u v v or 0. œ Ê œ œ Ê œ œ.. .
.
##
"
CASE 1: 0 x u, y v, and 0; then y x 1 v u 1 and v u v v 1.-œÊœ œ œ œÊœ œÊœ
##
v v 1 0 v no real solution.ʜʜ Ê
# Cœ
#
114
È
CASE 2: v and u v u ; x y and y x 1 x x 2xœ œ Ê œ œ œ Ê œ Ê œ
"""" """
## #
#
44 4 4
x y . Then f 2 the minimum distanceÊ œÊ œ ßßß œ œ Ê
"""""""
##
## #
88 884 848 8
77 73
ˆ ‰ˆ ‰ˆ‰ˆ‰
is 2. (Notice that f has no maximum value.)
3
8È
15. Let (x y ) be any point in R. We must show lim f(x y) f(x y ) or, equivalently that
!! !!ßßœß
Ðß ÑÄÐ ß Ñxy x y ÎÎ
lim f(x h y k) f(x y ) 0. Consider f(x h y k) f(x y )
Ðß ÑÄÐß Ñhk 00
kk!! !! !! !!ß ß œ ß ß
[f(x h y k) f(x y k)] [f(x y k) f(x y )]. Let F(x) f(x y k) and apply the Mean Valueœ ß ß ß ß œ ß
! ! !! !! !! !
Theorem: there exists with x x h such that F ( )h F(x h) F(x ) hf ( y k)00 0 0 !! ! ! !
w œ Ê ß
x
f(x h y k) f(x y k). Similarly, k f (x ) f(x y k) f(x y ) for some with"?"#?" ??"#?!! !! ! !! !! y((
y y k. Then f(x h y k) f(x y ) hf ( y k) kf (x ) . If M, N are positive real
!! !! !! ! !$$ " "? " # ? Ñ ? " " ?(0( kkkkkk
xy
numbers such that f M and f N for all (x y) in the xy-plane, then f(x h y k) f(x y )kk kk k k
xyÑÑ ? "?"#? !! !!
M h N k . As (h k) 0, f(x h y k) f(x y ) 0 lim f(x h y k) f(x y )Ñ" ? "?"#?` "?"#?kkkkkkkk !! !! !! !!
Ðß ÑÄÐß Ñhk 00

