Chapter 11 16 solucionario larson

respuestas del capitulo 11 al 16 9°edicion

1. © 2010 Brooks/Cole, Cengage Learning
CHAPTER 11
Vectors and the Geometry of Space
Section 11.1 Vectors in the Plane................................................................................2
Section 11.2 Space Coordinates and Vectors in Space ............................................13
Section 11.3 The Dot Product of Two Vectors.........................................................22
Section 11.4 The Cross Product of Two Vectors in Space ......................................30
Section 11.5 Lines and Planes in Space....................................................................37
Section 11.6 Surfaces in Space..................................................................................50
Section 11.7 Cylindrical and Spherical Coordinates................................................57
Review Exercises..........................................................................................................68
Problem Solving...........................................................................................................76

2. 2
© 2010 Brooks/Cole, Cengage Learning
CHAPTER 11
Vectors and the Geometry of Space
Section 11.1 Vectors in the Plane
1. (a)
51,4 2 4,2vπα απ
(b)
2. (a) 33,24 0,6vπαααπ α
(b)
3. (a)
θ〈
42,3 3 6,0vπαα ααα πα
(b)
4. (a) 12,31 3,2vπαα α πα
(b)
5. 53,62 2,4
31,8 4 2,4
u
v
uv
πα απ
πα απ
π
6.
θ〈
θ〈
14,805,8
72,7 1 5,8
u
v
uv
παα α π
πα ααπ
π
7.
60,23 6,5
93,510 6,5
u
v
uv
παααπα
πα α π α
π
8.
θ〈 θ〈
11 4 , 4 1 15, 3
25 0, 10 13 15, 3
u
v
uv
παααααπ α
πα απα
π

9. (b) 5 2,5 0 3,5vπα απ
(c)
35vijπ〉
(a), (d)

10. (b) θ〈
34,6 6 1,12vπα ααπα
(c) 12vi jπα 〉
(a), (d)
11. (b) 68,13 2,4vπαααπαα
(c) 2 4vijπα α
(a), (d)
5432
1
1
3
2
4
5
x
v
(4, 2)
y
x
−1 12345
−1
1
2
3
4
5
(3, 5)
(2, 0)
(5, 5)
v
y
x
−1−2
−6
−5
−4
−3
−2
−1
−3 231
v
(0, 6)−
y
4
2
−2
−2
−4
−4−6−8
x
v
(−6, 0)
y
x
−3 −1−2
3
2
1
( 3, 2)−
v
y
x
(8, 3)
(6, −1)
(−2, −4)
v
−2−4248
−6
2
4
6
y
−2−4−6−826810
−4
−6
2
4
6
8
x
v
y
(4, −6)
(3, 6)
(−1, 12)

3. Section 11.1 Vectors in the Plane 3
© 2010 Brooks/Cole, Cengage Learning
12. (b) θ〈50,1 4 5,3vπαα ααα πα
(c) 5 3vijπα 〉
(a) and (d).
13. (b) 66,62 0,4vπα απ
(c)
4vjπ
(a) and (d).

14. (b) θ〈
37,1 1 10,0vπαα ααα πα
(c) 10vi
πα
(a) and (d).
15. (b)
3514
22 3 3
,3 1,vπα απα
(c)
5
3
vijπα 〉
(a) and (d)
16. (b)
0.84 0.12, 1.25 0.60 0.72, 0.65vπα απ
(c) 0.72 0.65vijπ〉
(a) and (d).
17. (a) 223,56,10vππ

(b) 39,15vαπαα

(c)
735 21
222
,vπ

(d)
102
33
2,vπ

x
(−9, −15)
(3, 5)
v
−3v
y
−3−6−9−12−15 3 6
−6
−9
−12
−15
3
6
64
6
4
2
2
x
v
(6, 6)
(0, 4)
(6, 2)
y
x
v
8642−2−4−6−8
3
2
1
−2
−3
( 10, 0)−
(7, 1)−
(,1)− 3 −
y
21
3
2
−1−2
x
v
1
2
, 3((
3 24 3
, ((
5 3
−1, ((
y
x
v
1.000.750.25 1.250.50
1.25
1.00
0.75
0.50
0.25
(0.72, 0.65)
(0.84, 1.25)
(0.12, 0.60)
y
x
(6, 10)
(3, 5)
v2v
y
−2 246810
−2
2
4
6
8
10
x
v
−6−4−22
4
2
−2
( 5, 3)−
(0, 4)−
(5, 1)−−
y
x
(3, 5)
v
v
y
35
2
21
2
7 2
, ( (
−3 3 6 9 12 15 18
−3
3
6
9
12
15
18
x
−1 12345
−1
1
2
3
4
5
(3, 5)
v
v
y
10
3
2
3
2, ((

4. 4 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
18. (a) 442,3 8,12vπα πα

(b)
31
22
1,vαπα

(c)
00,0vπ

(d) 612,18uαπ α

19.
20. Twice as long as given vector u.

21.
22.
23. (a)
822
33 3
4, 9 , 6uππ
(b)
2, 5 4, 9 2, 14vuαπ αα παα
(c) 25 24,952,5 18,7uv〉π 〉 απ α
24. (a)
1622
33 3
3, 8 2,uπααπαα
(b)
8, 25 3, 8 11, 33vuαπ αααπ
(c) 2 5 2 3, 8 5 8, 25 34, 109uv〉παα〉 π
25. θ〈
333
222
23 3,vijijπαπαπα

26.
θ〈θ〈
2233,1vijijijπα〉〉π〉π

−2−4−6−8 246
4
6
8
10
12
x
v
4v
y
(−8, 12)
(−2, 3)
x
(−2, 3)
v
y
−1−2−33
−2
−3
2
3
1, −
3
2((
v
1
2
−
−1−2−31
−1
1
2
3
x
(−2, 3)
v
y
0v
−2−6261014
−6
−10
−14
−18
x
(−2, 3)
y
v
−6v
(12, −18)
x
−u
y
x
u
2u
y
x
−v
u
uv−
y
x
u
uv+ 2
2v
y
32
−1
1
−2
−3
3
2
x
u
u
y
x
2
1
−1
v
w
u
321
y

5. Section 11.1 Vectors in the Plane 5
© 2010 Brooks/Cole, Cengage Learning
27. θ〈θ〈222434,3vijijijπα〉〉π〉π

28. 53 52,131,2 7,11vuwπα π αα πα

29.
1
241
23
u
u
απα
απ

θ〈
1
23
5
3, 5u
u
Qπ
π
π

30.
θ〈
1
254
39
9, 6u
u
Qαπ
απα
πα
1
29
6
Terminal point
u
u
π
πα


31.
2
07 7vπ〉π

32.
θ〈
2
303vπα〉π

33.
22
43 5vπ〉π

34.
θ〈
2
2
12 5 13vπ〉απ

35.
θ〈
2
2
65 61vπ〉απ

36.
θ〈
2
2
10 3 109vπα 〉π

37.
3, 12vπ

22
312 153vπ〉π

v
u
v
ππ
3, 12 312
,
153 153 153
17 4 17
, unit vector
17 17
π
π

38.
5, 15
25 225 250 5 10
5, 15 10 3 10
, unit vector
10 10510
v
v
v
u
v
πα
π〉π π
α
ππ πα

39.
35
,
22
vπ

22
35 34
22 2
v

π〉π




35
,
22 35
,
34 34 34
2
334534
, unit vector
34 34
v
u
v



ππ π
π

40.
θ〈θ〈
22
6.2, 3.4
6.2 3.4 50 5 2
6.2, 3.4 31 2 17 2
, unit vector
50 5052
v
v
v
u
v
πα
πα 〉 π π
α
ππ πα
41.
1, 1 , 1, 2uvπα πα
(a) 11 2uπ〉π
(b) 14 5vπ〉π
(c) 0, 1
01 1
uv
uv
〉π
〉π 〉π

(d)
1
1, 1
2
1
u
u
u
u
πα
π

(e)
1
1, 2
5
1v
v
v
v
πα
π
(f )
0, 1
1
uv
uv
uv
uv〉
π
〉
〉
π
〉

4
2
4
−2
6
u + 2w
2w
x
u
y
−4−2
2
−6
−8
−10
−12
46810
5u
−3w
v
y
x

6. 6 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
42. 0, 1 , 3, 3uvππα
(a) 01 1uπ〉π
(b) 99 32vπ〉π
(c) 3, 2
94 13
uv
uv
〉π α
〉π 〉π

(d) 0, 1
1
u
u
u
u
π
π
(e)
1
3, 3
32
1v
v
v
v
πα
π
(f )
1
3, 2
13
1uv
uv
uv
uv〉
πα
〉
〉
π
〉

43.
1
1, , 2, 3
2
uvππ
(a)
15
1
42
uπ〉π
(b) 49 13vπ〉π
(c)
7
3,
2
49 85
9
42
uv
uv
〉π
〉π 〉 π

(d)
21
1,
25
1
u
u
u
u
π
π

(e)
1
2, 3
13
1v
v
v
v
π
π
(f )
27
3,
285
1
uv
uv
uv
uv
〉
π
〉
〉
π
〉

44. 2, 4 , 5, 5uvπα π
(a) 416 25uπ〉π
(b) 25 25 5 2vπ〉π
(c) 7, 1
49 1 5 2
uv
uv
〉π
〉π 〉π

(d)
1
2, 4
25
1u
u
u
u
πα
π
(e)
1
5, 5
52
1v
v
v
v
π
π
(f )
1
7,1
52
1uv
uv
uv
uv〉
π
〉
〉
π
〉

45.
2, 1
5 2.236
5, 4
41 6.403
7, 5
74 8.602
74 5 41u
u
v
v
uv
uv
uv u vπ
〈
π
〈
〉π



46.
3, 2
13 3.606
1, 2
5 2.236
2, 0
2
2135
u
u
v
v
uv
uv
u+v u v
πα
〈
πα
〈
〉πα
〉π



47.
1
0, 3 0, 1
3
660,10,6
0, 6
u
u
u
u
v
ππ

ππ


π

x
y
u
u + v
v
−1
1234567
1
2
3
4
5
6
7
−1−2−3 123
−1
−2
−3
1
2
3
x
y
u
u + v
v

7. Section 11.1 Vectors in the Plane 7
© 2010 Brooks/Cole, Cengage Learning
48.
1
1, 1
2
4221,1
22,22
u
u
u
u
v







49.
112
1, 2 ,
555
12
55, 5,25
55
5, 2 5
u
u
u
u
v







50.
1
3, 3
23
1
23,3
3
1, 3
u
u
u
u
v







51.

3cos0 sin0 3 3,0viji

52.

5 cos 120 sin 120
553 553
,
22 22
vij
ij




53.

2 cos 150 sin 150
33,1
vij
ij




54.

4 cos 3.5 sin 3.5
3.9925 0.2442
3.9925, 0.2442
vij
ij





55.


cos 0 sin 0
32 32
3 cos 45 3 sin 45
22
232 32 23232
,
22 22
uiji
vijij
uv i j






56.


4cos0 4sin0 4
2 cos 30 2 sin 30 3
535,3
uiji
vijij
uv i j



57.



2cos4 2sin4
cos 2 sin 2
2 cos 4 cos 2 2 sin 4 sin 2
2 cos 4 cos 2, 2 sin 4 sin 2
uij
vij
uv i j





58.




5 cos 0.5 5 sin 0.5
5cos0.5 5sin0.5
5cos0.5 5sin0.5
10 cos 0.5 10 cos 0.5, 0
uij
ij
vij
uv i





59. Answers will vary.
Sample answer: A scalar is a real
number such as 2. A vector is represented by a directed
line segment. A vector has both magnitude and direction.
For example
3,1has direction
6

and a magnitude
of 2.
60. See page 766:

61. (a) Vector. The velocity has both magnitude and
direction.
(b) Scalar. The price is a number.
62. (a) Scalar. The temperature is a number.
(b) Vector. The weight has magnitude and direction.
For Exercises 63–68,
au bw i 2j i j i 2 j.ababab
63. 2.vij So,
2, 2 1.ab ab Solving
simultaneously, you have
1, 1.ab
64. 3.vjSo,
0, 2 3.ab ab Solving
simultaneously, you have
1, 1.ab
65. 3.viSo,
3, 2 0.ab ab Solving
simultaneously, you have
1, 2.ab
66. 33.vij So,
3, 2 3.ab ab Solving
simultaneously, you have
2, 1.ab
67. .vij So,
1, 2 1.ab ab Solving
simultaneously, you have
21
33
,.ab
68.
7.vij So, 1, 2 7.ab ab Solving
simultaneously, you have
2, 3.ab
u
v
u + v
(u
1 + v
1
, u
2 + v
2
)
(v
1
,

v
2
)
(u
1
,

u
2
)
u
1
u
2
v
1
v
2
u
ku
(ku
1
, ku
2
)
(u
1
, u
2
)
u
1
ku
1
u
2
ku
2

8. 8 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
69. θ〈 θ〈 θ〈
2
,2,36fx x f x xfπππ
(a) 6.
mπLet
1, 6 , 37 ,wwππ then
1
1, 6 .
37
w
w

(b)
1
6
.mπαLet
6, 1 , 37,wwπα π then
1
6, 1 .
37
w
w

70.
θ〈 θ〈 θ〈
2
5, 2 , 1 2fx x f x xf πα 〉 πα πα
(a) 2.
mπαLet
1, 2 ,wπα 5,wπ then
1
1, 2 .
5
w
w

(b)
1
.
2
mπLet 2, 1 , 5,wwππ then
1
2, 1 .
5w
w


71. θ〈 θ〈
32
,33fx x f x xπππ at 1.xπ
(a) 3.mπLet
1, 3 ,wπ 10,wπ then
1
1, 3 .
10w
w

(b)
1
.
3
mπα Let
3, 1 , 10 ,wwπα π then
1
3, 1 .
10w
w


72. θ〈 θ〈
32
,312fx x f x xπππ at 2.xπα
(a) 12.mπLet
1, 12 ,wπ 145,wπ then
1
1, 12 .
145w
w

(b)
1
.
12
mπα Let
12, 1 , 145,wwπα π then
1
12, 1 .
145w
w


73. θ〈
2
25fxxπα

θ〈
2
3
425
x
fx
x
αα
ππ
α
at 3.xπ
(a)
3
.
4
mπαLet 4, 3 ,wπα 5,wπthen
1
4, 3 .
5w
w

(b)
4
.
3
mπ Let
3, 4 , 5,wwππ then
1
3, 4 .
5w
w


74. θ〈
θ〈
2
tan
sec 2 at
4
fx x
fx x x

π
πππ
(a) 2.mπLet
1, 2 ,wπ 5,wπ then
1
1, 2 .
5w
w

(b)
1
.
2
mπα
Let
2, 1 ,wπα 5,wπ then
1
2, 1 .
5w
w

x
y
−2 246810
2
4
6
8
10
(3, 9)
(a)
(b)
−1−3 123
−1
1
2
3
4 x
y
(1, 4)
(a)
(b)
x
y
12
1
2
(a)
(b)
(1, 1)
−2−4−624
−4
−6
−10
x
y
(a)
(b)
x
y
−1 12345
1
2
3
4
(3, 4)
(a)
(b)
4
π
2
π
4
π
−
2
π
−
x
y
−1.0
0.5
1.0
1.5
2.0
(a)
(b)

9. Section 11.1 Vectors in the Plane 9
© 2010 Brooks/Cole, Cengage Learning
75.
θ〈
22
22
2
22 22
,
22 22
uij
uv j
vuvu i jπ〉
〉π
π〉απα 〉 πα

76.
θ〈 θ〈 θ〈
23 2
333
323 332
uij
uv i j
vuvu i j
π〉
〉πα〉
π〉απαα 〉 α

323,332παα α
77. (a)–(c) Programs will vary.
(d) Magnitude 63.5
Direction 8.26
78. (a)
θ〈
93,1 4 6,5vπα ααπ
(b) 6 5vijπ〉
(c)
(d)
22
65 61vπ〉π
79.
1 1
2 2
3 3
123
1232, 33
3, 125
2.5, 110
1.33
132.5
F
F
F
RFFFF
F
F
RFFF



〉〉



π〉〉


80.
1 1
2 2
3 3
123
1232, 10
4, 140
3, 200
4.09
163.0
F
F
F
RFFFF
F
F
RFFF



〉〉



π〉〉


81.
θ〈θ〈θ〈 θ〈
θ〈 θ〈
θ〈 θ〈
12
22
12 500 cos 30 500 sin 30 200 cos 45 200 sin 45 250 3 100 2 250 100 2
250 3 100 2 250 100 2 584.6 lb
250 100 2
tan 10.7
250 3 100 2FF i j i j i j
FF



α

〉
82. (a)
θ〈180 cos 30 sin 30 275 430.88 90ijiij
Direction:
θ〈
90
arctan 0.206 11.8
430.88






Magnitude:
22
430.88 90 440.18〉 newtons
(b)
θ〈θ〈
22
275 180 cos 180 sin
180 sin
arctan
275 180 cos
M


π〉 〉

π

〉

(c)
(d)
(e) M decreases because the forces change from acting in the same direction to acting in the opposite direction as
increases
from 0to 180 .
0 30 60 90 120 150 180
M 455 440.2 396.9 328.7 241.9 149.3 95
0 11.8 23.1 33.2 40.1 37.1 0
−1 123456
1
2
3
4
5
6
x
(6, 5)
v
y
0 180
0
M
500
0 180
0
α
50

10. 10 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
83. θ〈θ 〈θ 〈
θ〈 θ〈
123
123
123
75 125 75 125
2222
75 cos 30 75 sin 30 100 cos 45 100 sin 45 125 cos 120 125 sin 120
3502 502 3
228.5 lb
71.3
RFFF
FF F i j i j i j
ij
RFFF

〉〉

π〉α〉〉〉
π〉〉


84.
θ〈 θ〈θ〈 θ〈 θ〈θ〈 θ〈 θ〈θ〈
θ〈 θ〈
123
22 400 cos 30 sin 30 280 cos 45 sin 45 350 cos 135 sin 135
200 3 140 2 175 2 200 140 2 175 2
200 3 35 2 200 315 2 385.2483 newtons
200 315 2
arctan 0.6908
200 3 35 2
R
FF F i j i j i j
ij
R



π〉α 〉α〉〉

πα〉α〉
α〉


α

39.6

85. (a) The forces act along the same direction. .

(b) The forces cancel out each other. 180 .

(c) No, the magnitude of the resultant can not be greater than the sum.
86.
1220, 0 , 10 cos sinFF
(a)
12
22 20 10 cos , 10 sin
400 400 cos 100 cos 100 sin
500 400 cos
FF

〉π 〉
π〉 〉 〉
π〉
(b)
(c) The range is
1210 30.FF
The maximum is 30, which occur at 0
πand 2 .π
The minimum is 10 at .
π
(d) The minimum of the resultant is 10.
87.
θ〈θ〈θ〈4, 1 , 6, 5 , 10, 3αα

88.
θ〈θ〈 θ〈
θ〈 θ〈 θ 〈1
2
1
3
71,52 6,3
2, 1
1, 2 2, 1 3, 3
1, 2 2 2, 1 5, 4
P
P
u
u
πα απ
π
π〉π
π〉 π
0
0
40
π2
x
8642
8
6
4
2
−4
(4, 1)−−
(1, 2)
(3, 1)
(8, 4)
y
x
8642
8
6
4
2
−4
−2
−4−2
(1, 2)
(3, 1)
(8, 4)
(6, 5)
y
x
81064
8
6
4
2
−4
−2
−2
(1, 2)
(3, 1)
(8, 4)
(10, 3)
y

11. Section 11.1 Vectors in the Plane 11
© 2010 Brooks/Cole, Cengage Learning
89. θ〈
θ〈
cos 30 sin 30
cos 130 sin 130
CB
CA
uuij
vvij
πππα
πππα




Vertical components: sin 30 sin 130 3000uv
Horizontal components: cos 30 cos 130 0uv
Solving this system, you obtain
1958.1u pounds
2638.2v pounds
90.
1
24
arctan 0.8761 or 50.2
20







2
24
arctan 1.9656 or 112.6
10




α


θ〈
θ〈
11
22cos sin
cos sin
uu i j
vv i j
π〉
π〉

Vertical components:
12sin sin 5000uv〉π
Horizontal components:
12cos cos 0uv〉π
Solving this system, you obtain
2169.4u and 3611.2.v

91. Horizontal component cos
1200 cos 6 1193.43 ft sec
vπ


Vertical component sin
1200 sin 6 125.43 ft sec
vπ


92. To lift the weight vertically, the sum of the vertical
components of u and v must be 100 and the sum of the
horizontal components must be 0.
θ〈
θ〈
cos 60 sin 60
cos 110 sin 110
uu i j
vv i j


So,
sin 60 sin 110 100,
3
sin 110 100.
2
uv
uv




or
And
cos 60 cos 110 0
1
cos 110 0.
2
uv
uv




or
Multiplying the last equation by
θ〈3and adding to the
first equation gives
θ〈sin 110 3 cos 110 100 65.27 lb.uv
Then,
1
65.27 cos 110 0
2
u




gives
44.65 lb.u
(a) The tension in each rope: 44.65 lb,
65.27 lb
u
v
π
π
(b) Vertical components:
sin 60 38.67 lb,
sin 110 61.33 lb
u
v




93.
θ〈
θ〈900 cos 148 sin 148
100 cos 45 sin 45
uij
vij



θ〈θ〈
θ〈θ〈
22
900 cos 148 100 cos 45 900 sin 148 100 sin 45
692.53 547.64
547.64
arctan 38.34 ; 38.34 North of West
692.53
692.53 547.64 882.9 km h
uv i j
ij
uv


α 〉



α


AB
C
30°
30°
50°
130°
uv
y
x
AB
C
v u
y
x
θ
1
θ
2
100 lb
20°
30°
u
v

12. 12 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
94.


400 plane
50 cos 135 sin 135 25 2 25 2 wind
400 25 2 25 2 364.64 35.36
35.36
tan 5.54
364.64
ui
vijij
uv i j i j






Direction North of East:
N 84.46 E
Speed:
336.35 mi h
95. True
96. True
97. True
98. False
0ab
99. False
2ab aij
100. True
101.
22
22
cos sin 1,
sin cos 1
u
v



102. Let the triangle have vertices at
0, 0 , , 0 ,aand
,.bcLet u be the vector joining 0, 0and ,,bcas
indicated in the figure. Then v, the vector joining the
midpoints, is


222
22
11
22
ab a c
bc
bc
vij
i+ j
ij u.






103. Let
uand vbe the vectors that determine the
parallelogram, as indicated in the figure. The two
diagonals are
uv and .vuSo,
,4 .xruvsvu But,

.
x yxyxy
urs
uv vu u v



So, 1
xy and 0. xy Solving you have
1
2
.xy


104.

cos sin cos sin
cos cos sin sin
2cos cos sin cos
22 22
vv uu
uv uv
uv uv uv uv
wuvvu
uv i v j vu i u j
uv i j
uv i j














sin cos
22
tan tan
2
cos cos
22
uv uv
uv
w
uv uv











So,

2
wuv and w bisects the angle between u and v.
105. The set is a circle of radius 5, centered at the origin.

22 22
,525xy xy xyu
x
ab+c
a
22
2
,( (
((
, 0
( , )b c
( , 0)a(0, 0)
u
v
y
u
s
r
v

13. Section 11.2 Space Coordinates and Vectors in Space 13
© 2010 Brooks/Cole, Cengage Learning
106. Let
0cosxvt π and
2
01
sin .
2
yvt gtπα

θ〈
2
0
000
22
2
0
2
2
2
0
222 2
00 2
22
00
222 2 4
000 2
22 22
00
22
0
2
0
1
sin
cos cos 2 cos
tan sec
2
tan 1 tan
2
tan tan
22 2 2
tan 2 tan
22 2
22xxx
tyv g
vvv
g
xx
v
gx
x
v
vgxgx v
x
gv v g
vgxgx v v
gv v gxgx
vgx
gv









πα
πα〉
πα α 〉 α

πα α α 〉

πα
2
22
0
2
0
tan
2
gx v
vgx


αα


If
22
0
2
0
,
22
vgx
y
g v
then can be chosen to hit the point θ〈,.xyTo hit θ〈0, :yLet 90 .Then

2
22
002
0
0
1
1,
222
vvg
yvt gt t
ggv

πα π α α

and you need
2
0
.
2
v
y g

The set H is given by 0 ,x0yand
22
0
2
0
22
vgx
y
g v


Note: The parabola
22
0
2
0
22
vgx
y
g v
πα is called the “parabola of safety.”
Section 11.2 Space Coordinates and Vectors in Space
1. θ〈
θ〈2, 3, 4
1, 2, 2A
B
αα


2. θ〈
θ〈2, 3, 1
3, 1, 4A
Bαα
α

3.
4.
5.
6.
7. 3,xπα 4,yπ 5:zπθ〈3, 4, 5α

8. 7,xπ 2,yπα 1:zπα

θ〈7, 2, 1αα

9. θ〈0, 12: 12, 0, 0yz xππ π

10. θ〈0, 3, 2: 0, 3, 2xyzπππ
x
y
4
3
2
4
1
2
3
3
4
5
6
z
(2, 1, 3)
(−1, 2, 1)
x y
(3, 2, 5)−
2
3
, 4, 2−( (
8
6
4
2
z
6
6
x
y
3
2
−3
1
4
1
2
3
3
2
1 −2
−3
z
(5, −2, 2)
(5, −2, −2)
x y
(4, 0, 5)
(0, 4, 5)−
8
z
6
2
−2
−4
−6
6
6
(x, y)
x
y
α

14. 14 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
11. The z-coordinate is 0.

12. The x-coordinate is 0.

13. The point is 6 units above the xy-plane.

14. The point is 2 units in front of the xz-plane.

15. The point is on the plane parallel to the -planeyz that
passes through 3.
x

16. The point is on the plane parallel to the xy-plane that
passes through
52.z

17. The point is to the left of the xz-plane.

18. The point is in front of the yz-plane.

19. The point is on or between the planes 3yand
3.
y

20. The point is in front of the plane 4.x

21. The point ,,
xyzis 3 units below the xy-plane, and
below either quadrant I or III.

22. The point ,,
xyzis 4 units above the xy-plane, and
above either quadrant II or IV.

23. The point could be above the xy-plane and so above
quadrants II or IV, or below the
xy-plane, and so below
quadrants I or III.

24. The point could be above the xy-plane, and so above
quadrants I and III, or below the
xy-plane, and so below
quadrants II or IV.

25.
222
40 20 70
16 4 49 69
d


26.
2 22
22 53 22
16 64 16 96 4 6d



27.
222
61 2 2 2 4
25 0 36 61
d



28.
222
42 52 63
4499 62d


29. 0, 0, 4 , 2, 6, 7 , 6, 4, 8ABC



222
2
22
22
2
222
263 497
6 4 12 196 14
4 2 15 245 7 5
245 49 196AB
AC
BC
BC AB AC




Right triangle

30.
3, 4, 1 , 0, 6, 2 , 3, 5, 6
941 14
0 1 25 26
9 1 16 26AB C
AB
AC
BC




Because
,ACBC the triangle is isosceles.

31. 1, 0, 2 , 1, 5, 2 , 3, 1, 1ABC

02516 41AB
419 14AC
4361 41BC
Because ,ABBC the triangle is isosceles.

32.
4, 1, 1 , 2, 0, 4 , 3, 5, 1
419 14
1360 37
1259 35
ABC
AB
AC
BC





Neither

33. The z-coordinate is changed by 5 units:

0, 0, 9 , 2, 6, 12 , 6, 4, 3

34. The y-coordinate is changed by 3 units:

3, 7, 1 , 0, 9, 2 , 3, 8, 6

35.

52 9373 3
,, ,3,5
222 2





36.

4808620
, , 6, 4, 7
22 2






37. Center: 0, 2, 5
Radius: 2


222
0254xyz

38. Center: 4, 1, 1
Radius: 5


222
41125xyz

39. Center:


2, 0, 0 0, 6, 0
1, 3, 0
2



Radius:
10


222
13010xyz

15. Section 11.2 Space Coordinates and Vectors in Space 15
© 2010 Brooks/Cole, Cengage Learning
40. Center: 3, 2, 4
3r

tangent to -planeyz


222
3249xyz

41.


222
22 2
222
26810
2 1 6 9 8 16 11916
13425
xyz xyz
xx yy zz
xy z




Center:
1, 3, 4
Radius: 5

42.


222
222
2
22
9 2 10 19 0
81 81
921102519125
44
9 109
15
24
xyz xy z
xx yy z z
xyz










Center:
9
,1, 5
2





Radius:
109
2


43.

222
222
222
2
22
21
39
21 11
39 99
1
3
99961810
20
21 1
101
xyzxy
xyz xy
xx yy z
xyz





Center:

1
3
,1,0
Radius: 1

44.



222
222
222
2311
444
1
2
4442448230
69 21 9 1
3116xyz xyz
xx yy zz
xyz


Center:

1
2
3, , 1
Radius: 4

45.
222
36xyz
Solid sphere of radius 6 centered at origin.

46.
222
4xyz
Set of all points in space outside the ball of radius 2 centered at the origin.

16. 16 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
47.
θ〈θ〈θ 〈
θ〈θ〈θ〈
222
222
222
46813
44 69 816491613
23416xyz xyz
xx yy zz
xyz



Interior of sphere of radius 4 centered at
θ〈2, 3, 4 .α

48.
θ〈θ〈θ 〈
θ〈θ〈θ〈
222
222
222
46813
44 69 816 134916
23416xyz xyz
xx yy zz
xyz



Set of all points in space outside the ball of radius 4 centered at
θ〈2, 3, 4 .αα

49. (a)
2 4, 4 2, 3 1 2, 2, 2vπα α απα
(b) 2 2 2
vijkπα 〉 〉
(c)

50. (a)
4 0, 0 5, 3 1 4, 5, 2vπα α απα
(b) 4 5 2
vijhπα〉
(c)

51. (a)
03,33,30 3,0,3vπαααπα
(b) 3 3
vikπα 〉
(c)

52. (a)
2 2, 3 3, 4 0 0, 0, 4vπα α απ
(b) 4
vkπ
(c)

53.
43,12,60 1,1,6
1, 1, 6 1 1 36 38
αα απα
απ〉〉π

Unit vector:
1, 1, 6 116
,,
38 38 38 38
α α
π

54. θ〈
14,7 5,32 5,12,5
5, 12, 5 25 144 25 194
αα αα αα πα α
ααπ 〉〉π
Unit vector:
5, 12, 5 512 5
,,
194 194 194 194
αα αα
π

55. θ〈
54,33,01 1,0,1
1, 0, 1 1 1 2
ααα α α πα α
ααπ 〉π
Unit vector:
1, 0, 1 11
,0,
222
αα αα
π

56. θ〈
21,4 2,2 4 1,6,6αααααπ α
1, 6, 6 1 36 36 73απ 〉 〉 π
Unit vector:
1, 6, 6 16 6
,,
73 73 73 73
α α
π


57. (b) θ〈
3 1 , 3 2, 4 3 4, 1, 1vπαα α απ
(c) 4
vijkπ〉〉
(a), (d)
x
y4
3
2
1
1
−3
−2
2
3
2
1
3
4
5
z
〈−2, 2, 2〉
x y
〈 −〉4, 5, 2
8
z
6
4
2
6
4
2
4
2
6
x
y4
3
2
1
1
−3
−2
2
3
2
1
3
4
5
z
〈−3, 0, 3〉
x y
〈〉0, 0, 4
4
3
2
1
1
3
2
1
2
3
z
x
y4
2
−2
2
4
2
3
4
5
z
(−1, 2, 3)
(3, 3, 4)
(0, 0, 0)
(4, 1, 1)
v

17. Section 11.2 Space Coordinates and Vectors in Space 17
© 2010 Brooks/Cole, Cengage Learning
58. (b) θ〈 θ〈4 2, 3 1 , 7 2 6, 4, 9vπα α αα αα πα
(c) 6 4 9
vijkπ〉〉
(a), (d)

59. θ〈θ〈θ〈123, , 0, 6, 2 3, 5, 6qq q απα

θ〈3, 1, 8Qπ

60. θ〈
θ〈 θ〈
θ〈
123
5 21
232
4
3
, , 0, 2, 1, ,
1, , 3
qq q
Q
απα
πα

61. (a)
2 2,4,4vπ

(b) 1, 2, 2vαπααα

(c)
33
22
,3,3vπ

(d)
0 0,0,0vπ


62. (a)
2, 2, 1vαπα α

(b) 24,4,2vπα

(c)
11
22
1, 1,vπα

(d)
55
22
5, 5,vπα


63.
1, 2, 3 2, 2, 1 1, 0, 4zuvπαπ α απα

64.
2
1, 2, 3 2, 2, 1 8, 0, 8 7, 0, 4zuv wπα〉
παα〉απα


65.
24
2, 4, 6 8, 8, 4 4, 0, 4 6, 12, 6zuvwπ〉α
π〉αααπ


66.
1
2
53
5, 10, 15 6, 6, 3 2, 0, 2
3, 4, 20zuvwπαα
παααα
πα


67.
12323 2,, 31,2,3 4,0,4zz zzuαπ α π α

11
22
33
7
2
5
2
75
22
234
260 3
294
,3,
zz
zz
zz
z



π
x y
12
( 4, 3, 7)−
( 6, 4, 9)−
(2, 1, 2)− −
z
9
6
3
9
9
x
y
2
1
1
−2
2
3
4
2
3
4
5
z
〈2, 4, 4〉
x
3
2
1
−3
−2
−2
−3
2
3
y
2
−2
−3
3
z
〈−1, −2, −2〉
x
1
−3
−2
−2
−3
2
3
y
2
−2
−3
3
z
3
2〈
, 3, 3〉
x
1
2
3
−3
−2
−2
−3
2
1
3
y
2
1
−2
−1
−3
3
z
〈0, 0, 0〉
x y
4
− − 〉2, 2, 1
z
3
2
1
3
3
〉
x y
8
−4, 4, 2
z
6
4
2
6
6
〉 〉
x y
2
2
1
〉 〉1, 1,−
z
1
x y
8
2
5
〈 〈5, 5,−
z
6
4
2
6

18. 18 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
68.
123
123
11
22
332 3 21,2,3 2,2, 1 4,0, 4 3 , , 0,0,0
0, 6, 9 3 , 3 , 3 0, 0, 0
03 0 0
63 0 2
93 0 3
0, 2, 3
zz z
zz z
zz
zz
zz
uvw z
z







69. (a) and (b) are parallel because
6, 4, 10 2 3, 2, 5 and
1042
33 3
2, , 3, 2, 5 .
70. (b) and (d) are parallel because

33412
32 234
2ijk ijk and

393 3 12
482234
.ij k i j k
71. 342zijk
(a) is parallel because 6 8 4 2 .ijk z
72.
7, 8, 3z
(b) is parallel because

14, 16, 6 .zz
73.
0, 2, 5 , 3, 4, 4 , 2, 2, 1PQR

3
2
3, 6, 9
2, 4, 6
3, 6, 9 2, 4, 6
PQ
PR






So, PQ

and PR

are parallel, the points are collinear.
74.

1
2
4, 2, 7 , 2, 0, 3 , 7, 3, 9
6, 2, 4
3, 1, 2
3, 1, 2 6, 2, 4
PQR
PQ
PR







So, PQ

and PR

are parallel. The points are collinear.
75.
1, 2, 4 ,P 2, 5, 0 , 0, 1, 5QR

1, 3, 4
1, 1, 1
PQ
PR





Because PQ

and PR

are not parallel, the points are not
collinear.
76.

0, 0, 0 , 1, 3, 2 , 2, 6, 4
1, 3, 2
2, 6, 4
PQ R
PQ
PR






Because PQ

and PR

are not parallel, the points are not
collinear.
77.


2, 9, 1 , 3, 11, 4 , 0, 10, 2 , 1, 12, 5
1, 2, 3
1, 2, 3
2, 1, 1
2, 1, 1
AB C D
AB
CD
AC
BD









Because
ABCD

and ,ACBD

the given points
form the vertices of a parallelogram.
78.

1, 1, 3 9, 1, 2 , 11, 2, 9 , 3, 4, 4
8, 2, 5
8, 2, 5
2, 3, 7
2, 3, 7AB C D
AB
DC
AD
BC








Because
ABDC
and ,
ADBC

the given points
form the vertices of a parallelogram.
79. 0, 0, 0
0v
v

80.
1, 0, 3
109 10
v
v



81. 35 0,3,5
0 9 25 34
vjk
v



82. 25 2,5,1
4251 30
vijk
v



83. 23 1,2,3
149 14
vi j k
v



84. 437 4,3,7vijk
16 9 49 74v

19. Section 11.2 Space Coordinates and Vectors in Space 19
© 2010 Brooks/Cole, Cengage Learning
85. 2, 1, 2
414 3
v
v
〈〉


(a)
1
2, 1, 2
3
v
v
〈〉
(b)
1
2, 1, 2
3
v
v
〉〈〉 〉
86. 6, 0, 8
36064 10
v
v
〈


(a)
1
6, 0, 8
10
v
v
〈
(b)
1
6, 0, 8
10
v
v
〉〈〉
87. 3, 2, 5
9 4 25 38
v
v
〈〉


(a)
1
3, 2, 5
38
v
v
〈〉
(b)
1
3, 2, 5
38
v
v
〉〈〉 〉
88. 8, 0, 0
8v
v〈
〈
(a)
1
1, 0, 0
8
v
v
〈
(b)
1
1, 0, 0
8
v
v
〉〈〉〉
89. (a)–(d) Programs will vary.
(e) 4, 7.5, 2
8.732
5.099
9.019
uv
uv
u
v





90. The terminal points of the vectors ,ttuu vand
stuvare collinear.

91.
222
2
2
7
3
22 4 4 7
97
949
vijkcc ccc
c
c
c
〈
〈


92.

222
2
2
8
7
23 4 9 4
14 4
14 16
uijkcc ccc
c
c
c
〈
〈


93.
0, 3, 3
10 10
32
1 1 10 10
10 0, , 0, ,
22 22
u
v
u
〈〈
〈〈

94.
1, 1, 1
33
3
111 333
3,, ,,
333 333
u
v
u
〈〈
〈〈

95.
2, 2, 133 3221 1
, , 1, 1,
223 2333 2
u
v
u
〉 〉
〈〈 〈 〈〉

96.
4, 6, 2 14 21 7
77 ,,
14214 14 14u
v
u 〉 〉
〈〈 〈

97.

2 cos 30 sin 30
30,3,1
vjk
jk





98.



52
5 cos 45 sin 45 or
2
52
5 cos 135 sin 135
2
vikik
vikik


99.

2
3
3, 6, 3
2, 4, 2
4, 3, 0 2, 4, 2 2, 1, 2
v
v
〈〉 〉
〈〉 〉


u
u + tv
su
su + tv
v
tv
x
−2
−2
−1
2
1
y
2
1
−2
−1
z
〈0, 3, 1〉
〈0, 3, −1〉
x y
8
2
5 2
(+ )ik 2
5 2
(+)−ik
z
6
4
2
6
6

20. 20 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
100.

102
33
10 13
33
5, 6, 3
,4, 2
1, 2, 5 , 4, 2 , 6, 3
v
v




101. (a)
(b)
0ab aab bwuvi jk

0, 0, 0aabb
So,
a and b are both zero.
(c)
2
1, 2, 1aabb
aab bijkijk
wuv




(d)
23
1, 2, 3aabb
aab bijkijk


Not possible
102. A sphere of radius 4 centered at
111,, .
xyz


211
222
111 ,
4v xxyyzz
xx yy zz



222
111
16xx yy zz
103.
0
xis directed distance to yz-plane.

0yis directed distance to xz-plane.

0zis directed distance to xy-plane.
104.

222
21 2 1 21
dxxyyzz
105.

222
2
000
xxyyzzr
106. Two nonzero vectors uand vare parallel if
cuvfor
some scalar
c.
107.

ABBC AC


So,
AB BC CA AC CA 0


108.


222
0
222
1112
1114xyz
xyzrr

This is a sphere of radius 2 and center

1, 1, 1 .
109. (a) The height of the right triangle is
22
18 .hL
The vector
PQ

is given by

0, 18, .PQ h


The tension vector T in each wire is
0, 18,chT where
24
8.
3
ch
So,
8
0, 18,
h
hT and


22
22 2
22
228
18
8
18 18
18
8
, 18.
18
TTh
h
L
L
L
L
L





(b)
(c)
18
xis a vertical asymptote and 8 yis a
horizontal asymptote.
(d)

22
18
22 2
8
lim
18
88
lim lim 8
18 118
L
LL
L
L
L
L L


! !
!



(e) From the table, 10
Timplies 30Linches.
110. As in Exercise 109(c),
xawill be a vertical
asymptote. So,
0
lim .
ra
T


!
L 20 25 30 35 40 45 50
T 18.4 11.5 10 9.3 9.0 8.7 8.6
1
1
1
v
u
yx
z
A
B
C
Q
P
L
(0, 0, 0)
(0, 18, 0)
18
(0, 0, )h
0100
0
30L = 18
T = 8

21. Section 11.2 Space Coordinates and Vectors in Space 21
© 2010 Brooks/Cole, Cengage Learning
111. Let be the angle between vand the coordinate axes.



cos cos cos
3cos 1
13
cos
33
33
1, 1, 1
33
vijk
v
vijk







112.


2
2
550 75 50 100
302,500 18,125
16.689655
4.085
4.085 75 50 100
306 204 409c
c
c
cij k
Fijk
ijk







113.
11
22
3
1230, 70, 115 , 0, 70, 115
60, 0, 115 , 60, 0, 115
45, 65, 115 , 45, 65, 115
0, 0, 500AB C
AC C
AD C
3
F
F
F
FFF F








So:

23
13
12360 45 0
70 65 0
115 500CC
CC
CCC


Solving this system yields
12
104 28
69 23
,,CC and
3
112
69
.C So:

1
2
3202.919N
157.909 N
226.521N
F
F
F




114. Let A lie on the
y-axis and the wall on the x-axis. Then 0,10,0, 8,0,6,AB 10, 0, 6C and
8, 10, 6 , 10, 10, 6 .AB AC


10 2, 2 59AB AC
Thus,
12420 , 650
ABAC
ABAC
FF




12 237.6, 297.0, 178.2 423.1, 423.1, 253.9 185.5, 720.1, 432.1FFF
860.0 lbF
115.
2dAP dBP






22 2 2
2 2
222 222
222
222
22
2
1121 2
2224 245
03 3 3 8 18 2 18
16 1 8 16 2 1
69 69
99 39 39
44 4 1
3
93 3xy z x y z
xyz yz xyz xy
xyzxyz
xx yy zz
xyz











Sphere; center:
41
,3, ,
33




radius:
211
3

y
x
0.2
0.2
0.4
0.6
0.4
0.4
0.6
333
333
,,( (
z

22. 22 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
Section 11.3 The Dot Product of Two Vectors
1. 3, 4 , 1, 5uv
(a)

31 45 17uv"
(b)

33 44 25uu"
(c)
2
22
34 25u
(d)

17 1, 5 17, 85uvv"
(e)

22 21734uv uv""
2. 4, 10 , 2, 3uv
(a)
42 103 22uv"
(b)

4 4 10 10 116uu"
(c)
22 2
4 10 116u
(d)

22 2, 3 44, 66uvv"
(e)

22 22244uv uv""
3. 6, 4 , 3, 2u= v
(a)

63 42 26uv"
(b)

66 4 4 52uu"
(c)
2
22
6452u
(d)

26 3, 2 78, 52uvv"
(e)

22 22652uv uv""
4. 4, 8 , 7, 5uv
(a)
47 85 12uv"
(b)

44 88 80uu"
(c)
2
22
4880u
(d)

12 7, 5 84, 60uvv"
(e)
22 21224uv uv""
5.
2, 3, 4 , 0, 6, 5uv
(a)
20 3 6 4 5 2uv"
(b)
22 3 3 44 29uu"
(c)

2
22 2
23429u
(d)

20,6,5 0,12,10uvv"
(e)
22 224uv uv""
6.
,uivi
(a) 1uv"
(b) 1uu"
(c)
2
1u
(d)
uvv i"
(e)
22 2uv uv""
7.
2 ,ui
jkv i k
(a)
21 1 0 1 1 1uv"
(b)
22 1 1 1 1 6uu"
(c)

2
22 2
2116u
(d)
uvv v i k"
(e)
22 2uv uv""
8.
22, 32uijkvijk
(a)
21 1 3 2 2 5uv"
(b)
22 11 2 2 9uu"
(c)

2
222
21 2 9u
(d)
5 3 2 5 15 10uvv i j k i j k"
(e)
22 2510uv uv""
9.
cos
uv
uv
"



85cos 20
3uv

"
10.

cos
5
40 25 cos 500 3
6
uv
uv
uv


"

"

11.
1, 1 , 2, 2uv

0
cos 0
28
2uv
uv



"



12.
3, 1 , 2, 1uv

51
cos
10 5 2
4uv
uv



"

23. Section 11.3 The Dot Product of Two Vectors 23
© 2010 Brooks/Cole, Cengage Learning
13. 3, 24uijvi j

21
cos
10 20 5 2
1
arccos 98.1
52uv
uv


"






14.


31
cos sin
6622
33 22
cos sin
4422
cos
3212 2
13
22 22 4
2
arccos 1 3 105
4
uijij
vijij
uv
uv










"








15.
1, 1, 1 , 2, 1, 1uv

22
cos
336
2
arccos 61.9
3
uv
uv


"



16.
32 , 23uijkvij

32 2 3 0
cos 0
2
uv
uv uv
=



"

17.
34, 23ui
jv=jk

8813
cos
65513
813
arccos 116.3
65
uv
uv


"






18.
23 , 2u i jkv i jk

99321
cos
1414 6 2 21
321
arccos 10.9
14
uv
uv


"






19.
4, 0 , 1, 1uv
not parallel
4 0 not orthogonal
Neither
cuv
uv
#
"#

20.
31
26
2, 18 , ,
not parallel
0 orthogonal
uv
uv
uv
c

#
"

21.
12
23
4, 3 , ,
not parallel
0 orthogonal
uv
uv
uv
c

#
"

22.

1
3
1
6
2, 2 4
paralleluijvij
uv

23. 6, 2
not parallel
8 0 not orthogonal
Neither
uj kvi jk
uv
uv
c

#
"#

24.
23 , 2ui
jkv i jk
not parallelcuv#
0 orthogonaluv"
25. 2, 3, 1 , 1, 1, 1
not parallel
0 orthogonal
uv
uv
uv
c

#
"

26. cos , sin , 1 ,
sin , cos , 0
not parallel
0 orthogonal
u
v
uv
uv
c


#
"

27. The vector
1, 2, 0 joining 1, 2, 0 and 0, 0, 0 is
perpendicular to the vector 2, 1, 0 joining
2, 1, 0 and 0, 0, 0 :1, 2, 0 2, 1, 0 0"
The triangle has a right angle, so it is a right triangle.
28. Consider the vector 3, 0, 0 joining 0, 0, 0 and
3, 0, 0 , and the vector 1, 2, 3 joining 0, 0, 0 and
1, 2, 3 :3, 0, 0 1, 2, 3 3 0"
The triangle has an obtuse angle, so it is an obtuse triangle.
29.


31
22
2, 0, 1 , 0, 1, 2 , , , 0ABC

53
22
11
22
2, 1, 1
,,1
,,2
AB
AC
BC







53
22
11
22
2, 1, 1
,,1
,,2
BA
CA
CB








3
2
1
2
53
44
510
120
20
AB AC
BA BC
CA CB
"
"
"




The triangle has three acute angles, so it is an acute
triangle.

24. 24 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
30. 2, 7, 3 , 1, 5, 8 , 4, 6, 1ABC

3, 12, 5
2, 13, 4
5, 1, 9
AB
AC
BC







3, 12, 5
2, 13, 4
5, 1, 9
BA
CA
CB








6 156 20 0
15 12 45 0
10 13 36 0
AB AC
BA BC
CA CB
"
"
"




The triangle has three acute angles, so it is an acute
triangle.
31.
22, 3ui j ku

222
1
3
2
3
2
3
144
999
cos
cos
cos
cos cos cos 1
$
%
$%





32.
222
5, 3, 1 35
5
cos
35
3
cos
35
1
cos
35
25 9 1
cos cos cos 1
35 35 35
uu

$
%
$%







33.
222
0, 6, 4 , 52 2 13
cos 0
3
cos
13
2
cos
13
94
cos cos cos 0 1
13 13
uu

$
%
$%






34.
222
222
222
222
222
222
222 222 222
,, ,
cos
cos
cos
cos cos cos 1abc a b c
a
abc
b
abc
c
abc
abc
abc abc abcuu

$
%
$%










35.
3, 2, 2 17
3
cos 0.7560 or 43.3
17
2
cos 1.0644 or 61.0
17
2
cos 2.0772 or 119.0
17
y
uu

$$
%






36.
4, 3, 5 50 5 2uu

4
cos 2.1721 or 124.4
52
3
cos 1.1326 or 64.9
52
51
cos or 45
452 2

$$

%%

25. Section 11.3 The Dot Product of Two Vectors 25
© 2010 Brooks/Cole, Cengage Learning
37. 1, 5, 2 30uu

1
cos 1.7544 or 100.5
30
5
cos 0.4205 or 24.1
30
2
cos 1.1970 or 68.6
30

$$
%%





38.
2, 6, 1 41uu

2
cos 1.8885 or 108.2
41
6
cos 0.3567 or 20.4
41
1
cos 1.4140 or 81.0
41

$$
%%





39.
11
1
22
2
50
: 4.3193
80
: 5.4183
C
C
F
F
F
F


12
4.3193 10, 5, 3 5.4183 12, 7, 5
108.2126, 59.5246, 14.1336
124.310 lb
FFF
F






108.2126
cos 29.48
59.5246
cos 61.39
14.1336
cos 96.53
F
F
F

$$
%%




40.
11
1
22
2
300
: 13.0931
100
: 6.3246
C
CF
F
F
F



12
13.0931 20, 10, 5 6.3246 5, 15, 0
230.239, 36.062, 65.4655
FFF




242.067 lbF

230.239
cos 162.02
36.062
cos 98.57
65.4655
cos 74.31
F
F
F

$$
%%






41.
222
222
0, 10, 10
0
cos 0 90
01010
10
cos cos
01010
1
45
2
OA


$%
$%






42.
11 0, 10, 10 .CF
111200 10 2 10 2CCF and
10, 100 2, 100 2F

22
33
123 4, 6, 10
4, 6, 10
0, 0,C
C
w
F
F
F
FF F F 0




23 23
23 23
2344 0
25 2
100 2 6 6 0 N
3
800 2
10 10 100 2
3CC CC
CC CC
WCC



43.
6, 7 , 1, 4uv
(a)
1w
2
22
proj
61 7 4
1, 4
14
34
1, 4 2, 8
17
v
uv
uv
v"








(b)
21 6, 7 2, 8 4, 1wuw
44. 9, 7 , 1, 3uv
(a)
1w
2
2
proj
91 73
1, 3
13
30
1, 3 3, 9
10
v
uv
uv
v"








(b)
21 9, 7 3, 9 6, 2wuw

26. 26 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
45. 23 2,3, 5 5,1uij vij
(a)
1w
2
2
proj
25 31
5, 1
51
13 5 1
5, 1 ,
26 2 2
v
uv
uv
v
"








(b)
21
51 15
2, 3 , ,
22 22
wuw
46. 23 2,3, 32 3,2uij vij
(a)
1w
2
22
proj
23 3 2
3, 2
32
03,2 0,0
v
uv
uv
v"








(b)
21 2, 3wuw
47. 0, 3, 3 , 1, 1, 1uv
(a)
1w
2
proj
01 31 31
1, 1, 1
111
6
1, 1, 1 2, 2, 2
3
v
uv
uv
v"








(b)
21 0,3,3 2,2,2 2,1,1wuw
48. 8, 2, 0 , 2, 1, 1uv
(a)
1w
2
2
proj
82 21 0 1
2, 1, 1
211
18
2, 1, 1 6, 3, 3
6
v
uv
uv
v"








(b)
21 8, 2, 0 6, 3, 3 2, 1, 3wuw
49. 222,1,2
34 0,3,4uijk
vjk

(a)
1w

2
22
proj
20 13 24
0, 3, 4
34
11 33 44
0, 3, 4 0, ,
25 25 25
v
uv
uv
v
"








(b)
21
33 44 8 6
2,1, 2 0, , 2, ,
25 25 25 25
wuw
50. 41,0,4
32 3,0,2ui k
vik

(a)
1w

2
22
,
proj
13 42
3, 0, 2
32
11 33 22
3, 0, 2 , 0
13 13 13
v
uv
uv
v
"








(b)
2w
1
33 22
1, 0, 4 , 0,
13 13
20 30
,0,
13 13
uw


51.
123 123 11 22 33,, ,,uu u vvv uv uv uvuv" "
52. The vectors u and v are orthogonal if 0.uv" The
angle
between u and v is given by
cos .
uv
uv
"

53. (a) and (b) are defined. (c) and (d) are not defined
because it is not possible to find the dot product of a
scalar and a vector or to add a scalar to a vector.
54. See page 786. Direction cosines of
123,,vv vv are
123
cos , cos , cos .
vvv
vvv
$% ,,$and %
are the direction angles. See Figure 11.26.
55. See figure 11.29, page 787.
56. (a)
2
c
uv
vu u v u
v
"


and vare parallel.
(b)
2
0
uv
v0 uv u
v"
"


and v are
orthogonal.

57. Yes,
22
22
11
uv vu
vu
vu
vu
uv vu
vu
vu
uv""

""




58. (a) Orthogonal,
2



(b) Acute,
0
2



(c) Obtuse,
2

27. Section 11.3 The Dot Product of Two Vectors 27
© 2010 Brooks/Cole, Cengage Learning
59. 3240, 1450, 2235
1.35, 2.65, 1.85u
v


3240 1.35 1450 2.65 2235 1.85
$12,351.25uv"

This represents the total amount that the restaurant
earned on its three products.

60.
3240, 1450, 2235
1.35, 2.65, 1.85u
v

Increase prices by 4%: 1.04
v
New total amount:
1.04 1.04 12,351.25
$12,845.30uv"


61. (a)–(c) Programs will vary.

62.
9.165
5.745
90u
v





63. Programs will vary.

64.
21 63 42
,,
26 26 13


65. Because u and v are parallel, proj
vuu

66. Because u and v are perpendicular, proj
vu0

67. Answers will vary. Sample answer:

13
.
42
uij Want 0.uv"

12 2vij and 12 2vij are orthogonal to u.

68. Answers will vary. Sample answer:
94.
uij Want 0.uv"

49vi
j and 4 9vij
are orthogonal to
u.

69. Answers will vary. Sample answer:

3, 1, 2 .u Want 0.uv"
0, 2, 1v and 0, 2, 1v are orthogonal to u.

70. Answers will vary. Sample answer:

4, 3, 6 .u Want 0uv"
0, 6, 3v and 0, 6, 3v
are orthogonal to
u.

71. (a) Gravitational Force 48,000Fj




1
2
1
cos 10 sin 10
48,000 sin 10
8335.1 cos 10 sin 10
8335.1 lbvij
Fv
wvFvv
v
v
ij
w
"
"



(b)

21
48,000 8335.1 cos 10 sin 10
8208.5 46,552.6
wFw
j ij
ij




247,270.8 lbw

72.
2
10, 5, 20 , 0, 0, 1
20
proj 0, 0, 1 0, 0, 20
1
proj 20
OA
OA
OA
v
v
v







73.
13
85
22
10
425 ft-lb
W
Fij
vi
Fv





"


74. 25 cos 20 sin 20
50
1250 cos 20 1174.6 ft-lbFij
vi
FvW

"

75. 1600 cos 25 sin 25
2000Fij
vi


1600 2000 cos 25
2,900,184.9 Newton meters (Joules)
2900.2 km-NFvW"



76. 40
100 cos 25PQ i
Fi



4000 cos 25 3625.2
WPQF

" Joules

77. False.
For example, let
1, 1 , 2, 3uv and
1, 4 .w Then 2 3 5uv" and
14 5.
uw"

78. True

00 0wuv wuwv" "" so, wand
uv are orthogonal.

28. 28 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
79. Let sθlength of a side.
,,
3
sss
sv
vθ
θ

1
cos cos cos
33
1
arcos 54.7
3s
s
$%
$%θθθθ






80.
1
1
2
2 ,,
3
,,0
2
22 6
cos
323
6
arcos 35.26
3
sss
s
ss
s
v
v
v
v


θ
θ
θ
θ
θθ


81. (a) The graphs
2
1
yxθ and
13
2
yxθ intersect at

0, 0 and 1, 1 .
(b)
12yxθ and
2
23
1
.
3
y
x
θ
At

0, 0 , 1, 0 is tangent to
1yand 0, 1 is
tangent to
2.y
At
11, 1 , 2yθ and
2
1
.
3
yθ


1
1, 2
5

is tangent to
1
1
,3,1
10
y
is tangent
to
2.y
(c) At
0, 0 , the vectors are perpendicular 90 .
At

1, 1 ,


11
1, 2 3, 1
51510
cos
11 50 2
45


"
θθθ




82. (a) The graphs
3
1
yxθ and
13
2
yxθ intersect at

1, 1 , 0, 0 and 1, 1 .
(b)
2
1
3yxθ and
2
23
1
.
3
y
x
θ
At
0, 0 ,
1, 0 is tangent to
1yand 0, 1 is
tangent to
2.y
At
1, 1 ,
13yθ and
2
1
.
3
yθ


1
1, 3
10

is tangent to
1
1
,3,1
10
y
is tangent
to
2.y
At
11, 1 , 3y and
2
1
.
3
yθ

1
1, 3
10

is tangent to
1
1
,3,1
10
y is tangent
to
2.y

(c) At
0, 0 , the vectors are perpendicular 90 .
At
1, 1 ,

11
1, 3 3, 1
6310 10
cos .
11 10 5
0.9273 or 53.13


"
θθθ


By symmetry, the angle is the same at
1, 1 .

83. (a) The graphs of
2
1
1yx and
22
1yx
intersect at
1, 0and 1, 0 .
(b)
1 2yx and
22.yxθ
At
11, 0 , 2y and
22.yθ
1
1, 2
5

is
tangent to
1
1
,1,2
5
y is tangent to
2.y
At
11, 0 , 2yand
2 2.y
1
1, 2
5

is
tangent to
1
1
,1,2
5
y is tangent to
2.y
(c) At
1, 0 ,
113
cos 1, 2 1, 2 .
555


"


0.9273 or 53.13
By symmetry, the angle is the same at
1, 0 .
y
x
v
s
s
s
z
y
x
v
1
v
2
( , , 0)s s
(, , )s s s
z
y
x
−112
−1
1
2
y = x
2
(1, 1)
(0, 0)
y = x
1/3
y
x
−1−2 1 2
−1
−2
1
2
(1, 1)
(0, 0)
(−1, −1)
y = x
3
y = x
1/3

29. Section 11.3 The Dot Product of Two Vectors 29
© 2010 Brooks/Cole, Cengage Learning
84. (a) To find the intersection points, rewrite the second
equation as
3
1.yx Substituting into the first
equation


2
6
10,1.yxxxx
There are two points of intersection,
0, 1and
1, 0 , as indicated in the figure.

(b) First equation:


2 1
1211
21
yxyyy
y


At
1
1, 0 , .
2
yθ
Second equation:
32
13.yx y x At
1, 0 , 3.yθ

1
2, 1
5

unit tangent vectors to first curve,
1
1, 3
10

unit tangent vectors to second curve
At
0, 1 , the unit tangent vectors to the first curve
are
0, 1 , and the unit tangent vectors to the
second curve are 1, 0 .
(c) At
1, 0 ,

11 51
cos 2, 1 1, 3 .
510 502
θ" θθ

4


or 45
At
0, 1the vectors are perpendicular, 90 .

85. In a rhombus,
.uvθ The diagonals are uvand
.
uv


22
0
uv uv uvu uvv
uu vu uv vv
uv" ""
""""

So, the diagonals are orthogonal.


86. If uand vare the sides of the parallelogram, then the
diagonals are
uv and ,uvas indicated in the
figure.
the parallelogram is a rectangle.


22
0
22
The diagonals are equal in length.uv
uv uv
uv uv uv uv
uv uv&"θ
&""
& " "
&
&


87. (a)
(b) Length of each edge:
222
02kk k
(c)

2
1
cos
222
1
arccos 60
2k
kk

θθ




(d)
1
2 ,,0 , , , ,
222 22 2
0, 0, 0 , , , ,
222 2 2 2
kkk kk k
rkk
kkk k k k
r
θ




2
2
1
4
cos
3
3
2
109.5
k
k





"





88.
cos , sin , 0 , cos , sin , 0uv $$θθ
The angle between
uand vis .$Assuming that
.$Also,



cos
cos cos sin sin
11
cos cos sin sin .
uv
uv
$
$ $
$ $
"


θ

y
x
−112
−2
1
(1, 0)
(0, −1)
y = x
3
−1
x = (y +1)
2
u
v
uv+
uv−
u
v
u + v
u − v
x
y
( , 0, )k k
( , , 0)k k
(0, , )k k
k
k
k
z

30. 30 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
89.

2
22
22
2
uv uv uv
uvu uvv
uu vu uv vv
uuvuvv
uvuv〉〈〉"〉
〈〉"〉〉"
""""
""
"

90.
cos
cos
cos
because cos 1.
uv u v
uv u v
uv
uv



"〈
"〈
〈


91.



2
22
2
22
2
2
uv uv uv
uvu uvv
=u u v u u v v v
uuvv
uuvv uv
"
""
""""
"


So,
.uv u v

92. Let
1
proj ,
vwu〈 as indicated in the figure. Because
1wis a scalar multiple of ,vyou can write
12 2 .cuw w vw
Taking the dot product of both sides with
vproduces

22ccuv vw v vvwv" " " "

2
,cv〈because
2wand vare orthogonol.
So,
2
2
cc
uv
uv v
v"
"
and
1
2
proj .c
v
uv
wuvv
v"
〈〈〈



Section 11.4 The Cross Product of Two Vectors in Space
1.
010
10 0
ijk
jik'〈 〈〉


2.
10 0
010
ijk
ij k'〈 〈


3.
010
00 1
ijk
jki'〈 〈


4.
00 1
010
ijk
kj i'〈 〈〉

θ
w
1
w
2
u
v
x y
i
j
−k
1
1
1
−1
z
x y
i
j
k 1
1
1
−1
z
x y
i
j
k
z
1
1
1
−1
x y
−i
j
k
1
1
1
−1
z

31. Section 11.4 The Cross Product of Two Vectors in Space 31
© 2010 Brooks/Cole, Cengage Learning
5. 10 0
00 1
ijk
ik j'〈 〈〉


6.
00 1
10 0
ijk
ki j'〈 〈


7. (a)
240 20 10 16
32 5
ijk
uv i j k'
(b)
20 10 16vu uv i j k''
(c)
vv 0'〈

8. (a)
30 5 15 16 9
23 2
ij k
uv i j k'
〉
(b)
15 16 9vu uv i j k'〈〉' 〈 〉 〉
(c)
vv 0'〈

9. (a)
73217 33 10
115
ijk
uv i j k'〈 〈 〉 〉
〉
(b)
17 33 10vu uv i j k''
(c)
vv 0'〈

10. (a)
322 8 5 17
15 1
ijk
uv i j k'
(b)
8517vu uv i j k''
(c)
vv 0'〈

11.
12, 3, 0 , 2, 5, 0uv〈〉 〈〉
12 3 0 54 0, 0, 54
250
ijk
uv k'〈 〉 〈 〈
〉

12 0 3 0 0 54
0uuv
uuv"'
('

20 50 054
0vuv
vuv"'
('

12.
1, 1, 2 , 0, 1, 0uv〈〉 〈
1 1 2 2 2, 0, 1
010
ijk
uv ik'〈〉 〈〉〉〈〉 〉


12 10 21
0
02 10 01
0uuv
uuv
vuv
vuv"'
('
"'
('

13.
2, 3, 1 , 1, 2, 1uv〈〉 〈〉
2 3 1 1,1,1
121
ijk
uv i jk'〈 〉 〈〉〉〉〈〉〉〉
〉

21 3 1 1 1
0uuv
uuv"'
('

11 2 1 1 1
0vuv
vuv"'
('

14.
10,0,6, 5, 3,0uv〈〉 〈 〉
10 0 6 18 30 30 18, 30, 30
530
ijk
uv i j k'
〉

10 18 0 30 6 30 0uuv
uuv"'
('

518 330 030 0vuv
vuv"'
('

15. ,2uijkv ijk

11 1 2 3 2,3,1
21 1
ij k
uv i jk'
〉

12 13 11
0uuv
uuv"'
('

22 13 1 1
0vuv
vuv"'
('

vu vu uv〉' 〈〉' 〈 '
x y
−1
i
k−j
z
1
1
1
−1
x y
i
j
k 1
1
1
−1
z

32. 32 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
16. 6, 2uijvi jk
16 0 6 13
211
ijk
uv i j k'


16 6 1 0uuv u uv"' ( '

26 1 1 113 0vuv v uv"' ( '

17.
18.
19.
20.
21.
4, 3.5, 7 , 2.5, 9, 3
73.5, 5.5, 44.75
2.94 0.22 1.79
, ,
11.8961 11.8961 11.8961
uv
uv
uv
uv

'
'

'

22.
8, 6, 4
10, 12, 2
60, 24, 156
1
60, 24, 156
36 22
5213
,,
322322322u
v
uv
uv
uv

'
'

'


23.
3, 2, 5 , 0.4, 0.8, 0.2
3.6, 1.4, 1.6
1.8 0.7 0.8
,,
4.37 4.37 4.37uv
uv
uv
uv
'
'

'

24.
7331
0, 0, , , 0,
10 2 5
21
0, , 0
20
0, 1, 0
uv
uv
uv
uv
'
'

'

25. Programs will vary.

26.
50, 40, 34
72.498uv
uv'
'

27.
010
01 1
uj
vjk
ijk
uv i

'

1uv iA'

28.
11 1
01 1
uijk
vjk
ijk
uv jk

'

2uv jkA'

29.
3, 2, 1
1, 2, 3
32 1 8,10,4
12 3u
v
ijk
uv

'

8, 10, 4 180 6 5Auv'
x
y
v
u
4
6
4
1
2
3
1
3
2
4
5
6
z
x
y
v
u
4
6
4
1
2
3
1
3
2
4
5
6
z
x
y
v
u
z
4
6
4
1
2
3
1
3
2
4
5
6
x
y
v
u
4
6
4
1
2
3
1
3
2
4
5
6
z

33. Section 11.4 The Cross Product of Two Vectors in Space 33
© 2010 Brooks/Cole, Cengage Learning
30. 2, 1, 0
1, 2, 0u
v


210 0,0,3
120
0, 0, 3 3
A
ijk
uv
uv
'

θ'θ θ


31. 0, 3, 2 , 1, 5, 5 , 6, 9, 5 , 5, 7, 2ABCD

1, 2, 3
1, 2, 3
5, 4, 0
5, 4, 0AB
DC
BC
AD
θθθ


θ
θ
θ
θ

Because
ABDC
θθθ
θ and ,BCAD
θ the figure ABCD is
a parallelogram.

AB
θθθand
AD
are adjacent sides

12 3 12,15,6
540AB AD
ijk
θθθ
'
144 225 36 9 5AABAD
θθθ'

32. 2, 3, 1 , 6, 5, 1 , 7, 2, 2 , 3, 6, 4ABCD

4, 8, 2
4, 8, 2
1, 3, 3
1, 3, 3AB
DC
BC
AD
θθθ








Because
ABDC
θθθ
θ and ,BCAD

θ the figure ABCD is
a parallelogram.
AB
θθθ
and AD

are adjacent sides
4 8 2 18, 14, 20
133
324 196 400 2 230AB AD
AABAD
ijk
θθθ
θθθ
'

'

33. 0, 0, 0 , 1, 0, 3 , 3, 2, 0ABC

11 11
22 2
1, 0, 3 , 3, 2, 0
10 3 6,9,2
32 0
36 81 4AB AC
AB AC
AABAC
ijk
θθθ
θθθ
θθθ

'

'


34. 2, 3, 4 , 0, 1, 2 , 1, 2, 0ABC

11
22
2, 4, 2 , 3, 5, 4
24 2 6 2 2
35 4
44 11AB AC
AB AC
AABAC
ij k
ijk
θθθ
θθθ
θθθ

'

θ'θ θ

35. 2, 7, 3 , 1, 5, 8 , 4, 6, 1ABC

11
22
3, 12, 5 , 2, 13, 4
3 12 5 113, 2, 63
213 4
16,742AB AC
AB AC
AABAC
ijk
θθθ
θθθ
θθθ

'

θ'θ


36. 1, 2, 0 , 2, 1, 0 , 0, 0, 0AB C

51
22
3, 1, 0 , 1, 2, 0
3105
120AB AC
AB AC
AABAC
ijk
k
θθθ
θθθ
θθθ

'

θ'θ


37. 20Fk


1
2
cos 40 sin 40
0 cos 40 2 sin 40 2 10 cos 40
0020
10 cos 40 7.66 ft-lbPQ
PQ
PQ jk
ijk
Fi
F




'

'


38.
2000 cos 30 sin 30 1000 3 1000
0.16
0 0 0.16
0 1000 3 1000
160 3
160 3 ft-lb
PQ
PQ
PQ
Fjkjk
k
ijk
F
i
F




θ
'θ

θ
'θ

x
y
40°
1
2
ft
PQ
F
z
x
y
60°
PQ
F
0.16 ft
z

34. 34 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
39. (a) Place the wrench in the xy-plane, as indicated in the figure.
The angle from AB

to Fis 30 180 210




18 inches 1.5 feet
33 3
1.5 cos 30 sin 30
44
56 cos 210 sin 210
33 3
0
44
56 cos 210 56 sin 210 0
ijij
Fij
ijk
F
OA
OA
OA


θθ




'θ

θθθ
θθθ
θθθ





42 3 sin 210 42 cos 210
42 3 sin 210 cos cos 210 sin 42 cos 210 cos sin 210 sin
13 31
42 3 cos sin 42 cos sin 84 sin
22 22 k
k
kk










84 sin , 0 180OAF
θθθ
'
(b) When
2
45 , 84 42 2 59.40
2
FOA '
θθθ

(c) Let
84 sin
84 cos 0 when 90 .T
dT
d

θ

This is reasonable. When 90 ,
the force is perpendicular to the wrench.

40. (a)

5
15 inches feet
4
12 inches 1 foot
5
4
180 cos sin
AC
BC
AB jk
Fjk
θθθ

θθ
θθ


(b)

5
4
01
0 180 cos 180 sin
225 sin 180 cos
225 sin 180 cos
ijk
F
F
AB
i
AB



'


'
θθθ
θθθ

(c) When
13
30 , 225 180 268.38
22
ABF



'


(d) If
225 sin 180 cos ,T 0 Tθfor
4
225 sin 180 cos tan 141.34 .
5

For 0 141.34,
225 cos 180 sin 0T
5
tan 51.34 .
4
AB
θθθ
and Fare perpendicular.
(e)
From part (d), the zero is 141.34 ,
when the vectors are parallel.
y
x
30
30
18 in.
AB
F
O


0
0
180
100
y = 84 sinθ
A
B
F
C 15 in.
12 in.

0
0
180
400

35. Section 11.4 The Cross Product of Two Vectors in Space 35
© 2010 Brooks/Cole, Cengage Learning
41.
100
010 1
001
uvw"'

42.
111
210 1
001
uvw"'

43.
201
030 6
001
uvw"'

44.
200
111 0
022
uvw"'

45.

110
011 2
101
2
V
uvw
uvw"'
"'


46.

13 1
06 6 72
40 4
72
uvw
uvw
V
"'

"'

47.


3, 0, 0
0, 5, 1
2, 0, 5
300
051 75
205
75
V
u
v
w
uvw
uvw


"'
"'

48.


0, 4, 0
3, 0, 0
1, 1, 5
040
3 0 0 4 15 60
115
60u
v
w
uvw
uvw
V



"'

"'

49. uv 0 u' and vare parallel.

uv 0 u" and vare orthogonal.
So,
uor v(or both) is the zero vector.

50. (a)


b
c
d
h
x
uvw vwu
wuv uvw
vwu u wv
vwu wuv"' ' "
"' '"
" ' "
" ' ' "

(e)

f
g
uwv wvu
wvu uvw"'"'
"' '"
So,
abcd h and efg

51.

123 123
23 32 13 31 12 21,, ,,uu u vv v
uv uv uv uv uv uvuv
ijk' "


52. See Theorem 11.8, page 794.

53. The magnitude of the cross product will increase by a
factor of 4.

54. From the vectors for two sides of the triangle, and
compute their cross product.

212 121 313131,, ,,xx y yz z x xy yz z'

55. False. If the vectors are ordered pairs, then the cross
product does not exist.

56. False. In general, uv vu''

57. False. Let
,01, 0, 0 , 1, 0 , 1, 0, 0 .uvw
Then,
, but .uv uw 0 v w'' #

58. True

59.
123 123 1 2 3,, , ,, , , ,uu u vvv wwwuvw




12 3
112 23 3
233322 133311 122211
23 32 13 31 12 21 2 3 3 2 1 3 3 1 1 2 2 1uu u
vwv wvw
uvwuvw uvwuvw uvwuvw
uv uv uv uv uv uv uw uw uw uw uw uw
ijk
uvw
ijk
ijk i j k
uv uw
'




''

36. 36 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
60.
123 123, , , , , , is a scalar:uu u vvv cuv



123
123
23 32 13 31 12 21
23 32 13 31 12 21ccucucu
vv v
cu v cu v cu v cu v cu v cu v
cuvuv uvuv uvuv c
ijk
uv
ijk
ijkuv
'

'



61.

123
123 23 32 13 31 12 21
123,,uu u
uuu uuuu uuuu uuuu
uu uu=
ijk
uu i j k 0
'


62.

123
123
123
123
123
123uu u
vv v
ww w
ww w
uu u
vv v
uvw
uvw wuv
"'
'""'



123 23 213 13 312 12
123 23 213 13 312 12wuv vu w uv vu wuv vu
uvw wv u vw wv uvw wv
uvw

"'


63.

23 32 13 31 12 21
23 32 1 31 13 2 12 21 3
23 32 1 31 13 2 12 21 3uv uv uv uv uv uv
uv uv u uv uv u uv uv u
uv uv v uv uv v uv uv v
uv i j k
uvu 0
uvv 0
'
'"
'"
So,
uv u'( and .uv v'(

64. If u and v are scalar multiples of each other, cuvfor some scalar c.

uv v v vv 0 0cc c' ' '
If ,
uv 0' then
sin 0.uv Assume , .u0v0## So, sin 0, 0, and u and v are parallel. So,

cuvfor some scalar c.

65.
sinuv uv '
If
u and v are orthogonal,
2 and sin 1.So, .uv uv'

66.
111 222 333,,, ,,, ,,cab a bc abcuv w



222 2332 2332 2332
333
11 1
23 32 32 23 23 32abc bc bc ac ac ab ab
abc
abc
bc bc ac ac ab ab
ijk
vw i j k
ijk
uvw
'
''





1 23 32 1 32 23 1 23 32 1 23 32
1 32 23 1 23 32
2131313 3121212 2131313 3121212
213 13 13 312 12babab cacac aabab cbcbc
aac ac bbc bc
a aa bb cc a aa bb cc b ab bb cc b aa bb cc
caa bb cc caa bb
uvw i j
k
ij
''









12
131313222 121212333
,, ,,
cc
aa bb cc a b c aa bb cc a b c
k
uwv uvw


""

37. Section 11.5 Lines and Planes in Space 37
© 2010 Brooks/Cole, Cengage Learning
Section 11.5 Lines and Planes in Space
1. 13, 2 , 25x ty tz t
(a)
(b) When
0, 1, 2, 2 .tP When 3, 10, 1, 17 .tQ

9, 3, 15PQ


The components of the vector and the coefficients of t are proportional because the line is parallel to .PQ


(c) 0ywhen 2.tSo, 7xand 12.z
Point:
7, 0, 12
0xwhen
1
3
.tSo,
7
3
y and
1
3
.z Point:

71
33
0, ,
0zwhen
2
5
.t So,
1
5
x and
12
5
.y Point:

112
55
,,0
2.
23, 2, 1
x ty z t
(a)
(b) When 0,t
2, 2, 1 .P When 2,t
4, 2, 1 .Q

6, 0, 2PQ


The components of the vector and the coefficients of
t are proportional because the line is parallel to .PQ


(c) 0zwhen 1.tSo, 1xand 2.y
Point:
1, 2, 0
0xwhen
2
3
.t So, 2yand
1
3
z
Point:

1
3
0, 2,
3.
2, 3, 4
x ty tz t
(a)
0, 6, 6 : For 0 2 ,xt you have
2.tThen
32 6y and
42 6.z Yes,
0, 6, 6 lies on the line.
(b)
2, 3, 5 : For 2 2 ,xt you have
4.tThen
3 4 12 3.y# No, 2, 3, 5 does
not lie on the line.
4.
37
2
28
xy
z


(a)
7, 23, 0 : Substituting, you have

73 237
02
28
222




Yes,
7, 23, 0 lies on the line.
(b)
1, 1, 3 : Substituting, you have

13 17
32
28
111




Yes,
1, 1, 3lies on the line.
5. Point:
0, 0, 0
Direction vector: 3, 1, 5
Direction numbers: 3, 1, 5
(a) Parametric:
3, , 5
xty tz t
(b) Symmetric:
35
x z
y

6. Point:
0, 0, 0
Direction vector:
5
2, , 1
2
v
Direction numbers:
4, 5, 2
(a) Parametric:
4, 5, 2
x ty tz t
(b) Symmetric:
452
x yz



y
x
z
x y
z

38. 38 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
7. Point: 2, 0, 3
Direction vector: 2, 4, 2v
Direction numbers:
2, 4, 2
(a) Parametric:
22, 4, 32
x ty tz t
(b) Symmetric:
23
24 2
xyz



8. Point:

3, 0, 2
Direction vector: 0, 6, 3v
Direction numbers: 0, 2, 1
(a) Parametric:
3, 2 , 2
x ytz t
(b) Symmetric: 2, 3
2
y
zx
9. Point:

1, 0, 1
Direction vector: 3 2vijk
Direction numbers:
3, 2, 1
(a) Parametric:
13, 2, 1
x ty tz t
(b) Symmetric:
11
321xyz



10. Point:

3, 5, 4
Directions numbers:
3, 2, 1
(a) Parametric:
33, 52, 4
x ty tz t
(b) Symmetric:
35
4
32xy
z




11. Points:

22
5, 3, 2 , , , 1
33





Direction vector:
17 11
3
33
vijk
Direction numbers:
17, 11, 9
(a) Parametric:
517, 311, 29
x ty tz t
(b) Symmetric:
532
17 11 9xyz



12. Points:

0, 4, 3 , 1, 2, 5
Direction vector: 1, 2, 2
Direction numbers:
1, 2, 2
(a) Parametric:
,42,32
xty tz t
(b) Symmetric:
43
22yz
x




13. Points:
7, 2, 6 , 3, 0, 6
Direction vector:
10, 2, 0
Direction numbers:
10, 2, 0
(a) Parametric:
710, 2 2, 6xty tz
(b) Symmetric: Not possible because the direction
number for z is 0. But, you could describe the
line as
72
, 6.
10 2
xy
z




14. Points: 0, 0, 25 , 10, 10, 0
Direction vector: 10, 10, 25
Direction numbers:
2, 2, 5
(a) Parametric:
2, 2, 25 5
xty tz t
(b) Symmetric:
25
22 5xyz



15. Point:
2, 3, 4
Direction vector: vk
Direction numbers: 0, 0, 1
Parametric:
2, 3, 4
x yz t
16. Point:
4, 5, 2
Direction vector: vj
Direction numbers:
0, 1, 0
Parametric:
4, 5 , 2xy tz
17. Point:
2, 3, 4
Direction vector: 3 2vijk
Direction numbers:
3, 2, 1
Parametric:
23, 32, 4
x ty tz t
18. Point
4, 5, 2
Direction vector: 2vijk
Direction numbers:
1, 2, 1
Parametric:
4, 52, 2
x ty tz t
19. Point:
5, 3, 4
Direction vector:
2, 1, 3v
Direction numbers:
2, 1, 3
Parametric:
52, 3 , 43
x ty tz t
20. Point:
1, 4, 3
Direction vector: 5vij
Direction numbers:
5, 1, 0
Parametric:
15, 4 , 3xtytz

39. Section 11.5 Lines and Planes in Space 39
© 2010 Brooks/Cole, Cengage Learning
21. Point: 2, 1, 2
Direction vector: 1, 1, 1
Direction numbers:
1, 1, 1
Parametric:
2, 1, 2
x ty tz t
22. Point:

6, 0, 8
Direction vector: 2, 2, 0
Direction numbers:
2, 2, 0
Parametric:
62, 2, 8xtytz
23. Let
0: 3, 1, 2tP other answers possible

1, 2, 0v any nonzero multiple of is correctv
24. Let
0: 0, 5, 4tP other answers possible

4, 1, 3v any nonzero multiple of is correctv
25. Let each quantity equal
0:

7, 6, 2P other answers possible

4, 2, 1v any nonzero multiple of is correctv
26. Let each quantity equal 0:

3, 0, 3P other answers possible

5, 8, 6v any nonzero multiple of is correctv
27.
1
2
3
4:3,2,4
:6,4,8
:6,4,8
:6,4,6
L
L
L
L
v
v
v
v







12 3
6, 2, 5 on line
6, 2, 5 on line
6, 2, 5 not online
not parallel to , , nor
LLL





12and
L Lare identical.
12LL and is parallel to
3.L
28.
1
2
3
4:2,6,2
:2,1,3
:2,10,4
:2,1,3
L
L
L
L
v
v
v
v







3, 0, 1 on line
1, 1, 0 on line
1, 3, 1 on line
5, 1, 8 on line




24and
L Lare parallel, not identical, because 1, 1, 0is
not on
4.
L
29.
1
2
3
4:4,2,3
:2,1,5
:8,4,6
: 2,1,1.5
L
L
L
L
v
v
v
v





8, 5, 9 on line
8, 5, 9 on line




13and
L Lare identical.
30.
1
2
3
4
1
2
:2,1,2
:4,2,4
:1,,1
:2,4,1
L
L
L
L
v
v
v
v







3, 2, 2 on line
1, 1, 3 on line
2, 1, 3 on line
3, 1, 2 on line






12,
LLand
3Lhave same direction.

3, 2, 2is not on
23nor
L L

1, 1, 3is not on
3
L
So, the three lines are parallel, not identical.
31. At the point of intersection, the coordinates for one line
equal the corresponding coordinates for the other line.
So,
(i) 422 2,ts (ii) 32 3,s and
(iii) 1 1.ts
From (ii), you find that 0sand consequently, from
(iii), 0.tLetting 0,st you see that equation (i)
is satisfied and so the two lines intersect. Substituting
zero for s or for t, you obtain the point
2, 3, 1 .



4First line
2 2 Second line
81 7 717
cos
5117 9 3 17uik
vijk
uv
uv



"

32. By equating like variables, you have
(i) 313 1,ts (ii) 412 4,ts and
(iii) 2 4 1.ts
From (i) you have ,
stand consequently from (ii),
1
2
t and from (iii), 3.tThe lines do not intersect.
33. Writing the equations of the lines in parametric form you
have

3
14xt
x s


2
2
yt
ys


1
33.zt
zs


For the coordinates to be equal, 3 1 4ts and
22.ts Solving this system yields
17
7
t and
11
7
.s When using these values for s and t, the z
coordinates are not equal. The lines do not intersect.

40. 40 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
34. Writing the equations of the lines in parametric form you
have

23
32
x t
x s


26
5
yt
ys


3
24.zt
zs


By equating like variables, you have
23 32,ts 26 5 ,ts 324.ts
So,
1, 1ts and the point of intersection is
5, 4, 2 .



3, 6, 1 First line
2, 1, 4 Second line
4 4 2 966
cos
48346 21 966u
v
uv
uv


θ
"
θθ θθ
35.
23
52
1
xt
yt
zt



27
8
21
xs
ys
zs




Point of intersection:
7, 8, 1

Note: 2and 0tsθθ


36. 21
410
xt
yt
zt


θ
512
311
24
xs
ys
zs




Point of intersection:
3, 2, 2


37. 436 6xyz
(a)

0, 0, 1 , 0, 2, 0 , 3, 4, 1
0, 2, 1 , 3, 4, 0PQ R
PQ PR



(b)
021 4,3,6
340PQ PR
ijk

'
The components of the cross product are
proportional to the coefficients of the variables in
the equation. The cross product is parallel to the
normal vector.

38. 234 4xyz

0, 0, 1 , 2, 0, 0 , 3, 2, 2PQ R
(a)
2, 0, 1 , 3, 2, 3PQ PR


(b) 20 1 2,3,4
32 3PQ PR
ijk

'

The components of the cross product are
proportional
for this choice of , ,PQ
and , they are the sameR to the coefficients of the
variables in the equation. The cross product is
parallel to the normal vector.

39. 2410xyz
(a)
7, 2, 1 : 7 2 2 4 1 1 0
Point is in plane
(b)
5, 2, 2 : 5 2 2 4 2 1 0
Point is in plane

40. 2360xy z
(a)
3, 6, 2 : 2 3 6 3 2 6 0
Point is in plane
(b)
1, 5, 1 : 2 1 5 3 1 6 6 0 #
Point is not in plane

41. Point: 1, 3, 7
Normal vector:
0, 1, 0njθθ

01130 7 0
30
xy z
y




42. Point: 0, 1, 4
Normal vector:
0, 0, 1nkθθ

0001140
40
xyz
z




43. Point: 3, 2, 2
Normal vector: 2 3
nijk

2332120
23 10
xyz
xyz




44. Point: 0, 0, 0
Normal vector: 3 2
nik

3000200
32 0
xyz
xz



x y
6
8
10
4
2
4
−8
−
4
6
8
10
(7, 8, −1)
z
z
x y
(3, 2, 2)
3
3
3
2
2
2
−2
−3
xt
yt
zt
= 2 1
= 4 + 10
=
−
−
xs
ys
zs
=5 12
= 3 + 11
=2 4
−−
−−

41. Section 11.5 Lines and Planes in Space 41
© 2010 Brooks/Cole, Cengage Learning
45. Point: 1, 4, 0
Normal vector: 2, 1, 2v

2114200
2260
xy z
xy z




46. Point: 3, 2, 2
Normal vector: 4 3
vijk

43 2320
438
xy z
xy z




47. Let u be the vector from 0, 0, 0to

2, 0, 3 : 2, 0, 3u
Let
u be the vector from 0, 0, 0 to

3, 1, 5 : 3, 1, 5v
Normal vectors: 203 3,19,2
315
ijk
uv'



30190200
319 2 0
xyz
xyz




48. Let u be the vector from 3, 1, 2to 2, 1, 5 :

1, 2, 3u
Let
ube the vector from 3, 1, 2to 1, 2, 2 :

2, 1, 4v
Normal vector:
1 2 3 5, 10, 5 5 1, 2, 1
214
ijk
uv'



1321 20
210
xyz
xyz




49. Let ube the vector from 1, 2, 3 to
3, 2, 1 : 2 2uik
Let
vbe the vector from 1, 2, 3 to
1, 2, 2 : 2 4vijk
Normal vector:

1
2
10 1 4 3 4
24 1
ijk
uv ijk'


413 2430
43410
xy z
xyz




50. 1, 2, 3 , Normal vector: ,1 1 0, 1xxvi

51. 1, 2, 3 ,Normal vector: ,1 3 0, 3zzvk

52. The plane passes through the three points

0, 0, 0 , 0, 1, 0 , 3, 0, 1 .
The vector from
0, 0, 0 to 0, 1, 0 :uj
The vector from
0, 0, 0 to
3, 0,1 : 3vik
Normal vector: 010 3
301
ijk
uv i k'


30xz

53. The direction vectors for the lines are
2,
uijk 34 . vijk
Normal vector:
21 1 5
34 1
ijk
uv i jk'


Point of intersection of the lines:
1, 5, 1

1510
5
xyz
xyz




54. The direction of the line is 2 .uijk Choose any
point on the line,
0, 4, 0 , for example ,
and let v be the
vector from
0, 4, 0 to the given point 2, 2, 1 :
22
vijk
Normal vector:
211 2
221
ijk
uv i k'



22 1 0
20
xz
xz




55. Let v be the vector from 1, 1, 1 to
2, 2, 1 : 3 2vijk
Let
n be a vector normal to the plane
23 3: 23xyz nijk
Because
v and n both lie in the plane P, the normal
vector to P is

312 7 11
231
ijk
vn i j k'



72121110
7115
xy z
xy z

42. 42 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
56. Let v be the vector from 3, 2, 1 to
3, 1, 5 : 6 vjk
Let
n be the normal to the given plane:
672
nijk
Because
v and n both lie in the plane P, the normal
vector to P is:


016 40 36 6
67 2
220 18 3
ijk
vn i j k
ijk
'



20 3 18 2 3 1 0
20 18 3 27
xyz
xyz




57. Let uiθand let v be the vector from 1, 2, 1 to
2, 5, 6 : 77 vi j k
Because
u and v both lie in the plane P, the normal
vector to P is:

10 0 7 7 7
17 7
ijk
uv j k jk'


210
1yz
yz




58. Let ukθand let vbe the vector from 4, 2, 1 to
3, 5, 7 : 736vijk
Because
u and v both lie in the plane P, the normal vector to P is:

001 3 7 3 7
736
ijk
uv i j i j'



347 20
37 26
xy
xy




59. xy-plane: Let 0.zθ


Then 0 4 4 1 2 4 7 and
2 3 4 10. Intersection: 7, 10, 0tt x
y


-plane:xz Let 0.yθ



221
333
10 1021
33 33
Then 0 2 3 1 2 and
4 . Intersection: , 0,
tt x
z



-plane:yz Let 0.xθ



111
222
7711
22 22
Then 0 1 2 2 3 and
4 . Intersection: 0, ,
tt y
z



60. Parametric equations: 2 3 ,xt 1,yt 32zt
-plane:xy Let 0.zθ



335
222
35 55
22 22
Then 3 2 0 2 3 and
1 . Intersection: , , 0
tt x
y



-plane:xz Let 0.yθ


Then 1 2 3 1 5 and
3 2 1 5. Intersection: 5, 0, 5
tx
z


-plane:yz Let 0.xθ


522
333
5552
33 33
Then 2 3 0 1 and
3 2 . Intersection: 0, ,
tt y
z



x
y
6
8
2
10
2
2
4
6
z
(−7, 10, 0)
0, − , −
7
2
1 2
))
)
)− , 0, −
10
31 3
−4
−6
−8
x
z
y
5
2
−
5
3
−
(
, ,
5
2
− 0 )
(5, 0, 5)
,
5
3
)0
,(
5
4
4
6
3
2
1
−1
−2
−3
−4
−1
2
−2
−3

43. Section 11.5 Lines and Planes in Space 43
© 2010 Brooks/Cole, Cengage Learning
61. Let ,,xyzbe equidistant from 2, 2, 0 and 0, 2, 2 .



222 222
222222
220 022
44 44 44 44
48 48
0Plane
xyz xyz
xx yy zxyy zz
xz
xz





62. Let ,,
xyzbe equidistant from 1, 0, 2 and 2, 0, 1 .



222 222
222 2 22
102 201
21 44 44 21
245 425
22 0
0Plane
xy z x y z
xx yzz xx yzz
xz xz
xz
xz






63. Let ,,xyzbe equidistant from 3, 1, 2 and 6, 2, 4 .



22 2 2 2 2
222 2 22
312 6 24
6 9 2 1 4 4 12 36 4 4 8 16
62414 124856
18 6 4 42 0
932210Plane
xyz xyz
xx yy zz x x yy zz
xyz xyz
xyz
xyz






64. Let ,,
xyzbe equidistant from 5, 1, 3 and 2, 1, 6



222 22 2
222222
513 216
10 25 2 1 6 9 4 4 2 1 12 36
10 2 6 35 4 2 12 41
14 4 18 6 0
72930Plane
xyz xyz
xx yy zz xx yy zz
xyz xy z
xy z
xyz






65. The normal vectors to the planes are

15, 3, 1 ,n
21, 4, 7 ,n
12
12
cos 0.
nn
nn

"


So,
2 and the planes are orthogonal.

66. The normal vectors to the planes are

13, 1, 4 ,n
2 9, 3, 12 .n
Because
213,nn the planes are parallel, but not equal.

67. The normal vectors to the planes are

1 36,nijk
25,nijk

12
12 536 4 138 2 138
cos .
414 20746 27nn
nn

"

So,
2 138
arccos 83.5 .
207







68. The normal vectors to the planes are

132 ,nijk
2 42,nijk

12
12 382 76 6
cos .
42 614 21nn
nn

"

So,
6
arccos 65.9 .
6







69. The normal vectors to the planes are
11, 5, 1n and
25, 25, 5 .n Because
215,nn the planes are
parallel, but not equal.

70. The normal vectors to the planes are

122, 0, 1 , 4, 1, 8 ,nn

1
12
cos 0
2nn
nn

"


So,
2


and the planes are orthogonal.

44. 44 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
71. 42612xyz


72. 362 6xyz


73. 234xy z


74. 24xyz


75. 6xz


76. 28xy


77. 5xθ


78. 8zθ


79. 26xyz


80. 33xz


81. 54680xyz


82. 2.1 4.7 3 0xyz

x
y
6
6
4
6
4
z
(0, 0, 2)
(0, 6, 0)
(3, 0, 0)
x
y2
3
2
3
3
z
(0, 0, 3)
(0, 1, 0)
(2, 0, 0)
x
y
−1
−4
3
3
2
z
(0, −4, 0)
(2, 0, 0)
4
3((0, 0,
x
y1
−4
3
4
1
2
z
(0, 0, 4)
(0, −4, 0)
(2, 0, 0)
yx
z
(0, 0, 6)
(6, 0, 0)
8
8
8
yx
z
(0, 8, 0)
(4, 0, 0)
8
8
8
x
y
5
5
3
z
(5, 0, 0)
x
y
5
5
8
z
yx
2
4
6
−6
2
4
6
Generated by Maple
z
x y
1
2
2
1
1
2
3
Generated by Mathematica
z
Generated by Maple
y
x
1
−2
−1
2
z
x
y
2
1
1
2
3
Generated by Mathematica
z

45. Section 11.5 Lines and Planes in Space 45
© 2010 Brooks/Cole, Cengage Learning
83.
1
2
3
4:15,6,24
:5,2,8
:6,4,4
: 3,2,2
P
P
P
Pn
n
n
n




0, 1, 1 not on plane
0, 1, 1 on plane


Planes
1Pand
2Pare parallel.

84.
1
2
3
4:2,1,3
:3,5,2
:8,4,12
:4,2,6
P
P
P
Pn
n
n
n




4, 0, 0 on plane
4, 0, 0 not on plane


1Pand
3Pare parallel.

85.
1
2
3
4:3,2,5
:6,4,10
:3,2,5
: 75, 50, 125
P
P
P
Pn
n
n
n





1, 1, 1 on plane
1, 1, 1 not on plane
1, 1, 1 on plane




1Pand
4Pare identical.

14PPand is parallel to
2.P

86.
1
2
3
4
12 3: 60,90,30 or 2,3,1
:6,9,3or2,3,1
:20,30,10or2,3,1
: 12, 18, 6 or 2, 3, 1
, , and are parallel.
P
P
P
P
PP Pn
n
n
n







9
10
2
3
5
6
0, 0, on plane
0, 0, on plane
0, 0, on plane



87. Each plane passes through the points

,0,0, 0, ,0,and 0,0, .cc c

88.
xyc
Each plane is parallel to the z-axis.

89. If 0, 0cz is xy-plane.
If
1
0, 0ccyz y z
c

#
is a plane parallel to
x-axis and passing through the points
0, 0, 0and
0, 1, .c

90. 0xcz
If
0, 0cz is the yz-plane.
If
0, 0cxcz# is a plane parallel to the y-axis.

91. (a)
132nijk and
2 42nijk

12
12 7 6
cos
614 21nn
nn

"

1.1503 65.91

(b) The direction vector for the line is

2 142 7 2.
321
1
ijk
nn j k
'

Find a point of intersection of the planes.

64214
42 0
714
2
xyz
xyz
x
x





Substituting 2 for x in the second equation, you have
42 2yz or 21.zy Letting 1,ya
point of intersection is
2, 1, 1 .

2, 1 , 1 2
x ytz t

92. (a)
163 ,nijk
2 5nijk

12
12 4 2 138
cos
20746 27
1.6845 96.52nn
nn


"



(b) The direction vector for the line is

12 631 16,31,3.
115
ijk
nn
'

Find a point of intersection of the planes.
63 5
55
xyz
xyz




63 5
6 6 30 30
33135
xy z
xy z
yz




Let
9, 2 4 4, 9, 2 .
416, 931, 23
yz x
x ty tz t




93. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:


13
,,12
22
13 3
22 1212,
22 2
xty tz t
tttt







Substituting 32tinto the parametric equations for
the line you have the point of intersection
2, 3, 2 .
The line does not lie in the plane.

94. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:


14, 2, 36
1
21 4 32 5,
2
x ty tz t
tt t




Substituting
1
2
t into the parametric equations for
the line you have the point of intersection
1, 1, 0 .
The line does not lie in the plane.

46. 46 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
95. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:

13, 12, 3
x ty tz t

2 1 3 3 1 2 10, 1 10,tt contradiction
So, the line does not intersect the plane.

96. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:


42, 13, 25
5 4 2 3 1 3 17, 0
x ty tz t
ttt


Substituting 0tinto the parametric equations for the
line you have the point of intersection
4, 1, 2 .
The line does not lie in the plane.

97. Point: 0, 0, 0Q
Plane: 2 3 12 0xyz
Normal to plane:
2, 3, 1n
Point in plane:
6, 0, 0P
Vector
6, 0, 0PQ



12614
714PQ
Dn
n

"



98. Point: 0, 0, 0Q
Plane: 5 9 0xyz
Normal to plane:
5, 1, 1n
Point in plane:

0, 9, 0P
Vector 0, 9, 0PQ



9
3
27PQ
Dn
n

"



99. Point: 2, 8, 4Q
Plane: 2 5xyz
Normal to plane:
2, 1, 1n
Point in plane: 0, 0, 5P
Vector: 2, 8, 1PQ



11 11 6
66PQ
Dn
n

"



100. Point:
1, 3, 1Q
Plane: 34560xyz
Normal to plane: 3, 4, 5n
Point in plane:

2, 0, 0P
Vector :1,3,1PQ



20
22
50PQ
Dn
n

"



101. The normal vectors to the planes are
11, 3, 4n and
21, 3, 4 .n Because
12 ,nn the planes are
parallel. Choose a point in each plane.


10, 0,0P is a point in 3 4 10.xyz


6, 0, 0Q is a point in 3 4 6.xyz
4, 0, 0 ,PQ


1
1 4226
1326PQ
Dn
n

"



102. The normal vectors to the planes are
14, 4, 9n and
24, 4, 9 .n Because
12 ,nn the planes are
parallel. Choose a point in each plane.


5, 0, 3P is a point in 449 7.xyz


0, 0, 2Q is a point in 4 4 9 18.xyz
5, 0, 1PQ



1
1 11 11 113
113113PQ
Dn
n

"



103. The normal vectors to the planes are
1 3, 6, 7n and
26, 12, 14 .n Because
212,nn the planes are
parallel. Choose a point in each plane.


0, 1, 1Pis a point in 367 1.xyz

25
,0,0
6
Q



is a point in 6 12 14 25.xyz

1
1
25
,1, 1
6
27 2 27 27 94
18894 2 94
PQ
PQ
D
n
n



"

47. Section 11.5 Lines and Planes in Space 47
© 2010 Brooks/Cole, Cengage Learning
104. The normal vectors to the planes are
12, 0, 4n and
22, 0, 4 .n Because
12 ,nn the planes are
parallel. Choose a point in each plane.


2, 0, 0P is a point in 2 4 4.xz


5, 0, 0Q is a point in 2 4 10.xz

1
1 635
3, 0, 0 ,
520PQ
PQ Dn
n


"



105.
4, 0, 1u is the direction vector for the line.
1, 5, 2Q is the given point, and 2, 3, 1P is on the
line.
3, 2, 3PQ


32 3 2,9,8
40 1PQ
ij k
u

'


149 2533
1717PQ
D u
u

'



106.
2, 1, 2u is the direction vector for the line.
1, 2, 4Q is the given point, and 0, 3, 2P is a point
on the line
let 0 .t

1, 1, 2PQ


11 2 0,2,1
212
ijk
u
PQ'



55
39u
uPQ
D'




107.
1, 1, 2u is the direction vector for the line.

2, 1, 3Q is the given point, and 1, 2, 0P is on the line
let 0 in the parametric equations for the line .t 3, 1, 3PQ


313 1,9,4
112PQ
ijk
u

'


18116 98 7 73
6311 4 3PQ
Du
u

'




108.
0, 3, 1u is the direction vector for the line.

4, 1, 5Qis the given point, and
3, 1, 1P is on the line.
1, 2, 4PQ


124 14,1,3
031PQ
ijk
u

'

2
14 1 9 206 103 515
10 5 591
PQ
Du
u

'






109. The direction vector for
1
Lis
1 1, 2, 1 .v
The direction vector for
2
Lis
23, 6, 3 .v
Because
213,vv the lines are parallel.
Let
2, 3, 4Q to be a point on
1
Land 0, 1, 4P a point
on
2.
L 2, 0, 0 .PQ



2uvis the direction vector for
2.
L

2220 6,6,18
363PQ
ijk
v

'


2
2
36 36 324 396 22 66
54 3 39369
PQ
D
v
v

'






110. The direction vector for
1
Lis
16, 9, 12 .v
The direction vector for
2
Lis
24, 6, 8 .v
Because
12
3
2
,vv the lines are parallel.
Let
3, 2, 1Qto be a point on
1
Land 1, 3, 0P a point
on
2.4,5,1.LPQ



2uvis the direction vector for
2.
L

2451 34,36,44
468PQ
ijk
v

'


2
2
222
34 36 44
16 36 64
4388 1097 31813
29 29116
PQ
D
v
v

'

48. 48 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
111. The parametric equations of a line L parallel to
,,,abcv and passing through the point
111,,Px y zare

1 ,xxat
1 ,yybt
1 .zzct
The symmetric equations are
111
.
xxyyzz
abc




112. The equation of the plane containing 111,,Px y zand
having normal vector
,,abcn is

111 0.ax x by y cz z
You need
nand P to find the equation.

113. Simultaneously solve the two linear equations
representing the planes and substitute the values back
into one of the original equations. Then choose a value
for t and form the corresponding parametric equations
for the line of intersection.

114. :
xaplane parallel to yz-plane containing ,0,0a

:ybplane parallel to xz-plane containing 0, , 0b
:zcplane parallel to xy-plane containing
0, 0,c

115. (a) The planes are parallel if their normal vectors are
parallel:

111 2 2 2,, , , , 0abc ta bc t#
(b) The planes are perpendicular if their normal vectors
are perpendicular:

111 2 2 2,, , , 0abc a bc"

116. Yes. If
1vand
2vare the direction vectors for the lines
1
Land
2,Lthen
12vvv' is perpendicular to both
1Land
2.L

117. An equation for the plane is

1
xyz
bcx acy abz abc
abc

For example, letting 0,yz the x-intercept is
,0,0.a

118. (a) Matches (iii)
(b) Matches (i)
(c) Matches (iv)
(d) Matches (ii)

119. Sphere


222
32516xyz

120. Parallel planes

4 3 10 4 10 4 26xyz n

121. 0.92 1.03 0.02 0.02 0.92 1.03
x yz z x y
(a)
The approximations are close to the actual values.
(b) According to the model, if x and z decrease, then so will y. (Answers will vary.)

122. On one side you have the points 0, 0, 0 , 6, 0, 0 , and 1, 1, 8 .

1 600 48 6
118
ijk
njk

On the adjacent side you have the points
0, 0, 0 , 0, 6, 0 , and 1, 1, 8 .

2 060 48 6
118
ijk
nik


12
12 36 1
cos
2340 65
1
arccos 89.1
65nn
nn


"



Year 1999 2000 2001 2002 2003 2004 2005
x 1.4 1.4 1.4 1.6 1.6 1.7 1.7
y 7.3 7.1 7.0 7.0 6.9 6.9 6.9
z 6.2 6.1 5.9 5.8 5.6 5.5 5.6
Model z 6.25 6.05 5.94 5.76 5.66 5.56 5.56
x
y
4
66
4
6
2
(0, 6, 0)
(6, 0, 0)
(0, 0, 0)
( 1, 1, 8)−−
z

49. Section 11.5 Lines and Planes in Space 49
© 2010 Brooks/Cole, Cengage Learning
123.
11:6;Lxt
18,yt
13zt

22:1,
Lxt
22,yt
22zt
(a) At 0,tthe first insect is at
16, 8, 3P and the second insect is at 21, 2, 0 .P
Distance

222
6 1 8 2 3 0 70 8.37 inches
(b) Distance


222 22
22
12 1 2 12
562 3 53070,0 10xx yy zz t t t t t
(c) The distance is never zero.
(d) Using a graphing utility, the minimum distance is 5 inches when 3tminutes.



124. First find the distance D from the point 3, 2, 4Q to the plane. Let 4, 0, 0P be on the plane.
2, 4, 3n is the normal to the plane.

7, 2, 4 2, 4, 3 14 8 12 18 18 29
294 16 9 29 29
n
n
PQ
D
" "




The equation of the sphere with center
3, 2, 4 and radius
18 29 29 is
222 324
324.
29
xyz


125. The direction vector vof the line is the normal to the
plane,
3, 1, 4 .v
The parametric equations of the line are
53,x t 4,yt 34.zt
To find the point of intersection, solve for t in the
following equation:


4
13
35 3 4 4 3 4 7
26 8
tt t
t
t



Point of intersection:


77 48 2344 4
13 13 13 13 13 13
53 ,4 ,34 ,,

126. The normal to the plane,
2, 1, 3n is perpendicular
to the direction vector
2, 4, 0v of the line because 2, 1, 3 2, 4, 0 0. "
So, the plane is parallel to the line. To find the distance
between them, let
2, 1, 4Q be on the line and
2, 0, 0P on the plane.
4, 1, 4 .PQ



4, 1, 4 2, 1, 3 19 19 14
14419 14
PQ
D
n
n
"

"




127. The direction vector of the line L through 1, 3, 1and
3, 4, 2 is
2, 1, 1 .v
The parametric equations for L are
12,xt 3,yt 1.zt
Substituting these equations into the equation of the
plane gives


3
4
12 3 1 2
43
.ttt
t
t


Point of intersection:


333 9 11
444244
12 ,3 ,1 , ,

128. The unknown line L is perpendicular to the normal vector
1, 1, 1n of the plane, and perpendicular to the direction
vector 1, 1, 1 .u So, the direction vector of L is
11 1 2,2,0.
11 1
ij k
v


The parametric equations for L are 1 2 ,
x t 2,yt
2.z

129. True

130. False. They may be skew lines.

See Section Project
15
0
0
15

50. 50 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
131. True

132. False. The lines ,
xtθ 0,yθ 1zθand 0,xθ
,ytθ
1zθare both parallel to the plane 0,zθbut
the lines are not parallel.

133. False. Planes 7 11 5xy z and 5 2 4 1xyz
are both perpendicular to plane 23 3,xyz but
are not parallel.

134. True.
Section 11.6 Surfaces in Space
1. Ellipsoid
Matches graph (c)

2. Hyperboloid of two sheets
Matches graph (e)

3. Hyperboloid of one sheet
Matches graph (f )

4. Elliptic cone
Matches graph (b)

5. Elliptic paraboloid
Matches graph (d)

6. Hyperbolic paraboloid
Matches graph (a)

7. 5yθ
Plane is parallel to the xz-plane.


8. 2zθ
Plane is parallel to the xy-plane.


9.
22
9yz
The x-coordinate is missing so you have a right circular
cylinder with rulings parallel to the x-axis. The generating
curve is a circle.


10.
22
25xz
The y-coordinate is missing so you have a right circular
cylinder with rulings parallel to the y-axis. The generating
curve is a circle.


11.
2
yxθ
The z-coordinate is missing so you have a parabolic
cylinder with rulings parallel to the z-axis. The generating
curve is a parabola.


12.
2
6yz
The x-coordinate is missing so you have a parabolic
cylinder with the rulings parallel to the x-axis. The
generating curve is a parabola.


13.
22
22
44
1
14
xy
xy



The z-coordinate is missing so you have an elliptic
cylinder with rulings parallel to the z-axis. The generating
curve is an ellipse.

x
3
2
1
−1
−2
−3
1
z
2
4
5
y
3
2
1
−2
−3
x
3
2
−1
−2
−3
2
3
z
y
3
2
x
y
4
7
6
4
z
x
y
8
8
6
4
z
x
y
4
4
3
3
2
4
z
x
y2
−2
2
z
x
y
2 3
3
2
3
z
−3

51. Section 11.6 Surfaces in Space 51
© 2010 Brooks/Cole, Cengage Learning
14.
22
22
16
1
16 16
yz
yz



The x-coordinate is missing so you have a hyperbolic
cylinder with rulings parallel to the x-axis. The generating
curve is a hyperbola.


15. sinzyθ
The x-coordinate is missing so you have a cylindrical
surface with rulings parallel to the x-axis. The generating
curve is the sine curve.


16.
y
zeθ
The x-coordinate is missing so you have a cylindrical
surface with rulings parallel to the x-axis. The generating
curve is the exponential curve.


17.
22
zx y
(a) You are viewing the paraboloid from the x-axis:
20, 0, 0
(b) You are viewing the paraboloid from above, but not
on the z-axis:
10, 10, 20
(c) You are viewing the paraboloid from the z-axis:
0, 0, 20
(d) You are viewing the paraboloid from the y-axis:
0, 20, 0

18.
22
4yz
(a) From
10,0,0:

(b) From
0, 10, 0 :

(c) From
10, 10, 10 :


19.
222
1
141
xyz

Ellipsoid
xy-trace:
22
1
14
xy
ellipse
xz-trace:
22
1xz circle
yz-trace:
22
1
41
yz
ellipse

20.
222
1
16 25 25
xyz

Ellipsoid
xy-trace:
22
1
16 25
xy
ellipse
xz-trace:
22
1
16 25
xz
ellipse
yz-trace:
22
25yz circle

21.
22 2
2
22
16 16 4
441
4
xy z
y
xz



Hyperboloid of one sheet
xy-trace:
2
2
41
4
y
x
hyperbola
xz-trace:

22
41xz circle
yz-trace:
2
2
41
4
y
z

hyperbola

x
y
46
6
6
4
2
−4
−4
−6
2
z
x
y
3
3
4
2
1
z
x y
20
15
10
5
1
2
3
4
3
z
y
z
y
z
y
x
3
3
3
z
x
y2
2
2
−2
z
x
y
z
3
4
5
2
1
3
4
5
1
2
3
4
5
2
1
x y
3
2
−2
−3
3
2
3
−3
−2
z

52. 52 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
22.
222
222
81818 2
9941
xyz
yzx



Hyperboloid of one sheet
xy-trace:
22
941yx hyperbola
yz-trace:
22
991yz circle
xz-trace:
22
94 1zx hyperbola


23.
222
41xyz
Hyperboloid of two sheets
xy-trace:
22
41xy hyperbola
yz-trace: none
xz-trace:
22
41xz hyperbola


24.
2
22
1
4
y
zx

Hyperboloid of two sheets
xy-trace: none
xz-trace:
22
1zx hyperbola
yz-trace:
2
2
1
4
y
z
hyperbola

10:z
22
1
936
xy
ellipse


25.
22
0xyz
Elliptic paraboloid
xy-trace:
2
yxπ
xz-trace:

22
0,
point 0, 0, 0
xz
yz-trace:
2
yzπ

1:yπ
22
1xz

26.
22
4zx y
Elliptic paraboloid
xy-trace: point
0, 0, 0
xz-trace:
2
zxπparabola
yz-trace:
2
4zyπ parabola

27.
22
0xyz
Hyperbolic paraboloid
xy-trace: yx
xz-trace:
2
zx
yz-trace:
2
zyπ

1:y
2
1zx

28.
22
3zyx
Hyperbolic paraboloid
xy-trace: yx
xz-trace:
21
3
zxπ
yz-trace:
21
3
zy

29.
2
22
9
y
zx

Elliptic cone
xy-trace: point
0, 0, 0
xz-trace:
zx
yz-trace:
3
y
z
When 1,z
2
2
1
9
y
x
ellipse

30.
222
22
x yz
Elliptic Cone
xy-trace: 2x y
xz-trace: 2x z
yz-trace: point:
0, 0, 0
x y
2
1
2
2
−2
1
z
x
y
3
3
2
−3
z
x
y
5
5
5
z
x
y
1
1
3
−3
−1
z
x
y
10
10
24
20
28
z
x
y
5
5
5
z
x
y2 2
3 3
3
z
x
y3
4
2
1
−3
3
2
1
3
−3
−2
z
x
y
z
3
2
1
4
2
1

53. Section 11.6 Surfaces in Space 53
© 2010 Brooks/Cole, Cengage Learning
31.



22 2
22 2
22
2
22
2
16 9 16 32 36 36 0
16 2 1 9 4 4 16 36 16 36
1619 216 16
12
1
11691xy z xy
xx yy z
xy z
xy z




Ellipsoid with center
1, 2, 0 .

32.


22 2
22 2
22 2
995445440
9 6 9 4 4 9 6 9 81 81
93 2930
xy z xy z
xx yy zz
xy z




Elliptic cone with center
3, 2, 3 .


33. 2coszxπ


34.
22
0.5zx y


35.
22 2
22
7.5
7.5
zx y
zxy





36.
22
22
2
3.25
3.25
3.25
yx z
zyx
zyx





37.
2
22
2
2
2
4
xy
z
yx
z








38.

22
22
ln
z
xye
xyz






39.
10zxy

x
y4
2
1
1
2
2
−2
z
x y
z
10
15
10
−15
5
15
20
25
y
x 2π
3
3
z
x
y
4
1
2
2
−2
1
z
yx
2 2
−2
−2
2
z
x
y
−1
2
3
1
−2
−1
2
1
z
2
x y
4
4
4
z
x y
4
−4−4
−3 −3
4
4
z
y
x
12
88
−8
−4
z

54. 54 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
40.
22
8
x
z
xy

π




41.
222
222
222
22
646 36
64636
32318
1
2318
3xyz
zyx
zyx
zyx






42.
222
22
9 1
82
948 72
9xyz
zxy




43.
22
22
22
2
2
22
1zxy
z
xy
xy
π




44.
2
2
4
4
0, 0, 0zx
yx
xyz

πππ



45.
22
1
2
0xy
xz
z

π



46.
22
4
2
0zxy
yz
z
π
π


47.
2
22
xzry

and 2;zry y so,
22
4.xzy

48.
2
22
xzry

and 3;zry yππ so,
22 2
9.xzy

49.
2
22
xyrz

and ;
2
z
yrz
ππ so,

2
22 2 2 2
,4 4 .
4
z
xyxyz

50.
2
22
yz rx

and
21
2
4;zrx x so,

22 2 1
4
4,yz x
222
444.xyz
yx
2 2
2
4 4
4
z
x
6
6
4
2
2
−6
−6
−4
−2
z
y
yx
10
10
20
10
20
20
z
x
y
1
2
2
32
−2
−2
z
yx 4 4
5
3
z
x
y
3
2
4
2
3
3
z
x
y
3
−3
3
3
z

55. Section 11.6 Surfaces in Space 55
© 2010 Brooks/Cole, Cengage Learning
51.
2
22
yz rx

and
2
;
yrx x
ππ so,
2
22 22
2
24
,.
yz yz
x x






52.
2
2
xyrz
)


and ;
z
yrz eππ so,
22 2
.
z
xye

53.
22
2
22
20
2xy z
xyz

Equation of generating curve: 2yzπ or 2x zπ

54.
22 2
cos
xzy
Equation of generating curve: cos or cosx yz yππ

55. Let C be a curve in a plane and let L be a line not in a
parallel plane. The set of all lines parallel to
L and
intersecting
C is called a cylinder.Cis called the
generating curve of the cylinder, and the parallel lines
are called rulings.

56. The trace of a surface is the intersection of the surface
with a plane. You find a trace by setting one variable
equal to a constant, such as 0
xπor 2.zπ

57. See pages 814 and 815.

58. In the xz-plane,
2
zxπis a parabola.
In three-space,
2
zxπis a cylinder.

59.
4
34
4
2
0
0
4 218
24 2
34 3xx
Vxxxdx




*



60.
+,
0
2
0
2sin
2sin cos 2Vyydy
yy y



π
*



61.
22
24
xy
z

(a) When
2zπwe have
22
2,
24
x y
or
22
1
48
xy

Major axis: 28 42π
Minor axis: 24 4π

2222
,4,2cabc c
Foci:
0, 2, 2
(b) When 8
zπwe have
22
8,
24
xy
or
22
1.
16 32
x y

Major axis: 232 82π
Minor axis: 216 8π

2
32 16 16, 4cc
Foci:
0, 4, 8

62.
22
24
xy
z

(a) When 4
yπyou have
2
4,
2
x
z


21
44.
2
zx





Focus:
9
0, 4,
2




(b) When 2
xπyou have


2
2
2,42 .
4
y
zzy

Focus:
2, 0, 3

63. If ,,
xyzis on the surface, then


22
22
2222
22
22
44 44
8yxyz
yy xyy z
xz y



Elliptic paraboloid
Traces parallel to
xz-plane are circles.

64. If ,,
xyzis on the surface, then


2
222
2222
22
22
4
816
816 2
88zxy z
zxyz z
xy
zx y z


Elliptic paraboloid shifted up 2 units. Traces parallel to
xy-plane are circles.
x
1234
4
3
2
1
px()
hx()
z
y
2
ππ
0.5
1.0
z

56. 56 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
65.
222
222
1
3963 3963 3950
xyz



66. (a)

2
22
2
22
21
220
xy rz
z
xy z






(b)
2
2
0
2
3
0
2
4
23
0 1
23 1
2
1
22
2
2 4 12.6 cm
8
Vxxdx
xxdx
x
x














*
*


(c)
2
2
12
2
3
12
2
4
2
12
3 1
23 1
2
1
22
2
2
8
31 225
4 11.04 cm
64 64
Vxxdx
xxdx
x
x
















*
*



67.
22
22
,
yx
z z bx ay
ba


22
22
42 24
22 2 2
22
22
22
22
22
11
44
22
22
yx
bx ay
ba
ab ab
xabx yaby
ab
ab ab
xy
ab
babab
yx
a







π



Letting ,
xatπyou obtain the two intersecting lines
,xatπ ,ybt 0zπand ,xatπ
222
,2 .ybtabz abtab

68. Equating twice the first equation with the second equation:

222 222
2644826432
48 32
34 6,a planexyzy xyzx
yx
xy



69. True. A sphere is a special case of an ellipsoid (centered
at origin, for example)

222
222
1
xyz
abc

having .
abcππ

70. False. For example, the surface
22 2 y
xze

can be
formed by revolving the graph of
y
xe

π about the
y-axis, as the graph of
y
ze

πabout the y-axis.

71. False. The trace 2xπof the ellipsoid
22
2
1
49
xy
z
is the point 2, 0, 0 .

72. False. Traces perpendicular to the axis are ellipses.

73. The Klein bottle does not have both an “inside” and an
“outside.” It is formed by inserting the small open end
through the side of the bottle and making it contiguous
with the top of the bottle.
x
y
4000
4000
4000
z
x
y
1
3
−2
2
2
3
4
z
x
3
2
1
123
y
x
3
2
1
123
y

57. Section 11.7 Cylindrical and Spherical Coordinates 57
© 2010 Brooks/Cole, Cengage Learning
Section 11.7 Cylindrical and Spherical Coordinates
1. 7, 0, 5 , cylindrical
cos 7 cos 0 7
sin 7 sin 0 0
5
xr
yr
z




7, 0, 5 , rectangular

2. 2, , 4 , cylindrical


cos 2 cos 2
sin 2 sin 0
4xr
yr
z




2, 0, 4 , rectangular

3.
3, , 1 ,
4



cylindrical

32
3cos
42
32
3sin
42
1
x
y
z






3232
,,1,
22



rectangular

4.
6, , 2 ,
4




cylindrical

6cos 3 2
4
6sin 3 2
4
2x
y
z












32,32,2, rectangular

5.
7
4, , 3 ,
6


cylindrical

7
4cos 2 3
6
7
4sin 2
6
3
x
y
z






23,2,3, rectangular

6.
4
0.5, , 8 ,
3



cylindrical

14 1
cos
234
14 3
sin
23 4
8
x
y
z





13
,,8,
44



rectangular

7. 0, 5, 1 ,rectangular


22
055
5
arctan
02
1r
z






5, , 1 ,
2



cylindrical

8.
22,22,4, rectangular


22
22 22 4
arctan 1
4
4
r
z






4, , 4 ,
4




cylindrical

9. 2, 2, 4 , rectangular


2
2
2222r


arctan 1
4




4z

22, ,4,
4




cylindrical

10. 3, 3, 7 , rectangular



2
2
33 1832
arctan 1
4
7r
z






32, ,7,
4




cylindrical

58. 58 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
11. 1, 3 , 4 ,rectangular

2
2
132
arctan 3
3
4
r
z



ππ
π

2, , 4 ,
3



cylindrical

12.
23,2,6, rectangular
12 4 4
15
arctan
63
6
r
z







π

4, , 6 ,
6




cylindrical

13. 4zπis the equation in cylindrical coordinates.
(plane)

14. 9, rectangular equation
cos 9
9 sec , cylindrical equationx
r
r


π
π
π

15.
222
22
17, rectangular equation
17, cylindrical equationxyz
rz


16.
22
2
11, rectangular equation
11, cylindrical equationzx y
zr


17.
2
,yxπ rectangular equation


2
2
sin cos
sin cos
sec tan , cylindrical equation
rr
r
r

π
π
π"

18.
22
2
8 , rectangular equation
8cos
8 cos , cylindrical equationxy x
rr
r



π
π

19.

22
2
2
22 2
10 , rectangular equation
sin 10
sin 10, cylindrical equationyz
rz
rz






20.
222
22
3 0, rectangular equation
3 0, cylindrical equationxyz z
rz z


21.
22
22
3
3
9
r
xy
xy
π




22. 2zπ
Same


23.
6
tan
6
1
3
3
30
y
x
y
x
x y
xy



π
π
π
π




24.
22
2
22
2
2
0
4
z
r
z
xy
z
xy
π




x
y3
4
4
3
3
−3
2
z
x
y
3
3
2
1
3
z
x
y
2
1
2
1
2
−2
−2
z
x
y2
2
−2
−2
4
z

59. Section 11.7 Cylindrical and Spherical Coordinates 59
© 2010 Brooks/Cole, Cengage Learning
25.
22
222
5
5
rz
xyz





26.
22
2
coszr
zxπ
π


27.

2
22
22
2
2
2sin
2sin
2
20
11
r
rr
xy y
xy y
xy
π
π






28.

2
22
22
2
2
2cos
2cos
2
20
11
r
rr
xy x
xy x
xy
π
π






29. 4, 0, 0 , rectangular

222
400 4-
tan 0 0
y
x

arccos 0
2

.
ππ

4, 0, , spherical
2





30. 4, 0, 0 , rectangular



2
22
4004
tan 0 0
arccos arccos 0
2
4, 0, , spherical
2
y
x
z
e-






πππ







31.
2, 2 3, 4 , rectangular



2
2
2
223442-

23
tan 3
2
2
3
y
x




π


1
arccos
42
2
4 2, , , spherical
34
.

ππ




32.
2, 2, 4 2 ,rectangular

2
22
22 42 210-

tan 1
4
y
x




2
arccos
5
2
2 10, , arccos , spherical
4 5
.
π




33.
3,1, 2 3 ,rectangular
3112 4-

1
tan
3
6
y
x


ππ
π


3
arccos
26
.
ππ

4, , ,
66



spherical
x
y
3
3
3
−3
z
x
y
z
1
2
3
9
6
5
4
3
2
1
x
y
1
2 2
1
2
−2
−2
−1
z
x
y
3
2
−2
2
z

60. 60 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
34. 1, 2, 1 , rectangular


2
22
121 6
tan 2 arctan 2
1
arccos
6
y
x
-

.







1
6, arctan 2 , arccos ,
6





spherical

35.
4, , ,
64



spherical


4sin cos 6
46
4sin sin 2
46
4cos 2 2
4
6, 2, 2 2 , rectangularx
y
z







36.

3
12, , , spherical
49
3
12 sin cos 2.902
94
3
12 sin sin 2.902
94
12 cos 11.276
9
2.902, 2.902, 11.276 , rectangular
x
y
z












37.

12, , 0 , spherical
4
12 sin 0 cos 0
4
12 sin 0 sin 0
4
12 cos 0 12
0, 0, 12 , rectangular
x
y
z

















38.

9, , , spherical
4
9sin cos 0
4
9sin sin 0
4
9cos 9
0, 0, 9 , rectangular
x
y
z















39.
3
5, , , spherical
44
35
5sin cos
442
35
5sin sin
442
352
5cos
42
55 5 2
, , , rectangular
22 2
x
y
z














40.

6, , , spherical
2
6sin cos 6
2
6sin sin 0
2
6cos 0
2
6, 0, 0 , rectangular
x
y
z















41. 2,yrectangular equation
sin sin 2
2 csc csc , spherical equation
-.
-.



42. 6, rectangular equationz
cos 6
6 sec , spherical equation
-.
-.



43.
222
2
49,
49
7,xyz
-
-



rectangular equation
spherical equation


44.

22 2
222 2
222
2
3 0, rectangular equation
4
4cos
14cos
1
cos
2
, cone spherical equation
3xy z
xyz z
--.
.
.

.

61. Section 11.7 Cylindrical and Spherical Coordinates 61
© 2010 Brooks/Cole, Cengage Learning
45.
22
16,xy rectangular equation


22 2 22 2
22 2 2
22
sin sin sin cos 16
sin sin cos 16
sin 16
sin 4
4 csc , spherical equation-.-.
-.
-.
-.
-.

π
π
π


46. 13,xπrectangular equation
sin cos 13
13 csc sec , spherical equation
-.
-.π
π


47.
22 2
22 2 22 2 2 2
22 2 2 2 2
22 2 2
2
2
2
2 , rectangular equation
sin cos sin sin 2 cos
sin cos sin 2 cos
sin 2 cos
sin
2
cos
tan 2
tan 2, spherical equation
xy z
-.-.-.
-. -.
-.-
.
.
.
.




π
π
π



48.
222
2
9 0, rectangular equation
9cos 0
9 cos , spherical equationxyz z
--.
-.


π


49.
222
5
25xyz
-π



50.
3
4

π

tan
1
0
y
x
y
x
xy
π





51.
6

.
π
222
222
2
222
222
cos
3
2
3
4
33 0,0
z
xyz
z
xyz
z
xyz
xyz z
.π

π

π

/

x
y3
−3
−3
3
z
x
y
2
1 1
2
2
−2
−1
− 1
−1
− 2
z
x
y
6
5
5
6
5
6
−6
z

62. 62 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
52.
2

.
π
222
222
cos
0
0
z
xyz
z
xyz
z
.π

π

π

xy-plane


53. 4cos
- .π


222
222
222
2
22 4
40
24,0z
xyz
xyz
xyz z
xy z z



/



54. 2sec
cos 2
- .
-.π
π


2zπ


55.
22
22
csc
sin 1
1
1
xy
xy
- .
-.π
π




56.
4csc sec
4
sin cos
sin cos 4
- ..
.
-.
π
π
π

4
xπ


57.
4, , 0 , cylindrical
4





22
40 4
4
arccos 0
2
4, , , spherical
42-



.

