Chapter 14

1. January 27, 2005 11:55 L24-ch14 Sheet number 1 Page number 603 black
603
CHAPTER 14
Partial Derivatives
EXERCISE SET 14.1
1. (a)f(2,1) =(2)
2
(1)+1=5 (b)f(1,2) =(1)
2
(2)+1=3
(c)f(0,0) =(0)
2
(0)+1=1 (d)f(1,−3) =(1)
2
(−3)+1=−2
(e)f(3a, a)=(3a)
2
(a)+1=9a
3
+1 (f)f(ab,a−b)=(ab)
2
(a−b)+1=a
3
b
2
−a
2
b
3
+1
2. (a)2t (b)2x (c)2y
2
+2y
3. (a)f(x+y, x−y)=(x+y)(x−y)+3=x
2
−y
2
+3
(b)f
J
xy,3x
2
y
3
a
=(xy)
J
3x
2
y
3
a
+3=3x
3
y
4
+3
4. (a)(x/y) sin(x/y) (b)xysin(xy) (c)(x−y) sin(x−y)
5.F(g(x),h(y)) =F
J
x
3
,3y+1
a
=x
3
e
x
3
(3y+1)
6.g(u(x, y),v(x, y)) =g
J
x
2
y
3
,πxy
a
=πxysin
n
J
x
2
y
3
a
2
(πxy)
u
=πxysin
J
πx
5
y
7
a
7. (a)t
2
+3t
10
(b)0 (c)3076
8.
√
te
−3ln(t
2
+1)
=
√
t
(t
2
+1)
3
9. (a)v=7 lies betweenv=5 andv=15, and 7 =5 + 2 =5 +
2
10
(15−5), so
WCI≈19 +
2
10
(13−19) =19−1.2=17.8
◦
F
(b)v=28 lies betweenv=25 andv=30, and 28 =25 +
3
5
(30−25), so
WCI≈19 +
3
5
(25−19) =19 + 3.6=22.6
◦
F
10. (a)AtT=35,14=5+9=5+
9
10
(15−5), soWCI≈31 +
9
10
(25−31) =25.6
◦
F
(b)Atv=15,32 =30 +
2
5
(35−30), soWCI≈19 +
2
5
(25−19) =21.4
◦
F
11. (a)The depression is 20−16 =4, so the relative humidity is 66%.
(b)The relative humidity≈77−(1/2)7 =73.5%.
(c)The relative humidity≈59+(2/5)4 =60.6%.
12. (a)4
◦
C
(b)The relative humidity≈62−(1/4)9 =59.75%.
(c)The relative humidity≈77+(1/5)(79−77) =77.4%.
13. (a)19 (b)−9 (c)3
(d)a
6
+3 (e)−t
8
+3 (f)(a+b)(a−b)
2
b
3
+3
14. (a)x
2
(x+y)(x−y)+(x+y)=x
2
J
x
2
−y
2
a
+(x+y)=x
4
−x
2
y
2
+x+y
(b)(xz)(xy)(y/x)+xy=xy
2
z+xy
15.F
J
x
2
,y+1,z
2
a
=(y+1)e
x
2
(y+1)z
2
16.g
J
x
2
z
3
, πxyz, xy/z
a
=(xy/z) sin
J
πx
3
yz
4
a

2. January 27, 2005 11:55 L24-ch14 Sheet number 2 Page number 604 black
604 Chapter 14
17. (a)f(
√
5,2,π,−3π)=80
√
π (b)f(1,1,...,1) =
n
r
k=1
k=n(n+1)/2
18. (a)f(−2,2,0,π/4) =1
(b)f(1,2,...,n)=n(n+ 1)(2n+1)/6, see Theorem 2(b), Section 5.4
19.
1
x
y
20.
2
x
y
21.
x
y
22.
x
y
23. (a)all points above or on the liney=−2
(b)all points on or within the spherex
2
+y
2
+z
2
=25
(c)all points in 3-space
24. (a)all points on or between the vertical linesx=±2.
(b)all points above the liney=2x
(c)all points not on the planex+y+z=0
25.
3
x
y
z
26.
(0, 3, 0)
z
y
x

3. January 27, 2005 11:55 L24-ch14 Sheet number 3 Page number 605 black
Exercise Set 14.1 605
27.
z
yx
28.
z
yx
29.
z
y
x
30.
(2, 0, 0)
(0, 2, 0)
(0, 0, 4)
z
yx
31.
z
y
x
(0, 0, 1)
32.
(0, 1, 0)
(1, 0, 0)
z
y
x
33.
z
y
x
(0, 0, 1)
(0, –1, 0)
34.
z
y
x
35. (a)f(x, y)=1−x
2
−y
2
, becausef=cis a circle of radius
√
1−c(providedc≤1), and the
radii in (a) decrease ascincreases.
(b)f(x, y)=
y
x
2
+y
2
becausef=cis a circle of radiusc, and the radii increase uniformly.
(c)f(x, y)=x
2
+y
2
becausef=cis a circle of radius
√
cand the radii in the plot grow like
the square root function.