943. Chapter 14 Practice Exercises 939
0 f is continuous at (x y ).œÊ ß !!
16. At extreme values, f and are orthogonal because f 0 by the First Derivative Theorem for™™† vœœœ
ddfd
dt dt dt
rr
Local Extreme Values.
17. 0 f(x y) h(y) is a function of y only. Also, 0 g(x y) k(x) is a function of x only.
` `
```
`f f
xyx
g
œÊ ßœ œ œÊ ßœ
Moreover, h (y) k (x) for all x and y. This can happen only if h (y) k (x) c is a constant.
`
``
` ww wwf
yx
g
œÊ œ œ œ
Integration gives h(y) cy c and k(x) cx c , where c and c are constants. Therefore f(x y) cy cœ œ ßœ
"#"# "
and g(x y) cx c . Then f(1 2) g(1 2) 5 5 2c c c c , and f(0 0) 4 c 4 cßœ ßœ ߜʜ œ ßœÊ œÊœ#"#"
"
#
c . Thus, f(x y) y 4 and g(x y) x . ? ?#
## ##
""99
18. Let g(x y) D f(x y) f (x y)a f (x y)b. Then D g(x y) g (x y)a g (x y)b? ? ? " ? ? ? " ?
uu xy x y
f (x y)a f (x y)ab f (x y)ba f (x y)b f (x y)a 2f (x y)ab f (x y)b .œßß ß ß œß ß ß
xx yx xy yy xx xy yy
####
19. Since the particle is heat-seeking, at each point (x y) it moves in the direction of maximal temperatureß
increase, that is in the direction of T(x y) e sin x 2e cos x . Since T(x y) is parallel to™™? " ?aba b
``2y 2y
ij
the particle's velocity vector, it is tangent to the path y f(x) of the particle f (x) 2 cot x.œÊœœ
w 2e cos x
e sin x
Î
Î2y
2y
Integration gives f(x) 2 ln sin x C and f 0 0 2 ln sin C C 2 ln lnœ œÊœ Êœœkk ˆ‰ ¸ ¸ Š‹
11
44
2 2
2
È
È#
#
ln 2. Therefore, the path of the particle is the graph of y 2 ln sin x ln 2.œœ kk
20. The line of travel is x t, y t, z 30 5t, and the bullet hits the surface z 2x 3y whenœœœ œ
##
30 5t 2t 3t t t 6 0 (t 3)(t 2) 0 t 2 (since t 0). Thus the bullet hits theœ Ê œÊ œÊœ
## #
surface at the point (2 2 20). Now, the vector 4x 6y is normal to the surface at any (x y z), so that?? " # ?? ijk
8 12 is normal to the surface at (2 2 20). If 5 , then the velocity of the particleni jk vijkœ ßß œ
after the ricochet is 2 proj ( 5 )wv vv nv n ij k i j kœ œ œ œ
n Š‹ ˆ‰ ˆ ‰
2 2 25 400 600 50
209 209 209 209
vn
n
††
kk

.œ
191 391 995
209 209 209
ijk
21. (a) is a vector normal to z 10 x y at the point ( 0 10). So directions tangential to S at ( 0 10) willk # # !? ? !? ?
##
be unit vectors a b . Also, T(x y z) (2xy 4) x 2yz 14 y 1uij i j k" ?? " " " " " "™ abab
##
T( 0 10) 4 14 . We seek the unit vector a b such that D T(0 0 10)Ê!ßßœ œ ßß™ ijk uij
u
(4 14 ) (a b ) (4 14 ) (a b ) is a maximum. The maximum will occur when a bœ œ ijkij ijij ij††
has the same direction as 4 14 , or (2 7 ).iju ijœ
"
È
53
(b) A vector normal to S at (1 1 8) is 2 2 . Now, T(1 1 8) 6 31 2 and we seek the unitßß œ ßßœ nijk i jk ™
vector such that D T(1 1 8) T has its largest value. Now write T , where is paralleluu vwv
ußß œ œ ™† ™
to T and is orthogonal to T. Then D T T ( ) . Thus™™™†††††w u vwuvuwuwu
uœœœœ
D T(1 1 8) is a maximum when has the same direction as . Now, T
ußß œ uwwn ™ Š‹
™†Tn
nkk

(6 31 2 ) (2 2 ) 6 31 2œ œ ijk ijk i j kˆ‰ ˆ‰ˆ‰ˆ‰
12 62 2 152 152 76
441 9 9 9


(98 127 58 ).œ Ê œ œ
98 127 58
999 29,097
ijku ijk
w
wkk È
"
22. Suppose the surface (boundary) of the mineral deposit is the graph of z f(x y) (where the z-axis points upϧ
into the air). Then is an outer normal to the mineral deposit at (x y) and points in##" ? "
`` ``
`` ``
ff ff
xy xy
ijk ij
the direction of steepest ascent of the mineral deposit. This is in the direction of the vector at (0 0)
``
``
ff
xy
ij"?
(the location of the 1st borehole) that the geologists should drill their fourth borehole. To approximate this
vector we use the fact that (0 0 1000), (0 100 950), and (100 1025) lie on the graph of z f(x y).ß ß ß ß ß!ß œ ß
The plane containing these three points is a good approximation to the tangent plane to z f(x y) at the pointϧ

944. 940 Chapter 14 Partial Derivatives
(0 0 0). A normal to this plane is 2500 5000 10,000 , or 2 4 . So at00050
00 0 25
ßß œ "
j#
ââ
ââ
ââ
ââ
ââ
ââ
ijk
ij kijk
( 0) the vector is approximately 2 . Thus the geologists should drill their fourth borehole!? " # "
``
``
ff
xy
ij ij
in the direction of ( 2 ) from the first borehole.
"
È
5
#"ij
23. w e sin x w re sin x and w e cos x w e sin x; w w , where c is theœÊœ œ Êœ œ
rt rt rt rt
t x xx xx t
c
111111
## "

positive constant determined by the material of the rod e sin x re sin xÊ œ11 1
# "rt rt
c
ab
r c e sin x 0 r c w e sin xÊ œÊœ Êœab
## ##
11 1 1
rt c t

1
24. w e sin kx w re sin kx and w ke cos kx w k e sin kx; w wœÊœ œ Êœ œ
rt rt rt rt
t x xx xx t
c
# "

k e sin kx re sin kx r c k e sin kx 0 r c k w e sin kx.Ê œ Ê œ Ê œ Ê œ
##### "rt rt rt c k t
c

abab
Now, w(L t) 0 e sin kL 0 kL n for n an integer k w e sin x .ßœ Ê œ Ê œ Ê œ Ê œ
œ œjckt cn t Lnn
LL

1
11 1
ˆ‰
As t , w 0 since sin x 1 and e 0.Ä_ Ä Ÿ Ä ¸¸ˆ‰
n
L
cn t L1 1œj

945. CHAPTER 15 MULTIPLE INTEGRALS
15.1 DOUBLE INTEGRALS
1. 4 y dy dx 4y dx dx 16'' ' '
00 0 0
32 3 3
ab ’“œœœ
#
#
!
y
33
16


2. (x y 2xy dy dx xy dx'' '
02 0
30 3
ab ’“
##
!
#
œ
xy
2
œœ
4x 2x dx 2x 0œ œœ'
0
3
ab ’“
##
$
!
2x
3


3. (x y 1) dx dy yx x dy'' '
11 1
01 0
"" $ " " ’“
x
2
œ
"
"
(2y 2) dy y 2y 1$"$"$'
1
0
cd
#
!
"

4. (sin x cos y) dx dy ( cos x) (cos y)x dy
'' '


22
0
œ cd
1
!
( cos y 2) dy sin y 2y 2$"$"$'


2
11 1 cd
#1
1

5. (x sin y) dy dx x cos y dx
'' '
00 0
x
x

œcd
!
(x x cos x) dx (cos x x sin x)œ œ '
0

’“
x
2
œ
1
!
2$"
1
œ
#

6. y dy dx dx sin x dx'' ' '
00 0 0
sin x sin x

œœ’“
y
2
œ
!
"
#
#
(1 cos 2x) dx x sin 2xœ œ œ
"""
!4424'
0

‘
1
1

946. 942 Chapter 15 Multiple Integrals
7. e dx dy e dy ye e dy'' ' '
10 1 1
ln 8 ln y ln 8 ln 8
xy xy y y ln y
œœ
œœcd a b
!
(y 1)e e 8(ln 8 1) 8 eœ œ cd
yy ln 8
1
8 ln 816eœ

8. dx dy y y dy'' '
1y 1
2y 2

œœab ’“
#
#
#
"
yy
3
œ
2œ œœˆ‰ˆ‰
8735
3336
""
##

9. 3y e dx dy 3y e dy
'' '
00 0
11
0
y
y


$#xy xy
œcd
3y e 3y dy e y e 2œœœ
'
0
1
Š‹’“
## $
"
!
yy

10. e dy dx x e dx'' '
10 1
4x 4
yx yx
x
0


33
2#""
œ ‘È
È
(e 1) x dx (e 1) x 7(e 1)œ œ œ
332
23#
$Î#
%
" '
1
4
È ‘ˆ‰

11. dy dx x ln y dx (ln 2) x dx ln 2'' ' '
1x 1 1
22x 2 2
2x
x
x3
y
œœœcd
#
12. dy dx (ln 2 ln 1) dx (ln 2) dx (ln 2)'' ' '
11 1 1
22 2 2
1
xy x
œœ œ
" #
13. x y dy dx x y dx x (1 x) dx x x dx'' ' ' '
00 0 0 0
11x 1 1 1 x
0

ab ’“ ’ “ ’ “
## # # #$
œœ œ
y (1x) (1x)
333

000œ œ œ’“ ˆ‰ˆ ‰
xx
34 1 34 1 6
(1 x)
" "

##
"
!
"" " "
14. y cos xy dx dy sin xy dy sin y dy cos y ( 1 1)'' ' '
00 0 0
111

œœœœœcd ‘
1
111!
""
"
!
11
2
15. v u dv du v u du u(1 u) du'' ' '
00 0 0
11u 1 1 u
0

ˆ‰ÈÈ È ’“ ’ “œ œ
v12uu
2
œœ

#
uuu du uuœ œ œœœ'
0
1
Š‹’ “
" """ " "
## # ## #
"Î# $Î# $Î# &Î#
"
!
u uuu2 2 22 2
2 6 3 5 635 5 10
œœ

947. Section 15.1 Double Integrals 943
16. e ln t ds dt e ln t dt (t ln t ln t) dt ln t t ln t t'' ' '
10 1 1
2lnt 2 2
ln t
0
ss tt
24
œœœcd ’“
œœ
#
"
(2 ln 212 ln 22) 1œœ ˆ‰
""
44
17. 2 dp dv 2 p dv 2 2v dv'' ' '

2v 2 2
0v 0 0
v
v
œœcd

2v 8œ œcd
#
0
2


18. 8t dt ds 4t ds'' '
00 0
11s 1
1s
0





œcd
#
41 s ds 4sœœœ'
0
1
ab ’“
#
"
!
s8
33


19. 3 cos t du dt (3 cos t)u
'' '

""

30 3
3sect 3
sec t
0
œ cd
3 dt 2œœ
'
"
"

3
3
1

20. dv du du'' '
01 0
342u 3
42u
1



42u 2u4
vv

œ œ‘
(3 u) du 3u uœ#œœ!
'
0
3
cd
#
$
!

21. dx dy
''
20
44y)2


22. dy dx
''

œ20
0x2

948. 944 Chapter 15 Multiple Integrals
23. dy dx''
0x
1x


24. dx dy''
01y
11y


25. dx dy
''
1lny
e1

26. dy dx
''
10
2lnx

27. 16x dx dy''
00
99y
1
2

28. y dy dx''
00
44x

949. Section 15.1 Double Integrals 945
29. 3y dy dx''

10
11x


30. 6x dx dy''

20
24y


31. dy dx dx dy sin y dy 2'' '' '
0x 00 0
y

sin y sin y
yy
œœœ

32. 2y sin xy dy dx 2y sin xy dx dy
'' ''
0x 00
22 2y
##
œ
2y cos xy dy 2y cos y 2y dyœ œ
''
00
22
y
0
cda b
#
sin y y 4 sin 4œ œ cd
##
#
!

33. x e dx dy x e dy dx xe dx
'' '' '
0y 00 0
11 1x 1
xy xy xy x
0
##
œœ cd
xe x dx eœœœ'
0
1
xx
ab ’“

"
##
"
!
2
xe2
œ œ

34. dy dx dx dy'' ''
00 00
24x 4 4y


xe xe
4y 4y
2y 2y

œ

dy dyœœœœ''
00
44 4y
0
’“ ’“
xe e e e
(4 y) 4 4
œ #2y 2y 2y
# #
%
!
"


35. e dx dy e dy dx'' ''
0y/2 00
2 ln 3 ln 3 ln 3 2x
xx


œ
œ
2xe dx e e 1 2œœœœ'
0
ln 3
xx ln3
ln 3
0



cd

950. 946 Chapter 15 Multiple Integrals
36. e dy dx e dx dy'' ''
0x3 00
31 13y
yy"
$$

œ
3y e dy e e 1œœœ
'
0
1
yy
#
"
!
$$
cd

37. cos 16 x dx dy cos 16 x dy dx
'' ''
0y 00
11612 12x
"" "
"Î%
%
ab ab11
&&
œ
x cos 16 x dxœœœ'
0
12
"
%&
"Î#
!
"
ab ’“1
sin 16 x
80 80
ab1
11
%

38. dy dx dx dy'' ''
0x 00
82 2y
$
$

""
ßßy1 y1
"" œ
dy ln y 1$$"$'
0
2
y
y1 44
ln 17

"
ß
" %
#
!
cdab

39. y 2x dA
''
R
ab
#
y 2x dy dx y 2x dy dxœ'' ''

1x1 0x1
0x1 11x
ab ab
##
y 2x y dx y 2x y dxœ ''


10
01
x1x
x1 x1
‘ ‘
""## ##
22

(x 1) 2x (x 1) ( x 1) 2x ( x 1) dxœ'
1
0
‘
""
##
## ##
(1 x) 2x (1 x) (x 1) 2x (x 1) dx '
0
1
‘
""
##
## ##

4 x x dx 4 x x dx 4 4œ œ ''

10
01 0
1
ab ab ’“’“
$# $#
"
!
xx xx
43 43
"
448œ œ œ œ’“ ˆ‰ˆ ‰
(1) (1)
4 3 4 3 12 12 12 3
34 8 2 ""

40. xy dA xy dy dx xy dy dx''
R
$"'' ''
0x 23x
23 2x 1 2 x
"
"
xy dx xy dx$"''
023
23 1
2x 2 x
xx
"
"

‘ ‘
""##
22

2x x dx x(2 x) x dxœ ''
023
23 1
"
"
ˆ‰ ‘
$$ #$"""
###
x dx 2x x dxœ''
023
23 1
"
"
3
#
$#
ab

xxx 1œœ œœ‘ ‘ˆ‰ˆ‰ˆ‰ ‘ˆ‰ˆ‰ˆ ‰
3 2 3 16 2 4 2 8 6 27 36 16 13
8 3 8 81 3 9 3 27 81 81 81 81 81
%#$
"
#Î$
23
0"
41. V x y dy dx x y dx 2x dxœœœœ'' ' '
0x 0 0
12x 1 1 2x
x

ab ’“ ’ “’ “
## # #
"
!
y (2x) (2x)
3 3 3 3 12 12
7x 2x 7x
"
00œ œˆ‰ˆ‰
27 16 4
31212 12 3
"

951. Section 15.1 Double Integrals 947
42. V x dy dx x y dx 2x x x dx x x xœœœœ'' ' '


2x 2 2
12x 1 1
2x
x


# # #%$ $ & %
"
#
cd a b ‘
211
354
œœ œœˆ‰ˆ ‰ˆ ‰ˆ ‰
2 16 32 16 40 12 15 320 384 240 189 63
3 5 4 3 5 4 60 60 60 60 60 60 60 20
""
43. V (x 4) dy dx xy 4y dx x 4 x 4 4 x 3x 12x dxœœœ'' ' '


43x 4 4
14x 1 1
4x
3x


cd c d abab
###
x 7x 8x 16 dx x x 4x 16x 12 64œ œ œ'
4
1
ab ‘ˆ ‰ˆ ‰
$# % $#
"
%
"17 7 64
43 43 3
œœ
157 625
3412
"
44. V (3 y) dy dx 3y dx 3 4 x dxœœœ'' ' '
00 0 0
24x 2 2 4x
0


’“ ’ “
È
Š‹
y
2
4x
œ œ
#
#
x4 x 6 sin 2x 6 4 3œ œœœ’“
È ˆ‰ ˆ‰
3xx81698
26663
# "
##
#
!


11
1
45. V 4 y dx dy 4x y x dy 12 3y dy 12y y 24 8 16œ œœœœœ
'' ' '
00 0 0
23 2 2
ab cd abcd
## #$
$
!!
#
46. V 4 x y dy dx 4 x y dx 4 x dx 8 4x dxœœœœ'' ' ' '
00 0 0 0
24x 2 2 2 4x

ß

a b ab ab ’“ Š‹
## ## "
##
#y
2
x
œ "
8x x x 16œ œœ œ ‘
4 32 32 480 320 96 128
3 10 3 10 30 15
$&"
#
!
47. V 12 3y dy dx 12y y dx 24 12x (2 x) dxœœœ'' ' '
00 0 0
22x 2 2

ab cd c d
#$ $
#
!
x
24x 6x 20œ œ’“
#
#
!
(2 x)
4
"
48. V (3 3x) dy dx (3 3x) dy dx 6 1 x dx 6 (1 x) dx 4 2 6œ œ œœ'' '' ' '

1x1 0x1 1 0
0x1 11x 0 1
ab
##
49. V (x 1) dy dx xy y dx 1 1 2 1 dxœœœœ'' ' ' '
11x 1 1 1
21x 2 2 2
1x
1x"
"
"

cd ‘ˆ‰ˆ‰
"" "
xx x
2 x ln x 2(1 ln 2)$" $"cd
#
"
50. V 4 1 y dy dx 4 y dx 4 sec x dx$"$"$"'' ' '
00 0 0
3 sec x 3 3 sec x
0
"""
ab ’“ Š ‹
# y
33
sec x

7 ln sec x tan x sec x tan x 7 ln 2 3 2 3$"" $""
22
33
cdkk ’“Š‹
ÈÈ
1Î$
!
51. dy dx dx dx lim lim 1 1'' ' '
1e 1 1
1
e
b
1
¸¸¸
ß
ßx
x
"""
"
xy x x x b
ln y x
œœœœœ’“ ˆ‰ ‘ˆ‰
bbÄ_ Ä_
52. (2y 1) dy dx y y dx dx 4 lim sin x'' ' '




11/1x 1 1
11/1x 1 1
1/ 1 x
1/ 1 x
b


$"
#!





"$" $ $ cd c dº
#"

!
1
1
2
1x
È
œ
b1Ä
4 lim sin b 0 2œœ
b1Ä

cd
"
153. -dx dy 2 lim tan b tan 0 dy 2 lim dy'' ' '

¸¸ ¸
" "
ßß ß ß
" "
ababx1y1 y1 y1
2
œœ œ œ œœ
0 0
b
Š‹Š ‹
bbÄ_ Ä_
1
2 lim tan b tan 0 (2 )œœœ111Š‹ ˆ‰
bÄ_
" " #
#
1
54. xe dx dy e lim xe e dy e lim be e 1 dy'' ' '
00 0 0
x2y 2y x x 2y b b b
0
¸¸ ¸ ¸

œœ
bbÄ_ Ä_
cd a b
e dy lim e 1œœœ'
0
2y 2b
¸

""
##
bÄ_
ab

952. 948 Chapter 15 Multiple Integrals
55. f(x y) dA f 0 f(0 0) f 0 f 0 f f f''
R
߸ßßßßß!ßß
""" """"""""""""
######488444884
ˆ ‰ ˆ‰ ˆ‰ ˆ ‰ ˆ‰ ˆ‰
00œ œ
""" " ""
## #484416
33
ˆ‰ˆ ‰
56. f(x y) dA f f ffff ff''
R
ß ¸ ß ß ß ß ß ß ß ß
"
44444444444 444444
79 99 511 711 911 1111 513 713
ˆ‰ˆ‰ˆ‰ˆ‰ˆ‰ˆ ‰ˆ‰ˆ‰
ff ff ‘ˆ‰ˆ ‰ˆ‰ˆ‰ß ßßß
913 1113 715 915
44 4 4 44 44
(252727293133313335373739) 24$ """"""""""" $ $
"
16 16
384
57. The ray meets the circle x y 4 at the point 3 1 the ray is represented by the line y .)œœßÊ œ
1
6
x
3
##
Š‹
È
È
Thus, f(x y) dA 4 x dy dx 4 x 4 x dx 4x''
R
ßœ œ œ'' '
0x3 0
34x 3
3
3
0





"

ÈÈ
’“ab ”• ## # xx
3
4x
33
œ
ab
È
œ
20 3
9
È
58. dy dx dx dx 6 '' ' ' '
20 2 2 2
2 2
0
¸¸¸¸
"


ab abxx(y1)
3(y 1)
xx xx xx x(x1)
33 dx
œœ

œœœ $$"$’“ ˆ‰
6 lim dx 6 lim ln (x 1) ln x 6 lim [ln (b 1) ln b ln 1 ln 2]œ œ œ
bb bÄ_ Ä_ Ä_
'
2
b
b
2
ˆ‰ cd
""
x1 x
6 lim ln 1 ln 2 6 ln 2œœ’“ ˆ‰
bÄ_
"
b
59. V x y dy dx x y dx$"$"'' '
0x 0
12x 1 2x
x

ab ’“
## # y
3

2x dxœ œ'
0
1
’“’“
#
"
!
7x 2x 7x
33 31212
(2 x) (2 x)
"
00œ œˆ‰ˆ‰
27 16 4
3121 12 3
"
#

60. tan x tan x dx dy dx dx dy dx dy'''''''
00x0y2y
22x2y22
ab
" " "" "
ßß ß
1œ œ


1y 1y 1y
œœ œ
""
dy dy ln 1 y 2 tan y ln 1 y$"$""""''
02
22
2
2
ˆ‰ ˆ‰1y 2
1y 1y 2
#
ßß #
" " #" #
!
¸
œœ

y
ˆ‰ ‘cd abab
1
11
ln 5 2 tan 2 ln 1 4 2 tan 2 ln 5œˆ‰ ab
1
11 1
" " " # "
#
1
22
11
2 tan 2 2 tan 2 ln 1 4œ
" " # "
#
11
2
ln 5
1
ab
61. To maximize the integral, we want the domain to include all points where the integrand is positive and to
exclude all points where the integrand is negative. These criteria are met by the points (x y) such thatß
4 x 2y 0 or x 2y 4, which is the ellipse x 2y 4 together with its interior.   Ÿ œ
## ## ##
62. To minimize the integral, we want the domain to include all points where the integrand is negative and to
exclude all points where the integrand is positive. These criteria are met by the points (x y) such thatß
x y 9 0 or x y 9, which is the closed disk of radius 3 centered at the origin.
## ##
Ÿ Ÿ
63. No, it is not possible By Fubini's theorem, the two orders of integration must give the same result.

953. Section 15.1 Double Integrals 949
64. One way would be to partition R into two triangles with the
line y 1. The integral of f over R could then be writtenœ
as a sum of integrals that could be evaluated by integrating
first with respect to x and then with respect to y:
f(x y) dA
''
R
ß
f(x y) dx dy f(x y) dx dy.œßß
'' ''
022y 1y1
12y2 22y2

Partitioning R with the line x 1 would let us write theœ
integral of f over R as a sum of iterated integrals with
order dy dx.

65. e dx dy e e dx dy e e dx dy e dx e dy
'' '' ' ' ' '

bb bb b b b b
bb bb b b b b
xy y x y x x y

œœ œ Œ $ Œ $Œ $
e dx 2 e dx 4 e dx ; taking limits as b gives the stated result.œœ œ Ä_Œ $Œ $ Œ $
'''

b0 0
bbb
xxx

###
66. dy dx dx dy dy '' '' ' '
00 00 0 0
13 31 3 3
xx x
(y 1) (y 1) (y 1) (y 1) 33
dy
$$


""
"
!
œœœ ’“
lim lim lim (y 1) lim (y 1)œ œ
""

"Î$ "Î$
33
dy dy
(y 1) (y 1)
b1 b1 b1 b1ÄÄ ÄÄ
$$
''
0b
b3
b3
0b
$
‘ ‘
lim (b 1) ( 1) lim (b 1) (2) (0 1) 0 2 1 2œœœ’“’“Š‹
ÈÈ
b1 b1Ä Ä
$
"Î$ "Î$ "Î$ "Î$ $$
67-70. Example CAS commands:
:Maple
f := (x,y) -> 1/x/y;
q1 := Int( Int( f(x,y), y=1..x ), x=1..3 );
evalf( q1 );
value( q1 );
evalf( value(q1) );
71-76. Example CAS commands:
:Maple
f := (x,y) -> exp(x^2);
c,d := 0,1;
g1 := y ->2*y;
g2 := y -> 4;
q5 := Int( Int( f(x,y), x=g1(y)..g2(y) ), y=c..d );
value( q5 );
plot3d( 0, x=g1(y)..g2(y), y=c..d, color=pink, style=patchnogrid, axes=boxed, orientation=[-90,0],
scaling=constrained, title="#71 (Section 15.1)" );
r5 := Int( Int( f(x,y), y=0..x/2 ), x=0..2 ) + Int( Int( f(x,y), y=0..1 ), x=2..4 );
value( r5);
value( q5-r5 );
67-76. Example CAS commands:
: (functions and bounds will vary)Mathematica
You can integrate using the built-in integral signs or with the command . In the command, theIntegrate Integrate
integration begins with the variable on the right. (In this case, y going from 1 to x).

954. 950 Chapter 15 Multiple Integrals
Clear[x, y, f]
f[x_, y_]:= 1 / (x y)
Integrate[f[x, y], {x, 1, 3}, {y, 1, x}]
To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done
with ImplicitPlot and all bounds involving both x and y can be plotted. A graphics package must be loaded. Remember to
use the double equal sign for the equations of the bounding curves.
Clear[x, y, f]
<<Graphics`ImplicitPlot`
ImplicitPlot[{x==2y, x==4, y==0, y==1},{x, 0, 4.1}, {y, 0, 1.1}];
f[x_, y_]:=Exp[x ]
2
Integrate[f[x, y], {x, 0, 2}, {y, 0, x/2}] Integrate[f[x, y], {x, 2, 4}, {y, 0, 1}]
To get a numerical value for the result, use the numerical integrator, . Verify that this equals the original.NIntegrate
Integrate[f[x, y], {x, 0, 2}, {y, 0, x/2}] NIntegrate[f[x, y], {x, 2, 4}, {y, 0, 1}]
NIntegrate[f[x, y], {y, 0, 1},{x, 2y, 4}]
Another way to show a region is with the FilledPlot command. This assumes that functions are given as y = f(x).
Clear[x, y, f]
<<Graphics`FilledPlot`
FilledPlot[{x , 9},{x, 0,3}, AxesLabels {x, y}];
2
Ä
f[x_, y_]:= x Cos[y ]
2
Integrate[f[x, y], {y, 0, 9}, {x, 0, Sqrt[y]}]
67. dy dx 0.603 68. e dy dx 0.558'' ''
11 00
3x 11
xy
"
xy
¸¸


69. tan xy dy dx 0.233 70. 3 1 x y dy dx 3.142'' ''
00 10
11 11x
" ##
¸¸


¸

È
71. Evaluate the integrals:
e dx dy
''
02y
14
x

e dy dx e dy dxœ'' ''
00 20
2x/2 41
xx

e 2 erfi 2 2 erfi 4œ
""
44
4
ˆ‰ ÈÈ ab ab11
1.1494 10¸‚
6
The following graphs was generated using
Mathematica.

955. Section 15.2 Areas, Moments, and Centers of Mass 951
72. Evaluate the integrals:
x cos y dy dx x cos y dx dy'' ''
0x 00
39 9 y
22
2
ab abœ
¸
0.157472œ¸
sin 81
4
ab
The following graphs was generated using
Mathematica.

73. Evaluate the integrals:
x y xy dx dy x y xy dy dx'' ''
0y 0x/32
242y 8 x
22 22
32
3
¸ ¸
ab abœ
97.4315œ¸
67,520
693
The following graphs was generated using
Mathematica.

74. Evaluate the integrals:
e dx dy e dy dx'' ''
00 00
24y 4 4x


xy xy
œ
¸
20.5648¸
The following graphs was generated using
Mathematica.

956. 952 Chapter 15 Multiple Integrals
75. Evaluate the integrals:
dy dx''
10
2x

1
xy¸
dx dy dx dyœ'' ''
01 1 y
12 42
11
xy xy¸¸
¸
1 ln 0.909543 ¸ˆ‰
27
4
The following graphs was generated using
Mathematica.

76. Evaluate the integrals:
dx dy dy dx'' ''
1y 11
28 8 x
3
3
11
xy xyÈÈ
22 22
¸¸
œ
¸
0.866649¸
The following graphs was generated using
Mathematica.

15.2 AREAS, MOMENTS, AND CENTERS OF MASS
1. dy dx (2 x) dx 2x 2,'' '
00 0
22x 2

$!$!$ ’“
x
2

#
!
or dx dy (2 y) dy 2'' '
00 0
22y 2

$!$

2. dy dx (4 2x) dx 4x x 4,
'' '
02x 0
24 2
2
0
$! $!$ cd
#
or dx dy dy 4'' '
00 0
4y2 4
¸
œœ
y
#

3. dx dy y y 2 dy
'' '

2y2 2
1y 1

œab
#
2yœ ’“
yy
3

#
"
#
224œ œˆ‰ˆ‰
""
##33
89

957. Section 15.2 Areas, Moments, and Centers of Mass 953
4. dx dy 2y y dy y'' '
0y 0
2yy 2


$!$!ab ’“
##
#
!
y
3
Ä
4$!$
84
33

5. dy dx e dx e 2 1 1
'' '
00 0
ln 2 e ln 2
xx ln 2
0
x
$ $ $!$ cd

6. dy dx ln x dx x ln x x
'' '
1ln x 1
e2 ln x e
e
1
$$! cd
(e e) (0 1) 1$!!!$

7. dx dy 2y 2y dy y y'' '
0y 0
12yy 1



$! $!ab ‘
##$
"
!
2
3
œ
"
3

8. dx dy y 1 2y 2 dy
'' '

12y2 1
1y1 1


œab
##
1 y dy y$! $!$'
1
1
ab ’“
#
"
"
y
33
4
Ä

9. dx dy 2y dy y'' '
0y3 0
62y 6

¸
$!$!Š‹’“
yy
39
Ä
#
'
!
36 12$! $
216
9

10. dy dx 3x x dx x x'' '
0x 0
32xx 3


$!$!ab ‘
##$ "
$
!
3
23
9$!$
27 9
##

958. 954 Chapter 15 Multiple Integrals
11. dy dx''
0sin x
4cos x
¸
(cos x sin x) dx sin x cos xœœ'
0
4
4
0


¸
¸
cd
(0 1) 2 1œœŠ‹
È
ÈÈ
22
##

12. dx dy y 2 y dy 2y'' '

#
1y 1
2y2 2 2
1

œ œab ’“
# yy
23
#
24 2 5œœœˆ‰ˆ‰
89
33
"" "
###

13. dy dx dy dx
'' ''
¸
12x 0 x2
01x 21x

(1 x) dx 1 dxœ''
10
02
ˆ‰
x
#
xx 1(21)œ œœ’“’“ ˆ‰
xx 3
24

02
10

"
##

14. dy dx dy dx'' ''
0x4 00
20 4 x



¸
4 x dx x dxœ ''
00
24
ab
# "Î#
4x x 8œ œœ’“ ‘ˆ ‰
x2 81632
33 333
#
2
0
4
0
$Î#

15. (a) average sin (x y) dy dx cos (x y) dx [ cos (x ) cos x] dxœœœ
"""
111

'' ' '
00 0 0
0


cd 1
sin (x ) sin x [( sin 2 sin ) ( sin sin 0)] 0œ œ œ
""
11
cd 1111

0
(b) average sin (x y) dy dx cos (x y) dx cos x cos x dxœœœ
" Î#
! #
Š‹




'' ' '
00 0 0
2
¸
22
11
1 1
cd ‘ˆ‰
sin x sin x sin sin sin sin 0œ œ œ
223 4
11 1
111

‘ ‘ˆ‰ ˆ ‰ˆ ‰
###

0
1
16. average value over the square xy dy dx dx dx 0.25;œœœœœ '' ' '
00 0 0
11 1 1
’“
xy
24
x
"
!
#
"
average value over the quarter circle xy dy dx dxœœ
"
ˆ‰

4
'' '
00 0
11x 1 1x
0
¸ ¸


4 xy
21
’“

x x dx 0.159. The average value over the square is larger.œœœ¸
22xx
24111'
0
1
ab ’“
$
"
!
"
#
4
17. average height x y dy dx x y dx 2x dxœœœœœ
"""" ## # #
##
!!
#44343333
y 8x4x8''''
0000
2222
ab ’“ ’“ ˆ‰
# #

959. Section 15.2 Areas, Moments, and Centers of Mass 955
18. average dy dx dxœœ
"""
(ln 2) xy (ln 2) x
ln y
ÄÄ
'' '
ln 2 ln 2 ln 2
2 ln 2 2 ln 2 2 ln 2 2 ln 2
ln 2
’“
(ln 2 ln ln 2 ln ln 2) dx ln xœœœ
"" " "
#(ln 2) x ln 2 x ln
dx
Ä
''
ln 2 ln 2
2 ln 2 2 ln 2
2 ln 2
ln 2
ˆ‰ ˆ‰ cd
(ln 2 ln ln 2 ln ln 2) 1œœˆ‰
"
#ln
19. M 3 dy dx 3 2 x x dx ; M 3x dy dx 3 xy dx$$!!$$ $'' ' '' '
0x 0 0x 0
12x 1 12x 1
y
2x
x
Ä Ä
Ä
Ä Ä
Ä
ab cd
#
#
7
3 2x x x dx ; M 3y dy dx y dx 4 5x x dxœ œœ œ œ œ'''''
00x00
112x11
x
2x
x
ab cd ab
$# # #%
##
53319
4 5
Ä
Ä
Ä
Ä
x and yÊœ œ
538
14 35
20. M dy dx 3 dx 9 ; I y dy dx dx 27 ; R 3;œœœœ œœœœ$$$$ $ $'' ' '' '
00 0 00 0
33 3 33 3
x x
3
0
#
’“ É È
y
3M
I
#
x
I x dy dx x y dx 3x dx 27 ; R 3
y y
00 0 0
33 3 3œœœœœœ$$$$'' ' '## #
$
!
cd É È
I
My
21. M dx dy 4 y dy ; M x dx dy x dy$$!!$$ $'' ' '' '
0y2 0 0y2 0
24y 2 24y 2
y
4y
y2
Ä Ä
Ä
_ _
Ä Ä
Ä
_
Š‹ cd
y 14
3
Ä
##
" #
16 8y y dy ; M y dx dy 4y y dyœœœ œœ
"
# #
##''''
00y20
224y2
x
Š‹ Š‹
yy
415 3
128 10
_ #

Ä
_
Ä
x and yÊœ œ
64 5
35 7
22. M dy dx (3 x) dx ; M x dy dx xy dx 3x x dx$$!$$ $$!$'' ' '' ' '
00 0 0 0 0 0
33x 3 3 3x 3 3
y
3x
0
ÄÄ
Ä
9 9
# #
#
cd a b
x 1 and y 1, by symmetryÊœ œ
23. M 2 dy dx 2 1 x dx 2 ; M 2 y dy dx y dx$$!$$$ $'' ' '' '
00 0 00 0
11x 1 11x 1
x
1x
0
_ _
_
Ä Ä
Ä Ä Ä
Ä
È ˆ‰ cd #
#
#11
4
1 x dx x y and x 0, by symmetry$! $!$Î$ $'
0
1
ab ’“
#
"
!
x2 4
33 3
#
1
24. M ; M x dy dx xy dx 5x x dx ;$$ $ $ !$
125 625
6 1
$ $
y
0x 0 0
56xx 5 5
6x x
x$$$'' ' '
Ä
Ä
Ä
Ä
cd a b
#$
#
M y dy dx y dx 35x 12x x dx x and y 5
x
0x 0 0
56xx 5 5
6x x
xœœœœÊœœ$'' ' '
Ä
Ä
Ä
Ä
$$ $
## #
##$%
cd a b
625 5
6
25. M dy dx ; M x dy dx xy dx x a x dx$$$ $$!$'' '' ' '
00 00 0 0
aax aax a a
y
ax
0
__
_ÄÄ ÄÄ
ÄÄ
ÄÄ
Ä
1a a
4 3
Ä #
cd
È
##
x y , by symmetryÊœœ
4a
31
26. M dy dx sin x dx 2; M y dy dx y dx sin x dxœœœœ œ œ'' ' '' ' '
00 0 00 0 0
sin x sin x
x
sin x
0
ÄÄ Ä Ä Ä
""
##
##
cd
(1 cos 2x) dx x and y$! $Î$ $
"
#448'
0
Ä
111
27. I y dy dx dx 4 x dx 4 ; I 4 , by symmetry;
x y
24x 2 2
24x 2 2 4x
4x$$$!$$'' ' '
ÄÄÄ Ä Ä
Ä Ä
ÄÄ
_
_ _
_
Ä
Ä Ä
Ä
##
$Î#
’“ ab
y
33
2
#
11
III8
oxyœœ 1
28. I x dy dx sin x 0 dx (1 cos 2x) dx
y
2sinxx 2 2
0$$!$!$'' ' '
ÄÄÄ
ÄÄÄ
ÄÄ
_
## "
##
ab
1
29. M dy dx e dx lim e dx 1 lim e 1; M x dy dx xe dx$$$ $!$$ $'' ' ' '' '
Ä
0e 0 0 0e 0
0b 0
xx b x
y
x x
bbÄ_ Ä_
lim xe dx lim xe e 1 lim be e 1; M y dy dx$$!$!!!$!$
bb bÄ_ Ä_ Ä_
'''
b 0
0 0e
xxx bb 0
b x
cd ab
Ä!
x

960. 956 Chapter 15 Multiple Integrals
e dx lim e dx x 1 and y$$ $Î$!$
"" " "
##''
Ä!
00
2x 2x
b
bÄ_
44
30. M x dy dx lim xe dx lim 1 1
y
00 0
eb
x2
b
0$$ $!!$'' '
!
Ä_
ĸ
Ä
Ä
Ä
_x2
x2
bbÄ_ Ä_
Ä ‘
"
e31. M (x y) dx dy xy dy 2y 2y dy ;œœœœœ'' ' '
0y 0 0
2yy 2 2 yy
yÄ
Ä Ä
Ä
Ä Ä
’“ Š ‹’ “
x 8
22 10315
yyy2y
Ä
$#
#
#
!
I y (x y) dx dy xy dy 2y 2y dy ;
x
0y 0 0
2yy 2 2 yy
yœœœœ'' ' '
Ä
Ä Ä
Ä
Ä Ä
#$&%
’“ Š ‹
xy y
2 2 105
64
ÄÄ $
R2
xœœœÉ ÉÉ
I
M7 7
82x
32. M 5x dx dy 5 dy 12 4y 16y dy 23 3$$$!!$'' ' '
Ä_Ä_Ä_
_Ä _ _ Ä
___
___ _ _32 4y 32 32
32 12 4y 32 32 12 4y
4y
Ä
Ä Ä
Ä
’“ ab
È
x5
2
Ä
#
#%
33. M (6x 3y 3) dy dx 6xy y 3y dx 12 12x dx 8;œœœœ'' ' '
0x 0 0
12x 1 1
2x
x
Ä
Ä
Ä ‘ ab
3
#
##
M x(6x 3y 3) dy dx 12x 12x dx 3; M y(6x 3y 3) dy dx
y x
0x 0 0x
12x 1 12xœœœœ'' ' ''
Ä Ä
ab
$
14 6x 6x 2x dx x and y$!!! $Î$ $'
0
1
ab
#$
#
17 3 17
816
34. M (y 1) dx dy 2y 2y dy ; M y(y 1) dx dy 2y 2y dy ;œœœœ œœ'' ' '' '
0y 0 0y 0
12yy 1 12yy 1
x
Ä Ä
Ä Ä
Ä Ä
ab a b
$#% "
#
4
15
M x(y 1) dx dy 2y 2y dy x and y ; I y (y 1) dx dy
y
0y 0 0y
12yy 1 12yyœœœÊœœœ '' ' ''
Ä Ä
Ä Ä
Ä Ä
ab
#% # 488
15 15 15
x
2 y y dy$!$'
0
1
ab
$& "
6
35. M (x y 1) dx dy (6y 24) dy 27; M y(x y 1) dx dy y(6y 24) dy 14;œœœœ œœ'' ' '' '
00 0 00 0
16 1 16 1
x
M x(x y 1) dx dy (18y 90) dy 99 x and y ; I x (x y 1) dx dy
y y
00 0 00
16 1 16œœœÊœœœ '' ' '' 11 14
327
#
216 dy 432; R 4œœœœ'
0
1
y
ˆ‰ É
y
36 M
11 I y
36. M (y 1) dy dx x dx ; M y(y 1) dy dx dxœœœœ œ'' ' '' '
ÄÄ ÄÄ1x 1 1x 1
11 1 11 1
x
Ä Ä
Š‹ Š‹
x332 5xx
15 6 3
_ œ¸
## #
#
; M x(y 1) dy dx x dx 0 x 0 and y ; I x (y 1) dy dxœœ œœÊœœœ
48 3x x 9
35 14
y y
1x 1 1x
11 1 11
'' ' ''
ÄÄ Ä
Ä Ä
Š‹
##
$#
!
x dx ; R$!!$$$'
Ä1
1
y
Š‹ É É
3x x 16 3
22 35 M 14
I
Ä $
% y
37. M (7y 1) dy dx x dx ; M y(7y 1) dy dx dx ;œœœœ œœ'' ' '' '
ÄÄ Ä Ä10 1 10 1
1x 1 1x 1
x
Ä Ä
Š‹ Š‹
7x 31 7x x 13
15 3 2 15
_ œ¸
#
#
M x(7y 1) dy dx x dx 0 x 0 and y ; I x (7y 1) dy dx
y y
10 1 10
1x 1 1xœœœÊœœœ '' ' ''
ÄÄ Ä
Ä Ä
Š‹
7x 13
31
!
#
$#
x dx ; Rœœœœ'
Ä1
1
y
Š‹ É É
7x 7 21
5M31
I
$
#
%
y
38. M 1 dy dx 2 dx 60; M y 1 dy dx 1 dx 0;œœœœ œ œ'' ' '' '
01 0 01 0
20 1 20 20 1 20
x Ä Ä
ˆ‰ ˆ‰ ˆ‰ ˆ‰ ’“ Š‹
xx x x
20 10 20 0
y
##
"
Ä"
Ä
M x 1 dy dx 2x dx x and y 0; I y 1 dy dx
y x
01 0 01
20 1 20 20 1œœœÊœœœ '' ' ''
Ä Ä
ˆ‰ ˆ‰ Š‹
x x 2000 100 x
20 10 3 9 20
Ä
#
1 dx 20; Rœœœœ
2x
320 M3
I'
0
20
x
ˆ‰ É É
x "

961. Section 15.2 Areas, Moments, and Centers of Mass 957
39. M (y 1) dx dy 2y 2y dy ; M y(y 1) dx dy 2 y y dy ;œœœœ œœ'' ' ''
0y 0 0y
1y 1 1y
x
1
0
ab ab '
#$# 57
36
M x(y 1) dx dy 0 dy 0 x 0 and y ; I y (y 1) dx dy 2y 2y dy
y x
0y 0 0y 0
1y 1 1y 1œœœÊœœœ œ'' ' '' '

7
10
#%$
ab
R ; I x (y 1) dx dy 2y 2y dy R ;œÊœ œ œ œ œÊœ œ
9 3
10 M 10 3 10 M 10
I 36 I 32
xy y
0y 0
1y 1É ab É
x
È È
'' '

#%$ " y
III R
oxyœœ Ê œ œ
6
5M5
I 32
!
É
o
È
40. M 3x 1 dx dy 2y 2y dy ; M y 3x 1 dx dy 2y 2y dy ;œœœœ œœ'' ' '' '
0y 0 0y 0
1y 1 1y 1
x
ab a b ab a b
#$ # %#
#
3 16
15
M x 3x 1 dx dy 0 x 0 and y ; I y 3x 1 dx dy 2y 2y dy
y x
0y 0y 0
1y 1y 1œœÊœœœ œœ'' '' '

ab ab a b
###&$ 32 5
45 6
R ; I x 3x 1 dx dy 2 y y dy R ;Êœ œ œ œ œÊœ œ
xy y
0y 0
1y 1É ab ˆ‰ É É
I
M3 5 3 30 M 45
5 311 11 Ix
È
'' '

## & $ " y
III R
oxy oœœ Ê œ œ
62
5M
I
5
É
o
È
41. dy dx 10,000 1 e 10,000 1 e '' ' ' '
$
$52 5 5
50 5 5
10,000e
11
dx dx dx
11
y
¸¸
# #

xx
xx


œ œ ab ab ’“
10,000 1 e 2 ln 1 10,000 1 e 2 ln 1œ ab ab ‘ ‘ˆ‰ ˆ‰
22 xx
##
!&
& !
10,000 1 e 2 ln 1 10,000 1 e 2 ln 1 40,000 1 e ln 43,329œ œ ¸ab ab ab ‘ ‘ˆ‰ˆ‰ ˆ‰
# # #
##
557
2
42. 100(y 1) dx dy 100(y 1)x dy 100(y 1) 2y 2y dy 200 y y dy'' ' ' '
0y 0 0 0
12yy 1 1 1
2y y
y






œ œ œ cd ab ab
#$
200 (200) 50$!$ $’“ ˆ‰
yy
24 4
¸ "
!
"
43. M dy dx 2a 1 x dx 2a x ; M y dy dx$$!$!$$'' ' ''

10 0 10
1a1x 1 1a1x
x


ab ’“
#
"
!
x4a
33
#
1 2x x dx a x y . The angle between theœœœÊœœœ
2a 2x x 8a 2a
35 15 M 5
M

#
#% #
"
!
'
0
1
ab ’“
x
Š‹
Š‹
8a
15
4a
3

)
x-axis and the line segment from the fulcrum to the center of mass on the y-axis plus 45° must be no more than
90° if the center of mass is to lie on the left side of the line x 1 tan a .œÊŸ Ê Ÿ ÊŸ)
11 1
454
2a 5
##
"
ˆ‰
Thus, if 0 a , then the appliance will have to be tipped more than 45° to fall over. Ÿ
5
#
44. f(a) I (y a) dy dx dx (2 a) a ; thus f (a) 0 4(2 a) 4aœœ œ œ œÊ
a
00 0
42 4
'' ' #$$w##
’“ cd
(2 a)
33 3
a4
# #
0 a (2 a) 0 4 4a 0 a 1. Since f (a) 8(2 a) 8a 16 0, a 1 gives aœÊ œÊ œÊœ œ œ œ
## w w
minimum value of I .
a
45. M dy dx dx 2 sin x 2 0 ; M x dy dx$$$$!$$'' ' ''
011x 0 011x
111x 1 111x
y

¸ ¸
¸ ¸


2
1x
È

"
"
! #

cd ˆ‰
1
1
dx 2 1 x 2 x and y 0 by symmetry$$!!$Î$$'
0
1
2x 2
1x
È

#
"Î#
"
!

’“ab
1
46. (a) I x dx RœœÊœœ'
¸
¸L2
L2
$
# "$$
$
LLL
12 12 L 23
# #
É †
È
(b) I x dx RœœÊœœ'
0
L
$
# "$$
$
LLL
33L 3
# #
É †
È
47. (a) M dx dy 2 y y dy 2 2
" "
# #
#
"
!
$$ $ ! $ ! $ $Î$'' '
0y 0
12yy 1



$$ $ $ $ ab ’“ ˆ‰
yy
23 6 3
3
#
$

962. 958 Chapter 15 Multiple Integrals
(b) average value , so the values are the sameœœœœ
''
''
0y
12yy
0y
12yy






(y 1) dx dy
dx dy
Š‹
Š‹



3
3
#
$
48. Let (x y ) be the location of the weather station in county i for i 1 254. The average temperature
iißœßáß
in Texas at time t is approximately , where T(x y ) is the temperature at time t at the! !
ß
!
254
i1

T(x y ) A
A
iiii i?
ß
weather station in county i, A is the area of county i, and A is the area of Texas.?
i49. (a) x 0 M x (x y) dy dx 0œœÊ œ ß œ
M
M
y
y
''
R
$
(b) I (x h) (x y) dA x (x y) dA 2hx (x y) dA h (x y) dA
Lœ ßœ ß ß ß'' '' '' ''
RRRR
## #
$$ $$
I 0 h (x y) dA I mhœ ßœ
ycm
##''
R
$

50. (a) I I mh I I mh 14 ; I I mh 12 14
cm L x5/7y y1114x œ Êœ œ œœ œ œ
## #
œ
# #
39 523 1147
5 7 35 14 14 ˆ‰ ˆ ‰
(b) I I mh 14 ; I I mh 14 24
x1 x57 y2 œ œ œœ œ œ
##
##
23 2 9 47 17
35 7 5 14 14ˆ‰ ˆ ‰
y1114œÎ
51. M y dA y dA M M x ; likewise, y ;
xpp


œ'' ''
R R 
"#

œ Êœ œ xx
MM MM
mm mm


xx yy
thus x y M M M M mx mx my mycij i j i jœœ œ
""

"" ## " # "#mm mm
xx yy


cdc dabab a ba b
mx y mx yœ œ
"

"" ## "#

mm mm
mm


cdababij ij
cc
52. From Exercise 51 we have that Pappus's formula is true for n 2. Assume that Pappus's formula is true forœ
n k 1, i.e., that (k 1) . The first moment about x of k nonoverlapping plates isœ œc
!
!
k1
i1
k1
i1



m
mii
ic
y dA y dA M M x ; similarly, y ;!

k1
i1

œ
'' ''
RRik
ikxx
MM MM
mm mm œ Êœ œ
c
cc
k1 k
xx yy
k1 k1 kk
ik ik



!!
k1 k1
i1 i1

thus (k) x y M M M Mcij i jœœ
"
!
k
i1
i
k1 k k1 k


m
‘ˆ‰ˆ‰
xx yy
cc
m x m x m y m yœ
"
!
k
i1
i

m
”•Œ  Œ  Œ   Œ  
!!
k1 k1
i1 i1

œœ
ikk ik kc c ij
mx y mx yœ œ
"

!!
!
k k1
i1 i1
k1
i1




m m
ikk k
m(k1)mi i
ikk
–—
abab
!
k1
i1

œ
c c
ccij ij

, and by mathematical induction the statement follows.œ
mm m m
mm m m   
 cc c c á
á k1k1 kk
k1 k
53. (a) x and ycœœÊœœ
8( 3 ) 2(3 3.5 ) 14 31
82 10 5 10
731ij i j i j

(b) x and ycœœÊœœ
8( 3 ) 6(5 2 ) 38 36
14 14 7 7
19 18ij i j i j
(c) x and ycœœÊœœ
2(3 3.5 ) 6(5 2 ) 36 19
8828
919ij ij ij
(d) x and ycœœÊœœ
8( 3 ) 2(3 3.5 ) 6(5 2 ) 44 43
16 16 4 16
11 43ij i j ij i j
54. cœœ œœ
15 7 48(12 )
15 48 4 63 4 63 4 7
15(3 28 ) 48(48 4 ) 2349 612 261 68
ˆ‰
3
4
ij ij ij ij i j ij


†††
x and yÊœ œ
261 17
87#

963. Section 15.3 Double Integrals in Polar Form 959
55. Place the midpoint of the triangle's base at the origin and above the semicircle. Then the center of
mass of the triangle is 0 , and the center of mass of the disk is 0 from Exercise 25. Fromˆ‰ ˆ ‰ßß
h4a
33 1
Pappus's formula, , so the centroid is on the boundarycœœ
(ah)
ah ah
ˆ‰ ˆ ‰Š‹ Š ‹
Š‹ Š‹
h a 4a ah 2a
323 3
aa
jj j







 
if ah 2a 0 h 2a h a 2 . In order for the center of mass to be inside T we must have
# $ ##
œÊ œ Êœ
È
ah 2a 0 or h a 2.
# $

È
56. Place the midpoint of the triangle's base at the origin and above the square. From Pappus's formula,
, so the centroid is on the boundary if 0 h 3s 0 h s 3.cœ œÊ œÊœ
Š‹Š ‹ ˆ‰
Š‹
sh h s
32
sh



jj

#
##
s
sh s
6



s
È
15.3 DOUBLE INTEGRALS IN POLAR FORM
1. dy dx r dr d d'' '' '

10 0 0 0
11x 1

œœœ

))
"
##
1
2. dy dx r dr d d'' '' '

11x 00 0
11x 21 2



œœœ

))1
"
#
3. x y dx dy r dr d d'' ' ' '
00 0 0 0
11y 21 2

ab
## $ "
œœœ

))
48
1
4. x y dx dy r dr d d'' '' '

11y 00 0
11y 21 2


ab
## $ "
#
œœœ

))
4
1
5. dy dx r dr d d a'' '' '

aax 00 0
aax 2a 2



œœœ

))1
a
2

#
6. x y dx dy r dr d 4 d 2'' ' ' '
00 0 0 0
24y 22 2

ab
## $
œœœ

))1
7. x dx dy r cos dr d 72 cot csc d 36 cot 36
'' ' ' '
00 40 4
6y 26 csc 2
2
4
œœœ œ








###
)) ) )) ) cd
8. y dy dx r sin dr d tan sec d'' ' ' '
00 0 0 0
2x 42 sec 4
œœœ

##
)) ) ))
84
33
9. dy dx dr d 2 1 dr d 2 (1 ln 2) d'' '' '' '

11x 0 0
0 0 32 1 32 1 32

22r
1xy 1r 1r
"
È
œœ œ


))) ˆ‰
(1 ln 2)œ 1
10. dx dy dr d 4 1 dr d 4 1 d'' '' '' '

11y 20 20 2
1 0 32 1 32 1 32
4x y
1x y 1 r 1 r 4
4r
È





"
œœ œ


))) ˆ‰ ˆ‰
1
4œ11
#
11. e dx dy re dr d (2 ln 2 1) d (2 ln 2 1)'' '' '
00 00 0
ln 2 (ln 2) y 2 ln 2 2
r



Èxy

œœ œ

))
1
#
12. e dy dx re dr d 1 d'' ' ' '
00 0 0 0
11x 21 2
xy





œœ œ

""
#
r
e4e
(e 1)

)) ˆ‰
1
13. dy dx r dr d 2 cos 2 sin cos d'' ' ' '
00 0 0 0
21(x1) 22 cos 2

x y r(cos sin )
xy r


#
œœ

))
)))))ab
sin 1œ œœ ‘))
sin 2 2
2
)11 #
##
2
0

964. 960 Chapter 15 Multiple Integrals
14. xy dx dy sin cos r dr d sin cos d sin'' ' ' '
01(y1) 20 2
2 0 2 sin
2



## %( )
œœœœ




)) ) ))) )
32 4 4
555
cd
15. ln x y 1 dx dy 4 ln r 1 r dr d 2 (ln 4 1) d (ln 4 1)'' ' ' '

11y 00 0
11y 21 2


ab ab
## #
œ œ œ

))1
16. dy dx 4 dr d 4 d 2 d'' ' ' ' '

11x 00 0 0
11x 21 2 2



22r
1x y 1r 1rab ab
"

"
!


œœ œœ

)))1 ‘
17. r dr d 2 (2 sin 2 ) d 2( 1)'' '
00 0
222sin 2 2


)))1œ œ
18. A 2 r dr d 2 cos cos dœœ œ '' '
01 0
21cos 2

))))ab
# 8
4
1
19. A 2 r dr d 144 cos 3 d 12œœœ '' '
00 0
6 12 cos 3 6

)))1
#
20. A r dr d dœœœ'' '
00 0
243 2

)))
864
927
# 1

21. A r dr d 2 sin d 1œœ œ '' '
00 0
21sin 2

)))
13 cos 2 3
28
ˆ‰
##
)1
22. A 4 r dr d 2 2 cos d 4œœ œ '' '
00 0
21cos 2

))) ˆ‰
3cos 23
2##
)1
23. M 3r sin dr d (1 cos ) sin d 4
x
00 0
1cos œœ œ'' '

# $
)) ) ))
24. I y k x y dy dx k r sin dr d d ;
x
aax 00 0
aax 2a 2œ œœœ'' '' '






### & #
#
cdab

)) )
ka 1 cos 2 ka
66

)1
I k x y dy dx k r dr d d
o
aax 00 0
aax 2a 2œ œœœ'' '' '






ab
## &
#

))
ka ka
63

1
25. M 2 dr d 2 (6 sin 3) d 6 2 cos 6 3 2œœ œ œ '' '







63 6
26 sin 2
2
6
)))))1 cd
È
26. I r dr d cos 2 cos d 2 sin 2
o
21 2
32 1cos 32
32
2œœ œ œ'' '







)))) )
""
###
#
ab ‘
sin 2
44
)) 1
27. M 2 r dr d (1 cos ) d ; M 2 r cos dr dœœ œœ '' ' ''
00 0 00
1cos 1cos
y

))) ))
##
#
31
2 cos 2 sin cos d x and y 0, by symmetryœ œÊœ œ'
0

ˆ‰
4 cos 15 cos 4 5 5
324 4 4 6
))1
))) )
#
28. I r dr d (1 cos ) d
o
00 0
21cos 2œœ œ'' '

$% "
)))
416
351
29. average r a r dr d a dœ œœ
442a
a3a311

'' '
00 0
2a 2


È## $
))
30. average r dr d a dœœœ
442a
a3a311

'' '
00 0
2a 2

# $
))
31. average x y dy dx r dr d dœ œœœ
"" ## #
111aa33
a2a

'' '' '

aax 00 0
aax 2a 2



È

))

965. Section 15.3 Double Integrals in Polar Form 961
32. average (1 x) y dy dx (1 r cos ) r sin r dr dœ œ
"" ## ###
11''
R
cd c d ''
00
21

)))
r 2r cos r dr d dœ œ œ œ
""" $ #
#111
))'' '
00 0
21 2
2
0


ab ˆ‰ ‘)) ))
3 2 cos 3 2 sin 3
43 4 3
33. r dr d 2 ln r dr d 2 r ln r r d 2 e 1 1 d 2 2 e'' '' ' '
01 01 0 0
2e 2e 2 2
e
1

Š‹ cd ÈÈ ‘ˆ‰ˆ‰
ln r
r

)) ) )1œœ œ œ

"
#
34. dr d dr d (ln r) d d 2'' '' ' '
01 01 0 0
2e 2e 2 2
e
1

Š‹ ˆ‰ cd
ln r 2 ln r
rr

))))1œœœœ
#
35. V 2 r cos dr d 3 cos 3 cos cos dœœ '' '
01 0
21cos 2

## $%
)) ) ) ))
2
3
ab
sin 2 3 sin sinœ œ
25 sin 4 4 5
38 32 3 8
‘
" $))1
)))
2
0
36. V 4 r 2 r dr d (2 2 cos 2 ) 2 dœ œ '' '
00 0
42 cos 2 4


È ‘ # $Î# $Î#
)))
4
3
1 cos sin d cos œ œ œ
22 22 6240264
33 333 9
32 32 cos111 )
ÈÈÈÈ '
0
4 4
0

ab ’“
#
))) )

37. (a) I e dx dy e r dr d lim re dr d
#
œœ œ'' ' ' ' '
00 0 0 0 0
22b

abxy


Š‹ ”•
rr

))
bÄ_
lim e 1 d d Iœ œœÊœ
""
###
''
00
22

bÄ_
Š‹
b
4

))
1 1È
(b) lim dt e dt 1, from part (a)
xÄ_
''
00
x
2e 2 2 t
 

t
ÈÈ È
È
11 1
1
œœœ
 

#
Š‹Š‹
38. dx dy dr d lim dr lim '' ' ' '
00 0 0 0
2b
b
0

" "
#a b ab ab1x y 1r 1r
rr
41r


œœ œ

)
11
bbÄ_ Ä_
‘
lim 1œ œ
11
41b4
bÄ_
ˆ‰
"


39. Over the disk x y : dA dr d ln 1 r d
## #""

Ÿœ œ
3r
41xy 1r 2''
R


'' '
00 0
232 2
32
0




)) ‘ab
ln d (ln 2) d ln 4œ œœ''
00
22

ˆ‰
""
# 4
))1
Over the disk x y 1: dA dr d lim dr d
## "

Ÿœœ ''
R
a1
1x y 1r 1r
rr

'' ' '
00 0 0
21 2 a

)) ’“
Ä

lim ln 1 a d 2 lim ln 1 a 2 , so the integral does not exist overœ œ œ_'
0
2

a1 a1ÄÄ

‘ ‘ab ab
""
##
##
)1 1††
xy1
##
Ÿ
40. The area in polar coordinates is given by A r dr d d f ( ) d r d ,œœœœ '' ' ' '

0
f f
0

)))))’“
r
2

""
##
##
where r f( )œ)
41. average (r cos h) r sin r dr d r 2r h cos rh dr dœ œ
"" ### $ ##
11aa

'' ''
00 00
2a 2a

cdab))) ))
d dœ œ œ
"""
###111
)))))
a43 43 43
a 2a h cos a h a 2ah cos h a 2ah sin h


''
00
22 2
0

Š‹Š‹’“ ))
a2hœ
"
#
##
ab

966. 962 Chapter 15 Multiple Integrals
42. (a) A r dr d 4 sin csc dœœ '' '



4csc 4
34 2 sin 34
))))
"
#
##
ab
2sin 2cot œ œ
"
##
cd)))
34
4


1 (b) V 2 yA 2œœ œ 11 1 ˆ‰ˆ‰
34 4
33
11 1
1

#
#

44-46. Example CAS commands:
:Maple
f := (x,y) -> y/(x^2+y^2);
a,b := 0,1;
f1 := x -> x;
f2 := x -> 1;
plot3d( f(x,y), y=f1(x)..f2(x), x=a..b, axes=boxed, style=patchnogrid, shading=zhue, orientation=[0,180], title="#43(a)
(Section 15.3)" ); # (a)
q1 := eval( x=a, [x=r*cos(theta),y=r*sin(theta)] ); # (b)
q2 := eval( x=b, [x=r*cos(theta),y=r*sin(theta)] );
q3 := eval( y=f1(x), [x=r*cos(theta),y=r*sin(theta)] );
q4 := eval( y=f2(x), [x=r*cos(theta),y=r*sin(theta)] );
theta1 := solve( q3, theta );
theta2 := solve( q1, theta );
r1 := 0;
r2 := solve( q4, r );
plot3d(0,r=r1..r2, theta=theta1..theta2, axes=boxed, style=patchnogrid, shading=zhue, orientation=[-90,0],
title="#43(c) (Section 15.3)" );
fP := simplify(eval( f(x,y), [x=r*cos(theta),y=r*sin(theta)] )); # (d)
q5 := Int( Int( fP*r, r=r1..r2 ), theta=theta1..theta2 );
value( q5 );
: (functions and bounds will vary)Mathematica
For 43 and 44, begin by drawing the region of integration with the command.FilledPlot
Clear[x, y, r, t]
<<Graphics`FilledPlot`
FilledPlot[{x, 1}, {x, 0, 1}, AspectRatio 1, AxesLabel {x,y}];ÄÄ
The picture demonstrates that r goes from 0 to the line y=1 or r = 1/ Sin[t], while t goes from /4 to /2.11
f:= y / (x y )
22

topolar={x r Cos[t], y r Sin[t]};ÄÄ
fp= f/.topolar //Simplify
Integrate[r fp, {t, /4, /2}, {r, 0, 1/Sin[t]}]11
For 45 and 46, drawing the region of integration with the ImplicitPlot command.
Clear[x, y]
<<Graphics`ImplicitPlot`
ImplicitPlot[{x==y, x==2 y, y==0, y==1}, {x, 0, 2.1}, {y, 0, 1.1}];
The picture shows that as t goes from 0 to /4, r goes from 0 to the line x=2 y. will find the bound for r.1 Solve
bdr=Solve[r Cos[t]==2 r Sin[t], r]//Simplify
f:=Sqrt[x y]
topolar={x r Cos[t], y r Sin[t]};ÄÄ
fp= f/.topolar //Simplify
Integrate[r fp, {t, 0, /4}, {r, 0, bdr[[1, 1, 2]]}]1

967. Section 15.4 Triple Integrals in Rectangular Coordinates 963
15.4 TRIPLE INTEGRALS IN RECTANGULAR COORDINATES
1. F x, y, z dy dz dx dy dz dx 1 x z dz dx
'' ' '' ' ''
00 xz 00 xz 00
11x1 11x1 11x


ab a b œœ
1x x1x dx dxœ œœœ ''
00
11
’“’“abab
ab ab ab1x 1x
22 6
1x
6
0
1
"
22
3
2. dz dy dx 3 dy dx 6 dx 6, dz dx dy, dx dy dz, dx dz dy,''' '' ' ''' ''' '''
000 00 0 000 000 000
123 12 1 213 321 231
œœœ
dy dx dz, dy dz dx
''' '''
000 000
312 132
3. dz dy dx'' '
00 0
1 22x 33x3y2

3 3x y dy dxœ ''
00
122x

ˆ‰
3
2
3(1 x) 2(1 x) 4(1 x) dxœ '
0
1
‘††
3
4
#
3 (1 x) dx (1 x) 1,œ œ œ '
0
1
# $
"
!
cd
dz dx dy, dy dz dx,
'' ' '' '
00 0 00 0
2 1y2 33x3y2 1 33x 22x2z3

dy dx dz, dx dz dy,'' ' '' '
00 0 00 0
31z322x2z3 233y21y2z3

dx dy dz'' '
00 0
322z31y2z3


4. dz dy dx 4 x dy dx 3 4 x dx x 4 x 4 sin 6 sin 1 3 ,''' '' '
000 00 0
23 4x 23 2 2
0


œ œ œ œœ
ÈÈÈ
’“
###
##
" "3x
1
dz dx dy, dy dz dx, dy dx dz, dx dy dz, dx dz dy''' '' ' '' ' ''' '''
0 00 00 0 00 0 000 000
3 2 4x 2 4x 3 2 4z 3 2 3 4z 3 2 4z


5. dz dy dx 4 dz dy dx'' ' '' '

24xxy 00 xy
2 4x 8xy 2 4x 8xy



œ
4 8 2 x y dy dxœ ''
00
24x


cdab
##
8 4 x y dy dxœ ''
00
24x


ab
##
8 4 r r dr d 8 2r dœ œ '' '
00 0
22 2

ab ’“
##
#
!
))
r
4

32 d 32 16 ,œœœ '
0
2

)1ˆ‰
1
#
dz dx dy,'' '

24yxy
24y8xy


dx dz dy dx dz dy,''' '' '

2y zy 24 8zy
2 4 zy 2 8y 8zy






dx dy dz dx dy dz, dy dz dx dy dz'' ' '' ' ''' '' '
0 z zy 4 8z 8zy 2 x zx 2 4 8zx
4 z zy 8 8z 8zy 2 4 zx 2 8x 8zx













dx,
dy dx dz dy dx dz'' ' '' '
0zzx 48z8zx
4z zx 88z 8zx

968. 964 Chapter 15 Multiple Integrals
6. The projection of D onto the xy-plane has the boundary
x y 2y x (y 1) 1, which is a circle.
## # #
œÊ œ
Therefore the two integrals are:
dz dx dy and dz dy dx'' ' '' '
0 2yy xy 1 1 1x xy
2 2y y 2y 1 1 1 x 2y








7. x y z dz dy dx x y dy dx x dx 1''' '' '
000 00 0
111 11 1
ab ˆ‰ ˆ‰
### ## # "
œ œ œ
33
2
8. dz dx dy 8 2x 4y dx dy 8x x 4xy dy''' '' '
00x3y 00 0
23y8xy 23y 2
3y
0





œ œ ab ‘
## $ #2
3
24y 18y 12y dy 12y y 24 30 6œ œ œœ'
0
2
2
0


ab ‘
$$ # %"5
2
9. dx dy dz dy dz dy dz dz dz 1''' '' '' ' '
111 11 11 1 1
eee ee ee e e ee
11
"""
xyz yz yz z z
ln x ln y
œœœœœ’“ ’“
10. dz dy dx (3 3x y) dy dx (3 3x) (3 3x) dx (1 x) dx'' ' '' ' '
00 0 00 0 0
1 33x 33xy 1 33x 1 1

œ œ œ ‘
## #"
##
9
(1 x)œ œ
33
##
$
"
!
cd
11. y sin z dx dy dz y sin z dy dz sin z dz (1 cos 1)''' '' '
00 0 00 0
111

œœœ 1
11

##
12. (x y z) dy dx dz xy y zy dx dz (2x 2z) dx dz x 2zx dz''' '' '' '
111 11 11 1
111 11 11 1
œ œ œ ‘ cd
"##
"
"
"
"2
4z dz 0œœ'
1
1
13. dz dy dx 9 x dy dx 9 x dx 9x 18'' ' '' '
00 0 00 0
39x 9x 39x 3


œ œ œ œ
È
ab ’“
# #
$
!
x
3

14. dz dx dy (2x y) dx dy x xy dy 4 y (2y) dy'' ' '' ' '
04y0 04y 0 0
24y 2xy 24y 2 2 4y
4y









œ œ œ cd ab
##
"Î#
4y (4)œ œœ’“ab
2216
333
# $Î#
$Î#
#
!
15. dz dy dx (2 x y) dy dx (2 x) (2 x) dx (2 x) dx'' ' '' ' '
00 0 00 0 0
12x2xy 12x 1 1

œ œ œ ‘
## #""
##
(2 x)œ œ œ ‘
"" $
"
!6666
87
16. x dz dy dx x 1 x y dy dx x 1 x 1 x dx x 1 x dx'' ' '' ' '
00 3 00 0 0
11x 4xy 11x 1 1


œ œ œ a b abab ab ’“
####
##""
##
1xœ œ’“ab
"" #
$
"
!
12 12
17. cos (u v w) du dv dw [sin (w v ) sin (w v)] dv dw''' ''
000 00

œ 1
[( cos (w 2 ) cos (w )) (cos (w ) cos w)] dwœ
'
0

11 1

969. Section 15.4 Triple Integrals in Rectangular Coordinates 965
sin (w 2 ) sin (w ) sin w sin (w ) 0œ œcd 11 1
1
!
18. ln r ln s ln t dt dr ds (ln r ln s) t ln t t dr ds (ln s) r ln r r ds s ln s s 1''' '' '
111 11 1
eee ee e
eee
111
œ œ œ œcd c dc d
19. e dx dt dv lim e e dt dv e dt dv e dv'' ' '' '' '
00 00 00 0
4 ln sec v 2t 4 ln sec v 4 ln sec v 4
x 2t b 2t 2 ln sec v


œ œœ
bÄ_
ab ˆ‰
""
##
dvœ œ œ'
0
4

Š‹ ‘
sec v tan v v
22 8

## #
""
Î%
!
1
1
20. dp dq dr dq dr 4 q dr dr''' '' ' '
000 00 0 0
72 4q 72 7 7

q
r1 r1 3(r1) 3 r1
q4 q 8

"" #
$Î#
#
!
œœ œ
È
’“ab
8 ln 2œœ
8 ln 8
3
21. (a) dy dz dx (b) dy dx dz (c) dx dy dz'' ' '' ' '' '

10 x 0 1z x 0 0 y
1 1x 1z 1 1z 1z 1 1z y







(d) dx dz dy (e) dz dx dy'' ' '' '
00 y 0 y 0
11yy 1y1y



22. (a) dy dz dx (b) dy dx dz (c) dx dy dz''' ''' '' '
00 1 00 1 0 1 0
11 z 11 z 1 z1


(d) dx dz dy (e) dz dx dy''' '''
10 0 10 0
0y 1 01y

23. V dz dy dx y dy dx dxœœœœ''' '' '
010 01 0
11 y 11 1

# 22
33
24. V dy dz dx (2 2z) dz dx 2z z dx 1 x dx xœœ œ œ œ œ '' ' '' ' '
00 0 00 0 0
11x22z 11x 1 1
1x
0


cd ab ’“
##
"
!
x2
33

25. V dz dy dx (2 y) dy dx 2 4 x dxœœ œ '' ' '' '
00 0 00 0
44x2y 44x 4


’“
È ˆ‰
4x
#
(4 x) (4 x) (4) (16) 4œ œ œœ ‘
4 4 32 20
34 3433
$Î## $Î#""
%
!
26. V 2 dz dy dx 2 y dy dx 1 x dxœœ œ œ '' ' '' '
01x0 0 1x 0
10 y 1 0 1



ab
# 2
3
27. V dz dy dx 3 3x y dy dx 6(1 x) 4(1 x) dxœœ œ '' ' '' '
00 0 00 0
1 22x 33x3y2 1 22x 1

ˆ‰ ‘
33
4#
##
†
3(1 x) dx (1 x) 1œ œ œ
'
0
1
# $
"
!
cd
28. V dz dy dx cos dy dx cos (1 x) dxœœœ '' ' '' '
00 0 00 0
11xcosx2 11x 1

ˆ‰ ˆ ‰
11xx
##
cos dx x cos dx sin u cos u du cos u u sin uœ œ œ'' '
00 0
11 2
2
0
ˆ‰ ˆ‰ ‘ cd
111
11 11
xx2x4 24
###
"
!





1œ œ
24 4
11 1
1

ˆ‰
#
29. V 8 dz dy dx 8 1 x dy dx 8 1 x dxœœ œ œ '' ' '' '
00 0 00 0
11x 1x 11x 1



È
ab # # 16
3
30. V dz dy dx 4 x y dy dx 4 x 4 x dxœœ œ '' ' '' '
00 0 00 0
24x 4xy 24x 2


a b abab ’“
###
##"
#
4 x dx 8 4x dxœ œ œ
"
##
##
#''
00
22
ab Š‹
x 128
15

970. 966 Chapter 15 Multiple Integrals
31. V dx dz dy (4 y) dz dy (4 y) dyœœ œ '' ' '' '
00 0 00 0
416y24y 416y2 4

È16 y
#
2 16 y dy y 16 y dy y 16 y 16 sin 16 yœ œ ''
00
44
ÈÈÈ ‘’“ab ### ""
#
" #
%
!
$Î#
%
!
y
46
16 (16) 8œ œˆ‰
1
#
" $Î#
63
32
1
32. V dz dy dx (3 x) dy dx 2 (3 x) 4 x dxœœ œ '' ' '' '

24x0 24x 2
24x3x 24x 2



È #
3 2 4 x dx 2 x 4 x dx 3 x 4 x 4 sin 4 xœ œ ''
22
22
ÈÈÈ
’“’“ ab ### " #
##
# #
$Î#x2
23
12 sin 1 12 sin ( 1) 12 12 12œ œ œ
" "
## ˆ‰ ˆ ‰
11
1
33. dz dy dx 3 dy dx'' ' ''
0 0 2xy 2 0 0
22x42x2y 22x


œ ˆ‰
3x 3y
##
3 1 (2 x) (2 x) dxœ '
0
2
‘ˆ‰
x3
4#
#
66x dxœ '
0
2
’“
3x 3(2 x)
4

#

6x3x (121240) 2œ œ œ’“
#
#
!
x2
24 4
(2 x)


34. V dx dy dz (8 2z) dy dz (8 2z)( z) dz 64 24z 2z dzœœ œ ) œ
''' '' ' '
0zz 0z 0 0
488z 48 4 4

ab
#
64z 12z zœ œ ‘
# $
%
!
2 320
33
35. V 2 dz dy dx 2 (x 2) dy dx (x 2) 4 x dxœœ œ '' ' '' '

20 0 20 2
24x2x2 24x2 2

È #
24 x dx x4 x dx x4 x 4 sin 4 xœ œ ''
22
22
ÈÈÈ
’“’“ ab ### " #
#
##
# #
" $Î#x
3
44 4œ œˆ‰ ˆ ‰
11
##
1
36. V 2 dz dx dy 2 x y dx dy 2 xy dyœœ œ '' ' '' '
00 0 00 0
11y xy 11y 1 1y
0


ab ’“
## # x
3

2 1 y 1 y y dy 2 1 y y y dy 1 y dyœ œ œ '''
000
111
abab ab ab’“ ˆ‰
### ## %'"""" #
33333
2
yœ œœ
2264
37 377
y
’“ ˆ‰ˆ‰
"
!
37. average x 9 dz dy dx 2x 18 dy dx 4x 36 dxœ œ œ œ
""" ###
8883
31''' '' '
000 00 0
222 22 2
ab ab ab
38. average (x y z) dz dy dx (2x 2y 2) dy dx (2x 1) dx 0œ œ œ œ
"""
222''' '' '
000 00 0
112 11 1
39. average x y z dz dy dx x y dy dx x dx 1œ œ œ œ ''' '' '
000 00 0
111 11 1
ab ˆ‰ ˆ‰
### ## # "
33
2
40. average xyz dz dy dx xy dy dx x dx 1œœœœ
"""
842''' '' '
000 00 0
222 22 2
41. dx dy dz dy dx dz dx dz z dz''' ''' '' '
002y 00 0 00 0
412 42 x2 42 4
4 cos x 4 cos x x cos x
2z 2z z
sin 4ab ab ab
ÈÈÈ

œœœ

ˆ‰
#
"Î#
(sin 4)z 2 sin 4œœ ‘
"Î#
%
!

971. Section 15.4 Triple Integrals in Rectangular Coordinates 967
42. 12xz e dy dx dz 12xz e dx dy dz 6yz e dy dz 3e dz''' ''' '' '
00x 000 00 0
111 11 y 11 1

zy zy zy zy

œœœ

’“
"
!
3ez dz3e1 3e6œ œ œ '
0
1
zz
ab cd
"
!
43. dx dy dz dy dz dz dy''' '' ''
0z0 0z 00
11 ln 3 11 1y



1 1 11 11e sin y 4 sin y 4 sin y
yyy
2x
ab ab ab


œœ
4 y sin y dy 2 cos y 2( 1) 2(1) 4œœ œ œ'
0
1
11 1ab c d ab
##
"
!
44. dy dz dx dz dx x dx dz (4 z) dz'' ' '' '' '
00 0 00 00 0
24x x 24x 4 4z 4



sin 2z x sin 2z sin 2z sin 2z
4z 4z 4z 4z#
"
œœ œ

ˆ‰ ˆ‰
cos 2z sin zœ œ œ ‘ ‘
"""
%%
! ##
#
!44
sin 4

45. dz dy dx 4 x y a dy dx'' ' ''
00 a 00
1 4ax 4x y 1 4ax


œÊ œ
44
15 15
ab
#
4 a x 4 a x dx 4 a x dx (4 a) 2x (4 a) x dxÊ œÊ œÊ '''
0 00
1 11
’“abab ab c d
## # ## %
## #""
##
44
15 15
(4 a) x x (4 a) (4 a) (4 a) 15(4 a) 10(4 a) 5 0œÊ œÊ œÊ œ
82x828
15 3 5 15 3 5 15
’“
# $ ##
"
!
"
 
3(4 a) 2(4 a) 1 0 [3(4 a) 1][(4 a) 1] 0 4 a or 4 a 1 a or a 3Ê œÊ œÊ œ œÊœ œ
# "
33
13
46. The volume of the ellipsoid 1 is so that 8 c 3.
x z 4abc
abc 3 3
y 4(1)(2)(c)



œœÊœ
1 1
1
47. To minimize the integral, we want the domain to include all points where the integrand is negative and to
exclude all points where it is positive. These criteria are met by the points (x y z) such thatßß
4x 4y z 4 0 or 4x 4y z 4, which is a solid ellipsoid centered at the origin.
### ###
Ÿ Ÿ
48. To maximize the integral, we want the domain to include all points where the integrand is positive and to
exclude all points where it is negative. These criteria are met by the points (x y z) such thatßß
1 x y z 0 or x y z 1, which is a solid sphere of radius 1 centered at the origin.  Ÿ
### ###
49-52. Example CAS commands:
:Maple
F := (x,y,z) -> x^2*y^2*z;
q1 := Int( Int( Int( F(x,y,z), y=-sqrt(1-x^2)..sqrt(1-x^2) ), x=-1..1 ), z=0..1 );
value( q1 );
: (functions and bounds will vary)Mathematica
Due to the nature of the bounds, cylindrical coordinates are appropriate, although Mathematica can do it as is also.
Clear[f, x, y, z];
f:= x y z
22
Integrate[f, {x, 1,1}, {y, Sqrt[1 x ], Sqrt[1 x ]}, {z, 0, 1}]
22
N[%]
topolar={x r Cos[t], y r Sin[t]};ÄÄ
fp= f/.topolar //Simplify
Integrate[r fp, {t, 0, 2 }, {r, 0, 1},{z, 0, 1}]1
N[%]

972. 968 Chapter 15 Multiple Integrals
15.5 MASSES AND MOMENTS IN THREE DIMENSIONS
1. I y z dx dy dz a y z dy dz a yz dz
x
c2 b2 a2 c2 b2 c2
c2 b2 a2 c2 b2 c2 b2
b2œœœ''' '' '



ab ab ’“
## ## # y
3
œ
a bz dz ab z ab b c b c ;œœœœœ'


c2
c2 c2
c2
Š‹ ’“ Š‹ abab
bbzbccabcM
12 12 3 12 12 1 12
œ
#####
#
R ; likewise R and R , by symmetry
xyzœœœÉÉÉ
bc ac ab
12 12 12

œœœ
2. The plane z is the top of the wedge I y z dz dy dxœÊœ
42y
3
##
x
3243
34 42y3
'''


ab
dy dx dx 208; I x z dz dy dxœœœœ '' ' '''

32 3 3243
34 3 34 42y3
y
’“ ab
8y 2y 8(2 y)
3 3 81 81 3
64 104
œœ
##
dy dx 12x dx 280;œœœ'' '
œœ œ32 3
34 3
’“ ˆ‰
(4 2y) x (4 2y)
81 3 3 81 3
4x 64 32 #
œ
I x y dz dy dx x y dy dx 12 x 2 dx 360
z
3243 32 3
34 42y3 34 3œœœœ''' '' '
œœœ

ab ab ab ˆ‰
## ## # 8
33
2y
3. I y z dz dy dx cy dy dx dx
x
000 00 0
abc ab aœœœœ''' '' 'ab Š‹ Š ‹
## # œccbcb
3333
abc b c
œœœ ab
b c where M abc; I a c and I a b , by symmetryœ œ œ œ
MMM
333
ab ab ab
## ## ##
yz
4. (a) M dz dy dx (1 x y) dy dx x dx ;œœœœ'' ' '' '
00 0 00 0
11x1xy 11x 1
œœœ œ
Š‹
x
6

##
""
M x dz dy dx x(1 x y) dy dx x 2x x dx
yz
00 0 00 0
11x1xy 11x 1œœœœ'' ' '' '
œœœ œ
""
#
$#
ab
24
x y z , by symmetry; I y z dz dy dxÊœœœ œ
" ##
4
x
00 0
11x1xy
'' '
œœœ
ab
y xy y dy dx (1 x) dx I I , by symmetryœ œœÊœœ'' '
00 0
11x 1
yx
œ
’“
##$ % """(1xy)
3 6 30 30
œ
(b) R 0.4472; the distance from the centroid to the x-axis is 0
xœœœ¸ œœÉ ÉÉÉ
I
M55 16 16 8 4
5 2
x " "" "#
È È
0.3536¸
5. M 4 dz dy dx 4 4 4y dy dx 16 dx ; M 4 z dz dy dxœœœœœ''' '' ' '''
004y 00 0 004y
114 11 1 114
xy
œ œ
ab
# 232
33
2 16 16y dy dx dx z , and x y 0, by symmetry;œœœÊœœœ'' '
00 0
11 1
ab
% 128 128 12
55 5
I 4 y z dz dy dx 4 4y 4y dy dx 4 dx ;
x
004y 00 0
114 11 1œœœœ''' '' '
œ
ab ’“ˆ‰ Š‹
## # % 64 1976 7904
3 3 105 105
64y
"
I 4 x z dz dy dx 4 4x 4x y dy dx 4 x dx
y
004y 00 0
114 11 1œœœ''' '' '
œ
ab ’“ˆ‰ ˆ ‰ Š‹
## # ## # 64 8 128
33 37
64y
"
; I 4 x y dz dy dx 16 x x y y y dy dxœœ œ
4832
63
z
004y 00
114 11
''' ''
œ
ab a b
## ####%
16 dxœœ '
0
1
Š‹
2x 2 256
315 45

6. (a) M dz dy dx (2 x) dy dx (2 x) 4 x dx 4 ;œœœœ'' ' '' '
œœœ
24x20 24x2 2
24x22x 24x2 2
""

""
œœ
œœ
Š‹
È
#
1
M x dz dy dx x(2 x) dy dx x(2 x) 4 x dx 2 ;
yz
24x20 24x2 2
24x22x 24x2 2œœœœ'' ' '' '
œœœ


""

""
œœ
œœ
Š‹
È
#
1
M y dz dy dx y(2 x) dy dx
xz
24x20 24x2
24x22x 24x2œœ'' ' ''
œœœ


""

""
œœ
œœ
(2 x) dx 0 x and y 0œ œÊœœ
" "
##'
œ2
2
’“
4x 4x
44

973. Section 15.5 Masses and Moments in Three Dimensions 969
(b) M z dz dy dx (2 x) dy dx (2 x) 4 x dx
xy
24x20 24x2 2
24x22x 24x2 2œœœ'' ' '' '
œœœ


""

""
œ œ
œ œ
""
##
## #
Š‹
È
5 zœÊœ1
5
4
7. (a) M 4 dz dy dx 4 r dz dr d 4 4r r dr d 4 4 d 8 ;œœœœœ '' ' ' '' ' ' '
00 xy 0 0r 0 0 0
2 4x 4 224 22 2
"
œ
% œ
œœ œ
œœ œ
)))1 ab
$
M zr dz dr d 16 r dr d d z , and x y 0,
xy
00r 00 0
224 22 2œœœœÊœœœ''' '' '
œœ œ
œ
)))
r32648
333#
%
ab
1
by symmetry
(b) M 8 4 r dz dr d cr r dr d d c 8 c 2 2,œÊœ œ œ œ ʜʜ11 ) ) ) '' ' '' '
00 r 0 0 0
2cc 2 c 2
œœ œ""

œ
ab
È
$#
#
cc
4

1
since c 0&
8. M 8; M z dz dy dx dy dx 0; M x dz dy dxœœ œ œœ
xy yz
13 1 13 13 1
151 15 151
''' '' '''
œœ œ œœ
’“
z
2

"
"
2 x dy dx 4 x dx 0; M y dz dy dx 2 y dy dx 16 dx 32œœœœ œ œœ'' ' ''' '' '
œœ œœœ œ13 1 13 1 13 1
15 1 151 15 1
xz
x 0, y 4, z 0; I y z dz dy dx 2y dy dx 100 dx ;Êœ œ œ œ œ œ œ
x
13 1 13 1
151 15 1
''' '' '
œœ œ œ
ab ˆ‰
## # 2 2 400
33 3
I x z dz dy dx 2x dy dx 3x 1 dx ;
y
13 1 13 1
151 15 1œœœœ''' '' '
œœ œ œ
ab a b ˆ‰
## # # 24 16
33 3
I x y dz dy dx 2 x y dy dx 2 2x dx R R
z xz
13 1 13 1
151 15 1œœœœÊœœ''' '' '
œœ œ œ
ab ab ˆ‰ É
## ## # 98 400 50
33 3
and R
yσ
2
3
9. The plane y 2z 2 is the top of the wedge I (y 6) z dz dy dxœ Êœ
L
221
24 2y2
'''
œœœ

cd
##
dy dx; let t 2 y I 4 5t 16t dt 1386;œœÊœœ'' '
œœ œ22 2
24 4
L
’“ Š‹
(y 6) (4 y) (2 y)
24 3 24 3
13t 49
#
" #
œ œ
M (3)(6)(4) 36 RœœÊœœ
"
##
LÉ É
I
M
77L
10. The plane y 2z 2 is the top of the wedge I (x 4) y dz dy dxœ Êœ
L
221
24 2y2
'''
œœœ

cd
##
x 8x 16 y (4 y) dy dx 9x 72x 162 dx 696; M (3)(6)(4) 36œ œ œ œ œ
" "
# #
## #'' '
œœ œ22 2
24 2
ab ab
RÊœ œ
LÉ É
I
M3
58L
11. M 8; I z (y 2) dz dy dx y 4y dy dx dxœœ œ œ œ
L
000 00 0
421 42 4
''' '' 'cd ˆ‰
## # 13 10 40
333
RÊœ œ
LÉ É
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L
000 00 0
421 42 4
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33
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13. (a) M 2x dz dy dx 4x 2x 2xy dy dx x 4x 4x dxœœœœ'' ' '' '
00 0 00 0
22x2xy 22x 2
œœœ œ
abab
#$# 4
3
(b) M 2xz dz dy dx x(2 x y) dy dx dx ; M by
xy xz
00 0 00 0
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# x(2 x)
31515
88
œ
symmetry; M 2x dz dy dx 2x (2 x y) dy dx 2x x dx
yz
00 0 00 0
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## #
#
ab
16
15
x , and y zÊœ œœ
42
55

974. 970 Chapter 15 Multiple Integrals
14. (a) M kxy dz dy dx k xy 4 x dy dx 4x x dxœœœœ '' ' '' '
00 0 00 0
2x4x 2x 2
"" œ
œ
ab a b
##%
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k 32k
15
(b) M kx y dz dy dx k x y 4 x dy dx 4x x dx
yz
00 0 00 0
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"" œ
œ
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ab a b
k8k
3
x ; M kxy dz dy dx k xy 4 x dy dx 4x x dxÊœ œ œ œ
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xz
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'' ' '' '
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ab ˆ‰
y ; M kxyz dz dy dx xy 4 x dy dxœÊœ œ œ
256 2k 40 2
231 77
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00 0 00
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'' ' ''
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ab
#
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16x 8x x dx zœœÊœ
k 256k 8
4 105 7'
0
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15. (a) M (x y z 1) dz dy dx x y dy dx (x 2) dxœœœœ''' '' '
000 00 0
111 11 1
ˆ‰
35
##
(b) M z(x y z 1) dz dy dx x y dy dx x dx
xy
000 00 0
111 11 1œœœœ''' '' ' ""
##
ˆ‰ ˆ‰
5134
363
M M M , by symmetry x y zÊœœœ Êœœœ
xy yz xz
48
315
(c) I x y (x y z 1) dz dy dx x y x y dy dx
z
000 00
111 11œœ''' ''ab ab ˆ‰
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x 2x x dx I I I , by symmetryœœÊœœœ'
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xyz
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16. The plane y 2z 2 is the top of the wedge.œ
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121 12
14 2y2 14
ˆ‰
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#
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121 12
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M y(x 1) dz dy dx y(x 1) 2 dy dx 0;
xz
121 12
14 2y2 14œœœ''' ''
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43
1

(c) I (x 1) y z dz dy dx (x 1) 2y dy dx 45;
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24x162xy 24x
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19. (a) Let V be the volume of the ith piece, and let (x y z ) be a point in the ith piece. Then the work done?
iiii ßß
by gravity in moving the ith piece to the xy-plane is approximately W m gz (x y z 1)g V z
iii iii iiœœ ?
the total work done is the triple integral W (x y z 1)gz dz dy dxʜ
'''
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111
g xz yz z z dy dx g x y dy dx g xy y y dxœœœ'' '' '
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11 11 1
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975. Section 15.5 Masses and Moments in Three Dimensions 971
(b) From Exercise 15 the center of mass is and the mass of the liquid is the work done byˆ‰
888 5
15 15 15
ßß Ê
#
gravity in moving the center of mass to the xy-plane is W mgd (g) g, which is the same asœœ œ ˆ‰ ˆ ‰
584
2153
the work done in part (a).
20. (a) From Exercise 19(a) we see that the work done is W g kxyz dz dy dxœ '' '
00 0
2x4x
" œ
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kg xy 4 x dy dx x 4 x dx 16x 8x x dxœœœ'' ' '
00 0 0
2x 2 2
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44
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4 3 5 7 105
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gravity in moving the center of mass to the xy-plane is W mgd (g)œœ œ ˆ‰ ˆ‰
32k 8
15 7 105
256k g†
21. (a) x 0 x (x y z) dx dy dz 0 M 0œœÊ ßß œÊ œ
M
M
yz '''
R
$
yz
(b) I h dm (x h) y dm x 2xh h y dm
Lœœ œ ''' ''' '''
DD D
kk k k abvi ij
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x y dm 2h x dm h dm I 0 h m I h mœ œœ''' ''' '''
DDD
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xcm &&
22. I I mh ma ma ma
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24. M dz dy dx (4 y) dy dx 4y dx 12 dx 72;œ œ œœœ''' '' ' '
œœœ
3243 32 3 3
34 42y3 34 3 3
22
332
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26. (a) 12 2 4 12 2 x , y , zcijk ijkœ œ Êœ œ œˆ‰ˆ ‰
3 13 13 13
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976. 972 Chapter 15 Multiple Integrals
27. (a) , where m and m ; ifcœœœœ
Š‹
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34 3 8
ah3a
34
œœœ
"
Š‹Š ‹Š ‹ Š‹Š ‹kkœ
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mm mm 3 3
ah 2 a

k
11
0, or h a 3, then the centroid is on the common base
h3a
4


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È
(b) See the solution to Exercise 55, Section 15.2, to see that h a 2.œ
È
28. , where m and m s ; if h 6s 0,cœœ œœœ
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sh h s
34
s
12
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mm mm 3
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or h 6s, then the centroid is in the base of the pyramid. The corresponding result in 15.2, Exercise 56, is h 3s.œ œ
È È
15.6 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES
1. dz r dr d r 2 r r dr d 2 r d''' '' '
00r 00 0
21 2r 21 2
œœ œ
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8. 4r dr d dz 2(1 cos ) d dz 6 d 12''' '' '
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12 1cos 12 1
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00 0 00 00
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0r2 0 0r2 0
24r2 24r 2œœ
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977. Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 973
11. (a) dz r dr d'''
000
21 4r
œ
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)
(b) r dr dz d r dr dz d'' ' '''
00 0 0 30
231 224z
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(c) r d dz dr'' '
00 0
14r2
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12. (a) dz r dr d
'''
00r
212r
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21z 22 2z
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13. f(r z) dz r dr d
'''
œ