π
ππ





58.
3, , 0 , cylindrical
4






22
30 3
4
0
arccos
92-



.


ππ




3, , ,
42




spherical

59.
4, , 4 ,
2



cylindrical

22
44 42
2
4
arccos
442
4 2, , , spherical
24-



.

π

ππ





x
y3
3
−3
−3
−3
−2
3
2
z
x
y
3
3
2
1
2
5
4
3
2
−2
−3
z
x
y
3
3
2
1
3
z
x y
1
2
1
2
1
2
− 2
−2 −2
− 1
z
x
y6
6
4
6
4
z

63. Section 11.7 Cylindrical and Spherical Coordinates 63
© 2010 Brooks/Cole, Cengage Learning
60.
2
2, , 2 ,
3



cylindrical


2
2
2222
2
3
13
arccos
42
23
2 2, , , spherical
34-



.










61.
4, , 6 , cylindrical
66






22
46 213
6
3
arccos
13
3
2 13, , arccos , spherical
6 13-


.









62.
4, , 4 ,
3




cylindrical


2
2
4442
3
1
arccos
42
4 2, , , spherical
34-



.








63. 12, , 5 ,cylindrical

22
12 5 13
5
arccos
13
5
13, , arccos , spherical
13-

.







64.
22
4, , 3 , cylindrical
2
43 5
2
3
arccos
5
3
5, , arccos , spherical
25

-


.











65.
10, , , spherical
62
10 sin 10
2
6
10 cos 0
2
10, , 0 , cylindrical
6
r
z
















66.
4, , , spherical
18 2
4sin 4
2
18
4cos 0
2
4, ,0 ,cylindrical
18
r
z
















67.

36, , , spherical
2
sin 36 sin 36
2
cos 36 cos 0
2
36, , 0 , cylindrical
r
z



-.


-.








68.
18, , , spherical
33
sin 18 sin 9
3
3
cos 18 cos 9 3
3
9, , 9 3 , cylindrical
3
r
z


-.



-.











69.
6, , , spherical
63
6sin 3 3
3
6
6cos 3
3
3 3, , 3 , cylindrical
6
r
z

64. 64 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
70.
5
5, , , spherical
6
5sin 0
5
6
5cos 5
5
0, , 5 , cylindrical
6
r
z


















71.
7
8, , , spherical
66
8sin 4
6
7
6
83
8cos
62
7
4, , 4 3 , cylindrical
6
r
z















72.
3
7, , , spherical
44
372
7sin
42
4
372
7cos
42
72 72
, , , cylindrical
24 2
r
z















Rectangular
Cylindrical Spherical

73. 4, 6, 3 7.211, 0.983, 3 7.810, 0.983, 1.177

74. 6, 2, 3 6.325, 0.322, 3 7.000, 0.322, 2.014

75. 4.698, 1.710, 8
5, , 8
9




9.434, 0.349, 0.559

76. 7.317, 6.816, 6 10, 0.75, 6 11.662, 0.750, 1.030

77. 7.071, 12.247, 14.142 14.142, 2.094, 14.142
2
20, ,
34




78. 6.115, 1.561, 4.052 6.311, 0.25, 5.052 7.5, 0.25, 1

79. 3, 2, 2 3.606, 0.588, 2 4.123, 0.588, 1.064

80.
32,32,3 6, 0.785, 3 6.708, 0.785, 2.034

81.
54 3
,,
23 2





2.833, 0.490, 1.5 3.206, 0.490, 2.058

82. 0, 5, 4 5, 1.571, 4 6.403, 1.571, 0.896

83. 3.536, 3.536, 5
3
5, , 5
4




7.071, 2.356, 2.356

84. 1.732, 1, 3
11
2, , 3
6




3.606, 2.618, 0.588

5
use the cylindrical coordinate 2, , 3
6
Note:

65. Section 11.7 Cylindrical and Spherical Coordinates 65
© 2010 Brooks/Cole, Cengage Learning
Rectangular Cylindrical Spherical

85. 2.804, 2.095, 6 3.5, 2.5, 6 6.946, 5.642, 0.528

Use the cylindrical coordinates 3.5, 5.642, 6
Note:

86. 2.207, 7.949, 4 8.25, 1.3, 4 9.169, 1.3, 2.022

87. 1.837, 1.837, 1.5 2.598, 2.356, 1.5
3
3, ,
43




88. 0, 0, 8 0, 0.524, 8
8, ,
6






89. 5r
Cylinder
Matches graph (d)

90.
4



Plane
Matches graph (e)

91. 5
-
Sphere
Matches graph (c)

92.
4

.

Cone
Matches graph (a)

93.
222
,rzxyz
Paraboloid
Matches graph (f )

94. 4sec , cos 4z-.-.
Plane
Matches graph (b)

95. Rectangular to cylindrical:
222
tan
rxy
y
x
zz




Cylindrical to rectangular: cos
sin
xr
yr
zz






96. cis a half-plane because of the restriction 0.r/

97. Rectangular to spherical:
2222
222
tan
arccos
xyz
y
x
z
xyz
-

.







Spherical to rectangular: sin cos
sin sin
cos
x
y
z
-.
-.
-.





98. (a) ra Cylinder with z-axis symmetry
b
Plane perpendicular to xy-plane
zc Plane parallel to xy-plane
(b)
a
- Sphere
b
Vertical half-plane
c
. Half-cone

99.
222
25xyz
(a)
22
25rz
(b)
2
25 5--

100.
22 2
4
xyz
(a)
22
42rz rz
(b)

22 2 22 2 2 2
4 sin cos sin sin cos
-.- .- .

22
2
4sin cos ,
1
tan ,
4
11
tan arctan
22..
.
..




101.
222
20xyz z
(a)

2
22 2
20 1 1rz z r z
(b)

2
2cos 0
2cos 0
2cos--.
-- .
- .

66. 66 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
102.
22
xyz
(a)
2
rzπ
(b)
22
2
2
sin cos
sin cos
cos
sin
csc cot-.-.
-. .
.
-
.
- ..
π
π
π
π


103.
22
4
xyy
(a)
2
4sin , 4sinrr rππ
(b)

22
sin 4 sin sin
sin sin 4 sin 0
4sin
sin
4sin csc
-.-.
-.-.

-
.
- .
π

π
π


104.
22
36xy
(a)
2
36 6rr
(b)
22 2 22 2
22
sin cos sin sin 36
sin 36
6csc-.-.
-.
- .

π
π


105.
22
9xy
(a)
22 22
2
22
cos sin 9
9
cos sinrr
r

π

(b)
22 2 22 2
22
22
2
2
22
sin cos sin sin 9
9
sin
cos sin
9csc
cos sin-.-.
-.

.
-

π

π


106. 4yπ
(a)
sin 4 4 cscrr
(b) sin sin 4,
4csc csc
-.
- .
π
π


107.
0
2
02
04
r
z








108.
22
03
0cos
r
zr








109. 02
0ra
rza





110.
22
02
24
68
r
zrr





111.
02
0
6
0sec
a


.
- .





112.
02
42
01

.
-



x
y2
3
1
2
3
5
3
2
z
y
x
4
−4
4
4
3
z
x
y
a a
−a
− a
a
z
y
x
5
−5
5
4
3
z
x
y
30 °
z
a
y
x
2
−2
−2
2
2
z

67. Section 11.7 Cylindrical and Spherical Coordinates 67
© 2010 Brooks/Cole, Cengage Learning
113. 0
2
0
2
02



.
-






114.
0
0
2
13

.
-




115. Rectangular
010
010
010
x
y
z






116. Cylindrical:

0.75 1.25
08 r
z



117. Spherical
46
-


118. Cylindrical

22
1
2
3
02
99
r
rz r







119. Cylindrical coordinates:

22
9,
3cos ,0
rz
r






120. Spherical coordinates:

2
3
0
4
-
-

.
/




121. False.
22 2
rz x y z is a cone.

122. True. They both represent spheres of radius 2 centered at
the origin.
x
y
z
2
2
2
4
1
2
3
−1
−2
−3
3
2
y
x
z
y
x
10
10
10
z
x y
8
22
−2 −2
z
y
x
8
−8
−8
8
z
y
x
4
−4
−4
4
z
2
2
11
3
2
x
y
z
−1
−2
x
y
3
2
3
4
3
−2−3
z

68. 68 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
123. False. , , 0, 0, 1rz and ,, 0,,1rz
represent the same point
, , 0, 0, 1 .xyz

124. True except for the origin .

125. sin , 1
sin
1
zr
yy
zy
r


The curve of intersection is the ellipse formed by the
intersection of the plane
zy and the cylinder 1.r

126. 2 sec cos 2 2 plane
4 sphere
z
- .-.
-


The intersection of the plane and the sphere is a circle.
Review Exercises for Chapter 11
1. 1, 2 , 4, 1 , 5, 4PQR
(a)
41,12 3,1
5 1,4 2 4,2
PQ
PRu
v




(b)
3uij
(c)
22
42 2025v
(d) 223,14,210,0uv

2. P2,1,Q5,1,R2,4
(a)
7, 0
4, 5u
vPQ
PR




(b) 7
ui
(c)
22
45 41v
(d) 18, 5227,04,5uv

3.


cos sin
8 cos 60 sin 60
13
84434,43
22vv i j
ij
iji j






4.
cos sin
11
cos 225 sin 225
22
22 22
,
44 44vv i+v j
ij
ij=



5.
0, 4, 5: 5, 4, 0zyx
6.
0, 7: 0, 7, 0 xz y
7.
Looking down from the positive x-axis towards the

yz-plane, the point is either in the first quadrant
0, 0yz or in the third quadrant 0, 0 .yz
The
x-coordinate can be any number.
8. Looking towards the xy-plane from the positive z-axis.
The point is either in the second quadrant
0, 0xy
or in the fourth quadrant
0, 0 .xy The

z-coordinate can be any number.

9.
2
222
15
326
2
xyz






10. Center:

04 06 40
,, 2,3,2
222





Radius:



22 2
222
20 30 24 494 17
23217xyz




11.

222
23
2
44 69 449
23 9
xx yy z
xyz


Center:
2, 3, 0
Radius: 3

12.


22
2
222
10 25 6 9
4 4 34 25 9 4
5324
xx yy
zz
xyz




Center:5, 3, 2
Radius:
2

x
y5
4
3
2
3
4
4
5
6
6
z
x
y2
8
6
4
2
6
4
z

69. Review Exercises for Chapter 11 69
© 2010 Brooks/Cole, Cengage Learning
13. (a), (d)

(b)
42,4 1,73 2,5,10v
(c) 2510vijk
14.
(a), (d)
(b)
36,32,80 3,5,8v
(c) 358vi jk
15. 1 3, 6 4, 9 1 4, 2, 10v
5 3, 3 4, 6 1 2, 1, 5 w

Because 2,wv the points lie in a straight line.
16.
85,54,57 3,1,2 v
11 5, 6 4, 3 7 6, 10, 4 w
Because vand ware not parallel, the points do not lie in
a straight line.
17. Unit vector:
2,3,5 235
,,
38 38 38 38
u
u

18.
6, 3, 28482416
86,3,2,,
777749



19.
5, 0, 0 , 4, 4, 0 , 2, 0, 6PQ R
(a) 1, 4, 0PQu


3, 0, 6PRv=


(b)
1 3 40 06 3" uv
(c) 9 36 45"
vv

20.
2, 1, 3 , 0, 5, 1 , 5, 5, 0PQR
(a) 2, 6, 2PQu


3, 6, 3PR v


(b)
23 66 2 3 36" uv
(c) 9369 54" vv
21.
7, 2, 3 , 1, 4, 5uv

Because 0," uv the vectors are orthogonal.

22.
4, 3, 6 , 16, 12, 24 uv

Because 4,vu the vectors are parallel.

23.
+,
3352
5cos sin
442



uijij

22
2cos sin 3
33



vijij


52
13
2
" uv
52uv



5221 3
26
cos
52 4

"

uv
uv

26 32
arcos 15 or, or 15
44312

"



24.
623, 5 ui
jkv i j

610 4
cos
49 26 7 26uv
uv

"


83.6

25.
10, 5, 15 , 2, 1, 3 uv

5uvu is parallel tovand in the opposite
direction.

26.

1, 0, 3
2, 2, 1
1
10
3
1
cos
310
u
v
uv
u
v
uv
uv



"


"


83.9

x
y
3
2
1
54
5
3
1
2
3
−2
−9
−10
−8
z
(2, −1, 3)
(4, 4, −7)
(2, 5, −10)
x
y
1
3
4
2
1
5
6
1
6
5
7
8
z
(3, −3, 8)
(−3, −5, 8)
(6, 2, 0)
v

70. 70 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
27. There are many correct answers.
For example:6, 5, 0 .v
28. cos 75 8 cos 30
300 3 ft-lbWPQ PQ "

FF


In Exercises 29–38,
u 3,2,1, v 2,4,3,
w1,2,2.
29.

2
2
33 2 2 1 1
14 14uu
u"

30.
11
cos
14 29
11
arccos 56.9
14 29
uv
uv


"






31.
2
proj
5
3, 2, 1
14
15 10 5
,,
14 14 14
15 5 5
,,
14 7 14
u
uw
wu
u

"







32.
Work 342 5uw "
33.
243 2
122
5
ijk
nvw ij
n


'







1
2,
5
n
ij
n
unit vector or
1
2
5
ij

34.
321 10118
243
ijk
uv i j k


'






243 10118
321
ijk
vu i j k


'





So,
''uv vu
35.

V"'uvw 3, 2, 1 2, 1, 0 4 4 "
36.

3, 2, 1 1, 2, 1
321 4 4 4
121
uvw
ijk
ijk
' '









321 10118
243
ijk
uv i j k


'





321 6 7 4
122
ijk
uw i j k


'





444'' 'uv uw i j k u vw
37. Area parallelogram

2
22
10, 11, 8 10 11 8
285
'
uv (See Exercises 34, 36)
38. Area triangle

2211 5
2 1 See Exercise 33
22 2
vw'
39.
cos 20 sin 20cFjk

2
00 2 2cos20
0 cos 20 sin 20
200 2 cos 20
100
cos 20
k
ij k
F= i
F
PQ
PQ c
cc
PQ c
c

'

'







2
100
cos 20 sin 20 100 tan 20
cos 20
100 1 tan 20 100 sec 20 106.4 lb
Fjkjk
F




x
y
70°
PQ
F
2 ft
z

71. Review Exercises for Chapter 11 71
© 2010 Brooks/Cole, Cengage Learning
40.
210
021 25 10
012
uvw
V


""





41.
93,110,62 6,11,4v
(a) Parametric equations:
36, 11, 24
x ty tz t
(b) Symmetric equations:
32
6114
xyz


42.
81,104,53 9,6,2v
(a) Parametric equations:
19, 46, 32
x ty tz t
(b) Symmetric equations:
143
962
xy z


43.
vj
(a)
1, 2 , 3xy tz
(b) None
(c)
44. Direction numbers: 1, 1, 1
(a)
1, 2, 3
x ty tz t
(b) 123
xy z
(c)
45. 337 4, 2 3xyz xyz

Solving simultaneously, you have 1.zSubstituting
1zinto the second equation, you have 1.yx
Substituting for
xin this equation you obtain two points
on the line of intersection,
0, 1, 1 , 1, 0, 1 . The
direction vector of the line of intersection is
vij .
(a)
,1,1xty tz
(b)
1, 1xy z
(c)
46.
251 211113
314
ijk
uv i j k
'

Direction numbers: 21, 11, 13
(a)
21 , 1 11 , 4 13
x ty tz t
(b)
14
21 11 13xy z


(c)
47.
+,
3, 4, 2 , 3, 4, 1 , 1, 1, 2
0, 8, 1 , 4, 5, 4
08 1 27 4 32
45 4
ij k
nijk
PQR
PQ PR
PQ PR

'





27 34 432 2 0
27 4 32 33xy z
xy z



x y
2
−2
−2
−4
−4
4
2
−4
4
4
z
x
y
3
4
5
6
4
2
3
1
5
6
1
6
5
4
3
z
y
x
1
2
3
4
2
3
4
324−2
−3
−4
−3
−2
−4
−3−4
z
x y
2
−2
−2
−4
−4
4
2
−4
4
4
z
x y
−2
−4
−4
−2
2
−2
2
4
4
z

72. 72 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
48. 3nijk

3213110
380xyz
xyz



49. The two lines are parallel as they have the same direction
numbers,
2, 1, 1. Therefore, a vector parallel to the
plane is
2.vi+
jk A point on the first line
is
1, 0, 1and a point on the second line is1, 1, 2 . The
vector 2 3uijk connecting these two points is
also parallel to the plane. Therefore, a normal to the
plane is


211
213
24 2 2.
ijk
vu
ij ij
'



Equation of the plane:
12 0
21xy
xy



50. Let
52,12,31 3,3,2v be the direction
vector for the line through the two points. Let
2, 1, 1n be the normal vector to the plane. Then
33 2 5,7,3
21 1
ijk
vn
'

is the normal to the unknown plane.

5571330
573270xyz
xyz



51.
1, 0, 2Q point
236 6xyz
A point
Pon the plane is3, 0, 0 .

2, 0, 2PQ


2, 3, 6n normal to plane

8
7
n
n
PQ
D
"



52.
3, 2, 4Q point
25 10xyz
A point
Pon the plane is5, 0, 0 .

2, 2, 4PQ


2, 5, 1n normal to plane
10 30
330
n
n
PQ
D
"



53. The normal vectors to the planes are the same,

5, 3, 1 .n
Choose a point in the first plane
0, 0, 2 .P Choose a
point in the second plane,
0, 0, 3 .Q

0, 0, 5
5 535
735 35
n
n
PQ
PQ
D

"




54.
5, 1, 3Q point

1, 2, 1u direction vector

1, 3, 5P point on line

6, 2, 2PQ


622 2,8,14
12 1
ijk
u
PQ'




264
211
6
PQ
Du
u

'


55. 236xyz
Plane
Intercepts:
6, 0, 0 , 0, 3, 0 , 0, 0, 2 ,

x
y
3
2
1
1
2
4
3
−2
(−2, 3, 1)
−2
−1
z
x y
2
−2
−4
−2
4
2
−2
−4
2
4
4
z
x
y
2
4
22
4
( 2, −2, 1)
( 5, 1, 3)
z
x
y
6
3
3
(0, 0, 2)
(6, 0, 0)
(0, 3, 0)
z

73. Review Exercises for Chapter 11 73
© 2010 Brooks/Cole, Cengage Learning
56.
2
yz
Because the x-coordinate is missing, you have a
cylindrical surface with rulings parallel to the x-axis.
The generating curve is a parabola in the yz-coordinate
plane.

57.
1
2
yz

Plane with rulings parallel to the x-axis.

58. cosyz
Because the x-coordinate is missing, you have a
cylindrical surface with rulings parallel to the x-axis.
The generating curve is cos .yz

59.
22
2
1
16 9
xy
z

Ellipsoid
xy-trace:
22
1
16 9
xy

xz-trace:
2
2
1
16
x
z
yz-trace:
2
2
1
9
y
z

60.
222
16 16 9 0xyz
Cone
xy-trace: point
0, 0, 0
xz-trace:
4
3
x
z
yz-trace:
4
3
y
z


22
4, 9zxy

61.
22
2
22
2
1
16 9
1
916
xy
z
yx
z



Hyperboloid of two sheets
xy-trace:
22
1
916
yx

xz-trace: None
yz-trace:
2
2
1
9
y
z

62.
22 2
1
25 4 100
xy z

Hyperboloid of one sheet
xy-trace:
22
1
25 4
xy

xz-trace:
22
1
25 100
xz

yz-trace:
22
1
4 100
yz


y
x
4
2
1
2
3
z
x
y
6
2
2
z
y
x
2
2
4
−2
z
x
y
5
4
2
−2
−4
z
x
y33
2
−3
−3
4
z
x
y
5 5
2
−2
z
x
y
5
−5
12
z

74. 74 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
63.
22
4.xz
Cylinder of radius 2 about y-axis

64.
22
16.yz
Cylinder of radius 4 about x-axis

65. Let

2yrx xand revolve the curve about the
x-axis.
66.
222
22 2
23
32
xyz y
xzyy



Let
22
32
x yy (Trace in xy-plane)
Then
2
32x yy is a generating curve. Revolve
the curve about the y-axis.

67.
2
2zyrevolved about y-axis


2
22
22
2
2
2
zy xzry y
xz y






68. 231xz revolved about the x-axis

12
3
x
z




2
2
22
12
,
3
x
yz rx



Cone

69.
22,22,2, rectangular
(a)

22
22 22 4,
3
arctan 1 , 2,
4
3
4, , 2 , cylindrical
4
r
z








(b)

22
2
22 22 2 25,
321
, arccos arccos ,
4 25 5
35
2 5, , arccos , spherical
45-

.








70.
333 3
,, ,
44 2




rectangular
(a)
2
2
33 3
,
442
arctan 3 ,
3
33 3 33
, , , , cylindrical
2222
r
z












(b)
22
2
3333 30
,,
44 2 2 3
330 3
arccos , , , arccos , spherical
2310 10
-

.








71.
100, , 50 , cylindrical
6






22
100 50 50 5
6
50 1
arccos arccos 63.4 or 1.107
50 5 5
50 5, , 63.4 , sperical or 50 5, , 1.1071
66-


.











72.
5
81, , 27 3 , cylindrical
6
6561 2187 54 3
5
6
27 3 1
arccos arccos
2354 3
5
54 3, , , spherical
63
-



.















x
y
z
2
2
−2
x
y
z
−2
2
2
−2

75. Review Exercises for Chapter 11 75
© 2010 Brooks/Cole, Cengage Learning
73.
2
2
3
25, , , spherical
44
3252
25 sin
42
4
3252
cos 25 cos
42
25 2 25 2
, , , cylindrical
242
rr
z




-.















74.
2
2
2
12, , , spherical
23
2
12 sin 6 3
3
2
2
cos 12 cos 6
3
6 3, , 6 , cylindrical
2
rr
z




-.



















75.
22
2
xyz
(a) Cylindrical:

22 22 2
cos sin 2 cos 2 2rr zr z
(b) Spherical:

22 2 22 2
2
2
sin cos sin sin 2 cos
sin cos 2 2 cos 0
2sec2 cos csc
-.- .-.
-. .
-..



76.
222
16xyz
(a) Cylindrical:
22
16rz
(b) Spherical: 4
-

77. 5 cos , cylindrical equationr

2
5cosrr

22
5
xyx

22 25 25
5
44
xx y

22
2
55
, rectangular equation
22
xy







78. 4, cylindrical equationz
4, rectangular equationz


79.
, spherical coordinates
4




tan tan 1
4




1
y
x


, 0, rectangular coordinates, half-planeyxx/


80. 3 cos , spherical coordinates-

222
222
222 3
30z
xyz
xyz
xyz z





22
22
33
,
22
xy z




rectangular coordinates, sphere


2
3
3
x
y
z
x
y
z
−2
4
2
2
34
4
3
2
1
−3
3
x
y
z
x y
2
2 1
1
4
3
−2
z

76. 76 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 11
1.


abc 0
babc 0
ba bc 0

'
''

sin
sin
ab bc
bc bc
ab ab
A
C
''
'
'

Then,

sin
sin
.
bc
aabc
ab
abc
c
A
C
'

'




The other case,
sin sin
ab
A B
is similar.

2.
4
0
1
xfxtdt*

(a)
(b)



4
1
01tan
4
122
,
222
uij
fx x
f







(c)
22
,
22

(d) The line is:,.yxxtyt


3. Label the figure as indicated.
From the figure, you see that

11 11
and .
22 22
SP RQ SR PQab ab


Because and ,SP RQ SR PQ



PSRQ is a parallelogram.

4. Label the figure as indicated.


22
0, because
ab
ba
ab ba b a
PR
SQ


"




ab in a rhombus.

5. (a) 0, 1, 1u is the direction vector of the line
determined by
1Pand
2.P

1
2, 0, 1 0, 1, 1
2
1, 2, 2 332
222
u
u
PQ
D
'

'






(b) The shortest distance to the line segment
is
1 2, 0, 1 5.PQ
x
y
4
3
2
4
1
2
3
3
4
5
6
z
P
1
P
2
Q
−4−2
−2
2
4
−4
24
y
x
a
a
bS
P
Q
R
1
2
b
1 2
+
a
1
2
b
1 2
−
a
b
S
P Q
R
b
c a

77. Problem Solving for Chapter 11 77
© 2010 Brooks/Cole, Cengage Learning
6. ′″′″ 00nnPP PP(
⎜⎜⎜′⎜⎜⎜ ′
Figure is a square.
So,
0PP n
⎜⎜⎜π
⎜ and the points P form a circle of radius
nin the plane with center at
0.P


7. (a)
′″
1
2
21
0
0
1
22z
Vzdz


⎜⎜⎜


*

Note:
1
2
(base)(altitude)
′″
11
1
22
⎜⎜
(b)
′″
′″ ′″
22
22
22
22
: slice at
1
xy
zzc
ab
xy
ca cb



At
,zc⎜figure is ellipse of area

′″′″ .ca cb abc ⎜

22
0
0
22
k
k
abc abk
V abc dc

⎜"⎜ ⎜


*

(c)
′″
11
22
V abk k
⎜⎜ (area of base)(height)

8. (a) ′″
3
22 2 3
0
0
4
22
33
r
r
x
Vrxdxrx r


⎜π⎜π⎜


*

(b) At height 0,
zd

222
222
22 2 22
22 2 2
1
1
xyd
abc
xydcd
ab c c

π



′″
′″
22
22 2 22 2
22
1.
xy
ac d bc d
cc
ππ

Area
′″
′″
′″
22 2 22 2
22
222
ac d bc d ab
cd
ccc
ππ
⎜⎜π


′″
22
2
0
3
2
2
0
2
24
33
c
cab
Vcddd
c
ab d
c d abc
c


⎜π

⎜π⎜

*

9. (a) 2sin
- .⎜
Torus

(b) 2 cos
Sphere
- .⎜


10. (a) 2cosr⎜
Cylinder
(b)
2
222
cos 2zr
zxy⎜
⎜π

Hyperbolic paraboloid

11. From Exercise 66, Section 11.4,

′″′ ″′″ ′″ .uv wz uvzw uvwz'' ''"'"


12.
′″
1
3, 1, 2 1; 4, 3,
2
x tytztQ s
(a) 2, 1, 4u⎜π direction vector for line

′″3, 1, 1P⎜π point on line

1, 2, 1PQ s
⎜⎜⎜′


′″′ ″
12 1
21 4
7625
PQ s
ss
ij k
u
ijk
⎜⎜⎜′
'
π



′″′ ″
22
76225
21u
uPQ ss
D
'
⎜⎜
⎜⎜⎜′


P
0
n + PP
0
P
nn
n − PP
0
x
y
z
3
−3
2
−2
3
x
y
z
3
2
1
−2
−2
−3
3
2
1

78. 78 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
(b)
The minimum is
2.2361 at 1.Ds⎜π
(c) Yes, there are slant asymptotes. Using
,
sx⎜you have
′″ ′″
2
221555
( ) 5 10 110 2 22 1 21 1
2121 21 21
Ds x x x x x x
′″
105
1
21
ys slant asymptotes.

13. (a)
′″cos 0 sin 0uu i j ui
Downward force
w
j⎜π
′″′″′″
′″
′″
cos 90 sin 90
sin cos
sin cos
TT i j
Ti j
0uwT uijT i j





sin
1cosuT
T
⎜
⎜
If
′″ ′″
212
30 , 1 2 and 1 3 2 1.1547 lb and 0.5774 lb
233
uT TT u






(b) From part (a),
tan and sec .uT⎜⎜
Domain: 0 90

(c)
(d)
(e) Both are increasing functions.
(f)
22
lim and lim .
Yes. As increases, both and increase.
T
T
u
u


ππ

⎜! ⎜!
0 10 20 30 40 50 60
T 1 1.0154 1.0642 1.1547 1.3054 1.5557 2
u 0 0.1763 0.3640 0.5774 0.8391 1.1918 1.7321
−11
−4
10
10
060
0
T
⎜⎜u⎜⎜
2.5

79. Problem Solving for Chapter 11 79
© 2010 Brooks/Cole, Cengage Learning
14. (a) The tension Tis the same in each tow line.


6000 cos 20 cos 20 sin 20 sin 20
2 cos 20
6000
3192.5 lb
2 cos 20
iij
i
TT
T
T




(b) As in part (a),
6000 2 cosiT

3000
cos
T


Domain: 0 90

(c)


(d)
(e) As
increases, there is less force applied in the direction of motion.
15. Let ,
$ the angle betweenuandv.Then


sin .
uv vu
uv uv$
''


For
cos , sin , 0u and cos , sin , 0 , 1vuv$$ and
cos sin 0 sin cos cos sin
cos sin 0
ijk
vu k.$$ $ $
'
So,

sin sin cos cos sin .vu$ $ $'
16. (a) Los Angeles:
4000, 118.24 , 55.95 Rio de Janeiro: 4000, 43.23 , 112.90
(b) Los Angeles:



4000 sin 55.95 cos 118.24
4000 sin 55.95 sin 118.24
4000 cos 55.95
, , 1568.2, 2919.7, 2239.7
x
y
z
xyz




Rio de Janeiro:


4000 sin 112.90 cos 43.23
4000 sin 112.90 sin 43.23
4000 cos 112.90
, , 2684.7, 2523.8, 1556.5
x
y
z
xyz





(c)


1568.2 2684.7 2919.7 2523.8 2239.7 1556.5
cos 0.02047
4000 4000
91.17 or 1.59 radiansuv
uv


"



(d)
4000 1.59 6360sr miles
10 20 30 40 50 60
T 3046.3 3192.5 3464.1 3916.2 4667.2 6000.0

090
0
10,000

80. 80 Chapter 11 Vectors and the Geometry of Space
© 2010 Brooks/Cole, Cengage Learning
(e) For Boston and Honolulu:
a. Boston:
4000, 71.06 , 47.64 Honolulu: 4000, 157.86 , 68.69
b. Boston:


4000 sin 47.64 cos 71.06
4000 sin 47.64 sin 71.06
4000 cos 47.64
959.4, 2795.7, 2695.1
x
y
z



Honolulu:

4000 sin 68.69 cos 157.86
4000 sin 68.69 sin 157.86
4000 cos 68.69
3451.7, 1404.4, 1453.7
x
y
z




c.


959.4 3451.7 2795.7 1404.4 2695.1 1453.7
cos 0.28329
4000 4000
73.54 or 1.28 radians
uv
uv


"



d.
5120 milessr 0 12)3

17. From Theorem 11.13 and Theorem 11.7 (6) you have


.
n
n
wuv uvw uvw
uv uv uvPQ
D
"

"' '" "'

'''



18. Assume one of,,,abcis not zero, say.aChoose a point
in the first plane such as
1,0,0.da The distance
between this point and the second plane is

12
222
12 12
222 22200
.
ada b c d
D
abc
dd dd
abc abc







19.
22
1xy cylinder
2zy
plane
Introduce a coordinate system in the plane 2 .zy

The new u-axis is the original x-axis.
The new v-axis is the line
2, 0.zyx
Then the intersection of the cylinder and plane satisfies
the equation of an ellipse:

22
2
2
2
2
1
1
2
1ellipse
4
xy
z
x
z
x







20. Essay.
x
y
z
−2
2
3
2
2
(0, 1, 2)
(0, −1, −2)

81. © 2010 Brooks/Cole, Cengage Learning
CHAPTER 12
Vector-Valued Functions
Section 12.1 Vector-Valued Functions .....................................................................82
Section 12.2 Differentiation and Integration of Vector-Valued Functions .............92
Section 12.3 Velocity and Acceleration..................................................................102
Section 12.4 Tangent Vectors and Normal Vectors ...............................................113
Section 12.5 Arc Length and Curvature..................................................................131
Review Exercises........................................................................................................153
Problem Solving.........................................................................................................161

82. 82
© 2010 Brooks/Cole, Cengage Learning
CHAPTER 12
Vector-Valued Functions
Section 12.1 Vector-Valued Functions
1.
1
3
12
rijk
t
tt
t



Component functions:



1
1
2
3
ft
t
t
gt
ht t





Domain:
,1 1,! 4 !
2.

22
46ttttrijk
Component functions:


2
2
4
6
ftt
gt t
ht t



Domain:
+
,2, 2

3. ln
t
ttetrijk
Component functions:


ln
t
ftt
gte
ht t




Domain:
0,!

4. sin 4 cosrijktt tt
Component functions:


sin
4cos
ftt
gtt
ht t




Domain:
,! !

5.
cos sin cos sin 2 costtt t tt t t ttrFG i jk i j ik
Domain:
+0,!

6. tr

22
22
ln 5 3 4 3
ln 1 5 4 3 3 ln 1Fijkijk
ij k ij
tGt ttt tt
ttttttt


Domain:
0,!

7.
22
sin cos 0 cos sin cos sin
0sincos
ttt tt t tt t
tt
ijk
rFG i j k'
Domain:
,! !

8.

33
33 3
3
22
111
2
1
ttttt
ttt tt tt tt tt
tt
tt
t
ijk
rFG i j k

'




Domain:
,1, 1,! !
9.


21
2
1tttri
j
(a)

1
2
1ri
(b)
0rj
(c)


22
11
22
1111 1
s ss ssrijij
(d)



22
2
11
22
11
22
222 21222 12
24
rr i jij ijij
ij ij
ttt ttt
tt t tt t
5 5 5 5 5 5
55 5 55 5

83. Section 12.1 Vector-Valued Functions 83
© 2010 Brooks/Cole, Cengage Learning
10. cos 2 sintt trij
(a)
0ri
(b)
2
2
42
rij




(c)
cos 2 sin cos 2 sinrijij
(d)
cos 2 sin cos 2 sin
666 6 66
tttrr i jij

5 5 5



11.

1
ln 3tt t
t
rijk

(a)

1
2ln2 6
2
rijk

(b)
3ris not defined. ln 3 does not exist.
(c)


1
4ln 4 3 4
4
rijktt t
t



(d)


1
11ln1 31 03ln1 3
11
rr ij kijk i jk
t
tt t t t
tt
5
5 5 5 5 5

5 5

12.

32 4 t
ttterijk


(a)
0rk
(b)

1
428 eri
j k


(c)


32 24
222rijk
c
ccc e



(d)


+,




32 94 94
32 94 94
9999 327
939 27
rr i jkijk
ijk
t
t
ttte e
ttee
5
5
5 5 5
55

13.



2
22
22
34
34
916 125
tttt
tttt
ttttt
rijk
r



14.



22
22
sin 3 cos 3
sin 3 cos 3 1
tttt
ttttt
rijk
r



15.
0, 0, 0 , 3, 1, 2PQ


3, 1, 2
32,01
v
rijk
PQ
tttt t





3, , 2,0 1,xtytzt t Parametric equation

Answers may vary
16.
0, 2, 1 , 4, 7, 2PQ

4, 5, 3vPQ



425 13,0 1ri j ktt t t t

4, 2 5, 1 3,
xty tz t
0 1, Parametric equationt

Answers may vary
17.
2, 5, 3 , 1, 4, 9PQ

1, 1, 12vPQ



25 312,01rijkttt tt

2, 5, 312,
x ty tz t
0 1, Parametric equationt

Answers may vary
18.
1, 6, 8 , 3, 2, 5PQ

4, 4, 3vPQ



14 64 83,0 1tt t ttrijk

14,
x t 64,yt 83,zt
0 1, Parametric equationt
(Answers may vary)

19.

23 3
32 3 3 321
4
31 8 4
3 2 4 5 , a scalar.
tt t t t t
tt t t tt
ru"

No, the dot product is a scalar-valued function.

84. 84 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
20. ′″′″′″′″′″′ ″′″ ′″
232
3 cos 4 sin 2 sin 6 cos 2 2 ,tt t t t tt t ttru" a scalar.
No, the dot product is a scalar-valued function.

21. ′″
2
2,2 2tt tt trijk

2
,2,
xty tz t⎜⎜ ⎜
So,
2
.zx⎜Matches (b)

22. ′″ ′″ ′″
′″ ′″
2
2
cos sin , 1 1
cos , sin ,rijkttttt
xtytzt

⎜⎜⎜

So,
22
1.xy Matches (c)

23. ′″
20.75
20.75
,2 2
,,
t
t
ttt e t
xtytze
rij k
⎜⎜ ⎜
So,
2
.yx⎜Matches (d)

24. ′″
2
ln , 0.1 5
3
2
,ln,
3
t
tt t t
t
xty tz
rijk
⎜⎜ ⎜

So,
2
3
zx⎜ and ln .yx⎜Matches (a)

25. (a) View from the negative x-axis: ′″20,0,0π
(b) View from above the first octant:
′″10, 20, 10
(c) View from the z-axis:
′″0, 0, 20
(d) View from the positive x-axis:
′″20, 0, 0

26. ′″ 2
,,2
ttt
xty tz x y
rijk


(a)
′″0, 0, 20 (b) ′″10, 0, 0 (c)
′ ″5, 5, 5


27.
4
4
1
41
t
x tx
yt
yx

⎜π
⎜π


28.
55
5
x tt x
yt
yx

⎜
⎜π

Domain: 0t/


29.
32
23
,
xty t
yx
⎜⎜
⎜



30.
22
,xt tyt t

2
3
31
1
2
y
x
x
y
−1−2−3−4 1234
−2
1
2
3
4
−1−2−3−4−5 12345
1
2
3
4
5
6
7
8
9
x
y
x
345
3
2
4
5
−1−4
−3
−2
y
21−2−5−3
6
7
−1
−1
1
2
3
4
5
12345
y
x
231
3
1
y
z
y
x
2
2
2
3
3
1
−1
−1
−2
−2
−2
−3
−3
−3
z

85. Section 12.1 Vector-Valued Functions 85
© 2010 Brooks/Cole, Cengage Learning
31.
2
2
cos , 3 sin
1, Ellipse
9
xy
y
x⎜⎜



32.
22
2cos
2sin
4, circle
x t
yt
xy
⎜
⎜



33.
22
3sec , 2tan
1, Hyperbola
94
xy
xy⎜⎜



34.
33
2cos , 2sin
x ty t⎜⎜

23 23
22
23 23 23
cos sin
22
1
2
xy
tt
xy




⎜



35. 1
42
23
xt
yt
zt




Line passing through
the points:
′″′″0, 6, 5 , 1, 2, 3

36.
25
3
xt
yt
yt
⎜
⎜π
⎜
Line passing through
the points:
′″
′″
515
22
0, 5, 0 , , 0,π

37. 2cos , 2sin ,
x ty tz t⎜⎜⎜

22
1
44
xy
zt

⎜

Circular helix

38. ,3cos,3sin
xty tz t⎜⎜ ⎜

′″′″
33
22
3cos 3sin 9yz t t
Circular helix along cylinder
22
9yz


39. 2sin , 2cos ,
t
x ty tz e
π
⎜⎜⎜

22
4
t
xy
ze
π

⎜


40.
2
2 3
,2,
2
3
,
44
xty tz t
y
xzy⎜⎜⎜
⎜⎜



t 2π 1π 0 1 2
x 4 1 0 1 4
y 4π 2π 0 2 4
z 3π
3
2
π 0
3 2
3

23−2−3
1
2
x
y
1
−1
−11
x
y
x
1296−6
−6
−3
−9
−12
−9−12
12
9
6
3
y
−3−2
−2
2
3
−3
23
y
x
x
y
(0, 6, 5)
(1, 2, 3)
(2, 2, 1)−
4
3
5
6
4
3
5
1
3
z
x
y
3
−3
3
7
z
x
y4
4
4
8
z
x
y
3
−3
3
6
z
1
1
−1
2
−2
3
23
4
5
−4
−3−2
−1
z
y
x
−3(0, 5, 0)−
y
x
6
6
44
4
−4
−4
−2
−6
−6 −6
2 2
2
5
2
15
2
, 0,()
z

86. 86 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
41.
23
23 2
3
2
3
,,
,
xty t z t
yxz x
⎜⎜ ⎜
⎜⎜



42. cos sin
sin cos
x tt t
yttt
zt
⎜π
⎜


22 2 2
11
xytz
or
222
1xyz

zt⎜
Helix along a hyperboloid of one sheet

43. ′″
2213
22
ttttrijk
Parabola

44. ′″
2231
22
tt t tri jk
Parabola

45. ′″
311 3
sin cos cos
2222
tt tt tri j k





Helix

46. ′″
2sin 2cos 2sinttttrijk
Ellipse


47. (a) (b) (c)
The helix is translated 2 The height of the helix The orientation of the
units back on the
x-axis. increases at a faster rate. helix is reversed.
(d) (e)
The axis of the The radius of the helix is
helix is the
x-axis. increased from 2 to 6.
t 2π 1π 0 1 2
x 2π 1π 0 1 2
y 4 1 0 1 4
z
16
3
π
2
3
π 0
2 3

16
3


x y
5
2
6
4
2
−2
−4
−6
2, 4,))
16
3
−2, 4, −))
16
3
z
x
y3
2
4
2
3
4
z
yx
2
3
−2
−3
−1
−2
−3
−2
−3
−4
−5
1
z
x y
33
22
1
1
1
−3
−3
−2
−2
−1
z
y
x
2
3
−1
2
1
−2
2
z
x y
2
2
1
1
1
−2
−1
−1
−1
z
y
x
π2
π
1
−3
1
2
−2
−2
z
yx
π8
π4
2
−2
−2
2
z
y
x
π2
π
−2
2
2
−2
z
yx
π2
π
2
−2
−2
2
z
yx
π2
π
2
−2
−2
2
z
z
yx
π
6
−6
6
−6

87. Section 12.1 Vector-Valued Functions 87
© 2010 Brooks/Cole, Cengage Learning
48.
231
2
rijhttt t

(a)
2ttur jπ′ is a
translation 2 units to the
left along the
y-axis.
(b)

23 1
2
tt t tuijk
has the roles of
x and y
interchanged. The graph
is a reflection in the
plane
.
xyπ
(c)
4ttur k is
an upward shift 4 units.
(d)

231
8
ttt tuijk
shrinks the
z-value by a
factor of 4. The curve
rises more slowly.
(e)
tturπ′ reverses
the orientation.

49. 5yx
Let ,
xtπthen 5 yt

5tttri j

50. 2350xy
Let ,
xtπthen
1
3
25.yt



1
3
25ttritj

51.
2
2yxπ′
Let ,
xtπthen
2
2.ytπ′


2
2tttri j

52.
2
4yxπ′
Let ,
xtπthen
2
4.ytπ′


2
4tt tri j

53.
22
25xy
Let
5cos ,
x tπthen 5sin .ytπ

5cos 5sintttrij

54.
2
2
24xy
Let
22cos, 2sin.
x ty t′π π

22cos 2sintttrij

55.
22
1
16 4
xy
′π
Let
4sec , 2tan .
x ty tππ

4sec 2tantttrij

56.
22
1
916
xy

Let
3cos
x tπand 4sinytπ
Then
22
22
cos sin 1
916
xy
tt

3cos 4sinrijttt

57.


22
1
2
3
,0 2
24,02
4,0 4
ttt t yx
tt t
rt t trij
rij
j


(Other answers possible)
−2
1
1
2
3
4
5
2
32
3
4
5
z
y
x
−2
1
1
2
3
4
5
2
32
3
4
5
z
y
x
1
1
1
2
3
4
5
2
3
4
5
2
3
4
5
z
y
x
1
1
2
3
4
5
5
2
3
4
5
z
y
x
1
1
2
3
4
5
5
3
2
1
2
3
4
5
z
y
x
1
1
1
2
3
4
5
2
3
4
5
2
3
4
5
z
y
x

88. 88 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
58. ′″ ′ ″ ′ ″ ′″
′″ ′ ″
′″
′″ ′″ ′″
′″ ′″′″
′″
111
2
22
3
33 , 0 10 0 , 10 10
10 cos sin ,
0010,5252
44
521 521 ,
0105252,1
Other answers possible
tt t
ttt
t
tt t
t
ri r0r i
rij
rir ij
rij
rijr0










59.
22
,0zx yxy
Let ,
xt⎜then yxt⎜π ⎜π
and
22 2
2.zx y t
So,
2
,,2.xty tz t⎜⎜π⎜

′″
2
2ttt trijk


60.
22
,4zx yz
So,
22
4xy or

2cos ,
x t⎜ 2sin , 4.ytz⎜⎜

′″2cos 2sin 4tttrijk

61.
22 2
22
4,
2sin , 2cos
4sin
xyzx
x ty t
zx t

⎜⎜
⎜⎜



′″
2
2 sin 2 cos 4 sinttt tri
j k


62.
222 2
44 16,
x yz xz
If ,
zt⎜then
2
xt⎜and
421
2
16 4 .ytt⎜ππ


′″
2421
2
16 4tt tt tri jk

63.
222
4, 2xyz xz
Let
1sin,
x tthen 21sinzx t⎜π⎜π and
222
4.xyz

′″ ′″
22
222
1 sin 1 sin 2 2 sin 4ty t ty

22
2cos , 2cos
1sin, 2cos
1sinyty t
x ty t
zt


⎜π


′″ ′ ″
′″
′″ ′ ″ ′″
1 sin 2 cos 1 sin and
1 sin 2 cos 1 sin
tttt
tttt
rijk
rijk



t 0
6


4


2


3
4

x 0 1 2 2 2 0
y 2 3 2 0 2π 2π
z 0 1 2 4 2 0
t 1.3π 1.2π 1π 0 1 1.2
x 1.69 1.44 1 0 1 1.44
y 0.85 1.25 1.66 2 1.66 1.25
z 1.3π 1.2π 1π 0 1 1.2

t
2

π
6

π 0
6


2


x 0
1
2
1
3 2
2
y 0
6
2
2
6
2
0
z 2
3
2
1
1
2
0

x
y
5
123
−
3
3
2
2, 2, 4− 2, 2, 4)(( )−
z
x
y2
2
6
z
x
y2
2
4
4
2
z
x
y
3
−3
−3
3
3
z
x
y
3
−3
3
4
z

89. Section 12.1 Vector-Valued Functions 89
© 2010 Brooks/Cole, Cengage Learning
64.
222
10, 4xyz xy
Let
2sin,
x tthen 2sinyt and
2
21 sin 2cos .ztt



2 sin 2 sin 2 costt ttri j k
65.
22 22
4, 4xz yz
Subtracting, you have
22
0xy or .yx
So, in the first octant, if you let ,
xtthen ,xt ,yt
2
4.zt


2
4ttt trij k
66.

222
16, 4 first octantxyz xy
Let ,
xtthen
4
y
t
and
2222 2
2 16
16.xyzt z
t


421
16 16
ztt
t


843 843t



4241
16 16tt t t
tt
rij k

67.
22
22 2 2
2cos 2sin 4 4yz tt tt t x

68.
22
22 2 2
cos sin
ttt
xye t ete z




69. lim cos sinijkij
t
ttt





70.
2
2
21 21
lim 3 6
132
t
t
ttijkijk






t
2


6

0
6


2


x 1
3
2
2
5
2
3 2
y 3
5
2
2
3 2
1 2
z 0
6
2
2
6
2
0 2

x
y
4
4
4
z
x
y
z
4
(2, 2, 0)
(0, 0, 2)
2
3
4
3
x
y
4
4
8
12
16
8
12
16
7
6
5
z
x
y
40
80
120
40
80
120
120
160
80
40
z
x
y
4
4
4
z
t 843 1.5 2 2.5 3.0 3.5 843
x 1.0 1.5 2 2.5 3.0 3.5 3.9
y 3.9 2.7 2 1.6 1.3 1.1 1.0
z 0 2.6 2.8 2.7 2.3 1.6 0

90. 90 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
71.
2
0 1cos
lim 3 0
t
t
tt
t
ij k

π




because
00
1cos sin
lim lim 0.
1
tt
tt
t

π
⎜⎜
′″L'Hôpital's Rule

72.
2
1
ln 1
lim
11
t
t
t
tt
ijk




ππ

does not exist because
1
1
lim
1
tt π
does not exist.

73.
0
sin
lim
tt
t t
ee
t
ijkijk
π





because

′″
00
sin cos
lim lim 1 L'Hôpital's Rule
1
tt
tt
t

⎜⎜

74.
2
1
lim 0
1
t
t t
e
tt
ij k
π
!




because

lim 0,
t
t
e
π
!
⎜
1
lim 0,
tt !
⎜and
2
lim 0.
1
t
t
t
!
⎜


75. ′″
1
tt
t
rij
Continuous on
′″′″,0, 0,π! !

76. ′″
1tttri j
Continuous on
+″1,!

77. ′″ ′ ″arcsin 1tt ttri j k
Continuous on
+
,1, 1π

78. ′″ ′ ″
2, ,ln 1
tt
teetr
ππ
⎜π
Continuous on 1 0t or
′″1: 1, .t!

79. ′″
2
,,tan
t
tet tr
π
⎜
Discontinuous at
2
tn



Continuous on ,
22
nn







80. ′″
3
8, ,tttr⎜
Continuous on
+″0,!

81. ′″ ′ ″
2
3tt t tri jk
(a)
′″ ′″ ′ ″ ′ ″
2
333tt tt tsr ki
j k
(b)
′″ ′″ ′″ ′″
2
223tt t t tsr i i jk
(c)
′″ ′″ ′ ″
2
52tt tt tsr ji jk

82. A vector-valued function ris continuous at ta⎜ if the
limit of
′″trexists as ta and ′″ ′ ″lim .
ta
tarr

⎜
The function
′″
2
2
t
t
tij
r
ij/6
⎜7
8
is not continuous at
0.t⎜

83. One possible answer is

′″
1
1.5 cos 1.5 sin , 0 2ttttt
rijk

Note that
′″21.52.rik


84. (a) 3cos 1,xt 5sin 2,yt 4z⎜

′″
′″
22
12
1, 4
925
xy
z
ππ


(b) 4,x⎜
3cos 1,yt 5sin 2zt

′″
′″
22
12
1, 4
925
yz
x
ππ


(c)
3cos 1,xt⎜π 5sin 2,yt⎜π π 4z⎜

′″
′″
22
12
1, 4
925
xy
z



(d)
3cos2 1,xt 5sin2 2,yt 4z⎜

′″
′″
22
12
1, 4
925
xy
z
ππ


(a) and (d) represent the same graph
1
1
−1
2
3
2
−1
−2
−1
−2
2
z
x y

91. Section 12.1 Vector-Valued Functions 91
© 2010 Brooks/Cole, Cengage Learning
85. Let 111txtytztrijk and 222 .txt yt ztuijk Then:

9
:


12 21 12 21 1 2 2 1
12 21 12 21
12 2lim lim
lim lim lim lim lim lim lim lim
lim lim lim lim
ru i j k
ij
tc tc
tc tc tc tc tc tc tc tc
tc tc tc
t t ytzt ytzt xtzt xtzt xtyt xtyt
yt zt yt zt xt zt xt zt
xt y t x t



'







1
111 2 2 2
lim lim lim lim lim lim
lim lim
k
ijk i jk
ru
tc
tc tc tc tc tc tc
tc tc
yt
xt yt zt x t y t z t
tt






'

'

86. Let
111txtytztrijk and 222 .txt yt ztuijk Then:




12 1 2 12
12 1 2 12
111 2 2 2lim lim
lim lim lim lim lim lim
lim lim lim lim lim lim
lim lim
tc tc
tc tc tc tc tc tc
tc tc tc tc tc tc
tc tc
tt xtxtytytztzt
xt x t yt y t zt z t
xtytzt xtytzt
tt
ru
ijk i jk
ru




"



"

"
87. Let
.txt yt ztrijk Because ris
continuous at ,tcthen
lim .
tc
tcrr



,,cxc yc zc xcyczcrijk
are defined at c.



222
222
lim
tc
xt yt ztxcyczc c
r
rr




So,
ris continuous at c

88. Let


1, if 0
1, if 0
t
ft
t
/6
7
8

and
.tftri Then ris not continuous at
0,cwhereas,
1ris continuous for all t.

89.

22
2
920
34 54.ri jk
uijktt t t
ss s s



Equating components:

2
2
2
34
920
54
ts
ts
ts




So,
2
3454 4
920 16 4.
sss
ts t



The paths intersect at the same time 4tat the point
16, 16, 16 . The particles collide.

90.

23
23 8 122
ttt t
ss ss
rijk
uijk


Equating components

2
3
23
8
12 2
ts
ts
ts






2
2
2
23 8
41298
42090
29210
s s
s ss
ss
ss





For
1
2
,s

1
2
232.t
For
9
2
,s

9
2
236t and

2 9
2
836t and

3 9
2
12 54.t Impossible.
The paths intersect at
2, 4, 8 , but at different times

1
2
2and .ts No collision.

91. No, not necessarily. See Exercise 90.

92. Yes. See Exercise 89.

93. True

94. False. The graph of
3
xyzt represents a line.

95. True. See Exercises 89 and 90.

96. True.
2222 22 2
sin cosyz t tt tt x

92. 92 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
Section 12.2 Differentiation and Integration of Vector-Valued Functions
1. ′″
′″ ′″
2
0
2
2
,2
,tt tt
xttytt
xy
rij
⎜⎜
⎜


′″
′″
′″242
2
24
tt
rij
rij
rij




′″0tris tangent to the curve at
0.t

2. ′″ ′″
2
0
1, 1ttt tri j

′″ ′″
′″
′″
′″
2
2
,1
1
1
2
12
xt t yt t
yx
tt
ri
rij
rij
⎜⎜π
⎜π
⎜



′″0rtis tangent to the curve at
0.t

3. ′″
′″ ′″
2
0
2
21
,2
1
,
1
tt t
t
xt t yt
t
x
y
rij
⎜⎜
⎜

′″
′″
′″
2
1
24
2
1
2
1
24
4
tt
t
rij
rij
rij
⎜π
⎜π


′″0tris tangent to the curve at
0.t

4. (a) ′″ ′ ″
3
0
1,1ttttrij

′″
3
3
1
1
xt
yt x

⎜⎜π
(b)
′″
′″
′″
2
12
3
13
tt
rij
rij
rij




′″0tris tangent to the curve at
0.t

5. ′″
′″ ′″
0
22cos sin ,
2
cos , sin
1
tttt
xt t yt t
xyrij


⎜⎜


′″
2
sin cos
2
ttt
rj
rij
ri


⎜




⎜π




′″0tris tangent to the curve at
0.t

6. ′″
′″ ′″
0
223sin 4cos ,
2
3sin , 4cos
1, ellipse
34
tttt
xt t yt t
xy
rij


⎜⎜






′″
3
2
3cos 4sin
4
2
i
ttt
r
rij
rj


⎜


⎜π

⎜π




′″0tris tangent to the curve at
0.t
7.
′″
2
0
,, 0
tt
teetr⎜⎜

′″ ′″ ′″
′″
′″
′″
2
2
2
2
,
,0
01,1
,2
01,2
r
r
r
ttt
tt
xteyte e
yxx
tee
⎜⎜⎜

⎜
⎜
⎜

′″0rtis tangent to the curve at
0.t

864
4
2
2
−2
−4
r′
x
r
(4, 2)
y
−1−2−3 123
−2
−3
1
2
3
r′
r
(1, 0)
x
y
r′
x
r
y
4,
1
2((
2
1
2
y
x
−1123
−1
1
2
3
r
r
(2, 1)
(1, 0)
−1−2−4124
−2
−3
1
2
3
r′
r
(3, 0)
x
y
123
1
2
3
(1, 1)
y
x
r
r′
x
r
1
(0, 1)
y
r′

93. Section 12.2 Differentiation and Integration of Vector-Valued Functions 93
© 2010 Brooks/Cole, Cengage Learning
8. ′″
′″ ′″
0,, 0
1
,
1
,0
tt
tt
t
teet
xte yte
e
yx
x
r
π
π
⎜⎜
⎜⎜ ⎜


′″
′″
′″
01,1
,
01,1
r
r
r
tt
tee
π
⎜
⎜π
⎜π

′″0rtis tangent to the curve at
0.t
9. (a) and (b)
′″
0
3
2cos 2sin ,
2
tttttrijk


22
4,xy zt

′″
2sin 2cos
33
2
22
3
2
2
tttrijk
rjk
rik












10.
′″
2
0 3
2
,2ttt trijk

2 3
2
,yxz⎜⎜

′″
′″
′″
3
2
2
224
24
ttrij
rijk
rij




11.
′″
′″
3
2
3
33
tt t
tt
rij
rij
⎜π
⎜π

12.
′″
′″
′″
3
2
1
1
3
2
tt t
tt
t
ri j
rij

⎜π
13.
′″
′″
2cos ,5sin
2sin ,5cos
ttt
ttt
r
r
⎜
⎜π
14.
′″
′″
cos , 2 sin
sin cos , 2 cos
ttt t
ttt t tr
r⎜π

15.
′″
′″
23
2
67
614 3
tttt
ttt
rijk
rijk



16.
′″
′″
2
2
1
16
2
1
16
t
tt
t
tt
t
rijk
rijk



17.
′″
′″
33
22
cos sin
3 cos sin 3 sin cos
ta ta t
tattatt
rijk
rij



18.
′″
22
4lntttttrijk

′″
2
32
22
2
2
25 2
2
t
ttt
ttt
t
tt
ri jk
ijk





19.
′″
′″
′″
45
55
tt
ttt
te te
te eterijk
ri k
π
π


20.
′″
′″
3
2
,cos3,sin3
3,3sin3,3cos3
tt t t
tt t t
r
r
⎜
⎜π

21.
′″
′″
sin , cos ,
sin cos , cos sin , 1
tttttt
tttttttr
r⎜

22.
′″
′″
22
arcsin , arccos , 0
11
,,0
11
r
r
ttt
t
tt
⎜
⎜π
ππ

23.
′″
321
2
tt t
rij
(a)
′″
2
3tttrij
(b)
′″6rijtt
(c)
′″ ′″ ′ ″
23
36 18tttttttrr"
24.
′″′″′″
22
ttt ttrij
(a)
′″ ′ ″ ′ ″21 21ttrt i j
(b)
′″22rijt
(c)
′″ ′″ ′ ″′ ″ ′ ″′ ″212 212 8rrtt t t t"
25.
′″4cos 4sintttrij
(a)
′″ 4sin 4costttrij
(b)
′″ 4cos 4sintttrij⎜π π
(c)
′″ ′″ ′ ″′ ″ ′ ″4sin 4cos 4cos 4sin 0tt t t t trr"
−112
−1
1
2
r′
x
r
(1, 1)
y
x y
))
3π
2
2
1
2
−2
2
π
π
r
r′
0, −2,
z
x
y
2
2
−2 −2
−4
−4
4
−6
−6
r
r'
z

94. 94 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
26. ′″8cos 3sintttrij
(a)
′″ 8sin 3costttrij
(b)
′″ 8cos 3sinrijttt⎜π π
(c)
′″ ′″ ′ ″′ ″ ′ ″8 sin 8 cos 3 cos 3 sin 55 sin costt t t t t ttrr"
27.
′″
2311
26
tttt
rijk
(a)
′″
21
2
tt trijk
(b)
′″ttrik
(c)
′″ ′″ ′″ ′ ″
′″
3
2
1
110
22
t
ttt ttt
rr"
28.
′″ ′ ″ ′ ″23 35tt t tri j k
(a)
′″ 23trijk
(b)
′″0tr⎜
(c)
′″ ′″ 0ttrr"⎜
29.
′″
cos sin , sin cos ,ttttttttr
(a)
′″
sin sin cos , cos cos sin , 1 cos , sin , 1t t tt t t tt t t tt tr
(b)
′″
cos sin , sin cos , 0rttttttt
(c)
′″ ′″ ′ ″′ ″ ′ ″′ ″cos cos sin sin sin costttttttttttttrr"
30.
′″ ′″
2
,,tan
t
tet tr
π
⎜
(a)
′″
2
,2,sec
t
tettr
π
⎜π
(b)
′″
2
,2,2sec tan
t
te ttr
π
⎜
(c)
′″ ′″
24
42sectan
t
tte t ttrr
π
"
31.
′″ ′ ″ ′ ″
′″ ′″ ′″
′″
′″
′″
′″ ′″ ′″
2
0
22
2
2
2
2
22 1
cos sin ,
4
sin cos 2
12 21
42 22
1221 141
4222 42
14 1
22
14 41
cos sin 2
1
4
ttttt
tttt
ttt
rijk
rijk
rijk
r
r
ijk
r
rijk
r
















π

π


π


′″
′″
′″
′″
22
22
22
2
4
22
4
22
2
22
122
24
422
14 1
224
14 24
ijk
r
r
ijk
r








π

π

x
y
r″
r′
⎜⎜r″⎜⎜
⎜⎜r′⎜⎜
z

95. Section 12.2 Differentiation and Integration of Vector-Valued Functions 95
© 2010 Brooks/Cole, Cengage Learning
32.




2
0
14
14
12 12
14
12 12 12
1431
,
24
3131
2,
24816
131
422
191 1
10 4
444 2
14 312
14 10 4 10 4 10 4
1
2, 2
4
rijk
rijkr ijk
rijk
ijk
rjkr jk
r
t
t
t
tttet
tte e
e
ree
r e
r eee
te e




























12
14
12 121
4
4
1
24
1 44
4
r
jk
r
e
e
ee
















33.



23
2
23
0
tt t
ttt
rij
rij
r0




Smooth on
,0, 0,! !
34.



2
1
3
1
1
3
1
tt
t
t
t
rij
rij





Not continuous when 1t
Smooth on
,1 , 1,! !
35.


33
22
2cos 3sin
6 cos sin 9 sin cos
2
rij
rij
r0
n








Smooth on
1
,,
22
nn
n


any integer.
36.




sin 1 cos
1cos sin
2 1 , any integer
rij
rij
r0nn





Smooth on
21,21nn
37.


2sin 1 2cos
12cos 2sin
for any value of
rij
rij
r0



#

Smooth on
,! !
38.



2
33
34
22
33
22
88
16 4 32 2
88
tt
t
tt
ttt
t
tt
rij
rij







tr0#for any value of .t
ris not continuous when 2.t
Smooth on
,2, 2,! !
39.


2
21
1
1
2
tt t
t
tt
t
rijk
rijk0

#

ris smooth or all
0: , 0 , 0,t#! !
40.

3
3
tt
tt
tee t
tee
rijk
rijk0



#

ris smooth for all t:
,! !
41.


2
3tan
3sec
tt t t
ttrij k
rij k0
#
ris smooth for all
21
.
22
n
tn

#

Smooth on intervals of form
,,
22
nn







nis an integer.
42.



2 1
1
4
11
2
42
ttt t
tt
t
ri jk
rijk0

#

ris smooth for all
0: 0,t!

96. 96 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
43.
2
3,tt ttrijk
23
4ttttuijk
(a)
32ttrijk
(b)
2trk
(c)

235
43tt tttru"


24
89 5
tDt t t t tru"

(d)


223
2
393
39263
ru i j k
ru i j k
t
ttttt tt
Dt t t tt




(e)


443 3 2
3232
24 12
8124 324
t
ttttt t t
Dt t t t t t t
ru i j k
ru i j k
'
'


(f )


24 2
2
2
10 10
10 2
10
r
r
t
ttttt
t
Dt
t






44.


2sin 2cos
1
2sin 2cosri j k
ui j ktt t t
ttt
t



(a)
2cos 2sintttri j k
(b)
2sin 2costttrjk
(c)


22
14sin 4cos 5
0, 0ru
ru
t
tt t t
Dt t t
"
" #


(d)


2
1
334sin4cos
1
334cos4sin
t
tt t t t
t
Dt t t t
t
ru i j k
ru i j k









(e)

112sin 2cos
2cos 2sin
1
2sin 2cos
tt t
tt tt tt
tt
tt
t
ijk
ru j k



'







22
11 11
2sin 2cos 1 2cos 2sin 1
tDt t t t t tt t
tt ttru j k





(f )


2
12
2
2
4
1
42
2 4
t
tt
t
Dt t t
tr
r





45.

23 4
2,tt t t ttri
jku k
(a)

7
tttru"
(i)

6
7
tDt t tru"

(ii) Alternate Solution:


23 3 2 4 6 6 6
2443 437
tDt t t t t t t t t t t t t t t tru ru r u i jk k i j k k" " " " "

97. Section 12.2 Differentiation and Integration of Vector-Valued Functions 97
© 2010 Brooks/Cole, Cengage Learning
(b) ′″ ′″
23 6 5
4
22
00
ttttttt
t
ijk
ru ij
'⎜ ⎜π
(i)
′″ ′″
54
12 5
tDt t t tru i
j'

(ii) Alternate Solution:
′″ ′″ ′″ ′″ ′″ ′″
23 2 5 4
34
2143125
004 00
ijkijk
ru ru r u i j
tDt t t t t t tt t tt t t
tt
' ' ' '

46.
′″cos sin ,ttttrijk ′″ttujk
(a)
′″ ′″
2
sintt ttru"
(i)
′″ ′″ cos 2
tDt t t tru"

(ii) Alternate Solution:

′″ ′″ ′″ ′″ ′″ ′″
′″′ ″′″
cos sin sin cos cos 2 cos
tDt t t t t t
ttt t t ttttttru ru r u
ijkk i jkjk
" " "

""

(b)
′″ ′″
′″′″cos sin sin cos cos
01
tt tttttttt t
t
ijk
ru i j k
'
(i)
′″ ′″ ′ ″ ′ ″cos sin 1 cos sin sin
tDt t t t t tt t tru i j k'

(ii) Alternate Solution:

′″ ′″ ′″ ′″ ′″ ′″
′″′″cos sin sin cos 1 sin cos 1 sin cos sin
001 0 1
tDt t t t t t
ttt t t ttt tt t t
t
ru ru r u
ijki jk
ijk
' ' '


47.
′″
′″
′″ ′″3sin 4cos
3cos 4sin
9 sin cos 16 cos sin 7 sin costtt
ttt
tt tt tt ttrij
rij
rr

⎜π
"⎜ π ⎜π


′″ ′″
′″
′″
′″ ′″
2222
2222
7sin cos
cos
9sin 16cos 9cos 16sin
7sin cos
arccos
9sin 16cos 9cos 16sin
rr
rrtt tt
tt tttt
tt
tttt


" π
⎜⎜


π

⎜




1.855
⎜ maximum at
5
3.927
4
t

⎜⎜


and
0.785 .
4t

⎜⎜



1.287
⎜ minimum at
3
2.356
4
t

⎜⎜


and
7
5.498 .
4
t

⎜⎜




′″1.571
2


⎜⎜ for
,0,1,2,3,
2
n
tn
″

⎜⎜
−17
0
⎜

98. 98 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
48.



2
3
42 2
3
42 2
3
42 2
2
2
,41
2
cos
41
2
arccos
41
rij
rij
rr
rr
tt t
tt
tttt
tttt t
tt
tt t
tt
tt t




"


π


π



0.340 19.47 maximum at
2
0.707 .
2
t

ππ




2


#for any t.
49.





2
2
00
2
00
321 321
lim lim
32
lim lim 3 2 3 2
ijij
rr
r
ij
ijij
tt
tt
tt tt t t
tt t
t
tt
tttt
tt t
t
5 5
5 5
55
5
ππ
55
5 55
5
5

50.




0
0 0
0 0
lim
33 33
22
lim lim 2
31313
lim 2 lim 2
2
rr
r
ij kijk
ijk
ijk ijki
t
t t
t t
tt t
t
t
tt tt t t
tt ttt t ttt
ttt
tt
tttt ttt tt t ttt t t
5
5 5
5 5
5
π
5

5 5 ′ 5 5 5

555



5′5

5 5 5 5 55

2
2jk
t
′

51.



0
22
2
000
lim
,0,2 ,0,2 2 ,0,2
lim lim lim 2 , 0, 2 2 , 0, 2
rr
r
t
ttt
tt t
t
t
tt tt t t tt t t
tt t
tt
5
5 5 5
5
π
5
5 5 5 5 5
5
55

52.




00
0 0
0, sin , 4 0, sin , 4
lim lim
0, sin cos sin cos sin , 4 sin cos 1 sin
lim lim 0, cos , 4
0, 0 cos , 4 0, cos , 4
rr
r
tt
t t
tttt tttt t
t
tt
tt tttt tt t
t
ttt
tt
5 5
5 5
5 5 5
ππ
55
"55 5 5′ 5

555



53.

2
2tdttttijk i j kC *

54.

34232 8
3
464 3ij k i j kCtt tdtttt *

55.
32 5212
ln
5
ij k i j kCtdt ttt
t




*

56.

1
ln ln lnijk i jkCtdtttttt
t




*

(Integration by parts)
0
−0.5
1.0
π

99. Section 12.2 Differentiation and Integration of Vector-Valued Functions 99
© 2010 Brooks/Cole, Cengage Learning
57.
32432
21 4 3 2ij k ij kCtttdttttt
*

58. sin cos cos sinijk i jkC
tt
ettdte tt
*

59.
2
2 1
sec tan arctan
1
ij i jCtdttt
t




*

60.
sin cos sin cos cos sin
22ij i jC
tt
tt
ee
ete tdt t t t t




*

61.

+,
1
2
1 1 1
2
000
0
1
84 4
22
ijk i j k i jk
t
tt dt t t



*

62.

11 1
24
1
33 43
1
111
3
244
tt
tt tdt t
ij k i j k 0






*

63.

+, +, +,
2 222
0000
cos sin sin cos
2
a t at dt at a t t aaijk i jkijk


*

64.


4 4
2
00
1
2
sec tan tan 2 sin cos sec ln sec sin 2 1 ln 2tt t ttdt t t tij k i jk ijk



*

65.


2
2
2 22
22
000
0
12 1 1
2
tt t t t
te tedt e t e e
ijk i j ek i j k




*

66.
224 2
1tt t t t tij for 0t/

3
33 32
22232
00
0
11
33
11 101t t dt t t dt tij



**
67.



22
2
43 23
023 2 3
231
rijijC
rijCiCj
ri j
tt tt
tt
teedtee
te e


*


68.




2332
332
36 4
02
24rjkjkC
rij
ri j ktt tdttt
C
ttt


*


69.






1
1
2
232 32
0 600 3 600
600 3 600 32
600 3 600 32
600 3 600 16
0
600 3 600 16
tdtt
tt
ttdt
ttt
ttttrjjC
rC ij
ri j
rij
ijC
rC0
ri j







*
*


70.





1
11
2
22
4cos 3sin
4sin 3cos
03 3
4cos 3sin
04 4 0
4cos 3sin
ttt
ttt
ttt
j
tttrjk
rjkC
rkkCC0
rjkC
rCjC
rjk







71.

22
1 1
2
11
02
22
ttt
ttee dteetrijk ijkC
rijCijkCijk


*





2
21
121
2
2
21
2
tt
t
t
tee t
e
et
rijk
ijk

100. 100 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
72.
22
111
1
1
arctan ln
tdt
ttt
tt
t
rijk
ij kC





*




122
44
1
2arctan1 ln
4
ttt
t
rijCiC ij
rijk










73. See “Definition of the Derivative of a Vector-Valued
Function” and Figure 12.8 on page 842.
74. To find the integral of a vector-valued function, you
integrate each component function separately. The
constant of integration Cis a constant vector.
75. At
0,ttthe graph of tuis increasing in the x, y,
and z directions simultaneously.
76. The graph of
tudoes not change position relative to
the xy-plane.
77. Let
.txtytztrijk Then
ct cxt cyt cztrijk and



.
tDc t cxt cyt czt
cxt yt zt c t
rijk
ijkr




78. Let
111txtytztrijk and 222 .txt yt ztuijk



12 1 2 12
12 1 2 12
111 2 2 2
tttxtxt ytyt ztzt
Dttxtxtytytztzt
xtytzt xtytzt t t
ru i j k
ru i j k
ijk i jkru







79. Let
,txtytztrijk then .rijkwt t wtxt wtyt wtzt


rijk
ijk ijk r r
tD wt t wt x t w t xt wt y t w t yt wt z t w t zt
wt x t y t z t w t xt yt zt wt t w t t




80. Let
111txtytztrijk and 222 .txt yt ztuijk




12 1 2 12 12 1 2 12
1 12 121212 12121212
12 1 2 12 1 2
12 1 2ru i j k
ru i j
k
t t ytzt ztyt xtzt ztxt xtyt ytxt
D t t y tz t y tz t z ty t z ty t x tz t x tz t z tx t z tx t
xty t x ty t ytx t y tx t
ytz t zty t
'

'




9 :
9 :

12 12 1 2 12
12 1 2 12 12 1 2 12ijk
ijk
ru r u
xtz t ztx t xty t ytx t
ytzt ztyt xtzt ztxt xtyt ytxt
tttt




' '

81. Let
.rijktxt yt zt Then wt xwt ywt zwtri
j kand




Chain Rule
.
rijk
ijkr
tD wt x wt w t y wt w t z wt w t
wt xwt ywt zwt wt wt





82. Let
.txtytztrijk Then .txt yt ztrijk





t
t t ytzt ztyt xtzt ztxt xtyt ytxt
D t t ytz t ytz t zty t z tyt xtz t xtz t ztx t z txt
xty t xtyt ytx t ytxt
ytzt ztyt xtzt ztx
rr i j k
rr i j
k
i
'

'





txtytytxt ttjkrr '

101. Section 12.2 Differentiation and Integration of Vector-Valued Functions 101
© 2010 Brooks/Cole, Cengage Learning
83. Let 111 ,txtytztrijk 222 ,txtytztuijk and 333 .txtytztvijk Then:






123 23 123 23 123 23
123 12 3 1 23 132
13 2 1 32 123 12 3 1 23
12 3
tt t t x t y tz t z ty t y t x tz t z tx t z t x ty t y tx t
D t t t xtytz t xty tzt xtytzt xtytz t
xty tzt xtytzt ytxtz t ytx tzt ytxtzt
ytz tx t
ruv
ruv
" '

"'

′′′′′




9 :

12 3 1 2 3 12 3 12 3
123123 123123
123 32 1 23 23 123 23
123 32 1 23 23ytz txt y tztxt ztxty t ztx tyt
z txtyt ztytx t zty txt z tytxt
xt ytzt ytzt y t xtzt ztxt z t xtyt ytxt
xt y tzt ytz t yt x tzt z txt






9 :
9 :

123 23
123 32 1 23 23 123 23zt x tyt y txt
xt y tz t y tz t y t x tz t z tx t z t x ty t y tx t
ttt t tt tt truv ruv ruv




"' " ' "'


84. Let
.txtytztrijk If ttrr"is constant, then:



+,



222
222
22 20
20
20.
tt
xt yt zt C
Dxt yt zt DC
xt x t yt y t zt z t
xt x t yt y t zt z t
ttrr






"

So,
0.ttrr"π
85.
sin 1 costt t trij
(a)
The curve is a cycloid.
(b)

1cos sin
sin cos
rij
rij
ttt
ttt




22
12cos cos sin
22cos
rtttt
t

π′
Minimum of

tris 0, 0.tπ
Maximum of

tris 2, .tπ


22
sin cos 1tttr
Minimum and maximum of

tris 1.
86.
2cos 3sinrt t t ij
(a) Ellipse

(b)



22
2sin 3cos
2cos 3sin
4sin 9cos
ttt
ttt
ttt
rij
rij
r

π′ ′


Minimum of

tris 2, 2.tπ
Maximum of

tris 3, 0.tπ

y
x
−1−313
−1
−2
1
2
40
0
0
5

102. 102 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
87.




22
sin cos
cos sin cos sin
sin cos sin cos cos sin sin cos 2 cos 2 sin
2sincos 2sincos 0
rij
rij
rijij
rr
tt
tt tt
tt tt t ttt t t
tt
te te t
tetet etet
t e te te te t e te te te t e t e t
ttettett



"π′π

So,
tris always perpendicular to .tr
88. (a)

2
4tt tri j
(b)


13
1.25 1.25 2.4375
1.25 1 0.25 0.5625
rij
rij
rr i j


′π ′

(c)



2
12
1.25 1 0.25 0.5625
2.25
1.25 1 0.25
ttrij
rij
rr ij
ij
π′
π′
′ ′
ππ′
′

This vector approximates
1.r
89. True
90. False. The definite integral is a vector, not a real number.
91. False. Let
cos sin .rijkttt





2
0
sin cos
1r
r
rij
rt
d
t
dt
ttt
t
π
π


π


92. False.

ru ru r uDt t t t t t " "
"
(See Theorem 2.2, part 4)

Section 12.3 Velocity and Acceleration
1.

31
3
ttt
tt
ttri j
vr ij
ar 0

ππ

3,xtπ 1,ytπ′

1
3
x
yπ′
At
3, 0 , 1.tπ

13 ,1vija0

2.

6
0
6, , 6
ttt
tt
tt
x ty ty x
rij
vr ij
ar


ππ
π′ π π′



3.


2
22
2
2
,,
tt t
ttt
tt
xty tx y
rij
vr ij
ar i

ππ
πππ

At
4, 2 , 2.tπ

24
22
vij
ai

π

4.


2
4
2
2
tt t
tt t
tt
ri j
vr ij
rj

ππ′
ππ′
a


22
,44
xty t x
At
1, 3 , 1,tπ 12,vijπ′ 12ajπ′

x
2
5
3
1
3−1−3
−1
r(1)
r(1.25)
rr(1.25) (1)−
y
64
2
−2
−4
v
x
(3, 0)
y
864
4
2
2
−4
−2
v
x
a
(4, 2)
y
−1−3
1
2
3
5
x
(1, 3)
y
v
a
2
2
4
4
x
(3, 3)
v
y

103. Section 12.3 Velocity and Acceleration 103
© 2010 Brooks/Cole, Cengage Learning
5.


23
2
23 23
23
26
,,
tt t
tttt
tt t
xty tx y
rij
vr ij
ar ij


πππ
At
1, 1 , 1.tπ

123
126vij
aij



6.



3
21
4
3
4
3
2
1tt t
ttt
trt trij
vr ij
ai

ππ


31
4
1,xt

3
31
4
1
41
yt x y
yx



At
3, 2 , 2, 2 3 , 2 3t vijai


7.


22
2cos 2sin
2sin 2cos
2cos 2sin
2cos , 2sin , 4
ttt
tt t t
tt t t
xtytxyrij
vr i j
ar i j

ππ′ ′


At
2, 2 , .
4
t

π
22
4
22
4vij
aij






π′ ′




8.


22
3cos 2sin
3sin 2cos
3cos 2sin
3 cos , 2 sin , 1 Ellipse
94
ttt
ttt
ttt
xy
xtyt
rij
vij
aij


π′ ′

At
α
3, 0 , 0.tπ

02
03
vj
ai
π
π′


9.



sin , 1 cos
1cos,sin
sin , cos
sin , 1 cos cycloid
tt t t
tt tt
tt tt
xt ty t
r
vr
ar
π′ ′
ππ′
ππ
π′ π′

At
α
,2, .tπ


2, 0 2
0, 1
vi
aj
ππ
π′π′
10.



,
,
,
11
,,
tt
tt
tt
tt
t
tee
ttee
ttee
xe yey
ex
r
vr
ar
′
′
′
′
π
ππ′
ππ
ππ π π
At
1, 1 , 0.tπ


01,1
01,1
vij
aij


11.



22
53
53
15 3 35
tt t t
t
st t
rijk
vijk
v





α
ta0π
12.




442
442
16 16 4 6
tttt
t
st t
t
rijk
vijk
v
a0



π

13.




2
2
22 2
2
2
14 15
2
rijk
vijk
v
ajk
t
ttt
ttt
stt tt t
t





14.




2
221
4
1
2
11
44
1
2
3
3
91 10
rijk
vijk
v
ak
tttt
tt
stt t t
t



π

(1, 1)
a
v
2
1
2
3
4
5
6
7
8
−1 345678
y
x
x
(3, 2)
y
v
a
−2−4−6 246
−4
−6
2
4
6
2, 2)(
3
3
−3
−3
x
v
a
y
v
x
2
a
4
2
π
π
π
( , 2)
y
v
a
(3, 0)
y
x
1
−1
−112−2
−3
3
1
1
2
2
x
(1, 1)
va
y

104. 104 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
15.




2
2
22
22
32
2
9
9
18
11
99
9
9
rij k
vij k
v
ak
ttt t
t
t
t
tt
st t
tt
t
t








16.




232
22
2
23
419 491
3
2
2
rijk
vijk
v
ai k
tt t t
tt t
st t t t t t
t
t




17.




22
4,3cos ,3sin
4, 3sin ,3cos 4 3sin 3cos
16 9 sin 9 cos 5
0, 3 cos , 3 sin 3 cos 3 sin
r
vijk
v
ajk
tt tt
ttt tt
st t t t
ttttt





18.




2
222 2
2cos ,2sin ,
2sin ,2cos ,2
4sin 4cos 4 2 1
2cos , 2sin ,2
r
v
v
a
tttt
tttt
stt t tt t
ttt





19.


cos , sin ,
cos sin sin cos
r
vijk
ttt
tt tt t
t e te te
tetet etete






22
222
cos sin cos sin
3
v
ttt
t
st t
ettet te
e





2sin 2cosaijk
ttt
tetete
20.




4
3
2
6210
24 2
2
231
ln , ,
11
,,4
11 1
16 1 16
12
,,12
r
v
v
a
ttt
t
tt
tt
stt t tt
tt t
tt
tt





21. (a)



3
2
0
2
,,, 1
4
3
1, 2 ,
4
3
11,2,
4
t
ttt t
t
tt
r
r
r





13
1, 12,
44
x ty tz t
(b)


13
1 0.1 1 0.1, 1 2 0.1 , 0.1
44
1.100, 1.200, 0.325
r


22. (a)



22
0
22
,25 ,25 , 3
1, ,
25 25
33
31,,
44
3
3, 4
4
tt t tt
tt
t
tt
xtyz t
r
r
r







(b)


33
30.1 30.1,4 0.1,4 0.1
44
3.100, 3.925, 3.925
r


23.







2
2
,0 ,0
0, ,
2
0, ,
2
22 222
t
tdtttt
tttttt
t
ttttdt
t
t
aijkv 0r 0
vijkijkC
vC0vijkv ijk
rijk ijkC
rC0r ijk
rijkijk






*
*

24.






22
22 3
2
3
2
23,0 4,0
23 2 3
04 243
243 4
00 4
2486
t
tdttt
tt t
tt tdtttt
tt t t
aikv jr0
vikikC
vCjv ijk
rijkijkC
rC r ijk
rijk






*
*

105. Section 12.3 Velocity and Acceleration 105
© 2010 Brooks/Cole, Cengage Learning
25.






22
22
22 3 3
3
,1 5,1
22
11 91
15
22 22
91
22 22
91 9 1
22 22 62 62
14 1 14 1
1
33 33
ttt
tt
tttdt
tt
t
tt t t
tdttt
t
t
ajkv jr0
vjk jkC
vjkCjCjk
vjk
rjkjkC
rjkC0Cjk
r











*
*

3
914 11
62 3 62 3
17 2
2
33
t
tt jk
rjk





26.







2
2
32
32 32
032 32132
32 132 3 2 16
052 352216
26 60
t
tdtt
tt
ttdttttt
tt t tt
ak
vkkC
v Cijk v ij k
rij kij kC
rCjkr i j k
rijk







*
*

27.







cos sin , 0 , 0
cos sin sin cos
0
sin cos
sin cos cos sin
0
cos sin
2cos2 sin22
ttt
tttdttt
ttt
tttdtttt
tttt
aijvjkri
vijijC
vjCjkCk
vijk
rijkijkC
r iCi C0
rijk
rijk








*
*

28.












2
2
8
88
023 3
1318
1318
34
0
13 4
aik
vikikC
v iC i jk C i jk
vijk
rijk
ij kC
riC0Ci
rijk
t
tt
t
t
t
t
te
te dtet
te t
te tdt
et t t t
tet ttt









*
*

29.


2
2
88 cos 30 10 88 sin 30 16
44 3 10 44 16
tt tt
ttt
ri j
ij





0
0
300
50

106. 106 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
30.
22
900 cos 45 3 900 sin 45 16 450 2 3 450 2 16ri ji jtt ttttt

The maximum height occurs when

450 2 32 0,yt t which implies that 225 2 16.t
The maximum height reached by the projectile is

2
225 2 225 2 50,649
3 450 2 16 6331.125
16 16 8
y




feet.
The range is determined by setting

2
3 450 2 16 0yt t t which implies that

450 2 405,192
39.779
32
t



seconds
Range:
450 2 405,192
450 2 25,315.500
32
x





feet
31.

0022
001
cos sin 3 16
2 22
vv
tv thv tgt t ttri ji j







0
300
2
v
twhen
0 2
3163.
2
v
tt

0
300 2
,t
v

2
0
00
300 2 300 2
16 0,
2
v
vv




2
2
0
300 32
300 0
v



2
0
300 32 ,v
0 9600 40 6,v
040 6 97.98 ft secv
The maximum height is reached when the derivative of the vertical component is zero.


0 222 40 6
3 16 3 16 3 40 3 16
22
tv
yt t t t t t


40 3 32 0
40 3 5 3
32 4
yt t
t


Maximum height:
2
53 53 53
3 40 3 16 78
444
y




feet
32.

2
220
50 mi h ft sec
3
220 220
cos 15 5 sin 15 16
33
ri jtt tt






The ball is 90 feet from where it is thrown when
220 27
cos 15 90 1.2706
3 22 cos 15
xtt

seconds.
The height of the ball at this time is
2
220 27 27
5 sin 15 16 3.286 feet.
3 22 cos 15 22 cos 15
y




33.

0
0cos or
cos
x
xt tv t
v



2
0
sin 16yt tv t h


2
22
0
22 2
00 0
16
sin 16 tan sec
cos cos
xx
yv hx xh
vv v

107. Section 12.3 Velocity and Acceleration 107
© 2010 Brooks/Cole, Cengage Learning
34.
2
0.005yx xπθ
From Exercise 33 we know that tan
is the coefficient
of x. So, tan 1,
π
4rad 45. Also

2
2
016
sec
v
πnegative of coefficient of
2
x

2
0
16
2 0.005
v
π
or
080 ft secvπ



2
Position function40 2 40 2 16ri j.tt tt
When 40 2 60,tπ



Direction
60 3 2
440 2
40 2 40 2 32
32
40 2 40 2 24 2
4
825 2.
32
Speed 8 2 25 4 8 58 ft sec
4
vi j
vi j
ij
v
t
tt
ππ











35.

2
0.004 0.37 6tt t tri j
(a)
2
0.004 0.37 6yxx
(b)
(c) 0.008 0.37 0 45.8375yx x
and
45.8375 14.56 fty
(d) From Exercise 33,
tan 0.3667 20.14

22
2
0
22
0
0
16 sec 16 sec 4000
0.004
0.004 cos
67.4 ft sec.
v
v
v




36.

2
140 cos 22 2.5 140 sin 22 16tt ttri j
When 375,xπ 2.889t
and 20.47y feet.
So, the ball clears
the 10-foot fence.

37.
miles feet sec 440
100 mi h 100 5280 3600 ft sec
hr mile hour 3

ππ



(a)

2
00440 440
cos 3 sin 16
33
tt ttri j






Graphing these curves together with 10yπshows that
020 .
(b)
(c) You want


440
cos 400
3
xt t


π/


and

2440
3 sin 16 10.
3
yt t t


/



From
,xtthe minimum angle occurs when
30 11 cos .t π Substituting this for tin ytyields:




2
2
2
2
2
1
440 30 30
3 sin 16 10
3 11 cos 11 cos
14,400
400 tan sec 7
121
14,400
1 tan 400 tan 7 0
121
14,400 tan 48,400 tan 15,247 0
48,400 48,400 4 14,400 15,247
tan
2 14,400
48,400 1,464,332,80
tan







θ



θπ



π
θ
π
0
19.38
28,800





0
0
18
120
450
0
0
60
5000
0
100
θ
0
= 10θ
0
= 15
θ
0
= 20θ
0
= 25

108. 108 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
38. 7hfeet, 35 , 30 yards 90 feet


2
00
cos 35 7 sin 35 16tv t v ttri j


(a)
0cos 35 90vt when
2
0
7 sin 35 16 4vtt



0
2
0
00
22
0
2
0
2
0
90
cos 35
90 90
7 sin 35 16 4
cos 35 cos 35
129,600
90 tan 35 3
cos 35
129,600
cos 35 90 tan 35 3
54.088 ft sec
t
v
v
vv
v
v
v











(b) The maximum height occurs when

0sin 35 32 0.yt v t

0sin 35
0.969 sec
32
v
t

At this time, the height is
0.969 22.0 ft.y
(c)

090 cos 35 90
90
2.0 sec
54.088 cos 35
xt v t
t




39.

2
cos sin 16ri jtv t v tt


(a) You want to find the minimum initial speed v as a function of the angle .
Because the bale must be thrown to the position
16, 8 , you have



2
16 cos
8sin 16.
vt
vtt





16 costv from the first equation. Substituting into the second equation and solving for ,vyou obtain:


2
22
22
22
2
2
2
16 16
8sin 16
cos cos
sin 1
1 2 512
cos cos
1sin
512 2 1
cos cos
1sincos2sincoscos
21
cos 512 512
512
2sin cos cos
v
vv
v
v
v
v
























You minimize

2
512
.
2sin cos cos
f









22
2
2
2cos 2sin 2sin cos
512
2sin cos cos
02cos2 sin2 0
tan 2 2
1.01722 58.28
f
f
















Substituting into the equation for
, 28.78 ft sec.vv

109. Section 12.3 Velocity and Acceleration 109
© 2010 Brooks/Cole, Cengage Learning
(b) If 45 ,


22
2
16 cos
2
2
8 sin 16 16
2
vtvt
vttvtt

ππ
π′π′
From part (a),

2
2 512 512
1024 32 ft sec.
12
222 22 22
vv
′

40. Place the origin directly below the plane. Then 0,
π
0792vπ and



22
00
cos 30,000 sin 16 792 30,000 16
792 32 .
ri ji j
vij
tv t v tt t t
tt
π′

At time of impact,
22
30,000 16 0 1875 43.3tt t seconds.





43.3 34,294.6
43.3 792 1385.6
43.3 1596 ft sec 1088 mi h
30,000
tan 0.8748 0.7187 41.18
34,294.6
ri
vij
v

π
π′
ππ


41.

2
00
cos sin 16tv t v ttri j




2
0
sin 16 0vtt′π when 0tπand
0sin
.
16
v
t
π
The range is


2
00
00
sin
cos cos sin 2 .
16 32
vv
xv tv

ππ π
So,


2
1200 1
sin 2 3000 sin 2 1.91 .
32 15
x

42. From Exercise 41, the range is

2
0
sin 2
32
v
xπ
So,

2
0
200 sin 24
32
v
x


2
0
6400 sin 24v

0125.4 ft secv
43. (a) 10 ,

066 ft secvπ




2
2
66 cos 10 0 66 sin 10 16
65 11.46 16
tt tt
tt tt
ri j
ri j




Maximum height: 2.052 feet
Range: 46.557 feet
(b) 10 ,

0146 ft secvπ




2
2
146 cos 10 0 146 sin 10 16
143.78 25.35 16
tt tt
tttt
ri j
ri j




Maximum height: 10.043 feet
Range: 227.828 feet
(c)
045 , 66 ft secv




2
2
66 cos 45 0 66 sin 45 16
46.67 46.67 16
tt tt
tt tt
ri j
ri j




Maximum height: 34.031 feet
Range: 136.125 feet
α
α
30,000
34,295(0, 0)
0
0
50
5
0
0
300
15
0
0
200
40

110. 110 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
(d) 45 ,
0146 ft secv




2
2
146 cos 45 0 146 sin 45 16
103.24 103.24 16
tt tt
tt tt
ri j
ri j




Maximum height: 166.531 feet
Range: 666.125 feet
(e)
060 , 66 ft secv




2
2
66 cos 60 0 66 sin 60 16
33 57.16 16
tt tt
tt tt
ri j
ri j




Maximum height: 51.047 feet
Range: 117.888 feet
(f )
060 , 146 ft secv




2
2
146 cos 60 0 146 sin 60 16
73 126.44 16
ri j
ri j
tt tt
tt tt




Maximum height: 249.797 feet
Range: 576.881 feet
44. (a)

2
00
cos sin 16ttv tv tri j

0sin 16 0tv t when
0sin
.
16
v
t

Range:
2
00
0
sin
cos sin 2
32 32
vv
xv




The range will be maximum when

2
0
2cos2 0
32
dx v
dt




or

2, rad.
24



(b)

2
0
sin 16yt tv t

0sin 32 0
dy
vt
dt when
0sin
.
32
v
t

Maximum height:

22 22 22
00 0 0
2
sin sin sin sin
16
32 32 32 64
vv v v
y




Minimum height when
sin 1,or
.
2



45.


2
00
2
cos sin 4.9
100 cos 30 1.5 100 sin 30 4.9
tv thv t t
ttt
ri j
ij




The projectile hits the ground when

2 1
2
4.9 100 1.5 0 10.234tt t seconds.
So the range is

100 cos 30 10.234 886.3 meters.
The maximum height occurs when 0.dy dt

100 sin 30 9.8 5.102 sectt
The maximum height is

2
1.5 100 sin 30 5.102 4.9 5.102
129.1 meters.
y


46.


2
00
2
00
cos sin 4.9
cos 8 sin 8 4.9
tv thv t t
vtvtt
ri j
ij




50xwhen

0
0
50
cos 8 50 .
cos 8
vt t
v


For this value of
,0:ty



2
0
00
2
00
22 2
0
50 50
sin 8 4.9 0
cos8 cos8
4.9 2500 4.9 50
50 tan 8 1777.698 42.2 m sec
cos 8 tan 8 cos 8
v
vv
vv
v






0
0
140
60
0
0
600
300
0
0
800
200

111. Section 12.3 Velocity and Acceleration 111
© 2010 Brooks/Cole, Cengage Learning
47.






222
2
sin 1 cos
cos sin 1 cos sin
sin cos sin cos
21cosrij
vijij
aijij
v
atbt t b t
tb tb tb tb t
tb t b tb t t
tb t
tb;; ;
;;;;;; ;;;
;; ;; ; ; ;
;;
;



π′
π
(a)

0tvπwhen 0,2,4, .t;π
(b)

tvis maximum when ,3 , ,t;π then 2.vtb;π
48.

sin 1 cos
1cos sin
rij
vij
tbt t b t
tb t t;; ;
;; ;



Speed

22
1 2 cos cos sin 2 1 cos .vtb t t t b t;;;;;;
The speed has a maximum value of 2b
;when ,3 ,t;π

α 60 mi h 88 ft sec 88 rad sec since 1 .bππ π
So, the maximum speed of a point on the tire is twice the speed of the car:

α
2 88 ft sec 120 mi hπ
49.


22
sin cos
sin cos sin cos 0vij
rvtb tb t
tt b t tb vt t;; ;;
;; ; ; ;
"
So,
trand tvare orthogonal.
50. (a) Speed
vπ
22 2 22 2 22 2 2
sin cos sin cosbtb tb ttb;; ;; ; ; ; ;

(b)
The graphing utility draws the circle faster for greater values of
.;
51.

22 2 2
cos sin cos sinaijijrtb tb t b t t t;; ;; ;; ; ;


tais a negative multiple of a unit vector from 0, 0 to cos , sintt;; and so tais directed toward the origin.

52.

22
cos sinaijtb t vt b;; ;

53.
2
,2atbb;ππ





22 1
32
132
210
410radsec
810ftsec
F
v
m
mb
tb
;;
;
;
π
ππ π
π
ππ

54.
30 mi h 44 ft sectvππ


44
rad sec
300vt
b
;ππ



2
2
2
3400 44 2057
300 lb
32 300 3
F
at b
mb
;
;π

ππ π


Let
n be normal to the road.

cos 3400
2057
sin
3n
n
π
π
Dividing,
121
tan
600
11.4

π

10−10
−10
10

112. 112 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
55. To find the range, set

2
01
sin 0
2
yt h v t gt
then

2
01
0sin.
2
gtv th

π′ ′


By the Quadratic
Formula,
αdiscount the negative value




2
00
22
00
sin sin 4 1 2
212
sin sin 2
second.vv gh
t
g
vv gh
g


π



π

At this time,


22
00
0
0 22
00
2
0
2
0 2
2
0
sin sin 2
cos
cos 2
sin sin
cos 2
sin sin feet.vv gh
xt v
g
vgh
vv
gv
vgh
gv







π














56. 6feet,hπ
045vπ feet per second, 42.5 . From
Exercise 55,


2
2
45 sin 42.5 45 sin 42.5 2 32 6
32
2.08 seconds.
t

π

At this time,
69.02 feet.xt

57. txtytztrijk Position vector

txt yt ztvijk Velocity vector

txtytztaijk Acceleration vector
Speed


222
, is a constant.
txtytzt
CC
v
π





222
0
22 2 0
20
0
d
xt yt zt
dt
xtx t yty t ztz t
xtx t yty t ztz t
ttva





"π

Orthogonal

58.





22
, and are constants.
, is a constant.
txt yt
yt mxt b m b
txt mxtb
txtmxt
txtmxtCC
rij
ri j
vij
s







So,



2
1
0
.
C
xt
m
xt
txtmxt
aij0
π

π


59. 6cos 3sintttrij
(a)
6 sin 3 cos vr i jtt t tππ′


22
22 2
36 sin 9 cos
34sin cos 33sin 1ttt
tt t
v


6cos 3sintt t tav i jππ′ ′
(b)
(c)
(d) The speed is increasing when the angle between
vand ais in the interval

0, .
2




The speed is decreasing when the angle is in the interval

,.
2





60.
cos sinrijta tb t;;
(a)
sin cosrv i jtta tb t;; ;;
Speed

22 2 22 2
sin cosvta tb t;;;;
(b)
ttav π


22
2
2
cos sin
cos sin
ij
ij
ratbt
atbt
t;; ;;
;; ;
;π′ ′
π′ ′
π′
61. The velocity of an object involves both magnitude and
direction of motion, whereas speed involves only
magnitude.
t 0
4


2


2
3

Speed 3
3
10
2
6
3
13
2
3
−2−4−82468
−2
−4
−6
−8
2
4
6
8 x
y

113. Section 12.4 Tangent Vectors and Normal Vectors 113
© 2010 Brooks/Cole, Cengage Learning
62. (a) The speed is increasing.
(b) The speed is decreasing.
63. (a)

1
21
2
txtytzt
tt
rijk
rr

Velocity:
22
21rrtt
Acceleration:
21 42ttrr
(b) In general, if
31 ,rrtt; then:
Velocity:
31rtrt;;
Acceleration:

2
31
rrtt;;
64.







1
11
2
22
sin cos
cos sin
0
cos sin
sin cos
0
sin cosaij
va ijC
viiCC0
vij
rv ijC
rjjCC0
rijttt
ttdt tt
ttt
ttdt tt
ttt







*
*

The path is a circle.
65. False. The acceleration is the derivative of the velocity.
66. True
67. True
68. False. For example,

3
6.ritt t Then
2
3vitt and
6.aitt vtis not orthogonal to .at
Section 12.4 Tangent Vectors and Normal Vectors
1.
2.
3.
4.
5.
2
2, 1tt ttrij

22,ttrij
22
442 1tt tr





1 122
1
221 2
r
Tijij
r



6.







32
2
42
2
34
916
1 134
134
551 916
tt t
tt t
ttt
rij
rij
r
r
Tijij
r







7.



22
4cos 4sin ,
4
4sin 4cos
16 sin 16 cos 4
224
422
4
rij
rij
r
r
Tij
r
tttt
ttt
ttt


















8.





22
6cos 2sin
6sin 2cos
36sin 4 cos
33 13
33
3 2836 3 4 1 4
3
rij
rij
r
r
ij
Tij
r
ttt
ttt
ttt

















x
y
y
x
T
T
T
N
N
N
y
x
x
y
T
TT
N
N
N

114. 114 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
9.





2
2
3ln,
1
3
1
3
1
3
3
0.9926 0.1217
1 91
9
tt tte
t
t
e
e
e
e
e
ee
e
e
rij
rij
rij
r
T
r
ij
ij
ij
πθ π
πθ
πθ

π

θ
θ




10.







cos , 0
cos sin
0
0 22
0
220 2
rij
rij
rij
r ij
Tij
r
tt
tt t
te tet
tetete







11.


2
,0,0,0
2
ttt tP
tt
rijk
rijk



When 0,tπ
0,rik 0at 0,0,0 .t






0 2
0
20
r
Tik
r




Direction numbers: 1,aπ 0,bπ 1cπ
Parametric equations: ,xtπ 0,yπ ztπ
12.

2 4
,
3
tt t
rijk
4
1, 1,
3
P





2ttrij
When 1,tπ

4
12 11,1,.
3
ttatrr ij










1 25
12
51 5
r ij
Tij
r




Direction numbers: 2,aπ 1,bπ 0cπ
Parametric equations: 21,xt 1,yt
4
3
zπ
13.
3cos 3sin , 3,0,0ttttPrijk

3sin 3costttrijk
0tπat
3, 0, 0P

03rjk




0 3
0
0 10r
jk
T
r
ππ


Direction numbers: 0,aπ 3,bπ 1cπ
Parametric equations: 3,xπ 3,ytπ
ztπ
14.


2
,, 4 , 1,1, 3ttt tPrπθ


2
1, 1,
4
t
t
t
rπθ
θ

When 1,tπ

1
11,1, ,
3
rπθ 1tπat 1, 1, 3 .






1 21 1
11,1,
71 3
r
T
r

ππ θ


Direction numbers:
1
1, 1,
3
abcπππθ

Parametric equations: 1,xt 1,yt
1
3
3
zt

15.



2 cos , 2 sin , 4 , 2, 2, 4
2sin ,2cos ,0
r
r
tttP
ttt
π
πθ

When ,
4
t

π 2, 2, 0 ,
4
r

πθ



at 2, 2, 4 .
4
t

π






1
2, 2, 0
42 4r4
T
r

ππθ



Direction numbers:
2, 2, 0abcπθ π π
Parametric equations: 22,xt 22,yt
4zπ
16.



2
2sin ,2cos ,4sin , 1, 3,1
2cos , 2sin ,8sin cos
r
r
ttttP
ttttt
π
πθ

When ,
6
t

π 3, 1, 2 3 ,
6
r

πθ



at 1, 3, 1 .
6
t

π






6 1
3, 1, 2 3
64 6r
T
r

ππθ



Direction numbers:
3,aπ 1,bπθ 23cπ
Parametric equations: 31,xt 3,yt
23 1zt
17.


23
22
,,
3
1, 2 , 2
r
rtttt
ttt
π
π
When 3,tπ

31,6,18,rπ

3at 3,9,18 .t





3 1
31,6,18
193r
T
r
ππ


Direction numbers: 1,aπ 6,bπ 18cπ
Parametric equations: 3,xt 69,yt 18 18zt
x
y
3
3
−3
6
6
9
9
12
12
15
15
18
18
z

115. Section 12.4 Tangent Vectors and Normal Vectors 115
© 2010 Brooks/Cole, Cengage Learning
18.

1
3cos 4sin
2
1
3sin 4cos
2
ttt
ttt
rijk
rijk



When ,
2
t


1
3,
22
rik





at 0, 4, .
24
t









2 211
36
22 2 37 37r
Tikik
r





Direction numbers: 6,a 0,b 1c
Parametric equations: 6,xt 4,y
4
zt


19.

0ln , 1tt t ttrijk


11
;
2
t
t t
rij k
1
1
2
rijk






12 221
1
33311 14
t
t
rijk
Tijk
r




Tangent line: 1 ,x t ,yt
1
1
2
zt



00.1 1.1 1.1 0.1 1.05
1.1, 0.1, 1.05
trrijk

20.
02cos 2sin , 0
t
te t ttri jk



2sin 2cos
t
te t trijk



02,rij 02,rik
05r




0 2
0
0 5r ik
T
r



Parametric equations:
1,
xss 2,ys2zs s



00.1 0 0.1
1 0.1,2,20.1 0.9,2,0.2
trr


21.


42,16,2
82,16,2r
u

So the curves intersect.


1
1, 2 , ,
2
ttr

1
41,8,
2
r


2311
,2, ,
43
ssu


11
8,2,
412
u




48 16.29167
cos 1.2
16.2951348ru
ru

"



22.


00,1,0
00,1,0r
u

So the curves intersect.


1, sin , cos ,tttr 01,0,1r


sin cos cos , sin cos cos ,
11
cos 2
22
s ss s ss s
s
u




01,0,1u




00
cos 0
200ru
ru
"



23.

21
,2
2
tt tt
rij

ttrij




2
1
t t
t
t t
r ij
T
r







32 32
22
1
11
t
t
tt
Tij






32 32
21
2
55
Tij







2 1255
22
552 5
T
Nijij
T




24.

6
,3tt t
t
rij


2
6
t
t
rij





2
4
2
2
4
16
136
6
36
t
ti
tt
t
t
ttr
Tj
r
ij












3
32 32
44
72 12
36 36
tt
t
tt
Tij




32 32
144 96
2
52 52
Tij





2 1
232
2 13T
Nij
T



4
4
4
1
1
2
3
5
5
5
yx
z

116. 116 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
25. ln 1 , 2ttttri j


1
t
t
rij




2
2
1
1 1
1
t t
t
t
t t
t
ij
r ij
T
r










32 32
22
1
11
t
t
tt
Tij






32 32
21
2
55
Tij







2 25 5
2
552
T
Nij
T




26.

cos sin ,
6
ttttrij




sin costttrij


tr




sin cos
t
ttt
t
r
Tij
r





cos sin , 1ttttTijT

31
622
Tij





316
622
6
T
Nij
T













27.

2
ln , 1ttt ttrijk


1
2tt
t
rijk




2
42
2
2
1
2
2
1 41
14
ijk
r ijk
T
r
t
t tt
t
t
t tt
t
t










43 3
32 32 32
42 42 42
14 2 4 8
414141
ttttt
t
tt tt tt
Tijk






+,
32 32 32 32
36 9 3
123
666 6
Tijkijk





2 3 14 2 14 3 14
1
14 14 1414
ijk
Nijk


28.

2,0rijk
tt
tteet




2
tt
teeri j k





2
22
2
t t tt tt
teeeeeer






2
tt
tt
t ee
t
eet
r ij k
T
r










222
2 22
tt
tt tt tt
ee
t
ee ee ee
Tijk








11
0
22
Tjk



22
0
22
Njk
29.

3
6cos 6sin ,
4
ttttrijk


6sin 6costttrij




sin cos
t
ttt
t
r
Tij
r





cos sin , 1ttttTijT



34322
422 34
T
Nij
T





30.
cos 3 2 sin 3 ,tt ttrijk

3sin3 6cos3tttrij




22
3sin3 6cos3
9sin 3 36cos 3
r ij
T
r
t tt
t
t tt




The normal vector is perpendicular to
tTand points
toward the z-axis:


22
6cos3 3sin3
9sin 3 36cos 3
tt
t
tt
ij
N






6
36
i
Ni

117. Section 12.4 Tangent Vectors and Normal Vectors 117
© 2010 Brooks/Cole, Cengage Learning
31. 4ttri

4tvi

ta0




4
4
t
t
t
v i
Ti
v



tT0




t
t
t
T
N
T



is undefined.
The path is a line and the speed is constant.
32.
42tttrij

42tvij

ta0





1
2
5
t
t
t
v
Tij
v



tT0




t
t
t
T
N
T



is undefined.
The path is a line and the speed is constant.
33.

2
4ttri

8ttvi

8tai




8
8
t t
t
tt
v i
Ti
v



tT0




t
t
t
T
N
T



is undefined.
The path is a line and the speed is variable.
34.

2
ttrjk

2ttvj

2taj




2
2
t t
t
tt
v j
Tj
v



tT0




t
t
t
T
N
T



is undefined.
The path is a line and the speed is variable.
35.

1
,tt
t
rij

2
1
,t
t
vij
1,vij



3
2
,1 2t
t
ajaj






2
2
2
44
11
11
t t
tt
tt tt
v
Tijij
v








12
1
22
Tijij







3
32 32
44
4
2
4
22
11
2
1
1
1
tt
ttt
t
tt
t
t
t
ij
T
N
T
ij











12
1
22
2
2
a
a
T
N
Nijij
aT
aN

"
"

36.

2
2, 1tt ttrij

22,122ttvijvij

2, 1 2taiai





22
11
22
44 1
v
Tijij
v
t
ttt
t tt






122
1
222
Tijij






32 32
22
2
2
1
11
1
1
1
1
t
ttt
t
t
t
t
t
ij
T
N
T
ij











22
1
22
Nij
2a
TaT"
2a
NaN"

118. 118 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
37.
32
2, 1ttt ttrij


2
13 4,1 2 4tttvijvij

64,1 64ttaijaij





2
42
13 4
9101
ttt
t
t tt
ijv
T
v







24 2 5
12
520 5
ij ij
Tij







2
4
32 32
42 42
16 3 1 436
9101 9101
tt t
t
tt tt
Tij






32 32
64 32
1
20 20
Tij





2
1
5
ij
N





1145
68
55
a
TaT"


185
12 4
55
a
NaN"
38.

32
41,ttt trij 0t


2
342,0 4tt tvijvi

62,02ttaijaj





2
42
342
92016
ttt
t
t tt
ijv
T
v






4
0
16
i
Ti








2
4
32 32
42 42
43 4 32 18
9 20 16 9 20 16
tt t
t
tt tt
Tij






32
32 1
1
16 2
Tjj


1Nj
0a
TaT"

2a
NaN"

39.
2
,0
tt
teetrij




2
2,0 2
tt
te evi
jvi j




2
4,0 4
tt
te eai
jai j






2
42
2
4
tt
tt
t ee
t
t ee
v ij
T
v








2
0
5
ij
T





2
0
5
ij
N





175
18
55
a
TaT

"


165
24
55
a
NaN"


40. ,0
tt
tee ttrijk



,0
tt
teevijkvijk



,0
tt
teeaijaij






22
1
v ijk
T
v
tt
tt
t ee
t
t ee








0
3
ijk
T










22 2 2 4
32 32 32
42 42 42
2211
111
Tijk
tt t t t t
tt tt tt
ee e e e e
t
ee ee ee






32 32
33
0
33
Tij



22
0
22
Nij
0a
TaT"

2a
NaN"

119. Section 12.4 Tangent Vectors and Normal Vectors 119
© 2010 Brooks/Cole, Cengage Learning
41. cos sin
tt
tet etrij

cos sin cos sin
tt
te t t e t tvij

2sin 2cos
tt
te t e taij
At
,
2
t


12
.
22v
Tijij
v

Motion along
ris counterclockwise. So,


12
.
22
Nij ij

2
2
2
2
ae
ae
T
NaT
aN


"
"

42. cos sinrijta tb t;;

sin cosvijta tb t;; ;;

0vj b;

22
cos sinai
jta tb t;; ;;


2
0ai a ;




0
0
0
v
Tj
v


Motion along tris counterclockwise . So, 0.Ni

2
0
T
NaT
aN
a
aa
;
"
"

43. 0000000cos sin sin cosrijttttttt;; ; ;; ;


22
000 00
cos sinvijttt tt;; ;;


2
0000000
cos sin cos sinaijttttttt;;;; ;; ;



000 cos sin
v
Tij
v
ttt ;;
Motion along
ris counterclockwise. So

000 sin cos .Nijttt ;;

2
TaTa ;"


23
00
NaNatt;; ;"

44. 00 0 0 sin 1 cosrijtt t t;; ;

000 1cos sinvijttt;; ;



2
000
sin cosai
jttt;; ;


00
01cos sin
21 cos
ijv
T
v
tt
t ;;
;



Motion along
ris clockwise. So,

00
0sin 1 cos
.
21 cos
ij
N
tt
t;;
;




22
0
0
0
sin
1cos
21 cos 2
TaT
t
at
t
;; ;
;
;
"


2
0
1cos
2
NaNat
;
;
"

45. cos sinrijta ta t;;

sin cosvijta ta t;; ;;


22
cos sinaijta ta t;; ;;




sin cos
v
Tij
v
t
ttt
t;;




cos sin
T
Nij
T
t
ttt
t;;




0a
TaT"

2
NaNaa ;"

46. tTpoints in the direction that ris moving. tNpoints
in the direction that
ris turning, toward the concave side
of the curve.


47. Speed:
vta ;
The speed is constant because 0.a
T

48. If the angular velocity;is halved,

2
2
.
24
N
a
aa;;




a
Nis changed by a factor of
1
.
4

x
a
a
N
T
y

120. 120 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
49. 0
1
,2tt t
t
rij


1
,1xty xy
t




2
1
t
t
rijπ′



2
4
1
t
t
t
ij
T′
π




2
4
1
t
t
t
ij
N
π




1
22
2
rij




17
24
17
Tijπ′



17
24
17
Nij

50.
3
0
,1tt ttrij

3313
,orxtyt x y yx


2
3ttri
j




2
4
3
91
t t
t
t t
r ij
T
r

ππ




3
1
10
310 10
10 10ij
T
ij
π




10 3 10
1
10 10
Nijπ′

51.
2
0 1
44,
4
tttt
rij
4,
xtπ

2
2
2
44
44
x x
yt

ππ π




11
44
r
ij






48ttrij


22
48 2
16 64 1 4
tt
t
ttij ij
T
ππ



1
2
1
4
12255
455 51 ij ij
Tij





1
4
N



is perpendicular to
1
:
4
T





12
4 5 ij
N
π




52.
2
0
21 , 2tt ttrij
21,xt

2
2
1
2
x
yt
′
π′ π′




254rijπ′

22ttrijπ′


22
22
44 1
tt
t
ttij ij
T′′
ππ




2
2
5ij
T′
π



2
2,
5ij
N′′
π
perpendicular to 2T

53.
02cos 2sin ,
4
ttttrij



2cos ,
x tπ
22
2sin 4ytxy

2sin 2costttrij



1
2sin 2cos sin cos
2
ttttt
Tijij

cos sintttNijπ′ ′

22
4rij







2
42
Tij







2
42
Nij

π′′




54. 03cos 2sin ,ttttrij

cos ,
3
x
tπ
22
sin 1, Ellipse
294
yxy
t

3sin 2costttrij

2rjTj

Niπ
N
x
32
T
3
2
1
1
2,
1
2
y
))
y
x
−112
−1
1
2
(1, 1)
T
N
−1−212
−1
1
2
3
y
x
T
N
1,
1
4
))
−2−4−6 246
−4
−6
−8
2
x
(5, −4)
y
N
T
1
−1
1−1
x
N
T
y
2, 2)(
y
x
−1−212
−1
−3
1
3
T
N

121. Section 12.4 Tangent Vectors and Normal Vectors 121
© 2010 Brooks/Cole, Cengage Learning
55. 23,1tt t ttrijk

23tvijk

ta0




1
23
14
14
23 1
14
t
v
T
v
ijk
ijkT






t
T
N
T


is undefined.

,aa
TNare not defined.

56. 442t tttrijk

442tvijk

ta0



1
22
3
tv
Tijk
v



t
T
N
T


is undefined.

,aa
TNare not defined.

57.
cos sin 2 ,
3
tttttrijk



sin cos 2tttvijk


cos sin , 1tttaija


22
sin cos 4 5tttv





1
sin cos 2
5
t
ttt
tv
Tijk
v


0att
TaT"

22
10 1aa
NT a

131
2
322 5
Tijk




13
322
TnaijTNN aa





58.

2
3,1t ttttrijk

32,ttvijk
2
10 4ttv


2, 2taka




2
32
10 4
t t
t
t tv ij k
T
v




12,ak

1
132
14
Tijk


4
11
14
a
TaT

"


22 16 20 2 35
4
14 7 7
aa
NT a


41 235
232
714 14
aa
TNak T N ijk N







7 2 35 6 2 10
23 2
710777235
Nkijk ijk

122. 122 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
59.
2
2
,1
2
t
ttt trijk









2
32
2
2
2
2
12
2
1
2
15
6
12
6
52
15 52
5 51 5
15ttt
t
ttt
t
t
t t
t
t
tvijk
vijk
ajk
v
Tijk
v
Tijk
ijk
Tijk
N
T















30
152
30
Nijk

56
6
a
TaT"

30
6
a
NaN"
60.

2
21 4, 2tt tttrijk



22
22 4,
20 4 2 5tt
tttvijk
v



2, 2taja



22
22 4 1
2
25 5t
tt
ttijk
Tijk







1
22,2 22
3
ajT ijk


4
22
3
a
TaT"

22 16 2
45
93
aa
NT a


42
2225
93
TNaj T N ijk Naa


34
222
925
25 5 45
15 3 15
Njijk
ij k







61.
sin cos
tt t
tete terijk

cos sin sin cos
tt tt t
t e te t e te t evijk

0vijk

2cos 2sin
ttt
teteteaijk

02aik



1
cos sin sin cos
3
t
tt t t
v
T
v
ijk






+,
1
0
3
Tijk



1
sin cos cos sin
2
tttttNij



22
0
22
Nij
3a
TaT"
2a
NaN"
62.
2,0
tt
te tetrijk




22
2, 4
tt tt
te e t e evijkv




22
,
tt t t
tee t e eaika



02,0vijkaik


2
0
6ijk
T


0
a
TaT"

22
2
NT aaa


2
1
2aa
TNaik T N N
Nik

123. Section 12.4 Tangent Vectors and Normal Vectors 123
© 2010 Brooks/Cole, Cengage Learning
63. 43cos 3sin,
2tt t ttri j k



43sin 3costttvi j k

43
2
vij

πθ




3cos 3sintttajkπθ θ

3
2
ak

πθ






1
43sin 3cos
5
ttt
v
Tijk
v



1
43
25
Tij
πθ





cos sinttt
T
Njk
T

ππθθ



2
Nk

πθ



0
a
TaTπ" π
3
a
NaNπ" π
64.

2cos 1sin ,
3
t
tttt
rijk



110
sin cos ,
33
ttt tvijkv


1
3
vjk

cos sintttaij

aiπ


310 1
sin cos
10 3
ttt
v
Tijk
v







310 1
10 3
T
jk








310
cos sin
10
cos sin
310
10
ij
T
Nij
Ttt
ttt






Niπ

0, 1aa
TNaT aNπ" π π" π

65.

2
2
3
2
t
tt t
rijk

6tttvijk

2122vijk

6tajk



2
1
6
137
ttt
t
v
Tijk
v






1
2122
149
Tijk



+,
+,
32
2
2
2
1
37 6
137
37
137
1
37 6
37 1 37
t
t
t
t
t
T
N
T
ijk
ijk

π



π






+,

1
2746
37 149
1
74 6
5513
Nijk
ijk




74
149
a
TaTπ" π

37 37
5513 149
a
NaNπ" π π
66.

2
2, 1tt ttrijk

22ttvik

122vik

2taiπ



22
1
22
44 1 t
tt
ttvik
Tik
v







1
1
2
Tik




32
2
32
22
1
111
tt it
t
tt
2
ikTk
N
T t
θ
ππ π






1
1
2
Nikπθ


2
2
2
aa
T Tπ" π π

2
2
2
aa
N Nπ" π π
x
y
2π
4π
3
3
z
T
N
x
y
z
2
2
3
1
1
2
1
T
N
x
y
z
2
3
2
3
4
5
1
3
4
T
N
y
x 4
6
8
2
4
2
4
N
T
z

124. 124 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
67. Let C be a smooth curve represented by r on an open
interval
I. The unit tangent vectorTtat t is defined as




, 0.
r
Tr
r
t
tt
t

#

The principal unit normal vector
Ntat t is defined as




, 0.
T
NT
T
t
tt
t

#

The tangential and normal components of acceleration
are defined as
.
TNaTNtatat
68. The unit tangent vector points in the direction of motion.
69. (a) If 0,
a
Nthen the motion is in a straight line.
(b) If 0,
a
Tthen the speed is constant.
70.
34tttrij

34, 9165tt tvr ijv

ttav 0




34
55
t
t
t
v
Tij
v



0ttTN does not exist.
The path is a line. The speed is constant (5).
71.

sin , 1 cosrtt t t
The graph is a cycloid.
(a)

sin , 1 cosrtt t t


cos , sinvttt


22
sin , cosattt





1
1cos,sin
21 cos
v
T
v
t
ttt
t t









1
sin , 1 cos
21 cos
T
N
T
t
ttt
t t










2
22
1sin
sin 1 cos cos sin
21 cos 21 cos
TaT
t
atttt
tt



"









22
22 2
21 cos1cos1
sin cos 1 cos
221 cos 21 cos
NaN
tt
attt
tt



"



When
22 2
122
:,
222 2
ta a
TN


When
2
1: 0 ,ta a
TN
When
3
:
2
t
22
22
,
22
aa
TN


(b) Speed:



2
21 cos
sin
21 cos
T
v
s tt
ds t
a
dt t




When
2
12
:0
22
ta
T

the speed in increasing.
When
1: 0ta
T the height is maximum.
When
2
32
:0
22
ta
T

the speed is decreasing.
x
t= 1
t=
2
1
t=
2
3
y

125. Section 12.4 Tangent Vectors and Normal Vectors 125
© 2010 Brooks/Cole, Cengage Learning
72. (a) cos sin , sin cosrttttttt


2222
sin sin cos , cos cos sin cos , sinvttttttttttttt


2323
cos sin , sin cosattttttt




cos , sin
v
T
v
t
ttt
tππ


23 23 2
cos cos sin sin sin cos
TaTatttttttt "


22 4 22 4 3
1aa tt
NT a
When 1,tπ
2
,a
Tπ
3
.a
NπWhen 2,tπ
2
,a
Tπ
3
2.a
Nπ
(b) Because
2
0a
T for all values of t, the speed is increasing when 1tπand 2.tπ
73.

02cos 2sin ,
22
t
ttttrijk




1
2sin 2cos
2
tttrijk



217 1
2sin 2cos
17 2
tttTijk






cos sintttNij

2
24
rjk






217 1 17
24
217 2 17
Tikik




2
Nj







4 17 17 17 4 17 17
04
222 17 171717 17
010
ijk
BTN i k ik
'




74.

3
2
0
,1
3
t
ttt trijk


2
2tttrijk



2
241
2
14
ttt
tt
Tijk





34 3
24 241
21 2
14 1
tttttt
tt tt
Nijk





1
1
3
rijk



1
12
6
Tijk




12
133
263
Nikik



666
333 3
111 636
333 3
22
0
22
ijk
BTN i j k ijk'


1
1
2
−1
−2
2
3 −2
yx
N
B
T
0, 2,
π
2
()
z
1
1
2
1
2
−
1
2
N
B
T
1
3
1, 1,()
x
y
z

126. 126 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
75. 0sin cos ,
4
ttttri j k



cos sin ,tttrjk


1tr

22
4422
rT jk





sin cos ,tttTjk

22
422
Njk





22
0
444 22
22
0
22
ijk
BTN i
'




76.
02cossin,0
tt t
teetettri j k

2cossin sincos
tt t t t
t e e tet ete tri j k



1
02 0 2
6
rijkT ijk




222222 2 2222 2
222 2 2
4 cos sin 2 cos sin sin cos 2 sin cos
42cossin 6
tt t t t t t
tt t
teetetettetetett
ee t te
r




6
t
ter





1
2cossin sincos
6t
ttttt
tr
Tijk
r








1
sin cos cos sin
6
tttttTjk



+,
122
00
226
TjkNjk




211
333
000 666
333
22
0
22
ijk
BTN i j k'

127. Section 12.4 Tangent Vectors and Normal Vectors 127
© 2010 Brooks/Cole, Cengage Learning
77. 04sin 4cos 2 ,
3tttttrijk



4cos 4sin 2 ,tttrijk


22
16 cos 16 sin 4 20 2 5tttr
2232
3
rijk






151555
2232 3
3555525
Tijkijkijk







1
4sin 4cos
25
tttTij

31
322
Nij






5155
51545 5
34555
333 10101010
31
0
22
ijk
BTN i j k ijk

'




78.

3cos2 3sin2 ,
4tttttrijk



6sin2 6cos2tttrijk


37tr
6
4
rik










1
6sin2 6cos2
37rt
ttt
rt
Tijk






1
12 cos 2 12 sin 2
37
tttTij


1
6
4 37
Tik








cos 2 sin 2
t
ttt
tT
Nij
T





4
jN






6116
0
444 37 37 37 37
010
ijk
BTN i k
'

128. 128 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
79. From Theorem 12.3 you have:


2
00
cos sin 16tvt hvt tri j

00cos sin 32tv v tvi j

32taj




00
2
22
00cos sin 32
cos sin 32
ij
T
vvt
t
vvt









00
2
22
00
Motion is clockwise.
sin 32 cos
cos sin 32
ij
N
vtv
t
vvt







0
2
22
0032 sin 32
cos sin 32
TaT
vt
a
vvt


"




0
2
22
0032 cos
cos sin 32
NaN
v
a
vvt

"

Maximum height when
0sin 32 0;vt vertical component of velocity
At maximum height, 0a
Tand 32.a
N
80.
045 , 150v

0
2
cos 150 75 2
2
v
"

0
2
sin 32 150 32 75 2 32
2
vt t t
"




2 2
32 75 2 32 16 32 75 2
256 1200 2 5625
11250 75 2 32
tt
a
tt
t
T








2 2
32 75 2
1200 2
256 1200 2 5625
11250 75 2 32
a
tt
t
N



At the maximum height, 0a
Tand 32.a
N
81. (a)


2
00
221
2
cos sin
120 cos 30 5 120 sin 30 16 60 3 5 60 16
tv thv tgt
tttttt
ri j
ijij




(b)
Maximum height
61.25 feet
range
398.2 feet
(c)

60 3 60 32ttvi j
Speed


2
2
3600 3 60 32 8 16 60 225ttttv

32taj
0
0
400
70

129. Section 12.4 Tangent Vectors and Normal Vectors 129
© 2010 Brooks/Cole, Cengage Learning
(d)
(e) From Exercise 79, using
0120vπ and 30 ,





2
2
2
2
32 60 32
60 3 60 32
32 60 3
60 3 60 32
T
N
t
a
t
a
t

π

π



At
1.875, 0ta
Tππ and the projectile is at its maximum height. When
Taand
Nahave opposite signs, the speed is
decreasing.
82. (a)


2
00
2
21
2
cos sin
220 cos 45 4 220 sin 45 16
110 2 4 110 2 16
tv thv tgt
ttt
ttt
ri j
ij
ij





(b)
Maximum height 382.125 at 4.86t
Range 1516.4
(c)



22
110 2 110 2 32
110 2 110 2 32
32
tt
tt
t
vi j
v
aj





(d)
83.