4. January 27, 2005 11:55 L24-ch14 Sheet number 4 Page number 606 black
606 Chapter 14
36. (a)III, because the surface has 9 peaks along the edges, three peaks to each edge
(b)IV, because the center is relatively flat and the deep valley in the first quadrant points in the
direction of the positivex-axis
(c)I, because the deep valley in the first quadrant points in the direction of the positivey-axis
(d)II, because the surface has four peaks
37. (a)A (b)B (c)increase
(d)decrease (e)increase (f)decrease
38. (a)Medicine Hat, since the contour lines are closer together near Medicine Hat than they are
near Chicago.
(b)The change in atmospheric pressure is about ∆p≈999−1010 =−11, so the average rate of
change is ∆p/1400≈−0.0079.
39.
1234k = 0
x
y
40.
k = –1
k = –2 k = 2
k = 1
k = 0x
y
41.
x
y
k = 2
k = 1
k = 0
k = –1
k = –2
42.
1234k = 0
x
y
43.
x
y
k = 2
k = 2
k = 1
k = 1
k = 0
k = 0
k = –1
k = –1
k = –2
k = –2
–2
–2
2
2
44.
2
co
k = -2
k = 2
k = –1
k = 0
k = 1
x
y

5. January 27, 2005 11:55 L24-ch14 Sheet number 5 Page number 607 black
Exercise Set 14.1 607
45.
(0, 4, 0)
(0, 0, 2)
(2, 0, 0)
z
y
x
46.
z
yx
47.
(0, 0, 3)
z
x y
48.
(0, 0, 1)
( , 0, 0)
1
4
(0, − , 0)
1
2
z
x
y
49.concentric spheres, common center at (2,0,0)
50.parallel planes, common normal 3i−j+2k
51.concentric cylinders, common axis they-axis
52.circular paraboloids, common axis thez-axis, all the same shape but with different vertices along
z-axis.
53. (a)f(−1,1) =0;x
2
−2x
3
+3xy=0 (b)f(0,0) =0;x
2
−2x
3
+3xy=0
(c)f(2,−1) =−18;x
2
−2x
3
+3xy=−18
54. (a)f(ln 2, 1) =2;ye
x
=2 (b)f(0,3) =3;ye
x
=3
(c)f(1,−2) =−2e;ye
x
=−2e
55. (a)f(1,−2,0) =5;x
2
+y
2
−z=5 (b)f(1,0,3) =−2;x
2
+y
2
−z=−2
(c)f(0,0,0) =0;x
2
+y
2
−z=0
56. (a)f(1,0,2) =3;xyz+3=3,xyz=0 (b)f(−2,4,1) =−5;xyz+3=−5,xyz=−8
(c)f(0,0,0) =3;xyz=0
57. (a)
4
4
x
y
T = 1
T = 2
T = 3
(b)At (1,4) the temperature is
T(1,4) =4 so the temperature
will remain constant along
the pathxy=4.

6. January 27, 2005 11:55 L24-ch14 Sheet number 6 Page number 608 black
608 Chapter 14
58.V=
8
y
16 +x
2
+y
2
x
2
+y
2
=
64
V
2
−16
the equipotential curves are circles.
10 20
20V = 2.0
V = 1.0
V = 0.5
x
y
59. (a)
5–5
–3
2 (b)
4–4
–3
1
60. (a) 10
–10
–10 10
(b) 40
–40
–55
61. (a)
z
x y
(b)
–2 –1012
–2
-1
0
1
2
62. (a)
0
1
2
3
4
x
0
y
–5
0
5
z
6
c
i
o
(b)
0 1 2 3 4
0
3
6
9
c
f
i
l
o