œ
œ 20 0
2 cos 3r
œ
ßß))
14. r dz dr d r cos dr d cos d''' '' '
œ

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œ 20 0 20 2
21r cos 21 2
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55
2
15. f(r z) dz r dr d 16. f(r z) dz r dr d'' ' ' ' '
00 0 20 0
2 sin 4 r sin 2 3 cos 5 r cos
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ßß ßß)) ))
17. f(r z) dz r dr d 18. f(r z) dz r dr d
'' ' '' '
œ

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21 0 2cos 0
21cos 4 2 2 cos 3 r sin
ßß ßß)) ))
19. f(r z) dz r dr d 20. f(r z) dz r dr d
''' '''
00 0 40 0
4 sec 2 r sin 2 csc 2 r sin

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21. sin d d d sin d d sin d d''' '' ' '
000 00 0 0
2 sin
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242 24 2 2
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978. 974 Chapter 15 Multiple Integrals
25. 3 sin d d d 8 sec sin d d 8 cos sec d'' ' '' '
00 sec 00 0
232 23 2
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00 0 00 0
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04 0 0 0
20 2 20 20 2œ

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6 csc 0 6 csc 6
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00 0 00 0
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30. 5 sin d d d 32 csc sin d d 32 sin csc d d''' '' ''
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## ##
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sin d d d sin d d d
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00 1 00
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979. Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 975
35. V sin d d d (1 cos ) sin d d dœœœ''' '' '
000 00 0
21cos 2 2
œœ Êœœ œœ
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040 04 0
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39. (a) 8 sin d d d (b) 8 dz r dr d''' '''
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00 0
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980. 976 Chapter 15 Multiple Integrals
45. V dz r dr d r sin dr d 9 cos (sin ) dœœœ'' ' '' '
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21r 21 2
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57. V dz r dr d 4r r sin dr d 8 1 d 16œœœœ''' '' '
000 00 0
224r sin 22 2
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000 00 0
2 2 4 r cos r sin 2 2 2
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59. The paraboloids intersect when 4x 4y 5 x y x y 1 and z 4
# # ## ##
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004r 00 0 0
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010 01 0 0
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given by the triple integral V 4 dz r dr d 4 r 2 r r dr dœœ ''' ''
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64. average r dz dr d 2r 1 r dr dœœ
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sin r r 1 r 1 2r d 0 d d (2 )œœœœœ
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000 00 0 000
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xy
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33
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r dr d d z , and x y 0, by symmetryœœœÊœœœœœ
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000 00 0 000
22r 22 2 22 r
yz
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# 84
33
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r cos dr d 4 cos d 4; M y dz r dr d r sin dr dœœœœ œ'' ' ''' ''
00 0 000 00
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982. 978 Chapter 15 Multiple Integrals
69. M ; M z sin d d d cos sin d d d 4 cos sin d dœœ œ œ
8
3
1
xy
030 030 03
222 222 22
'' ' '' ' ''
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70. M sin d d d sin d d d ;œœœœ'' ' '' '
00 0 00 0
24a 24 2
œœ œœ œ
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#

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333
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M sin cos d d d sin cos d d d
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z , and x y 0, by symmetryÊœ œ œ œ œœ
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000 00 0 000
24 r 24 2 24 r
xy
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r dr d d z , and x y 0, by symmetryœœœÊœœœœ
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œ

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31 1r 31 3
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73. I x y dz r dr d 4 r dr d 15 d 30 ; M dz r dr d
z
010 01 0 010
224 22 2 224œœ œœœ''' '' ' '''
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z
00 1 00 0
211 21 2œœœœ''' '' '
œœœ
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(b) I r sin z dz r dr d 2 2r sin dr d d
x
00 1 00 0
211 21 2œœœ''' '' '
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1
4
which is through the vertex and parallel to the base of the cone I r sin z dz r dr dʜ
x
00r
211
'''
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ab
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rr sin sin 2
3 3 20 10 40 80 10 0 5 4

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z
00 00 0
2a 2a 2
ar
ar
aœœœ'' '' '
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z
00 r 00 00
2ah 2a h 2a hœœ œ''' '' ' ''
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a a
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983. Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 979
78. (a) M z dz r dr d r dr d d ; M z dz r dr dœœœœœ ''' '' ' '''
000 00 0 000
21r 21 2 21r
xy
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r dr d d z , and x y 0, by symmetry;œœœÊœœœ
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000 00 0 000
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xy
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r dr d d z , and x y 0, by symmetry; I r dz dr dœœœÊœœœ œ
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00r 00 0 00r
211 21 2 211
xy
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ab
84
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r r dr d d z , and x y 0, by symmetry; I zr dz dr dœœœÊœœœ œ
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z
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(b) M z dz r dr d from part (a); M z dz r dr d r r dr dœœœœ ''' ''' ''
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211 211 21
xy
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ab
d z , and x y 0, by symmetry; I z r dz dr d r r dr dœœÊœœœ œ œ
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000 00 0
2a 2 2
œœ œœ œ
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555
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I sin d d d 1 cos sin d d cos d
z
000 00 0
2a 2 2œœœ''' '' '
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000 00 0
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00 0 00 0
2a ar 2a 2
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h
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hh
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h a 2ha h
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xy
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82. Let the base radius of the cone be a and the height h, and place the cone's axis of symmetry along the z-axis
with the vertex at the origin. Then M and M z dz r dr d h r r dr dœœ œ
1ah h
3 a


xy
00 r 0 0
2ah 2 a
''' ''
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h
a ))
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h r r h a a ha ha ha 3 3
24a 4 8 4 M 4 ah 4
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22 2
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1 xy
x y 0, by symmetry the centroid is one fourth of the way from the base to the vertexœœ Ê
83. M (z 1) dz r dr d h r dr d d ;œœœ œ''' '' '
000 00 0
2ah 2a 2
œœ œ
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h ah 2h ah 2h
4

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M z z dz r dr d r dr d d
xy
000 00 0
2ah 2a 2œœœ œ''' '' '
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a 2h 3h a 2h 3h
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a2h 3h
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ab
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984. 980 Chapter 15 Multiple Integrals
I (z 1)r dz dr d r dr d d ;
z
000 00 0
2ah 2a 2œœ œ œ''' '' '
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$$ œœ
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h2h h2h a
44
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ah 2h 2a
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z 1
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ab È
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84. The mass of the plant's atmosphere to an altitude h above the surface of the planet is the triple integral
M(h) e sin d d d e sin d d dœœ
''' '''
00R R00
2h h2
œœ œœ
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Ð Ñ # Ð Ñ #cR cR33
e ( cos ) dd 2 e e dd 4 e e dœœ œ'' '' '
R0 R0 R
h2 h2 h
œœ
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!!!
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ccc
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h
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ccc c c cŠ‹

ch ch ch cR cR cR
The mass of the planet's atmosphere is therefore M lim M(h) 4 .œœ
hÄ_
1.!Š‹
R2R2
cc c

œ
85. The density distribution function is linear so it has the form ( ) k C, where is the distance from the$3 3 3œ
center of the planet. Now, (R) 0 kR C 0, and ( ) k kR. It remains to determine the constant$$33œÊ œ œ
k: M (k kR) sin d d d k kR sin d dœ œ''' ''
000 00
2R 2 R
œœ œœ
3 3 939) 99)
#
’“
33
œ
43

k sin d d R cos d R d kœ œœœÊœ'' ' '
00 0 0
222
œœ œ œ
Š‹ cd
RR k k kR 3M
43 1 6 3 R


99) 9 ) )
#
%%
!
1 1
1
( ) R . At the center of the planet 0 (0) R .Êœ œÊœ œ$3 3 3 $
3M 3M 3M 3M
RR R R11 1 1
œ
ˆ‰
86. x y a sin cos sin sin a sin cos sin a sin a
222 2 22 2 2 2 22 2 22
œÊ œÊ œÊ œabababa b39) 39) 39 ) ) 39
sin a or sin a sin a or a csc , since 0 and 0.Êœ œÊœœ ŸŸ  39 39 39 3 9 91 3
87. (a) A plane perpendicular to the x-axis has the form x a in rectangular coordinates r cos a rœÊœÊœ )
a
cos )
r a sec , in cylindrical coordinates.ʜ )
(b) A plane perpendicular to the y-axis has the form y b in rectangular coordinates r sin b rœÊœÊœ )
b
sin )
r b csc , in cylindrical coordinates.ʜ )
88. ax by c a r cos b r sin c r a cos b sin c r .œÊ œÊ œÊœabab a b)) ))
c
a cos b sin ))œ
89. The equation r f z implies that the point r, , zœab a b )
f z , , z will lie on the surface for all . In particularœabab))
f z , , z lies on the surface whenever f z , , z doesab abab ab)1 )
the surface is symmetric with respect to the z-axis.Ê

90. The equation f implies that the point , , f , , lies on the surface for all . In particular, if39 39) 99) )œœab a b a b ab
f , , lies on the surface, then f , , lies on the surface, so the surface is symmetric wiith respect to theab a bab ab99) 99) 1
z-axis.
15.7 SUBSTITUTIONS IN MULTIPLE INTEGRALS
1. (a) x y u and 2x y v 3x u v and y x u x (u v) and y ( 2u v);œ œ Ê œ œ Ê œ œ
""
33


`ß
`ß
""
"
""(x y)
(u v) 9 9 3
33
2
33
2
œ œœ
$
»»

985. Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 981
(b) The line segment y x from ( 0) to (1 1) is x y 0œ!ßßœ
u 0; the line segment y 2x from (0 0) toÊœ œ ß
(1 2) is 2x y 0 v 0; the line segment x 1ß œ Ê œ œ
from (1 1) to ( 2) is (x y) (2x y) 3ß"ß œ
u v 3. The transformed region is sketched at theʜ
right.
2. (a) x 2y u and x y v 3y u v and x v y y (u v) and x (u 2v);œ œÊœ œÊœ œ
""
33


`ß
`ß
"
"
""(x y)
(u v) 9 9 3
33
2
1
33
2
œœœ
$
»»
(b) The triangular region in the xy-plane has vertices (0 0),ß
(2 0), and . The line segment y x from (0 0)ßß œßˆ‰
22
33
to is x y 0 v 0; the line segmentˆ‰
22
33
ߜʜ
y 0 from (0 0) to ( 0) u v; the line segmentœß#ßÊœ
x 2y 2 from to (2 0) u 2. Theœ ß ßÊœ ˆ‰
22
33
transformed region is sketched at the right.

3. (a) 3x 2y u and x 4y v 5x 2u v and y (u 3x) x (2u v) and y (3v u);œ œÊœ œ Êœ œ
"" "
# 510



`ß
`ß
"(x y)
(u v) 50 50 10
21
55
13
10 10
61
œ œœ
$
$
»»
(b) The x-axis y 0 u 3v; the y-axis x 0œÊœ œ
v 2u; the line x y 1Êœ œ
(2u v) (3v u) 1ʜ
""
510
2(2u v) (3v u) 10 3u v 10. Theʜʜ
transformed region is sketched at the right.

4. (a) 2x 3y u and x y v x u 3v and y v x x u 3v and y u 2v;œœÊœœÊœœ
23 1
3
12
`ß
`ß
(x y)
(u v)
œ œœ
$Î $
$$
ºº
(b) The line x 3 u 3v 3 or u 3v 3;œ Ê œ œ
x0 u3v0; yx v0; yx1œÊ œ œÊœ œ
v 1. The transformed region is the parallelogramʜ
sketched at the right.

5. x dx dy dy 1 1 y y dy'' ' '
0y2 0 0
4y21 4 4
¸"
ˆ‰ ˆ‰ˆ‰ˆ‰ˆ‰ ’“ ’ “œ œ
yxyyyyy x
2#######
"
##

y
2
y
2
(y 1 y) dy dy (4) 2œœ œœ
"""
###''
00
44

986. 982 Chapter 15 Multiple Integrals
6. 2x xy y dx dy (x y)(2x y) dx dy'' ''
RR
ab
##
œ
uv du dv uv du dv;œœ'' ''
GG
¹¹
`ß
`ß
"(x y)
(u v) 3
We find the boundaries of G from the boundaries of R,
shown in the accompanying figure:
xy-equations for Corresponding uv-equations Simplified
the boundary of R for the boundary of G uv-equations
y2x4œ
"
3
(2uv) (uv)4 v4
y 2x7 (2uv) (uv)7 v7
y x 2 ( 2u v) (u v) 2 u 2
y x 1 ( 2u v) (u v) 1 u 1
œ œ
œ œ œ
œ œ œ
œ œ œ
2
3
33
2
33
1
33
1
"
"
"
uv du dv uv dv du u du u du (4 1)Êœ œ œœœœ
"" "
(#
%"
##33 32 244
v 11 11 u 11 33''
G
'' ' '
œœœ14 1 1
27 2 2
’“ ’“ ˆ‰ ˆ‰

7. 3x 14xy 8y dx dy''
R
ab
##

(3x 2y)(x 4y) dx dyœ
''
R
uv du dv uv du dv;œœ'' ''
GG
¹¹
`ß
`ß
"(x y)
(u v) 10
We find the boundaries of G from the boundaries of R,
shown in the accompanying figure:

xy-equations for Corresponding uv-equations Simplified
the boundary of R for the boundary of G uv-equations
yx1œ
3
#
"
10 10
3
33
10 10
41020
1
41020
1
(3v u) (2u v) 1 u 2
y x 3 (3v u) (2u v) 3 u 6
y x (3v u) (2u v) v 0
y x 1 (3v u) (2u v) 1 v 4
œ œ
œ œ œ
œ œ œ
œ œ œ
#
"
""
""
uv du dv uv dv du u du u du (18 2)Êœ œœœœœ
"" "
%'
!#
10 10 10 2 5 5 2 5 5
v4 4u4 64''
G
'' ' '
20 2 2
64 6 6
’“ ’“ ˆ‰ ˆ‰

8. 2(x y) dx dy 2v du dv 2v du dv; the region G is sketched in Exercise 4'' '' ''
R GG
œ œ ¹¹
`ß
`ß
(x y)
(u v)
2v du dv 2v du dv 2v(3 3v 3v) dv 6v dv 3v 3Ê œ œ œ œ œ''
G

'' ' '
03v 0 0
133v 1 1œ
œ
cd
#
"
!
9. x and y uv v and xy u ; J(u v) v u v u ;
vuv
vu
œ œÊœ œ œßœ œœ
$
u 2u
vx (uv) v
y(xy)# # " " `ß
`ß
" #
ºº
y x uv v 1, and y 4x v 2; xy 1 u 1, and xy 9 u 3; thusœÊ œÊœ œ Êœ œÊœ œÊœ
u
v
xy dx dy (v u) dv du 2u dv du 2uv 2u ln v du''
R
Š‹ Š‹É È ˆ‰ cd
y
xvv
2u 2u
œ œœ '' '' '
11 11 1
32 32 3

#
#
"
2u 2u ln 2 du u u ln 2 8 (26)(ln 2) 8 (ln 2)œ œ œ œ'
1
3
ab œ ‘
###
$
"
22 52
33 3

987. Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 983
10. (a) J(u v) u, and
0
vu
`ß
`ß
(x y)
(u v)
œßœ œ
"
ºº
the region G is sketched at the right

(b) x 1 u 1, and x 2 u 2; y 1 uv 1 v , and y 2 uv 2 v ; thus,œÊœ œÊœ œÊ œÊœ œÊ œÊœ
"
uu
2
dy dx u dv du uv dv du u du u du'' '' '' ' '
11 11u 11u 1 1
22 22u 22u 2 2 2u
1u
y
xu 2u2u
uv v 2
œœœœ



ˆ‰ ˆ ‰ ’“


"
u du ln u ln 2; dy dx dx ln x ln 2œœœ œ†œœœ
333 13dx33
uxx2x
yy
### ###
" # #
" "'''''
11111
22222 2
1
ˆ‰ cd cd ’“

2
11. x ar cos and y ar sin J(r ) abr cos abr sin abr;
a cos ar sin
b sin br cos
œœÊœßœ œœ
$
)) ) ))
))
))
`ß
`ß
##(x y)
(r ))
ºº
I x y dA r a cos b sin J(r ) dr d abr a cos b sin dr d
!
## ### ## $## ##œœ ß œ ''
R
ab a b kk ab '' ''
00 00
21 21
œ œ
)))) )))
a cos b sin dœœœ
ab ab a a sin 2 b b sin 2
4424244
ab a b'
0
2 2
œ œ
ab ’“
## ##
!
œ
)))

)))) 1ab
12. J(u v) ab; A dy dx ab du dv ab dv du
a0
0b
`ß
`ß
(x y)
(u v)
œßœ œ œ œ œºº '' ''
R G
''
œœœ
œ11u
11u
"
" œ
œ
2ab 1 u du 2ab 1 u sin u ab sin 1 sin ( 1) ab abœœ œœœ'
œ1
1
ÈÈ
’“ cd œ ‘ˆ‰ ## "
###
" " "
"
"
u
2
11
1
13. The region of integration R in the xy-plane is
sketched in the figure at the right. The
boundaries of the image G are obtained as
follows, with G sketched at the right:

xy-equations for Corresponding uv-equations Simplified
the boundary of R for the boundary of G uv-equations
xy (u2œ
"
3
v) (u v) v 0
x 2 2y (u 2v) 2 (u v) u 2
y0 0 (uv) vu
œ œ
œ œ œ
œœ œ
1
3
33
2
1
3
"
Also, from Exercise 2, J(u v) (x 2y) e dx dy ue dv du
`ß
`ß
""(x y)
(u v) 3 3
œßœÊ œ '' ''
0y 00
23 2 2y 2 u
yx v
œ

¸¸
u e du u1 e du uu e e 22 e 2 e 1œœ œœ
"" " "
#
#
!
33 3 3
u''
00
22
vuuu22u
cd ab ab c d ’“ ab
œœœœœœ


3e 1 0.4687œ¸
"
3
ab
œ2

988. 984 Chapter 15 Multiple Integrals
14. x u and y v 2x y (2u v) v 2u andœ œ Ê œ œ
v
#
J(u v) 1; next, u x
0
`ß
`ß #
"
#
(x y)
(u v)
v
œßœ œ œ
"
"
ºº
x and v y, so the boundaries of the region ofœ œ
y
#
integration R in the xy-plane are transformed to the
boundaries of G:

xy-equations for Corresponding uv-equations Simplified
the boundary of R for the boundary of G uv-equations
xuœœ
y v
##
v
y vv
#
###
u0
x2 u 2 u2
y0 v0 v0
y2 v2 v2
œ
œ œ œ
œœ œ
œœ œ y (2x y) e dx dy v (2u) e du dv v e dv v e 1 dvÊœ œ œ'' '' ' '
0y2 00 0 0
2y22 22 2 2
16

$ÐÑ $ $ $ ""
#
!
2x y 4u 4u
44

’“ ab
e1 e1œ œ
"
#
!
44
v
ab ’“
16 16

15. (a) x u cos v and y u sin v u cos v u sin v u
cos v u sin v
sin v u cos v
œœÊœ œœ
$
`ß
`ß
##(x y)
(u v)
ºº
(b) x u sin v and y u cos v u sin v u cos v u
sin v u cos v
cos v u sin v
œœÊœ œœ
$
`ß
`ß
##(x y)
(u v)
ºº
16. (a) x u cos v, y u sin v, z w u cos v u sin v u
cos v u sin v 0
sin v u cos v 0
00
œœœÊœ œœ
$
"
`ßß
`ßß
##(xyz)
(uvw)
ââ
ââ
ââ
ââ
ââ
ââ
(b) x 2u 1, y 3v 4, z (w 4) (2)(3) 3
200
030
00
œ œ œ Ê œ œ œ
""
#`ßß #
`ßß
"
#
(xyz)
(uvw)
ââ
ââ
ââ
ââ
ââ
ââ
ˆ‰
17.
sin cos cos cos sin sin
sin sin cos sin sin cos
cos sin 0
ââ
ââ
ââ
ââ
ââ
ââ
9)39)39)
9)3 9)39 )
939
$
$
(cos ) ( sin )
cos cos sin sin sin cos sin sin
cos sin sin cos sin sin sin cos
œ
$$
939
39)39) 9)39)
39)39) 9)39)
ºººº
cos sin cos cos sin cos sin sin sin cos sin sinœaba baba b3 9 99 ) 99)39 9 ) 9)
########
sin cos sin sin cos sin sin œœ œ3 9 93 93 9 9 93 9
###$# ###
aba b
18. Let u g(x) J(x) g (x) f(u) du f(g(x))g (x) dx in accordance with Theorem 6 inœÊœœ Ê œ
du
dx
ww ''
ag(a)
bg(b)
Section 5.6. Note that g (x) represents the Jacobian of the transformation u g(x) or x g (u).
w "
œœ
19. dx dy dz dy dz (y 1) dy dz''' '' ''
00y2 00 00
341y2 34 34 y2
y2


ˆ‰ œ ‘’“
2x y xy yzxxz z
323 3

####
ÊœÄ
"
œ œ


dz dz 2 dz 2z 12œœœœœ'''
000
333
’“ ’“ ˆ‰ˆ‰
(y 1) y yz
443 434 3 3
94z 4z 2zœ
% $
! !
"

20. J(u v w) abc; the transformation takes the ellipsoid region 1 in xyz-space
a00
0b0
00c
ßß œ œ Ÿ
ââ
ââ
ââ
ââ
ââ
ââ
xz
abc
y



into the spherical region u v w 1 in uvw-space which has volume V
## #
Ÿ œ ˆ‰
4
3
1

989. Chapter 15 Practice Exercises 985
V dx dy dz abc du dv dwÊœ œ œ ''' '''
R G
4abc
3
1
21. J(u v w) abc; for R and G as in Exercise 19, xyz dx dy dz
a00
0b0
00c
ßß œ œ
ââ
ââ
ââ
ââ
ââ
ââ
kk '''
R
a b c uvw dw dv du 8a b c ( sin cos )( sin sin )( cos ) sin d d dœœ'''
G
### ### # '''
000
221
œœ
39 )39)3 93 939) ab
sin cos sin cos d d sin cos dœœœ
4abc abc abc
336

'' '
00 0
22 2
œœ œ
)) 999) )))
$
22. u x, v xy, and w 3z x u, y , and z w J(u v w) ;
100
0
00
œœ œÊœœ œ Êßßœ œ $
v
u3 3u
v
uu
3
"" "
"
ââ
ââ
ââ
ââ
ââ
ââ

x y 3xyz dx dy dz u 3u J(u v w) du dv dw v du dv dw''' '''
R G
ab kkœ ‘ˆ‰ˆ‰ ˆ‰ˆ ‰
## "
œ ßßœ
vvw vw
uu3 3 u '''
001
322
(v vw ln 2) dv dw (1 w ln 2) dw (1 w ln 2) dw w ln 2œ œ œ œ
""
#$
!!
332332
v2 2w'' ' '
00 0 0
32 3 3
’“ ’ “

3 ln 2 2 3 ln 2 2 ln 8œ œ œ
29
3
ˆ‰
#
23. The first moment about the xy-coordinate plane for the semi-ellipsoid, 1 using the
xz
abc
y



œ
transformation in Exercise 21 is, M z dz dy dx cw J(u v w) du dv dw
xyœœßß''' '''
D G
kk
abc w du dv dw abc M of the hemisphere x y z 1, z 0 ;œ œ œ  œ
## ###'''
G
aba b†
xy
abc
4

1
the mass of the semi-ellipsoid is z c
2abc abc 3 3
3 4 2abc 8
11
1
Êœ œŠ‹ ˆ‰

24. A solid of revolutions is symmetric about the axis of revolution, therefore, the height of the solid is solely a function of r.
That is, y f x f r . Using cylindrical coordinates with x r cos , y y and z r sin , we haveœœ œ œ œab ab ))
V r dyddr r dyddr ddr r fr ddr dr r y r f rœœ œ œ œ
'''
G
)))) )'' ' '' '' '
a0 0 a0 a0 a
b2 b2 b2 b
2
œœœ
œfr
00
fr
ab
ab
cd ab c d ab
rf r dr. In the last integral, r is a dummy or stand-in variable and as such it can be replaced by any variable name.
'
a
b
21ab
Choosing x instead of r we have V xf x dx, which is the same result obtained using the shell method.œ
'
a
b
21ab
CHAPTER 15 PRACTICE EXERCISES
1. ye dx dy e dy
'' '
10 1
10 1 y 10
y



xy xy
œ cd
"
(e 1) dy 9e 9œœ'
1
10

990. 986 Chapter 15 Multiple Integrals
2. e dy dx x e dx'' '
00 0
1x 1
x
&
&
yx yxÎÎ
œ œ‘

xe x dx eœœœ'
0
1
Š‹’ “
xx
2
xe2

"
##
"
!

3. t ds dt ts dt'' '
094t 0
32 9 4t 32
94t
94t

œœ
œ
œœ
"
"
"
"
œ
œ
œ
œ
œ cd
2t 9 4t dt 9 4tœœ'
0
32

È
’“ab # " #
$Î#
$Î#
!
6
09œ œ œ
"$Î# $Î#
#66
27 9
ˆ‰

4. xy dx dy y dy'' '
0y 0
12y 1 2y
y"
" "
"
œ œ
œ ’“
x
2

y44yyy dyœ
"
#'
0
1
ˆ‰È
2y 2y dy yœ œœ'
0
1
ˆ‰ ’“
$Î# #
"
!
"4y
55
$

5. dy dx x 2x dx
'' '
œ ¸œ
œ22x4 2
04x 0
œ
œab
#
x4œ œ œ’“ ˆ‰
x84
333
œ
#
!
#
dx dy dy'' '
04y 0
4y4)/2 4
y4
2œœ
œ
œ
"
(
œˆ‰ È4y
48 4 2y 4 yœœ†$$ $’“ ab
y3/2
23
2 3/2
4
0
2
3
2
4œ œ
16 4
33

6. x dx dy x dy'' '
0y 0
1y 1
y
y
"
"
È œ ‘œ
2
3
$Î#
y y dy y yœœ
2242
3375'
0
1
ˆ‰ œ ‘
$Î% $Î# (Î% &Î#
"
!
œœ
24 2 4
37 5 35
ˆ‰
x dy dx x x x dx x x dx'' ' '
0x 0 0
1x 1 1
2
È ab ˆ‰œœ
1/2 2 3/2 5/2
xxœ œœœ ‘
22 224
57 5735
5/2 7/2
0
1

7. y dy dx dx'' '
œœ

30 3
3129x 3 12 9 x
" " œ œ
œ’“
y
2
Ð

9 x dxœœ'
œ3
3
" #
$
$
8824
9x x
ab ’“
œ
œœœˆ‰ˆ ‰
27 27 27 27 27 9
824 824 6 #
y dx dy 2y dy'' '
094y 0
32 9 4y 32

œ
*
"
" œ
œ
œ È94y
9œ † œ † œ œ
""
#43 6 6
2279 3/2
0
3/2
3/2
ab º94y
*

991. Chapter 15 Practice Exercises 987
8. 2x dy dx 2xy dx'' '
00 0
24x 2
4x
0
œ
œ
2
2
œcd
2x 4 dx 8x 2 dxœœ
''
00
22
23
ababab xx 4168œ œœ’“ x
2
2x16
22
4
!
2x dx dy x dy'' '
00 0
44y 4
4y
0
"
"
œ
œ
œcd
2
4 y dy 16 8 4yœœ œœ $'
0
4
ab ’“
y
2
0
4
16
2
2

9. 4 cos x dx dy 4 cos x dy dx 2x cos x dx sin x sin 4
'' '' '
02y 00 0
12 2x/2 2
ab ab ab c d ab
####
#
!
œœœœ
10. e dx dy e dy dx 2xe dx e e 1
'' '' '
0y2 00 0
21 12x 1
xxxx

œœœœ
œœœœ cd
"
!
11. dy dx dx dy dy'' '' '
0x 00 0
82 2y 2
&
&
"
"""
œœœy1 y1 y1 44
4y ln 17

œ œœœ
12. dx dy dy dx 2 x sin x dx cos x ( 1) ( 1) 2'' '' '
0y 00 0
11 1x 1
&
&
"

2 sin x 2 sin x
xx
11 11ab ab


œœœœœ 11 1ab c d ab
##
"
!
13. A dy dx x 2x dx 14. A dx dy y 2 y dyœœœ œœœ'' ' '' '
œ ¸œ œ
œ2 2x4 2 1 2y 1
04x 0 4 y 4
œ
ab ˆ‰È
# 4 37
3 6
"
15. V x y dy dx x y dx 2x dxœ œ œœ'' ' '
0x 0 0
12x 1 1 2x
x
œ œ
ab ’“ ’ “’ “
## # #
"
!
y (2x) (2x)
3 3 3 3 12 12
7x 2x 7x
œœ œœ  
œ œˆ‰
2724
31212 12 3
"

16. V x dy dx x y dx 6x x x dxœœœœ'' ' '
œœœ
œ
œ3x 3 3
26x 2 2
6x
x
œ
œ
# # #%$
cd a b
125
4
17. average value xy dy dx dx dxœœœœ '' ' '
00 0 0
11 1 1
’“
xy
224
x
"
!
"
18. average value xy dy dx dx x x dxœœœœ
" "$
#ˆ‰
œ
4
'' ' '
00 0 0
11x 1 1 1x
" "
œ œ
œ œ

42 xy
2111
’“ ab

19. M dy dx 2 dx 2 ln 4; M x dy dx x 2 dx 1;œœœœ œœ'' ' '' '
12x 1 12x 1
22 2 22 2
y
ˆ‰ ˆ‰
22
xx

M y dy dx 2 dx 1 x y
x
12x 1
22 2œœœÊœœ'' '

ˆ‰
2
xln 4

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#
20. M dx dy 4y y dy ; M y dx dy 4y y dy ;œ œœ œ œ œœ'' ' '' '
02y 0 02y 0
42yy 4 42yy 4
xœ œ
œ œ
œ œ
ab a b ’“
##$
%
!
32 64
3343
4y y
œ
M x dx dy 2y dy x and y 2
y
02y 0
42yy 4œœœœÊœœœœ'' '
œ
œ
œ
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ab2y y yy
10 2 5 M 5 M
128 12 M M
#
#
%
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21. I x y (3) dy dx 3 4x dx 104
o
02x 0
24 2œ œ œ'' 'ab Š‹
## # 64 14x
33
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22. (a) I x y dy dx 2x dxo
240
33œœœ'' '
œœ œ21 2
21 2
ab ˆ‰
## #
(b) I y dy dx dx ; I x dx dy dy I I Ixy oxy
2b 4ab 2a 4a b
33 33œœœœ œœÊœ'' ' '' '
œœ œ œœ œab a ba b
ab a ba b
##
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œœ
4ab 4a b
33 3
4ab b a
œœ abœ

992. 988 Chapter 15 Multiple Integrals
23. M dy dx dx 3 ; I y dy dx x dx 2 Rœœœœ œœœÊœ$$$$ $'' ' '' '
00 0 00 0
32x3 3 32x3 3

2x 8 8 3 2
3818143
x x
#$ $$
ˆ‰Š‹ É

24. M (x 1) dy dx x x dx ; M y(x 1) dy dx xxxx dx ;œ œ œ œ œ œ'' ' '' '
0x 0 0x 0
1x 1 1x 1
œ œ
ab a b
$$&#% """
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x
3
M x(x 1) dy dx x x dx x and y ; I y (x 1) dy dxy x
2813
15 15 30œœœÊœœœ '' ' ''
0x 0 0x
1x 1 1x
œ œ
ab
#% #
xxxx dx R ; I x(x1) dydx xx dxœ œ Ê œ œ œ œ
" %($' # $&
3 280 M 70
17 17
xy
I''''
0 0x 0
1 1x 1
ab ab É É
x
œ
RœÊ œ œ
11
12 M 3
y
I
É É
y
25. M x y dy dx 2x dx 4; M y x y dy dx 0 dx 0;œ œœœ œœ'' ' '' '
œœ œ œœ œ11 1 11 1
11 1 11 1
ˆ‰ ˆ‰ ˆ‰
## # ## " "
33 3
4
x
M x x y dy dx 2x x dx 0y
33
4œœœ'' '
œœ œ11 1
11 1
ˆ‰ ˆ‰
## $ "
26. Place the ABC with its vertices at A(0 0), B(b 0) and C(a h). The line through the points A and C is? ßß ß
y x; the line through the points C and B is y (x b). Thus, M dx dyœœœ
hh
aab ''
0ayh
habyhb

$
b 1 dy ; I y dx dy b y dy ; Rœœœ œ œœœ$$$ ''''
00ayh0
hhabyhbh
x
ˆ‰ Š‹ É
y y
h h1 M
bh bh h
x
I
6
$ $
# #
##


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x
È
27. dy dx dr d d d'' '' ' '
œœœ
œ11x 00 0 0
11x 21 2 2
"
" œ
œ

22r
1x y 1r 1rab abœœ œ # œ
""
"
!


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œœœ
)))1œ ‘
28. ln x y 1 dx dy r ln r 1 dr d ln u du d u ln u u d'' '' '' '
œœœ
œ11y 00 01 0
11y 21 22 2"
" œ
œ
ab ab cd
## # ""
##
#
"
œ œ œ
œœœ
)) )
(2 ln 2 1) d [ln (4) 1]œœ
"
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2
œ
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29. M r dr d d 3 ; M r cos dr d 9 cos d 9 3 x ,œœœœ œ œÊœ'' ' '' '
œ

œœ œ œ
œœ œ œ30 3 30 3
33 3 33 3
y
))1 )) ))
9 33
#
# È
È
1
and y 0 by symmetryœ
30. M r dr d 4 d 2 ; M r cos dr d cos d x , andœœœœ œ œÊœ'' ' '' '
01 0 01 0
23 2 23 2
y
œœ œ œ
))1 )) ))
# 26 26 13
333 1
y by symmetryœ
13
31
31. (a) M 2 r dr d œ ''
01
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)
2 cos d ;œœ '
0
2
œ
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1cos 2 8
4
œœ
#
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M (r cos ) r dr d
y
21
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œ

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œ
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cos cos dœ '
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2
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3

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x , andœÊœ
32 15 15 32
24 6 48
œœ
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11
1
y 0 by symmetryœ
(b)
32. (a) M r dr d d a ; M (r cos ) r dr d dœœœœ œœ'' ' '' '
œœ œ œ%% % %
%% % %00
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a a cos 2a sin
33
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2a sin 2a sin
33
!!
!!
!1 !1ÄÄ
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993. Chapter 15 Practice Exercises 989
(b) x and y 0œœ
2a
51

33. x y x y 0 r r cos 2 0 r cos 2 so the integral is dr dabab
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24

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R

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!
ab ab ab1x y 1r
r
21 r


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00 0
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sec du


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R
b
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r
21 r
0
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00
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35. cos (x y z) dx dy dz [sin (z y ) sin (z y)] dy dz''' ''
000 00
œœœ œœ
œ 1
[ cos (z 2 ) cos (z ) cos z cos (z )] dz 0œ œ
'
0
œ
11 1
36. e dz dy dx e dy dx e dx 1
''' '' '
ln 6 0 ln 4 ln 6 0 ln 6
ln 7 ln 2 ln 5 ln 7 ln 2 ln 7
Äœœ_ Äœ_xyz xy x
œœœ37. (2x y z) dz dy dx dy dx dx''' '' '
00 0 00 0
1x xy 1x 1
œ œ
%
œ œ œ Š‹ Š‹
3x 3x x 83y
35
'"
## ##
38. dy dz dx dz dx ln x dx x ln x x 1''' '' '
110 11 1
exz ex e
e
1
2y
zz
œ œœœœ
"
cd
39. V 2 dz dx dy 2 2x dx dy 2 cos y dy 2œœœœœ'' ' '' '
0cos y0 0cos y 0
20 2x 20 2 2
œœœ œ
œœ



#
#
’“
ysin 2y
24
1
40. V 4 dz dy dx 4 4 x dy dx 4 4 x dxœœœ'' ' '' '
00 0 00 0
24x4x 24x 2
""
œœ œœœ œ
ab ab
##
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x 4 x 6x 4 x 24 sin 24 sin 1 12œ œ œ’“ab
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x
2
1
41. average 30xz x y dz dy dx 15x x y dy dx 15x x y dx dyœœœ
""" ###
333''' '' ''
000 00 00
131 13 31
ÈÈÈ
5 x y dy 5(1 y) 5y dy 2(1 y) 2y 2(4) 2(3) 2œœœœ
"" "" # $Î# $Î# &Î# &Î# &Î# &Î#
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00
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994. 990 Chapter 15 Multiple Integrals
42. average sin d d d sin d d dœœœœ
33a3a3a
4a 16 8 4111
œ
''' '' '
000 00 0
2a 2 2
œœ œœ œ
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43. (a) 3 dz dx dy'' '
œœœ %
œœœ
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"
""
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22y 4xy
œœœ
œœœ
(b) 3 sin d d d'' '
00 0
242
œœ
3939)
#
(c) 3 dz r dr d 3 r 4 r r dr d 3 4 r d'' ' '' '
00 r 00 0
224r 22 2
œœ œ
"" "
œ
œ
)))œœ’“’ “ab ab
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r
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224 d842d2842œ œ œ ''
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22
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44. (a) 21(r cos )(r sin ) dz r dr d 21r cos sin dz r dr d
''' '''
œ

œœ
œœ20 r 20 r
21r 21r
œ œ
œ œ
)) ) )) )
#$#
œ
(b) 21r cos sin dz r dr d 84 r sin cos dr d 12 sin cos d 4
''' '' '
œ

œ
œœœ20 r 0 0 0
21r 21 2
œ
œ
$# '# #
)) ) ))) )))œœœ
45. (a) sin d d d
'' '
00 0
2 4 sec
œœ "
3939)
#
(b) sin d d d (sec )(sec tan ) d d tan d d'' ' '' ' '
00 0 00 0 0
24sec 24 2 2
4
œœ Êœœ œœ

3939) 9 999) 9 ) )
# #""""
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33263
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1
1
46. (a) (6 4y) dz dy dx (b) (6 4r sin ) dz r dr d'' ' ' ''
00 0 000
1 1x xy 21r
" "œ œœœ

œ
))
(c) (6 4 sin sin ) sin d d d
'''
040
2 2 csc
œœ "
œ


39)3 939)ab
#
(d) (6 4r sin ) dz r dr d 6r 4r sin dr d 2r r sin d''' '' '
000 00 0
21r 21 2
œœœ
œœ)) )) )) abcd
#$ $%
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0
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47. z yx dz dy dx z yx dz dy dx'' ' ' ' '
01x1 10 1
13x 4xy 33x 4xy
"
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œ
œœœ œœœœ
œœœ œœœ
##

48. (a) Bounded on the top and bottom by the sphere x y z 4, on the right by the right circular
###
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cylinder (x 1) y 1, on the left by the plane y 0œ œ
##
(b) dz r dr d'' '
00 4r
22 cos 4r
œ œ
œœ
"
"
œ
œ
)
49. (a) V dz r dr d r 8 r 2r dr d 8 r r dœœœ''' '' '
002 00 0
22 8r 22 2
œœ œ
"
œ
œ
))) Š‹’ “
È
ab
# " ##
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#
!
3
(4) 4 (8) d 2328 d 425 dœ œ œ œ'' '
00 0
22 2
œœ œ
œ ‘ Š‹Š‹
È È
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33 3 3 3
44
8425
)))
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(b) V sin d d d 2 2 sin sec sin d dœœ '' ' ''
00 2 sec 00
248 24
œœ œœ
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3
Š‹
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2 2 sin tan sec d d 2 2 cos tan dœœ
88
33'' '
00 0
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88
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1
50. I ( sin ) sin d d d sin d d d
z
00 0 00 0
232 232œœ'' ' '' '
œœ œœ
393 939) 3 939)
## % $
ab
sin cos sin d d cos dœ œœ
32 32 8
5533
cos'' '
00 0
23 2
œœ œ
ab ’“9999) 9 )
#
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9
1
1

995. Chapter 15 Additional and Advanced Exercises 991
51. With the centers of the spheres at the origin, I ( sin ) sin d d d
z
00a
2bœ'''
œœ
$3 9 3 9 3 9 )
##
ab
sin d d sin cos sin d dœœ
$$ab abba ba
55
$$ $$
$#'' ''
00 00
22
œœ œœ
99) 9 9 9 9) ab
cos d dœœ œ
$$1$ 9
1
ab ab abba 4ba 8 ba
5 3 15 15
cos
$$ $$ $$ œ

!''
00
22
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52. I ( sin ) sin d d d sin d d d
z
000 000
21cos 21cos œœ''' '''
œœ œ œ
393 939) 3 939)
## % $
ab
(1 cos ) sin d d (1 cos ) (1 cos ) sin d d ;œ œ
" &$ '
5'' ''
00 00
22
œœ œœ
999) 9 999)
u(2u) dud d 2 d
u1cos
du sin d
”• ’“ ˆ‰
œ
œ
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9
99
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00 0 0
22 2 2
œœœ

d dœœœ
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556 35 35
2 2 32 64''
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53. x u y and y v x u v and y vœ œ Ê œ œ
J(u v) 1; the boundary of the
0
Êßœ œ
""
"
ºº
image G is obtained from the boundary of R as
follows:

xy-equations for Corresponding uv-equations Simplified
the boundary of R for the boundary of G uv-equations
yx vuv uœœ œ
œœ œ
0
y0 v0 v0
e f(x y y) dy dx e f(u v) du dvÊßœ ß
'' ''
00 00
x

ÐÑsx s u v
54. If s x y and t x y where ( ) ac b , then x , y ,œ œ œ œ œ!" #$ !$"#
##

$"
!$ "# !$ "#
#!st st
and J(s t) e ds dtßœ œ Ê
$
$
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()
st
ac b
!$ "# !$ "#


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ab
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re dr d d . Therefore, 1 ac b .œœœ œÊœ
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ÈÈÈÈ
acb acb acb acb
r


'' '
00 0
22
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)) 1
11
CHAPTER 15 ADDITIONAL AND ADVANCED EXERCISES
1. (a) V x dy dx (b) V dz dy dxœœ
'' '' '
œ œ
œ œ3x 3x 0
26x 26x x
œ œœ
#
(c) V x dy dx 6x x x dx 2xœœœœ'' ''
œœ
œœ3x 3x
26x 26x
œœ
# #%$ $
#
$
ab ’“
x x 125
54 4
$
2. Place the sphere's center at the origin with the surface of the water at z 3. Thenœ
9 25 x y x y 16 is the projection of the volume of water onto the xy-planeœ Ê œ
## ##

996. 992 Chapter 15 Multiple Integrals
V dz r dr d r 25 r 3r dr d 25 r r dÊœ œ œ ''' '' '
00 25r 00 0
24 3 24 2
œœ œ
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œ
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26 52
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3. Using cylindrical coordinates, V dz r dr d 2r r cos r sin dr dœœ ''' ''
000 00
2 1 2 r cos sin 2 1
œ œ"%$
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##
1 cos sin d sin cos 2œ œ œ'
0
2
œ
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"" ""
#
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))))) )1
1
4. V 4 dz r dr d 4 r 2 r r dr d 4 2 r dœœœ''' '' '
00r 00 0
21 2r 21 2
œœ œ
œ
œ
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È
ab
# $# " $Î#
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r

4 d dœ œ œ''
00
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34 3 3 6
22 827
82 7ÈÈ Š‹
È
))
1
5. The surfaces intersect when 3 x y 2x 2y x y 1. Thus the volume isœ Ê œ
## # # ##
V 4 dz dy dx 4 dz r dr d 4 3r 3r dr d 3 dœœœœœ'' ' ' '' ' ' '
0 0 2x 2y 0 0 2r 0 0 0
1 1x 3xy 213r 21 2
"
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œœ œ

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))) ab
$
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31
6. V 8 sin d d d sin d dœœ ''' ''
000 00
222 sin 22
œœ Êœœ
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3
sin d d 16 d 4 d 2œ œ œ œ
64 3
344 24
sin cos sin 2''' '
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99 9 9
1
1
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7. (a) The radius of the hole is 1, and the
radius of the sphere is 2.

(b) V 2 r dr dz d 3 z dz d 2 3 d 4 3œœœœ '' ' '' '
00 1 00 0
234z 23 2
œœ œ
"" "
œ
œ
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ÈÈ
#
8. V dz r dr d r 9 r dr d 9 r dœœœ'' ' '' '
00 0 00 0
3 sin 9 r 3 sin 3 sin
œ
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# " #
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3
9 9 sin (9) d 9 1 1 sin d 9 1 cos dœ œ œ '''
000
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33
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9. The surfaces intersect when x y x y 1. Thus the volume in cylindrical
## ## œœ
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xy1

coordinates is V 4 dz r dr d 4 dr d 4 dœ œ œ ''' '' '
00r 00 0
21 r12 21 2
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rr r r
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10. V dz r dr d r sin cos dr d sin cos dœœœ''' '' '
010 01 0
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997. Chapter 15 Additional and Advanced Exercises 993
11. dx e dy dx e dx dy lim e dx dy'''''''
00aa0a0
bb bt
!! !
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x
xy xy xy
ax bx

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tÄ_
lim dy lim dy dy ln y lnœœ œœœ'' '
aa a
bb b t
b
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ee b
yyyya
xy yt
!
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12. (a) The region of integration is sketched at the right
ln x y dx dyÊ ''
0y cot
a sin a y
&
&
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ab
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r ln r dr d ;œ ''
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a
&
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ur
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(b) ln x y dy dx ln x y dy dx'' ''
0 0 a cos 0
a cos (tan )x a a x
&&
&
ab ab
## ##

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13. e f(t) dt du e f(t) du dt (x t)e f(t) dt; also'' '' '
00 0t 0
xu xx x
mx t mx t mx tÐÑ ÐÑ ÐÑ
œœ
e f(t) dt du dv e f(t) du dv dt (v t)e f(t) dv dt
''' ''' ''
000 0t t 0t
xvu xxv xx
mx t mx t mx tÐÑ ÐÑ ÐÑ
œœ
(v t) e f(t) dt e f(t) dtœ œ''
00
xx
x
t
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" #ÐÑ ÐÑ
#2
mx t mx t (x t)

14. f(x) g(x y)f(y) dy dx g(x y)f(x)f(y) dy dx'' ''
00 00
1x 1x
Š‹ œ
g(x y)f(x)f(y) dx dy f(y) g(x y)f(x) dx dy;œ œ
'' ' '
0y 0 y
11 1 1
Œ
g x y f(x)f(y) dx dy g(x y)f(x)f(y) dy dx g(y x)f(x)f(y) dy dx
'' '' ''
00 00 0x
11 1x 11
abkkœ
g(x y)f(x)f(y) dx dy g(y x)f(x)f(y) dy dxœ
'' ''
0y 0x
11 11
g(x y)f(x)f(y) dx dy g(x y)f(y)f(x) dx dyœ '' ''
0y 0y
11 11 ðóóóóóóóóóóóóñóóóóóóóóóóóóò
simply interchange x and y
variable names
2 g(x y)f(x)f(y) dx dy, and the statement now follows.œ
''
0y
11
15. I (a) x y dy dx x y dx dx
o
00 0 0
axa a a xa aœœœœ'' ' '


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## #
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xx x x
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a ; I (a) a a 0 a a . Since I (a) a 0, theœ œ œÊ œÊœ œ œ
a
41 6 3 3 3


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o o
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value of a does provide a for the polar moment of inertia I (a).minimum
o
16. I x y (3) dy dx 3 4x dx 104
o
02x 0
24 2œ œ œ'' 'ab Š‹
## # 14x 64
33
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998. 994 Chapter 15 Multiple Integrals
17. M r dr d sec dœœ'' '
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b sec
a
)))Š‹
ab

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#
a b tan a cos bœ œ
## #" #
)) ˆ‰ Š‹
b
ab
ab
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a
o
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a
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b
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18. M dx dy 1 dy y ; M x dx dyœœœœœ'' ' ''
œœ
21y4 2 21y4
22y2 2 22y2
y $% $%
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yy
4123
8
œ #
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dy 4 y dy 16 8y y dy 16yœœœœ'' '
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22 2 2y2
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x33 3
23232 1635
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32 x , and y 0 by symmetryœœ œÊœœ œ œ
36432332848 4836
16 3 5 16 15 15 M 15 8 5
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† y
19. e dy dx e dy dx e dx dy'' '' ''
00 00 00
ab abxa bayb
max b x a y b x a yab

ß
œ

x e dx y e dy e e e 1 e 1œœœ''
00
ab ab
ˆ‰ ˆ‰ ’“’“Š‹Š‹
ba
a b 2ab 2ba ab ab
bx ay bx ay ba ab

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!!
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ab
ab
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20. dx dy dy dx F(x y) F(x y)'' ' '
yx y y
yx y y x
x
y
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xy y y y
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F(x y ) F(x y ) F(x y ) F(x y )œßßßß
"" !" "! !!
21. (a) (i) Fubini's Theorem
(ii) Treating G(y) as a constant
(iii) Algebraic rearrangement
(iv) The definite integral is a constant number
(b) e cos y dy dx e dx cos y dy e e sin sin 0 (1)(1) 1'' ' '
00 0 0
ln 2 2 ln 2 2
xx ln 20
œœ
œœœœŒ Œ ab ˆ‰
1
#
(c) dx dy dy x dx 1 0'' ' '
11 1 1
21 2 1œœ
xx
yy y2

œœœœŒ Œ ’“’“ ˆ‰ˆ‰
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22. (a) f x y D f u x u y; the area of the region of integration is ™œ Ê œ ij
u "#
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average 2 (u x u y) dy dx 2 u x(1 x) u (1 x) dxÊœ œ '' '
00 0
11x 1
œ
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#
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23 3 6 6 3
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(b) average (u x u y) dA x dA y dA u u u x u yœœ œœ
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area area area M M
uu M M'' '' ''
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23. (a) I e dx dy e r dr d lim re dr d
#
œœœ'' ' ' ' '
00 0 0 0 0
xy r r
22b

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ab ”•))
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lim e 1 d d Iœ œ œ Ê œ
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00 0
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999. Chapter 15 Additional and Advanced Exercises 995
24. Q kr (1 sin ) dr d (1 sin ) d cos œœœœ'' '
00 0
2R 2
œœ
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kR kR 2 kR
333
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cd
1 1
25. For a height h in the bowl the volume of water is V dz dy dxœ '' '
œœœ %
œ
"
"
"
"hhxxy
hhx h
œœœ
œ
h x y dy dx h r r dr d d d .œœœœœ'' '' ' '
œœœ
œ
"
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"
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hhx 2 h 2 2 h
œ
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ab ab ’“
## #
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)))
hr r h h
24 4

È
1
Since the top of the bowl has area 10 , then we calibrate the bowl by comparing it to a right circular cylinder1
whose cross sectional area is 10 from z 0 to z 10. If such a cylinder contains cubic inches of water1 œœ
h

1
#
to a depth w then we have 10 w w . So for 1 inch of rain, w 1 and h 20; for 3 inches of1œÊœ œ œ
hh
20

1
# È
rain, w 3 and h 60.œœ
È
26. (a) An equation for the satellite dish in standard position
is z x y . Since the axis is tilted 30°, a unitœ
""
##
##
vector 0 a b normal to the plane of thevijkœ
water level satisfies b cosœœ œvk† ˆ‰
1
6
3
È
#
a1b Êœ œÊœ
È
# ""
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vjk
È
3
(y 1) z 0Ê œ
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È
3
ˆ‰
zyʜ
"""
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33
Š‹
is an equation of the plane of the water level. Therefore

the volume of water is V dz dy dx, where R is the interior of the ellipseœ '''
R
11
22
111
332
xy
y
œœ
%
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x y y 1 0. When x 0, then y or y , where
##

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3 3
41
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24 2
39 3
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and V 1 dz dx dy"œÊœ
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39 3

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22 11
322 3
22 111
32 333
y1 y x y
y1 y y
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(b) x 0 z y and y; y 1 1 the tangent line has slope 1 or a 45 ° slantœÊœ œ œÊ œÊ
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# dz dz
dy dy
at 45 ° and thereafter, the dish will not hold water.Ê
27. The cylinder is given by x y 1 from z 1 to z r z dV
## ##
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D
ab
dz r dr d lim dz dr dœœ''' '''
001 001
21 21a
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lim dr d lim dr dœ œ
aaÄ_ Ä_
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21 21 a
1
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rrr
rz ra r1ab ab ab


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lim r a r 1 d lim 1 a 2 a dœœ
aaÄ_ Ä_
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lim 2 1 a 2 .œœ
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28. Let's see?
The length of the "unit" line segment is: L 2 dx 2.œœ
'
0
1
The area of the unit circle is: A 4 dy dx .œœ ''
00
11x
"
œ 2
1
The volume of the unit sphere is: V 8 dz dy dx .œœ '' '
00 0
11x 1xy
" "œœœ 222
4
3
1
Therefore, the hypervolume of the unit 4-sphere should be:
V 16 dw dz dy dx.hyperœ'' ' '
00 0 0
11x 1xy 1xyz
" ""œœœœœœ222222
Mathematica is able to handle this integral, but we'll use the brute force approach.

1000. 996 Chapter 15 Multiple Integrals
V 16 dw dz dy dx 16 dz dy dxhyperœœ'' ' ' '' '
00 0 0 00 0
11x 1xy 1xyz 11x 1xy
222
" """ "œœœœœœ œœœ222222 222
È1xyz
16 dz dy dx
cos
dz 1 x y sin d
œœ
œ
œ '' '
00 0
11x 1xy
22 z
1x y
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2
22
È É1x y1
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z
1x y
22
È
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)
))
16 1 x y 1 cos sin d dy dx 16 1 x y sin d dy dxœ œ'' ' '' '
00 /2 00 /2
11x 0 11x 0
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œ œ 2 2
ab ab
È
22 22 22
œ œ
))) ))
16 1 x y dy dx 4 1 x x 1 x 1 x dxœ œ '' '
00 0
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22 2 2 22
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4 1 x dx 1 x dx sin d1x
x cos
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33
24
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d 1 2 cos 2 cos 2 d 2 cos 2 dœ œ œ œ
81cos 2 2 23 cos 4
32 3 3 22
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00 0
ˆ‰ ˆ ‰ ab

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2

1001. CHAPTER 16 INTEGRATION IN VECTOR FIELDS
16.1 LINE INTEGRALS
1. t ( t) x t and y 1 t y 1 x (c)ri jœ" Ê œ œÊ œÊ
2. t x 1, y 1, and z t (e)rij kœ Ê œ œ œ Ê
3. (2 cos t) (2 sin t) x 2 cos t and y 2 sin t x y 4 (g)rijœ Êœ œÊœÊ
##
4. t x t, y 0, and z 0 (a)riœÊœ œ œÊ
5. t t t x t, y t, and z t (d)rijkœ Ê œ œ œÊ
6. t (2 2t) y t and z 2 2t z 2 2y (b)rj kœ Ê œ œ Ê œ Ê
7. t 1 2t y t 1 and z 2t y 1 (f)rjkœ Êœ œÊœÊab
## z
4
œ
8. (2 cos t) (2 sin t) x 2 cos t and z 2 sin t x z 4 (h)rikœ Êœ œÊœÊ
##
9. (t) t (1 t) , 0 t 1 2 ; x t and y 1 t x y t ( t) 1ri j ij jœ ŸŸ Ê œÊ œ œ œÊ œ"œ
dd
dt dt
rr
¸¸È
f(x y z) ds f(t 1 t 0) dt (1) 2 dt 2 t 2Êßßœßß œ œ œ'' '
C0 0
11
¸¸ Š‹ ’ “
ÈÈÈ
d
dt
r
"
!
10. (t) t (1 t) , 0 t 1 2; x t, y 1 t, and z 1 x y z 2ri jk ijœ ŸŸ Ê œÊ œ œ œ œ Ê
dd
dt dt
rr
¸¸È
t (1 t) 1 2 2t 2 f(x y z) ds (2t 2) 2 dt 2 t 2t 2œ œ Ê ßß œ œ œ''
C0
1
ÈÈ È
cd #
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!
11. (t) 2t t (2 2t) , 0 t 1 2 2 4 1 4 3; xy y zrij k ijkœ ŸŸÊ œ Ê œ œ
dd
dt dt
rr
¸¸È
(2t)t t (2 2t) f(x y z) ds 2t t 2 3 dt 3 t t 2t 3 2œÊ ßß œ œ œœ''
C0
1
ab œ ‘ˆ ‰
#$# ""
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2213
33
12. (t) (4 cos t) (4 sin t) 3t , 2 t 2 ( 4 sin t) (4 cos t) 3rijk ijkœŸŸÊœ 11
d
dt
r
16 sin t 16 cos t 9 5; x y 16 cos t 16 sin t 4 f(x y z) ds (4)(5) dtÊœ œ œ œÊ ßßœ¸¸ÈÈ È
d
dt
r ## ## ## ''
C2
2 œœ
œ
20t 80œœcd
#
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1
1
1
13. (t) ( 2 3 ) t( 3 2 ) (1 t) (2 3t) (3 2t) , 0 t 1 3 2r ijk ijk i j k ijkœ œ ŸŸÊ œ
d
dt
r
194 14; xyz(1t)(23t)(32t)66t f(xyz) dsÊ œ œ œ œ Ê ß߸¸È È
d
dt
r '
C
(6 6t) 14 dt 6 14 t 6 14 3 14œ œ œ œ'
0
1
ÈÈ È È
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2
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!
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14. (t) t t t , 1 t 3 ; rijk ijkœ ŸŸ_Ê œ Ê œ œ œ
dd
dt dt x y z t t t 3t
333rr
¸¸È
ÈÈÈ
œ œ œ œœœ œ

f(x y z) ds 3 dt lim 1 1Êßßœ œœ œ''
C1

Š‹
È œ‘ˆ‰
È
3
3t t b
1
œ
_
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bÄ_

1002. 998 Chapter 16 Integration in Vector Fields
15. C : (t) t t , 0 t 1 2t 1 4t ; x y z t t 0 t t 2t"
# ###rij ijœ ŸŸ Ê œ Ê œ œ œ œ
dd
dt dt
rr
¸¸È È
È
kk
since t 0 f(x y z) ds 2t 1 4t dt 4t (5) 5 5 1 ; Ê ßß œ œ " œ œ ''
C0
1œ
È
’“ Š‹ab
È # """" # $Î#
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!
6666
C : (t) t , 0 t 1 1; x y z 1 1 t 2 t#
###rijk kœ ŸŸ Ê œ Ê œ œ œ
dd
dt dt
rr
¸¸ È È
f(x y z) ds 2 t (1) dt 2t t 2 ; therefore f(x y z) dsÊßßœ œœœ ßß'' '
C0 C
1
ab œ ‘
#$ ""
"
!333
5
f(x y z) ds f(x y z) ds 5œßßßßœ''
CCœ
53
6È
#
16. C : (t) t , 0 t 1 1; x y z 0 0 t t"
###rk kœŸŸÊœÊœœœ
dd
dt dt
rr
¸¸ È È
f(x y z) ds t (1) dt ;Êßßœ œœ''
C0
1œ
ab ’“
#
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33

C : (t) t , 0 t 1 1; x y z 0 t 1 t 1#
#rjk jœ ŸŸ Ê œ Ê œ œ œ
dd
dt dt
rr
¸¸ È ÈÈ
f(x y z) ds t 1 (1) dt t t 1 ;Êßßœ œœœ''
C0
1
ˆ‰ œ ‘È
22
333
$Î#
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!
"
C : (t) t , 0 t 1 1; x y z t 1 1 t$
#rijk iœ ŸŸÊ œÊ œ œ œ
dd
dt dt
rr
¸¸ È È
f(x y z) ds (t)(1) dt f(x y z) ds f ds f ds f dsÊ ßßœ œ œÊ ßßœ œ' ' ' '''
C 0 C CCC
1_
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t
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6
17. (t) t t t , 0 a t b 3 ; rijk ijkœ ŸŸ Ê œ Ê œ œ œ
d d ttt 1
dt dt x y z t t t t
xyzrr
¸¸È



œ œ œ œœœ
f(x y z) ds 3 dt 3 ln t 3 ln , since 0 a bÊßßœ œ œ Ÿ''
Ca
b b
a
ˆ‰ ˆ‰ÈÈ È
’“ kk
1b
ta
18. (t) (a cos t) (a sin t) , 0 t 2 ( a sin t) (a cos t) a sin t a cos t a ;rjk jkœ ŸŸÊœ Êœ œ 1
dd
dt dt
rr
¸¸È
kk ## # #
x z 0 a sin t f(x y z) ds a sin t dt a sin t dt
a sin t, 0 t
a sin t, t 2
œ œ Ê ßß œ
Î&&
ŸŸ
ÈÈ
œ
kk
kk
kk kk
## ##
##
1
11
'' '
C0
2
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œ
a cos t a cos t a ( 1) a a a ( 1) 4aœ œ œcdcdc dc d
## #### #
!
#11
1
19. (x) x y x , 0 x 2 x 1 x ; f(x y) f x 2x f dsrijij ijœœ ŸŸÊ œ Ê œ ßœ ß œ œ Ê
xdd xx
dx dx
œ œ
# #
#rr
¸¸È
Š‹
Š‹
x
œ
œ
'
C
(2x) 1 x dx 1 x 5 1œœœœ'
0
2
È
’“ab ˆ‰ # # $Î#
$Î#
#
!
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333
10 5 2
È
20. (t) 1 t 1 t , 0 t 1 1 1 t ; f(x y) f 1 t 1 trijœ ŸŸÊ œ ßœ ß œabab ab abab ¸¸É Š‹
1d 1 2 2
dt
1t 1t
11t
# #
#

r abab
É ab
1
4
4
œ
f ds 1 1 t dt 1 t 1 t dt 1 t 1 tÊœ œœ'' '
C0 0
11
abab
É ab
1t 1t
11t
11
420
425

# "
#
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1
4
4
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0œ œˆ‰
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## #00
11
21. (t) (2 cos t) (2 sin t) , 0 t ( 2 sin t) (2 cos t) 2; f(x y) f(2 cos t 2 sin t)rij ijœ ŸŸÊœ ʜߜß
1
#
dd
dt dt
rr
¸¸
2 cos t 2 sin t f ds (2 cos t 2 sin t)(2) dt 4 sin t 4 cos t 4 ( 4) 8œÊ œ œ œœ
''
C0
2
œ_
cd
1Î#
!
22. (t) (2 sin t) (2 cos t) , 0 t (2 cos t) ( 2 sin t) 2; f(x y) f(2 sin t 2 cos t)rij ijœ ŸŸÊœ ʜߜß
1
4dt dt
ddrr
¸¸
4 sin t 2 cos t f ds 4 sin t 2 cos t (2) dt 4t 2 sin 2t 4 sin tœÊ œ œ
## Î% ''
C0
4
0
œ_
abc d
1
21 2œ 1Š‹
È

1003. Section 16.1 Line Integrals 999
23. (t) t 1 2t , 0 t 1 2t 2 2 t 1; M (x y z) ds (t) 2 t 1 dtrjk jkœ ŸŸÊ œÊ œ œ ßß œ ab ¸¸ ÈÈ
Š‹
# ##dd
dt dt
rr ''
C0
1
$$
t2t1 dt t1 2 1221œœœœ'
0
1
ˆ‰Š‹’ “
ÈÈ
ab
3 3/2
#
# # $Î#
"
!
24. (t) t 1 2t , 1 t 1 2t 2rjk jkœ ŸŸÊ œab
# d
dt
r
2 t 1; M (x y z) dsÊœ œ ß߸¸ È '
d
dt
r #
C
$
15 t 1 2 2 t 1 dtœ'
œ1
1
ˆ‰Èab Š‹
È ##
30 t 1 dt 30 t 60 1 80;œœœœ'
œ1
1
ab ’“Š‹ ˆ‰
#
"
œ"
"t
33

M y (x y z) ds t 1 30 t 1 dt
xz
C1
1œßßœ ''$
œ
abc d ab
##
30 t 1 dt 30 t 60 1œœœ'
œ1
1
ab ’“Š‹ ˆ‰
%
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œ"
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55
"

48 y ; M x (x y z) ds 0 ds 0 x 0; z 0 by symmetry (since isœ Ê œ œ œ œ ß ß œ œ Ê œ œ
M
M805
48 3
yzxz ''
CC
$$ $
independent of z) (x y z) 0Êßßœ!ß߈‰
3
5
25. (t) 2t 2t 4 t , 0 t 1 2 2 2t 2 2 4t 2 1 t ;rij k ijkœ ŸŸÊœÊœœ
ÈÈ ÈÈ È È
ab ¸¸
# ##dd
dt dt
rr
(a) M ds (3t) 2 1 t dt 2 1 t 2 2 1 4 2 2œœ œ œœ''
C0
1
$ Š‹’ “
ÈÈ
ab ˆ‰
# # $Î#
$Î#
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!
(b) M ds (1) 2 1 t dt t 1 t ln t 1 t 2 ln 1 2 (0 ln 1)œ œ œ œ ''
C0
1
$ Š‹’ “’ “
ÈÈ ÈÈÈ
Š‹ Š‹
## #
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!
2ln1 2œ
ÈÈ
Š‹
26. (t) t 2t t , 0 t 2 2 t 1 4 t 5 t;rij k ijkœ ŸŸ Ê œ Ê œ œ
2d d
3dt dt
$Î# "Î# rr
¸¸È È
M ds 3 5 t 5 t dt 3(5 t) dt (5 t) 7 5 (24) 36;œœ œœœœœ'' '
C0 0
22
$ ˆ‰ˆ‰ œ ‘ÈÈ ab
333
2
###
#
! ##
M x ds t[3(5 t)] dt 15t 3t dt t t 30 8 38;yz
5
2œœœœœœ'' '
C0 0
22
$ ab œ ‘
##$ "
#
!
M y ds 2t[3(5 t)] dt 2 15t 3t dt 76; M z ds t [3(5 t)] dtxz xy
2
3œœ œ œœ œ '' ' ''
C0 0 C 0
22 2
$$ ab
# $Î#
10t 2t dt 4t t 4(2) (2) 16 2 2 2 xœœœœœÊœ'
0
2
ˆ‰ œ ‘ ÈÈ È
$Î# &Î# &Î# (Î# &Î# (Î#
#
!
4 4 32 144
7777M
M yz
, y , and z 2œœ œ œœ œ œ œ
38 19 76 19 4
36 18 M 36 9 M 7 36 7
M M 144 2 xz xy
È
†
È
27. Let x a cos t and y a sin t, 0 t 2 . Then a sin t, a cos t, 0œœŸŸœœœ 1
dx dz
dt dt dt
dy
dt a dt; I x y ds a sin t a cos t a dtÊœœœ ʈ‰ ˆ‰Š‹ ab a b
dx dz
dt dt dt
dy
z
##
#
## ## ## ''
C0
2
$$
œ
a dt 2 a ; M (x y z) ds a dt 2 a R a.œœœßßœœÊœœœ'''
0C0
22
œœ
$$
$1$ $ $ 1$ z
I
M2a
2aÉ É
z 1$
1$

28. (t) t (2 2t) , 0 t 1 2 5; M ds 5 dt 5;rj k jkœ ŸŸ Ê œ Ê œ œ œ œ
dd
dt dt
rr
¸¸ÈÈÈ ''
C0
1
$$ $
I y z ds t (2 2t) 5 dt 5t 8t 4 5 dt 5 t 4t 4t 5 ;x
55
33œ œ œ œ œ'' '
C0 0
11
ab c d a b ÈÈÈ È œ ‘
## # # # $ #
"
!
$$ $$ $
I x z ds 0 (2 2t) 5 dt 4t 8t 4 5 dt 5 t 4t 4t 5 ;y
44
33œ œ œ œ œ'' '
C0 0
11
ab c d a b
ÈÈÈ È œ ‘
## # # # $ #
"
!
$$ $$ $
I x y ds 0 t 5 dt 5 5 R , R ,
z
C0
1œ œ œ œ Êœœ œœœ''ab ab
ÈÈ È
’“ É ÉÉ É
## ##
"
!
"
$$$$
t542
33 M3 M3
xy
I I
3

x y
È
and Rz
I
M 3œœÉ
z "
È

1004. 1000 Chapter 16 Integration in Vector Fields
29. (t) (cos t) (sin t) t , 0 t 2 ( sin t) (cos t) sin t cos t 1 2;rijk ijkœŸŸÊœÊœ œ 1
dd
dt dt
rr
¸¸ÈÈ ##
(a) M ds 2 dt 2 2; I x y ds cos t sin t 2 dt 2 2œœ œ œœ œ'' ' '
C0 C 0
2 2
z
$ $ 1$ $ $ 1$
œ œ
ÈÈ ÈÈ
ab a b ## # #
R1 Êœ œ z
I
MÉ
z
(b) M (x y z) ds 2 dt 4 2 and I x y ds 2 dt 4 2œßßœ œ œ œ œ'' ' '
C0 C 0
4 4
z
$ $ 1$ $ $ 1$
œ œ
ÈÈ ÈÈ
ab ##
R1 Êœ œ z
I
MÉ
z
30. (t) (t cos t) (t sin t) t , 0 t 1 (cos t t sin t) (sin t t cos t) 2t rijk i jkœ ŸŸÊœ
22
3dt
d
È
$Î# r È
(t 1) t 1 for 0 t 1; M ds (t 1) dt (t 1) 2 1 ;Êœœ ŸŸœ œœ œ œ¸¸ œ ‘È ab
d 3
dt 2
r # "" ###
"
! ## ''
C0
1
$
M z ds t (t 1) dt t t dt t txy
22 22 22
33 375
22œœ œ œ '' '
C0 0
11
$ Š‹ ˆ‰ œ ‘
ÈÈ È
$Î# &Î# $Î# (Î# &Î#
"
!
z ; I x y dsœœ œÊœœ œ œ
22 22 162 162 322
3 7 5 3 35 35 M 35 3 105
22 24 2 M
z
ÈÈÈ ÈÈ
ˆ‰ ˆ‰ ˆ‰ Š‹ ab
xy '
C
##
$
t cos t t sin t (t 1) dt t t dt Rœ œ œ œœ Ê œ œ''
00
11
abab ’“ É É
## ## $#
"
!
""tt 7 7
43 4312 M 18
z
I

z
31. (x y z) 2 z and (t) (cos t) (sin t) , 0 t M 2 2 as found in Example 4 of the text;$11ßß œ œ Ÿ Ÿ Ê œ rjk
also 1; I y z ds cos t sin t (2 sin t) dt (2 sin t) dt 2 2 R¸¸ ab a b É
d
dt M
x x
Ir
œœ œ œ œÊœ'' '
C0 0
## # #
$1
œœ
x
1œ
32. (t) t t , 0 t 2 2 t t 1 2t t (1 t) 1 t forri jk i jkœ ŸŸ Ê œ Ê œ œ œ
22
3dtdt
td d
È
$Î# "Î#
#
##
œ
rr ÈÈ ¸¸ È
0 t 2; M ds (1 t) dt dt 2; M x ds t (1 t) dt 2;ŸŸ œ œ œ œ œ œ œ œ'' ' ' '
C0 0 C 0
22 2
yz
$$ ˆ‰ ˆ‰ ’“
""

#
!
t1 t1 2
t
œ
M y ds t dt t ; M z ds dt x 1,xz
22 42
31515 63M
32 t t Mœœ œ œœœœœÊœœ'' ''
C0 C0
2 2
xy
$$
ÈÈ
$Î# &Î#
##
!!
#
%
’“ ’“
œ
yz
y , and z ; I y z ds t t dt t ;œ œ œ œ œ œ œ œœ
M
M15 M3 9 4 9 20 92045
16 8 2 t 32 32 232M
xxz xy #" ## $ % %
#
! ''
C0
2
ab ˆ‰ ’“$
"
I x z ds t t dt ; I x y ds
y z
C0 C
2œ œ œœœœ'' 'ab ab ˆ‰ ’“
## # % ## "
#
!
$$
432032015
tt 83264
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t t dt t R , R , andœ œ œœÊœœ œœ'
0
2
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#$ %
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!
8 t 2 8 32 56 2 29 32
939399 M35 M15
xy
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I
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z
33-36. Example CAS commands:
:Maple
f := (x,y,z) -> sqrt(1+30*x^2+10*y);
g := t -> t;
h := t -> t^2;
k := t -> 3*t^2;
a,b := 0,2;
ds := ( D(g)^2 + D(h)^2 + D(k)^2 )^(1/2): # (a)
'ds' = ds(t)*'dt';
F := f(g,h,k): # (b)
'F(t)' = F(t);
Int( f, s=C..NULL ) = Int( simplify(F(t)*ds(t)), t=a..b ); # (c)
`` = value(rhs(%));
: (functions and domains may vary)Mathematica
Clear[x, y, z, r, t, f]
f[x_,y_,z_]:= Sqrt[1 30x 10y]
2

1005. Section 16.2 Vector Fields, Work, Circulation, and Flux 1001
{a,b}= {0, 2};
x[t_]:= t
y[t_]:= t
2
z[t_]:= 3t
2
r[t_]:= {x[t], y[t], z[t]}
v[t_]:= D[r[t], t]
mag[vector_]:=Sqrt[vector.vector]
Integrate[f[x[t],y[t],z[t]] mag[v[t]], {t, a, b}]
N[%]
16.2 VECTOR FIELDS, WORK, CIRCULATION, AND FLUX
1. f(xyz)xyz xyz (2x) xxyz ; similarly,ß ß œ Ê œ œ ab ab ab
### ### ###
"Î# $Î# $Î#`"
`#
f
x
yxyz and zxyz f
``
``
### ###
$Î# $Î# œœ œ

ff
yz
xyz
xyz
œ œ Ê œab ab ™
ijk
ab
œœœ

2. f(x y z) ln x y z ln x y z (2x) ;ßßœ œ Ê œ œÈ ab Š‹
### "`""
#`#
### fx
x xyz x y z
œœœ œ œ œ
similarly, and f
``
` `
ffz
y xyz z xyz xyz
yxyz
œœÊœ
œœœ œœœ œœœ ™
ijk
3. g(x y z) e ln x y , and eß ß œ Ê œ œ œ
z zgg2yg
xxyyxy z
2x
ab
## `` `
```
œœ œœ
geÊœ™ Š‹Š‹
œ