1
20
10 cos 10 , 10 sin 10 , 4 4 , 0rttttt
(a)



22
22
2
2
100 sin 10 , 100 cos 10 , 4
100 sin 10 100 cos 10 16
100 16 4 625 1 314 mi h
r
r
ttt
ttt




(b) 0a
Tπand
2
1000a
N π
0a
Tπbecause the speed is constant.
T 0.5 1.0 1.5 2.0 2.5 3.0
Speed 112.85 107.63 104.61 104.0 105.83 109.98
0
−20
4
40
a
N
a
T
0 1800
−10
450
t 0.5 1.0 1.5 2.0 2.5 3.0
Speed 208.99 198.67 189.13 180.51 172.94 166.58

130. 130 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
84. 600 mi h 880 ft sec


2
880 16 36,000tt tri j

880 32ttvij

32taj


22
880 32 55 2
16 4t 3025 4t 3025
ij ij
T
tt
t




Motion along ris clockwise, therefore


2
255
4 3025
ij
N
t
t
t





2
64
4 3025
t
a
t
TaT"



2
1760
4 3025
NaNa
t
"


85.
cos sinrijta t a t;;
From Exercise 45, we know 0aT" and
2
.aN a;"
(a) Let
02.w; Then


2
22
0
24aN aa a;;;"
or the centripetal acceleration is increased by a factor
of 4 when the velocity is doubled.
(b) Let
0 2.aaThen

22 2
0 1
22
aN
a
aa
;; ;

"



or the centripetal acceleration is halved when the
radius is halved.
86.
cos sinrijtr t r t;;

sin cosvijtr tr t;; ;;


1vtr rv;;


2
cos cosaijtr tr t;; ;;


2
atr;
(a)


2
222
a
mmv
Fmt mr r
rr ;;
(b) By Newton’s Law:

2
2
2
,,
mv GMm GM GM
vv
rrrr

87.
4
9.56 10
4.82 mi sec
4000 115
GM
r
v
'



88.
4
9.56 10
4.75 mi sec
4000 245
GM
r
v
'



89.
4
9.56 10
4.67 mi sec
4385
v
'

90. Let xdistance from the satellite to the center of the earth 4000 .xr Then:






4
22 4
22
22
4
3
2
2 2 9.56 10
24 3600
49.5610
24 3600
9.56 10 24 3600
26,245 mi
4
2 26,245
1.92 mi sec 6871 mi h
24 3600xx
v
tx
x
x
xx
v


'

'

'



91. False. You could be turning.
92. True. All the motion is in the tangential direction.
93. (a)



22 2 2
cosh sinh , 0
cosh , sinh
cosh sinh 1, hyperbolatbtbtb
xbtybt
x y bt btrij


(b)
sinh coshtb bt b btvij


22 2
cosh sinhtb bt b bt btai
j r

131. Section 12.5 Arc Length and Curvature 131
© 2010 Brooks/Cole, Cengage Learning
94. Let cos sintTij. . be the unit tangent vector.
Then



dt
sin cos .
T
T
T
ijM
d
t
dd d d
ddt dt dt
. ..
..
.
π
φφππ φ


sin cos
cos 2 sin 2
Mij
ij
. .
. .



and is rotated counterclockwise through an angle of
2from .T
If 0,ddt.then the curve bends to the left and
Mhas the same direction as .T
So,p
Mhas the same
direction as

,
T
N
T

π


which is toward the
concave side of the curve.
If 0,ddt.then the curve bends to the right and
Mhas the opposite direction
as .TThus,

T
N
T

π


again points to the
concave side of the curve.
95.
txt ytrij

,yt mxt bm and b are constants.

txt mxtbri j



txtmxtvij



22
2
1t xt mxt xt mv







2
, constant
1
tm
t
t m
vij
T
v

ππ

So,
.tT0π
96. Using
,aa
TNaTN ,TT 0'π and
1,TN'π
you have:



aa
aa
a
aa
TN
TN
N
NNva vT T N
vTTvTN
vTN
va v TN v
' '
''
π'
'π ' π
So,
.a
N
va
v
'
π


97.

2
22 2 2
22
222
2
aa aa
aaaa
aa
aa
TN TN
TTNN
TN
NT
aaa
TN TN
TTNN
a
π"
"
"

φπ
Because
0,a
Nwe have
22
.aa
NT aφπ

98.
1
dv dv
Fma
dt dt
ππ π Force

23
xat bt ct

2
23
dx
vabtct
dt

26
dv
bct
dt




22 22
22
2
424 36
41223
412
Fb bctct
bcbtct
bcva






2
41212Ffv b ac cv
The sign of the radical is the sign of 2 6 ,bct which
cannot change.
Section 12.5 Arc Length and Curvature
1. +
,3,0,3tttrijφπ
3, 1, 0
dx dy dz
dt dt dt
φφπφ


33 2
2
0 0
31 10310sdtt
*

y
x
69
−3
−6
3
(0, 0)
(9, −3)
x
M
T
φ
y
x
M
N
T
φ
y

132. 132 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
2.
2
tttrij
1,
dx
dt
π 2,
dy
t
dt
π 0
dz
dt
π


4
2
0
4
2
0
14
11
2 1 4 ln 2 1 4 8 65 ln 8 65 16.819
44
stdt
tt t t


*


3. +
,
32
,0,1tt trij

2
3, 2, 0
dx dy dz
tt
dt dt dt
πππ


11
42 2
00
1
1 12 32
2232
0
0
94 94
111
9 4 18 9 4 13 8 1.4397
18 27 27
sttdtttdt
ttdtt



**
*


4.
2
1,0 6tt t trij

1, 2
dx dy
t
dt dt
ππ

θ
6
2
0
6
22
0
14
111
ln 4 1 2 4 1 ln 145 12 3 145
424
stdt
tttt




*


5.
33
cos sinta ta tri
j

22
3cos sin, 3sin cos
dx dy
att att
dt dt
φπ φ

+,
2 22
22
0
22 2
000
4 3 cos sin 3 sin cos
12 sin cos 3 2 sin 2 3 cos 2 6
sattattdt
a t tdt a tdt a t a




φφφπφ
*
**


6. cos sintatatrij

sin , cos
dx dy
at a t
dt dt
φπ φ
+,
22 2
22 2 2
000
sin cos 2
s ata tdt adtat a

**


7. (a)

2
00
221
cos sin
2
1
100 cos 45 3 100 sin 45 32 50 2 3 50 2 16
2
tv thv tgt
tttttt
ri j
ijij










(b)

50 2 50 2 32ttvi j

25 2
50 2 32 0
16
tt
Maximum height:
2
25 2 15 2
3 50 2 16 81.125 ft
16 16





x
12
8
16
4
4
213(0, 0)
(4, 16)
y
y
x
(0, 0)
(1, 1)
1
1
y
x
12345678−1
5
10
15
20
25
30
35
40
(1, 0)
(7, 36)
x
a
a
−a
−a
y
x
a
a
y

133. Section 12.5 Arc Length and Curvature 133
© 2010 Brooks/Cole, Cengage Learning
(c)
2
3 50 2 16 0 4.4614tt t
Range: 50 2 4.4614 315.5 feet
(d)
224.4614
0
50 2 50 2 32 362.9 feetstdt*

8. (a)

2
001
cos sin
2
tv t v tgtri j








2
01
sin
2
yt v t gt
φπ

0sin 0yt v gtφπφ when
0sin
.
v
t
g
π
Maximum height when
sin 1, or .
2


ππ
(b)

02
012sin
sin 0
2
v
yt v t gt t
g


Range:

2
00 2
0
2sin
cos sin
vv
xt v
gg

ππ


The range

xtis a maximum for sin 2 1, or .
4


ππ
(c)

0
0cos
sin
xt v
yt v gt
π
φπ


22 2
22 22 22 2 22 2 22
00 000 00
cos sin cos sin 2 sin 2 sin
xtytv v gtv v vgtgtv vgtgt


2sin 12
0 222
00
0
2sin
vg
dt
svvgtgt



*

Because
096 ft sec, you havevπ


6sin 12
22
0
96 6144 sin 1024 .
dt
stt



*

Using a computer algebra system,

sis a maximum for 0.9855 56.5 .
9.
+
,43,0,1ttttrijk
1, 4, 3
dx dy dz
dt dt dt
φπ φ φ

11
0 0
1 16 9 26 26sdtt
*

10.
+
,
23
,0,2tttrijk

2
0, 2 , 3
dx dy dz
tt
dt dt dt
ππ π


2
24
0
2
2 32
2 2 32 32
0
0
49
111
49 49 40 4 80108
27 27 27
sttdt
ttdt t




*
*

z
x
y
(0, 0, 0)
(−1, 4, 3)
1
1
2
−2
−3
3 −1
3
2
4
−2
2
3
4
5
x
y
z
563
4
1
2
3
4
5
6
7
8
−4
(1, 0, 0)
(1, 4, 8)

134. 134 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
11.
3
4, cos ,sin , 0,
2
ttttr
φπ




4, sin , cos
dx dy dz
tt
dt dt dt
ππ π

3232 32
22
000 3
16 sin cos 17 17 17
2
sttdtdtt



**


12.

+,2sin ,5,2cos , 0,ttttr π
2cos , 5, 2sin
dx dy dz
tt
dt dt dt
φφφπ

22
00
4 cos 25 4 sin 29 29sttdtdt

**

13.
cos sintatatbtrijk

sin , cos ,
dx dy dz
at a t b
dt dt dt
φπ φ φ

2
22 2 2 2
0
2
2
22 22 22
0
0
sin cos
2
satatbdt
abdt abt ab







*
*

14.

2
cos sin , sin cos ,ttttttttr
cos ,
dx
tt
dt
π sin ,
dy
tt
dt
π 2
dz
t
dt
π


2 222
0
2
22
2
2
0
0
cos sin 2
5
55
28
s tt tt tdt
t
tdt






πππ

*
*

15.

2
lntt t tri
j k
2,
dx
t
dt
π 1,
dy
dt
π
1dz
dt t
π



2
42 42
333 22
2
111
14141
2 1 8.37
tt tt
s t dt dt dt
tt t




***

16.

3
sin costtttrijk

cos ,
dx
t
dtπ
sin ,
dy
t
dtφπ
2
3
dz
t
dt
π


22 222
224
00
cos sin 3 9 11.15stttdttdt **

x
y
(6 , 0, −1)
(0, −1, 0)π
21
18
15
12
9
6
−12
−9
−6
−3
6
6
−6
−9
−12
9
z
4
−4
−4
−6
−8
−10
−6
6
8
10
12
(0, 0, 2)
(0, 5
π, −2)
x
y
z
101214
468
x
y
(a, 0, 2 b)
2b
b
( , 0, 0)a
π
π
π
z
1
2
3
2
3
2
3
z
y
x
(1, 0, 0)
, 1,
π
2
π
4()
2

135. Section 12.5 Arc Length and Curvature 135
© 2010 Brooks/Cole, Cengage Learning
17.
23
4,02tt t t tri jk
(a)

0 0, 4, 0 , 2 2, 0, 8rr
distance
222
248 84
2 21 9.165


(b)

00,4,0r


0.5 0.5, 3.75, 0.125r


11,3,1r


1.5 1.5, 1.75, 3.375r


22,0,8r
distance



22 2
22 2
22 2
22 2
0.5 0.25 0.125
0.5 0.75 0.875
0.5 1.25 2.375
0.5 1.75 4.625
0.5728 1.2562 2.7300 4.9702
9.529







(c) Increase the number of line segments.
(d) Using a graphing utility, you obtain 9.57057.

18.
6cos 2sin ,0 2
44
tt
tttrijk




(a)


0 6 6, 0, 0
2 2 2 0,2,2ri
rjk

distance
222
622 44
2 11 6.633


(b)

06,0,0r


0.5 5.543, 0.765, 0.5r


1.0 4.243, 1.414, 1.0r


1.5 2.296, 1.848, 1.5r


2.0 0,2,2r
distance 6.9698
(c) Increase the number of line segments.
(d) Using a graphing utility, you obtain

2
0
7.0105.stdtr*


19.
2cos ,2sin ,ttttr
(a)


222 222
00 0 0
2sin 2cos 1 5 5 5
ttt t
s x u y u z u du u u du du u t

***

(b) 5
s
t



2cos , 2sin ,
555
2cos 2sin
555
s ss
xyz
sss
srijk









(c) When
5:s

2 cos 1 1.081
2 sin 1 1.683
1
1.081, 1.683, 1.000
x
y
z



When 4:s

4
2 cos 0.433
5
4
2 sin 1.953
5
4
1.789
5
0.433, 1.953, 1.789
x
y
z





(d)

2
2 2
22 141
sin cos 1
5555 5 5 5
r
ss
s




20.

23
4sin cos ,4cos sin ,
2
ttttttttr
(a)


222
0
222
22
000
5
4sin 4cos 3 16 9 5
2
t
ttt
sxuyuzudu
uu u u udu u udu udu t


*
***

136. 136 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
(b)
2
5
s
t

22 2
4sin cos
55 5
s ss
x






22 2
4cos sin
55 5
s ss
y





2
32 3
25 5
s s
z






22 2 22 23
4sin cos 4cos sin
55 5 55 55
s ss ssss
s
rijk





(c) When
5:s

25 25 25
4 sin cos 1.030
55 5
x






25 25 25
4 cos sin 5.408
555
y






35
1.342
5
z

1.030, 5.408, 1.342
When
4:s

88 8
4 sin cos 2.291
55 5
x






88 8
4 cos sin 6.029
55 5
12
2.4
5
y
z







2.291, 6.029, 2.400
(d)

22
2
42 4 2 3 169
sin cos 1
55 5 5 5 2525ss
s
r





21.
22
11
22
s ssrij








22 11
and 1
22 22
ssrijr





s
s s
s
r
Tr
r





The curve is a line.0T0 TsKs
22.
3ssrij

sriand
1sr


s sTr


The curve is a line.0T0 TsKs
23.

2cos 2sin
555
s ss
s
rijk







22 1
sin cos
55 5 5 5
Tr
ijkss
ss








22
cos sin
55 55 ss
s
Tij







2
5
Ks
T

137. Section 12.5 Arc Length and Curvature 137
© 2010 Brooks/Cole, Cengage Learning
24.
22 2 22 23
4 sin cos 4 cos sin
55 5 55 55
s ss ssss
s
rijk







424 23
sin cos
555 55ss
ss
Tr i jk


45 2 45 2
cos sin
25 2 5 25 2 5
s s
s
ss
Tij


45 210
25 2 25
s
Ks
s s
T
25.
42tttrij

42tvij



1
2
5
t
Tij




0
t
t
K
t
T0
T
r





(The curve is a line.)
26.

2
ttrij

2ttvi

tTi

tT0



0
t
K
tT
r




27.

1
tt
t
rij



2
1
t
t
vij


1vij


3
2
t
taj

12aj


2
1
t
t
t
4
ij
T








2
12
41
1
tt
t
Nij





1
1
2
Nij


2
2
2
K
aN
v
"

28.

31
9
tt trij



21
3
ttvi j



4
2,
2
vij

16 5
21
93
v


2
3
ttaj


4
2
3
aj



2
2
441
33
9
1
9
t
t
t
tt
ij
ij
T








34
2
5
ij
T





43
2
ij
N
5



2
45 36
25 9 125
aN
v
K
"

29.

,sintttr



2
1, cos , 1 costtt trr
1, 0 , 1
22
rr







0, sin , 0, 1
2
ttaa







2
1
1, cos
1cos
tt
t
T



1, 0
2
T





0, 1
2
N






2
1
1
1
K
aN
v
"

138. 138 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
30. 5cos ,4sin ,
3
ttttr



5cos , 4sin
xttytt



32
22
32
22
32
22
5sin 4sin 4cos 5cos
25 sin 16 cos
20
25 sin 16 cos
xy yx
K
xy
tt t t
tt
tt
=













32
20 160 91
3 828125 3 4 16 1 4
K





31.
4cos2 4sin2tttrij

8sin2 8cos2tttrij

sin 2 cos 2tttTij

2cos2 2sin2tttTij



21
84
t
K
t
T
r




32.
2cos sintttrij

2sin costttrij


22
4sin costttr


22
2sin cos
4sin cos
tt
t
tt
ij
T







32
22
2cos 4sin
4sin cos
tt
t
tt
ij
T








22
22
32
22
2
4sin cos
4sin cos
2
4sin cos
t tt
K
t tt
tt
T
r










33.
cos sinrijta ta t;;

sin cosrijta ta t;; ;;

sin cosTijttt;;




cos sin
1Tij
T
rttt
t
K
aat;; ;;
;
;



34.
cos sinrijta tb t;;

sin cosrijta tb t;; ;;




22 2 2
sin cos
sin cos
ij
T
atb t
t
atb t;;
;;







22
32
22 2 2
cos sin
sin cos
ij
T
ab t a b t
t
atb t;; ;;
;;










22 2 2
22 2 2
32
22 2 2
sin cos
sin cos
sin cos
T
r
ab
t atb t
K
t atb t
ab
atb t
;
;;
;; ;
;;







35.

sin , 1 cosrtat ta t;; ;
From Exercise 44, Section 12.4, we have:




2
2
2
22
1cos
2
1cos
22
21cos 41cos
aN
aN
v
a
t
tt
K
t
a
t
at at
;
;
;
;
;; ;
" "
"







36.

cos sin , sin cosrttttttt;; ; ;; ;
From Exercise 43, Section 12.4, we have:


3
3
242
1
aN
aN
v
t
tt t
K
tt
;
;
;;"
"


37.

2
2
2
t
ttt
rijk

2tttrijk


2
2
15
tt
t
t
ijk
T







32
2
52
15
t
t
t
ijk
T









2
32
2 2
5
15 5
15 15
tt
K
t t t
T
r

139. Section 12.5 Arc Length and Curvature 139
© 2010 Brooks/Cole, Cengage Learning
38.
22 1
2
2
ttttrijk


4tt trijk


2
4
117
tt
t
tij k
T






32
2
417
117
t
t
t
ijk
T









2
12
2
32
2
32
2
289 17
117
117
17
117T
rt t
Kt
t
t
t








39. 43cos 3sintt t tri j k

43sin 3costttri j k


+,
1
43sin 3cos
5
ttt
Tijk


+,
1
3cos 3sin
5
tttTjk




35 3
525
t
K
tT
r




40.
22 2
cos sin
tt t
te e te tri j k


22 2 2 2 2
2 2 cos sin 2 sin cos 2 2 cos sin 2 sin cosri j ki j k
tt t t t t
te e tet ete t e t t t t




+,
12
12
22 22 222
4 4 cos 4 cos sin sin 4 sin 4 sin cos cos 9 3
t tt
te t tt t t tt t e er






22 1 2 1
cos sin sin cos
33 3 3 3
t
ttttt
t
r
Ti j k
r







21 2 1
sin cos cos sin
33 3 3
ttttt
Tjk







12
22 22
41 4 4 14 5
sin cos sin cos cos sin cos sin
99 9 9 99 3
ttttttttt
T








22
5
53
39
tt
t
K
eet
T
r




41.

2
32,3,2 1tttP trij
3,
xt 3,x 0x

2
2,yt 4,yt 4y




32 32
22 2
34 0
94
xy yx
K
xy t





At 1,t

32
12 12
125916
K


42.
4, 1,0 0
t
te tP trij
,
t
xe ,
t
xe
t
xe
4,yt 4,y 0y


32 32 32
22
04 4
17116
xy yx
K
xy





43.


3
2
, 2,4,2 2
4
t
ttt P trijk


23
2
4
ttt
rijk


243,226rijkr


3
2
2
tt
rjk

223rjk


2 2 143632
023
ijk
rr ijk'


22 497rr'

332
7726
26 676
K
rr
r
'

140. 140 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
44. cos sin , 1, 0, 1 0
ttt
te teteP trijk

cos sin sin cos
tt tt t
tetet eteterijk



222 2
cos 2 cos sin sin sin 2 sin cos cos 1 3
t t
te t tt t t tt t er



1
cos sin sin cos
3
tttttTijk




1
sin cos cos sin
3
tttttTij



003rijkr



12
00
33
TijT



0 23 2
30 3
K
T
r




45. 32yx
Because 0,y 0,Kand the radius of curvature is
undefined.
46. ymxb
Because 0,y 0,Kand the radius of curvature is
undefined.
47.
2
23, 1yx x
4yx
4y


32 32
2
44
0.057
17
14
K




32
radius of curvature
117
17.523
4
K

48.
4
2,1
yx x
x


2
4
2,12
yy
x


3
8
,1 8
yy
x

32 32 32
2
88
514
1
y
K
y






32
radius of curvature
15
8
K

49.
cos 2 , 2yxx

2sin2yx

4cos2yx
At 2 ,x
1,y 0,y 4y

32
2
4
4
10
K






11
4K

50.
3
,0
x
yex

33
3, 9
x x
yey e
At 0,x 1,y 3,y 9y

32 32
2
99
10
13
K




11010
9K

51.
22
,0yaxx

22
x
y
ax






2
32
22
a
y
ax



At
0:x


32
2
radius of curvature
0
1
11
10
1
y
y
a
a
K
a
a
K

141. Section 12.5 Arc Length and Curvature 141
© 2010 Brooks/Cole, Cengage Learning
52.

2
23
16
4
9
16
916
16
yx
x
y
y
y
y
y


π


π

At
0:xπ


32
2
radius of curvature
0
3
16
316 3
16
10
116
3
y
y
K
K
π


ππ

π


53.
3
,2yxxππ

2
3, 6yxy xππ
At 2,xπ 8,yπ 12,yπ 12yπ



32 32
2
12 12
145
112
Kππ




1 145 145
12K
π

54. ,
n
yxπ 1,xπ 2n/

1n
ynx

π


2
1
n
ynnx


At 1,xπ 1,yπ ,ynπ
1ynn


32
2
1
1
nn
K
n

π




55. (a) Point on circle:
,1
2




Center:
,0
2




Equation:
2
2
1
2
xy





(b) The circles have different radii because the curvature
is different and
1
.r
K
π

56. (a)
2
2
4
3
x
y
x
π



2
2
24
3
x
y
x
π




2
3
2
72 1
3
x
y
x

π


At
0:xπ

32
2
0
72 8
27 3
83 8
3
10
13
8
y
y
K
r
K
π
ππ
ππ

ππ

Center:
3
0,
8




Equation:
2
2
39
864
xy





(b) The circles have different radii since the curvature is
different and
1
.r
K
π

57.
23
11 2
,,yx y y
x xx





32
2
2
2at 1,2
10
Kππ


Radius of curvature
12.πBecause the tangent line is
horizontal at
1, 2 , the normal line is vertical. The center
of the circle is
12unit above the point 1, 2at 1, 5 2 .
Circle:

2
2
51
1
24
xy





(1, 2)
−6
−4
6
4

142. 142 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
58.

ln ,
1
,
11,
yx
y
x
y
π
π
π


2
1
1
11
x
y
x
y
π
φπ
φπ



32 32
2
1 1
,
2
11
K
π
ππ

321
222r
K
ππ π

The slope of the tangent line at
1, 0 is 11.yπ
The slope of the normal line is 1.π
Equation of normal line:
11yx x
The center of the circle is on the normal line
22units
away from the point
1, 0 .





22
22
2
2
1022
118
2428
2230
23 10
3or 1
xy
xx
xx
xx
xx
xx



ππφ

φφπ

Because the circle is below the curve, 3xπand 2.yφπ
Center of circle:
3, 2π
Equation of circle:

22
328xy

59.

,
,
01,
x
x
ye
ye
y
π
π
π

0
01
x
y
ye
y
π
π
π



32 32
2
1111
,22
2 2211
Kr
K
πππππ


The slope of the tangent line at
θ
0, 1 is 01.yπ
The slope of the normal line is 1.π
Equation of normal line: 1yxπφπ or 1yx
The center of the circle is on the normal line 22units
away from the point
0, 1 .


22
22
2
0122
8
4
2
xy
xx
x
x


π


Because the circle is above
the curve, 2xφπand 3.yπ
Center of circle:
2, 3π
Equation of circle:

22
238xy

60.

3
21
,
3
,
11,
yx
yx
y
π
π
π

1
2
12
x
y
y
π
π
π


32
211
,2
211
Kr
K
ππππ


The slope of the tangent line at
1
1,
3



is
11.yπ
The slope of the normal line is 1.π
Equation of normal line:

1
1
3
yxπφππ
or
4
3
yx

The center of the circle is on the normal line
2units
away from the point
1
1, .
3








2
2
22
2
1
12
3
112
11
0or 2
xy
xx
x
xx





πφ
ππ

Because the circle is above the curve, 0xπand
4
.
3
yπ

Center of circle:
,
4
0
3




Equation of circle:
2
2
4
2
3
xy






61.
62.
63.
2
13,yx 21,yxφπ 2yπ

θ

32 32
2 2
22
14 112 1
K
xx
ππ



(a) K is maximum when 1xπor at the vertex
1, 3 .
(b)
lim 0
x
K
!
π
(0, 1)
−6
0
3
6
−2
x
y
π
π
π
A
B
−3
−5
(1, 0)
9
3
(0, 1)
−3
−1
3
3
2−2−4
−3
3
4
−4
4
y
x
A
B

143. Section 12.5 Arc Length and Curvature 143
© 2010 Brooks/Cole, Cengage Learning
64.
3
,yx
2
3,yx 6yx


32
4
6
19
x
K
x



(a) K is maximum at 44 4433
11 11
,,,.
45 4545 45




(b)
lim 0
x
K
!


65.
23
,yx
132
,
3
yx


432
9
yx





43
32 32
23 13 2 3
29 6
149 94
x
K
x xx







(a) K !as 0.x No maximum
(b)
lim 0
x
K
!


66.
1
,y
x

2
1
,y
x

3
2
.y
x
Assume 0.x


3
3
32 32 32
2 44
2 2
11 11
xy x
K
xxy








24
52
4
61
1
x xdK
dx
x




(a) K has a maximum at 1x
and 1 by symmetry .x
(b)
lim 0
x
K
!


67. ln ,yx
1
,y
x

2
1
y
x



2
32 32
2 2
1
111
xx
K
xx







2
52
2
21
1
dK x
dx
x




(a) K has a maximum when
1
.
2
x

(b)
lim 0
x
K
!


68. ,
x
ye
x
yye


32 32
2 2
11
x
x
y e
K
ey








2
52
2
12
1
x x
x
eedK
dx
e




(a)
22 1111
12 0 ln ln2
2222
xx
ee x





K has maximum curvature at
1
ln 2.
2
x

(b)
lim 0
x
K
!


69. sinh ,yx cosh ,yx sinhyx


32
2
sinh
1cosh
x
K
x




(a) By symmetry, consider 0,x/so
sinh sinh :xx




32 12
22
3
2
3
2
1 cosh cosh sinh 1 cosh 2 sinh cosh
1cosh
x xx x xx
K
x





Setting
0:K







22
22
22
2
1 cosh cosh sinh 3 cosh
1 cosh 3 sinh
11sinh 3sinh
sinh 1
arcsinh 1 0.8814, Maximum
Also, arcsinh 1
xx x x
xx
xx
x
x
x







Maximum curvature at
arcsinh 1 , 1
(b)
lim 0
x
K
!

144. 144 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
70. cosh ,yx sinh ,yx coshyx

32 32 2
22
cosh cosh 1
cosh
1sinh cosh
xx
K
x
xx




(a)
3
2sinh
0
cosh
x
K
x

when 0, Maximumx
(b)
lim 0
x
K
!


71.
3
1,yx
2
3,yx 6yx

32
4
6
19
x
K
x





Curvature is 0 at
0: 0, 1 .x

72.
3
13,yx
2
31,yx 61yx




32 32
24
61
0
1' 191
xy
K
yx




at 1.x
Curvature is 0 at

1, 3 .

73. cos ,yx sin ,yx cosyx


32 32
2 2
cos
0
1sin1
yx
K
xy




for
.
2
x K



Curvature is 0 at
,0 .
2
K







74. sin ,yx cos ,yx sinyx


32
2
sin
0
1cos
x
K
x


for
.
xn
Curvature is 0 for
:,0xnn

75. (a)

222bb
aa
s x t y t z t dt r t dt **

(b) Plane:
T
TdK s
dS

Space:




3
Trr
rrttt
K
tt
'



Answers will vary

76. The curve is a line.

77.

32
2
1
y
K
y





At the smooth relative extremum 0,yso
.K y
Yes, for example,
4
yxhas a curvature of 0 at its
relative minimum
0, 0 . The curvature is positive at any
other point on the curve.

78.
2
rijttt
(a)
1, 2
dx dy
t
dt dt


2
22
22 2 2
00
0
111
14 142 2 214 ln2 14 Theorem8.2
222
1
4 17 ln 4 17 4.647
4
s t dt t t dt u t t t t t

"



**

(b) Let
2
,yx 2,yx 2y
At 0,t 0,x 0,y 0,y 2,y
2K

+,
32
10 2
At 1,t 1,x 1,y 2,y 2y


32 32
2
22
0.179
5
12
K



At 2,t 2,x 4,y 4y 2y

+,
32 32
22
0.0285
17116
K


(c) As t changes from 0 to 2, the curvature decreases.

145. Section 12.5 Arc Length and Curvature 145
© 2010 Brooks/Cole, Cengage Learning
79. Endpoints of the major axis: θ2, 0
Endpoints of the minor axis:
0, 1





22
2
222
233
3
32 32
2 22
32 32
22
44
28 0
4
41 4
16
44 1
16 16 4
14 16
1614
16 16
12 4 16 3
xy
xyy
x
y
y
yxy
y
y
yxy y x
yyy
y
K
yxxy
yx


φπ
πππ
π
π
πππ
π π
ππ


ππ


So, because
22,
x K is largest when 2xand
smallest when 0.xπ

80.
12 ,
2
x
yaxbxy
x
φπ φ

You observe that
0, 0 is a solution point to both
equations. So, the point P is origin.

1 ,yaxbxφπ 1 2,yabxφπ
1 2yaφπ

2 ,
2
x
y
x
π


2 2
2
,
2
y
x
π


2 3
4
2
y
x
π
π


At P,
10yabπand

2 2
21
0.
202
yππ


Because the curves have a common tangent at ,P
1200yyπ or
1
.
2
abπ
So,
1
1
0.
2
yπ
Because the
curves have the same curvature at P,
1200.KKπ






1
1 32 32
22
10 2
0
11210
y a
K
y
π
ππ









2
2 32 32
22
20 12
0
11210
y
K
y
π
ππ



So,
1
2
2
aor
1
.
4
aIn order that the curves
intersect at only one point, the parabola
must be concave downward. So,

1
4
aπ
and
1
2.
2
b
a
ππ


1
1
2
4
yxxφπ
and
2
2
x
y
x
π


81.
42
fxxxφπ
(a)
2
64232
26 1
16 16 4 1
x
K
xxx
π
π


(b) For 0,xπ 2.Kπ
00.fπAt 0, 0 , the circle
of curvature has radius
1
.
2
Using the symmetry of
the graph of ,
fyou obtain
2
2
11
.
24
xy





For
1, 255.xKππ 10.fπAt 1, 0 , the
circle of curvature has radius
51
.
2 K
π
Using the graph of f, you see that the center of
curvature is
1
0, .
2



So,

2
2
15
.
24
xy





To graph these circles, use

211
24
yx and
215
.
24
yx
(c) The curvature tends to be greatest near the extrema
of f, and K decreases as
.
x !f and K,
however, do not have the same critical numbers.
Critical numbers of f :
0,xπ
2
0.7071
2

Critical numbers of K:
0,xπ 0.7647, 0.4082

82.
851
,0 5
4
yx x

(a)
(rotated about y-axis)
(b)
85 85 35
5
3
0
5 125 5
2 114.6 cm
44 9
x
Vx dx

φπφ

*

(c)
352
,
5
yxπ
25
2566
25 25
yx x
π
ππ

25
32
65
32
25 65
6
25
4
1
25
6
4
25 1
25
x
K
x
xx
π




π





(d) No, the curvature approaches !as
0.x

So, any
spherical object will hit the sides of the goblet before
touching the bottom
0, 0 .
x
y
2
4
4
2
2
4
z
50
0
1
−33
−2
f
2
3−3
−2
5
2
4
−4
−2−424
x
y
2
y
2
y
1
P
y

146. 146 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
83. (a) Imagine dropping the circle
2
2
16xyk
into the parabola
2
.yxφThe circle will drop to the point where the tangents
to the circle and parabola are equal.

2
yxφand
22
222
16 16xyk xxk
Taking derivatives,
22 0xyky and 2 .yxφSo,


.
x
yky x y
yk
π

π
So,


22 1
22 12 .
2
x
xxxyk xkxk
yk
π

π

So,


2
2
22 2 2
1
16 15.75.
2
xxk x x





Finally,
21
16.25,
2
kx
and the center of the circle is 16.25 units from the vertex of
the parabola. Because the radius of the circle is 4, the circle is 12.25 units from the vertex.
(b) In 2-space, the parabola
2
zyφ
2
orzxφ has a curvature of 2Kφat 0, 0 . The radius of
the largest sphere that will touch the vertex has radius
1
1.
2
Kφφ


84.
c
s
K
φ

31
3
yxφ


2
yxφ
2yxφ


32
4
2
1
x
K
x
φ


When
1:xφ
4
4
4
1
2
2
12
30
30 2
2
K
c
s c
cc
φ
φφ


At
3
,
2
xφ

32
4
3
0.201
18116
3302
56.27 mi h
2
K
c
s
KK
φ







x
510−5−10
10
15
y

147. Section 12.5 Arc Length and Curvature 147
© 2010 Brooks/Cole, Cengage Learning
85. 00,Px ypoint on curve .yfx Let ,$be the center of curvature. The radius of curvature is
1
.
K


.yfx Slope of normal line at 00,xyis

0
1
.
fx



Equation of normal line:

00
0
1
yy xx
fx





,$is on the normal line: 00 0
fxy x$ Equation 1

00,xylies on the circle:


2
32
2
2
0
22
00
0
1
1
fx
xy
K fx
$








Equation 2
Substituting Equation 1 into Equation 2:









2
2 2
00 0
3
2
0
22
00 2
0
2
2
0
2
0 2
0
1
1
1
1
fx y y
K
fx
yfx
fx
fx
y
fx
$$
$
$













When
00,fx
00,y
$ and if 00,fxthen
00.y$
So




2
0
0
0
2
0
00
0
1
1
fx
y
fx
fx
yyz
fx
$
$







Similarly,
00 .
xfxz

86. (a) ,
x
yfx e ,0,1
x
fx f x e



2
10
2
0
f
z
f





, 0 2, 1 2 2, 3$
(b)
2
1
,,1,1,
22
x
yyxy








2
11
2
1
f
z
f






15
,12,21,
22
$





(c)

2
,2, 2,0,0yxy xy



2
10 1
02
f
z
f






11
,0,0 0,
22
$





y
x
(x
0
, y
0
)P
1
K
(,)

148. 148 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
87. cos sin cos sinrr r f fij i j

cosxf

sinyf

sin cosxf f

cos sinyf f

cos sin sin cos cos 2 sin cosxf f f f f f f

sin cos cos sin sin 2 cos sinyffff f ff






2 22 2
32 32 32
22 2 2
22
2 2fff f rrr r
xy yx
K
xy rr ff







88. (a) 1sinr

cosr

sinr







2 2
22
32
332
2 2
2
2 2 cos 1 sin sin 1 sin
31 sin 3
221 sin81 sin
cos 1 sin
rrrr
K
rr








(b)


2
2
2
32 32
2 2
2
1
0
2
2
1
r
r
r
rrrr
K
rr











(c)
sinra

cosra

sinra



2
2
22 22 22
2
32 3
32
2 22 22
2 2cos sin sin 22
,0
cos sin
rrrr aaa a
Ka
aa
rr aa





(d)
re



re



re





2
2
2
32 32
2 2
2
2
21
22
rrrr
e
K
eerr







89. ,0
a
rea



a
rae



2a
rae





2
2
22 22 2
32 32
22222
2
2 2 1
1
aaa
aaarrrr ae ae e
K
eaae err






(a) As
,0.K !
(b) As
,a ! 0.K

149. Section 12.5 Arc Length and Curvature 149
© 2010 Brooks/Cole, Cengage Learning
90. At the pole, 0.rφ




22
2
32 3
2
2
22
2
rrrr r
K
rr
rr

φφφ





91. 4sin2rφ

8cos2rφ
At the pole:

221
840
K
r
φφφ



92. 6cos3rφ

18 sin 3rφπ
At the pole,

,
6


φ
18,
6
r

φπ



and


221
.
18 96
K
r

φφφ
π


93. ,
xft y gtφφ



dy
gtdy
dt
y
dxdx f t
dt

φφ φ










2
3
gtd
dt f t
y
dx
dt
ftgt gtf t
ft
ftg t g t f t
ft ft



φ
π
π

φφ















3
32 32
22
3
32
3 22
22
2
1
1
ftgt gtf t
fty
K
y gt
ft
ftgt gtf t
ft
ftg t g t f t
ft gtft gt
ft
π

φφ




π
π
φφ
6< ==
7>
==8?


94.
3
,
xttφ θ
2
3,xttφ 6xt tφ



21
,,1
2
yt t y t t y tφφφ




2
32
2 2
2
2
32 32
32 2
31 6
3
33
91 91
ttt
K
tt
t
tt tt
π
φ



φφ


0K as t !


95.

sin
1cos
sin
xa
xa
xa

φπ
φπ
φ


1cos
sin
cos
ya
ya
ya

φπ
φ
φ






+,
+,

32
22
222
32
2
222
32
32
1cos cos sin
1cos sin
cos 11
22cos
11cos
1cos 0
221 cos
11
csc
42222cos
xy yx
K
xy
aa
aa
a
a
aa









π
φ


ππ
φ


π
φ
π
π
φπ/
π

φφ

π

Minimum:
1
4
a
φ
Maximum: none
as 0K !

96. (a)

23
2
33
633
tt tt
tt tri j
vi j




2
2
2
31 , 6
ds d s
tt t
dt dt
v


2
2
2
31
K
t
φ


2
2
6
ds
at
dt
Tφφ



2
2
2
2
2
2
91 6
31ds
aK t
dt
t
N

"



4−4
0
5

150. 150 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
(b)
221
2
ttt trijk

2tttvijk


2
51
ds
tt
dt
v

2
2
2
5
51
ds t
dt t




2tajk


12 2
021
rr va
ijk jk
tttt
tt
''





323
2
5
51rr
rtt
K
t
t
'





2
2
2
5
51
ds t
a
dt t
T





2
2
32
2
2
5
51
51
5
51
N
ds
aK t
dt
t
t








97.

2
2
2
30 5280 ft5500 lb 1
3327.5 lb
32 ft sec 100 ft 3600 sec
N
ds
Fma mK
dt









98.

2
2
2
35 5280 ft6400 lb 1
32 ft sec 250 ft 3600 sec
94864
2108.1 lb
45
N
ds
Fma mK
dt










99.
cosh
2
x x
ee
yx





sinh
2
xx
ee
yx





cosh
2
xx
ee
yx






32 32 22
2 2
cosh cosh 1 1
cosh
cosh1sinh
x x
K
xy
xx






100. (a)




, by the Chain Rule
dddt
Ks
ds dt ds
ttddt
ds dt tt
TT
T
TTT
vr
"



(b)



tt
t
ds dttrr
T
r






ds
tt
dt
rT



2
2
ds ds
ttt
dt dt
rTT







2
2
2
ds d s ds
tt tt tt
dt dt dt
r r TT TT

' ' '


Because
ttTT 0' and
,
ds
t
dtr you have:




2
222
1 from ( )
rr r TT
rr r TT r TT r r
tt ttt
tt ttt ttt tKt a
' '

' '

So,


3
.
tt
K
t
rr
r
'



(c)










2223
tt tt
tt t t tt
K
ttttrr va
rr r v aN
rrrr''
' "

151. Section 12.5 Arc Length and Curvature 151
© 2010 Brooks/Cole, Cengage Learning
101. False
102. False
Curvature
1
radius


103. True
104. True

2
ds
aK
dt
N





105. Let
.
xtytztrijk Then
222
rxtytztr
and .rxt yt ztijk Then,
9 :

12
222 222
1
22 2
2
.
dr
r xt yt zt xt yt zt xtxt ytyt ztz t
dt
xt x t yt y t zt z t
rr


"


"


106.
3
3
GmM
mm
r
GM
r
Fa a r
ar




Because
ris a constant multiple of ,athey are parallel. Because ar is parallel to ,r .rr 0' Also,

.
d
dt
rr r r rr 00 0

'''


So, rr' is a constant vector which we will denote by .L

107. Let xyzrijk where x, y, and z are function of t, and
.rr






22
2
3
222
3
22 2 2 2 2
3
3
using Exercise 105
1
1
rrrrrrr
rrrr
ijk ijk
ijk
ijk
rrrdrdtd
dt r r r
r
r
xyzxyz xxyyzzxyz
r
xy xz xyy xzz x y z y xxy zzy x z y z xxz yyz
r
yz y z x
r
"



"







+,9:
3
1
rr rzxzxyxy
r
xyz
''


108.
+, +,9:
+, +,9:
+, +,9:
+, +,9:
3
33
33
3
11
11
1
1rr
Lr0rLrrr
r
0rrrrr
r
rr rr r
rr r rr r 0d
dt GM r GM r
GM
GM r r
rr
r

' ' ' ' '



''''


' ' ' '
''''

So,
rr
L
GM r

'


is a constant vector which we will denote by .e

152. 152 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
109. From Exercise 106, you have concluded that planetary motion is planar. Assume that the planet moves in the
xy-plane with the sum at the origin. From Exercise 108, you have
.GM
r
r
rL e
'



Because
rL'and rare both perpendicular to ,Lso
is
.eSo, elies in the xy-plane. Situate the coordinate system so that elies along the positive x-axis
and
is the angle between eand .rLet
.eeφ Then cos cos .rere re "φ φ Also,



+,
2
cos .
LLL
rr L
rr L
r
re
rr
re
GM
r
GM
r
GM re r
φ"
φ'"
φ" '

"


"
"




So,
2
1cos
GM
r
e
L

φ

and the planetary motion is a conic section. Because the planet returns to its initial position periodically,
the conic is an ellipse.
110.
Lrr φ'
Let:


cos sin
sin cos
r
dd dd
r
dt dt d dt
rij
rr
rij




"



Then:
22cos sin 0
and .
sin cos 0
ijk
rr k L rr
ddrr
rr
dt dtdd
rr
dt dt


'φ φ φ' φ
π

111.
21
2
A rd
$

φ*

So,

211
22
dA dA d d
r
dt d dt dt
L


φφ φ
and rsweeps out area at a constant rate.
112. Let
Pdenote the period. Then

0
1
.
2
LPdAA dt P
dt
φφ*

Also, the area of an ellipse isab
where 2aand 2bare the lengths of the major and minor axes.

1
2
2
L
L
ab P
ab
P

φ
φ



22
22 22 24 2 2
22222 3 333
22 222
444 44 4
1
L
LL LLL
GMaa aeded
Pacae a aaKa
aGM

φπφ πφ φφ φφ



x
θ
r
e
Sun
Planet
y

153. Review Exercises for Chapter 12 153
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 12
1. () tanrijkttt
(a) Domain: ,
2
tn


# n an integer
(b) Continuous for all
,
2
tn


# n an integer
2.
1
()
4
rijktt
t



(a) Domain:
[0, 4) and 4,!
(b) Continuous except at 4t
π
3. () Inri
jktttt
(a) Domain:
0,!
(b) Continuous for all 0
t
4.
2
21ri
jktt tt
(a) Domain:
,! !
(b) Continuous for all
t
5.
2
21 2rijktt t t
(a)

02ri k
(b)
234rij
(c)

2
121 1 1rijkccc c
(d)



2
2
112111 333
2233
rr i j kijk
ij k
tttt
ttt t
5 5 5 5

5 5 5 5

6. (a)
03rij
(b)
22
rk





(c)
3cos 1 sinri
j kss ss
(d)

3cos( 1 sin ) 3
3cos 3 sin
rr i jkijk
ik
tttt
ttt 5 5 5 5
5 55

7.

cos , sinrtttπ

cos , sin
x ty tππ

22 2
,xy circle
8.

2
2, 1rttt
22
xt t x


2
2
1 2 1, parabolayt x

9.

2
rijkttt

22
1x
yt
zt z y
π
π

10.

23
3rijktt tt

x
y
−11235 4
−2
−1
1
2
4
3
x
y
2
1
2
1
z
x
y
z
2
2
4
4
2
−2
−4
−4
y
x
−1−2−4124
−2
−4
1
2
4

154. 154 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
11. sin
1, sin , 1
ri jktt
xy tz
ππ π

12.

22
2cos 2sin
2cos , , 2sin
4
rijktttt
x ty tz t
xz

πππ


13.

21
ln
2
rijktt t t


14.

311
24
rijktttt

15. One possible answer is:



1
2
3 34,0 1
34 ,0 4
3,0 3rij
ri j
rittt t
ttt
ttt



16. One possible answer is:




1
2
3 4,0 1
4cos 4sin ,0
2
4,0 4tt t
tttt
tttri
rij
rj





17. The vector joining the points is
7, 4, 10 .One path is


27,34,810.rtttt
18. The -xand y-components are 2cost and 2sin .t At

3
,
2
t
π
the staircase has made
3
4
of a revolution and is 2 meters
high. So, one answer is


4
2cos 2sin .
3
rijktttt


19.
22
2
,0,
,,2
zx yxy tx
xty tz t





2
2ri
j+kttt t
20.
22
2
4, 0,
,, 4
xzxytx
xtytz t






2
2
4
4rij k
rij kttt t
ttt t



21.

4
lim 4 4ij+kik
t
tt



22.
00
sin2 2 cos 2
lim lim
1
2
ijk ijk
ijk
tt
tttt
ee
t








23.

232
31,
3
ri juijkttt ttt t

(a)
3rijt
(b)
0rtπ
(c)


22 3 2
2
312
34
ru
rt u
t
ttttt tt
Dttt
"
"


x
y
2
3
2
π
z
y
x
1
21
2
3
3
2
1
z
x
y
5
123
−3
3
2
z
x
y
3
4
5
z
yx
−1
1
1
1
2 2
2
3
3
3
4
5
6
−1
−2−2
−3−3
z
y
x
121
2
3
−2
2
3
1
z

155. Review Exercises for Chapter 12 155
© 2010 Brooks/Cole, Cengage Learning
(d)

23
2 2
25 22
3
25222
ur i jk
ur i jk
t
ttttt t
Dt t t t




(e)


2
2
10 2 1
10 1
10 2 1
r
r
t
ttt
t
Dt
tt






(f )



43 4 32
32 3 22
23
3
8
28921
3
ru i j k
ru i j k
t
tt tt t ttt
Dt t t t t t t
'

'



24.

1
sin cos , sin cosrijkuijkttttttt
t

(a)
cos sinrijkttt
(b)
sin cosrijttt
(c)

2
0
ru
ru
t
tt
Dt t
"
"


(d)


2
1
2sincos 2
1
2cossin 2
ur i j k
ur i j k
t
tt t t t
t
Dt t t t
t









(e)


2
2
1
1
r
r
t
tt
t
Dt
t





(f )


22
11
cos cos sin sin
11 1 1
sin cos sin cos cos sin cos sin
ru i j
ru i j
t
tt ttt ttt
tt
Dt t t tt t t t tt t t
tt t t

'



'



25.
and xt yt are increasing functions at
0,ttand zt is a decreasing function at
0.tt
26. The graph of
uis parallel to the yz-plane.
27.
cos cos sin sin cosij i jCttdttttt*

28.


2
ln ln ln 1 2 ln
4
ijk i jkC
t
ttt dtttt t t *

29.
22 21
cos sin 1 1 ln 1
2
ijk Ctttdt tdttt t t



**

30.

34 3 2
2232

34 3 2
jkijk i jk i j kC
tt t t
tt ttdt t t t tdt

'

**

31.

2
234
2
23
2
2
32 32
32
234 3
ijk ijk j
ttt
tttdt






*

156. 156 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
32.
1
1
32
0
0
22
sin sin cos sin 1 cos 1
33
j kj kj ktttdt t ttt



*

33.

2 2
22 23
00
32 2282ijk ijk ijk
tt
et dtett e
*

34.


1
43
1
32 2
1
1
2
arcsin arcsin 1
433
tt
tttdt tttijki jkk






*
35.



2
2
2
03524
12 4
rijkijkC
rjkCijkCijk
ri+jk
tt tt
tt
tteedttee
tt e e




*

36.



2
3
3
sec tan
ln sec tan ln cos
3
03
ln sec tan ln cos 3
3
ttttdt
t
tt t
t
tttt
rijk
ijkC
rCk
rijk





*

37.



3
2
44
4
43
Speed 16 9 1 17 9
6
tttt
tt t
tt
tt
rijk
rv ijk
v
aj




38.



2
2
32
52
1
54
2
1
Speed 25 16
4
1
4
4
tttt
tt t
t
t
t
t
t
rijk
rv ijk
v
aik





39.


33
22
cos ,sin ,3
3cos sin ,3sin cos ,3
r
vr
tttt
tt tt tt






42 4 2 22 2 2 22
9 cos sin 9 sin cos 9 3 cos sin cos sin 1 3 cos sin 1vttttt tttt tt



22 22
22 22
6 cos sin 3 cos cos , 6 sin cos 3 sin sin , 0
3 cos 2 sin cos , 3 sin 2 cos sin , 0
avtt tt tttt tt
tt tt t t


40.




2
42
2
,tan,
1, sec ,
1sec
0, 2 sec tan ,
r
rv
v
ra
t
t
t
t
tt te
tt te
tte
tt tte



"
41.



2
01
ln 3 , , , 4
2
11
,2,
32
1
4 1, 8, direction numbers
2
r
r
r
ttttt
tt
t







Because 4 0, 16, 2 , the parametric equations are
1
,168,2 .
2
r
xty tz t





00.1 4.1 0.1, 16.8, 2.05rrt
42.



03 cosh , sinh , 2 , 0
3sinh ,cos , 2
0 0, 1, 2 direction numbers
r
r
r
ttttt
ttht






Because 0 3, 0, 0 , the parametric equations are
3, , 2 .
r
xytzt





00.1 0.1 3, 0.1, 0.2rrt
43.

2
00
21
cos , sin
2
42 3 , 42 16
rtv tv tgt
tt t



2 21
42 16 0,
8
21 441 3
Range 42 3 190.96 ft
84
ttt

157. Review Exercises for Chapter 12 157
© 2010 Brooks/Cole, Cengage Learning
44.
2
00 0
0 6
16 6 0
4
616
4
4 6
86
6.532 ft/sec
3
yt t
xvt v v
v







45. Range
2
0
sin 2 95
9.8
v
x



2
0
0
9.8 95
sin 40
38.06 m/sec
v
v




46.

2
001
cos sin 9.8
2
ri jtv t v t t





(a)

2
20 cos 30 20 sin 30 4.9ri jtttt



Maximum height 5.1 m; Range 35.3 m
(b)

2
20 cos 45 20 sin 45 4.9ri jtttt



Maximum height 10.2 m; Range 40.8 m
(c)

2
20 cos 60 20 sin 60 4.9ri jtttt



Maximum height 15.3 m; Range 35.3 m
(Note that 45gives the longest range)
47.




23
3
19 10
0
1
3
10
is not defined
0
does not exist
rij
vij
v
a
Tij
N
aT
aN
ttt
t
t
t
t





"
"
(The curve is a line)
48.




14 23
43
5
1
43
5
does not exist.
0
does not exist.rij
vij
v
a0
Tij
N
aT
aNtt t
t
t
t
t





"
"

49.







1
2
41
2
1
4
12
2
41412
2
41
1
441
1
24 1
rij
vi j
v
aj
ij
ij
T
ij
N
aT
aNtt t
t
t
t
t
t
t
tt
t
t
t
ttt
t
t
t
tt t
tt













"

"


450
0
20
450
0
20
450
0
20

158. 158 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
50.














2
4
2
3
2
4
2
4
34
2
21
1
2
2
1
211
1
4
1
1
11
1
11
4
111
tt
t
t
t
t
t
t
t
t
t
t
t
t
t
t
tt
rij
vi j
v
aj
ij
T
ij
N
aT















"






2
34 4
41 4
111 111
aN=
t
tt tt

"




51.






22
22
22
rij
vij
v
aij
ij
T
ij
N
tt
tt
tt
tt
tt
tt
tt
tt
tee
tee
tee
tee
ee
t
ee
ee
t
ee



















22
22
22
2
aT
aN
tt
tt
tt
ee
ee
ee




"

"


52.











22
2
2
2
cos sin
sin cos cos sin
speed sin cos cos sin
1
cos 2 sin sin 2 cos
sin cos cos sin
1
cos sin sin cos
1
rij
vr i j
v
ar i j
vij
T
v
ij
N
tt ttt
tttttttt
ttttttt
t
tttt ttt t
ttttttt
t
t t
tttttt
t
t















2
2
2
1
2
1
aT
aN
t
tt
t
t
tt
t
"


"


53.





22
2
2
21
2
2
15
2
2
15
52
51 5
rijk
vijk
v
ajk
ijk
T
ijk
N
ttt t
ttt
t
t
tt
t
t
t
t
t












2
22
5
15
55
51 5 1 5
aT
aN
t
t
tt
"

"


54.






2
4
2
3
22
4
2
4
1
1
1
21
2
21
2
22 1
rijk
vijk
v
ak
ijk
T
ij k
N
tt t
t
t
t
t
t
t
t
t
tt
t
t
t
t
t













34
4
2
21
4
22 1
aT
aN
tt
tt

"

"


55.




2cos 2sin , 3
33. Point: 1,3,3
3
2sin 2cos
33
rijk
rijk
rijk
rijk
ttttt
ttt






Direction numbers:
3,1,1
13, 3, 3x ty tz t
56.

23 2 322
33
,, ,rijkttt txtytz t
When
16
3
2, 2, 4, .txyz


2
22ri
j kttt
Direction numbers when
2, 1, 4, 8tabc

16
3
2, 4 4, 8xt y t z t

159. Review Exercises for Chapter 12 159
© 2010 Brooks/Cole, Cengage Learning
57.
4
9.56 10
4.58
4000 550
GM
v
r
'


mi/sec
58. Factor of 4
59.
23,0 5
23
ttt t
t
rij
rij





55
0 0
49 13 513
b
a
s t dt dt tr
**


60.


2
2,0 3
22
rik
rik
tt t t
tt






3
2
0
3
22
0
44
ln 1 1
ln 10 3 3 10 11.3053
r
b
a
s tdt t dt
tttt




**


61.


33
22
10 cos 10 sin
30 cos sin 30 sin cos
ttt
ttttt
rij
rij





42 4 2
30 cos sin sin cos
30 cos sin
rttttt
tt

π


2
2
2
0
0
sin
4 30 cos sin 120 60
2
t
sttdt



π"π π


*


62.



10 cos 10 sin
10 sin 10 cos
10
rij
rij
r
ttt
ttt
t


π


2
0
10 20sdt

ππ*


63.

324,0 3
324
tttt t
trijk
rijk



3
0
3
0
9416
29 3 29
r
b
a
s tdt dt
dt

ππ**
*


64.


2
2
2,0 2
22, 54
rijk
rijkr
ttt t t
tt t t






2
2
0
55
44
54
21 ln 5 ln 105 4 5 6.2638r
b
a
stdt tdt
**


65.


8cos ,8sin , ,0
2
8sin ,8cos ,1, 65r
rrttttt
tttt





2
0 65
65
2
rb
a
s t dt dt

πππ
**


2−2−4
−4
2
−6
−8
−10
−12
−14
−16 4 6 8 10 12 14
x
y
(0, 0)
(10, −15)
z
y
(9, 0, 6)
(0, 0, 0)
x
2
2
11
2
3
4
5
6
7
9
−2
3
4
5
6
−10
10
2
−10 10−22
x
y
−8−6−4
−4
2
4
6
8
−6
−8
−2 2468
y
x
x
y
6
8
2
4
10
2
2
6
4
8
10
12
z
(0, 0, 0)
(−9, 6, 12)
4
3
2
1
1
3
1
2
2
4
z
y
x
(2, 4, 4)
x
y
z
π
4
4
6
8
6
8
(8, 0, 0)
0, 8,
2
π
2
))

160. 160 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
66.

2
2sin cos ,2cos sin , ,0
2
2sin,2cos,1, 4 1
r
rr
ttttttttt
ttttt t t






2
2
0
22
41
1
ln 1 1 3.055
44
r
b
a
stdt tdt





**

67.
32rijttt
Line
0K
68.



32
23
1119
3, 9
1
2
rij
rijr
ri
ttt
t
tt
ttt
tt








32
32
32
32
32 323 3213 3
30 ;
22
1
00
2
3/2 3
19 219
t
tt
t
tt t
K
t tt t
ijk
rr krr
rr
r


' '

'



69.



22
21
2
2
22,54
2
rijk
rijkr
rjk
tttt
ttt t
t






32 323
22
22 42, 20
01 2
20 2 5
54 45
ijk
rr j krr
rr
r
tt
K
tt
' '
'



70.



25cos 5sin
25sin 5cos, 29
5cos 5sin
ri j k
ri j kr
rjk
tt t t
tttt
ttt




25sin 5cos
05cos 5sin
25 10 sin 10 cos
ij k
rr
ijk
tt
tt
tt
'





323
725
725 25 29 5
2929 2929
rr
rr
r
K
'
' "


71.



23
21111
,,1, 1
2323
,1
2, 1 2
rijk
rijkrijk
rikrik
ttttP t
tt t
tt









33
11 1 2
10 2
411 6 2
333
3
ijk
rr ijk
rr
r
K
'
'


72.


4cos 3sin , 4,0,
4sin 3cos , 3
4cos 3sin , 4
rijk
rijkrjk
rijri
ttttP t
ttt
ttt







323
031 4 12
400
16 144 2
591
ijk
rr j k
rr
r
K
'
'



73.
21
2
2
1
yx
yx
y






32 32
2 2
32
32
1
11
1
At 4, and 17 17 17.
17
y
K
xy
xK r






74.

2
22
2
32 32
2
11
,
24
1
4
1
1
1
4
x
xx
x
x
ye
yeye
e
y
K
y
e













At

32 32
14 2 2 2 5
0, ,
5255554
xK

55
.
2
r

161. Problem Solving for Chapter 12 161
© 2010 Brooks/Cole, Cengage Learning
75.

2
2
32 32
22
ln
1
1
1
111
yx
y
x
y
x
y x
K
yx
π
π


ππ



At
32
11 2
1, and 2 2.
24 22
xK rπππ π π
76.
2
2
tan
sec
2sec tan
yx
yx
yxx
π
π
π



2
32 32
24
2sec tan
1sec1
xxy
K
xy

ππ



At
32
4445 55
, and .
45 25 4 55
xK r
ππππ π
77. The curvature changes abruptly from zero to a nonzero
constant at the points B and C.
78.

53
42
3
4
32
2
42
53
20 6
20 6
15 3
yax bx cx
yaxbxc
yaxbx
ax bx
K
ax bx c




π




At
1:xπ

02060
053 0
11 1
kab
yabc
y abc




Solving these 3 equations for
,,,abcyou obtain
3
8
,aπ
515
48
,.bc By symmetry, the same holds
at 1.x

533515
84 8
yx x x



Problem Solving for Chapter 12
1.


22
00
22
cos , sin
22
cos , sin
22
tt uu
xtduytdu
tt
xt yt


ππ


ππ

**

(a)

22
00
aa
s x t y t dt dt a**

(b)


22
22
22
sin , cos
22
cos sin
22
1
At , .
tt
xt t yt t
tt
tt
K t
taK a











ππ
ππ

(c) Ka
ππ (length)
2.
23 23 23
x ya


13 13
13
1322
0
33
Slope at , .
xyy
y
yPxy
x



π









33
22
cos sin
3 cos sin 3 sin cos
3cos sin
cos sin
sin cos
ttt
ttttt
ttt
t
ttt
t
tttrij
rij
ri
r
Tij
r
Tij

π






0, 0, 0Q origin


33
cos , sin , 0Pttπ on curve.


33 3 3
cos sin 0 cos sin sin cos
cos sin 0
PQ t t t t t t
tt
ijk
Tk

'





cos sin
1
3cos sin
T
T
T
r
PQ
Dtt
t
K
ttt
'
ππ

ππ


So, the radius of curvature,
1
,
K
is three times the
distance from the origin to the tangent line.
y
x
−11
−1
1
(1, 1)
(−1, −1)

162. 162 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
3. Bomb:
2
1
5000 400 , 3200 16rttt
Projectile:

2
20 0
cos , sin 16rtv tv tt
At 1600 feet: Bomb:

2
3200 16 1600 10tt seconds.
Projectile will travel 5 seconds:

0
05 sin 16 25 1600
sin 400.
v
v



Horizontal position:
At
10, bomb is at 5000 400 10 1000.t
At
05, projectile is at 5 cos .tv
So,
0cos 200.v
Combining,

0
0sin 400
tan 2 63.43 .
cos 200
v
v




0
200
447.2
cos
v

ft/sec
4. Bomb:

2
1
5000 400 , 3200 16rttt
Projectile:

2
20 0
cos , sin 16rtv tv tt
At 1600 feet: Bomb:

2
3200 16 1600 10tt
Projectile will travel 5 seconds:

0
05 sin 16 25 1600
sin 400.
v
v



Horizontal position:
At
10, bomb is at 5000 400 10 9000.t
At
05, projectile is at cos 5.tv
So,

0
05 cos 9000
cos 1800.
v
v



Combining,

0
0sin 400 2
tan 12.5 .
cos 1800 9
v
v




0
1800
1843.9
cos
v

ft/sec
5.
1cos, sin,0xy )



22 2
2
2
1cos sin
22cos 4sin
2
xy







2sin 4cos 4cos
222
sin , cos
t
t
t
st d
xy









*



3
3
1 cos cos sin sin
2 sin
2
cos 1
8 sin
2
1
4 sin
2
K













So,
1
4sin
2 t
K
- and
22 2 2
16 cos 16 sin 16.
22
tt
s-





6. 1cos
sin
r
r






2
2
1 cos sin 2 2 cos
2sin 4cos 4cos
222
tt
t
t
stdd
t
d









**
*




2
2
32
2
2
2
2
3
2
33
2
2 sin 1 cos cos 1 cos
8sin
2
sin
33cos 33
2
4
8 sin sin 4 sin
222
4sin
1
2
3
rrrr
K
rr
K






-











22 2 2
9 16 cos 16 sin 16
22
s

-


7.




2
2
2
rrr
rr
rrrrrrrr
r
ttt
ttdd d
tttttttt
dt dt dt t
"
"
""

163. Problem Solving for Chapter 12 163
© 2010 Brooks/Cole, Cengage Learning
8. (a)
2
22 2
22 2
2
2
position vector
cos sin
cos sin sin cos
cos sin sin cos sin
sin cosrij
rij
r
ij
r
a ixy
rr
ddr d dr d
rr
dt dt dt dt dt
ddr drddrd d d
rr
dt dt dt dt dt dt dt dt
dr dr d
dt dt d


















2
2
2
cos sin cos
dr d d d
rr
tdt dt dt dt








2
2 2
22
2 2
22
2 22
22
2 22
cos sin
cos 2 sin cos cos cos sin
sin 2 sin cos sin cos sinau a i j
rra
dr dr d d d
rr
dt dt dt dt dt
dr dr d d d dr d
rr r
dt dt dt dt dt dt dt



" "









2
2
sin cos 2aa i j
dr d d
au r
dr dt dt



" "


2
22
22
2aauu au u u
rr r
dr d drd d
ur r
dt dt dt dt dt


" "


(b)
2
2
42,000 cos 42,000sin
12 12
42,000, 0, 0rij
r
tt
dr d r
dt dt



πππ


2
2
,0
12
dd
dt dt

ππ
So,
2
2
875
42,000 .
12 3
auu
rr






Radial component:
2875
3

Angular component: 0
9.


4cos 4sin 3 ,
2
4sin 4cos 3 , 5
4cos 4sin
ttttt
ttt t
tttrijk
rijkr
rij






443
sin cos
555
44
cos sin
55
cos sin
33
sin cos
55 5
Tijk
Tij
Nij
4
BTN i j ktt
tt
tt
tt



'

At
,
2t

π
43
255
2
34
255
Tik
Nj
Bik















x
y
z
π3
12
3
4
4
π6
T
T
B
B
N
N

164. 164 Chapter 12 Vector-Valued Functions
© 2010 Brooks/Cole, Cengage Learning
10.

cos sin ,
4
sin cos , 1
sin cos
cos sin
cos sinrijk
rijr
Tij
Tij
Nij
BTNktttt
tttt
tt
tt
tt






π' π

At
,
4t

π
22
422
22
422
4
Tij
Nij
Bk










π



11. (a)
1 constant length
B
BTN Bd
ds
' (



0
B
TN TN T N
B
TTTNTTN
T
TTN T T
T
dd
ds ds
d
ds
'''
""'"'

'"" '



So, and
BBB
BT N
ddd
ds ds ds
@((
for some scalar .
@
(b) .BTNπ' Using Section 11.4, exercise 66,


BN TN N N TN
NNT NTN
T
' ' '' '
" "






.
BT TN T T TN
TNT TTN
N
' ' '' '
" "

π

Now,



TTT
NT
T
sdd
Ks
ds dss

πππ


Finally,




.
NBTBTBT
BN NT
TB
d
s
ds
K
K
@
@
'''
' '

12.
52
32
121
32
5
64
15
128
yx
yx
yx
π
π
π


12
32
315
128
25
1
4096
x
K
x
π





At the point



32
32
89120 1
4,1 , 7.
12089
Kr
K

13.

cos , sin , 0 2rtt ttt t
(a)
(b) Length


2
0
2
22
0

1
6.766 graphing utility
rtdt
tdt

π


*
*

(c)






22
32
22
2
32
2
2
1
02
2
11.04
1
20.51
t
K
t
K
K
K




π


π





(d)
(e)
lim 0
t
K
!
π
(f ) As ,t !the graph spirals outward and the
curvature decreases.
14. (a) Eliminate the parameter to see that the Ferris wheel has a radius of15 meters and is centered at
16 .
jAt 0,tπ
the friend is located at
10,rjπwhich is the low point on the Ferris wheel.
(b) If a revolution takes
t5seconds, then


2
10 10
tt t

5


and so
20t5πseconds. The Ferris wheel makes three revolutions per minute.
−33
−2
2
0 5
0
5
−1
−2
z
y
x
2
2
2
1
−1
−1
−2
−2

165. Problem Solving for Chapter 12 165
© 2010 Brooks/Cole, Cengage Learning
(c) The initial velocity is20 8.03 11.47 .ijrt The speed is
22
8.03 11.47 14 m/sec. The angle of inclination
is
11.47
arctan 0.96
8.03




radians or 55 .
(d) Although you may start with other values,
00tis a fine choice. The graph at
the right shows two points of intersection. At 3.15t sec the friend is near the
vertex of the parabola, which the object reaches when


0
11.47
1.17
24.9
tt

sec.
So, after the friend reaches the low point on the Ferris wheel, wait
02tsec
before throwing the object in order to allow it to be within reach.
(e) The approximate time is 3.15 seconds after starting to rise from the low point on
the Ferris wheel. The friend has a constant speed of

1 15rt m/sec. The speed
of the object at that time is



2
2
2
3.15 8.03 11.47 9.8 3.15 2 8.03r

m/sec.
0
0
20
30

166. © 2010 Brooks/Cole, Cengage Learning
CHAPTER 13
Functions of Several Variables
Section 13.1 Introduction to Functions of Several Variables.................................167
Section 13.2 Limits and Continuity.........................................................................174
Section 13.3 Partial Derivatives ..............................................................................183
Section 13.4 Differentials ........................................................................................197
Section 13.5 Chain Rules for Functions of Several Variables...............................203
Section 13.6 Directional Derivatives and Gradients ..............................................212
Section 13.7 Tangent Planes and Normal Lines.....................................................221
Section 13.8 Extrema of Functions of Two Variables ...........................................236
Section 13.9 Applications of Extrema of Functions of Two Variables.................246
Section 13.10 Lagrange Multipliers ..........................................................................256
Review Exercises........................................................................................................269
Problem Solving.........................................................................................................281

167. © 2010 Brooks/Cole, Cengage Learning 167
CHAPTER 13
Functions of Several Variables
Section 13.1 Introduction to Functions of Several Variables
1. No, it is not the graph of a function. For some values of x
and y
for example, , 0, 0 ,xy there are 2 z-values.
2. Yes, it is the graph of a function.
3.
22
22
2
2
310
10 3
10 3
xz y xy
xzxyy
xyy
z
x





Yes, z is a function of x and y.
4.
22
24xz xy y
No, z is not a function of x and y. For example,
,1,0xy corresponds to both 2.z

5.
22
2
1
49
xy
z
No, z is not a function of x and y. For example,
,0,0xy corresponds to both 1.z

6.

ln 8 0
18 ln
ln
81
zxy yz
zyxy
xy
z
y





Yes, z is a function of x and y.

7. ,
fxy xy
(a)
3, 2 3 2 6f
(b)
1,4144f
(c)
30, 5 30 5 150f
(d)
5, 5
fyy
(e)
,2 2
fxx
(f )
5, 5
ftt

8.
22
,4 4
fxy x y
(a)
0, 0 4f
(b)
0, 1 4 0 4 0f
(c)
2, 3 4 4 36 36f
(d)

22
1, 4 1 4 3 4
fyyy
(e)

22
,0 4 0 4
fxx x
(f )

22
,1 4 4
fttt

9. ,
y
fxy xe
(a)

0
5, 0 5 5fe
(b)

2
3, 2 3
f e
(c)

12
2, 1 2fe
e


(d)
5, 5
y
fye
(e)

2
,2
fxxe
(f )
,
t
ftt te

10.
,lngxy x y
(a)

1, 0 ln 1 0 0g
(b)

0, 1 ln 0 1 ln 1 0g
(c)

0, ln 0 1ge e
(d)

1,1 ln1 1 ln2g
(e)
3
,ln ln ln3lnln2
222
1ln3ln2
eee
ge e e






(f )

2, 5 ln 2 5 ln 7g

11.
,,
xy
hx y z
z


(a)


23 2
2, 3, 9
93
h

(b)


10
1, 0, 1 0
1
h

(c)


23 3
2, 3, 4
42
h



(d)


54 10
5, 4, 6
63
h



12.
,,fxyz x y z
(a)

0, 5, 4 0 5 4 3f
(b)

6, 8, 3 6 8 3 11f
(c)

4, 6, 2 4 6 2 12 2 3f
(d)

10, 4, 3 10 4 3 3f

168. 168 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
13. ,sinfxy x y
(a) 2, 2 sin 2
44
f





(b)
3, 1 3 sin 1f
(c)
333
3, 3 sin 3
3322
f



(d)
4, 4 sin 4
22
f






14.
2
,Vrh rh
(a)

2
3, 10 3 10 90V
(b)

2
5, 2 5 2 50V
(c)

2
4, 8 4 8 128V
(d)

2
6, 4 6 4 144V

15.
222
,23
333
y
x
y
x
gxy t dt
tt yyxx


*

(a)
4, 0 0 16 12 4g
(b)
4, 1 1 3 16 12 6g
(c)

399 25
242 4
4, 16 12g
(d)

3999
2424
,0 0g

16.
1
,lnlnlnln
y
y
x
x
y
gxy dt t y x
tx




*

(a)

1
4, 1 ln ln 4
4
g

(b)

3
6, 3 ln ln 2
6
g

(c)

5
2, 5 ln
2
g

(d)
1
2
17
,7 ln ln14
2
g






17.
2
,2
fxy x y
(a)


22
22,, 2
2, 0
xxy xyfx xy fxy x
x
xxx
5 5 5
5#
555

(b)


2 22
2,,2 2
2,0
yy yfxy y fxy x y y x y
yyy
yyy
555 5
55#
555


18.
2
,32
fxy x y
(a)


2 22
3232,, 63
63, 0
xx y x yfx xy fxy xx x
xxx
xxx
5 5 5 5
55#
555

(b)


22
32 32,, 2
2, 0
xyyxyfxy y fxy y
y
yyy
5 5 5
5#
555


19.
22
,
fxy x y
Domain:
9
:, : is any real number, is any real numberxy x y
Range: 0z/

20. ,
xy
fxy e
Domain: Entire xy-plane
Range: 0z

21.
,gxy x y
Domain:
9
:,: 0xy y/
Range: all real numbers

22.
,
y
fxy
x

Domain:

9 :,: 0xy x
Range: all real numbers

23.
xy
z
xy



Domain:

9 :,: 0and 0xy x y##
Range: all real numbers

169. Section 13.1 Introduction to Functions of Several Variables 169
© 2010 Brooks/Cole, Cengage Learning
24.
xy
z
xy



Domain:
9
:,:xyx y#
Range: all real numbers

25.
22
,4fxy x y
Domain:
9:
22
22
22
40
4
,: 4
xy
xy
xy x y
/


Range: 0 2z

26.
22
,44fxy x y
Domain:
22
22
22
440
44
1
41
xy
xy
xy
/





22
,: 1
41
xy
xy
6<
7>
8?
Range: 0 2z

27. , arccos
fxy x y
Domain:
9
:,:1 1xy x y
Range: 0 z


28.
,arcsin
y
fxy
x





Domain:

,:1 1
y
xy
x
6<
7>
8?

Range:
22
z



29. ,ln4
fxy x y
Domain:
9:
40
4
,: 4
xy
xy
xy y x




Range: all real numbers

30. ,ln6fxy xy
Domain:
60
6
xy
xy




9
:,: 6xy xy
Range: all real numbers

31.
22
4
,
1
x
fxy
xy




(a) View from the positive x-axis:
20, 0, 0
(b) View where x is negative, y and z are positive:
15, 10, 20
(c) View from the first octant:
20, 15, 25
(d) View from the line
yx in the xy-plane:
20, 20, 0

32. (a) Domain:


9 :, : is any real number, is any real numberxy x y
Range:
22z
(b) 0zwhen 0xwhich represents points on the
y-axis.
(c) No. When x is positive, z is negative. When x is
negative, z is positive. The surface does not pass
through the first octant, the octant where y is
negative and x and z are positive, the octant where y
is positive and x and z are negative, and the octant
where x, y and z are all negative.

33. ,4fxy
Plane:
4z

34. ,623
fxy x y
Plane
Domain: entire xy-plane
Range: z! !

35.
2
,
fxy y
Because the variable x is missing, the surface is a
cylinder with rulings parallel to the x-axis. The
generating curve is
2
.zyThe domain is the entire
xy-plane and the range is 0.z/

x
y
4
2
3
1
3
5
2
1
2
3
5
z
x
y
2
33
44
6
z
x
y2
3
1
4
4
5
z

170. 170 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
36.
1
,
2
gxy y
Plane:
1
2
zy


37.
22
zxy
Paraboloid
Domain: entire xy-plane
Range: 0z

38.
221
2
zxy

Cone
Domain of f : entire xy-plane
Range: 0z/

39. ,
x
fxy e


Because the variable y is missing,
the surface is a cylinder with
rulings parallel to the y-axis. The
generating curve is
.
x
ze


The domain is the entire xy-plane
and the range is 0.z

40.
,0,0
,
0, else where
xy x y
fxy
//6
7
8

Domain of
:
fentire xy-plane
Range: 0z/

41.
22
1zy x
Hyperbolic paraboloid
Domain: entire xy-plane
Range: z! !

42.
221
, 144 16 9
12
fxy x y
Semi-ellipsoid
Domain: set of all points
lying on or inside the ellipse

22
1
916
xy


Range: 0 1z

43.
22
,
xy
fxy xe




44. ,sin
fxy x y


45.
22
1
xy
ze


Level curves:

22
1
22
22
ln 1
1ln
xy
ce
cxy
xy c





Circles centered at
0, 0
Matches (c)

46.
22
1
xy
ze


Level curves:

22
1
22
22
ln 1
1ln
xy
ce
cxy
xy c





Hyperbolas centered at
0, 0
Matches (d)
x
y
2
2
−2
1
z
x
y
44
2
4
6
8
z
x
y
5
25
20
15
10
5
z
x
y
z
4 4
4
−2
−4
yx
z
x
y
z
4
4
−4
−4
−4
yx
z
x
y
1
1
2
2
33
2
z
x
y 1
2
2
1
−1
−1
−2
z

171. Section 13.1 Introduction to Functions of Several Variables 171
© 2010 Brooks/Cole, Cengage Learning
47.
2
lnzyx
Level curves:

2
2
2
ln
c
c
cyx
eyx
yx e




Parabolas
Matches (b)

48.
2
2
cos
4
xy
z




Level curves:

22
22
1
22 1
2
cos
4
2
cos
4
24cos
x y
c
xy
c
xy c









Ellipses
Matches (a)

49. zxy
Level curves are parallel
lines of the form
.
xyc

50. ,623
fxy x y
The level curves are of the
form
62 3
xyc or
23 6.xyc So, the
level curves are straight
lines with a slope of
2
3
.

51.
22
4zx y
The level curves are ellipses of the form
22
4
x yc

22
except 4 0 is the point 0, 0 .xy


52.
22
,9fxy x y
The level curves are of the form

22
22 2
9
9,circles.
cxy
xy c





22
0 is the point 0, 0 .xy

53. ,
fxy xy
The level curves are
hyperbolas of the form
.
xyc

54.
2
,
xy
fxy e
The level curves are of the form

2
,
xy
ecor ln .
2
xy
c

So, the level curves are
hyperbolas.

55.
22
,
x
fxy
xy



The level curves are of the form

22
22
22
2
0
11
.
22
x
c
xy
x
xy
c
xy
cc








So, the level curves are circles passing through the origin
and centered at 12 ,0.c

56. ,ln
fxy x y
The level curves are of the form

ln
.
c
c
cxy
exy
yxe




So, the level curves are parallel
lines of slope 1 passing through
the fourth quadrant.

57.
22
,2fxy x y

4
4
2
2
−2
−2
x
c = −1
c = 0
c = 2
c = 4
y
x
−2
3
c= 0
c= 2
c= 4
c= 6
c= 8
c= 10
y
c = 0
c = 1
c = 2
c = 3
c = 4
x
y
2
−22
−2
−11
−1
1
c = 0
c = 1
c = 2
c = 3
x
y
1
1
−1
−1
x
c = 6
c = 5
c = 4
c = 3
c = 2
c = 1
c = −1
c = −2
c = −3
c = −4
c = −5
c = −6
y
x
2
2
c = 1
c = −1
c = −2
c = 2
y
1
2
−c =
1
2
c =
3
2
c =
3
2
−c =
−6
−4
6
x
c=2−
c=1−
c= 2
c=
c=−
c=±
c= 1
c= 0
1
1
3
2
2
2
y
−9
−6
9
6
−2−1
−1
1
2
−2
1
1
2
2
c = 3
c = 2
c =
c = 4
1
3
c =
1
4
c =
y
x

172. 172 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
58. ,fxy xy

59.
22
8
,
1
gxy
xy




60.
,3sinhx y x y

61. The graph of a function of two variables is the set of all
points
,,xyzfor which ,zfxy and ,xyis in
the domain of f. The graph can be interpreted as a
surface in space. Level curves are the scalar fields
,,
fxy cwhere c is a constant.

62. No, the following graphs are not hemispheres.


22
22
xy
ze
zx y





63.
,
x
fxy
y

The level curves are the lines
x
c
y
or
1
.yx
c


These lines all pass through the origin.

64. ,,0,0fxy xyx y//
(a)
(b) g is a vertical translation of f three units downward.
(c) g is a reflection of f in the xy-plane.
(d) The graph of g is lower than the graph of f. If
,zfxy is on the graph of f, then
1
2
zis on the
graph of g.
(e)

65. The surface is sloped like a saddle. The graph is not
unique. Any vertical translation would have the same
level curves.
One possible function is


,.fxy xy

66. The surface could be an ellipsoid centered at 0, 1, 0 .
One possible function is



2
2
1
,1.
4
y
fxy x




67.

10
10.061
, 1000
1
R
VIR
I







68. , 5000
rt
Art e


69. ,, , 1fxyz x y zc

1,Planexyz

70. ,, 4 2
fxyz x y z
4c
44 2xy z
Plane
−6
−4
6
4
−6
−4
6
4
−1
−1
1
1
x
y
5
25
20
15
10
5
z
x
y
5
25
20
15
10
5
z
Inflation Rate
Tax Rate 0 0.03 0.05
0 1790.85 1332.56 1099.43
0.28 1526.43 1135.80 937.09
0.35 1466.07 1090.90 900.04
Number of Year
Rate 5 10 15 20
0.02 5525.85 6107.01 6749.29 7459.12
0.03 5809.17 6749.29 7841.56 9110.59
0.04 6107.01 7459.12 9110.59 11,127.70
0.05 6420.13 8243.61 10,585.00 13,591.41

x
y
2
−1
−2
1
2
1
2
1
z
y
x
1
2
3
2
3
4
z

173. Section 13.1 Introduction to Functions of Several Variables 173
© 2010 Brooks/Cole, Cengage Learning
y
z
x
71.
222
,,fxyz x y z
9c

222
9
xyz
Sphere

72.
221
4
,,
fxyz x y z
1c

221
4
1
x yz
Elliptic paraboloid
Vertex:
0, 0, 1

73.
222
,, 4 4
fxyz x y z
0c

222
04 4
x yz
Elliptic cone

74. ,, sin
fxyz x z
0c

0sin
xzor sinzx

75.
2
4
,
4
d
NdL L




(a)


2
22 4
22, 12 12 243
4
N




board-feet
(b)


2
30 4
30, 12 12 507
4
N




board-feet

76.
1
,wyx
xy



(a)

11
15, 9 h 10 min
15 9 6
w


(b)

11
15, 13 h 30 min
15 13 2
w


(c)

11
12, 7 h 12 min
12 7 5
w


(d)

11
5, 2 h 20 min
52 3
w



77.
22
600 0.75 0.75Txy
The level curves are of the form

22
22
600 0.75 0.75
600
.
0.75
cxy
c
xy




The level curves are circles
centered at the origin.

78.
22
5
,
25
Vxy
xy




79.




0.6 0.4
0.6 0.4
0.6 0.4
0.6 0.4
0.6 0.4
0.6 0.4
0.6 0.4
, 100
2 , 2 100 2 2
100 2 2
100 2 2
2 100 2 ,
fxy x y
fxy x y
xy
xy
xyfxy








80.
1aa
zCxy



ln ln ln 1 lnzCax ay

ln ln ln ln lnzyCaxay

ln ln ln
zx
Ca
yy


81.

base front and back 2 ends
1.20 2 0.75 2 0.75
1.20 1.50
Cxy xz yz
xy xz yz




82.

2
23
4
34
33
r
Vrl r lr



83. PV kT
(a)
520
26 2000 300
3
kk

(b)
520
3
kT T
P
VV





The level curves are of the form

520
,
3
T
c
V




or
520
.
3
VT
c


These are lines through the origin with slope
520
.
3c

x
y
−4
−4
4
4
4
z
x
y5
3
5
z
x
y
−2
−2
2
2
12
z
y
x
2
8
4
z
30
30
y
x
−30
c = 600
c = 500
c = 400
c = 300
c = 200
c = 100
c = 0
−30
y
x
−25 25
−25
25
c =
1
4
c =
1
3
c =
1
2
r
l

174. 174 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
84. (a) , 0.026 0.316 5.04zfxy x y

(b) y has the greater influence because its coefficient (0.316) is greater than x’s coefficient (0.026).
(c)
, 95 0.026 0.316 95 5.04 0.026 35.06fx x x
This gives the shareholder’s equity in terms of net sales x, assuming total assets of
95 billion .y

85. (a) Highest pressure at C
(b) Lowest pressure at A
(c) Highest wind velocity at B

86. Southwest

87. (a) No; the level curves are uneven and sporadically
spaced.
(b) Use more colors.

88. (a) The different colors represent various amplitudes.
(b) No, the level curves are uneven and sporadically
spaced.

89. False. Let


,2
1, 2 2, 1 , but 1 2.
fxy xy
ff

#

90. False. Let

,5.fxy
Then,

2
2,2 5 2 , .
fxy fxy#

91. True

92. False. If there were a point ,xyon the level curves
1,
fxy Cand 2,,fxy C then
12 .CC
Section 13.2 Limits and Continuity
1.
,1,0
lim 1
xy
x



,,1fxy xL
We need to show that for all 0,
Athere exist a
-neighborhood
B about 1, 0such that


,1fxy L x A
Whenever
,1,0xy# lies in the neighborhood.
From

22
010,xy B it follows that

222
11 10.xx xy B
So, choose
BAand the limit is verified.

2. Let 0Abe given. We need to find 0Bsuch that


,4fxy L x A
whenever



22 22
041.xa yb x y B
Take .
BA
Then if

22
041 ,xy BA we have


2
4
4.
x
xA
A


3.
,1,3
lim 3.
xy
y

,,
fxy y 3L
We need to show that for all 0,
Athere exists a
-neighborhood
B about 1, 3such that


,3fxy L y A
whenever
,1,3xy# lies in the neighborhood.
From

22
013xy B it follows that

222
33 13.yy xy B
So, choose
BAand the limit is verified.

4. Let 0Abe given. We need to find 0Bsuch that

,fxy L y b A
whenever

22
0. xa yb B Take
.
BA
Then if

22
0, xa yb BA we have


2
.
yb
yb
A
A

Year 2002 2003 2004 2005 2006 2007
z 35.2 39.5 43.6 49.4 53.2 61.6
Model 35.3 39.4 44.0 49.4 53.3 61.8

175. Section 13.2 Limits and Continuity 175
© 2010 Brooks/Cole, Cengage Learning
5.






,, ,, ,,
lim , , lim , lim , 4 3 1
xy ab xy ab xy ab
fxy gxy fxy gxy




6.







,,
,,
,,
5lim ,
5, 54 20
lim
,lim,33
xy ab
xy ab
xy ab
fxy
fxy
gxy gxy










7.






,, ,, ,,
lim , , lim , lim , 4 3 12
xy ab xy ab xy ab
fxygxy fxy gxy







8.









,, ,,
,,
,,
lim , lim ,
,, 43 7
lim
,lim,44
xy ab xy ab
xy ab
xy ab
fxy gxy
fxy gxy
fxy fxy









9.


2
,2,1
lim 2 8 1 9
xy
xy


Continuous everywhere

10.


,0,0
lim 4 1 0 4 0 1 1
xy
xy


Continuous everywhere

11.


12 2
,1,2
lim
xy
xy
ee e


Continuous everywhere

12.

22
,2,4
24 6
lim
1215
xy
xy
x





Continuous everywhere

13.

,0,2
0
lim 0
2
xy
x
y


Continuous for all 0y
#

14.

,1,2
12 1
lim
12 3
xy
xy
xy





Continuous for all
.
xy#

15.

22
,1,1
1
lim
2
xy
xy
xy



Continuous except at
0, 0

16.

,1,1
12
lim
211
xy
x
xy



Continuous for 0xy


17.



,4,2
lim cos 2 cos 0
2
xy
yxy




Continuous everywhere

18.

,2,4
2
lim sin sin 1
4
xy
x
y




Continuous for all 0y
#

19.

,0,1
arcsin arcsin 0
lim 0
11
xy
xy
xy



Continuous for 1,xy
#
1xy

20.

,0,1
arccos
arccos 0
lim
112
xy
x
y
xy








Continuous for 1,xy
# 0,y#
0
x
y

21.

,, 1,3,4
lim 1 3 4 2 2
xyz
xyz


Continuous for 0xyz
/

22.


10
,, 2,1,0
lim 2 2
yz
xyz
xe e


Continuous everywhere

23.

,1,1
111
lim 0
111
xy
xy
xy






24.

2
2
,1,1
11
lim
1112
xy
xy
xy






25.

,0,0
1
lim
xy
xy
does not exist
Because the denominator xyapproaches 0 as
,0,0.xy

26.

22
,0,0
1
lim
xy
xy
does not exist because the denominator
xy approaches 0 as
,0,0.xy

27.





22
,2,2 ,2,2
,2,2
lim lim
lim 4
xy xy
xy
xyx yxy
xy xy
xy








28.





2222
44
22 22
,0,0 ,0,0
22
,0,0
224
lim lim
22
lim 2 0
xy xy
xy
x yx yxy
xy xy
xy

176. 176 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
29.
,0,0
lim
xy
xy
x y




does not exist because you can’t approach 0, 0from
negative values of x and y.

30.






,2,1
,2,1
,2,1
11
lim
11
11
lim
1
lim 1 2
xy
xy
xy
xyxy
xy xy
xy xy
xy
xy




"







31. The limit does not exist because along the line 0y
you have



22
, 0,0 ,0 0,0 ,0 0,0
1
lim lim lim
xy x x
xy x
xyx x




which does not exist.

32. The limit does not exist because along the line
xyyou have



22 22
,0,0 ,0,0 ,0,0
lim lim lim .
0
xy xx xx
x xx
xy xx



Because the denominator is 0, the limit does not exist.

33.


2
22
,0,0
0
lim 0
1111
xy
x
xy





34.


22
,0,0
lim ln
xy
xy

does not exist
because

22
lnxy ! as ,0,0.xy

35. The limit does not exist because along the path
0,x 0,yyou have



222 2
, , 0, 0, 0 0, 0, 0, 0, 0
0
lim lim 0
xyz z
xy yz xz
xyz z





whereas along the path
,
xyzyou have



222
2 22 222
, , 0, 0, 0 , , 0, 0, 0
lim lim
1
xyz xxx
xyyzxz xxx
xyz xxx






36. The limit does not exist because along the path
0,yz you have



22
222 2
, , 0, 0, 0 , 0, 0 0, 0, 0
0
lim lim 0
xyz x
xy yz xz
xyz x





However, along the path 0,z
,
xyyou have



22 2
222 22
, , 0, 0, 0 , , 0 0, 0, 0
lim lim
1
2
xyz xx
xy yz xz x
xyz xx






37.
,0,0
lim 1
xy
xy
e


Continuous everywhere

38.


22
22
,0,0
cos
lim 1
xy
xy
xy


!



The limit does not exist.
Continuous except at
0, 0

39.
22
,
xy
fxy
xy



Continuous except at
0, 0
Path: 0y

Path:
yx

The limit does not exist because along the path 0ythe function equals 0, whereas
along the path
yx the function equals
1
2
.
,
xy 1, 0 0.5, 0 0.1, 0 0.01, 0 0.001, 0
,fxy 0 0 0 0 0

,xy 1, 1 0.5, 0.5 0.1, 0.1 0.01, 0.01 0.001, 0.001
,fxy
1
2

1
2

1
2

1
2

1
2

177. Section 13.2 Limits and Continuity 177
© 2010 Brooks/Cole, Cengage Learning
40.
22
,
y
fxy
xy



Continuous except at
0, 0
Path:
yx

Path: 0y

The limit does not exist because along the path 0ythe function equals 0, whereas along the path
yx the function tends
to infinity.

41.
2
24
,
xy
fxy
xy



Continuous except at
0, 0
Path:
2
xy

Path:
2
x y

The limit does not exist because along the path
2
xythe function equals
1
2
,whereas along the path
2
x ythe function
equals
1
2
.

42.
2
2
2
,
2
xy
fxy
xy




Continuous except at
0, 0
Path: 0y

Path:
yx

The limit does not exist because along the 0ythe function tends to infinity, whereas along the line
yx the function
tends to 2.
,xy 1, 1 0.5, 0.5 0.1, 0.1 0.01, 0.01 0.001, 0.001
,fxy
1
2
1 5 50 500

,xy 1, 0 0.5, 0 0.1, 0 0.01, 0 0.001, 0
,fxy 0 0 0 0 0

,xy 1, 1 0.25, 0.5 0.01, 0.1 0.0001, 0.01 0.000001, 0.001
,fxy
1
2

1
2

1
2

1
2

1
2


,xy 1, 1 0.25, 0.5 0.01, 0.1 0.0001, 0.01 0.000001, 0.001
,fxy
1
2

1
2

1
2

1
2

1
2


,xy 1, 0 0.25, 0 0.01, 0 0.001, 0 0.000001, 0
,fxy 1 4 100 1000 1,000,000

,xy 1, 1 0.25, 0.25 0.01, 0.01 0.001, 0.001 0.0001, 0.0001
,fxy
1
3
1.17 1.95 1.995 2.0

178. 178 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
43.




2222
44
22
22 22
,0,0 ,0,0 ,0,0
lim lim lim 0
xy xy xy
xyxyxy
xy
xy xy





So, f is continuous everywhere, whereas g is continuous everywhere except at 0, 0 .g has a removable discontinuity at 0, 0 .

44.





2222
44
22
22 22
,0,0 ,0,0 ,0,0
224
lim lim lim 2 0
22
xy xy xy
xyxyxy
xy
xy xy





So, g is continuous everywhere, whereas f is continuous everywhere except
0, 0 . f has a removable discontinuity at 0, 0 .

45.

22
22
,0,0
4
lim 0
xy
xy
xy




So,




,0,0 ,0,0
lim , lim , 0.
xy xy
fxy gxy


f is continuous at
0, 0 ,whereas g is not continuous at
0, 0 .

46.




222
22
,0,0 ,0,0
2
22
,0,0
2
lim , lim
2
lim 1 1
xy xy
xy
x xy y
fxy
xy
xy
xy









(same limit for g)
So, f is not continuous at
0, 0 , whereas g is continuous
at
0, 0 .

47.
,0,0
lim sin sin 0
xy
xy



48.

,0,0
11
lim sin cos
xy
x x

Does not exist

49.

2
42
,0,0
lim
2
xy
xy
x y

Does not exist. Use the
paths 0xand
2
.yx

50.

22
2
,0,0
lim
xy
xy
xy


Does not exist

51.

22
,0,0
5
lim
2
xy
xy
x y

Does not exist. Use the paths 0xand
.
xy


52.

22
,0,0
6
lim 0
1
xy
xy
xy





53.



22
2
22 2
,0,0 0
2
0
cos sin
lim lim
lim cos sin 0
xy r
r
rrxy
xy r
r








54.



33 3
33
22 2
,0,0 0
33
0
cos sin
lim lim
lim cos sin 0
xy r
r
rxy
xy r
r









55.

22 4 2 2
22 2
,0,0 0
22 2
0
cos sin
lim lim
lim cos sin 0
xy r
r
xy r
xy r
r







x
y
z
x
6
4
y
6
2
z
x
y
2
−2
2
2
z
x y
4
4
18
z
x
y
−2
−3
−3
3
3
z
2
x
−4
−2
−4
−4
y
z
4
2
−2
4
2
4

179. Section 13.2 Limits and Continuity 179
© 2010 Brooks/Cole, Cengage Learning
56. cos ,xr sin ,yr
22 22 22 2
,cossinxy rxy r




22 2
22
22
22,0,0 0 0
cos sin
lim lim lim cos sin 0
xy r r
rxy
r
rxy







57.


22 2
,0,0 0
lim cos lim cos cos 0 1
xy r
xy r



58.


22
,0,0 0
lim sin lim sin sin 0 0
xy r
xy r



59.
22
xyr



22
22,0,0
0
sinsin
lim lim 1
xy
r
rxy
rxy







60.


22
22
2
22 2
,0,0 0 0 0
sin sin 2 cos
lim lim lim lim cos 1
2
xy r r r
xy rrr
r
xy r r






61.
22 2
xyr




22 2
22 2
,0,0 0
1 cos 1 cos
lim lim 0
xy x
xy r
xy r






62.
22 2
xyr



22 22 22 2
,0,0 0 0
lim ln lim ln lim 2 ln
xy r r
xyxy rr rr


By L’Hôpital’s Rule,



22
23
0000
2ln 2
lim 2 ln lim lim lim 0
12
rrrr
r r
rr r
rr






63.
222
1
,,fxyz
xyz



Continuous except at

0, 0, 0

64.
22
,,
4
z
fxyz
xy


Continuous for
22
4.xy#

65.
sin
,,
x y
z
fxyz
ee



Continuous everywhere

66. ,, sin
fxyz xy z
Continuous everywhere

67. For 0,xy#the function is clearly continuous.
For 0,xy#let
.zxyThen

0
sin
lim 1
z
z
z


implies that f is continuous for all x, y.

68. For
22
,
x y# the function is clearly continuous.
For
22
,
x y# let
22
.zx y Then


0
sin
lim 1
z
z
z


implies that f is continuous for all x, y.

69.
2
,
ftt ,23gxy x y


2
,2323
fgxy fxy xy
Continuous everywhere

70.



22
22
22
1
,
1
,
ft
t
gxy x y
fgxy fx y
xy





Continuous except at (0, 0)

71.

1
,, 23ftgxyxy
t



1
,23
23
fgxy f x y
xy



Continuous for all
2
3
yx#

180. 180 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
72.
221
,,
1
ftgxyxy
t




22
22 1
,
1
fgxy fx y xy



Continuous for
22
1xy#

73.
2
,4
fxy x y
(a)



2
2 2
00 00
44
,, 2
lim lim lim lim 2 2
xx xx
xx y x y
fx xy fxy xx x
xxx
xxx
5 5 5 5
5
5 5 5

5
555

(b)



22
00 00
44,, 4
lim lim lim lim 4 4
yy yy
xyyxyfxy y fxy y
yyy
5 5 5 5
55 5

555


74.
22
,
fxy x y
(a)



2
222 2
00 00
,, 2
lim lim lim lim 2 2
xx xx
xx y xy
fx xy fxy xx x
xxx
xxx
5 5 5 5
5
5 5 5

5
555

(b)



2
222 2
00 00
,, 2
lim lim lim lim 2 2
yy yy
xyy xy
fxy y fxy yy y
yy y
yyy
5 5 5 5
5
5 5 5

5
555


75.
,
x
fxy
y

(a)

0000
,, 11
lim lim lim lim
xxxx
xxx x
fx xy fxy yy y
x xxyy5 5 5 5
5 5

5

555

(b)



2
00000
,,
lim lim lim lim lim
yyyyy
xx
fxy y fxy xy xy xy
xyxxyyy
y y yyyy yyyy yyyy
5 5 5 5 5

5 5 5 5

5555555


76.
1
,fxy
xy



(a)




000
2
00
11
,,
lim lim lim
11
lim lim
xxx
xx
fx xy fxy x y x x y xxyxy
xxxxyxyx
x
xxyxyx xxyxy
xy
5 5 5
5 5

5 5 5

5555
5

5 5 5

(b) By symmetry,


2
0
,, 1
lim .
y
fxy y fxy
y
xy
5
5

5


77. ,3 2
fxy x xy y
(a)



00
00
,, 3 232
lim lim
3
lim lim 3 3
xx
xx
fxxyfxy xxxxyy xxyy
xx
xyx
yy
x
5 5
5 5
5 5 5

55
55

5

(b)



00
00
,,3 232
lim lim
2
lim lim 2 2
yy
yy
fxy y fxy xxy y y y xxy y
yy
xy y
xx
y
5 5
5 5
5 5 5

55
55

5

181. Section 13.2 Limits and Continuity 181
© 2010 Brooks/Cole, Cengage Learning
78. ,1fxy yy
(a)


00
,, 1 1
lim lim 0
xx
fx xy fxy yy yy
xx
5 5
5

55

(b)




32 12
32 12
00
32 12
32 12
00
12 12
,,
lim lim
lim lim
31
L Hôpital's Rule
22
31
2
yy
yy
yy yy y yfxy y fxy
yy
yy y yy y
yy
yy
y
y
5 5
5 5

5 5 5

55
5 5

55





79. True. Assuming ,0fxexists for 0.x#

80. False. Let
22
,.
xy
fxy
xy



See Exercise 39.

81. False. Let

22
ln , , 0, 0
,.
0, 0, 0
xyxy
fxy
xy
6 #=
7
=8


82. True

83.
22
,0,0
lim
xy
xy
xy


(a) Along :yax





2 222
2
,0,0 0
2
1
lim lim
1
,0
xax x
x axax
xax ax
a
a
a




#

If 0,athen 0yand the limit does not exist.
(b) Along



2
22
2
2
22 0
,0,0
1
:lim lim
x
xx
xx
x
yx
xxx




Limit does not exist.
(c) No, the limit does not exist. Different paths result in
different limits.


84.
2
42
,
xy
fxy
xy



(a)



2
2 224
:,
xax ax
yaxfxax
xaxax



If 0,a#

22
,0,0
lim 0.
xax
ax
xa



(b)
2
:yx


22
4
2
2 4
42
,
2
xx
x
fxx
x
xx




4
4
2
,
1
lim
22
xx
x
x

(c) No, the limit does not exist. f approaches different numbers along different paths.

85.


222 2
,, 0,0,0 0
2
0
sin cos sin sin cos
lim lim
lim sin cos sin cos 0
xyz
xyz
xyz
-
-
-.-.-.
-
-..








86.

11
222 2
,, 0,0,0
0 11
lim tan lim tan
2
xyz xyz -

-

182. 182 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
87. As ,0,1,xy
2
11x and
2
2
10.xy
So,

2
1
2
2,0,1
1
lim tan .
21
xy
x
xy









88.



22 22
222
2
,0,0 0 0
cos sin
lim , lim cos sin lim cos sin cos sin 0
xy r r
rr
fxy r r r
r






So, define

0, 0 0.f

89. Because

1
,,lim , ,
xy ab
fxy L

then for 20,Athere corresponds
10Bsuch that 1,2fxy LA whenever

22
1
0. xa yb B
Because

2
,,lim , ,
xy ab
gxy L

then for 20,Athere corresponds
20Bsuch that 2,2gxy L A whenever


22
2
0. xa yb B
Let
Bbe the smaller of
1Band
2.BBy the triangle inequality, whenever
22
,xa ybB we have


12 1 2 1 2,, , , , , .
22
fxygxy LL fxyL gxyL fxyL gxyL
AA
A

So,

12
,,lim , , .
xy ab
fxy gxy L L





90. Given that ,
fxyis continuous, then


,,
lim , , 0,
xy ab
fxy fab

which means that for each 0,Athere corresponds
a 0
Bsuch that
,,fxy fab A whenever


22
0. xa yb B
Let

,2,fabA then ,0fxyfor every point in the corresponding Bneighborhood because






,, ,
,, ,,
22 2
31
,, ,0.
22
fab f ab f ab
f x y f ab f x y f ab
fab fxy fab




91. See the definition on page 899. Show that the value of


,,
00
lim ,
xy x y
fxy

is not the same for two different paths to 00,.xy

92. See the definition on page 902.

93. (a) True
(b) False. The convergence along one path does not imply convergence along all paths.
(c) False. Let



2
22
22
23
,4 .
23
xy
fxy
xy




(d) True

94. (a) No. The existence of 2, 3f has no bearing on the existence of the limit as ,2,3.xy
(b) No,
2, 3f can equal any number, or not even be defined.

183. Section 13.3 Partial Derivatives 183
© 2010 Brooks/Cole, Cengage Learning
Section 13.3 Partial Derivatives
1. 4, 1 0
xf

2. 1, 2 0
yf

3. 4, 1 0
yf

4. 1, 1 0
xf

5. No, y only occurs in the numerator.

6. Yes, x occurs in both the numerator and denominator.

7. Yes, x occurs in both the numerator and denominator.

8. No, y only occurs in the numerator.

9. ,253fxy x y

,2
xfxy

,5
yfxy

10.
22
,24fxy x y

,2
x
fxy x

,4
y
fxy y

11.
23
,
fxy xy


3
,2
x
fxy xy


22
,3
y
fxy xy

12.
32
,4
fxy xy




22
,12
x
fxy xy




33
,8
y
fxy xy



13.
2
zxy
z
y
x
zx
y y

C

C
C

C

14.
2
2zyx

2
zy
x x
C

C


4
z
yx
y
C

C

15.
22
43
24
46
zx xy y
z
xy
x
z
xy
y

C

C
C

C

16.
32
21zy xy

2
2
z
y
x
C

C

2
34
z
yxy
y
C

C

17.
xy
xy
xy
ze
z
ye
x
z
xe
y

C

C
C

C


18.
1
2
1
xyxy
xy
xy
ze e
z
e
xy
zx
e
yy


C

C
C

C

19.
22
2
22
2
2
y
y
y
zxe
z
xe
x
z
xe
y

C

C
C

C


20.
1
yx yx
zye ye



2
1
2
2
yx y xzy
ye yx e
xx
C


C


1
1
yx yx yxzy
eyee
yx xC


C


21.
ln ln ln
x
zxy
y


1z
xx
C

C


1z
yy
C

C


22.

1
ln ln
2
zxy xy


11
22
zy
x xy x
C

C


11
22
zx
yxyy
C

C

184. 184 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
23.
22
lnzxy

22
2zx
xxy
C

C


22
2zy
yx y
C

C


24.
ln ln ln
xy
zxyxy
xy




11 2zy
xxy xy xyxy
C

C



11 2zx
yxyxy xyxyC

C


25.
22
3
2
x y
z
yx


23 3
22
23 3
2
zxyxyx yx xy
C

C


233
22
612
22
zx y yx
yyx xyC

C


26.
22
,
xy
fxy
xy











22
32
22
22 22
22
32
22
22 22
2
,
2
,
x
y
xyy xyx yxy
fxy
xy xy
xyx xyy
xxy
fxy
xy xy








27.





22
22
22
,
,2
,2
xy
xy
x
xy
y
hx y e
hxy xe
hxy ye







28.

22 22 1
,ln ln
2
gxy x y x y


22 22
12
,
2
x
xx
gxy
xyxy





22 22
12
,
2
y
yy
gxy
xyxy




29.
22
,fxy x y



12
22
221
,2
2
x
x
fxy x y x
xy






12
22
221
,2
2
y
y
fxy x y y
xy




30.
3
,2fxy x y

12
3
311
22
2 2
f
xy
x
xy
C

C



2
12
32
3
13
23
2 22
fy
xy y
y
xy
C

C


31.
cos
sin
sin
zxy
z
yxy
x
z
xxy
y

C

C
C

C

32.


sin 2
cos 2
2cos 2
zxy
z
xy
x
z
xy
y

C

C
C

C


33.


2
2
tan 2
2sec 2
sec 2
zxy
z
xy
x
z
xy
y

C

C
C

C


34.
sin 5 cos 5
5 cos 5 cos 5
5 sin 5 sin 5
zxy
z
x y
x
z
x y
y

C

C
C

C

35.

sin
cos
sin cos
cos sin
y
y
yy
y
ze xy
z
ye xy
x
z
exyxe x
y
ex xy xy
C

C
C

C


36.


22
22
22
cos
2sin
2sin
zxy
z
xxy
x
z
yxy
y

C

C
C

C


37.


sinh 2 3
2cosh 2 3
3cosh 2 3
zxy
z
xy
x
z
xy
y

C

C
C

C

185. Section 13.3 Partial Derivatives 185
© 2010 Brooks/Cole, Cengage Learning
38.
2
22
2
cosh
sinh
2sinh
zxy
z
yxy
x
z
xyxy
y

C

C
C

C


39.


2
333
22
2
,1
333
,11
,1
y
x
y
x
x
y
fxy t dt
tyx
tyx
fxy x x
fxy y





*


+You could also use the Second Fundamental Theorem
of
,Calculus.

40.

+,
,2121
21 21
2222
yx
xy
yy
xx
y y
xx
fxy t dt t dt
tdt tdt
dt t y x


**
**
*



,2
,2
x
xfxy
fxy




41. ,32
fxy x y



0
0
0
,,
lim
3232
lim
3
lim 3
x
x
x
fx xy fxyf
xx
xxyxy
x
x
x
5
5
5
5 C

C5
5

5
5

5




0
0
0
,,
lim
32 32
lim
2
lim 2
y
y
y
fxy y fxyf
yy
x yy xy
y
y
y
5
5
5
5 C

C5
5

5
5

5


42.
2
22
,2
fxy x xy y x y




0
2
22 2
0 0
,,
lim
22
lim lim 2 2 2
x
x x
fx xy fxyf
xx
xx xxyyx xyy
xxy xy
x
5
5 5
5 C

C5
5 5
5
5





0
2
222
0 0
,,
lim
22
lim lim 2 2 2
y
y y
fxy y fxyf
yy
x xyy yy x xyy
xyy yx
y
5
5 5
5 C

C5
55
5
5


43.
,fxy x y




0
0
0 0
,,
lim
lim
11
lim lim
2
x
x
x x
fx xy fxyf
xx
xxy xy
x
xxy xyxxy xy
xxy xy xyxx xy xy
5
5
5 5
5 C

C5
5

5
5 5

5 55


f
y
C

C



00
0
0
,,
lim lim
lim
11
lim
2
yy
y
y
fxy y fxy xy y xy
yy
xyy xyxyy xy
yxy y xy
xy y xy xy
5 5
5
5
5 5

55
5 5

55

5

186. 186 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
44.
1
,fxy
xy






2
000
11
,, 11
lim lim lim
xxx
fx xy fxyf xxyxy
x x x x xyxy
xy
5 5 5

5 C 5

C5 5 5




2
000
11
,, 11
lim lim lim
yyy
fxy y fxyf xy xy
yy yxyyxy
xy
5 5 5

5 C 5

C5 5 5



45. ,sin
y
fxy e x

,cos
y
x
fxy e x
At
,0, ,0 1.
xf

,sin
y
y
fxy e x
At
,0, ,0 0.
yf

46. ,cos
x
fxy e y



,cos
x
x
fxy e y


At
0, 0 , 0, 0 1.
xf

,sin
x
y
fxy e y


At
0, 0 , 0, 0 0.
yf

47. ,cos2
fxy x y

,2sin2
x
fxy x y
At ,,
43



,2sin 1.
43 2 3
xf






,sin2
y
fxy x y
At ,,
43



1
,sin .
43 2 3 2
yf






48. ,sin
fxy xy

,cos
x
fxy y xy
At 2, ,
4



2, cos 0.
442
xf






,cos
y
fxy x xy
At 2, ,
4



2, 2 cos 0.
42
yf






49.
,arctan
y
fxy
x




22222
1
,
1
x
yy
fxy
x xyyx





At

1
2, 2 : 2, 2
4
xf



2222
11
,
1
y
x
fxy
x xyyx





At

1
2, 2 : 2, 2
4
yf

50. ,arccos
fxy xy


22
,
1
x
y
fxyxy




At
1, 1 ,
xfis undefined.


22
,
1
y
x
fxyxy




At
1, 1 ,
y
fis undefined.

51.
,
xy
fxy
xy








2
22
,
x
yx y xy y
fxy
xyxy




At
2, 2 :
1
2, 2
4
xf





2
22
,
y
xx y xy x
fxy
xyxy




At

1
2, 2 : 2, 2
4
yf

187. Section 13.3 Partial Derivatives 187
© 2010 Brooks/Cole, Cengage Learning
52.
22
2
,
45
xy
fxy
x y






3
32
22
10
,
45
x
y
fxy
xy



At

32
10 10
1, 1 , 1, 1 .
927
xf



3
32
22
8
,
45
y
x
fxy
xy



At

1, 1 ,
32
88
1, 1 .
927
yf

53.

22
,4
,2
x
gxy x y
gxy x



At
1, 1 : 1, 1 2
xg

,2
y
gxy y
At
1, 1 : 1, 1 2
yg

54.
22
,hx y x y

,2
xhxy x
At

2, 1 : 2, 1 4
xh

,2
yhxy y
At
2, 1 : 2, 1 2
yh

55.

22
49 , 2, 2, 3, 6zxyx
Intersecting curve:
2
45zy

2
45
zy
y y
C

C

At

31
2, 3, 6 :
245 9
z
y
C

C



56.
22
4, 1,2,1,8zx yy
Intersecting curve:
2
4zx

2
z
x
x
C

C

At

2, 1, 8 : 2 2 4
z
x
C

C

57.
22
9,zxy 3,y 1, 3, 0
Intersecting curve:
2
99zx

18
z
x
x
C

C

At

1, 3, 0 : 18 1 18
z
x
C

C

58.
22
9,1,1,3,0zxyx
Intersecting curve:
2
9zy

2
z
y
y
C

C
At

1, 3, 0 : 2 3 6
z
y
C

C

59.


,, sin 2 3
,, cos 2 3
,, 2cos 2 3
,, 3cos 2 3
x
y
z
Hxyz x y z
Hxyz x y z
Hxyz x y z
Hxyz x y z





60.



22
22
,, 3 5 10
,, 6 5
,, 3 5 10
,, 5 20
x
y
z
fx y z x y xyz yz
f x y z xy yz
fxyz x xz z
f x y z xy yz






61.
222
222
222
222
wxyz
wx
x
xyz
wy
y xyz
wz
z xyz

C

C
C

C
C

C


62.




1
22
27
7
77 7
7
7xz
wxzxy
xy
xyz xzwyz
x
xyxy
wxz
y xy
wx
zxy



C

C
C

C
C

C
yx
x= 2
10
8
8
z
x y
20
4
4
z
z
x
y
160
2
4
3
4
y= 3
x
y
40
4
4
z

188. 188 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
63.



222 222
222
222
222 1
, , ln ln
2
,,
,,
,,
x
y
z
Fxyz x y z x y z
x
Fxyz
xyz
y
Fxyz
xyz
z
Fxyz
xyz








64.






222
32
222
32
222
32
222
1
,,
1
,,
1
,,
1
,,
1
x
y
z
Gx yz
xyz
x
Gxyz
xyz
y
Gxyz
xyz
z
Gxyz
xyz










65.






32
22
32
3
,,
,, 3
1, 1, 1 3
,,
1, 1, 1 1
,, 2
1, 1, 1 2
x
x
y
y
z
z
fxyz xyz
fxyz xyz
f
fxyz xz
f
fxyz xyz
f









66.






23
3
22
,, 2 3
,, 2 2
2, 1, 2 4 4 0
,, 3 2 3
2, 1, 2 12 8 6 2
,, 2 3
2, 1, 2 4 3 7
x
x
y
y
z
z
fx y z x y xyz yz
fxyz xy yz
f
fxyz xy xz z
f
fxyz xy y
f









67.






2
2
,,
1
,,
1, 1, 1 1
,,
1, 1, 1 1
,,
1, 1, 1 1
x
x
y
y
z
z
x
fxyz
yz
fxyz
yz
f
x
fxyz
yz
f
x
fxyz
yz
f











68.












2
22
2
2
22
2
22
,,
,,
11
3, 1, 1 0
3
,,
93 2
3, 1, 1
33
0
,,
31
3, 1, 1
93
x
x
y
y
z
z
xy
fxyz
xyz
xyzyxy yyz
fxyz
xyz xyz
f
xyzxxy xxz
fxyz
xyz xyz
f
xyz xy xy
fxyz
xyz xyz
f



















69.



,, sin
,, cos
0, , 4 4 cos 0
22
,, cos
0, , 4 4 cos 0
22
,, sin
0, , 4 sin 1
22
x
x
y
y
z
z
fxyz z x y
fxyz z x y
f
fxyz z x y
f
fxyz x y
f





















70.
22 2
32xyz







22 2 22 2
22 2 22 2
22 2 22 2
63
,,
23 2 3 2
6335
1, 2, 1
523 4 2 5
2
,,
23 2 3 2
225
1, 2, 1
55
42
,,
23 2 3 2
225
1, 2, 1
55
x
x
y
y
z
z
xx
fxyz
xyz xyz
f
yy
fxyz
xyz xyz
f
zz
fxyz
xyz xyz
f















71.
2
3zxy

2
3,
z
y
x
C

C
2
2
0,
z
x
C

C
2
6
z
y
yx
C

CC


6,
z
xy
y
C

C
2
2
6,
z
x
y
C

C
2
6
z
y
xy
C

CC

189. Section 13.3 Partial Derivatives 189
© 2010 Brooks/Cole, Cengage Learning
72.
22
3zx y
2,
z
x
x
C

C
2
2
2,
z
x
C

C
2
0
z
yx
C

CC


6,
z
y
y
C

C
2
2
6,
z
y
C

C
2
0
z
xy
C

CC


73.
22
2
2
2
2
2
2
23
22
2
2
26
6
2
zx xy y
z
xy
x
z
x
z
yx
z
xy
y
z
y
z
xy

C

C
C

C
C

CC
C

C
C

C
C

CC


74.
4224
32
2
22
2
2
23
2
22
2
2
3
46
12 6
12
64
612
12
zx xy y
z
xxy
x
z
xy
x
z
xy
yx
z
xy y
y
z
xy
y
z
xy
xy

C

C
C

C
C

CC
C

C
C

C
C

CC


75.




22
22
22
322
22
2
32
22
22
22
322
22
2
32
22
zxy
zx
x
xy
zy
x
xy
zxy
yx
xy
zy
y xy
zx
y
xy
zxy
xy
xy

C

C
C

C

C

CC

C

C
C

C

C

CC



76.




2
22
2
2
2
22
2
2
ln
1
1
1
11
1
1zxy
z
xxy
z
x xy
z
yx xy
z
yxyyx
z
y xy
z
xy xy

C

C
C

C
C

CC
C

C
C

C
C

CC

So,
22
.
zz
yx xy
CC

CC CC


77.
2
2
2
2
2
2
2
2
2
2
tan
tan
tan
sec
sec
2sec tan
sec
x
x
x
x
x
x
x
ze y
z
ey
x
z
ey
x
z
ey
yx
z
ey
y
z
eyy
y
z
ey
xy

C

C
C

C
C

CC
C

C
C

C
C

CC


78.
2
2
2
2
2
2
23
23
3
23
23
2
23
yx
yx
x
yx
yx
y
yx
zxe ye
z
eye
x
z
ye
x
z
eye
yx
z
xe e
y
z
xe
y
z
ee
xy







C

C
C

C
C

CC
C

C
C

C
C

CC

190. 190 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
79. coszxy
sin ,
z
yxy
x
C

C
2
2
2
cos
z
yxy
x
C

C


2
cos sin
z
yx xy xy
yx
C

CC

sin ,
z
xxy
y
C

C
2
2
2
cos
z
x xy
y
C

C


2
cos sin
z
xyxy xy
xy
C

CC


80.







22222
2
22
22
22
222
22
22 22
2222
2
22
22
22
222
22
22 22
arctan
1
1
2
2
11
1
2
2
y
z
x
zyy
xxxy yx
zxy
x
xy
xy yyzyx
yx
xyxy
zx
yxxy yx
zxy
y
xy
xy xxzyx
xy
xy xy

C


C
C

C

C

CC

C


C
C

C

C

CC



81. ,2 20
xfxy x y

,220
yfxy x y
220 22
xyyx

22 2 2 0 3 6 0 2,
2
xx x x
y



Point:

2, 2

82. ,2 50
xfxy x y

,210
yfxy x y
250 25xy y x

22 5 1 0 3 9 0 3,
1
xx x x
y



Point:

3, 1

83. ,244,
xfxy x y ,4216
yfxy x y
0:
xyff 24 4
42 16
xy
xy



Solving for x and y,
6xand 4.y

84.

,2 0
,20
202
22 0 0, 0
x
yfxy x y
fxy x y
xy y x
xx xy




Point:
0, 0

85.

22
11
,,,
xy
fxy yf xy x
xy

0:
xyff
2
1
0y
x

and
2
1
0x
y



2
1
y
x
and
2
1
x
y



4
1yy y x
Points:
1, 1

86.
2
,912,
x
fxy x y
2
,123
yfxy x y
0:
xyff
22
22
91203 4
3120 4
x yxy
yx yx



Solving for x in the second equation,
2
4,xy you
obtain
2
2
34 4.yy

4
364 0yyy or
13
4
3
y

0 x or
23
116
43
x





Points:

23 13
44
0, 0 , ,
33





87.

22
22
,2 0
,2 0
xxyy
x
xxyy
y
fxy x ye
fxy x ye







20 2
22 0 0 0
xy y x
xx x y



Point:
0, 0

88.

22
22
2
,00
1
2
,00
1
x
y
x
fxy x
xy
y
fxy y
xy





Points:
0, 0

191. Section 13.3 Partial Derivatives 191
© 2010 Brooks/Cole, Cengage Learning
89.

2
2
2
2
22
2
2
sec
sec
0
sec tan
sec tan
sec sec tan
sec tan
zx y
z
y
x
z
x
z
yy
yx
z
xyy
y
z
xyy y
y
z
yy
xy

C

C
C

C
C

CC
C

C
C

C
C

CC

So,
22
zz
yx xy
CC

CC CC

There are no points for which 0 ,
x yzz because
sec 0.
z
y
x
C
#
C

90.
22
25zxy





22
22
322
22
2
32
22
22
22
322
22
2
32
22
25
25
25
25
25
25
25
25
zx
x
xy
zy
x
xy
zxy
yx
xy
zy
y xy
zx
y
xy
zxy
xy
xy
C

C
C

C

C

CC

C

C
C

C

C

CC



0
zz
xy
CC

CCif 0xy

91.

22
22
ln ln ln
x
zxxy
xy










22
22 22
24224
22
22 2
2
2
22
22
22
2
22
22
2
2
22
12
4
4
2
2
4
zxyx
xxx y
xxy
zx xy y
x
xx y
zxy
yx
xy
zy
yxy
yxz
y
xy
zxy
xy
xy
C

C
C

C

C

CC

C

C
C

C

C

CC


There are no points for which 0.
xyzz

92.










2
22
22
32
2
22
43
2
22
22
32
2
22
43
2
22 1 2
2
22 2
xy
z
xy
yx y xyzy
x xy xy
zy
x xy
xy y y xyzxy
yx
xyxy
xx y xyzx
y xy xy
zx
y xy
xy x x xyzxy
xy xy xy


C

C
C

C
C

CC
C

C
C

C
C

CC

There are no points for which 0.
xyzz

93.







,,
,,
,,
,, 0
,,
,,
,, 0
,, 0
,, 0
x
y
yy
xy
yx
yyx
xyy
yxy
fxyz xyz
fxyz yz
fxyz xz
fxyz
fxyz z
fxyz z
fxyz
fxyz
fxyz









So, 0.
xyy yxy yyxfff

192. 192 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
94.








23
,, 3 4
,, 2 3
,, 3 4
,, 0
,, 3
,, 3
,, 0
,, 0
,, 0
x
y
yy
xy
yx
yyx
xyy
yxy
fxyz x xy yz z
fxyz x y
fxyz x z
fxyz
fxyz
fxyz
fxyz
fxyz
fxyz









So, 0.
xyy yxy yyxfff

95.








2
2
2
2
,, sin
,, sin
,, cos
,, sin
,, cos
,, cos
,, sin
,, sin
,, sin
x
x
x
x
y
x
yy
x
xy
x
yx
x
yyx
x
xyy
x
yxy
fxyz e yz
fxyz e yz
fxyz ze yz
fxyz ze yz
fxyz ze yz
fxyz ze yz
fxyz ze yz
fxyz ze yz
fxyz ze yz



















So, .
xyy yxy yyzf ff

96.















2
2
3
3
3
4
4
4
2
,,
2
,,
2
,,
4
,,
4
,,
4
,,
12
,,
12
,,
12
,,
x
y
yy
xy
yx
yyx
xyy
yxy
z
fxyz
xy
z
fxyz
xy
z
fxyz
xy
z
fxyz
xy
z
fxyz
xy
z
fxyz
xy
z
fxyz
xy
z
fxyz
xy
z
fxyz
xy

























97.
2
2
2
2
5
5
0
5
0
zxy
z
y
x
z
x
z
x
y
z
y

C

C
C

C
C

C
C

C

So,
22
22
00 0.
zz
xy
CC

CC


98.
2
2
2
2
sin
2
cos
2
sin
2
sin
2
sin
2
yy
yy
yy
yy
yy
ee
zx
zee
x
x
zee
x
x
zee
x
y
zee
x
y








C

C
C

C
C

C
C

C
So,

22
22
sin sin 0.
22
yy yy
z z ee ee
xx
xy

CC

CC


99.
2
2
2
2
sin
sin
sin
cos
sin
x
x
x
x
x
ze y
z
ey
x
z
ey
x
z
ey
y
z
ey
y

C

C
C

C
C

C
C

C
So,
22
22
sin sin 0.
xxzz
eyey
xy
CC

CC


100.
arctan
y
z
x

From Exercise 80, we have


22
2222
22 22
22
0.
zz xy xy
xy
xy xy
CC

CC

193. Section 13.3 Partial Derivatives 193
© 2010 Brooks/Cole, Cengage Learning
101.




2
2
2
2
2
sin
cos
sin
cos
sin
zxct
z
cxct
t
z
cxct
t
z
xct
x
z
xct
x

C

C
C

C
C

C
C

C

So,
22
2
22
.
zz
c
tx
CC

CC


102.





2
2
2
2
2
22
22
22
cos 4 4
4sin4 4
16 cos 4 4
4sin 4 4
16 cos 4 4
16 cos 4 4
zxct
z
cxct
t
z
cxct
t
z
xct
x
z
xct
x
zz
cxctc
tx

C

C
C

C
C

C
C

C
CC

CC


103.



22
22
2
22
22 2
2
222
ln
1
1zxct
zc
txct
zc
t xct
z
xxct
z
x xct
zc z
c
txxct

C

C
C

C
C

C
C

C
C C

CC


104.




2
22
2
2
2
2
sin sin
cos sin
sin sin
sin cos
sin sin
zwctwx
z
wc wct wx
t
z
wc wct wx
t
z
wwct wx
x
z
wwctwx
x

C

C
C

C
C

C
C

C

So,
22
2
22
.
zz
c
tx
CC

CC


105.
2
22
cos
cos
1
sin
1
cos
t
t
t
tx
ze
c
zx
e
tc
zx
e
x cc
zx
e
x cc





C

C
C

C
C

C
So,
2
2
2
.
zz
c
tx
CC

CC

106.
2
22
sin
sin
1
cos
1
sin
t
t
t
tx
ze
c
zx
e
tc
zx
e
xcc
zx
e
x cc





C

C
C

C
C

C
So,
2
2
2
.
zz
c
tx
CC

CC

107. Yes. The function
,cos32fxy x y satisfies
both equations.

108. A function
,fxywith the given partial derivatives
does not exist.

109.
,, 0
xfxyz
x
C


C
There are no xs in the expression.

110.


2
21
,, sinh
yyz
y
fxyz
xz

C


C


111. If
,,zfxy then to find
xfyou consider y constant
and differentiate with respect to x. Similarly, to find
,
y
fyou consider x constant and differentiate with
respect to y.

112.

f
x
C
C
denotes the slope of surface in the x-direction.

f
y
C
C
denotes the slope of the surface in the y-direction.
x y
Plane: x = x
0
(x
0
, y
0
, z
0
)
z
x
Plane: y = y
0
y
(x
0
, y
0
, z
0
)
z

194. 194 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
113. The plane ,zxyfxy satisfies
0
f
x
C

Cand
0.
f
y
C

C

114. The plane
,zxy fxy satisfies
0
f
x
C

Cand
0.
f
y
C

C

115. In this case, the mixed partials are equal, .
xyyxf f
See Theorem 13.3.

116.





,sin2
,cos2
,sin2
,2sin2
,2cos2
,4sin2
,2sin2
x
xx
xy
y
yy
yx
fxy x y
fxy x y
fxy x y
fxy x y
fxy x y
fxy x y
fxy x y








So,
,,.
xy yx
fxy f xy

117.
22
121122
200 200 4 8 4
R x xxxxx
(a)
12
1200 8 8
r x x
x
C

C
At
12,4,12,xx
1
200 32 96 72.
R
x
C

C
(b)
12
2200 8 8
R x x
x
C

C
At
12,4,12,xx
2
72.
R
x
C

C

118. (a)
32 175 205 1050Cxyxy
16 175
Cy
xx
C

C


80, 20
1
16 175 183
4
C
x
C


C

16 205
Cx
yy
C

C

80, 20
16 4 205 237
C
y
C


C
(b) The fireplace-insert stove results in the cost
increasing at a faster rate because
.
CC
yx
CC

CC

119.
, 100
M
IQ M C
C


100
, 12, 10 10
MMIQ IQ
C



2
100
, 12, 10 12
cc
M
IQ IQ
C



When the chronological age is constant, IQ increases at a
rate of 10 points per mental age year.
When the mental age is constant, IQ decreases at a rate
of 12 points per chronological age year.