7. January 27, 2005 11:55 L24-ch14 Sheet number 7 Page number 609 black
Exercise Set 14.2 609
63. (a)The graph ofgis the graph offshifted one unit in the positivex-direction.
(b)The graph ofgis the graph offshifted one unit up thez-axis.
(c)The graph ofgis the graph offshifted one unit down they-axis and then inverted with
respect to the planez=0.
64. (a)
z
y
x
(b)Ifais positive and increasing then the graph ofgis more pointed, and in the limit asa→+∞
the graph approaches a ’spike’ on thez-axis of height 1. Asadecreases to zero the graph of
ggets flatter until it finally approaches the planez=1.
EXERCISE SET 14.2
1.35 2.π
2
/2 3.−8 4.e
−7
5.0 6.0
7. (a)Alongx=0 lim
(x,y)→(0,0)
3
x
2
+2y
2
=lim
y→0
3
2y
2
does not exist.
(b)Alongx=0,lim
(x,y)→(0,0)
x+y
2x
2
+y
2
=lim
y→0
1
y
does not exist.
8. (a)Alongy=0: lim
x→0
x
x
2
=lim
x→0
1
x
does not exist because
2
2
2
2
1
x
2
2
2
2
→+∞asx→0 so the original
limit does not exist.
(b)Alongy=0: lim
x→0
1
x
2
does not exist, so the original limit does not exist.
9.Letz=x
2
+y
2
, then lim
(x,y)→(0,0)
sin
J
x
2
+y
2
a
x
2
+y
2
=lim
z→0
+
sinz
z
=1
10.Letz=x
2
+y
2
, then lim
(x,y)→(0,0)
1−cos
J
x
2
+y
2
a
x
2
+y
2
=lim
z→0
+
1−cosz
z
=lim z→0
+
sinz
1
=0
11.Letz=x
2
+y
2
, then lim
(x,y)→(0,0)
e
−1/(x
2
+y
2
)
=lim
z→0
+
e
−1/z
=0
12.Withz=x
2
+y
2
, lim
z→+∞
1
√
z
e
−1/
√
z
; letw=
1
√
z
,limw→+∞
w
e
w
=0
13. lim
(x,y)→(0,0)
J
x
2
+y
2
aJ
x
2
−y
2
a
x
2
+y
2
=lim
(x,y)→(0,0)
J
x
2
−y
2
a
=0
14. lim
(x,y)→(0,0)
J
x
2
+4y
2
aJ
x
2
−4y
2
a
x
2
+4y
2
=lim
(x,y)→(0,0)
J
x
2
−4y
2
a
=0

8. January 27, 2005 11:55 L24-ch14 Sheet number 8 Page number 610 black
610 Chapter 14
15.alongy=0: lim
x→0
0
3x
2
=lim
x→0
0 =0; alongy=x: lim
x→0
x
2
5x
2
=lim
x→0
1/5=1/5
so the limit does not exist.
16.Letz=x
2
+y
2
, then lim
(x,y)→(0,0)
1−x
2
−y
2
x
2
+y
2
=lim
z→0
+
1−z
z
=+∞so the limit does not exist.
17.8/3 18.ln 5
19.Lett=
y
x
2
+y
2
+z
2
, then lim
(x,y,z)→(0,0,0)
sin
J
x
2
+y
2
+z
2
a
y
x
2
+y
2
+z
2
=lim
t→0
+
sin
J
t
2
a
t
=0
20.Witht=
y
x
2
+y
2
+z
2
, lim
t→0
+
sint
t
2
=lim
t→0
+
cost
2t
=+∞so the limit does not exist.
21.yln(x
2
+y
2
)=rsinθlnr
2
=2r(lnr) sinθ, so lim
(x,y)→(0,0)
yln(x
2
+y
2
) =lim
r→0
+
2r(lnr) sinθ=0
22.
x
2
y
2
y
x
2
+y
2
=
(r
2
cos
2
θ)(r
2
sin
2
θ)
r
=r
3
cos
2
θsin
2
θ, so lim
(x,y)→(0,0)
x
2
y
2
y
x
2
+y
2
=0
23.
e
√
x
2
+y
2
+z
2
y
x
2
+y
2
+z
2
=
e
ρ
ρ
, so lim(x,y,z)→(0,0,0)
e
√
x
2
+y
2
+z
2
y
x
2
+y
2
+z
2
=lim
ρ→0
+
e
ρ
ρ
does not exist.
24. lim
(x,y,z)→(0,0,0)
tan
−1
7
1
x
2
+y
2
+z
2