2x
xy xy
2y
œœ œœijk
z
4. g(x y z) xy yz xz y z, x z, and y x g (y z) ( z) (x y)ßß œ Ê œ œ œ Ê œ B
`` `
`` `
gg g
xy z
™ ijk
5. inversely proportional to the square of the distance from (x y) to the origin (M(x y)) (N(x y))kk ÈF ßÊßß
##
, k 0; points toward the origin is in the direction of œ Ê œ
k x
xy xy xy
y
œœ œœ œœ

œ

FFnij
ÈÈ
a , for some constant a 0. Then M(x y) and N(x y)ʜ ߜ ߜFn
œ

œax
xy xy
ay
ÈÈ
œœ œœ
(M(x y)) (N(x y)) a a , for any constant k 0Êßߜʜʜ È
##

œ

kkx
xy
xy xy
ky
œœ
œœ œœ
Fij
ab ab
6. Given x y a b , let x a b cos t and y a b sin t. Then
#### ## ##
œ œ œ
ÈÈ
a b cos t a b sin t traces the circle in a clockwise direction as t goes from 0 to 2rijœ Š‹Š‹
ÈÈ
## ##
1
a b sin t a b cos t is tangent to the circle in a clockwise direction. Thus, letÊœ vijŠ‹Š‹
ÈÈ
## ##
y x and (0 0) .Fv F i j F 0œÊ œ ßœ
7. Substitute the parametric representations for (t) x(t) y(t) z(t) representing each path into the vectorrijkœ
field , and calculate the work W .FF œ '
C
†
d
dt
r
(a) 3t 2t 4t and 9t W 9t dtFijk ijkFœ œÊ œÊ œ œ
dd 9
dt dt
rr
† '
0
1 #
(b) 3t 2t 4t and 2t 4t 7t 16t W 7t 16t dt t 2tFijk ijkFœ œ Ê œ Êœ œ
#% $ #( #($)
"
!
dd 7
dt dt 3
rr
† '
0
1
ab œ ‘
2 œœ
713
33
(c) t t and t ; 3t 2t and 5t W 5t dt ;r i j r ij kF i j ij F"# " " "
#œ œ œ œ Ê œ Ê œ œ
dd
dt dt
5rr&&
† '
0
1
3 2 4t and 4t W 4t dt 2 W W WFijk kF###"#
#œ œÊ œ Ê œ œÊ œ œ
dd
dt dt
9rrœœ
† '
0
1

1006. 1002 Chapter 16 Integration in Vector Fields
8. Substitute the parametric representation for (t) x(t) y(t) z(t) representing each path into the vectorrijkœ
field , and calculate the work W .FF œ '
C
†
d
dt
r
(a) and W dt tan tFjijkFœ œ Ê œ Ê œ œ œˆ‰ cd
"""

œ"
"
!t1 dt dt t1 t1 4
dd
œœœ
rr
† '
0
1
1
(b) and 2t 4t W dt ln t 1 ln 2FjijkFœœÊœÊœœœˆ‰ cdab
"

$#
"
!t1 dt dt t1 t1
d d 2t 2t
œ œœ
rr
† '
0
1
(c) t t and t ; and ; and rijrijkF j ijF F j k"# " " #
"""
#œ œ œ œ Ê œ œ œ ˆ‰
t 1 dt dt t 1 dt
dd d
œœ
ŸŸ œ
rr r
†
0 W dtʜʜ œF #
"
†
d
dt t 1 4
rœ
œ '
0
1
1
9. Substitute the parametric representation for (t) x(t) y(t) z(t) representing each path into the vectorrijkœ
field , and calculate the work W .FF œ '
C
†
d
dt
r
(a) t 2t t and 2 t 2t W 2 t 2t dt t tFijk ijkFœ œ Ê œ Ê œ œ œÈÈ È È ˆ‰ œ ‘
dd 4
dt dt 3 3
rr
† '
0
1
$Î# #
"
!
"
(b) t 2t t and 2t 4t 4t 3t W 4t 3t dt t tFijk ij kFœ œ Ê œ Ê œ œ œ
# $ %# %# &$
"
!
"dd 4
dt dt 5 5
rr
† '
0
1
ab œ ‘
(c) t t and t ; 2t t and 2t W 2t dtr i j r ij kF j k ij F"# " " "œ œ œ œ Ê œ Ê œ È
dd
dt dt
rr&&
† '
0
1
1; t 2 and 1 W dt 1 W W W 0œ œ œÊ œÊ œ œÊœœFijk kF ###"#
È
dd
dt dt
rrœœ
† '
0
1
10. Substitute the parametric representation for (t) x(t) y(t) z(t) representing each path into the vectorrijkœ
field , and calculate the work W .FF œ '
C
†
d
dt
r
(a) t t t and 3t W 3t dt 1Fijk ijkFœ œÊ œ Ê œ œ
### # # dd
dt dt
rr
† '
0
1
(b) t t t and 2t 4t t 2t 4t W t 2t 4t dtFijk ij kFœ œ Ê œ Ê œ
$'& $ $() $() dd
dt dt
rr
† '
0
1
ab
t œ œ’“
tt4 17
449 18
Î(
*
"
!
(c) t t and t ; t and t W t dt ;rijrijkF i ijF"# " " "
### "œ œ œ œ Ê œ Ê œ œ
dd
dt dt 3
rr&&
† '
0
1
t t and t W t dt W W WFijk k F###"#
"
#œ œÊ œÊ œ œÊœœ
dd
dt dt 6
5rrœœ
† '
0
1
11. Substitute the parametric representation for (t) x(t) y(t) z(t) representing each path into the vectorrijkœ
field , and calculate the work W .FF œ '
C
†
d
dt
r
(a) 3t 3t 3t and 3t 1 W 3t 1 dt t t 2FijkijkFœ œ Ê œ Ê œ œ œab abcd
###$
"
!
dd
dt dt
rr
† '
0
1
(b) 3t 3t 3t and 2t 4t 6t 4t 3t 3tFijkijkFœ œ Ê œab
# % $ &$# dd
dt dt
rr
†
W 6t4t3t3t dt ttt tÊ œ œ œ'
0
1
ab œ ‘
&$# '%$ #
##
"
!
33
(c) t t and t ; 3t 3t and 3t 3tr i j r ij kF ik ij F"# " "
##œ œ œ œ Ê œ ab
dd
dt dt
rr&&
†
W 3t 3t dt t t ; 3t and 1 W dt 1Êœ œ œ œ œÊ œÊœ œ"###
#$#
"
!
"
#
''
0 0
1 1
ab œ ‘
3
2dtdt
dd
Fjk kF
rrœœ
†
WW WÊœœ "#
1
2
12. Substitute the parametric representation for (t) x(t) y(t) z(t) representing each path into the vectorrijkœ
field , and calculate the work W .FF œ '
C
†
d
dt
r
(a) 2t 2t 2t and 6t W 6t dt 3t 3Fijk ijkFœ œÊ œÊ œ œ œ
dd
dt dt
rr
† '
0
1
cd
#
"
!
(b) t t t t t t and 2t 4t 6t 5t 3tFijkijkFœ œ Ê œa babab
#% % # $ &%# dd
dt dt
rr
†
W 6t5t3t dtttt 3Êœ œ œ
'
0
1
abcd
&%# '&$
"
!

1007. Section 16.2 Vector Fields, Work, Circulation, and Flux 1003
(c) t t and t ; t t 2t and 2t W 2t dt ;rijrijkFijk ijF"# " " "œ œ œ œÊ œ Ê œ œ"
dd
dt dt
rrŸŸ
† '
0
1
(1 t) (t 1) 2 and 2 W 2 dt 2 W W W 3FijkkF ###"#œ œÊ œÊ œ œÊ œ œ
dd
dt dt
rrœœ
† '
0
1

13. t t t , 0 t 1, and xy y yz t t t and 2tri jk F ij k F i j k i jkœ ŸŸ œ Êœ œ
#$#$ d
dt
r
2t work 2t dtʜʜ œF†
d
dt
r $$ "
# '
0
1
14. (cos t) (sin t) , 0 t 2 , and 2y 3x (x y)rijk Fij kœ ŸŸ œ
t
6
1
(2 sin t) (3 cos t) (cos t sin t) and ( sin t) (cos t) Êœ œ ÊFij k ijkF
dd
dt 6 dt
rr "
†
3 cos t 2sin t cos t sin t work 3 cos t 2 sin t cos t sin t dtœÊœ
## "" ""22
66 66 '
0
2
œ
ˆ‰
t sin 2t t sin t cos tœ œœ ‘
3 3 sin 2t
24 2 6 6
""
#
!
1
1
15. (sin t) (cos t) t , 0 t 2 , and z x y t (sin t) (cos t) andrijk FijkFijkœ ŸŸ œÊœ 1
(cos t) (sin t) t cos t sin t cos t work t cos t sin t cos t dt
dd
dt dt
rr
œÊœÊœ ijkF †
## '
0
2
œ
ab
cos t t sin t sin tœ œœ ‘
tsin 2t
24
#
!
1
1
16. (sin t) (cos t) , 0 t 2 , and 6z y 12x t cos t (12 sin t) andrijk FijkFi j kœ ŸŸ œÊœ
t
6
1
##
ab
(cos t) (sin t) t cos t sin t cos t 2 sin t
dd
dt 6 dt
rr
œÊœ ijkF
" #
†
work t cos t sin t cos t 2 sin t dt cos t t sin t cos t 2 cos t 0Êœ œ œ'
0
2
œ
ab œ ‘
#$
#
!
1
3
1
17. x t and y x t t t , 1 t 2, and xy (x y) t t t andœœœÊœŸŸ œÊœ
## # $ #
ri j F i j F i j ab
2t t 2t 2t 3t 2t xy dx (x y) dy dt 3t 2t dt
dd d
dt dt dt
rr r
œ Ê œ œ Ê œ œ ijF F ††
$#$ $# $#
ab ab '''
CC œÎ
Ê
tt 12œ œ œœœ ‘ˆ ‰ˆ‰
3 2 16 3 2 45 18 69
43 3 43434
%$
#
œ"
18. Along (0 0) to (1 0): t , 0 t 1, and (x y) (x y) t t and t;ßßœŸŸ œÊœ œÊœri F i j Fij i F
dd
dt dt
rr
†
Along (1 0) to (0 1): (1 t) t , 0 t 1, and (x y) (x y) (1 2t) andß ß œ ŸŸ œ Êœ rij F ijF ij
2t;
dd
dt dt
rr
œ Ê œij F †
Along (0 1) to (0 0): (1 t) , 0 t 1, and (x y) (x y) (t 1) (1 t) andß ß œ ŸŸ œ Êœ rj F ijFij
t 1 (x y) dx (x y) dy t dt 2t dt (t 1) dt (4t 1) dt
dd
dt dt
rr
œ Ê œ Ê œ œ jF † '''''
C0000
11 1 1
2t t 2 1 1œ œœcd
#
"
!
19. x y y y , 2 y 1, and x y y y 2y and 2y yrij ij F ij ij ij Fœœ    œ œ Ê œ œ
##% & dd
dy dy
rr
†
ds dy 2y y dy y yÊœ œœœœœ'' '
C2 2
11
FT F††
œœ
d 64 4 3 63 39
dy 3 3 3 3
r
ab œ ‘ˆ‰ˆ ‰
&'# "" ""
#####
œ"
#
20. (cos t) (sin t) , 0 t , and y x (sin t) (cos t) and ( sin t) (cos t)rij FijFij ijœ ŸŸ œÊœ œ
1
#
d
dt
r
FFrʆ †
d
dt
r
œ œ Ê œ œsin t cos t 1 d ( 1) dt
##
# ''
C0
2
œ"
1
21. ( ) t( 2 ) (1 t) (1 2t) , 0 t 1, and xy (y x) 1 3t 2t t andrij ij i j F i j F ijœ œ ŸŸ œ Ê œ ab
#
2 1 5t 2t work dt 1 5t 2t dt t t t
dd d 5225
dt dt dt 2 3 6
rr r
œ Ê œ Ê œ œ œ œij F F ††
###$
"
! ''
C0
1
ab œ ‘
22. (2 cos t) (2 sin t) , 0 t 2 , and f 2(x y) 2(x y)rij F ijœ ŸŸ œœ 1 ™
4(cos t sin t) 4(cos t sin t) and ( 2 sin t) (2 cos t) Êœ œ ÊFij ijF
dd
dt dt
rr
†

1008. 1004 Chapter 16 Integration in Vector Fields
8 sin t cos t sin t 8 cos t cos t sin t 8 cos t sin t 8 cos 2t work f dœ œ œ Ê œaba bab
## ## '
C
™†r
dt 8 cos 2t dt 4 sin 2t 0œœ œœ''
C0
2
F†
d
dt
r
œ
cd
#
!
1
23. (a) (cos t) (sin t) , 0 t 2 , x y , and y x ( sin t) (cos t) ,rij FijFij ijœ ŸŸœ œÊœ 1 "#
d
dt
r
(cos t) (sin t) , and ( sin t) (cos t) 0 and sin t cos t 1FijF ijF F"# "#
##œ œ Ê œ œœ ††
dd
dt dt
rr
Circ 0 dt 0 and Circ dt 2 ; (cos t) (sin t) cos t sin t 1 andÊœ œ œ œœ Ê œœ"# "
##
''
00
22
œœ
1nijFn †
0 Flux dt 2 and Flux 0 dt 0Fn
#" #†œÊ œ œ œ œ ''
00
22
œœ
1
(b) (cos t) (4 sin t) , 0 t 2 ( sin t) (4 cos t) , (cos t) (4 sin t) , andri j i jF i jœ ŸŸÊœ œ 1
d
dt
r
"
( 4 sin t) (cos t) 15 sin t cos t and 4 Circ 15 sin t cos t dtFijF F#"#"œ Ê œ œ Ê œ ††
dd
dt dt
rr '
0
2
œ
sin t 0 and Circ 4 dt 8 ; cos t sin t œœœœœ Êœ ‘ Š‹Š‹
" "#
#
!
# "
54
2 17 17
1 '
0
2
œ
1nijFn
ÈÈ
†
cos t sin t and sin t cos t Flux ( ) dt 17 dtœ œ Êœ œ
44 15 4
17 17 17 17
ÈÈ È È
##
#""
Fn Fnv†† ''
00
22
œœ
kk Š‹
È
8 and Flux ( ) dt sin t cos t 17 dt sin t 0œœ œ œœ1 ##
#
#
!
''
00
22
œœ
Fnv†kk Š‹
È œ ‘
15 15
17 2È
1
24. (a cos t) (a sin t) , 0 t 2 , 2x 3y , and 2x (x y) ( a sin t) (a cos t) ,rij FijFij ijœ ŸŸ œ œ Êœ 1 "#
d
dt
r
(2a cos t) (3a sin t) , and (2a cos t) (a cos t a sin t) (a cos t) (a sin t) ,FijFi jnvij"#œ œÊœ kk
2a cos t 3a sin t, and 2a cos t a sin t cos t a sin tFnv Fnv
"#
## ## ## # ##††kk kkœ œ
Flux 2a cos t 3a sin t dt 2a 3a a , andÊœ œ œ"
## ## # # #
##
!!
'
0
2
œ
ab œ ‘ œ ‘
tsin 2t tsin 2t
24 24
11
1
Flux 2a cos t a sin t cos t a sin t dt 2a sin t a a#
## # ## # # # #
##
!! #
#
!œœœ'
0
2
œ
abcd œ ‘ œ ‘
t sin 2t a t sin 2t
24 24
11 1
œ
1
25. (a cos t) (a sin t) , ( a sin t) (a cos t) 0 Circ 0; M a cos t,Fij ijF" """œ œ ʜʜœ
dd
dt dt
rrŸŸ
†
N a sin t, dx a sin t dt, dy a cos t dt Flux M dy N dx a cos t a sin t dt
""""
## ##œœ œÊœœ ''
C0
œ
ab
a dt a ;œœ
'
0
œ
##
1
t , t Circ t dt 0; M t, N 0, dx dt, dy 0 FluxFi iF##### #œœÊ œÊœ œœœœœÊ
dd
dt dt
rrœœ
† '
œa
a
M dy N dx 0 dt 0; therefore, Circ Circ Circ 0 and Flux Flux Flux aœ œ œ œœ œœ''
Ca
a
## "# "#
#
œ
1
26. a cos t a sin t , ( a sin t) (a cos t) a sin t cos t a cos t sin tFij ijF" "
#### $#$#œ œÊœ abab
dd
dt dt
rrŸŸ
†
Circ a sin t cos t a cos t sin t dt ; M a cos t, N a sin t, dy a cos t dt,Êœ œ œ œ œ"""
$#$# ## ##
'
0
œ
ab
2a
3

dx a sin t dt Flux M dy N dx a cos t a sin t dt a ;œ Ê œ œ œ """
$$ $$ $
''
C0
œ
ab
4
3
t , t Circ t dt ; M t , N 0, dy 0, dx dtFi iF### ##
####œ œÊ œÊ œ œ œ œœœ
dd
dt dt 3
2arrœœ

† '
œa
a
Flux M dy N dx 0; therefore, Circ Circ Circ 0 and Flux Flux Flux aÊ œ œ œœ œœ### "# "#
$
'
C
4
3
27. ( a sin t) (a cos t) , ( a sin t) (a cos t) a sin t a cos t aFij ijF" "
## # # #œ œ Ê œ œ
dd
dt dt
rrŸŸ
†
Circ a dt a ; M a sin t, N a cos t, dx a sin t dt, dy a cos t dtÊœ œ œ œ œ œ
"""
##
'
0
œ
1
Flux M dy N dx a sin t cos t a sin t cos t dt 0; t , 0Êœ œ œœœÊ œ""" # #
##
''
C0
œ
ab Fj iF
dd
dt dt
rrœœ
†
Circ 0; M 0, N t, dx dt, dy 0 Flux M dy N dx t dt 0; therefore,ÊœœœœœÊœ œœ
### # ##
''
Ca
a œ
Circ Circ Circ a and Flux Flux Flux 0œœ œœ "# " #
# 1

1009. Section 16.2 Vector Fields, Work, Circulation, and Flux 1005
28. a sin t a cos t , ( a sin t) (a cos t) a sin t a cos tFij ijF" "
## # # $$ $ $œ œ Ê œ abab
dd
dt dt
rr&&
†
Circ a sin t a cos t dt a ; M a sin t, N a cos t, dy a cos t dt, dx a sin t dtÊ œ œ œ œ œ œ"""
$$ $ $ $ ## # #
'
0
œ
ab
4
3
Flux M dy N dx a cos t sin t a sin t cos t dt a ; t , 0Êœ œ œ œ œÊ œ""" # #
$#$# $#
''
C0
œ
ab
2
3dt dt
dd
Fj iF
rrœœ
†
Circ 0; M 0, N t , dy 0, dx dt Flux M dy N dx t dt a ; therefore,ÊœœœœœÊœ œœ### # ##
# #$
''
Ca
a œ
2
3
Circ Circ Circ a and Flux Flux Flux 0œœ œœ "# " #
$4
3
29. (a) (cos t) (sin t) , 0 t , and (x y) x y ( sin t) (cos t) andrij F i j ijœ ŸŸ œÊœ 1 ab
## d
dt
r
(cos t sin t) cos t sin t sin t cos t sin t cos t dsFi jF FTœ Ê œ Êab
## #
††
d
dt
r '
C
sin t cos t sin t cos t dt sin t sin tœ œ œ'
0
œ
ab œ ‘
## "
## !24
tsin 2t
1
1
(b) (1 2t) , 0 t 1, and (x y) x y 2 and (1 2t) (1 2t) ri Fi j iFi jœ ŸŸ œ Ê œ œ Ê ab
## # d
dt
r
4t 2 ds (4t 2) dt 2t 2t 0FFT††
d
dt
r
œÊ œ œ œ''
C0
1
cd
#
"
!
(c) (1 t) t , 0 t 1, and (x y) x y and (1 2t) 1 2t 2trij Fi j ijFi j"
## #œ ŸŸ œ Ê œ œ ab a b
d
dt
r&
(2t 1) 1 2t 2t 2t Flow 2t dt ; t (t 1) ,ÊœœÊœ œ œœFFrij††
d d
dt dt 3
2r r&
ab
## #
"# ''
C0
1œ
œ
0 t 1, and (x y) x y and t t 2t 1ŸŸ œ Ê œ œ Fi j ijFi jab a b
## ## d
dt
r œ
2t 2t 1 1 2t 2t 1 2t 2t Flow 2t 2t dtœ Ê œ œ Ê œ œ ijF Fab ab ab
### #
#
††
d d
dt dt
r rœ œ ''
C0
1
t t Flow Flow Flow 1œ œ Ê œ œœœ ‘
#$
"
!
""
"#
22
33 33
30. From (1 0) to (0 1): (1 t) t , 0 t 1, and (x y) x y ,ß ß œ ŸŸ œ Ê œrij Fi j ij "
## ab
d
dt
r&
1 2t 2t , and 2t 2t Flux 2t 2t dtFi j nv ij Fnvœ œ Ê œ Ê œ ab kk kk ab
###
"" " "
† "
'
0
1
tt ;œ œœ ‘
#$
"
!
"2
33
From (0 1) to ( 1 0): t (1 t) , 0 t 1, and (x y) x y ,ßß œŸŸ œÊœri j F i j ij #
## ab
d
dt
rœ
(1 2t) 1 2t 2t , and (2t 1) 1 2t 2t 2 4t 2tFi jnvijFnvœ œ Ê œ œ ab kk kk ab
# ##
## ##
†
Flux 2 4t 2t dt 2t 2t t ;Êœœœ#
##$
"
!
'
0
1
ab œ ‘
22
33
From ( 1 0) to (1 0): ( 1 2t) , 0 t 1, and (x y) x y 2 ,ß ß œ Ÿ Ÿ œ Ê œri Fiji $
## ab
d
dt
r
( 1 2t) 1 4t 4t , and 2 2 1 4t 4tFi jnvjFnvœ œ Ê œ ab kk kk ab
##
$$ $$
†
Flux 2 1 4t 4t dt 2 t 2t t Flux Flux Flux FluxÊ œ œ œ Ê œ œœ$ "#$
##$
"
!
""
'
0
1
ab œ ‘
42 22
3 3 333 3
31. on x y 4;Fijœ œ
y
xy xy
x
ÈÈ
œœ œœ

##
at (2 0), ; at (0 2), ; at ( 2 0),ßœ ßœßFj F i
; at ( 2), ; at 2 2 , ;Fj Fi F ijœ !ß œ ß œ Š‹
ÈÈ
È
3
##
"
at 2 2 , ; at 2 2 ,Š‹ Š‹
ÈÈ ÈÈ
ß œ ßFij
È
3
##
"
; at 2 2 , Fij Fijœ ß œ
ÈÈ
33
## ##
""
Š‹
ÈÈ

1010. 1006 Chapter 16 Integration in Vector Fields
32. x y on x y 1; at (1 0), ;Fij Fiœ œ ß œ
##
at ( 1 0), ; at (0 1), ; at (0 1),ß œ ß œ ßFi Fj
; at , ;Fj Fi jœ ß œ Š‹
""
## # #
ÈÈ
33
at , ;Š‹ß œ
""
## # #
ÈÈ
33
Fij
at , ;Š‹
""
## # #
ß œ
ÈÈ
33
Fi j
at , .Š‹ß œ
""
## # #
ÈÈ
33
Fij

33. (a) P(x y) Q(x y) is to have a magnitude a b and to be tangent to x y a b in aGijœßß œ
È
## ####
counterclockwise direction. Thus x y a b 2x 2yy 0 y is the slope of the tangent
#### w w
œ Ê œÊ œ
x
y
line at any point on the circle y at (a b). Let b a a b , with in aÊœ ß œÊ œ
w ##a
b
vijv v kk
È
counterclockwise direction and tangent to the circle. Then let P(x y) y and Q(x y) xߜ ߜ
y x for (a b) on x y a b we have b a and a b .Ê œ Ê ß œ œ œ Gij GijG
#### ##
kk
È
(b) x y a b .GF Fœœˆ‰È Š‹
È
## ##
34. (a) From Exercise 33, part a, y x is a vector tangent to the circle and pointing in a counterclockwiseij
direction y x is a vector tangent to the circle pointing in a clockwise direction Ê Êœij G
y x
x y
i jœ
È
œœ
is a unit vector tangent to the circle and pointing in a clockwise direction.
(b) GFœ
35. The slope of the line through (x y) and the origin is x y is a vector parallel to that line andßÊœ
y
x
vij
pointing away from the origin is the unit vector pointing toward the origin.ÊœF
x y
x y
i j
È
œœ
36. (a) From Exercise 35, is a unit vector through (x y) pointing toward the origin and we wantÎ(
x y
x y
i j
È
œœ
to have magnitude x y x y x y .kk ÈÈ Š‹FFij
## ##

Êœ œ
x y
x y
i j
È
œœ
(b) We want where C 0 is a constant C .kk Š‹Š‹FFœ Á Êœœ
CC
x y x y x y
x y x y
x yÈÈÈ
œœ œœ œœ œœ



i j i j
37. 4t 8t 2 and 2t 12t Flow 12t dt 3t 48Fijk ijFœ œ Ê œ Ê œ œ œ
$# $ $ %
#
!
dd
dt dt
rr
† '
0
2
cd
38. 12t 9t and 3 4 72t Flow 72t dt 24t 24Fjk jkFœ œÊ œÊ œ œ œ
## # # $
"
!
dd
dt dt
rr
† '
0
1
cd
39. (cos t sin t) (cos t) and ( sin t) (cos t) sin t cos t 1Fik ikFœ œ Ê œ
dd
dt dt
rr
†
Flow ( sin t cos t 1) dt cos t t 0Êœ œ œœ'
0
œ
œ ‘ˆ ‰ˆ ‰
"""#
! ##2
1
11
40. ( 2 sin t) (2 cos t) 2 and (2 sin t) (2 cos t) 2 4 sin t 4 cos t 4 0Fijk ijkFœ œ Ê œ œ
dd
dt dt
rr
†
##
Flow 0ʜ
41. C : (cos t) (sin t) t , 0 t (2 cos t) 2t (2 sin t) and ( sin t) (cos t)"
#rijk F ij k ijkœŸŸÊœ œ
1 d
dt
r
2 cos t sin t 2t cos t 2 sin t sin 2t 2t cos t 2 sin tÊœ œ F†
d
dt
r