120.
0.7 0.3
, 200
fxy x y
(a)
0.3
0.3 0.3
140 140
fy
xy
xx
C


C

At
, 1000, 500 ,xy

0.3 0.3
500 1
140 140 113.72.
1000 2
f
x
C


C

(b)
0.7
0.7 0.7
60 60
fx
xy
xy

C

C

At
, 1000, 500 ,xy


0.7
0.7
1000
60 60 2 97.47.
500
f
x
C


C


121. An increase in either price will cause a decrease in demand.

122.










10
109
211
99
1
10.061
, 1000
1
10.06110.061 10.061
, 10,000 10,000
1 11
0.03, 0.28 11,027.20
10.06110.061 0.06
, 10,000 600
11 1
I
I
R
R
VIR
I
RRR
VIR
I II
V
RR
VIR
II I











0






0.03, 0.28 653.26
RV
The rate of inflation has the greater negative influence.
y
x
2
4
4
2
4
z
x
y
−6
6
8
z

195. Section 13.3 Partial Derivatives 195
© 2010 Brooks/Cole, Cengage Learning
123.
22
500 0.6 1.5Txy
1.2 , 2, 3 2.4 m
TT
x
xx
CC

CC

32,39m
TT
y
yy
CC

CC

124. 0.885 22.4 1.20 0.544Athth
(a)
0.885 1.20
A
h
t
C

C

30 , 0.80 0.885 1.20 0.80 1.845
A
t
C

C

22.4 1.20
A
t
h
C

C

30 , 0.80 22.4 1.20 30 13.6
A
h
C

C
(b) The humidity has a greater effect on A because its
coefficient 22.4 is larger than that of t.

125.
2
2
1
n
PV RT
xB
PV T V
T
nn P
RR
xB xB
nn
RTRT
P
xB xB
P
VVV
nn
RT R
VxB xB
V
PTP
nn
RT R
TPV V
xB xB
nPV T V P
R
xB
nn
RT RT
xB xB
nVP
RT
xB

C

C
C

C
C

C

CCC
""
CC C





126.
22
53Uxxyy
(a) 10
xUxy
(b) 6
yUxy
(c)
2, 3 17
xU and 2, 3 16.
yU The person
should consume one more unit of y because the rate
of decrease of satisfaction is less for y.
(d)

127. 0.92 1.03 0.02zxy
(a)
0.92
1.03
z
x
z
y
C

C
C

C

(b) As the consumption of flavored milk (x) increases,
the consumption of plain light and skim milk (z)
decreases. As the consumption of plain reduced-fat
milk (y) decreases, the consumption of plain light
and skim milk decreases.

128.
22
1.2225 0.0096 71.381 4.121 354.65zxyxy
(a)
2
2
2
2
2.445 71.381
2.445
0.0192 4.121
0.0192
z
x
x
z
x
z
y
y
z
y
C

C
C

C
C

C
C

C

(b) Concave downward
2
2
0
z
x
C

C
The rate of increase of Medicare expenses
zis declining with respect to worker’s compensation expenses
.x
(c) Concave upward
2
2
0
z
y
C

C
The rate of increase of Medicare expenses
zis increasing with respect to public assistance expenses ( y).
x
y
z
−2
2
2 1
1

196. 196 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
129. False
Let 1.
zxy

130. True

131. True

132. True

133.



22
22
,, 0,0
,
0, , 0, 0
xy x y
xy
fxy xy
xy
6
= #
7
=

8

(a)










222 3 3 3 4 224
22
22 22
223 2 3 3 4 224
22
22 22
324
,
324
,
x
y
xyxyy xyxyx yx xy y
fxy
xy xy
xyx xy xy xy y xx xy y
fxy
xy xy







(b)



2
00
00
,0 0,0
0, 0 lim lim 0
x
xx
x
fx f
f
xx
5 5
5
5


55





2
00
00
0, 0, 0
0, 0 lim lim 0
y
yy
y
fyf
f
yy
5 5
5
5


55

(c)





4
2
000 2
0, 0
0, 0, 0
0, 0 lim lim lim 1 1
xx
xy
yyy
yy
fyff
f
yx y
yy
5 5 5
55
5CC


CC 5
55






4
2
000 2
0, 0
,0 0,0
0, 0 lim lim lim 1 1
yy
yx
xxx
xx
fx ff
f
xy x
xx
5 5 5
55
5CC

CC 5
55

(d)
yx
for
xyfor both are not continuous at 0, 0 .


134.
3
,1
y
x
fxy t dt*

By the Second Fundamental Theorem of Calculus,

333
111
yx
xyfd d
tdt tdt x
xdx dx
C

C
**


33
11.
y
xfd
tdt y
ydy
C

C
*


135.
13
33
,fxy x y
(a)


0
0
0,0 0,0
0, 0 lim
lim 1
x
x
x
fxf
f
x
x
x
5
5
5

5
5

5




0
0
0, 0 0, 0
0, 0 lim
lim 1
y
y
y
fyf
f
y
y
y
5
5
5

5
5

5

(b)
,
x
fxyand ,
yfxyfail to exist for
,0.yxx #
136.
23
22
,fxy x y
For
,0,0,xy#


13
22
13
2224
,2.
3
3
x
x
fxy x y x
xy


For
,0,0,xy use the definition of partial derivative.





43
13
000
00,0
0, 0 lim lim lim 0
x
xxx
fxf x
fx
xx
5 5 5
5 5
5
55

197. Section 13.4 Differentials 197
© 2010 Brooks/Cole, Cengage Learning
Section 13.4 Differentials
1.
23
322
2
46zxy
dz xy dx x y dy



2.
2
2
2
2
x
z
y
xx
dz dx dy
yy




3.
22
1
z
xy









22
22 22
2
22
22
2xy
dz dx dy
xy xy
xdx ydy
xy






4.
3
xy
w
zy






22
13
3 33
xz xy
dw dx dy dz
zy zy zy





5. cos coszx yy x


cos sin sin cos
cos sin sin cos
dz y y x dx x y x dy
yy xdx x y xdy



6.

22 221
2
xy xy
zee







22 22
22 22
22 22
2
2
2
2
xy xy
xy xy
xy xy
ee
dz x dx
ee
ydy
eexdxydy















7.

sin
sin cos
x
xx
ze y
dz e y dx e y dy



8.
2
cos
sin cos 2
y
yy
we xz
dw e x dx e x dy z dz



9.
3
2sinwzyx

332
2 cos 2 sin 6 sindw z y xdx z xdy z y xdz

10.


22
222
2
sin
2cos
2cos
w x yz yz
dw xyz dx x z z yz dy
xyz y yzdz




11. ,23
fxy x y
(a)


2, 1 1
2.1, 1.05 1.05
2.1, 1.05 2, 1 0.05f
f
zf f


5
(b)
2 3 2 0.1 3 0.05 0.05dz dx dy

12.
22
,
fxy x y
(a)


2, 1 5
2.1, 1.05 5.5125
2.1, 1.05 2, 1 0.5125f
f
zf f


5
(b)

22
22 0.1 21 0.05 0.5dz x dx y dy


13.
22
,16
fxy x y
(a)


2, 1 11
2.1, 1.05 10.4875
2.1, 1.05 2.1 0.5125f
f
zf f


5
(b)

22
22 0.1 21 0.05 0.5dz x dx y dy


14.
,
y
fxy
x

(a)


2, 1 0.5
2.1, 1.05 0.5
2.1, 1.05 2, 1 0f
f
zf f


5
(b)

2
111
0.1 0.05 0
42
y
dz dx dy
xx




15. ,
x
fxy ye
(a)



2
2.1
2, 1 7.3891
2.1, 1.05 1.05 8.5745
2.1, 1.05 2, 1 1.1854fe
fe
zf f


5
(b)

22
0.1 0.05 1.1084
xx
dz ye dx e dy
ee



16. ,cos
fxy x y
(a)


2, 1 2 cos 1 1.0806
2.1, 1.05 2.1 cos 1.05 1.0449
2.1, 1.05 2, 1 0.0357f
f
zf f


5
(b)

cos sin
cos 1 0.1 2 sin 1 0.05 0.0301dz ydx x ydy

198. 198 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
17. Let
2
,zxyΔ 2,xΔ 9,yΔ 0.01,dxΔ 0.02.dyΔ
Then:
2
2dz xy dx x dy


2
22
2.01 9.02 2 9 2 2 9 0.01 2 0.02 0.44"

18. Let
22
,5,zxyx 3,yΔ 0.05,dxΔ 0.1.dyΔ
Then:

22 22
xy
dz dx dy
xy xy






22
22
22 22 5 3 0.55
5.05 3.1 5 3 0.05 0.1 0.094
3453 53



19. Let
22
1,zxyΔπ 3,xΔ 6,yΔ 0.05,dxΔ 0.05.dyΔπ Then:


2
23
212 xx
dz dx dy
yy
ππ









2 2
2
2 22 3
21 313.05 2313
0.05 0.05 0.012
66 65.95ππ π



20. Let
22
sin ,zxy 1,xyΔΔ 0.05,dxΔ 0.05.dyΔπ Then:
22 22
2 cos 2 cosdz x x y dx y x y dy


22
22 22
sin 1.05 0.95 sin 2 2 1 cos 1 1 0.05 2 1 cos 1 1 0.05 0


21. See the definition on page 918.

22. In general, the accuracy worsens as
x5and
y5increases.

23. The tangent plane to the surface ,zfxyΔ at the point
Pis a linear approximation of z.

24. If ,,zfxyΔ then zdz5 is the propagated error,
and
zdz
zz
5

is the relative error.

25.
1
Alh
dA l dh h dl
A dl h dh lh
h dl l dh dl dh
Δ

5



AdA dl dh5π Δ

26.
2
2
2
Vrh
dV rh dr r dh
Δ



27.
2
,4,8
3
rh
Vrh

ΔΔΔ

2
24
2164
333 3
rh r r
dV dr dh h dr r dh dr dh



22
2
48 128
33
Vrrhhrh r h

5 5 5 5 5


dAΔAAd
ll
Δ
Ad
h
Δ
h
Δr
r
Δh
2rhdr
π
Δ−VdVrdh
2
π
r5 h 5 dV V5 VdV5π
0.1 0.1 8.3776 8.5462 0.1686
0.1 0.1π 5.0265 5.0255 0.0010π
0.001 0.002 0.1005 0.1006 0.0001
0.0001π 0.0002 0.0034π 0.0034π 0.0000

199. Section 13.4 Differentials 199
© 2010 Brooks/Cole, Cengage Learning
28.
22
,6,16Srrhr h

22
12 12
22 222
22
2dS r h
rh rrh
dr rh






22
dS rh
dh rh



+,
22
22
2 328 96
292dS r h dr rh dh dr dh
rh





6, 16 322.101353S



22 2 2
6 6 16 322.101353Srrrrhh r r h5 5 5 5 5 5 5


29. 0.92 1.03 0.02zxy
(a)
0.92 1.03dz dx dy
(b)
+
, + ,0.92 0.25 1.03 0.25 0.23 0.2575dz
Maximum error: 0.4875

Relative error:

0.4875
0.081 8.1%
0.92 1.9 1.03 7.5 0.02
dz
z





30.
, 7.2, 2.5 , 0.05, 0.05x y dx dy

22
rxydr
22 22 2 2 2 2
7.2 2.5
7.2 2.5 7.2 2.5
0.9447 0.3280
xy
dx dy dx dy
xy xy
dx dy





1.2727 0.05 0.064dr
arctan
y
d
x





22 22
0.0430 0.1239
yx
dx dy dx dy
xy xy



Using the worst case scenario, 0.05
dx and 0.05,dy you see that
0.0083.d

31.
22
2
2 2 0.04 0.02 0.10 10%
VrhdV rhdrrdh
dV dr dh
Vrh


32.


2
1
2
1
2
11 1
216 16
sin
sin sin cos
4 sin 45 3 sin 45 12 cos 45 0.02 0.24 in.
AabC
dA b C da a C db ab C dC






r5 h 5 dS S 5 SdS5
0.1 0.1 7.7951 7.8375 0.0424
0.1 0.1 4.2653 4.2562 0.0091
0.001 0.002 0.0956 0.0956 0.0000
0.0001 0.0002 0.0025 0.0025 0.0000

200. 200 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
33.
0.16 0.16
0.16
0.84 0.84
35.74 0.6215 35.75 0.4275
0.6215 0.4275
5.72 0.0684
CTvTv
C
v
T
C
vTv
v
ππ

C

C
C

C


θ θ
0.16 0.84 0.84
0.6215 0.4275 23 1 5.72 23 0.0684 8 23 3
1.3275 1.1143 2.4418 Maximum propagated error
CC
dC dT dv
Tv
ππCC

CC



2.4418
0.19 or 19%
12.6807
dC
C


π


34.

2
2
2
2
2 2 0.03 0.02 0.08 8%
v
a
r
vv
da dv dr
rr
da dv dr
avr
Δ
Δπ
ΔπΔ ππΔ Δ


Note: The maximum error will occur when dv and dr differ in signs.

35. (a)

331
18 sin 18 cos 16 12 31,104 sin in. 18 sin ft
222
V bhl

ΔΔ Δ Δ



V is maximum when
sin 1Δor
2.Δ
(b)
2
sin
2
s
VlΔ




22
22
33
sin cos sin
22
118 18 1
18 sin 16 12 16 12 cos sin 1809 in. 1.047 ft
222 290222
ss
dV s l ds l d dl






36. (a) Using the Law of Cosines:


222 2 2
2 cos 330 420 2 330 420 cos 9
107.3 ft.
abc bcA
a



(b)

22
420 2 420 cosab b


+, +,
12
22
12
22
121
420 840 cos 2 840 cos 840 sin
2
1
330 420 840 330 cos 2 330 840 cos 6 840 330 sin
2 20 20 20 180
1
11512.79 1774.79 8.27 ft
2
da b b b db b d


π
π
π








37.
2
,
E
P
R
Δ 3% 0.03,
dE
E
ΔΔ 4% 0.04
dR
R
ΔΔ

2
2
2EE
dP dE dR
RR
Δπ



22
2
22
2122dP EE EE
dE dR P dE dR E R dE dR
RR RRPER

Δπ Δπ Δπ



Using the worst case scenario,
0.03
dE
E
Δand 0.04:
dR
R
Δπ 2 0.03 0.04 0.10 10%.
dP
P

18 18
h
θ
2
b
2
330 ft
420 ft
9°

201. Section 13.4 Differentials 201
© 2010 Brooks/Cole, Cengage Learning
38.
12
12
12
11
22
111
0.5
2
RRR
RR
R
RR
dR R
dR R



5
5



22
21
212 22
12 12 12
RR R R
RdR dR dR R R
RR RR RR
CC
5 5 5
CC

When
110Rand
215,Rwe have




22
22
15 10
0.5 2 0.14 ohm.
10 15 10 15
R5



39.



6
46
2
0.00021 ln 0.75
1 100 1 16
0.00021 0.00021 6.6 10
100 2
0.00021 ln 100 0.75 8.096 10 6.6 10 micro henrys
h
L
r
dh dr
dL
hr
LdL







'

''


40.
2
32.23 32.09 0.14
2.48 2.50 0.02
L
T
g
dg
dL
TT L
TdT dg dL dg dL
gLgg Lg



CC
5
CC

When 32.09g and 2.50,L



2.5
0.14 0.02 0.0108 seconds.
32.09 32.09 2.5 32.09
T
5


41.




2
2
2 2
2
12 1 2
,2
,,2 22 2
2222 0
, , where and 0.
xy
zfxy x xy
zfx xy yfxy x xx x x xy y x xy
xx x x y x x y xx y
fxyx fxyy x y x
AA A A

5 5 5 5 5 5 5
5 5 5 5 55 55 5
5555 5

As
12,0,0,0and0.xy AA55

42.



22
22
22 22
12 1 2
,
,,2 2
22 , , where and .
xy
zfxy x y
zfx xy yfxy x xx x y yy y x y
xxyyxxyyfxyxfxyy x y x y AA A A

5 5 5 5 5 5 5
5 5 5 5 5 5 5 5 5 5 5 5

As
12,0,0,0and0.xy AA55

43.




2
2
22
22 2
22
2
12 1 2
,
,,2
222 2
,, whereand2.
xy
zfxy xy
zfx xy yfxy x xx xy yxy
xyx yx xy xx y x y xyx xy yxx xx x y
fxyx fxyy x y yx xx x
AA A A

5 5 5 5 5 5
55555555555555

5555 5 55

As
1,0,0,0xy A55 and
2 0.A

202. 202 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
44.





3
23
32 3
2
2
2
12 1 2
,510
,,
5 5 10 10 3 3 5 10
531003
,, where0and3 .
xy
zfxy x yy
zfx xy yfxy
x x y yy yy yy y x yy
xy y xyyyy
fxy x f xy y x y y y yAA A A

5 5 5
5 5 555
5 55 55 5
5555 55
As
1,0,0,0xy A55 and
2 0.A

45.


2
42
3
,, 0,0
,
0, , 0, 0
xy
xy
xyfxy
xy
6
#
=
7
=

8





4
00
0
0
,0 0,0
0, 0 lim lim 0
x
xx
fx f x
f
xx
5 5

5 5

55





2
00
0
0
0, 0, 0
0, 0 lim lim 0
y
yy
fyf y
f
yy
5 5

5 5

55

So, the partial derivatives exist at
0, 0 .
Along the line :yx


3
42 2
,0,0 0 0
33
lim , lim lim 0
1
xy x x
xx
fxy
xx x



Along the curve
2
:yx


4
4
,0,0
33
lim ,
22
xy
x
fxy
x



fis not continuous at 0, 0 .So, fis not differentiable at 0, 0 .See Theorem 12.5

46.


2
33
5
,, 0,0
,
0, , 0, 0
xy
xy
xyfxy
xy
6
#
=
7
=

8




00
,0 0,0 00
0, 0 lim lim 0
x
xx
fx f
f
xx
5 5
5

55




00
0, 0, 0 00
0, 0 lim lim 0
y
yy
fyf
f
yy
5 5
5

55

So, the partial derivatives exist at
0, 0 .
Along the line :yx


3
3
,0,0 0
55
lim , lim .
22
xy x
x
fxy
x


Along the line 0,x


,0,0
lim , 0.
xy
fxy


So, f is not continuous at
0, 0 .Therefore f is not differentiable at 0, 0 .

47. If f is differentiable at 00,,
xythen the partial derivative
00,
xfxyexists, which implies that 0,fxyis
differentiable at
00,.
xyIf
22
,,fxy x y then

2
,0 .fxxx Because xis not differentiable
at 0,x

22
,fxy x y is not differentiable.

48.
22
,fxy x y
(a)

22
1, 2 1 4 5 2.2361
1.05, 2.1 1.05 2.1 2.3479
f
f


(b)
1.05, 2.1 1, 2 0.1118zf f5
(c)

22 22
12
0.5 0.1 0.1118
55
xy
dz dx dy
xy xy




The results are the same.

203. Section 13.5 Chain Rules for Functions of Several Variables 203
© 2010 Brooks/Cole, Cengage Learning
Section 13.5 Chain Rules for Functions of Several Variables
1.

22
2, 3
22 23
46 81826
wx y
xtyt
dw w dx w dy
xy
dt x dt y dt
xyt t t


CC

CC



2.
22
22 22
2
22 2 2
cos ,
sin
sin cos sin
cos
t
t
tt
t
wxy
xtye
dw w dx w dy
dt x dt y dt
xy
te
xy xy
xtye t te
xy te


CC

CC







3. sinwx y
,
t
xey t


sin cos 1
sin cos sin cos
t
tt t tdw w dx w dy
ye x y
dt x t y dt
te e t e t e t

CC

CC C



4.

ln
cos
sin
11
sin cos
1
tan cot
sin cos
y
w
x
xt
yt
dw
tt
dt x y
tt
tt









5.
2
,,
tt
wxyxeye


(a)

22 2
22
ttttttt
dw w dx w dy
dt x dt y dt
ye x e e e e e e

CC

CC


(b)
2tt t
t
wee e
dw
e
dt





6.
2
cos , , 1wxyxty
(a)



2
sin 2 sin 0
2sin 2sin 1
dw
xyt xy
dt
txy tt


(b)


22
cos 1 , 2 sin 1
dw
wt tt
dt


7.
222
,wx y z cos ,xt sin ,yt
t
ze
(a)

22
2sin 2cos 2
2 cos sin 2 sin cos 2 2
t
tt
dw w dx w dy w dz
dt x dt y dt z dt
xtytze
tt t t e e
CCC

CCC



(b)
222 2
2
cos sin 1
2
tt
t
wtte e
dw
e
dt



8.
2
cos
arccos
wxy z
xt
yt
zt




(a)

222
22
333 311
cos 1 cos 2 sin 2 1
11
24
dw
yz xzt xyz ttttttt t
dt tt
ttt t





(b)
43
,4
dw
wt t
dt


9. ,wxyxzyz 1,xt
2
1,ytzt
(a)


222
2
11211321
dw w dx w dy w dz
yz xzt xy
dt x dt y dt z dt
ttttttt t
CCC

CCC

(b)


22
222
11 1 1
21 12131321
wt t t tt t
dw
tt t t t t
dt

204. 204 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
10.
22 2
,wxy xzyz
2
,2,2xty tz
(a)

2222233
22 2 2 20 4 42 4 42 24 8
dw w dx w dy w dz
dt x dt y dt z dt
yxzt xyz xyz ttt t t
CCC

CCC


(b)

22 4 4
3
422468
24 8
wtt t t t t
dw
t
dt



11. Distance

22 2 2
12 1 2
10 cos 2 7 cos 6 sin 2 4 sin
ftxxyy t t t t




12
22
1
10 cos 2 7 cos 6 sin 2 4 sin
2
2 10 cos 2 7 cos 20 sin 2 7 sin 2 6 sin 2 4 sin 12 cos 2 4 cos
ft t t t t
tt tt tt tt







12
212
2
1 1 22 11 29
10 4 2 10 7 2 4 12 116 44 2.04
22 2 29 229
f








12. Distance


22
22
21 2 1
48 3 2 48 1 2 48 8 2 2 2 6
48 8 2 2 2 6 1
ft x x y y t t
ft f





13. ln , ,
tt
wxyxeye




11
tt
tt
tt
dw w dx dw dy
dt x dt y dt
ee
ee
xyxy ee



C

CC








2
22
2
4
tttt tttt
tt
tt
eeee eeeedw
dt
ee
ee









At 0,t

2
22
4
1.
11
dw
dt




14.
2
2
1
1
x
w
y
xt
yt
t










2
2
342 443
22 2
2
21
1422 34
1 11 1
dw w dx w dy x x
t
dt x dt y dt y y
ttttt ttt
t tt t
CC

CC







2
32 43
2
42
112 12 3 4 2 1
1
ttttttdw
dt t




At
1:t

2
2
424 7 4 68
4.25
16 16
dw
dt




15.
22
,
wx y
xsty st



21 21 2 2 4
w
x yststs
s
C

C

21 2 1 2 2 4
w
x yststt
t
C

C
When 1sand 0,t
4
w
s
C

Cand
0.
w
t
C

C

16.
32
3
,
s t
wy xy
xey e




22 2
63306 6
s sts stw
xy e y x e e e e
s
C

C



22 2 2
32
60 3 3 3 3
33
ttst
tstw
xyyxeeee
t
ee

C

C


When 1sand 2,t
6
w
s
C

C
and
6
33.
w
e
t
C

C


17. sin 2 3wxy
xst
yst





2 cos 2 3 3 cos 2 3
5cos 2 3 5cos 5
w
xyxy
s
xyst
C

C



2 cos 2 3 3 cos 2 3
cos 2 3 cos 5
w
xyxy
t
xyst
C

C


When 0sand ,
2
t

0
w
s
C

Cand
0.
w
t
C

C

205. Section 13.5 Chain Rules for Functions of Several Variables 205
© 2010 Brooks/Cole, Cengage Learning
18.
22
cos
sin
wx y
xst
yst





22
2cos 2sin
2cos 2sin 2cos2
w
xt yt
s
s ts t s t
C

C



2
2 sin 2 cos 2 sin 2
w
xst ys t s t
t
C

C

When 3sand ,
4
t

0
w
s
C

Cand
18.
w
t
C

C

19.
,
yz
w
x

2
,x ,yr zr
(a)
22
2
011
wyz z y zyr
rx x x x

C

C







2
42
22
2
33
211
2
2 22
wyz z y
xxx
rr r r
r r






C

C





(b)

2
22
2
2
3
1
2
2
rryz r
w
x
wr
r
wr





C

C
C

C


20.
22
2,wx xyy ,xr yr
(a)
221 2210
w
xy xy
r
C

C

221 22144 4 4 8
w
xy xy xy xy r r

C


C

(b)

22
2222222
22224
0
8
wr r r r r r r r r
w
r
w


C

C
C

C


21.
arctan ,
y
w
x
cos ,xr sinyr
(a)


22 22 2 2
22 22 2 2
sin cos cos sin
cos sin 0
sin sin cos cos
sin cos 1
wy x r r
rxy xy r r
rr rrwy x
rr
xy xy r r



C

C
C

C

(b)

sin
arctan arctan tan
cos
0
1
r
w
r
w
r
w




C

C
C

C

22.
22
25 5 5 ,wxy cos , sinxr yr
(a)
22
22 22 22 2
555cos5sin5
cos sin
25 5 5 25 5 5 25 5 5 25 5
wx y rr r
r
x yxy xyr

C

C



22 2
22 22 22
555sincos5sincos
sin cos 0
25 5 5 25 5 5 25 5 5
wx y r r
rr
xy xy xy

C

C

206. 206 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
(b)
2
2
25 5
5
;0
25 5
wr
wrw
r r


CC

CC

23. ,wxyz ,
xst ,yst
2
zst


2
2 2 2 22 224 224 222
11
233
w
yz xz xy t
s
stst s tst s t s tt st st t st t t s t
C

C




22 33333
22
112 222224
22
w
yz xz xy st s t st s t st s t s t st st s t st s t st
t
st s t
C

C


24.
222
,wx y z
2
sin , cos ,
xtsyt szst

2
22 44
2cos 2 sin 2
2sincos 2sincos 2 2
w
xsytszt
s
t sst ssst st
C

C




2 2 23 23
2sin 2cos 2 2
2sin 2cos 4 2 4
w
xs y s zst
t
tst sst tst
C

C



25. ,
xy
wze ,,xsty stz st





22
22
11
221
xy xy xy
stst
stst stw
yze xze e t
s
eststststt
estttes


C

C











22
22
11
212
xy xy xy
stst
stst stw
yze xze e s
t
eststststs
estsset


C

C






26. cos ,wx yz
2
,
xs
2
,yt 2zs t


23 22 23
cos 2 sin 0 sin 1
cos 2 2 sin 2
w
yz s xz yz xy yz
s
st t s s t st t
C

C






2232223
22 3 2 3
cos 0 sin 2 sin 2
22sin22sin2
62sin2
w
yz xz yz t xy yz
t
sts t st t st st t
st st st t
C

C




27.
22
0xxyyxy



, 21 21
,2121
x
yFxydy x y y x
dx F x y x y y x





28. sec tan 5 0xy xy

dy
dx




2
2
2
2
, sec tan sec
,sectansec
sec tan sec
sec tan sec
x
yFxy yxyxyy xy
F x y x xy xy x xy
yxyxy xy y
xxxyxy xy








29.

22
22
ln 4
1
ln 4 0
2
xyxy
xy xy






2222
22
22
1
,
,
1
x
y
x
Fxydy x x y xy
ydx F x y y x y
xy








30.
2
22
60
x
y
xy








2
222 2
2
22
22
2
22
22
4235
,
,
22
22
22 4 2
x
yFxydy
dx F x y
yxxy
xyx y y
yx
xy y x y
yx
xyyx xy y












31.
222
,, 1Fxyz x y z
2, 2, 2
xyzFxFyFz

zx
xz
y
z Fx
Fz
Fzy
yF z
C

C
C

C

207. Section 13.5 Chain Rules for Functions of Several Variables 207
© 2010 Brooks/Cole, Cengage Learning
32. ,,F x y z xz yz xy

x
y
z
x
z
y
zFzy
Fzx
Fxy
zF yz
x Fxy
Fzxz
yF xy



C

C
C

C


33.
22
,, 2 1 0Fxyz x yz z





,, 2
,, 2 2
,, 2
,, 2 2
x
z
y
zFxyzzxx
x Fxyz y z y z
Fxyzzzz
yFxyz yzyz
C

C
C

C


34. sin 0xyz
(i)



1cos 0implies
1
sec .
cos
z
yz
x
z
yz
xyz
C

C
C

C
(ii)
1cos 0
z
yz
y
C

C implies
1.
z
y
C

C

35. ,, tan tan 1Fxyz x y y z










2
22
2
2
2
22
2
2
2
sec
sec sec
sec
sec
sec
sec sec
sec
sec
1
sec
x
y
z
x
z
y
zFxy
Fxyyz
Fyz
xyzF
xF yz
F
xyyzz
yF yz
xy
yz



C

C
C

C






36. ,, sin
x
Fxyz e y z z








sin
cos
cos 1
sin
1cos
cos
1cos
x
x
x
y
x
z
x
x
x
z
x
y
x
z
Fe yz
Fe yz
Fe yz
eyzzF
x Feyx
F eyzz
yF eyz



C

C
C

C


37. ,, 0
xz
Fxyz e xy





,,
,,
,, 1
,,
xz
x
xz
z
y
xz
xz xz
z
Fxyzzzey
xFxyz xe
Fxyzzx
e
yFxyzxee

C

C
C

C


38.
22
ln 8 0xyyzz
(i)


2
,
,, ln
,2
x
z
z
Fxyzzy
x Fxy y z
C

C

(ii)


2
23
2
,, 2
,, 2 2
y
z
x
yz
Fxyzzxyz y
yFxyz y z y yz

C

C


39. ,,,Fxyzw xy yz wz wx s

x
y
z
wFyw
Fxz
Fyw
Fzx






x
w
y
w
z
wwF ywyw
x Fzxzx
Fwxzxz
yF zxzx
w F yw yw
zF zxzx
C

C
C

C
C

C


40.
222 2
5102 ,,,
xyz yw w Fxyzw
2, 2 5 ,
xyFxFyw 2, 5 20
zwFzF yw

22
520 520
52
20 5
2
520
x
w
y
w
z
wwF x x
x Fywyw
Fwwy
yF wy
wF z
zF yw
C

C
C

C
C

C


41. ,,, cos sin 20Fxyzw xy yz wz

sin
sin cos
cos
x
w
y
w
z
wwFyxy
xF z
Fwxxyzyz
yF z
wF yzyw
zF z
C

C
C

C
C

C

208. 208 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
42. ,,, 0Fxyzw w x y y z



12
12 12
11
21 2
11 11
22 22
1
2
x
w
y
w
z
w
xywF
xF xy
Fw
xy yz
yF
xyyz
wF
zF yz


C

C
C

C
C

C


43. (a)




22
22 22
,
,,
xy
fxy
xy
tx ty xy
ftx ty t tf x y
xytx ty







Degree: 1
(b)




33
32 32
2222 22
,, 1,
xy
yxxy
xfxy yfxy x y fxy
xyxy xy








44. (a)
323
,3
fxy x xy y


323
33 2 3 3
,3 3 ,
ftx ty tx tx ty ty t x xy y t f x y
Degree: 3
(b)

22 2 3 23
,,33633933,
xy
xfxy yfxy xx y y xy y x xy y fxy

45. (a)

,
,,
xy
tx ty x y
fxy e
ftx ty e e f x y



Degree: 0
(b)

2
1
,, 0
xy xy
xy x
xf x y yf x y x e y e
yy





46. (a)




2
22
2
2
22 22
,
,,
x
fxy
xy
tx x
ftx ty t tf x y
xytx ty







Degree: 1
(b)






22 2
32 2 422 2
32 32 32 32
2222 22 22 22
2
,, ,
xy
xx yxxy xy xxy x
xfxy yfxy x y fxy
xyxy xy xy xy









47.
dw w dx w dy f dg f dh
dt x dt y dt x dt y dt
CC CC

CC CC

At 2,t 4,x 3,y

4, 3 5
xf and 4, 3 7.
yf
So, 51 76 47
dw
dt


48.
53 75 50
wwxwy
sxsys
fg fh
xs ys
CCCCC

CCCCC
CC CC

CC CC


52 78 66
wwxwy
txtyt
fg fh
xt yt
CCCCC

CCCCC
CC CC

CC CC

209. Section 13.5 Chain Rules for Functions of Several Variables 209
© 2010 Brooks/Cole, Cengage Learning
xx
h
θ
2
b
2
49. Page 925
dw w dx w dy
dt x dt y dt
CC

CC

50.
Page 925
wwxwy
sxsys
wwxwy
txtyt
CCCCC

CCCCC
CCCCC

CCCCC


51.






,
,
,,
,,
,,
,,
x
y
x
z
y
zfxydy
dx f x y
fxyzz
x fxyz
fxyzz
yfxyz
Δπ
C
Δπ
C
C
Δπ
C


52. θ
,, ,fxyz xyzΔ
2
,xtΔ 2,ytΔ
t
ze
π
Δ
(a)


223 2
22
422 23
t
ttt t
df f dx f dy f dz
dt x dt y dt z dt
yz t xz xy e
te te te te t
π
πππ π
CCC

CCC



(b)

23
22
tt
ftte te
ππ
ΔΔ

32 2
26 23
tt tdf
te te te t
dt
ππ π

The results are the same.

53.
2
VrhΔ




2 3
2
2 2 12 2 36 6 12 4 4608 in. min
2
2 2 2 24 36 6 12 4 624 in. min
dV dr dh dr dh
rh r r h r
dt dt dt dt dt
Srrh
dS dr dh
rh r
dt dt dt













54.


2
2
23
22 21
3
11
2 2 12 36 6 12 4 1536 in. min
33
Surface area includes base.
Vrh
dV dr dh
rh r
dt dt dt
Srrh r


Δ












2
22
22 22
22
22 22
22
2
36 12144
12 36 2 12 6 4
12 36 12 36
12 36 648 36
12 10 6 144 4 144 in. min 20 9 10 in. min
510 10 10
dS r dr rh dh
rh r
dt dt dt rh rh















55.

1
1
pV mRT
TpV
mR
dT dp dV
Vp
dt mR dt dt
Δ
Δ






56.
2
2
2
22
1
sin cos sin
2222
sin cos
2
16 32 2
6 sin cos m hr 2.566 m h
42 2 490 2 10
x
Abhx x
dA A dx A d dx x d
x
dt x dt dt dt dt





ΔΔ Δ


CC

CC

210. 210 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
57.

22
12
12 2
121
2
1
22 628228cmsec
2
Imrr
dI dr dr
mr r m m
dt dt dt







58.




22
22
22
3
3
22
3
2 15 25 10 4 15 2 25 10 4 15 15 25 25 12
3
19,500
3
6,500 cm min
VrrRRh
dV dr dR dh
rRh r Rh r rRR
dt dt dt dt































2
2
2 2
2 2
2 2
2 2
2
2
2
2
2
2
2
2
2
2
25 15
25 15 10 25 15 4
25 15 10
25 15
25 15 10 25 15 4
25 15 10
SRrRrh
Rr RrdS dr dR
Rr h Rr Rr h Rr
dt dt dt
Rr h Rr h
hdh
Rr
dt
Rr h



6
=

7

=
8
<
=
>
=
?
6
=

7

=
8










2
2
2
10
25 15 12
25 15 10
320 2 cm min

<
=

>

=
?



59.
,
0
wfxy
xuv
yvu
wwdxwdyww
uxduyduxy
wwdxwdy ww
vxdvydv xy
ww
uv



CC C CC

CC C CC
CC C CC

CC C CC
CC

CC

60.


sin
cos sin
cos sin
0
wxy yx
w
xyyx yx
x
w
xyyx yx
y
ww
xy

C

C
C

C
CC

CC

211. Section 13.5 Chain Rules for Functions of Several Variables 211
© 2010 Brooks/Cole, Cengage Learning
61. ,,wfxy cos ,xr sinyr
cos sin
ww w
rx y
CC C

CC C


sin cos
ww w
rr
xy

CC C

CC C

(a)



2
2
22
2
cos cos sin cos
sin sin sin cos
cos sin cos sin
cos sin
sin
cos First Formula
sin sin cos sin
co
ww w
rrr
rx y
ww w
rr
xy
www
rrr
rx
ww w
rr
xr
ww w
xr r
ww w
rr r
rx y










CC C

CC C
CC C

CC C
CCC

CCC
CC C

CC C
CC C

CC C
CC C

CC C

2
ssincoscos
ww w
rr
xy

CC C

CC C




22
sin cos sin cos
sin cos
cos
sin Second Formula
www
rrr
ry
ww w
rr
yr
ww w
yr r






CCC

CCC
CC C

CC C
CC C

CC C

(b)
2222 2
222
2
22 2
2
1
cos 2 sin cos sin sin
2sincos cos
www ww w w
rr x xy y x
ww w w w
xy y x y



CCC CC C C

CCC CC C C
CC C C C

CC C C C


62.

arctan , cos , sin
sin
arctan arctan tan for
cos 2 2
y
wxryr
x
r
r








22
,
wy
xxy
C

C
22
,
wx
yxy
C

C
0, 1
ww
r

CC

CC




22
22
22 22 2
22 22
22
222
11
111
01
ww y x
x yxyr
xy xy
ww
rr r r

CC

CC
CC


CC

So,
2222
2
1
.
ww w w
xy rr

CC C C

CC C C

212. 212 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
63. Given
uv
x y
CC

CC
and ,
uv
yx
CC

CC cosxr and sin .yr

cos sin cos sin
uu u v v
rx y y x
CC C C C

CC C C C


sin cos cos sin
vv v v v
rrr
xy yx

CC C C C


CC C C C

So,
1
.
uv
rr

CC

CC


cos sin cos sin
vv v u u
rx y y x
CC C C C

CC C C C


sin cos cos sin
uu u u u
rrr
xy yx

CC C C C


CC C C C

So,
1
.
vu
rr

CC

CC

64. Note first that

22
22
.
uv x
xyxy
uv y
yxxy
CC

CC
CC

CC


22
22 22 2
cos sin 1
cos sin
ux y r r
rxy xy r r
C

C



22 2 2
22 22 2
sin cos
sin cos 1
vy x r r
rr
xy xy r

C

C

So,
1
.
uv
rr

CC

CC


22 22 2
sin cos sin cos
cos sin 0
vy x r r
rxy xy r
C

C



22
22 22 2
sincos sincos
sin cos 0
ux y r r
rr
xy xy r

C

C

So,
1
.
vu
rr

CC

CC

Section 13.6 Directional Derivatives and Gradients
1.
,349
34
55
fxy x xy y
vij





,3494
1, 2 5 5
34
55
ij
ij
v
uij
v
fxy y x
f
D
D



1, 2 1, 2 3 4 1Df f
u uD "
2.




33
22 2
,,
2
,33
4, 3 48 27
22
22
fxy x y
fxy x y
f
vij
ij
ij
v
uij
v

D
D




4, 3 4, 3
27 21
24 2 2 2
22
Df f
u uD "

213. Section 13.6 Directional Derivatives and Gradients 213
© 2010 Brooks/Cole, Cengage Learning
3. ,fxy xy

1
3
2
vi j


,
0, 2 2
fxy y x
f
ij
i
D
D

13
22
v
uij
v


0, 2 0, 2 1Df f
u uD "
4.




2
,
1
,
1, 1
1, 1 1, 1 1
x
fxy
y
x
fxy
yy
f
Df f
u
vj
ij
ij
v
uj
v
u


D
D

D "

5.

,sin
sin cos
1,
2
1, 1,
22
x
xx
hx y e y
he ye y
he
Dh h e
u
vi
ij
i
v
ui
v
u




D

D




D "


6.
,arccos
gxy xy

vj




22
,
11
yx
gxy
xyxy
ij

D



1, 0g
jD

v
uj
v



1, 0 1, 0 1Dg g
u uD "

7.



22
22 22
,
34
34
3, 4
55
34
55
7
3, 4 3, 4
25
gxy x y
xy
g
xyxy
g
Dg g
u
vij
ij
ij
v
uij
v
u


D

D

D "

8.





22
22 22
,
22
0, 0
0, 0 0, 0 0
xy
xy xy
hx y e
hxe ye
h
Dh h
u
vij
ij
0
u




D
D
D "
9.




222
,,
3
3
,, 2 2 2
1, 1, 1 2 2 2
333
333
2
1, 1, 1 1, 1, 1 3
3
fxyz x y z
fxyz x y z
f
Df f
u
vijk
ijk
ijk
v
uijk
v
u


D
D

D "

10.
,,
fxyz xy yz xz

2vijk




,,
1, 2, 1 3
1
2
6
fxyz y z x z y x
f ijk
ik
v
uijk
v
D
D




1, 2, 1 1, 2, 1
23 6
666
Df f
u uD "


11.



,,
2, 1, 2
2, 1, 1 2 2
212
333
8
2,1,1 2,1,1
3
hx y z xyz
hyzxzxy
h
Dh h
u
v
ijk
ijk
v
uijk
v
u


D
D

D "

12.






22
, , arctan
1, 2, 1
, , arctan
11
4, 1, 1 2 2
4
12 1
,,
66 6
868
4,1,1 4,1,1
2446hx y z x yz
xz xy
hx y z yz
yz yz
h
Dh h
u
v
ijk
ijk
v
u
v
u




D

D


D "

214. 214 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
13.

22
,
11
22
22
22
2
22
fxy x y
fxy
Df f x y x y
u
uij
ij
u


D
D "

14.





22
22
2
,
31
22
3
22
1
3
2
y
fxy
xy
yx
f
xy xy
yx
Df f
xyxy
yx
xy
u
uij
ij
u



D

D "




15.




,sin2
13
22
2 cos 2 cos 2
3
cos 2 cos 2
2
23
cos 2
2fxy x y
u
f xy xy
Df f x y x y
xy
u
ij
ij
u


D
D "





16.


,
13
22
13
31
22 2
y
yy
y
yy
gxy xe
ge xe
e
Dg e xe x
u
ui j
ij


D


17.



22
,3
34
26,1,126
34
55
624
1, 1 1, 1 6
55
u
vij
ij ij
v
uij
v
u
fxy x y
fxyf
Df f


D D

D "

18.




,cos
2
sin sin
12
55
12
sin sin
55
15
sin sin
55
fxy x y
fxyxy
Df xy xy
xyxy
u
vij
ij
v
uij
v




D




At
0, , 0.Df
u
19.
,,
24
z
z zz
gxyz xye
gye xe xye
vij
ij k


D

At

2, 4, 0 ,428.gijkD

12
55
44 8
55 5
Dg g
u
v
uij
v
u

D "


20.


,, ln
33
1
hx y z x y z
h
xyz
vijk
ijk


D


At
1, 0, 0 , .h ijkD


1
33
19
7719
1919
Dh h
u
v
uijk
v
u

D "

21.


2
,351
,310
2, 1 3 10
fxy x y
fxy y
f
ij
ij

D
D

22.



,2
2
,22
2, 0 2 2
yx
yx yx yx
gxy xe
y
gxy e e e
x
g
ij
ij


D


D

23.



2
22
ln
21
,
2, 3 4
zxy
x
zx y
xyxy
z
ij
ij

D

D

215. Section 13.6 Directional Derivatives and Gradients 215
© 2010 Brooks/Cole, Cengage Learning
24.


22
22 22
cos
,2sin 2sin
3, 4 6 sin 25 8 sin 25 0.7941 1.0588
zxy
zx y x x y y x y
z
ij
ijij

D
D

25.


222
35 2
,, 6 10 4
1, 1, 2 6 10 8
wx y z
wx y z x y z
w
ijk
ijk

D
D

26.




2
2
22
tan
,, tan sec
sec
4, 3, 1 tan 2 4 sec 2 4 sec 2
ij
k
ijk
wx yz
wx y z y z x y z
xyz
w

D

D
27.

22
,
22
,22,1,224
222 32
PQ
gxy x y g
Dg g
u
iju i j
ij ij
u


DD
D "
28.

21
42,
55
62, 1,4 68
12 8
45
55
u
iju i j
ij ij
u
PQ
fxyf
Df f

DD
D "

29.

21
2,
55
cos sin
0, 0
225
55
yy
PQ
fe xe x
f
Df f
u
iju i j
ij
i
u


D
D
D "
30.

12
,
2 55
2 cos 2 cos sin 2 sin
,0 2
225
55
PQ
f xy xy
f
Df f
u
iju i j
ij
i
u





D
D
D "

31.




2
,2
,222
1, 0 2 2
1, 0 2 2
fxy x xy
fxy x y x
f
f
ij
ij

D
D
D

32.





2
,
1
11
,
1 1
11
0, 1
24
11 1
0, 1 5
416 4
xy
fxy
y
x
fxy
y y
f
f
ij
ij




D

D
D

33.


2
,tan
,tan sec
2, 4
4
2, 17
4
hx y x y
hx y y x y
h
h
ij
ij



D

D



D



34.




22
2
,cos
,sin
cos sin
333
0,
36 6
39633
0,
336 36
32 2 3 3
6
i
j
ij
hx y y x y
hx y y x y
xyyxy
h
h




D



D


D





35.




,
,
0, 5 5
0, 5 26
x
x x
gxy ye
gxy ye e
g
g
ij
ij



D
D
D

36.





22 223
22 22 1
,ln ln
3
12 2
,
3
12 4 2
1, 2 2
35 5 15
25
1, 2
15
gxy x y x y
xy
gxy
xy xy
g
g
ij
ij ij


D



D


D

37.




222
222
,,
1
,,
1
1, 4, 2 4 2
21
1, 4, 2 1
fxyz x y z
fxyz x y z
xyz
f
f
ijk
ijk

D

D
D

216. 216 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
38.




222
3
222
1
1
1
1
0, 0, 0
0, 0, 0 0
w
xyz
wxyz
xyz
w
w
ijk
0


D

D
D

39.
22
22 2 2
22ijk
wxyz
wyz xyz xyz

D




2, 1, 1 4 4
2, 1, 1 33
ijkw
w
D
D

40.




,,
,,
2, 0, 4 8
2, 0, 4 65
yz
yz yz yz
fxyz xe
fx y z e xze xye
f
f
ijk
ij

D
D
D

For exercises 41–46, ,3
32
xy
fxy and





11
,cossin.
32
Dfxy
41.

,3
32
xy
fxy


42. (a)
4
12 12 52
3, 2
32 22 12
Df






(b)
23
11 13233
3, 2
32 22 12
Df






(c)
43
11 1 3
3, 2
32 2 2
233
12
Df







(d)
6
13 11
3, 2
32 2 2
323
12
Df








43. (a)

1
2
11 11 52
32 1222
Df f
u
uij
u





D "



(b)
34
916 5
34
55
12 3
55 5
Df f
u
vij
v
uij
u


D "

(c)
34
916 5
34
55
12 1
55 5
u
vi j
v
uij
u
Df f



D "

(d)
3
10
13
10 10
11 11 10
60610
u
vi j
v
uij
u
Df f




D "


44.
11
32
f
ij

D




45.
11 1
13
94 6
fD

46.

11
32
1
23
13
f
f
f
ij
ijD
D

D
So,
1133 2ui j and
3, 2 0.Df f f
u uD " D is the direction of greatest
rate of change of f. So, in a direction orthogonal to
,fDthe rate of change of f is 0.

47. (a) In the direction of the vector 4ij
(b)



11
10 10
1121
10 10 5 10
23 32
1, 2 4 1
Same direction as in part a
fxy xy
f ij
ijijD
D

(c)
21
510
,f ijD the direction opposite that of the
gradient
x
y
3
6
9
(3, 2, 1)
z

217. Section 13.6 Directional Derivatives and Gradients 217
© 2010 Brooks/Cole, Cengage Learning
48. (a) In the direction of the vector ij
(b)


11 1 1
22 224
11
1, 2
22
Same direction as in part
y
f yx x
xx
f
a
ij ij
ij
D
D
(c)
11
,
22
f ijθD π θ θ
the direction opposite that of the
gradient

49.
22
,,4,3,7fxy x yπθ θ
(a)
(b)

,,2cos2sin
4, 3 8 cos 6 sin
Dfxy fxy x y
Df
u
u u
πD " π θ


(c) Zeros:
2.21, 5.36
These are the angles
for which 4, 3Df
u equals
zero.
(d)

4, 3 8 cos 6 sin
8sin 6cos
gDf
g
u


Critical numbers:
0.64, 3.79
These are the angels for which
4, 3Df
u θis a
maximum
0.64and minimum 3.79 .
(e)

4, 3 2 4 2 3 64 36 10,f ijD
the maximum value of
4, 3 ,Df
u θat 0.64.
(f )

22
,7fxy x yπθπ

4, 3 8 6f ijD is perpendicular to the level
curve at
4, 3 .θ


50. (a)
22
8
,2
1
y
fxy
xy
ππ




22
22
2
2
41
4441
23
yxy
yy x
yx




Circle: center:
0, 2 , radius:
3
(b)



22
22
22 22
16 8 8 8
11
3
3, 2
2
xyxy
f
xy xy
f
ij
i

D

θ
Dπ


(c) The directional derivative of f is 0 in the direction
.
j
(d)

51.

,623
6, 0, 0
,23
fxy x y
cP
fxy
ij
πθ θ
ππ
Dπθθ

62 3 6
02 3
xy
xy
θθπ



0, 0 2 3f ijDπθθ

52.



22
22
,
25, 3, 4
,22
25
3, 4 6 8
fxy x y
cP
fxy x y
xy
f ij
ij

ππ
D

D

53.


,
3, 1, 3
,
3
1, 3 3
fxy xy
cP
fxy y x
xy
f ij
ij
π
πθ π θ
D
πθ
Dθ π θ

x
y
z
4
2
−4
−8
−12
8
12
ππ
Generated by Mathematica
θ
D
u
f
x
y
2
−2
−4246−6
−4
−6
4
6
Generated by Mathematica
x
2
1
3
4
12−1−2
y
−6
−6
6
6
yx
z

218. 218 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
54.





22
22
22
22 22
22
22
,
1
,1,1
2
2
,
1
2
20
1
1, 1
2
x
fxy
xy
cP
yx xy
fxy
xy xy
x
xy
xy x
f
ij
j
π

ππ
θ
Dπ θ

π


Dπθ


55.
2
,4
fxy x yπθ
(a)

,8
2, 10 16
fxy x
f
ij
ij
Dπθ
Dπθ
(b)
16 257ijθπ

1
16
257
ijθis a unit vector normal to the level
curve
2
46xyθπ at 2, 10 .
(c) The vector 16
ijis tangent to the level curve.
Slope
16
16
1
ππ


10 16 2
16 22 Tangent line
yx
yx
θπ θ
πθ

(d)

56.
2
,
fxy x yπθ
(a)

,2
4, 1 2
fxy y
f ij
ijDπθ
D
(b)

4, 1 5fDθπ

1
2
5
ijis a unit vector normal to the level
curve
2
3xyθπ at 4, 1 .θ
(c) The vector 2
ijθis tangent to the level curve.
Slope
1
.
2
πθ



1
14
2
1
1 Tangent line
2
yx
yx



(d)

57.
22
,32
fxy x yπθ
(a)

64
1, 1 6 4
f xy
fij
ijDπ θ
Dπθ

(b)

1, 1 36 16 2 13fD

1
32
13
ijθis a unit vector normal to the level
curve
22
32 1xyθπ at 1, 1 .
(c) The vector 2 3
ijis tangent to the level curve.
Slope
3
.
2
π



3
2
3 1
22
11
tangent line
yx
yx
θπ θ
πθ
(d)

58.
22
,94
fxy x y
(a)

18 8
2, 1 36 8
fxy
fij
ijD
Dθπ θ

(b)

2, 1 1360 4 85fDθπ π

1
92
85
ijθis a unit vector normal to the level
curve
22
94 40xy at 2, 1 .θ
(c) The vector 2 9
ijis tangent to the level curve.
Slope
9
.
2
π



9
12
2
9
10 Tangent line
2
yx
yx

πθ

(d)
y
x
15105−15−10−5
−10
−5
−226
−2
−4
2
4
y
x
(4, −1)
y
x
321−3−2−1
−3
−2
1
2
3
−44
−2
−4
2
4
y
x
(2, −1)

219. Section 13.6 Directional Derivatives and Gradients 219
© 2010 Brooks/Cole, Cengage Learning
59. See the definition, page 934.

60. Let ,
fxybe a function of two variables and
cos sinui j a unit vector.
(a) If 0 ,
then
.
f
Df
x
u
C

C

(b) If 90 ,
then
.
f
Df
y
u
C

C


61. See the definition, pages 936 and 937.

62.
63. The gradient vector is normal to the level curves. See
Theorem 13.12.

64.
22
,9
fxy x y and

,2cos2sin
2cos sinDfxy x y
xy



(a)

22
,9
fxy x y

(b) 4
2
1, 2 2 2 2
2
Df






(c)
3
1
1, 2 2 3 1 2 3
2
Df






(d)


1, 2 2 4
1, 2 4 16 20 2 5
f
f
ijD
D

(e)




1, 2 2 4
1, 2 1
2
1, 2 5
f
f
f
ij
ij
D
D

D

Therefore, 152uij and
1, 2 1, 2 0.Df f
u uD "
65.


22
22
22
22 22
2
724 1
3, 4 7 24
625 625 625
x
T
xy
yx xy
T
xy xy
T
ij
ij ij



D

D

66.



22
, 5000 0.001 0.004
0.002 0.008
500, 300 2.4 or
5512
ij
ij
ij
hx y x y
hxy
h
h

D
D
D

67.
68. The wind speed is greatest at B.
69.

22
, 400 2 ,Txy x y 10, 10P

4
dx
x
dt
2
dy
y
dt



4
1
t
xtCe


2
2
t
yt Ce



110 0xC 210 0yC


4
10
t
xte


2
10
t
yt e



2
10
y
x

24
100
t
yt e



2
10yx
70.

22
, 100 2 ,Txy x y 4, 3P

2
dx
x
dt
4
dy
y
dt



2
1t
xtCe


4
2t
yt Ce



140xC 230yC


2
4
t
xte


4
3
t
yt e



2
42
33
16 16
tx
eyux


x
y
5
3
3
z
P
x
y
9
3
3
(1, 2, 4)
z
1800
1800
A
B
1994
1671

220. 220 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
71. (a)
(b)




2
2
7
1
2
1
2
, 400
3, 5 400 3
xy
Txy e x
Te
ij
ij


D

D


There will be no change in directions perpendicular
to the gradient:
6ij
(c) The greatest increase is in the direction of the
gradient:
1
2
3i
j
72. (a)
(b) The graph of
2
250 30 50 sin 2Dxy
would model the ocean floor.
(c)

1, 0.5 250 30 1 50 sin 315.4 ft
4D


(d)
60
D
x
x
C

C
and
1, 0.5 60
D
xC

C

(e)
25 cos
2
Dy
y
C

C
and
1, 0.5 25 cos 55.5
4
D
y
C

C

(f )

60 25 cos
2
1, 0.5 60 55.5
y
Dx
D
ij
ij


D


D

73. True
74. False


,21Dfxy
u when cos sin .
44
uij





75. True
76. True
77. Let

2
,, cos .
2
x zfxyz e y C Then ,, cos sin .
xx
fxyz e y e y zi jkD
78. We cannot use Theorem 13.9 because f is not a differentiable function of x and y. So, we use the definition of directional
derivatives.





0
2
222
0000
cos , sin ,
,lim
0,0 0,0 4
112222 22
0, 0 lim lim lim lim which does not exist.
22
t
tttt
fx t y t fxy
Dfxy
t
tt tt
ff
t
Df
ttttt tt
u
u














If
0, 0 2,f then
2
2
00
0,0 2
1222
0, 0 lim lim 2 0
tt
tt
f
t
Df
ttt
u









which implies that the directional derivative exists.

79. (a)
3
,fxy xy is the composition of two continuous functions, ,hx y xyand
13
,gzz and
therefore continuous by Theorem 13.2.
6
500
yx
6
z
x
y
300
400
1
2
1
2
D

221. Section 13.7 Tangent Planes and Normal Lines 221
© 2010 Brooks/Cole, Cengage Learning
(b)

13
00
0,0 0,0 0 0
0, 0 lim lim 0
x
xx
fxf x
f
xx
5 5
5 "5
πππ
55





13
00
0, 0 0, 0 0 0
0, 0 lim lim 0
y
yy
fyf y
f
yy
5 5
5 "5
πππ
55

Let
cos sin ,iu
j 0, , , .
22

E
#
Then



23 3
13
000
0 cos ,0 sin 0,0 cos sin cos sin
0, 0 lim lim lim , does not exist.
ttt
ft t f t
Df
ttt
u



πππ

(c)

Section 13.7 Tangent Planes and Normal Lines
1. ,, 3 5 3 15 0
35315Plane
Fxyz x y z
xyz




2.
222
222
,, 25 0
25Fxyz x y z
xyz

Sphere, radius 5, centered at origin.

3.
222
22 2
,, 4 9 4 0
4 9 4 Elliptic coneFxyz x y z
xy z


4.
22
,, 16 9 36 0Fxyz x y z

22
16 9 36 0xy z Hyperbolic paraboloid

5. ,, 3 4 12 0Fxyz x y z
3412,
F ij kD
9 16 144 13FD

3412
13 13 13F
F
nijk
D

D


6. ,, 4Fxyz x y z
F
ijk

D



13
33F
F
nijkijk
D

D


7.
222
,, 6Fxyz x y z



222
1, 1, 2 2 2 4
1, 1, 2 4 4 16 2 6Fxyz
F
F ijk
ijkD
D
D


112
666F
F
nijk
D

D


8.


22
22 22
,,
,,
34
3, 4, 5
55Fxyz x y z
xy
Fxyz
xy xy
F
ijk
ijk

D

D


F
F
n
D
π
D


53 4
5552
1
345
52
2
345
10
ijk
ijk
ijk








9.



3
2
,,
3
2, 1, 8 12
2, 1, 8 145
1
12
145
Fxyz x z
Fx
F
F
F
F
ik
ik
nik
πθ
Dπ θ
Dθπθ
Dθ π
D
ππ θ
D

x
y
−2
−1
2
2
z
3

222. 222 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
10.


24
423
,,
,, 2 4
1, 2, 16 32 32
Fxyz xy z
Fxyz xy xy
F
ijk
ijk

D
D


F
F
n
D

D


1
32 32
2049
2049
32 32
2049
ijk
ijk




11.


23
2
,, 3 9
,, 2 3 3
2, 1, 2 4 3 12
Fxyz x y z
Fxyz x z
F
ij k
ij k

D
D



1
4312
13
F
F
nijk
D

D


12.




23 2 3
33 22 22
,, 2 4
22 3 2 6
1, 1, 1 4 5 7
1, 1, 1 3 10
1
457
310
Fxyz xy yz xz
Fxyz xyyz xzy
F
F
F
F
ijk
ijk
nijk

D
D
D
D

D


13.



, , ln ln ln
11 1
,,
1, 4, 3
x
Fxyz x y z
yz
Fxyz
xyzyz
F
ijk
ijk



D

D



13
33
F
F
nijkijk
D

D


14.


22
22 22 22
,, 3
,, 2 2
2, 2, 3 12 12
xy
xy xy xy
Fxyz ze
Fxyz xze yze e
F
ijk
ijk



D
D


1
12 12
17
F
F
nijk
D

D


15.

,, sin 4
,, sin cos
1
6, , 7 3 3
62
Fxyz x y z
Fxyz y x y
F
ijk
ijk


D

D



F
F
n
D

D


21
33
2113
1
63 2
113
113
63 2
113
ijk
ijk
ijk








16.

,, sin 2
,, cos cos
333
,,
36 2 2 2
Fxyz x y z
Fxyz xy xy
F
ijk
ijk


D

D




F
F
n
D

D


23 3
2210
1
332
10
10
332
10
ijk
ijk
ijk








17.

22
22
,3,2,1,8
,, 3
fxy x y
Fxyz x y z




,, 2
xFxyz x ,, 2
yFxyz y ,, 1
zFxyz

2, 1, 8 4
xF 2, 1, 8 2
yF 2, 1, 8 1
zF

4221180
42 2
xyz
xyz




18.


,,1,2,2
,,
y
fxy
x
y
Fxyz z
x




2
,,
x
y
Fxyz
x

1
,,
yFxyz x
,, 1
zFxyz

1, 2, 2 2
xF 1, 2, 2 1
yF 1, 2, 2 1
zF

21 2 20
220
22
xy z
xyz
xyz

223. Section 13.7 Tangent Planes and Normal Lines 223
© 2010 Brooks/Cole, Cengage Learning
19.

22
22
,,3,4,5
,,
fxy x y
Fxyz x y z





22
,,
x
x
Fxyz
xy



22
,,
y
y
Fxyz
xy


,, 1
zFxyz


3
3, 4, 5
5
xF
4
3, 4, 5
5
yF 3, 4, 5 1
zF



34
3450
55
334 4550
345 0
xyz
xyz
xyz





20.


,arctan,1,0,0
, , arctan
y
gxy
x
y
Gx yz z
x






2
2222
,,
1
x
yx y
Gxyz
xyyx






2222
1
,,
1
y
xx
Gxyz
xyyx


,, 1
zGxyz

1, 0, 0 0
xG 1, 0, 0 1
yG 1, 0, 0 1
zG
0yz

21.

22
22
,,1,1,2
,,
gxy x y
Gx yz x y z




,, 2
xGxyz x ,, 2
yGxyz y ,, 1
zGxyz

1, 1, 2 2
xG 1, 1, 2 2
yG 1, 1, 2 1
zG

2121120
22 2
xyz
xyz




22.
22
2,1,2,1zx xyy


22
,, 2
Fxyz x xy y z

,, 2 2
xFxyz x y ,, 2 2
yFxyz x y ,, 1
zFxyz

1, 2, 1 2
xF 1, 2, 1 2
yF 1, 2, 1 1
zF

212 2 10
22 10
22 1
xyz
xyz
xyz




23.


2
3
2
3
,2 ,3,1,1
,, 2
fxy x y
Fxyz x y z




2
3
,, ,
xFxyz ,, 1,
yFxyz ,, 1
zFxyz


2
3
2
3
3110
20
233 6
xyz
xyz
xyz

224. 224 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
24.

sin 1 , 0, , 2
2
,, sin 1
x
x
ze y
Fxyz e y z







,, sin 1
x
x
Fxyz e y ,, cos
x
y
Fxyz e y ,, 1
zFxyz
0, , 2 2
2
xF





0, , 2 0
2
yF





0, , 2 1
2
zF





22xz

25.


22
22 22
,ln ,3,4,ln5
1
, , ln ln
2
hx y x y
Hxyz x y z x y z





22
,,
3
3, 4, ln 5
25
x
x
x
Hxyz
xy
H






22
,,
4
3, 4, ln 5
25
y
y
y
Hxyz
xy
H





,, 1
3, 4, ln 5 1
z
zHxyz
H






34
34ln50
25 25
334 425ln50
3 4 25 25 1 ln 5
xyz
xy z
xy z





26.

2
,cos,5,,
42
,, cos
hx y y
Hxyz y z







,, 0
2
5, , 0
42
x
xHxyz
H






,, sin
22
5, ,
42 2
y
yHxyz y
H






,, 1
2
5, , 1
42
z
zHxyz
H








22
0
24 2
222
0
282
42 8 2 4
yz
yz
yz








27.

222
222
4 36, 2, 2, 4
,, 4 36
xyz
Fxyz x y z





,, 2
2, 2, 4 4
x
xFxyz x
F


,, 8
2, 2, 4 16
y
yFxyz y
F


,, 2
2, 2, 4 8
z
zFxyz z
F




42162840
24 22 4 0
4218
xyz
xyz
xyz

225. Section 13.7 Tangent Planes and Normal Lines 225
© 2010 Brooks/Cole, Cengage Learning
28.

222
22 2
2,1,3,2
,, 2
xzy
Fxyz x y z





,, 2
1, 3, 2 2
x
xFxyz x
F


,, 2
1, 3, 2 6
y
yFxyz y
F


,, 4
1, 3, 2 8
z
zFxyz z
F




216 3820
13 34 2 0
34 0
xyz
xy z
xyz





29.
22
38,1,3,2xy x z


22
,, 3 8Fxyz xy x z



2
,, 3
1, 3, 2 12
x
xFxyz y
F



,, 2
1, 3, 2 6
y
yFxyz xy
F


,, 2
1, 3, 2 4
z
zFxyz z
F



12 1 6 3 4 2 0
12 6 4 22
63211
xy z
xyz
xyz




30. 2 3, 4,4,2xyz

,, 2 3Fxyz x yz y


,, 1
4, 4, 2 1
x
xFxyz
F


,, 2 3
4, 4, 2 1
y
yFxyz z
F


,, 2
4, 4, 2 8
z
zFxyz y
F



41 48 2 0
816
816
xyz
xy z
xy z




31.
9, 3, 3, 3
,, 9
xyz
Fxyz x y z




,, 1
3, 3, 3 1
x
xFxyz
F


,, 1
3, 3, 3 1
y
yFxyz
F


,, 1
3, 3, 3 1
z
zFxyz
F




3330
9 same plane!
xyz
xyz


Direction numbers: 1, 1, 1
Line: 333xyz

32.

222
222
9, 1, 2, 2
,, 9
xyz
Fxyz x y z





,, 2
1, 2, 2 2
x
xFxyz x
F


,, 2
1, 2, 2 4
y
yFxyz y
F


,, 2
1, 2, 2 4
z
zFxyz z
F


Direction numbers: 1, 2, 2
Plane:
12 22 2 0, 2 2 9xy z xyz
Line:
122
122
xy z

226. 226 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
33.

22
22
9, 1, 2, 4
,, 9
xyz
Fxyz x y z




,, 2
1, 2, 4 2
x
xFxyz x
F



,, 2
1, 2, 4 4
y
yFxyz y
F



,, 1
1, 2, 4 1
z
zFxyz
F



Direction numbers: 2, 4, 1
Plane:
214 2 40,xyz 24 14xyz
Line:
124
241
xy z



34.

22
22
16 , 2, 2, 8
,, 16
zxy
Fxyz x y z




,, 2
2, 2, 8 4
x
xFxyz x
F



,, 2
2, 2, 8 4
y
yFxyz y
F



,, 1
2, 2, 8 1
z
zFxyz
F




4242 80
44 24
44 24
xyz
xyz
xyz




Direction numbers: 4, 4, 1
Line:
22
8
44
xy
z




35.

22
22
,3,2,5
,,
zx y
Fxyz x y z




,, 2
3, 2, 5 6
x
xFxyz x
F



,, 2
3, 2, 5 4
y
yFxyz y
F



,, 1
3, 2, 5 1
z
zFxyz
F




6342 50
64 5
xyz
xyz



Direction numbers:
6, 4, 1
Line:
325
641
xyz




36. 0, 2, 3, 6xy z

,,Fxyz xy z


,,
2, 3, 6 3
x
xFxyz y
F



,,
2, 3, 6 2
y
yFxyz x
F



,, 1
2, 3, 6 1
z
zFxyz
F



Direction numbers: 3, 2, 1
Plane:
3223 60,32 6xyz xyz
Line:
236
321
xyz



37. 10, 1, 2, 5xyz

,, 10F x y z xyz


,,
1, 2, 5 10
x
xFxyz yz
F



,,
1, 2, 5 5
y
yFxyz xz
F



,,
1, 2, 5 2
z
zFxyz xy
F



Direction numbers: 10, 5, 2
Plane:
10 1 5 2 2 5 0, 10 5 2 30xy z xyz
Line:
125
10 5 2
xy z

227. Section 13.7 Tangent Planes and Normal Lines 227
© 2010 Brooks/Cole, Cengage Learning
38.
2
, 0,2,2
xy
zye


2
,,
xy
Fxyz ye z



22
,, 2
0, 2, 2 8
xy
x
x
Fxyz ye
F




2
,2,2
,, 1 2
01
xy
y
y
Fxyz xye
F



,, 1
0, 2, 2 1
z
zFxyz
F



80 2 20
80
xyz
xyz



Direction number:
8, 1, 1
Line:
22
81 1
xy z




39.
arctan , 1, 1,
4
y
z
x







, , arctan
y
Fxyz z
x



22
,,
1
1, 1,
42
x
x
y
Fxyz
xy
F










22
,,
1
1, 1,
42
y
y
x
Fxyz
xy
F








,, 1
1, 1, 1
4
z
zFxyz
F







Direction numbers:
1, 1, 2
Plane:

112 0, 2
42
xy z xyz





Line:
411
11 2
zxy




40.
2
ln 2, , 2, 1yxz e

+
,,, ln 2ln 2Fxyz y x z



,,
2
,2,1
x
x
y
Fxyz
x
Fe
e



,, ln 2ln
,2,1 1
y
yFxyz x z
Fe




2
,,
,2,1 4
z
z
y
Fxyz
z
Fe





2
24 10
2
48
xe y z
e
xy z
e



Direction numbers:
2
,1, 4
e



21
214
xe y z
e



41.
22
,, 5Fxyz x y ,,Gx yz x z

,, 2 2Fxyz x yijD ,,Gx yzikD

1, 1, 1 2 2F ijD 1, 1, 1G ikD
(a) FGD'D
220 2 2 2 2
10 1
ijk
ijk ijk


Direction numbers:
1, 1, 1
Line:
1
11
1
y
xz

228. 228 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
(b)

21
cos
222 2
FG
FG

D"D

DD

Not orthogonal

42.
22
,,Fxyz x y z ,, 4Gx yz y z

,, 2 2Fxyz x yijkD ,,Gx yz
jkD

2, 1, 5 4 2F ijkD 2, 1, 5G
jkD
(a) 421 4 4
011
FG
ij k
ijk
D'D

Direction numbers:
1, 4, 4.
215
14 4
xyz



(b)
3342
cos ;
1421 2 42
FG
FG

D"D

DD
not orthogonal

43.

22
,, 25
22
3, 3, 4 6 8
Fxyz x z
Fxz
F
ik
ik

D
D

22
,, 25
22
3, 3, 4 6 8
Gx yz y z
Gyz
G
jk
jk

D
D
(a)
608 484836 12443
068
FG
ijk
ijk ijkD 'D
Direction numbers:
4, 4, 3.
334
44 3
xyz



(b)

64 16
cos ;
10 10 25
FG
FG

D"D

DD
not orthogonal

44.
22
,,Fxyz x y z ,, 5 2 3 22Gx yz x y z


22 22
,,
xy
Fxyz
xy xy ijkD
,, 5 2 3Gx yzijkD


34
3, 4, 5
55
F
ijkD 3, 4, 5 5 2 3G ijkD
(a)
23426
35 45 1
55 5
523
FG
ijk
ijk
D'D

Direction numbers:
1, 17, 13

345
11713
xyz


Tangent line
(b)
85 8
cos
238 576
FG
FG

D"D

DD
Not orthogonal

45.
222
,, 14Fxyz x y z ,,Gx yz x y z

,, 2 2 2Fxyz x y zijkD ,,Gx yzijkD

3, 1, 2 6 2 4F ijkD 3, 1, 2G ijkD

229. Section 13.7 Tangent Planes and Normal Lines 229
© 2010 Brooks/Cole, Cengage Learning
(a) FGD'D +,62 4 2 10 8 2 5 4
111
ij k
ijkijk


Direction numbers:
1, 5, 4
Line:
312
15 4
xyz



(b)
cos 0 orthogonal
FG
FG
D"D

DD


46.
22
,,Fxyz x y z ,, 6 33Gx yz x y z

,, 2 2Fxyz x yijkD ,, 6Gx yzij kD

1, 2, 5 2 4F ijkD 1, 2, 5 6G ij kD
(a)
24 1 25 13 2
116
FG
ijk
ijk
D'D
Direction numbers:
25, 13, 2.
125
25 13 2
xy z



(b)
cos 0;
FG
FG
D"D

DD
orthogonal

47.
22
,, 3 2 15,2,2,5Fxyz x y z

,, 6 4Fxyz x yijkD

2, 2, 5 12 8F ijkD



2, 2, 5 1
cos
2, 2, 5 209
1
arccos 86.03
209
F
F k


D"

D






48.
3
, , 2 , 2, 2, 2Fxyz xy z

2
223Fyxzij kD

2, 2, 2 4 4 12F ij kD



2, 2, 2 12 311
cos
112, 2, 2 176
F
F k

D"

D


311
arccos 25.24
11







49.
22
,, ,1,2,3Fxyz x y z

,, 2 2Fxyz x yijkD

1, 2, 3 2 4ijkFD



1, 2, 3 1
cos
1, 2, 3 21
1
arccos 77.40
21
F
F k


D"

D



50.
22
,, 5,2,1,3Fxyz x y

,, 2 2Fxyz x yijD

2, 1, 3 4 2F ijD



2, 1, 3
cos 0
2, 1, 3
arccos 0 90
F
F k


D"

D



51.

22
,, 3 6
,, 2 2 6
Fxyz x y y z
Fxyz x y
ijk

D


20, 0xx

260, 3yy


22
3036312z

0, 3, 12 vertex of paraboloid

52.
22
,, 3 2 3 4 5Fxyz x y x y z

,, 6 3 4 4Fxyz x y ijkD

1
2
630,xx

440, 1yy


2
2
3111
22 4
3213415z


311
24
,1,

230. 230 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
53.
22
,, 2 2Fxyz x xy y x y z

,, 2 2 2 2Fxyz x y x y ijkD
220
220
xy
xy



22 2222
360 2yx x x
xx



2, 4yz
Point:
2, 2, 4

54.
22
,, 4 4 2 8 5 4Fxyz x xy y x y z

,, 8 4 8 4 4 5Fxyz x y x y ijkD
8480
4450
xy
xy


Adding,
31
42
12 3 0 ,xxy and
5
4
z
Point:

351
424
,,

55. ,, 5Fxyz xy z

,, 5 5Fxyz y x ijkD

50
50
y
x


0
xyz
Point:
0, 0, 0

56.
11
,,Fxyz xy z
xy




22
11
,,Fxyz y x
xy
ijk

D



2
1
y
x


4
21
1, 1, 3xxxyz
y


Point:
1, 1, 3

57.

222
, , 2 3 3, 1, 1, 0Fxyz x y z


,, 2
1, 1, 0 2
x
xFxyz x
F


,, 4
1, 1, 0 4
y
yFxyz y
F


,, 6
1, 1, 0 0
z
zFxyz z
F



2141000
24 6
23xyz
xy
xy





222
,, 6 10 14, 1,1,0Gx yz x y z x y


,, 2 6
1, 1, 0 4
x
xGxyz x
G


,, 2 10
1, 1, 0 8
y
yGxyz y
G


,, 2
1, 1, 0 0
z
zGxyz z
G



4181000
48120
23xyz
xy
xy



The tangent planes are the same.

58.
222
,, 8 12 4 42,2,3,3Fxyz x y z x y z


,, 2 8
2, 3, 3 4
x
xFxyz x
F


,, 2 12
2, 3, 3 6
y
yFxyz y
F


,, 2 4
2, 3, 3 2
z
zFxyz z
F



4263230
462200
23 10
xyz
xyz
xyz






22
,, 2 7,2,3,3Gx yz x y z


,, 2
2, 3, 3 4
x
xGxyz x
G


,, 2
2, 3, 3 6
y
yGxyz y
G


,, 2
,, 2
z
zGxyz
Gxyz



4263230
462200
23 10xyz
xyz
xyz



The tangent planes are the same.

231. Section 13.7 Tangent Planes and Normal Lines 231
© 2010 Brooks/Cole, Cengage Learning
59. (a)
2
,, 2 ,Fxyz xy z 1, 1, 2 2 2 0F


22
,, 8 5 8 13,Gx yz x y z 1, 1, 2 8 5 16 13 0G
So,
1, 1, 2 lies on both surfaces.
(b)
2
24 ,Fy xyijkD 1, 1, 2 2 4F ijkD
16 10 8 ,
Gxy ijkD 1, 1, 2 16 10 8G ijkD

216 4 10 1 8 0FGD"D
The tangent planes are perpendicular at
1, 1, 2 .

60. (a)
222
,, 2 4 4 12Fxyz x y z x y z

1, 2, 1 0F



22 2
,, 4 16 24
1, 2, 1 0
Gx yz x y z
G



So,
1, 2, 1lies on both surfaces.
(b)
22 24 24ijkFx y zD

1, 2, 1 4 8 2ijkFD
8232
Gx y zij kD

1, 2, 1 8 4 32G ij kD
32 32 64 0
FGD"D
The planes are perpendicular at
1, 2, 1 .

61.
222
,, 4 9Fxyz x y z
282
FxyzijkD
This normal vector is parallel to the line with direction number
4, 8, 2.
So, 2 4 2
88
22
x tx t
ytyt
ztzt




222 222
4944901xyz ttt t
There are two points on the ellipse where the tangent plane is perpendicular to the line:


2, 1, 1 1
2, 1, 1 1
t
t



62.
222
,, 4 1Fxyz x y z
282
FxyzijkD
The normal to the plane, 4
ni jk
must be parallel to .
FD
So,
2
2
84
2
2
2
t
xt x
t
yty
t
ztz




22
222 2 2
411.
44
tt
xyz t t t
Two points:
111
,,
222




1t and
111
,,
222




1t

232. 232 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
63. 000 0 000 0 000 0,, ,, ,, 0
xy zFxyz x x Fxyz y y Fxyz z z

Theorem 13.13

64. For a sphere, the common object is the center of the sphere. For a right circular cylinder, the common object is the axis of the
cylinder.

65. Answers will vary.

66. (a)
222
0, 5, 13, 12xyz


222
,,Fxyz x y z


,, 2
5, 13, 12 10
x
xFxyz x
F
π
θπ
,, 2
5, 13, 12 26
y
yFxyz y
F
πθ
θπθ
,, 2
,, 24
z
zFxyz z
Fxyz
π
πθ
Direction numbers:
5, 13, 12θθ
Plane:
5 5 13 13 12 12 0
51312 0xy z
xyz
θθπ

(b) Line:
51312
51312
xy z
ππ
θθ


67.

22
4
,,22,03
11
xy
zfxy x y
xy



(a) Let


22
4
,,
11
xy
Fxyz z
xy
πθ









22
22 2 2
22 2222
222222
41 41412412
,,
11
111111
ijkijk
yx xyyx x xy y
Fxyz
yx
xyyxxy

θθ

D




1, 1, 1F kDπθ
Direction numbers:
0, 0, 1θ
Line:
1, 1, 1
x yz tπππθ
Tangent plane:
0101110 1xyz z
(b)


2
4346
1, 2, 0
525 25
F ijkjk
θθ
D



Line:
64
1, 2 ,
25 5
x ytzt
Plane:


64
01 21 0
25 5
61225200
625320
xyz
yz
yz




θθ θ π
θθπ

(c)
x
y
−1
3
2
1
z
x
y
1
2
2
3
−2
−1
z

233. Section 13.7 Tangent Planes and Normal Lines 233
© 2010 Brooks/Cole, Cengage Learning
68. (a)
sin
,,33,02
y
fxy x y
x

Let

sin
,,
y
Fxyz z
x




2
sin cos
,,
yy
Fxyz
xx
ijk

D


11
2, ,
22 4
F ik

D



Direction numbers:
1
,0, 1
4
or 1, 0, 4
Line:
1
2, , 4
22
x ty z t


Tangent plane:

1
120 4 0 440
22
xy z xz





(b)
23 3 9
,,
322 4
F ik

D



Direction numbers:
9
,0, 1
4

or 9, 0, 4
Line:
233
9, , 4
322
x ty z t


Tangent plane:
23 3
90 4 094120
322
xy z xz





(c)

69.

2
2
,6 ,,2
4
yfxy x gxy x y
(a)

2
2
,, 6
4
y
Fxyz z x ,, 2Gx yz z x y


1
,, 2
2
Fxyz x y ijkD ,, 2Gx yz ijkD

1, 2, 4 2F ijkD 1, 2, 4 2G ijkD
The cross product of these gradients is parallel to the curve of intersection.


1,2,4 1,2,4 2 1 1 2 4
211
ijk
ij
FGD'D

Using direction numbers
1, 2, 0, you get 1, 22, 4.xty tz

411 4
cos 48.2
666FG
FG

D"D

DD

(b)
x
y
−3
3
2
π
3
z
x
y
6
8
8
(1, 2, 4)
z

234. 234 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
70. (a)

22
22
,16 24
2
,1364
2
fxy x y x ygxy x y x y



(b)




22 22
22 22
22
22
22
,,
1
16 2 4 1 3 6 4
2
32 2 2 4 8 1 3 6 4
2313 12
21423 4 4
1423 2fxy gxy
xyxy xyxy
xyxy xyxy
xx y y
xx yy
xy







To find points of intersection, let 1.
xThen



2
2
3242
214
214y
y
y




1, 2 14 2 ,f jD 1, 2 14 1 2 .g jD The normals to f and g at this point are 2jk and
12 ,jk which are orthogonal.
Similarly, 1, 2 14 2f jD and 1, 2 14 1 2g jD and the normals are 2jkand
12 ,jk which are also orthogonal.
(c) No, showing that the surfaces are orthogonal at 2 points does not imply that they are orthogonal at every point of
intersection.

71.



222
222
2
2
2
,, 1
2
,,
2
,,
2
,,
x
y
z
xyz
Fxyz
abc
x
Fxyz
a
y
Fxyz
b
z
Fxyz
c




Plane:

000
000
22 2
222
000 000
22222222 2
0
1xyz
xx yy zz
ab c
xx yy zz x y z
abcabc




72.
222
222
,, 1
xyz
Fxyz
abc



2
2
,,
x
x
Fxyz
a


2
2
,,
y
y
Fxyz
b



2
2
,,
z
z
Fxyz
c

Plane:

000
000
22
2
222
000 000
22222222 2
0
1xyz
xx yy zz
ab c
xx yy zz x y z
abcabc



73.
22 2 2 2
,,Fxyz ax by z


2
,, 2
xFxyz ax


2
,, 2
yFxyz by

,, 2
zFxyz z
Plane:

22
00 0 000
22 22222
000 000
22 20
0ax x x by y y z z z
axx byy zz ax by z


So, the plane passes through the origin.
x
y
5
5
5
f
g
z

235. Section 13.7 Tangent Planes and Normal Lines 235
© 2010 Brooks/Cole, Cengage Learning
74.




2
,,
,,
1
,,
,, 1
x
y
x
y
zxf
x
y
Fxyz xf z
x
yyyyyy
Fxyz f xf f f
x xx x xx
yy
Fxyz xf f
xx x
Fxyz

π



πθ







ππ


πθ

Tangent plane at
000,, :xyz


000 0
000
000 0
000 0 0 0 0 0
00 0 0
000 0 0 0 0 0
000
000
0
0
yyy y
ffxxfyyzz
xxx x
yyy y y y y y
ffxxfyfyfyfzxf
xxx x x x x x
yyy
ff
xxx







θ

0
0
0
y
xf yz
x




So, the plane passes through the origin
, , 0, 0, 0 .xyzπ

75. ,
xy
fxy e
θ
π

,,
xy
x
fxy e
θ
π ,
xy
yfxy e
θ
πθ

,,
xy
xx
fxy e
θ
π ,,
xy
yy
fxy e
θ
π ,
xy
xyfxy e
θ
πθ
(a)
1, 0,0 0,0 0,0 1
xyPxy f f x f y x y
(b)


2222
21111
2222
, 0,0 0,0 0,0 0,0 0,0 0,0 1
x y xx xy yyPxy f f x f y f x f xy f y x y x xy y
(c) If 0,xπ

2
21
2
0, 1 .Py y y This is the second-degree Taylor polynomial for .
y
e
θ

If 0,yπ

2
21
2
,0 1 .Px x x This is the second-degree Taylor polynomial for .
x
e
(d) (e)

76. ,cos
fxy x y

,sin,
x
fxy x y ,sin
yfxy x y

,cos,
xx
fxy x y ,cos,
yyfxy x y ,cos
xyfxy x y
(a)
1, 0,0 0,0 0,0 1
xyPxy f f x f y
(b)


22
2
2211
22
11
22
, 0, 0 0, 0 0, 0 0, 0 0, 0 0, 0
1
x y xx xy yyPxy f f x f y f x f xy f y
xxy y
πθ θ θ
(c) If 0,xπ

2
21
2
0, 1 .Py yπθ This is the second-degree Taylor polynomial for cos y.
If 0,yπ

2
21
2
,0 1 .Px xπθ This is the second-degree Taylor polynomial for cos x.
x y
,
fxy 1,Pxy 2,Pxy
0 0 1 1 1
0 0.1 0.9048 0.9000 0.9050
0.2 0.1 1.1052 1.1000 1.1050
0.2 0.5 0.7408 0.7000 0.7450
1 0.5 1.6487 1.5000 1.6250
z
f
P
1
P
2
y
x
4
2
2
1
−2
−2
−4
−2

236. 236 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
(d) (e)
77. Given
,,zfxy then:







000 00 00
000
000
22 2
00 00
22
00 00
,, , 0
,, , ,
,,
cos
,,
1
,,1
1
,,1
ijk
k
k
xy
xy
xy
Fxyz f xy z
Fx y z f x y f x y
Fx y z
Fx y z
fxy fxy
fxy fxy


D
D"

D








78. Given ,,wFxyz where F is differentiable at

000,,xyz and 000,, ,Fx y z0D#
the level surface of F at
000,,xyz is of the form ,,Fxyz Cfor some constant C. Let

,, ,, 0.Gxyz Fxyz C
Then
000 000,, ,,Gxyz FxyzDD where 000,,Gx y zD is normal to ,, 0Fxyz C at 000,, .xyz So,
000,Fx yzD is normal to the level surface through 000,, .xyz
Section 13.8 Extrema of Functions of Two Variables
1.
22
,130gxy x y /
Relative minimum:1, 3, 0

Check:
210 1
230 3
x
ygx x
gy y



2, 2, 0, 2 2 0 4 0
xx yy xygggd
At critical point
1, 3 , 0dand 0
xxg relative
minimum at
1, 3, 0 .
2.
22
,325gxy s x y
Relative maximum:3, 2, 5

Check:

230 3
220 2
2, 2, 0
22 0 4 0
x
y
xx yy xygx x
gy y
ggg
d





At critical point
3, 2 , 0d and 0
xxg relative
maximum at
3, 2, 5 .
x y
5
5
5
z
x y ,fxy 1,Pxy 2,Pxy
0 0 1 1 1
0 0.1 0.9950 1 0.9950
0.2 0.1 0.9553 1 0.9950
0.2 0.5 0.7648 1 0.7550
1 0.5 0.0707 1 0.1250

237. Section 13.8 Extrema of Functions of Two Variables 237
© 2010 Brooks/Cole, Cengage Learning
3.
22
,11fxy x y/
Relative minimum:0, 0, 1

Check:



22
22
2
32
22
2
32
22
32
22
00
1
00
1
1
1
1
1
1
x
y
xx
yy
xy
x
fx
xy
y
fy
xy
y
f
xy
x
f
xy
xy
f
xy














At the critical point0, 0 , 0
xxf and

2
0.
xx yy xyff f
So,
0, 0, 1is a relative minimum.
4.
2
2
,252 5fxy x y
Relative maximum: 2, 0, 5

Check:


2
2
2
2
2
02
25 2
00
25 2
x
y
x
fx
xy
y
fy
xy












2
32
2
2
2
32
2
2
32
2
2
25
25 2
25 2
25 2
2
25 2
xx
yy
xy
y
f
xy
x
f
xy
yx
f
xy














At the critical point
2, 0 , 0
xxf
and

2
0.
xx yy xyff f
So,
2, 0, 5 is a relative maximum.
5.
22
22
,2661344fxy x y x y x y /
Relative minimum:1, 3, 4

Check: 220 1
260 3
2, 2, 0
x
y
xx yy xyfx x
fy y
fff



At the critical point
1, 3 , 0
xxf and
2
0.
xx yy xyff f So, 1, 3, 4 is a relative minimum.
6.

22
22
22
,101264
10 25 12 36 25 36 64 5 6 3 3
fxy x y x y
xx yy x y



Relative maximum:5, 6, 3

Check:

2100 5
2120 6
2, 2, 0, 2 2 0 4 0
x
y
xx yy xyfx x
fy y
fffd




At critical point5, 6 , 0dand 0
xxf relative maximum at 5, 6, 3 .
7.
22
,326416fxy x y x y


660
1, 1
440
6, 4, 0, 6 4 0 24 0.
x
y
xx yy xyfx
xy
fy
fffd
<
>

?



At the critical point1, 1 , 0d

and 0 1,1,11
xxf is a relative minimum.
8.
22
,32345fxy x y x y
630
xfx when
1
2
.x
440
yfy when 1.y
6, 4, 0
xx yy xyfff
At the critical point

1
2
,1, 0
xxf
and

2
0.
xx yy xyff f
So,

311
24
,1, is relative maximum.

238. 238 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
9.
22
, 5 10 10 28fxy x y x y

2100
5, 1
10 10 0
x
yfx
xy
fy
<
>

?


2, 10, 0, 2 10 0.
xx yy xyff fd

At the critical number5, 1 , 0d
and
05,1,2
xxf is a relative maximum.
10.
22
,22 23fxy x xy y x
4220
Solving simultaneously

22 0 yields 1 and 1.
x
yfxy
fxy xy
<
>

?

4, 2, 2
xx yy xyfff

At the critical point1, 1 , 0
xxf
and

2
0.
xx yy xyff f
So,

1, 1, 4 is a relative minimum.
11.
22 1
2
,2
fxy z x xy y x y

220 Solving simultaneously
yields 3, 410
x
yfxy
xyfxy
<
>

?


2, 1, 1, 2 1 1 1 0.
xx yy xyfffd
At the critical point
3, 4 , 0d
and
03,4,5
xxf is a relative minimum.
12.
22
,54 1610fxy x xy y x

10 4 16 0 Solving simultaneously
yields 8 and 16.42 0
x
yfxy
xyfxy
<
>

?


10, 2, 4
xx yy xyfff

At the critical point 8, 16 , 0
xxf
and

2
0.
xx yy xyff f
So,
8, 16, 74 is a relative maximum.
13.
22
,fxy x y

22
22
0
0
0
x
y
x
f
xy
xy
y
f
xy
<

=

=
>
=

=

?


Because,0fxy/for all ,xyand
0,0 0, 0,0,0f is a relative minimum.
14.
13
22
,2hx y x y



23
22
23
22
2
0
3
0, 0
2
0
3
x
y
x
h
xy
xy
y
h
xy
<

=
==
>
=

=

=?


Because,2hx y/for all ,,xy0, 0, 2is a relative
minimum.
15.
,4gxy x y
0, 0 is the only critical point. Because,4gxyfor
all
, , 0, 0, 4xy is a relative maximum.
16.
,2fxy x y
Since
,2fxy/for all ,,
xythe relative minima
offconsist of all points,xysatisfying 0.xy
17.
22
4
1
x
z
xy




Relative minimum:1, 0, 2
Relative maximum:
1, 0, 2
18.
3222
,3331fxy y yx y x
Relative maximum:0, 0, 1
Saddle points:


0, 2, 3 , 3, 1, 3
19.
22
221
4
xy
zx ye


Relative minimum:0, 0, 0
Relative maxima:
0, 1, 4
Saddle points:
1, 0, 1
20.
xy
ze
Saddle point:0, 0, 1

x
y
5
−4
−4
4
4
z
z
x
y
3
3
20
40
−4
44
−4
5
6
yx
z
yx
33
100
z

239. Section 13.8 Extrema of Functions of Two Variables 239
© 2010 Brooks/Cole, Cengage Learning
21.
22
,8080hx y x y x y

80 2 0
40
80 2 0
x
yhx
xy
hy
<
>

?



2, 2, 0,
22 0 4 0
xx yy xyhhh
d



At the critical point
40, 40 , 0d and
0 40, 40, 3200
xxh is a relative maximum.
22.
22
,
gxy x y x y

12210
210 12
x
y xgx
gy y
<
>
?


2, 2, 0, 2 2 0 4 0
xx yy xygg gd

At the critical point
12, 12, 0d
12, 12,0 is a saddle point.
23. ,gxy xy
0 and 0
x
ygy
xy
gx
<
>

?


0, 0, 1
xx yy xyggg

At the critical point
2
0, 0 , 0.
xx yy xygg g
So,
0, 0, 0 is a saddle point.
24.
22
,3hx y x xy y

2 3 0 Solving simultaneously
32 0 yields 0 and 0.
x
yhxy
hxy xy
<
>
?


2, 2, 3
xx yy xyhh h

At the critical point
2
0, 0 , 0.
xx yy xyhh h
So,0, 0, 0 is a saddle point.
25.
22
,3
fxy x xy y x y
230
210
x
yfxy
fxy



Solving simultaneously yields 1, 1.xy


2
2, 2, 1
22 1 5 0
xx yy xyff f
d



At the critical point
1, 1 , 0 1, 1, 1d is a saddle point.
26.

421
,2 1
2
fxy xy x y



3
3
22 Solving by substitution yields 3 critical points:
0, 0 , 1, 1 , 1, 122
x
yfyx
fxy
<
=
>
=?


22
6, 6, 2
xx yy xyfxfyf
At

2
0, 0 , 0 0, 0, 1
xx yy xyff f saddle point.
At

2
1, 1 , 0
xx yy xyff f and 01,1,2
xxf relative maximum.
At

2
1, 1 , 0
xx yy xyff f and 01,1,2
xxf relative maximum.
27. ,sin
x
fxy e y



sin 0 Because 0 for all and sin and cos are never
cos 0 both zero for a given value of , there are no critical points.
x x
x
x
y
fey ex yy
fe y y


< =
>
=?

240. 240 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
28.
22
2211
,
2
xy
fxy x y e









22
32 1
22
231
22 3 0
26
Solving yields the critical points 0, 0 , 0, , , 0 .
22
22 0
xy
x
xy
y
fxxyxe
fxyyye

<

=
>

=
?





22
422 22 1
22
422221
22
33 1
44 1223
4 4 281
442
xy
xx
xy
yy
xy
xy
fxxyxye
fyxyxye
fxyxyxye







At the critical point
2
0, 0 , 0.
xx yy xyff f So,
0, 0, 2eis a saddle point. At the critical
points0, 2 2 , 0
xxf and
2
0.
xx yy xyff f So, 0, 2 2, e are relative maxima. At the critical
points62,0, 0
xxf and
2
0.
xx yy xyff f So, 62,0, ee are relative minima.
29.

4
22
0. 0
xy
zz
xy

/

if 0.xy#
Relative minimum at all points,, 0.xx x#

30.


2
22
22
0. 0
xy
zz
xy

/
if
22
0.xy#
Relative minima at all points
,
xxand,, 0.xxx#

31.

2
2
94 6 0
xx yy xyff f
Insufficient information.
32. 0
xxfand
2
2
38 2 0
xx yy xyff f

fhas a relative maximum at00,xy
33.

2
2
96 10 0
xx yy xyff f

fhas a saddle point at00,.xy

34. 0
xxfand
2
2
25 8 10 0
xx yy xyff f

fhas a relative minimum at00,xy
35.
222
28 16 0
xx yy xy xy xydff f f f

2
16 4 4
xy xyff
36.

2
0
xx yy xydff f if
xxfand
yyfhave opposite
signs. So,
,, ,ab f abis a saddle point. For example,
consider

22
,
fxy x y and,0,0.ab
37.
33
,fxy x y
(a)
2
2
30
0
30
x
yfx
xy
fy
<
=
>
=?

Critical point:
0, 0
(b) 6 , 6 , 0
xx yy xyfxfyf


2
At 0, 0 , 0.
xx yy xyff f

0, 0, 0 is a saddle point.
(c) Test fails at
0, 0.
(d)
z
yx
3
3
40
60
x y5
5
2
z
x
y
−2
−2
−2
2
2
1
z
2
Saddle point
(0, 0, 0)

241. Section 13.8 Extrema of Functions of Two Variables 241
© 2010 Brooks/Cole, Cengage Learning
38.
33 2 2
,69122719fxy x y x y x y
(a)
2
2
312120 Solving yields
2 and 3.318270
x
yfx x
xyfy y
<
=
>
=?

(b) 6 12, 6 18, 0
xx yy xyfxfyf
At

2
2, 3 , 0.
xx yy xyff f

2, 3, 0is a saddle point.
(c) Test fails at
2, 3 .
(d)
39.
22
,140fxy x y /
(a)



2
2
critical points:21 4 0
1, and , 4
21 40
x
yfxy
ab
fxy
<
=
>

=
?

(b)



2
2
24
21
41 4
xx
yy
xyfy
fx
fxy




At both
1,aand
2
,4, 0.
xx yy xybfff
Because
,0,fxy/there are absolute minima
at
1, , 0aand ,4,0.b
(c) Test fails at
1,aand ,4.b
(d)
40.
22
,120fxy x y/
(a)


22
22
1
0
12 Solving yields
1 and 2.2
0
12
x
y
x
f
xy
xyy
f
xy
<

=
==
>

=

=

=?

(b)






2
32
22
2
32
22
32
22
2
12
1
12
12
12
xx
yy
xy
y
f
xy
x
f
xy
xy
f
xy













At

2
1, 2 ,
xxyy xyfff is undefined.

1, 2, 0 is an absolute minimum.
(c) Test fails at
1, 2 .
(d)
41.
23 23
,0fxy x y/
(a)

13
13
2
and are undefined
3
at 0 and 0.
2
Critical point: 0, 0
3
xy
x
yfff
x
xy
f
y
<

=
=
>
=

=
?

(b)
43 43
22
,,0
99
xx yy xyfff
xy


At

2
0, 0 ,
xx yy xy
fff is undefined.

0, 0, 0is an absolute minimum.
(c) Test fails at
0, 0 .
(d)
x
y
2
2
4
−4
−2
(2, −3, 0)
z
4
2
6
x
y
−4
−2
4
z
6
Absolute
minimum
(1, a, 0)
Absolute
minimum
(b, −4, 0)
x
y
22
4
−4
−6(1, −2, 0)
z
4
4
2
6
x
y
2
4
6
z
6
Absolute
minimum (0, 0, 0)

242. 242 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
42.
23
22
,0fxy x y /
(a)


13
22
13
22
4
3 and are undefined at 0, 0.
Critical Point: 0, 04
3
x
xy
y
x
f
xy ff xy
y
f
xy
<

=
=
>
=
=

?

(b)





22
43
22
22
43
22
43
22
43
9
43
9
8
9
xx
yy
xy
x y
f
xy
xy
f
xy
xy
f
xy










At

2
0, 0 ,
xx yy xy
fff is undefined.

0, 0, 0 is an absolute minimum.
(c) Test fails at
0, 0 .
(d)
43.
22
2
,, 3 1 0fxyz x y z /



20
230 Solving yields the critical point 0, 3, 1 .
210
x
y
zfx
fy
fz
<
=
>
=

?


Absolute minimum: 0 at0, 3, 1
44.

2
,, 9 1 2 9fxyz xy z


The absolute maximum value of
fis 9, and realized at all points where120.xy z
So, the critical points are of the form
0, , , , 1, , , ,ab c d e f z
where
,,, ,,abcde f are real numbers.
45.
9
:
2
,45,,:14,02fxy x xy R xy x y

24 0
0
40
x
yfxy
xy
fx
<
>

?
(not in region R)
Along

2
0, 1 4: 5, 1, 0 6, 4, 0 21.yxfxf f
Along

2
2, 1 4: 8 5, 2 8 0
1, 2 2, 4, 2 11.yxfxxfx
ff


Along
1, 0 2: 4 6, 1, 0 6, 1, 2 2.xyfyf f
Along
4, 0 2: 21 16 , 4, 0 21, 4, 2 11.xyf yf f
So, the maximum is
4, 0, 21 and the minimum is4, 2, 11 .
2
3
3
1
2
4
4
−4
−2
−3
z
y
x
(0, 0, 0)
5
−1 1234
−1
1
2
3
4
y
x

243. Section 13.8 Extrema of Functions of Two Variables 243
© 2010 Brooks/Cole, Cengage Learning
46. 9 :
2
,,,:2,1fxy x xyR xy x y

20
0
0
x
yfxy
xy
fx
<
>

?


0, 0 0f

Along
2 1
2
1, 2 2, , 2 1 0 .yxfxxfx x

Thus,

11
24
2, 1 2, , 1ff and2, 1 6.f
Along
2 1
2
1, 2 2, , 2 1 0 .yxfxxfxx
Thus,


11
24
2, 1 6, , 1 , 2, 1 2.fff
Along
2, 1 1, 4 2 2 0.xyfyf #
Along
2, 1 1, 4 2 2 0.xyfyf #
So, the maxima are
2, 1 6f and 2, 1 6f and the minima are

11
24
,1f and

11
24
,1 .f
47. ,1232
fxy x y has no critical points. On the line 1, 0 1,yx x
,()12321510fxy fx x x x

and the maximum is10, the minimum is 5. On the line24,1 2,yx x

,1232244fxy fx x x x
and the maximum is6,the minimum is 5. On the line
1
2
1, 0 2,yx x



1
2
,12321210fxy fx x x x
and the maximum is10, the minimum is 6.
Absolute maximum:10 at
0, 1
Absolute minimum: 5 at
1, 2
48.
2
,2
fxy x y

42 0 2
22 0 2
x
y
f xy x y
f xy x y




On the line 1, 0 1,yx x


2 2
,211fxy fx x x x

and the maximum is1,the minimum is 0. On the line
1
2
1, 0 2,yx x



2 2
51
22
,211fxy fx x x x
and the maximum is16, the minimum is 0. On the line
24,1 2,yx x


2 2
,22444fxy fx x x x
and the maximum is16, the minimum is 0.
Absolute maximum:16 at
2, 0
Absolute Minimum: 0 at
1, 2and along the line 2 .yx
2
−1
−2
1
x
y
2
2
31
3
1
x
yx=+ 1
yx= 2 + 4−
(0, 1)
(1, 2)
(2, 0)
1
2
yx=+ 1−
y
x
1
2
312
3
(0, 1)
(1, 2)
(2, 0)
yx= 2
y

244. 244 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
49.
22
,324fxyxyy

60 0
0, 1 2
440 1
x
yfx x
f
fy y
<
>

?


On the line4, 2 2,yx


22
, 3 32 16 3 16fxy fx x x
and the maximum is 28, the minimum is16. On the curve
2
,2 2,yx x


2
22 24222
,324221fxy fx x x x x x x x
and the maximum is 28, the minimum is
1
8
.
Absolute maximum: 28 at
2, 4
Absolute minimum:
2at0, 1
50.
2
,22
fxy x xy y

22 0 1
1, 1 1
22 0 1
x
yfyy
f
fyx yxx
<
>

?


On the line1, 1 1,yx

,2211.fxy fx x x
On the curve
2
,1 1yx x


2
22 43
,22 22
fxy fx x xx x x x x
and the maximum is1, the minimum is
11
16
.
Absolute maximum:
1at1, 1 and on 1y
Absolute minimum:
11
16
0.6875 at

11
24
,
51.
9 :
22
,2,,:2,1fxy x xy y R xy x y

22 0
22 0
x
yfxy
yx
fxy
<
>

?



222
,20fx x x x x
Along
1, 2 2,yx


2
2 1, 2 2 0 1, 2, 1 1, 1, 1 0, 2, 1 9.fx x f x x f f f
Along
1, 2 2,yx


2
21, 220 1,2,19,1,10,2,11.fx x f x x f f f
Along
2
2, 1 1, 4 4 , 2 4 0.xyfyyfy #
Along
2
2, 1 1, 4 4 , 2 4 0.xyfyyfy #
So, the maxima are
2, 1 9f and 2, 1 9,f and the minima are ,0,11.fx x x
x
2
1
3
12−1−2
( 2, 4)− (2, 4)
y
x
2
1−1
( 1, 1)− (1, 1)
y
2
−1
−2
1
x
y

245. Section 13.8 Extrema of Functions of Two Variables 245
© 2010 Brooks/Cole, Cengage Learning
52. 9:
22 22
,2,,:8fxy x xy y R xyx y

22 0
22 0
x
yfxy
yx
fxy
<
>

?



222
,20fx x x x x

On the boundary
22
8,xy we have
22
8yx and
2
8.yx Thus,


222 2
2
12 12
22 2
2
28 8 828
16 4
8228 .
8
fx x x x x x
x
fxxx
x






Then, 0fimplies
2
16 4
x or 2.x

2, 2 2, 2 16ff and 2, 2 2, 2 0ff
So, the maxima are
2, 2 16f and 2, 2 16,f and the minima are
,0,2.fx x x
53.

9:
22
4
,,,:01,01
11
xy
fxy R xy x y
xy






2
2
22
41
01
11
x
xy
fx
yx



or 0y




2
2
22
41
0
11
y
yx
fx
xy



or 1y


For 0, 0,xy also, and0, 0 0.f

For 1, 1, 1, 1 1.xyf
The absolute maximum is
11,1.f
The absolute minimum is
00,0.In fact, 0, ,00.ffyfx
54.

9:
22
224
, , , : 0, 0, 1
11
xy
fxy R xyx y x y
xy
//





2
2
22
41
01
11
x
xy
fx
yx



or 0y



2
2
22
41
01
11
y
yx
fy
xy



or 0x
For
0, 0,xy also, and0, 0 0.f
For 1xand 1,ythe point
1, 1is outside .
R
For


2
22 2
24
41
1, , , 1 ,
2
x x
xy fxy fx x
xx



and the maximum occurs at
22
,.
22
xy
Absolute maximum is
822
,.
922
f





The absolute minimum is
00,0.In fact,0, ,00ffyfx

2
4
−2
−4
−2−424
x
y
x
1
1
R
y
x
1
1
R
y

246. 246 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
55. In this case, the pointAwill be a saddle point. The
function could be
,.
fxy xy
56. AandBare relative extrema. Cand Dare saddle points.
57.
No extrema
58.
Extrema at all

,xy
59.
Saddle point

60.

22 22
,,,
fxy x y gxy x y
(a)
20, 20 0,0
xyfx f y is a critical
point.

20, 20 0,0
xygxgy is a critical
point.
(b)

2, 2, 0
2 2 0 0 0, 0 is a saddle point.
xx yy xyff f
d





2, 2, 0
2 2 0 0 0, 0 is a relative minimum.
xx yy xyggg
d



61. False.
Let

,1 .fxy x y

0, 0, 1is a relative maximum, but 0, 0
xf and
0, 0
yf do not exist.
62. False. Consider
22
,.
fxy x y
Then
0, 0 0, 0 0,
xyff but 0, 0, 0 is a saddle
point.
63. False. Let
22
,
fxy xy (See Example 4 on page 958).
64. False.
Let
422
,2.fxy x x y

Relative minima:1, 0, 1

Saddle point:0, 0, 0
Section 13.9 Applications of Extrema of Functions of Two Variables
1. A point on the plane is given by

,, ,,3 .
xyz xy x y The square
of the distance from

0, 0, 0 to this point is




2
22
3.
223
223
x
y
Sx y xy
Sx xy
Sy xy




From the equations
0
xSand 0
ySwe obtain

42 6
24 6.
xy
xy



Solving simultaneously, we have
1, 1, 1.xy z
So, the distance is

2
22
111 3.

2. A point on the plane is given by

,, ,,3 .
xyz xy x y The square of
the distance from 1, 2, 3 to this point is



22 2
222
123 3
12 .
Sx y xy
xy yx





212
222
x
ySx yx
Sy yx



From the equation
0
xSand 0
ySwe obtain

42 2
24 4.
xy
xy


Solving simultaneously, we have
43, 53, 103.xyz
So, the distance is

22 2
45 5413
12 .
33 333





x
y
2
30
45
60
75
2
z
x
y
2
3
3
4
4
4
z
x
y
6
7
−3
36
z

247. Section 13.9 Applications of Extrema of Functions of Two Variables 247
© 2010 Brooks/Cole, Cengage Learning
3. A point on the surface is given by,, ,, 1 2 2 .xyz xy x y The square of the distance
from

2, 2, 0 to a point on the surface is given by



2
22 22
221220 22122.Sx y xy x y xy


222
222
x
ySx
Sy


From the equations 0
xSand 0,
ySwe obtain
220
1, 5.
220
x
xy z
y
<
>
?

So, the distance is


2
22
12 12 5 7.
4. A point on the surface is given by ,, ,, 1 2 2 .xyz xy x y The square
of the distance from
0, 0, 2 to a point on the surface is given by




2
22
2122.
22 1 2 2
4
222
12 2 12 2
22 1 2 2
4
222
12 2 12 2
x
y
Sx y x y
xy
Sx x
xy xy
xy
Sy y
xy xy








From the equation 0
xSand 0,
ySwe obtain .
xy
So,
4
220.
14
x
x



Using a graphing utility, 0.3221.xy
The minimum distance is

12
2
22
2 1 2 2 0.667.Sxy xy





5. Let, , and
xyz be the numbers. Because 27,xyz

22
27
.
27
.
27 27
10,10.
xy
Z
xy
Sxyzxy
xy
SS
xy xy




2
2
27
3
27
xy
xy
xy
<=
>
=?

So, 3.xyz
6. Because 32, 32 .
xyz z xy So,



22223
2232
22 2
32
32 2 32 2 0
64 2 3 64 2 3 0.
x
y
P xy z xy x y xy
Pyxyyy xy
Pxyxyxyyxxxy




Ignoring the solution 0yand substituting
32 2yx into 0,
yPwe have



2
64 2 3 32 2 0
480.
xx x x
xx



So,
8, 16,xy and 8.z
7. Let,,and
xyz be the numbers and let

222
.Sx y z Because
30,xyz we have


2
22
30Sx y xy


2230 10 230
2230 10 2 30.
x
ySx xy xy
Sy xy xy
< =
>
=?

Solving simultaneously yields 10,x
10,yand 10.z
8. Let, , and
xyz be the numbers. Because
1, 1 .xyz z xy

222 22
22 1
Sx y z x yxy



32 23
22
20,20
xySx Sy
xy xy




32
42 24
23
1
1
xxy
xyxy xy
yxy
<
=
>
=
?
So, 1.xyz

248. 248 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
9. The volume is
668.25
668.25 .xyz z
xy






2
2
668.25 668.25
0.06 2 2 0.11 0.12 0.11
80.19 80.19
0.11
80.19
0.11 0
8.19
0.11 0
x
y
Cyzxzxy xy
xy
Cxy
xy
Cy
x
Cx
y









Solving simultaneously, 9xy and 8.25.z
Minimum cost:

80.19 80.19
0.11 $26.73
99
xy

10. Let x, y, and z be the length, width, and height, respectively. Then
01.5 2 2Cxyyzxz and

01.5
.
2
Cxy
z
xy




The volume is given by






22
0
22
0
2
22
0
2
1.5
2
236
4
236
.
4
x
y
Cxy xy
Vxyz
xy
yC x xy
V
xy
xCyxy
V
xy










In solving the system 0
xVand 0,
yVwe note by the symmetry of the equations that .yx
Substituting
yx into 0
xVyields


22
0
2
0000
2
29 11 1
0, 2 9 , 2 , 2 , and 2 .
16 3 3 4
xC x
Cxx Cy C z C
x



11. Let .abc k Then

2244 4
33 3
abc
V abkab kababab





2 2
2 24
20 20
3
4
20 20
3
a
bVkbabb kb ab b
Vkaaab ka a ab

<

=
=
>
=

=?

Solving this system simultaneously yieldsaband substitution yields
3.bkSo, the solution is 3.abck
12. Consider the sphere given by
222 2
xyz r and let a vertex of the rectangular box be
22 2
,, .xyr x y
Then the volume is given by





22 2 22 2
22 2 2 2 2
22 2 22 2
22 2 22 2
22 2 22 2
222 8
8
820
8
820.
x
y
Vxyrxy xyrxy
xy
Vxy yrxy rxy
rxy rxy
yx
Vxy xrxy rxy
rxy rxy














Solving the system

22 2
222
2
2
xyr
x yr



yields the solution 3.xyzr
y
z
x

249. Section 13.9 Applications of Extrema of Functions of Two Variables 249
© 2010 Brooks/Cole, Cengage Learning
13. Let , ,xyandzbe the length, width, and height, respectively and let
0Vbe the given volume.
Then
0V xyz and
0.zVxyThe surface area is

00
0 2
0
2
0 2
02
222 2
020
20 0.
x
y
VV
Sxyyzxz xy
xy
V
xy VSy
x
V
Sx xy V
y



<

=
=
>

=

=
?

Solving simultaneously yields
33 3
00 0
,,and.
x Vy V z V
14.

22
2221
30 2 30 2 2 cos sin 30 sin 2 sin sin cos
2
30 sin 4 sin 2 sin cos 0
30 cos 2 cos 2 cos 1 0
Axxxxxxx
A
xx
x
A
xx





C

C
C

C

From
0
A
x
C

Cwe have
215
15 2 cos 0 cos .
x
xx
x




From
0
A

C

C
we obtain

2
22
2
2
2
215 215 215
30 2 2 1 0
3021522152215 0
3300
10.
xx x
xxx
xx x
xxx xx
xx
x










Then
1
cos 60 .
2

15.
22
12 1 2 12 1 2
, 5 8 2 42 102
Rxx x x xx x x

12 121
21 12210 2 42 0, 5 21
16 2 102 0, 8 51
x
xRxx xx
Rxx xx



Solving this system yields
13xand
26.x

11
12
22
10
2
16
xx
xx
xxR
R
R





11
0
xxRand
2
11 2 2 1 2
0
xx x x xxRR R
So, revenue is maximized when
13xand
26.x
16.

12 1 2 1 2
22 22
12 11 22 1 2 12
11 1
22 2
11
12
22,15
15 15 0.02 4 500 0.05 4 275 0.02 0.05 11 11 775
0.04 11 0, 275
0.10 11 0, 110
0.04
0
0.10
x
x
xx
xx
xx
Px x x x C C
xx xx x x x x xx
Px x
Px x
P
P
P









110
xxPand
2
11 2 2 1 2
0
xx x x xxPP P
So, profit is maximized when
1275x and
2110.x

250. 250 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
17. ,, 2 2 2 .P p q r pq pr qr
1pqr implies that 1 .rpq


22 22
,221 21
2222222 22222
Ppq pq p p q q p q
pq p p pq q pq q pq p q p q




224; 2 24
PP
qp pq
pq
CC

CC
Solving
0
PP
pq
CC

CCgives 21
21
qp
pq



and so
1
3
pq
and
11 1 1 1 1 1 6 2
,22222 .
33 9 3 3 9 9 9 3
P





18. (a)

ln ln ln , 1 ln ln 1 ln 1
1ln 1 ln1 0
1ln 1 ln1 0
x
y
H xxyyzzxyz xxyy xy xy
Hx xy
Hy xy





ln 1 ln ln .xy x y x y
So,

1
3
ln12 ln 12 .xx xxxyz
(b)

11 11 11
3 3 33 33 1
3
ln ln ln ln ln 3H
19. The distance fromPtoQis
2
4.xThe distance from QtoRis
2
1.yxThe distance from RtoSis10 .y




2
2
22
22
342 110
32 0
4 1
1
20
2
11
x
y
Ckx kyx k y
yxx
Ck k
x yx
yx yx
Ck k
yx yx
















2
2
2
22
2
1
320
24
1
34
34
94
1
2
2
2
x
kk
x
x
x
xx
xx
x
x














2
22
2
21
41
1
3
112332
632
yx yx
yx yx
yx
y





So,
2
0.707
2
x km and
23 32
1.284
6
y

km.

251. Section 13.9 Applications of Extrema of Functions of Two Variables 251
© 2010 Brooks/Cole, Cengage Learning
20.

22 22 22
123
2
00 0 02 2 02 2
24 2
Sdd d y y y
yy






2
22
10
42
ydS
dy
y



when
23 6 23
2.
33
y


The sum of the distance is minimized when
23 3
0.845.
3
y


21. (a)



123
22 22 22
22 22
22,
00 22 42
22 42
Sxy d d d
xy xy xy
xy x y x y




From the graph we see that the surface has a minimum.
(b)




22 2 2 2 2
22 2 2 2 2
24
,
22 42
22
,
22 42
x
y
xx x
Sxy
xy xy xy
yy y
Sxy
xy xy xy







(c)

112
1, 1 1, 1 1, 1
2210
xySSS

D

iji j

210 12
2
tan 1 186.027
12 5





(d)

22 1 11 1 1,1
121
,,, 1,1
2102
xy
xyxSxytySxyt t t




2
2
2121 210 252
1,1 2 221
5552102
210 25 2
10 2 2 1
555
210 25 2
10 4 2 1
555
St t t t
tt
tt












Using a computer algebra system, we find that the minimum occurs when 1.344.t So
22, 0.05, 0.90 .xy
(e)
33 2 22 2 22, , , , 0.05 0.03 , 0.90 0.26
xy
xyxSxytySxyt t t




22 2 2
22
0.05 0.03 , 0.90 0.26 0.05 0.03 0.90 0.26 2.05 0.03 1.10 0.26
3.95 0.03 1.10 0.26
Stt t t t t
tt



Using a computer algebra system, we find that the minimum occurs when 1.78.t So
33, 0.10, 0.44 .xy

44 3 33 3 33, , , , 0.10 0.09 , 0.44 0.01
xy
xyxSxytySxyt t t




22 2 2
22
0.10 0.09 , 0.45 0.01 0.10 0.09 0.45 0.01 2.10 0.09 1.55 0.01
3.90 0.09 1.55 0.01
Stt t t t t
tt



Using a computer algebra system, we find that the minimum occurs when 0.44.t So,
44, 0.06, 0.44 .xy

Note: The minimum occurs at , 0.0555, 0.3992xy
(f )
,SxyD points in the direction thatS decreases most rapidly. You would use ,SxyD for maximization problems.
x
y
4
6
8
24
20
4
2
2
4
6
8
S

252. 252 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
22. (a)
22222
2
416122Sx y x y x y
The surface appears to have a minimum near
,1,5.xy
(b)



22222
2
22222
2
41 12
416122
62
416122
x
y
xx x
S
xyx y x y
yy y
S
xyx y x y







(c) Let
11,1,5.xy Then 1, 5 0.258 0.03SD ij. Direction 6.6
(d) 0.94,t
21.24,x
25.03y
(e) 3.56,t
31.24,x
35.06,y 1.04,t
41.23,x
45.06y

Note: Minimum occurs at , 1.2335, 5.0694xy
(f )
,SxyD points in the direction thatS decreases most rapidly.
23. Write the equation to be maximized or minimized as a function of two variables. Set the partial derivatives equal to zero (or
undefined) to obtain the critical points. Use the Second Partials Test to test for relative extrema using the critical points. Check
the boundary points, too.
24. See pages 964 and 965.
25. (a)




2
36 04 313 4 34
,40,
38 0 4 3 4 3 4 3
ab yx





(b)
22 2
34 4 34 1
01 3
23 3 23 6
S





26. (a)




2
46 04 313 3
,401, 1
10 4 10 10420 0
abyx





(b)
222 2
1713191
0112
10 10 10 10 5
S





2
2
30
−2
−4
4
4
6
8
x
y
z
x y xy
2
x
2 0 0 4
0 1 0 0
2 3 6 4
0
ixF 4
iyF 6
iixyF
2
8
ixF
x y xy
2
x
3 0 0 9
1 1 1 1
1 1 1 1
3 2 6 9
0
ixF 4
iyF 6
iixyF
2
20
ixF

253. Section 13.9 Applications of Extrema of Functions of Two Variables 253
© 2010 Brooks/Cole, Cengage Learning
27. (a)




2
44 48 1
2, 8 2 4 4, 2 4
46 4 4
abyx





(b)

2222
44 23 21 00 2S
28. (a)




2
837 28 8 72 1 1 1 3 1 3
,828 ,
144 2 8 2 4 2 48 116 28
abyx





(b)
2 2222222
3 11355793
0 0011222
4 44444442
S





29. 0,0, 1,1, 3,4, 4,2, 5,5
13, 12,
iixyFF

2
46, 51
ii ixy xFF



2
546 1312 74 37
86 43551 13
a






137 7
12 13
543 43
37 7
43 43
b
yx







30.
1, 0 , 3, 3 , 5, 6
9, 9,
iixyFF

2
39, 35
ii ixy xFF




2
339 99 36 3
24 2335 9
13 93
99
32 62
33
22
a
b
yx










x y xy
2
x
0 4 0 0
1 3 3 1
1 1 1 1
2 0 0 4
4
ixF 8
iyF 4
iixyF
2
6
ixF

x y xy
2
x
3 0 0 9
1 0 0 1
2 0 0 4
3 1 3 9
4 1 4 16
4 2 8 16
5 2 10 25
6 2 12 36
28
ixF 8
iyF 37
iixyF
2
116
ixF

−2 10
−1
7
(0, 0)
(1, 1)
(4, 2)
(3, 4)
(5, 5)
yx=+
37 7
43 43
6
7
−1
−1

254. 254 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
31. 0,6, 4,3, 5,0, 8, 4, 10, 5
27, 0,
iixyFF

2
70, 205
ii ixy x FF




2
570 270 350 175
296 1485 205 27
1 175 945
027
5 148 148
a
b









175 945
148 148
yx


32.
6,4, 1,2, 3,3, 8,6, 11,8, 13,8; 6n
42 31
iixyFF

2
275 400
ii ixy xFF



2
6 275 42 31 29
0.5472
536 400 42
1 29 425
31 42
6 53 318
1.3365
a
b










29 425
53 318
yx

33.
(a) 1.6 84yx
(b)
(c) For each one-year increase in age, the pressure
changes by approximately1.6, the slope of
the line.
34. (a) Using a graphing utility, 300 832.yx
(b) When
1.59, 300 1.59 832 355.xy
35. 1.0, 32 , 1.5, 41 , 2.0, 48 , 2.5, 53

2
7, 174, 322, 13.5
ii ii ixy xy x FF F F

14, 19, 14 19abyx
When
1.6, 41.4xy bushels per acre.
36. (a) 2.09 60.2yx
(b) For each one-point increase in the percent ,
xy
increases by about 2.09 million.
37.
2
2
1
,,
n
ii i
i
Sabc y ax bx c

F




22
1
2
1
2
1
20
20
20
n
ii i i
i
n
ii i i
i
n
ii i
i
S
x y ax bx c
a
S
x y ax bx c
b
S
yax bxc
c



C

C
C

C
C

C
F
F
F

4322
1111
32
1111
2
11 1
nnnn
iiiii
iiii
nnnn
iiiii
iiii
nn n
ii i
ii i
ax bx cx xy
ax bx cx xy
ax bx cn y





FFFF
FFFF
FF F
38. Let , ,
xyandzbe the length, width, and height,
respectively. Then the sum of the length and girth is
given by
2 2 108xyz
or 108 2 2 .
x yz The volume is given by



22
2
2
108 2 2
108 4 2 108 4 2 0
108 2 4 108 2 4 0.
y
z
V xyz zy zy yz
Vzyzzz yz
Vyyyzy yz




Solving the system 4 2 108yz
and 2 4 108,yz we obtain the solution
36xinches, 18yinches, and 18zinches.
39. 2, 0 , 1, 0 , 0, 1 , 1, 2 , 2, 5

2
3
4
2
0
8
10
0
34
12
22
i
i
i
i
i
ii
iix
y
x
x
x
xy
xy






F
F
F
F
F
F
F

34 10 22, 10 12, 10 5 8ac b ac

23626 3626
7535 7535
,, ,abc yxx
−4 18
−6
(0, 6)
(4, 3)
(5, 0)
(8, 4)−
(10, 5)−
y = − x +
175
148
945
148
8
−1
−1
14
9
0
50
250
80
−2
6−9
( 2, 0)−
( 1, 0)−
(0, 1)
(1, 2)
(2, 5)
8

255. Section 13.9 Applications of Extrema of Functions of Two Variables 255
© 2010 Brooks/Cole, Cengage Learning
40. 4, 5 , 2, 6 , 2, 6 4, 2

2
3
4
2
0
19
40
0
544
12
160
i
i
i
i
i
ii
iix
y
x
x
x
xy
xy






F
F
F
F
F
F
F

544 40 160, 40 12, 40 4 19ac b ac

253 53 41 41
24 10 6 24 10 6
,,,abcyxx
41.
0,0, 2,2, 3,6, 4,12

2
3
4
2
9
20
29
99
353
70
254
i
i
i
i
i
ii
iix
y
x
x
x
xy
xy






F
F
F
F
F
F
F

353 99 29 254
99 29 9 70
29 9 4 20
abc
abc
abc




2
1, 1, 0,ab c yxx
42.
0,10, 1,9, 2,6, 3,0

2
3
4
2
6
25
14
36
98
21
33
i
i
i
i
i
ii
iix
y
x
x
x
xy
xy






F
F
F
F
F
F
F

98 36 14 33
36 14 6 21
14 6 4 25
abc
abc
abc




25 9 199 5 9 199
420 20 42020
,, ,abcyxx
43.
(a)
0, 0 , 2, 15 , 4, 30 ,
6, 50 , 8, 65 , 10, 70

2
3
4
2
30
230
220
1800
15,664
1670
13,500
i
i
i
i
i
ii
iix
y
x
x
x
xy
xy






F
F
F
F
F
F
F

15,664 1800 220 13,500
1800 220 30 1670
220 30 6 230
abc
abc
abc




2225 541 25
112 56 14
0.22 9.66 1.79yxx x x
(b)
44. (a) 0.075 5.3yx
(b)
2
0.002 0.12 5.1yxx
(c)
(d) For
2014, 24x and

0.075 24 5.3 7.1y (billion)


2
0.002 24 0.12 24 5.1 6.8y (billion).
The quadratic model is less accurate because of the
negative
2
xcoefficient.
−4
9−9
(2, 6)
(4, 2)
( 2, 6)−
( 4, 5)−
8
−5
−2
7
(0, 0)
(2, 2)
(3, 6)
(4, 12)
14
−1
9−9
(0, 10)
(3, 0)
(2, 6)
(1, 9)
11
−114
−20
120
5 18
5.5
7

256. 256 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
45. (a) ln 0.1499 9.3018Ph
(b)
0.1499 9.3018 0.1499
ln 0.1499 9.3018
10,957.7
hh
Ph
Pe e


αα
(c)
(d) Same answers
46. (a)
1
0.0074 0.445
1
0.0074 0.445
ax b x
y
y
x

α


(b)
(c) No. For
70, 14,xyαβ which is nonsense.

1000yα which seems inaccurate.
47.
2
1
,
n
ii
i
Sab ax b y
α
F






2
111
11
2
1
1
,2 2 2
,2 22
,2
,2
,2
nnn
aiiii
iii
nn
bii
ii
n
aa i
i
bb
n
ab i
i
Sab a x b x xy
Sab a x nb y
Sab x
Sab n
Sab x
ααα
αα
α
α


α
α
αFFF
FF
F
F


,0
aaSab as long as 0
ix#for all .i(Note: If 0
ixαfor all ,ithen 0xαis the least squares regression line.)

22
22 2
11 11
44 4 0
nn nn
aa bb ab i i i i
ii ii
dSS S nx x nx x
αα αα


αβα β α β /


FF FF since
2
2
11
.
nn
ii
ii
nx x
αα

/

FF
As long as
0,d#the given values foraandbyield a minimum.
Section 13.10 Lagrange Multipliers
1. Maximize,.
fxy xyα
Constraint: 10xy
f gGDαD

ij ijyx G

10 5
y
xy
x
xy x y
G
Gα<
α>
α?