=lim
ρ→0
+
tan
−1
1ρ
2
=
π
2
25. (a)No, since there seem to be points near (0,0) withz=0 and other points near (0,0) with
z≈1/2.
(b)lim
x→0
mx
3
x
4
+m
2
x
2
=lim
x→0
mx
x
2
+m
2
=0 (c)lim
x→0
x
4
2x
4
=lim
x→0
1/2=1/2
(d)A limit must be unique if it exists, sof(x, y) cannot have a limit as (x, y)→(0,0).
26. (a)Alongy=mx: lim
x→0
mx
4
2x
6
+m
2
x
2
=lim
x→0
mx
2
2x
4
+m
2
=0;
alongy=kx
2
: lim
x→0
kx
5
2x
6
+k
2
x
4
=lim
x→0
kx
2x
2
+k
2
=0.
(b)lim
x→0
x
6
2x
6
+x
6
=lim
x→0
1
3
=
1
3
7=0
27. (a)lim
t→0
abct
3
a
2
t
2
+b
4
t
4
+c
4
t
4
=lim
t→0
abct
a
2
+b
4
t
2
+c
4
t
2
=0
(b)lim
t→0
t
4
t
4
+t
4
+t
4
=lim
t→0
1/3=1/3
28.π/2 because
x
2
+1
x
2
+(y−1)
2
→+∞as (x, y)→(0,1)
29.−π/2 because
x
2
−1
x
2
+(y−1)
2
→−∞as (x, y)→(0,1)

9. January 27, 2005 11:55 L24-ch14 Sheet number 9 Page number 611 black
Exercise Set 14.2 611
30.withz=x
2
+y
2
, lim
z→0
+
sinz
z
=1=f(0,0)
31.The required limit does not exist, so the singularity is not removeable.
32. lim
(x,y)→(0,0)
f(x, y) =0 so the limit exists, andfis not defined at (0,0), thus the singularity is
removable.
33.
–1
x
y
34.
y = x
x
y
35.
5
x
y
36.
1
y = 2x + 1
x
y
37.
x
y
38.
x
y
39.
x
y
xy = –1
xy = 1
xy = 1
xy = –1
40.
x
y

10. January 27, 2005 11:55 L24-ch14 Sheet number 10 Page number 612 black
612 Chapter 14
41.all of 3-space
42.all points inside the sphere with radius 2 and center at the origin
43.all points not on the cylinderx
2
+z
2
=1 44.all of 3-space
EXERCISE SET 14.3
1. (a)9x
2
y
2
(b)6x
3
y (c)9y
2
(d)9x
2
(e)6y (f)6x
3
(g)36 (h)12
2. (a)2e
2x
siny (b)e
2x
cosy (c)2 siny (d)0
(e)cosy (f)e
2x
(g)0 (h)4
3. (a)
∂z
∂x
=
3
2
√
3x+2y
; slope =
3
8
(b)
∂z
∂y
=
1
√
3x+2y
; slope =
1
4
4. (a)
∂z
∂x
=e
−y
; slope =1 (b)
∂z
∂y
=−xe
−y
+ 5; slope =2
5. (a)
∂z
∂x
=−4 cos(y
2
−4x); rate of change =−4 cos 7
(b)
∂z
∂y
=2ycos(y
2
−4x); rate of change =2 cos 7
6. (a)
∂z
∂x
=−
1
(x+y)
2
; rate of change =−
1
4
(b)
∂z
∂y
=−
1
(x+y)
2
; rate of change =−
1
4
7.∂z/∂x=slope of line parallel toxz-plane =−4;∂z/∂y=slope of line parallel toyz-plane =1/2
8.Moving to the right from (x
0,y0) decreasesf(x, y), sof x<0; moving up increasesf,sof y>0.
9. (a)The right-hand estimate is∂r/∂v≈(222−197)/(85−80) =5; the left-hand estimate is
∂r/∂v≈(197−173)/(80−75) =4.8; the average is∂r/∂v≈4.9.
(b)The right-hand estimate is∂r/∂θ≈(200−197)/(45−40) =0.6; the left-hand estimate is
∂r/∂θ≈(197−188)/(40−35) =1.8; the average is∂r/∂θ≈1.2.
10. (a)The right-hand estimate is∂r/∂v≈(253−226)/(90−85) =5.4; the left-hand estimate is
(226−200)/(85−80) =5.2; the average is∂r/∂v≈5.3.
(b)The right-hand estimate is∂r/∂θ≈(222−226)/(50−45) =−0.8; the left-hand estimate is
(226−222)/(45−40) =0.8; the average is∂r/∂v≈0.
11.III is a plane, and its partial derivatives are constants, so III cannot bef(x, y). If I is the graph
ofz=f(x, y) then (by inspection)f
yis constant asyvaries, but neither II nor III is constant as
yvaries. Hencez=f(x, y) has II as its graph, and as II seems to be an odd function ofxand an
even function ofy,f
xhas I as its graph andf yhas III as its graph.
12.The slope atPin the positivex-direction is negative, the slope in the positivey-direction is
negative, thus∂z/∂x <0,∂z/∂y <0; the curve throughPwhich is parallel to thex-axis is
concave down, so∂
2
z/∂x
2
<0; the curve parallel to they-axis is concave down, so∂
2
z/∂y
2
<0.