1011. Section 16.2 Vector Fields, Work, Circulation, and Flux 1007
Flow ( sin 2t 2t cos t 2 sin t) dt cos 2t 2t sin t 2 cos t 2 cos t 1 ;Êœ œ œ"
"
Î#
!
'
0
2
œ"
œ ‘
2
1
1
C : (1 t) , 0 t 1 (1 t) 2 and #
# #rj k F j k k Fœ ŸŸ Ê œ œ Ê œ
1 1
11
dd
dt dt
rr
†
Flow dt t ;Êœœœ
#
"
!
'
0
1
111cd
C : t (1 t) , 0 t 1 2t 2(1 t) and 2t$ri j F i k ij Fœ ŸŸ Ê œ œÊ œ
dd
dt dt
rr
†
Flow 2t dt t 1 Circulation ( 1 ) 1 0Êœ œœÊ œœ
$
#
"
!
'
0
1
cd 11
42. x y z , where f(x y z) x y x f tF Fr† †
d dx dz f dx f f dz d d
dt dt dt dt x dt y dt z dt dt dt
dy dyr r
œœ ßßœ Ê œ
``` "
``` #
###
ab ab abab
by the chain rule Circulation dt f t dt f b f a . Since C is an entire ellipse,Êœœ œ ''
Ca
b
Frrr†
dd
dt dt
r
abababab ab abab
b a , thus the Circulation 0.rrab abœœ
43. Let x t be the parameter y x t and z x t t t t , 0 t 1 from (0 0 0) to (1 1 1)œ Êœœ œœÊœ ŸŸ ßß ßß
## #
ri jk
2t and xy y yz t t t t 2t t 2t Flow 2t dtÊ œ œ œ Ê œ œ Ê œ
d d
dt dt
r r
ijkF ij kijkF
$#$ $ $$ $ $
† '
0
1
œ
"
#
44. (a) xy z , where f(x y z) xy z dtFF FœÊœœ ßߜʙ† †ab
#$ #$```
```
d f dx f z dz df d
dt x dt y dt z dt dt dt
dyr r )
C
f t dt f b f a 0 since C is an entire ellipse.œœœ '
a
b
d
dt
abababab ab ababrrr
(b) xy z dt xy z (2)(1) ( 1) (1)(1) (1) 2 1 3''
C 111
21 1
F†
dd
dt dt
r
œœœœœ
Ÿßß_
Ÿßßœ_
abcd
#$ #$ # $ # $
#Á"Áœ"#
Ð"ß"ß"Ñ
45. Yes. The work and area have the same numerical value because work d y dœœ ''
CC
Fr ir††
[f(t) ] dt [On the path, y equals f(t)]œ'
b
a
ii j†œ ‘
df
dt
f(t) dt Area under the curve [because f(t) 0]œœ '
a
b
46. x y x f(x) f (x) ; (x y ) has constant magnitude k and points awayriji j i jF ijœœ Ê œ œ
dk
dx x y
r w
È
œœ
from the origin k x [f(x)] , by the chain ruleÊœœ œ F†
dkx d
dx dxx y x y x [f(x)]
k y f (x) kx k f(x) f (x)r
ÈÈ È
œœ œœ œ œ
__

†† † † ##
È
ds dx k x [f(x)] dx k x [f(x)]Êœ œ œ'' '
CC a
b
b
a
FT F††
dd
dx dx
r ÈÈ œ ‘## ##
k b [f(b)] a [f(a)] , as claimed.œˆ‰ÈÈ
####
47-52. Example CAS commands:
:Maple
with( LinearAlgebra );#47
F := r -> < r[1]*r[2]^6 | 3*r[1]*(r[1]*r[2]^5+2) >;
r := t -> < 2*cos(t) | sin(t) >;
a,b := 0,2*Pi;
dr := map(diff,r(t),t); # (a)
F(r(t)); # (b)
q1 := simplify( F(r(t)) . dr ) assuming t::real; # (c)
q2 := Int( q1, t=a..b );
value( q2 );
: (functions and bounds will vary):Mathematica
Exercises 47 and 48 use vectors in 2 dimensions
Clear[x, y, t, f, r, v]
f[x_, y_]:= {x y , 3x (x y 2)}
65

1012. 1008 Chapter 16 Integration in Vector Fields
{a, b}={0, 2 };1
x[t_]:= 2 Cos[t]
y[t_]:= Sin[t]
r[t_]:={x[t], y[t]}
v[t_]:= r'[t]
integrand= f[x[t], y[t]] . v[t] //Simplify
Integrate[integrand,{t, a, b}]
N[%]
If the integration takes too long or cannot be done, use NIntegrate to integrate numerically. This is suggested for exercises
49 - 52 that use vectors in 3 dimensions. Be certain to leave spaces between variables to be multiplied.
Clear[x, y, z, t, f, r, v]
f[x_, y_, z_]:= {y y z Cos[x y z], x x z Cos[x y z], z x y Cos[x y z]}
2
{a, b}={0, 2 };1
x[t_]:= 2 Cos[t]
y[t_]:= 3 Sin[t]
z[t_]:= 1
r[t_]:={x[t], y[t], z[t]}
v[t_]:= r'[t]
integrand= f[x[t], y[t],z[t]] . v[t] //Simplify
NIntegrate[integrand,{t, a, b}]
16.3 PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE FIELDS
1. x , y , z Conservative
``` `` `
``````
PNMPNM
yzzxxy
œœ œœ œœ Ê
2. x cos z , y cos z , sin z Conservative
``` ```
``````
PNMPNM
yzzxxy
œœ œœœœÊ
3. 1 1 Not Conservative 4. 1 1 Not Conservative
`` ` `
`` ` `
PN NM
yz x y
œ Á œ Ê œ Á œ Ê
5. 0 1 Not Conservative
``
``
NM
xy
œÁœ Ê
6. 0 , 0 , e sin y Conservative
``` `` `
````` `
PNMPN M
yzzxx y
x
œœ œœ œ œ Ê
7. 2x f(x y z) x g(y z) 3y g(y z) h(z) f(x y z) x h(z)
``
```##
# #`ff
xyy
g3y 3y
œÊßßœßÊœœÊßœ Êßßœ
œœ
h (z) 4z h(z) 2z C f(x y z) x 2z CÊœ œÊ œÊßßœ
`
` #
w###f
z
3y
œ
8. y z f(x y z) (y z)x g(y z) x x z z g(y z) zy h(z)
``
````
``ff
xyyy
gg
œ Ê ßß œ ß Ê œ œ Ê œ Ê ß œ
f(x y z) (y z)x zy h(z) x y h (z) x y h (z) 0 h(z) C f(x y z)Êßßœ Êœ œÊ œÊ œÊßß
`
`
wwf
z
(yz)xzyCœ
9. e f(x y z) xe g(y z) xe xe 0 f(x y z)
``
````
``ff
xyyy
y 2z y 2z y 2z y 2z gg
œÊßßœßÊœœ ÊœÊßß
xe h(z) 2xe h (z) 2xe h (z) 0 h(z) C f(x y z) xe CœÊœ œ ʜʜÊßßœ
y 2z y 2z y 2z y 2zf
z
ww `
`
10. y sin z f(x y z) xy sin z g(y z) x sin z x sin z 0 g(y z) h(z)
``
````
``ff
xyyy
gg
œÊßßœ ßÊœœÊœÊßœ
f(x y z) xy sin z h(z) xy cos z h (z) xy cos z h (z) 0 h(z) C f(x y z)Êßßœ Êœ œ Ê œÊ œÊßß
`
`
wwf
z
xy sin z Cœ

1013. Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1009
11. f(x y z) ln y z g(x y) ln x sec (x y) g(x y)
`" `
` # ``
## # `fz f
zyz x x
g
œÊßßœ ßÊœœÊß
œœ ab
(x ln x x) tan (x y) h(y) f(x y z) ln y z (x ln x x) tan (x y) h(y)œ Êßßœ
"
#
##
ab
sec (x y) h (y) sec (x y) h (y) 0 h(y) C f(x y z)Êœ œ Ê œÊ œÊßß
`
`
#w# wf
yyz yz
yy
œœ œœ
ln y z (x ln x x) tan (x y) Cœ
"
#
##
ab
12. f(x y z) tan (xy) g(y z)
``
` ` `
œ" `
œ
ffxxz
x1xy y1xy y1xy
y g
1yz
œ Êßßœ ßÊœœ
œœ œœ œœ œœ È
g(y z) sin (yz) h(z) f(x y z) tan (xy) sin (yz) h(z)Êœ Êßœ Êßßœ
`
` œ
œ" œ" œ"g
y
z
1yzÈ
œœ
h (z) h (z) h(z) ln z CÊ Êœ œ Ê œÊ œ
`""
` œœ
wwf
zzz
yy
1yz 1yzÈÈ
œœ œœ
kk
f(x y z) tan (xy) sin (yz) ln z CÊßßœ
œ" œ"
kk
13. Let (x y z) 2x 2y 2z 0 , 0 , 0 M dx N dy P dz isFijkßßœ Êœœ œœ œœ Ê
``` `` `
````` `
PNMPNM
yzzxx y
exact; 2x f(x y z) x g(y z) 2y g(y z) y h(z) f(x y z) x y h(z)
``
```
#### `ff
xyy
g
œ Ê ßßœ ßÊ œ œ Ê ßœ Ê ßßœœ
h (z) 2z h(z) z C f(x y z) x y z C 2x dx 2y dy 2z dzÊ œ œ Ê œÊ ßßœÊ
`
`
w# ###f
z '
&((`
Ÿßßœ_000
23 6
f(2 3 6) f( ) 2 3 ( 6) 49œ ß ß !ß!ß! œ œ
## #
14. Let (x y z) yz xz xy x , y , z M dx N dy P dz isFijkßßœ Ê œœ œœ œœ Ê
``` `` `
``````
PNMPNM
yzzxxy
exact; yz f(x y z) xyz g(y z) xz xz 0 g(y z) h(z) f(x y z)
``
````
``ff
xyyy
gg
œÊßßœ ßÊœœÊœÊßœ Êßß
xyz h(z) xy h (z) xy h (z) 0 h(z) C f(x y z) xyz Cœ Êœ œÊ œÊ œÊßßœ
`
`
wwf
z
yz dx xz dy xy dz f(3 5 0) f(1 1 2) 0 2 2Ê œ ßß ßß œœ'
&((`
&((`112
350
15. Let (x y z) 2xy x z 2yz 2z , 0 , 2xFijkßß œ Ê œ œ œ œ œ œab
## ``````
``````
PNMPNM
yzzxxy
M dx N dy P dz is exact; 2xy f(x y z) x y g(y z) x x z zÊ œÊßßœßÊœœÊœ
``
````
##### ``ff
xyyy
gg
g(y z) yz h(z) f(x y z) x y yz h(z) 2yz h (z) 2yz h (z) 0 h(z) CÊ ß œ Ê ß ß œ Ê œ œ Ê œ Ê œ
### ww `
`
f
z
f(x y z) x y yz C 2xy dx x z dy 2yz dz f( ) f( ) 2 2(3) 16Ê ß ß œ Ê œ "ß#ß$ !ß!ß! œ œ
## ## # '
&((`
&((`000
123
ab
16. Let (x y z) 2x y 0 , 0 , 0Fijkßß œ Ê œœ œœ œœ
#
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h(z) ln z C f(x y z) 2x cos y ln y ln z CÊœ Êßßœ kk kk kk
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M dx N dy P dz is exact is conservative path independenceÊ Ê Ê F
27. 0 , 0 , is conservative there exists an f so that f;
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3x 3
#
8. M x y, N x y 1, 1, 2x, 2y Flux (1 2y) dx dyœ œ Ê œ œ œ œ Ê œ ab
## ``` `
``` `
MMN N
xyx y ''
R
(1 2y) dy dx x x dx ; Circ ( 2x 1) dx dy ( 2x 1) dy dxœ œœœœ'' ' ''
00 0 00
1x 1 1x
ab
# "
6 ''
R
2x x dxœ œ'
0
1
ab
# 7
6
9. M x e sin y, N x e cos y 1 e sin y, e cos y, 1 e cos y, e sin yœ œ Ê œ œ œ œ
xx xx xx MMN N
xyxy
````
````
Flux dx dy r dr d cos 2 d sin 2 ;Êœ œ œ œ œ''
R
'' '
œ"œ"
__
œœ
œ 40 4
4cos2 4
œ
))))ˆ‰ œ ‘
"""
##
Î%
Î%4
1
1
Circ 1 e cos y e cos y dx dy dx dy r dr d cos 2 dœ œ œ œ œ'' ''
RR
ab ˆ‰
xx '' '
œ"œ"
__
œœ
œ 40 4
4cos2 4
œ
)))
""
##
10. M tan , N ln x y , , , œ œÊœœœœ
œ" # # ````
````
œyy 2y
x x xy y xy x xy y xy
MMxN2xN
ab
œœ œœ œœ œœ
Flux dx dy r dr d sin d 2;Êœ œ œ œ''
R
Š‹ ˆ‰
œ

y2y
xy xy r
r sin
œœ œœ œ
'' '
01 0
2
œœ
)
)))
Circ dx dy r dr d cos d 0œ œ œ œ''
R
Š‹ ˆ‰
2x x r cos
xy xy r
œœ œœ œ

'' '
01 0
2
œœ
)
)))
11. M xy, N y y, x, 0, 2y Flux (y 2y) dy dx 3y dy dxœœÊœœœœÊœ œ
# ````
````
MMNN
xyxy ''
R

''
0x
1x

dx ; Circ x dy dx x dy dx x x dxœ œ œœœœ''''
0 0x 0
1 1x 1
Š‹ ab
3x 3x
5 1
œ Î
## #
" "#$
''
R


12. M sin y, N x cos y 0, cos y, cos y, x sin yœ œ Ê œ œ œ œ
`` ` `
`` ` `
MM N N
xy x y
Flux ( x sin y) dx dy ( x sin y) dx dy sin y dy ;Êœœœœ''
R
'' '
00 0
22 2
œœ œ__ _
Š‹
11
œœ
88
Circ [cos y ( cos y)] dx dy 2 cos y dx dy cos y dy sin yœ œ œ œ œ''
R
'' '
00 0
22 2
œœ œ__ _
111 cd
1Î#
!
13. M 3xy , N e tan y 3y , œ œ Êœ œ
xMN
1y x 1y y 1y
x
``
œ" `"`"
œ œœ
Flux 3y dx dy 3y dx dy (3r sin ) r dr dÊœ œ œ '' ''
RR

Š‹
""
1y 1y
œœ
''
00
2a1cos
œ &) `
))
a (1 cos ) (sin ) d (1 cos ) 4a 4a 0œ œ œœ'
0
2
œ
$$ %$$
#
!
))) ) ’“ ab
a
4

1
14. M y e ln y, N 1 , Circ 1 dx dy ( 1) dx dyœ œ Ê œ œ Ê œ œ
x eM eNe e e
yy yxy y y
xxx xx
``
`` '' ''
RR
’“Š‹
dy dx 3 x x 1 dx x x 2 dxœ œ œ œ'' ' '
œ  œ œ
œ1x 1 1 1
13x 1 1
Î

cdababab
#% %# 44
15
15. M 2xy , N 4x y 6xy , 8xy work 2xy dx 4x y dy 8xy 6xy dx dyœœÊœ œÊœ œ
$## # # $## ## ``
``
MN
yx )
C
''
R
ab
2xy dy dx x dxœœœ'' '
00 0
1x 1
_
#"! 22
333

1019. Section 16.4 Green's Theorem in the Plane 1015
16. M 4x 2y, N 2x 4y 2, 2 work (4x 2y) dx (2x 4y) dyœ œ Ê œ œÊ œ
``
``
MN
yx )
C
[2 ( 2)] dx dy 4 dx dy 4(Area of the circle) 4( 4) 16œ œ œ œ œ'' ''
RR
11†
17. M y , N x 2y, 2x y dx x dy (2x 2y) dy dxœœÊœ œÊ œ
## ## ``
``
MN
yx )
C
''
R
(2x 2y) dy dx 3x 4x 1 dx x 2x x 1 2 1 0œœœœœ'' '
00 0
11x 1
œ
abcd
#$#
"
!
18. M 3y, N 2x 3, 2 3y dx 2x dy (2 3) dx dy 1 dy dxœœÊœœÊ œ œ
``
``
MN
yx )''
C00
sin x
''
R
œ
sin x dx 2œ œ'
0
œ
19. M 6y x, N y 2x 6, 2 (6y x) dx (y 2x) dy (2 6) dy dxœ œ Ê œ œÊ œ
``
``
MN
yx )
C
''
R
4(Area of the circle) 16œ œ 1
20. M 2x y , N 2xy 3y 2y, 2y 2x y dx (2xy 3y) dy (2y 2y) dx dy 0œ œ Ê œ œ Ê œ œ
# #``
``
MN
yx )
C
ab ''
R
21. M x a cos t, N y a sin t dx a sin t dt, dy a cos t dt Area x dy y dxœœ œœ Ê œ œ Ê œ
"
#)
C
a cos t a sin t dt a dt aœ œœ
""
##
## ## # #''
00
22
œœ
ab 1
22. M x a cos t, N y b sin t dx a sin t dt, dy b cos t dt Area x dy y dxœœ œœ Ê œ œ Ê œ
"
#)
C
ab cos t ab sin t dt ab dt abœ œœ
""
##
##''
00
22
œœ
ab 1
23. M x a cos t, N y sin t dx 3 cos t sin t dt, dy 3 sin t cos t dt Area x dy y dxœœ œœ Ê œ œ Ê œ
$$ # # "
# )
C
3 sin t cos t cos t sin t dt 3 sin t cos t dt sin 2t dt sin u duœœœœ
""
##
## # # ## # #''''
0000
2224
œœœœ
aba bab
33
816
œ œ
3 u sin 2u 3
16 2 4 8
œ ‘
%
!
1
1
24. M x t , N y t dx 2t dt, dy t 1 dt Area x dy y dxœœ œœ Ê œ œ Ê œ
## "
#
t
3

ab )
C
t t 1 t (2t) dt t t dt t t 9 3 15 3œœœœ
""" "
##
## % # & $
$
œ$''
œœ
œœ
œœ33
33
’“ Š‹ab Š‹ ˆ‰ œ ‘ ÈÈ
t111
33215315
È
È
3œ
8
5È
25. (a) M f(x), N g(y) 0, 0 f(x) dx g(y) dy dx dyœœÊœœÊ œ
`` ``
`` ``
MN NM
yx xy )
C
''
R
Š‹
0 dx dy 0œœ
''
R
(b) M ky, N hx k, h ky dx hx dy dx dyœœÊœœÊ œ
`` ``
`` ``
MN NM
yx xy )
C
''
R
Š‹
(h k) dx dy (h k)(Area of the region)œ œ
''
R 26. M xy , N x y 2x 2xy, 2xy 2 xy dx x y 2x dy dx dyœœÊœ œÊ œ
## # # `` ``
`` ``
MN NM
yx xy )
C
ab Š‹ ''
R
(2xy 2 2xy) dx dy 2 dx dy 2 times the area of the squareœ œ œ'' ''
RR

1020. 1016 Chapter 16 Integration in Vector Fields
27. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Green's
Theorem, with M 4x y and N x , 4x y dx x dy x 4x y dx dyœœ œ
$%$% %$ ``
`` )
C
''
R
’“ab a b
xy
4x 4x dx dy 0.œ œ''
R
ðóóñóóò
ab
$$
0
28. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Green's theorem, with
M x and N y , y dy x dx y x dx dy 0.œœœ œ
$$$$ $$ ``
`` )
C
''
R
”•ðñò
ab ab
ï
xy
0 0
29. Let M x and N 0 1 and 0 M dy N dx dx dy x dyœœÊœ œÊ œ Ê
`` ``
`` ``
MN MN
xy xy ))
C C
''
R
Š‹
(1 0) dx dy Area of R dx dy x dy; similarly, M y and N 0 1 andœ Ê œ œ œ œÊœ'' ''
RR
)
C
`
`
M
y
0 M dx N dy dy dx y dx (0 1) dy dx y dx
```
```
NNM
xxy
œÊ œ Ê œ Ê)))
C CC
'' ''
RR
Š‹
dx dy Area of Rœœ
''
R
30. f(x) dx Area of R y dx, from Exercise 29')
aC
b
œœ
31. Let (x y) 1 x Ax x dA (x 0) dx dy$ߜʜ œ œ œ Ê œ œ
M
MA
x (x y) dA x dA x dA
(x y) dA dA
y
'' '' ''
'' ''
RRR
RR
$
$
ß
ß
'' ''
RR
dy, Ax x dA (0 x) dx dy xy dx, and Ax x dA x x dx dyœœœœ œœ)
C
C
x 2
33
œ
#
"
'' '' ''
RR R R
)'' ˆ‰
x dy xy dx x dy xy dx x dy xy dx AxœÊ œœœ))))
CCCC
"" " "## #
#33 3
32. If (x y) 1, then I x (x y) dA x dA x 0 dy dx x dy,$$ßœ œ ß œ œ œ
y
C
'' '' ''
RRR
### $ "
ab
3
)
x dA 0 x dy dx x y dx, and x dA x x dy dx'' '' '' ''
RR RR
###### "
œ œ œ ab ˆ‰ )
C
3
44
x dy x y dx x dy x y dx x dy x y dx x dy x y dx IœœÊ œœœ)))))
CCCCC
y
"" " " "$ # $# $ # $#
44 4 3 4
33. M , N , dx dy dx dy 0 for suchœœÊœ œÊ œ œ
```````` ``
```````` ``
ffMfNfff ff
yxyyxxyx xy
œœ œœ
œœ œœ
)
C
''
R
Š‹
curves C
34. M x y y , N x x y , 1 Curl 1 x y 0 in the interior ofœ œÊ œ œÊ œ œ
"" ` ` `` "# $ ## ##
`` ``43 y4 x xy 4
M1 N N M
ˆ‰
the ellipse x y 1 work d 1 x y dx dy will be maximized on the region
""## ##
44
œÊ œ œ '
C
Fr† ''
R
ˆ‰
R {(x y) | curl } 0 or over the region enclosed by 1 x yœß   œ F
"##
4
35. (a) f M , N ; since M, N are discontinuous at (0 0), we™œ Êœœ ߊ‹Š‹
2x 2x
xy xy xy xy
2y 2y
œœ œœ œœ œœ

ij
compute f ds directly since Green's Theorem does not apply. Let x a cos t, y a sin t dx a sin t dt,
'
C
™†n œœÊœ dy a cos t dt, M cos t, N sin t, 0 t 2 , so f ds M dy N dxœœœŸŸ œ
22
aa
1''
CC
™†n
cos t a cos t sin t a sin t dt 2 cos t sin t dt 4 . Note that this holds for anyœœœ''
0 0
2 2
œ œ
œ ‘ˆ‰ ˆ‰ab a b a b
22
aa
22
1

1021. Section 16.4 Green's Theorem in the Plane 1017
a 0, so f ds 4 for any circle C centered at 0, 0 traversed counterclockwise and f ds 4œ œ ''
C C
™† ™†n n1 1ab
if C is traversed clockwise.
(b) If K does not enclose the point (0 0) we may apply Green's Theorem: f ds M dy N dxߜ
''
CC
™†n
dx dy dx dy 0 dx dy 0. If K does enclose the pointœ œ œ œ'' '' ''
RR R
Š‹ Š ‹
``
``
œœ

MN
xy
2y x 2x y
xy xy
ˆ‰ˆ‰
abab
22 22
22 22
22
(0 0) we proceed as in Example 6:ß
Choose a small enough so that the circle C centered at (0 0) of radius a lies entirely within K. Green's Theoremß applies to the region R that lies between K and C. Thus, as before, 0 dx dy
R
œ ''Š‹
``
``
MN
xy
M dy N dx M dy N dx where K is traversed counterclockwise and C is traversed clockwise.œ ''
KC
Hence by part (a) 0 4 4 f ds. We have shown: M dy N dx M dy N dxœÊœœ ÎÎ’“ '' '
KK K
11 ™†n
f ds
0 if (0 0) lies inside K
4 if (0 0) lies outside K
'
K
™†nœ
ß
ß
œ
1
36. Assume a particle has a closed trajectory in R and let C be the path C encloses a simply connected region
"" Ê
R C is a simple closed curve. Then the flux over R is ds 0, since the velocity vectors are
"" "ʜ )
Cœ
Fn F†
tangent to C . But 0 ds M dy N dx dx dy M N 0, which is a"
``
`` œœœÊœ))
CC
xyœœ
Fn† ''
RÎ
Š‹
MN
xy
contradiction. Therefore, C cannot be a closed trajectory."
37. dx dy N(g (y) y) N(g (y) y) dx dy [N(g (y) y) N(g (y) y)] dy''''
gy cgy c
gy dgy dœ œ
&` &`
&` &`
``
``
#" #"
NN
xx
œßßÊ œ ßß ˆ‰
N(g (y) y) dy N(g (y) y) dy N(g (y) y) dy N(g (y) y) dy N dy N dyœßßœßßœ
''''''
cc cd CC
dd dc
#" #"
œ
dy N dy dx dy
R
œÊ œ))
CC
''
`
`
N
x
38. dy dx [M(x d) M(x c)] dx M(x d) dx M(x c) dx M dx M dx.'' ' ' ' ' '
ac a a a C C
bd b b b
`
`
M
y
œßßœßßœ
3 œ
Because x is constant along C and C , M dx M dx 0#%
''
CC
œœ
M dx M dx M dx M dx M dx dy dx M dx.Ê œÊ œŠ‹ '''' ) '' )
CCCC C ac C
bdœ

`
`
M
y
39. The curl of a conservative two-dimensional field is zero. The reasoning: A two-dimensional field M NFijœ
can be considered to be the restriction to the xy-plane of a three-dimensional field whose k component is zero,
and whose and components are independent of z. For such a field to be conservative, we must haveij
by the component test in Section 16.3 curl 0.
`` ``
`` ``
NM NM
xy xy
œÊœœ F
40. Green's theorem tells us that the circulation of a conservative two-dimensional field around any simple closed
curve in the xy-plane is zero. The reasoning: For a conservative field M N , we have Fijœ œ
``
``
NM
xy
(component test for conservative fields, Section 16.3, Eq. (2)), so curl 0. By Green's theorem,Fœœ
``
``
NM
xy
the counterclockwise circulation around a simple closed plane curve C must equal the integral of curl over theF
region R enclosed by C. Since curl 0, the latter integral is zero and, therefore, so is the circulation.Fœ
The circulation ds is the same as the work d done by around C, so our observation that
))
CC
FT F r F††
circulation of a conservative two-dimensional field is zero agrees with the fact that the work done by a
conservative field around a closed curve is always 0.

1022. 1018 Chapter 16 Integration in Vector Fields
41-44. Example CAS commands:
:Maple
with( plots );#41
M := (x,y) -> 2*x-y;
N := (x,y) -> x+3*y;
C := x^2 + 4*y^2 = 4;
implicitplot( C, x=-2..2, y=-2..2, scaling=constrained, title="#41(a) (Section 16.4)" );
curlF_k := D[1](N) - D[2](M): # (b)
'curlF_k' = curlF_k(x,y);
top,bot := solve( C, y ); # (c)
left,right := -2, 2;
q1 := Int( Int( curlF_k(x,y), y=bot..top ), x=left..right );
value( q1 );
: (functions and bounds will vary)Mathematica
The command will be useful for 41 and 42, but is not needed for 43 and 44. In 44, the equation of the lineImplicitPlot
from (0, 4) to (2, 0) must be determined first.
Clear[x, y, f]
<<Graphics`ImplicitPlot`
f[x_, y_]:= {2x y, x 3y}
curve= x 4y ==4
22

ImplicitPlot[curve, {x, 3, 3},{y, 2, 2}, AspectRatio Automatic, AxesLabel {x, y}];ÎÎ ‚ ‚
ybounds= Solve[curve, y]
{y1, y2}=y/.ybounds;
integrand:=D[f[x,y][[2]], x] D[f[x,y][[1]], y]//SimplifyÎ
Integrate[integrand, {x, 2, 2}, {y, y1, y2}]Î
N[%]
Bounds for y are determined differently in 43 and 44. In 44, note equation of the line from (0, 4) to (2, 0).
Clear[x, y, f]
f[x_, y_]:= {x Exp[y], 4x Log[y]}
2
ybound = 4 2xÎ
Plot[{0, ybound}, {x, 0,2. 1}, AspectRatio Automatic, AxesLabel {x, y}];ÄÄ
integrand:=D[f[x, y][[2]], x] D[f[x, y][[1]], y]//SimplifyÎ
Integrate[integrand, {x, 0, 2}, {y, 0, ybound}]
N[%]
16.5 SURFACE AREA AND SURFACE INTEGRALS
1. , f 2x 2y f (2x) (2y) ( 1) 4x 4y 1 and f 1;pk i jk pœ œÊ œ œ œ™™ ™† kk k kÈÈ
## # ##
z 2 x y 2; thus S dA 4x 4y 1 dx dyœÊ œ œ œ
## ## '' ''
RR
kk
kk
™
™†
f
fp
È
4r cos 4r sin 1 r dr d 4r 1 r dr d 4r 1 dœ œ œ ''
R
ÈÈ
’“ab## ## # " #
$Î#
#
!
))) ) ) '' '
00 0
22 2
œœ
œ
12
È
dœœ'
0
2
œ
13 13
63
)1
2. , f 2x 2y f 4x 4y 1 and f 1; 2 x y 6pk i jk pœ œÊ œ œ ŸŸ™™ ™† kk k kÈ
## ##
S dA 4x 4y 1 dx dy 4r 1 r dr d 4r 1 r dr dÊœ œ œ œ '' '' ''
RR R
kk
kk
™
™†
f
fp
È ÈÈ## # #
))''
02
26
œ
œ
œ
4r 1 d dœ œœ''
00
22
œœ
’“ab
" #
$Î#
'
#
12 6 3
49 49
È
È
))1

1023. Section 16.5 Surface Area and Surface Integrals 1019
3. , f 2 2 f 3 and f 2; x y and x 2 y intersect at (1 1) and (1 1)pk i j k pœœÊœ œœ œ ß ß™™™† kk k k
##
S dA dx dy dx dy 3 3y dy 4Êœ œ œ œ œ'' ''
RR
kk
kk
™
™†
f
f
33
p ##
# '' '
œœ
œ1y 1
12y 1


ab
4. , f 2x 2 f 4x 4 2 x 1 and f 2 S dApk i k pœ œÊ œœ œÊœ™™ ™† kk k k
ÈÈ
## ''
R
kk
kk
™
™†
f
fp
dx dy x 1 dy dx x x 1 dx x 1 (4)œœœœœœ''
R
2x 1
3333
7
È
È
œ

#
## """# $Î#
$Î#
$
! '' '
00 0
3x 3
œœ
ÈÈ
’“ab
5. , f 2x 2 2 f (2x) ( 2) ( 2) 4x 8 2 x 2 and f 2pk i j k pœ œ Ê œ œ œ œ™™ ™† kk k kÈ ÈÈ
### # #
S dA dx dy x 2 dy dx 3x x 2 dx x 2Êœ œ œ œ œ '' ''
RR
kk
kk
È™
™†
f
f
2x 2
p
œ

#
## #
$Î#
#
! '' '
00 0
23x 2
ÈÈ
’“ab
66 22œ
È È
6. , f 2x 2y 2z f 4x 4y 4z 8 2 2 and f 2z; x y z 2 andpk i j k pœ œ Ê œ œ œ œ œ™™ ™† kk k kÈ È È
### ###
z x y x y 1; thus, S dA dA 2 dAœÊœ œ œ œÈ È
## ##
#
" '' '' ''
RR R
kk
kk
È™
™†
f
fz z
22
p
2 dA 2 2 1 2 d 2 2 2œœœœ
ÈÈÈÈÈ
Š‹Š‹ ''
R
"
œ
ÈÈab2xy
r dr d
2r
œœ œ
'' '
00 0
21 2
œœ
)
)1
7. , f c f c 1 and f 1 S dA c 1 dx dypk ik pœ œÊ œ œÊœ œ ™™ ™† kk k k
ÈÈ
# # '' ''
RR
kk
kk
™
™†
f
fp
c 1 r dr d d c 1œœ œ'' '
00 0
21 2
œœ
ÈÈ##
#
))1
È
c1œ
8. , f 2x 2z f (2x) (2z) 2 and f 2z for the upper surface, z 0pk i j pœœÊœœ œ  ™™ ™† kk k kÈ
##
S dA dA dy dx 2 dy dx dxÊœ œ œ œ œ'' '' ''
RRR
kk
kk ÈÈÈ
™
™†
f
fz
21
1x 1x 1x
p #
""
œœœ
œœœ
'' '
œ"œ"
__ _12 0 12
12 12 12
sin xœœœcd ˆ‰
œ"
"Î#
"Î#
111
663
9. , f 2y 2z f 1 (2y) (2z) 1 4y 4z and f 1; 1 y z 4pi i j k pœ œÊ œœ œŸŸ™™ ™† kk k kÈÈ
### ## ##
S dA 1 4y 4z dy dz 1 4r cos 4r sin r dr dÊœ œ œ '' ''
RR
kk
kk
™
™†
f
fp
È È ## ## ## ''
01
22
œ
)))
14r r drd 14r d 171755 d 171755œœ œ œ'' ' '
01 0 0
22 2 2
œœ œ
È
’“Š‹Š‹ab
ÈÈ ÈÈ # "" #
$Î#
#
"
#
)) )
12 1 6
1
10. , f 2x 2z f 4x 4z 1 and f 1; y 0 and x y z 2 x z 2;pj ij k pœœÊœ œœ œÊœ™™ ™† kk k k
È
## ## ##
thus, S dA 4x 4z 1 dx dz 4r 1 r dr d dœœœ œœ'' ''
RR
kk
kk
™
™†
f
f 63
13 13
p ÈÈ## # '' '
00 0
22 2
œœ
œ
))1
11. , f 2x 15 f 2x 15 ( 1) 4x 8 2xpk i jkœœÊœ œœ™™ˆ‰ ˆ‰ ˆ‰ÈÈ
kk Ê Š‹ É É
22 42
xx xx
##
#
##
œ
2x , on 1 x 2 and f 1 S dA 2x 2x dx dyœ ŸŸ œÊœ œ
2
xf
f
kk ab™†p '' ''
RR
kk
kk
™
™†p
œ"
2x 2x dx dy x 2 ln x dy (3 2 ln 2) dy 3 2 ln 2œ œ œœ'' ' '
01 0 0
12 1 1
ab c d
œ" #
#
"
12. , f 3 x 3 y 3 f 9x 9y 9 3 x y 1 and f 3pk i j k pœ œÊ œœ œ™™ ™†È È kk k kÈÈ
S dA x y 1 dx dy x y 1 dx dy (x y 1) dyÊœ œ œ œ '' ''
RR
kk
kk
™
™†
f
f 3
2
p
ÈÈ œ ‘ '' '
00 0
11 1
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!
(y 2) (y 1) dy (y 2) (y 1) (3) (2) (2) 1œ œ œ '
0
1
œ ‘ œ ‘ œ ‘
22 4 4 4
3 3 15 15 15
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1024. 1020 Chapter 16 Integration in Vector Fields
93 82 1œ
4
15
Š‹
È È
13. The bottom face S of the cube is in the xy-plane z 0 g(x y 0) x y and f(x y z) z 0 ÊœÊ ßßœ ßßœœÊœ pk
and f f 1 and f 1 d dx dy g d (x y) dx dy™™ ™†œÊ œ œÊ œ Ê œ kpkk k k 55
'' ''
S R
(x y) dx dy ay dy a . Because of symmetry, we also get a over the face of the cubeœœœ'' '
00 0
aa a
Š‹
a
œ
#
$$
in the xz-plane and a over the face of the cube in the yz-plane. Next, on the top of the cube, g(x y z)
$
ßß
g(x y a) x y a and f(x y z) z a and f f 1 and f 1 d dx dyœßßœ ßßœœÊœ œÊ œ œÊœ pk k p™™ ™† kk k k 5
g d (x y a) dx dy (x y a) dx dy (x y) dx dy a dx dy 2a .
'' ''
S R
5œ œ œ œ '' '' ''
00 00 00
aa aa aa
$
Because of symmetry, the integral is also 2a over each of the other two faces. Therefore,
$
(xyz) d 3a 2a 9a.''
cube
œ œ5 ab
$$ $
14. On the face S in the xz-plane, we have y 0 f(x y z) y 0 and g(x y z) g(x 0 z) z andœ Ê ßß œ œ ßß œ ßß œ Ê œ pj
f f 1 and f 1 d dx dz g d (y z) d z dx dz 2z dz™™ ™†œÊ œ œ Ê œ Ê œ œ œjpkk k k 555
'' ''
SS
'' '
00 0
12 1
1.œ
On the face in the xy-plane, we have z 0 f(x y z) z 0 and g(x y z) g(x y 0) y andœ Ê ßß œ œ ßß œ ßß œ Ê œ pk
f f 1 and f 1 d dx dy g d y d y dx dy 1.™™ ™†œÊ œ œÊ œ Ê œ œ œkpkk k k 555
'' ''
SS
''
00
12
On the triangular face in the plane x 2 we have f(x y z) x 2 and g(x y z) g(2 y z) y z andœ ßßœœ ßßœ ßßœÊ œ pi
f f 1 and f 1 d dz dy g d (y z) d (y z) dz dy™™ ™†œÊ œ œ Ê œ Ê œ œ ipkk k k 555
'' ''
SS
''
00
11y
œ
1 y dy .œœ'
0
1
""
#
#
ab
3
On the triangular face in the yz-plane, we have x 0 f(x y z) x 0 and g(x y z) g(0 y z) y zœ Ê ßß œ œ ßß œ ßß œ
and f f 1 and f 1 d dz dy g d (y z) dÊœ œÊ œ œÊ œ Ê œ pi i p™™ ™† kk k k 555
'' ''
SS
(y z) dz dy .œœ''
00
11y
œ
"
3
Finally, on the sloped face, we have y z 1 f(x y z) y z 1 and g(x y z) y z 1 andœ Ê ßß œœ ßß œœ Ê œ pk
f f 2 and f 1 d 2 dx dy g d (y z) d™™™†œ Ê œ œ Ê œ Ê œ jk p kk k k
ÈÈ
555 '' ''
SS
2 dx dy 2 2. Therefore, g(x y z) d 1 1 2 2 2 2œ œ ßß œ œ''
00
12
ÈÈ ÈÈ ''
wedge
5
""
33 3
8
15. On the faces in the coordinate planes, g(x y z) 0 the integral over these faces is 0.ßß œ Ê
On the face x a, we have f(x y z) x a and g(x y z) g(a y z) ayz and f f 1œ ßßœœ ßßœ ßßœ Ê œ œÊ œ pi i ™™ kk
and f 1 d dy dz g d ayz d ayz dy dz .kk™†pœÊ œ Ê œ œ œ555 '' ''
SS
''
00
cb
ab c
4
œœ
On the face y b, we have f(x y z) y b and g(x y z) g(x b z) bxz and f f 1œ ßßœœ ßßœ ßßœ Ê œ œÊ œ pj j ™™ kk
and f 1 d dx dz g d bxz d bxz dx dz .kk™†pœÊ œ Ê œ œ œ555 '' ''
SS
''
00
ca
abc
4
œœ
On the face z c, we have f(x y z) z c and g(x y z) g(x y c) cxy and f f 1œ ßßœœ ßßœ ßßœ Ê œ œ Ê œ pk k ™™ kk
and f 1 d dy dx g d cxy d cxy dx dy . Therefore,kk™†pœÊ œ Ê œ œ œ555 '' ''
SS
''
00
ba
abc
4
œœ
g(xyz) d .''
S
ßß œ5
abc(ab ac bc)
4

1025. Section 16.5 Surface Area and Surface Integrals 1021
16. On the face x a, we have f(x y z) x a and g(x y z) g(a y z) ayz and f f 1œ ßßœœ ßßœ ßßœ Ê œ œÊ œ pi i ™™ kk
and f 1 d dz dy g d ayz d ayz dz dy 0. Because of the symmetrykk™†pœÊ œ Ê œ œ œ555
'' ''
SS
''
œœbc
bc
of g on all the other faces, all the integrals are 0, and g(x y z) d 0.''
S
ßß œ5
17. f(x y z) 2x 2y z 2 f 2 2 and g(x y z) x y (2 2x 2y) 2 x y ,ßßœ œÊ œ ßßœ œÊœ™ ijk pk
f 3 and f 1 d 3 dy dx; z 0 2x 2y 2 y 1 x g d (2 x y) dkk k k™™†œ œ Ê œ œ Ê œÊœÊ œ p 555
'' ''
SS
3 (2 x y) dy dx 3 (2 x)(1 x) (1 x) dx 3 2x dx 2œœœœ'' ' '
00 0 0
11x 1 1
œ
œ ‘ Š‹
"
###
# 3x
œ
18. f(x y z) y 4z 16 f 2y 4 f 4y 16 2 y 4 and f 4ßß œ œ Ê œ Ê œ œ œ Ê œ
# ##
™™ ™†jk pk pkk k kÈÈ
d dx dy g d x y 4 dx dy dx dyÊœ Ê œ œ55
2y 4 y 4
4
xy 4ÈÈ ab
œ œ œ

#
##

''
S
'' ''
œœ40 40
41 41
ˆ‰È Š‹
y 4 dy 4y 16œœœœ'
œ4
4
"""#
##
%
!
4333
y 64 56
ab ’“ ˆ‰

19. g(x y z) z, g g 1 and g 1 Flux d ( ) dAßß œ œ Ê œ Ê œ œ Ê œ œp k k p Fn Fk™™™† † †kk k k '' ''
S R
5
3 dy dx 18œœ
''
00
23
20. g(x y z) y, g g 1 and g 1 Flux d ( ) dAßß œ œ Ê œ Ê œ œ Ê œ œ pj j p Fn Fj™™ ™† † †kk k k '' ''
S R
5
2 dz dx 2(7 2) dx 10(2 1) 30œœœœ
'' '
œœ12 1
27 2
21. g 2x 2y 2z g 4x 4y 4z 2a; ;™™ †œ Ê œ œ œ œ Ê œijk n Fn kk È
###
#
2x 2y 2z x y z
xyz aa
zijk ijk
È
œœœ
œ
g 2z d dA Flux dA z dA a x y dx dykk abŠ‹ˆ‰ È™†kœÊ œ Ê œ œ œ 5
2a z a
2z a z '' '' ''
RRR
œ
###
a r r dr dœœ''
00
2a
œ"
È##
)
1a
6

22. g 2x 2y 2z g 4x 4y 4z 2a; ™™ †œ Ê œ œ œ œ Ê œ ijk n Fn kk È
###
#
2x 2y 2z x y z xy xy
xyz aaa
ijk ijk
È
œœœ
0; g 2z d dA Flux d 0 d 0œœÊœÊœ œœkk™† †kFn 555
2a
2z '' ''
SS
23. From Exercise 21, and d dA Flux dAnFnœœÊœœÊœ
xyz xy xy
az aaaa az
azzzaijk
5 † ''
R
ˆ‰ˆ‰
1 dAœœ''
R
1a
4
œ
24. From Exercise 21, and d dA z aznFnœœÊœœœ
xyz zy xyz
az aaaa
azxzijk
5 †
œ œœœœ
Š‹
Flux (za) dx dy a dx dy a (Area of R) aÊœ œ œ œ '' ''
RR
ˆ‰
a
z4
## % "
1
25. From Exercise 21, and d dA a FluxnFnœœÊœœÊ
xyz y
az aaa
axzijk
5 †
œœ œ
a dA dA dA r dr dœœœ œ'' '' ''
RRR
ˆ‰
aa a a
zz axy ar
œœ œ
œœœ œœ
ÈÈab œ
''
00
2a
œ"
)
aar dœœ'
0
2 a
œ"
# ##
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)
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1026. 1022 Chapter 16 Integration in Vector Fields
26. From Exercise 21, and d dA 1nFnœœÊœœœ
xyz
az a
a
xyz
ijk


5 †
Š‹ Š‹Œœ
È
Š‹
xz
aaa
y
a
a
œœ œ
œ
œœœ Flux dx dy dx dy r dr dÊœ œ œ œ '' ''
RR
aa aa
z axy ar
ÈÈab
œœœ œœ
œ
œ
#
''
00
2a
œ"
)
1
27. g(x y z) y z 4 g 2y g 4y 1 ßß œ œ Ê œ Ê œ Ê œ
# #

™™ jk n kk È
2y
4y 1
jk
È
œ
; g 1 d 4y 1 dA FluxÊœ œÊ œÊœ ÊFn p k p†™†
2xy 3z
4y 1
œ

#
È
œ
kk È5
4y 1 dA (2xy 3z) dA; z 0 and z 4 y y 4œœœœÊœ''
R R
Š‹ È ''
2xy 3z
4y 1
œ

# ##
È
œ
Flux 2xy 3 4 y dA 2xy 12 3y dy dx xy 12y y dxÊœ œ œ ''
R
cdabcdab
###$
#
œ# '' '
02 0
12 1œ
32 dx 32œ œ'
0
1
28. g(x y z) x y z 0 g 2x 2y g 4x 4y 1 4 x y 1ßßœœÊ œÊ œ œ
## ## ##
™™ ijk kk abÈ È
; g 1 d 4 x y 1 dAÊœ Ê œ œÊ œÊ œ nFnpkp
2x 2y 8x 8y 2
4x y 1 4x y 1
ijk

##
ÈÈab ab
œœ œœ
œœ †™† kk abÈ5
Flux 4 x y 1 dA 8x 8y 2 dA; z 1 and x y zÊœ œ œ œ'' ''
RR
Š‹ Èab a b
8x 8y 2
4x y 1
œœ
œœ


## ## ##
Èab
x y 1 Flux 8r 2 r dr d 2ÊœÊ œ œ
## # ''
00
21
œ
ab )1
29. g(x y z) y e 0 g e g e 1 ; ßß œ œ Ê œ Ê œ Ê œ Ê œ œ
xx 2x e2e2y
e1 e1
™™ † ij n Fn pikk
È
xx
2x 2x
ijœœœ

ÈÈ g e d dA Flux dA dAʜʜ Êœ œkk Š‹Š‹™†p
x e1 e 1
eee
2e 2y
e1
2e 2e
5
ÈÈ
È2x 2x
x xx
x
2x
xx
œœ

œœ '' ''
RR
4 dA 4 dy dz 4œ œ œ''
R
''
01
12
30. g(x y z) y ln x 0 g g 1 since 1 x eßß œ œ Ê œ Ê œ œ Ÿ Ÿ™™
""
xxx
1x
ij kk É
œ
œ
È
; g 1 d dAÊœ œ Ê œ œÊ œÊ œnFnpjp
Š‹
Œ œ
ÈÈ
È



"
œ
x
1x
x
ij
ij
œ
x 2xy 1 x
1x 1x x
œœ
œ
†™† kk 5
Flux dA 2y dx dz 2 ln x dz dx 2 ln x dxÊœ œ œ œ ''
R
Š‹Š‹
2xy 1 x
1x xÈ
È


œ
œ
'' '' '
01 10 1
1e e1 e
2 x ln x x 2(e e) 2(0 1) 2œ œœcd
e
"
31. On the face z a: g(x y z) z g g 1; 2xz 2ax since z a;œßßœÊœÊ œœÊœœ œ ™™ †knkFnkk
d dx dy Flux 2ax dx dy 2ax dx dy a .5œÊœ œ œ
''
R
''
00
aa
%
On the face z 0: g(x y z) z g g 1; 2xz 0 since z 0;œßßœÊœÊ œœÊœœ œ™™ †knkFnkk
d dx dy Flux 0 dx dy 0.5œÊœ œ
''
R
On the face x a: g(x y z) x g g 1; 2xy 2ay since x a;œßßœÊœÊ œœÊœœ œ ™™ †iniFnkk
d dy dz Flux 2ay dy dz a .5œÊœ œ
''
00
aa
%
On the face x 0: g(x y z) x g g 1; 2xy 0 since x 0œßßœÊœÊ œœÊœœ œ™™ †iniFnkk
Flux 0.ʜ
On the face y a: g(x y z) y g g 1; 2yz 2az since y a;œßßœÊœÊ œœÊœœ œ ™™ †jnjFnkk
d dz dx Flux 2az dz dx a .5œÊœ œ
''
00
aa
%
On the face y 0: g(x y z) y g g 1; 2yz 0 since y 0œßßœÊœÊ œœÊœœ œ™™ †jnjFnkk
Flux 0. Therefore, Total Flux 3a .Êœ œ
%

1027. Section 16.5 Surface Area and Surface Integrals 1023
32. Across the cap: g(x y z) x y z 25 g 2x 2y 2z g 4x 4y 4z 10ßßœœ Ê œ Ê œ œ
### ###
™™ ijk kk È
; g 2z since z 0 d dAÊœ œ Ê œ œÊ œ  Ê œnFnpkp
™
™
g
g5 555 2z
xyz yz xz z 10
kk
ijk
†™†
œ œ
kk 5
Flux d dA x y 1 dx dy r 1 r dr dÊœ œ œ œ
cap
00
24
'' '' ''
cap RR
Fn†5 )Š‹ ˆ‰ ab ab
xz z 5
555z
yz
œ œ
## # ''
œ
72 d 144 .œœ'
0
2
œ
)1
Across the bottom: g(x y z) z 3 g g 1 1; g 1ßß œ œ Ê œ Ê œ Ê œ Ê œ œ Ê œ™™ † ™†knkFnpkpkk k k
d dA Flux d 1 dA 1(Area of the circular region) 16 . Therefore,Ê œ Ê œ œ œ œ55 1
bottom
'' ''
bottom R
Fn†
Flux Flux Flux 128œ œ
cap bottom 133. f 2x 2y 2z f 4x 4y 4z 2a; f 2z since z 0 d dA™™ ™†œ Ê œ œ œÊ œ  Êœijk pk p kk k kÈ
###
5
2a
2z
dA; M d (surface area of sphere) ; M z d z dAœœ œ œ œ œ
aaa
z8 z
xy '' '' ''
SS R
$5 $5 $
$$1
œ
#
ˆ‰
a dA a r dr d z . Because of symmetry, x yœœ œÊœœ œ œ$$ )'' Š‹ˆ‰
R
''
00
2a
œ"
$1 $1
$1
aa2a
4M4a
M