5, 5 25f α

2. Maximize
,.
fxy xyα
Constraint: 2 4xy
f gGDαD

2ij ijyxGG

2y
xG
Gα
α

2444
1, 1, 2
xy
xy G
G
ααα


1, 2 2f α
2
2
4
4
6
6
8
8
10
10
12
12
x
Constraint
Level curves
y
600
0
40
−2 24
−2,000
14,000

257. Section 13.10 Lagrange Multipliers 257
© 2010 Brooks/Cole, Cengage Learning
3. Minimize
22
,.fxy x y
Constraint: 4xy

22
ijij
fg
xyG
GGDαD



2
2
42
x
xy
y
xy x yG
Gα<
α>
α?



2, 2 8f α

4. Minimize
22
,.
fxy x y
Constraint: 2 4 5xy

22 24
ijij
fg
xyG
GGDαD



11
22
22
24 2
24 510 5
,,1
xx
yy
xy
xyGG
GG
G
G


ααα



51
24
,1f α
5. Minimize
22
,.
fxy x y
Constraint: 250xy
f gGDαD

22 2ijijxy G

2
22
x
y
G
Gα<
=
>
=α?
2x
yG
Gα
α

250xy

25 2,1, 2
2
xy
G
GG


1, 2 5f α
6. Maximize
22
,.
fxy x yαβ
Constraint:
2
20yxβα
f gGDαD

22 2 2
22 0 or 1ij ijxy x
xxx GG
GG

If 0,xαthen 0yαand
0, 0 0.f α
If
1,Gαβ

2
22 2 1 2 2.yyxxG
2,1 2 1 1,f αβα Maximum
7. Maximize,22 .fxy x xy y
Constraint: 2 100xy
f gGDαD

22 2 1 2ijijyx GG

22 2 1
21
21
2
yy
yx
xx GG
G
G <
=
α>β

=
?

2 100 4 100xy x

25, 50xyαα

25, 50 2600f α
8. Minimize, 3 10.fxy x y
Constraint:
2
6xyα

f gGDαD

2
32ij i jxy xGG

2
2
2
3
32
2 3
32
21
1
0
xy
xy
x
xxyy
x
xx
GG
GG
<

=
=
>
=

= #?


22
3
3
3 3
66
2
4
34
4,
2
x
xy x
x
xy




α
αα


33
3
34 94 20
4,
22
f

α



x
4
4−4
−4
Constraint
y
Level curves

258. 258 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
9. Note:
22
,6fxy x y is maximum when
,gxyis maximum.
Maximize

22
,6 .
gxy x y
Constraint: 2xy

2
2
x
xy
y
G
G <
>
?

21xy x y


1, 1 1, 1 2fg
10. Note:
22
,fxy x y is minimum when ,gxy
is minimum.
Minimize

22
,.
gxy x y
Constraint: 2 4 15xy

22
2
24
x
yx
y
G
G<
>
?


2 4 15 10 15
3
,3
2
3335
,3 ,3
222
xy x
xy
fg







11. Minimize
222
,, .fxyz x y z
Constraint: 9xyz

2
2
2
x
yxyz
zG
G
G<
=
>
=

?

93xyz x y z

3, 3, 3 27f
12. Maximize ,, .
fxyz xyz
Constraint: 3xyz

yz
xz yzxzxy xyz
xy
G
G
G<
=
>
=

?

31xyz x y z

1, 1, 1 1f
13. Minimize
222
,, .
fxyz x y z
Constraint: 1xyz

2
2
2
x
yxyz
zG
G
G<
=
>
=

?



1
3
111 1
333 3
1
,,
xyz x y z
f



14. Minimize
22
, 10 14 28.fxy x x y y
Constraint: 10xy



1
2
1
2
10210
214 14
xx
y yGG
G G <
>
?

10xy


11
22
10 14 10GG
2G

4, 6xy

4, 6 16 40 36 84 28 44f
15. Maximize or minimize
22
,3.
fxy x xy y
Constraint:
22
1xy
Case 1: On the circle
22
1xy

22
23 2
32 2
xy x
x y
xy y
G
G <
>
?


22 22
1,
22
xy x y
Maxima:
225
,
22 2
f





Minima:
22 1
,
22 2
f





Case 2: Inside the circle


2
23 0
0
32 0
2, 2, 3, 0
x
y
xx yy xy xx yy xyfxy
xy
fxy
ffffff
<
>

?


Saddle point:
0, 0 0f
By combining these two cases, we have a maximum
of
5
2
at
22
,
22




and a minimum of

1
2

at
22
,.
22

259. Section 13.10 Lagrange Multipliers 259
© 2010 Brooks/Cole, Cengage Learning
16. Maximize or minimize
4
,.
xy
fxy e


Constraint:
22
1xy
Case 1: On the circle
22
1xy



4
22
4
42
42
xy
xy
ye x
x y
xe y
G
G


< =
>
=?


22 2
1
2
xy x

Maxima:
1822
, 1.1331
22
fe





Minima:
1822
, 0.8825
22
fe






Case 2: Inside the circle



4
4
40
0
40
xy
x
xy
y
fye
xy
fxe


< =
>
=?


22
44
11
,,
16 16 16 4
xy xy xy
xx yy xyyx
fefefexy





At

2
0, 0 , 0.
xx yy xyff f
Saddle point:
0, 0 1f
Combining the two cases, we have a maximum
of
18
eat
22
,
22




and a minimum
of
18
e

at
22
,.
22






17. Maximize,, .
fxyz xyz

Constraints: 32
0
xyz
xyz




f ghGHDDD

yz xz xy GH ijk ijk ijk

yz
xzyzxyxz
xy
GH
GH
GH <
=
>
=

?


32
22 32 8
0
xyz
xz xz
xyz
<
>
?

16 y

8, 16, 8 1024f

18. Minimize

222
,, .fxyz x y z
Constraints: 2 6
12
xz
xy



f ghGHDDD
222 2xyz GH ijkikij

2
222
22
x
yxyz
zGH
H
G <
=
>
=
?


6
26 3
22
x x
xz z


12 12xy y x


9
2212 3 27 6
22
x
xx xx






6, 0xz

6, 6, 0 72f
19.
22
22
,00fxy x y x y
Constraint: 1xy

22
22
xx
xy
yy
GG
GG<<
>>
??


1
1
2
xy
xy



Minimum distance:
22
11 2
22 2




20. Minimize the square of the distance
22
,
fxy x y
subject to the constraint23 1.xy

22 3
223
x x
y
y
G
G<
>
?


23
23 1 ,
13 13
xy x y
The point on the line is
23
,
13 13



and the desired
distance is
22
2313
.
13 13 13
d

260. 260 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
21.
2
2
,2fxy x y
Constraint: 4xy


22
4
22
2
xx
yyGG
G
G<
=
>
=
?

4xy

4
4
22
6GG
G






3, 1xy
Minimum distance:

2
2
31232
22.
2
2
,1
fxy x y
Constraint: 43 xy


2
21
2
224
xx
yyG
G
GG<
=
>
=
?

43xy


2
42 3
2G
G


216 6
17 4
4
17GG
G
G



19 8
,
17 17
xy
Minimum distance:
22
19 8 5 17
17 17 17




23.
2
2
,3fxy x y
Constraint:
2
0yx
22
x xG
23y G

2
yx
If
0, 0,xy and
0, 0 9f distance 3.
If 0, 1, 5 2, 5 2xyxG#

2
111
52,52 52 3
24
f




Minimum distance:
11
2


24.
2
2
,3
fxy x y
Constraint:
2
0yx


2
23 2
2
x x
y
yxG
G




2
22yxG


3
23 22x x

3
4260xx


2
2 1 2 2 3 0 1, 1,xxx x y
Minimum distance:

2
2
11 2
25.
22
,121fxy x y
Constraint:
2
10yx



1
2
2
22
21
1
x x
y
yxG
G





22
222 122yx xG

231
2
2224
x xx x

3
4210xx
0.3855
1.1486
x
y



Minimum distance:

0.3855, 1.1486 0.188f

261. Section 13.10 Lagrange Multipliers 261
© 2010 Brooks/Cole, Cengage Learning
26. Minimize the square of the distance
2
2
,10fxy x y subject to the constraint
2
2
44.xy



22 4 10 5
10
422102
xx xy
yx
xyyy G
G< =
>
=?



2
22 2
2 25
4 4 8 16 50 100 4
4
29
58 112 0
4
xy xx xx
xx






Using a graphing utility, we obtain 3.2572x and 4.7428x or by the Quadratic Formula,



2
58 58 4 29 4 112 58 2 29 4 29
4.
2294 292 29
x


Using the smaller value, we have
29
41
29
x




and
10 29
1.8570.
29
y
The point on the circle is
29 10 29
41 ,
29 29



and the desired distance is
22
29 10 29
16 1 10 8.77.
29 29
d





The larger x-value does not yield a minimum.

27. Minimize the square of the distance


222
,, 2 1 1fxyz x y z
subject to the constraint 1.xyz



22
21
21
x
yyz
z G
G
G<
=
>
=

?
and 1yx

1211
1, 0
xyz x x
xyz



The point on the plane is
1, 0, 0 and the desired distance
is

222
12 01 01 3.d
28. Minimize the square of the distance


2
22
,, 4
fxyz x y z
subject to the constraint
22
0.xyz



22
22
24
24 2
2
22
2
xx
x
zxy
x xyy
y
z yyxy
z
GG
GG
G
<

=

=
==
>
=
=

=
=?


22
0, 2, 0, 2xyz x y z
The point on the plane is

2, 0, 2 and the desired distance
is

2
22
24 0 2 22.d
29. Maximize ,,fxyz zsubject to the constraints

222
0xyz and 2 4.xz

222
02
02 0
12 2
0
24 42
x
yy
z
xyz
xzxz
GH
G
GH








2
22
2
4
3
42 0 0
316160
34 40
or 4
zz
zz
zz
zz





The maximum value of
foccurs when4zat the point
of 4, 0, 4 .
30. Maximize ,,fxyz zsubject to the constraints

222
36xyz and 2 2.xyz

02 2
02 2
12x
yxy
zGH
GH
GH <
=
>
=

?


222
36xyz
222252xyz z xy y


22
2
2
2
25236
30 20 32 0
15 10 16 0
5 265
15
yy y
yy
yy
y






Choosing the positive value for y we have the point

10 2 265 5 265 1 265
,, .
15 15 3

262. 262 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
31. Optimization problems that have restrictions or
constraints on the values that can be used to produce the
optimal solution are called contrained optimization
problems.
32. See explanation at the bottom of page 971.
33. Minimize
222
,, .
fxyz x y z
Constraint:
,, 3gx y z x y z

22
22
22
xx
yy
zzGG
GG
GG



3xyz

3
222
3
3
2
2
GGG
G
G






1, 1, 1xy z
Minimum distance

2
22
111 3
34. Minimize
222
,, 1 2 3.fxyz x y z
Constraint:
,, 3gx y z x y z




2
21
2
4
22
2
6
23
2
xx
yy
zzG
G
G
G
G
G






3
xyz

246
3
222GGG


346
2
3G
G



4510
,,
333
xyz

Minimum distance
222
1113
3333




35. Minimize
,, .fxyz x y z
Constraint: ,, 27gx y z xyz

1
1
1
yz x xyz
xzy xyz xyz
xy z xyz
GG
GG
GG <
=
>
=
?

27
xyz

3
27 3xxyz
36. Maximize

2
,, .Px y z xyz
Constraint: ,, 32gx y z x y z

2
2
2
yz
xyz
xyG
G
G



32
xyz


22
0xy y z x z y #

22 2
22 2
xyz xy x y xy x y

232
8
16
8
xxx
x
y
z





37. Minimize
, , 0.06 2 2 0.11 .fxyz yz xz xy
Constraint: , , 668.25gx y z xyz


0.12 0.11
0.12 0.11
0.12
668.25
zyyz
zxxz
yx xy
xyz G
G
G




0.12 0.11 0.12 0.11xz yx xyz yz xy x y
G


2 0.24
0.12 2xx
x
GG

0.24 0.11 11
0.12 0.11 0.24
0.12 12
x x
zxxz zz
x





211 33
668.25 9,
12 4
xyz x x x y z





33
9, 9, $26.73
4
f

263. Section 13.10 Lagrange Multipliers 263
© 2010 Brooks/Cole, Cengage Learning
38. Maximize,,fx y z xyz (volume).
Constraint:
,, 1.5 2 2
gxyz xy xz yz C
1.5 2
1.5 2
22
yz y z
xzxz
xyx y
GG
GG
GG



1.5 2 2
xyxzyzC

+
,+ ,
2
1.5 2 1.5 2
22
(also by symmetry)
22 4.
xyz x y z y x z
xz yz
xy
xxx xGG GG
GG
GGG





3
1.5 2
44 4
xx
xzx z zx





22 33 2 2
1.5 2 2 ,
44 9 3
C
xxxxxCx Cx






22
,
34
CC
yz
39. Maximize
4
,, .
3
abc
fxyz

Constraint: ,abc K constant

4
443
33 34
3
4
3
bc
bc ac K
ab abc
ac
ab
G


G

G <

=
=
>
=

=?


So, ellipse is a sphere:

222 2
xyz a
40. Consider the sphere given by
222 2
xyz r

and let a vertex of the rectangular box be,, .
xyz
Maximize
,, 2 2 2 8 .V x y z x y z xyz
Constraint:

222 2
,,
gxyz x y z r

2
2
2
82 4
82 4
82 4
yz x xyz x
xz y xyz y x y z
xy z xzy z
GG
GG
GG <
=
>
=

?


222 2
xyz r
So, the rectangular box is a cube.

41. Maximize,, 2 2 2 .P p q r pq pr qr
Constraint:
,, 1gpqr pqr

22
22
22
qr
pr pqr
pq G
G
G <
=
>
=
?


1
3
31pqr p p and


111 2 2
333 9 3
,, 3 .P
42. Maximize
, , ln ln ln .Hxyz x x y y y z
Constraint: ,, 1gx y z x y z
(a)
ln 1
ln 1
ln 1
x
yxyz
z G
G
G <
=
>
=
?

1
3
31xyz x x y z
(b)

11
33111
333
,, 3 ln ln3H


y
z
x

264. 264 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
43. Maximize ,, 2 2 2 8Vxyz x y z xyz subject to the constraint
222
222
1.
xyz
abc


2
222
2222
2
2
8
2
8
2
8
x
yz
a
yx y z
xz
babc
z
xy
c
G
G
G
<

=
=
=
>
=
=
=
?


222 2 2 2
222 2 2 2
333
11,1,1
,,
333
xyz x y z
abc a b c
abc
xyz



So, the dimensions of the box are
23 23 23
.
333
abc
''
44. (a) Yes. Lagrange multipliers can be used.
(b) Maximize
,,Vxyz xyz subject to the constraint 2 2 108.xyz

2 and 2
2
yz
xzyzxy
xy
G
G
G<
=
>
=
?


2 2 108 6 108, 18xyz y y

36, 18xyz
Volume is maximum when the dimensions are 36 18 18'' inches.
45. Minimize ,, 5322C x y z xy xz yz xy subject to the constraint 480.xyz

86
86 ,4 3
66
yzyz
xzxzxyy z
xyxy
G
G
G <
=
>
=
?


34
3
33 4
3
480 480
360, 360
xyz y
xy z



Dimensions:
33 3 4
3
360 360 360'' feet.
46. (a) Maximize,,Px y z xyz subject to the
constraint .
xyz S

yz
xzxyz
xy
G
G
G<
=
>
=
?


3
S
xyz S x y z
So,
3
3
3
,,, 0
333
27
3
.
3
SSS
xyz x y z
S
xyz
S
xyz
xyz
xyz








x
y
z
x
y
z

265. Section 13.10 Lagrange Multipliers 265
© 2010 Brooks/Cole, Cengage Learning
(b) Maximize
123 n
Pxxxx subject to the constraint
1
.
n
i
i
xS

F

23
13
12312
123 1
n
n
nn
nxx x
xx x
xxx xxx x
xx x x
G
G
G
G

<
=

=
=
>
=
=
=?







123
1
n
in
i
S
xS xx x x
n

F
So,

123 ,0
ni
SSS S
xx x x x
nnn n

/




123
n
n
S
xx x x
n





123
n
n
S
xx x x
n



123
123
.
nn
n
xxx x
xx x x
n





47. Minimize
2
,2 2
Arrhr subject to the constraint
2
0
.rh V

2
24 2
2
2
hr rh
hr
rr G
G <
>
?


23
00
2rh V r V
Dimensions:
0
3
2
V
r

and
0
3
2
2
V
h



48. Maximize
22
, , 100Txyz x y subject to the constraints
222
50xyz and 0.xz

22
22
02
xx
yy
z GH
G
GH <
=
>
=

?

If 0,
y# then 1Gand 0, 0.z
H
So, 0
xz and
50.y
0, 50, 0 100 50 150T
If
0ythen
22 2
250xz x and
50 2.xz

50 50 50
, 0, 100 112.5
22 4
T




So, the maximum temperature is150.

266. 266 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
49. Using the formula
Distance
Time ,
Rate
θ
minimize
2222
21
12
,
dydx
Txy
vv


subject to the constraint .
xya

22
11
22 2 2
11 2 2
22
22
x
vd x xy
y vd x v d y
vd y
G
G
<
θ
=

=
θ>
=
θ
=

?


xya
Because
1
22
1sin
x
dxθ

and
2
22
2sin ,
y
dyθ

we have
2222
21
12
yd yxd x
vv

θ
or

12
12sin sin
.
vv
θ


50. Case 1: Minimize
,2
2
l
Pl h h l


subject to
the constraint
2
.
8
l
lh A






1
24
22
2,1
22
2
l
h
h
l
ll
lh

G

GG



θ

Case 2: Minimize
2
,
8
l
Al h lh




subject to
the constraint
2.
2
l
hl P




42
2,
2424
or 2
2
l
h
llll
lh
l
hlh

G

GG
(



θθ


51. Maximize

0.25 0.75
, 100Px y x yθ subject to the
constraint 72 60 250,000.
xy

0.75
0.75 0.75
0.25
0.25 0.25
0.75 0.25
72
25 72
25
60
75 60
75
72 75
25 60y
xy
x
x
xy
y
yy
xx G
G
G
G
G
G
π
π







θ



18
5
18
5
y
x
yx
θ
θ


18 15,625
72 60 288 250,000
518
3125
xxx x
y



θ


15625
, 3125 226,869
18
P



Medium 2
Q
d
2
d
1
x
y
a
1
θ
2
θ
Medium 1P
l
h

267. Section 13.10 Lagrange Multipliers 267
© 2010 Brooks/Cole, Cengage Learning
52. Maximize
0.4 0.6
, 100Px y x y subject to the
constraint72 60 250,000.
xy

0.6
0.6 0.6
0.4
0.4 0.4
72
40 72
40
60
60 60
60y
xy
x
x
xy
y G
G
G
GG










0.6 0.4
72 1
40
99
55
9 125,000
72 60 180 250,000
59
yy
xx
y
yx
x
xxx xG
G
"







2500y

125,000
, 2500 496,399
9
P




53. Minimize
,7260Cx y x y subject to the
constraint
0.25 0.75
100 50,000.xy

0.75
0.75 0.75
0.25
0.25 0.25
72
72 25
25
60
60 75
75
y
xy
x
x
xy
y
G
G
G
G










0.75 0.25
72 75
25 60
18 18
3.6
55
yy
xx
y
yx x
x G
G
"





0.75
0.25
100 3.6 50,000xx

0.75
500
191.3124
3.6
3.6 688.7247
x
yx




191.3124, 688.7247 55,097.97C
54. Minimize,7260Cx y x y subject to the constraint
0.6 0.4
100 50,000.xy

0.4
0.4 0.4
0.6
0.6 0.6
72
72 60
60
60 3
60 40
40 2
y
xy
x
x
xy
y
G
G
G
GG










0.4 0.6
0.4
0.6
72 2
60 3
44
55
4
100 50,000
5
yy
xx
y
yx
x
xxG
G
"










0.4
500
45
x



0.4
400
45
y



0.4 0.4
500 400
, $65,601.72
45 45
C




55. (a) Maximize,, cos cos cosg$G $ % subject to the constraint . $%

sin cos cos
cos sin cos tan tan tan
cos cos sin$%G
$%G $ % $%
$%G <
=
>
=

?


3

$% $%


1
,
33 3 8
g
"

268. 268 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
(b) $% %$




cos cos cos
cos cos cos cos sin sin
cos cos cos
g$ $ $
$ $ $
$$





56.
Letrαradius of cylinder, and hαheight of cylinder αheight of cone.

22
22Srhrhr constant surface area

22
2
25
33
rh rh
Vrh

volume
We maximize

2
,
frh rhα subject to
22
,.grh rh r h r C


2
22 2
Crh rh r

24
24
2
2
CCrhr
Cr
h
Cr
βα
β
α



24 5
2
,
222
Cr Crr
frh Fr r
Cr C
β
αα αβ





4
24
2
5
0
22
5
5
Cr
Fr
C
Cr
C
r
αβ α
α
α



3
10r
Fr
Cβ
α






24 2 2
14
2
14
2
2
5
2
25
45
25
2
5
2
5
5
25
5
Cr CC
h
Cr
CC
C
C
C
r
r
r
r
ββ
αα
α
α
α
α

So,
25
.
5
h
r
α
By the Second Derivative Test, this is a maximum.
α β 3
3
2
3
γ

269. Review Exercises for Chapter 13 269
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 13
1.
22
22
,, 2
2
fxyz x y z
yx z

Elliptic paraboloid


2.
22 2
,, 4 4 0fxyz x y z
Elliptic cone.

3.
22
,
fxy x y
(a)
(b)
,,2gx y f x y is a vertical translation
of
ftwo units upward.
(c)
,,
gxy fxy zαβ is a horizontal translation
offtwo units to the right. The vertex moves
from
0, 0, 0 to0, 2, 0 .
(d)

1,zfyα ,1zfxα
4.
22
,1fxy x yαββ
(a)
(b) γ γ ,2,gxy fx y is the graph offshifted
2units back on the x-axis. The hemisphere has
equation


2
22
21,0.xyzz /
(c)
γ γ,4 ,gxy fxyαβ is the graph offreflected in
the xy-plane, and shifted
4units upward.
(d)

γ0,fy ,0fx
22
1zxyαββ
5.
22
,
xy
fxy e

α
The level curves are of the form

22
22
ln .
xy
ce
cx y

α


The level curves are circles centered at the origin.

6.
γ,lnfxy xyα
The level curves are of the form
ln
.
c
cxy
exy
α
α

The level curves are hyperbolas

x
y3
2
2
−2
−2
z
x
y
3
3
1
2
3
−1
−2
−3
z
x
y
22
1
−2
4
5
z
x
y2
2
4
5
z
z = f(1, y)
x
y
2
2
4
5
z
z = f(x, 1)
x
y
1
1
1
−1
z
x
y
1
1
1
−1
z
x
y
1
1
1
−1
z
x
y
Generated by Mathematica
2−2
c = 10
c = 1
−2
2
c= 0
c= 1
c=
2
c=
2
−33
−2
2
13

270. 270 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
7.
22
,fxy x yαβ
The level curves are of the form

22
22
1.
cx yx y
cc
αβ
αβ

The level curves are hyperbolas.

8.
,
x
fxy
xy
α

The level curves are of the form

1
.
x
c
xy
c
yx
c
α

β
α


The level curves are passing through the origin with
slope
1
.
c
cβ


9.
γ

22
,
xy
fxy e

α

10.
1
,
x
gxy y

α

11.
γγ
22
,1,1
1
lim
2
xy
xy
xy

α

Continuous except at
0, 0 .
12.
γγ
22
,1,1
lim
xy
xy
xy β

Does not exist.
Continuous except when .yx
13.
γγ
2
2
,0,0
00
lim 0
110
y
xy
yxe
x
β


αα


Continuous everywhere.
14.
γγ
2
42
,0,0
lim
xy
xy
xy


For
24
2
42 44
1
,.
2
xy x
yx
xy xx
αα



For
2
42
0, 0
xy
y
xy
αα

for 0.x#
The limit does not exist.
Continuous to all
γ,0,0xy#
15. γ,cos
cos
sin
x
x
x
x
y
fxy e y
fey
f ey
α
α
αβ

16.



2
22
2
2
,
x
y
xy
fxy
xy
yx y xy y
f
xyxy
x
f
xy
α


αα

α


17.
,
yx
x y
ze e
zz
ee
xy
ββ
ββ

CC
αβ αβ
CC
18.
γ
22
22
22
ln 1
2
1
2
1
zxy
zx
xx y
zy
yx y

C
α
C
C
α
C

x
y
Generated by Mathematica
1
4
4
c = 12
c = −12c = −2
c = 2
−41
−1
−4
c= 2
c=2−
c=1−
c= 1
c=
1
2
c=−
3
2
c=−
2
c=
3
2
−33
−2
2
1
x
y3 3
3
−3
−3
−3
z
y
x
5
5
60
z

271. Review Exercises for Chapter 13 271
© 2010 Brooks/Cole, Cengage Learning
19.






22
22 22
22
22 22
22
2
22
,
2
x
y
xy
gx y
xy
yx y xy x yy x
g
xy xy
xx y
g
xy
α


αα

β
α

20.

222
12
222
222
222
222
1
2
2
wxyz
wx
xyz x
x
xyz
wy
y xyz
wz
z xyz
β
αββ
C
αββ α
C ββ
Cβ
α
C ββ
Cβ
α
C ββ

21.
, , arctan
y
fxyz z
x
α


22222
2222
1
1
1
arctan
x
y
z
zy yz
f
x xyyx
zxz
f
x xyyx
y
f
x
β
βα



αα


α

22.

222
12
222
1
,,
1
1
fxyz
xyz
xyz
β
α








32
222
32
222
32
222
32
2221
12
2
1
1
1
f
xyz x
x
x
xyz
fy
y
xyz
fz
z
xyz
βC

C
β
α

Cβ
α
C

Cβ
α
C


23.


2
2
2
2
,sin
cos
sin
nt
nt
nt
uxt ce nx
u
cne nx
x
u
cn e nx
t
β
β
β
α
C
α
C
C
αβ
C

24.
γ γ


,sincos
cos cos
sin sin
uxt c akx kt
u
akc akx kt
x
u
kc akx kt
t
α
C
α
C
C
αβ
C

25.
26.

γ

2
ln 1
2 ln 1 . At 2, 0, 0 , 0.
zx y
zz
xy
xx

CC

CC
Slope in x-direction.

2
.
1
zx
yy
C
αα
C
At2, 0, 0 , 4.
z
y
C
α
C
Slope in y-direction.
27.
γ
23
2
,3 2
6
6
6
12
1
1
x
y
xx
yy
xy
yx
fxy x xy y
fxy
fxy
f
fy
f
f

αβ

α
α
αβ
αβ

28. ,
x
hx y
xy
α






2
2
3
3
2
2
x
y
xx
yy
y
h
xy
x
h
xy
y
h
xy
x
h
xy
α

β
α

β
α

α








2
43
2
43
2
2
xy
yx
xy yxy
xy
h
xy xy
xy yxy xy
h
xyxy
β
αα

β
αα


x
y
3
3
−1
3
z

272. 272 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
29. ,sincoshx y x y y x

sin sin
cos cos
cos
sin
cos sin
cos sin
x
y
xx
yy
xy
yxhyyx
hxy x
hyx
hxy
hyx
hyx







30. ,cos2
gxy x y




sin 2
2sin 2
cos 2
4cos 2
x
y
xx
yy
g xy
g xy
g xy
g xy







2cos 2
2cos 2
xy
yx
g xy
g xy



31.
22
zx y

2
2
2
2
z
x
x
z
x
C

C
C

C


2
2
2
2
z
y
y
z
yC

C
C

C

So,
22
22
0.
zz
xy
CC

CC

32.
32
3zx xy

22
2
2
2
2
33
6
6
6
z
x y
x
z
x
x
z
xy
y
z
x
y
C

C
C

C
C

C
C

C

So,
22
22
0.
zz
xy
CC

CC

33.
22
y
z
xy





2
22
22 22
32 32
22 22 22
2
413
22
zxy
x
xy
zx xy
yy
x
xyxy xy
C

C


C

C






22
22
22
22 22
2xy yzxy
y
xyxy
C

C





2
22 2222
2 22
432
22 22
22 2 3
2
xy y xyxyyzxy
y
y
xyxy
C

C


So,
22
22
0.
zz
xy
CC

CC

34.
2
2
2
2
sin
cos
sin
sin
sin
y
y
y
y
y
ze x
z
ex
x
z
ex
x
z
ex
y
z
ex
y

C

C
C

C
C

C
C

C

So,
22
22
sin sin 0.
yyzz
exex
xy
CC

CC

273. Review Exercises for Chapter 13 273
© 2010 Brooks/Cole, Cengage Learning
35.

2
sin
cos sin cos
zxxy
zz
dz dx dy xy xy xy dx x xy dy
xy
θ
CC

CC
36.
22
xy
z
xy
θ





22 22 22 22
33
32 3222 22
22 22
zz
dz dx dy
xy
xyyxyxxy xyxxyyxy
yx
dx dy dx dy
xy xy
xy xy
CC

CC










37.
222
222
51 121 17
0.654 cm
13 2 13 2 26
zxy
zdz xdx ydy
xy
dz dx dy
zz







Percentage error:
17 26
0.0503 5%
13
dz
z


38. From the accompanying figure we observe

2
tan or tan
tan sec .
111
Letting 100, , , and .
2 60 180
h
hx
x
hh
dh dx d dx x d
x
xdx d



θθ
CC

CC


(Note that we express the measurement of the angle in radians.) The maximum error is approximately

211 1 11
tan 100 sec 0.3247 2.4814 2.81
60 2 60 180
dh


feet.
39.

21
3
2
2 3521 2111 1
33 383866
25 2 in.
Vrh
dV rh dr r dh

θ

40.

22
2
22
22 22
22
22 22
2 825 1101 43
8829 29 8 29
Arrh
rrh
dA r h dr dh
rh rh
rh rh
dr dh
rh rh













41.
2
ln , 2 , 4wxyxty t
(a) Chain Rule:

22
2
21
21
81
44
dw w dx w dy
dt x dt y dt
x
xy xy
t
tt
CC

CC


π
θ


(b) Substitution:

22
ln ln 4 4wxy t t


2
1
81
44
dw
t
dt t t
θπ


42.
2
,cos,sinuy xx ty tθπ θ θ
(a) Chain Rule:


1sin 2cos
sin 2 sin cos
sin 1 2 cos
du u x u y
dt x t y t
tyt
ttt
tt
CC CC

CC CC




(b) Substitution:
2
sin cosuttθπ


2sin cos sin
sin 1 2 cos
du
tt t
dt
tt


x
h
θ

274. 274 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
43. ,2, ,2
xy
w x rty rtz rt
z

(a) Chain Rule:






2
2
223
2
2
223
2
22
2222
22 2
44
2
11
44
2
wwxwywz
rxryrzr
yxxy
t
zzz
rtt rtrtrt
rt rt rt
rt rt t
rt
wwxwywz
txtytzt
yx xy
r
zz z
rt rt r
rt
CCCCCCC

CCCCCCC







CCCCCCC

CCCCCCC





(b) Substitution:



22
223
2
223
2
2 2
22
44
2
44
2
rtrt
xyrtrt
w
zrt rt
wrtrtt
r rt
wrtrt r
t rt



C

C
C

C

44.
222
,cos,sin,ux y zxr tyrtzt
(a) Chain Rule:



22
22
2cos 2sin 20
2cos sin 2
2 sin 2 cos 2 2 sin cos sin cos 2 2
uuxuyuz
rxryrzr
xt yt z
rtrt r
uuxuyuz
txtytzt
xrt yr t z r t tr t t t t
CCCCCCC

CCCCCCC


CCCCCCC

CCCCCCC


(b) Substitution:

22 22 2 2 2
,cossin
2
2
ur t r t r t t r t
u
r
r
u
t
t

C

C
C

C

45.
222
0
220
xxyyyzz
zz
xyy z
xx

CC

CC


2
2
zxy
x yz
C

C

220
zz
xyy zz
yy
CC

CC

2
2
zxyz
yyz
C

C

46.
2
2
2
sin 0
2cos0
cos 2
2cossin0
sin
2cos
xz y z
zz
xz z y z
xx
zz
xyzxz
zz
xz y z z
yy
zz
yxzyz

CC

CC
C

C
CC

CC
C

C

275. Review Exercises for Chapter 13 275
© 2010 Brooks/Cole, Cengage Learning
47.

2
2
,
2
5, 5 50 25
fxy xy
fxyx
f
α
D
D
ij
ij


34
55
uijαβ unit vector

5, 5 5, 5
30 20 50
Df fβαDβ"
αβ β αβ
u u
48.


221
,
4
1
2
2
1, 4 2 2
1255
555
1, 4 1, 4
45 25 25
55 5
fxy y x
fxy
f
Df f
αβ
Dαβ
D

αD "

u
i+ j
ij
uv ij
u

49.


2
1212
3333
442 2
333 3
2
1, 2, 2 2 4
1,2,2 1,2,2
wy xz
wz y x
w
Dw w

D
D

D "
u
ijk
ijk
uvijk
u

50.



22
2
523
10 2 2 6 3
1, 0, 1 10 2
1
3
wx xyyz
wxy xyzy
w

D
D

ijk
ij
uijk



1, 0, 1 1, 0, 1
10 2 12
43
33 3
Dw w αD "

u u
51.


2
2
2
2, 1 4 4
2, 1 4 2
z
zxy
zxyx
z
α
D
D
Dα
ij
ij

52.
cos
cos sin
22 22
0, ,
422 22
0, 1
4
ij
ij
x
xx
ze y
ze ye y
z
z


β
ββ
α
Dαβ β

Dαββαββ



Dα



53.



22
22
22
22 22
2
11
1, 1 , 0
22
1
1, 1
2
ij
i
y
z
xy
xy x y
z
xy xy
z
z
α

β
D

Dαβαβ
Dα

54.



2
22
22
2
2, 1 4
2, 1 4
ij
j
x
z
xy
xxy x
z
xy xy
z
z
α
β
β
D
ββ
Dα
Dα

55.
γ
22
,94,65,3,2fxy x y c Pαβ α
(a) γ
,188
3, 2 54 16
fxy x y
f
Dαβ
Dαβ ij
ij
(b) Unit normal:

54 16 1
27 8
54 16 793
β
αβ
βij
ij
ij

(c) Slope
27
.
8
α



27
3
8
yz xβα β


27 65
88
yxαβ
Tangent line
(d)
56.
,4sin,3,,1
2
fxy y x yc P

αβα



(a)

,4cos 4sin1
,1 3
2 ij
jfxy y x x
f

D

Dα


(b) Unit normal vector:
j
(c) Tangent line horizontal: 1yα
(d)
y
x
64−6−4
−6
−4
−2
4
2
6
Tangent line
Unit normal
vector
2
3
2
π
2
π
π
2
π
−
y
x
((, 1

276. 276 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
57.

2
2
,, 0
2
2, 1, 4 4 4
Fx yz xy z
Fxyx
F

D
D
ijk
ijk

So, the equation of the tangent plane is

4241 40xyz or
44 8,xyz
and the equation of the normal line is

42, 41, 4.xt yt z t
58.

22
,, 25 0
22
2, 3, 4 6 8 2 3 4
Fx yz y z
Fyz
F

D
D
jk
jkjk

So, the equation of the tangent plane is

33440yz or 3 4 25,yz
and the equation of the normal line is

34
2, .
34
yz
x



59.


22
,, 4 6 9 0
24 26
2, 3, 4
Fxyz x y x y z
Fx y
F

D
D
ijk
k

So, the equation of the tangent plane is
40z or 4,z
and the equation of the normal line is

2, 3, 4 .
x yz t
60.

222
,, 9 0
22 2
1, 2, 2 2 4 4 2 2 2
Fx yz x y z
Fx y z
F

D
D
ijk
ijk ijk

So, the equation of the tangent plane is

12 22 2 0xy z or
229,xyz
and the equation of the normal line is

122
.
122
xy z

61.


2
,, 9 0,2,2,5
,,
Fx yz y z
Gx y z x y


2Fy
G
D
D
jk
ij

2, 2, 5 4FD jk
041 4
110
FGD'D

ijk
ij k

Tangent line:
225
11 4
xyz



62.




22
,, 0
,, 3 0
22
2, 1, 3 4 2
4212 2
00 1
Fx yz x y z
Gx y z z
Fx y
G
F
FG


D
D
D
D'D

ijk
k
ijk
ijk
ij

So, the equation of the tangent line is

21
,3.
12
xy
z


63.


222
,, 14
,, 2 2 2
fxyz x y zfxyz x y z

D ijk

2, 1, 3 4 2 6fD ijk Normal vector to plane.

6314
cos
1456
36.7


"

nk
n

64. (a)



1
,cossin,0,01
sin , 0, 0 0
cos , 0, 0 1
,1
xx
yy
fxy x yf
fxf
fyf
Pxy y





(b)


cos , 0, 0 1
sin , 0, 0 0
0, 0, 0 0
xx xx
yy yy
xy xyfxf
fyf
ff






21
2 2
,1Pxy y x

277. Review Exercises for Chapter 13 277
© 2010 Brooks/Cole, Cengage Learning
(c) If 0,yθyou obtain the 2
nd
degree Taylor polynomial for cos .x
(d)
(e)
The accuracy lessens as the distance from
0, 0 increases.
65.
22
,269814
4680
618 0, 3
x
y
fxy x xy y x
fxy
fxyxy






4
3
43 6 8 , 4yy y x

4
18
6
xx
yy
xyf
f
f
θ
θ
θ


2 2
4 18 6 36 0.
xx yy x yff f
So,

4
3
4, , 2ππ is a relative minimum.
66.
22
,35
fxy x xy y x
2350
3
32 0
2
x
yfxy
f xy y x




3
23 5
2
xx




49 10
2, 3
xx
xy
πθ
θπ θ

2, 2, 3, 4 9 0
xx yy xyfffd

2, 3 is a saddle point.

67.
2
2
2
2
11
,
1
0, 1
1
0, 1
x
y
fxy xy
xy
fy xy
x
fx xy
y

θπ θ θ
θπ θ θ

So,
22
xyxyθ orxyθand substitution yields the
critical point1, 1 .

3
3
2
1
2
xx
xy
yyf
x
f
f
y
θ
θ
θ

At the critical point1, 1 , 2 0
xxf and

2
30.
xx yy x yff f
So, 1, 1, 3 is a relative minimum.

1
2
1
−1
y
x
z
1
2
1
−1
y
x
z
2
2
3
21
1
−1
−1
−2
−2
x
y
z
x y ,fxy 1,Pxy 2,Pxy
0 0 1.0 1.0 1.0
0 0.1 1.0998 1.1 1.1
0.2 0.1 1.0799 1.1 1.095
0.5 0.3 1.1731 1.3 1.175
1 0.5 1.0197 1.5 1.0

x
y
−24
−20
3
4
4
20
(1, 1, 3)
z

278. 278 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
68.
33
2
2
50 0.1 20 150 0.05 20.6 125
50 0.3 20 0, 10
50 0.15 20.6 0, 14
x
y
zxy xx y y
zxx
zyy



Critical Points:
10, 14 , 10, 14 , 10, 14 , 10, 14
0.6 , 0.3 , 0
xx yy x yzxzyz
At

2
2
10, 14 , 6 4.2 0 0, 0.
xx yy x y xxzz z z

10, 14, 199.4 is a relative maximum.
At

2
2
10, 14 , 6 4.2 0 0.
xx yy x yzz z

10, 14, 349.4 is a saddle point.
At

2
2
10, 14 , 6 4.2 0 0.
xx yy xyzz z

10, 14, 200.6 is a saddle point.
At

2
2
10, 14 , 6 4.2 0 0, 0.
xx yy x y xxzz x z

10, 14, 749.4 is a relative minimum.
69. The level curves are hyperbolas. There is a critical point at
0, 0 , but there are no relative extrema. The gradient is normal to
the level curve at any given point
00,.xy
70. The level curves indicate that there is a relative extremum at ,
Athe center of the ellipse in the second quadrant, and that there is
a saddle point at ,Bthe origin.
71. 12 1 2,Px x R C C
22
1212 1 1 2 2
22
121212
225 0.4 0.05 15 5400 0.03 15 6100
0.45 0.43 0.8 210 210 11,500
xxxx x x x x
xxxxxx





1210.9 0.8 210 0
xPxx

120.9 0.8 210xx

2120.86 0.8 210 0
xPxx

120.8 0.86 210xx
Solving this system yields
194x and
2157.x

11
12
22
0.9
0.8
0.86
xx
xx
xxP
P
P





11
2
11 2 2 1 2
0
0
xx
xx x x xxP
PP P



So, profit is maximum when
194x and
2157.x

279. Review Exercises for Chapter 13 279
© 2010 Brooks/Cole, Cengage Learning
72. Minimize
22
12 1 1 2 2
,0.25100.1512Cx x x x x x
subject to the constraint
12 1000.xx

1
12
20.50 10
53 20
0.30 12
x
xx
x G
G <
θπ>
?


12 1 2
12
1
1
2 1000 3 3 3000
53 20
8 3020
377.5
622.5
xx x x
xx
x
x
x

θπ
π
π
π


377.5, 622.5 104,997.50C π
73. Maximize
,4 2fxy x xy y subject to the
constraint 20 4 2000.xy

420
56
24
y
xy
x G
G <
θπθ>
?


20 4 2000 5 500
56
10 494
49.4
253
xy xy
xy
x
x
y

θπθ
π
π
π

49.4, 253 13,201.8f π
74. Minimize the square of the distance:




222
22
22 32
22 32
,, 2 2 0.
222 20 22 2 0
222 20 22 2 0
x
y
fxyz x y x y
fx xyx xxxy
fy xyy yyxy

< =
>
=?



3
2
2222
Clearly and hence: 4 2 0. Using a computer algebra system, 0.6894.
So, distance 0.6894 2 0.6894 2 2 0.6894 4.3389.
Distance 2.08
xy x x x




75. (a)
2
0.004 0.07 19.4yxx
(b) When

2
80, 0.004 80 0.07 80 19.4 50.6 Kg.xy
76. (a) 2.29 2.0yt

(b)
Yes, the data appear linear.
(c)
1.24 8.37 lnyt
(d)

0
0
20
9
−1
−5
3
20

280. 280 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
77. Optimize,,fxyz xy yz xz subject to the
constraint 1.xyz

yz
xzxyz
xy
G
G
G <
=
>
=
?


1
3
1xyz x y z
Maximum:

111 1
333 3
,,f π
78. Optimize
2
,
fxy xyπ subject to the
constraint 2 2.xy

2
2
2
40 or4
2
xy
x xy x x y
x
G
Gπ<
>
π?

22xy
If
0, 1.xyππ If
14
33
4,then , .xy y xπππ
Maximum:

1641
33 27
,f π
Minimum:
0, 1 0f π
79.
2
2
4,
1,
;10
PQ x
QR y
RS z x y z





22
3421Cx y z
Constraint: 10xyz
Cg
GDπD

+,
22
32
41
xy
xy
G
ijkijk

2
2
22 2
22 2
34
21
1
1
94
2
1
41
3
xx
yy
xx x
yy yG
G
G

π



So,
2
0.707 km,
2
3
0.577 km,
3
23
10 8.716 km.
23
x
y
z




80. ,,,0fxy ax byxy
Constraint:
22
1
64 36
xy

(a) Level curves of
,43
fxy x y are lines of form
4
.
3
yxC
Using
4
12.3,
3
yx
you obtain
7, 3, and 7, 3 28 9 37.xy f

Constraint is an ellipse.
(b) Level curves of
,49
fxy x y are lines of form
4
.
9
yxC

Using
4
7,
9
yx
you obtain
4, 5.2, and 4, 5.2 62.8.xy f π
−10 10
−8
8

281. Problem Solving for Chapter 13 281
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 13
1. (a) The three sides have lengths5, 6, and 5.
Thus,
16
2
8sππ and
8323 12.Aππ
(b) Let
2
, , area ,fabc ss a s b s cππθθθ
subject to the constraint
constant (perimeter).abc
Using Lagrange multipliers,




.
ss b s c
ss a s c
ss a s b
G
G
Gθθ θπ
θθ θπ
θθ θπ

From the first
2equations
.
sbsa ab
Similarly,bcπand henceabcππ which is an
equilateral triangle.
(c) Let
,, ,
fabc a b c subject
to

2
Area
ssasbscπθ θ θ constant.
Using Lagrange multipliers,


1
1
1
ssbsc
ssasc
ssasb
G
G
Gπθ θ θ
πθ θ θ
πθ θ θ

So,
sasb ab and .abcππ

2.
24
3
Vrrh

E

Material
2
42Mr rh


3
2
1000 4 3
1000
r
Vh
r
θ

So,

3
2
2
22
2
2
3
13
3
1000 4 3
42
2000 8
4
3
2000 16
80
3
16 2000
8
3
8
2000
3
750 6
5.
r
Mr r
r
rr
r
dM
rr
dr r
rr
r
r
rr






θ




πθ θ π
θπ

π







Then,

2
1000 4 3 750
0.h
r
θ
ππ

The tank is a sphere of radius
13
6
5.r


π



3. (a) ,, 1 0
,,
xyz
F x y z xyz
FyzFxzFxy
πθπ
πππ

Tangent plane:

00 0 00 0 0 0 0 0yz x x xz y y xy z z

00 00 0 0 0 00 33yzx xzy xyz xyz
(b)

00 00 0 0
1
base height
3
11 3 3 3 9
32 2
V
yz xz xy
π

ππ




4. (a) As
13
3
,1
x fx x x ! and
hence
lim lim 0.
xx
fx gx fx gx
! !



(b) Let


13
3
00
,1xx θ be a point on the graph of .
f
The line through this point perpendicular
togis
33
00
1.yxx x
This line intersects
gat the point

3333
00 0011
1, 1 .
22
xx xx







The square of the distance between these two points
is


2
33
000
1
1.
2
hx x xπθ θ

his a maximum for
0
3
1
.
2
xπ
So, the point
on
ffarthest fromgis
33
11
,.
22

θ




x
y
3
3
3
z
3
x
0
y
0
3
y
0
z
0
3
x
0
z
0
Tangent plane
Base
−(
−6
−4
6
4

282. 282 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
5. (a)
Maximum value of fis2, 2 2 2.f θπ
Maximize
,.
fxy x yπθ
Constraint:

22
,4gxy x y
:
f gGDπD
22
12
12
4
x
y
xy
G
Gπ
θπ

22
x yx yGG

2
24 2, 2xxy
2, 2 2 2, 2, 2 2 2ffθπ θ πθ
(b)
,
fxy x yπθ
Constraint:

22
0, 0,0xy xy
Maximum and minimum values are 0.
Lagrange multipliers does not work:

12
0,
12
x
xy
y
G
Gπ <
πθ π>
θπ ?
a contradiction.
Note that
0, 0 .gDπ 0

6. Heat Loss 53333H k xy xy xz xz yz yz

666kxy xz yz

1000
1000 .Vxyz z
xy


Then 1000 1000
6.Hkxy
yx




Setting 0,
xyHHππ you obtain 10.xyzπππ
7. 566H k xy xz yz

1000 6000 6000
5.zHkxy
xy y x





2
26000
5 0 5 6000
xHy yx
x


By symmetry,
33
1200.xy x y

So,
3
2 150xyππ and
35
150.
3
zπ

` 8. (a)
22
22
2
2
, 2 10 10
11
2
44
11
2
24
Txy x y y
xyy
xy








2
2
12
1
18 14
yx θ

ellipse
(b) On

22 2 2 2
1, , 2 1 10 12
xyTxyTy yyy yy


13
210 , .
22
Ty y y x
Inside:
1
40, 210 0,
2
xyTxTy






139
0,
24
T

π


minimum

31 49
,
22 4
T




maximum
−
1
2
1 2
−
1
2
1
2
y
x
3
−1
3
−22( (,
−2 2( (,
y
x
c = 1
c = −1
c = 0

283. Problem Solving for Chapter 13 283
© 2010 Brooks/Cole, Cengage Learning
9. (a)
11
,1
aa aaff
Cax y C a x y
xy
θθ θCC
ππθ
CC




11
1
1
1
1
aa aa
aa
aaff
x yCaxyCaxy
xy
Ca C a x y
Cx y f
θθ
θ
θCC

CC


ππ

(b)
1
11 1
,,
aa
aa a a a a
ftx ty C tx ty Ct x t y Cx y t tf x y
θ
θθ θ
ππ ππ
10. cos , sin ,xryrzzπππ


2222 222
22 2
sin cos Similarly,
cos sin .
sin cos cos
uuxuyuzu u
rr
xyz x y
uu u
rx y
u ux uy uz u ux uy uz u
rrr r
xxyxz x yxyyz




CCCCCCCC C

CCCCCCCC C
CC C

CC C
CCCCCCCC CCCCCCC


C CCCCCCCC C CCCCCCCC
22 2
22 2 2 2
22
sin
sin cos 2 sin cos cos sin
y
uu u uu
rr r rr
xy xy xy

C
CC C CC

CC CC CC


Similarly,
22 2 2
22
22 2
cos sin 2 cos sin .
uu u u
rx y xy
CC C C

CC C CC

Now observe that

222222
22
222222
22 2 2
22
22 2
222
222
11 1
cos sin 2 cos sin cos sin
11
sin cos 2 sin cos cos sin
uu uu u u u u u
rrrr z x y xy rx y
uu u u uu
x yxy rxryz
uuu
xyz



CC CCC C C C C


CCCCC C CC C C
CC C CCC


CC CC C CC
CCC

CCC
.

So, Laplace’s equation in cylindrical coordinates, is
222
2222
11
0.
uu uu
rrrr z

CC CC

CCCC

11. (a)

22
64 cos 45 32 2
64 sin 45 16 32 2 16
xtt
ytttt



(b)
2
tan
50
32 2 16
arctan arctan
50 32 2 50
y
x
ytt
x t
π

θ
ππ

(c)



22
22 432
2
64 8 2 25 25 2 16 8 2 25 25 2
1
64 256 2 1024 800 2 625
32 2 16 32 2 50
1
32 2 50
tt tt
d
dt tttt
tt t
t


ππ
θ




"
(d)
No. The rate of change of
is greatest when the projectile is closest to the camera.
0 4
−5
30

284. 284 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
(e) 0
d
dt

when



2
2
8 2 25 25 2 0
25 25 4 8 2 25 2
0.98 second.
28 2
tt
t




No, the projectile is at its maximum height when
32 2 32 0dy dt t or 21.41t seconds.
12. (a)
22
22 2 2 3 4 2
32 2 32 2 16 4096 1024 2 256 16 4 2 16dxy t tt t t t tt t
(b)

2
2
32 3 2 8
42 16
tt
dd
dt
tt




(c) When
2:t

32 12 6 2
38.16
20 8 2
dd
dt



ft/sec
(d)


32
2
322
2
32 6 2 36 32 12
0
42 16
ttt
dd
dt
tt



when 1.943t seconds. No. The projectile is at its maximum height
when
2.t
13. (a) There is a minimum at 0, 0, 0 , maxima at 0, 1, 2e and saddle point at 1, 0, 1 :e








22 22
22
22 22
22 32 32
222
222 24 2 0 2 0
xy xy
x
xy xy
fxye x xe
e xyxxe xxyx xxyx













22 22
22
22 22
22 32 32
224
224 42 4 02 20
xy xy
y
xy xy
fxye y ye
exyyye yxyy yxyy





Solving the two equations
32
20xxyx and
32
220,yxy y you obtain the following critical points:
0, 1 , 1, 0 , 0, 0 . Using the second derivative test, you obtain the results above.

(b) As in part (a), you obtain





22
22
22
22
212
22 2
xy
x
xy
y
fe xx y
feyxy







The critical numbers are
0, 0 , 0, 1 , 1, 0 .
These yield

1, 0, 1e minima
0, 1, 2e maxima

0, 0, 0 saddle
x
y
1
2
−1
1
z
x
y
1
2
2
z

285. Problem Solving for Chapter 13 285
© 2010 Brooks/Cole, Cengage Learning
(c) In general, for 0you obtain

0, 0, 0minimum

0, 1,e$ maxima
1, 0,e saddle
For 0,
you obtain

1, 0,e minima
0, 1,e$ maxima

0, 0, 0saddle

14. Given that
fis a differentiable function such that
00,,fx yDπ 0then0, 0 0
xfxy π and00,0.
yfxy π
Therefore, the tangent plane is
00zzθθ π or
000 ,zz fxyππ which is horizontal.

15. (a)
(b)
(c) The height has more effect since the shaded region
in (b) is larger than the shaded region in (a).
(d)
Ahl dA l dh h dl
If 0.01dlπ and 0,dhπthen
1 0.01 0.01.dAππ
If 0.01dhπ and 0,dlπthen
6 0.01 0.06.dAππ

16.
,5,
18
r


π




0.05, 0.05
cos 5 cos 4.924
18
sin 5 sin 0.868
18dr d
xr
yr







(a)
dxshould be more effected by changes in.r


cos sin
0.985 0.868
dx dr r d
dr d

θ

dxis more effected by changes inrbecause
0.985 0.868.
(b)
dyshould be more effected by changes in .
sin cos
0.174 4.924
dy dr r d
dr d




dyis more effected bybecause 4.924 0.174.
17. Let
,.
x
gxy yf
y

π




2
,
1
,
y
x
x xx x xx
gxy f yf f f
yyyyyy
xx
gxy yf f
yy y
θ



ππ


Tangent plane at
000,,
xyzis
0000 0
000
0000 0
0000
0000
10
0.
xxxx x
fxxf f yyzyf
yyyy y
xxxx
fxf f yz
yyyy






This plane passes through the origin, the common point of intersection.
1 cm
6 cm
1 cm
6 cm
x
1
4
5
2
314 52
3 5
18
π
=θ
18
π
θ
(r, ) = 5,()
y

286. 286 Chapter 13 Functions of Several Variables
© 2010 Brooks/Cole, Cengage Learning
18.
22
2xyx

2
2
11xy Circle

22
22
1
xy
ab
Ellipse
The circle and ellipse intersect at
,
xyand ,xyθfor a unique value of .x

2
222
2
b
yax
a
πθ
Ellipse


2
222
2
2
b
xaxx
a
Circle

2
22
2
120
b
xxb
a


Quadratic
For these to be a unique
x-value, the discriminant must be 0.

2
2
2
2224
441 0
0
b
b
a
aabb

θθ π



We use lagrange multipliers to minimize the area
,
fab abπ of the ellipse subject to the constraint

2224
,0.gab a ab b
f gGDπD

22 3
,22,24ba a ab ab b G



2
23
22
24
baab
aabbG
Gπθ



422 222 42 2
232
42 22 2
22 4 2 2
ba a
bab aab bab
aab b ab

G

θθ
Using the constraint,
2224
0,aabb
2
22
0
22
3
2 2
36
2, .
22
aa
aa
a
ab

π
ππ
Ellipse:

22
1
92 32
xy

19.



2
2
2
2
1
cos cos
2
1
sin sin
2
1
cos cos
2
1
sin sin
2
u
xtxt
t
u
xtxt
t
u
xtxt
x
u
xtxt
x
C


C
C


C
C


C
C


C

Then,
22
22
.
uu
tx
CC
π
CC
20.


1
,
2
uxt f x ct f x ct


Let rxctπθ and .
sxct

Then

1
,.
2
ur s f r f s







22 2 222
22
22 2 22
22 2 22
22
22 2 22
11
22
11
222
11
11
22
111
11
22 2 2
uurus df df
cc
trtst dr ds
u df df c df df
cc
tdr ds drds
uurus df df
xrxsx dr ds
udf df dfdf
xdr ds drds
CCCCC

CCCCC
C


C
CCCCC

CCCCC
C


C

So,
22
2
22
.
uu
c
tx
CC
π
CC
y
x
−11
−1
−2
1
2
(x, y)
(x, −y)

287. © 2010 Brooks/Cole, Cengage Learning
CHAPTER 14
Multiple Integration
Section 14.1 Iterated Integrals and Area in the Plane.............................................288
Section 14.2 Double Integrals and Volume ............................................................297
Section 14.3 Change of Variables: Polar Coordinates ...........................................308
Section 14.4 Center of Mass and Moments of Inertia............................................316
Section 14.5 Surface Area .......................................................................................325
Section 14.6 Triple Integrals and Applications.......................................................332
Section 14.7 Triple Integrals in Cylindrical and Spherical Coordinates ...............344
Section 14.8 Change of Variables: Jacobians.........................................................350
Review Exercises........................................................................................................358
Problem Solving.........................................................................................................367

288. 288 © 2010 Brooks/Cole, Cengage Learning
CHAPTER 14
Multiple Integration
Section 14.1 Iterated Integrals and Area in the Plane
1.
2222
00
22
x x
xydy xy y x x x
*

2.
2
2422
2
11
1
22 2
x
x
x
x
yy xxx
dy x
xx xx



*

3.
+,

2 2
11
ln
ln 2 0 ln 2 , 0
y yy
dx y x
x
yy yyy


*

4. +,
cos cos
00
cos
y y
ydx yx y y*

5.
2
4
242
4
222
0
0
14
22
x
x
xx
xydy xy






*

6.

3 3
22 23 2 233 523259
33
3
x x
xx
x ydy xy y x x x xx x x x x x


*

7.

2
222 2ln 1 1
ln ln ln ln , 0
22 2
y
y
y
y
e
y
e
yx y
dx y x y y e y y y
x




*

8.

2
1
22
1 32 12
22 32 2 2 2 2
2
1
2
1
2111
21 1 12
33 3
y
y
y
y
y
xydx x yx y y y y








*

9.

33 33
222
42 2 2
0000
1
,, ,
xx xx
yx yx yx x yx x x
yx yx
ye dy xye x e dy x e x e x e x e
u y du dy dv e dy v xe





**

10.


22
32
2
33
11
33
sin cos 1 cos sin cos
cos cos cos cos cos cos
yy
y
xydx x xydx
x xy y yy




**

11.

212 1 1 1
22
000 0 0 0
1
2
22 2 3x y dy dx xy y dx x dx x x
** * *

12.
2
3
12 1 1
22 2 2 2
12 1 1
2
1
3
1
2
1
1
88
22
333
16 4 16 4 16 4 16
48
3333333
y
x y dy dx x y dx x x dx
x
xdx x










** * *
*

13.
4 2
3
24 2 2
22 2 2 3
10 1 1
10
64 64 8 128 64 64 8 8
228
333333333
x
x y dxdy xy dy y dy y y




** * *

14.

3
2
23 2 2
2222
11 1 1
1
2
2
23
1
1
91
3
222
2162
42 4 8 4 18
333
x
x y dx dy xy dy y y dy
ydy y y











** * *
*

289. Section 14.1 Iterated Integrals and Area in the Plane 289
© 2010 Brooks/Cole, Cengage Learning
15.
1 2
2
21 2 2
00 0 0
00
111
cos cos cos sin
2222
y
y x dy dx x dx x dx x






** * *

16.

ln 4 ln 3 ln 4 ln 3
000 0
ln 4 ln 4
ln3 ln3 ln4 ln3 ln4 ln3
00
1124316
xy xy
xx xx
edydx e dx
eedxee e ee






** *
*

17. +,
sin sin
2
000 0 0 0
1
2
1 cos cos sin sin cos cos sin 1 1 2
x x
xdydx y y x dx x x xdx x x



** * *

18.

444 4
2 41
41111 1 1 14
24x x
xxxxx
ye dy dx y e dx xe e dx xe e e
ee



** * *

19.

1
111 32
2222
00 0 0
0 0
12 1
23 3
1111
x
x
xdydx y x dx x xdx x


** * *

20.

2 42
444 32 32
33233
40 4 4
0 4
204822
999
64 64 64 64 0 128 2
x
x
xdydx y x dx x xdx x





** * *


21.

3
3
53 5
22 2
10 1
0
5
55
33 3 2
11
1
11
33
434
3 39 9 39 9 39 9 39
9 9 9 25 625 1629
44216216216
y
y
x
x y dx dy x xy dy
y y ydy y ydy y y

0










** *
**

22.
2
3
22 2 2
22 2 33 33
000
2
4
2
32
0
0
2162
10 2 2 10 2 20 4 10 2
333
20 5 80 140
10 5 20
3333
y
y
y
y
x
x y dx dy x y x dy y y y y y y dy
y
yydyy







** * *
*

23.

2
12 1
11 1 1 32
22232
00 0 0
0 0
11 11122
22 26233
11 1
y
y
x y dx dy x xy dy y y y dy y y y




** * *

24. +,

2 2222 2 2 2
23 34
22 3603 6 0 0 0
8
3
33 3843 16
yy yy
yyyy
ydxdy xy dy y y dy y y




** * *

25.
+,
2
4
2
24 2 2 2
000 0 0 22
0
22
224
44
y
y
x
dx dy dy dy y
yy






** * *

26.
+,
33 33 3
22 110 1 1 1
0 44 4
arctan ln ln 3
4
y
y
x
dx dy dy dy dy y
xy y y y y



** * * *

27.
2cos 2
2
22cos 2 2
2
00 0 0
00
1
2cos sin2
222
r
rdrd d d






** * *


28.

3cos
22
43cos 4 4 4
03 0 0 0
3
4
4
0
0
3cos 3 3 3
1cos2
22242
333333
cos 2 sin 2
4484816
r
rdrd d d d
d














** * * *
*

290. 290 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
29.

sin
2
2sin 2 2
2
00 0 0
0
2
22
2
0
0
1
sin
22
111 1
cos 2 cos 2 sin 2
44242328
r
rdrd d d
d








ππ





** * *
*

30.
4 4
4
4cos 4 4 cos
23 3
000 0 0
0
cos 1 1 3
3 sin sin cos sin 1
44 16 2
rdrd r d d





πππθπθθπ

** * *

31.
1
2
1
2
10 1 1
10
11 1 11
0
22 2 22
x
x
y
ydydx dx dx
xx
!
!!!



** * *

32.
3
2 3
33 3
22
2 000 0 0
0
9
arctan
12232
xx
dy dx x y dx x dx
y ! !
πππ"π

** * *

33.

11 1 1
1
11 11
ln 0
Diverges
dx dy x dy dy
xy y y y
!
!! ! !

ππ!θ


** * *

34.


22 22
22
00 0 0
0
0
1111
2244
xy xy
yy
xye dx dy ye dy ye dy e
!
!
!! ! !
θθ


πθ π πθ π


** * *

35. +, + ,
+, + ,
83 8 8 38
0000 0 0
38 3 3 83
0000 0 0
3324
8824
A dy dx y dx dx x
A dx dy x dy dy y
πππππ
πππππ** * *
** * *

36. +, + ,
+, +,
23 2 2 32
1111 1 1
32 3 3 23
1111 1 1
222
2
A dy dx y dx dx x
A dx dy x dy dy y
πππππ
πππππ** * *
** * *

37. +,
+,
2
32 224 2 2 4
2
000 0 0
0
4
44 4 4 4 4 12 32
000 0 0 0
0
16
44
33
2216
44148
333x x
y y x
Adydxydxxdxx
A dx dy x dy y dy y dy yθ θ
θ θ
πππθπθπ



πππθπθθθπθθππ


** * *
** * * *


x
4
6
8
2468
2
y
2
2
31
3
1
x
y
x
−11
1
2
2
3
4
3
yx= 4−
2
y

291. Section 14.1 Iterated Integrals and Area in the Plane 291
© 2010 Brooks/Cole, Cengage Learning
38. +,

+, +,

+,
551 1 5 5 11
020 2 2 2
2
12 5 1 1 1
02 122
1
212 1 12 1511 12
222 0012 012
121
212
1
11
313 2x x
y
y
A dy dx y dx dx x
x
A dxdy dxdy
x dy x dy dy dy y y
yy
θ θ


ππππθπ

θ




** * *
** **
** **

39.

+,



42 2 2
42 4 4 2
000 0 0
0
2
42
00
88
44 4
323
8
3x x
y x
A dy dx y dx x x dx x x x
Adxdyθ θ
θ



ππ
** * *
**

Integration steps are similar to those above.

40.
+,


42
32
0
442
32
32
00
4
252
0
23
88
23
02 0
8
2
53
0
2
2216
16 32
555
2
3316
32 16
545 5
x
x
x
x
y
y
Adydx
ydx xxdx
xx
y
A dx dy y dy
y
y
π
ππθ

πθ πθ π



ππθ



πθπθπ


**
**
** *


41.
+, +,

+,
323 55
00 30
3523 5
0003
35
03
35
22
03
25
032
2 5
320
2
0
2
2
2
0
0
2
5
3
11
55
32
3
5
2
55
555
24
xx
xx
y
y
y
y
Adydxdydx
ydx ydx
x
dx x dx
xxx
Adxdy
xdy
y
ydy
y
dy y y
θ
θ
θ
θ







π
π

πθθ



πθ πθ π

** **
**
**
**
*
*
*

42.
+, +,
+,


+, +,

399
00 30
39 39 9
0003 03
3
9
2
3
0
19 39
01
1399
01
13
01
13
22
01
9
1
9ln
2
9
9ln9 ln3
2
9
1ln9
2
9
9
11
99ln
22
xx
xx
y
yy
y
yy
A dydx dydx
ydx y dx xdx dx
x
xx
A dxdy dxdy
xdy x dy
ydy ydy
y
yy yy
















** **
** **
** **
**
**

9
1ln9
2


43.

+,



22 22
000 0
2
22 2
00
2
0
2
0
4
cos
sin , cos
1cos2
2
1
sin 2
22
4
abaax a ba a x
aA
dy dx y dx
b
a x dx ab d
a
xa dxa d
ab
d
ab
ab








θ θ
ππ
πθπ
ππ





π** *
**
*

So, .
A abπ


22
00
44
babby
A ab
dx dyθ
ππ**

So, .
A abπ Integration steps are similar to those
above.
x
25
3
3
5
4
2
41
1
1
x−1
y=
y
x
2
3
4
12 34
1
yx= (2 )−
2
y
−11
1
2
3
4
5
6
7
8
2345678
y
x
(4, 8)
y = x
3/2
y = 2x
x
2
3
4
1234 5
1
−1
yx= 5−
yx=
2
3
y
x
4
6
24 68
2
−2
(3, 3)
(9, 1)
yx=
y=
9
x
y
a
b
x
b
a
yax=
22
−
y

292. 292 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
44.


242
02 22
24
02
24
22
02
2
2
22 2
00
0
2
22
2
44
143 2
22
2
y
yy
x
x
Adxdydxdy
yy
dy dy
yy
y
x
Adydxxxdx











ππθππ

** **
**
** *

45.
+,


2
14
22
21 4
22
1
2
2
1
2
2
1
23
2
911
23 2
42
2
2
x
x
x
x
Adydx
ydx
xxdx
xxdx
xx x
θ

θ
θ
θ
θ
θ
π
π
πθθθ
πθθ
πθθ π
**
*
*
*


+, +,


32 44
04 30
34 24
4003
34
03
34
32 32
2
03
912 4
23 3 2
2
2
24 24
24 4
yy
y
yy
y
A dxdy dxdy
xdy xdy
yydyydy
yy y y
θθ
θθ
θθ
θθ



πθθθ θθ π
** **
**
**

46.





2
24
00
2
2
0
2
2
0
2
0
2
0
2
2
24 2
2
00 0
22
2
00
2
0
2
1
2
1
2
4
4cos
21cos2
2sin2
2sin , 2cos , 4 2cos
4
4cos 2 1cos2
2sin2
2sin , 2cos , 4 2
x
y
Adydx
xdx
d
d
xdxdx
Adxdyydy
dd
ydydy












θ
θ
π
πθ
π



ππ θπ
ππθ



ππ θπ**
*
*
*
** *
**

cos

47.

4
00
44
0
,,0 ,04
,
y
x
fxydxdy x y y
fxydydx

π**
**


48.


42
0
2
2
00
,, 2,04
,
y
x
fxydxdy y x y
fxydydx

π**
**


49.

2
24
2
20
2
24
2
04
,,0 4,22
x
y
y
fxydydx y x x
dx dy
θ
θ
θ
θθ

π**
**


50.

2
24
2
00
44
00
,,04,02
,
x
y
fxydydx y x x
fxydxdy
θ
θ

π**
**


x
2
3
4
12 34
1
yx= 2
yx=
y
x
−2−11
1
2
2
3
yx=+ 2
yx= 4−
2
(1, 3)
y
x
2
12
1
yx= 4−
2
y
1234
1
2
3
x
y
x
2
3
4
12 34
1
yx=
2
y
−2−112
−1
3
1
y
x
−1
−1
1
2
3
4
1234
y
x

293. Section 14.1 Iterated Integrals and Area in the Plane 293
© 2010 Brooks/Cole, Cengage Learning
51.

10 ln
10
ln 10 10
0
,,0ln,110
,
y
x
e
fxydxdy x y y
fxydydx

π**
**


52.

2
10
2
2ln
2
01 1
,,0 ,12
,,
x
e
x
eey
e
fxydydx y e x
fx y dx dy f x y dx dy
θ
θ
θ
θ
θ
θ
θθ

**
** **


53.

11
2
2
1
1
0
,, 1,11
,
x
y
y
fxydydxx y x
fxydxdy
θ
θ

π**
**


54.


2cos
20
1 arccos
0 arccos
,,0cos,
22
,
x
y
y
fxydydx y x x
fxydxdy


θ
θ

π**
**



55.
12 21
00 0 0
2dy dx dx dyππ** **


56.
24 42
12 21
2dx dy dy dxππ** **


57.
22
11 1 1
2
01 10
2
yx
y
dx dy dy dx
θθ
θθ θ
ππ** * *


58.

2
24 2
22
2
24 2
2
24
2
24
44 4
4
x
x
y
y
dy dx x x dx
dx dy

θ
θθθ θ
θ
θθθ

π** *
**


59.
24424
00 20 0
4
xx y
y
dy dx dy dx dx dy
θθ
** ** **


1
2
4
6
8
23
y
x
−112
2
3
y
x
x
−2−112
2
3
4
y
2
x
2
1
2
3
4
π
4
π
−
y
312
3
2
1
x
y
x
2
3
4
12 34
1
y
x
1
1
−1
y
−1
−1
1
1
y
x
x
2
3
12 34
1
−1
y

294. 294 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
60.

42 66 4 6
00 40 0 4
2
2
26 2
02 0
0
6426
2
3
63 6 6
2
xx
y
y x
dy dx dy dx dx x dx
y
dx dy y dy yθ
θ


πθ πθ π
** ** * *
** *

61.
21 12
02 00
1
y
x
dy dx dx dyππ** **


62.

93 9
00
9
32
0
3
32
33
2
00 0
0
3
2
327189
3
9
3
x
y
dy dx x dx
xx
y
dx dy y dy
πθ

πθ πθπ



πππ

** *
** *


63.
3
11
23
00
5
12
yx
yx
dx dy dy dxππ** **


64.
2
24 4 4
20 0 4
32
3
yx
x
dx dy dy dx
θθ
θθθ
ππ** **



65.
The first integral arises using vertical representative rectangles. The second two integrals arise using horizontal representative
rectangles.




2
550 5 32
22 2 2 5
00
2
55250 552 32
22 22 5 2 2
00 5 0 0 5
15,62511
3324
11
33
15,625 15,625 15,625 15,625
18 18 18 24
50
50
x
x
yy
x y dy dx x x x dx
x y dx dy x y dx dy y dy y y dy

θ
θ

πθθπ



** *
** * * * *

66. (a)
22
2
0
y
y
A dx dyπ** +,
2
3
222
22
2
00
0
84
24
333y
y y
xdy yydy y

ππθπθπθπ


**

(b)
4
02
x
x
A dy dxπ** +,
44
200
4
2
32
0
2
2164
4
343 3
x
x x
ydx x dx
x
x

ππθ



πθπθπ


**

Integrals (a) and (b) are the same.

y =
x
2
y = 6 − x
y
x
−1
−1
1
1
2
3
4
5
6
23456
(4, 2)
21
2
1
x
y
−1
−2
1
2
3
4
5
−3
−4
−5
123456789
y
x
y =x
2
2
1
1
3
x
(1, 1)
y
x = y
x = y
2
x
2
12 3
1
−1
−2
xy= 4−
2
y
x
5
5
(5, 5)
(0, )5 2
yx=50−
2
yx=
y
1234
1
2
3
4
x
(4, 2)
y = x
x
= y
2
y =
x
2
x = 2y
y

295. Section 14.1 Iterated Integrals and Area in the Plane 295
© 2010 Brooks/Cole, Cengage Learning
1
1
x
y
1234
1
2
3
4
(4, 2)
x
y
67.


22 2
33
000
2
2
3
0
0
2
32
0
2
32
3
0
11
1
2
1
1
2
112
1
233
1126
27 1
999
y
x
y
x ydydx x ydxdy
x
ydy
yydy
y


"



""


** **
*
*


68.
2
42 2
33
000
2
2
3
0
0
2
22
3
3
0 0
33
22
3
2
3
ln 2
2
ln 10 ln 2 ln 5y
x
y
dy dx dx dy
yy
x
dy
y
y
dy y
y









** **
*
*


69.
12 2 2 22
02 0 0
2
22 22
00
0
2
2
4
0
44
42
1
y
yy
x
y
yy
y
e dy dx e dx dy
xedy yedy
ee







** **
**



70.

22 2 22
000
2 2
0
0
2 2
0
2
2
0
40
4
1
2
11
22
11
1 0.4908
2
y
yy
x
y
y
y
y
e dy dx e dx dy
xe dy
ye dy
e
ee
e


















** **
*
*


71.





11 1
22
000
1
2
0 0
1
2
0
1
2
0
sin sin
sin
sin
1
cos
2
11
cos 1 1
22
1
1 cos 1 0.2298
2
x
y
x
x dx dy x dy dx
yxdx
xxdx
x









** **
*
*


72.
+,
24 4
2
000
4
0 0
4
0
4
0
sin sin
sin
sin
sin cos
sin 4 4 cos 4 1.858
x
y
x
xxdxdy x xdydx
yx x dx
xxdx
xx x





** **
*
*


(4, 2)
x
y
y = x
1234
1
2
3
4
321
3
2
1
x
y
321
3
2
1
x
y
12
1
2
(2, 2)
x
y

296. 296 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
73.
22
32
2
0
1664
105
3 15.848
x
x
x y dy dx**

74.
12
0 sin 2 sin 3
sin 0.408
23y
y
x y dx dy**

75.


4 2
00 2
ln 5 2.590
11y
dx dy
xy

**

76.
4
22
00
6
aax a
x y dy dxθ
**

77. (a)
313
2
2
42 32
32
xy yx
x
xyxyy
π&π
π&π&π

(b)
13
8
22
2
032
x
x
xyxydydxθ**

(c) Both integrals equal
67,520
97.43.
693



78. (a)
22
2
44
4164
4
yxxy
x
yxy
πθ&πθ
πθ & π θ
(b)
22 32 4 164
222 22 22
04 20 30
11 1
y
yxy xy xy
dx dy dx dy dx dy
xy xy xy θ
θ

** ** **


(c) Both orders of integration yield 1.11899.
79.
2
24
00
20.5648
x
xy
edydx
θ
**

80.
22
33
0
16 6.8520
x
x y dy dx **

81.
21cos
2
00 15
6cos
2
rdrd

π**

82.
2
21sin
00
45 135
15 30.7541
32 8
rdrd

**

83. An iterated integral is integration of a function of several
variables. Integrate with respect to one variable while
holding the other variables constant.
84. A region is vertically simple if it is bounded on the left
and right by vertical lines, and bounded on the top and
bottom by functions of x. A region is horizontally simple
if it is bounded on the top and bottom by horizontal lines,
and bounded on the left and right by functions of y.
85. The region is a rectangle.
86. The integrations might be easier. See Exercises 59–62.
87. True
88. False, let ,.
fxy xπ
x
1
1
2
2
3
4
y
x
2468
−2
2
4
(8, 2)
xy=
3
xy= 4 2
y

297. Section 14.2 Double Integrals and Volume 297
© 2010 Brooks/Cole, Cengage Learning
Section 14.2 Double Integrals and Volume
For Exercises 1–4, 55 1
iixy and the midpoints of the squares are

35 33353311 1 1 71 1 7
22 22 22 22 22 22 22 22
,,,,,,,,,,,,,,,.

1.

8
1
,
, 12342345 24
ii i i
i
fxy x y
fx y x y


55
F




2
2
42 4 4 4
2
000 0 0
0
22 2 24
2
y
x y dy dx xy dx x dx x x


** * *

2.

2
11
,
2
1 9 25 49 3 27 75 147
,21
16 16 16 16 16 16 16 16
ii i i
i
fxy xy
fx y x y
!


55
F


24
22 3
42 4 4
22
00 0 0
00
164
21.3
24 33
xy x
x y dy dx dx x dx



** * *

3.

22
8
1
,
210265010183458
,52
44 4 4 4 4 4 4
ii i i
i
fxy x y
fx y x y


55
F



24
33
42 4 4
22 2 2
00 0 0
00
8 2 8 160
2
33333
yxx
x y dy dx x y dx x dx



** * *

4.


8
1
1
,
11
4 4 4 4 4 4 4 4 7936
, 1.680
9 15 21 27 15 25 35 45 4725
ii i i
i
fxy
xy
fx y x y



55
F





2
42 4
00 0
0
4 4
00
11
ln 1
11 1
ln 3
ln 3 ln 1 ln 3 ln 5 1.768
1
dy dx y dx
xy x
dx x
x




"


** *
*

5.
44
00
, 32312823 3130 27 22 2827 2419 23221914
400
fxydydx
**

Using the corner of the ith square furthest from the origin, you obtain 272.
6.
22
00
,428620fxydydx **

x
2
3
4
12 34
1
y

298. 298 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
7.
21 2 2 12
22
0000 0 0
12 2 2 22 2 8x y dy dx y xy y dx x dx x x
** * *


8.

2
2
22 2
00 0
0
2
2
00
0
11
sin cos sin sin 2
22
11
sin 1 cos 2 sin 2
22 8 82 8
x y dy dx x y y dx
xdx xdx x x










ππθπθπ


** *
**


9.

3 6
63 6 6
2223
02 0 0
2 0
195935
336
2282224
y
y
x y dx dy x xy dy y y dy y y y



** * *


10.


4
32 72 5 92 6
444
22
012 0 0
12 0
2 1024 256 256
3 3 24 27 144 27 9 27
y
y
y
y
xy y y y y
x y dx dy dy dy

ππθπθπθπ

** * *


11.

22
22
32
22222
22
22
12
20
23
ax a
aax a a
aax a a
ax a
x y dy dx xy y dx x a x dx a x
θ
θ
θθ θ θ θ
θθ θ




** * *


312
3
1
2
x
y
2
2
31
3
1
x
y
642
6
4
2
x
(3, 6)
y
1
1
2
3
4
234
y
x
(2, 4)
a
x
a−a
−a
y

299. Section 14.2 Double Integrals and Volume 299
© 2010 Brooks/Cole, Cengage Learning
12.

10 11 1 1 01
1001 00 0 0
11
21 21 1
00
11
22
y y
xy xy xy xy
yy
yy
e dx dy e dx dy e dy e dy
ee dy ey e ee
θ θ

θθ
θθθ




** ** * *
*

13.
53 35
00 00
333 5
22
00 0 0
25 25 2251
2244
xy dx dy xy dy dx
xy dx x dx x
π
ππππ
** **
**

14.
+,
22
00
2
0
sin sin sin sin
sin cos sin 0
x y dx dy x y dy dx
xydx xdx




θθ
θθ
π
πθ π π** **
**

15.


24222
22 22 22
11 2 2 1
22
22
1
2
22
1
2
2
1
1
1
ln
2
1
ln 5 ln 2
2
15 1 5 15
ln ln ln
22 2 2 22
yx
yx
x
xyyy
dx dy dx dy dy dx
xy xy xy
xy dx
xxdx
dx x




πθ

ππ π


** ** **
*
*
*

16.
44 44
00 00
xy
yy
xedydx xedxdy
θθ
π** **

For the first integral, you obtain:




44 4
4
000
4
2
444
0
1 5 8 13.
2
x
yx
x
xe dx xe x dx
x
ex ee
θ
θ
θ
πθ





**

17.


2
44 14
34 04
2
1 4
2
40
1 2 2
2
0
1
24 2
0
1
5
32
0
22
44
16 8 16 8
6
34
55
yx
yx
x
x
y dx dy y dy dx
ydx
xxdx
xxxxdx
x
xx
θθ
θθ
θ
θ
θπ θ
πθ


πθ θ θ θ






** **
*
*
*

x
−11
2
yx=+ 1−yx=+ 1
y
x
25
3
3
5
4
2
41
1
y
x
y
22
π
2
3π
2
3π
2
5π
π
ππ
2π
2
π
−
2
3π−−
x
2
3
4
12 34
1 yx=
yx= 2y
x = 1
x= 2
1
1
2
3
4
234
y
x
134
1
2
3
4
(1, 3)
y = 4 − x
y = 4 − x
2

y
x

300. 300 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
18.

24 4
2 22
000
4
2
44
2
22
00
00
11
1111
ln 1 ln 17
21 21 4 4
x
y
xyy
dx dy dy dx
xx
yx
dx dx x
xx
π





** **
**

19.

22
434 5 25 3 25
00 40 043
2
325
2
043
33
23
0 0
1
2
25 25 1
18 18 3
9925
xx y
y
y
y
xdydx xdydx xdxdy
xdy
ydy y y
θθ
θ

π

πθπθπ
** ** **
*
*

20.




2
24
22
2
04
2
24
22
20
2
4
2
23
2
0
2 32
22 2
2
2
32 32
22 22
2
1
3
1
44
3
11
4 4 4 arcsin 4 6 4 24 arctan 4
42 212 2
y
y
x
x
x y dx dy
x y dy dx
xy y dx
xx xdx
xx x
xxx xxxx

θ
θθ
θ
θ
θ
θ
θ
θ













**
**
*
*

21.
2
2
42 4 4
00 0 0
0
4
24
yy
Vdydx dxdx

ππππ

** * *


22.
42
00
44 2
2
000
62
6832
V y dy dx
y y dx dx
πθ
πθ π π
**
**


23.
2
00
2
2
0
0
2
2
2
0
2
33
2
0
4
4
2
4
2
2
63
88
84
63
y
y
Vxydxdy
x
x xy dy
y
yydy
yy
y
πθθ

πθθ



πθθ


πθθ


πθθ π
**
*
*

24.
2
22
2
00 0
0
4428
x
Vdydxxdxx

ππππ
** *



x
2
3
4
12 34
1
yx=
y
x
25
3
3
5
4
2
41
1
xy= 25−
2
(4, 3)
xy=
4
3
y
x
−2−11
1
2
3
4
xy= 4−
2xy=4− −
2
y
x
2
3
4
12 34
1
y
x
2
3
4
12 34
1
y
1
1
2
2
y = x
y
x
1
1
2
2
y = x
y
x

301. Section 14.2 Double Integrals and Volume 301
© 2010 Brooks/Cole, Cengage Learning
25.

23 4
6234 6
2
00 0
0
6
6
232
0
0
12 2 3 1 3
12 2
442
11
26 6 12
618
x
x
xy
Vdydxyxyydx
xx dx xxx


θθ
ππθθ






** *
*

26.

2
2
22 2
00 0
0
2
2 23
0
0
22
2
114
22
263
x
x
y
Vxydydxyxydx
xdx x
θ
θ

πθθπθθ



πθπθθπ


** *
*

27.
1
00
1
2324
11
00
00
1
3
22288
y
y
Vxydxdy
xy y y y
xdyydy
πθ

πθ π θ πθπ
**
**

28.

2
2
00
2
4
2
32
0
0
4
424
4
y
V y dx dy
y
yydy y
πθ

πθπθπ

**
*

29.


22 2 2
00 0 0
00
1111
1
111 11 1
Vdydxdxdx
xxy xy x
! !
!! ! !

ππθππθπ

** * *

30.

22 22
000 0 0 0
2244
xy xy xx
V e dy dx e dx e dx e
!!! ! ! !
θθ
ππθππθπ

** * *

31.
2
22 8
22
00
32 2
48
3x
Vxydydx
θ
πθθπ**

32.
1
2
00
1
3
1
x
V x dy dxπθπ**

33.
111 1
234
00 0 0 0 0
11 11
22 88
x x
V xydydx xy dx x dx x πππππ
** * *


x
1
1
−1
2
2
3
3
4
4
5
5
6
2
3
x+ 4y=−
y
1
1
2
2
y = 2 − x
y
x
x
1
1
yx=
y
x
1
1
yx=
y
x
2
12
1
yx=
y

302. 302 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
34.
+,
5
00
55
2
000
5
3
0
1251
33
x
x
V x dy dx
xydx xdx
x
π
ππ
ππ

**
**

35.
24
2
00
22 4
22
000
2
3
0
4
432
33
V x dy dx
xydx xdx
x
π
ππ


ππ

**
**


36.


22
22 2
00
22
22 2 22
0 22
0
3
22 2 3
0
0
8
4arcsin
14
42
233
rrx
rx
r
r
r
V r x y dy dx
y
yr x y r x dx
rx
r
rxdx rx x


θ
θ
πθθ


θ

πθπθπ

**
*
*

37. Divide the solid into two equal parts.


11
22
00 0
0
1
1 32
22
0
0
22
33
21 21
21 1
x
x
V x dy dx y x dx
xxdx x

πθπθ


πθπθθπ
** *
*

38.


2
24
2
00
2
22
0
2
35
2
24
0
0
4
44
8 64 32 256
16 8 16 32
35 3515
x
V x dy dx
xxdx
xx
xxdx x
θ
πθ
πθθ



**
*
*


39.



2 2
24 2 4
2
00 0 0
2
22
0
2
32
23
0
1
2
1
2
1611
363
42
42
xx
Vxydydxxyydx
xx xdx
xxx
θθ






** *
*

40.
+,
22
2 000 0
2
2
0
0 1
arctan
1
22
Vdydxydx
y
x
dx


! !
ππ


πππ

** *
*


x
−11
1
2
2
3
4
3
y
x
r
r
yrx=
22
−
y
x
1
1
yx=
y
12
1
2
3
4
34
y
x
y = 4 − x
2
x
2
12
1
yx= 4−
2
y
x
2
12
1
y
x
25
3
3
5
4
2
41
1
yx=
y

303. Section 14.2 Double Integrals and Volume 303
© 2010 Brooks/Cole, Cengage Learning
41.

+,


2
211
22
00
2
211
22
00
24 42
22
x
x
V x y x dy dx
xx ydydx
θθ
θθ
πθθθθ

πθθ
**
**


42.


2
211
22
00
22
x
V x x y dy dx
θθ


**


43.

2
24
22
00
4
x
Vxydydx
θ
**


44.
5
2
00
sinV x dx dy

π**


45.



2
2221
22
2
0221
42
y
y
V y x y dx dy
θθ
θθθ


**


46.


2
39
22 22
2
39
2
39
22
00
18
41822
x
x
x
Vxyxydydx
x y dy dx
θ
θθθ
θ


πθθ
**
**


47.

22
2
39
22
00
9,0
81
49
2
x
zxyz
Vxydydx
θ
πθ θ π
πθθπ
**

48.
99
00
81
2
9
y
V y dx dy
θ
πθπ**

49.
20.51
22
00 2
1.2315
1x
Vdydx
xy


**

50.

16 4
00
ln 1 38.25
y
Vxydxdy
θ
**

51. f is a continuous function such that
0,1fxy over a region R of area 1. Let
,fmnπthe minimum value of f over R and
,fMN πthe maximum value of f over R. Then

,,,.
RR R
fm n dA f x y dA f M N dA** ** **

Because
1
R
dAπ**
and
0, ,1,fmn fM N you have
0,1 , ,11.
R
fmn fxydA fM N **

So,
0,1.
R
fxydA**

1
−1
1
x
y
(x − 1)
2
+ y
2
= 1
1
−1
1
x
y
(x − 1)
2
+ y
2
= 1
1
−1
−11
x
xy
22
+= 4
y
x
25
3
3
5
4
2
41
1
y
−1−212
−1
−2
1
2
x
y
x
2
+ y
2
= 9
−1−212
−1
1
3
x
y
+ (y − 1)
2
= 1
x
2
2

304. 304 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
52. 1
1
xyz
abc
xy
zc
ab


πθθ






1
2
1
00 0
0
2
2
0
23
23
2
0
,1
2
111
2
11
2236 23
bxa
ab xa a
R
a
a
xy xyy
VfxydA c dydxcy dx
ab a b
xxb xb x
cb dx
aa a ba
ab x x b x b ab x ab ab
cc
aaa a
θ

θ


ππ θθπθθ




πθθθθθ





** ** *
*
26 6
ab ab abc




53.
112 12 222
02 00
12
12 22
14 14
0
0
2 1 1 0.221
x
xx
y
xx
e dx dy e dy dx
xe dx e e e
θθ
θθθ θ
π


** * *
*

54.
+,
ln 10 10 10 ln
010
ln
10
1
0
10 10
11 11
ln ln
ln
9 y
x
e
y
dy dx dx dy
yy
x
dy
y
dy y
π

π


πππ** **
*
*

55.

2
22 4
24 24 2
222
22
224 24 2
4
2
3
2
2
2
2
444
21616
24 8 16 16
333
64
3
y
xy
xy
y
y dy dx y dx dy x y dy
y
ydy y
θ
θθ
θθθ θθθ θ
θθ
θ
θ

θπ θπ θ





π
** ** *
*

56.

3
31 13 1
444
03 00 0
0
1
1
2
4
0
0
11
111
33
arctan
12
33
24 8
x
x
y
y
dx dy dy dx dx
xxx
x
dx x
x


ππ



ππ



ππ


** ** *
*

57.

1 arccos
2
00
2cos
2
00
2
2 12 32
22
0
0
12 1
23 3
sin 1 sin
sin 1 sin
1sin sincos 1sin 221
y
x
x x dx dy
x x dy dx
xxxdx x





"
**
**
*

x
1
1
1
2
1
2
yx= 2
y
x
25
6
3
10
8
4
41
2
ye=
x
y
−11
−1
1
x
y
x
2
+ y
2
= 4
23
1
2
3
x
y
y

= 3x
(1, 3)
x
2
1
2
yx= cos
ππ
y
x
y
b
c
a
R
z

305. Section 14.2 Double Integrals and Volume 305
© 2010 Brooks/Cole, Cengage Learning
58.

+, +,
22 2 2
2
012 00
22
00
2
0
cos cos
2 cos 2 cos
2cos sin 2cos2 2sin2 1
y
x
y y dy dx y y dx dy
yy ydy y ydy
yy y


** **
**

59.
4
2
42 4
00 0
0
11
Average 2 2
888
x
xdydx xdx



** *
60.
53 5 3
2
000 0
5
2
5
0
011
Average 2
15 15
11915
9
15 15 2 2
xydydx xy dx
x
xdx





** *
*

61.

22
22
00
2
3
22
22
00
0
2
3
01
Average
4
118
2
43 43
18 2 8
43 3 3
x y dx dy
x
xydy ydy
yy








**
**

62.

+,
+,
1
00
11
000
1 1
0011
Average
12
2ln 2ln2ln
2 ln2 2 ln2 2ln2x
x
dy dx
xy
xydx x xdx
dx x





**
**
*


63.

11 1
12
00
1
12 2 2
0
2
21
Average 2
12
111
22
222
21 1
xy x x
x
xx
edydx e edx
ee eee
ee e







** *

64.



+,
2
00
2 00
2
0
22 00
1
Average sin
1
cos
1
cos cos
11
2cos 2sin 0
x y dy dx
xydx
xxdx
xdx x












**
*
*
*

65.

325 250
0.6 0.4
300 200
250
1.6
325
0.4
300
200
325
0.4
300
325
1.4
3001
Average 100
1250
1
100
1250 1.6
128,844.1
1250
103.0753 25,645.24
1.4
xydxdy
x
ydy
ydy
y









**
*
*

66.


24
22
00
1
8
281 224
83 3
Average 20 4
C
xydydx
**

67. See the definition on page 994.
68. The value of ,
R
fxydA**
would be kB.
69. (a) The total snowfall in the county R.
(b) The average snowfall in R.
70. Part (b) is invalid. You cannot have the variable of
integration
yas a limit of integration.
71. No, the maximum possible value is Area 6 6 .
72. The second is integrable. The first contains
2
sinydy*
which does not have an elementary
antiderivation.
73.
,0fxy/for all ,xyand



52
00
5
0
22
01
2
0
1
10
1
5
1
10
11
10 5
,
1
02,12
.
fx y dA dy dx
dx
Px y dydx
dx
!!
! !



** **
*
**
*

(2, 2)
y =x
21
2
1
1
2
2
y
x
1
(1, 1)
y = x
x
y

306. 306 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
74. ,0fxy/for all ,xyand



22 2
00 0
12 1
01 01
,1
42
133
01,12 .
4816
x
fxydA xydydx dx
x
Px y xydydx dx!!
! !

** ** *
** *

75.
,0fxy/for all ,xyand




63
22
36 3 3
03 0 0
30
16 1
04 0
11 11
,9 9 1
27 27 2 2 9 2 18
127
01,4 6 9 4 .
27 27 27
yxx
fxydA x ydydx y xy dx xdx
Px y xydydx xdx!!
! !




** ** * *
** *

76. ,0fxy/for all ,
xyand





0000 0 0
11 1 1 1
21
000
1
2
21 21 1
0
,limlim1
01, 1
1111
1 0.1998.
2222
bb
xy xy x x
bb
xy xy x x
xx
xx
f x y dA e dy dx e dx e dx e
P x x y e dy dx e dx e e dx
ee ee e
!! !! ! !

! ! ! !










** ** * *
** * *


77. Divide the base into six squares, and assume the height
at the center of each square is the height of the entire
square. So,


3
4 3 6 7 3 2 100 2500 m .V

78. Sample Program for T1-82:
Program: DOUBLE
: Input A
: Input B
: Input M
: Input C
: Input D
: Input N
: 0 V
:

BAM G
:

DCN H
: For (I, 1, M, 1)
: For (J, 1, N, 1)
:

A0.5G2I1 X
:
C0.5H2J1 Y
:
VsinXY GH V''
: End
: End
: Disp V
79.
12
00
sin 4, 8x y dy dx m n**

(a) 1.78435
(b) 1.7879
80.
24 3
8
00
20 10, 20
x
edydx m n

**

(a) 129.2018
(b) 129.2756
81.
62
40
cos 4, 8y x dx dy m n **

(a) 11.0571
(b) 11.0414
82.
42
33
11
6, 4x y dx dy m n**

(a) 13.956
(b) 13.9022
83. 125V
Matches d.
84. 50V
Matches a.

x
y
(15, 5, 6)
(15, 15, 7)
(5, 5, 3)
(5, 15, 2)
(25, 5, 4)
(25, 15, 3)
7
20
30
z
x
y
5
5
16
(4, 0, 16)
(4, 4, 16)
(0, 4, 0)
(4, 0, 0)
(4, 4, 0)
z
x
y
3
4
3
3
z

307. Section 14.2 Double Integrals and Volume 307
© 2010 Brooks/Cole, Cengage Learning
85. False

2
11
22
00
81
y
Vxydxdy
θ
πθθ**

86. True
87.


11 11 22
001 0
11 22
00 0
1
2
0
111
222
Average
11
x
tt
x
t
tt
t
fx dx e dt dx e dt dx
edxdt tedt
ee e
ππ πθ
πθ πθ

πθ πθ θ π θ
*** **
** *


88.
2
2
2
1
1
1
x x
xy xy
ee
edy e
xx
θθ
θθ
θ
πθ π


*

So,

+,
2
2
001
2
10
2
1
0
2 2
11
1
ln ln 2.
xx
xy
xy
xy
ee
dx e dx dy
x
edxdy
e
dy
y
dy y
y
θθ
!!
θ
!
θ
!
θ
θ
π
π

πθ


πππ
***
**
*
*

89.
22
9zxyπθ θ is a paraboloid opening downward
with vertex

0, 0, 9 .The double integral is maximized if
0.z/That is,


9 :
22
,: 9.Rxyxy


22 81
The maximum value is 9 .
2
R
xydA

θθ π


**

90.
22
4zx y is a paraboloid opening upward with vertex

0, 0, 4 .θThe double integral is minimized if 0.zThat is,
9 :
22
,: 4.Rxyxy
[The minimum value is
8.]θ

91.


22222
11
22 2
0002
2
22 2 2
2 2 22 22
02 0 2
2
21
0 11 1
tan tan
11 1
2
1 1 11 11
11 1
1ln1 2tan
22xy
xy y
y
yy
x x dx dy dx dx dy dx dy
yy y
xxyy y
dy dy dy dy
yyyyyy
yy







θθ
θ









*******
** * *



2
2
2
121
11 2
ln 1
11 1 1
1ln52tan2 ln142tan2 ln5
222
11
ln 5 2 tan 2 2 tan 2 ln 1 4 0.8274
22y





θθ
θθ










92.
2
39
22
00
9
9
2y
x y dx dy
θ
θθ π**

because this double integral represents the portion of the sphere
222
9xyz in the first octant.


314 9
3
83 2
V


π" π
93. Let
9:
22 22
max ,
00
.
ab bx ay
I e dy dxπ**

Divide the rectangle into two parts by the diagonal line
.ay bxπ On lower triangle,
22 2 2
bx ay/ because
.
b
yx
a


22 22 22 22
00 0 0 0 0
22
22 22 22 22
00
11 1 1
11
22 2
abxa b ayb a b
bx ay bx ay
abab
bx ay ba ab bx ay
I e dy dx e dx dy e dx e dy
ab
e
ee ee
ab ab ab ab

θ


** ** * *

x
1
1
t
y = x
y = x
x
y
−1−2 12345
1
2
3
4
5
6
7
π
y
x
b
a
(a, b)
ay = bx

308. 308 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
94. Assume such a function exists.



1
1111
000 1
1;,01
2
x
x
ux uyuy x dy x
u x dx dx u y u y x dy dxGG
G
*
****

Change the order of integration.


11 1
000 00
11
yy
u x dx u y u y x dx dy u y u y x dx dyG G



*** **

Hold y fixed and let
,.zyxdz dxπθ πθ


10 1
000
11
y
y
u y u z dz dy u y u z dz dyG G



** **

Let

0
.
y
fyuzdzπ*
Then ,0 0,1 .fy uy f f πππ




1
2
1 22
2
0
0
11 1
111101
222 2
fy
fyfydy f f
G G G G






*


2
220.G
For
to exist, the discriminant of this quadratic must be nonnegative.

2 1
4480
2
bac GG/
But,
1
,
2
Ga contradiction.
Section 14.3 Change of Variables: Polar Coordinates
1. Rectangular coordinates
2. Polar coordinates
3. Polar coordinates
4. Rectangular coordinates
5. 9
:,:0 8,0Rr r
6. 9 :,:0 4sin,0Rr r
7. 9 :,:0 33sin,0 2Rr r Cardioid
8. 9 :,:0 4cos3,0Rr r
9.

cos
00
cos
2
0
0
2
0
0
0
2
1
cos
2
1
1cos2
4
11
sin 2
42 4
rdrd
r
d
d
d













π


π





**
*
*
*

10.

sin
3
sin
2
00 0
0
32
00
3
0
3
11
sin 1 cos sin
33
1cos
cos
33
11 1
11
33 3
4
9
r
rdrd d
dd









π


ππθ








π
** *
**

21
0
2
π
1
0
2
π

309. Section 14.3 Change of Variables: Polar Coordinates 309
© 2010 Brooks/Cole, Cengage Learning
11.
+,
26 2 6
23
000 0
2
0
2
0
3sin sin
216 sin
216 cos 0rdrd r d
d





π

π
πθ π
** *
*


12.
4
3
44 4
2
00 0
0
4
2
0
sin cos sin cos
3
64 sin 16
32 3
r
rdrd d





π



ππ


** *


13.

3
23 2 32
22
02 0
2
2
0
1
99
3
55 55
36
rrdrd r d





θπθθ



ππ

** *


14.

3
23 2 22
00 0
0
2
9
0
9
1
2
1
1
2
1
1
4
rr
re dr d e d
e
e





θθ
θ
πθ



πθ θ



πθ


** *



15.


1sin
2
21sin 2
00 0
0
2 2
0
2
2 2
0
2
2
1
1sin
2
11111
sin cos cos sin sin
82228
39
32 8
r
rdrd d
d












π




"



** *
*

16.

1cos 2
2
21cos 2 2 32
00 0 0
00
sin 1 1
sin sin 1cos 1cos
22 66r
rdrd d d




θ
θ

ππθπθπ


** * *


17.

2
33322
22
2
00 0 0 0
0
sin sin cos
333
aay a aaa
ydxdy r drd d



θ
πππθπ

** * * *

4

0
2
π
0
2341
π
2
321
0
2
π
0
231
π
2
21
0
2
π
0
1
( , ) = (0, 1)x y
π
2

310. 310 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
18.
2
33322
22
2
00 0 0 0
0
cos cos sin
333
aax a aaa
xdydx r drd d



θ
ππππ

** * * *

19.

2
42
24 2
22 2
20 0 0 0 0
0
44
4
x r
x y dy dx r r dr d d d

θ
θ



** ** * *

20. Note that
2
22
111 1
.
444 2
xx x x x




So
2
22
11
.
24
yxxyx








cos
42
12cos2
22 2
2
0202
0
2
44
20
4
11
cos cos
42
11 3 3
Wallis's Formula
22 4 2 32
xx
xx r
xydydx rrdrd d
dd








θ
θθ θ θ
θ



ππ

π""π


** * * *
**

21.

2
39 23 2 32
22 4
00 0 0 0
243 243
510x
x y dy dx r dr d d

θ
** * * *

22.

4
33 3
2
28 422 4
22 2
0000
0
22 22 22
42
33 343
y
y
x y dx dy r dr d d



θ


"



** * * *

23.
2
62
22 22cos 2
35
00 0 0 0
0
4cos 2
cos sin 4 cos sin
63xx
xy dy dx r dr d d



θ
πππθπ

** * * *

24.

2
44 24sin 2
232 42
00 0 0 0
2
3
2
46 5
0
0
cos 64 sin cos
64 sin cos 3
64 sin sin sin cos sin cos 2
648
yy
xdxdy r drd d
d






θ
ππ


** * * *
*

25.

1
2
11 1
22 2 2
10 0 0 0 0
0
11
cos cos sin sin 1 sin 1 1.3218
222x
x y dy dx r r dr d r d d

θ
θ



** ** * *

26.
+,+ ,

2
24 22 2 2
22
000 0 0 0
2
0
sin sin sin cos Integration by parts
sin 2 2 cos 2 sin 2 2 cos 2 2.7357
2
x
x y dy dx r r dr d r r r d
d





θ

πθ πθ** * * *
*

27.
2
2228 422
22 22 2
00 2 0 0 0
4
0
16 2 4 2
33
xx
x y dy dx x y dy dx r dr d
d





θ

ππ** * * * *
*

28.


2
522 5 25 4 5
3
00 5220 00
44
2
0 0
625 625 625
4816
sin cos
sin cos sin
xx
xy dy dx xy dy dx r dr d
d




θ

πππ
** ** **
*

2
r = cosθ
0
π
2
y = x − x
2
0
231
π
2

311. Section 14.3 Change of Variables: Polar Coordinates 311
© 2010 Brooks/Cole, Cengage Learning
29.

2
24 22 22
2
00 00 00
22
0 0
8816
333
cos sin cos sin
cos sin sin cos
x
xy dy dx r r r dr d r dr d
d




θ


** * * * *
*

30.


5
25 2 22
22
20 2
0
2
25 2
2
2
25 2 25 2
2
1
11
rr
erdrd e d
ed
ee









θθ
θθ
θ
θ
θθ
θ
πθ

πθ
πθ π θ

** *
*

31.
22
12 4 2 4
2
01 12
42
01
4
22
4
0
0
arctan arctan
333
2464
yy
yy yy
dx dy dx dy
xx
rdrd
d






θθ
θ

π

πππ

** **
**
*

32.


2
39 23
22 2
00 0 0
3
23 2 2
324
00 0 0
0
99
9 1 81 81
9
24 4 8
x
x y dy dx r r dr d
rrdrd r r d d





θ
θθ π θ

πθπθππ

** * *
** * *

33.

21
00
221 2
3
00 0 0
1111
28168
cos sin
sin2 sin2 cos2
Vrrrdrd
rdrd d



π
πππθπ
**
** *

34.

1
42
21 2 2
2
00 0 0
0
377
434 4
42 4 2
rr
V r rdrd d d




** * *

35.
25 2
2
00 0 125 250
33
Vrdrd d

πππ** *

36.

22 22
22 2
01 01
2
2
22
00
1
ln ln 2 ln
33
212ln 2ln4 4ln4
444
R
VxydA rrdrd rrdrd
r
rd d







** * * * *
**

37.


4cos
3
24cos 2 2
223
00 0 0
0
2
3
2
2
0
0
12
2 16 2 16 64 sin 64
33
128 128 cos 64
1sin1cos cos 3 4
3339
V r rdrd r d d
d








πθπθθπθθ






** * *
*

38.

4
3
24 2 2
22
01 0 0
1
1
16 16 5 15 10 15
3
V r rdrd r d d


πθπθθπ π


** * *

−5−4−3−2
−2
1
2
3
4
5
−3
−4
−5
−1 1234
y
x
0
21
22(),
2
1
2
1
,( (
π
2

312. 312 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
39.
4
33
24 2
222
00
11
16 16 16 2
33
a
a
V r rdrd r d a



πθπθθπθ


** *

One-half the volume of the hemisphere is

64 3.



32
2
32
2
223
3223
33264
16
33
16 32
16 32
16 32 16 8 2
4 4 2 2 2 4 2 2 2.4332
a
a
a
a
a
θπ
θπ
θπ
πθ πθ
πθ πθ
40.
222 2 2 22 22
xyz a z a xy ar



2
22
00
2
33 3
22 32
22
00
0 0
8 8 times the volume in the first octant
12 8 4
88
23 3 3 3
a
a
Varrdrd
aa a
ar d d





πθ

πθ"θ π π π

**
**

41.

4
24 2 2 22
4444
00 0 0
0
Total volume 25 50 50 1 1 100 308.40524
rr
Verdrd ed ede


θθθθ


** * *

Let c be the radius of the hole that is removed.


22 22
44
00 0
0
2 22
44
0
2
4
2
21
25 50
10
50 1 30.84052 100 1
0.90183
0.10333
4
0.41331
0.6429
diameter 2 1.2858
c
c
rr
cc
c
Verdrd ed
ed e
e
c
c
c
c




θθ
θθ
θ
ππθ



θπθ
π
π

** *
*

42.



2
22 229911
;1cos
424949
zr
xy xy

θ



(a)
22
99
436 436
z
rr
θ



(b)
2
2
Perimeter .
dr
rd
d
$







*



22111
1cos cos
222
cos sin
r
dr
d



πθ


2
222
01
Perimeter 2 1 cos cos sin 5.21
4
d
*

(c)

2
2121cos
2
014
9
2 0.8000
436
Vrdrd
r



**

−0.7
−1 1
0.7
x
y
1
1
1
z

313. Section 14.3 Change of Variables: Polar Coordinates 313
© 2010 Brooks/Cole, Cengage Learning
43.
6cos
2
00 0 0
0 1
18 cos 9 1 cos 2 9 sin 2 9
2
Ardrd d d







** * *

44.
24 2
02 0
612Ardrd d

πππ** *

45.

21cos 2
2
00 0
2
2
0
0 1
12cos cos
2
11cos21113
12cos 2sin sin2
222222
Ardrd d
d










** *
*

46.

+,
22sin 2 2 2
2
00 0 0
2
0
2
0 11
2 sin 4 4 sin sin
22
11cos2
44sin
22
1111 9
44cos sin2 84 4
2242 2
Ardrd d d
d









θ






** * *
*

47.

332sin3 3 3
2
00 0 0 0
3 1
26
34sin331cos63sin6Ardrd d d

πππθπθπ

** * *

48.

4
43cos2 4 4
2
00 0 0
0
19
8 4 9 cos 2 18 1 cos 4 18 sin 4
42
Ardrd d d







** * *

49.
2cos
2
32cos 3 3
2
01 0 0
1
3
3
0
0
12cos
3
1
2222cos
22
11sin2 3 3
21cos2 2 2
222 6432
r
r
Ardrd d d
d











πππθ





** * *
*

50.
124
22cos 1 cos ,
233
r



22cos
2
23 22cos 23
01 0
1
22
2
r
A rdrd d






ππ

** *




+,
23 2
0
23 23
2
00
23
0
22cos 1
3 8 cos 4 cos 3 8 cos 2 1 cos 2
10 3 10 7 3
58sin sin2 43
3232
d
dd











*
**

51.
1
3 cos 1 cos cos
23
r





+,
3cos
2
33cos 3 3 2
2
01cos 0 0
1cos
33
2
00
3
0
2 2 9 cos 1 cos
2
8 cos 2 cos 1 4 1 cos 2 2 cos 1
32sin22sin 3 3 3
3
r
A rdrd d d
dd























** * *
**

3
r = 2 cosθ
r = 1
0
2
π
23
1
r = 2 + 2 cos
r = 1
θ
0
π
2
1
r = 3 cosθ
r = 1 + cosθ
0
2
π

314. 314 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
52.
1
1 cos 3 cos cos
23
r





2 1 cos 1 cos
33cos 20
1 cos 1 cos 22
222
22
323 2
3cos 0
2
321
2
1 cos 9 cos 1 cos
22 2 2
12cos 41cos2 11cos2
cos
224
3
sin si
2
A r dr d r dr d
rr
dd d d
dd
























** **
*** *
**
2
32
3sin23 3333
n2 sin 1 1
484222488



θθ



So,
.
4
A

π
53.
15 5
4sin3 2 sin3 3 , ,
2661818
r





4sin3
2
518 4sin3 518 518 2
18 2 18 18
2
518
518
18
18
3
33 4sin34
22
334
81 cos6 4 4 sin6
223
310 4 3 2 4 3 4
23
29 3 2 9 32 3
r
A rdrd d d
d












πππθ







θ



** * *
*
54.
222cos cos 0
2
r







+,
22
022cos
2
2
2
0
22cos
2 2
0
2
2
0
2
0
2
0
2
2
2
422cos
8 cos 4 cos
8cos 21 cos2
8 sin 2 sin 2 8
Ardrd
r
d
d
d
d














θ
θ
π

π


πθθ

πθ

πθθ πθ**
*
*
*
*

55. Let R be a region bounded by the graphs of
1rgπ and 2,rgπ and the lines a πand
.b
π
When using polar coordinates to evaluate a double
integral over R, R can be partitioned into small polar
sectors.
56. See Theorem 14.3.
57. r-simple regions have fixed bounds for 2

-simple regions have fixed bounds for r.
58. (a) Horizontal or polar representative elements
(b) Polar representative element
(c) Vertical or polar
59. (a)

2
39
2
39
,
x
x
fx y dy dx
θ
θθθ
**

(b)

23
00
cos , sin
fr r rdrd

**

(c) In general, the integral in part (b) is easier to
evaluate. The endpoints of the region of integration
are constants.
60.
222
04,0 2, .rxyr
Answer (c)
61. You would need to insert a factor of r because of the
rdrdnature of polar coordinate integrals. The plane
regions would be sectors of circles.
1
r = 3 cosθ
r = 1 + cosθ
0
2
π
r = 4 sin 3θ
r = 2
134
0
2
π
1
r = 2 − 2 cos
r = 2
θ
0
π
2

315. Section 14.3 Change of Variables: Polar Coordinates 315
© 2010 Brooks/Cole, Cengage Learning
62. (a) The volume of the subregion determined by the point 5, 16, 7 is base height 5 10 8 7 .'""
Adding up the 20 volumes, ending with 45 10 8 (12),"" you obtain



+, +,
3
10 5 7 9 9 5 15 8 10 11 8 25 10 14 15 11
8
35 12 15 18 16 45 9 10 14 12
55
150 555 1250 2135 2025 6115 24,013.5 ft
44
V


"





(b)
57 24,013.5 1,368,769.5 pounds
(c)
7.48 24,103.5 179,621 gallons
63.
25
3
40
1 sin 56.051rr drd


**



25
3
40
This integral equals sin 1 .Note: dr rdr







**

64.
44
00
5 87.130
r
erdrd


**

65. Volume base height
8 12 300

'
'
Answer (c)
66.
9
4
Volume base height 3 21' '
Answer (a)

67. False
Let
,1fr r where R is the circular sector
06r and 0 .
Then,

1 0 but 1 0
R
rdA r I**
for all r.
68. True
69. (a)

22
2222 22
222
00 0 0
0
4442
xy
rr
I e dA e r dr d e d d


!
!! !

! !



** ** * *

(b) So,
2.I
70. (a) Let 2,ux then
22
2 11
2.
22
xu
edx e du
!!

! !
**

(b) Let 2 ,
ux then
22
4 11
.
22
xu
edx e du
!!

! !
**

71.


2 22 7
749 27 2 0.01 22
0.01 0.01
2
749 00 0
0
0.49 0.49
4000 4000 200,000
2 200,000 1 400,000 1 486,788
x xy
rr
x
e dydx e rdrd e d
ee











** ** *

x
y4
6
4
6
16
z
x
y
4
4
2
2
4
6
z

316. 316 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
72.
≤
22
2 2
00 0 0
2 2
0
0
2
0
2
24
xy
r
r
ke dy dx ke r dr d
k
ed
kk
d







!! !
θ
!
θ
π

πθ


ππ** * *
*
*

For
,
fxyto be a probability density function,
1
4
4
.
k
k

π
π

73. (a)
4
23y
y
fdx dy**

(b)
23
232
43 3 4 4
243
x
x
xx
fdydx
fdy dx f dy dx
**
** **

(c)
34csc
42csc
frdrd


**


74. (a)
2
22 4
02
4
y
fdx dy

**

(b)

2
242
00
4
x
fdy dx
θθ
**

(c)
24cos
00
2
frdrd

**


75.

22
21 12
21
22 2
rr rr
A rr rr

55
πθπ5 θπ55



Section 14.4 Center of Mass and Moments of Inertia
1.
2
2
22 2 2 2
2
000 0 0
0
24
2
xy
m xydydx dx xdx x

πππππ


** * *


2.


3
2 23
9 22
22
39 3 3
00 0 0
0
0
99 1 1 243
0 243
224344
x
x
xx xxy
mxydydx dx dx
θ
θ
θθ






** * *

3.

12
42
21 2 2
00 0 0
00
11sin1
cos sin cos sin cos sin
44 428r
mrrrdrd d d




ππππ"π


** * *

4.
≤

2
39
22 2
33 9 3 3 3
223
03 0 0 0
3
3
24
32
2
0
1
39 9 69 9
22 2
1 9 1 81 81 297
29 54
2242248
x
x
yx
mxydydxxdx xdxxxxxdx
xx
x












** * * *

x
25
3
3
5
41
1
(4, 4)
(2, 2)
yx= 3
yx=
, 2
2
3( (
, 4
4
3( (
y
x
2
1
−2
−1
3
1
(2)+
2
= 4xy−
2
y

317. Section 14.4 Center of Mass and Moments of Inertia 317
© 2010 Brooks/Cole, Cengage Learning
5. (a)

2
00
23
00 0
3
00
22
2
,
22
, , center of square
22
aa
aa a
x
aa
y
y x
m k dy dx ka
ka ka
Mkydydxdx
ka
Mkxdydx
M aMa
xy
mm
aa
xy







**
** *
**

(b)

3
00
24
00
4
001
2
1
3
1
4
2
,
23
2
,,
23aa
aa
x
aa
y
y x
m kydydx ka
Mkydydxka
M kyx dy dx ka
M aMa
xy
mm
aa
xy







**
**
**

(c)

3
00
4
00
23
001
2
1
4
1
3
2
,
32
2
,,
32aa
aa
x
aa
y
y x
mkxdydxka
M kxy dy dx ka
Mkxdydxka
M aMa
xy
mm
aa
xy







**
**
**

6. (a)

22
00
23
2
00
32
2
00
32
22
23
22
4
6
6
62
,
43
62
43
22
,,
33
ab
ab
x
ab
y
y
x ka b
m kxy dy dx
ka b
M kxy dy dx
ka b
M kx y dy dx
M ka b a
x
mkab
Mkab
yb
mkab
ab
xy









**
**
**

(b)












22 22
00
2
23 22
00
2
32 2 2
00
222 22
22 22
222 22
22 22
3
23
12
32
12
12 3 2 3 2
34
12 2 3 2 3
34
ab
ab
x
ab
y
y
x kab
m k x y dy dx a b
kab
Mkxyydydxab
ka b
Mkxxydydxab
ka b a b a a bM
x
m kabab ab
kab a b b a bM
y
m kabab ab









**
**
**







22 22
22 22
32 23
,,
44
aa b ba b
xy
ab ab





7. (a)

2
00
3
00
3
001
2
1
3
1
6
2
33
2
,,
33ay
ay
x
ay
y
y x
m k dx dy ka
Mkydxdyka
M kxdxdy ka
M aMa
xy
mm
aa
xy







**
**
**
(b)

3
00
24
00
4
001
3
1
4
1
8
33
,
84
33
,,
84ay
ay
x
ay
y
y x
mkydxdyka
Mkydxdyka
M kxy dx dy ka
M aMa
xy
mm
aa
xy







**
**
**

(c)

3
00
4
00
24
001
6
1
8
1
12
3
24
3
,,
24ay
ay
x
ay
y
y x
m kxdxdy ka
M kxy dx dy ka
Mkxdxdyka
M aMa
xy
mm
aa
xy







**
**
**

318. 318 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
8. (a)

2
2
02
2
3
02
2
3
02 1
2
1
6
1
4
,,,,
2323aay
y
aay
x
y
aay
y
y
y x
m k dx dy ka
Mkydxdyka
Mkxdxdyka
M aMa aa
xy xy
mm
θ
θ
θ
ππ
ππ
ππ

ππ ππ π

**
**
**

(b)

2
4
02
2
25
02
2
25
02
,
1
12
1
24
11
240
11 11
,,,
20 2 20 2aay
y
aay
x
y
aay
y
y
y x
y
m kxy dx dy ka
M kxy dx dy ka
M kx y dx dy ka
M aMa aa
xyx
mm
θ
θ
θ
ππ
ππ
ππ

ππ ππ π

**
**
**

9. (a) The x-coordinate changes by 5:
,5,
22
aa
xy





(b) The x-coordinate changes by 5:
2
,5,
23
aa
xy





(c) ≤









5 2
50
5 2
2
50
5 3
2
50
3
2
2
2 1
525
2
1
525
4
1
5 125
3
2 5 125 21575
3103525
2
21575
,,
310 2aa
aa
x
aa
y
y
x
mkxdydxkaa
M kxy dy dx ka a
Mkxdydxkaa
a aaM
x
ma a
Ma
y
m
aa a
xy
a








ππ π


ππ

π

**
**
**

10. The x-coordinate changes by c units horizontally and d
units vertically. This is not necessarily true for variable
densities. See Exercise 9.
11.

1
00
1
2
00
1
00 1
4
2
15
1
6
2
3
8
15
28
,,
315x
x
x
x
y
y
x
m kydydx k
Mkydydxk
M kxy dy dx k
M
x
m
M
y
m
xy
ππ
ππ
ππ
ππ
ππ

π

**
**
**

12.

2
2
00
2
2
2
00
2
2
2
00
16
3
32
3
64
7
12
7
2
12
,,2
7x
x
x
x
y
y
x
m kxy dy dx k
M kxy dy dx k
M kx y dy dx k
M
x
m
M
y
m
xy
ππ
ππ
ππ
ππ
ππ

π

**
**
**

(0, 0) ( a, 0)
( , a)
a
2
a 2
y = 2x
y = −2x + 2a
a
x
y
1
1
(1, 1)
y = x
2
3
8
15
,( (
x
y
1234
1
2
3
4 (2, 4)y = x
2
x
y

319. Section 14.4 Center of Mass and Moments of Inertia 319
© 2010 Brooks/Cole, Cengage Learning
13.
44
2
10
44
2
10
44
3
10
30
24
84
84 14
30 5
24 4
30 5
x
x
x
x
y
y
y
mkxdydxk
M kx y dy dx k
Mkxdydxk
M k
x
mk
M k
y
mk
ππ
ππ
ππ
πππ
ππ π**
**
**



14 4
,,
55
xy

π



14.
≤
≤


2
111
10
2
111
10
0by symmetry
2
2
8
22
2
84
x
x
x
x
x
k
m k dy dx
k
Mkydydx
Mk
y
mk





θ

θ
π
ππ


"
**
**



2
,0,
4
xy

π



15. (a)


1
00
1
2
00
1
00
2
1
1
1
4
1
,
1
11
,
41 4
11
,,
14
x
e
x
e
x
x
e
y
y
x
m k dy dx k e
Mkydydxke
Mkxdydxk
M
x
me
Me e
y
me
e
xy
e
ππθ
ππθ
ππ
ππ
θ

ππ π
θ

π

θ**
**
**

(b)








2
1
00
3
1
2
00
2
1
00
3
2
22
3
2
22
1
4
1
9
1
8
411
,,
21 91
411
,,
2191
x
e
x
e
x
x
e
y
y x
e
mkydydx k
e
Mkydydxk
e
M kxy dy dx k
eM eM
xy
mm ee
ee
xy
ee
θ
ππ
θ
ππ

ππ
θ
ππ ππ
θθ
θ
π
θθ

**
**
**

16. (a)











1
2
00
1
23
00
1
2
00
2
2
3
2
3
2
22 1
1
4
1
1
9
1
13
8
13
21
41
91
4113
,,
21 91
x
e
x
e
x
x
e
y
y
x
mkydydxek
Mkydydxek
M kxy dy dx e k
M e
x
m e
eM
y
m e
ee
xy
ee
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θθ
ππθ
ππθ
ππθ
θ
ππ
θ
θ
ππ
θ
θθ
π
θθ
**
**
**

(b)










1
23
00
1
34
00
1
23
00
3
3
4
3
4
3
33 1
1
9
1
1
16
1
14
27
14
31
91
16 1
9114
,,
31 161
x
e
x
e
x
x
e
y
y
x
mkydydxek
Mkydydxek
M kxy dy dx e k
M e
x
m e
eM
y
m e
ee
xy
ee
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θθ
ππθ
ππθ
ππθ
θ
ππ
θ
θ
ππ
θ
θθ
π
θθ
**
**
**

17.


2
24
20
2
24
20
256
15
4096
105
0by symmetry
16
7
16
,0,
7x
x
x
x
mkxdydxk
M kxy dy dx k
x
M
y
m
xy
θ
θ
θ
θ
ππ
ππ
π
ππ

π

**
**

x
2
3
4
12 34
1
y=
4
x
y
x
−11
2
y=
1
1 +x
2
y
−1−212
1
2
3
y = 4 − x
2
x
y

320. 320 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
L
L
L
2
x
x
y
y = sin
π
18.


2
39
30
2
39
30
2
39
2
30
648
5
0 by symmetry
23,328
35
36
,0
7
36
,,0
7y
y
x
y
y
y k
m kxdxdy
M kxy dx dy
k
Mkxdxdy
M
xy
m
xyθ
θ
θ
θ
θ
θ
ππ
ππ
ππ
ππ π

π

**
**
**
19.



sin 2
00
sin
00
by symmetry
2
2
4
Lx
LxL
x
L
x
kL
m k dy dx
kL
Mkydydx



π
ππ
ππ
**
**



8
,,
28
xM
y
m
L
xy

ππ

π



20.






2cos
00
2cos
2
00
22
2cos
2
00
2
2
2
2
8
2
9
4
32
4
4
16
9
416
,,
49
LxL
LyL
x
LxL
y
y
x kL
m kydydx
kL
Mkydydx
Lk
M kxy dy dx
LM
x
m
M
y
m
L
xy










ππ
ππ
θ
ππ
θ
ππ
ππ
θ
π

**
**
**

21.


2
3
4
2
00
3
4
2
00
3
2
3
2
8
22
sin
6
2
cos
6
28 42
63
22 422
8
63
a
x
R
a
y
R
y
x
ak
m
ka
MkydA krdrd
ka
MkxdA krdrd
M ka a
x
mak
ka a
M
y
mak







π
θ
ππ π
ππ π
ππ " π
θθ
ππ " π
** * *
** * *



42 2
42
,,
33
a
a
xy

θ

π



22.


422
2
22 3
00 0 0
522
2
22 4
00 0 0
5
4
8
sin
5
by symmetry
88
55
aax a
aax a
x
yx
y ka
m k x y dy dx kr dr d
ka
Mkxyydydxkrdrd
MM
M ka a
xy
mka



θ
θ


π
ππ π " π** * *
** * *



88
,,
55
aa
xy


π


246 10
−2
−4
−6
2
4
6
x
y
x = 9 − y
2
0
a
ra=
yx=
π
2

321. Section 14.4 Center of Mass and Moments of Inertia 321
© 2010 Brooks/Cole, Cengage Learning
23.





4
2
00
6
2
2
00
4
2
2
00
4
4
6
62
6
4
4 62
15
8
17
27
113
8
13
5
87
27 5
8713
,,
527 5
x
e
x
e
x
x
e
y
y
x
e
m kxy dy dx k
e
M kxy dy dx k
e
M kx y dy dx k
M e
x
me
eM
y
m ee
ee
xy
e ee
θ
θ
θθ
θθ
θ
ππ
θ
ππ
θ
ππ
θ
ππ
θ
θ
ππ
θ
θθ
π
θ θ

**
**
**

24.

ln
10
ln
10
ln
10
2
6
2
2
1
21
63
1
,2,
2
ex
ex
x
ex
y
y
x kk
mdydx
x
kk
M y dy dx
x
k
M x dy dx k
x
M k
x
mk
Mk
y
mk
xy
ππ
ππ
ππ
π π"π
π π"π

π


**
**
**

25.
62cos3
60
62cos3
2
60
0by symmetry
3
cos
27 3
1.17
40
81 3
1.12
40
R
y
R
y
y
k
m k dA kr dr d
MkxdA
kr dr d
kk
M
x
m







θ
θ
π
ππ π
π
π


** * *
**
**


,1.12,0xy

26.



21cos
00
21cos 2
223
00 0
2 2
22
0
0by symmetry
3
2
cos cos 1 3 cos 3 cos cos
3
315
cos 1 cos 3 cos 1 sin 1 cos 2
32 4 4
525
43 6
5
,,0
6
R
y
R
y
y
k
m k dA kr dr d
k
MkxdA krdrd d
kk
d
M k
x
mk
xy










π
ππ π





ππ"π

π


** * *
** * * *
*
27.
3
2
00
3
2
00
32
32
3
3
13
33 3 3
13
33 3 3
bh
x
bh
y
y
x
mbh
bh
I y dy dx
bh
I x dy dx
I bh b b
x b
mbh
Ibh hh
yh
mbh
π
ππ
ππ
ππ "πππ
ππ "π ππ
**
**

28.



00
3
2
00
3
2
00
3
3 2
12
12
12 6
26 6
12 6
26 6
bhhxb
bhhxb
x
bhhxb
y
y
x bh
mdydx
bh
I y dy dx
bh
I x dy dx
I bh b
x b
mbh
Ibh h
yh
mbh
θ
θ
θ
ππ
ππ
ππ
ππ ππ
ππ ππ**
**
**

12
1
2
x
y
y = e
−x
2
2
3e1
3
1
x
yx= ln
y
0
1
r= 2 cos 3θ
θ
θ
π
6
π
6
=
=−
π
2
0
1
r= 1 + cosθ
π
2

322. 322 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
29.
2
4
2
232
00
4
2
232
00
44 4
0
4
2
sin
4
cos
4
44 2
1
42
a
x
R
a
y
R
xy
x
ma
a
IydA rdrd
a
IxdA rdrd
aa a
III
Ia a
xy
ma













"
** * *
** * *

30.
2
4
232
00
4
232
00
44 4
0
4
2
2
sin
8
cos
8
88 4
2
82
a
x
R
a
y
R
xx
x
a
m
a
IydA rdrd
a
IxdA rdrd
aa a
III
Ia a
xy
ma














"
** **
** **

31.
2
4
2
232
00
4
2
232
00
44 4
0
4
2
4
sin
16
cos
16
16 16 8
4
16 2
a
x
R
a
y
R
xy
x
a
m
a
IydA rdrd
a
IxdA rdrd
aa a
III
Ia a
xy
ma














"
** * *
** * *

32.




3322
32
222222222
33
00 0 0
32 3
22 2 2222 4
3
0
322
2
00
3
0
4
44
33
41
arcsin 2 arcsin
32 8 4
4
4
4abaax a a
x
a
babby
y
yx
mab
bb
I y dy dx a x dx a a x x a x dx
aa
ba x x ab
xaxa xxaaxa
aa a
ab
I x dx dy
ab a
III















** * *
**

3
22
3
3
44
1
42
1
42
y
x
bab
ab
I ab a
x
mab
Iab b
y
mab





"
"

33.
2
00
4
3
00
32
2
00
423
0
32 2
2
42
2
2
4
6
32
12
63
23 3 3
42
22 2 2
ab
ab
x
ab
y
xy
y
x
ky
kab
mk ydydx
kab
Ik ydydx
ka b
Ik xyydydx
kab kb a
III
I ka b a a
x a
mkab
Ikab bb
yb
mkab
-







**
**
**

34.

322
22
00 0
522
3
0
522
2
0
5
0
52
3
52
3
2
2
3
4
15
2
15
2
5
215 5
23 5 5
415 2 10
23 5 5aax a
aax
x
a
aax
y
a
xy
y
x
ky
ka
m k ydydx k a x dx
ka
Ik ydydx
ka
I k x y dy dx
ka
III
I ka a a
x
mka
Ika aa
y
mka
-












** *
**
**

323. Section 14.4 Center of Mass and Moments of Inertia 323
© 2010 Brooks/Cole, Cengage Learning
35.
2
24
00
2
24
2
00
2
24
3
00
0
4
32
3
16
3
16
16 3 4 2 2 3
43 3 3
32 3 8 4 2 6
43 3 6
x
x
x
x
y
xy
y
x
kx
mk xdydx k
k
I k xy dy dx
k
Ik xdydx
III k
I k
x
mk
Ik
y
mk
-










**
**
**

36.



11
35
2
00
11
359
2
00
11
357
2
00
0
224
460
248
93
240 80
48 2
24 2
60 10
24 5
x
x
x
x
x
x
y
x
xy
y
x
kxy
kk
m k xy dy dx x x dx
kk
I k xy dy dx x x dx
kk
I k x y dy dx x x dx
kk
III
I k
x
mk
Ik
y
mk
-






** *
** *
** *

37.
4
00
4
3
00
4
3
00
0 32
3
16
512
5
592
5
512 3 48 4 15
532 5 5
16 3 3 6
132 2 2x
x
x
x
y
xy
y
x
kxy
k
m kxy dy dx
I kxy dy dx k
k
I kx y dy dx
k
III
I k
x
mk
Ik
y
mk
-




"
"
**
**
**

38.



22
1
22
2
0
1
222
2
0
1
222
2
0
0
6
35
158
2079
158
2079
316
2079
158 35 395 351,945
2079 6 891 891
351,945
891x
x
x
x
x
x
y
x
xy
y
x
xy
mxydydx
Ixyydydx
I x y x dy dx
III
I
x
m
I
yx
m-




"

**
**
**

39.
1
2
0
1
2
2
0
1
3
2
0
0 3
20
3
56
18
55
504
20 30
18 3 9
320 70
56 3 14x
x
x
x
x
x
y
x
xy
y
x
kx
k
mkxdydx
k
I kxy dy dx
k
Ikxdydx
k
III
I k
x
mk
Ik
y
mk
-




"
"
**
**
**

40.
24
3
0
24
3
3
0
24
2
3
0
0 512
2
21
32,768
2
65
2048
2
45
321,536
585
2048 21 28 2 105
45 512 15 15
32,768 21 8 1365
65 512 65x
x
x
x
x
x
y
x
xy
y
x
ky
k
mkydydx
k
Ikydydx
k
I kx y dy dx
k
III
I k
x
mk
Ik
y
mk
-




"
"
**
**
**

41.


22
22
22
0
4222
22 2 2 2 2 2 2 2 2
22
22 20 4
824
bbx b
bb
bbb
bbb
I k x a dy dx k x a b x dx
babkb
k xbxdxaxbxdxa bxdx k b a










** *
***

42.

4
42 4 223
00 0
0
2 416
626 6
33
kk
Ikxdydxkxdxx




** *

324. 324 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
43.
4
44 2
2927252
00 0
0
2 24 72 42,752
61236
9 7 5 315x k
I kx x dy dx kx x x x dx k x x x




** *

44.



22
2
0
22
4322
0
2
4 224 222 222 22
23 5 2
4222
2222 4
2
43 2
12
2
43 2
12 2
arcsin
43532
1
2arcsin
8
aax
a
ax
a
a
a
a
Ikyyadydx
yayay
kdx
aa
ka ax x aa x xa x a xdx
ax x a a x
kax xa x a
a
x
xx a a x a
a


















**
*
*
23
2
44 2 3 55
555 3 5
23
1212 7 5615
22
435 34162 3 158 60
a
a
ax
ax
aa a a a a a
ka a a a k ka










45.



22
22 22
234
00 00 0
0
22
43 22 34
00
4 322 222 2222 4 224 4
0
4224322 2
4
46 4
4
464 2
4
78 8 4
4
ax
aax aax a
a ax
a
k
I k a y y a dy dx k a y dy dx a y dx
k
a ayay ayy dx
k
a aax aax aaxax a axx adx
k
aaxxaax ax













** ** *
*
*

22
0
2
4 3 3 22 2 2222 4
0
5555 5 5
8
74arcsin2 arcsin
435 2
81 1 717
72
4 3 5 4 16 15
a
a
axdx
kax xa x
ax x axaxa xxaaxa
aa
k
aaaa a ak


J











*

46.


2
4
2
24 2 2 23
2
20 2 2
0
2
2
246 3 5 7
2
2
2128
33
6 1 2 192 128 1408
16 12 6 16 4 32 32
3 3 5 7 3 5 7 105
x
x
kk
Ikydydxydx xdx
kkkk
xxxdx xx x x













** * *
*

47. Let ,xy- be a continuous density function on the
planar lamina R.
The movements of mass with respect to the x- and y-axes
are

,and ,.
xy
RRM y xydA M x xydA--** **

If m is the mass of the lamina, then the center of mass is

,,.
y x
MM
xy
mm




48.

2
2
, , Moment of inertia about -axis
, , Moment of inertia about -axis
x
R
y
RIyxydA x
IxxydA y-
-
**
**

49. See the definition on page 1017.
50. (a) ,
xyky-
ywill increase.
(b)

,2xyk x-
,xywill be the same.
(c)
,
xykxy-
Both xand ywill increase.
(d)
,44
xyk x y-
Both xand ywill decrease.

325. Section 14.5 Surface Area 325
© 2010 Brooks/Cole, Cengage Learning
51.


2
00
3
3
0
0
3
,,
22
2
2
312
12
22 3
bL
y
L
b
y
a
LL
yAbLh
L
I y dy dx
yL
Lb
dx
I L Lb L
yy
hA L bL
ππ π

πθ





ππ


πθ π θ π
**
*

52.



2
3
00
3
,,
22
212
3212
232 2
ba
y
a
aa
yAabhL
aab
I y dy dx
aL aaab
y
LaLaab
ππ πθ

πθ π


θ
πθ π
θ

**


53.
2
2
02
3
2
0
2
3
2
0
2
4
33
0
3
2
2
,,
323
2
2
3
22
33
222
327 3
222
327 8 3 36
236
362
bL
y
Lx b
L
b
Lx b
b
b
a
LbLL
yAh
L
I y dy dx
L
ydx
LLxL
dx
b
LxbLx L Lb
Lb
LLb L
y
Lb
πππ

πθ




πθ



πθθ



πθ θ π

πθ π
**
*
*

54.

2
22
2
232
22
00
44
2
2
0
4
2
2
0, ,
sin
sin
44
4
4
aax a
y
aax
a
yAahL
I y dy dx r dr d
aa
d
a a
y
La L








θ
θθ θ
ππ π
ππ
ππ
πθ πθ
** **
*

55. Orient the xy-coordinate system so that L is along the y-
axis and R is the first quadrant. Then the volume of the
solid is

2
2
2
2.
R
R
R
R
R
VxdA
xdA
xdA
dA
dA
xA


π
π


π



π**
**
**
**
**

By our positioning,
.xrπSo, 2.VrAπ

Section 14.5 Surface Area
1.


22
,22
2
11443
xy
xy
fxy x y
ff
ff

ππ



44 4
00 0
4
2
0
334
34 24
2
x
Sdydxxdx
x
x
θ
ππθ

πθπ
** *


2.


22
33 3
00 0
,1523
2, 3
114
14 3 14 9 14
xy
xy
fxy x y
ff
ff
Sdydxdx

ππθ

πππ
** *


x
L
R
(, )x y
y
1234
1
2
3
4
y = 4 − x
x
y
2
2
31
3
1
x
R
y

326. 326 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
3.


22
22 2
00 0
,722
2
11443
3612
xy
xy
fxy x y
ff
ff
Srdrdd



ππ

πππ
** *


4.


22
23 2
00 0
,1223
2, 3
114914
914
14 9 14
2
xy
xy
fxy x y
ff
ff
Srdrd d



ππθ

πππ
** *


5.


2
22
2
,9
2, 0
114
xy
xy
fxy x
fxf
f fx
πθ
πθ π



≤


22 2
22
00 0
2
22
0
14 2 14
1
2ln14 2 14
42
1
2ln17 4 17
4
1
217 ln4 17
2
S x dy dx x dx
x
xx x









** *


6.



2
22
2
,
square with vertices 0, 0 , 3, 0 , 0, 3 , 3, 3
0, 2
114
xy
xy
fxy y
R
ffy
ff y
π
π
ππ


≤

33 3
22
00 0
3
22
0
3
4
3
4
14 314
214 ln2 14
637 ln6 37
S y dx dy y dy
yy y y




** *


7.

32
12
22
,3
3
,0
2
949
11
42
xy
xy
fxy x
fxf
x
ff x

ππ





34 3
00 0
3
32
0 49 49
4
22
44
4 9 31 31 8
27 27
xx
Sdydxdx
x

ππ




** *


8.

32
12
222
3
,2
0,
11
xy
xy
fxy y
ffy
f fy

ππ




22 2
00 0
2
32 52
0
32 52
2
5
82212
5555
112
21 1
23 3 2 3
y
S y dx dy y y dy
yy
θ



" " ** *


−11
−1
1
x
2
+ y
2
= 4
x
y
−1−212
−1
−2
1
2
x
2
+ y
2
= 9
x
y
12
1
2
x
y
2
2
31
3
1
x
R
y
x
2
12
1
R
y = 2 − x
y
1234
1
2
3
4
x
y

327. Section 14.5 Surface Area 327
© 2010 Brooks/Cole, Cengage Learning
9.


22
2
,lnsec
,:0 ,0 tan
4
tan , 0
11tansec
xy
xy
fxy x
Rxyx yx
fxf
f fxx

π
6<
7>
8?
ππ



+,
4tan
00
4
0
4
0
sec
sec tan
sec 2 1
x
S x dy dx
xxdx
x



π
π
ππθ**
*

10.


22
22
22
,13
2, 2
1144
xy
xy
fxy x y
fxf y
f fxy

ππθ






22
2
00
2
2 32
2
0
0
2
32
0
1
12
1
12
14
14
17 1
17 17 1
6
Srrdrd
rd
d










πθ
πθ**
*
*


11.

9:
22
22 22
22 22
,
,:0 , 1
01,1
,
xy
fxy x y
Rxy fxy
xy xy
xy
ff
xyxy



ππ




22
22
22 22
2
11 21
2
11 00
11 2
222
xy
x
x
xy
ff
xy xy
Sdydxrdrd


θ
θθθ


πππ
** **


12.

9:

22
22
22
,
,: 16
,
11
xy
xy
fxy xy
Rxyxy
fyfx
f fyx
π

ππ




2
416
22
2
416
24
2
00
1
2
117171
3
x
x
Syxdydx
rrdrd


θ
θθ θ

**
**



13.

9:

≤
22 2
22 2
22 2 22 2
22
22
22 2 22 2
22 2
22
2
22
22
0022 2 22
,
,: ,0
,
11
2
xy
xy
bbx b
bbx
fxy a x y
Rxyxyb ba
xy
ff
axy axy
xy a
ff
axy axy axy
aa
Sdydxrdrdaaab
axy ar


θ
θθ θ
πθθ

θθ
ππ
θθ θθ

θθ θθ θθ
πππθθ
θθ θ
** **
x
2
1
4
yx= tan
π
2
π
R
y
−1−212
−1
−2
1
2
x
y
x
2
+ y
2
= 2
1
1
x
xy
22
+= 1
y
2
−2
−22
x
xy
22
+ = 16
y
ab
b
a
−b
−b
x
xyb
222
+≤
y

328. 328 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
14. See Exercise 13.

22
22
22 2
2
2
00 22
2
aax
aax
a a
Sdydx
axy
a
rdrd a
ar


θ
θθ θ
π
θθ
ππ
θ**
**


15.


22
83212
00
24 3 2
114
14 48 14
xy
x
zxy
ff
Sdydx

πθθ

ππ
**



16.

22
22
22
16
1144
yy
zxy
f fxy
πθθ





2
416
22
00
24
2
00
14
1 4 65 65 1
24
x
Sxydydx
rrdrd


θ

**
**



17.

22
22
22
22 22
22
25
5
11
25 25 25
xy
zxy
xy
ff
xy xy
xy
πθθ

θθ θθ θθ



2
39 23
2
39 00 222
55
2220
2525x
x
Sdydxrdrd
rxy


θ
θθθ
πππ
θ** **



18.

22
22
22
22 22
2
44
11 5
xy
zxy
xy
ff
xy xy





22
00
545Srdrd

ππ**



x
8
12
16
124 168
4
y
42
4
2
6
6
x
yx= 16−
2
y
1
2
2
−1
−2
−2−11
x
xy
22
+= 9
y
1
−1
−11
x
y
x
2

+ y
2

= 4

329. Section 14.5 Surface Area 329
© 2010 Brooks/Cole, Cengage Learning
19.



2
22
2
,2
triangle with vertices 0, 0 , 1, 0 , 1, 1
154
xy
fxy y x
R
ff x

π




1
2
00 1
54 2755
12x
S x dy dx **



20.



2
22
2
,2
triangle with vertices 0, 0 , 2, 0 , 2, 2
154
xy
fxy x y
R
ff y

π



2
2
00
54
5 8 21 37 21 5 5
ln
45 412
x
S y dy dx



**



21.

9:

22
22 22
22
22
,9
,:0 ,
09 9
2, 2
1144
xy
fxy x y
Rxy fxy
xy xy
fx x fy y
f fxy
πθ θ


πθ πθ




2
39
22
2
39
23
2
00
14 4
14
37 37 1 117.3187
6
x
x
Sxydydx
rrdrd



θ
θθθ


πθ**
**


22.

9:

22
22
22
22
,
,:0 , 16
016
2, 2
1144
xy
xy
fxy x y
Rxy fxy
xy
fxfy
f fxy



ππ




2
416
22
2
416
24
2
00
14 4
65 65 1
14
6
x
x
Sxydydx
rdrd


θ
θθ θ

θ
**
**



23.

9:

22
22
22
,4
,:0 1,0 1
2, 2
1144
xy
xy
fxy x y
Rxy x y
fxfy
f fxy
πθ θ

πθ πθ




11
22
00
1 4 4 1.8616Sxydydx **


24.
9:

32
12
2222
3
,cos
,:0 1,0 1
sin , 0
11sin
xy
xy
fxy x x
Rxy x y
fx xf
f fxx


πθ π



211
00
1 sin 1.02185Sxxdydx **


25. Surface area 4 6 24"
Matches (e)

x
1
1
yx=
R
y
2
2
31
3
1
x
R
yx=
y
2
−2
−22
x
xy
22
+ = 16
y
x
y
5
5
10
z

330. 330 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
26. Surface area 9
Matches (c)


27.

9:

22
2
,
,:0 1,0 1
,0
11
x
x
xy
x
xy
fxy e
Rxy x y
fef
ff e






11
2
00
1
2
0
1
1 2.0035
x
x
S e dy dx
e

**
*


28.
9:

52
32
22
32
5
,
,:0 1,0 1
0,
11
xy
xy
fxy y
Rxy x y
ffy
ff y





11 1
33
00 0
1 1 1.1114S y dx dy y dy** *


29.



33
22
22
,3
square with vertices 1, 1 , 1, 1 , 1, 1 , 1, 1
333 ,
33 3
x
y
fxy x xy y
R
fxyxy
fxyyx







11 22
22
11
19 9Sxyyxdydx

**


30.

9:

22
,3
,:0 4,0
23, 32 32
xy
fxy x xy y
Rxy x yx
f xyf xy xy








2 22
22
112332
113
xy
f fxyxy
xy





4
22
00
113
x
Sxydydx**


31. ,sin
sin , cos
x
xx
xy
fxy e y
f eyfe y







22 22 22 2
11sincos1
x xx
xyf feyeye



2
24
2
2
24
1
x
x
x
S e dy dx



**


32.



22
22
22 22
22
222 2 222 2 22 2 2 2
,cos
,:
2
2sin , 2sin
114sin4sin14sin
xy
xy
fxy x y
Rxyxy
fxxyfyxy
f fxxyyxy xyxy


6<
7>
8?







2
22
22222
2
22
14 sin
x
x
Sxyxydydx



**


33.

9:


22
22 22 2 2 2
,
,:0 4,0 10
,
11 1
xy
xy xy
xy
xy xy xy
xy
fxy e
Rxy x y
fyefxe
f fyexeexy







410
22 2
00
1
xy
S e x y dy dx**

x
y
3
3
3
2
z

331. Section 14.5 Surface Area 331
© 2010 Brooks/Cole, Cengage Learning
34.

9:
,sin
,:0 4,0
sin , cos
x
xx
xy
fxy e y
R xy x y x
fe yfe y







22
22 2 2 2
11sincos1
x xx
xyf feyeye



4
2
00
1
x
x
Sedydx

**


35. See the definition on page 1021.

36.
22
,
fxyxy is a paraboloid opening upward.
Using the figure below, you see that the surface areas
satisfy:


b ca


37. No, the surface area is the same.

,and ,z fxy z fxy k
have the same partial derivatives.

38. (a) Yes. For example, let R be the square given by

01,01,xy
and S the square parallel to R given by

01,01,1.xyz
(b) Yes. Let R be the region in part (a) and S the surface
given by
,.
fxy xy
(c) No.

39.
2
22
,1; ,0
1
xy
x
fxy x f f
x






22
1
00 2
1
1 12
2
0 2
0
1
1
16
1
16 16 1 16
1
xy
R
xSffdA
dy dx
x
x
dx x
x





**
**
*


40.

22
22 2 2
22
2
22 22
,
11 1
xy
fxy k x y
kx ky
ff k
xy xy






22
2222
11111
xy
RRRSffdAkdAkdAAkrk
)
** ** **


41. (a)

22
25
225 2 32
22
04 0
4
3
,
8 625 where is the region in the first quadrant
2
8 625 4 625
3
8
0 609 609 812 609 cm
32
R
R
Vfxy
xydA R
rrdrd r d










"
**
**
** *

(b)

22
22
22 22
225
0422 2
22
425
181
625 625
25 25
88
625 625
lim 200 625 100 609 cm
2
xy
RR
R
b
b
xy
A f f dA dA
xy xy
dA rdrd
xy r
r











"
** **
** * *

x
y
12
1
2
x
2
+ y
2
= 4
x
82024
12
12
20
24
16
8
164
4
R
y

332. 332 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
42. (a)
321416
25
75 25 15
zyyy

(b)


15
32
01416
2 50 25 100 266.25 26,625 cubic feet
75 25 15
Vyyydy




*

(c)

32
2
50 15
22
001416
,25
75 25 15
1816
0,
25 25 15
2 1 3087.58 sq ft
xy
yx
fxy y y y
ff yy
Sffdydx



**

(d)

Arc length 30.8758
Surface area of roof 2 50 30.8758 3087.58 sq ft



Section 14.6 Triple Integrals and Applications
1.

1 2
2 2
321 32 32 3
000 00 00 0
0 0
3 3
2
00
11
2222
12 2 3 18
xz
xy z dx dz dy xy xz dz dy y z dz dy z yz dy
ydyyy





*** ** ** *
*

2.
1
111 11
222 322
111 11
1
111 1 1 1
22 32 2 3
111 1 1 1
1
3
82244
3992727
x y z dx dy dz x y z dy dz
y z dy dz y z dz z dz z







*** **
** * *

3. +,
1
22 4 5
11111
2
000 0 00 00 0 0
00
1
2 2 10 10
x
xxy x x xy
xy x x
xdzdydx xz dydx x ydydx dx dx




*** ** ** * *

4.

22
993 9 93 9 9 3
22 23 3 4
000 0 00 0 0 0
7291121
2 2 18 36 4
93
yyx y y
zdzdxdy y x dxdy xy x dy y dy y



** * ** * *

5.

41 41 41 22 2
100 10 10
0
4
21
44 2
11
11
0
1
222
15 1
11 1
22
x
x
xx x
x
ze dy dx dz ze y dx dz zxe dx dz
z
ze dz z e dz e
e








*** ** **
**

6.


2
2
222
41 4 4 4 4 41
0 111 0 1 1 1 1 1 1
1
lnln 2
ln ln 2 ln 2 ln 4
2
e
exz e e xz
zz
zdydzdx z y dzdx dzdx dx dx x
xz x x





** * ** ** * *

7.


421 42 42 1
000 0 00 00
4
23
44 2
000
0
cos cos 1 cos
64 40
1sin 1 8
23 3 3
x x
x y dz dy dx x y z dy dx x x y dy dx
xx
xxydx xxdx







** * ** **
**

8.
2
221 22 2
00 0 00 0
0
sin 1 1 1
sin sin cos
222yy y y
ydzdxdy dxdy ydy y
y






*** ** *

9.
22
39
2
09 0
324
5
yy
y
ydzdxdy


** *

333. Section 14.6 Triple Integrals and Applications 333
© 2010 Brooks/Cole, Cengage Learning
10.
22 2
22 4 22
23
22
00 2 00
16 2
42 2
15xy x
xy
ydzdydx y x y y dydx


** * **

11.

222
24 4 24 4
2
100 1 00
2
22 4
222
000
sin
sin ln
ln 4 cos ln 4 1 cos 4 2.44167xx
xxy
dz dy dx x y z dy dx
z
x y dx x x dx





** * **
**

12.


3223 62 3 6 6 2 6 23 22 22
00 0 00 0
2
63 2 22
00
16 2
2.118
23
yyz xxy
xy xy
x
xy
ze dx dz dy ze dz dy dx
xy
edydx








** * ** *
**

13.
55 5
00 0
xxy
Vdzdydx

** *

14.
2
32 9
00 0
xx
Vdzdydx

** *

15.
222 222
66 6 66 6
22
66 0 66 0
xxy yxy
xy
Vdzdydxdzdxdy


** * ** *

16.
222
416 16
2
416 0
xxy
x
Vdzdydx


** *

17.



222
222 2 2 22
222
416 80
222
416 121
2
2
2 80 2 80 0 8 10 0 8 2 16
xxy
xxy
zxy zxy
xyz zz z z z z z xy z
Vdzdydx
)





** *

18.


2 2
42 324
22
223
42 3
x x
xy
x
Vdzdydx



** *


22 2
22
22
43
42 3
1ellipse
243
zxxy
xy
xy




19.

22
24 24
20 0 20
222 2
22435
20 0
82561 1
23515
416816
yx y
V dz dx dy x dx dy
ydy y ydy y y y





** * **
**

20.
222 22 2 2 2
2
0000 00 0 0
248
xy
V dzdydx xydydx xy dx xdx
*** ** * *

21.


22 222 22
22 2
00 0 00
22
22 2 22
0 22
0
22 2 3 3
0
0
88
4arcsin
14
42
233
aax axy aax
ax
a
a
a
V dz dy dx a x y dy dx
y
ya x y a x dx
ax
axdx ax x a











** * **
*
*

334. 334 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
x
y
6
3
6
6
3
z
22.


2
36
3222 2
636 36 636 6
22 2
00 0 00 0
0
6 32
22 2 2
0
6
32
22
0
4436436
3
1
43636 36 36
3
1
4 9 36 324 arcsin 36 4 162 648
66
x
xxy x
y
Vdzdydxxydydxyxydx
xx x x dx
x
xx xx

θ
θθθ θ

ππθθπθθ



πθθθθθ






** * ** *
*

23.
22
224 4 2 2 2
22435
00 0 0 0 0
8 2561
35 15
416816
xx
V dz dy dx x dx x x dx x x x
θθ


** * * *


24.


23
22 9
00 0
2
22
3
00
2
32
0
2
235
0
2
6
34
0
9
92
18 9 2
1
18 3
26
42
18 2 6 2 2 12 2
33
xx
x
Vdzdydx
xdydx
xxdx
xxxdx
x
xx x
θθ
θ
π
πθ
πθθ





** *
**
*
*

25.
2
32 4
00 2
32
2
00
2
32
3
0
0
3
0
3
0
42
2
32
8
42
3
10
10
3
y
y
Vdzdydx
y y dy dx
yy
ydx
dx
x
θ
θ
π










ππ

***
**
*
*

26. The region in the xy-plane is:

+,
22 22
22
00 0
22 2
3
2
00
2
34
2
0
2
88
44
343 3
xx x
xx
x
x
V dz dy dx x dy dx
xydx xx xdx
xx
x


ππ



** * **
**

27.

11
00 1
z
dy dz dx
θ
θ
***

28.

2
11 1
2
10
yz
y
dx dz dy
θθ
θ
** *

29. Plane: 36412xyz


3124312436
00 0
zzx
dy dx dz
θθθ
** *

30. Top plane: 6xyz
Side cylinder:
22
9xy

2
39 6
00 0
yxy
dz dx dy
θθθ
** *

y
x
2
1
33
4
z
1234
1
2
3
4
x
y
y = x +2
y = x
2
x
y
1
1
−1
z
y
x
2
1
1
−1
z
x
y2
3
3
4
z

335. Section 14.6 Triple Integrals and Applications 335
© 2010 Brooks/Cole, Cengage Learning
31. Top cylinder:
22
1yz
Side plane: xy

2
11
00 0
xy
dz dy dx

***

32. Elliptic cone:
22 2
4
xzy

22
44 2
00
yz
z
dx dy dz

***

33. 9
:,, :0 1,0 ,0 3Qxyz x yx z


311 31
00 000
131 13
00 00 0
11 3 1 3
00 000
9
16
x
y
Q
x
y
x
y
xyz dV xyz dx dy dz xyz dy dx dz
xyz dx dz dy xyz dy dz dx
xyz dz dx dy xyz dz dy dx


*** * * * * * *
*** ***
*** ***

34. 9
:
2
, , : 0 2, 4, 0 2Qxyz x xy z x




24 2
2
00
42
00 0
22 4
2
00
22 4
2
00
2
22 24 2
2
00 0 0 2 0
42 42 2
00 0 02 0
104
21
x
x
Q
yx
x
x
z
x
zy z
z
yy z
y
xyz dV xyz dz dy dx
xyzdzdxdy
xyzdydzdx
xyz dy dx dz
xyz dx dy dz xyz dx dy dz
xyzdxdzdy dxdzdy













*** * * *
** *
** *
** *
** * ** *
** * ** *

35. 9
:
22
,, : 9,0 4Qxyzxy z


22
43 9 43 9
22
03 9 03 9
22
34 9 3 9 4
22
30 9 3 9 0
22
34 9 3 9 4
22
30 9 3 9 0
0
xy
xy
Q
yy
yy
xx
xx
xyz dV xyz dy dx dz xyz dx dy dz
xyz dx dz dy xyz dz dx dy
xyz dy dz dx xyz dz dy dx








*** * * * * * *
*** ** *
*** ** *

36. 9
:
2
,, :0 1, 1 ,0 6Qxyz x y x z

2
11 6 1 1 6
00 0 00 0
16 1 61 1
00 0 0 00
22
161 611
00 0 0 00
3
2
xy
Q
yy
xx
xyz dV xyz dz dy dx xyz dz dx dy
xyz dx dz dy xyz dx dy dz
xyz dy dz dx xyz dy dx dz





*** * * * * * *
*** ***
*** ***


x
1
1
yx=
R
y
x
y
44
2
4
2
(2, 4)
z
x
y
4
3
5
4
3
z
x
y2
1
6
2
1
z
x
y
1
1
1
z
x
y
5
5
4
3
3
2
2
1
1
z

336. 336 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
37. 9:
2
,, :0 1,0 1 ,0 1Qxyz y x y z y

2
11 1 1 1 1
00 0 00 0
2
12 1 11 1
2
00 0 02 0
11 1 11 1 1
01 1 0 0 0 0
22
11 1 11 1
00 0 00 0
5
12
yy xy
zz z x
zz
zxx
x
yy zy
dz dx dy dz dy dx
dy dx dz dy dx dz
dy dz dx dy dz dx
dx dz dy dx dy dz
θθ θθ
θθ θ
θ
θθθθ
θθ
θθ θθ
π


πππ** * ** *
** * ** *
** * ** *
** * ** *

38. 9
:
2
,, :0 3,0 ,0 9Qxyz x yx z x

22
3 9 339
000 0 0
2
39 9 9
00 0 00 0
2
99 9 39 9
00 00
81
4
xx x
y
xx zx
zz y z
yy
dz dy dx dz dx dy
dy dz dx dy dx dz
dx dy dz dx dz dy
θθ
θθ
θθ θ θ
π
ππ
πππ*** ***
** * ** *
** * ** *


39.



6423 2 2 3
00 0
6423 2 2 3
00 0
8
12
12 3
82
xyx
xyx
yz
yz
m k dz dy dx k
Mk xdzdydxk
M k
x
mk
θθθ
θθθ
ππ
ππ
πππ** *
** *

40.


55 151533
00 0
55 151533
2
00 0 125
8
125
4
2xxy
xxy
xz
xz
mk ydzdydz k
Mk ydzdydx k
M
y
m
θθθ
θθθ
ππ
ππ
ππ** *
** *

41.



444 44
000 00
4
2
0
2
444 44
000 00
4
23
0
4
128
44
3
4
2
128
2168
3
1
x
x
xy
xy
m k x dz dy dx k x x dy dx
k
kxxdx
x
M k xz dz dy dx k x dy dx
k
k xxxdx
M
z
m
θ
θ
ππθ
πθπ
θ
ππ

ππ*** **
*
*** **
*

42.


11
00 0
2
11
00 0
2
6
24
24
64
ba yb c yb xa
ba yb c yb xa
xz
xz kabc
m k dz dx dy
kab c
Mk ydzdxdy
Mkabc b
y
m kabcθθθ

θθθ

ππ
ππ
ππ π** *
** *


43.
5
000
6
2
000
6
2
000
6
000
6
5
6
5
6
5
4
6
6
8
62
43
62
43
8
42
bbb
bbb
yz
bbb
xz
bbb
xy
yz
xz
xy kb
m k xy dz dy dx
kb
M k x y dz dy dx
kb
M k xy dz dy dx
kb
M k xyzdzdydx
M kb b
x
mkb
Mkb b
y
mkb
M kb b
z
mkb
ππ
ππ
ππ
ππ
ππ π
ππ π
ππ π
***
***
***
***


44.
2
000
3
2
000
22
000
22
000
22
2
22
2
3
2
2
3
4
4
4
22
4
22
32
23
abc
abc
xy
abc
yz
abc
xz
yz
xz
xy kabc
m k zdzdydx
kabc
Mk zdzdydx
ka bc
M k xz dz dy dx
kab c
M k yz dz dy dx
M ka bc a
x
m kabc
Mkabc b
y
m kabc
M kabc c
z
m kabc
ππ
ππ
ππ
ππ
ππ π
ππ π
ππ π
***
***
***
***

y + z = 1
z = 1 − 1 − x
11
1
z
y
x
x = 1 − y
2
3
9
3
z
y
x
z = 9 − y
2
z = 9 − x
2
y = x

337. Section 14.6 Triple Integrals and Applications 337
© 2010 Brooks/Cole, Cengage Learning
45. xwill be greater than 2, whereas yand zwill be
unchanged.
46.
zwill be greater than 85, whereas x and ywill be
unchanged.
47.
ywill be greater than 0, whereas xand zwill be
unchanged.
48.
,xyand zwill all be greater than their original values.

49.


2
22
22
00
2 22
22 2
2
00
222
32
22
2
0
22
21
3
0 by symmetry
4
2
4
34
43
34
rrxh
xy
hx y r
rrx
r
xy
mkrh
xy
Mk zdzdydx
kh
rxydydx
r
kh k r h
rx dx
r
M krh h
z
mkrh












** *
**
*



3
,, 0,0,
4
h
xyz






50.
2
39
00 0
2
39
30 0
2
39
30 0
2
39
30 0
218
0
81
8
81
16
0
9
16
9
32
xy
xy
yz
xy
xz
xy
xy
yz
xz
xy
m k dz dy dx k
Mk xdzdydx
M k ydzdydx k
Mk zdzdydx k
M
x
m
M
y
m
M
z
m

















** *
** *
** *
** *



99
,, 0, ,
16 32
xyz




51.



22 2
22 222
44 4
00 0
22
4
22
44 4
22 2 2 3
00 0
0
4 32
22
0
2
4
0
128
3
0 by symmetry
4
4
1
24 216
3
4
4
3
1024
cos let 4 sin
3
64 by Wallis's Formula
6
xxy
xy
x
x
xy
k
m
xy
zxy
M k zdzdydx
k xydydxk yxyy dx
k
xdx
k
dx
k
M
z
m



















** *
** *
*
*

433
1 128 2
3
, , 0, 0,
2
k
k
xyz

"

338. 338 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning

52.
≤
≤

≤

2
211 1 21 2
2
000 00 0
2
211 1 21 2
2
000 00 0
2
211 1
000
1
21 2
2 2
00 0 2
0
0
1
222
14
22ln2ln4
1
2
11
arctan
2( 1) 2
1
y
y
xz
y
xy
x
m k dz dy dx k dy dx k dx k
y
y
M k y dz dy dx k dy dx k dx k
y
M k zdzdydx
y
kdydxk ydx
y
y




π

ππππ


ππππ

π




*** ** *
*** ** *
***
** *

2
011
48 24
ln 4 ln 4
21
24 4
ln 4 2
,, 0, ,
4
xz
xy
kdxk
Mk
y
mk
M
zkk
m
xyz










ππ π





π


*

53.
5
,
12
fxy yπ





20 3 5 12 5 12
00 0
20 3 5 12 5 12
00 0
20 3 5 12 5 12
00 0
20 3 5 12 5 12
00 0
200
1000
1200
250
1000
5
200
1200
6
200
250 5
200 4
xy
xy
yz
xy
xz
xy
xy
yz
xz
xy
m k dz dy dx k
Mk xdzdydx k
M k ydzdydx k
M k zdzdydx k
M k
x
mk
Mk
y
mk
M k
z
mk




ππ
ππ
ππ
ππ
ππ π
ππ π
ππ π** *
** *
** *
** *



5
,, 5,6,
4
xyz

π



54.

1
,601220
15
fxy x yπθθ





5353115601220
00 0
5353115601220
00 0
5353115601220
00 0
5353115601220
00 0
10
25
2
15
2
10
25 2 5
10 4
15 2
10
xxy
xxy
yz
xxy
xz
xxy
xy
yz
xz
m k dz dy dx k
k
Mk xdzdydx
k
Mk ydzdydx
M k zdzdydx k
M k
x
mk
Mk
y
mk




ππ
ππ
ππ
ππ
ππ π
ππ π** *
** *
** *
** *
3
4
10
1
10
xyM k
z
mk
πππ



53
,, ,,1
44
xyz

π



x
820
12
12
20
16
8
164
4
yx= + 12−
3
5
y
x
25
3
3
5
4
2
41
1
yx= (5 −)
3
5
y

339. Section 14.6 Triple Integrals and Applications 339
© 2010 Brooks/Cole, Cengage Learning
55. (a)
22 22
000 00
5
32 3 2 3 3
00
0 0
5
11112
33333
2
by symmetry
3
aaa aa
x
aa
aa
xyz
I k y z dx dy dz ka y z dy dz
ka
ka y z y dz ka a az dz ka a z az
ka
III





*** **
**

(b)



2
22 3 3
000 00
24 23 4 422 4 8
23
00
0 0
8
2
2
2
242 8 824 8
by symmetry
8
aaa aa
x
aa
aa
xyz ka
I k y z xyzdxdydz y z yz dydz
ka y z y z ka ka a z z ka
dz a z z dz
ka
III





*** **
**

56. (a)
5
222
2
222
5
55 5
12
by symmetry
12
12 12 6
aaa
xy
aaa
xz yz
xyz ka
Ik zdzdydx
ka
II
ka ka ka
III



***

(b)




37
222 22
22 2 2 2
222 22
7
222 22
22 2 22 4
222 22
7
7
12 72
7
360
by symmetry
30
30
aaa aa
xy
aaa aa
aaa aa
xy
aaa aa
yz xz
xxyxz
yxyyz
zyzxz ak ak
Ik zxydzdydx xydydx
ka
I k y x y dz dy dx ka x y y dy dx
II
ak
II I
ak
III
II I







*** **
*** **
7
7
180
ka


57.
(a)




44 4 44 3
22 2
00 0 00
4 4
3
44 3324
00
00
44 4 44 3
22 2
00 0 00 1
44
3
64 4 32 1
4 4 4 4 4 4 256
33 33 3 3
1
44
3
44x
x
x
y
I k y z dz dy dx k y x x dy dx
yy
kxxdxkxxdxkxxk
I k x z dz dy dx k x x x dy dx
k














*** **
**
*** **



4
4 34
23 3 4
0
0
44 4 44
22 22
00 0 00
4
3
44
22
00
0
1 4 1 1 512
44 4
334123
4
64
4 4 4 256
33
x
z
k
xx xdx kx x x
I k x y dz dy dx k x y x dy dx
y
kxy xdxk x xdx k









*
*** **
**

340. 340 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
(b)




44 4 44 3
22 3
00 0 00
4 4
42
44 3324
00
00
44 4 4 3
22 2
00 0 00 1
44
3
8 2 2048
4 4 64 4 4 32 4 4
46 3 3 3
1
44
3x
x
x
y
I k y y z dz dy dx k y x y x dy dx
yy k
kxxdxkxxdxkxx
Ik yxzdzdydxk xy x y xdydx













*** **
**
*** *




4
4
4 34
23 3 4
0
0
44 4 44
22 2 3
00 0 00
4
22 4
44
2
00
0
4
23 2 3 4
0
1 4 1 1 1024
84 4 8 4
334123
4
48644
24
41
83284 8324
34
x
z
k
kxx xdxkx x x
Ik yxydzdydxk xyy xdx
xy y
kxdxkxxdx
kxxxdxkxxxx











*
*
*** **
**
*
4
0
2048
3
k





58.
(a)



2
424 42 42 4
32 2468
000 00 00
2
3579
4 4
0 0
0
2
424 42 2
222
000 00
1
4 256 256 96 16
44
256 96 16 16,384 65,536
256
4 3 5 7 9 945 315
1
4
2
1
2y
xy
y
xz k
I k z dzdydx k y dydx y y y y dydx
kyyyy k
ydxkdx
I k y z dz dy dx k y y dy dx
k







*** ** **
**
*** **



2
357
42 4 4
246
00 0 0
0
2
424 42 2
222
000 00
2
35
42 4 4
224 2 2
00 0 0
0
16 8 1024 2048
16 8
2 3 5 7 2 105 105
1
4
2
1 8 256 8
16 8 16
2235215
y
yz
kyyy k k
yyydydx dx dx
I k x z dz dy dx k x y dy dx
kyyk
k x y y dy dx x y dx x dx









** * *
*** **
** * *
192
45
2048 8192 63,488
,,
9 21 315
x xz xy y yz xy z yz xz
k
kk k
III II I II I

(b)



2
424
2
000
22
424 424
23
000 000
2
424
2
000
22
424 24
22
000 00
4
32,768 65,536 32,768
4
105 315 315
4
1024 2048 1024
4
15 105 21
y
xy
yy
y
xz
yy
Izzdzdydx
kk k
k z dz dy dx k z dz dy dx
Iyzdzdydx
kkk
k y dz dy dx k y z dz dy dx







***
*** ***
***
*** **

4
0
2
424
2
000
22
424 424
22
000 000
4
4096 8192 4096
4
94515
48,128 118,784 11,264
,,
315 315 35
y
yz
yy
x xz xy y yz xy z xz yz
I k x z dz dy dx
kk k
k x dz dy dx k x z dz dy dx
kkk
II I II I II I





*
***
*** ***

341. Section 14.6 Triple Integrals and Applications 341
© 2010 Brooks/Cole, Cengage Learning
59.
≤
22
22
22222
22
22
2
2
22 2 2222 4
2
44 4
2
2
2
3
21
arcsin 2 arcsin
32 8
2
2
3416 4La ax La
xy
La ax La
a
L
L
a
L
L
I k z dz dx dy k a x a x dx dy
ax x
k xaxa xxaxaa dy
aa
kaa aLk
dy

θ
θθθθ θθ
θ
θ
θ
ππθθ





πθπ
*** **
*
*

Because
22
,4.
xymaLkI maππ

22
22
2222
22
22
2 23
2 2
22 2 2 2 2
2 2
22
2
2222
22
22
2
21
2arcsin
23812
2
La ax La
xz
La ax La
a
L L
L L
a
La ax La
yz
La ax La
Ik ydzdxdyk yaxdxdy
yx kaL
kxaxa dykaydy mL
a
I k x dzdxdy k x a x dxdy


θ
θθθθ θθ
θ θ
θ
θ
θθθθ θθ
ππθ



ππθ*** **
**
*** *



2
442
2 2
2222 4
2 2
22
22
22 2
22
22
1
22 arcsin
8444
3
41212
44 2
3
12 4 12
a
L L
L L
a
xxyxz
yxyyz
zxzyz
xka kaLma
kxxaaxa dy dy
a
ma mL m
II I aL
ma ma ma
III
mL ma m
II I aL
θ θ
θ







*
**


60.


3
222 22
222
222 22
33
222 22 2
22 22
222 22 2
3
22 2
22
22 2
11
12 12 12
11
12 12 12 12
1
12cab ca
xy
cab ca
cab ca c
xz
cab ca c
ab c
yz
ab c b
I z dz dy dx dy dx b abc mb
ba ba c
I y dz dy dx b y dy dx dx a abc ma
abc
I x dz dy dx ab x dx
θθ θ θθ
θθ θ θθ θ
θθ θ
ππππ
ππππππ
ππππ*** **
*** ** *
** *




2
22
2
22
22
22 1
12 12
1
12
1
12
1
12c
c
xxyxz
yxyyz
zxzyz
c abc mc
III mab
III mbc
III mac
θ
π


*

61.
111
22222
110
x
xy x y z dz dy dx
θ
θθ
***


62.
222
11 4
22 2
2
11 0
xxy
x
kx x y dz dy dx
θθθ
θθθ
** *

63. kz
-π
(a)

222
24 4
2
24 0
32
3xxy
x k
mkzdzdydxθθθ
θθθ
ππ


** *

(b)
0xyππ by symmetry

222
24 4
2
2
24 0
1
2 xxy
xy
xM
zkzdzdydx
mm θθθ
θθθ
ππ π ** *

(c)

222
24 4
22
2
24 0
32
3xxy
z
x k
Ixykzdzdydxθθθ
θθθ



** *

2
4
2
z
y
x
z = 4 − x
2
− y
2
x
2
+ y
2
= 4

342. 342 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
64. kxy-π
(a)
222
525 25
00 0
625
3xxy
m kxy dz dy dx k
θθθ
ππ


** *

(b)

222
525 25
00 0
125
32 xxy
yzM
x x kxy dz dy dx
mm θθθ
ππ π


** *


yxπby symmetry

222
525 25
00 0
125
16 xxy
xyM
zzkxydzdydx
mm θθθ
ππ π


** *

(c)

222
525 25
22
00 0
62500
21xxy
z
I x y kxy dz dy dx k
θθθ



** *


65. See the definition, page 1027.
See Theorem 14.4, page 1028.
66. Because the density increases as you move away from
the axis of symmetry, the moment of inertia will
increase.
67. (a) The annular solid on the right has the greater density.
(b) The annular solid on the right has the greater
moment of inertia.
(c) The solid on the left will reach the bottom first. The
solid on the right has a greater resistance to
rotational motion.
68. The region of integration is a cube:

Answer: (a)
69. 1 unit cubeVπ




111
2
000
11 1
22
00 0
1
3
0
1
Average value , ,
4
44
113
44
333
Q
fxyzdV
V
zdxdydz
zdydz zdz
z
z
π





***
***
** *

70. 64 cube with sides of length 4Vπ


444
000
44
00
44
00
1
Average value , ,
1
64
1
8
64
1
88
8
Q
fxyzdV
V
xyz dx dy dz
yz dy dz
zdz zdz
π
π
π
πππ
***
***
**
**
71.

1
base height
3
11 4
22 2
32 3
Vπ'

ππ




,,
fxyz x y z
Plane: 2xyz



22 2 22
00 0 00
2 2
0
1
Average value , ,
331
22
442
31 3 3
42 2
46 4 2
Q
xxy x
fxyzdV
V
xy z dz dy dx x y x y dy dx
xx dx
θθθ θ
π

***
** * **
*

5
5
5
z
y
x
x
2
+ y
2
+ z
2
= 25
x
2
+ y
2
= 25
y
x
3
2
1
z
2
2
2
z
y
x
x + y

+ z = 2
(0, 0, 2)
(0, 2, 0)
(2, 0, 0)

343. Section 14.6 Triple Integrals and Applications 343
© 2010 Brooks/Cole, Cengage Learning
72.
34
343
3
Vππ


222
33 3
222
33 3
11
Average value , , 0, by symmetry
43 xxy
xxy
Q
f x y z dV x y dz dy dx
V

θθθ
θθθθθθ
*** * * *

73.
22 2
22 2
12 3 0
231
xy z
xy z
θθθ/



9
:
22 2
, , : 2 3 1 ellipsoidQxyzxyz


≤
≤
222
12 312 12
22 2
2
2212 12
12 3
1 2 3 0.684
xyx
x
xy
xy zdzdydx
θθθ
θθθ
θθ θ
** *

Exact value:
46
45

74.
9:

222
222
222
222
11 1
222
222
11 1
10
1
, , : 1 sphere
1 1.6755
xxy
xxy
xyz
xyz
Qxyzxyz
xyzdzdydx
θθθ
θθθ θθθ
θθθ/



** *

Exact value:
8
15


75.



22 2
13 4 13
2
00 00
2
3 2
2
11
222 2
00
0
14
4
15
3 94 11 1
443
221532 ay xy ay
a
ay
dz dx dy x y a dx dy
ayxa
yax dy ya ay dy a
θθ θθ θθ
θθ
ππθθθ
θθ



** * **
**

So,

2
322320
23 16 0
16
2, .
3
aa
aa
a

θθπ
π

76.
22
2
2
1
9
yz
x
b


By symmetry, the volume in the first octant is


1
16 2 .
8
π

2222
11 31
00 0
21
bx xyb
dz dy dz
θθθ
π** *

By trial and error, 4.bπ

222
222
4
Volume at ellipsoid 1 is .
3xyz
abc
abc





Note:
1
1
−1
1
z
y
x
2x
2
+ y
2
+ 3z
2
= 1

344. 344 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
77. Let 1 .
kkyx
112 1
22 22
nnnx xnyy y yy
nn n


So,



11 1
2
1112
00 0
00 0 11 1
2 2
112 1122
11 1 00 0
cos
2
sin sin
22
nn
nn nnIxxdxdxdx
n
y y dy dy dy x x dx dx dx I
nn

6<
7>
8?
6< 6<
7> 7>
8? 8?
** *
** * ** *




12 1
1
1.
2
II I
Finally,
1
1
lim .
2
n
I
!

Section 14.7 Triple Integrals in Cylindrical and Spherical Coordinates
1.

25523 52 5 5
10 0 10 1 1 01
99999
22222
cos cos sin 5 1 27rdrddz ddz dz dz z





** * ** * *


2.


6
2
466 46 46
32
000 00 00
0
6
4
44
32
00
0
1
12 36
22
1127
4 18 108 54
24 2 4 2
r
r
rz
rz dz dr d dr d r r r dr d
r
rrd d














*** ** **
**


3.

2
2cos
422 2
2 2 cos 4 2 2 cos 2
22
00 0 00 0
0
2
59
2
48
0
0
sin 4 sin 2 sin
4
8 cos 4 cos 52
8 cos 4 cos sin
945
r r
rdzdrd rr drd r d
d












J
** * ** *
*


4.

2
2
22 2 2 33
2 88
000 00 00
0
11
11
336
e ddd e dd e dd e
--
- -. . .




*** ** **


5.
24cos 24 2 4
234
000 0 00 0 11
sin cos sin cos
3128
ddd dd d .
-.-. ... .


*** ** *


6.

44cos 44
23
000 00
44
2
00
4
3
4
0
0
4
2
4
0
0 1
sin cos cos sin cos
3
1
sin cos cos 1 sin
3
1sin
sin cos sin
33
52 52sin 52
sin cos
36 36 2 144
ddd dd
dd
d
d





-..-. ...
.. .

.. .
.
...








*** **
**
*
*


7.

42
4
000
3
z
r
re d dr dz e

***


8.
2sin
2
000
8
9
2cos ddd

.- - .***

345. Section 14.7 Triple Integrals in Cylindrical and Spherical Coordinates 345
© 2010 Brooks/Cole, Cengage Learning
9.
2 3
23 23 2 2 22
99
000 00 0 0
0
11
11
224
r
e
rr
rdzdrd re drd e d e d e








*** ** * *



10.

5
242
255 25 2 2
3
00 2 00 0 0
0
525252525
52
24 24 4 2r rr
rdzdrd r r drd d d

"


*** ** * *



11.
+,
224 22 2 2 2
2
6060 06 0 0 64 64 32 3 64 3
sin sin cos
3333
ddd dd d d

-.-. .. .
*** ** * *



12.
+,
25 2 2
2
0000 00 0 117 117 468
sin sin cos 156
333
ddd dd d

-.-. .. . *** ** *



13.


224
2
2
00
2arctan124sec 2 2 cotcsc
32 32
0 0 0 0 arctan 1 2 0
cos 0
sin cos sin cos 0
r
rdzdrd
ddd ddd

. ..

-.-. -.-.
***
** * ** *


14.
22
22 16
2
000
2
264 222csc
32 32
000 0 64
8
23
3
8
sin sin 2 3
3r
rdzdrd
ddd ddd

.



-.-. -.-.

***
*** ***


15.
22
2
2
00
42 2cos
32
00 sec
cos 0
sin cos 0
aa ar
a
a
a
rdzdrd
ddd

.
.

-.-.


***
***

x
y
z
1
2
2
3
3
3
1
y
x
3
3
5
−1
−3
z
x
y
4
4
4
z
y
x
7
7
7
r= 5
r= 2
z

346. 346 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
16.
2
23 9
22
000
223
3
000
81
8
81
sin
8r
r z rdzdrd
ddd




-.-.

***
***


17.

22
2cos 2cos
22
00 0 00
2
3
2
33 3 2 3
0
0
44
441422
1 sin cos sin 2 3 4
3333239
aar a
V r dz dr d r a r dr d
a
ada a











** * **
*


18.


2
222 24 163
00 0 0220
2
44
3
Volume of lower hemisphere 4 Volume in the first octantrr
V r dz dr d r dz dr d






** * ***




222 24
22
00 022
4
2 32
2
0
22128
416
3
128 8 2 1
416
33 3
128 8 2 8 2 128 64 2 64
422
333333
Vrdrdrrdrd
rd
















** **
*


19. In the xy-plane,
22
22 2xx y



222 2
22
2
01414
12 12
xxy xx y
xy



In polar coordinates, use
cosr for this circle.


cos 2 cos
2
00 2
cos
23
00
cos
34
0
0
4
4
0
4
0
2cos 2
2
cos
32
2cos
cos
32
1
cos
616
r
r
V r dz dr d
rrdrd
rr
d
d
d





















** *
**
*
*
*


20.
22 22 22
21xy xy xy



2
212
2
00
21
2
00
1
4
2
2
0
0
22
1
2
22
r
r
V r dz dr d
rrdrd
r
rd












***
**
*


21.



22
cos
00 0
cos
2
00
cos
32
22
0
0
3
3
0
333
0
2
2
1
2
3
2
1sin
3
2cos2
cos 3 4
339
aar
a
a
V r dz dr d
ra rdrd
ar d
a
d
aa













)










** *
**
*
*


22.


2
224
00
22
22
00
2
3
2 32
2
0
0
4
18
422
333
r
r
V r dz dr d
rrrdrd
r
rd













***
**
*


23.

+,
+,
+,
229cos2sin
000
22
2
00
2
44
2
3
0
0
2
0
2
0
9cos 2sin
3cossin
42
24 4 cos 8 sin
24 4sin 8cos
48 8 8 48
rr
m kr r dz dr d
kr r r dr d
rr
kr d
kd
k
kk




















***
**
*
*

y
x
7
7
7
z

347. Section 14.7 Triple Integrals in Cylindrical and Spherical Coordinates 347
© 2010 Brooks/Cole, Cengage Learning
24.


2
2212 22 2
000 00
2
2 2
0
0
2
4
0
4
12
6
66
31
r
e
r
r
k r dz dr d ke r dr d
ke
ke k d
ke
















*** **
*
*


25.



22
0
00
2
0 00
000
2
0 2
0
00
0
3
2
0
0
0
3
0 2
0
0
4
4
4
6
41
62 3
rhrrr
r
hh
zh x y r r
rr
V r dz dr d
h
rr r drd
r
rh
d
r
rh
rh
r
















***
**
*


26.
0xy by symmetry

2
01
3
mrhk
from Exercise 25



2000
000
2
20
223
00
2
00
0
24 22
00
2
0
22
0
2
0
4
2
2
2
12 2 12
3
12 4
rhrrr
xy
r
xy
M k zr dz dr d
kh
rr rr r drd
r
kh r kr h
r
M kr h h
z
mrhk















***
**


27.
22
kx y kr-
0xy by symmetry

2000
23
0
000 1
4
6rhrrr
mk rdzdrd krh



***


2000
2
000
32
0
32
0
3
0
4
1
30
30
65
rhrrr
xy
xy
M k r z dz dr d
krh
M krh h
z
mkrh








***


, , 0, 0,
5
h
xyz






28. kz
-
0xy by symmetry

2000
22
0
000 1
4
12rhrrr
m k zr dz dr d k r h



***


2000
2
000
23
0
23
0
22
0
4
1
30
30 2
12 5
rhrrr
xy
xy
M k z r dz dr d
krh
M krh h
z
mkrh








***



2
, , 0, 0,
5
h
xyz






29.


2000
3
000
20
34
0
00
0
5
0 4
0
0
4
4
41
20 2 10
rhrrr
z
r
Ik rdzdrd
kh
rr r drd
r
kh r
krh
r











***
**

Because the mass of the core is
2
01
3
mkV k rh






from Exercise 25, we have
2
0
3.kmrh So,

442
000
2
0113 3
.
10 10 10
z
m
I k rh rh mr
rh







30.

22
2000
4
000
20
0 4
00
0
0
56 55
22
00
00
0
0
2
555
000
0
,,
4
4
44
56 5 6
111
44
30 30 2 15
z
Q
rhrrr
r
rIxyxyzdV
k r dz dr d
rr
kh r dr d
r
rr rr
kh d kh d
r
kh r d kh r r kh




-














***
***
**
**
*


31.



22 22
2
3
00
22
344
00
44 2222
22
4
4
22
1
2
bh
z
a
b
a
mkbh ah khb a
Ik rdzdrd
kh r dr d kh b a d
kb ah kb a b ah
ma b











***
** *


32.
2
22sin
342
00 0
33
2
22ah
z
mkah
Ik rdzdrd kahma




** *

348. 348 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
33.
+,
223
2
040
22
04
2 2
40
2
0
sin
9sin
9cos
22
91892
22
Vddd
dd
d
d







-.-.
..
.






***
**
*
*


34.
222
22 2
2
22
11
44
11
24
xyz z
xy zz
xy z










Sphere with center
1
0, 0, : cos
2
- .






4
34
24cos 24 2 2
2
00 0 00 0 0
0
cos cos 1 1
sin sin 1
3121248
V d dd dd d d

.
. .
-.-. ..



*** ** * *


35.
24sin
22
000
sin 16Vddd
.
-.-. ***


36.




42
2
00
42
33
00
4
33
0
4
33 33 33
0
8 sin includes upper and lower cones
8
sin
3
4
sin
3
4242
cos 1 2 2
3233
b
a
Vddd
ba dd
ba d
ba ba ba




-.-.
..

..

.





***
**
*


37.

22
3
000
22
4
00
2 2
444
00
8sin
2sin
sin cos
a
mk ddd
ka d d
ka d ka ka



-.-.
..
...


***
**
*


38.
22
32
000
22
42
00
2
42
0
2
4424
0
8sin
2sin
sin
11 1
sin 2
24 44
a
mk ddd
ka d d
ka d
ka ka k a




-.-.
..
..

. .





***
**
*


39.
32
3
mkr


0xy by symmetry


22
3
000
22
4
00
4
2
0
2
44
0
4
3
4cossin
1
sin 2
2
sin 2
4
11
cos 2
84
43
238
3
, , 0, 0,
8
r
xy
xy
Mk ddd
kr d d
kr
d
kr kr
M kr r
z
mkr
r
xyz




-..-.
..

..
.












***
**
*

y
x
3
3
2
3
z
y
x
1
1
1
−1
−1
z
y
x
a
a
b
b
b
z

349. Section 14.7 Triple Integrals in Cylindrical and Spherical Coordinates 349
© 2010 Brooks/Cole, Cengage Learning
40. 0xy by symmetry

33 3322 2
33 3
mk R r kR r


















22
3
00
22
44
00
2
44
0
2
44 44
0
44 44
33 33
44
33
4cossin
1
sin 2
2
1
sin 2
4
11
cos 2
84
43
238
3
, , 0, 0,
8
R
xy
r
xy
Mk ddd
kR r d d
kR r d
kR r kR r
kR r R rM
z
m kR r R r
Rr
xyz
Rr




-..-.
..
..
.














***
**
*


41.

22cos
43
40 0
22
53
40
2
52
4
2
68
4
4sin
4
cos sin
5
2
cos 1 cos sin
5
21 1
cos cos
5 6 8 192
zIk ddd
kdd
kd
k
k
.







-.-.
...
....

..





***
**
*


42.




22
43
00
22
55 3
00
2
55 2
0
2
3
55
0
55
4sin
4
sin
5
2
sin 1 cos
5
2cos
cos
53
4
15
R
z
r
Ik ddd
k
Rr dd
k
Rrd
k
Rr
k
Rr




-.-.
..

. ..
.
.







***
**
*


43.
222
cos
sin tan
xrxyr
y
yr
x
zz zz






44.
2222
222
sin cos
sin sin tan
cos cos
xxyz
y
y
x
z
z
xyz
-. -
-.
-. .





45.





cos , sin2
2 2
cos , sin11 1
cos , sin ,
ghrr
ghrr
fr r z r dz dr d


** *


46.
22 2 2
111
sin cos , sin sin , cos sin
f ddd
.-
.-
-.-.-.- .-.***


47. (a)
0:rrright circular cylinder about z-axis

0:plane parallel to z-axis

0:zzplane parallel to xy-plane
(b)
0:
-- sphere of radius
0-

0:plane parallel to z-axis

0:
..cone

48. (a)
22
2
22
000
aar
r z rdzdrd



***

(b)
22
3
000
sin
a
ddd

-.-.***

Integral (b) appears easier. The limits of integration are all constants.

y
x
a
a
a
z

350. 350 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
49.



222222222 22222
22 22
00 0 0 00 0
22
2
22 2
000
22
2
22 2 22
00 22
0
22
2
22
00
16 16
16
1
16 arcsin
2
84
22
aax axy axyz aax axy
aar
ar
a
a
dw dz dy dx a x y z dz dy dx
ar zdzrdrd
z
za r z a r rdrd
ar
ar
a r rdrd















** * * ** *
***
**
**
442
22
4
00
0
42
a
ra
da d




**

50.




222
222
2 2
2
000
2 2
3
000
2 2
3
000
2
2
0
sin
lim sin
lim sin
1 1
22 lim 4 2
22
xyz
k
k
k
k
k
k
x y z e dx dy dz
eddd
eddd
dd ed
e

-

-

-
-
-- .-.
-.-.
.. - -
-

!!!
! ! !
!


!

!

!










***
***
***
***


51.
2
222 22
836
xyz xy
In cylindrical coordinates,



2
22 2
22
22
2
2
836
86
69 10
31.
rz r
rz r
rr z
rz





This is a torus: rotate

2
2
31xz about the
z-axis. By Pappus' Theorem,


2
23 6.V
Section 14.8 Change of Variables: Jacobians
1.


1
2
1
2
x uv
yuv



11 11 1
22 22 2
xy yx
uv uv
CC CC


CC CC


2.
xau bv
ycudv




xy yx
ad cb
uv uv
CC CC

CC CC


3.
2
xuv
yuv



11 1 2 1 2
xy yx
vv
uv uv
CC CC

CC CC

4. 2
xuv u
yuv



22
xy yx
vuvu u
uv uv
CC CC

CC CC

5. cos sin
sin cos
xu v
yu v




22
cos sin 1
xy yx
uv uv
CC CC

CC CC


6.
xua
yva



11 0 0 1
xy yx
uv uv
CC CC

CC CC

7. sin
cos
u
u
xev
ye v




2
sin sin cos cos
uu uu uxy yx
eveve ve v e
uv uv
CC CC

CC CC


8.
u
x
v
yuv




222
11
11
xyyx u uuv
uv uv v v v v v
CC CC


CC CC

351. Section 14.8 Change of Variables: Jacobians 351
© 2010 Brooks/Cole, Cengage Learning
9.

32
3
3
2322
3339
xuv
yv
y
v
xyxvxy
u

π
π
θθ
ππ πθ


10.


1
3
1
3
4
4
x uv
yuv
uxy
vx y
πθ
πθ
πθ
πθ









11.


1
2
1
2
x uv
yuv

πθ

uxy

vxyπθ








12.


1
3
1
3
2
x vu
yvu
πθ


2uy x
vxy
πθ












13.
84
33
8 4
33
5 2
33
21
32
34
20
,4
38
24
,4
35
24
,1
31
20
,1
y
xy
yxxy
y
xy
yxxy
y
xy
yxxy
y
xy
xyxy
π
θπ <
>
ππ ?
π
θπθ <
>
ππ ?
π
θπθ <
>
ππθ ?
π
θπ <
>
ππ ?




1
3
1
3
3
2 23 2
uv y y uvuxy
vx y uv x x uv
<
>
πθ ?


23 42 43 42 83 4
23 1 23 2 43 2
32 164 296 164
27 27 27 9
33 33
xx x
xx x
R
xy dA xy dy dx xy dy dx xy dy dx

θθ
** * * * * * *

,xy ,uv
(0, 0) (0, 0)
(3, 0) (1, 0)
(2, 3) (0, 1)
,xy ,uv
(0, 0) (0, 0)
(4, 1) (3, 0)
(2, 2) 0, 6θ
(6, 3) 3, 6θ
,xy ,uv

11
22
, (1, 0)
(0, 1) 1, 1θ
(1, 2) 3, 1θ

33
22
, (3, 0)
,xy ,uv

14
33
,θ (2, 1)

12
33
, (0, 1)

84
33
, (0, 4)

102
33
, (2, 4)
u
1
1
(0, 1)
(1, 0)
v
(0, 0) (3, 0)
(3, −6)(0, −6)
−1
−1
1
−2
−3
−4
−5
−6
12 4 56
v
u
2
1
(1, 0)
(1, −1)
(3, 0)
(3, −1)
−1
−2
u
v
1234
1
2
3
4
u
v
(0, 1)
(0, 4)
(2, 1)
(2, 4)
123
−1
3
x
y
x + y = 1
x + y = 4
x − 2y = 0
x − 2y = −4
4
3
8 3
,((
2
3
1 3
,(( 8
3
4 3
,((
2 35 3
,((
−

352. 352 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
14.



11 232
2
01 1 1
2
22
sin
cos 1515178
sin 1 cos 1 sin 1 cos 1 sin 1 cos 1 7 3
16 3 16 16 3
13 13 13 13
sin 1 cos 1 sin 2 2.363
33 36
xx
xx
R
x y x y dA f x dy dx f x dy dx

θθ

θ



** * * * *


15.


1
2
1
2
x uv
yuv

πθ


11 11 1
22 22 2
xy yx
uv uv
CC CC
θπθθ πθ

CC CC





11 22
22
11
1
3
11 1
22 2
11 1
1 111
44
442
18
22
3333
R
x y dA u v u v dv du
uu
u v dv du u du
θθ
θθ θ
θ







** * *
** *


16.


1
,
2
1
,
2
x uv u xy
yuvvxy


11 1 1 1
22 2 2 2
xy yx
uv uv
CC CC
θπθθπ

CC CC




13
11
13
22
11
3 1
3
11
22 3
11
11 111
60 60
222
15
2
15 15 26 15 2 26 2 26
2 15 120
23 2 3 23 3 3 3
R
xy dA u v u v dv du
v u dv du
v
uv du u du u u
θ
θ
θθ
θ




πθθ

πθ θ π θ π θ π θ πθ

** * *
**
**



17.
xuv
yu
π

10 11 1
xy yx
uv uvCC CC
θπθπθ
CC CC



34
00
3
0
1
836
R
yx y dA uv dvdu
udu
θπ
ππ** * *
*



18.


1
2
1
2
x uv
yuv

πθ


1
2xy yx
uv uvCC CC
θπθ
CC CC





20
02
2
2
0
2
2
22
0
2 1
44
2
21
2
2
21
xy v
Ru
u
uu
xy e dA ue dv du
uedu
u
ue e
e
θ
θ
θ
θθ
θ



πθ




πθ
** * *
*

,
xy ,uv
(0, 1) 1, 1θ
(2, 1) (1, 3)
(1, 2) 1, 3θ
(1, 0) (1, 1)

12
1
2
y = x + 1
y = x − 1
y = −x + 3
y = −x + 1
x
y
(−1, 1) (1, 1)
(1, −1)(−1, −1)
u
v
(1, 3)(−1, 3)
(−1, 1) (1, 1)
−2−1
−1
2
12
v
u
u
2
3
4
12 34
1
v
u
12
−1
−2
vu=2−
v

353. Section 14.8 Change of Variables: Jacobians 353
© 2010 Brooks/Cole, Cengage Learning
19.
2
14
:,2,,
4
,,
xy
R
edA
x
Ry y xy y
xx
y
x vu y uv u v xy
x
′
ππ ππ

**


θΔ
θΔ
12
32 12 12
12 12
12 12
111
, 11 1 122
,42 11
22
xx v
xy uv u uv
uv u u uyy vu
uv uv
CC
′
C CC
ππ π′φπ′

C CC
CC

Transformed Region:

1
11
4
44
222
11
444
yyxv
x
yuxv
x
y
yx u
x
xy
yu
x






θΔ
θΔ θΔ θΔ
4
2
24 2 2
22 212
14 1 14 14
1
2
212 212 12 2
14
11
2
1
ln ln 2 ln ln 8 0.9798
4
v
xy v
R
e
e dA e dv du du e e du
uu u
ee u ee e e
′
′′ ′′
′′ ′′ ′ ′

ππ′π′′





** * * * *

20.
u
x
v
yv
π
π


1xy yx
uv uv v
CC CC
′π
CC CC


θΔ
+, θΔ
44 4 4
111 1 1
sin sin 3 sin 3 cos 3 cos 1 cos 4 3.5818
R
yxydA vudvdu udu u
v
πππ′π′** * * *

21.
1, 1
1, 3
uxy vxy
uxy vxy
π′π πφπ
π′π′ πφπ


θΔ
θΔ
1
2
1
2
x uv
yvu
πφ
π′


θΔ
θΔ
, 11 1 1 1
,22222
xy xy yx
uv u v u v
C CC CC
π′π′′π

CCCCC


θΔ θΔ θΔ
1
3
31 31 3
22 2
11 11 1
1
3
3
3
2
1
1
111
48 48 6 6
222 3
222 22
62 6 6182 96
333 33
R
u
xydA u v v u dudv v u dudv uv dv
v
vdv v
′′
′

πφ′π′π′



π′π′π′′φπ


** ** ** *
*

y
x
2
3
4
12 34
1
yx= 2
yx=
1
4
y=
1
x
y=
4
x
R
u
2
3
134
S
v
12
1
2
(1, 2)
(2, 1)
y = x + 1
y = x − 1
y = −x + 3
x
y
y = −x + 1
R

354. 354 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
22. 20,320
28,3216
uyx vxy
uyx vxy
π′π πφπ
π′π πφπ


θΔ
θΔ
1
4
1
3
8
x vu
yvu
π′
πφ


θΔ
θΔ
, 11 31 1
,48848
xy
uv
C
π′ ′ π′

C


θΔ
8
2
16 8 16 162
12 32
00 0 0
0
232163240962
32 2 2 2
33233
R
x y y x dA v u du dv vu dv v dv

φ′π π π π π

** * * * *


23.
θΔ θΔ
4, 0
8, 4
11
22
uxy vxy
uxy vxy
x uv y uv
πφπ π′π
πφπ π′π
πφ π′


θΔ
θΔ
, 1
,2
xy
uv
C
π′
C


θΔ
θΔ θΔ θΔ
8
84 8
4244
40 4
4
11 1
11121
22 4
xy v
R
xye dA ue dvdu ue du ue e
′
φπ π′π′π′


** ** *


24.
θΔ θΔ
,0
2,
11
,
22
uxy vxy
uxy vxy
x uv y uv

πφπ π′π
πφπ π′π
πφ π′

θΔ
θΔ
, 1
,2
xy
uv
C
π′
C


θΔθΔ
2
334
22
222
00
0
111cos2717
sin sin sin 2
2232 12212
R
uv
x y x y dA u v du dv dy v v




′
φ′π π π′ π

** * * *


25.
θΔ θΔ
40, 0
45, 5
11
4,
55
ux y vxy
ux y vxy
x uv y uv
πφ π π′π
πφ π π′π
πφ π′


11 14 1
55 55 5
xy yx
uv uv
CC CC
′π′′ π′

CC CC


θΔθ Δ
55
55 5
32 32
00 0
0 0
1 1 2 2 5 2 100
4
553 339
R
x y x y dA uv du dv u v dv v

′φ π π π π
** ** *


26.
θΔ θΔ
32 0, 2 0
32 16, 2 8
11
,3
48
uxy vyx
uxy vyx
x uv y u v
πφ π π ′π
πφ π π ′π
π′ πφ


13 1 1 1
48 8 4 8
xy yx
uv uv
CC CC
′π′′π

CC CC


θΔθΔ
8
816 832
32 32 52
00 0
0
1 2 4096
322 16 16 2
855
R
x y y x dA u v du dv v dv v

φ′ π π π π

** ** *

−1−2 234
−1
2
3
5
(0, 0)
(4, 2)
(2, 5)
(−2, 3)
y =x + 4
1
2
y = −x + 8
3
2
y = −x
3
2
y =x
1
2
x
y
R
42
4
2
6
6
x
xy−= 0
xy−= 4
xy+= 8
xy+= 4
y
2
x
xy−= 0
xy−=
xy+=
π
π
π
π
xy+= 2π
π
2
π
2
π3
2
π3
y
1
2
−1
−2
−134
x
xy+ 4 = 0
xy+ 4 = 5
xy−= 5
xy−= 0
y
x
2
3
3
5
2
41
−1
−1−2
2= 0yx−
2= 8yx−
3 + 2 = 16xy
3 + 2 = 0xy
y
(0, 0)
(−2, 3)
(2, 5)
(4, 2)

355. Section 14.8 Change of Variables: Jacobians 355
© 2010 Brooks/Cole, Cengage Learning
27. θΔ θΔ
11
,, ,
22
uxyvxyx uvy uvπφπ′πφ π′


1
2
xy yx
uv uv
CC CC
′π′
CC CC


52 52
00
0122
255
a
au a
Ru
xy dA u dv du u u du u a
′

φπ π π π


** * * *



28.
1, 1
4, 4
,
ux vxy
ux vxy
u
xu y
v
ππ π π
ππ π π
ππ

1xy yx
uv uv u
CC CC
′π
CC CC

θΔ +, θΔ
44
44 4
2
22 2
11 1
11
1111 117
ln 1 ln 17 ln 2 ln ln ln 4
11 2 2 22
R
xy v
dA dv du v du u
xy v u u

ππφπ′π

φφ
** * * *


29. 2uxy
vxy
π′
πφ


θΔ
1
3
3
xuv x uv
Then θΔ θΔ
11
33
2.yvxv uv vuπ′π′ φ π ′








One side is parallel to the u-axis.

30.

0, 3 6
2, 3 2
yx yx
yx yx
′π ′π
′π′ ′π

Let
and 3 .uyx v yxπ′ π ′
Then
θΔ θΔ
11
22
3and .
x vu y vuπ′ π′
So,
θΔθ Δ
θΔ θΔθΔ
11
22
,, 3, .Tuv x y v u v uππ′ ′
Note that:

θΔ,
xy θΔ,uv
(0, 0) (0, 0)
(6, 3) (9, 9)
(2, 7) θΔ3, 9′ θΔ,xy θΔ,uv
(1, 1) (0, 2)
(4, 2) (–2, 2)
(6, 4) (–2, 6)
(3, 3) (0, 6)
x
a
a
xya+=
y
u
a
−a
2a
v = −u
vu=
v
x
2
3
4
12 34
1
x= 4
x= 1
xy= 4
xy= 1
y
2468
2
4
6
8
R
(6, 3)
(2, 7)
7
2
y =
x
1
2
y =
x
y = −x + 9
y
x
26
2
4
6
(1, 1)
(3, 3)
(6, 4)
(4, 2)
1
3
y =
x + 2
1
3
y =
2
3
x +
y = x
y = x − 2
R
y
x
−2−4 246810
6
8
12
R
v
u
(9, 9)(−3, 9)

356. 356 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
31.

22
22
22
22
22
1, ,
1
1
xy
xau y bv
ab
au bv
ab
uv



(a)
22
22
22
11
xy
uv
ab


(b)



,
00
,
xy xy yx
ab ab
uv u v u v
C CC CC

CCCCC

(c)


2
1
S
A ab dS ab ab**


32. (a)

22
22
,16
:1
16 9
,
R
fxy x y
xy
R
VfxydA



**

Let 4
xuand 3.yv



2
11
22 2 2
2
11
21
22 22
00
1
22
242 42 2 2
00
0
16 16 16 9 12 Let cos , sin .
16 16 cos 9 sin 12
12 8 4 cos sin 12 8 4 cos sin
1 cos 2 9 1 cos 2
12 8 4
24
u
Ru
x y dA u v dv du u r v r
r r rdrd
rr r d d










KK


00



** * *
**
**
2 2
0 0
2
0 39 7
12 cos 2
288
39 7 39
12 sin 2 12 117
816 4
dd












**

(b)

22
22
,cos
2
x y
fxy A
ab






22
22
:1
xy
R
ab

Let
xauand .ybv


2
11
22
2
11
,cos
2
u
Ru
fxydA A u v abdvdu





** * *

Let
cos , sin .ur vr



1
21
2
00
0
2
24
cos sin cos 2
222
4224
200
rr r
Aab r rdrd Aab
Aab
Aab













**
x
a
R
b
y
u
1
S
1
v

357. Section 14.8 Change of Variables: Jacobians 357
© 2010 Brooks/Cole, Cengage Learning
33.
θΔ
θΔ
,
Jacobian
,
xy xyyx
uv u v u v
C CC CC
ππ′
CCCCC


34. See Theorem 14.5.

35. θΔ θ Δ1, 1 ,
xuvyuvwzuvwπ′ π ′ π

θΔ
θΔ
θΔθΔ θΔθΔ θΔ θΔ θΔ θΔ
2222 222
10
,,
11 11 1 1
,,
vu
xyz
v w u w uv v u v w u vw u uv w uv w v u v u uv u v
uvw
vw uw uv
′′
C
π′ ′ ′π′ ′φ φ ′φ π′ φ π

C


36. 4,4 ,
xuvy vwz uwπ′ π′ πφ

θΔ
θΔ
410
,,
04117
,,
10 1
xyz
uvw
′
C
π′π
C


37.
θΔ θΔ
11
,,2
22
x uvy uvz uvwπφ π′ π

θΔ
θΔ
+,
12 12 0
,,
12 12 0
,,
222
21414
xyz
uvw
vw uw uv
uv uv
C
π′
C
π′′π′


38. ,2,
xuvwy uvz uvwπ′φ π πφφ

θΔ
θΔ
θΔ θΔ θ Δ
111
,,
220
,,
111
12 12 12 2 4
xyz
vu
uvw
uvvuv
′
C
π
C
πφφ′π


39. sin cos , sin sin , cosxyz
-. -. -.πππ

θΔ
θΔ
θΔ
222 2 2222
222 22
sin cos sin sin cos cos
,,
sin sin sin cos cos sin
,,
cos 0 sin
cos sin cos sin sin cos cos sin sin cos sin sin
cos sin cos sin cos sin sin cos
xyz.-.-.
.-.- .
-.
.-.
.-..-..-.-.-.
.- . . - .- .
′
C
π
C
′
π′ ′ ′ φ

π′ φ ′ φ

θΔ
θΔ
2
22232 22 2
sin
sin cos sin sin cos sin sin
-..- .-. . . -.

π′ ′ π′ φ π′


40. cos , sin ,
xryrzzπππ

θΔ
θΔ
22
cos sin
,,
sin cos 0 1 cos sin
,,
001
r
xyz
rrrr
rz


C
ππφπ

C


41. Let
θΔ
θΔ
30, 3
,3,.
3, 22 01
xyx xu
uvy y v
uv
C

C


Region A is transformed to region ,Aand region B is transformed to region .B

22
3
39
AB m m


Note: You could also calculate the integrals directly.
−1−313
−2
−3
2
3
x
y =
x
2
x
2
9
A
y
+ y
2
= 1
−1−313
−2
−3
2
3
B
y = mx
x
y
x
2
9
+ y
2
= 1
−11
−1
1
A′
u
v =
3
2
u
u
2
+ v
2
= 1
v −11
−1
1
B′
v = 3mu
u
v
u
2
+ v
2
= 1

358. 358 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 14
1. θΔ θΔ
2 2
32 332
11
ln 1 ln 1 ln ln
x x
xydy xy y x x x x x x x
*

2. θΔ
2
33
2
22 2
10
33
y
y
y
y
x y
xydx xy

φπφπ

*


3. θΔ
θΔ
11 1 1
2
000 0
1
1
232
0
0
5294
32 6
32 3
451
x x
x y dy dx xy y dx
xxdx xxx
φ φ
φπφ

πφφπφφπ

** *
*


4. θΔ
θΔ
22 2 2
222
22
00
2
234
0
2
345
0
88412
325 15
2
422
x x
xx
x y dy dx x y y dx
x xxdx
xxx
φπφ

πφ′
πφ′ π

** *
*

5.
θΔ
2
39 3
2
00 0
3
32
2
0
4
3
449
936
x
xdydx x x dx
x
′
π′

π′ ′ π
** *


6.
2
32 4 3
2
2
024 0
3
2
0
24
44arcsin
2
4
3
3
y
y
dx dy y dy
y
yy

φ′
′′
π′

π′φ


πφ** *

7.
θΔ
3(3)3 133
00 00
1133 1
2
00 0 0
33
22
33 3
xy
y
dy dx dx dy
Adxdy ydyyy
′′
′
π
ππ′π′π
** **
** *

8.
θΔ
23622(6)2
00 20 0
22(6)2 2
2
00 0
311
222
(6 3 ) 6 3
xx y
y
y
y
dy dx dy dx dx dy
Adxdy ydyyy
′′
′
φπ
ππ′π′π
** ** **
** *

9.
22 2
325 425 43 525
2222
525 5 25 425 4 25
3
2
325 3
22
50 5
5
25 3
2 2 25 25 25 arcsin 12 25 arcsin 67.36
52 5
xy y
xyyy
x
dy dx dx dy dx dy dx dy
x
Adydxxdxxx

′′′ ′
′′ ′ ′ ′ ′ ′′ ′ ′ ′
′
′′
′
πφφ

ππ′π′φπφφ

** ** ** **
** *

1234
1
2
3
4
y
x
(2, 4)
y = 2x
y = x
2
4321
4
3
2
1
x
y
yx=9−
2
123
1
2
(2, 2)y = 3
(x − 2)
2
+ y
2
= 4
y
x
321
3
2
1
x
y
y = x + 1

359. Review Exercises for Chapter 14 359
© 2010 Brooks/Cole, Cengage Learning
10.

2
46 0 1 1 81 1 9 3 9
2
02 111 039 839
2
446 4
223
2
02 0 0
642
33
82 4
xx y y y
xx y y y
xx
xx
dy dx dy dx dx dy dx dy
Adydxxxdxxx






** * * ** **
** *

11.

1
2
11 1 32
22
00 0
0
2
12 (1 1 4 ) 2
02
(1 1 4 ) 2
44
4411
33
4xx
y
y
Adydxxxdxx
Adxdy







** *
**

12.
2
21 12 52
00 00 1 1
14
3
y
x
Adxdydydxdydx


** ** **

13.
51 21 23
2
23 10 1 1
9
2
2
xxy
xy
Adydx dydxdxdy


** ** **

14.
2
32 0 1 1 11 1
03011
9
2
yy x x
yx x
A dxdy dydx dydx


** * * **

15. Both integrations are over the common region R shown in the figure. Analytically,



2
122
02
2
22 22 8 2
00 2 0
44
33
5 4144
33 3 33
2
22
y
y
xx
x y dx dy
x y dy dx x y dy dx



**
** * *

16. Both integrations are over the common region R shown in the figure. Analytically,


25
5
032
323 55
53 53 5
00 30
82
55
38 22
55 55
y
xy
y
xx
xy xy
edxdy e
e dy dx e dy dx e e e e e





**
** **


17.

2
44
2
00
2
4
4
22
0
0
4
42
0
4
53
0
1
2
1
2
329614
10 3 15
4
4
48
8
x
x
Vxydydx
xyy y dx
xx dx
xxx







**
*
*

18.
3
00
3
2
0 0
3
2
0
3
3
0
1
2
3
2
271
22
x
x
Vxydydx
xyydx
xdx
x






**
*
*

19. Area 16R


22
22
22
2
3
2
2
2
2
2
2
2
2
3
21
Average Value 16
16
1
16
16 3
116
64 4
16 3
1416
64
16 3 3
1646440
256
16 3 3 3
xydydx
y
yxy dx
xdx
x
xx





















**
*
*

(2, 1)
yx=
1
2
yx=8 −
2
1 2 3
−1
1
2
x
1
2
y
R
x
1
1
2
2
3
3
4
4
5
5
(3, 2)
y

360. 360 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
20. Area 9R

3
3
33 3 3 3
22 2 2 3
000 0 0
0
1111
Average Value 2 2 6 9 2 9 9
99399
y
x y dy dx x y dx x dx x x




** * *

21. Area 35 15R


+,
5
3
35 3
22 2
00 0
0
3
3
23
0
0
11
Average temperature 40 6 40 6
15 15 3
1 125 1 125 1 2
200 30 200 10 600 270 125 13
15 3 15 3 15 3
y
x y dy dx y x y dx
x
x dx x x C








** *
*

22.
60 50
22
45 40
1
150
Average 192 576 5 2 5000 13,246.67x y x y xy dx dy
**

23.



00 0 0
0
0
1
1
xy xy
x
x
kxye dy dx kxe y dx
kxe dx
kx e k
!!! !

!

!






** *
*

So, 1.k


11
00
0.070
xy
Pxyedydx

**

24.
2
11
00 0
0
1
34
1
0
0
2
288
x
x
kxy
kxydydx dx
kx kx k
dx








** *
*

Because
1,
8
k
you have 8.k

0.25 0.5
00
0.25 0.5
2
00
0.25
0
0.25
2
0
8
4
0.125
2
P xydydx
xydx
xdx
x








**
*
*

25.

927
22
Volume base height
3



Matches (c)
26. Matches (c)


27. True
28. False,
11 2 2
00 1 1
xdy dx x dy dx#** **


29. True
30. True,
11 11
22 2
00 00 11
114
dx dy dx dy
xy x

** **

31.

4sec
22 2
00 0 0
33 3
4 4
3
00
sec sec tan ln sec tan 2 ln 2 1
36 6
hx h
x y dy dx r dr d
hh h
d





** * *
*

32.
4
42
416 24 2 2
22 3
00 0 0 0 0
0
64 32
4
y r
xydxdy rdrd d d





** * * * *

33.
2cos 2
00 0 0
0 1cos2 1 sin2 9
22cos44cos44sin
2242
Ardrd d d







** * *

34.

2
22sin2 2 2 2
00 0 0
0
1cos4 sin4
422sin2842
24
Ardrd d d







** * *

x
y
(3, 3)
(3, 0)
2
2
4
4
4
6
z
x
y
2
2
1
2
3
z

361. Review Exercises for Chapter 14 361
© 2010 Brooks/Cole, Cengage Learning
35. θΔ
32
21 2
223
00 1 00 0
0
1
4211
33
h
hz h h
h
V rdrddz z ddz zdz z
φ
ππφ′πππ


** * ** *

36.
θΔ θΔ θΔ
22 2 32 32 32
22 22 22 22
00 0
88 4
333
8
R
R
b
b
VRrrdrdRrdRbdRb



π′π′′π′π′

** * *

37. (a) θΔθΔ
2
22 22
9
xyxyφπ′

θΔ θ Δ
θΔ
2
22222
222
9cos sin
9cos sin
9cos2
3cos2
rr r
r
r


π′
π′
π
π

(b)
43cos2
00
49Ardrd

ππ**

(c)
43cos2
2
00
4 9 20.392V r rdrd

π′**

38.
12 13 3
tan 0.9828
2813

The polar region is given by 0 4r and
0 0.9828.
So,

θΔθΔ
θΔarctan 3 2 4
00 288
cos sin .
13
r r rdrd
π**


39. (a)
12
3
02
12
2
3
02
12
2
3
02
4
16
55
8
45
32
45
64
55
x
x
x
x
x
x
y
x
y
x k
m k xy dy dx
k
M k xy dy dx
k
M k x y dy dx
M
x
m
M
y
m
ππ
ππ
ππ
ππ
ππ
**
**
**


θΔ
32 64
,,
45 55
xy

π



(b)
θΔ
θΔ
θΔ
12
22
3
02
12
22
3
02
12
22
3
02 17
30
392
585
156
385
936
1309
784
663x
x
x
x
x
x
y
x
y
x k
m k x y dy dx
k
Mk yxydydx
k
Mk xxydydx
M
x
m
M
y
m
πφπ
πφπ
πφπ
ππ
ππ
**
**
**


θΔ
936 784
,,
1309 663
xy

π





40.
θΔ θΔθ Δ
θΔ θΔθ Δ
222
22
2
00 0
2
2222
22
2
00 0
2234223452
234 234
0
0
7
2
212
2
8
43 2 2 17
44
88258Lh xLxL L
Lh xLxL L
x
L
L kh x x khL
m k dy dx dx
LL
kh x x
Mk ydydx dx
LL
kh x x x x kh x x x x kh
dx x
LL L L LL L L

′′


′′


ππ′′π


ππ′′


π ′′φφ π ′′φφ π"


** *
** *
*
θΔ θΔθ Δ
2
22
22
00
23 3 4 2 2
2
22
0
0
2
2
17
10 80
55
2
223421224
5125
24 7 14
17 12 51
80 7 140
Lh xLxL
y
L
L
y
x
L kh L
Mk xdydx
kh x x kh x x kh L khL
xdxx
LL L L
M khL L
x
mkhL
MkhL h
y
mkhL

′′

π
π

π′′π′′π"π

ππ " π
ππ " π
**
*

−4
−66
4
x
1
1
2
3
4
4
, )(12/ 138/ 13
8/ 13
xy
22
+ =16
2
3
xy=
θ
y
x
2
12
1
yx= 2
3
yx= 2
y
x
h
L
y= 2− −
h
2
x
L
x
2
L
2( (
y

362. 362 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning

41.








2232
00
234
00
2
32 4 2 2
0
2
00
4 2
2
32 2
2 1
,
6
1
,
4
11
23
64 12
1
,
2
14 2
12 2 2
16 3
12 3 3ab
x
R
ab
y
R
xy
ab
R
y
x
IyxydAkxydydxkba
I x x y dA kx dy dx kba
ka b
III kba kba b a
mxydAkxdydxkba
Ikba aa
x
mkba
kb aIbb
y
mkba -
-
-




** **
** **
** **


42.



2
24
23
00
2
24
22
00
0
2
24
00
16,384
,
315
512
,
105
16,384 512 17,920 512
315 105 315 9
128
,
15
512 105 4 2 7
128 15 7 7
16,384 3x
x
R
x
y
R
xy
x
R
y
x
I y x y dA ky dy dx k
IxxydA kxydydxk
kk
III k k
m x y dA kydydx k
I k
x
mk
Ik
y
m -
-
-








** **
** **
** **
15 128 8 42
128 15 21 21k


43.
22
,25
2, 2
xy
fxy x y
f xf y







22
22
25
2
00
5
2 32
2
0
0
2 32
0
1
14 4
414
1
14
3
1
101 1
3
101 101 1
6
xy
R
RSffdA
x ydA
rrdrd
rd
d
















**
**
**
*
*

44.

9:

2
22
2
,16
,:0 2,0
1, 2
124
xy
xy
fxy x y
R xy x y x
ffy
ff y








22 2
222
00
2
32
222
0
24 224 24
11
224 2ln2 24 24
212
1122
418 2ln4 18 1818 ln 2
21212
92 2 52
62 ln4 32 ln2 ln22 3
263
y
S y dx dy y y y dy
yy y y y










** *

45.
2
,9
0, 2
xy
fxy y
f fy






22
3
333 32 32
22 22
000
0
12 1
43 6
1
14 14 2 14 14 37 1
xy
R
y
y
y
ySffdA
ydxdy yx dy y ydy y




**
** * *

363. Review Exercises for Chapter 14 363
© 2010 Brooks/Cole, Cengage Learning
x
y
50
50
50
R
z
46. θΔ
2
,4, 2,0
xyfxy x f xfπ′ π′ π

02 22
22 2
20
14 14 14
Rxx
S x dA x dy dx x dy dx
′′
πφπ φ φ φ** * * **

These integrals are equal by symmetry.
θ Δ θΔ θΔ
22 2
222
00
2
32
222
0
171111
21221212
2 14 214 14
2 ln 1 4 2 1 4 1 4 2 ln 17 4 2 17 17 7.0717
x
S x dy dx x x x dx
xxx x x

πφπφ′φ

πφφφφ′φπ φφ′φ
** *

47. (a)
θΔ
θΔ
2222
50 50 50
22 22 22
00 0
50
43
32
22 2 22 3
0
50
20 20 50 50 50
100 5 200 5 10
25 1
10 50 50 arcsin 50 250 30,415.74 ft
50 4 800 15 30x xy x y x x x
Vdydxxxxdy
xx x
xx x x x′ φ′
πφ′π′φ′′′′



π′φ φ′φ′′φ


** *

(b)
θΔ θΔ
22 222
22
22
20
100
100
11
100 100 100
xy
xy
z
xyyx
ff
πφ
φφ
φφ πφφ π


22
50 50 2 50
22 2 22 2
00 0 0
11
100 100 2081.53 ft
100 100 x
Sxydydxrrdrd


′
πφφπφ** * *

48. (a) Graph of θΔ
θ Δ
2222
1000
2
,251 cos
1000
xy xy
fxy z e′φ φ
ππ φ


over region R

(b)
θΔ θΔ
22
Surface area 1 , ,
xy
R
fxy f xy dAπφ φ**

Using a symbolic computer program, you obtain surface
area 4540 sq. ft.
49.
θΔ
2
39 9 239
22 2
222 2
39 00
3
5
23 2 2
24 3
00 0 0
0
162 324
93
55 5
x
xxy r
x y dz dy dx r dz dr d
r
rrdrd r d d





′
′′′ φ
φπ

π′π′ππ

** * ***
** * *

50. θΔ
θ Δ
222 2
24 2 222 22 2
22 3 5
2
24 0 000 00 0
11632
233xxy r
x
x y dz dy dx r dz dr d r dr d d

′φ
′′′
φπ π ππ** * *** ** *

51. θΔ
θΔ
θΔ θΔ
222 3 2 2
000 00
33 2 3 3 3 222
0 1
3
11 1 1 1 1
33 3 3 3 3
abc ab
a
x y z dx dy dz c cy cz dy dz
bc b c bcz dz abc ab c a bc abc a b c
φφ π φ φ
πφφπφφπφφ*** **
*

52.
+,
θΔθΔ θΔ
2222
525 25 225
222 2
00 0 000
22 5
000
2 2
00
1
sin
11
arctan sin
5 arctan 5 cos 5 arctan 5
2xxy
dz dy dx d d d
xyz
dd
d


-
.-.
-
--..

.′′′
π
φφφ φ
π′

** * ***
**
*

−1−212
1
3
4
y = x
y = −x
(−2, 2) (2, 2)
y
x

364. 364 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
53.
222 2
11 1 211
22 3
222 2
11 1 00 1
8
15xxy r
xxy r
x y dz dy dx r dz dr d



** * * **

54.
222
24 4
00 0
4
3
xxy
xyzdzdydx

** *

55.

2
2 2 cos 4 2 2 cos
2
00 0 0 0
2cos 2
22 32
23 3
00
0 0
444
432 321322
41sincoscos
33 33323
r
V r dz dr d r r dr d
rd d











** * **
**

56.


2
22sin 16 2 2sin
2
00 0 0 0
22
24 24
00
2
3
0
2216
2 32sin 4sin 8 8sin sin
131129
8 4 2 sin 2 sin cos sin 2
4424 2
r
V r dz dr d r r dr d
dd













** * **
**


57.
22cos
2
40 0
2
22 2
334
40 4
4
22cos
3
40 0
22 2
55
40 4
4sin
4221
cos sin cos sin cos
333424
4cossin
11
cossin cossin
212
xy
mk ddd
k
kddkdk
Mk ddd
kddkdk
.





.



-.-.

... ... .
-..-.
... ...





***
** *
***
** *
2
6
4
cos
96
96 1
24 4
0by symmetry
xy
k
M k
z
mk
xy



.










1
, , 0, 0,
4
xyz






58.
2sin 2 2
23 3
000 00 0
2sin 2 2
232424
000 00 0
2sin
232
000 00 22
22sinsin
33
11
2 sin 2 sin sin
28
2sinacr a
acr a
xz
acr a
xy
m k r dz dr d kc r dr d kca d kca
M k r dzdrd kc r drd kca d kca
M k rz dz dr d kc r dr d







*** ** *
*** ** *
*** *
22
24 2 24
0
4
3
24
3 11
sin
416
0
83
2316
16 3
23 32
xz
xy
kc a d kc a
x
M kca a
y
m kca
M kc a ca
z
mkca






**



33
,, 0, ,
16 32
aca
xyz

365. Review Exercises for Chapter 14 365
© 2010 Brooks/Cole, Cengage Learning
x
ya
a
a
ah−
h
z
59.
θΔ
3
22
2
000
4
22
2
000
4
3
sin
6
cos sin
16
63
16 8
a
a
xy
xy ka
mk ddd
ka
Mk ddd
M ka a
xyz
mka

-.-.

-.- .-.


ππ
ππ

πππ π π

***
***


θΔ
333
,, , ,
888
aaa
xyz

π




60. θ Δ
θΔ
2
32 25 32
2
00 00
3
32
22
0
2
234 25 25
22
00 25 03 25
500 500
25 4
33
500 1 500 64 125 500 14
2 25 2 2 18 162
33 3 3333
0by symmetry r
r
xy
rr
m rdzddr r r rddr
rr
xy
M zr dz dr d zr dz dr d





′
′
′′ ′′
π′ π′ ′′

π′′′′π′′′φπ′π


ππ
πφ ** * **
*** ***
θΔ
23
2
00
2
3
23 2
342
00 0
0 0 1
825 0
2
1 9 1 9 81 81
22 84 8 4
81 1 1
4162 8
xy
rrdrd
rrdrd rrd
M
z
m








π′′ φ



π′π′π′π′


ππ′ π′
**
** *


θΔ
1
, , 0, 0,
8
xyz

π′



61.
θΔ
2
2416 24
335
030 03
833
4416
3r
z k
I k r dzdrd k r r drd
′
ππ′π*** **

62. θΔ
6
2
22 2
00 0
4
sin sin
9a
z ka
Ik ddd
-.--.-.
ππ** *

63. θΔ
22 2 22
2
,
02
zfxy ax y ar
rahh
ππ′′π′


(a) Disc Method

θΔ
θΔ
θΔ
+,
3
33
22 2 3 2
33 33
332 22 2 2
33 3
1
3
33 333
a
a
ah
ah
ahya
Vaydyay a aah
aa hh
a a ah ah ah ah h a h


′
′
′
π′π′π′′′′



π′′φφ′φ′π ′π ′


*

Equivalently, use spherical coordinates.

θΔ
θΔ
1
2cos
2
00 sec
sin
aha a
ah
Vddd

.
-.-.
′
′
′
π** *

(b)
θΔ
θΔ
θΔ
θΔ
θΔ
θΔ
θΔ
1
2cos 2
22
00 sec
2
2
2
2
1
cos sin 2
4
1
2
23
4
1 43
3
3aha a
xy
ah
xy
Mdddhah
hah
Mah
z
Vah
hah

.
-.- .-.


′
′
′
ππ′
′
′
ππ π
′
′** *

Centroid:
θΔ
θΔ
2
32
0, 0,
43
ah
ah
′

′

x
yax=
22
−
2a
a
a−a
ah−
y

366. 366 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
(c) If
θΔ
θΔ
2
3 3
,.
42 8
a
haz a
a
ππ π

Centroid of hemisphere:
3
0, 0,
8
a




(d)
θΔ
θΔ
θΔ
2 2
00
3432
lim lim
43 12
hh
aah
za
ah a

′
πππ
′

(e)
θΔ
θΔ
θΔ
θΔ
22 22
31
2cos
22 2 2 2
00 sec
sin
sin sin 20 15 3
30
aha a
z
ah
xy
h
Idddaahh

.
-.
-.- . - .
′
′
′
φπ
ππ′φ
** *

(f ) If
θΔ
3
222 5
4
,20153 .
30 15
z
a
haI a a a a

ππ ′φπ
64.
2
22
2
1
z
xy
a
φφ π

θΔ θΔ
22 222
11
22 22
22 222
11
8
15aza yza
z
Qazayza
I x ydV x ydxdydz a
′′ ′ ′′
′′′′ ′′′′
πφπ φ π** * * * *

65.
26sin
2
000
sinddd
.
-.-.***

Because
6sin
- .π represents (in the yz-plane) a circle of radius 3 centered at θΔ0, 3, 0 , the integral represents the volume of
the torus formed by revolving
θΔ02 this circle about the z-axis.
66.
2
21
000
r
rdzdrd


φ
***

Because
2
1zrπφ represents a paraboloid with vertex θΔ0, 0, 1 , this integral represents the volume of the solid below the
paraboloid and above the semi-circle
2
4yxπ′ in the xy-plane.
67.
θΔ
θΔ
θΔ θΔ
,
13 23 9
,
xy xy yx
uv u v u v
C CC CC
π′π′′π′
CCCCC

68.
θΔ
θΔ
θΔθ Δ θΔθΔ
,
22 22 8
,
xy xy yx
uv uv uv
uv u v u v
C CC CC
π′π′′ π′
CCCCC

69.
θΔ
θΔ
θΔ θΔ
, 11 11 1
,22222
11
,,
22
xy xy xy
uv u v v u
x uvy uv u xyv xy
C CC CC
π ′ π′′ π′

CCCCC



Boundaries in -plane Boundaries in -plane
33
55
11
11
xy uv
xy u
xy u
xy v
xy v
φπ π
φπ π
′π′ π′
′π π


θΔ
θΔ θΔ +,
θΔθΔ
51 51 5 5
331 31 3111 1
ln ln ln ln ln
222 2
5 ln 5 5 3 ln 3 5 ln 5 3 ln 3 2 2.751
R
xy dA u v u v dv du u dv du u du u u u
u
′′

φπ φφ′ π π π ′


π′′′π′′
** * * * * *

y = −x + 5y = x + 1
y = −x + 3 y = x − 1
1
1
2
3
23
y
x

367. Problem Solving for Chapter 14 367
© 2010 Brooks/Cole, Cengage Learning
70.
θΔ
θΔ
, 11
10
,
,,
xy xy xy
uv u v v u u u
v
xuy uxvxy
u
C CC CC
π′π′π

CCCCC



Boundary in -plane Boundary in -plane
11
55
11
55
xy uv
xu
xu
xy v
xy v
ππ
ππ
ππ
ππ


θΔ
,
55 55 5
222 22211 11 1
5
1 114
111 1
4 arctan 4 arctan 5
R
xu
dA du dv du dv dv
xy u v v uvu
v


πππ

φφφ φ
ππ′
** * * * * *

Problem Solving for Chapter 14
1.
θΔ
+, θΔ
2
41 4 32
22 2
2
00 0
4
0
16 1
16 1
16 1 cos 1 cos 1
3cos
16
sec cos tan 8 2 2 4.6863
3
R
VxdA
rrdrd d





π′

π′ π′ ′′

**
** *

2.
θΔ
θΔ θΔ
θΔ
22
22
22
22 222 222
22
1
Plane
,
11
1
xy
xy
RR
z d ax by
c
ab
ff
cc
ab
ff
cc
a b abc abc
SdAdAAR
cc c c
π′′
π′ π′
φφ πφφ
φφ φφ
πφφπ π
** **

3. (a)
θΔ
22
22
22 221
arctan . Let 2 , .
11
Then arctan .
2 22
du u
ca uuv
au a a
v
dv C
uv uu
πφπ′π
φ
πφ
′φ ′′
*
*

(b)
22
1
0 22
22 22
0 022 2 222
arctan
22
24
arctan arctan arctan
22 2 22
u
u
v
Idu
uu
uu u
du du
uu u uu
′

π
′′
′
π′π
′′ ′ ′′
*
**

Let
222
2 sin , 2 cos , 2 2 2 sin 2 cos .ududu ππ ′π′π
θΔ
6 2
22
6 6
1
0 0
0
12sin 4
4 arctan 2 cos 4 arctan tan 2
26182cos 2cos
Idd





π"ππππ

**

y =
1
x
y =
1
x
x = 5
x = 1
1
1
2
3
4
5
45
y
x
x
y
1
1
1
z
y = x
R

368. 368 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
(c)
2
2
2
22 22
2
2 2
22 2222 2 22
2
arctan
22
22242
arctan arctan arctan
22 2 22
u
u
v
Idu
uu
uu u
du du
uu u uu
′φ
′

π
′′
′φ ′ ′
π′π

′′ ′ ′′
*
**

Let
2sin .uπ

2 2
2
6 61 2 2 sin 1 sin
4 arctan 2 cos 4 arctan
cos2cos 2cos
Idd


′′
π"π


**

(d)
θΔθΔ
θΔθΔ
θΔ
θΔθΔ
θΔ
22
2
1cos 2 1sin 1sin1 1 sin 1 sin
tan
2 2 1 cos 2 1 sin 1 sin 1 sin cos cos

′′ ′′
′π π π π π

φ′φ φ′

(e)
22 22
2
66 66
2
22222 2
22
6 1sin 1 1
4 arctan 4 arctan tan 4 2
cos 2 2 2 2 2
18 9 6 1 4
22 2
2 2 4 8 12 72 72 36
Id ddd






′
ππ′π′π′

′′φ
π ′ π ′′′ π π π

** **
9

(f) θΔ θΔ
21
11
1
xy xy xy
xy

′


θΔ θΔ θΔ
θΔ θΔ
1
1
11 11 11 1 2
00 00 00 0
00 0
1
1
1
22 2
0
0001
0
1
1
1 1
11
1 11
KK
K
KK
KK
KKKn
xy
dx dy xy xy dx dy xy dx dy dy
xy K
yy
dy
Kn KK
!! φ
ππ
!!!! φ
ππππ
πφφφ π π
′ φ
ππππ
φ φφ
FF** ** ** *
FFFF*

(g)
θΔ
θΔ
,
22
2
22
2
22
, 12 12
1
, 12 12
xy yx
uv
x uv
uv x
yuv
uv y
xy
uv
φ′
ππ
′

φ

C ′
ππ
C


θΔ θΔ
θΔ
θΔ
θΔ θΔ
RS
0, 0 0, 0
11
1, 0 ,
22
11
0, 1 ,
22
1, 1 2 , 0
L

L′



L


L


22 2
11 22 2 2
1222 22
00 0 22 2
11 1
1 18 9 6
11
22 22 uu
uu
dx dy dv du dv du I I
uv uvxy
′φ
′′
πφ πφπφπ
′
′φ ′φ** * * * *

R
y
x
−1
1
1234
2
−2
2
1
2
2, 0
1
,
()
2
1
2
1
,()
−
()
S
v
u

369. Problem Solving for Chapter 14 369
© 2010 Brooks/Cole, Cengage Learning
4.
2
25
3
04
2
210
3
09
1333
:4.36ft
16 160 960
523
1.71 ft
16 160 960
rr
Ardrd
rr
Brdrd









**
**

The distribution is not uniform. Less water in region of
greater area.
In one hour, the entire lawn receives

2
210
3
00
125
32.72 ft .
16 160 12
rr
rdrd



**

5.
2
2
Boundary in -plane Boundary in -plane
1
22
1
3
3
1
4
4
xy uv
yx u
yx u
yx v
yx v
ππ
ππ
ππ
ππ

θΔ
θΔ
θΔ
θΔ
23 13
13 23
12
33, 1
,3
21
33
, 1
11
,3
RS
vu
uvxy
uv
vu
uv
xy
AdA dA
uv


C
ππ′
C



C
ππ π
C
** **

6. (a)
θΔ
2
22 8
002
8
42 5
3r
V r dz dr d

′
ππ′***

(b)
θΔ
2422
00 2sec
sin
8
42 5
3
Vddd

.
-.-.

)
π
π′***

7.
32 6
00
18
xx
x
Vdydzdx
′
ππ** *

8. θΔ θΔ
1
1
11 1
00 0
0
1
0
1
1
22
0
1
1
1
1
11
n
nn n
n
n
x
x y dx dy y dy
n
ydy
n
y
nn
φ
φ

π

φ
π
φ




** *
*


θΔ
11
2
00 1
lim lim 0
1
nn
nn
x y dx dy
n
! !
ππ
φ**

9. From Exercise 69, Section 14.3,

2
2
2.
x
edx
!
′
′!
π*

So,
22
2
00 2
and
22
xx
edx edx
!!
′′
ππ**


222
2
00
0 11
22
1
22 4
xxx
xedx xe edx

!
!!
′′′

π′ φ


π" π
**

10. Let
1
ln , .dx
vdv x x

ππ′




θΔ θΔ
10
00
1
,,
ln 1
vvv
vv
exedxedv
x
xdx v e dv ve dv
′′
!
′′
!
ππ π′
π′π
** *

Let
2
,,2 .uvuvududvππ π
θΔ θΔ
1 2
00
2
2
0
ln 1 2
22
42
u
u
xdx ue udu
ue du

!
′
!
′
π

πππ

**
*

(See Problem Solving #9.)
11. θΔ
θΔ
0, 0
,
0elsewhere
xya
ke x y
fxy
′φ
6 //=
π7
=8


θΔ
θΔ
00
00
,
xya
xa y a
fxydA ke dxdy
ke dx e dy
!! !!
′φ
′! ′!
!!
′′
π
π"** **
**

These two integrals are equal to

θΔ
00
lim .
b
xa xa
b
edx ae a
!
′′
!
π′ π

*

So, assuming
,0,ak you obtain

2 1
1or .ka a
k
ππ

12. By the shell method,

22
0
0
lim 2 lim .
b
b
xx
bb
Vxedxe
′′
! !
ππ′π

*

This same volume is given by

θ Δ
θ Δ
22
22
00
22
00
2
2
0
4
4
4.
xy
xy
xy
x
edydx
edydx
edxedy
edx
!! ′φ
′! ′!
!! ′φ
!!
′′
!
′
π
π
π

π
**
**
**
*

So,
2
0
.
2
x
edx
!
′
π*

x
y
(0, 0, 0)
(3, 3, 6)
(3, 3, 0)
(0, 6, 0)
2
4
5
6
6
3
z
b
y = e
−x
2
y
x

370. 370 Chapter 14 Multiple Integration
© 2010 Brooks/Cole, Cengage Learning
13. Essay
14. sec
cos
x
Alw y xy

5
π" π 5π 55



Area in
xy-plane:
xy55
15. The greater the angle between the given plane and the xy-plane, the greater the surface area. So:
214 3zzzz
16. Converting to polar coordinates,

θΔ
θΔ
θΔ
θΔ θΔ
2
22
00 0 0 22 2
2
0 2
2
2
0
2
011
11
2
1
lim 1 2
4
1
lim
41 4
t
t
t
t
dx dy r d dr
xy r
r
dr
r
rrdr
r





!! !
!
′
!
!
π
φφ φ

π

φ
πφ
′
π" π

φ** **
*
*

17.
θΔ
θΔ
θΔ
11
3
00
11
3
00 1
2
1
2xy
dx dy
xy
xy
dy dx
xy
′
π′
φ
′
π
φ
**
**

The results are not the same. Fubini's Theorem is not valid because
f is not continuous on the region 01,01.xy
18. The volume of this spherical block can be determined as follows. One side is length .
-5Another side is .-.5Finally, the
third side is given by the length of an arc of angle
5in a circle of radiussin .
-.Thus:

θΔθ Δθ Δ
2
sin
sin
V
--. - .
-.-.55 55
π555


θ
P
Δx
Δx
θcos
Δy
Δy
x
y
θφρ
i i i
sinΔ
ρΔ
i
φρ
ii
Δ
z

371. © 2010 Brooks/Cole, Cengage Learning
CHAPTER 15
Vector Analysis
Section 15.1 Vector Fields.......................................................................................372
Section 15.2 Line Integrals ......................................................................................383
Section 15.3 Conservative Vector Fields and Independence of Path ....................395
Section 15.4 Green’s Theorem................................................................................402
Section 15.5 Parametric Surfaces............................................................................410
Section 15.6 Surface Integrals .................................................................................418
Section 15.7 Divergence Theorem..........................................................................427
Section 15.8 Stokes’s Theorem ...............................................................................433
Review Exercises........................................................................................................440
Problem Solving.........................................................................................................449

372. 372
© 2010 Brooks/Cole, Cengage Learning
CHAPTER 15
Vector Analysis
Section 15.1 Vector Fields
1. All vectors are parallel to x-axis.
Matches (d)
2. All vectors are parallel to y-axis.
Matches (c)
3. Vectors are in rotational pattern.
Matches (e)
4. All vectors point outward.
Matches (b)
5. Vectors are parallel to x-axis for
.yn
Matches (a)
6. Vectors along x-axis have no x-component.
Matches (f )
7.

,
2
xy

Fij
F

8.

,2
2
xy

Fi
F
9.

22
,xyyx
yx


Fij
F

10.

22
,2
4
xyy x
yx


Fij
F

11.

,, 3
3
xyz y
yc


Fj
F
12.

,xyx
xc


Fi
F

13.

22
22
22
,4
16
1
16
xy x y
xyc
xy
cc



Fij
F

14.


22
2
22
,
1
xy x y
xy


Fij
F
15.

,,
3
xyz

Fijk
F

16.

222
2222
,,
xyz x y z
xyz c
xyzc



Fijk
F

1
−4
−4
x
y
x
−2
−2
−1
1
21
y
5
−5
5
−5
x
y
x
−2
−2
2
2
y
y
x
4
4
2
z
x
−2−3 −1
1
2
2
3
3
1
y
x
−2
−2
−112
2
y
−22
4
6
8
10
y
x
x
y
4
4
4
−4
−4
z
x y
2
2
2
−2
−2
−2
z

373. Section 15.1 Vector Fields 373
© 2010 Brooks/Cole, Cengage Learning
17.
21
,2
8
ijFxy xy y


18.
,2323
xyyx yxFij

19.

222
,,
ijkxyz
Fxyz
xyz





20.
,,
xyz x y zFijk

21.




22
,2
,2
,4
,24
x
y
fxy x y
fxy x
fxy y
xyxy



Fij

Note that
.fDF
22.




221
4
1
2
1
2
,
,2
,
,2
x
y
fxy x y
fxy x
fxy y
xyxy



Fij

23.




22
,53
,103
,32
, 103 32
x
y
gx y x xy y
gxy x y
gxy x y
xyxyxy



Gij

24.



,sin3cos4
,3cos3cos4
, 4 sin 3 sin 4
,3cos3cos44sin3sin4
x
y
gx y x y
gxy x y
gxy x y
xyxyxy



ijG

25.




,, 6
,, 6
,, 6
,, 6
,, 6 6 6
x
y
z
fxyz xyz
fxyz yz
fxyz xz
fxyz xy
xy z yz xz xy




Fijk


26.


222
222
222
222
222 222 222
,, 4
4
4
4
4
4
,,
444
Fijk
x
y
z
fxyz x y z
x
f
xyz
y
f
xyz
z
f
xyz
xyz
xyz
x yz x yz x yz










x
−2−112
2
1
−1
−2
y
−6
−6
2
4
6
−4−2246
y
x
y
x
2
2
2
1
1
1
z
y
x
2
2
2
1
1
1
z

374. 374 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
27.




2
2
2
22
,,
,, 2
,,
,, 1
,, 2Gijk
x
x
x
x
y
z
xx
gxyz z ye
g x y z xye
gxyz e
gxyz
x y z xye e






28.





2
2
2
222
,,
,,
1
,,
1
,,
11
,,
x
y
z
yzxz
gx y z
zx y
zz
gxyz
xy
xz
gxyz
zy
yx
gxyz
zxy
zz xz y x
xyz
x yzy zxy







Gijk

29.





,, ln
,, ln
,, ln
,, 0
,, ln ln
x
y
z
hx y z xy x y
xy
hxyz y x y
xy
xy
hxyz xxy
xy
hxyz
xy xy
xyz y x y x x y
xy xy










Hij

30.





22
22
22 22
,, arcsin
,, arcsin
,,
1
,,
1
,, arcsin
11
x
y
z
hx y z x yz
hxyz yz
xz
hxyz
yz
xy
hxyz
yz
xz xy
xyz yz
yz yz








Hijk

31.

22
,
xyxy xyFi j

22
andMxy Nxy have continuous first partial
derivatives.

2 conservative
NM
xy
xy
CC

CC F
32.

22
11
,
y
xy y x
x xx
Fijij


2
Myx and 1N x have continuous first partial
derivatives for all 0.x#

2
1
is conservative.
NM
xx y
CC

CC
F
33. ,sin cos
xyyxyFij

sinMyand cos
Nxy have continuous first
partial derivatives.
cos is conservative.
NM
y
xy
CC

CC
F
34.


111
,xy y x
xyxy
Fijij
1Mxand 1N yhave continuous first partial
derivatives for all
,0.xy#

0 is conservative.
NM
xy
CC

CC
F

375. Section 15.1 Vector Fields 375
© 2010 Brooks/Cole, Cengage Learning
35.
2
,5 3xyyyxFi j

32
5, 15MyN xy

2
15 Conservative
NM
y
xy
CC

CC

36.
22
222
,
xyxy x
MeN e
yy




2
3
22
Conservative
xy
yxNM
e
xy y
CC

CC

37.

22 22
32 32
22 22
11
,
Not conservative
MN
xy xy
NxMy
xy
xy xy


CC
#
CC



38.

32
,
11
2
Conservative
21
yx
MN
xy xy
Nxy M
xy xy


CC

CC

39.
+, +,

,
1 Conservative
,,, ,
xy
xy y x
yx
yx
fxy yf xy x fxy xy k

CC

CC

Fij
40.
22 3
22 2
32
,3 2
36
26
xyxy xy
xy xy
y
xy xy
x

C


C
C


CFij
Conservative




22
3
32
,3
,2
,
x
yfxy xy
fxy xy
fxy xy K




41.
+,
2
2
,2
22
2
xyxyx
xy x
y
xx
x

C

C
C


CFij
Conservative


22
,2,, ,,
xy
fxy xyf xy x fxy xy K
42.
2
22 2
3
22 2
23
,2
222
22
xy
xyxy xy
xyxy xy
xy xe y x
xye xe x ye
y
xe xe xye
x

C


C
C


C
Fij

Conservative




2
2
2
2
,2
,
,
xy
x
xy
y
xy
f x y xye
fxy xe
fxy e K




43.
32
32 2 2
,155
15 45 5 5
xy y xy
yy xyy
yx

CC
#

CC
Fij
Not conservative
44.

2
2
1
,2
12
xyyx
y
x
yy

Fij
ij

2
22
11
22
yy y
x
xyy
C


C
C


C

Not conservative
45.
2
2
2
22
2
,
22
2
yx
xy
x y
y
yx x
xx
xy y

C


C
C


C
Fij
Not conservative
46.



22 22
222
22
222
22
,
2
2
xy
xy
xyxy
xxy
yx y
xy
yxy
xx y
xy


C


C
C


C
Fij
Conservative




22
22
22
,
,
1
,ln
2
x
y
x
fxy
xy
y
fxy
xy
fxy x y K

376. 376 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
47. ,cossin
cos sin
sin sin
x
xx
xx
xye y y
ey ey
y
ey ey
x

C


C
C


CFij
Conservative



,cos
,sin
,cos
x
x
x
y
x
fxy e y
fxy e y
fxy e y K




48.



22
22 22
23
22 22
23
22 22
22
,
28
28
xy
xy
xy xy
xxy
y
xy xy
yxy
x
xy xy



C


C



C


C


Fij
Conservative






2
22
2
22
22
2
,
2
,
1
,
x
y
x
fxy
xy
y
fxy
xy
fxy K
xy







49. ,, ,Fijk
xy z xyz xyz xyz 2, 1, 3


xyz
xyz xyz xyz
xzxy yzxy yzxz
CCC

CCC

ijk
curl F
ijk


2, 1, 3 6 2 3 2 3 6
43

curl F i j k
ij k

50.
2
,, 2 ,2,1,3Fijkxyz xz xz yz



2
2
2
2
20 20
22
xyz
x z xz yz
zx x z
zxx z
CCC

CCC



ijk
curl F
ij k
ijk


2,1,3746curl F i j k
51.
,, sin cos ,0,0,1
xx
xyz e y e yFij

cos cos 2 cos
sin cos 0
ijk
curl F k k
xx x
xx
eyey ey
xyz
eye y
CCC

CCC


0, 0, 1 2curl F k
52. ,, ,3,2,0
xyz
xyz e

Fijk


ijk
curl F i j k
xyz xyz xyz
xyz xyz xyz
xz xy e yz xy e yz xz e
xyz
eee

CCC

CCC


3, 2, 0 6 6curl F i j
53.
22
, , arctan ln
x
xyz x y
y



Fijk




2
222 22
22
2
1
1
arctan ln 1
2
ijk
curl F k k
xyxx
x y z xy xy xy
x
xy
y
CCC

CCC

377. Section 15.1 Vector Fields 377
© 2010 Brooks/Cole, Cengage Learning
54. ,,
yz xz xy
xyz
yz xz xy


Fijk



22 22 22
22 22 22
222
22 2 2 22
11 11 11
ijk
curl F i j k
ijk
xx yy zz
xyz xy xz xy yz xz yz
yz xz xy
yzxzxy
xyz
xy xz xy yz yz xz

CCC

CCC







55.
, , sin sin sinxyz x y y z z xFijk


cos cos cos
sin sin sin
yz zx xy
xyz
xy yz zx
CCC

CCC

ijk
curl F i j k
56.


222
,,xyz x y zFijk


222
222 222 222
yz zx xy
xyz xyz
xyz xyz xyz
CCC

CCC

ijk
ijk
curl F

57.

22 2 2 2 2
,,
xy z xy z x yz x y zFijk

22 2 2 2 2
xyz
xy z x yz x y z
CCC

CCC
ijk
curl F 0

Conservative





22
22
22
222
,,
,,
,,
1
,,
2
x
y
zfxyz xyz
fxyz xyz
fxyz xyz
fxyz xyz K





58.
23 3 22
,, 2 3
xy z yz xyz xyz Fijk

23 3 22
23
xy z
y z xyz xy z
CC C

CC C
ij k
curl F 0

Conservative


23
,,
fxyz xyz K
59. ,, sin sin sinxyz z x yFijk

sin sin sin
cos cos cos
ijk
curl F
ijk0
xyz
zxy
yzx
CCC

CCC
#

Not conservative

60. ,,
z xy
xy z ye ze xeFijk

ijk
curl F i j k 0
yx y z xz
zx y
xe e e ye ze e
xyz
ye ze xe
CCC
#
CCC

Not conservative

378. 378 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
61.
2
,,
zxzx
xyz
yy y
Fijk

22 22
2
11
ijk
curl F i j k 0
xx zz
xyz y y yy y y
zxzx
yyy
CCC

CCC


Conservative





2
,,
,,
,,
,,
x
y
z
z
fxyz
y
xz
fxyz
y
x
fxyz
y
xz
fxyz K
y





62.
22 22
2222
,,
1
xy
xyz
xy xy
xyz
xy
xyxy


CCC

CCC
Fijk
ijk
curl F 0
Conservative










22
22
22
22
1
22
22
2
3
22
,,
,,
,, 1
,,
1
ln ,
2
,,
1
ln ,
2
,, ,
1
,, ln
2
x
y
z
x
fxyz
xy
y
fxyz
xy
fxyz
x
fxyz dx
xy
xygyzK
y
fxyz dy
xy
xyhxzK
fxyz dz z pxy K
fxyz x y z K













*
*
*

63.


22
22
,2
div , 2 2 4
Fij
F
xy x yxyx yxy
xy

CC

CC

64.


,
div ,
Fii
F
xy
xyx xyy
xy xe ye
xy xe ye
xy
xeeyee

CC

CC


65.
+, +,
2
2
,, sin cos
div , , sin cos
cos sin 2
Fijk
F
xyz x y zxyz x y z
xy z
xyz

CC C


CC C


66.


+,
22 22
22 22
22 22
,, ln ln
22
div , , ln ln
xyz x y xy y z
x z
xyz x y xy y z x
x yz xy yz

CCC


CCC
Fijk
F

379. Section 15.1 Vector Fields 379
© 2010 Brooks/Cole, Cengage Learning
67.

,,
div , , 1
div 2, 1, 1 1 2 1 4xyz xyz xy z
xyz yz x



Fijk
F
F

68.



2
,, 2
div , , 2
div 2, 1, 3 11
xy z x z xz yz
xyz xz y



Fijk
F
F

69.



2
,, sin cos
div , , sin sin 2
div 3, 0, 0 0
xx
xx
xyz e y e y z
xyzeyeyz



Fijk
F
F

70.



,, ln
111
div , ,
11 11
div 3, 2, 1 1
32 6
xyz xyz
xyz
xyz



Fijk
F
F
71. See the definition, page 1058. Examples include velocity
fields, gravitational fields, and magnetic fields.
72. See the definition of Conservative Vector Field on page
1061. To test for a conservative vector field, see
Theorem 15.1 and 15.2.
73. See the definition on page 1064.
74. See the definition on page 1066.
75.

,, 3 2
,,
xyz x y
xyz x y z


Fijk
Gijk



22
13 2
32 2 3
xy
xyz
xz y z xy y x
'


ijk
FG
ij k




22
32 2 3
11 6 3 2 4
92
ijk
curl F G
ijk
jk
xyz
xz y z xy y x
xxyy
xy
CCC
'
CCC




76.


22
,,
,,
xyz x z
xyz x y z


Fik
Gijk


22
22
0
x zyz xz xz xy
xyz
'
ijk
FG i j k



22
22
222121
ijk
curl F G i j k i k xxzx yy z xzz xxz zzx
xyz
yz xz x z xy
CCC
'
CCC


77.
,,
xy z xyz y zFijk
xyxz
xyz
xyz y z
CCC

CCC
ijk
curl F j k


0
zy
xy z
xy xz
CC C

CC C

ijk
curl curl F j k

380. 380 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
78.
2
,, 2xy z x z xz yzFi j k

2
2
22
2
zxx z
xyz
xz xz yz
CCC

CCC

ijk
curl F i j k


2
2
22
x
xyz
zxx z
CCC

CCC

ijk
curl curl F j k
79.

,, 3 2
,,
xyz x y
xyz x y z


Fijk
Gijk



22
13 2
32 2 3
xy
xyz
xz y z xy y x
'


ijk
FG
ij k


div 3 2 zx'FG
80.


22
,,
,,
xyz x z
xyz x y z


Fik
Gijk


22
22
0
x zyz xz xz xy
xyz
'
ijk
FG i j k

div 0'FG
81.


,,
div 0
xyz xyz y z
xyxz
xyz
xyz y z
xx

CCC

CCC

Fijk
ijk
curl F j k
curl F

82.



2
2
2
,, 2
22
2
div 2 2 0
xyz xz xz yz
zxx z
xyz
xz xz yz

CCC

CCC


Fijk
ijk
curl F i j k
curl F

83. Let
MNPFijk and QRSGijk where ,,,,,MNPQR and S have continuous partial derivatives.

MQ NR PSFG i j k



xyz
MQNRPS
PS NR PS MQ NR MQ
yz xz x y
PN PM NM SR SQ RQ
yz x z x y yz xz xy
CCC

CCC

CC CC C C


CC CC C C
CC CC C C CC CC CC

CC C C C C CC CC CC

ijk
curl F G
ij k
ijkijk
curl Fcurl G

84. Let
,,fxyzbe a scalar function whose second partial derivatives are continuous.


22 22 22
fff
f
xyz
ff ff ff
f
xyz yz zy xz zx xy yx
fff
xyz
CCC
D
CCC
CCC C C C C C C
D
CCC CC CC CC CC CC CC
CCC
CCC
ijk
ijk
curl i j k 0

381. Section 15. 1 Vector Fields 381
© 2010 Brooks/Cole, Cengage Learning
85. Let MNPFijk and Gijk.RST


div
div div FG
FG
MRNSPT
MR NS PT
xyz xxyyzz
MNP RST
x yz xyz
CCCCCCCCC

CCC CCCCCC
CCC CCC


CCC CCC


86. Let MNPFijk and Gi
jk.RST
MNP NTPS MTPR MSNR
RST
'
ijk
FG i j k



div NT PS PR MT MS NR
xy z
TNSPRP TM SM RN
NT PSPRMT MS NR
xxxxyy yy zz zz
PN M P NM TS RT S
RS TMNP
yz z x x y yz zx
CC C
'
CC C
CC CCCC CC CC CC

CCCCCC CC CC CC
CC C C C C CC CC C

CC CC C C CC CC
FG

R
x y
C

CC
""curl F G F curl G

87. MNPFijk




Exercise 83
Exercise 84
ff
f
'' '

'
'
' '
Fcurl F
curl curl F
curl F
F
DD D D D
D D
D
DD

88. Let MNPFijk.


+,
ijk
F
ij k
ijk
ijk F
Df
xyz
fM fN fP
fPf NfPf Mf Nf M
Pf Nf Pf M f Nf Mf
yyz zxxz zx xy y
PN PM NM fff
ff
yz xz x y xyz
MNP
CCC
'
CCC
CCCCCCCCCCCC

CCC CCCC CC CC C
CC CC C C CCC
D'D
CC CC C C CCC
Ff'

89. Let , MNPFijk then .Fijk
f fM fN fP


div
div
F
FF
MfNfPf
ffMfNfPfMfNfP
xyz xxyyzz
MNN f f f
fMNPff
xyz x y z
C C C C CC CCC

CCC CCCCCC
CCC C C C
D"
CCC C C C

382. 382 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
90. Let MNPFijk.