œ
xy
#
the centroid is .œÊ ßß
a aaa
2
ˆ‰
###
34. f 2y 2z f 4y 4z 4 y z 6; f 2z since z 0 d dA™™ ™†œ Ê œ œ œ œÊ œ  Êœjk pk kkk ab kkÈ È
## ##
5
6
2z
dA; M 1 d dx dy dx dy 9 ; M z dœœ œ œ œœ
333
zz 9y
xy '' ''
S S
515'' ''
œœ30 30
33 33 Èœ œ
z dx dy 54; M y d y dx dy dx dy 0;œœœœœ œ'' '' ''
œœœ30 30 30
33 33 33
ˆ‰ ˆ‰
33
zz
xz
3y
9y ''
S
5
Èœ œ
M x d dx dy . Therefore, x , y 0, and zyz
3x 27 3 54 6
9y 99œœ œ œœœœœ''
S
51''
œ30
33 È
Š‹
œ ## œ
27
œ
1
111
35. Because of symmetry, x y 0; M d d (Area of S) 3 2 ; f 2x 2y 2zœœœœœ œ œ '' ''
SS
$5 $ 5 $ 1 $
È
™ ijk
f 4x4y4z2xyz; f 2z d dAÊ œ œ œÊ œÊœkk k kÈÈ™™†
### ###

#
pk p 5
2xyz
z
È
œœœ
dA dA M z dAœœÊœ
ÈÈ È ab
ÈÈ
xy xy
zz z
2x y 2x y
xyœœ œœ œœ œœ

$''
R
Š‹
2 x y dA 2 r dr d zœœ œÊœœ$$)$''
R
ÈÈÈ ## # ''
01
22
œ
14 2
39 32
141
$
1$
È
Œ œ
È
14 2
3œ
œ
x y z 0 0 . Next, I x y d x y dAÊßßœßß œ œ ab ab abˆ‰ Š‹
14
9 z
z
2x y '' ''
S R
## ##
$5 $
ÈÈ œœ
2 x y dA 2 r dr d Rœœ œÊœœ$$)$
ÈÈ
ab É ''
R
## $
## ''
01
22
z
œ
15 2 I
M
101
ÈÈ
z
36. f(x y z) 4x 4y z 0 f 8x 8y 2z f 64x 64y 4zßßœ œÊ œ Ê œ
### ###
™™ ijk kk È
2 16x 16y z 2 4z z 2 5 z since z 0; f 2z d dA 5 dAœ œ œ   œÊ œ Ê œ œÈ È ÈÈ kk
### ##
pk p ™† 5
25z
2z
È
I x y d 5 x y dx dy 5 r dr dÊœ œ œ œz
35
'' ''
S R
ab ab ÈÈ
## ## $
#
$5 $ $ ) ''
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œ
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22cos
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1$
37. (a) Let the diameter lie on the z-axis and let f(x y z) x y z a , z 0 be the upper hemisphereßßœœ  
####
f 2x 2y 2z f 4x 4y 4z 2a, a 0; f 2z since z 0Ê œ Ê œ œ œÊ œ  ™™ ™†ijk pk p kk k kÈ
###
d dA I x y d a dA a r dr dÊœ Êœ œ œ5$5$ $)
aa r
zz
xy
axy ar
z
00
2a
''ab ˆ‰
S R
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œ
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ÈÈab
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a r a r a r d a a d a the moment of inertia is a forœ œ œÊ$)$)$$ ''
0 0
2 2a
œ œ
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È
ab
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224 8
333 3
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1028. 1024 Chapter 16 Integration in Vector Fields
the whole sphere
(b) I I mh , where m is the mass of the body and h is the distance between the parallel lines; now,
Lc.m.œ
#
I a (from part a) and d dA a dy dxc.m.
8ma
3z axyœœœœ
1%
#
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$$5$$ '' '' ''
S RR
ˆ‰
È abœœœ
a r dr d a a r d a a d 2 a and h aœœœœœ$)$ )$)1$'' ' '
00 0 0
2a 2 2 a
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ar
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38. (a) Let z x y be the cone from z 0 to z h, h 0. Because of symmetry, x 0 and y 0;œ œœ œ œ
h
aÈ##
z x y f(x y z) x y z 0 f 2zœ Êßßœ œÊœ
h h 2xh
aa aa
2yhÈ ab## ## #
œœ
œœœ
œ
™ ijk
f 4z2 xy xy 2 xy 1Êœœ œ kk abab abÉ ÉÉ ˆ‰ ˆ ‰™
4x h h h h h
aa a
4y h
aaa
œ
ÎÎ Î

œœœ
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2 z h a since z 0; f 2z d dAœœ œÊœÊœÉˆ‰ˆ‰ È
kk
### ha 2z
aa 2z
ha
œœ
œ
œœ
pk p ™† 5
ˆ‰È
2z
a
dA; M d dA a a h a ;œœœ œœ
ÈÈÈ
ha ha ha
aaaœœ œœ œœ
# ## '' ''
S R
511 ab
È
M z d z dA x y dx dy r dr d
xy
00
2aœœ œ œ'' '' ''
S RR
5 )Š‹ È
ÈÈ È
ha ha hha
aaa a
hœœ œœ œœ
œ
## # ''
œ
z the centroid is 0 0œÊœœÊ ßß
2ah h a
3M3 3
M 2h 2h1
È
œœ
xy
ˆ‰
(b) The base is a circle of radius a and center at (0 0 h) (0 0 h) is the centroid of the base and the mass isßß Ê ßß
M d a . In Pappus' formula, let , h , m a h a , and m aœœ œ œœ œ''
S
51 1 1
# #
"#" #
##
ckck
2h
3 È
the centroid is Ê œ œ Ê !ß !ßck
11
11
ah a ah
ah a a
2h h a 3ah 2h h a 3ah
3haa 3haa
È
Š‹
È
ÈÈ
Š‹ Š‹
ÈÈ
œœ œ
œœ œ
œœ œœ
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2h
3
kk

(c) If the hemisphere is sitting so its base is in the plane z h, then its centroid is 0 0 h and its mass isœßß ˆ‰
a
#
2 a . In Pappus' formula, let , h , m a h a , and m 2 a111
# #
"# " #
#
##
ckc kœœ œ œ
2h a
3
ˆ‰ È
the centroid isÊœ œ Êck
11
11
ah a 2ah
ah a 2a
2h h a 6ah 3a
3ha2a
È ˆ‰ ˆ ‰
È
È
Š‹
È
œœ œ
œœ œ
œœ œ
œœ




2h
3
kk
a
œ
. Thus, for the centroid to be in the plane of the bases we must have z h

!ß !ß œ
2h h a 6ah 3a
3ha2a
È
Š‹
È
œœ œ
œœ

h 2h h a 6ah 3a 3h h a 6ah 3a h h aʜʜʜ
2h h a 6ah 3a
3ha2a
È
Š‹
È
œœ œ
œœ


## ## ## ## ÈÈ È
9a h h a h a h 9a 0 h (the positive root) h aʜ ʜʜ ʜ
% ### %## % #
œ"
##
œ
ab
Š‹ É
È È37 a 237 2
œ
39. f (x y) 2x, f (x y) 2y f f 1 4x 4y 1 Area 4x 4y 1 dx dy
xy xyßœ ßœ Ê œ Ê œ É ÈÈ
## ## ## ''
R
4r 1 r dr d 13 13 1œœ''
00
23
œ
œ
È
Š‹
È#
)
1
6
40. f (y z) 2y, f (y z) 2z f f 1 4y 4z 1 Area 4y 4z 1 dy dz
yz yzßœ ßœÊ œ Ê œ É ÈÈ
## ## ## ''
R
4r 1 r dr d 5 5 1œœ''
00
21
œ
È
Š‹
È#
)
1
6
41. f (x y) , f (x y) f f 1 1 2xy
xx
xy xy
yy
xy xy xyßœ ßœ Ê œ œ
ÈÈœœ œœ
œ
œœ œœ
œ

##

É É È
Area 2 dx dy 2(Area between the ellipse and the circle) 2(6 ) 5 2Êœ œ œ œ ''
Rxy
ÈÈ È È
11 1

1029. Section 16.6 Parameterized Surfaces 1025
42. Over R : z 2 x 2y f (x y) , f (x y) 2 f f 1 4 1xy x y
22 47
33 93 xyœ Ê ß œ ß œÊ œ œ É É
##
Area dA (Area of the shadow triangle in the xy-plane) .Êœ œ œ œ ''
Rxy
77 737
33 3
ˆ‰ˆ‰
##
Over R : y 1 x z f (x z) , f (x z) f f 1 1xz x z
11 17
33 946 xzœ Ê ß œ ß œ Ê œ œ
"""
##
## È É
Area dA (Area of the shadow triangle in the xz-plane) (3) .Êœ œ œ œ ''
Rxz
77 7 7
66 6
ˆ‰
#
Over R : x 3 3y z f (y z) 3, f (y z) f f 1 9 1yz y z
3397
yz 4œ Ê ß œ ß œ Ê œ œ
## #
##
É É
Area dA (Area of the shadow triangle in the yz-plane) (1) .Êœ œ œ œ ''
Ryz
77 7 7
22 2
ˆ‰
#
43. y z f (x z) 0, f (x z) z f f 1 z 1 ; y z z 4œ Ê ßœ ßœ Ê œ œ Ê œ Êœ
2 16 16 2
3 333
xz xz
$Î# "Î# $Î# ## È È
Area z 1 dx dz z 1 dz 5 5 1Êœ œ œ '' '
00 0
41 4
ÈÈ Š‹
È 2
3
44. y 4 z f (x z) 0, f (x z) 1 f f 1 2 Area 2 dA 2 dx dzœ Ê ß œ ß œÊ œ Ê œ œxz xz
È ÈÈÈ## ''
Rxz
00
24z
''
œ

2 4 z dzœœ
È
ab '
0
2
# 16 2
3
È
16.6 PARAMETRIZED SURFACES
1. In cylindrical coordinates, let x r cos , y r sin , z x y r . Thenœœœœ)) ˆ‰È
##
#
#
(r ) (r cos ) (r sin ) r , 0 r 2, 0 2 .rijkßœ ŸŸ ŸŸ))) )1
#
2. In cylindrical coordinates, let x r cos , y r sin , z 9 x y 9 r . Thenœœœœ))
## #
(r ) (r cos ) (r sin ) 9 r ; z 0 9 r 0 r 9 3 r 3, 0 2 . Butrijkßœ  Ê Ê ŸÊŸŸ ŸŸ))) )1 ab
###
3 r 0 gives the same points as 0 r 3, so let 0 r 3.Î&& && &&
3. In cylindrical coordinates, let x r cos , y r sin , z z . ThenœœœÊœ))
Èxy rœœ

##
(r ) (r cos ) (r sin ) . For 0 z 3, 0 3 0 r 6; to get only the first octant, letrijkßœ ŸŸ ŸŸÊŸŸ))) ˆ‰
rr
##
0.ŸŸ)
1
#
4. In cylindrical coordinates, let x r cos , y r sin , z 2 x y z 2r. ThenœœœÊœ)) È
##
(r ) (r cos ) (r sin ) 2r . For 2 z 4, 2 2r 4 1 r 2, and let 0 2 .rijkßœ ŸŸ Ÿ ŸÊŸŸ ŸŸ))) )1
5. In cylindrical coordinates, let x r cos , y r sin since x y r z 9 x y 9 rœœ œÊœœ))
## # ## #2
ab
z 9 r , z 0. Then (r ) (r cos ) (r sin ) 9 r . Let 0 2 . For the domainÊœ   ßœ ŸŸ
ÈÈ
# #
rijk))) )1
of r: z x y and x y z 9 x y x y 9 2 x y 9 2r 9œ œÊ œÊ œÊ œÈÈ ˆ‰ ab
## ## ### ## ## #
#
r 0 r .Êœ Ê ŸŸ
33
22
ÈÈ
6. In cylindrical coordinates, (r ) (r cos ) (r sin ) 4 r (see Exercise 5 above with x y z 4,rijkßœ œ)))
È
# ###
instead of x y z 9). For the first octant, let 0 . For the domain of r: z x y and
###
#
##
œ ŸŸ œ )
1 È
x y z 4 x y x y 4 2 x y 4 2r 4 r 2. Thus, let 2 r 2
### ## ## # ##
#
œÊ œÊ œÊ œÊœ ŸŸˆ‰È ab
ÈÈ
(to get the portion of the sphere between the cone and the xy-plane).

1030. 1026 Chapter 16 Integration in Vector Fields
7. In spherical coordinates, x sin cos , y sin sin , x y z 3 3œœœÊœÊœ39 ) 39)3 3 3 È È
### #
z 3 cos for the sphere; z 3 cos cos ; z 3 cos Êœ œ œ Ê œ Ê œ œ Ê œ
ÈÈ È
9999 9
ÈÈÈ
333
3####
" 1
cos . Then ( ) 3 sin cos 3 sin sin 3 cos ,ʜʜ ߜ 99 9) 9) 9) 9
"
#
2
3
1
rijkŠ‹Š‹Š‹
ÈÈÈ
and 0 2 .
11
33
2
ŸŸ ŸŸ9)1
8. In spherical coordinates, x sin cos , y sin sin , x y z 8 8 2 2œœœÊœÊœœ39 ) 39)3 3 3 È È È
### #
x 2 2 sin cos , y 2 2 sin sin , and z 2 2 cos . Thus letÊœ œ œ
ÈÈ È
9) 9) 9
( ) 2 2 sin cos 2 2 sin sin 2 2 cos ; z 2 2 2 2 cos rijk9) 9 ) 9 ) 9 9ßœ œÊœŠ‹Š‹Š‹
ÈÈÈ È
cos ; z 2 2 2 2 2 2 cos cos 1 0. Thus 0 andʜʜœÊœ ʜʜ ŸŸ99 999 9
"
È
2
3 3
4 4
1 1 ÈÈÈ
02.ŸŸ)1
9. Since z 4 y , we can let be a function of x and y (x y) x y 4 y . Then z 0œ Ê ßœ œ
# #
rrijk ab
0 4 y y 2. Thus, let 2 y 2 and 0 x 2.Êœ Êœ„ ŸŸ ŸŸ
#
10. Since y x , we can let be a function of x and z (x z) x x z . Then y 2œÊßœœ
# #
rrijk
x 2 x 2. Thus, let 2 x 2 and 0 z 3.ʜʜ„ ŸŸ ŸŸ
# ÈÈÈ
11. When x 0, let y z 9 be the circular section in the yz-plane. Use polar coordinates in the yz-planeœœ
##
y 3 cos and z 3 sin . Thus let x u and v (u,v) u (3 cos v) (3 sin v) whereÊœ œ œ œÊ œ )) ) ri j k
0 u 3, and 0 v 2 .ŸŸ ŸŸ 1
12. When y 0, let x z 4 be the circular section in the xz-plane. Use polar coordinates in the xz-planeœœ
##
x 2 cos and z 2 sin . Thus let y u and v (u,v) (2 cos v) u (3 sin v) whereÊœ œ œ œÊ œ )) ) rijk
2 u 2, and 0 v (since we want the portion the xy-plane).Î& & & & 1 above
13. (a) x y z 1 z 1 x y. In cylindrical coordinates, let x r cos and y r sin œÊœ œ œ ))
z 1 r cos r sin (r ) (r cos ) (r sin ) (1 r cos r sin ) , 0 2 andÊœ Ê ß œ ŸŸ))))) )))1rij k
0r3.ŸŸ
(b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let
y u cos v, z u sin v where u y z and v is the angle formed by (x y z), (x 0 0), and (x y 0)œ œ œ ßß ßß ßß È
##
with (x 0 0) as vertex. Since x y z 1 x 1 y z x 1 u cos v u sin v, then is aßß œ Ê œ Ê œ r
function of u and v (u v) (1 u cos v u sin v) (u cos v) (u sin v) , 0 u 3 and 0 v 2 .Êßœ ŸŸ ŸŸrijk 1
14. (a) In a fashion similar to cylindrical coordinates, but working in the xz-plane instead of the xy-plane, let
x u cos v, z u sin v where u x z and v is the angle formed by (x y z), (y 0 0), and (x y 0)œ œ œ ßß ßß ßß
È
##
with vertex (y 0 0). Since x y 2z 2 y x 2z 2, then (u v)ßß œ Ê œ ß r
(u cos v) (u cos v 2u sin v 2) (u sin v) , 0 u 3 and 0 v 2 .œ ŸŸ ŸŸijk
È
1
(b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let
y u cos v, z u sin v where u y z and v is the angle formed by (x y z), (x 0 0), and (x y 0)œ œ œ ßß ßß ßß È
##
with vertex (x 0 0). Since x y 2z 2 x y 2z 2, then (u v)ßß œ Ê œ ß r
(u cos v 2u sin v 2) (u cos v) (u sin v) , 0 u 2 and 0 v 2 .œ ŸŸ ŸŸ ijk
È
1
15. Let x w cos v and z w sin v. Then (x 2) z 4 x 4x z 0 w cos v 4w cos v w sin vœ œ œÊ œÊ
## # # ## ##
0 w 4w cos v 0 w 0 or w 4 cos v 0 w 0 or w 4 cos v. Now w 0 x 0 and y 0,œÊ œÊ œ œ Ê œ œ œ Ê œ œ
#
which is a line not a cylinder. Therefore, let w 4 cos v x (4 cos v)(cos v) 4 cos v and z 4 cos v sin v.œÊœ œ œ
#
Finally, let y u. Then (u v) 4 cos v u (4 cos v sin v) , v and 0 u 3.œ ß œ ŸŸ ŸŸrijkab
#
##
11

1031. Section 16.6 Parameterized Surfaces 1027
16. Let y w cos v and z w sin v. Then y (z 5) 25 y z 10z 0œœ œÊœ
## ##
w cos v w sin v 10w sin v 0 w 10w sin v 0 w(w 10 sin v) 0 w 0 orʜʜʜʜ
## ## #
w 10 sin v. Now w 0 y 0 and z 0, which is a line not a cylinder. Therefore, let w 10 sin vœœÊœœ œ
y 10 sin v cos v and z 10 sin v. Finally, let x u. Then (u v) u (10 sin v cos v) 10 sin v ,Êœ œ œ ßœ
# #
ri j k ab
0 u 10 and 0 v .ŸŸ ŸŸ 1
17. Let x r cos and y r sin . Then (r ) (r cos ) (r sin ) , 0 r 1 and 0 2œœ ßœ ŸŸŸŸ) ) ))) )1 rijk ˆ‰
2r sin œ
#
)
(cos ) (sin ) and ( r sin ) (r cos )Êœ œ rijkr i jkr
sin r cos )) ) ) ˆ‰ ˆ ‰
))
)
##
cos sin
r sin r cos
Ê‚œ Î
ÎÎ
rr
ijk r
sin
r cos
)
)
)
ââ
ââ
ââ
ââ
ââ
ââ
))
))
#
#
r cos r sin rœ œŠ‹Š‹ ab
œ
## ## #
##r sin cos r sin r cos r(sin )(r cos ))) ) ) ))
ij kjk
œœ
))
r A dr d d dÊ‚œ œ Êœ œ œ œkk É ’“rr r
r
44
5r 5r 5r 5
)
1
œ œ
#
## #
"
!
ÈÈÈ È '' ' '
00 0 0
21 2 2
œœœ
)))
18. Let x r cos and y r sin z x r cos , 0 r 2 and 0 2 . ThenœœÊœœŸŸŸŸ)) ) )1
(r ) (r cos ) (r sin ) (r cos ) (cos ) (sin ) (cos ) andrijkrijkßœ Ê œ ))) ) )))
r
( r sin ) (r cos ) (r sin )rijk)œ )))
cos sin cos
r sin r cos r sin
Ê‚œ Î
Î
rr
ijk
r )
ââ
ââ
ââ
ââ
ââ
ââ
)) )
)))
r sin r cos (r sin cos r sin cos ) r cos r sin r rœ œab ab
## ##
) ) )) )) ) )ijkik
r r r 2 A r 2 dr d d 2 2 d 4 2Ê‚œ œ Êœ œ œ œkk
ÈÈ È ÈÈ
’“rr r
r2
2
)
##
#
!
'' ' '
00 0 0
22 2 2
œœœ
)))1
œÈ
19. Let x r cos and y r sin z 2 x y 2r, 1 r 3 and 0 2 . ThenœœÊœœŸŸŸŸ)) )1 È
##
(r ) (r cos ) (r sin ) 2r (cos ) (sin ) 2 and ( r sin ) (r cos )rijkrijkrijßœ Ê œ œ ))) )) ) ) r )
( 2r cos ) (2r sin ) r cos r sincos sin 2
r sin r cos 0
Ê‚œ œ
Î
rr i j k
ijk r )
ââ
ââ
ââ
ââ
ââ
ââ
ab))
))
))))
##
( 2r cos ) (2r sin ) r 4r cos 4r sin r 5r r 5œ Ê ‚ œ œ œ)) ))ijkrr kk
È È È r )
## ## # #
A r 5 dr d d 4 5 d 8 5Êœ œ œ œ'' ' '
01 0 0
23 2 2
œœœ
ÈÈÈ
’“)))1
r5
2
œÈ
$
"
20. Let x r cos and y r sin z , 3 r 4 and 0 2 . ThenœœÊœœŸŸŸŸ)) )1
Èxy
33
rœœ

(r ) (r cos ) (r sin ) (cos ) (sin ) and ( r sin ) (r cos )r i jkr i jkr i jßœ Ê œ œ ))) )) ) ) ˆ‰ ˆ‰
r
33
r
"
)
cos sin r cos r sin r cos r sin
r sin r cos 0
Ê‚œ œ
Î
rr i j k
ijk r
33 3
)
ââ
ââ
ââ
ââ
ââ
ââ
ˆ‰ˆ‰ ab)) ) ) ) )
))
"" " ##
r cos r sin r r cos r sin rœ Ê ‚ œ œ œˆ‰ˆ‰ kk É É
"" "" ## ## #
33 99 93
r
10r r10
)) ))ijkrr )
œ È
A dr d d dÊœ œ œ œ'' ' '
03 0 0
24 2 2
œœœ
r 10 r 10 7 10 7 10
3663
ÈÈÈÈ
)))’“
œ %
$
1
21. Let x r cos and y r sin r x y 1, 1 z 4 and 0 2 . Thenœ œ Ê œœ ŸŸ ŸŸ)) )1
###
(z ) (cos ) (sin ) z and ( sin ) (cos )rijkrkrijßœ Ê œ œ ))) ) ) z )
(cos ) (sin ) cos sin 1sin cos 0
001
Ê‚œ œ Ê‚œ œÎrr i j rr
ijk ) )z z
ââ
ââ
ââ
ââ
ââ
ââ
kk
È
)) )) ))
##

1032. 1028 Chapter 16 Integration in Vector Fields
A 1 dr d 3 d 6Êœ œ œ'' '
01 0
24 2
œœ
))1
22. Let x u cos v and z u sin v u x z 10, 1 y 1, 0 v 2 . ThenœœÊœœŸŸŸŸ
###
1 (y v) (u cos v) y (u sin v) 10 cos v y 10 sin vrijk ijkßœ œ Š‹Š‹
ÈÈ
10 sin v 10 cos v and 10 sin v 0 10 cos v
010
Êœ œÊ‚œ Îrikrjrr
ijk vyvyŠ‹Š‹
ÈÈ
ââ
ââ
ââ
ââ
ââ
ââ
ÈÈ
10 cos v 10 sin v 10 A 10 du dv 10u dvœ Ê ‚ œ Ê œ œŠ‹Š‹ ’“
ÈÈ È È È
kkikrr
vy
01 0
21 2
'' '
œœ
œ "
œ"
2 10 dv 4 10œœ'
0
2
œ
ÈÈ
1
23. z 2 x y and z x y z 2 z z z 2 0 z 2 or z 1. Since z x y 0,œ œ Ê œ Ê œ Ê œ œ œ  
## # # ## ## ÈÈ
we get z 1 where the cone intersects the paraboloid. When x 0 and y 0, z 2 the vertex of theœœœœÊ
paraboloid is (0 0 2). Therefore, z ranges from 1 to 2 on the “cap" r ranges from 1 (when x y 1) to 0ßß Ê œ
##
(when x 0 and y 0 at the vertex). Let x r cos , y r sin , and z 2 r . Thenœœ œ œ œ ))
#
(r ) (r cos ) (r sin ) 2 r , 0 r 1, 0 2 (cos ) (sin ) 2r andrijk rijkßœ ŸŸ ŸŸ Ê œ ))) )1 )) ab
#
r
( r sin ) (r cos ) cos sin 2r
r sin r cos 0
rijrr
ijk))œ Ê ‚ œ Î
Î
)) ))
)) r
ââ
ââ
ââ
ââ
ââ
ââ
2r cos 2r sin r 4r cos 4r sin r r 4r 1œÊ‚œœabab kk
È È
## %# %# # #
)) ))ijkrr r )
A r 4r 1 dr d 4r 1 d d 5 5 1Êœ œ œ œ '' ' '
00 0 0
21 2 2
œœ œ
È
’“Š‹Š‹ab
È# " #
$Î#
"
!
œ"
#
)))
12 1 6
55
È
1
24. Let x r cos , y r sin and z x y r . Then (r ) (r cos ) (r sin ) r , 1 r 2,œ œ œœ ßœ ŸŸ) ) )))
### #
rijk
0 2 (cos ) (sin ) 2r and ( r sin ) (r cos )ŸŸ Ê œ œ )1 ) ) ) )rijkr i j
r )
2r cos 2r sin r cos sin 2r
r sin r cos 0
Ê‚œ œ Ê ‚
Î
rr i jk rr
ijk r r) )
ââ
ââ
ââ
ââ
ââ
ââ
abab kk))
))
))
##
4r cos 4r sin r r 4r 1 A r 4r 1 dr d 4r 1 dœœÊœ œ
È ÈÈ
’“ab
%# %# # # # " #
$Î#
#
"
)) ) ) '' '
01 0
22 2
œœ
12
d171755œœ'
0
2
œ
Š‹Š ‹
È È
17 17 5 5
16
ÈÈ
œ
#
)
1
25. Let x sin cos , y sin sin , and z cos x y z 2 on the sphere. Next,œ œ œ Êœ œ39 ) 39) 3 9 3 È È
###
x y z 2 and z x y z z 2 z 1 z 1 since z 0 . For the lower
### ## # ##
œ œ Ê œÊ œÊœ  ÊœÈ 9
1
4
portion of the sphere cut by the cone, we get . Then91œ
( ) 2 sin cos 2 sin sin 2 cos , , 0 2rijk9) 9 ) 9 ) 9 9 1 ) 1ßœ ŸŸ ŸŸŠ‹Š‹Š‹
ÈÈÈ
1
4
2 cos cos 2 cos sin 2 sin and 2 sin sin 2 sin cos Êœ œ rijkr ij9 )Š ‹Š‹Š‹Š ‹Š‹
ÈÈÈ ÈÈ
9) 9) 9 9) 9)
2 cos cos 2 cos sin 2 sin
2 sin sin 2 sin cos 0
Ê‚œ Î
Î
rr
ijk 9)
ââ
ââ
ââ
ââ
ââ
ââ
ÈÈ È
ÈÈ
9) 9) 9
9) 9 )
2 sin cos 2 sin sin (2 sin cos )œabab
##
9) 9) 99ijk
4 sin cos 4 sin sin 4 sin cos 4 sin 2 sin 2 sin Ê‚œ œ œ œkk kk È Èrr 9)
%# %# ## # 9) 9) 99 9 9 9
A 2 sin d d 2 2 d 4 2 2Êœ œ œ'' '
04 0
22
œœ œ
œ"
99) ) 1 Š‹Š ‹
ÈÈ
26. Let x sin cos , y sin sin , and z cos x y z 2 on the sphere. Next,œ œ œ Êœ œ39 ) 39) 3 9 3 È
###
z 1 1 2 cos cos ; z 3 3 2 cos cos . Thenœ Ê œ Ê œ Ê œ œ Ê œ Ê œ Ê œ99 9 99 9
"
# #
2
3 6
31 1 ÈÈ
È

1033. Section 16.6 Parameterized Surfaces 1029
( ) (2 sin cos ) (2 sin sin ) (2 cos ) , , 0 2rijk9) 9 ) 9 ) 9 9 ) 1ßœ ŸŸ ŸŸ
11
63
2
(2 cos cos ) (2 cos sin ) (2 sin ) andʜ rijk 9 9) 9) 9
( 2 sin sin ) (2 sin cos )rij
)œ 9) 9 )
2 cos cos 2 cos sin 2 sin
2 sin sin 2 sin cos 0
Ê‚œ Î
Î
rr
ijk
9)
ââ
ââ
ââ
ââ
ââ
ââ
9) 9) 9
9) 9 )
4 sin cos 4 sin sin (4 sin cos )œabab
##
9) 9) 99ijk
16 sin cos 16 sin sin 16 sin cos 16 sin 4 sin 4 sin Ê‚œ œ œ œkk kk È Èrr 9)
%# %# ## # 9) 9) 99 9 9 9
A 4 sin d d 2 2 3 d 4 4 3Êœ œ œ'' '
06 0
223 2
œœ œ
œ"
"
99) ) 1 Š‹Š‹
ÈÈ
27. Let the parametrization be (x z) x x z 2x and 12x0
00
rijkrijrkrr
ijk
ßœ Ê œ œ Ê ‚œ
"
#
xzxz
ââ
ââ
ââ
ââ
ââ
ââ
2x 4x 1 G(x y z) d x 4x 1 dx dz 4x 1 dzœÊ‚œ Ê ßß œ œ ij r rkk ab
ÈÈ
’“
xz
00 0
32 3
## " #
$Î#
#
! ''
S
5'' '
12
17 17 1 dzœœ'
0
3
"
#
œ"
14
17 17
Š‹
È
È
28. Let the parametrization be (x y) x y 4 y , 2 y 2 and rij k rirj kßœ ŸŸÊ œ œÈ
#
œ
xy
y
4yÈ
œ
1 10 0
01
Ê‚œ œ Ê ‚œ œ
Î
rr jk rr
ij k
xy xy
ââ
ââ
ââ
ââ
ââ
ââ
kk É
y
4y
yy
4y 4y 4y
2
È
ÈÈ
œ
œœ œ
œ
œ œ
œ
œ
G(x y z) d 4 y dy dx 24Êßßœ œ''
S
5''
12
42œ
È Š‹ #
œ
2
4yÈ
œ
29. Let the parametrization be ( ) (sin cos ) (sin sin ) (cos ) (spherical coordinates with 1rijk9) 9 ) 9 ) 9 3ßœ œ
on the sphere), 0 , 0 2 (cos cos ) (cos sin ) (sin ) andŸŸ ŸŸ Ê œ 91 ) 1 9 ) 9 ) 9rijk
9
( sin sin ) (sin cos ) cos cos cos sin sin
sin sin sin cos 0
rijrr
ijk)9)œ Ê ‚ œ Î
Î
9) 9 ) 9) 9) 9
9) 9 )
ââ
ââ
ââ
ââ
ââ
ââ
sin cos sin sin (sin cos ) sin cos sin sin sin cosœÊ‚œabab kk È
## %# %# ##
9) 9) 99 9 ) 9 ) 9 9ijkrr 9)
sin ; x sin cos G(x y z) cos sin G(x y z) d cos sin (sin ) d dœœ Êßßœ Êßßœ99) )9 5 )999)
## ## ''
S
''
00
2
œœ
ab
cos 1 cos (sin ) d d ; cos u 1 du d
u cos
du sin d
œ Ä
œ
œ
'' ''
00 01
2 21
œœ œ
aba b aba b ”•
## ##
)999) ) )
9
99
œ
cos u d cos dœœœœ''
00
22
œœ
ab ’“ œ ‘
##
œ"
"
#
!
))))
u4 4sin 24
33 3243

)) 1
1
30. Let the parametrization be ( ) (a sin cos ) (a sin sin ) (a cos ) (spherical coordinates withrijk9) 9 ) 9 ) 9ߜ
a, a 0, on the sphere), 0 (since z 0), 0 239)1œ  ŸŸ   ŸŸ
1
#
(a cos cos ) (a cos sin ) (a sin ) andʜ rijk 9 9) 9) 9
( a sin sin ) (a sin cos ) a cos cos a cos sin a sin
a sin sin a sin cos 0
rijrr
ijk
)9)œ Ê ‚ œ Î
Î
9) 9 ) 9) 9) 9
9) 9 )
ââ
ââ
ââ
ââ
ââ
ââ
a sin cos a sin sin (a sin cos )œabab
## ## #
9) 9) 99ijk
a sin cos a sin sin a sin cos a sin ; z a cos Ê‚œ œ œkk Èrr 9)
%% # %% # %# # # 9) 9) 99 9 9
G(x y z) a cos G(x y z) d a cos a sin d d aÊßßœ Ê ßß œ œ
## ## # %
95 999)1''
S
''
00
22
œœ"
abab
2
3

1034. 1030 Chapter 16 Integration in Vector Fields
31. Let the parametrization be (x y) x y (4 x y) and rij krikrjkßœ Ê œ œ xy
3 F(x y z) d (4 x y) 3 dy dx10 1
01
Ê ‚ œ œ Ê ‚ œ Ê ßß œ Î
Î_
rr ijk rr
ij k xy xy
ââ
ââ
ââ
ââ
ââ
ââ
kk
ÈÈ
''
S
5''
00
11
3 4y xy dx 3 x dx 3 x 3 3œœœœ''
00
11
ÈÈÈÈ
’“ ’“ ˆ‰
y
22
77x
œ œ""
!!
##
32. Let the parametrization be (r ) (r cos ) (r sin ) r , 0 r 1 (since 0 z 1) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1
(cos ) (sin ) and ( r sin ) (r cos ) cos sin
r sin r cos 0
Êœ œ Ê‚œ "
Î
rijkr i jrr
ijk
r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
( r cos ) (r sin ) r ( r cos ) ( r sin ) r r 2; z r and x r cos œ Ê ‚ œ œ œ œ)) ) ) )ijkrr kk È È r )
###
F(x y z) r r cos F(x y z) d (r r cos ) r 2 dr d 2 (1 cos ) r dr dÊßßœ Ê ßß œ œ )5 )) )) ''
S
'' ''
00 00
21 21
œœ
Š‹
ÈÈ
#
œ
22
3
1
È
33. Let the parametrization be (r ) (r cos ) (r sin ) 1 r , 0 r 1 (since 0 z 1) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1 ab
#
(cos ) (sin ) 2r and ( r sin ) (r cos ) cos sin 2r
r sin r cos 0
Êœ œ Ê‚œ Î
Î
r i jkr i jrr
ijk r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
2r cos 2r sin r 2r cos 2r sin r r 1 4r ; z 1 r andœÊ‚œ œœabab kk ababÉ È
## #####
#
)) ))ijkrr r )
x r cos H(xyz) r cos 1 4r H(xyz) dœÊßßœ Ê ßß)) 5 ab
È
## # ''
S
r cos 1 4r r 1 4r dr d r 1 4r cos dr dœœœ'' ''
00 00
21 21
œ œ
ab abŠ‹Š ‹
ÈÈ
## $ # # ##
))))
11
12
1
34. Let the parametrization be ( ) (2 sin cos ) (2 sin sin ) (2 cos ) (spherical coordinates withrijk9) 9 ) 9 ) 9ߜ
2 on the sphere), 0 ; x y z 4 and z x y z z 4 z 2 z 2 (since39œ ŸŸ œ œ Ê œÊ œÊœ
1
4
### ## # ## È È
z 0) 2 cos 2 cos , 0 2 ; (2 cos cos ) (2 cos sin ) (2 sin ) Ê œ Ê œ Êœ ŸŸ œ 999)19)9)9
È
È
2
4#
1
9
rijk
and r ( 2 sin sin ) (2 sin cos ) 2 cos cos 2 cos sin 2 sin
2 sin sin 2 sin cos 0
)9)œ Ê ‚ œ Î
Î
9) 9 ) 9) 9) 9
9) 9 )
ijrr
ijk
ââ
ââ
ââ
ââ
ââ
ââ
4 sin cos 4 sin sin (4 sin cos )œabab
##
9) 9) 99ijk
16 sin cos 16 sin sin 16 sin cos 4 sin ; y 2 sin sin andÊ‚œ œ œkk Èrr 9)
%# %# ## 9) 9) 99 9 9)
z 2 cos H(x y z) 4 cos sin sin H(x y z) d (4 cos sin sin )(4 sin ) d dœ Ê ßß œ Ê ßß œ9 99) 5 99) 99)
''
S
''
00
24
œœ"
16 sin cos sin d d 0œœ''
00
24
œœ"
#
99)9)
35. Let the parametrization be (x y) x y 4 y , 0 x 1, 2 y 2; z 0 0 4 yrij kßœ ŸŸŸŸ œÊœab
##
y 2; and 2y 2y d 10 0
01 2y
Êœ„ œ œ Ê ‚œ œ Ê
Î
ri rj k rr jk Fn
ij k xy xy
ââ
ââ
ââ
ââ
ââ
ââ
†5
dy dx (2xy 3z) dy dx 2xy 3 4 y dy dx dœ‚œœÊFrr Fn††
rr
rrxy
xy‚
‚
#
kk
kk cd abxy
''
S
5
2xy 3y 12 dy dx xy y 12y dx 32 dx 32œœœœ
'' ' '
02 0 0
12 1 1œ
abcd
##$
#
œ#

1035. Section 16.6 Parameterized Surfaces 1031
36. Let the parametrization be (x y) x x z , 1 x 1, 0 z 2 2x and rijk rijrkßœ ŸŸ ŸŸÊ œ œ
#
xz
2x d dz dx x dz dx12x0
001
Ê‚œ œ Ê œ ‚ œrr ij Fn F rr
ijk xz xz
ââ
ââ
ââ
ââ
ââ
ââ
kk††5
rr
rrxz
xz‚
‚
#
kk
d x dz dxÊœœ''
S
Fn†5 ''
œ10
12
# 4
3
37. Let the parametrization be ( ) (a sin cos ) (a sin sin ) (a cos ) (spherical coordinates withrijk9) 9 ) 9 ) 9ߜ
a, a 0, on the sphere), 0 (for the first octant) 0 (for the first octant)39)œ  ŸŸ ߟŸ
11
##
(a cos cos ) (a cos sin ) (a sin ) and ( a sin sin ) (a sin cos )Êœ œ rijkr ij9 ) 9) 9) 9 9) 9)
a cos cos a cos sin a sin
a sin sin a sin cos 0
Ê‚œ Î
Î
rr
ijk
9)
ââ
ââ
ââ
ââ
ââ
ââ
9) 9) 9
9) 9 )
a sin cos a sin sin a sin cos d d dœ Êœ‚ababab kk
## ## # ‚
‚
9) 9) 99 5 )9ijkFnFrr ††
rr
rrœ
œ
kk
9)
a cos sin d d since z (a cos ) d a cos sin d dœœœÊœ œ
$# $#
99)9 9 5 999)Fk k Fn ''
S
† ''
00
22
œœ""
1a
6

38. Let the parametrization be ( ) (a sin cos ) (a sin sin ) (a cos ) (spherical coordinates withrijk9) 9 ) 9 ) 9ߜ
a, a 0, on the sphere), 0 , 0 2391)1œ  ŸŸ ŸŸ
(a cos cos ) (a cos sin ) (a sin ) and ( a sin sin ) (a sin cos )Êœ œ rijkr ij
9 ) 9) 9) 9 9) 9)
a cos cos a cos sin a sin
a sin sin a sin cos 0
Ê‚œ Î
Î
rr
ijk
9)
ââ
ââ
ââ
ââ
ââ
ââ
9) 9) 9
9) 9 )
a sin cos a sin sin a sin cos d d dœ Êœ‚ababab kk
## ## # ‚
‚
9) 9) 99 5 )9ijkFnFrr ††
rr
rrœ
œ
kk
9)
a sin cos a sin sin a sin cos d d a sin d d since x y zœ œ œab
$$ # $$ # $ # $
99 9) 99)9 9)9 Fijk
(a sin cos ) (a sin sin ) (a cos ) d a sin d d 4 aœÊœ œ9) 9) 9 5 99)1ijkFn
''
S
† ''
00
2
œœ
$$
39. Let the parametrization be (x y) x y (2a x y) , 0 x a, 0 y a and rij k rikrjkßœ ŸŸ ŸŸÊ œ œ xy d dy dx 10 1
01 1
Ê ‚ œ œ Ê œ ‚Î
Î
rr ijk Fn F rr
ij k xy xy
ââ
ââ
ââ
ââ
ââ
ââ
kk††5
rr
rrxy
xy‚
‚kk
[2xy 2y(2a x y) 2x(2a x y)] dy dx since 2xy 2yz 2xzœ œ Fijk
2xy 2y(2a x y) 2x(2a x y) dœ ÊijkFn
''
S
†5
[2xy 2y(2a x y) 2x(2a x y)] dy dx 4ay 2y 4ax 2x 2xy dy dxœ œ
'' ''
00 00
aa aa
ab
##
a 3a x 2ax dx aœœœ'
0
a
ˆ‰ˆ‰
4 432 13a
3336
$# # %
#
Î
40. Let the parametrization be ( z) (cos ) (sin ) z , 0 z a, 0 2 (where r x y 1 onrijk))) )1ßœ ŸŸ ŸŸ œ œ È
##
the cylinder) ( sin ) (cos ) and (cos ) (sin ) sin cos 0
001
Êœ œÊ‚œ œ Îrijrkrr ij
ijk )) )) )) )) zz
ââ
ââ
ââ
ââ
ââ
ââ
d dz d cos sin dz d dz d , since (cos ) (sin ) zÊœ ‚œ œ œFn F r r F i j k††5)))))))
rr
rr
‚
‚
##z
z
kk
kk ab) z
d 1 dz d 2 aÊœ œ''
S
Fn†5)1 ''
00
2a
œ

1036. 1032 Chapter 16 Integration in Vector Fields
41. Let the parametrization be (r ) (r cos ) (r sin ) r , 0 r 1 (since 0 z 1) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1
(cos ) (sin ) and ( r sin ) (r cos ) r sin r cos 0
cos sin 1
Êœ œ Ê‚œ Îrijkr i jrr
ijk
r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
(r cos ) (r sin ) r d d dr r sin cos r d dr sinceœÊœ ‚œ )) 5 ) )))ijkFnFrr ††
rr
rr
‚
‚
$##r
r
kk
kk ab) r
FikFnœÊœ œ ab a b ˆ‰r sin cos r d r sin cos r dr d sin cos d
#$### ""
)) 5 ) ) ) ) ) )''
S
† '' '
00 0
21 2
œœ
43
cosœ œœ ‘
" $
#
!12 3 3
2
)
)1
1
42. Let the parametrization be (r ) (r cos ) (r sin ) 2r , 0 r 1 (since 0 z 2) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1
(cos ) (sin ) 2 and ( r sin ) (r cos ) r sin r cos 0
cos sin 2
Êœ œ Ê‚œ Îrijkr i jrr
ijk
r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
(2r cos ) (2r sin ) r d d drœÊœ ‚)) 5 )ijkFnFrr ††
rr
rr
‚
‚r
r
kk
kk) r
2r sin cos 4r cos sin r d dr sinceœab
$# $
)) )) )
Fi jkFnœ Ê œ aba b a br sin 2r cos d 2r sin cos 4r cos sin r dr d
## # $# $
)) 5 ))))) ''
S
† ''
00
21
œ sin cos cos sin d sin sinœœœ'
0
2
œ
ˆ‰ œ ‘
""""#$#
#
#
!2262
1
)) )) ) ) ) ) 1
1
43. Let the parametrization be (r ) (r cos ) (r sin ) r , 1 r 2 (since 1 z 2) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1
(cos ) (sin ) and ( r sin ) (r cos ) r sin r cos 0
cos sin 1
Êœ œ Ê‚œ Îrijkr i jrr
ijk
r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
(r cos ) (r sin ) r d d dr r cos r sin r d drœÊœ ‚œ)) 5 ) )))ijkFnFrr ††
rr
rr
‚
‚
## ## $r
r
kk
kk ab) r
r r d dr since ( r cos ) (r sin ) r d r r dr dœ œ Ê œ œab ab
#$ # #$
))) 5 )FijkFn ''
S
† ''
01
22
œ
73
6
1
44. Let the parametrization be (r ) (r cos ) (r sin ) r , 0 r 1 (since 0 z 1) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1
#
(cos ) (sin ) 2r and ( r sin ) (r cos ) r sin r cos 0
cos sin 2r
Êœ œ Ê‚œ Îr i jkr i jrr
ijk r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
2r cos 2r sin r d d dr 8r cos 8r sin 2r d drœÊœ ‚œabab kk ab
## $# $#‚
‚
)) 5 ) )))ijkFnFrr ††
rr
rr
r
r
kk
) r
8r 2r d dr since (4r cos ) (4r sin ) 2 d 8r 2r dr d 2œ œ Ê œ œab ab
$ $
))) 5 )1FijkFn ''
S
† ''
00
21
œ
45. Let the parametrization be ( ) (a sin cos ) (a sin sin ) (a cos ) , 0 , 0rijk9) 9 ) 9 ) 9 9 )ßœ ŸŸ ŸŸ
11
##
(a cos cos ) (a cos sin ) (a sin ) and ( a sin sin ) (a sin cos )Êœ œ rijkr ij9 ) 9) 9) 9 9) 9)
a cos cos a cos sin a sin
a sin sin a sin cos 0
Ê‚œ Î
Î
rr
ijk
9)
ââ
ââ
ââ
ââ
ââ
ââ
9) 9) 9
9) 9 )
a sin cos a sin sin a sin cos œababab
## ## #
9) 9) 99ijk
a sin cos a sin sin a sin cos a sin a sin . The mass isÊ‚œ œ œkk ÈÈrr 9)
%% # %% # %# # %# # 9) 9) 99 9 9
M d a sin d d ; the first moment is M x dœœ œ œ'' ''
S S
599) 5''
00
22
yz
œœ""
ab
#
#
a
œ
1
(a sin cos ) a sin d d x the centroid is located at , , byœœÊœœÊ''
00
22
œœ""
9) 99)ab ˆ‰
#
# ###
aa aaa
4


œ
œ
1
Š‹
Š‹
a
4
a"
"
symmetry

1037. Section 16.6 Parameterized Surfaces 1033
46. Let the parametrization be (r ) (r cos ) (r sin ) r , 1 r 2 (since 1 z 2) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1
(cos ) (sin ) and ( r sin ) (r cos ) r sin r cos 0
cos sin 1
Êœ œ Ê‚œ Îrijkr i jrr
ijk
r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
(r cos ) (r sin ) r r cos r sin r r 2. The mass isœÊ‚œ œ)) ))ijkrr kk
ÈÈ ) r
## ## #
M d r 2 dr d 3 2 ; the first moment is M z d r r 2 dr dœœ œ œ œ'' ''
S S
$5 $ ) 1$ $ 5 $ )'' ''
01 01
22 22
y
œ œ
ÈÈ È
Š‹ Š‹
x
z the center of mass is located at 0 0 by symmetry. TheœÊœ œÊ ßß
Š‹
È
Š‹
È
14 2
39 9
32
14 14
1$
1$

œ14 2
3 œ
ˆ‰
moment of inertia is I x y d r r 2 dr d the radius of gyration is
z
01
22œœ œ Ê''
S
$5$ )ab Š‹
È
## #
# ''
œ Š‹
È
15 21$
R z
I
M
5œœÉ É
z
#
47. Let the parametrization be ( ) (a sin cos ) (a sin sin ) (a cos ) , 0 , 0 2rijk9) 9 ) 9 ) 9 9 1 ) 1ßœ ŸŸ ŸŸ
(a cos cos ) (a cos sin ) (a sin ) and ( a sin sin ) (a sin cos )Êœ œ rijkr ij
9 ) 9) 9) 9 9) 9)
a cos cos a cos sin a sin
a sin sin a sin cos 0
Ê‚œ Î
Î
rr
ijk
9)
ââ
ââ
ââ
ââ
ââ
ââ
9) 9) 9
9) 9 )
a sin cos a sin sin a sin cos œababab
## ## #
9) 9) 99ijk
a sin cos a sin sin a sin cos a sin a sin . The moment ofÊ‚œ œ œkk ÈÈrr 9)
%% # %% # %# # %# # 9) 9) 99 9 9
inertia is I x y d (a sin cos ) (a sin sin ) a sin d d
z
00
2œœ ''
S
$5$9)9)99)ab c dab
## # ## ''
œœ
a sin a sin d d a sin d d a cos sin 2 dœœœœ'' '' '
00 00 0
222
œœ œœ œ
$999) $99)$ 99 )abab a b œ ‘ˆ‰
## # %$ % # "
!33
8a
1
$1
Î
48. Let the parametrization be (r ) (r cos ) (r sin ) r , 0 r 1 (since 0 z 1) and 0 2rijkßœ ŸŸ ŸŸ ŸŸ))) )1
(cos ) (sin ) and ( r sin ) (r cos ) r sin r cos 0
cos sin 1
Êœ œ Ê‚œ Îrijkr i jrr
ijk
r r )) ) ) ))
)) ))
ââ
ââ
ââ
ââ
ââ
ââ
(r cos ) (r sin ) r r cos r sin r r 2. The moment of inertia isœÊ‚œ œ)) ))ijkrr kk
ÈÈ ) r
## ## #
I x y d r r 2 dr d
z
00
21œœ œ''
S
$5$ )ab Š‹
È
## #
# ''
œ
1$
È
2
49. The parametrization (r ) (r cos ) (r sin ) rrijkߜ )))
at P 2 2 2 , r 2,!œßßÊœœŠ‹
ÈÈ
)
1
4
(cos ) (sin ) andrijkijkr
22œ œ))
ÈÈ
##
( r sin ) (r cos ) 2 2rijij)œ œ ))
ÈÈ
2/2 2/2 1
220
Ê‚œ
Î
rr
ijk r )
ââ
ââ
ââ
ââ
ââ
ââ
ÈÈ
ÈÈ
2 2 2 the tangent plane isœ Ê
ÈÈ
ijk

0 2 2 2 x 2 y 2 (z 2) 2x 2y 2z 0, or x y 2z 0.œ Êœ œŠ‹’ “
ÈÈ È È È È È
Š‹Š‹ijk i j k†
The parametrization (r ) x r cos , y r sin and z r x y r z the surface is z x y .rßÊœ œ œÊ œœ Ê œ )))
#### ## È

1038. 1034 Chapter 16 Integration in Vector Fields
50. The parametrization ( )r9)ß
(4 sin cos ) (4 sin sin ) (4 cos )œ 9) 9) 9ijk
at P 2 2 2 3 4 and z 2 3!œßß Êœ œŠ‹
ÈÈ ÈÈ
3
4 cos ; also x 2 and y 2œÊœ œ œ99
1
6 ÈÈ
. Then ʜ)
1
9
4
r
(4 cos cos ) (4 cos sin ) (4 sin )œ 9) 9) 9ijk
662 andœ
ÈÈ
ijk
( 4 sin sin ) (4 sin cos )rij
)œ 9) 9 )
22 at P 662
220
œ Ê ‚ œ Î
Î
ÈÈ
ââ
ââ
ââ
ââ
ââ
ââ
ÈÈ
ÈÈ
ij rr
ijk ! 9)
2 2 2 2 4 3 the tangent plane isœ Ê
ÈÈ È
ijk

22 22 43 x 2 y 2 z 23 0 2x 2y 23z 16,Š‹’ “
ÈÈ È È ÈÈ ÈÈÈ
Š‹Š‹Š ‹ijk i j k œÊ œ†
or x y 6z 8 2. The parametrization x 4 sin cos , y 4 sin sin , z 4 cos œ Ê œ œ œ
È È
9) 9) 9
the surface is x y z 16, z 0.Ê œ  
###
51. The parametrization ( z) (3 sin 2 ) 6 sin zrijk)))ߜ ab
#
at P 0 and z 0. Then!
##œßßÊœ œŠ‹
33 9
3
È
)
1
(6 cos 2 ) (12 sin cos )ri j)œ )))
3 3 3 and at Pœ œijrk
È z !
33 3 3330
001
Ê‚œ œ Îrr ij
ijk ) z
ââ
ââ
ââ
ââ
ââ
ââ
È È
the tangent plane isÊ
33 3 x y (z 0) 0Š‹’ “
È
Š‹ ˆ‰ij i j kœ†
33 9
È
##
3x y 9. The parametrization x 3 sin 2ʜ ʜ
È
)
and y 6 sin x y 9 sin 2 6 sinœÊœ
#### #
#
))) ab

9 4 sin cos 36 sin 6 6 sin 6y x y 6y 9 9 x (y 3) 9œœœÊœÊœab ab
## % # ## # #
)) ) )
52. The parametrization (x y) x y x atrijkߜ
#
P (1 2 1) 2x 2 and at P! !œßß Ê œ œ œri kik rj xy
2 the tangent plane10 2
00
Ê‚œ œÊ Î
"
rr ik
ij k xy
ââ
ââ
ââ
ââ
ââ
ââ
is (2 ) [(x 1) (y 2) (z 1) ] 0ik i j kϠ
2x z 1. The parametrization x x, y y andÊœ Êœ œ
z x the surface is z xœ Ê œ
##

53. (a) An arbitrary point on the circle C is (x z) (R r cos u, r sin u) (x y z) is on the torus withßœ Ê ßß
x (R r cos u) cos v, y (R r cos u) sin v, and z r sin u, 0 u 2 , 0 v 2œ œ œ ŸŸ ŸŸ 11

1039. Section 16.7 Stoke's Theorem 1035
(b) ( r sin u cos v) (r sin u sin v) (r cos u) and ( (R r cos u) sin v) ((R r cos u) cos v)rijkr i juvœ œ
r sin u cos v r sin u sin v r cos u
(R r cos u) sin v (R r cos u) cos v 0
Ê‚œ ÎÎ

rr
ijk
uv
ââ
ââ
ââ
ââ
ââ
ââ
(R r cos u)(r cos v cos u) (R r cos u)(r sin v cos u) ( r sin u)(R r cos u)œ ijk
(R r cos u) r cos v cos u r sin v cos u r sin u r(R r cos u)Ê‚ œ Ê‚œkk ab kkrr rr
uv uv
# ## # # # # # # #
A rR r cos u du dv 2 rR dv 4 rRÊœ œ œ '' '
00 0
22 2
œœ œ
ab
##
11
54. (a) The point (x y z) is on the surface for fixed x f(u) when y g(u) sin v and z g(u) cos vßß œ œ œ ˆ‰ ˆ‰
11
##
x f(u), y g(u) cos v, and z g(u) sin v (u v) f(u) (g(u) cos v) (g(u) sin v) , 0 v 2 ,Êœ œ œ Ê ßœ ŸŸ ri j k 1
aub ŸŸ
(b) Let u y and x u f(u) u and g(u) u (u v) u (u cos v) (u sin v) , 0 v 2 , 0 uœœÊœ œÊßœ ŸŸŸ
## #
ri j k 1
55. (a) Let w 1 where w cos and sin cos cos cos and cos sin
# #
œ œ œ Ê œ Êœ œ
zzxx
ccabab
yy
œ œ
œ œœ
œ
99 9 9) 9)
x a cos cos , y b sin cos , and z c sin Êœ œ œ )9 )9 9
( ) (a cos cos ) (b sin cos ) (c sin )Êßœ rijk)9 )9 )9 9
(b) ( a sin cos ) (b cos cos ) and ( a cos sin ) (b sin sin ) (c cos )rijrijk
)9œ œ )9 )9 )9 )9 9
a sin cos b cos cos 0
a cos sin b sin sin c cos
Ê‚œ Î
ÎÎ
rr
ijk
)9
ââ
ââ
ââ
ââ
ââ
ââ
)9 )9
)9 )9 9
bc cos cos ac sin cos (ab sin cos )œabab)9 )9 99
##
ijk
b c cos cos a c sin cos a b sin cos , and the result follows.Ê‚ œ kkrr
)9
# ## # % ## # % ## # #)9 )9 99
A d d d d a b sin cos b c cos cos a c sin cosÊ‚œ
'' ''
00 00
22
1/2
œœ œœ
kk cdrr)9 9) 9) 99 )9 )9
## # # ## # % ## # %
56. (a) ( u) (cosh u cos ) (cosh u sin ) (sinh u)rijk)))ߜ
(b) ( u) (a cosh u cos ) (b cosh u sin ) (c sinh u)rijk)))ߜ
57. ( u) (5 cosh u cos ) (5 cosh u sin ) (5 sinh u) ( 5 cosh u sin ) (5 cosh u cos ) andrijkrij))) ))ßœ Ê œ
)
(5 sinh u cos ) (5 sinh u sin ) (5 cosh u)rijkuœ ))
r 5 cosh u sin 5 cosh u cos 0
5 sinh u cos 5 sinh u sin 5 cosh u
Ê‚œ Î
)r
ijku
ââ
ââ
ââ
ââ
ââ
ââ
))
))
25 cosh u cos 25 cosh u sin (25 cosh u sinh u) . At the point (x y 0), where x y 25œ ßßœabab
## ##
!!
))ijk
00
we have 5 sinh u 0 u 0 and x 25 cos , y 25 sin the tangent plane isœÊœ œ œ Ê !! ))
5(x y ) [(x x) (y y) z ] 0 xx x yy y 0 xx yy 25
!! ! ! ! ! ! !
##ij i jkœÊœÊœ†
00
58. Let w 1 where cosh u and w sinh u w w cos and w sin
zz xx
cc abab
yy
œ œ
œ œœ
œ
œ œ œ Ê œÊœ œ
##
))
x a sinh u cos , y b sinh u sin , and z c cosh uÊœ œ œ ))
( u) (a sinh u cos ) (b sinh u sin ) (c cosh u) , 0 2 , uÊßœ ŸŸ__rijk))) )1
16.7 STOKES' THEOREM
1. curl 0 0 (2 0) 2 and curl 2 d dx dy
x2xz
FF ijkknkFn
ijk
œ ‚œ œ œ œÊ œÊ œ™†
ââ
ââ
ââ
ââ
ââ
ââ
```
```
##
xyz
5
d 2 dA 2(Area of the ellipse) 4Êœ œ œ
)
C
Fr† ''
R
1

1040. 1036 Chapter 16 Integration in Vector Fields
2. curl 0 0 (3 2) and curl 1 d dx dy
2y 3x z
F F i j kk nk Fn
ij k
œ ‚œ œ œ œÊ œÊ œ
Î
™†
ââ
ââ
ââ
ââ
ââ
ââ
`` `
`` `
#
xy z
5
d dx dy Area of circle 9Êœ œ œ
)
C
Fr† ''
R
1
3. curl x 2x (z 1) and curl
yxzx
FF ijkn Fn
ijk
œ‚œ œ œ Ê™†
ââ
ââ
ââ
ââ
ââ
ââ
```
```
#

xyz
3
ijk
È
( x 2x z 1) d dA d ( 3x z 1) 3 dAœÊœ Ê œ
" "
È È
È
3 3
3
1
5 )
C
Fr† ''
R
È
[ 3x (1 x y) 1] dy dx ( 4x y) dy dx 4x(1 x) (1 x) dxœ œ œ '' '' '
00 00 0
11x 11x 1
œœ
œ ‘
"
#
#
3x x dxœ œ'
0
1
ˆ‰
"
##
#75
6
4. curl (2y 2z) (2z 2x) (2x 2y) and
yzxzxy
FF i j kn
ijk
œ‚œ œ œ

™
ââ
ââ
ââ
ââ
ââ
ââ
```
```
######

xyz
3
ijk
È
curl (2y 2z 2z 2x 2x 2y) 0 d 0 d 0Ê œ œÊ œ œFn F r††
"
È
3 )
C
''
S
5
5. curl 2y (2z 2x) (2x 2y) and
yzxyxy
FF i j knk
ijk
œ‚œ œ œ

™
ââ
ââ
ââ
ââ
ââ
ââ
```
```
######
xyz
curl 2x 2y d dx dy d (2x 2y) dx dy x 2xy dyʜʜʜ œFn F r†† 5 )'' '
C11 1
11 1 œœ œ
cd
#
"
œ"
4y dy 0œ œ'
œ1
1
6. curl 0 0 3x y and
xy 1 z
FF ijkn
ijk
œ ‚œ œ œ œ™
ââ
ââ
ââ
ââ
ââ
ââ
```
```
#$
##

xyz
2x 2y 2z x y z
2xyz 4
ijk ijk
È
œœœ
curl x y z; d dA (Section 16.5, Example 5, with a 4) dÊœ œ œÊFn F r††
34
4z
##
5 )
C
x y z dA 3 r cos r sin r dr d 3 (cos sin ) dœ œ œ''
R
ˆ‰ˆ‰ abab ’“
34 r
4z 6
## # # # # #
#
! '' '
00 0
22 2
œœ
))) )))

32 sin 2 d 4 sin u du 4 8œ œ œ œ''
00
24
œœ
" ##
%
!424
usin 2u
)) 1 œ ‘
1
7. x 3 cos t and y 2 sin t (2 sin t) 9 cos t 9 cos t 16 sin t sin e at theœœÊœ Fi j k aba b
##% œŸ_Ÿ_6 sin t cos t 0
base of the shell; (3 cos t) (2 sin t) d ( 3 sin t) (2 cos t) 6 sin t 18 cos trijr ijFœ Êœ Êœ †
d
dt
r #$
d 6 sin t 18 cos t dt 3t sin 2t 6(sin t) cos t 2 6Ê‚œ œ œ''
S
™†Fn5 1 '
0
2
œ
ab ab œ ‘
#$ #
#
!
3
2
1
8. curl 2 ; f(x y z) 4x y z f 8x 2z
ztanyx
FF j ijk
ijk
œ‚œ œ ßßœÊ œ

™ ™
ââ
ââ
ââ
ââ
ââ
ââ
ââ
```
```
""
#
œ"
##
xyz
x4z
and f 1 d dA f dA; ( 2 f)Êœ œÊ œÊ œ œ ‚ œ œnpjp Fnj
™
™™†™™
™f
ffff
f 2
kk k k kk kk
kk
kk kk™† ™ ™ † †™ 5
p
"œ
d 2 dA d 2 dA 2(Area of R) 2( 1 2) 4 , where RÊ ‚ œ Ê ‚ œ œ œ œ™† ™† ††Fn Fn55 11 '' ''
S R
is the elliptic region in the xz-plane enclosed by 4x z 4.
##
œ

1041. Section 16.7 Stoke's Theorem 1037
9. Flux of d d , so let C be parametrized by (a cos t) (a sin t) ,™™† †‚œ ‚ œ œ FFnFr rij ''
S
5)
C
0 t 2 ( a sin t) (a cos t) ay sin t ax cos t a sin t a cos t aŸŸ Ê œ Ê œ œ œ1
dd
dt dt
rr
ijF †
## # # #
Flux of d a dt 2 aÊ‚œœœ ™†FFr )'
C0
2
œ
##
1
10. (y ) ; x y z
y00
™‚ œ œ œ œ œikn ijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
```
```

xyz
f
f
2x 2y 2z
2xyz
™
™kk È
ijk
œœœ
(y ) z; d dA (Section 16.5, Example 5, with a 1) (y ) dÊ‚ œ œ œÊ ‚™† ™†in in 55
"
z ''
S
( z) dA dA , where R is the disk x y 1 in the xy-plane.œ œ œ Ÿ'' ''
RR
ˆ‰
" ##
z
1
11. Let S and S be oriented surfaces that span C and that induce the same positive direction on C. Then
"#
d d d'' ''
SSÎ Ê
™† † ™†‚œœ‚Fn F r Fn"" ##55)
C
12. d d d , and since S and S are joined by the simple'' '' ''
SSS
™† ™† ™†‚œ‚‚Fn Fn Fn555
Î Ê
"#
closed curve C, each of the above integrals will be equal to a circulation integral on C. But for one surface
the circulation will be counterclockwise, and for the other surface the circulation will be clockwise. Since the
integrands are the same, the sum will be 0 d 0.Ê‚œ
''
S
™†Fn5
13. 5 2 3 ; (cos ) (sin ) 2r and ( r sin ) (r cos )
2z 3x 5y
™‚œ œ œ œ Fijkrijkrij
ijk
ââ
ââ
ââ
ââ
ââ
ââ
```
```xyz r
)) ) ) )
2r cos 2r sin r ; and d dr dcos sin 2r
r sin r cos 0
Ê‚œ œ œ œ‚ Î
Î
rr i jkn rr
ijk r r) )
ââ
ââ
ââ
ââ
ââ
ââ
abab kk))
))
)) 5 )
## ‚
‚
rr
rr r
r

kk
d ( ) ( ) dr d 10r cos 4r sin 3r dr d dÊ‚ œ‚ ‚ œ Ê ‚™† ™ † ™†Fn F r r Fn5))))5 r ) ab
## ''
S
10r cos 4r sin 3r dr d r cos r sin r dœœ'' '
00 0
22 2
œœ
ab œ ‘
## $ $ # "
#
#
!
))) ) ) )
043
33
cos sin 6 d 6(2 ) 12œœœ'
0
2
œ
ˆ‰
80 32
33
)))11
14. 2 2 ; 2r cos 2r sin r and
yzzxxz
™‚œ œ ‚ œ

Fijkrrijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
abab
```
```
##
xyz r )
))
d ( ) ( ) dr d (see Exercise 13 above) d™† ™ † ™†‚œ‚‚ Ê‚Fn F r r Fn5) 5
r )
''
S
2r cos 4r sin 2r dr d r cos r sin r dœ œ '' '
00 0
23 2
œœ
ab œ ‘
## $ $#
$
!
))) ) ))
24
33
18 cos 36 sin 9 d 9(2 ) 18œ œ œ'
0
2
œ
ab )))11
15. 2y 0 x ;
x y 2y z 3z
cos sin 1
r sin r cos 0
™‚œ œ ‚ œ
Î
Fijkrr
ijk ijk
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
```
```
#$
$#
xyz r )
))
))
( r cos ) (r sin ) r and d ( ) ( ) dr d (see Exercise 13 above)œ ‚ œ ‚ ‚)) 5 )ijkFn Frr ™† ™ †
r )
d 2ry cos rx dr d 2r sin cos r cos dr dÊ‚œ œ '' ''
S R
™†Fn5)) ))))aba b
$# % $# ''
00
21
œ
3

1042. 1038 Chapter 16 Integration in Vector Fields
sin cos cos d sinœœœ'
0
2
œ
ˆ‰ œ ‘ˆ‰
2sin 2
5410444
34
)) )) )
"""#
#
#
!
)) 1
1
16. ;
xyyzzx
cos sin 1
r sin r cos 0
™‚ œ œ ‚ œ
ÎÎÎ
Î
Î
Fijkrr
ijk ijk
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
ââ â â
```
```xyz r )
))
))
(r cos ) (r sin ) r and d ( ) ( ) dr d (see Exercise 13 above)œ ‚œ‚‚)) 5 )ijkFn Frr ™† ™ †
r )
d (r cos r sin r) dr d (cos sin 1) d (2 ) 25Ê‚œ œ œ œ''
S

™†Fn5))))))11'' '
00 0
25 2
œœ
’“ ˆ‰
r25
œ
##
&
!
17. 0 0 5 ;
3y 5 2x z 2
™‚œ œ
ÎÎ
Fijk
ij k
ââ
ââ
ââ
ââ
ââ
ââ
`` `
`` `
#
xy z
3 cos cos 3 cos sin 3 sin
3 sin sin 3 sin cos 0
rr
ijk9)‚œ Î
Î
ââ
ââ
ââ
ââ
ââ
ââ
ÈÈ È
ÈÈ
9) 9) 9
9) 9 )
3 sin cos 3 sin sin (3 sin cos ) ; d ( ) ( ) d d (see Exerciseœ ‚œ‚‚abab
##
9) 9) 99 5 9)ijkFnFrr ™† ™ † 9)
13 above) d 15 cos sin d d cos d d 15Ê‚œ œ œœ''
S
™†Fn5999)9))1'' ' '
00 0 0
2/2 2 2
œœ œ œ
œ ‘
15 15
2
#
Î#
! #
1
18. 2z 2y ;
yz x
™‚œ œ Fijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
```
```
##
xyz
2 cos cos 2 cos sin 2 sin
2 sin sin 2 sin cos 0
rr
ijk9)‚œ Î
Î
ââ
ââ
ââ
ââ
ââ
ââ
9) 9) 9
9) 9 )
4 sin cos 4 sin sin (4 sin cos ) ; d ( ) ( ) d d (see Exerciseœ ‚œ‚‚abab
##
9) 9) 99 5 9)ijkFnFrr ™† ™ † 9)
13 above) d 8z sin cos 4 sin sin 8y sin cos d dÊ‚œ '' ''
S R
™†Fn59)9)9)9)ab
##
16 sin cos cos 4 sin sin 16 sin sin cos d dœ ''
00
2/2
œœ
ab
###
9 9 ) 9) 9) )9)
sin cos 4 (sin ) 16 (sin cos ) dœ '
0
2
œ
œ ‘ˆ‰ ˆ‰
" $
##
Î#
!
6
344
sin 2 sin 2
9) ) )) )
99 99
1
cos sin 4 sin cos d sin cos 2 sin 0œ œ œ'
0
2
œ
ˆ‰ œ ‘
16 6
33
)1 ) 1 ) )) )1 ) 1 )
" #
#
!
1
19. (a) 2x 2y 2z curl d d 0 d 0Fijk F0 Fr Fnœ Ê œÊ œ ‚ œ œ )
C
†™†'' ''
SS
55
(b) Let f(x y z) x y z f curl d dßß œ Ê ‚ œ ‚ œ Ê œ Ê œ ‚
##$
™™™ † ™†F0F0FrFn )
C
''
S
5
0 d 0œœ
''
S
5
(c) (x y z ) d d 0 d 0Fijk0F0Fr Fnœ‚œÊ ‚œÊ œ ‚ œ œ™™†™†
)
C
'' ''
SS
55
(d) f f d d 0 d 0FF0FrFnœÊ‚œ‚œÊ œ ‚ œ œ™™ ™™ † ™†
)
C
'' ''
SS
55
20. f x y z (2x) x y z (2y) x y z (2z)Fijkœœ ™
"""
###
### ### ###
$Î# $Î# $Î#
ab ab ab
xxyz yxyz zxyzœ ababab
### ### ###
$Î# $Î# $Î#
ijk (a) (a cos t) (a sin t) , 0 t 2 ( a sin t) (a cos t)rij ijœ ŸŸÊœ 1
d
dt
r
xxyz (a sin t)yxyz (a cos t)Êœ F†
d
dt
r
ab ab
### ###
$Î# $Î#
( a sin t) (a cos t) 0 d 0œ œ Ê œˆ‰ ˆ‰
a cos t a sin t
aa

)
C
Fr†

1043. Section 16.8 The Divergence Theorem and a Unified Theory 1039
(b) d d f d d 0 d 0)
C
Fr Fn n 0n†™† ™™† †œ‚ œ‚ œ œ œ'' '' '' ''
SS SS
5555
21. Let 2y 3z x 3 2 ;
2y 3z x
F i jk F ijkn
ijk
œÊ ‚œ œ œ
Î
™
ââ
ââ
ââ
ââ
ââ
ââ
`` `
`` `

xy z
22
3
ijk
2 2y dx 3z dy x dz d d 2 dÊ™ † † ™ † Fn F r Fn‚œÊ œ œ ‚ œ ))
CC
'' ''
SS
55
2 d , where d is the area of the region enclosed by C on the plane S: 2x 2y z 2œ œ
'' ''
SS
55
22. 0
xyz
™‚œ œF
ijk
ââ
ââ
ââ
ââ
ââ
ââ
```
```xyz
23. Suppose M N P exists such that Fijk F i j kœ ‚œ ™ Š‹ Š ‹ ˆ‰
`` ` ` ` `
`` `` ` `
PN MP NM
yz zx x y
x y z . Then (x) 1. Likewise, (y)œ œ Ê œ œijk
`` ` ` ` ` `` ` `
` ` ` ` `` `` ` ` ` `x y z x xy xz y z x y
PN P N MP
Š‹ ˆ‰
œœ
1 and (z) 1. Summing the calculated equationsÊœ œ Êœ
`` ``` ` ``
`` `` ` ` ` ` `` ``
œœ œœ
MP NM NM
yz yx z x y z zx zy
Š‹
3 or 0 3 (assuming the second mixed partials areÊ œœŠ‹Š‹Š‹
`` `` ``
`` `` `` `` `` ``
œœ œœ œœ
PP NN MM
xy yx zx xz yz zy
equal). This result is a contradiction, so there is no field such that curl x y z .FFijk œ
24. Yes: If , then the circulation of around the boundary C of any oriented surface S in the domain of™‚œF0 F
is zero. The reason is this: By Stokes's theorem, circulation d d dFFrFn0n œœ‚œ
)
C
†™† †'' ''
SS
55
0.œ
25. r x y r x y r 4x x y 4y x y M NœÊœÊœ œ œÈ ab ababab
## % ## % ## ##
#
Fijij™
r ds ds M dy N dx dx dyÊœœœ)))
CCC
™† †ab Š‹
% ``
``
nFn ''
R
MN
xy
4 x y 8x 4 x y 8y dA 16 x y dA 16 x dA 16 y dAœœ œ ''cdabab ab
R RRR
## # ## # ## # # '' '' ''
16I 16I .œ
yx
26. 0, 0, 0, 0, , curl .
`` ` `` `
`` ``` `
œœ œœ

PNMPN M
yz zxx y
yx yx yx yx
xy xy xy xy
œœœœœ œ Ê œ œ
œœ œœ œœ œœ
œœ œœ œœ œœ
œœ œœ
ab ab abab
Fk0’“ However, x y 1 (cos t) (sin t) ( sin t) (cos t)
##
œÊœ Ê œ rij ij
d
dt
r
sin t cos t sin t cos t 1 d 1 dt 2 which isÊœ Ê œ œÊ œ œFijF Frabab ††
d
dt
r ## ))
C
œ

_
1
not zero.
16.8 THE DIVERGENCE THEOREM AND A UNIFIED THEORY
1. div 0 2. x y div 1 1 2FF FijFœ Ê œ œ œ Ê œœ


y x xy xy
xy xy
ij
È
ab
œœ œœ

3. div GMFFœ Ê œ
GM(x y z )
xyz
xyz 3xxyz
xyz
ijk


ab
abab
ab
œœœ

œœœ œœœœ

œœœ
”•
GM GMÎΔ•”•
abab abab
ab ab
xyz 3yxyz xyz 3zxyz
xyz xyz
œœœ œœœœ œœœ œœœœ

œœœ œœœ

1044. 1040 Chapter 16 Integration in Vector Fields
GM 0œ œ’“
3xyz 3xyz xyz
xyz
ababab
ab
œœœ œœœ œœœ
œ
œœœ
!ʜ


4. z a r in cylindrical coordinates z a x y a x y div 0œ Êœ Êœ Ê œ
## # ## ###
ab a b vkv
5. (y x) 1, (z y) 1, (y x) 0 2 Flux 2 dx dy dz 2 2
```
```
$
xyz
œ œ œÊ œÊ œ œ ™†F '''
œœœ111
111
ab
16œ
6. x 2x, y 2y, z 2z 2x 2y 2z
```
```
###
xyx
ab ab abœœœÊœ ™†F
(a) Flux (2x 2y 2z) dx dy dz x 2x(y z) dy dz (1 2y 2z) dy dzœœœ
''' '' ''
000 00 00
111 11 11
cd
#
"
!
y(1 2z) y dz (2 2z) dz 2z z 3œœœœ''
00
11
cd cd
##
""
!!
(b) Flux (2x 2y 2z) dx dy dz x 2x(y z) dy dz (4y 4z) dy dzœœœ''' '' ''
œœœ œœ œœ111 11 11
111 11 11
cd
#
"
œ"
2y 4yz dz 8z dz 4z 0œ œ œœ''
œœ11
11
cd cd
##
""
œ" œ"
(c) In cylindrical coordinates, Flux (2x 2y 2z) dx dy dzœ '''
D
(2r cos 2r sin 2z) r dr d dz r cos r sin zr d dzœœ''' ''
000 00
122 12

œ œ
)) ) ) ) ) œ ‘
22
33
$$#
#
!
cos sin 4z d dz sin cos 4z dz 8 z dz 4 z 4œœœœœ'' ' '
00 0 0
12 1 1
œ
ˆ‰ œ ‘ cd
16 16 6 16
33 33
))) ) )) 111
"
#
!
#
"
!
1
7. (y) 0, (xy) x, ( z) 1 x 1; z x y z r in cylindrical coordinates
`` `
`` `
## #
xy z
œœœÊœœÊœ ™†F
Flux (x 1) dz dy dx (r cos 1) dz r dr d r cos r r dr dÊœ œ œ
'''
D
''' ''
000 00
22r 22
œœ

)) )) ab
$#
cos d cos 4 d sin 4 8œœ œœ''
00
22
œœ
’“ ˆ‰ œ ‘
r r 32 32
54 5 5
_ Î
)) )) )) 1
#
!
#
!
1
8. x 2x, (xz) 0, (3z) 3 2x 3 Flux (2x 3) dV
```
```
#
xyz
abœœœÊœÊœ ™†F '''
D
(2 sin cos 3) sin d d d sin cos sin d dœœ''' ''
000 00
22 2
œœ œœ
3 9 ) 3 9 39) 9 ) 3 99)ab ’“
#$
#
!
3
Î
2
(8 sin cos 8) sin d d 8 cos 8 cos d (4 cos 16) dœœœ'' ' '
00 0 0
22 2
œœ œ œ
9) 99) ) 9) 1) ) œ ‘ˆ‰
99
1
24
sin 2
!
32œ1
9. x 2x, ( 2xy) 2x, (3xz) 3x Flux 3x dx dy dz
`` `
`` `
#
xy z
abœœ œÊœ '''
D
(3 sin cos ) sin d d d 12 sin cos d d 3 cos d 3œœœœ''' '' '
000 00 0
222 22 2
œœ œœ œ__ __ _
3 9 ) 3 9 39) 9 )9) 1 )) 1ab
##
10. 6x 2xy 12x 2y, 2y x z 2, 4x y 0 12x 2y 2
```
```
###$
xyz
ab ababœ œ œÊ œ ™†F
Flux (12x 2y 2) dV (12r cos 2r sin 2) r dr d dzÊœ œ
'''
D
'' '
00 0
322
œ_
)) )
32 cos sin 4 d dz 32 2 dz 112 6œœœ'' '
00 0
32 3
œ_
ˆ‰ˆ‰))) 1 1
16 16
33
11. (2xz) 2z, ( xy) x, z 2z x Flux x dV
`` `
`` `
#
xy z
œ œ œ Ê œ Ê œ ab ™†F '''
D
x dz dy dx (xy 4x) dy dx x 16 4x 4x 16 4x dxœœœ'' ' ' ' '
00 0 0 0 0
2 16 4x 4 y 2 16 4x 2
œœ
œœ œ
’“ab
È
1
2
# #
4x x 16 4xœ œ’“ ab
#% #"" $Î#
#
!
23 3
40

1045. Section 16.8 The Divergence Theorem and a Unified Theory 1041
12. x 3x , y 3y , z 3z 3x 3y 3z Flux 3 x y z dV
```
```
$#$#$# ### ###
xyz
ab ab ab a b '''œœœÊœÊœ ™†F
D
3 sin d d d 3 sin d d 3 dœœœœ''' '' '
000 00 0
2a 2 2
œœ œœ œ
33 9 39) 99) )
##
ab
a2a12a
555
"""
1
13. Let x y z . Then , , ( x) x , ( y) y333333œ œœœÊ œ œ œ È Š‹ Š‹
###
``` ` `
``` ` ` ` `
``333 3 3
333 3xyz x x y y
xz x y
œ
, ( z) z 3 4 , since x y zœ œ œÊ œ œ œ
y xyz
zz
z
œ œœœœ
333
3
33 3 3 33 3
`
``
` ###
Š‹ È™†F
Flux 4 dV (4 ) sin d d d 3 sin d d 6 d 12Êœ œ œ œ œ'''
D
333939)99))1''' '' '
001 00 0
22 2 2
œœ œœ œ
œ
ab
#
14. Let x y z . Then , , . Similarly,3œ œœœÊ œ œÈ Š‹ Š ‹
###
``` `
``` ` `
`"333 3
333 333 33xyz x x
xzxx1x y
œ
œ
and
`" `"
``

yz
y y xyz zz 3 2
Š‹ Š‹
333 333 3 3 3
œ œ Ê œ œ
œ œœœ

œ
™†F
Flux dV sin d d d 3 sin d d 6 d 12Êœ œ œ œ œ'''
D
22
33''' '' '
001 00 0
22 2 2
œœ œœ œ
Š‹ab3939) 99) ) 1
#
15. 5x 12xy 15x 12y , y e sin z 3y e sin z, 5z e cos z 15z e sin z
`` `
`` `
$###$ # $ #
xy z
yy y y
ab ab a bœ œ œ
15x 15y 15z 15 Flux 15 dV 15 sin d d dÊ œœ Ê œ œ™†F
### # # ##
33 33939)'''
D
'''
001
22
œœ
œ
aba b
12 2 3 sin d d 24 2 6 d 48 2 12œœœ'' '
00 0
22
œœ œ
Š‹ Š‹Š ‹
ÈÈÈ
99) ) 1
16. ln x y , tan , z x y x y
`` `
`` `
## œ"

## ##
xxyyxxx xyz
2x 2z 2z 2z y
1
cdab ˆ‰ˆ‰ ˆ‰
–—
ÈÈœ œ œ œ
œœ œœœ
Š‹
ˆ‰
œ
x
y
x x y Flux x y dz dy dxʜʜ ™†F
2x 2z 2x 2z
xy xy xy xy
œœ œœ œœ œœ