22 22 22
div
0 because the mixed partials
PN PM NM
yz xz x y
PN PM NM
xy z yx z zx y
PNPMNM
xy xz yx yz zx zy
CC CC C C

CC CC C C
CC C CC C CC C


CC C CC C CC C
CC CC CC

CC CC CC CC CC CC
curl F i j k
curl F
are equal

In Exercises 91-93, F,, =i+j+kx
yzx yzand .
222
,, =F ,, = + +fxyzx yzx yz
91.



222
222 222 222 222 21
ln ln
2
ln
fxyz
xyzxyz
f
xyz xyz xyz xyz f


D

ijk F
ijk

92.



222
32 32 32 33
222 222 222
222
11
1 ijk F
ijk
f xyz
xyzxyz
f f
xyz xyz xyz
xyz



D



93.
222
n
n
f xyz


11 1
222 222 222
222 222 222
2
222 2
ijk
ijk F
nn n
n
n
n
xyz
f nxyz nxyz nxyz
xyz xyz xyz
nx y z x y z nf



D


94. (a)

22
,
xy
xy
xy



ij
F

(b)

22
,
xy
xy
xy



ij
G

(c) All the vectors are unit vectors. Those of F point away from origin. Answers will vary.
95. True.


24
16 0 as , 0, 0 .xy x y xy F
96. True. If

,xyis on the positive y-axis, then
0xand 0.ySo,


2
,0,
xyyyFF j.
97. False. Curl is defined on vector fields, not scalar fields.
98. False. See Example 7.
−2−3−4 234
−2
−3
−4
1
2
3
4
x
y
−2−3−4 234
−2
−3
−4
1
2
3
4
x
y

383. Section 15.2 Line Integrals 383
© 2010 Brooks/Cole, Cengage Learning
Section 15.2 Line Integrals
1.

,01
22,12
tt t
t
ttt
6
=
7
=8
ij
r
ij

2.


2
,02
44,24
8, 4 8
tt t
tt t
tt
6
=
7
=

8
ij
rij
j

3.




,03
33,36
93,69
12 , 9 12
tt
tt
t
tt
tt
6
=

=
7

=
=

8
i
ij
r
ij
j

4.



4
5
,0 5
59 ,5 9
14 , 9 14
tt t
ttt
tt
6
=
=
7
=

=8
ij
ri j
i

5.

22
22
22
2
2
2
2
9
1
99
cos sin 1
cos
9
sin
9
3cos
3sin
3cos 3sin
02
rij
xy
xy
tt
x
t
y
t
xt
yt
ttt
t











6.

22
22
2
2
2
2
1
16 9
cos sin 1
cos
16
sin
9
4cos
3sin
4cos 3sin
02
xy
tt
x
t
y
t
xt
yt
ttt
t









rij

7.

43,0 1
43
ttt t
t


rij
rij


1
22
0
1 1
23
00
43 4 3
60 20 20
C
xy ds t t dt
tdt t


**
*

8.
2,0 2tt t t
t


ri j
rij



2 2
2
0
2
0
2
2
0
33211
32 2 2
32 2 0
C
xyds t t dt
tdt
tt



**
*

9.

sin cos 2 , 0
2
cos sin
rijk
rij
ttt t
ttt






22
222 2 2 2 2
00 5
sin cos 4 cos sin 5
2
C
xyzds t t t tdt dt

** *

10.
12 5 84 , 0 1
12 5 84
tttt t
t


rijk
rijk



1
4
11 22
23
00
0
2 2 12 5 84 12 5 84 10,080 85 856,800 214,200
4
C
t
xyz ds t t t dt t dt




** *

11. (a) ,0 1ttt t rij
(b)

,2rijrtt


1
22 22
0
1
3
0
2
22
22
33
C
xyds t t dt
t




**

12. (a) 2, 0 2tt t t rij
(b)

2, 5tt rijr


2
22 2 2
0
2
3
0
45
5405
5
33
C
xyds t t dt
t




**

384. 384 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
13. (a) cos sin , 0
2
tttt

rij
(b)


22 22
22 2 2
00
cos sin sin cos
2
C
x y ds t t t t dt dt



** *

14. (a)
2cos 2sin , 0
2
tttt

rij
(b)


2 222
22 2 2
0 0
4cos 4sin 2sin 2cos 8 4
C
x y ds t t t t dt dt



** *

15. (a) ,0 1tt tri
(b)


1
2
1
0
0
,1
1
4
22rir
C
tt
t
xydstdt





**

16. (a) ,1 9tt trj
(b)

,1ttrjr

99
32
1 1
88 208
33 3
44 271
C
xyds tdtt
**

17. (a)

,01
21,12
3, 2 3
tt
ttt t
tt
6
=
7
=

8i
rij
j

(b)





1
01
2
2
2 32
12
1
3
3 32
23
2 1
4
2
8192
424111221
23 6
88
443 3
33
19 1 2
1192 8 19192
4
263 6 6
C
C
C
C
xydstdt
t
x y ds t t dt t t
xyds tdt t
xyds











**
**
**
*

18. (a)



,02
22,24
62,46
8, 6 8
tt
tt
t
tt
tt
6
=

=
7

=
=

8i
ij
r
ij
j

(b)





2
01
4
22
6
43
8
64
42
16 2
42424
3
4642282
16 2
448
3
16 2 16 2 56
424 282 82
333
C
C
C
C
C
xydstdt
xyds tdt
xyds t dt
xyds tdt
xyds




**
**
**
**
*

x
C
1
C
2
C
3
(1, 0)
(0, 1)
y
C
1
C
2
C
3
C
4
1
1
2
2
y
x
(2, 2)

385. Section 15.2 Line Integrals 385
© 2010 Brooks/Cole, Cengage Learning
19. (a)

1
1
1
22
01
0: 0,0,0 to 1,0,0 : ,0 1, , 1
221 ri rir
C
Cttttt
xy zds tdt t




**




2
1
2
1
2
02
0: 1,0,0 to 1,0,1 : ,0 1, , 1
3
222
22 rik r kr
C
Cttttt
t
xy zds tdt t





**




3
1
3
1
22
03
0: 1,0,1 to 1,1,1 : ,0 1, , 1
4
221
33 rijk r jr
C
Cttttt
t
xy zds t dt t





**

(b) Combining,

2 34 23
21.
23 6
C
xy zds *

20. (a)

1
1
3
1
22
01
0: 0,0,0 to 0,1,0 : ,0 1, , 1
1
2
33 rj rjr
C
Cttttt
t
xy zds tdt





**




2
1
2
1
2
02
0: 0,1,0 to 0,1,1: ,0 1, , 1
1
21
22 rjk r kr
C
Cttttt
t
xy zds tdt t





**




3
1
32
11 2
22
003
0: 0,1,1 to 0,0,0 : 1 1 ,0 1, , 2
2
211222
32 6 rjkrjkr
C
Ctttttt
tt
xy zds t t dt t t dt






** *

(b) Combining,

2 11 2 5 2
2.
32 6 6
C
xy zds

*

21.

2221
,,
2
xyz x y z-




22
2cos 2sin ,0 4
2sin 2cos
4sin 4cos 1 5rijk
rijk
rttttt
ttt
ttt








4
222
0
4
3
4
2
0
0
3
2
Mass , ,
1
4cos 4sin 5
2
55
44
223
56485
16 4 3 795.7
233
C
xyzds
tttdt
t
tdt t



-










*
*
*

22. ,,
xyz z-




22
2cos 2sin ,0 4
2sin 2cos
4sin 4cos 1 5
ttttt
ttt
ttt

rijk
rijk
r


4
2
4
2
0
0
Mass , ,
5585
2
C
xyzds
t
tdt


-




*
*

23.

cos sin , 0
sin cos , 1
tttt
tttt
rij
rijr



+,
0
0
Mass ,
cos sin
sin cos
11 2
CC
xyds x yds
ttdt
tt


-


**
*

(0, 0, 0)
(1, 0, 1)
(1, 0, 0)
(1, 1, 1)
C
1
1
x
y
z
x
y
z
C
1
1
(0, 0, 0)
(0, 1, 0)
(0, 1, 1)

386. 386 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
24.

2
2
2, 0 1
22, 4 4
tt t t
tt t t

rij
rijr





1
2
0
1 12
2
0
1
32
2
0
3
4
3
4
Mass ,
24 4
31
1221
CC
xyds yds
tt dt
tt dt
t-




**
*
*

25.

2
2
2,1 3
22 , 4 5
rijk
rijkr
tt tt t
tt t t





3
2
1
3
32
2
1
Mass , ,
45
45
12
41 41 27
12
CC
xyzds kzds
kt t dt
kt
k-








**
*

26.


2cos 2sin 3 , 0 2
2sin 2cos 3
49 13
ttttt
ttt
t

rijk
rijk
r




2
0
2
2
0
2
Mass , ,
313
3
13
2
13 2 6
CC
xyzds k zds
kt dt
t
kt
k


-





**
*

27.




1 1
2
00
,
:,01
1Fij
rij
Fij
rij
Fr
C
xy x y
Ct t t t
ttt
t
dttdtt




"

**

28.



,
:4cos4sin,0
2
16 sin cos 4 sin
4sin 4cos
Fij
rij
Fij
rij
xy xy y
Ct t t t
tttt
ttt








2
2
0
2
32
0
64 sin cos 16 sin cos
64 40
sin 8 sin
33
Fr
C
dttttdt
tt


"



**

29.




,34
:cossin,0 2
3cos 4sin
sin cos
xy x y
Ct t t t
ttt
ttt





Fij
rij
Fij
rij



2
0
2
2
0
3 cos sin 4 sin cos
sin
12
2
C
dttttdt
t


"


**
Fr
30.




2
2
2
,34
:4,22
344
4
xy x y
Ct t t t
tt t
t
t
t




Fij
ri j
Fi j
ri j


2
2
2
2
2
34 0
2
C
t
dttdt



"

**
Fr
31.




2
323
,,
:2,01
22
22
xyz xy xz yz
Ct t t t t
tt t t
tt



Fijk
rijk
Fijk
rijk


1
4
1
333
0
0
99
44
44
Fr
C
t
dtttdt

"


**

32.




222
2
224
,,
1
:2sin2cos ,0
2
1
4sin 4cos
4
2cos 2sin
xyz x y z
Ct t t t t
tttt
tttt





Fijk
rijk
Fijk
rijk

225
0
6
33
0
66 1
8sin cos 8cos sin
4
88
sin cos
33 24
8816
3243 24 3
Fr
C
dtttttdt
t
tt




"







**

33.



22
2
2222
,, 6
ln , 1 3
ln 6 ln
1
2
xyz xz y yz
ttt t t
ttt t t t
dtdt
t







Fijk
rijk
Fijk
ri jk


3 2
23
1
ln 12 ln 249.49Fr
C
dtttttdt"
**

387. Section 15.2 Line Integrals 387
© 2010 Brooks/Cole, Cengage Learning
34.



222
22
,,
,0 2
2
t
t
t
t
xyz
xyz
xyz
ttte t
tte
t
te
dedt








ijk
F
rijk
ij k
F
rij k


2
2
0 22 1
26.91
2
t
C t
dtedt
te
"
**
Fr
35.


3
,2
:,02
xy x y
Ct t t t

Fij
rij



2
3
3
2
tt
tt t


rij
Fij



2
2
2
56
0
0
Work 6 66
2 Fr
C
t
dttdt t

"


**

36.


2
33
,
: cos , sin from 1, 0 to 0, 1
xy x xy
Cx ty t


Fij






33
22
633
2
845444442
442 8 6
cos sin , 0
2
3 cos sin 3 sin cos
cos cos sin
3 cos sin 3 cos sin 3 cos sin cos sin 3 cos sin cos 1 cos
3 cos sin 2 cos 2 cos 1 6 cos sin 6 cos
rij
rij
Fi j
Fr
tttt
ttttt
tttt
tt t t tt t t tt t t
tt t t tt





"


4
sin 3 cos sintt tt


2
864
0
2
975
0
Work 6 cos sin 6 cos sin 3 cos sin
2 cos 6 cos 3 cos 43
3 7 5 105
C
dttttttdt
ttt


"




**
Fr
37.


,
01
:2 1,12
323
xy x y
tt
Ct t t t
tt

6
=
7
=

8
Fij
i
rij
j
On
1,,Ct tt Firi

1
01 1
Work
2
Fr
C
dtdt" **

On
2,2 1,Ct t t t Fijrij



2
12
2
2
1
Work 2 1
3
46 13 0
Fr
C
dttdt
tt
"



**

On
3,3,Ct t t Fjrj



3
2
3
23
2
Work 3 3
2
91
926
22
Fr
C
t
dtdt t

"






**


11
Total work 0 0
22


38.
,
xyyx Fij
C: counterclockwise along the semicircle
2
4yx from 2, 0 to 2, 0




22
2cos 2sin , 0
2sin 2cos
2sin 2cos
4sin 4cos 4cos2
tttt
ttt
ttt
tt t


"
rij
rij
Fij
Fr


+,
0
0
Work
4cos2
2sin2 0
Fr
C
d
tdt
t


"

*
*

39.

,, 5
:2cos2sin,02
xyz x y z
Ct t tt t



Fijk
rijk


2sin 2cos
2cos 2sin 5
5
ttt
tttt
t


"
rijk
Fijk
Fr

2
2
0
Work 5 10
C
dtdt

" **
Fr

388. 388 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
40.
,,
: line from 0, 0, 0 to 5, 3, 2
xy z yz xz xy
C
Fijk




222
2
1
2
0
532,0 1
532
61015
90
Work 90 30
C
tttt t
t
tt t t
t
dtdt



"
"
**
rijk
rijk
Fijk
Fr
Fr

41. Because the vector field determined by F points in the
general direction of the path C, 0"FT and work will
be positive.
42. Because the vector field determined by F points for the
most part in the opposite direction of the path C,
0"FT and work will be negative.
43. Because the vector field determined by F is
perpendicular to the path, work will be 0.
44. Because the vector field is perpendicular to the path,
work will be 0.
45.

2
,
xyx xyFi j
(a)





1
1
2
3
2
11
236
3
21,13
2
421
821
C
ttt t
t
tt tt
dtttdt



"
**
ri j
rij
Fi j
Fr

Both paths join
2, 0 and 6, 2 . The integrals are
negatives of each other because the orientations are
different.
(b)




2
2
2
2 2
02
236
3
23 2 , 0 2
2
43 23 2
83 23 2
C
tttt
t
tt tt
dtttdt



"


**
rij
rij
Fi j
Fr

46.

232
,xyxyxyFi j
(a)




2
1
1
2
23
2 2
24
01
1,0 2
2
11
256
121
5
rij
rij
Fij
Fr
C
tt t t
tt
tt t tt
dttttdt



"

**

(b)





2
2
2
2
23
2 2
23
02
1 2 cos 4 cos , 0
2
2sin 8cos sin
12cos 4cos 12cos8cos
256
1 2 cos 4 cos 2 sin 8 cos sin 1 2 cos 8 cos
5
C
tttt
tttt
ttt tt
dtttttttdt





"

**
rij
rij
Fij
Fr

Both paths join
1, 0and 3, 4 .The integrals are negatives of each other because the orientations are different.
47.



,
:2
2
2
22 0
xy y x
Ct t t
t
ttt
tt




"
Fij
rij
rij
Fij
Fr

So,
0.
C
d"*
Fr
48.




3
2
3
33
,3
:
3
3
33 0
xy y x
Ct t t
tt
ttt
tt




" Fij
rij
rij
Fij
Fr
So,
0.
C
d"*
Fr

389. Section 15.2 Line Integrals 389
© 2010 Brooks/Cole, Cengage Learning
49.





32
2
2
32
2
32
,2
2
:
2
2
2
22 0
2
y
xy x x x
Ct t t
tt
t
ttt t
t
tt tt










"
Fij
rij
rij
Fij
Fr
So,
0.
C
d"*
Fr
50.


,
:3sin3cos
3cos 3sin
3sin 3cos
9sin cos 9sin cos 0
xy x y
Ct t t
ttt
ttt
tt tt




" Fij
rij
rij
Fij
Fr
So,
0.
C
d"*
Fr
51.
2 , 10 , 0 1 5 or , 0 10
5
y
xty t t yxx y


10
2
10
223
0
0
3 3 1010
510
C
yy
xydy ydy y



**

52. 2, 10, 0 1 5 , 0 2xty t t yx x


2
2
2
223
0
0
37525202
2
C
x
x y dx x x dx x



**

53.
1
2, 10, 0 1 , 0 10,
55
y
xty t t x y dx dy

10
232
10
0
0
190
or
25 75 2 3
C
yyy
xy dx y dy y dy



**


5, 5 ,0 2y x dy dx x


2
32
2
2
0
0
5 25 190
525
32 3
C
xx
xy dx y dy x x dx



**

54. 2, 10, 0 1 5 , 5 , 0 2xty t t yxdy dx x



22 2
2 2
00
2
23
0
125 125 1084
333
3 3 5 5 5 14 125
7288
C
y x dx y dy x x dx x dx x x dx
xx


** *


55.
,0 5
,0
,0
tt t
xt t yt
dx dt dy



ri


5
0
23225
C
x y dx x y dy t dt **


56.

,0 2
0,
0,
tt t
xtytt
dx dy dt



rj


22
2
0 0
3
2
2336
C
x y dx x y dy t dt t
**


x
12 345
1
2
3
−1
−2
y
x
1
2
1
−1
y

390. 390 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
57.

,03
33,36
tt
t
tt
6=
7
=8
i
r
ij



1
3
01:,0,
,0
2329
C
Cxt tyt
dx dt dy
x y dx x y dy t dt



**





2
6
2
6
32
3:3, 3
0,
345
233336
22
45 63
239
22
C
C
Cxt yt t
dx dy dt
t
x y dx x y dy t dt t
x y dx x y dy







**
*

58.

,03
33,3 5
tt
t
tt
6=
7
=8j
r
ij







1
3
01
2
55 2
32 3
27
2
27 47
22
:0,
0,
233
:3,3
,0
232333310
2310
C
C
C
Cxt yt t
dx dy dt
x y dx x y dy t dt
Cxt t yt
dx dt dy
x y dx x y dy t dt t t
x y dx x y dy








**
**
*

59.
2
,1,01, ,2
xt t y t t t dx dt dy t dt



1
22
0
1
43
1
32 2
0
0
2321332
311
641 2
23 6
C
x y dx x y dy t t t t t dt
tt
tt t dt tt




**
*

60.
32 12 3
2
,,04,,
xt t y t t t dx dt dy t dt




4
32 32 12
0
44
232 3522
0 0
3
2
93 592111
22 25 5 5
2323
2963216
C
x y dx x y dy t t t t t dt
tt tdtttt



**
*

61.
2
,2,02
,4
xt t yt t t
dx dt dy t dt





222
22 32 432
00 0
3162
33
23226424226
C
x y dx x y dy t t dt t t t dt t t t dt t t t
** *

62.
4sin , 3cos , 0
2
4cos , 3sin
xt t yt t t
dx t dt dy t dt








2
0
2
2
22 2
0
0
2 3 8 sin 3 cos 4 cos 4 sin 9 cos 3 sin
55
5 sin cos 12 cos 12 sin sin 12 6
22
C
x y dx x y dy t t t dt t t t dt
tt t tdt t t








**
*

x
32
2
1
1
3
(3, 3)
C
1
C
2
y
x
321
C
1
C
2
−1
−2
−3 (2, 3)−
y

391. Section 15.2 Line Integrals 391
© 2010 Brooks/Cole, Cengage Learning
63. ,fxy h
C: line from
0, 0 to 3, 4



34,0 1
34
5
tt t
t
t



rij
rij
r
Lateral surface area:


1
0
,55
C
fxyds hdt h**

64. ,fxy y
C: line from
0, 0 to 4, 4




,0 4
2
ttt t
t
t



rij
rij
r
Lateral surface area:



4
0
,282
C
fxyds t dt**

65.

22
,
: 1 from 1, 0 to 0, 1
fxy xy
Cx y







cos sin , 0
2
sin cos
1rij
rij
rtttt
ttt
t





Lateral surface area:


2
2
2
0
0
sin 1
,cossin
22
C
t
fxyds t tdt






**

66.

22
,
: 1 from 1, 0 to 0, 1
fxy x y
Cx y







cos sin , 0
2
sin cos
1rij
rij
rtttt
ttt
t





Lateral surface area:



+,
2
0
2
0
,cossin
sin cos 2
C
fx y ds t t dt
tt



**
67.

2
,
: 1 from 1, 0 to 0, 1
fxy h
Cy x








2
2
111,01
21
141
ri j
ri j
r
tt t t
tt
tt




Lateral surface area:




1
1 222
0
0
, 1 41 21 1 41 ln 21 1 41
4
2 5 ln 2 5 1.4789
4
C
h
fxyds h t dt t t t t
h
h






**

68.

2
,1
: 1 from 1, 0 to 0, 1
fxy y
Cy x







2
2
111,01
21
141ri j
ri j
rtt t t
tt
tt




Lateral surface area:







111 22 2 22
000
1
22
0
1
22 2
0
1
2
1
64
11
264
23 33 1
32 64
, 2 1 1 41 2 1 41 1 1 41
21 1 41 ln 21 1 41
21 2 4 1 1 1 41 ln 21 1 41
25 ln2 5 185 ln2 5
5ln25
C
fx y ds t t dt t dt t t dt
tt t t
tt t t t












** * *
64
46 5 33 ln 2 5 2.3515

392. 392 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
69.

2
,
: 1 from 1, 0 to 0, 1
fxy xy
Cy x
π


You could parameterize the curve C as in Exercises 67 and 68. Alternatively, let cos ,
x tπ then:

22
1cos sinytt




2
222 2
cos sin , 0
2
sin 2 sin cos
sin 4 sin cos sin 1 4 cosrij
ri j
rtttt
tttt
ttttt t





Lateral surface area:



22 12
2222
00
, cos sin sin 1 4 cos sin 1 4 cos sin cos
C
fx y ds t t t t dt t t t t dt




** *

Let
2
sinutπ and
12
2
1 4 cos sin cos ,dv t t t then 2sin cosdu t t dtπ and
32
21
14cos .
12
vt





2
232 32
22 2
0
0
2
32 52 52
22 2
0
11
, sin 1 4 cos 1 4 cos sin cos
12 6
11 1111
sin 1 4 cos 1 4 cos 5 25 5 11 0.3742
12 120 12 120 120 120
C
fxyds t t t t tdt
tt t











**

70.
22
22
,4
:4
fxy x y
Cx y






2cos 2sin , 0 2
2sin 2cos
2
tttt
ttt
t

π
rij
rij
r

Lateral surface area:



222
22
00 0
1
2
, 4 cos 4 sin 4 2 8 1 cos 2 8 sin 2 16
C
fxyds t t dt tdt t t



** *

71. (a)




2
,1
2cos 2sin , 0 2
2sin 2cos
2
fxy y
tttt
ttt
t




π
rij
rij
r



2 2
2 2
00
, 1 4 sin 2 2 4 sin cos 12 37.70 cm
C
Sfxyds tdttttt


**

(b)

312
0.2 12 7.54 cm
5


(c)
x
y
33
−3
4
5
z

393. Section 15.2 Line Integrals 393
© 2010 Brooks/Cole, Cengage Learning
72.
32
1
,20
4
:,040
fxy x
Cy x x







32
12
,0 40
3
2
9
1
4
ttt t
tt
tt





rij
ri j
r
Lateral surface area:

40
0 19
,201
44
C
fx y ds t t dt



**

Let
9
1,
4
ut




then
24
1
9
tu and
8
.
9
dt u du

40 91 91
242
01 1
91
53
1 19 1 8 8
20 1 20 1 179
44 9 981
8 179 850,304 91 7184
6670.12
81 5 3 1215
ttdt uuudu uudu
uu







** *
73.

cos sin , 0 2
sin cos ,
tatat t
tatatta
rij
rijr




2
222
0
2
32 3
0
2
222
0
2
32 3
0
,sin1
sin
,cos1
cos
x
C
y
CIyxyds atadt
atdta
Ixxyds atadt
atdta




-

-



**
*
**
*

74.

cos sin , 0 2
sin cos ,
tatat t
tatatta
rij
rijr




2
2222
0
2
43
0
2
2222
0
2
42
0
,sinsin
sin 0
,cossin
cos sin 0
x
C
y
CI y x y ds a t t a dt
atdt
I x x y ds a t t a dt
attdt




-
-


**
*
**
*

75. (a) Graph of:
2
3 cos 3 sin 1 sin 2 , 0 2ttt tt rij k
For ybconstant,
3sin sin
3
b
tb t and



2
2
22
2
22 2
1 sin 2 1 2 sin cos
14sincos
4
14sin1sin 1 1 .
99
ttt
tt
b
tt b






(b) Consider the portion of the surface in the first quadrant. The curve
2
1sin2zt is over the curve

1 3cos 3sin ,0 2.tttt rij So, the total lateral surface area is



2
22
0 3
4, 41sin2312 9cm.
4
C
fxyds t dt





**

(c) The cross sections parallel to the xz-plane are rectangles of height

2
2
1431 9yy and base
2
29 .y So,

22
3
23
0
27
Volume 2 2 9 1 4 1 42.412 cm .
99 2
yy
ydy



*

x
y
3
4
3
3
2
1
4
z

394. 394 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
76.

2
,0 1
2
ttt t
tt

rij
rij



10
544244611
34
16
3
Fr
C
d

"

*

77.
2
2
0
0
10
3sin 3cos , 0 2
2
175
10
3cos 3sin
2
1750 1750
1750 ft lb
22
rijk
Fk
rijk
Fr
C
ttttt
dttdt
ddtt












" "


**

78.




222
2
1
22 2 2
1
15 4 60 15
15 15
,2
60 15 15 2 120 4 8 parabola
Fr
CC
WdMdxNdy
Mxy xccx
Nxyxccx
dx dx dy cx dx
Wxccxxccxcxdxcc

"




**
*


1
4
16 4 0Wc c yields the minimum work, 119.5. Along the straight line path, 0,ythe work is 120.
79. See the definition of Line Integral, page 1070. See
Theorem 15.4.
80. See the definition, page 1074.
81. The greater the height of the surface over the curve, the
greater the lateral surface area. So,
312 4 .zzzz

82. (a) Work 0
(b) Work is negative, because against force field.
(c) Work is positive, because with force field.
83. False

1
2
0
2
C
xyds t dt**

84. False, the orientation of C does not affect the form.

,.
C
fxyds*

85. False, the orientations are different.
86. False. For example, see Exercise 32.
87.

,
1,01
12
Fij
rij
rij
xy y x xy
tkttt t
tk t









1
2
0
1
2
0
1
23 3 2 22 2
0
Work 1
1112
112 1
23
12
12
Fr
ijij
C
d
tkt t kt t k t dt
tkt tk t kt tdt
k
k t kt kt k t k t kt dt
k
"
"





*
*
*
*

,
xy 0, 0
11
,
416




11
,
24




39
,
416




1, 1
,xyF 5i 3.5 ij 22ij 1.5 3 ij 5ij
tr i 0.5ij ij 1.5 ij 2ij
"Fr 5 4 4 6 11
x
−11
c
yc x=1−
2
))
y
x
2
3
4
12 34
1
y
y
x
−1 12
−1
1
2
(0, 1)
(0, 0)

395. Section 15.3 Conservative Vector Fields and Independence of Path 395
© 2010 Brooks/Cole, Cengage Learning
Section 15.3 Conservative Vector Fields and Independence of Path
1.
2
,xyx xyFi j
(a)




2
1
1
23
1
24
0
,0 1
2
11
2
15
rij
rij
Fij
Fr
C
ttt t
tt
tt t
dttdt



"
**

(b)



2
2
2
23
sin sin , 0
2
cos 2 sin cos
sin sin
rij
ri j
Fijt









2
35
2
24
0
0
sin 2 sin 11
sin cos 2 sin cos
35 15
Fr
C
dd




"


**

2.
22
,
xyxy x Fij
(a)


1
1
2 ,0 4
1
2
tt t t
t
t
tttt



rij
ri j
Fij

4
2
0
4
3232
0 1
2
80
32 3 3
C
dtttdt
ttt

"





**
Fr
(b)




2
2
2
42 2
,0 2
2
ww w w
ww
www w



rij
rij
Fij


2
42 2
0
2
643
0
2
80
323 3
C
dwwwwdw
www
"



**
Fr
3. ,xy y xFij
(a)



1
2
1 sec tan , 0
3
sec tan sec
tan secrij
rij
Fij












3
23
0
3
23
0
3
0
3
0
sec tan sec
sec sec 1 sec
sec
ln sec tan
ln 2 3 1.317
Fr
C
dd
d
d







"





**
*
*

(b)



2
2 1,03
11
212
1
rij
rij
Fi j
tt t t
t
tt
ttt









3
0
3
0
3
0 2
3
0 2
3
2
0 1
212
11
2 1
11
2 14 14
11
2
12 14
11
ln
22
17 1
ln 2 3 ln
22 2
1
ln 7 4 3 1.317
2
Fr
C
tt
ddt
tt
dt
tt
dt
tt
dt
t
ttt

"

















**
*
*
*

396. 396 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
4.
2
,xyyxFi j
(a)






1
1
2
3
23
3 2
0
0 23,03
32
32 69
32
23 2
rij
rij
Fij
Fr
C
tttt
t
tt t
tt
dttdt




"



**

(b)





3
2
2
2
3
23
3
2
1
1
2ln 3ln ,1
11
3ln 2ln
3ln 2ln11 69
3ln 2ln
23 2
rij
rij
Fij
Fr
e
e
C
wwwwe
w
ww
ww w
ww
dwwdw
ww




"



**

5.

,sincos
cos cos
Because , is conservative.
xx
xx
xy e y e y
NM
ey ey
xy
NM
xy

CC

CC
CC

CC
Fij
F
6.

22 3
22
,15 10
30 30
Because , is conservative.
xy xy xy
NM
xyxy
xy
NM
xy

CC

CC
CC

CC
Fij
F

7.

2
22
1
,
11
Because , is not conservative.
x
xy
yy
NM
xy y y
NM
xy

CC

CC
CC
#
CC
Fij
F

8.

,, ln ln
so is not conservative.
xy
xyz y z x z
z
Px x N
yz z z

#
CC
#
CC
Fijk
curl F 0 F

9.

22
,, 2
is conservative.
xyz yz xyz xy

Fijk
curl F 0 F
10.
, , sin cos sin
, so is not conservative.
xy z yz xz yz xy yz
#
Fijk
curl F 0 F

11.

2
,2
xyxyxFij
(a)



2
1
1
32
,0 1
2
2
ttt t
tt
ttt



rij
rij
Fij


1
3
0
41
C
dtdt" **
Fr
(b)



3
2
2
2
42
,0 1
3
2
ttt t
tt
ttt



rij
rij
Fij


1
4
0
51
C
dtdt" **
Fr
12.
,
xyxy
xyye xeFij
(a)


1
1
22
33 3,0 3
3
tt tt
ttt t
t
tte te




ri j
rij
Fij




3 22
33
0
3 2
3
0
3
2
300
0
3
32
0
tt tt
C
tt
tt
dtetedt
etdt
eee



"





**
*
Fr

(b)
,
xyF is conservative because
.
xy xyMN
xye e
yx
CC

CC

The potential function is
,.
xy
fxy e K
By Theorem 15.7,
0.Fr
C
d"*

397. Section 15.3 Conservative Vector Fields and Independence of Path 397
© 2010 Brooks/Cole, Cengage Learning
13. ,xyyxFij
(a)


1
1 ,0 1
0
C
ttt t
t
ttt
d



"
*
rij
rij
Fij
Fr

(b)



2
2
2
2
1
2
0
1
3
,0 1
2
C
ttt t
tt
tt t
dtdt



"
**
rij
rij
Fij
Fr
(c)



3
3
2
3
3
1
3
0
1
2
,0 1
3
2
C
ttt t
tt
tt t
dtdt



"
**
rij
rij
Fij
Fr
14.

22
,2Fi
jxyxy xy
(a)


1
1
2
1
,1 3
1
1
2
tt t
t
t
t
tt
t



rij
rij
Fij


3 3
111
ln ln 3
Fr
C
ddtt
t
"
**

(b)




2
2
22
1
13,02
3
1
3
12
13 1 3
93
tt t t
t
ttt tt



rij
rij
Fij




2 22
0
2
32
0
2
432
012
13 1 3
99
1
3773
9
13 7 7 44
3
94 3 2 27
Fr
C
dtt ttdt
tttdt
ttt
t

"






**
*

15.
2
2
C
ydx xydy*

Because

2
2, , 2My Nx y xy y xyCCCC Fi jis conservative. The potential function is
2
,.fxy xy k So, you
can use the Fundamental Theorem of Line Integrals.
(a)


4, 4
22
0, 0
264
C
ydx xydy xy
*

(b)


1, 0
22
1, 0
20
C
ydx xydy xy



*

(c) and (d) Because C is a closed curve,
2
20.
C
ydx xydy*

16.
231 3 5
C
xydxxy dy *

Because 3, , 2 3 1 3 5My Nx xy x y x yCCCC Fij is conservative. The potential function is

22
,325.fxy x xy y x y k
(a) and (d) Because C is a closed curve,
231 3 5 0.
C
x y dx x y dy *

(b)



0,1
2
2
0, 1
231 3 5 3 5 10
2
C
y
x y dx x y dy x xy x y




*

(c)




2
2,
2
224
0,1
1
231 3 5 3 5 32
22
e
C
y
xydxxy dyxxy xy ee


*

398. 398 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
17.
22
2
C
xydx x y dy*

Because 2,My Nx xCCCC
22
,2 xyxyxyFi j is conservative.
The potential function is

3
2
,.
3
yfxy xy k
(a)


0, 4
3
22 2
5, 0
64
2
33
C
y
xy dx x y dy x y



*

(b)


0, 4
3
22 2
2, 0
64
2
33
C
y
xy dx x y dy x y



*

18.
22
2
C
xydx xydy*

Because 2,My Nx yCCCC
22
,2 xyxy xy Fij is conservative. The potential function is

32
,3 .fxy x xy k
(a)


8, 4
3
22 2
0, 0
896
2
33
C
x
x y dx xy dy xy




*

(b)


0, 2
3
22 2
2, 0
8
2
33
C
x
x y dx xy dy xy




*

19. ,,xyz yz xz xyFijk
Because
,,,curl F 0 F
xyz is conservative. The
potential function is
,, .
fxyz xyz k
(a)
1 2,0 4tt t t rijk

+,


4, 2, 4
0, 2, 0
32
C
dxyz" *
Fr
(b)

22
2
,0 2tt tt t rijk

+,


4, 2, 4
0, 0, 0
32
C
dxyz" *
Fr
20.
,,xyz z y Fijk
Because
,,,curl F 0 F
xyz is conservative. The
potential function is
,, .
fxyz x yz k
(a)

2
1
cos sin , 0tttt t rijk

+,


2
1, 0,
1, 0, 0
2
C
dxyz

" *
Fr
(b)

2
2
12 ,0 1tttt rik

+,


2
1, 0,
1, 0, 0
2
C
dxyz

" *
Fr

21.

2
,, 2 2 4
xyz y x x z y zFijk

,,xyzF is not conservative.
(a)





2
1
1
22 2
1
32
0
,0 1
2
2124
2
22
3rijk
rij
Fijk
Fr
C
ttt t
tt
tttt t
dtttdt



"
**

(b)



2
2
2
22
2
21,0 1
42 1
3212421
rij k
rij k
Fi j k
ttt t t
tt
ttt t t t









1 23
2
0
1
3
32
1 23 4
2
0
0
3 21 8211621
2117 5 17
17 5 2 1 16 2 1 2 2 1
32 6 6
Fr
C
dttt tt tdt
ttt
ttt t dt t
"





**
*

399. Section 15.3 Conservative Vector Fields and Independence of Path 399
© 2010 Brooks/Cole, Cengage Learning
22.
2
,, 3xyz y x xz Fijk

,,xyzF is not conservative.
(a)


1
1
2 cos sin , 0
sin cos
sin cos 3 cos
tttt t
ttt
ttttt


rijk
rijk
Fijk

+,

222
0
2
0
2
0 0 0
2
0
sin cos 3 cos
13cos
3sin 6 sin
3sin 6sin cos
5Fr
C
dttttdt
ttdt
ttt ttdt
tt t ttt





"








**
*
*

(b)


2
2
22 12 ,0 1
2
12 3 12
tttt
t
tttt




rik
rik
Fj k



1
32
0
1
32 3
0
1
34 3
3
0
312
32
3
32 2Fr
C
dttdt
ttdt
tt


"




**
*

23. ,,
z
xyz e y x xyFijk

,,
xyzF is conservative. The potential function is
,, .
z
fxyz xye k
(a)
1 4cos 4sin 3 , 0ttt t rijk



4, 0, 3
4, 0, 3
0
z
C
d xye

"

*
Fr
(b)
2 48 3,0 1tt t rik



4, 0, 3
4, 0, 3
0
z
C
d xye

"

*
Fr
24. ,, sin sin cosxyz y z x z xy xFijk
(a)



22
1
1
42
,0 2
22
cos
tt t t
ttt
tt t


rij
rij
Fk

2
0
00
C
ddt" **
Fr
(b)


2
2
2 44,0 1
44
16 cos 4
ttt t
t
tt t


rij
rij
Fk

1
0
00
C
ddt" **
Fr
25. +,


3, 8
0, 0
33 3 72
C
yxd xy" *
ijr
26.



3, 2
2
2
1, 1
22 5025
C
xy xy d xy

"

*
ijr
27.
+,

32, 2
0,
cos sin sin cos sin sin 1
C
x y dx x y dy x y



*

28.

23,2
22
1, 1
arctan
3412
C
ydx xdy x
xy y


*

29.


2,0
0, 0
sin cos sin 0
xx x
C
eydxe ydyey



*

30.


1, 5
22 22
22 22
7, 5
22 1112
26 74 481
C
xy
dx dy
xy
xy xy
1



*

31.
22
C
z y dx x z dy x y dz*


,,xyzF is conservative and the potential function is ,, 2
fxyz xz xy yz
(a)
+,


1, 1, 1
0, 0, 0
2202xz xy yz
(b)
+,


+,


0, 0,1 1,1,1
0, 0, 0 0, 0,1
22022xz xy yz xz xy yz
(c)
+,


+,


+,



1, 0 , 0 1, 1, 0 1, 1, 1
0,0,0 1,0,0 1,1,0
22202222xz xy yz xz xy yz xz xy yz

400. 400 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
32.
C
zy dx xz dy xy dz*

Note: Because
,,
xy z yz xz xyFijk is conservative and the potential function is ,, ,fx y z xyz k the integral
is independent of path as illustrated below.
(a)
+,


1, 1, 1
0, 0, 0
1xyz
(b)
+,


+,


0, 0,1 1,1,1
0, 0, 0 0, 0,1
011xyz xyz
(c)
+,


+,


+,


1, 0 , 0 1, 1, 0 1, 1, 1
0,0,0 1,0,0 1,1,0
0011xyz xyz xyz
33.
+,

2, 3, 4
0, 0, 0
sin cos 12 1 11
C
xdx zdy ydz x yz

*

34.
,,xyzF is conservative:
22
,, 3 4 10
fxyz x yz z



3, 4, 0
22
0, 0, 0
3 4 10 27
C
dxyzz"
*
Fr
35.

22 3
,9 61xy xy xyFij is conservative.



5, 9
32
0, 0
Work 3 30,366xy y

36.
,xyF is conservative.
2
,
x
fxy
y



3, 2
2
1, 1
97
Work 1
22
x
y






37.





22
2
2
3
2cos2 2sin2
4sin2 4cos2
8cos2 8sin2
1
cos 2 sin 2
32 4
cos 2 sin 2 4 sin 2 cos 2 0 0
4
rij
rij
aij
Fa a i j
Fr i j i j
CC C
ttt
ttt
ttt
tmt t t t
W d t t t t dt dt









" "
** *

38.
12 3,,xyz a a aFijk
Because
,,xyzF is conservative, the work done in moving a particle along any path from P to Q is

+,



,,
123
12 3 111222333
,,
123
,, . F
Qqq q
Ppp p
fxyz ax ay az aq p a q p a q p PQ


"


39. 175Fj
(a)


50
0
50 , 0 50
175 8750 ft lbs
C
tt t t
ddt
ddt


" "
**
ri j
rij
Fr

(b)



2
1
50
1
25
50 , 0 50
50
tt t t
dt

ri j
ri j



50
0
50
2
0
1
25
175 50
7 50 8750 ft lbs
2
Fr
C
dtdt
t
t
"

"

**

40. No. The force field is conservative.
41. See Theorem 15.5.
42. A line integral is independent of path if
C
d"*
Frdoes
not depend on the curve joining P and Q. See Theorem
15.6.

401. Section 15.3 Conservative Vector Fields and Independence of Path 401
© 2010 Brooks/Cole, Cengage Learning
43. (a) For the circle cos sin , 0 2 ,tatat t rij you have
22 2
,xya and

2 2
22
22
0 0 sin cos
sin cos sin cos 2 .Fr i j i j
C
at a t
datatdtttdt
aa


" "


** *

(b) For this curve, the answer is the same, 2 .

(c) For the opposite overtation, the answer is 2 .

(d) For the curve away from the origin, the answer is 0.
44. (a) The direct path along the line segment joining

4, 0 to 3, 4 requires less work than the path
going from
4, 0 to 4, 4 and then to 3, 4 .
(b) The closed curve given by the line segments joining
4, 0 , 4, 4 , 3, 4 , and 4, 0 satisfies
0.Fr
C
d"#*

45. Conservative.
C
d"*
Fris independent of path.
46. Not conservative. The value of
C
d"*
Frfrom
1, 0 to
1, 0 is positive if the path is above the x-axis, and
negative if the path is below the x-axis.
47. False, it would be true if F were conservative.
48. True
49. True
50. False, the requirement is .My NxCCCC
51. Let
.
f f
MN
yx
CC

CC
Fij i j
Then
22
22
and .
Mff N f f
yyy y xxx x
CCCC CCC C

CCC C CCC C Because
22
22
0
ff
xy
CC

CC
you have
.
MN
yx
CC

CC

So,
F is conservative. Therefore, by Theorem 15.7, you have
0
CCC
ff
dx dy M dx N dy d
yx
CC
"
CC
***
Fr
for every closed curve in the plane.
52. Because the sum of the potential and kinetic energies remains constant from point to point, if the kinetic energy is decreasing
at a rate of 15 units per minute, then the potential energy is increasing at a rate of 15 units per minute.
53.
22 22
,
yx
xy
xyxy


Fij
(a)




22
22
22
22
22 22
22
22
22
22
22 22
12
12
y
M
xy
xy yyMxy
y
xy xy
x
N
xy
xy xx
N xy
x
xyxy


C

C



C

C


So, .
N M
x y
CC

CC

(b)


+,
22
00
cos sin , 0
sin cos
sin cos
sin cos
C
tttt
tt
dttdt
dttdtt





"
**
rij
Fij
rij
Fr

402. 402 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
(c)
cos sin , 0
sin cos
sin cos
tttt
tt
dttdt

rij
Fij
rij

+,
22
00
sin cosFr
C
dttdtt

" **

(d)

cos sin , 0 2
sin cos
sin cos
rij
Fij
rij
tttt
tt
dttdt




+,
2 2
22
00
sin cos 2Fr
C
dttdtt

" **

This does not contradict Theorem 15.7 because F is not continuous at
0, 0 in R enclosed by curve C.
(e)

2
22 22 22
1
arctan
11
ijijF
xy xy y x
yxyxy xy xy

D


Section 15.4 Green's Theorem
1.

2
,01
22,12
tt t
t
ttt
6=
7
=8ij
r
ij





12 22
22 4 2
01
2
1 3
54
12 2
43
01
0
1
22 2
22 72 1
222
52 3 10330
C
y dx x dy t dt t t dt t dt t dt
ttt
ttdt tdt






** *
**

By Green's Theorem,



11
2
22
00
1
345
1
234
0
0
22 2
1
2
325 30
x x
xRxNM
dA x y dy dx xy y dx
xy
xxx
xxxdx
CC

CC



** ** *
*


2.

,01
22,12
tt t
t
ttt
6=
7
=8ij
r
ij





12 2
22 2 2
01
12 32
2
01
2
1 252
3
0
1 1
22
22
1
222
2
222
325
211 1
325 30
C
y dx x dy t dt t dt t dt t dt
t
tdt t t dt
ttt













** *
**

By Green's Theorem,



11
2
00
1
23
1
32 2 52
0
0
22 2
41
2
523 30
x x
xRxNM
dA x y dy dx xy y dx
xy
xx
xxxdx x
CC

CC




** ** *
*

1
1
(1, 1)
C
1
C
2
y = x
y = x
2
y
x
1
1
(1, 1)
C
1
C
2
y = x
y =
y
x
x

403. Section 15.4 Green’s Theorem 403
© 2010 Brooks/Cole, Cengage Learning
3.



01
11 2
323
434
tt
tt
t
tt
tt
6
=

=
7

=
=

8i
ij
r
ij
j



12 3 4 222
22 2
01 2 3
23
12
00 101 1 30 400
11 0
C
y dx x dy dt t t dt dt t t dt
dt dt


*** * *
**

By Green's Theorem,



11
00
11 11
22
0000
22
221 0
R
NM
dA x y dy dx
xy
xy y dx x dx x x
CC

CC


** **
**


4.



03
3337
10 4 7 10
14 10 14
tt
tt
t
tt
tt
6
=

=
7

=
=

8
i
ij
r
ij
j




3 7 10 14 222
22 2
03 7 10
710
37
00 309 161000140
9 16 9 7 3 16 10 7 12
C
y dx x dy dt t t dt dt t dt t
dt dt


** * * *
**

By Green's Theorem,


+,
34 3 4
2
000 0
3 3
2
00
22 2
8 16 4 16 36 48 12
R
NM
dA x y dy dx xy y dx
xy
xdxxx
CC

CC


** ** *
*


5.
22
:4Cx y
Let
2cos
x tand 2sin ,0 2 .ytt



2
2sin 2cos
0
2
24 2 22
24 4
2
24 2
2 cos 2 sin 2 cos 19.99
2 4 19.99
yx t t
C
x
xy x x x
Rx
xe dx e dy te t e t dt
NM
dA exedydx xexe xe dx
xy






CC


CC
**
** * * *


6. C: boundary of the region lying between the graphs of yx and
3
yx



10 3
2
01
11 3
3
3
00
3 2.936 2.718 0.22
0.22
yx x x xx
C
x
xy x x
Rx
xe dx e dy xe x e dx xe e dx
NM
dA e xe dy dx xe x e dx
xy

CC

CC** *
** ** *

In Exercises 7–10,
1.
NM
xy
C C

C C


7.

3
2
02
3
32
3
2
0
0
2
3279
29
32 22
x
Cxx
y x dx x y dy dy dx
xx
xx xdx





***
*

1
1
(1, 1)
C
1
C
3
C
2
C
4
y
x
4
4 (3, 4)
C
1
C
3
C
2
C
4
y
x
1234
−1
1
2
3
(3, 3)
y = x
y = x
2
− 2x
y
x

404. 404 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
8. Because C is an ellipse with 2aand 1,bthen R is an ellipse of area 2 .ab So, Green's Theorem yields

2 1 Area of ellipse 2 .
CR
y x dx x y dy dA ***


9. From the accompanying figure, we see that R is the shaded region. So, Green's Theorem yields

2 1 Area of 6 10 2 2 56.
CR
y x dx x y dy dA R ***



10. R is the shaded region of the accompanying figure.


+,
1
2
21
Area of shaded region
25 9 8
CR
y x dx x y dy dA



***

11.
+,
2 211 1 1 1
22
010 1 1
1
34
1
23 2
1
1
2
12 2 1 21
17 4
122
32663
CR
x x
NM
xy dx x y dy dA
xy
xdy dx y xy dx x x x dx
xx
xxxdxx x




CC

CC






***
** * *
*


12. The given curves intersect at 0, 0and 9, 3 .So, Green's Theorem yields


2
9
22
999
00 0 0
00
2
81
.
2244
CR
x
x
ydx xydy y ydA
yxx
ydydx dx dx



***
** * *


13.

2 2
416 4 16
22 2
22
16416 4
22220
x x
xCR xNM
x y dx xy dy dA y y dy dx y dx
xy
CC

CC
******


14. In this case, let sin , cos .yr xr Then dA rdrd and Green's Theorem yields




21cos
22
00
2
4
21cos 2 3
2
00 0
0
244sin
1cos4
4sin sin1cos 0.
33
CR
x y dx xy dy y dA r r dr d
rdrd d












*****
** *


15. Because
2sin2
xMN
ey
yx
CC

CC
you have
0.
R
NM
dA
xy
CC

CC
**


16. Because
22
2
,
MxN
yxy x
CC

CC

you have path independence and
0.
R
NM
dA
xy
CC

CC
**
x
2
2
4
4
−2
−4
(1, 1)
(5, 3)( 5, 3)−
( 1, 1)−
(5, 3)−
(1, 1)−
(5, 3)−−
(1, 1)−−
y
−5−4−3−2
−2
1
2
4
−3
−4
−5
−1 12345
y
x

405. Section 15.4 Green’s Theorem 405
© 2010 Brooks/Cole, Cengage Learning
17. By Green's Theorem,


2
11
00
1
223
1
0
0
cos sin sin sin
2
11 1
22 4 6 4612
x
x
CRx
x
y
ydx xy x y dy y y y dA ydydx dx
xx x x
dx







******
*


18. By Green's Theorem,


22 2
22
2 2 Area of 2 6 2 3 60 .
xy
CR
e y dx e x dy dA R


***


19. By Green's Theorem,


+ ,+ ,3 1 3 4 Area Large Circle Area Small Circle 4 9 32
CR
x y dx x y dy dA ***


20. By Green's Theorem,


22
22 1 2
22
12 11
12 1 1
22
22 12
22 2 22 12
22 1
33
33
33
7272
16 16 2 2 .
yy y
CR
yy
yy
xedx edy xedA
x e dy dx x e dy dx
x e dy dx x e dy dx
ee ee ee e e
eeee









***
** **
** **


21.
22
,
:1
xyxy xy
Cx y

Fij



21
00
1 2
22
22
00
00
Work 1 1 cos
111
cos 1 cos sin
22 2 2 2
CR
xy dx x y dy x dA r r dr d
rr
dd








*****
**


22. ,36
:2cos
xy
xye y e x
Cr


Fij

Work 3 6 9 9
xy
CR
e y dx e x dy dA ***
because 2cosr is a circle with a radius of one.

23.


32
,365
: boundary of the triangle with vertices 0, 0 , 5, 0 , 0, 5
xy x y x y
C
Fij

32 2251
22
Work 3 6 5 9 9 5 5
CR
x y dx x y dy dA ***


24.
22
,3 4
: boundary of the region bounded by the graphs of , 0, 9
xy x y xy
Cyxyx

Fij



99
22 2 3212
00 0
5584
35
Work 3 4 4 1
x
C
x y dx xy dy y dy dx x x dx ****


25. C: let cos , sin , 0 2 .xa tya t t By Theorem 15.9, you have


2
2
22
22
00
0
11 1
cos cos sin sin .
22 22
C
a
A xdyydx atatatatdt adt t a






** *

x
(1, 1)
(2, 2)
(2, 2)−− (2, 2)−
(1, 1)−(, )−−
(,)−
(,)−
y

406. 406 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
26. From the figure you see that

1
2
3
33
:, ,02
22
1
:4,
22
:0, 0.
Cy xdy dx x
x
C y dy dx
Cx dx







20
02
20
0213 3 1 1 1
40
22 2 2 2 2 2
1
42 4
2
x
Axxdx xdx
dxdx





**
**


27.
2
1
2
:1,2
:53, 5
Cy x dy xdx
Cy x dy dx



So, by Theorem 15.9 you have


+, +, +,
41
2
14
4
3
1
4
111
21 553
22
11119
3189.
23 2 2 2 2
A xx x dx x x dx
x
xx




**


28. Because the loop of the folium is formed on the interval 0 ,t!



3
2
3
31 2
1
t
dx dt
t



and

4
2
3
32
,
1
tt
dy dt
t




you have










43
2
2233
0 33
23
52
2
23
33 300 033
0
32 31 213 3
21 1
11
199 3 33
31 .
22 2 2 2111
tt ttt
Adt
tt
tt
tttt
dt dt t t dt
ttt!
!
!! !










*
** *


29. See Theorem 15.8, page 1093.

30. See Theorem 15.9:
1
2
.
C
A xdy ydx*


31. For the moment about the x-axis, .
x
RMydA**
Let 0Nand
2
2.My By Green's Theorem,

2
22
11
and .
22 22
x
x
CC CyM
M dx y dx y y dx
AA

** *

For the moment about the y-axis,
.
y
RMxdA**
Let
2
2Nx and 0.MBy Green's Theorem,

2
22
11
and .
22 22 y
y
CC CMx
M dy x dy x x dy
AA

** *


32. By Theorem 15.9 and the fact that cos , sin ,xr yr you have


211 1
22 2
cos cos sin sin .
C
A xdy ydx r r d r r d r d ** *

(2, 3)
3x − 2y = 0
x 2y = 8
1
1
2
3
4
234
y
x
C
1
C
2
C
3
1234
5
10
15
(4, 17)
(1, 2) y = x
2
+ 1
y = 5x
− 3
y
x
C
1
C
2

407. Section 15.4 Green’s Theorem 407
© 2010 Brooks/Cole, Cengage Learning
33.
2
3
2
2
2
2
22
12
32
44
33
11
22
CC
x
Axdxx
xxdyxdy
AA






*
**

For
1,2Cdy xdx and for
2,0.Cdy So,


2
4
2
2
2
2
13
20.
2323 64 2
x
xxxdx





*

To calculate
,ynote that 0yalong
2.CSo,




2
35
22 2
224
22
2
13 388
416816 .
2 32 3 64 64 3 5 5
8
,0,
5
xx
y x dx x x dx x
xy








**


34. Because
2
area of semicircle ,
2
a
A

you have
2
11
.
2A a
Note that 0yand 0dyalong the boundary 0.y
Let
cos , sin , 0 ,xa tya t t then




3
22 3 2
2
000
0
3
22 3
2
00
0
1 sin
cos cos cos 1 sin cos sin 0
3
1cos4
sin sin sin cos .
33
4
,0,
3
aa at
x a t a t dt t dt t t dt t
a
aata
yatatdt tdt t
a
a
xy



















***
**


35. Because
1
24
1
3
0
0
1
,
24 4
xx
Axxdx



*
you have
1
2.
2A

On
1Cyou have
32
,3yxdy xdx and on
2Cyou
have
,.y x dy dx So,


10
222 242
0112
10
262
01 62 8
2232 62
5315
22 8
222 .
73 21
CC C
C
x x dy x x dx x dx x dx x dx
y y dx x dx x dx

** * **
***



88
,,
15 21
xy






36. Because

1
2,
2
A ac ac you have
11
,
22A ac





1
2
3:0, 0
:,
:,.
Cy dy
cc
C y x a dy dx
ba ba
cc
C y x a dy dx
ba ba






So,



222
22 22
22
211 12
00
22 233
11 1
0 .
22 2333 ab a
Caab
ba
Ca b c c abc b
x x dy x dx x dx
ac ac b a b a ac
cb a cb acc c
y y dx x a dx x a dx
ac ac b a b a ac








****
** *


,,
33
bc
xy





x
12
1
2
3
−1−2
C
1
C
2
yx= 4−
2
y
x
1
1
C
1
C
2
(1, 1)
y
x
a
2a
−aC
1
C
3
C
2
(b, c)
y

408. 408 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
37.
2
2222
22 2
2
00
0
11cos2313
1cos 12cos 2sin sin2 3
222222422
aaaa
Aa d d







**


38.
222
22
00
0
11cos6sin6
cos 3
222464
aaa
Aa d d







**

Note: In this case R is enclosed by
cos 3ra where 0 .

39. In this case the inner loop has domain
24
.
33

So,

+,
43 43 43
2
2323 2311 133
1 4cos 4cos 3 4cos 2cos2 3 4sin sin2 .
22 22
Ad d

**


40. In this case, 0 2 and you let
2
22
sin 1 2
,cos , .
1cos 1 1
udu
ud
uu





Now u!as
and you have





2
2
222
00 0 22
2
22
2
22 22
00 2
0
0
2
0
2
19 1
1
29 18
2 2cos 1131
44
1
1
13 23 6 12 1 3
18 18 arctan 3
13 213 13 3313
66 6
a
21333 3
du
u
u
Ad du
uuu
u
u
u
du du u du
uuu
u
u
u




!!
!
!
!!
!












** *
** *
0
33
rctan 3 0 2 3 .
33
u

!






41. (a)

33 22
1
1
24
21 2 2
2
00 0 0
0
27 27 3 3
27 3 51 51
27 3
24 4 2
CR
ydx x xdy x y dA
rr
rrdrd d d








***
** * *

(b) You want to find c such that

2
0
27 3
C
rrdrd*
is a maximum:



2
4
2
27 3
24
27 3 3
c
fc c
fc c c c





23
2
00 243
Maximum Value: 27 3
2
rrdrd

**


42. (a)

,04
44, 48
12 12 , 8 12
tt
tt t
ttt
6
=
7
=

8i
ri j
ij



48 12 222
22 2
04 8
0 0 4 0 16 12 12
128 64
064
33
C
y dx x dy dt t t dt t dt t dt

** * *

By Green's Theorem,

44
2
00 0 64
22 .
3x
RNM
dA x y dy dx x dx
xy
CC

CC
** ** *

x
1
1
2
2
3
3
4
4
(4, 4)
yx=
y

409. Section 15. 4 Green’s Theorem 409
© 2010 Brooks/Cole, Cengage Learning
(b) cos sin , 0 2tttt rij



22
22 2 2 3 3
00
2
33
2
22
0
0
sin sin cos cos cos sin
sin cos
cos 1 sin sin 1 cos sin cos 0
33
C
y dx x dy t t dt t t dt t t dt
tt
ttt tdtt t









** *
*

By Green's Theorem,




2
11
2
11
21
00
2
0
22
2cos 2sin
22
cos sin 0 0.
33
x
RxNM
dA x y dy dx
xy
rrrdrd
d





CC

CC


** **
**
*


43.
22
C
ydx xdy
I
xy



*

(a) Let
22 22
.
yx
xyxy


Fij

F is conservative because

22
2
22
.
NM xy
xy
xy
CC

CC



F is defined and has continuous first partials everywhere except at the origin. If C is a circle (a closed path) that does not
contain the origin, then

0.
CC R
NM
dMdxNdy dA
xy
CC
"
CC
** **
Fr
(b) Let
cos sin , 0 2atat t rij be a circle
1Coriented clockwise inside C (see figure). Introduce line segments
2Cand
3C as illustrated in Example 6 of this section in the text. For the region inside C and outside
1,CGreen's Theorem
applies. Note that since
2Camd
3Chave opposite orientations, the line integrals over them cancel. So,
412 3CCCCC and

41
0.
CCC
ddd" " "***
Fr Fr Fr
But,



+,
2
22 22 22 22
01
2 2
22
00 sin sin cos cos
cos sin cos sin
sin cos 2 .
C
atat a ta t
ddt
atatatat
ttdtt



"



**
*
Fr

Finally,
1
2.
CC
dd " "**
Fr Fr

Note: If C were oriented clockwise, then the answer would have been 2 .
44. (a) Let C be the line segment joining 11,
xyand 22,.xy


21
11
21
21
21yy
yxxy
xx
yy
dy dx
xx










22
21 21 21
11 1 1
11
21 21 21
2
21 21
1 1 1 121 12 1 121 12 21
21 21
1xx
Cx x
x
xyy yy yy
y dx x dy x x y x dx x y dx
xx xx xx
yy yy
x yx x yxx xyy yxx xyxy
xx xx






** *

x
1
1
−1
−1
xy
22
+=1
y
x
−3
−2
2
3
4
C
1
C
2
C
3
C
y

410. 410 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
(b) Let C be the boundary of the region
11
11 .
22
CRR
A y dx x dy dA dA *****

So,

12
1
2
RC C C n
dA y dx x dy y dx x dy y dx x dy


** * * *

where
1Cis the line segment joining 11,
xyand 22 2,,xyCis the line segment joining 22,xyand 33,,,xyand
nC is the line segment joining ,
nnxyand 11,.xySo,
12 21 23 32 1 1 1 1
1
.
2
nn nn n n
RdA xy xy xy xy x y xy xy xy

**

45. Pentagon: 0,0, 2,0, 3,2, 1,4, 1,1


191
22
00 40 122 14 00A


46. Hexagon: 0,0, 2,0, 3,2, 2,4, 0,3, 1,1


1 21
2 2
0040124600300A


47. Because div ,
CR
ds dA"***
FN F then


2
div div .
N N
CC R R R
fD g ds f g ds f g dA f g f g dA f g f g dA D" D D D"D D D"D* * ** ** **

48.

22 22
NN N N
CCC
RR R
fD g gD f ds fD g ds gD f ds
fg f gdA g f g fdA f g g fdA

D D "D D D "D D D
***
** ** **

49.
00 0
CR R
f x dx g y dy g y f x dA dA
xy
CC


CC*** **

50.

0
00
CC R R
MN
NM NM
xy xy
NM
dMdxNdy dA dA
xy

CC CC

CC CC
CC
"
CC
* * ** **
Fij
Fr
Section 15.5 Parametric Surfaces
1. ,uv u v uv
zxy

rijk
Matches (e)
2.
22 2
,cossin
, cone
uv u v u v u
xy z

rijk
Matches (f )
3.

1
2
,
2,planeuv u u v v
yxz

ri jk
Matches (b)
4.
3
31
4
,
4 , cylinder
uv u v v
yz

rijk
Matches (a)
5. , 2 cos cos 2 cos sin 2 sinuv v u v u vrijk

222 2 2 2 2 2 2 2
4 cos cos 4 cos sin 4 sin 4 cos 4 sin 4, spherexyz v u v u v v v
Matches (d)
6.
22
,4cos4sin
4, circular cylinder
uv u u v
xy

rijk
Matches (c)

411. Section 15.5 Parametric Surfaces 411
© 2010 Brooks/Cole, Cengage Learning
7. ,
2
20
v
uv u v
yz

π≤
rijk
Plane

8.


2
22 2 2 2 21
2
11
28
,2cos2sin
,4
uv u v u v u
zuxy u z xy


rijk
Paraboloid

9.

22
,2cos 2sin
4
uv u v u
xz


rijk
Cylinder

10.

22 2 2 2 2 2
22 2
22
222
, 3 cos cos 3 cos sin 5 sin
9 cos cos 9 cos sin 9 cos
cos sin 1
925
1
9925
rijkuv v u v u v
xyvuvuv
xy z
vv
xyz






Ellipsoid

11.


4
2
22
,2cos2sin ,
01,02
16
uv u v u v u
uv
xy
z




π
rijk
12.

222
, 2 cos cos 4 cos sin sin ,
02,02
1
4161
uv v u v u v
uv
xyz




rijk
13.

22 2
, 2 sinh cos sinh sin cosh ,
02,02
1
141
uv u v u v u
uv
zxy



ππ ≤
rijk
14.

,2cos2sin ,
01,03
tan
uv u v u v v
uv
y
z
x



π
rijk
15.
, sin cos 1 cos sin ,
0,02
uv u u v u v u
uv



rijk

16.

33
,coscossinsin ,
0,02
2
uv u v u v u
uv




rijk

x
y
43
5
5
−4
3
2
z
y
4
4
4
x
z
x
y
5
5
−3
3
z
y
5
4
3
4
3
x
z
yx
2
2
3
2
1
z
x y
3
−4
−5
−5
−5
−4
−3
5
5
5
4
3
4
z
x
y
3
6
9
9
6
9
6
z
yx
2
−2
−4 −4
4
2
4
4
8
z
x
y3
3
5
2
2
1
−2
−2
−3
−3
−1
4
3
z
x
y
−1
−1
1
1
2
z

412. 412 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
For Exercises 17–20,
2
r, cos sin ,uv u vi u v juk 02,02.uv
Eliminating the parameter yields
22
,0 4.zx y z


17.


2
22
,cossin ,0 2,02uv u v u v u u v
zxy

sijk
The paraboloid is reflected (inverted) through the xy-plane.
18.
2
22
,cos sin,0 2,02uv u v u u v u v
yx z

sijk
The paraboloid opens along the y-axis instead of the z-axis.
19.
2
,cossin ,0 3,02uv u v u v u u v sijk
The height of the paraboloid is increased from 4 to 9.
20.
2
22
,4cos4sin ,0 2,0 2
16
uv u v u v u u v
xy
z


sijk
The paraboloid is "wider." The top is now the circle
22
64.xy It was
22
4.xy
21. zy

,uv u v vrijk
22.
6zxy

,6uv u v u vrij k
23.

22
22
49
,49
yxzxyx x z z

ri jk

or,


11
23
,cos sin,
0, 0 2
uv u v u u v
uv


/
rijk
24.

22
22
16
,16
xyz
yz y z y z

rijk

or,


1
4
,cossin,
0, 0 2
uv u u v u v
uv


/
ri jk
25.

22
25
,5cos5sin
xy
uv u u v

rijk
26.

22
416
,2cos4sin
xy
uv u u v

rijk
27.

2
2
,
zx
uv u v u

rijk

28.

222
1
941
, 3 cos cos 2 cos sin sin
xyz
uv v u v u v

rijk

29.
4zinside
22
9.xy

,cossin4,0 3uv v u v u vrijk
30.

22 22
2
inside 9.
,cossin ,0 3
zx y x y
uv v u v u v v

rijk
31. Function:
,0 6
2
x
yx
Axis of revolution: x-axis

,cos,sin
22
06,02
uu
xuy vz v
uv




x
y
2
5
2
z

413. Section 15.5 Parametric Surfaces 413
© 2010 Brooks/Cole, Cengage Learning
32. Function: ,0 4yx x
Axis of revolution: x-axis

,cos,sin
04,02
xuy u vz u v
uv




33. Function:
sin , 0xzz
Axis of revolution: z-axis

sin cos , sin sin ,
0,02
x uvy uvzu
uv




34. Function:
2
1, 0 2zy y
Axis of revolution: y-axis


22
1cos , , 1sin
02,02
xuvyuzuv
uv




35.





, , 1, 1, 1
,,,
At 1, 1, 1 , 0 and 1.
0, 1 , 0, 1
0, 1 0, 1 1 1 0 2
111
uv
uv
uv
uv u v u v v
uv uv
uv




'

rijk
rijrijk
rijrijk
ijk
Nr r ij k

Tangent plane:
11210
20
xy z
xy z



(The original plane!)

36.







, , 1, 1, 1
,,,
22
At 1, 1, 1 , 1 and 1.
11
1, 1 , 1, 1
22
1
1110
1, 1 1, 1 2
22
1
01
2
rijk
rikrjk
rikrjk
ijk
Nr r i jk
uv
uv
uv
uv u v uv
vu
uv uv
uv uv
uv




'

Direction numbers:
1, 1, 2
Tangent plane:
11210
20
xy z
xy z



37.




2
, 2 cos 3 sin , 0, 6, 4
,2cos3sin2
,2sin3cos
At 0, 6, 4 , 2 and 2.
2, 3 4 , 2, 4
22
2, 2, 0 3 4 16 12
22
40 0
rijk
rijk
rij
rjkr i
ijk
Nr r j k
u
v
uv
uv
uv u v u v u
uv v v u
uv u v u v
uv












'



Direction numbers:
0, 4, 3
Tangent plane:
46340
4312
yz
yz



38.





21
2
,2cosh2sinh ,
,2cosh2sinh
, 2 sinh 2 cosh
At 4, 0, 2 , 2 and 0.
2, 0 2 2 , 2, 0 4
88
rijk
rijk
rij
rikr j
Nrr i k
u
v
uv
uv
uv u v u v u
uv v v u
uv u v u v
uv





'

Direction numbers:
1, 0, 1
Tangent plane:
420
2
xz
xz



39.


12
00
,4 ,0 2,0 1
,4,,
400 4 4
011
16 16 4 2
42 4221 82
uv
uv
uv
uv u v v u v
uv uv
Adudv


'

'

**
rijk
rirjk
ijk
rr j k
rr

414. 414 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
40.


2
,2cos2sin ,0 2,0 2
,2cos2sin2
,2sin2cos
u
v
uv u v u v u u v
uv v v u
uv u v u v


rijk
rijk
rij


22
42 42 2 2
2cos 2sin 2 4 cos 4 sin 8
2sin 2cos 0
16 cos 16 sin 64 4 4
ijk
rr i j k
rr
uv
uv vvuuvuvu
uvu v
uvuvuuu
'

'


2
22 2 2 32
22
00 0 0
0
44464
44 4 888 16282 221
3333
A u u du dv u dv dv




** * *

41.



2
00
,cossin ,0 2,0
,sincos
,
sin cos 0 cos sin
001
2
u
v
uv
uv
b
uv a u a u v u v b
uv a u a u
uv
auau auau
a
A a du dv ab





'
'

**
rijk
rij
rk
ijk
rr i j
rr
42.



22 22 2
, sin cos sin sin cos , 0 , 0 2
, cos cos cos sin sin
,sinsinsincos
cos cos cos sin sin sin cos sin sin sin cos
sin sin sin cos 0
u
v
uv
uv a u v a u v a u u v
uv a u v a u v a u
uv a u v a u v
auvauvaua uva uva uu
auvau v


'

rijk
rijk
rij
ijk
rr i j k
r
2
2
22
00
sin
sin 4
uv au
A a ududv a


'

**
r

43.



2
2
2
222
00
,cossin,0 ,02
,cossin
,sincos
cos sin 1 cos sin
sin cos 0
1
11
u
v
uv
uv
b
uv au v au v u u b v
uv a v a v
uv au v au v
avav auvauvau
au v au v
au a
A a a u du dv ab a





'

'

**
rijk
rijk
rij
ijk
rr i j k
rr

415. Section 15.5 Parametric Surfaces 415
© 2010 Brooks/Cole, Cengage Learning
44.

, cos cos cos sin sin , , 0 2 , 0 2
, cos sin cos cos
, sin cos sin sin cos
rijk
rij
rijk
u
v
uv a b v u a b v u b v a b u v
uv a b v u a b v u
uv b v u b v u b v






cos sin cos cos 0
sin cos sin sin cos
cos cos cos sin cos cos sin cos
uv ab v uab v u
bv u bvu b v
b u va b v b u va b v b va b v
'


ijk
rr
ijk




22
2
00
cos
cos 4
uv ba b v
A b a b v du dv ab


'

**
rr

45.





24
00
,cossin,04,02
cos sin
,
22
,sincos
cos sin 1
1cossin
222
sin cos 0
1
4
1
17 17 1 36.177
46
rijk
rijk
rij
ijk
rr i j k
rr
u
v
uv
uv
uv u v u v u u v
vv
uv
uu
uv u v u v
vv
uv uv
uu
uv u v
u
A u du dv





'

'

**

46.



2
2
2
00
, sin cos sin sin , 0 , 0 2
, cos cos cos sin
, sin sin sin cos
sin cos cos sin sin sin
sin 1 cos
21
sin 1 cos 2 2 ln
21
rijk
rijk
rik
rr i j k
rr
u
v
uv
uv
uv u v u u v u v
uv u v u v
uv u v u v
uv uu uv
uu
Auududv





'
'



**

47. See the definition, page 1102.
48. See the definition, page 1106.
49. Function:
zx
Axis of revolution: z-axis


cos , sin ,
,cossin
0, 0 2
xuvyuvzu
uv u v u v u
uv




rijk

50. (a) From
10, 10, 0
(b) From
10, 10, 10
(c) From
0, 10, 0
(d) From
10,0,0
51.

33 33 3
, sin cos sin sin cos
0,02
uv a u v a u v a u
uv



rijk

3 3 23 23 2 2
33 232322
323232
sin cos sin cos
sin sin sin sin
cos cos
xau vx a u v
ya u v y a u v
za u z a u




23 23 23 23 2 2 2 2 2 23 2 2 23
sin cos sin sin cos sin cosx yzauvuvuauua

416. 416 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
52. Graph of ,cossinuv u v u v vrijk

0,0uv from
(a)
10,0,0 (b) 0, 0, 10 (c) 10, 10, 10

53. (a)
,4coscos
4cossin sin,
02,02
uv v u
vu v
uv




ri
j k
(b)

,42coscos
42cossin 2sin,
02,02
uv v u
vu v
uv




ri
j k


(c)

,8coscos
8cossin sin,
02,02
uv v u
vu v
uv




ri
j k
(d)

,83coscos
8 3 cos sin 3 sin ,
02,02
ri
j k
uv v u
vu v
uv






The radius of the generating circle that is revolved about the z-axis is b, and its center is a units from the axis of revolution.
54.
,2cos2sin ,0 1,0 3uv u v u v v u v rijk
(a) If
1:u≤ Helix


22
1, 2 cos 2 sin
4
03
vvvv
xy
z



rijk
(b) If
2
:
3
v
≤ Line

22
,3
33
3
2
3
rijkuuu
yx
z




≤π
≤

(c)If one parameter is held constant, the result is a curve in 3-space.
y
3
−3
z
x y
3
3
3
z
x y
6
6
4
−6
−4
−6
z
x y
66
4
z
1
−1−1
−2−2
1
1
2
22
yx
z
y
x 22
2
4
8
10
−2
−2
z
yx
3
9
−9
3
z
yx
12
12
12
−12
z
x
y
3
3
−3
−3

417. Section 15.5 Parametric Surfaces 417
© 2010 Brooks/Cole, Cengage Learning
55. , 20sin cos 20sin sin 20cos ,0 3, 0 2
20 cos cos 20 cos sin 20 sin
20 sin sin 20 sin cos
rijk
rijk
rij
u
v
uv u v u v u u v
uv uv u
uv u v




22 2 2
22
20 cos cos 20 cos sin 20 sin
20 sin sin 20 sin cos 0
400 sin cos 400 sin sin 400 cos sin cos cos sin sin
400 sin cos sin sin cos sin
ijk
rr
ij k
ijk
uv uv uv u
uv u v
uv uv uu v uu v
uv uv uu
'






42 42 22 4 22 2
400 sin cos sin sin cos sin 400 sin cos sin 400 sin 400 sinrr
uv uv uv uu u uu u u'
+,
23 2 2 3
2
000 0 0
400 sin 400 cos 200 400 m
S
S dS u du dv u dv dv

** * * * *

56.
222
1xyz
Let
cos , sin ,
xuvyuvππ and
2
1.zu Then,


2
,cossin
1
,sincos.
u
v
u
uv v v
u
uv u v u v



rijk
rij

At
1, 0, 0 , 1uπand 0. 1, 0
uvπr is undefined and 1, 0 .
v πrj The tangent plane at 1, 0, 0 is 1.xπ

57.




2
23
2
00
,cossin2,0 3,0 2
,cossin
,sincos2
cos sin 0 2 sin 2 cos
sin cos 2
4
313
43134ln
2
u
v
uv
uv
uv u v u v v u v
uv v v
uv u v u v
vv v vu
uvu v
u
A u du dv





'

'




**
rijk
rij
rijk
ijk
rr i j k
rr

58.


,cossin,,02
,cossin
,sincos
u
v
uv u fu v fu v a u b v
uv f u v f u v
uv fu v fu v


ri j k
ri j k
rjk




1 cos sin cos sin
0sin cos
ij k
rr i j k
uv fuvfuvfufufuvfuv
fu v fu v
'



2
1
uv fu f u '

rr



2 22
0
1 2 1 because
bb
aa
A f u f u du dv f x f x dx u x



** *

59. Answers will vary.
60. Answers will vary.

x
y
4
2
4
−4
−2
π2
π4
z

418. 418 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
Section 15.6 Surface Integrals
1. :4,0 4,0 3, 1, 0
zz
Sz x x y
xy
CC

CC




43 43 4 2
2
00 00 0
22411024223122
S
xyzdS xy x dydx ydydx dx ** ** ** *

2.
: 15 2 3 , 0 2, 0 4, 2, 3, 1 4 9 14
zz
S z x y x y dS dy dx dy dx
xy
CC

CC




24 24
00 00
2 2 15 2 3 14 14 15 128 14
S
xyzdS xy xy dydx xydydx ** ** **

3.
22
:2, 1, 0
zz
Sz x y
xy
CC

CC




2
11 21
2
2
11 00
2
2
0
0
222100cos2sin2
12 12 2 2
cos sin 1 sin cos 2 2
33 33 3 3
x
Sx
xy z dS x y dy dx r r r dr d
d





π
)
πππ




** ** * *
*

4.
32 122
:,01,0,,0
3
zz
Sz x x y x x
xy
CC

CC






1 2 2
32 12
00
1
32
00
1
52
0
1
132
52 32
0
0
1
1 132 32
52 32 12
00
0 2
2210
3
2
21
3
2
1
3
21 5
11
34 12
151 5
111
6123 24
252 5
31824x
S
x
xyzdS xy x x dydx
x y x x dy dx
xxdx
xx x xdx
xx xx x xdx
x

















** **
**
*
*
*
1
2
0
2
1
0
1
22
0
25 1 1
18 24 2 4
251 1 1 1
ln
18 24 2 2 4 2
253 13 11
2ln 2 ln
18 48 2 4 2 4 2
2152 5 1 612 5
ln ln 3 2 2 0.2536
18 96 192 288 192322
xdx
xdx
xxx x xx














*
*

5.

: 3 first octant , 1, 1
zz
Sz x y
xy
CC
≤ππ ≤π ≤π
CC




3
2
33 3 22
00 0
0
3
42
3 2
3
0
0
11 1 3
2
3 3 9 3 27 27 3
32
2242248
x
x
S
y
xy dS xy dy dx x
xx
xxdx x
π
π








** ** *
*

y
x
3
3
3
z
y = 3 − x

419. Section 15.6 Surface Integrals 419
© 2010 Brooks/Cole, Cengage Learning
6.

2
2
42
24 2
22
00 0
0
:,0 2,0 4, 0
11
422
224
x
S
zz
Sz h x y x
xy
x
xy dS xy dy dx x x dx x
π
CC

CC

≤≤π≤π≤

** ** *

7.
2
22
2
0
:9,0 2,0 ,
2, 0
391 17 1
14
240
Sy
Sz x x y x
zz
x
xy
xy dS xy x dx dy

CC
≤π ≤
CC


** **

8.
22
44
00
111
:,04,04,,
222
3904 160 5
1
44 15 3
S
zz
Sz xy x y y x
xy
yx
xy dS xy dy dx
CC

CC

** **

9.

22
22
2222
00
:10 ,0 2,0 2
2 2 1 4 4 11.47
S
Sz x y x y
x xy dS x xy x y dy dx


** **

10.

3
22 2
222 2
00 0
:cos,0 ,0
22
2 2 1 sin 1 sin 0.52
4
x
S
x
Sz x x y
x
xxydS xxy xdydx xdx




** * * *

11.

22
11
32
: 2 3 6 12 first octant 2
,,
Sx y z z x y
xyz x y
-









2 2 6423
22 22
00
6
346
234
0
0
711
32 6
7 7 364 212 4 1 12
6333636833
1
44 4
x
R
mxy dA xydydx
xx xdxxx x
π



** **
*

12.

22 2
:
,,
Sz a x y
xyz kz
-
≤ππ
≤



22
22 2
22 2 22 2
22 2 23
22 2
1
22
SR
RRR
xy
mkzdS kaxy dA
axy axy
a
k a x y dA ka dA ka dA ka a ka
axy


ππ


ππ ππ


≤ππ ≤≤≤≤

ππ

** **
** ** **

13.
:, 2,0 1,0 2
,2
100 2
012
5rijk
rirjk
ijk
rr jk
rr
uv
uv
uv
Suv uv v u v

'
'≤



2
2
21 2
00 0
0
5555555125
2
S
v
y dS v du dv v dv v




** ** *

x
−1
1
2
2
34
4
5
5
6
3
1
2
3
yx= 4−
R
y
y
x
a
a a
z

420. 420 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
14.
22
,2cos2sin ,0 2,0 1
2sin 2cos ,
2sin 2cos 0 2cos 2sin
001
4cos 4sin 2
uv
uv
uv
uv u u v u v
uu
uu u u
uv

'
' rijk
rijrk
ijk
rr i j
rr



2
2
12 1
00 0
0
sin
2cos 2sin 2 8 4
2
S
u
xy dS u u du dv dv






** ** *

15.
22
:, 2cos 2sin ,0 2,0 1
2sin 2cos ,
2sin 2cos 0 2cos 2sin
001
4cos 4sin 2rijk
rijrk
ijk
rr i j
rr
uv
uv
uv
Suv u uv u v
uu
uu u u
uu

'
'



+,
12 1 1 2
000 0 0
2 cos 2 sin 2 4 sin cos 4 2 8
S
x y dS u u du dv u u dv dv

** ** * *

16.

4
00
:, 4cos 4sin 3,0 4,0
12 cos 12 sin 16 20
10,240
4 cos 4 sin 20
3
uv
S
Suv u v u v u u v
uv uv u u
x y dS u v u v ududv


'

** **
rijk
rr i j k
17.
222
22
,,
:, 1, 1
fxyz x y z
zz
Sz x y x y
xy

CC

CC




2
11 2
22 22
2
11
2
11 21
22 2
2
11 00
1 2
4 2
22
00
0 0
,, 1 1 1
3222 322cossin
33sin
23 cossin 1 cossin 3
44 2 2 2
x
Sx
x
x
f x y z dS x y x y dy dx
x y xy dy dx r r r r dr d
rr
dd










0








** **
** * *
**

18.
22 22
,,
:,4 16
xy
fxyz
z
Sz x y x y






2
24
22 2
22 2
02
4
24 2 32
22
02 0
2
2
2
0
sin cos
,, 1 4 4 1 4
1
14sincos 14 sincos
12
65 65 17 17 sin
0
12 2
SS
xy r
fx y z dS x y dydx r rdrd
xy r
rr drd r d














** ** * *
** *

421. Section 15.6 Surface Integrals 421
© 2010 Brooks/Cole, Cengage Learning
19.
222
2222
,,
:,4
fxyz x y z
Sz xyxy






22
2 2
24
22 22
2
24 22 22
22222
24
22
2 22
24
2 2
32
24 22 2
22 2
2
24 00 0
00
,, 1
2
16 32
222
333
x
Sx
x
x
x
x xy
f x y z dS x y x y dy dx
xy xy
xyxy
x y dy dx
xy
r
x y dy dx r dr d d





















** * *
**
** ** *

20.

222
2
22 2
,,
:,11
fxyz x y z
Sz x y x y








22
2
22 22
22 22
22
2cos
22 22 2
22
00
3
32
00
0
,, 1
2
222
16 16 16 sin
cos 1 sin cos sin 0
33 33
SS
SS
xy
f x y z dS x y x y dy dx
xy xy
xy
xydydxxydydxrdrd
xy
dd

















** **
** ** **
**

21.
222
22
,,
:9,03,03,09
fxyz x y z
Sx y x y z



Project the solid onto the yz-plane;
2
9,0 3,0 9.xyy z




2
39 2
222
00 2
9
3
39 3
2
00 0 22
0
3
3
0 2
0
,, 9 1 0
9
33
99
399
3
324 972 arcsin 972 0 486
329
S
y
fxyzdS y y z dzdy
y
z
zdzdy zdy
yy
y
dy
y
















** **
** *
*

22.
222
22
,,
:9,03,0
fxyz x y z
Sx y x z x


Project the solid onto the xz-plane;
2
9.yx





2
3 2
222
00 2
3
33
2
00 0 22
0
3
333 12 12
232
000 2
,, 9 1 0
9
33
99
399
3
9279 9
39
x
S
x
x x
fxyzdS x x z dzdx
x
z
zdzdx zdx
xx
x
x dx x x dx x x dx
x











** **
** *
***




12
22 2
3
33
3 32
222 2 2
0
00
0
Let , 9 , then 2 , 9 .
2
27 9 9 2 9 81 9 81 18 99
3
u x dv x x dx du x dx v x
xxx xxdx x







*

422. 422 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
23.


,, 3 4
: 1 first octant
,, 1
,,
xyz z y
Sz x y
Gx y z x y z
Gx y z

≤π π

D
Fijk
ijk







11
00
11
00
11 1 1
2
000 0
11 2
2
00
4
3
34
31 4
13 2 3
131 1 22FN F
x
SR
x
x x
dS G dA z y dy dx
xy ydydx
x y dy dx y xy y dx
xxx xdx xdx
π
π
π π
" "D


≤πππ ≤πππ


** ** **
**
** *
**


24.


,,
: 6 3 2 , first octant
,, 3 2 6
,, 3 2
xyz x y
Sz x y
Gx y z x y z
Gx y z

≤π π

DFij
ijk



3
23
2
00
3
2 3
2
2
00
2
2
0
2
2
0
2
3
0
32
3
33
33 3
22
9
4
4
9916
412
43 4 3
FN F
SR
x
x
dS G dA
x y dy dx
xy y dx
x xxdx
xdx
x
x
π
π
"≤ "D







π
≤π
πππ
≤π≤ ≤

** **
**
*
*
*


25.



22
22
,,
:1 , 0
,, 1
,, 2 2
xyz x y z
Sz x y z
Gx y z x y z
Gx y z x y

≤π π /

D
Fijk
ijk







22
22 22
22
21
2
00
1
42
22
00
0
22
22 1
1
1
33
42 4 2
FN F
SR
R
R
R
dS G dA
xyzdA
x yxydA
xydA
rrdrd
rr
dd





"≤ "D







** **
**
**
**
**
**


26.




222
22
22
22 22
,,
: 36 first octant
36
,, 36
,,
36 36
xyz x y z
Sx y z
zxy
Gx y z z x y
xy
Gx y z
xy xy


≤ππ
≤π π π
D
ππ ππ
Fijk
ijk

22
22 22 22
36
36 36 36
xy
Gz
xyxy xy
"D
ππ ππ ππ
F

26
0022 236 36
improper 108
36 36
FN F
SR R
dS G dA dA r dr d
xy r

"≤ "D≤ ≤ ≤
ππ π** ** ** * *

x
1
1
yx=+ 1−
R
y
y
x
6
6
6
3
3
2
z
R
y = 3 −
3
2
x
1
1
2
3
4
5
6
23456
y
x
R
x
2
+ y
2
= 6
2
1
R
x
2
+ y
2
≤ 1
y
x

423. Section 15.6 Surface Integrals 423
© 2010 Brooks/Cole, Cengage Learning
27.


2222
22
,, 4 3 5
:,4
,,
,, 2 2
xyz
Sz x y x y
Gx y z x y z
Gx y z x y



D
Fijk
ijk



+,
22
00
22
332
0 0
2
0
2
0
85
32
64
3
64
3
865
8cos 6sin 5
cos 2 sin
cos 16 sin 10
sin 16 cos 10 20
FN F
SR R
dS G dA x y dA
rr rdrd
rrrd
d








" "D






** ** **
**
*
*

28.



22 2
22 2
22 2 22 2
22 222
22 2
22 2 22 2 22 2
,, 2
:
,,
,,
332
2
xyz x y z
Sz a x y
Gx y z z a x y
xy
Gx y z
axy axy
x yxya
Gaxy
axy axy axy

≤ππ
≤π π π
D
ππ ππ

"D
ππ ππ ππ
Fijk
ijk
F



222 22
2
0022 2 22
3
22
2
00 00 22 22
22 32
22 2 2 2 2 2 2
00
0
0
22
32
00
332 32
32
2
32
3
2
32 0
3
FN F a
SR R
aa
a
axya ra
dS G dA dA r dr d
axy ar
rr
dr d a dr d
ar ar
ra r a r d a a r d
ad a ad








"≤ "D≤ ≤
ππ π
≤π
ππ



≤ππππ π ππ


≤π≤
** ** ** * *
** **
**
**

29.



22
22
,,
:16 , 0
,, 16
,, 2 2
xyz x y y z
Sz x y z
Gx y z z x y
Gx y z x y

≤ππ ≤

D
Fijk
ijk



22 22222
2222216 216FGxxy yzx xyy xyxy xy"D


+,
24
22
00
4
44
22 2
2 2
000
0
2cossin 16
cos sin 8 192 128 cos sin 192 64 sin 384
42
FN F
SR
dS G dA
rr rdrd
rr
rd d




"≤ "D




** **
**
**

(The flux across the bottom 0z≤is 0)
1
−1
−11
x
xy
22
+4≤
R
y
a
a
−a
−a
x
xya
222
+≤
y

424. 424 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
30.
2
,, 4xyz xy z yzFijk
S: unit cube bounded by

0, 1, 0, 1, 0, 1xxyyzz

1:SThe top of the cube


11
001
1
2
,1
1
S
z
dS y dy dx

"
** **
Nk
FN


2:SThe bottom of the cube


11
002
,0
00
S
z
dS y dy dx

"
** **
Nk
FN


3:SThe front of the cube


11
003
,1
41 2
S
x
dS y dy dz

"
** **
Ni
FN


4:SThe back of the cube


11
004
,0
40 0
S
x
dS y dy dx

"
** **
Ni
FN


5:SThe right side of the cube

11
2
005
1
3
,1
S
y
dS z dz dx

"
** **
Nj
FN


6:SThe left side of the cube

11
2
006
1
3
,0
S
y
dS z dz dx

"
** **
Nj
FN

So,

5111
2332
020 .
S
dS" **
FN
31.
22
:1
yz xz xy
Sz x y


Eijk




22 22
2
11
2
1122
,,
11
2
330
1
xy
SR
R
x
RRxdS g x y g x y dA
xy
yz xz xy dA
xy xy
xyz
xy dA xy dA xy dy dx
xy


" "

"







** **
**
** ** **
EN E i j k
ijk i jk
y
1
1
1
x
z

425. Section 15.6 Surface Integrals 425
© 2010 Brooks/Cole, Cengage Learning
32.

22
2
:1 ,
xy z
Sz x y gxy

Eij k




22 22
22
22 22
22 22
22 2
21
0022 22 2
,,
2
11
2
11
21 228
3111
EN E i j k
ij k i jk
xy
SR
R
R
RRdS g x y g x y dA
xy
xy z dA
xy xy
xy
zdA
xy xy
xy xy xy r
dA dA r dr d
xy xy r


" "

"










** **
**
**
** ** * *

33.
22
,0zxy za


22
2
22 22
4422
2
22 22 3 2
00
122
22
22 2 2
422 2
SR R
a
z
SR
xy
m k dS k dA k dA k a
xy xy
ka k a a a m
IkxydSkxydAkrdrd ka










** ** **
** ** * *

34.
222 2
22 2
xyza
zaxy





22
22 2 22 2
2
22 2
0022 2 22
0
221
22 224
SR
a
a
R
xy
mkdSk dA
axy axy
ar
kdAkadrdkaarka
axy ar












** **
** * *




3
2
22 22
0022 2 22
2 2 2 use integration by parts
a
z
SRar
IkxydSkxy dAka drd
axy ar

** ** * *

Let

12
222 22
,,2,.u r dv r a r dr du r dr v a r



32
22 2 2 2 3 2 2 2
02222
22224
3333
a
ka r a r a r ka a a ka a m





35.

22 2
22
,0
,, 1
xya zh
xyz
yax
-




Project the solid onto the xz-plane.




2
2
22 2 22
00 22
3333
00 0 22
0
414 1 0
1
44arcsin42
2
ha
z
S
a
ha h x
I x y dS x a x dx dz
ax
x
adxdzadzahah
aax










** **
** *

x
y
a a
h
z

426. 426 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
36.
22
,0zx y zh
Project the solid onto the xy-plane.





2
22 22 2 2
2
2 32 52
22
00
32
32
1144
12
14 2 14 14
12 120 120
14
10 1 4 1 4 6 1 1
60 60 60
hhx
z
Shhx
h
IxydS xyxydydx
h
rrrdrd h h
h
hh hh












** * *
**

37.

22
:16 ,0
,, 0.5
Sz x y z
xyz z
/
π
Fk





22
24 2
2
00 0
,, 0.522
0.5 0.5 16
0.5 16 0.5 64 64FN F i j k k i j k
xy
SR R
RRdS g x y g x y dA z x y dA
zdA x y dA
rrdrd d

-- -
--
---" " "

** ** **
** **
** *

38.

22
:16
,, 0.5
Sz x y
xyz z

π
Fk



22 22
22
24 2
2
00 0
,,
0.5
16 16
0.5 0.5 16
64 64
0.5 16 0.5
33
xy
SR
R
RRdS g x y g x y dA
xy
zdA
xy xy
zdA x y dA
rrdrd d

--
-
--
-
--" "

"



** **
**
** **
** *
FN F i j k
kijk

39.
The surface integral of f over a surface S, where S is
given by
,,zgxyπ is defined as



0
1
,, lim , , .
n
iii i
S
i
fxyzdS fx y z S
5
π
π5F** (page 1112)
See Theorem 15.10, page 1112.
40. A surface is orientable if a unit normal vector N can be
defined at every nonboundary point of S in such a way
that the normal vectors vary continuously over the
surface S.
41. See the definition, page 1118.
See Theorem 15.11, page 1118.
42. Orientable
43. (a)
(b) If a normal vector at a point P on the surface is
moved around the Möbius strip once, it will point in
the opposite direction.
(c)
,0 4cos 2 4sin 2uuurij
This is circle.

(d) (construction)
(e) You obtain a strip with a double twist and twice as
long as the original Möbius strip.
z
y
x
h
x
y6
4
6
−6
−4
−6
z
y
x
4
−4
2
2
−2
z

427. Section 15.7 Divergence Theorem 427
© 2010 Brooks/Cole, Cengage Learning
44. (a)

2
2
112 2 4 12
210
u
v
uv u
v
uuuv v
v


'

rij k
rij
ijk
rr i j k


uv'rris a normal vector to the surface.
(b)

22
,uv u u v u v Fi jk


32 323 2
24 1224 4 22Fr r
uv uuvuv uv vuuvuvvuvuv"'
(c)


2
2
3
2 2 not in domain
1
1
4
xuv
uu
yuv
v
zu
<
=
>

=

?

(d) Calculate
uv
uv'
"
'rr
F
rr
at P.




3, 1, 4 4 3
2, 1 4 8 3
89
1373789
16 24 3
8989 89Fijk
rr i j k
rr
rr
F
rr
uv
uv
uv
uv

'
'
'
"
'

(e)


12 1
32 3 2 32
10 1 26 44
24 4 22 84 6
33
FN F r r
uv
SRdS dA
v
u u v uv v u v uv du dv v v dv

" "'



** **
** *

Section 15.7 Divergence Theorem
1. Surface Integral: There are six surfaces to the cube, each with
1.dS dA

2
1
22 24
002
3
3
004
5
3
00
0, , , 0 0
,, ,
0, , 2 , 0 0
,, 2, 2 2 2
0, , 2 , 0 0
,, 2, 2 2 2
S
aa
S
S
aa
S
S
aa
S
zzdA
za z adA adxdya
xxdA
xa x a dy dz a dy dz a
yydA
ya y adA adzdx a
"
"
"
"
"
"
**
** **
**
** **
**
***
NkFN
Nk FN
Ni FN
Ni FN
NjFN
Nj FN
6*

So,
4334
22 .FN
s
dS a a a a"**


Divergence Theorem: Because div 2 ,z F the Divergence Theorem yields

24
000 00
div 2 .
aaa aa
Q
dV z dz dy dx a dy dx a*** *** **
F

428. 428 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
2. Surface Integral: There are three surfaces to the cylinder.
Bottom:
2
0, ,zz"NkFN

1
00
S
dS**

Top:
2
,,zh z"NkFN


22 2
2
Area of circle 4
S
hdS h h **

Side:
,2cos2sin ,0 2,0rijkuv u u v u v h

2sin 2cos ,ri
j k
uv uur

2cos 2sinrr i j
uv uu'


22
8cos 8sin
uv uu"' Fr r


2
22
003
8cos 8sin 0FN
h
S
dS u u du dv

" ** **

So,
22
04 0 4 .FN
s
dS h h"**


Divergence Theorem: div 222 2zz F

22
000
224.
h
Q
zdV zrdzdrd h


)
*** * * *

3. Surface Integral: There are four surfaces to this solid.

0, ,zz"NkFN

1
00
S
dS**


0, , 2 ,NjFNyyzdSdAdxdz"


66 6
2
00 02
636
z
S
z dS z dx dz z z dz

** ** *


0, , 2 ,
x y x dS dA dz dy" NiFN


362 3
2
00 03
62 9
y
S
y dS y dz dy y y dy

** ** *


2253
26, , ,6
66
xyz
xy z dS dA

"
ijk
NFN


362 3
2
00 04
253 18 11 909020 45
y
S
xyzdzdy x ydxdy y ydy

** ** *

So,
0 36 9 45 18.
s
dS"**
FN

Divergence Theorem: Because div 1, F you have



11
Volume of solid Area of base Height 9 6 18.
33
Q
dV'***

y
x
h
2
2
z
x
y
6
3
6
z

429. Section 15.7 Divergence Theorem 429
© 2010 Brooks/Cole, Cengage Learning
4. ,,xyz xy z x yFijk
S: surface bounded by the planes
4, 4yz x≤≤π and the coordinate planes

Surface Integral: There are five surfaces to this solid.

0, ,NkFNzxy"


44 4
00 01
48 64
S
x y dS x y dy dx x dx ** ** *

0, ,yz≤≤π"≤π
NjFN


2
44 4
00 02
4 32
23x
S x
z dS z dz dx dxπ π
π≤ π ≤π ≤π
** ** *

4, ,yz≤≤"≤
NjFN


2
44 4
00 03
4 32
23x
S x
z dS z dz dx dxπ π
≤≤≤
** ** *

0, ,xxy≤≤π"≤π
NiFN

44
004
00
S
xy dS dSπ≤ ≤** **


+,
1
4, , , 2
22
xzxyxydSdA

"
ik
NFN


+,
44
00
51
2 128
2
S
xy x y dA xy x y dy dx ** **

So,
32 32
64 0 128 64.
33
FN
S
dS"**


Divergence Theorem: Because div ,y≤ F you have

444
000
div 64.
x
Q
dV ydzdydx
π
≤≤*** ***
F
5.
2
,, 2
Fxyz xz yz z ij k

Surface Integral: There are two surfaces.
Bottom:
2
0, , 2zz≤≤π"≤πNkFN

2
1
200
SR
dS z dA dA"≤π ≤ ≤** ** **
FN
Side: Outward unit normal is

22
22
441
xy
xy

≤

ijk
N


222
221
222
441
xzyzz
xy
"


FN



22 2
22
21 21 2 2
22 2 3
00 00 0
22
1
21 21 2 2
2FN
SS
dS x y z z dA
rr rrdrd rrdrd d


"




** **
** ** *


Divergence Theorem: div 4 6zz z z F


2
211
000
21 21 2 2
224
00 00 0
div 6
331
31 3 6 3
222F
r
Q
dV z r dz dr d
r r dr d r r r dr d d




π
≤



*** * * *
** ** *

x
y
1
1
−1
−1
1
z

430. 430 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
6.
22
,,xyz xy yx eFijk
S: Surface bounded by
22
zxy and 4z

Surface Integral: There are two surfaces.
Top: 4, ,
NkFNze"


1
area circle 16
S
dS e e " **
FN
Side:
22
22 22
(, ) , ,
xy
xy
zgxy x yg gxyxy




22 22 2 2 2 2
24
0022 2222
22
2
0
2cos sin
2048 sin cos 512
816
55
FN
SS
xy xy r r
dS e dA e r dr d
rxy xy
ed e





"





** ** **
*

So,
512
.
5
S
dS "**
FN

Divergence Theorem: div
22
yxF

4
5
244 24 2 2
2344
00 00 0 0
0
256 512
div 4
555
F
r
Q
r
dV r r dz dr d r r dr dz r dz dz




*** * * * * * * *

7. Because div 2 2 2 ,
xyzF you have



000
2232234
000 0
div 2 2 2
22 2 2 2 3.F
aaa
Q
aa a a
dV x y z dz dy dx
ax ay a dy dx a x a dx a x a x a


*** ***
** *

8. Because div
2
223,
xzxyF you have




23
000 00
42 3 635
0 2
3
3321
3234
div 2 2 3 2 3
22.F
aaa aa
Q
a
dV xz xy dz dy dx xa a xya dy dx
xa a xa dx a a a

*** *** **
*

9. Because div 222 2,xxxyz xyz F




2/2
2
00 0
2/2
53
00 0
2
52
2
00 0
0
div 2 2 sin cos sin sin cos sin
2 sin cos sin cos
1sin
sin cos 0.
222
a
QQ
a
aa
dV xyz dV d d d
ddd
dd d




-.-.-.- ..-
- ...-
-
-- -
J




*** *** * * *
** *
** *
F

10. Because div ,yzy z F you have

22 222 22
2
22
0000
23 224 4 4
222
00 0 0
0
div
.
22 4 8 8 4
aax axy aar
aax
Q
a
a
dV zdzdydx zrdzdrd
ar r ar r a a
dr d d d











*** * * * * * *
** * *
F

11. Because div 3, F you have



34
3
3 3 Volume of Sphere 3 3 108 .
Q
dV

***

431. Section 15.7 Divergence Theorem 431
© 2010 Brooks/Cole, Cengage Learning
12. Because div ,xzF you have

2
4
22
52 4 52 52
2
02 4 02 02
2
4
2
00.
2
y
y
y
Q y
x
xz dV xz dx dy dz dy dz dy dz








*** ** * ** **

13. Because div 1 2 1 2 ,yy F you have


5
2
75 25 75 7 3/2
22
2
05 25 05 0
5
4
22425250.
3y
y
Q
y dV y dx dy dz y y dy dz y dz






*** * * * * * *

14. Because div
22
z
yxeF , you have





2
16 256 8
22 22
222
02561/2
2168 216
2384/2
00 /2 00
2
88
0
1
2
131,052 262,104
55
8
100 200 .
x
zz
xxy
Q
zr
r
x y e dV x y e dz dy dx
rerdzdrd rre rredrd
ed e








*** * * *
*** **
*

15. Because div
22 2
304,
xxxF you have


644 64 6
222 2
000 00 0
4 4 4 4 32 2304.
y
Q
x dV x dz dy dx x y dy dx x dx

*** * * * * * *

16. Because div 3 ,
z zz z
eee eF you have


644 64 6
444
000 00 0
333135185.
y
zz y
Q
e dV e dz dy dx e dy dx e dx e




*** * * * * * *

17. div 4 .yxF Use spherical coordinates.



42
2
00 0
42
32 32 2
00 0
44
22
00 0
4 sin sin sin cos 4 sin
sin sin sin cos 4 sin
1024
8sin 16
3
Q
yxdV ddd
ddd
dd d



-.-. - ..-
-.- . - ..-

- . . - - -


*** * * *
***
** *

18. div 2F
div 2 .
S
QQ
dS dV dV" * * *** ***
FN F
The surface S is the upper half of a hemisphere of radius 2. Because the volume is

314
23
216/3, you have


32
2 Volume .
3
S
dS

" **
FN
19. Using the Divergence Theorem, you have





22
div
,, 6 2 2 4 4 6
4262
div 0.
curl F N curl F
ijk
curl F i j k i
curl F
S
Q
dS dV
xyz y z z x x y
xyz
xy z x yz xz
"
CCC

CCC


** ***

So,
div 0.
Q
dV***
curl F

432. 432 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
20. Using the Divergence Theorem, you have




div
,, sin sin cos cos .
cos sin
curl F N curl F
ijk
curl F i j k
S
Q
dS dV
xy z xz y x yz xy z yz x x z
xyz
xy z yz x xyz
"
CCC

CCC
** ***

Now, div
,, cos sin cos sin 0.xyz z y x z x z y x x z curl F So,

div 0.curl F N curl F
S
Q
dS dV" ** ***

21. See Theorem 15.12.
22. If
div , , 0,xyzF then source.
If
div , , 0,xyzF then sink.
If
div , , 0,xyzF then incompressible.
23. (a) Using the triple integral to find volume, you need
F so that

div 1.
MNP
xyz
CCC

CCC
F
So, you could have
,,or .
x yz FiFj Fk
For
dA dy dz consider ,(,),
xxfyzFi then
22
1
yz
yz
f f
f f



ijk
N and
22
1.
yzdS f f dy dz
For
dA dz dx consider ,(,),yy fxzFj then
22
1
xz
x z
f f
f f



ij k
N
and
22
1.
xzdS f f dz dx
For
dA dx dy consider ,(,),zz fxyFk then
22
1
xy
x y
ff
f f



ijk
N and
22
1.
xydS f f dx dy
Correspondingly, you then have
.
SSSS
dS x dy dz y dz dx z dx dy" ** ** ** **
VFN
(b)
23
00 0 0 0
aa a a a
v x dy dz a dy dz a dz a ** ** *

Similarly,
3
00 0 0
.
aa a a
y dz dx z dx dy a** **

24.
,,
xyz x y zFijk
Divergence Theorem: div 1 1 1 3F

3 3 Volume of cube 3
Q
dV***

Surface Integral: There are six surfaces.

1
2
3
4
5
6
0: , , 0 0
1: , , 1 1
0: , , 0 0
1: , , 1 1
0: , , 0 0
1: , , 1 1
NiFN
Ni FN
NjFN
Nj FN
NkFN
Nk FN
S
S
S
S
S
S
xxdS
xxdS
yydS
yydS
zzdS
zzdS
"
"
"
"
"
" **
**
**
**
**
**

So,
111 3.
S
dS"**
FN

433. Section 15.8 Stokes’s Theorem 433
© 2010 Brooks/Cole, Cengage Learning
25. Using the Divergence Theorem, you have div .
S
Q
dS dV"* * ***
curl F N curl F Let


22 22 22
,,
div 0.
xyz M N P
PN PM NM
yz xz x y
PNPMNM
xy xz yx yz zx zy

CC CC C C

CC CC C C
CC CC CC

CC CC CC CC CC CC
Fijk
curl F i j k
curl F

So,
00.
S
Q
dS dV" * * ***
curl F N
26. If
12 3,, ,
xyz a a aFijk then div 0.F
So,

div 0 0.
S
QQ
dS dV dV" * * *** ***
FN F
27. If
,, ,
xyz x y zFijk then div 3.F

div 3 3 .
S
QQ
dS dV dV V" * * *** ***
FN F
28. If
,, ,xyz x y zFijk then div 3.F

1113
div 3
S
QQQ
dS dV dV dV" * * *** *** ***
FN F
FFFF
29.

2
N
div divN
SS
QQ Q
fD g dS f g dS f g dV f g f g dV f g f g dVD" D DD"D DD"D* * * * *** *** ***

30.


NN N N
22 22
SSS
QQ QfDg gDfdS fDgdS gDfdS
fg f gdV g f g fdV f g g fdV

D D "D D D "D D D
** ** **
*** *** ***

Section 15.8 Stokes's Theorem
1. ,, 2
z
xyz y z e xyzFijk


2
12
z
z
xyz
y z e xyz
xz e yz
CCC

CCC


ijk
curl F
ijk

2.

222
,,
xyz x y xFijk

222
2x
xyz
xyx
CCC

CCC
ijk
curl F j
3.

2
2
2
, , 2 4 arctan
1
28
1
2 4 arctan
Fijk
ij k
curl F j k
xyz z x x
x
xy z x
zx x

CC C


CC C


4.


2
2
2
,, sin cos
sin cos
sin cos
Fijk
ijk
curl F
ik
xyz x y y x yz
xyz
xyy xyz
zyxxy

CCC

CCC

434. 434 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
5.
22 22
,,
xy yz
xyz e e xyz

Fijk



22 22
22 2 2
22 2 2
22
22
xy yz
yz xy
yz xy
xyz
eexyz
xz ze yz ye
zx e yz ye



CCC

CCC


ijk
curl F
ij k
ij k

6.

22
,, arcsin 1xyz y x yFijk

22
22
22
arcsin 1
1
2
11
1
2
11
xyz
yxy
x
y
xy
x
y
xy
CCC

CCC










ijk
curl F
ik
ik

7.
22
:9,0,0Cx y z dz
Line Integral:

CC
dydxxdy" **
Fr

3cos , 3sin , 3sin , 3cosx t dx t dt y t dy t dt


2
0
2
0
3sin 3sin 3cos 3cos
918
Fr
C
dttttdt
dt



"

**
*

Double Integral:

22
,9 , 2, 2
xy
gxy x y g xg y


2
2 2 area circle 18
SR
dS dA

"
** **
curl F k
curl F N



8. In this case,
,,MyzNxzPxy and C is the circle
22
1, 0, 0.xy z dz


CC C
dyzdxxzdyxydzydxxdy" ** *
Line Integral: F r
Letting
cos , sin ,
x ty t you have sin , cosdx t dt dy t dt and
2
22
0
sin cos 2 .
C
y dx x dy t t dt

**

Double Integral: Consider

222
,, 1.Fx yz x y z
Then
222
222
.
2
Fxyz
xyz
F xyz
D

D
ijk
Nijk

Because
222 2
1, ,
2
x
x x
zxyz
zz


and
22
22
1
,1 .
y
yxy
zdS dAdA
zzzz


Now, because 2 ,
curl F kyou have

1
2 2 2 Area of circle of radius 1 2 .
curl F N
SRR
dS z dA dA
z

"


** ** **

9. Line Integral:
From the figure you see that

1
2
3:0, 0
:0, 0
:0, 0
Cz dz
Cx dx
Cy dy





20120
0201212 3
0Fr
CC CC C
d xyz dx y dy z dz y dy y dy zdz z dz y dy y dy z dz z dz" ** ** *****


Double Integral:
xyxzcurl F j k
Letting
12 6 6 , , 6 .
x yzxygxyg g

+
,


22 22
2
00 00
2 2
22
00
66 6
61266 12126
6126 0curl F N curl F i j k
SR R
xx
x
dS dA xy xz dA
xy x x y dy dx xy x x dy dx
xy xy x y dx


" "



** ** **
** **
*

x
y 4 4
9
z
C
x y
z
12
4 4
C
2 C
3
C
1
(2, 0, 0) (0, 2, 0)
(0, 0, 12)

435. Section 15.8 Stokes's Theorem 435
© 2010 Brooks/Cole, Cengage Learning
10. Line Integral: From the figure you see that

1
2
2
2
3
2
4:0,0, 0
: , 0, 0, 2
:, , 0
:,,0,2.Cy z dy dz
C z y x dx dz y dy
Cy az ady dz
Cz yx adx dz ydy




So,


22 2 3 4 2 3
12 3 4
00 0
34 2342533
00
02 2
22 1.
Fr
CC CC CC
a a
aaa
d z dx x dy y dz dx y dy a dx a dy y dy
y dy a dx a y dy a x a y a a a a
"


** ****
***


Double Integral: Because S is given by
2
0,yz you have

2
2
14
y
y



jk
N
and
2
14 .dS y dA
Furthermore, 2 2 2 .
yzxcurl F i j kSo,



3
00 00
4425332
00
42 42 42
21.
aa aa
x
SR
a a
dS yz x dA yz x dA y dydx
aaxdx axax aaaa
"

** ** ** **
*
curl F N

11. These three points have equation:

2.xyz
Normal vector: Nijk
32 curl F i j k


6 6 area of triangle in -plane
62 12
SR
dS dA xy"
** **
curl F N

12. Let
0, 0, 0 , 1, 1, 1 ,AB and 0, 0, 2 .C Then ,ABUijk

and 2 ,ACVk

and

22
.
22 2
'

'UV i j i j
N
UV

So,
,,Fx yz x y and
2.dS dA Because
22
2
,
x
xy


curl F k you have 00.
SR
dS dS" ** **
curl F N
13.

22
22
222
2
,1 , 2, 2
xy
yz
xyz
zxy
zGxy x yG xG y
CCC

CCC

ijk
curl F i j k


22
2
11
23
2
11
1
1
22
1
1
22222 4 41 2
444 42
41 2arcsin 1 2curl F N i j k i j k
SR R
x
x
dS y z x y dA xy y x y dA
xy y x y y dy dx
xdx x x x





" "






** ** **
**
*

y
x
a
a
a
2
C
1
C
2
C
3
C
4
z

436. 436 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
14.



22
22
,, 4 4 , :9 , 0
444
,, 9
,, 2 2
xyz xz y xy S x y z
xxy
Gx y z x y z
Gx y z x y



D
Fijk
curl F i j
ijk



2
39
222
2
39
3 32
22 2
3
16
3
8244 888
16 9 9 0
x
SR x
dS x y x y dA x xy y dy dx
xx xdx



"


** ** * *
*
curl F N
15.

2
22
22 22
2
,4 , ,
44
xy
z
xyz
zyz
xy
zGxy x yG G
xyxy
CCC

CCC



ijk
curl F j



22 22
22 2
24
2
2422 22
2
44
242
20
44curl F N j i j k
SR
x
RR x
xy
zdA
xy xy
yxyyz
dA dA y dy dx
xy xy



" "






** **
** ** * *

16.
22 22
,, , : 4xy z x z xyz S z x y Fijk



22
22
22 22
2
,, 4
,,
44
xz z yz
xy z
xz xyz
Gx y z z x y
xy
Gx y z
xy xy
CC C

CC C


D

ij k
curl F i j
ijk






2
22 22
2
24
222
2
24
2
4
3
2
2
2
2
4
2
22 2 2 2
2
22 2 2
2
44
22
2
3
2
24 44 4 4
3
88
444 4
33
SR
x
Rx
x
x
zx x yz
dS dA
xy xy
x x y dA x x y dy dx
y
xy xy dx
xxxx x xdx
xxxx x







"















** **
** * *
*
*
curl F N


2
2
2
32
22 2 2
2
81 4 81
244 16arcsin 4 4 4arcsin
38 2 3 32 2
141 4
82 8 20
333 3
dx
xx
xx x x x x











*

437. Section 15.8 Stokes's Theorem 437
© 2010 Brooks/Cole, Cengage Learning
17.
22
, , ln arctan
x
xyz x y
y
Fijk




22 2222
22
1 2
1
1 2 ln arctan 1
ijk
curl F k k
y yy
xyz xyxy xy
xy xy

CCC


CCC




:923Sz x y over one petal of 2sin2r in the first octant.


,, 2 3 9
,, 2 3
Gx y z x y z
Gx y z

D ijk



22sin2
22 2
00
2
3
24sincos 2
2
00 0
022sin
8sin 8
2 sin 8 sin cos
33
curl F N
SR
yr
dS dA r dr d
xy r
dr d d





"




** ** * *
** *

18.

22
,, 2 3xyz yz y x y Fijk


22
22
23
yyxz
xy z
yz y x y
CC C

CC C

ij k
curl F i j k

:Sthe first octant portion of
22
16xzover
22
16xy



2
2
,, 16
,,
16
ik
Gx yz z x
x
Gx yz
x

D





2
22
2
16
2
416 4
222
00 0 22
0
4
3
4 32
22 2
0
022
16
16 16
2
16 16
16 16
16464
16 16 16 16 64
3333
curl F N
SR R
x
x
xy xy
dS z dA x dA
xx
xy x
xdydx y xy dx
xx
x
xx xdx x x



"










** ** **
** *
*
64
3





19.

2
,,2,0
curl F j k
xy
xy xz
zGxy xG xG




5
2
00
2
4
curl F N j k i k
aa
SR R a
xy xz x dA xz dA x x dy dx" "
** ** ** **

20.

,,Fijk
ijk
curl F j k
xyz xyz y z
xyxz
xyz
xyz y z

CCC

CCC

S: the first octant portion of
2
zx over
22 2
.
xya You have
2
22
and 1 4 .
14
x
dS x dA
x



ik
N



22
33
00
32 52
322 222 22 5
0
0
12 2
315 15
curl F N
aax
SRR
a
a
dS xz dA x dA x dy dx
xaxdx xax ax a

"



** ** ** **
*

438. 438 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
21. ,, 2
11 2
xyz
xyz

CCC

CCC

Fijk
ijk
curl F 0
Letting ,Nk you have
0.
S
dS"**
curl F N
22.

22
,,
:1
0
xyz z y
Sx y
xyz
zy


CCC

CCC

Fik
ijk
curl F i j
Letting
, 0"N k curl F N and
0.
S
dS"**
curl F N
23. See Theorem 15.13.
24. curl F measures the rotational tendency. See page 1135.
25. (a)
+
,rcurl N
CS
fgd fg dSD" D "***
(Stokes's Theorem)



22
22
ijk
ijk
curl
i
ggg
fg f f f
xyz
fg
xyz
fgx fgy fgz
gfg gfg
ff
yz y z zy z y
gfg g
ff
xz x z
CCC
D
CCC
CCC
D
CCC
CC CC CC
CCC CCC

CC C C CC C C
CCC C

CC C C
22
j
k
i
fg
zx z x
gfg gfg
ff
xy x y yx y x
fg fg f
yz zy x
CC

CC C C
CCC CCC

CC C C CC C C
CC CC CC

CC CC C
j k
ijk
gfg fgfg
zzx xyyx
fff
fg
xyz
ggg
xyz
CC CC CC

CCC CCCC
CCC
D'D
CCC
CCC
CCC

So,
+
, + , .rcurl N N
CS S
fg d f g dS f g dSD" D " D'D "*** **

(b)
using part a
0 because .
rN
0
CS
ff d f f dS
ff
D" D'D"
D'D***

(c)



using part a
0
rrr
NN
NN
CCC
SS
SS
fg gf d fg d gf d
f g dS g f dS
f g dS f g dS
DD " D " D "
D'D" D'D"
D 'D " D 'D " ***
** **
** **

439. Section 15.8 Stokes's Theorem 439
© 2010 Brooks/Cole, Cengage Learning
26.
22
,, , ,, , : 4fxyz xyz gxyz z S z x y
(a)



,,
,, ,,
2cos 2sin 0 , 0 2
,, ,, 0
k
k
rijk
r
C
gxyz
fxyz gxyz xyz
ttt t
fxyz gxyz d

D
D

D"

*

(b)


,,
,,
001
f x y z yz xz xy
gx y z
fg yzxzxy xz yz
D
D
D'D
ijk
k
ijk
ij


22 22
22
22 22 22
44
2
1
44 4
xy
xy xy
xy
dS dA dA
xy xy xy








Nijk





22
22 22 22
22
22
222 2
3
22 2
00 0 22
0
2
,, ,,
444
2
4
2cos sin 21
sin 2 0
244
N
SS
S
xz yz
f x y z g x y z dS dA
xy xy xy
xy
dA
xy
r r
r d dr dr
rr





D ' D "









** **
**
** *

27. Let ,abcCijk then

11 1
2
22 2
C r r curlC r N CN CN
CS SS
ddSdSdS'" '" " "*** ****

because
abc bzcy azcx aybx
xyz
'
ijk
Cr i j k
and

22.abc
xyz
bz cy cx az ay bx
CCC
'
CCC

ijk
curl C r i j k C
28. (a)
,,
yz
xyz e

Fi
From the figure you have




11 1
22 2
33 3
44 4:, 01,
:,01,
:1,01,
:1,01,
Crt t t
Crt t t
Crt t t
Crt t t




iri
ij r j
ij r i
j rj


11
010
1234
0012 3 4
001F r Fr Fr Fr Fr
CC C C C
d dtdtdtdtedtedte

" " " " "
** * * * * *

Double Integral:
yz yz
ee

curl F j k

,0,Gx y z Nk

11 1 1
000 0
1curl F N
yyyy
SR
dS e dA e dx dy e dy e e "
** ** ** *

x y
z
(0, 1, 0)
C
2
C
3
C
1
C
4
(1, 1, 0)
(0, 0, 0)
(1, 0, 0)
1
1 1

440. 440 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
(b)
22 2
,,xyz z x yFi j k
From the figure you have

:2cos2sin4,02.Ct t t t rijk


22
23
00
16 2 sin 4 cos 2 cos 0 32 sin 8 cos 0Fr
CC
d M dx N dy P dz t t t dt t t dt

"

** * *

Double Integral: 222yzxcurl F i j k


22
,,2,2
xyzgxy x y g xg y




22
22
2
00
2
0
16 128
35
222 22
44 2
4sin cos 4sin 2cos
16 sin cos sin 0
SR
R
dS y z x x y dA
xy y x y x dA
rr rr r rdrd
d




" "


≤π π π

≤π π π ≤
** **
**
**
*
curl F N i j k i j k

29. Let S be the upper portion of the ellipsoid

222
44,0xyz z/
Let

:2cos,sin,0,02,Ct t t t r be the boundary of S.
If ,,MNP≤F exists, then






22
22
00
0by i
Stokes's Theorem
by iii
sin 2 cos 1
, , 0 2 sin , cos , 0 2 sin 2 cos
44 4
curl F N
Fr
Gr
S
C
C
dS
d
d
tt
t t dt t t dt


≤"
≤"
≤"
π
"**
*
*
**

So, there is no such F.
Review Exercises for Chapter 15
1.
22 2 2
,, 2
12 5
xyz x
xx


Fijk
F


2.

2
,2
14
xyy
y
≤π

Fij
F


3.


22
,, 2
,, 4 2
fxyz x xy z
xyz f x y x z

D Fijk

4.



2
22
,,
,, 2
2
yz
yz yz yz
yz
fxyz xe
xyz xe xze xye
xe xz xy
≤


Fijk
ijk

5. Because
2
1,My x NxCC≤π ≤CC Fis conservative.
From
2
MUxyx≤C C ≤π and
1,N Uy x≤C C ≤ partial integration yields
Uyxhy and Uyxgx which
suggests that

,.Uxy yx C
6. Because
2
1,My y NxC C ≤π #C C Fis not
conservative.
x
y 2 2
4
z
C
−2 −2
1
2
1
2
z
y
x
S: z = 4 − x
2
− 4y
2
C
x
y
3
2
3
4
2
z
x
5
4
3
24
2
−2
−1
−3
−4
−5
y

441. Review Exercises for Chapter 15 441
© 2010 Brooks/Cole, Cengage Learning
7. Because 2
M
xy
y
C

C
and 2,
N
xy
x
C

C
Fis conservative.
From
22U
Mxyx
x
C

C
and
22
,
U
N xy y
y
C

C
partial integration yields

3
22
1
23
x
Uxy hy

and


3
22
1
.
23
y
Uxy gx

So,

3
3hy yand
3
3.gx xSo,


33
22
1
,.
233
xy
Uxy xy C

8. Because
2
6sin2 ,My y x NxCC CC Fis
conservative. From
3
2sin2MUx y xC C and

2
31cos2,N Uy y xC C you obtain

3
cos 2Uy xhy and

3
1cos2Uy xgx which suggests that

3
,,hy y gx C and

3
,1cos2.Uxy y x C
9. Because 8
M
xy
y
C

C
and 4,
N
x
x
C

C
,
MN
yx
CC
#
CC
so F is
not conservative.
10. Because

4,
2,
6,
MN
x
yx
MP
z
zx
N P
y
zy
CC

CC
CC

CC
CC
#
CC

F is not conservative.
11. Because

22
22
11
,,
,
MNMP
yyz xzyz x
Nx P
zyz y
CCCC

CCCC
CC

CC

F is conservative. From

22
1
,,
UUxUx
MNP
xyz y y z z yz
CCC

CCC

you obtain



,, ,,
,,, .
xx
UfyzUgxz
yz yz
xx
Uhxyfxyz K
yz yz



12. Because

sin , cos ,
MNM P
zyz
yxz x
CCC C
#
CCC C

F is not conservative.
13. Because

222
,, :
xyz x xy xz Fijk
(a)
2
div 2 2
xxy x F
(b)
2
222
2
ijk
curl F j k
xzy
xy z
xxyxz
CC C

CC C

14. Because

22
,, :xyz y zFjk
(a) div 2 2yzF
(b)
22
0
xy z
yz
CC C

CC C

ijk
curl F 0
15. Because
cos cos sin sin :yy x xx y xyz Fijk
(a)
div sin cosyxx yxy F
(b)
cos sin sin cosxz yz x y y x xz yz curl F i j k i j
16. Because
323:xy y z z x Fijk
(a) div 3 1 1 5F
(b) 2 3curl F i j k
17. Because
22
arcsin :xxy yzFijk
(a)
2
1
div 2 2
1
xyyz
x


F
(b)
22
zycurl F i k

442. 442 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
18. Because
22
sin :xyx yFi j
(a)
div 2 2 sin cos
x yy≤πF
(b) ≤curl F 0
19. Because

22 22
ln ln :
xyxyzFijk
(a)
22 22 22
22 22
div 1 1
F
xy xy
xy xy xy




(b)
22
22
curl F k
xy
xy
π
≤


20. Because
2
:
zz
z
xy
Fijk
(a)
22 22
11
div 2 2
zz
zz
x yxy



F
(b)
11
yx
curl F i j
21. (a) Let
3, 4, 0 1,xtyt t
then
916 5.ds dt dt


1
22 2 2
0
1
3
0
9165
125
125
33
C
xyds t t dt
t


≤≤

**

(b) Let
cos , sin , 0 2 ,xtyt t
then

22
sin cos 1.ds t t


2
22
0
2
C
x y ds dt

**

22. (a) Let
5, 4,0 1,xtyt t then
41 .ds dt≤

1
2
0 20 41
20 41
3
C
xy ds t dt≤≤**

(b)
1
2
3:,0,0 4,
:44,2,0 1, 25
:0,2,0 2,Cx ty t ds dt
C x t y t t ds dt
Cx y t t ds dt



So,

1
23
41 2
2
00 0
0
85
088250165 .
23 3
C
tt
xy ds dt t t dt dt




*** *

23.
1sin, 1cos,0 2xty tt


22
cos , sin , cos sin
dx dy
t t ds t t dt dt
dt dt



+,+ ,
22 22
22 22
00
2 2
00
1sin 1cos 12sin sin 12cos cos
32sin 2cos 3 2cos 2sin 6
C
xyds t t dt t t t tdt
ttdtttt






** *
*

24.
cos sin , sin cos , 0 2 , cos , sin
dx dy
x tt ty tt t t t t t t
dt dt



2 222
22 22 22 3 2 2
0 0
cos sin sin cos cos sin 2 1 2
C
x y ds t t t t t t t t t t dt t t dt



** *

25. (a) Let
3, 3, 0 1.xty t t


11 1
2
000
226333632791818
C
x y dx x y dy t t t t dt t t dt t
** *

(b) Let
3cos , 3sin , 3sin , 3cos ,0 2 .x t y t dx t dt dy t dt t


2 2
0 0
2 2 6 cos 3 sin 3 sin 3 cos 6 sin 3 cos 9 18
C
x y dx x y dy t t t t t t dt dt


** *

26.
cos sin , sin sin , 0 , cos , cos cos sin
2
x t t t y t t t t dx t t dt dy t t t t dt




2
222
0
2 3 sin cos 5 6 2 cos 1 sin 2 3 1.01
C
x y dx x y dy t t t t t t t t dt



**

x
2
3
4
34
(0, 2)
(4, 0)
C
2
C
1
C
3
y=x+ 2
1
2
y
−

443. Review Exercises for Chapter 15 443
© 2010 Brooks/Cole, Cengage Learning
27.




33
2
2
2
2 22
33
0
2,cossin,0
2
3cossin
3sincos
9
22cossin
5
C
C
xydst a t a t t
xt a t t
yt a t t
a
x y ds a t a t x t y t dt



"
"
" " *
**
rij
28.