## ##
ÈÈ Š‹ '''
D
r dz r dr d 6 cos 3r dr dœœ'' ' ''
01 1 0 1
222 2 2
œœ
œœ
œ
ˆ‰ ˆ ‰
2r cos 2z 3
rr r
)
œœ )))
#
6 21 cos 3 ln 2221 d 2 ln 2221œœ'
0
2
œ
’“Š‹Š‹
ÈÈÈ È
))1
3
#
17. (a) M N P curl Gijk G G i k k Gœ Ê ‚œ œ Ê ‚™ ™†™ Š‹ Š ‹ ˆ‰
`` ` ` ` `
`` `` ` `
PN MP NM
yz zx x y
div(curl )œœG
`` ` `` ` `` `
`` ` `` ` `` `xy z yz x zx y
PN MP NM
Š‹ Š ‹ ˆ‰
0 if all first and second partial derivatives are continuousœœ
``` ```
`` `` `` `` `` ``
œœœ œœœ
PNMPNM
xy xz yz yx zx zy
(b) By the Divergence Theorem, the outward flux of across a closed surface is zero because™‚G
outward flux of ( ) d™™†‚œ ‚GGn
''
S
5
dV [Divergence Theorem with ]œ‚ œ‚
'''
D
™†™ ™ GFG
(0) dV 0 [by part (a)]œœ
'''
D
18. (a) Let M N P and M N P a bFijkFijkFF"""" #### " #œ œ Ê
(aM bM ) (aN bN ) (aP bP ) (a b )œ Ê
"# "# "# "#ijkFF ™†
a b a b a bœ ˆ‰ ˆ‰ Š‹
`` `` ``
`` `` ``
MM NN PP
xx yy zzŸ œŸœŸœ
aba()b()œœ Š‹Š‹
``` ```
``` ```
"#
MNP MNP
xyz xyzŸŸŸ œœœ
™† ™†FF
(b) Define and as in part a (a b )FF FF
"# "# Ê‚™
ab ab a b abœ ’“Š‹ ˆ‰ œ ‘ˆ‰ˆ‰
`` ` ` ` ` ``
`` ` ` ` ` ``
PP NN MM PP
yy z z z z xxŸ œŸœ ŸœŸœ
ij

1046. 1042 Chapter 16 Integration in Vector Fields
ab ab a œ’“’ “ˆ‰ ˆ‰ Š‹Š‹ Š‹
`` ` ` `` `` ``
`` ` ` `` `` ``
NN MM PN MP NM
xx y y yz zx xyŸœ Ÿ œ ŸŸ ŸŸ ŸŸ
kijk
bab œ‚‚’“Š‹ Š ‹ ˆ‰
`` ` ` ` `
`` `` ` `
"#
PN MP NM
yz zx x yœœ œœ œ œ
ij kFF ™™
(c) (N P P N ) (M P P M ) (M N N M ) ( )MNP
MNP
FF i j k FF
ijk
" # "# " # "# " # " # " # " #"""
###‚œ œ Ê ‚
ââ
ââ
ââ
ââ
ââ
ââ
™†
[(N P P N ) (M P P M ) (M N N M ) ]œ™†
"# " # "# " # " # " #ij k (N P P N ) (M P P M ) (M N N M ) P N N Pœ œ
`` `
`` ` ````
"# " # "# " # " # " # # " # "
````
xy z xxxx
NPPN
ˆ‰
& œŸœ
MPPM MNNM Š‹ ˆ‰ "#"# "#"#
`` ` ` ` ` ` `
`` ` ` ` ` ` `
PMMP NMMN
yy y y z z z z
œŸœ Ÿ œ Ÿ œ Ÿ
MNP MNœ### ""
`` ` ` `` `` ``
`` `` ` ` `` ``Š‹ Š ‹Š‹ ˆ‰ ˆ‰
PN MP NM NP PM
yz zx x y zy xzŸ Ÿ ŸŸ Ÿ Ÿ œœ œ œ
P œ‚‚ "#""#
``
``Š‹
MN
yxœœ
FFFF†™ †™
19. (a) div(g ) g (gM) (gN) (gP) g M g N g PFFœœœ™†
``` ` ` `
``` ``````
```
xyz xxyyzz
MNP ggg
Š‹Š‹Š‹
MNP g g gœœ Š‹Š‹
```
``` ```
```ggg
xyz xyz
MNP
™† ™ †FF
(b) (g ) (gP) (gN) (gM) (gP) (gN) (gM)™‚œ Fijk’“ ’ “ œ ‘
`` ` ` ` `
`` ` ` ` `yz z x x y
PgNg Mg Pg Ng Mgœ Š‹Š ‹Š ‹
`` ` ` ` `
`` ` ` ` ` `` `` ` `
`` ` ` ` `gg g g g g
yy z z z z xx xx y y
PN MP N M
ij k
PN gg MP g g NMœ Š‹Š‹Š ‹ Š ‹ ˆ‰
`` `` ` `
`` `` `` `` ` `
`` ` `gg gg g g
yz yz zx zx x y
PN MP
ii jj k
gg g g œ‚‚Š‹
``
``
NM
xy
kFF™™
20. Let M N P and M N P .FijkFijk
"""" ####œ œ
(a) (N P P N ) (P M M P ) (M N N M ) ( )FF i j k FF
" # "# " # " # "# " # " # " #‚œ Ê ‚ ‚ ™
(M N N M ) (P M M P ) (N P P N ) (M N N M )œ’“ œ ‘
`` ``
`` ``
"# "# "# "# "# "# "# "#
yz zx
ij
(P M M P ) (N P P N )’“
``
``
" # "# "# " #
xy
k
and consider the -component only: (M N N M ) (P M M P )i
``
``
"# "# "# "#
yz
ÎÎ Î
NMMNMPPMœ #"#"#"#"
````````
````````
MNNMPMMP
yyyyzzzz
& œ Ÿ œ ŸœŸ œ
NP NP M Mœ Š‹Š‹Š‹Š‹ ## "" " #
`` `` `` ``
`` ```` ``
MM MM NP NP
yz yzyz yz
ŸŸ œœœœ ŸŸ
MNP MNP MœŠ‹Š‹Š‹ ### """ "
``` ``````
``` ``````
MMM MMMMNP
xyz xyzxyz
ŸŸŸ œœœœœœ
M . Now, -comp of ( ) M N P M œ Š‹ Š ‹
```
``` ` ` `
##"###"
```MNP
xyz x y z&&&
iFF †™
M N P ; likewise, -comp of ( ) M N P ;œ œŠ‹ Š‹ ### "#"""
``` ```
``` ```
MMM MMM
xyz xyz
&&& œœœ
iFF †™
-comp of ( ) M and -comp of ( ) M .iFF iFF™† ™† #" " "# #
``` ```
``` ```œ œŠ‹ Š‹
MNP MNP
xyz xyzœœœ &&&
Similar results hold for the and components of ( ). In summary, since the correspondingjk FF ™‚‚ "#
components are equal, we have the result
()()()()()™ †™†™™†™†‚‚œFFFFFF FF FF
"##""# #" "#
(b) Here again we consider only the -component of each expression. Thus, the -comp of ( )iiFF ™† "#
(MM NN PP) M M N N P Pœ œ
`
```````
"# "# "# " # " # " #
``````
xxxxxxx
MMNNPP
ˆ‰
œ Ÿ œ ŸœŸ
-comp of ( ) M N P ,iFF "#" " "
```
```†™ œŠ‹
MMM
xyzœœœ
-comp of ( ) M N P ,iFF #"# # #
```
```†™ œŠ‹
MMM
xyz&&&
-comp of ( ) N P , andiF F "#" "
`` ` `
`` ``‚‚œ ™ Š‹ ˆ‰
NM MP
xy zxœœ œœ

1047. Section 16.8 The Divergence Theorem and a Unified Theory 1043
-comp of ( ) N P .iF F #"# #
`` ` `
`` ``‚‚œ ™ Š‹ ˆ‰
NM MP
xy zxŸŸ ŸŸ
Since corresponding components are equal, we see that
( ) ( ) ( ) ( ) ( ), as claimed.™ † †™ †™ ™ ™FFFFFFF FF F
"#"##"" ## "œ‚‚‚‚
21. The integral's value never exceeds the surface area of S. Since 1, we have (1)(1) 1 andkk k k kkkkFFnFnŸœŸœ †
d d [Divergence Theorem]
''' ''
D S
™† †FFn55œ
d [A property of integrals]Ÿ
''
S
kkFn†5
(1) d 1 ŸŸ
''
S
5 cdkkFn†
Area of S. œ
22. Yes, the outward flux through the top is 5. The reason is this: Since (x 2y (z 3)™† ™†Fijkœ
1 2 1 0, the outward flux across the closed cubelike surface is 0 by the Divergence Theorem. The fluxœœ
across the top is therefore the negative of the flux across the sides and base. Routine calculations show that
the sum of these latter fluxes is 5. (The flux across the sides that lie in the xz-plane and the yz-plane are 0, while the flÎ ux
across the xy-plane is 3.) Therefore the flux across the top is 5.Î
23. (a) (x) 1, (y) 1, (z) 1 3 Flux 3 dV 3 dV
```
```xyz
œ œ œÊ œÊ œ œ ™†F ''' '''
DD
3(Volume of the solid)œ
(b) If is orthogonal to at every point of S, then 0 everywhere Flux d 0.F n Fn Fn ††œÊœœ
''
S
5
But the flux is 3(Volume of the solid) 0, so is not orthogonal to at every point.Á Fn
24. 2x 4y 6z 12 Flux ( 2x 4y 6z 12) dz dy dx™†Fœ Ê œ
'''
000
ab1
( 2x 4y 9) dy dx 2xb 2b 9b dx a b 2ab 9ab ab( a 2b 9) f(a b);œ œ œ œœß'' '
00 0
ab a
ab
###
2ab 2b 9b and a 4ab 9a so that 0 and 0 b( 2a 2b 9) 0 and
`` ``
`` ``
##ff ff
ab ab
œ œ œ œ Ê œ
a( a 4b 9) 0 b 0 or 2a 2b 9 0, and a 0 or a 4b 9 0. Now b 0 or a 0 œ Ê œ œ œ œ œ œ
Flux 0; 2a 2b 9 0 and a 4b 9 0 3a 9 0 a 3 b so that f 3 is theÊ œ œ œ Ê œ Ê œ Ê œ ß œ
3327
###
ˆ‰
maximum flux.
25. d dV 3 dV d dV Volume of D'' ''' ''' '' '''
SS DD D
Fn F Fn†™† †55œœÊœœ
"
3
26. 0 Flux d dV 0 dV 0FC F Fn FœÊ œÊ œ œ œ œ™† † ™† '' ''' '''
S DD
5
27. (a) From the Divergence Theorem, f d f dV f dV 0 dV 0
'' ''' ''' '''
S DDD
™ † ™†™ ™n5œœœœ
#
(b) From the Divergence Theorem, f f d f f dV. Now,'' '''
S D
™† ™†™n5œ
ff f f f ff f f f™™†™œ Ê œ ˆ‰ ˆ‰ ˆ‰ ˆ‰Š‹ ’ “ Š‹ ’ “ ”•
``` `` `` ``
``` `` `` ``
##
#
fff ff ff ff
xyz xx yy zz
ijk
œœ œ
œœ œ
f f f 0 f since f is harmonic f f d f dV, as claimed.œ œ Ê œ™™ ™ ™† ™
# ## #
kk kk kk '' '''
S D
n5
28. From the Divergence Theorem, f d f dV dV. Now,'' ''' '''
S DD
™ † ™†™n5œœ Š‹
```
```
œœœ
œœœ
fff
xyz
f(x y z) ln x y z ln x y z , , ßßœ œ Ê œ œ œÈ ab
### "```
#```
### fxf fz
x xyz y xyz z xyz
y
œœœ œœœ œœœ

1048. 1044 Chapter 16 Integration in Vector Fields
, , , Êœ œ œ Ê
``` ```
``` ```


œœœ œœœ
œœœ œœœ
œœœ œœœ œœœ
œœœ œœœ œœœ
œœœ
fff fff
xyz xyz
x y z xyz xyz
xyz xyz xyzab ab ab f d dddœœÊ œ œ
xyz
xyz xyz xyz
dV sin
œœœ
œœœ
œ œœœ œœœ œ
œ


"
ab '' '''
S D
™†n539) '''
000
22a
œœ__
39
3
a sin d d a cos d a dœœœœ'' ' '
00 0 0
22 2 2
œœ œ œ__ _ _
99) 9 ) ) cd
1 1Î#
! #
a
29. f g d f g dV f f f dV'' ''' '''
S DD
™† ™†™ ™†nijk5œœ Š‹
```
`` `
ggg
xy z
f f f dVœ '''
D
Š‹
``````
``` ``` ```
```
œœœ
œœœ
gggggg
xxx yyy zzz
fff
f dV f g f g dVœœ ''' '''
D D
’“Š‹Š ‹ ab
``` ` ` `
` ` ` ``````
``` #
œœœ
œœœ
ggg g g g
x y z xxyyzz
fff
™™†™
30. By Exercise 29, f g d f g f g dV and by interchanging the roles of f and g,
'' '''
S D
™† ™ ™†™n5œ ab
#
g f d g f g f dV. Subtracting the second equation from the first yields:'' '''
S D
™† ™ ™ †™n5œ ab
#
f g g f d f g g f dV since f g g f'' '''
S D
ab a b™™† ™ ™ ™†™™†™œ œÞn5
##
31. (a) The integral p(t x y z) dV represents the mass of the fluid at any time t. The equation says that'''
D
ßßß
the instantaneous rate of change of mass is flux of the fluid through the surface S enclosing the region D:
the mass decreases if the flux is outward (so the fluid flows out of D), and increases if the flow is inward
(interpreting as the outward pointing unit normal to the surface).n
(b) dV p dV p d p dV p''' ''' '' '''
DD D S
`
` `
`p
tdt t
d
œœœ Êœ vn v v†™† ™†5
3
Since the law is to hold for all regions D, p 0, as claimed™†vœ
`
`
p
t
32. (a) T points in the direction of maximum change of the temperature, so if the solid is heating up at the™
point the temperature is greater in a region surrounding the point T points away from the pointÊ™
T points toward the point T points in the direction the heat flows.%Î %Ι™
(b) Assuming the Law of Conservation of Mass (Exercise 31) with k T p and c T p, we haveœ œ™ v 3
c T dV k T d the continuity equation, ( k T) (c T) 0
d
dt t''' ''
D S
35 3œ Ê œ™† ™† ™n
`
`
c ( k T) k T T K T, as claimedÊœ œ Êœ œ3
``
``
###TTk
ttc
™† ™ ™ ™ ™
3
CHAPTER 16 PRACTICE EXERCISES
1. Path 1: t t t x t, y t, z t, 0 t 1 f(g(t) h(t) k(t)) 3 3t and 1, 1,rijkœ Ê œ œ œ ŸŸ Ê ß ß œ œ œ
# dx
dt dt
dy
1 dt 3 dt f(x y z) ds 3 3 3t dt 2 3
dz dx dz
dt dt dt dt
dy
œÊ œ Ê ßß œ œÊˆ‰ ˆ‰Š‹
ÈÈÈ
ab
##
#
#
''
C0
1
Path 2: t t , 0 t 1 x t, y t, z 0 f(g(t) h(t) k(t)) 2t 3t 3 and 1, 1,rij "
#œ ŸŸ Ê œ œ œ Ê ß ß œ œ œ
dx
dt dt
dy
0 dt 2 dt f(x y z) ds 2 2t 3t 3 dt 3 2 ;
dz dx dz
dt dt dt dt
dy
œÊ œ Ê ßß œ œÊˆ‰ ˆ‰Š‹
ÈÈÈ ' ab
##
#
#
C
0
1œ
'
t x 1, y 1, z t f(g(t) h(t) k(t)) 2 2t and 0, 0, 1rijk#œ Ê œ œ œ Ê ß ß œ œ œ œ
dx dz
dt dt dt
dy
dt dt f(x y z) ds (2 2t) dt 1ÊœÊßßœœÊˆ‰ ˆ‰Š‹
dx dz
dt dt dt
dy##
# ''
C0
1
f(xyz) ds f(xyz) ds f(xyz) 3 2 1Êßßœßßßßœ'''
CCC œ œ
È

1049. Chapter 16 Practice Exercises 1045
2. Path 1: t x t, y 0, z 0 f(g(t) h(t) k(t)) t and 1, 0, 0ri "
#œÊœ œ œÊ ß ß œ œ œ œ
dx dz
dt dt dt
dy
dt dt f(x y z) ds t dt ;ÊœÊßßœœÊˆ‰ ˆ‰Š‹
dx dz
dt dt dt 3
dy##
#
# " ''
C0
1œ
t x 1, y t, z 0 f(g(t) h(t) k(t)) 1 t and 0, 1, 0rij#œ Ê œ œ œ Ê ß ß œ œ œ œ
dx dz
dt dt dt
dy
dt dt f(x y z) ds (1 t) dt ;ÊœÊßßœœÊˆ‰ ˆ‰Š‹
dx dz 3
dt dt dt 2
dy##
# ''
C0
1
t x 1, y 1, z t f(g(t) h(t) k(t)) 2 t and 0, 0, 1rijk$œ Ê œ œ œ Ê ß ß œ œ œ œ
dx dz
dt dt dt
dy
dt dt f(x y z) ds (2 t) dtÊœÊßßœœÊˆ‰ ˆ‰Š‹
dx dz 3
dt dt dt 2
dy##
# ''
C0
1_
f(xyz) ds f(xyz) ds f(xyz) ds f(xyz) dsÊ ßß œ ßß ßß ßß œ''''
Path 1 C C C œ œ"
10
3
Path 2: t t x t, y t, z 0 f(g(t) h(t) k(t)) t t and 1, 1, 0rij %
#œ Ê œ œ œ Ê ß ß œ œ œ œ
dx dz
dt dt dt
dy
dt 2 dt f(x y z) ds 2 t t dt 2;ÊœÊßßœ œÊˆ‰ ˆ‰Š‹
ÈÈÈ
ab
dx dz 5
dt dt dt 6
dy##
#
# ''
C0
1Î
t (see above) f(x y z) dsrijk$œ Ê ßß œ '
C_
3
2
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## 33
2
3. (a cos t) (a sin t) x 0, y a cos t, z a sin t f g(t) h(t) k(t) a sin t a sin t andrjkœ ÊœœœÊßßœ œ ab kk
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1050. 1046 Chapter 16 Integration in Vector Fields
6. , 0 , 0 M dx N dy dz is exact; 1 f(x y z)
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yyzzz xx y x
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z
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7. y cos z y cos z is not conservative; 2 cos t 2 sin t , 0 t 2
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œ Á œ Ê œ Ÿ Ÿ Frijk abab 1
d 2 sin t 2 cos t d 2 cos t sin 1 2 cos t dt 2 sin t sin 1 2 sin tÊœ Ê œ rijFrabab c d ababab ab aba ba bab
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4 sin 1 sin t cos t dt 8 sin 1œœab a b ab'
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8. 0 , 0 , 3x is conservative d 0
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9. Let M 8x sin y and N 8y cos x 8x cos y and 8y sin x 8x sin y dx 8y cos x dyœœÊœ œÊ
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10. Let M y and N x 2y and 2x y dx x dy (2x 2y) dx dyœœÊœ œÊœ
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11. Let z 1 x y f (x y) 1 and f (x y) 1 f f 1 3 Surface Area 3 dx dyœ Ê ß œ ß œÊ œ Ê œ
xy xy É
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## ''
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12. f 3 2y 2z , f 9 4y 4z and f 3™™™†œ œ Ê œ œijkpi p kk k kÈ
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1051. Chapter 16 Practice Exercises 1047
15. f(x y z) 1 f f and fßßœœÊ œ Ê œ œÊ œ
xz
abc a b c a b c c
y
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ijk pk p
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since c 0 Surface Area dA c dA abc ,Ê œ œ œ '' ''
RR
É
Š‹
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c

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abc abc
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since the area of the triangular region R is ab. To check this result, let a c and a b ; the area can be
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vik w ijœ œ
found by computing .
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##
5
g(x y z) d 4y 1 dx dy y y 1 dx dy y y dx dyÊßßœ œ œ '' '' ''
S RR
5
yz
4y 1È
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13
3y y dy 3 0œœœ'
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z
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17. f 2y 2z , f 4y 4z 2 y z 10 and f 2z since z 0™™ ™†œ œÊ œœœ œ  jkpk p kk k kÈÈ
## ##
d dx dy dx dy g(x y z) d x y y z dx dyÊœ œ œ ßß œ 55
10 5 5
2z z z '' ''
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x y (25) dx dy x dx dy dy 50œœœœ''
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ab Š‹
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18. Define the coordinate system so that the origin is at the center of the earth, the z-axis is the earth's axis (north
is the positive z direction), and the xz-plane contains the earth's prime meridian. Let S denote the surface
which is Wyoming so then S is part of the surface z R x y . Let R be the projection of S ontoœab
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180 180
11
97,751 sq. mi.¸
19. A possible parametrization is ( ) (6 sin cos ) (6 sin sin ) (6 cos ) (spherical coordinates);rijk9) 9 ) 9 ) 9ߜ
now 6 and z 3 3 6 cos cos and z 3 3 3 3 6 cos 3999 9œœÊœ ʜʜ œÊœ
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also 0 2ŸŸ)1
21. A possible parametrization is (r ) (r cos ) (r sin ) (1 r) (cylindrical coordinates);rijkߜ )))
now r x y z 1 r and 1 z 3 1 1 r 3 0 r 2; also 0 2œ Êœ ŸŸÊŸŸÊŸŸ ŸŸÈ
##
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22. A possible parametrization is (x y) x y 3 x for 0 x 2 and 0 y 2rij kßœ ŸŸ ŸŸˆ‰
y
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1052. 1048 Chapter 16 Integration in Vector Fields
23. Let x u cos v and z u sin v, where u x z and v is the angle in the xz-plane with the x-axisœœ œ
È
##
(u v) (u cos v) 2u (u sin v) is a possible parametrization; 0 y 2 2u 2 u 1Ê ßœ ŸŸÊ ŸÊ Ÿrijk
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0 u 1 since u 0; also, for just the upper half of the paraboloid, 0 vÊŸŸ   ŸŸ 1
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1
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ijku v uv uvœ œ Ê ‚ œ œ Ê ‚ œ ""
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ââ
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00
11
kk
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26. xy z d (u v)(u v) v 6 du dv 6 u 2v du dv''
S
ab c d a b
ÈÈ
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11 11
6 2uv dv 6 2v dv 6 v vœ œ œ œ œ
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ÈÈ r r) )
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29. 0 , 0 , 0 Conservative
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yzzxx y
œœ œœ œœ Ê
30. , , Conservative
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P N M 3xz P N M
yzzxx y
3zy 3xy
xyz xyz xyz
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ab ab ab
œœœ œœœ œœœ
B"ʜ B"ʜ B"ʜ
31. 0 ye Not Conservative
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PN
yz
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z
32. , , Conservative
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œPx NM PN z M
y (xyz) z z (xyz) x x (xyz) y
y
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33. 2 f(x y z) 2x g(y z) 2y z g(y z) y zy h(z)
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xyy
g
œÊ ßßœ ß Ê œ œ Ê ßœ
f(x y z) 2x y zy h(z) y h (z) y 1 h (z) 1 h(z) z CÊ ßß œ Ê œ œ Ê œ Ê œ
#ww `
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f
z
f(xyz) 2x y zy z CÊßßœ
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34. z cos xz f(x y z) sin xz g(y z) e g(y z) e h(z)
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g yy
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f(x y z) sin xz e h(z) x cos xz h (z) x cos xz h (z) 0 h(z) CÊßßœ Êœ œ Ê œÊ œ
y f
z
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ww
f(x y z) sin xz e CÊßßœ
y

1053. Chapter 16 Practice Exercises 1049
35. Over Path 1: t t t , 0 t 1 x t, y t, z t and d ( ) dt 2t trijk rijk F ij kœ ŸŸ Ê œ œ œ œ Ê œ
##
d 3t 1 dt Work 3t 1 dt 2;ʜʜ œFr† ab ab
## '
0
1
Over Path 2: t t , 0 t 1 x t, y t, z 0 and d ( ) dt 2t trij rij F ijk"""
##œ ŸŸ Ê œ œ œ œ Ê œ
d 2t 1 dt Work 2t 1 dt ; t , 0 t 1 x 1, y 1, z t andÊ œ Ê œ œ œ ŸŸ Ê œ œ œFr rijk"" " #
##† ab ab '
0
1
5
3
d dt 2 d dt Work dt 1 Work Work Work 1rk F ijk Fr## ## # "#œÊœÊ œÊ œœÊœœœ † '
0
1
58
33
36. Over Path 1: t t t , 0 t 1 x t, y t, z t and d ( ) dt 2t tr i j k r ijk F i jkœ ŸŸ Ê œ œ œ œ Ê œ
##
d 3t 1 dt Work 3t 1 dt 2;ʜʜ œFr† ab ab
## '
0
1
Over Path 2: Since f is conservative, d 0 around any simple closed curve C. Thus consider)
C
Fr†œ
d d d , where C is the path from (0 0 0) to (1 1 0) to ( ) and C is the path
'''
curve C C
Fr Fr Fr†††œ ß ß ß ß "ß "ß "
œ œ
"#
from (1 1 1) to ( ). Now, from Path 1 above, d 2 0 d d ( 2)ß ß !ß !ß ! œ Ê œ œ '''
C curve Cœ
Fr Fr Fr†††
d2ʜ
'
Cœ
Fr†
37. (a) e cos t e sin t x e cos t, y e sin t from (1 0) to e 0 0 t 2rijœ Êœœ ßßÊŸŸabab ab
tt tt 2 1
1
e cos t e sin t e sin t e cos t and Êœ œ œ
d
dt
tt tt xy
xy
e cos t e sin t
e cost e sint
r ij ij
abab ijF



ab
abab
ab
œœ

œœ
tt
2t 2t
eœ Êœ œˆ‰ˆ‰ Š‹
cos t sin t d cos t sin t cos t sin t sin t cos t
ee dteeee
t
2t 2t t t t tijF †
r
œœ
œ
Work e dt 1 eÊœ œ '
0
2
œ
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(b) f(x y z) x y g(y z) FœÊœÊßßœßÊœ
xy yg
xy xy xy
fx f
x yy
ij `

` `
` ``
##
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ab ab ab
œœ œœ œœ

ab
g(y z) C f(x y z) x y is a potential function for dœ Ê ßœ Ê ßßœ Ê
y
xyab
œœ


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†
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2211 œ
38. (a) x ze is conservative d 0 for closed path CFF FrœÊ Êœ™†ab
#y )
C
any
(b) d x ze d x ze x ze 2 0 2''
C 100
102
Fr r†™†œœœœ
Ÿßß_
Ÿßß _œ
ab kkab ab
## #
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yy y
1
11
39. 2y ; unit normal to the plane is
yy3z
™‚œ œ œ œ
Î
Fk nijk
ijk
ââ
ââ
ââ
ââ
ââ
ââ
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xy z
263
4369
263
777
ijk
È
y; and f(x y z) 2x 6y 3z f 3 d dA dAÊ‚ œ œ ßßœÊ œÊœ œ™† ™†Fn p k p
6 7
7 f3
f
kk 5
kk
kk
™
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d y d y dA 2y dA 2r sin r dr d sin d 0Êœ œ œ œ œ œ)'''
C 00 0
21 2
Fr† '' '' ''
RR R
667 2
773 3
5))))ˆ‰ˆ ‰
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40. 8y ; the circle lies in the plane f(x y z) y z 0 with unit normal
x y x y 4y z
™‚œ œ ßßœœ

Fi
ijk
ââ
ââ
ââ
ââ
ââ
ââ
`` `
`` `
##
xy z
0 d d 0 d 0njk Fn Fr Fnœ Ê‚ œÊ œ ‚ œ œ
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C
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55
41. (a) 2t 2t 4 t , 0 t 1 x 2t, y 2t, z 4 t 2, 2, 2trij kœ ŸŸÊœ œ œÊœ œ œ
ÈÈ È È È È
ab
## dx dz
dt dt dt
dy
dt 4 4t dt M (x y z) ds 3t 4 4t dt (4 4t)ʜʜßßœ œÊˆ‰ ˆ‰ œ ‘Š‹
ÈÈ
dx dz
dt dt dt 4
dy##
#
## " $Î#
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1
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42 2œ
È

1054. 1050 Chapter 16 Integration in Vector Fields
(b) M (x y z) ds 4 4t dt t 1 t ln t 1 t 2 ln 1 2œßßœœœ''
C0
1
$
ÈÈ ÈÈÈ
’“Š‹ Š‹
## #
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42. t 2t t , 0 t 2 x t, y 2t, z t 1, 2, tri j kœ ŸŸ Ê œ œ œ Ê œ œ œ
2 2 dx dz
3 3 dt dt dt
dy$Î# $Î# "Î#
dt t 5 dt M (x y z) ds 3 5 t t 5 dtʜʜßߜʈ‰ ˆ‰Š‹ ÈÈÈ
dx dz
dt dt dt
dy##
# ''
C0
2
$
3(t 5) dt 36; M x ds 3t(t 5) dt 38; M y ds 6t(t 5) dt 76;œœœ œœœ œœ
'''''
0C0 C0
22 2
yz xz $$
M z ds 2t (t 5) dt 2 x , y , zxy
144 38 19 76 19
7 M 36 18 M 36 9 M 36
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2
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144
7
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43. t t , 0 t 2 x t, y t , z 1, 2 t , tri j kœ ŸŸ Ê œ œ œ Ê œ œ œŠ‹Š‹
È
22 22 dy
3 3 dt dt dt
t t dx dz
ÈÈ
$Î# $Î# "Î#
##
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dt 1 2t t dt (t 1) dt t 1 dt (t 1) dt on the domain given.ÊœœœœÊˆ‰ ˆ‰Š‹
È È kk
dx dz
dt dt dt
dy##
#
##
Then M ds (t 1) dt dt 2; M x ds t (t 1) dt t dt 2;œœ œœœ œ œœ'' ' ' ' '
C0 0 C 0 0
22 2 2
$$ ˆ‰ ˆ‰
""
t1 t1
yz
M y ds t (t 1) dt t dt ; M z dsxz xy
22 22
3t1 3 15
32œœ œ œœ'' ' '
C0 0 C
22
$$ Š‹ ˆ‰
ÈÈ
$Î# $Î#"

(t 1) dt dt x 1; y ; zœœœÊœœœœœœœ''
00
22
Š‹ˆ‰
tt4216
t1 3 M M 15 M
MM M
œœ
# # # #
"
yz xyxz
32
15
ˆ‰
; I y z ds t dt ; I x z ds t dt ;œœ œ œ œ œ œ œ
ˆ‰
4
3
#
## $ ## #2 8 t 232 t 64
39445 415
xy
'' ''
C0 C0
2 2
ab ab Š‹ Š‹$$
Î Î
I y x ds t t dt ; R ; R ;zxy
856
99 M M
I 229
35 15
I 42œ œ œœœ œ œœ œ''
C0
2
ab ˆ‰ É ÊÊ É
## # $
##
$
x y
Š‹ Š‹ ÈÈ
ÈÈ
232 64
45 15
R z
I
M3
27œœ œÉ Ê
z
Š‹ È
56
9
#
44. z 0 because the arch is in the xy-plane, and x 0 because the mass is distributed symmetrically with respectœœ
to the y-axis; (t) (a cos t) (a sin t) , 0 t ds dtrijœ ŸŸÊœ 1 ʈ‰ ˆ‰Š‹
dx dz
dt dt dt
dy##
#
( a sin t) (a cos t) dt a dt, since a 0; M ds (2a y) ds (2a a sin t) a dtœ œ   œ œ œ È
## '' '
CC 0
$
œ
2a 2a ; M y dt y(2a y) ds (a sin t)(2a a sin t) dt 2a sin t a sin t dtœ œ œ œ œ
2
1$
# ###
xz
CC 0 0
'' ' '
œœ
ab
2a cos t a 4a y x y z 0 0œ œ Ê œ œ Ê ßß œ ß ßœ ‘ˆ‰ˆ‰ ab
## #
! #œœ œ
œ
œœtsin 2t a 8 8
2 4 2a 2a 4 4 4 4
4a1
111
11 1
œ
œ
œœ Š‹
a
œ
œ
œ
45. (t) e cos t e sin t e , 0 t ln 2 x e cos t, y e sin t, z e e cos t e sin t ,rijkœŸŸÊœœœÊœabab a b
ttt ttt tt dx
dt
e sin t e cos t , e dt
dy dy
dt dt dt dt dt
tt t dz dx dz
œ œÊ ab ʈ‰ ˆ‰Š‹
##
#
e cos t e sin t e sin t e cos t e dt 3e dt 3 e dt; M ds 3 e dtœœœ œœÉababab
È ÈÈ
tt tt t 2t tt### ''
C0
ln 2
$
3; M z ds 3 e e dt 3 e dt z ;œœœ œ œÊœœœ
ÈÈÈ
Š‹ ab xy
tt 2t 33 M
M 3
3
'' '
C0 0
ln 2 ln 2
$
È
Œ œ
È##
xy
33
œ
œ
I x y ds e cos t e sin t 3 e dt 3 e dt Rz z
2t 2t t 3t 73
3M
Iœ œ œ œÊœ'' '
C0 0
ln 2 ln 2
ab a b Š‹
ÈÈ É
## # #
$
È
z
œœÊ É
73
33
7
3
È
È
46. (t) (2 sin t) (2 cos t) 3t , 0 t 2 x 2 sin t, y 2 cos t, z 3t 2 cos t, 2 sin t,rijkœ ŸŸÊœœ œÊœ œ 1
dx
dt dt
dy
3 dt 4 9 dt 13 dt; M ds 13 dt 2 13;
dz dx dz
dt dt dt dt
dy
œÊ œ œ œ œ œÊˆ‰ ˆ‰Š‹ È ÈÈÈ
##
#
''
C0
2
$$ 1$
œ

1055. Chapter 16 Practice Exercises 1051
M z ds (3t) 13 dt 6 13; M x ds (2 sin t) 13 dt 0;xy yzœœ œ œœ œ'' ''
C0 C 0
2 2
$$$1 $ $
œ œ
Š‹ Š‹
ÈÈ È
#
M y ds (2 cos t) 13 dt 0 x y 0 and z 3 ( 3 ) is thexz
M
M
613
213œ œ œ Ê œ œ œ œ œ Ê !ß !ß''
C0
2
$$ 11
œ
Š‹
È
xy $1
$1
œÈ
È
center of mass
47. Because of symmetry x y 0. Let f(x y z) x y z 25 f 2x 2y 2zœœ ßßœœ Ê œ
###
™ ijk
f 4x 4y 4z 10 and f 2z, since z 0 M (x y z) dÊ œ œ œÊ œ  Êœ ßßkk k kÈ™™†
###
pk p ''
R
$5
z dA 5 dA 5(Area of the circular region) 80 ; M z d 5z dAœœœ œœœ'' '' '' ''
RR RR
ˆ‰
10
z
xy
#
1$5
5 25 x y dx dy 5 25 r r dr d d zœœ œœÊœœ''
R
È Š‹
È ## # '' '
00 0
24 2
œœ
))1
490 980 49
3 3 80 12
Š‹
980
3
1
1
x y z 0 ; I x y d 5 x y dx dy 5r dr d 320 d 640 ;Êßßœß!ß œ œ œ œ œab ab abˆ‰
49
12
z '' ''
RR
## ## $
$5 ) ) 1 '' '
00 0
24 2
œœ
R22
zœœ œÉ É È
I
M80
640z 1
1
48. On the face z 1: g(x y z) z 1 and g g 1 and g 1 d dAœßßœœ œÊœÊ œ œÊœ pk k p™™™† kk k k 5
I x y dA 2 r dr d ; On the face z 0: g(x y z) z 0 g and Êœ œ œ œ ßßœœ Ê œ œ''
R
ab
## $ ''
00
/4 sec
œ
)
2
3
™ kpk
g 1 g 1 d dA I x y dA ; On the face y 0: g(x y z) y 0ÊœÊ œÊœÊœ œ œßßœœkk k k ab™™† p 5 ''
R
## 2
3
g and g 1 g 1 d dA I x 0 dA x dx dz ;Ê œ œÊ œÊ œÊ œ Êœ œ œ™™™†jpj p kk k k ab5 ''
R
## " ''
00
11 3
On the face y 1: g(x y z) y 1 g and g 1 g 1 d dAœßßœœÊœ œÊ œÊ œÊœ ™™™†jpj p kk k k 5
I x 1 dA x 1 dx dz ; On the face x 1: g(x y z) x 1 g and Êœ œ œ œ ßßœœ Ê œ œ''
R
ab ab
## # ''
00
11
4
3
™ ipi
g 1 g 1 d dA I 1 y dA 1 y dy dz ; On the faceÊœÊ œÊœÊœœ œkk k k ab ab™™† p 5 ''
R
## # ''
00
11
4
3
x 0: g(x y z) x 0 g and g 1 g 1 d dAœßßœœÊœ œÊ œÊ œÊœ ™™™†ipi p kk k k 5
I 0 y dA y dy dz IÊœ œ œ Ê œœ''
R
ab
## # "" ''
00
11
z
1224414
3 333333 3
49. M 2xy x and N xy y 2y 1, 2x, y, x 1 Flux dx dyœ œÊœ œ œ œÊ œ
```` ``
```` ``
MMNN MN
xyxy xy ''
R
Š‹
(2y1x1) dydx (2yx) dydx ; Circ dxdyœ œ œ œ '' ''
R R
''
00
11
3NM
xy#``
``
Š‹
(y 2x) dy dx (y 2x) dy dxœ œ œ''
R
''
00
11
"
#
50. M y 6x and N x y 12x, 1, 1, 2y Flux dx dyœ œ Ê œ œ œ œ Ê œ
## ```` ``
```` ``
MMNN MN
xyxy xy ''
R
Š‹
( 12x 2y) dx dy ( 12x 2y) dx dy 4y 2y 6 dy ;œ œ œ œ''
R
'' '
0y 0
11 1
ab
# 11
3
Circ dx dy (1 1) dx dy 0œ œœ'' ''
RR
Š‹
``
``
NM
xy
51. M and N ln x sin y and ln x sin y dy dxœ œ Ê œ œ Ê
cos y sin y sin y cos y
xyxxxx
MN``
`` )
C
dx dy dx dy 0œ œ œ'' ''
RR
Š‹ Š ‹
``
``
NM
xy x x
sin y sin y

1056. 1052 Chapter 16 Integration in Vector Fields
52. (a) Let M x and N y 1, 0, 0, 1 Flux dx dyœ œÊœœœœÊœ
```` ``
```` ``
MMNN MN
xyxy xy ''
R
Š‹
(1 1) dx dy 2 dx dy 2(Area of the region)œ œ œ
'' ''
RR
(b) Let C be a closed curve to which Green's Theorem applies and let be the unit normal vector to C. Letn
x y and assume is orthogonal to at every point of C. Then the flux density of at every pointFij F n Fœ
of C is 0 since 0 at every point of C 0 at every point of CFn†œÊœ
``
``
MN
xy
Flux dx dy 0 dx dy 0. But part (a) above states that the flux isÊœ œ œ '' ''
RR
Š‹
``
``
MN
xy
2(Area of the region) the area of the region would be 0 contradiction. Therefore, cannot beÊÊ F
orthogonal to at every point of C.n
53. (2xy) 2y, (2yz) 2z, (2xz) 2x 2y 2z 2x Flux (2x 2y 2z) dV
```
```xyz
œœœÊœÊœ ™†F '''
D
(2x 2y 2z) dx dy dz (1 2y 2z) dy dz (2 2z) dz 3œœœœ''' '' '
000 00 0
111 11 1
54. (xz) z, (yz) z, (1) 0 2z Flux 2z r dr d dz
```
```xyz
œœœÊœÊœ ™†F '''
D
)
2z dz r dr d r 16 r dr d 64 d 128œœœœ''' '' '
003 0 0 0
24 25r 2 4 2
œœœ
œ
œ

)))1 ab
#
55. ( 2x) 2, ( 3y) 3, (z) 1 4; x y z 2 and x y z z 1
```
```
### ##
xyz
œ œ œ Ê œ œ œ Ê œ ™†F
x y 1 Flux 4 dV 4 dz r dr d 4 r 2 r r dr dÊ œ Ê œ œ œ
## $# '''
D
''' ''
00r 00
21 2r 21
œœ


œ
œ
)) Š‹
È
42 d782œ œ '
0
2
œ
Š‹Š‹
ÈÈ
72 2
12 3 3
)1
56. (6x y) 6, ( x z) 0, (4yz) 4y 6 4y; z x y r
`` `
`` `
##
xy z
œ œ œ Ê œ œ œ ™†F È
Flux (6 4y) dV (6 4r sin ) dz r dr d 6r 4r sin dr dÊœ œ œ
'''
D
''' ''
000 00
/21r /21
œœ
)) )) ab
#$
(2 sin ) d 1œ œ'
0
/2
œ
)) 1
57. y z x 0 Flux d dV 0Fijk F Fn Fœ Ê œÊ œ œ œ™† † ™†
'' '''
S D
5
58. 3xz y z 3z 1 3z 1 Flux d dVFijk F Fn Fœ Ê œœÊ œ œ
#$ ##
™† † ™† '' '''
S D
5
1 dz dy dx dx xœœœœ'' ' '
00 0 0
416x2y2 4
œ
œ__
Š‹ ’ “
16 x x 8
16 48 3
œ
%
!
œ
59. xy x y y y x 0 Flux d dVFijk F Fn FœÊ œÊ œ œ
## ##
™† † ™† '' '''
S D
5
x y dV r dz r dr d 2r dr d dœœ œ œœ'''
D
ab
## # $ "
# ''' '' '
00 1 00 0
211 21 2
œœœ
œ
)))1
60. (a) (3z 1) 3 Flux across the hemisphere d dVFkF Fn Fœ Ê œÊ œ œ™† † ™†
'' '''
S D
5
3 dV 3 a 2 aœœ œ '''
D
ˆ‰ˆ ‰
"
#
$$4
3
11
(b) f(xyz)xyza0 f2x2y2z f 4x4y4z 4a2a sinceßßœœÊ œ Ê œ œ œ
#### ### #
™™ ijk kk È È
a0 (3z1) ; f f 2z Êœ œ Ê œ œÊ œ œnFnpkpk
2x 2y 2z x y z
aa a
zijk ijk
#
†™†™† ˆ‰

1057. Chapter 16 Additional and Advanced Exercises 1053
f 2z since z 0 d dA dA d (3z 1) dAÊœ ÊœœœÊ œkk ˆ‰ˆ‰™† †pFn 55
kk
kk
™
™†
f
f2z z az
2a a z a
p '' ''
S R xy
(3z 1) dx dy 3 a x y 1 dx dy 3 a r 1 r dr dœ œ œ '' ''
RRxy xy
00
2a
ˆ‰È Š‹
È ### ## ''
œ
)
a d a 2 a , which is the flux across the hemisphere. Across the base we findœœ '
0
2
œ
Š‹
a
œ
#
$#$
)1 1
[3(0) 1] since z 0 in the xy-plane (outward normal) 1 Flux acrossFkk nk Fnœœ œ Êœ Ê œÊ †
the base d 1 dx dy a . Therefore, the total flux across the closed surface isœœœ
'' ''
S R
Fn†51
xy
#
a2a a2a.ab1111
#$# $
œ
CHAPTER 16 ADDITIONAL AND ADVANCED EXERCISES
1. dx ( 2 sin t 2 sin 2t) dt and dy (2 cos t 2 cos 2t) dt; Area x dy y dxœ œ œ
"
#)
C
[(2 cos t cos 2t)(2 cos t 2 cos 2t) (2 sin t sin 2t)( 2 sin t 2 sin 2t)] dtœ
"
#'
0
2
œ
[6 (6 cos t cos 2t 6 sin t sin 2t)] dt (6 6 cos t) dt 6œ œ œ
""
##''
0 0
2 2
œ œ
1
2. dx ( 2 sin t 2 sin 2t) dt and dy (2 cos t 2 cos 2t) dt; Area x dy y dxœ œ œ
"
#)
C
[(2 cos t cos 2t)(2 cos t 2 cos 2t) (2 sin t sin 2t)( 2 sin t 2 sin 2t)] dtœ
"
#'
0
2
œ
[2 2(cos t cos 2t sin t sin 2t)] dt (2 2 cos 3t) dt 2t sin 3t 2œ œ œ œ
"""
###
#
!''
0 0
2 2
œœ
œ ‘
2
3
1
1
3. dx cos 2t dt and dy cos t dt; Area x dy y dx sin 2t cos t sin t cos 2t dtœœœœ
"""
###)'
C0
œ
ˆ‰
sin t cos t (sin t) 2 cos t 1 dt sin t cos t sin t dt cos t cos t 1œœœœœ
"""""
###
## # $
!''
0 0
œ œ
cdab ab œ ‘
333
2
1
4. dx ( 2a sin t 2a cos 2t) dt and dy (b cos t) dt; Area x dy y dxœ œ œ
"
#)
C
2ab cos t ab cos t sin 2t 2ab sin t 2ab sin t cos 2t dtœ
"
#
##'
0
2
œ
cdaba b
2ab 2ab cos t sin t 2ab(sin t) 2 cos t 1 dt 2ab 2ab cos t sin t 2ab sin t dtœ œ
" "
# #
## #''
0 0
2 2
œ œ
cdab ab
2abt ab cos t 2ab cos t 2 abœ œ
"
#
$
#
!
œ ‘
2
3
1
1
5. (a) (x y z) z x y is only at the point (0 0 0), and curl (x y z) is never .Fijk0 Fijk0ßßœ ßß ßßœ
(b) (x y z) z y is only on the line x t, y 0, z 0 and curl (x y z) is never .Fik0 Fij0ßßœ œ œ œ ßßœ
(c) (x y z) z is only when z 0 (the xy-plane) and curl (x y z) is never .Fi0 Fj0ßß œ œ ßß œ
6. yz xz 2xyz and , so is parallel to when yz , xz ,Fij kn F nœ œ œ œ œ
## ##

xyz xyz cy
xyz RRR
cxijk ijk
È
œœœ
and 2xyz 2xy y x y x and z 2x z 2x. Also,œÊ œœ ʜʜ„ œ„œ Êœ„
cz xz c
Rxy R
yz
œ œ
## # # È
x y z R x x 2x R 4x R x . Thus the points are: ,
#### ## ## ##
####
œ Ê œ Ê œ Êœ„ ßß
RRR 2R
Š‹
È
, , , , ,Š‹Š ‹Š ‹Š‹Š ‹
RR RR RR RR RR2R 2R 2R 2R 2R
## # # ## # # # # ## # # #
( (Î Î (Î ( Î (Î (Î (Î ( (Î (Î
ÈÈ ÈÈÈ
, Š‹Š ‹Î(( Î((Î
RR RR 2R 2R
## # ## #
ÈÈ
7. Set up the coordinate system so that (a b c) (0 R 0) (x y z) x (y R) zßß œ ß ß Ê ßß œ $ È
###
x y z 2Ry R 2R 2Ry ; let f(x y z) x y z R and œœ ßßœ œÈÈ
### # # ### #
pi
f 2x 2y 2z f 2 x y z 2R d dz dy dz dyÊ œ Ê œ œ Ê œ œ™™ ijk kk È
###
5
kk
kk
™
™†
f
f2x
2R
i

1058. 1054 Chapter 16 Integration in Vector Fields
Mass (x y z) d 2R 2Ry dz dy R dz dyÊœ ßßœ œ'' '' ''
S RR
$5
yz yz
È ˆ‰#
œ
œœ
R
x
2R 2Ry
Ryz
È
È
œ
œœœ
4R dz dy 4R 2R 2Ry sin dyœœ '' '
œœ
œ
œR0 R
RRy R
0
Ry
œ
œ

È
ÈÈ
2R 2Ry
Ryz
2 1 z
Ryœ
œœœ
œ
œœ
œ
œ
É Š‹
»
22
2 R 2R 2Ry dy 2 R 2R 2Ryœœœ11 '
œ
œR
R
3/2
R
R
É
»
2 116R
3R 3
2
ˆ‰ab
œ 1

8. (r ) (r cos ) (r sin ) , 0 r 1, 0 2 cos sin 0
r sin r cos
rijk rr
ijk
ßœ ŸŸ ŸŸ Ê ‚œ
Î_
)))) )1 ))
)) r )
ââ
ââ
ââ
ââ
ââ
ââ
(sin ) (cos ) r 1 r ; 2 x y 2 r cos r sin 2rœ Ê‚œœœ œ)) $ ) )ijkrr kk
ÈÈ È r )
#######
Mass (x y z) d 2r 1 r dr d 1 r d 2 2 1 dÊœ ßßœ œ œ ''
S
$5 ) ) )'' ' '
00 0 0
21 2 2
œœœ
ÈÈ
’“ Š‹ab # #
$Î#
"
!
22
33
22 1œ
4
3
1
Š‹
È
9. M x 4xy and N 6y 2x 4y and 6 Flux (2x 4y 6) dx dyœ œ Ê œ œÊ œ
# ``
``
MN
xx ''
00
ba
a 4ay 6a dy a b 2ab 6ab. We want to minimize f(a b) a b 2ab 6ab ab(a 2b 6).œ œ ßœœ'
0
b
ab
### ##
Thus, f (a b) 2ab 2b 6b 0 and f (a b) a 4ab 6a 0 b(2a 2b 6) 0 b 0 or
abßœ œ ßœ œÊ œÊœ
##
b a 3. Now b 0 a 6a 0 a 0 or a 6 (0 0) and (6 0) are critical points. On the otherœ œ Ê œ Ê œ œ Ê ß ß
#
hand, b a 3 a 4a( a 3) 6a 0 3a 6a 0 a 0 or a 2 (0 3) and ( ) are alsoœÊ œÊ œÊœ œÊß #ß"
##
critical points. The flux at (0 0) 0, the flux at (6 0) 0, the flux at (0 3) 0 and the flux at (2 1) 4.ߜ ߜ ߜ ߜ
Therefore, the flux is minimized at (2 1) with value 4.(Î
10. A plane through the origin has equation ax by cz 0. Consider first the case when c 0. Assume the plane is givenœ Á
by z ax by and let f(x y z) x y z 4. Let C denote the circle of intersection of the plane with the sphere.œ ßßœœ
###
By Stokes's Theorem, d d , where is a unit normal to the plane. Let
S
)''
C
Fr Fn n†™†œ‚ 5
(x y) x y (ax by) be a parametrization of the surface. Then a b 0a
0b
rij k rr ijk
ijk
ßœ ‚œ œ "
"
xy
ââ
ââ
ââ
ââ
ââ
ââ
d dx dy a b 1 dx dy. Also, and
zxy
Ê œ ‚ œ ‚ œ œ œ5kk
È
ââ
ââ
ââ
ââ
ââ
ââ
rr F ijk n
ijk xy xyz
ab
ab1
## ```
```

™
ijk
È
œœ
d a b 1 dx dy (a b 1) dx dy (a b 1) dx dy. NowÊ‚œ œœ'' '' '' ''
S RRR
™†Fn5
xy xy xy
ab1
ab1


##
È
œœ
È
x y (ax by) 4 x y xy 1 the region R is the interior of the
## # # #
#
œÊ œÊ Š‹ Š‹ ˆ‰
a1 b1 ab
44
œœ
xy
ellipse Ax Bxy Cy 1 in the xy-plane, where A , B , and C . By Exercise 47 in
##
#
œ œ œ œ
a1 ab b1
44
œœ
Section 10.3, the area of the ellipse is d h(a b) .
24
4AC B ab1 ab1
4(a b 1)11 1
ÈÈ È
œ

œ œœ œœ
œÊœßœ )
C
Fr†
Thus we optimize H(a b) : 0 andßœ œ œ
(a b 1)
ab1 a
H 2(a b 1) b 1 a ab
ab1

`
`

œ
œœ
œ
œœ
ab
ab
2
0 a b 1 0, or b 1 a ab 0 and a 1 b ab 0
`
`


##H
b
2(a b 1) a 1 b ab
ab1
œ œ Ê œ œ œ
ab
ab
œ
œœ
2
a b 1 0, or a b (b a) 0 a b 1 0, or (a b)(a b 1) 0 a b 1 0 or a b.Êœ œÊœ œÊœ œ
##
The critical values a b 1 0 give a saddle. If a b, then 0 b 1 a ab a 1 a a 0œ œ œ Ê œ
###
a 1 b 1. Thus, the point (a b) ( 1 1) gives a local extremum for d z x yÊ œ Ê œ ß œ ß Ê œ )
C
Fr†
x y z 0 is the desired plane, if c 0.Ê œ Á
: Since h( 1 1) is negative, the circulation about is , so is the correct pointing normal forNote
clockwiseÎ(Î Î nn

1059. Chapter 16 Additional and Advanced Exercises 1055
the counterclockwise circulation. Thus ( ) d actually gives the circulation.''
S
™†‚Fn 5 maximum
If c 0, one can see that the corresponding problem is equivalent to the calculation above when b 0, which does notœ œ
lead to a local extreme.
11. (a) Partition the string into small pieces. Let s be the length of the i piece. Let (x y ) be a point in the?
iii
th ß
i piece. The work done by gravity in moving the i piece to the x-axis is approximately
th th
W (gx y s)y where x y s is approximately the mass of the i piece. The total work done by
i iii i iii
thœ ??
gravity in moving the string to the x-axis is W gx y s Work gxy dsDD?
ii
iii i
CœÊœ
## '
(b) Work gxy ds g(2 cos t) 4 sin t 4 sin t 4 cos t dt 16g cos t sin t dtœœ œ '' '
C0 0
/2 /2
## # ##
œ œ
ab
È
16g gœœ’“Š‹
sin t 16
33


1Î#
!
(c) x and y ; the mass of the string is xy ds and the weight of the string isœœ
''
''
CC
CC
x(xy) ds y(xy) ds
xy ds xy ds
'
C
g xy ds. Therefore, the work done in moving the point mass at x y to the x-axis is'
C
abß
W g xy ds y g xy ds g.œœœŠ‹ ''
CC
# 16
3
12. (a) Partition the sheet into small pieces. Let be the area of the i piece and select a point (x y z ) in?5
i iii
th ßß
the i piece. The mass of the i piece is approximately x y . The work done by gravity in moving the
th th
ii i
?5
i piece to the xy-plane is approximately (gx y )z gx y z Work gxyz d .
th
ii i i iii i
?5 ?5 5œÊœ ''
S
(b) gxyz d g xy(1 x y) 1 ( 1) ( 1) dA 3g xy x y xy dy dx'' ''
S R

5œœ
xy
00
11x
È È
ab ## ## ''
œ
3g xy x y xy dx 3g x x x x dxœ œ
ÈÈ œ ‘ œ ‘ ''
00
11
x
" " " """"### $ #$%
###
"œ
!23 6 6
3gxxx x 3gœ œ œ
ÈÈœ ‘ˆ‰
"""" ""#$% &
"
! #126630 13020
3g
È
(c) The center of mass of the sheet is the point x y z where z with M xyz d andabßß œ œ
M
M
xy
xy
''
S
5
M xy d . The work done by gravity in moving the point mass at x y z to the xy-plane isœßß ''
S
5 ab
gMz gM gM gxyz d .œœœ œŠ‹
M
M20
3g
xy
xy
''
S
5
È
13. (a) Partition the sphere x y (z 2) 1 into small pieces. Let be the surface area of the i piece
## #
œ ?5
i
th
and let (x y z ) be a point on the i piece. The force due to pressure on the i piece is approximately
iii
th thßß
w(4 z ) . The total force on S is approximately w(4 z ) . This gives the actual force to beÎÎ
ii ii?5 D ?5
i
w(4 z) d .''
S
Î 5
(b) The upward buoyant force is a result of the -component of the force on the ball due to liquid pressure.k
The force on the ball at (x y z) is w(4 z)( ) w(z 4) , where is the outer unit normal at (x y z).ßß œ ßßnnn
Hence the -component of this force is w(z 4) w(z 4) . The (magnitude of the) buoyant forceknkkn œ††
on the ball is obtained by adding up all these -components to obtain w(z 4) d .kkn
''
S
Î †5
(c) The Divergence Theorem says w(z 4) d div(w(z 4) ) dV w dV, where D
'' ''' '''
S DD
œ œkn k†5
is x y (z 2) 1 w(z 4) d w 1 dV w, the weight of the fluid if it
## #
ŸÊ œ œ '' '''
S D
kn†51
4
3
were to occupy the region D.

1060. 1056 Chapter 16 Integration in Vector Fields
14. The surface S is z x y from z 1 to z 2. Partition S into small pieces and let be the area of theœ œ œÈ
##
?5
i
i piece. Let (x y z ) be a point on the i piece. Then the magnitude of the force on the i piece due to
th th th
iii
ßß
liquid pressure is approximately F w(2 z ) the total force on S is approximately
iiiœ Ê ?5 F w(2 z ) the actual force is w(2 z) d w 2 x y 1 dADD ?5 5
i
S R
iii
xyœ Ê œ '' '' ˆ‰È É ##

x
xy xy
y
œ
œœ œœ
œ
2 w 2 x y dA 2w(2 r) r dr d 2w r r d dœœ œœ''
Rxy
01 0 0
22 2 2
ÈÈȈ‰ œ ‘È ## #$"
#
" '' ' '
œœœ
)))
33
22w
È
œ
42w
3
È
1
15. Assume that S is a surface to which Stokes's Theorem applies. Then d ( ) d)
C
Er En†™†œ‚ ''
S
5
d d . Thus the voltage around a loop equals the negative of the rateœ œ'' ''
SS
``
``
B
tt
††nBn55
of change of magnetic flux through the loop.
16. According to Gauss's Law, d 4 GmM for any surface enclosing the origin. But if
''
S
Fn F H†™51œœ‚
then the integral over such a closed surface would have to be 0 by the Divergence Theorem since div 0.Fœ
17. f g d (f g) d (Stokes's Theorem)
)
C
™† ™ ™ †rnœ‚ ''
S
5
(f g f g) d (Section 16.8, Exercise 19b)œ‚‚
''
S
™™ ™ ™† n5
[(f)( ) f g] d (Section 16.7, Equation 8)œ‚
''
S
0n™™† 5
( f g) dœ‚
''
S
™™† n5
18. ( ) is conservative f; also, ™ ™ ™ ™ ™† ™†‚œ ‚Ê ‚ œÊ Êœ œFF FF0FF FF FF
" # #" #" #" " #
( ) 0 f 0 (so f is harmonic). Finally, on the surface S, f ( )ʜʜ œ™† ™ ™ † †FF nFFn #" #"
#
0. Now, (f f) f f f f so the Divergence Theorem givesœœ œ FnFn #"
#† † ™† ™ ™ †™ ™
f dV f f dV (f f) dV f f d 0, and since f 0 we have
''' ''' ''' ''
DDD S
kk™™™†™™† ™
# # #
œ œœ œ n5
f dV 0 0 dV 0 , as claimed.
''' '''
DD
kk k k™
##
#" #" # "
œ Ê œ Ê œ Ê œ FF FF0 FF
19. False; let y x (y) (x) 0 and 0 0 0
xy0
Fij0 F F ijk0
ijk
œÁÊ œ œ ‚œ œ œ™† ™
``
``
```
```xy xyz
ââ
ââ
ââ
ââ
ââ
ââ
20. sin 1 cos cos ( )k k kkkk kkkk ab kkkk kkkk kkkkrr rr rr rr rr rr rr
uv uv uv uv uv uv uv‚œ œ œ œ
### ## #### ## ## # #
)) ) †
EG F d du dv EG F du dvÊ‚ œ Êœ‚ œ kk kk
ÈÈ
rr rr uv uv
#
## 5
21. x y z 1 1 1 3 dV 3 dV 3V V dVrijk r r rœ Ê œœÊ œ œ Ê œ™† ™† ™† ''' ''' '''
DD D
"
3
d , by the Divergence Theoremœ
"
3''
S
rn†5

1061. Chapter 16 Additional and Advanced Exercises 1057
NOTES:

1064. to accompany
THOMAS’CALCULUS
ELEVENTHEDITION
AND
THOMAS’CALCULUS:EARLYTRANSCENDENTALS
ELEVENTHEDITION
BASED ON THEORIGINALWORK BY
George B. Thomas, Jr.
Massachusetts Institute of Technology
ASREVISED BY
Maurice D. Weir
Naval Postgraduate School
Joel Hass
University of California, Davis
Frank R. Giordano
Naval Postgraduate School
INSTRUCTOR’S
SOLUTIONSMANUAL
PARTTWO
ARDIS• BORZELLINO• BUCHANAN• MOGILL• NELSON
weir_22650_ISM_TTL_CPY.qxd 10/15/04 11:02 AM Page 1

1065. Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors.
Copyright © 2005 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted,
in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior
written permission of the publisher. Printed in the United States of America.
ISBN 0-321-22650-X
1 2 3 4 5 6 BB 08 07 06 05 04
This work is protected by United States copyright laws and is provided solely
for the use of instructors in teaching their courses and assessing student
learning. Dissemination or sale of any part of this work (including on the
World Wide Web) will destroy the integrity of the work and is not permit-
ted. The work and materials from it should never be made available to
students except by instructors using the accompanying text in their
classes. All recipients of this work are expected to abide by these
restrictions and to honor the intended pedagogical purposes and the needs of
other instructors who rely on these materials.
weir_22650_ISM_TTL_CPY.qxd 10/15/04 11:02 AM Page 2

1066. TABLE OF CONTENTS
11 Infinite Sequences and Series 697
11.1 Sequences 697
11.2 Infinite Series 708
11.3 The Integral Test 713
11.4 Comparison Tests 717
11.5 The Ratio and Root Tests 722
11.6 Alternating Series, Absolute and Conditional Convergence 725
11.7 Power Series 731
11.8 Taylor and Maclaurin Series 739
11.9 Convergence of Taylor Series; Error Estimates 743
11.10 Applications of Power Series 750
11.11 Fourier Series 757
Practice Exercises 761
Additional and Advanced Exercises 773
12 Vectors and Analytic Geometry in Space 779
12.1 Three-Dimensional Coordinate Systems 779
12.2 Vectors 781
12.3 The Dot Product 786
12.4 The Cross Product 792
12.5 Lines and Planes in Space 797
12.6 Cylinders and Quadric Surfaces 804
Practice Exercises 812
Additional Exercises 819
13 Vector-Valued Functions and Motion in Space 825
13.1 Vector Functions 825
13.2 Modeling Projectile Motion 833
13.3 Arc Length and the Unit Tangent Vector 837T
13.4 Curvature and the Unit Normal Vector 840N
13.5 Torsion and the Unit Binormal Vector 845B
13.6 Planetary Motion and Satellites 850
Practice Exercises 852
Additional Exercises 858
14 Partial Derivatives 863
14.1 Functions of Several Variables 863
14.2 Limits and Continuity in Higher Dimensions 870
14.3 Partial Derivatives 875
14.4 The Chain Rule 880
14.5 Directional Derivatives and Gradient Vectors 887
14.6 Tangent Planes and Differentials 891
14.7 Extreme Values and Saddle Points 897
14.8 Lagrange Multipliers 909
14.9 Partial Derivatives with Constrained Variables 917
14.10 Taylor's Formula for Two Variables 919
Practice Exercises 922
Additional Exercises 936

1067. 15 Multiple Integrals 941
15.1 Double Integrals 941
15.2 Areas, Moments, and Centers of Mass 952
15.3 Double Integrals in Polar Form 959
15.4 Triple Integrals in Rectangular Coordinates 963
15.5 Masses and Moments in Three Dimensions 968
15.6 Triple Integrals in Cylindrical and Spherical Coordinates 972
15.7 Substitutions in Multiple Integrals 980
Practice Exercises 985
Additional Exercises 991
16 Integration in Vector Fields 997

16.1 Line Integrals 997
16.2 Vector Fields, Work, Circulation, and Flux 1001
16.3 Path Independence, Potential Functions, and Conservative Fields 1008
16.4 Green's Theorem in the Plane 1012
16.5 Surface Area and Surface Integrals 1018
16.6 Parametrized Surfaces 1025
16.7 Stokes's Theorem 1035
16.8 The Divergence Theorem and a Unified Theory 1039
Practice Exercises 1044
Additional Exercises 1053

1068. PREFACE TO THE INSTRUCTOR
This Instructor's Solutions Manual contains the solutions to every exercise in the 11th Edition of THOMAS' CALCULUS
by Maurice Weir, Joel Hass and Frank Giordano, including the Computer Algebra System (CAS) exercises. The
corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the
CAS exercises (because the CAS command templates would give them all away).
In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or
rewritten every solution which appeared in previous solutions manuals to ensure that each solution
conforms exactly to the methods, procedures and steps presented in the textì
is mathematically correctì
includes all of the steps necessary so a typical calculus student can follow the logical argument and algebraì
includes a graph or figure whenever called for by the exercise, or if needed to help with the explanationì
is formatted in an appropriate style to aid in its understandingì
Every CAS exercise is solved in both the MAPLE and computer algebra systems. A template showingMATHEMATICA
an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within
the text grouping require a change only in the input function or other numerical input parameters associated with the problem
(such as the interval endpoints or the number of iterations).
Acknowledgments
Solutions Writers
William Ardis, Collin County Community College-Preston Ridge Campus
Joseph Borzellino, California Polytechnic State University
Linda Buchanana, Howard College
Tim Mogill
Patricia Nelson, University of Wisconsin-La Crosse
Accuracy Checkers
Karl Kattchee, University of Wisconsin-La Crosse
Marie Vanisko, California State University, Stanislaus
Tom Weigleitner, VISTA Information Technologies
Thanks to Rachel Reeve, Christine O'Brien, Sheila Spinney, Elka Block, and Joe Vetere for all their guidance and help at
every step.