232
12
4
222 243 2
0
,0 4
3
1, 2 ,
2
9
1 4 2080.59
4
rijk
C
ttt t t
xt yt tzt t
xyzds ttt t tdt



**

29.
,3sin
fxy x y

: 2 from 0, 0 to 2, 4Cy x




2,0 2
2
5
tt t t
t
t


rij
rij
r

Lateral surface area:


+,

2
0
2
0
2
0
,3sin25
53sin3
1
53 cos3
3
11
56 cos6
33
5
19 cos 6 13.446
3
C
fx y ds t t dt
tdt
tt











**
*

30.
,12
fxy x y


2
: from 0, 0 to 2, 4Cy x




2
2
,0 2
2
14
ttt t
tt
tt



rij
rij
r

Lateral surface area:


2
22
0
, 12 1 4 41.532
C
fxyds t t t dt**

31.
,2
xyxy xyFij



22
,0 1
22
tt t t
ttt

rij
rij



1
22 2 2
0
1 1
56
00
22 2
61
C
dttttttdt
tdt t
"


**
*
Fr

32.



2
2
2
0
0
4sin 3cos
4cos 3sin 4cos 3sin ,0 2
7sin
12 7 sin cos 12 24
2
C
dttdt
tt ttt
t
dttdtt








"

**
rij
Fij
Fr

33.


2
2
0
2sin 2cos
2cos 2sin ,0 2
2
C
dttdt
tttt
dtdt






"
**
rijk
Fijk
Fr
34.


2
2
22
2
2
2
0
0
2, 2, 4 ,0 2
2
4
42 4 4 2 0
222
2
C
xtytzttt
t
ddt
tt
ttt tt t
t
dtdt t






"


**
rij k
Fijk
Fr

444. 444 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
35. ,,xyz y z x z x y Fijk
Curve of intersection:
22 2
,, 2xtytzt t t



2
2, 0 2
4
ttt t t
tt

rijk
rijk



22 2
22 2 32
000
2 2 2 4 12 2 4 36Fr
C
dttttttdt ttdttt "
** *

36. Let
2
2sin , 2cos , 4sin ,0 .xty tz tt




23
0 0 16
3
2 cos 2 sin 8 sin cos
04 2sin
8 sin 16 sin cos 8 cos sin 16
C
dttttdt
t
dtttdttt




"

**
rij k
Fij k
Fr

37.


22
1
2
For , , 0 2
For 2 , 2 4 2 , 0 2
yx t tt t
yxt t t t


rij
rij




22 22 22
12
100 4
33
32
CC C
xydx x y dy xydx x y dy xydx x y dy
***

38.


2
22
2 cos 2 sin 2 sin 2 cos , 0
44
CC
C
dxydxyxdy
t ttt ttt t
d


"

" **
*
Fr
rij
Fr

39.
xyFi j is conservative.




4, 8
232 32
0, 0
812 1 2
23 2 3 3
Work 16 8 3 4 2xy


40.

2000 5280 25
10 sin 10 cos 10 sin 10 cos , 0
2332
rij kijkttt t tttt



2
0
20
25
10 cos 10 sin
33
500 250
mi ton
33 33
C
dtt
ddt








" "
**
Fk
rijk
Fr

41.


1, 3, 2
22 2
0, 0, 0
26
C
xyzdx x zdy x ydz x yz
*

42.


4, 4, 4
0, 0,11
ln 16 ln 4
C
y dx x dy dz xy z
z

*

43. (a)


1 2
2
0
11 1
22 2 32
000
2132131
3 2 1 2 3 4 1 9 14 5 3 7 5 15
C
ydx xydy t t t dt
tt tt dt t tdtttt



**
**

(b)


44 4
2 2
111 1
212 15
2
C
y dx xy dy t t t dt t t dt t
t



** *

(c)

2
,2
xyy xy f DFij where
2
,.fxy xy
So,

22
4 2 1 1 15.
C
d" *
Fr
x
2
3
4
12 34
1
(2, 4)
C
1
C
2
yx=
2
yx= 2
y

445. Review Exercises for Chapter 15 445
© 2010 Brooks/Cole, Cengage Learning
44. sin , 1 cos , 0 2xa ya
(a)
1
.
2
C
A xdy ydx≤π*

Because these equations orient the curve backwards, you will use




2 2
22
0 0
2
2
22
0
2
2
0
2
2
1
2
11
1cos 1cos sin sin 00
22
12cos cos sin sin
2
22cos sin
2
63.
2
a
Aydxxdy
aadd
a
d
a
d
a
a







≤π




≤ππ
≤≤*
**
*
*

(b) By symmetry,
.x a≤ From Section 15.4,



2 2
23 3
2011 15
1cos 1cos 5 .
22 6 23
C
yydxa d aa
AA a


≤π ≤ π π ≤ ≤**


45.

11
00
11
00
2
21 1
C
NM
y dx x dy dy dx
xy
dy dx
CC

CC
≤π≤
***
**


46.
22
22
00
2
0
2
24
C
xydx x y dy x x dydx
xdx

≤≤***
*


47.

22
22 0
CR
R
NM
xy dx x y dy dA
xy
xy xy dA
CC

CC
≤π≤
***
**


48.
22
22
22
24
00
aax
Caax
a
a
xy dx xy dy y dy dx
dx
π
ππ π
π

≤≤***
*


49.

+,

2
11
2
1
1 1
2
1
1
3
1
1
24
1
2
0
24
CR
x
x
NM
xy dx x dy dA
xy
xxdydx
xy dx
xxdx
xx
π
π
π
π
CC

CC
≤π
≤≤
≤π

≤π ≤


***
**
*
*


50.






32 32
23 23
1111
243 13 132
32 32
11 23 23
11
1
1 32 52 52
13 23 23 23 23
1
1
44
33
8816
3735
2
1110
xx
C
xx
y dx x dy x y dy dx x y y dx
xxdx x x x
ππ
ππ
ππ ππ
π
π



≤π≤ππππ≤

*** *
*


51. , sec cos 1 2 tan sin 2uv u v u v u ri jk

0,02
3
uv





52.
44
,cossin
6
uu u
uv e v e v
ππ
rijk
04,02uv


x
C
1
C
2
2a
π
y
2
2
4
6
4
−4
−2
y
x
z
2 2
2
yx
z

446. 446 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
53. (a) (b) (c)
(d)
The space curve is a circle:
32 32 2
,cossin
42 2 2
uuu



rijk
(e) 3 cos sin 3 cos cos
3 sin cos 3 sin sin cos
u
v vu v u
vu vu v

rij
rijk




22 2 2
22
3 cos sin 3 cos cos 0
3sin cos 3sin sin cos
3 cos cos 3 cos sin 9 cos sin sin 9 cos sin cos
3 cos cos 3 cos sin 9 cos sin
uv vu v u
vu vu v
vu vu vv u vv u
vu vu vv
'



ijk
rr
ij k
ijk


4 2 42 22 4 22
9 cos cos 9 cos sin 81 cos sin 9 cos 81 cos sinrr
uv vu vu vv v vv'
Using a Symbolic integration utility,
22
40
14.44.
uv du dv


'**
rr
(f ) Similarly,
42
00
4.27.
uv dv du

'**
rr

54. :, sin,0 2,0Suv uv uv v u v rijk



,
,cos
11 0 cos cos 2
11cos
u
v
uvuv
uv v
vv
v


'
rij
rijk
ij k
rr i j k


2
2cos 4
uv v' rr

2
2
00 62
sin 2 cos 4 2 6 2 ln
62
S
zdS v v dudv




** **


55. :, cos sin 12 ,0 2,0 2Suv u vu v u u u v rij k




,cossin 32
,sincos
cos sin 3 2 2 3 cos 2 3 sin
sin cos 0
u
v
uvuv v v u
uv u v u v
vv uuuvuuvu
uvu v


'
rijk
rij
ijk
rr i j k



2
23 1
uv uu' rr



22 22 22
2
00 00
cos sin 2 3 1 cos sin 2 3 1 0
S
x y dS u v u v u u du dv v v u u dv du

** * * **

x
y4
3
−4
−3
−2
4
−4
z
y
x
2
2
3
3
4
−4
−4
−3
−2
−1
−3
3
4
z
y
x
2
4
3
−3
−2
−3
−2
−3
−4
−4
3
2
3
4
z
y
x
2
4
3
−3
−2
−3
−2
−4
−4
3
1
3
4
z
x
y3
2
3
−3
−3
−2
z

447. Review Exercises for Chapter 15 447
© 2010 Brooks/Cole, Cengage Learning
56. (a)
22 2
22 2
,0
0
zaa x y za
zxya



(b)


222
22
:,
,
Sgxy z a a x y
xy k x y
-






22 22
22 2 2
22 2 22
22 22
3
22
22 2 23
00 0
,, 1
11
2
111
33
xy
SR
RR
amexyzdS kxyggdA
ax ay
kxy dAka xydA
xy xy
a
ka rdrd ka d ka a





** **
** **
** *


57.
2
,,
xyz x xy zFijk
Q: solid region bounded by the coordinates planes and the plane 23412xyz

Surface Integral: There are four surfaces for this solid.

1
2
2
3
0, , , 0 0
0, , , 0 0
0, , , 0 0
S
S
S
zzdS
yxydS
xxdS
"
"
" **
**
**
NkFN
NjFN
NiFN

234 1 9 29
23412, , 1
416 429
xy z dS dA dA




ijk
N






2
4
6423
2
00
22
6
2
0
6
32
0
43
21
234
4
1
231223
4
1 122 3122 122 122 3122
2122
4323 3 323
1
24 36
6
1
12 36
64 3
FN
SR
x
dS x xy z dA
xxy xydydx
xx x x x x
x xdx
xx x dx
xx
xx

"








** **
**
*
*
6
0
66



Divergence Theorem: Because div 2 1 3 1,xx xF Divergence Theorem yields










6122312234
00 0
61223
00
12 2 3
6
2
0
0
2
6
0
6
32
0
div 3 1
12 2 3
31
4
13
3112 2
42
1 122 3122
314122 2
4323
12
3359636
43
1
6F
xxy
Q
x
x
dV x dz dy dx
xy
xdydx
xyxyy dx
xx
x xx dx
xxx dx
















*** * * *
**
*
*
*
6
43
2
0
335
48 36 66.
43
xx
xx





ya
a
a
2
x
z

448. 448 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
58. ,,xyz x y zFijk
Q: solid region bounded by the coordinate planes and the plane 23412xyz

Surface Integral: There are four surfaces for this solid.

1
2
3
0, , , 0 0
0, , , 0 0
0, , , 0 0
S
S
S
zzdS
yydS
xxdS
"
"
" **
**
**
NkFN
NjFN
NiFN

234 1 9 29
23412, , 1
416 429
xy z dS dA dA




ijk
N




6
2
61223 6
00 04
0
11 2
234 12 34 34 36
44 33
NF
x
SR xx
dS x y z dydx dydx dx x
"


** ** ** *


Triple Integral: Because div 3,F the Divergence Theorem yields



11
div 3 dV 3 Volume of solid 3 Area of base Height 6 4 3 36.
32
F
QQ
dV



*** ***


59. ,, cos cos sin sinxyz y y x x x y xyz Fijk
S: portion of
2
zyover the square in the xy-plane with vertices 0, 0 , , 0 , , , 0,aaaa

Line Integral: Using the line integral you have:

1
2
2
2
3
2
4:0, 0
:0, 0, , 2
:, 0, , 0
:, 0, , 2
Cy dy
C x dx z y dz y dy
Cy ady z a dz
Cx adx z y dz ydy









+,+ ,
3
123 4
0
4
000
5
0
0
0
cos cos sin sin
0coscos sinsin 2
cos cos sin sin 2
cos sin sin cos 2
5
cos sin s
CC
CCC C
aaa
a
a
a
a
d y y x dx x x y dy xyz dz
dx a a x dx a a y dy ay y dy
dx a a x dx a a y dy ay dy
y
a x aa x y aa y a
aa aa aa
"






**
*** *
** * *
Fr
66
22
in cos
55
aa
aa aa


Double Integral: Considering
2
,, ,
fxyz z y you have:

2
22
,14,
14
fy
dS y dA
f y
D

D jk
N
and .curl F i
jxzyz
So,

56
24
00 00 0
22
22 .
55aa aa a
S aa
dS y z dy dx y dy dx dx"
** ** ** *
curl F N

y
x
(0, 4, 0)
(6, 0, 0)
(0, 0, 3)
z
y
x
1
aa
C
1
C
2
C
3
C
4
z

449. Problem Solving for Chapter 15 449
© 2010 Brooks/Cole, Cengage Learning
60.
2
,,xyz x z y z x Fijk
S: first octant portion of the plane 3 2 12xy z

Line Integral:

1
2
3
12 3 3
:0, 0, ,
22
12 1
:0, 0, ,
22
:0, 0, 123, 3
x
Cy dy z dz dx
y
Cx dx z dz dy
C z dz y x dy dx










2
2
123
0124
2
400
12 3 3 12
12 3 3
22 2
35 3
6 6 10 36 8
22 2
Fr
CC
CCC
d x zdx y zdy xdz
xy
x x dx y dy x x dx
x x dx y dy x dx
"








**
***
***


Double Integral:




4123 4
2
00 0
12 3
,,
2
31
,,
22
21
1 3 15 12 8
x
S
xy
Gx y z z
Gx y z
x
dS x dy dx x x dx



D

"
** ** *
ijk
curl F i j
curl F N


61. If ,
xyzcurl F i j k then div 1 1 1 3,curl F contradicting Theorem 15.3.
Problem Solving for Chapter 15
1. (a)

+,
32
222
25
Txyz
xyz

D

iik

2
1x x Ni k

2
1
1
dS dA
x








2
32 12 32
222 2 222
22
12 1
32 12 32 12
12 0 222 2 222 2
12 1 1 12
32 12 32 12
12 0 0 12 22 2 2
Flux 25
1
1
25
11
1112
25 25 25
2
11 1 1
N
SR
xz
kT dS k dA
xyz x xyz
xx
kdydx
xyz x xyz x
k dy dx k dy dx k
yx y x




D"














** **
**
** * *
2
25
36
k



(b)

,cos,,sinuv uv ur
sin , 0, cos , 0, 1, 0
uv uu rr
cos , 0, sin
uv uu' rr


+,

+,





32 32
222 2
22
32 32
22
123
32
03 2
25 25
cos sin
1
25 25
cos sin
11
25 2
Flux 25
6
1
ijk ij k
rr
uv
Txyzuvu
xyz v
Tuu
vv
k
du dv k
v



D


D" '



**

y
x
(0, 12, 0)
(0, 0, 6)
(4, 0, 0)
z

450. 450 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
2. (a)
22
22 22
1, ,
11
zxzy
zxy
xy
xyxy
CC

CC




22
22
32
222
1
1
1
25
25ijk ijk
zz
dS dA
xy xy
Txyzxyz
xyz
CC

CC

D



22
222222
22
1
11
1
1
zz
xyxy
xy
xy xy
zz
xy
xy x y xyz
CC


CC



CC

CC

ijk
Nijk
ij k ijk



22
21
0022 2
1
Flux 25
1
25 1
25 50
11
Nijkijk
SR
R
kT dS k x y z x y z dA
xy
k dA k rdrd k
xy r


D" "


** **
** * *

(b)

, sin cos , sin sin , cosuv u v u v ur

cos cos , cos sin , sin
sin sin , sin cos , 0
uv uv u
uv u v


u
vr
r


22 2 2
sin cos , sin sin , sin cos sin sin cos cosuv uvuu v uu v'
uvrr

22
00
sin
Flux 25 sin 50
u
k u du dv k


'

**
uvrr

3.


3cos ,3sin ,2
3sin ,3cos ,2, 13
tttt
tttt


r
rr






2
22 2 2 2
0
2
22 2 2 2
0
2
22 2 2
0 1
9sin 4 13 13 32 27
3
1
9cos 4 13 13 32 27
3
9 cos 9 sin 13 18 13
x
C
y
C
z
CIyzds ttdt
Ixzds ttdt
Ixyds t tdt



-
-
-

**
**
**


4.


232
12
22
,,
23
,1, 2 , 1
1
11
1
tt
tt
tt t tt
ds t dt
t
-




r
rr




4
1
22 3
0
1
22 2 3
0
4
1
22 2
0
849
4 9 180
85
99
23
460
y
C
x
C
z
C
t
Ixzds tdt
Iyzdsttdt
t
Ixyds tdt
-
-
-










**
**
**

451. Problem Solving for Chapter 15 451
© 2010 Brooks/Cole, Cengage Learning
5.




2222
522
222 222
2222
32 52 2
222 222
2222
32 52 2
222 222
222
2
32 222
222
11 2
2
2
0
wxyz
w
fx xyz xyz
wx wyxz
xy
xyz xyz
wy wzxy
yz
xyz xyz
wz www
w
zxyz
xyz
Cππ
≤≤ ≤
C
CCππ
≤π ≤
CC

CCππ
≤π ≤
CC

C CCC
D
CCCC


Therefore
1
w
f
≤is harmonic.

6.
nn
CR NM
y dx x dy dA
xy
CC

CC
***

For the line integral, use the two paths




11
22
22
1
2
22
222
22
11
0:,
:,to
0ri
ri j
nn
C
a n
nn n
Ca
aax
nn
Ra
Cxxaxa
Cxx axxaxa
ydx xdy
x
ydx xdy a x x dx
ax
NM
dA nx ny dy dx
xy
π
π
ππ
π



π

π
CC
π≤ π
CC
*
**
** * *

(a) For
1, 3, 5, 7,n≤ both integrals give 0.
(b) For n even, you obtain

35794 16 32 256
2: 4: 6: 8:
3 15 35 315
nan an an a≤π ≤π ≤π ≤π

(c) If n is odd then the integral equals 0.

7.


2
0
2 2
22 22 2
0 0
11
22
11
22
sin sin 1cos 1cos
sin sin 1 2 cos cos sin 2 cos 2 3
C
xdy ydx a a d a a d
adada



π≤ π ππ π



**
**

So, the area is
2
3.a

8.

2
0
11 2
22 3
2 sin 2 cos sin cos 2 2
C
x dy y dx t t t t dt

π≤ π ≤

**

So, the area is
4
3
.

9. (a)
,0 1rj
rjtt t
t

≤



11
00
1
C
W dr t dt dt" " ** *
Fijj
(b)


2
,0 1
12ri+j
rijttt t t
tt






1 2
22
0
11
2432 42
00
2112
13
122 2 1 4 2 1
15Fr i j ijWd tttt tdt
ttt t t t dt t t t dt

" "


*
**

x
−aa
2a
yax=
22
−
C
1
C
2
y

452. 452 Chapter 15 Vector Analysis
© 2010 Brooks/Cole, Cengage Learning
(c)

2
,0 1
12
tctt t t
tc t

rij
rij


2
222
24 22 2 2
12 11
221
d ctt tc t ctt
ct ct ct ct ct
"

Fr

2
2
211
1
30 6
11 5
0
15 6 2
15
0 minimum.
15 2
C
Wdcc
dW
cc
dc
dW
c
dc
"

*
Fr
10.

22 3
,3 2Fx y xy xy ij is conservative.


32
,
fxy xy potential function.

Work 2, 4 1, 1 8 16 1 127ff
11.
123
23131 2,, ,,
,,
aa a xyz
az ay az ax ay ax
' '
vr


1232,2 ,2 2aa a' curl v r v
By Stokes's Theorem,

2.v r r curl v r N v N
CS S
ddSdS' '" "*** **

12.


Area
cos sin , 0 2
sin cos
ab
tatbt t
tatbt



rij
rij


22
11
22
11 1
22 2
sin cos
sin cos
bt a t
dabtabtdtab

"

Fij
Fr



2
0
1
2
2W d ab ab

" *
Fr
Same as area.
13.



22
52
22
,,, 32
m
xy Mxy Nxy xy y x
xy


Fij ij












52
22
52
22
72 52
22 22
22
72
22 2 22
72
22
22
52
222 2
52
22
72 52
22 22 22
72
22 23
3
5
323
2
34
35
2
2
5
222
2
2
mxy
Mmxyxy
xy
M
mxy x y y x y mx
y
mx x y
mx x y y x y
xy
my x
Nmyxxy
xy
N
my x x y x x y mx
x
mx x y y







C


C







C


C






222
22
72
22 2 2
72
22
52
34
312
xxy
mx x y
mx x y x y
xy







So,
N M
x y
CC

CC
and F is conservative.

453. © 2010 Brooks/Cole, Cengage Learning
CHAPTER 16
Additional Topics in Differential Equations
Section 16.1 Exact First-Order Equations...............................................................454
Section 16.2 Second-Order Homogeneous Linear Equations................................464
Section 16.3 Second-Order Nonhomogeneous Linear Equations .........................472
Section 16.4 Series Solutions of Differential Equations ........................................480
Review Exercises........................................................................................................491
Problem Solving.........................................................................................................500

454. 454 © 2010 Brooks/Cole, Cengage Learning
CHAPTER 16
Additional Topics in Differential Equations
Section 16.1 Exact First-Order Equations
1.
22
230xxydx xydy

2
2
Exact
M
xy
y
N
xy
x
MN
yx
C

C
C

C
CC

CC
2.
10xy dx y xy dy

Not exact
M
x
y
N
y
x
MN
yx
C

C
C

C
CC
#
CC


3. sin cos 0x y dx x y dy

cos
cos
Not exact
M
xy
y
N
y
x
MN
yx
C

C
C

C
CC
#
CC

4. 0
xy xy
ye dx xe dy

Exact
xyxy
xyxy
M
exye
y
N
exye
x
MN
yx
C

C
C

C
CC

CC


5. 23 23 0x y dx y x dy

3 Exact
MN
yx
CC

CC




2
,,
23
3
fxy Mxydx
xydx
x xy g y



*
*



,3
23 2
yfxy x gy
yx gy y




2
1
gyyC


22
1
22
,3
3
fxy x xy y C
xxyyC



6. 0
xx
ye dx e dy

Exact
xMN
e
yx
CC

CC



,,
,0
xx
xx
x
fxy Nxydy edy ye gx
fxy ye gx ye gx

**


1
gxC

1,
x
x
fxy ye C
ye C



7.
22 2
310 6210 0yxydxxy xy

620 Exact
MN
yxy
yx
CC

CC



22
222
,,310
35
fxy Mxydx y xy dx
xy x y g y

**




22
1
, 6 10 6 2 10
22
y
fxy xy xy g y xy xy
gy gy y C




222
1
,35 2
fxy xy xy y C

222
35 2
xyxyyC

455. Section 16.1 Exact First-Order Equations 455
© 2010 Brooks/Cole, Cengage Learning
8. 2 cos 2 cos 2 0xydx xydy



2sin 2
2sin 2 Exact
MN
xy
yx
xy
CC

CC




,,
2cos 2 sin 2
fxy Mxydx
xydx x y gy


*
*




1
,cos2
cos 2 0
yfxy x y gy
xygy gyC





1,sin2
sin 2
fxy x y C
xy C



9.
32 3
46 46 0
12
6
xxydxyxydy
M
xy
y
N
y
x

C

C
C

C

Not exact

10.
22
2
22 0
xy xy
ye dx xye dy



2
3
2
3
4
2
xy
xy
M
xyye
y
N
xyye
x
C

C
C

C

Not exact

11.
22 22
0
yx
dx dy
xy xy





22
2
22
Myx N
yx
xy
CC

CC

Exact


, , arctan
x
fxy Mxydx gy
y


*





22
1
22
,
0
y
x
fxy gy
xy
x
gygyC
xy







1,arctan
arctan
x
fxy C
y
x
C
y







12.

22 22
0
xy xy
xe dx ye dy




22
2
xyMN
xye
yx CC

CC
Exact





22 22
,,
1
2
xy xy
fxy Mxydx
xedxe gy


*
*








22 22
1
,0
xy xy
y
fxy ye gy ye gy
gy C







22
1
22
1
,
2xy
xy
fxy e C
eC






13.
22
0
yx
dx dy
xy xy







3
3
2
2
Mxy
y
xy
N xy
x xy
C

C
C

C

Not exact

14. cos cos sin 0
yy
ye xy dx e x xy xy dy

cos cos sin
yy yM
exyyexyxyexy
y
C

C


+,cos sin cos
yN
exyxyxyyxy
x
C

C


MN
yx
CC

CC
Exact


,,
cos sin
yy
fxy Mxydx
ye xy dx e xy g y

*
*




1
,sin cos
0
yy
y
fxy e xyxe xygy
gy gy C




1,sin
y
fxy e xy C
sin
y
exyC

456. 456 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
15. (a) and (c)

(b)

22 1
2tan 5 sec 0,
24
x y dx x y dy y





2
2 sec Exact
MN
xy
yx
CC

CC




2
,,2tan5
tan 5
fxy Mxydx x y dx
xyxgy

**




22
22
,sec
sec
0
yfxy x y gy
xy
gygyC






2
,tan5
fxy x y x C

11511
,
24 4 2 4
f C





Answer:
2 11
tan 5
4
xyx

16. (a) and (c)

(b)

2
2cos0,1xy dx x y dy y

2
MN
x
yx
CC

CC
Exact


2
,,2
fxy Mxydx xydx xy gy**


,
yfxy


22
1
cos
cos
sin
xgyx y
gy y
gyyC






2
1
,sin
fxy xy y C

2
sin
xyyC

1,
f C
Answer:
2
sinxy y

17. (a) and (c)

(b)

22 22
0, 4 3
xy
dx dy y
xy xy




32
22
Exact
MxyN
yx
xy
CC

CC



,fxy


22
12
22
22
,
1
2
2
x
Mxydx dx
xy
xyxdx
xygy




**
*





22
22
,
0
y
y
fxy gy
xy
y
gygyC
xy







22
,fxy x y C


22
3, 4 3 4 25 5f C
Solution:
22
5xy or
22
25xy

18. (a) and (c)
(b)

22
0, 4 5x y dx y x dy y

1
MN
yx
CC

CC
Exact






3
2
2
3
2
1
,,
3
,
3
y
fxy Mxydx
x
xydx xy gy
fxy x gy y x
y
gyy gy C




*
*




33
1
33
,
33
33
4, 5 43
xy
fxy xy C
xy
xy C
fC




Solution:
33
3 129xy xy
4
4
x
y
−4
−
−6
−4
6
4
y
x
4−4
4
2 −6
−4
6
4
x
y
−4−224
−4
−2
2
4
6
−6
−99
−12 12
−8
8

457. Section 16.1 Exact First-Order Equations 457
© 2010 Brooks/Cole, Cengage Learning
19. ln 1 2 0
1
y
dx x y dy
x





1
Exact
1MN
yx x
CC

CC


,,ln1
fxy Mxydx y x gy*




2
1
,ln1
2
yfxy x gy
gyygyyC





2
1
,ln1
fxy y x y C


2
ln 1yx y C

24:y 4ln 2 1 16 16CC
Solution:

2
ln 1 16yx y

20.
22 22
0
xy
dx dy
xy xy




2
22
2
ExactMxyN
yx
xy
C C

CC





22
22
,,
1
ln
2
fxy Mxydx
x
dx x y g y
xy


*
*





22
1
,
0
y
y
fxy gy
xy
gygyC







22
11
,ln
2
fxy x y C


22
ln
xyC

04:ln16yC


22
ln ln 16xy
Solution:
22
16xy

21.
33
sin 3 cos 3 0
xx
eydxe ydy

3
3 cos 3 Exact
xMN
ey
yx
CC

CC




33
,,
1
sin3 sin3
3
xx
fxy Mxydx
eydxeygy

*
*




3
1
,cos3
0
x
y
fxy e y gy
gygyC





3
11
,sin3
3
x
fxy e y C

3
sin 3
x
eyC

0: 0yC
Solution:
3
sin 3 0
x
ey

22.
22
20x y dx xy dy

2Exact
MN
y
yx
CC

CC





3
22 2
,,
3fxy Mxydx
x
xydx xy gy


*
*


1,2 0
y
fxy xy gy gy gy C


3
2
1
,
3
x
fxy xy C

3
2
3
xxyC

31:y9312 C

3
2
12
3
x
xy

Solution:
32
336xxy

23.
22
29 2 1 0xy x dx y x dy

2Exact
MN
x
yx
CC

CC




2
23
,,29
3
fxy Mxydx xy x dx
xy x gy

**





22
2
1
,21
21
yfxy x gy y x
gy y
gy y y C






222
1
222
,3
3
fxy xy x y y C
xy x y y C



03:936yC
Solution:
222
36xy x y y

24.
22
24 26 0
4Exact
xy dx x y dy
MN
xy
yx

CC

CC




222
,,
24 4
fxy Mxydx
xydxxyxgy


*
*




22
1
,2 2 6 6
6
yfxyxygyxy gy
gy y C





22
1
22
,46
46
fxy xy x y C
xy xyC



1 8:16444813yC
Solution:

22
46 13xy xy

458. 458 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
25.


2
60
2
ydx x y dy
Nx My
ky
My

CCC C


Integrating factor:

2
ln
2 1kydy
y
ee
y

*


Exact equation:
2
1
60 x
dx dy
yy








1
,
6
6
6
x
fxy gy
y
gy
gy y C
x
yC
y






26.


3
20
2xydxxdy
My Nx
hx
Nx

CCCC

Integrating factor:

2
ln
21hx dx
x
ee
x

*

Exact equation:
2
1
20
y
xdxdy
xx










2
1
2
,
0
y
fxy x gy
x
gy
gy C
y
xC
x






27.


2
50
2xydxxdy
My Nx
hx
Nx

CCCC

Integrating factor:

2
ln
21hx dx
x
ee
x

*


Exact equation: 2
1
50
y
dx dy
xx









1
,5
0
5
y
fxy x gy
x
gy
gy C
y
xC
x






28.


22
520
1xydxydy
My Nx
hx
N

CCCC

Integrating factor:
hx dx
x
ee
*


Exact equation:

22
520
xx
xyedxyedy




22
, 5 10 10
xxxx
fxy xe xe e ye gy



0gy

1
gyC

22
51010
xxxx
ye xe xe e C



29.


tan 0
tan
xydx xdy
My Nx
xhx
N

CCCC


Integrating factor:
ln cos
cos
hx dx
x
ee x
*

Exact equation:
cos sin 0xy xdx xdy

,sincossin
fxy x x x y x gy



1
0gy
gyC




sin cos sin
xxxyxC

30.


23
21 0
1
xy dx x dy
My Nx
hx
Nx

CCCC

Integrating factor:

ln 1 1hx dx x
ee
x
*

Exact equation:
21
20xy dx x dy
x










2
1
2
,ln
0
ln
fxy xy x gy
gy
gy C
xy x C






31.


2
10
1
ydx xy dy
Nx My
ky
My

CCC C

Integrating factor:

ln 1 1kydy y
ee
y
*

Exact equation:
1
0ydx x dy
y







1
,
1
ln
ln
fxy xy gy
gy
y
gyyC
xy y C

459. Section 16.1 Exact First-Order Equations 459
© 2010 Brooks/Cole, Cengage Learning
32.


2
220
1
2
xxydxdy
My Nx
hx
N

CCCC

Integrating factor:

2
hx dx
x
ee
*

Exact equation:

222
220
xx
xxyedxedy






22
1
22
,224
0
24
x
x
fxy x x ye gy
gy
gy C
xx ye C






33.



2sin0
1
2
ydx x y dy
Nx My
ky
My

CCC C


Integrating factor:

ln 1 1ykydy
ee
y
*

Exact equation:
sin
20
yx
ydy dy
yy









1
,2
sin
2cos
cos
fxy yx gy
y
gy
y
gy y C
yx y C






34.


323
21 3 0
3
ydxxyxdy
My Nx
hx
Nx

CCCC

Integrating factor:


3
ln 1
3
1xhx dx
ee
x
*


Exact equation:
32
33 2
21 3
10
yy
dx dy
xx x









3
22
1
3
22
1
,
2
1
1
2
y
fxy gy
xx
gy
gy y C
y
yC
xx






35.
22 3
42 34 0xy y dx x xydy
Integrating factor:
2
xy
Exact equation:


33 4 42 23
42 34 0x y xy dy x y x y dy




43 24
1
43 24
,
0
fxy xy xy gy
gy
gy C
xy xy C






36.
22 3
35 32 0yxydxxyxdy
Integrating factor:
2
xy
Exact equation:


23 42 32 5
35 32 0x y x y dx x y x y dy




33 52
1
33 52
,
0
fxy xy xy gy
gy
gy C
xy xy C






37.
52 4 3
22 0yxydx xy xdy
Integrating factor:
23
xy


Exact equation:
2
22 3
1
22 0
yyx
dx dy
xy xy









2
2
1
2
2
,
0
yx
fxy gy
xy
gy
gy C
yx
C
xy






38.
322
0ydx xy x dy
Integrating factor:
22
xy


Exact equation:
22
11
0
y
dx dy
xxy









2
1
,
1
1
1
y
fxy gy
x
gy
y
gy C
y
y
C
xy

460. 460 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
39. 0ydx xdy
(a)
22 2
11 1
,0,
yMN
dx dy
xxx yxx
CC

CC

(b)
22
11
,0,
x
dx dy
yy y

2
1MN
yy x
CC

CC

(c)
11 1
,0,dx dy
xy x y
0
MN
yx
CC

CC
(d)

2222 22
22
2
22
1
,0,
yx
dx dy
xyxy xy
Mxy N
yx
xy


CC

CC



40.
22
0axy by dx bx y ax dy
Exact equation:
2,
M
axy b
y
C

C
2,
N
bxy a
x
C

C
MM
yx
CC

CC
only if .ab
Integrating factor:
mn
xy


12 1 21 1
0
mn mn m n mn
ax y bx y dx bx y ax y dy







11
11 2221
21 11
mn mn
mn mnM
an bman x y bn x y
y
N
bm x y am x y bn am
x

C <

=
C =
>
C =

=
C ?





2
2
22
22
2
2
abn b m b b aan bm b a
abn a m a a bbn am a b
abm baab
ba
m
ab
< =
>
=?







222
2
22
2
ba
bn a a b
ab
ba bab a a b
bn
ab ab
ab
n
ab













41.
22 22
22
,
0
yx
xy
xyxy
dy x
dx y
ydy xdx
yxC
Fij





Family of circles

42.
22 22
,
xy
xy
xyxy
Fij


0
dy y
dx x
xdy ydx
xyC




Family of hyperbolas
4
−4
−66
c= 4
c= 1
c= 9
4
−4
−6 6

461. Section 16.1 Exact First-Order Equations 461
© 2010 Brooks/Cole, Cengage Learning
43.
22
2
,4 2
x
xy xy xy
yFi j






3
3
4
4
2
4
2
24 2
1
24
82
21
1
ln 2 1 ln ln
21
2
dy y
dx x xy
y
dy dx
yx
yC
x
C
y
x
xy x C













44.
2
2
2
2
2
,1 2
2
1
12
1
1
ln ln ln
1
1
xyxxy
dy xy
dx x
x
dy dx
yx
yC
x
C
y
x
Fij













45.




2
2
1
22
3
30
1
,
2
3
3
2
23
dy y x
dx y x
x y dx y x dy
MN
yx
x
fxy xy gy
gy y
y
gy C
xxyyC




CC

CC





Initial condition:
21,443 , 3yCC
Particular solution:
22
23 3xxyy

46.




22
22
2
2
3
1
23
2
20
2
,
3
3
dy xy
dx x y
xy dx x y dy
MN
x
yx
fxy xy gy
gy y
y
gy C
xy y C




CC

CC




Initial condition:
02,8yC
Particular solution:
23
38xy y

47.




22
22
1
22
20
210
20 10 2 0
20 2
,10
0
10
xy xdy
Ex
yxydx
xy y dx x xy dy
MN
xy
yx
fxy xy xy gy
gy
gy C
xy xy K




CC

CC





Initial condition:
100 500, 100 , 25,000,000CxK



22
22
24
24
10 25,000,000
10 25,000,000 0 Quadratic Formula
5 1,000,000
10 100 4 25,000,000
2
xy xy
xy x y
x xx
xxx
y
xx






4
−4
−6 6
2
−2
−33
c= 4
c= 2
c= 6

462. 462 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
48.


0
dy
Pxy Qx
dx
dy P x y dx Q x dx
Pxy Qx dx dy






,Mxy Pxy Qx and ,1Nxy



+,

1
,,
,1
yx
dd
Pxy Qx
Mxy Nxy dy dx
Nxy
Px






By Theorem 16.2,


Pxdx
ux e
*
is an integrating factor.

49. (a) 4 0.5231y

(b)
22
dy xy
dx x y






22
0xy dx x y dy

+,
111
2
xyNM xx
Mxyy

function of y
alone.
Integrating factor:
1
lnydy
y
eey
*







223
22
2
4
2
1
22 4
0
,
2
,
4
,
24
y
xy dx x y y dy
xy
fxy xydx gy
y
fxy xy g y gy C
xy y
fxy C




*

Initial condition:

41 9
21,
24 4
yC
Particular solution:
22 4
9
244xy y

or
22 4
29.xy y
For
4,x
24
32 9 4 0.528yy y
(c)

50. (a) 5 6.6980y

(b)


2
22
6
32
6230
2Exactdy x y
dx y y x
xydx xy ydy
MN
y
yx




CC

CC





222
3
1
223
,6 3
2
,3
y
fxy x y dx x xy gy
f xy g y g y y C
fxy x xy y C



*

Initial condition:
01 1yC
Particular solution:
223
31xxyy
For
23
5, 75 5 1 0 6.695.xyy y
(c)
2
−1
−15
16
−2
−36
16
−2
−36
2
−1
−15

463. Section 16.1 Exact First-Order Equations 463
© 2010 Brooks/Cole, Cengage Learning
51. (a) 4 0.408y

(b)

22
22
0
dy xy
dx x y
xy dx x y dy






+,
11
2
1
function of alone.
xyNM xx
Mxy
y
y



Integrating factor:
1
lnydy
y
eey
*







223
22
2
4
2
1
22 4
0
,
2
,
4
,
24
y
xy dx x y y dy
xy
fxy xydx gy
y
fxy xy g y gy C
xy y
fxy C




*

Initial condition:

41 9
21,
24 4
yC
Particular solution:
22 4
22 4
9
244
29.
xy y
xy y


or
For
4,x
24
32 9 4 0.528yy y
(c)
The solution is less accurate. For #49, Euler’s
Method gives
4 0.523,y whereas in #51, you
obtain
4 0.408.y The errors are
0.528 0.523 0.005 and
0.528 0.408 0.120.

52. (a) 6.708ys

(b)


2
22
6
32
6230
2Exact
dy x y
dx y y x
xydx xy ydy
MN
y
yx




CC

CC





222
3
1
223
,6 3
2
,3
y
fxy x y dx x xy gy
f xy g y g y y C
fxy x xy y C



*

Initial condition:
01 1yC
Particular solution:
223
31xxyy
For
23
5, 75 5 1 0 6.695.xyy y
(c)
The solution is less accurate. For #50, Euler’s
Method gives
5 6.698,y whereas in #52, you
obtain
5 6.708.y The errors are
6.695 6.698 0.003 and
6.695 6.708 0.013.

53. If M and N have continuous partial derivatives on an
open disc R, then
,,0Mxydx Nxydy is exact
if and only if
.
MN
yx
CC

CC

54. See Theorem 16.2.

55. False

2
M
x
y
C

C
and 2
N
x
x
C

C


56. False

0ydx xdyis exact, but
2
0xy dx x dy is not
exact.

57. True


M
fx M
yy
CC


CC
and

N
gy N
x x
CC


CC

2
−1
−15
2
−1
−15
16
−2
−36
16
−2
−36

464. 464 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
58. True
0fx
y
C


Cand
0gy
x
C


C

59.
22 3
,Mxykxyx
32 2
Nxxyy

2
2,
M
xykx
y
C

C
2
32
N
x xy
x
C

C

3
MN
k
yx
CC

CC

60.
2
2,
xy
Mye x
2
2
xy
Nkxe y

22 2 2
2, 2
xyxy xy xyMN
e xye ke kxye
yx
CC

CC


1
MN
k
yx
CC

CC

61. sin ,Mgy x
2
Nyfx
sin ,
M
gyx
y
C

C

2N
yf x
x
C

C



2
:sin
MN
gyxfxy
yx
CC

CC


3
2
1
3
y
gyy gy C

2sin cos
fxxfx xC

62. ,
y
MgyeNxy

,
yyMN
gye g ye y
yx
CC

CC


yy
gye g ye y


y
gye y





2
2
yy
gye C


2
2
yy
gye C






Section 16.2 Second-Order Homogeneous Linear Equations
1.

33
12
33 3
12 2
33 3
12 2
33 3 33 3 3 3
12 2 12 2 12
33
96 9
6996 9 18618 99 0
xx
xx x
xx x
xx x xx x x x
yCe Cxe
yCeCe Cxe
yCeCeCxe
yyyCeCeCxe CeCe Cxe CeCxe









y approaches zero as .
x !


2.
22
12
22
12
22
12
22
44 4
4440
xx
xx
xx
yCe Ce
yCe Ce
yCe Ce y
yyyy







The graphs are different combinations of the graphs of
2
x
eand
2
.
x
e




3.
12
12
12cos 2 sin 2
2sin2 2 cos2
4cos2 4sin2 4
4440yC xC x
yCxCx
yCxCxy
yy yy




The graphs are basically the same shape, with left and
right shifts and varying ranges.

−2−4246
6
y
x
−2−424
−2
2
y
x
2
−1
−2
1
2
y
x

465. Section 16.2 Second-Order Homogeneous Linear Equations 465
© 2010 Brooks/Cole, Cengage Learning
4. 12 12cos 3 sin 3 cos 3 sin 3
xxx
yCe xCe xeC xC x




+,
12 1 2
12 1 2cos 3 sin 3 3 sin 3 3 cos 3
cos3 sin3 3 sin3 3 cos3
xx
x
yeC xC xeC xC x
eC xC xC xC x





+
,
+,
12 1 2
12 1 2cos 3 sin 3 3 sin 3 3 cos 3
3sin33cos39cos39sin3
x
x
yeCxCxCxC x
eC x C x C x C x






+
,
+,
+,12 1 2
12 1 2
12 1 2
12210 cos3 sin33sin33cos3
3sin3 3 cos3 9cos3 9 sin3
2 cos 3 sin 3 3 sin 3 3 cos 3
10 cos 3 sin 3 0
x
x
x
yy yeC xC xCxC x
CxC xCxC x
eC xC xC xC x
eC x C x









y approaches zero as .
x !The graphs are the same only reflected.

5. 0yy
Characteristic equation:
2
0mm
Roots:
0, 1m

12
x
yCCe

6. 20yy
Characteristic equation:
2
20mm
Roots:
0, 2m

2
12
x
yCCe



7. 60yy y
Characteristic equation:
2
60mm
Roots:
3, 2m

32
12
x x
yCe Ce



8. 650yyy
Characteristic equation:
2
650mm
Roots:
1, 5m

5
12
x x
yCe Ce



9. 2320yyy
Characteristic equation:
2
2320mm
Roots:
1
2
,2m

12 2
12x
x
yCe Ce



10. 16 16 3 0yyy
Characteristic equation:
2
16 16 3 0mm
Roots:
31
44
,m

14 34
12
x x
yCe Ce

11. 690yyy
Characteristic equation:
2
690mm
Roots:
3, 3m

33
12
x x
yCe Cxe



12. 10 25 0yyy
Characteristic equation:
2
10 25 0mm
Roots:
5, 5m

55
12
x x
yCe Cxe

13. 16 8 0yyy
Characteristic equation:
2
16 8 1 0mm
Roots:
11
44
,m

14 14
12
x x
yCe Cxe

14. 91240yyy
Characteristic equation:
2
91240mm
Roots:
22
33
,m

23 23
12
x x
yCe Cxe

15. 0yy
Characteristic equation:
2
10m
Roots: ,
mii

12cos sinyC xC x

16. 40yy
Characteristic equation:
2
40m
Roots:
2,2mii

12cos 2 sin 2yC xC x
234
1
y
x

466. 466 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
17. 90yy
Characteristic equation:
2
90m
Roots:
3, 3m

33
12
x x
yCe Ce



18. 20yy
Characteristic equation:
2
20m
Roots:
2, 2m

22
12
x x
yCe C



19. 240yyy
Characteristic equation:
2
240mm
Roots:
13,13mii
12cos 3 sin 3
x
yeC xC x

20. 4210yy y
Characteristic equation:
2
4210mm
Roots:
217,217mii

2
12
cos 17 sin 17
x
yeC xC x

21. 30yyy
Characteristic equation:
2
310mm
Roots:
3535
,
22
m



352 352
12
x x
yCe Ce




22. 34 0yyy
Characteristic equation:
2
3410mm
Roots:
2727
,
33
m



273 273
12
x x
yCe Ce




23. 912110yyy
Characteristic equation:
2
912110mm
Roots:
2727
,
33 ii
m




23
12 77
cos sin
33
x
ye C x C x





24. 2 670yyy
Characteristic equation:
2
2670mm
Roots:
3535
,
22 ii
m




32
12 55
cos sin
22
x
ye C x C x





25.
4
0yy
Characteristic equation:
4
10m
Roots:
1, 1, ,mii

12 3 4 cos sin
xx
yCe Ce C xC x



26.
4
0yy
Characteristic equation:
42
0mm
Roots:
0, 0, 1, 1m

12 3 4
x x
yCCxCe Ce



27. 61160yy yy
Characteristic equation:
32
61160mm m
Roots:
1, 2, 3m

23
12 3
x xx
yCe Ce Ce

28. 0yyyy
Characteristic equation:
32
10mmm
Roots:
1, 1, 1m

12 3
x xx
yCe Cxe Ce



29. 3750yyyy
Characteristic equation:
32
3750mm m
Roots:
1, 1 2 , 1 2mii

12 3 cos 2 sin 2
xx
yCe eC xC x

30. 33 0yyyy
Characteristic equation:
32
3310mmm
Roots:
1, 1, 1m

2
12 3
x xx
yCe Cxe Cxe

31.
12
12
100 0
cos 10 sin 10
10 sin 10 10 cos 10yy
yC xC x
yCxCx



(a)

1
2202:2
00:010 0
yC
yCC


Particular solution:
2 cos 10yx
(b)


1
22
1
5
00:0
02:210yC
yCC


Particular solution:
1
5
sin 10yx
(c)


1
22
3
10
01:1
03:310yC
yCC


Particular solution:
3
10
cos 10 sin 10yx x

467. Section 16.2 Second-Order Homogeneous Linear Equations 467
© 2010 Brooks/Cole, Cengage Learning
32. sin 3
3cos 3
3sin 3yC t
yCt
yC t




3 sin 3 sin 3
03
yyC t t
C;;
;




53
05:53
3
yCC and
53;

33. 30 0,yy y 01,y 04y
Characteristic equation:
2
30 0mm
Roots:
6, 5m

65
12
,
x x
yCe Ce


65
12
65
x x
yCe Ce


Initial conditions:
01,y 04,y
121,CC
1246 5CC
Solving simultaneously:
12
110
,
11 11
CC

Particular solution:

651
10
11
x x
ye e



34. 7120,yy y 03,y 03y
Characteristic equation:
2
7120mm
Roots:
3, 4m

34 3 4
12 1 2
,3 4
x xxx
yCe Cey Ce Ce
Initial conditions:
03,03,yy
12 1 2 3, 3 4 3CC C C
Solving simultaneously:
129, 6CC
Particular solution:
34
96
x x
ye e

35. 16 0,yy
00,y 02y
Characteristic equation:
2
16 0m
Roots: 4mi

12cos 4 sin 4yC xC x

124sin4 4 cos4yCxCx
Initial conditions:



1
22
1
2
00
024
yC
yCC



Particular solution:
1
2
sin 4yx

36. 230,02,yyy y 01y
Characteristic equation:
2
230mm
Roots:
2412
12
2
mi


12cos 2 sin 2
x
yeC xC x


12 122sin2 2cos2 cos2 sin2
xx
yeC xC xeC xC x


Initial conditions:


1
212 202
3
01 2 22
2
yC
yCCC C



Particular solution:
3
2cos 2 sin 2
2
x
ye x x






37. 96 0,yyy 02,y 01y
Characteristic equation:
2
9610mm
Roots:
11
33
,m

13 13
12
x x
yCe Cxe

13 13 13
12211
33
x xx
yCe Cxe Ce
Initial conditions:
02,y 01y

1
12
12
1
1 3
3
2
2,
1
C
CC
CC
<=
>

=?

Particular solution:
331
3
2
x x
ye xe

38. 44 0,yyy 03,01yy
Characteristic equation:

2
2
4410
21 0
mm
m



Roots:
11
22
,m

12 12
12
x x
yCe Cxe



12 12 12
12211
22
x xx
yCe Cxe Ce


Initial conditions:



1
12 2
51
22
03
01
yC
yCCC



Particular solution:
225
2
3
x x
ye xe

468. 468 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
39. 430,yyy 01,y 13y
Characteristic equation:
2
430mm
Roots:
1, 3m

3
12
x x
yCe Ce

01:y
12 1CC

13:y
3
12
3Ce C e
Solving simultaneously,
3
12
33
33
,ee
CC
ee ee




Solution:
3
3
33
33
x xee
ye e
ee ee





40. 40,yy
02,y 5y
Characteristic equation:
2
410m
Roots:
1
2
mi

12
11
22
cos sinyC xC x

02:y
12C

5:y
2 5C
Solution:
11
22
2cos 5sinyxx

41. 90,yy 03,y 5y
Characteristic equation:
2
90mm
Roots: 3mi

12cos 3 sin 3yC xC x

03:y
13C

5:y
15C
No solution

42. 420210,yyy 03,y 20y
Characteristic equation:
2
420210mm
Roots:
37
,
22
m

32 72
12
x x
yCe Ce



03:y
12 3CC

20:y
37 4
12 12
00Ce C e C C e


Solving simultaneously,
1
4
3
,
1
C
e



4
2
4
3
1e
C
e



Solution:

4
32 72
44
33
11
x xe
ye e
ee





43. 428490,yyy 02,y 11y
Characteristic equation:
2
428490mm
Roots:
77
,
22
m

72 72
12
x x
yCe Cxe

02:y
12C

11:y
72
72 72
12 2
72
12
1e
Ce C e C
e



Solution:

72
72 72
72
12
2
x xe
ye xe
e





44. 6450,yy y 04,y 8y
Characteristic equation:
2
6450mm
Roots:
636180
36
2
mi



33
12
cos 6 sin 6
xx
yCe xCe x



04:y
14C

8:y
3
1
8Ce


No solution

45. No, it is not homogeneous because of the nonzero term
sin x.

46. Answers will vary. See Theorem 16.4.

47. Two functions
1yand
2yare linearly independent if the
only solution to the equation
11 2 2 0Cy C y is the
trivial solution
12 0.CC

48. 20ykyky
Characteristic equation:
2
20mkmk

2
2
244
1
2
kkk
mkk


(a) For 1kand 1,k
2
10k and there are 2
distinct real roots.
(b) For 1,k
2
10k and the roots are repeated.
(c) For 1 1,k the roots are complex.

49. By Hooke’s Law,
32
48.
23
Fkx
F
k
x



Also, ,Fma and
32
1.
32
F
m
a


So,
1
cos 4 3
2
yt

469. Section 16.2 Second-Order Homogeneous Linear Equations 469
© 2010 Brooks/Cole, Cengage Learning
50. By Hooke’s Law,
32
48.
23
Fkx
F
k
x



Also, ,Fma and
32
1.
32
F
m
a

So,
2
cos 4 3 .
3
yt


51.
12cos sin ,y C kmt C kmt
48 4 3km
Initial conditions:


21
0,0
32
yy







12
1
12
22cos 4 3 sin 4 3
2
0
3
43 sin43 43 cos43
113
043
224 83
23
cos 4 3 sin 4 3
324
yC t C t
yC
yt C t C t
yCC
yttt







52.
12cos 4 3 sin 4 3yC t C t
Initial conditions:


11
0,0
22
yy






1
12
22
1
0
2
43 sin43 43 cos43
11
043
2 83
11
cos 4 3 sin 4 3
2 83
yC
yt C t C t
yC C
yt t t






53. By Hooke’s Law,
32 2 3 ,k so 48.kMoreover, because the weight w is given by mg, it follows that
32 32 1.mwg Also, the damping force is given by 18 .dy dt So, the differential equation for the
oscillations of the weight is

2
2
2
2
1
48
8
1
48 0.
8
dy dy
my
dt dt
dy dy
my
dt dt







In this case the characteristic equation is
2
8 384 0mm with complex roots
1 16 12,287 16 .mi
So, the general solution is

16
12 12,287 12,287
cos sin .
16 16
t tt
yt e C C






Using the initial conditions, you have

1
1
0
2
yC



2116
1212,287 12,287 12,287 12,287
sin cos
16 16 16 16 16 16
t Ct Ct
yt e C C








1
2212,287 12,287
00
16 16 24,574
C
yCC
and the particular solution is


16
12,287 12,287 12,287
cos sin .
2 16 12,287 16
t
et t
yt

470. 470 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
54. By Hooke’s Law, 32 2 3 ,k so 48.kAlso, 32 32 1.mwg The damping force is given by 14 .dy dt So,

2
2
2
2
1
48
4
1
48 0.
4
dy dy
my
dt dt
dy dy
my
dt dt







The characteristic equation is
2
4 192 0mm with complex roots
1 8 3071 8 .mi So, the general solution is


8
12 3071 3071
cos sin .
88
t tt
yt e C C






Using the initial conditions, you have


1
1
0
2
yC


2218
13071 3071 3071 3071
sin cos
888 888
t CtCCt
yt e C








1
223071 3071
00
8 8 6142
C
yC C
and the particular solution is


8
3071 3071 3071
cos sin .
2 8 3071 8
t
et t
yt







55. 90yy
Undamped vibration
Period:
2
3

Matches (b)

56. 25 0yy
Undamped vibration
Period:
2
5

Matches (d)

57. 2100yy y
Damped vibration
Matches (c)

58.
37
4
0yy y
Damped vibration
Matches (a)

59. Because
2mais a double root of the characteristic equation, you have

2
2
2
0
24
aa
mmam




and the differential equation is

2
40.yay ay The solution is






2
12
212
2
22
212
2
2222222
2 121212
22
22
44
0.
4442244
ax
ax
ax
ax
yCCxe
Ca C a
yCxe
Ca C a
yaCxe
aCaCaCaCaCaCa
yay ye Ca x Ca x x

471. Section 16.2 Second-Order Homogeneous Linear Equations 471
© 2010 Brooks/Cole, Cengage Learning
60. Because mi $ are roots to the characteristic equation, you have


222
20mimimm$ $ $


and the differential equation is

22
20.yy y$
2
:1.iNoteThe solution is






12
12 21
22 2 2
11 2 2 2 1
22
12 21
22 2 2 2 2
11 2 2cos sin
cos sin
2cos 2sin
222cos22sin
cos sin
x
x
x
x
x
yeC xC x
yeC C xC C x
yeC C C xC C C x
ye C C x C C x
yeC C xC C x





$$
$$ $$
$ $$ $ $$
$ $ $ $
$ $ $ $ $









So,

22
20.yy y$

61. False. The general solution is
33
12
.
x x
yCe Cxe

62. True

63. True

64. False. The solution
2
x
yxe requires that 1mis a triple root of the characteristic equation. Because the characteristic
equation is quadratic, 1mcan be at most a double root.

65.
1 ,
ax
ye
2 ,
bx
ye ab#



12,
0 for any value of .
ax bx
ax bx
ax bx
ee
Wy y
ae be
bae x


#


66.
12 ,
ax ax
yey xe


12
2,
0 for any value of .
ax ax
ax ax ax
ax
exe
Wy y
ae e axe
ex


#


67.
1 sin ,
ax
ye bx
2 cos , 0
ax
ye bxb#


12
22 2 2 2
sin cos
,
sin cos cos sin
sin cos 0 for any value of .
ax ax
ax ax ax ax
ax ax ax
ebx e bx
Wy y
ae bx be bx ae bx be bx
be bx be bx be x


#


68.
1 ,yx
2
2
yx


2
2
12
,0
12
xx
Wy y x
x
# for 0.x#

69.
2
0, 0xy axy by x
Let
.
t
xe
(a)
tdy dy dt dy
e
dx dx dt dt




222
2
222
t
tt t t
t d dt e dy dtdy dy dy dy dy
ee e e
dx e dt dt dt dt








2
0xy axy by

2
22
2
0
tt ttd y dy dy
ee aee by
dt dt dt







2
2
10
dy dy
aby
dt dt

472. 472 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
(b)
2
660xy xy y
Let
.
t
xeFrom part (a), you have:

2
2
560
dy dy
y
dt dt


2
560mm

320mm

123, 2mm


32
ln 1 ln 1
1232 3ln 2ln
12 1 2 1 2
32
.
xx
tt x x CC
yCe Ce Ce Ce Ce Ce
xx



70. 0yAy
If 0,Athen
12cos sin .yC AxC Ax

100 0yC and
2sinyC Ax


200 sin 1,2,3, 1,4,9,16,yCAA A
So, A is a perfect square integer.
If 0,Athen
12 .
Ax Ax
yCe Ce





12
12
1200 0
00
00
AA
yCC
CC y
yCeCe



<
=
>
=?

If
0,Athen 0yand the initial condition gives 0.y
Section 16.3 Second-Order Nonhomogeneous Linear Equations
1.

2
2
2
222
22cos
42sin
82cos
8 2 cos 2 cos 10
x
x
x
x xx
ye x
ye x
ye x
yy e x e x e





2.

1
2
11
22
1
2
11
22
2sin sin
'2cos cos sin
2sin sin cos
2 sin sin cos 2 sin sin cos
yxxx
yxxxx
yxxxx
yy xxx x x x x




3.

3sin cos ln sec tan
3cos 1 sin ln sec tan
3sin tan cos ln sec tan
3 sin tan cos ln sec tan 3 sin cos ln sec tan tan
yxxxx
yx xxx
yxxxxx
yy xxxxx xxxx x




4.
5 ln sin cos sinyxxxx
5 ln sin sin cos cot sin cos 6 sin sin ln sin cos cotyxxxxxxxxxxxxx

2
6 cos cos cos ln sin cos csc 1 sin cot
5 cos cos ln sin csc cot sin
yxxxxxxxxx
xxx xxxx



cos 5 ln sin csc cot sin 5 ln sin cos sin csc coty y x x xxxx x xxx xx

473. Section 16.3 Second-Order Nonhomogeneous Linear Equations 473
© 2010 Brooks/Cole, Cengage Learning
5. 71231
7120
yy yx
yy y




2
712 3 40mm m m when 3, 4m

34
12
01
1,
0
x x
h
p
pp
yCe Ce
yAAx
yAy





101712 712 31
pp pyy y A AAxx

1
1
10 16
10
1
4
12 3
,
712 1
A
AA
AA
<=
>
=?

Solution:
11
16 4
pyx
6. 64
60
yy y
yy y




2
6320mm m m when 3, 2m

32
12
,0
x x
h
ppp
yCe Ce
yAyy




2
3
664
pp pyy y A A
Solution:
2
3
py
7.
3
816
8160
x
yy ye yy y




2
2
816 4 0mm m when 4m

44
12
333
,3,9
xx
h
x xx
pp p
yCe Cxe
yAey Aey Ae





33 3
3
816 9 83 16
x xx
pp p
x
yy y Ae Ae Ae
e



924161 1AAA A
Solution:
3
x
pye
8.
2
3
30
x
yy ye yy y



2 111
30
2
mm m




222
,2,4
x xx
pp p
yAey Aey Ae


22 22
34 2 3
x xxx
yy y Ae Ae Ae e

1
4231
9
AAA A


Solution:
21
9
x
pye
9. 215sin
2150
yy y x
yy y



2
215 5 30 5,3mm m m m

sin cos
cos sin
sin cos
p
p
pyAxBx
yAxBx
yAxBx




2 15 sin cos 2 cos sin 15 sin cos sin
pp py y y A xB x A xB x A xB x x

215sin 215cos sin
ABAxBABx x

16 2 1 41
,
216 0 65 130
AB
AB
AB
<
>
?


Solution:
41
sin cos
65 130
pyxx

10.
2
45 cos
450
44
450 2
2
x
yyyex
yyy
mm m i





cos sin cos sin
xxx
p
yAe xBexeAxBx

cos sin sin cos sin cos
xx x
p
yeAxBxeAxBxeBAxAB x

sin cos cos sin 2 sin 2 cos
xx x
p
y eBA xAB xeBA xAB xeAxBx

4 5 2 sin 2 cos 4 sin cos 5 cos sin cos
xx xxx
ppp
yyyeAxBxeBAxABx AexBexex

24 5sin 24 5cos cos
A BA B x B AB A x x

69 0 32
,
961 39 39
AB
AB
AB
<
>
?


Solution:
32
cos sin
39 39
xx
p
yexex

474. 474 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
11. 322
320
yyyx
yyy




2
320mm when 1, 2.m

2
12
01
1
0
x x
h
p
p
p
yCeCe
yAAx
yA
y





01 132 23 2 2
pp pyyy AA Axx

01
3
10 2
1230
1,
22
AA
AA
A
<=
>
=?


2 3
12 2xx
yCe Ce x
12.
2
23 1
230
yyyx
yyy




2
230mm when 1, 3.m

3
12
2
01 2
12
2
2
2
xx
h
p
p
p
yCe Ce
yAAxax
yAAx
yA









2
212
2
012
23 3 34
322 1
pppyyy Ax AAx
AAAx




2
5 41
12 0 1 2 27 9 3
01231
34 0 , ,
322 1
A
AA A A A
AAA
<
=
>
=

?


32 514
12 3927xx
yCe Ce x x


13. 22
20
x
yye yy




2
20mm when 0, 2.m

2
12x
h
x
ppp
yCCe
yAeyy





23 2
x x
pp
yy Aee or
2
3
A

2 2
12 3
x x
yCCe e


14.
3
95
90
x
yye yy




2
90mwhen 3, 3.m



33
12
3
3
3
31
96
x x
h
x
p
x
p
x
p
yCe Ce
yAxe
yAex
yAex







33
96 5
x x
pp
yy Ae e or
5
6
A


33 5
12 6
x x
yCe C xe


15. 10 25 5 6
10 25 0
x
yyy e yyy



2
10 25 0mm when 5, 5.m

55
12
01
1
x x
h
x
p
x
pp
yCe Cxe
yAAe
yyAe




0110 25 25 16 5 6
x x
pp p
yy y AAe e
or
31
01 58
,AA


5 3 1
12 85
xx
yCCxe e
16. 16 8 4
16 8 0
x
yyy xe yyy



2
16 8 1 0mm when
11
44
,.m


14
12
01 2
12
2
x
h
x
p
x
p
x
p
yCCxe
yAAxAe
yAAe
yAe





011 216 8 8 9
44
x
ppp
xyyyAAAxAe
xe



or
4
2109
,4,32AAA


14 4
12 9
32 4
x
x
yCCxe xe
17. 9sin3
90
yy x
yy




2
90m when 3,3.mii



12
01 2 3
03 1
12 3cos 3 sin 3
sin3 sin3 cos3 cos3
96sin39sin3
69cos39cos3
h
p
pyC xC x
yA xAx xA xAx x
yAAxAxx
AA xAx x





31
1
13 696sin36cos3sin3,
0,
ppyy AxA x x
AA




1
126
cos 3 sin 3yC x xC x

475. Section 16.3 Second-Order Nonhomogeneous Linear Equations 475
© 2010 Brooks/Cole, Cengage Learning
18.
2
322
320
x
yyye yyy





3
320mm when 1, 1, 2.m




2
12 3
22
01
22
01 1
22
01 1
22
01 1
22
44 4
812 8
xx x
h
xx
p
xx
p
xx
p
x x
p
yCeCxeCe
yAe Axe
yAAeAxe
yAAe Axe
yAAeAxe












22 2
11 9
22
12 3 9
32 9 2 or
xx
pp p
xx x
yyy Ae e A
yCe Cxe C xe





19.
3
,0 1, 0 0
0
yyxy y
yy




2
10mwhen , .mii

12
23
01 2 3
2
12 3
23cos sin
23
26
h
p
p
pyC xC x
yAAxAxAx
yAAxAx
yAAx







32
32 13 02
3
62
ppyy AxAx A AxA A
x



or
32 1 01, 0, 6, 0AA A A

3
12
2
12
cos sin 6
sin cos 3 6
yC xC xx x
yCxCxx



Initial conditions:
12201,00,1 ,0 6, 6yy CCC
Particular solution:
3
cos 6 sin 6yx xxx
20. 44,01,06
40
yy y y
yy




2
40mwhen 2, 2.mii

12
0cos 2 sin 2
0
h
p
pyC xC x
yA
y





0444
ppyy A or
01A

12
12cos 2 sin 2 1
2sin2 2 cos2
yC xC x
yCxCx



Initial conditions:
1
12201,06,1 1,
0, 6 2 , 3
yy C
CCC


Particular solution:
3sin2 1yx
21. 2sin , 0 0, 0 3
0
yy xy y
yy




2
0mmwhen 0, 1.m

12
cos sin
sin cos
cos sin
x
h
p
p
p
yCCe
yAxBx
yAxBx
yAxBx







cos sin 2 sin
ppyy AB x AB x x

0
1, 1
2
AB
AB
AB
<
>
?



12
2 cos sin
sin cos
x
x
yCCe x x
yCe x x



Initial conditions:

12 2
21
00,0 3,
01,31,
2, 1
yy
CC C
CC



Particular solution:
12 cos sin
x
yexx


22. 23cos2,0 1,02
20
yy y xy y
yy y



2
20mmwhen 1, 2.m

2
12
cos 2 sin 2
2sin2 2cos2
4cos2 4sin2
xx
h
p
p
p
yCeCe
yA xBx
yAxBx
yAxBx







262cos226sin2
3cos2
pp pyy y AB x AB x
x




93
20 20
62 3
,
26 0
AB
AB
AB
<
>
?


2 93
12 20 20
2 93
12 10 10
cos 2 sin 2
2sin2cos2
xx
xx
yCe Ce x x
yCe Ce x x




Initial conditions:


9
12 20
3
12 10
01,02,1 ,
22
yy CC
CC




11
12 20 31
12 5417
12 10
,
2
CC
CC
CC
< =
>
=?

Particular solution:


21
20
415 9cos23sin2
xx
yee x x

476. 476 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
23.
4 1
3
'4 , 0
'4 0
xx
y y xe xe y
yy



40m when 4.m




4
24
01 2 3
01 1
24 4
23 2 3
42
x
h
xx
p
xx
p
x x
yCe
yAAxeAxAxe
yAAxeAe
AxAxe A Axe







4
01 1 2
44
3
433
2
x xx
pp
xx x
yy AAxeAeAe
Axe xe xe




11 1
0123 93 2
2411
29
,,0,
13
xx
AAAA
yCxe xe



Initial conditions:

11 1 4
33 9 9
0, ,yCC
Particular solution:

2441 1
92 9
13
x x
yxe xe
24.
2
2sin,
25
20
yy xy
yy





20m when 2.m

2
cos sin
sin cos
x
h
p
p
yCe
yAxBx
yAxBx







2
2 sin cos 2 cos sin
2sin2cossin
21
21,20 ,
55
12
cos sin
55
pp
x
hpyy AxBxAxBx
BAx AB x x
BA AB B A
yy y Ce x x






Initial conditions:
22 2
,,0
255 5
yCeC






Particular solution:
21
sin cos
55
yxx

25. sec
0
yy x
yy




2
10mwhen , .mii

12
12cos sin
cos sin
h
pyC xC x
yvxvx




12
12cos sin 0
sin cos sec
vxvx
vxvx x




1
0sin
sec cos
tan
cos sin
sin cos
x
xx
vx
xx
xx




1 tan ln cosvxdxx *


2
cos 0
sin sec
1
cos sin
sin cos
x
xx
v
xx
xx





2vdxx*


12ln cos cos sinyC x xCx x
26. sec tan
0
yy xx
yy




2
10mwhen , .mii

12
12cos sin
cos sin
h
pyC xC x
yv xv x



12cos sin 0vxvx

12sin cos sec tanvxvxxx

2
1
0sin
sec tan cos
tan
cos sin
sin cos
x
xx x
vx
xx
xx





22
1
tan sec 1
tan
vxdxxdx
xx

**


2
cos 0
sin sec tan
tan
cos sin
sin cos
x
xx x
vx
xx
xx





2 tan ln cos ln secvxdx x x*




12
12cos sin tan cos
ln sec sin
tan cos ln sec sin
hpyy y
CxCxx x x
xx
Cx x xC x x

477. Section 16.3 Second-Order Nonhomogeneous Linear Equations 477
© 2010 Brooks/Cole, Cengage Learning
27. 4csc2
40
yy x
yy




2
40mwhen 2,2.mii

12
12cos 2 sin 2
cos 2 sin 2 0
h
pyC xC x
yv xv x




12
12cos 2 sin 2 0
2sin2 2cos2 csc2
vxvx
vxvxx




1
0sin2
csc 2 2 cos 2 1
cos 2 sin 2 2
2sin2 2cos2
x
xx
v
xx
xx




1
11
22
vdxx
*


2
cos 2 0
2sin2 csc2 1
cot 2
cos 2 sin 2 2
2sin2 2cos2
x
xx
vx
xx
xx





2
11
cot 2 ln sin 2
24
vxdxx
*


12
11
cos 2 ln sin 2 sin 2
24
yC x xC x x





28.
22
44
440
x
yyyxe yyy




2
440mm when 2, 2.m



2
12
2
12
x
h
x
p
yCCxe
yvvxe





22
12
2222
12
0
221
xx
x xx
ve v xe ve v x e xe





2
22 2 34
3
1
422
22
0
21
221
x
xx x
xxx
xx
xe
xe x e xe
vx
eexe
exe





34 1
1 4
vxdxx *


2
222 24
2
2
44
0
2
x
xx x
xx
e
exe xe
vx
ee



23 1
2 3
vxdxx*


42
121
12
x
yCCx xe






29. 2ln
20
x
yyyex
yyy




2
210mm when 1, 1.m



12
12
x
h
x
p
yCCxe
yvvxe





12
12 0
1ln
xx
xxx
ve v xe
ve v x e e x




1
22
1
2
2 ln
ln ln
24
ln
ln ln
vxx
x x
vxxdx x
vx
vxdxxxx




*
*




2
2
12
ln 3
4
x
x
xe
yCCxe x

30.
2
44
440
x
e
yyy
x
yyy




2
440mm when 2, 2.m



2
12
2
12
x
h
x
p
yCCxe
yvvxe





22
12
2
22
12
0
221
xx
x
xx
ve v xe
e
ve v x e
x



1
1
2
2 1
1
1
1
ln
v
vdxx
v
x
vdxx
x




*
*



2
12
ln
x
yCCxxxxe
31. 12 0yy y

2
12 4 3 0 4, 3mm m m m

Let
2
.
pyAxBxC This is a generalized form of

2
.Fx x
32. 12 0yy y

2
12 4 3 0 4, 3mm m m m

Because
43
12
,
x x
h
yCe Ce

let

4
.
x
pyAxe

478. 478 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
33. Answers will vary. See the “Variation of Parameters”
box on page 1163.
34. (a) Because 0
pyand 3 3 4 12.
py
(b) 2
py
(c) 4
py
35. 10 25 6 sin 5 , 0 0, 0 0qqq tq q

2
10 25 0mm when 5, 5.m


5
12
cos 5 sin 5
5sin5 5cos5
25 cos 5 25 sin 5
t
h
p
p
p
qCCte
qAtBt
qAtBt
qAtBt








3
25
5 3
12 25
10 25 50 cos 5 50 sin 5
6sin5, , 0
cos 5
pp p
tqqq BtAt
tA B
qCCte t




Initial conditions:


3
11225
33
12 25 5
0 0, 0 0, 0, 5 0,
,qq C CC
CC



Particular solution:

553
25
5cos5
tt
qete t


36. 20 50 10 sin 5qqq t

2
20 50 0mm when
10 5 2.m

1052 1052
12
cos 5 sin 5
5cos5 5sin5
25 cos 5 25 sin 5
tt
h
p
p
p
qCe Ce
qAtBt
qBtAt
qAtBt







20 50 25 100 cos 5
25 100 sin 5 10 sin 5
pppq qq ABt
B At t




25 100 0 28
,
85 8525 100 10
AB
BA
BA
<
>
?


1052 1052
12 82
cos 5 sin 5
85 85tt
qCe Ce t t


Initial conditions:



12
12
12
8
00,00, 0,
85
2
10 5 2 10 5 2 0,
17
872 872
,
170 170
qq CC
CC
CC




Particular solution:

10 5 2 10 5 2872 872
170 170
82
cos 5 sin 5
85 85 tt
qe e
tt





37.

24 24 1
32 32 4
48 48 sin 4 , 0 , 0 0yy ty y

224
32
48 0m when 8.mi

12cos 8 sin 8
sin 4 cos 4
4cos4 4sin4
16 sin 4 16 cos 4
h
p
p
pyC tC t
yAtB t
yAtBt
yAtBt







24
32
24
32
48 36 sin 4 36 cos 4
48 sin 4 , 0, 1
ppyy AtBt
tB A



12cos 8 sin 8 sin 4
hpyy y C tC t t
Initial conditions:


11
144
1
22 2
0,00, ,
08 4yy C
CC


Particular solution:
11
42
cos 8 sin 8 sin 4yttt
38.

22 1
32 32 4
44sin8,0,00yy ty y

22
32
40m when 8.mi


12cos 8 sin 8
sin 8 cos 8
64 16 sin 8 16 64 cos 8
h
p
pyC tC t
yAttBt t
yAtBtABtt






2
32
21
32 4
4sin8cos8
4sin8 , 0,
ppyy BtAt
tA B



1
12 4
cos8 sin8 cos8yC tC t t t
Initial conditions:


11 11
12 244 432
0,00, ,08yy CC C
Particular solution:
111
4324
cos 8 sin 8 cos 8ytttt

39.

221
32 32 4
44sin8,0,03yy y ty y

21
16
40mm when 8, 8.m


8
12
sin 8 cos 8
8cos8 8sin8
64 sin 8 64 cos 8
t
h
p
p
p
yCCte
yAtB t
yAtBt
yAtBt








2
32
2
32
11
432
48sin88cos8
4sin8 8
,8 0 0
pp pyy y B tA t
tB
BAA




Initial conditions:


9111
11443232
3
12 2 4
0,03, ,
38yy C C
CC C



Particular solution:

893 1
32 4 32
cos 8
t
yte t


2
−2
0
2
0.3
−0.05
04
2
−2
06.28

479. Section 16.3 Second-Order Nonhomogeneous Linear Equations 479
© 2010 Brooks/Cole, Cengage Learning
40.
4125 1
0, 0 , 0 4
32 2 2 2
yy yy y

21125
0
822
mm

2
4 100 0mm when
246.mi

22
12
cos 4 6 sin 4 6
tt
yCe t Ce t


Initial conditions:


112
2
11
0,04, ,4246,
22
36
846
yy C CC
C



Particular solution:

2216
cos 4 6 sin 4 6
28
tt
ye t e t


41. In Exercise 37,


11 5 15 15
cos 8 sin 8 sin 8 arctan sin 8 arctan sin 8 2.6779 .
42 4 24 24
hytt t t t






42. (a)


254
32 2
12
11
122
2
22 5
12
25
0
cos 10 sin 10
0:
04:410
cos 10 sin 10
yy
yC xC x
yC
yCC
yxx






The motion is undamped.
(b) If 0,bthe motion is damped.
(c) If
5
2
,bthe solution to the differential equation is
of the form
12
12 .
mx m x
yCe Ce There would be
no oscillations in this case.
43.
2
4lnxy xy y x x

1yxand
2 lnyxx

12 1 2ln 0 lnux u x x u u x


12 2
44
1ln ln lnuu x x u x
xx


and

2
14
lnux
x




3
14
ln
3
ux
and
2
2
2lnux


33 342
ln 2 ln ln
33
pyxxxxxx


3
122
ln ln
3
hpy y y CxCxx x x
44. Let sin ln cos ln .
pyA xB x



111
cos ln sin ln cos ln sin ln
pyA x B x A xB x
xxx



2 22
111111
cosln sinln sinln cosln sinln cosln
pyAxBx AxBx BAx ABx
xxxxxx








2
4sinln coslncoslnsinln
4 sin ln cos ln sin ln
p
xyxy yBA x AB x A xB x
AxB x x



4sinln 4cosln sinln
BAB A x AB A B x x

1
31,30
3
AB A

So,

1
sin ln
3
pyx and

22
12 1
sin ln cos ln sin ln .
3
hpyy y C x C x x
0.5
−0.5
01.57

480. 480 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
45. True.


2
22 2
22
cos
sin 2 cos sin 2 cos
cos sin 2 sin 4 cos
xx
p
xxx x xxx x x
p
x xxxx xxx x x
p
yee
yeee ee eeee
yeeeee eeeee








So,

222
222
3 2 cos sin 2 sin 4 cos 3 sin 2 cos 2 cos
23sin 14 6 2cos cos.
x xxxxxx xxxx xx
pp p
xxx x x x x x x
y y y eee ee e e ee e e e e
eee e e e e e e







46. True.


222111
842
22211
24
,,
66
x xx
ppp
x xx
pp
yeyeye
yy e e e


47. 22
x
yyye

2
210 1,1mm m

2
12
,,
x xx
hp
yCeCxey xe particular solution
General solution:

22
12 12
x xx
fx C Cxe xe C Cx x e


22
212 2 12
22
x x
fxC xCCxxe xC xCCe
(a) No. If
0fxfor all ,
xthen
22
21 2 1
040xCxC C C& for all .
x
So, let
12 1.CC Then
2
32
x
fxxxe and
1
43
2
0.f
(b) Yes. If
0fxfor all ,
xthen


2
212
2
21
2
21
2
21
24 0
440
44
40
CCC
CC
CC
CC






0fx for all .
x
Section 16.4 Series Solutions of Differential Equations
1. 0.yy Letting
0
:
n
n
n
yax
!

F

1
1
100 0
10
nn nn
nn nn
nnn n
yy nax ax n ax ax
!!! !



FFF F


1
11
1
nn
n
nna a
a
a
n





10 2 0 0
102 3
0
0
0
,, ,,
22 3123 !
!
n
nx
n
aa a a a
aaa a a
n
a
yxae
n
!


""

F


Check: By separation of variables, you have:

1ln
x
dy
dx
y
yxC
yCe



**

481. Section 16.4 Series Solutions of Differential Equations 481
© 2010 Brooks/Cole, Cengage Learning
2. 0.yky Letting
0
:
n
n
n
yax
!

F



1
1
100 0
1
1
23
10 2 0
102 3 0
00 0
00
10
1
1
,, ,,
22 3123 !
!!
nn nn
nn nn
nnn n
nn
n
n
n
n
n
n
nkx
nn
yky nax kax n ax kax
na ka
ka
a
n
ka k a ka k a k
akaa a a a
n
kxk
yaxa ae
nn
!!! !





!!






""
FFF F
FF



Check: By separation of variables, you have:

1ln
kx
dy
kdx
y
ykxC
yCe



**

3. 90.yy Letting
0
:
n
n
n
yax
!

F




2
2
200 0
2
2
919 2190
21 9
9
21
nn nn
nn nn
nnn n
nn
n
n
y y nn ax ax n n a x ax
nna a
a
a
nn
!!! !








FFF F


00 11
01
23
22
20 3 1
45
01
221
99
232
99 9 9
43 4321 54 54321
99
2! 2 1!
nn
nn
aa aa
aa
aa
aa a a
aa
aa
aa
nn



"

"""" """""












221
01 1221
001
00 0 0 00
33 3399
2! 21! 2!3 21! ! !
nn nn
nn
nn
nn n n nn
x xxxaa a
yx xa C C
nn nn nn

!! ! ! !!





FF FF FF


33
01
x x
Ce Ce

where
01 0CCa and
1
01
.
3
a
CC
Check: 90yy is a second-order homogeneous linear equation.

2
1
90 3mm and
2 3m

33
12
x x
yCe Ce


4.
01 .
kx kx
yCe Ce

Follow the solution to Exercise 3 with 9 replaced by
2
.k

482. 482 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
5. 40.yy Letting
0
:
n
n
n
yax
!

F




2
2
200 0
2
2
414 2140
21 4
4
21
nn nn
nn nn
nnn n
nn
n
n
y y nn ax ax n n a x ax
nna a
a
a
nn
!!! !









FFF F






00 11
01
23
22
0123
45
20211
44
232
4444
43 4! 54 5!
14 14
2! 2 1!
nn
nn
nn
aa aa
aa
aa
aaaa
aa
aaa a
nn




"


""














221
01 1221
001
00 0 0
14 14 1 2 1 2
cos 2 sin 2
2! 21! 2! 2 21!
nn nnnn
nn
nn
nn n n
aa x x a
yx xa CxCx
nn n n

!! ! !





FF FF

Check: 40yy is a second-order homogeneous linear equation.

2
12
40 2
cos 2 sin 2
mmi
yC xC x



6.
01cos sin .yC kxC kx Follow the solution to Exercise 5 with 4replaced by
2
.k
7. 30.yxy Letting
0
:
n
n
n
yax
!

F

11
10
330
nn
nn
nn
y xy nax ax
!!


FF


11
21
10
230
nn
nn
nn
nax ax a
!!



FF and
2
3
2
n
na
a
n








00 1
01
23
2
01
40 5
3
23 2
01
60 7
33
34 3
00 1
89
34 0
33
0
23
33 3 33
0
42 2 53
33 3 33
0
62 232 735
33 3 33
0
8 232 2432 9 357
aa a
aa
aa
aa
aa a
aa
aa a
aa a
aa








""



""" ""



2
0
0
3
2!
n
n
n
n
x
ya
n
!



F



1
22 2
1
1 2
3 2! 3
lim lim lim 0
21! 21 3
n
n n
n
nn nnn n
n
xunx
un n x




! ! !

"


The interval of convergence for the solution is
,.! !

483. Section 16.4 Series Solutions of Differential Equations 483
© 2010 Brooks/Cole, Cengage Learning
8. 20.yxy Letting
0
:
n
n
n
yax
!

F


11
10
11
21
10
2
220
2 2 0 and
2
2
nn
nn
nn
nn
nn
nn
n
n
yxy nax ax
nax ax a
a
a
n
!!


!!







FF
FF

01
01
20 3
2
000 1
45
2
23 2
000 1
67
23
3
00 1
8900
22
0
23
22 2 22
0
42 2 2 2 53
22 2 22
0
62 2 2 3 2 3! 73 5
222
0
83! 4! 935 7
aa
aa
aa a
aaa a
aa
aaa a
aa
aa a
aa





"


""" "


""

2
0
0
!
n
n
x
ya
n
!

F


22 2
1
!
lim lim lim 0
1! 2 1
n
n
n
nn n
n
uxnx
un n


! ! !
"


The interval of convergence for the solution is ,.! !
9. 0.yxy Letting
0
:
n
n
n
yax
!

F




21
21
2
20
2
00
2
10
1
21
21
nn
nn
nn
nn
nn
nn
nn
nn
nn
n
n
yxy nn ax xnax
nn ax nax
nnax nax
na
a
nn
!!


!!


!!







FF
FF
FF

00 11aa aa

1
23
0
32
a
aa
"

There are no even powered terms.
31
5
51
733
54 5!
553
76 7!
aa
a
aa
a

"
"

"






21 21 21
01 01 01
000
1357 2 1 2!
21! 2!21! 2!21
nn n
nn
nnn
nx nx x
ya a a a a a
nnnnn
!!!

"""


FFF






223
1
121
2!21 21
lim lim lim 0
21!23 2123
nn
n
nn
nn n
n
nn n xux
unnx nn



! ! !

"


Interval of convergence:
,! !

484. 484 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
10. 0.yxyy Letting
0
:
n
n
n
yax
!

F


21
210
2
00
2
10
21 1
2
nnn
nnn
nnn
nn
nn
nn
n
n
y xy y nn ax x nax ax
nnax nax
a
a
n
!!!


!!






FFF
FF

00 11
01
23
20 0 3 1
45
2
40 5 1
67
3
6
071
89
4
23
4822! 535
623! 7357
824! 93579
aa aa
aa
aa
aa a a a
aa
aa a a
aa
aa a a
aa



"

""

"""



221
01
00
2! 1357 2 1
nn
n
nn
xx
ya a
nn
!!


""" FF






22 2
1
12
23
1
21
2!
lim lim lim 0
21! 21
1357 2 1 2
lim lim lim 0
135723 23
nn
n
nn
nn n
n
n
n
n
nn n
n
uxnx
unxn
nux x
unxn



! ! !



! ! !
"

"""
"
"""



Because the interval of convergence for each series is ,,! !the interval of convergence for the solution is
,.! !
11.
2
40.xyy Letting
0
:
n
n
n
yax
!

F




2 2
22 0
2
2
00
2
2
4141
14210
1
42 1
nnn
nnn
nn n
nn
nn
nn
n
n
x yy nnax nnax ax
nn ax n n ax
nn a
a
nn
!! !


!!







FF F
FF




00 11
00 11
23
20 3 1
45
24 3 5
01
42 1 8 43 2 24
377
4 4 3 128 4 5 4 1920
7
1
8 128 24 1920
aa aa
aa aa
aa
aa a a
aa
xx x x
ya ax

485. Section 16.4 Series Solutions of Differential Equations 485
© 2010 Brooks/Cole, Cengage Learning
12.
2
0.yxy Letting
0
:
n
n
n
yax
!

F



222
20
22
4
20
4
10
43
43
nn
nn
nn
nn
nn
nn
n
n
yxy nn ax ax
nnax ax
a
a
nn
!!


!!








FF
FF

Also:



23
01 2 3
2
23
223
230415
23 40 51
232 1
232 43 54 0
2 0, 6 0, 12 0, 20 0
n
n
n
n
ya axax ax ax
ya ax nnax
yxy a axa ax a ax
aa aa aa


"
" " "



So,
20aand
3671000,0,0,aaaa and
110.aTherefore,
42 0
na
and
43 0.
na


00 11
01
45
40 51
89
80 91
12 13
43 54
87 8743 98 9854
12 11 12 11 8 7 4 3 13 12 13 12 9 8 5 4
aa aa
aa
aa
aa aa
aa
aa aa
aa


""

" """ " """

" " """" " " """"


48 12
2
0
57 9
1
1
43 8743 12118743
54 9854 13129854
xx x
yxya
xx x
ax


" """ " """"


"""" """""

13. 21 0,0 2yxy y







44
234
12 0 0
12 2 0 2
12 4 0 10
12 6 0 2
22 10 2
2
1! 2! 3! 4!
yxy y
yxyyy
yxyyy
yxyyy
yx x x x x








Using the first five terms of the series,
1163
2.547.
264
y





Using Euler’s Method with 0.1x5 you have
12 .yxy

i x
i y
i
0 0 2
1 0.1 2.2
2 0.2 2.376
3 0.3 2.51856
4 0.4 2.61930
5 0.5 2.67169

486. 486 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
14. 20,01yxy y

















44
54 5
654 6
24 6
246
200
202
22 00
23 012
24 00
2 5 0 120
2 12 120
1
2! 4! 6!
11
1
26
yxy y
yxyy y
yxyy y
yxyy y
yxyy y
yxyy y
yx x x x
xxx













Using the first four terms of the series,

8
1 2.667.
3
y
Using Euler’s Method with 0.1x5 you have 2 .yxy

So,
1 2.335.y
15. Given a differential equation, assume that the solution
is of the form
0
.
n
n
n
yax
!

F Then substitute y and its
derivatives into the differential equation. You should
then be able to determine the coefficients
01,, .aa
16. A recursion formula is a formula for determining
the next term of a sequence from one or more of the
preceding terms. See Example 1.

17. (a) From Exercise 9, the general solution is


21
01
0
.
2!2 1
n
n
n
x
ya a
nn
!


F





0
2 2
11
2
00
100 0
21
2!2 1 2!
02
n n
n
nn
ya
nx
x
ya a
nn n
ya
!!






FF



21
0
2
2!2 1
n
n
n
x
y
nn
!


F
(b)


33
3
35 35
5
22
23 3
22 2
3425 320
xx
Px x x
x xxx
Px x x



"

""


(c) The solution is symmetric about the origin.
18. (a)
2
90 3mmi

12cos 3 sin 3yC xC x


1
12
2202
3sin3 3 cos3
063 2
yC
yCxCx
yCC




2cos3 2sin3
pyxx
i x
i y
i
0 0 1
1 0.1 1
2 0.2 1.02
3 0.3 1.0608
4 0.4 1.1244
5 0.5 1.2144
6 0.6 1.3358
7 0.7 1.4961
8 0.8 1.7056
9 0.9 1.9785
10 1.0 2.3346
−4
−12
12
4
P
3
(x)
P
5
(x)

487. Section 16.4 Series Solutions of Differential Equations 487
© 2010 Brooks/Cole, Cengage Learning
(b) Let
12
01 2
,,1
nn n
nn n
nn n
yaxy naxy nnax
!! !


FF F


2
20
9190
nn
nn
nn
y y nn ax ax
!!


FF




2
00
2
221 9 0
21 9
9
21
nn
nn
nn
nn
nn
nnax ax
nna a
aa
nn
!!








FF
For
neven, For nodd,



20
2
42 0
3
60
9
2
99
43 4!
9
6!
aa
aa a
aa




"





31
2
53 1
3
71
9
32
99
54 5!
9
7!
aa
aa a
aa


"


"



and in general,


20
9
2!
n
n
aa
n

and in general,


21 1
9
21!
n
n
aa
n





So,









221
221
00 0
221
221
0101
00 0 0
9 9 13 13
2! 2 1! 2! 2 1!
nn n
nn n
nn n
n n nn nn
nn
nn n n
yax ax ax
xx
ax ax a a
nn n n
!! !




!! ! !





FF F
FF F F
Applying the initial conditions,
01 2,aa and




221
00
13 13
2.
2! 2 1!
nn nn
nn
xx
y
nn

!!





FF
(c)

19. 20,01,03yxy y y



















44
55
64 6
754 7
200
202
22 012
23 00
24 016
2 5 0 120
yxy y
yxyy y
yxyy y
yxyy y
yxyy y
yxyy y









346 73 2 12 16 120
1
1! 3! 4! 6! 7!
yxxxxx


Using the first six terms of the series,
1
0.253.
4
y





20. 20,01,02yxyy y y










44
54 5
654 6
765 7
201
202
23 03
25 010
27 021
29 090
yxyy y
yxyy y
yxyy y
yxyy y
yxyy y
yxyy y









234 5 6 72 1 2 3 10 21 90
1
1! 2! 3! 4! 5! 6! 7!
y xxxx x x x


Using the first eight terms of the series,
1
1.911.
2
y





x
y
y
12
3
P
3
P
5
2

488. 488 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
21.
2
cos 0, 0 3,yxy xy y 02y

2
cosyxy xy 03y


22
2sincosyxyxy xyxy 02y

2323 2
3
1! 2! 3!
yxxx

Using the first four terms of the series,
1
3.846.
3
y





22. sin 0, 0 2, 0 1
x
yey xy y y
sin ,
x
yey xy 01y


cos sin
cos sin
xx
x
yeyey xy xy
ey y xy xy


01122y

2311 2
2
1! 2! 3!
yxxx

Using the first four terms of the series,
1
1.823.
5
y





23. ,,0.
xx
fx e f x e y y
Assume
0
,
n
n
n
yax
!

F then:


1
1
1
10
1
00
1
1
,0
1
n
n
n
nn
nn
nn
nn
nn
nn
n
n
ynax
na x a x
nax ax
a
an
n
!


!!


!!






/
F
FF
FF





10
10
2
20
3
30
4
40
5
00
10,
1,
22
2,
323
3,
4234
4,
52345
1! !
nn
naa
aa
na
aa
na
aa
na
aa
na
aa
aa
nn











0
0
!
n
n
x
ya
n
!

F which converges on ,.! !When
01,ayou have the Maclaurin Series for .
x
fxe
24. cos , sin , cos ,
0.
fxxfx xfx x
yy



Assume
0
,
n
n
n
yax
!

F then:




2
2
22
20
2
00
2
1
10
21
,0
12
n
n
n
nn
nn
nn
nn
nn
nn
n
n
ynnax
nn ax ax
nnax ax
a
an
nn
!


!!


!!






/
F
FF
FF








00 11
01
23
20 31
45
01
221
12 23
34 4! 45 5!
11
2! 2 1!
nn
nn
aa aa
aa
aa
aa aa
aa
aa
aa
nn













221
01
00
11
2! 2 1!
nn
nn
nn
xx
ya a
nn
!!




FF which
converges on
,! !
When
01aand
10,ayou have the Maclaurin
Series for
cos .
fxx

489. Section 16.4 Series Solutions of Differential Equations 489
© 2010 Brooks/Cole, Cengage Learning
25.




2
2
2
2
2
arctan
1
1
2
1
2
1
120
fxx
fx
x
x
fx
x
x
yy
x
xy xy











Assume
0
,
n
n
n
yax
!

F then:






1
1
2
2
22
200
2
200
2
00
2
2
1
12 1 120
112
21 1
21 1
,0
2
n
n
n
n
n
n
nnn
nnn
nnn
nnn
nnn
nnn
nn
nn
nn
nn
nn
ynax
ynnax
x y xy nn ax nn ax nax
nn ax nn ax nax
nnax nnax
nna nna
n
aan
n
!


!


!!!


!!!


!!










/
F
F
FFF
FFF
FF


200na all the even-powered terms have a coefficient of 0.

31
531
751
971
1
1,
3
31
3,
55
51
5,
77
71
7,
99
naa
naaa
naaa
naaa







1
211
21
n
n
a
a
n







21
1
0
1
21
n
n
n
x
ya
n
!




F which converges on 1, 1 .When
11,ayou have the Maclaurin Series for arctan .
fxx

490. 490 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
26.




2
32
2
22
2
2
arcsin
1
1
1
1
111
10
fx x
fx
x
x
fx
x
xx
yy
xxx
xy xy





"



Assume
0
,
n
n
n
yax
!

F then:


2
200
2
2
00
2
2
110
21
,0
12
nnn
nnn
nnn
nn
nn
nn
nn
ann x ann x anx
nnax nax
n
aan
nn
!!!


!!





/
FFF
FF

200na all the even-powered terms have a coefficient of 0.




















11
31
53 1 1
75 1 1
97 1 1
11 9 1 1
21 1 2
1
1,
23
99 3
3,
45 2345 245
925 3525
5,
67 234567 2467
92549 35749
7,
89 23456789 2468
9254981 357981
9,
1011 234567891011 24681011
2!
2! 2 1
n
n
aa
naa
naaaa
naaaa
naaaa
naa aa
n
aa
nn














21
1 2
0
2!
2! 2 1
n
n
n
n
ya x
nn
!



F which converges on 1, 1 .When
11,ayou have the Maclaurin Series for arcsin .
fxx
27. 0.yxy Let
0
.
n
n
n
yax
!

F


211
3
201 0
1
23
0
1320
232 0
nn nn
nn n n
nnn n
n
nn
n
yxy nn ax xax n n ax ax
annaax
!!! !



!





FFF F
F

So,
20aand

3
32
n
na
a
nn


for 0, 1, 2,n
The constants
0aand
1aare arbitrary.

00 11
01
34
30 41
67
32 43
65 6532 76 7643
aa aa
aa
aa
aa a a
aa


""

"""" """"

So,
01 0 134 6 7
01
.
6 12 180 504
aa a a
ya ax x x x x

491. Review Exercises for Chapter 16 491
© 2010 Brooks/Cole, Cengage Learning
y
x
42−4
4
−4
Review Exercises for Chapter 16
1.
32
0
12 1
yx xydxxdy
MN
xy
yx

CC
#
CC

Not exact
2. 550
1
x y dx y x dy
MN
yx

CC

CC

Exact
3. 10 8 2 8 5 2 0x y dx x y dy
Exact: 8
MN
yx
CC

CC






2
2
1
22
1
22
,1082582
,8 852
52
5
2
2
5
,582 2
2
5
582 2
2
y
fxy x y dx x xy x gy
fxy x gy x y
gy y
gy y y C
fxy x xy x y y C
xxyxy yC





*

4.
32
22 6 0xyydxxxydy
Exact:
2
61
MN
y
yx
CC

CC







3
23
22
1
23
1
23
,22
2
,6 6
0
,2
2
y
fxy x y ydx
xxyxygy
fx y xy x g y x xy
gy
gy C
fxy x xy xy C
xxyxyC







*


5. 53 0xy dx x yzdy

1
MN
yx
CC

CC Exact






2
2
1
,5 5
2
,32
32
3
2
2
y
x
fxy x y dx xy x gy
fxy x gy x y
gy y
gy y y C




*


2
2
1
3
520
22
x
xy x y y C


22
2103 4
x xy x y y C
6.
22 32
35 25 0xxydxyxydy

2
10 5
MN
xyy
yx
CC
#
CC

Not exact
7.
2
0
xx
dx dy
yy


22
1MxN
yy xy
CC
#
CC


Not exact
8. sin sin 0yxydxxxyydy

cos sin
MN
xy xy xy
yx
CC

CC Exact

,sin cos
fxy y xydx xy gy*




2
1
,sin sin
2
y
fxy x xy g y x xy y
y
gy y gy C




2
cos
2
y
xyC
9. (a)
(b)
220x y dx y x dy

1
MN
yx
CC

CC Exact


2
,2
fxy xydx x xygy*





2
1
22
,2
2
y
fxy x g y y x
gy y
gy y C
xxyy C






22:4444yC
Particular solution:
22
4xxyy
(c)
−66
−4
4

492. 492 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
10. (a) and (c)
(b)

322
64330xy y dx y x xy dy

2
63
MN
xy
yx
CC

CC Exact


323
,6 3
fxy xy y dx xy xy gy*




22 22
2
1
,33 433
42
y
fxyxxygy yxxy
gy y gy y C




232
32
xyxy y C

01:2yC
Particular solution:
232
322xy xy y
11. 23 310x y dx x y dy
Exact:
1
MN
yx
CC

CC







2
2
1
2
2
1
22
,23
3
,
31
31
3
2
,3
3
2
22632
y
fxy x y dx
xxy x g y
fxy x gy
xy
gy y
gy y y C
fxy x xy x
yyC
xxyxyyC









*

Initial condition:

20
801200 4
y
CC



Particular solution:

22
22632 4xxyxyy
12.
22 3 2
3230,12xy dx xy y dy y

2
6
MN
xy
yx
CC

CC Exact



22 32
332
,3
,2 23
y
fxy xy dx xy gy
fx y xy g y xy y


*




2
3
1
3gy y
gyyC




32 3
xyyC
Initial condition:
12:48yC
Particular solution:
32 3
4xy y
13.
22
320x y dx xy dy



22 2
2
My Nx yy
hx
Nxyx
CCCC



Integrating factor:

2
ln
21hx dx
x
ee
x

*


Exact equation:
2
2
2
30
yy
dx dy
xx








22
2
,3 3
22
,
y
yy
fxy dx x gy
xx
yy
fxy gy
xx



*


10
gygyC

2
3
y
x C
x


14.
22
20xy dx y x dy



22 2
2
Nx My xx
ky
Mxyy
CCC C

Integrating factor:

2
ln
2 1kydy
y
ee
y

*


Exact equation:
2
2
2
10
xx
dx dy
yy







2
2
,
xx
fxy dx gy
yy
*





22
22
1
,1
1
y
x x
fxy gy
yy
gy gy y C




2
x
yC
y

−6 6
−4
4

493. Review Exercises for Chapter 16 493
© 2010 Brooks/Cole, Cengage Learning
15.
2
30
y
dx x e dy





30
3
1
Nx My
ky
M
CCC C



Integrating factor:
3
kydy
y
ee
*

Exact equation:

33
30
yyy
edx xe edy



33
33
,
,3 3
yy
yyy
y
fxy e dx xe gy
fx y xe g y xe e


*




1
y
y
gy e
gyeC




3yy
xeeC

16. cos 2 sin cos 0ydx x y y y dy




2sin sin
cos
tanNx My yy
My
yky
CCC C



Integrating factor:

cos
kydy
ey
*

Exact equation:


22
cos 2 sin cos cos 0ydx x y y y y




22
,cos cos
fxy ydx x y gy*



2
,2cossin
2 sin cos 2 sin cos cos
yfxy x y y gy
xyyyyy y






2
2
1
2sin cos cos
cos
gyyyy y
gyyyC




22
cos cos
x yy yC
17.
22
12
22
12
22
12
22
44
xx
x x
x x
yCe Ce
yCe Ce
yCe Ce









22
12
22
12
44 4
40
xx
xx
yyCe Ce
Ce C e





18.

12
12
12cos 2 sin 2
2sin2 2 cos2
4cos2 4sin2yC xC x
yCxCx
yCxCx





12
1244cos24sin2
4cos2 sin2 0
yy C xC x
CxCx




19.


2
20
2210,2,1
yy y
mm m m m






2
12
2
12
12
12
2
00
032
x x
x x
yCe Ce
yCeCe
yCC
yCC







Adding these equations,
1133 1CC and
2 1.C

2
x x
ye e


20.
2
450
41620
450 2
2
yyy
mm m i





+,
+,+,

22
12
2
12
22
22
22
cos sin
02 2cos sin
2sin cos 2 2cos sin
07 22 3
xx
x
xx
yCe xCe x
yCyexCx
y e xC x e xC x
yCC








22
2cos 3sin
xx
ye xe x



21.


2
230
23 3 10 3,1
yyy
mm m m m






3
12
3
12
12
12
3
02
00 3
xx
x x
yCe Ce
yCeCe
yCC
yCC







Subtracting these equations,
1
11 2
24CC and
3
2 2
.C

33 1
22
x x
ye e


22.

2
2
12 36 0
12 36 6 0, 6, 6yyy
mm m m



66
12
66 6
12 2
66
xx
x xx
yCe Cxe
yCeCe Cxe






1
2202
01
62 13
yC
y
CC





66
213
x x
ye xe


−5
−1
5
7
y
1
y
3
y
2
−33
−2
2
−12 12
−8
8
3
−1
−1 5
−66
0
8
−1
−1
2
2

494. 494 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
23.
2
250
2420
250 12
2
yyy
mm m i






12
1
12
2
12cos 2 sin 2
14 cos2 sin2
20 cos4 sin4
x
yeC xC x
yeCC
yeCC







Solving this system, you obtain
19.0496,C
27.8161.C

9.0496 cos 2 7.8161 sin 2
x
ye x x


24.
2
0
10yy
mmi




12
1
2cos sin
02
1
2
yC xC x
yC
yC









2cos sinyxx
25. yis always positive according to the graph (concave
upwards), but yis negative when 0x(decreasing),
so .yy #
26. yis positive for 0x(concave upwards), but
1
2
0y for 0x(increasing). So,
1
2
.yy #

27.

3
2
12
23
01 2 3
2
12 3
23
23
02 13 2 3
3
01 23
3
12
10 when ,.
cos sin
23
26
26
0, 5, 0, 1
cos sin 5
h
p
p
p
pp
yyxx
mmii
yC xC x
yAAxAxAx
yAAxAx
yAAx
yyAA AAxAxAx
xx
AA AA
yC xC x xx











28.

2
2
12
2
01
2
1
2
22
01
11
0162
211
12 62
2
20 when 2,2.
cos 2 sin 2
2
4
26 22
,0,
cos 2 sin 2
x
h
x
p
x
p
x
p
xx
pp
x
yyex
mmii
yC xC x
yAeBBx
yAeB
yAe
yyAeBBxex
AB B
yC xC x e x











29. 2cosyy x

2
10m when ,.mii




12
12cos sin
cos sin
cos sin
2cos 2sin
2 cos 2 sin 2 cos
0, 1
cos sin
h
p
p
p
ppyC xC x
yAxxBxx
yBxAxBAxx
yBAxxBxAx
yy BxAx x
AB
yC x C x x







30.

2
54 sin2yyyx x

2
540mm when 1, 4.m


4
12
2
01 2 0 1
12 0 1
20 1
22
01 2 1 2 2 1 0
521 1
01 20132 8 4
sin 2 cos 2
22cos22sin2
24sin24cos2
5 4 4 5 2 4 10 4 10 sin 2 10 cos 2 sin 2
,,,0,
xx
h
p
p
p
ppp
yCe Ce
yAAxAxB xB x
yAAxB xB x
yABxBx
yyy AAA AAxAxBxBxx x
AA ABB







1
10
42 521 1 1
12 32 8 4 10
cos 2
xx
yCe Ce x x x



4
−2
−1 5
−4
−3
5
3

495. Review Exercises for Chapter 16 495
© 2010 Brooks/Cole, Cengage Learning
31. 22
x
yyyxe

2
210mm when 1, 1.m


12
12
x
h
x
p
yCCxe
yvvxe






12
12
2
1
23 2
1 3
2
2
2
31
12 3 0
12
2
2
2
2
xx
x xx
x
ve v xe
ve v x e xe
vx
vxdxx
vx
vxdxx
yCCxxe







*
*


32.
2
1
2
x
yyy
xe



2
210mm when 1, 1.m



12
12
x
h
x
p
yCCxe
yvvxe









12
12
2 0
1
1
xx
xx
x
ve v xe
ve v x e
ex






1
1
2
2
2
2
1
1
ln
1
11
v
x
vdxx
x
v
x
vdx
xx




*
*


12 ln 1
x
yCCx x e


33. 654,02,00yy y y y


2
12
32
12
60
320
3, 2
x x
h
mm
mm
mm
yCe Ce






9
pyby inspection

32
12
9
xx
hp
yy y Ce Ce


Initial conditions:



12 12
3322
12 1 2 5502:2 9 11
00:03 2 ,yCCCC
yCCCC





3211
5
23 9
xx
yee


34. 25 , 0 0, 0 0
x
yyey y

12cos 5 sin 5
,
h
x x
ppp
yC xC x
yAey y Ae



12
1
25 26 1
26
1
cos 5 sin 5
26
xxx
x
hp
Ae Ae e A A
yy y C xC x e





11
22
11
00:0
26 26
11
00:05
26 130
yCC
yCC




111
cos 5 sin 5
26 130 26
x
yxxe
35.
2
4cos
40 2
yy x
mmi





12cos 2 sin 2
cos sin
sin cos
cos sin
4cossin4cossincos
h
p
p
p
ppyC xC x
yAxBx
yAxBx
yAxBx
yy AxBxAxBx x






1
3
3cos 3sin cosAxBx x A and 0B

1
3
1
12 3
cos
cos 2 sin 2 cos
p
hpyx
yy y C xC x x


Initial conditions:


171
1133
22
06:6
06:62 3
yCC
yCC



Particular solution:
17 1
33
cos 2 3 sin 2 cosyxxx

496. 496 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
36.

2
12
3
12
32
2
22 2
3
2 2
3
32 3 2
12 32 3
36
3 0 0 and 3
32
62
3 6 2 3 3 2 9 6 6 2 3 6 , 0, 1, and
x
h
p
p
p
pp
p
xx
hp
yyx
mm m m
yCCe
yAxBxCxD
yAxBxC
yAxB
y y Ax B Ax Bx C Ax A B x B C x A B C
yx xD
yy y CCe x xDCCe











2 2
3
xx


Initial conditions:

32
10 10 10 24
22 333 3 3 302:2
0 : 3 and yCC
yCCC


Particular solution:

32 3 210 421
33 3 3
10 4 3 2
xx
yexx exx


37.

2
21 ,01,03
2210 2,1
x
yy y xey y
mm m m m







2
12
2
22
22
xx
h
x
p
x xx
p
yCe Ce
yABxCxe
yBxCxeBCxeBCBxCxe









22
222422
x xx
p
y B CBxCxe CB Cxe Cx B Cx C Be





222
2224 2 2
1
2 6 2 3 1 , 6 1 and 2 3 0.
2
x xx
pp p
xx
y y y C B C Bx Cxe B C Bx Cxe A Bx Cxe
ACxCBe xe A C CB





So,
1
6
C
and
1
.
9
B


22
12 111
296
x xx
hp
yy y Ce Ce x xe





Initial conditions:


12 12
12 12
13
01
22
128
032 2
99
yCCCC
yCCCC



Adding,
11
83 83
3.
18 54
CC

So,
2
1
.
27
C

Particular solution:
283 1 1 1 1
54 27 2 9 6
x xx
ye e xxe

497. Review Exercises for Chapter 16 497
© 2010 Brooks/Cole, Cengage Learning
38.
2
4,01,01,01
0
yy xy y y
yy




32
0mmwhen 0, 0, 1.m


12 3
23 4
012
23
01 2
2
01 2
12
22 41
01 1 2 2 0 1 2 33
23441
12 3 33
23 4
26 12
624
26 624 12 4 or 4, ,
4
x
h
p
p
p
p
pp
x
yCCxCe
yAxAxAx
yAxAxAx
yAAxAx
yAAx
yy AA A AxAxxA A A
yC CxCe x x x









234
23 3
2
3
'84
88 4
x
x
yCCe x x x
yCe xx


Initial conditions:
13 23 3 1 2 30 1, 0 1, 0 1, 1 , 1 , 1 8, 8, 8, 9y y y CCCCCC C C
Particular solution:
23441
33
88 4 9
x
yxxxxe
39. By Hooke’s Law, , 64 4 3 48.FkxkFx Also, Fma and 64 32 2.mFa So,

2
2
22
48
0
2
cos 2 6 sin 2 6 .
dy
y
dt
yC t C t






Because

1
2
0yyou have
1
1 2
Cand 00yyields
20.CSo,

1
2
cos 2 6 .yt
40. From Exercise 39 you have 48kand 2.mAlso, the damping force is given by
18 .dy dt

2
2
1
248
8
1
24 0
16
16 384 0
dy dy
y
dt dt
yyy
yy y






The characteristic equation
2
16 384 0mm has complex roots

1 24,575 1 5 983
.
32 32 32 32
i
mi
So,

32
12 5 983 5 983
cos sin .
32 32
t
yt e C t C t






Initial conditions:



1
1
22
11
0
22
5 983 983
00 0
32 32 9830
yC
C
yCC



Particular solution:

321 5 983 983 5 983
cos sin
2 32 9830 32
t
yt e t t

498. 498 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
41. (a) (i)
22
112 24
cos 2 sin 2 sin
244
yt t t





(ii)

1
162cos22 3sin22
2
yttt



(iii)
5
199 199
199 cos 199 sin
398 5 5
t
et t
y






(iv)

21
cos 2 sin 2
2
t
ye t t



(b) The object comes to rest more quickly. It may not
even oscillate, as in part (iv).
(c) It would oscillate more rapidly.
(d) Part (ii). The amplitude becomes increasingly large.
42.
11 1
44 4
cos , sin , cos
pp pyxy xy x


111
444
45 cos4sin5cos
cos sin
pppyyy x x x
xx



False.
43. (a) sin
pyAx and 3 3 sin .
pyAx
So,
3 sin 3 sin
2sin 12sin
ppyyAxAx
Axx



(b)
5
cos
2
pyx
(c) If cos sin ,
pyAxBx then
cos sin ,
pyAxBx and solving for
Aand
Bwould be more difficult.
44. 5,ybecause 0yy and 65 30
45. 40.xyy Letting
0
:
n
n
n
yax
!

F



1
01 0
1
1
010 1
1
1
001 02 1 0 0
2
0
0
44
14 1 410
141
1
4
111 1
,, ,,
444 4
4
nnn
nnn
nn n
nn n n
nn n n
nnnn
nn
nn
n
n
n
n
n
xy y y na x na x a x
nax nax nax nax
na na
aa
aaa aa a a a a
x
ya
!! !


!!! !





!






FFF
FFFF
F


12
−12
010
60
−60
0 14
1
−1
08
0.6
−0.2
03

499. Review Exercises for Chapter 16 499
© 2010 Brooks/Cole, Cengage Learning
46. 330.yxyy Letting
0
:
n
n
n
yax
!

F




21
210
2
00
2
33 1 3 3 0
21 33
31
21
nnn
nnn
nnn
nn
nn
nn
n
n
y xy y nn ax x nax ax
nnax nax
na
a
nn
!!!


!!







FFF
FF

00 11aa aa

20 3
3
0
21
aa a
"


There are no odd-powered terms for 1.n







0
40
3
00
6
34
00
8
45
00
10
1
22
00 0
23333
4321 4!
33 3 333
6 5 4! 6!
33 35 335
87 6! 8!
35 3 37 5 337
10 9 8! 10!
133572 33
22!
n
n
n
n
a
aa
aa
a
aa
a
aa
a
n
ya ax a x
n

!




""


"

"


"

"""


"

""


F


47.











44
54 5
0, 0 2, 0 0
02
0220
204
33 0 484
x
x
x
x
x
yyey y y
yyey y
yyeyy y
yyeyyy y
yyeyyyy y













45
23 4 5245
00 0 0 11
00 2
2! 3! 4! 5! 6 30
yy y y
yy yx x x x x x x x




Using the first four terms of the series,
1
2.063.
4
y





48.











4 4
5 5
64 4 6
754454 7
0, 0 1, 0 1
00
01
202
23 00
34 04
45 010
yxy y y
yxy y
yxyy y
yxyyyxyy y
yxyyyxyy y
yxyyyxyy y
yxyyyxyy y












7 34 6 7
27
00
00 1
2! 7! 6 12 180 504
yy
xxx x
yy yx x x x




1
1.474
2
y

500. 500 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 16
1.
22 2 2
350x kxy dx x y ky dy
2
10
5
M
kxy
y
N
xy
x
MN
k
yx
C

C
C

C
CC

CC

22 2 2
35 5 5 0xxydxxyydy Exact



22 322 5
,35
2
fxy x xy dx x xy gy *



222
,5 55
y
fxy xy g y xy y


23
1 5
5
3
gyygy yC

3223
2
322 355
23
615 10
x xy y C
x xy y C



2.
22
0kx y dx kxy dy
(a)

22
22
2
2
2211
0
0
2
2
kx y dx kxy dy
xx
yky
kdxdy
xx
MyNky
k
yx xx




CC

CC

(b)
2
2
2
20
yy
dx dy
xx



Exact



22
2
,2 2
yy
fxy dx x gy
xx


*



1
22
,
y
yyfxy g y gy C
xx



2
2
y
x C
x

3.
2
0, 0yay y

22
0ma mama m a

12 12
12
12
12
22
22
cosh sinh
ax ax ax ax
ax ax ax ax CC CC
yBe Be e e
ee ee
CC
CaxCax








4.

2
0yy$

22
0mmi
$ $

12cos sinyC xC x$ $
Let.be given by
2
1
cot , 0 2 .
C
C
. .
Then
12cos sin .CC
. .
Let
12
.
sin cos
CC
C
. .
Then


12cos sin
sin cos cos sin sin .
yC xC x
CxCxCx
$ $
.$.$ $.


Note that if
10,Cthen0
.and

sin .yC x
$ And if
20,Cthen
sin cos .
2
yC x C x

$ $





5. The general solution to 0yayby is


12 .
rsx rsx
yBe Be


Let
112CBB and
212 .CBB
Then
12
1
2
CC
B

and
12
2
.
2
CC
B


So

+,
12 12
12
12
22
22
cosh sinh .
rsx rsx
sx sx sx sx
rx
rxCC CC
ye e
ee ee
eC C
eC sx C sx











6. The roots of the characteristic equation
2
0, 0mamb ab are
2
4
.
2
aa b
m

You consider three cases:
(i) If the roots are equal, then
2
40ab and
2
12 0
a
x
yCCxe

as .
x !
(ii) If the roots are complex, ,
2
a
mi $


then
22
12cos sin 0
aa
xx
yCe xCe x$$

as
x !
(because
cos
x$andsinx$are bounded).
(iii) If the roots are real and distinct, then
2
24
4
2
2
12
.
aa b
aa b
x
x
yCe Ce


The second term clearly tends to 0 as .
x !
For the first term, note that
2
4ab
2
4
1.
b
aa
a
So
4
1
222
1
0
aa b
xa
yCe




as .
x !

501. Problem Solving for Chapter 16 501
© 2010 Brooks/Cole, Cengage Learning
7. 0, 0 0yay y yL
(a) If
0, 0 .ay ycxd 00yd
and
00.yL cL c So 0yis the
solution.
(b) If
0, 0ayay has characteristic equation
2
0.ma m a

12
ax ax
yCe Ce



12 1 200yCCCC

0yL

12
11
1
112
2
2
2sinh 0
aL aL
aL aL
aL aL
Ce C e
Ce Ce
ee
C
CaLCC










So, 0yis the only solution.
8. 0, 0, 0 0yay a y yL

2
0ma m ai




12
1
2
2cos sin .
00
sin
0sin
yC ax C ax
yC
yC ax
yL C aL




So
aL n

2
,
n
an
L




an integer.
9.
2
2
0, 0
dg g
dt L L



(a)

12sin cos
g g
tC tC t
L L






Let.be given by
1
2
tan , .
22
gC
LC

..




Then
21sin cos .
gg
CC
LL
. .





Let
21
cos sin
CC
A
gg
LL
. .








12sin cos sin sin cos cos cos
gg ggggg
tC tC t A tA t A t
LL LLLLL...





(b)





cos , 9.8, 0.25
0 cos 39.2 0.1
sin
4
0 39.2 sin 39.2 0.5
g
tA t g L
L
A
gg
tA t
L
A.
.
.
.











Dividing,
5
tan 39.2 0.1076 0.128.
39.2
A
..






0.128 cos 39.2 0.108tt

(c) Period
2
1
39.2
sec
(d) Maximum is 0.128.
(e)
0tat 0.359t sec, and at 0.860t sec.
(f )
0.359 0.801, 0.860 0.801

502. 502 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
10. (a)
21
2,0
2
Ay Wx Wx A



22
3
2
1
34
122
4
22
2
23
2
2312
WW W
yxx xx
AA A
Wx
yx C
A
Wx x
yCxC
A









200 0yC


1
11
16 16
20 2 0
2312
2
2
W
yC
A
WW
CC
A A








34
2
2
2312
Wx x
yx
A




(b) Using a graphing utility, the maximum deflection is
at 1.1074,x and the deflection is

1.43476 0.7174 .
2
WW
A A



11. 8160,01,01yy y y y
(a)
22
4, 4, 0,G;G; critically damped
(b)
12 4mm



4
12
44
12 2
,
4
t
tt
yCCte
yCCteCe








1
22
401
01 4 5
15
t
yC
yCC
yte





(c)
The solution tends to zero quickly.

12. 2260,01,04yy y y y
(a)
1, 26 ,G;

22
25 0,G; underdamped
(b)
1215, 15mimi

12cos 5 sin 5
tt
yCe t Ce t



101yC


12
12cos 5 sin 5
5sin5 5 cos5
t
t
yt e C t C t
eC tC t





12 204 5 1yCCC

cos 5 sin 5
t
ye t t


(c)
The solution oscillates.
13. 20 64 0, 0 2, 0 20yyyy y
(a)
22
10, 8, 36 0,G;G; overdamped
(b)



12
416
12
12
416
12
1210 6 4, 10 6 16
02
416
020416
tt
tt
mm
yCe Ce
yCC
yt Ce Ce
yCC








12
12
12 2
1, 1
45
CC
CC
CC
<
>
?


416tt
ye e


(c)
The solution tends to zero quickly.
14. 20,02,01yyy y y
(a)
22
1, 1, 0,G;G; critically damped
(b)

12
12 12 2 1
,
ttt
mm
yCCtey CCte Ce







1
2202
012 1
2
t
yC
yCC
yte





(c)
The solution tends to zero quickly.
05
−2
2
0
0
2
2
0
02
2
05
−2
2

503. Problem Solving for Chapter 16 503
© 2010 Brooks/Cole, Cengage Learning
15. Airy’s Equation: 0yxy

10yxyyyy x yy
Let

1
01
1, 1 ,
nn
nn
nn
yaxy nax
!!


FF
2
2
11.
n
n
n
ynnax
!


F

10yxyy



2
200
11 1
31
101
11 1 1 10
32 1 1 1 0
nnn
nnn
nnn
nn n
nnn
nnn
nn a x x a x a x
nnax ax ax
!!!


!!!




FFF
FFF


1
20 3 1
0
232 10
n
nnn
n
aa n n a aa x
!




F

20 2 001
1
20 ;,
2
aa a aaa
arbitrary
In general,

1
3
.
32
nn
naa
a
nn






01
3
10
12 10
4
01
0
23 01
5
6
1
22
12 12 24
1
4
26
20 20 120
aa
a
aa
aa aa
a
aa
a
aa aa
a













01 10
34 0 1
6
10 01
45 0 1
7 2
56624
30 30 720
24
91124 120
42 42 5040
aa aa
aa a a
a
aa a a
aa a a
a











So, the first eight terms are




23 4 5001 10 01
01
6701 0 1 24
11 1 1 1
2 6 24 120
56 911
11.
720 5040
aaa aa aa
ya ax x x x x
aa a a
xx





16. (a) 11 2
nnnTx xTxTx





01
2
2
23
3
3242
41,
2121
22 1 4 3
24 3 2 1 8 8 1
TTx
Txx x
Txx xxx
Txxxx xx




(b)
22
10xy xy ky
Substituting
04,,TTinto this equation shows that the polynomials
satisfy Chebyshev’s equation. For example, for
4,T


22 3 42
1 9616 3216168810xx xx x xx


(c)


42 3 5 3
5
53 42 642
6
28 8 1 4 3 16 20 5
216 20 5 8 8 1 32 48 18 1
Txxx xx x xx
Txx xxxx x x x




642 53 7 53
7
2 32 48 18 1 16 20 5 64 112 56 7Txx x x x xx x x xx

504. 504 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
17.
22
0xy xy xy Bessell equation of order zero
(a) Let

12
01 2
,,1.
nn n
nn n
nn n
yaxy naxy nnax
!! !


FF F

22
0xy xy xy




2212
210
2
210
222
22
010
10
10
21 2 0
nnn
nnn
nnn
nnn
nnn
nnn
nnn
nnn
nnn
xnnax xnax xax
nn ax nax ax
nnax nax ax
!!!


!!!


!!!





FFF
FFF
FFF


2
122
0
21 2 0
n
nnn
n
ax n n a n a a x
!




F

10aand

2 2 .
2
n
na
a
n





All odd terms
iaare 0.



0
2
2
20
40 2222 4
40
60 22222 6
2
1
424 21 2
1
6246 23!
a
a
aa
aa
aa
aa




" "


""




2
0 2
2
0
1
2!
n
n
n
n
x
ya
n
!



F
(b) This is the same function (assuming
01a).
18. (a) Let
12
01 2
,,1
nn n
nn n
nn n
yaxy naxy nnax
!! !


FF F

22
10xy xy x y


2212
210
110
nn n
nn n
nnn
xnnax xnax x ax
!!!


FFF


2
2100
10
nnn n
nnn n
nnnn
nn ax nax ax ax
!!!!


FFFF


2222
22 2
0102
21 2 0
nnnn
nnnn
nnnn
nnax nax ax ax
!!!!



FFFF


2
011 2 2 2
0
21 2 0
n
nnnn
n
aaax n na n a aax
!




F



2
0222
0 and 2 1 2 1 4 3
13
n
nn nnna
annnaannaaa
nn







All even terms
iaare 0.

11
3
3
31 1
5
55
51
7
7
24 2
2
46 23! 2 2!3!
2
68 23!4!
aa
a
aa a
a
aa
a


"


""


"




21
1
21
0
1
2
2!1!
n
n
n
n
x
ya
nn
!





F
(b) This is the same function (assuming
121a).

505. Problem Solving for Chapter 16 505
© 2010 Brooks/Cole, Cengage Learning
19. (a) Let
12
01 2
,,1.
nn n
nn n
nn n
yaxy naxy nnax
!! !


FF F
280yxyy




21
210
2
000
2
0
12 8 0
21 2 8 0
21 2 8 0
nnn
nnn
nnn
nnn
nnn
nnn
n
nnn
n
nn ax x nax ax
nnax nax ax
nna naax
!!!


!!!


!





FFF
FFF
F






2
4222
2000
24
21
22 1
16 48
43 3
24
48 4 12
2
nn
n
aa
nn
aaaa
aaaa











42
4
16 48 12Hx x x
(b)







0
0
1
1
22 2
1
2
2
0
2
1
0!
2
2
1!
12!2 22 2
42
!2 2 ! 2! 1
nn
n
x
Hx
x
Hx x
xx
Hx x
nn







F










32 3 1
1
3
3
0
42 4 2
2
42
4
0
13!2 3!2 3!2
812
!3 2 ! 3! 1
1 4!2 4!2 4!2 4!
16 48 12
! 4 2 ! 4! 2! 2!
nn
n
nn
n
xxx
Hx x x
nn
xxx
Hx x x
nn










F
F

506. 506 Chapter 16 Additional Topics in Differential Equations
© 2010 Brooks/Cole, Cengage Learning
20. (a) 10xy x y ky

Let
12
01 2
,,1.
nn n
nn n
nn n
yaxy naxy nnax
!! !


FF F




21
210
11
2110
11
1010
11 0
10
11 0
nnn
nnn
nnn
nnnn
nnnn
nnnn
nnnn
nnnn
nnnn
xnn ax x nax kax
nn ax nax nax kax
nnax nax nax kax
!!!


!!!!


!!!!




FFF
FFFF
FFFF

10 1 1
1 11 0
n
nnnn
n
a ka n na n a na ka x
!



F

10 1 0 0aka a ka



2
11 2
10
1
nnn n
nk
na kna a a
n






Let
01.a

For 12 00, 0 1.kaa Lx

For 123 11, 1, 0 1 .ka aa Lx x
For

2
121 2
211 1
2, 2, 1 2 .
22 2
ka a a Lx xx



In general, for a given integer 0,k/
12 0.
kkaa
Furthermore, in the given formula for

,
k
Lxyou can verify that

1 2 .
1
nn
nk
aa
n




Finally, you can see that for ,kn/












11 2 222 2
2
2022
2222
2
022
11111 2
1
112 1120
1121
112 ! 1!
!! !!
nn n n
n
n
n
knnk kn nk
aa a a
nn n n
kn kn kn kn k
aa
nn nn
kn kn kkn k
a
nkn knn



"



"








(b)









0
0 2
0
1
1 2
0
2 2
2 2
0
10!
1
0!!
11!
1
1!!
12!
12
22!!
n
n
n
n
n
n
n
n
n
x
Lx
nn
x
Lx x
nn
x
x
Lx x
nn












F
F
F







33
2
3 2
0
4
23 4
4 2
0
13! 3
13
263!!
14! 21
14 3
3244!!
n
n
n
n
n
n
x x
Lx x x
nn
x
Lxxxxx
nn








F
F