11. January 27, 2005 11:55 L24-ch14 Sheet number 11 Page number 613 black
Exercise Set 14.3 613
13.∂z/∂x=8xy
3
e
x
2
y
3
,∂z/∂y=12x
2
y
2
e
x
2
y
3
14.∂z/∂x=−5x
4
y
4
sin(x
5
y
4
),∂z/∂y=−4x
5
y
3
sin
J
x
5
y
4
a
15.∂z/∂x=x
3
/(y
3/5
+x)+3x
2
ln(1 +xy
−3/5
),∂z/∂y=−(3/5)x
4
/(y
8/5
+xy)
16.∂z/∂x=ye
xy
sin(4y
2
),∂z/∂y=8ye
xy
cos(4y
2
)+xe
xy
sin(4y
2
)
17.
∂z
∂x
=−
y(x
2
−y
2
)
(x
2
+y
2
)
2
,
∂z
∂y
=
x(x
2
−y
2
)
(x
2
+y
2
)
2
18.
∂z
∂x
=
xy
3
(3x+4y)
2(x+y)
3/2
,
∂z
∂y
=
x
2
y
2
(6x+5y)
2(x+y)
3/2
19.f x(x, y)=(3/2)x
2
y
J
5x
2
−7
aJ
3x
5
y−7x
3
y
a
−1/2
fy(x, y)=(1/2)x
3
J
3x
2
−7
aJ
3x
5
y−7x
3
y
a
−1/2
20.f x(x, y)=−2y/(x−y)
2
,fy(x, y)=2x/(x−y)
2
21.f x(x, y)=
y
−1/2
y
2
+x
2
,fy(x, y)=−
xy
−3/2
y
2
+x
2
−
3
2
y
−5/2
tan
−1
(x/y)
22.f
x(x, y)=3x
2
e
−y
+(1/2)x
−1/2
y
3
sec
√
xtan
√
x,fy(x, y)=−x
3
e
−y
+3y
2
sec
√
x
23.f
x(x, y)=−(4/3)y
2
sec
2
x
J
y
2
tanx
a
−7/3
,fy(x, y)=−(8/3)ytanx
J
y
2
tanx
a
−7/3
24.f x(x, y)=2y
2
cosh
√
xsinh
J
xy
2
a
cosh
J
xy
2
a
+
1
2
x
−1/2
sinh
√
xsinh
2
J
xy
2
a
f
y(x, y)=4xycosh
√
xsinh
J
xy
2
a
cosh
J
xy
2
a
25.f
x(x, y)=−2x, f x(3,1) =−6;f y(x, y)=−21y
2
,fy(3,1) =−21
26.∂f/∂x=x
2
y
2
e
xy
+2xye
xy
,∂f/∂x
2
2
(1,1)=3e;∂f/∂y=x
3
ye
xy
+x
2
e
xy
,∂f/∂y
2
2
(1,1)=2e
27.∂z/∂x=x(x
2
+4y
2
)
−1/2
,∂z/∂x
2
2
(1,2)=1/
√
17 ;∂z/∂y=4y(x
2
+4y
2
)
−1/2
,∂z/∂y
2
2
(1,2)=8/
√
17
28.∂w/∂x=−x
2
ysinxy+2xcosxy,
∂w
∂x
(1/2,π)=−π/4;∂w/∂y=−x
3
sinxy,
∂w
∂y
(1/2,π)=−1/8
29. (a)2xy
4
z
3
+y (b)4x
2
y
3
z
3
+x (c)3x
2
y
4
z
2
+2z
(d)2y
4
z
3
+y (e)32z
3
+1 (f)438
30. (a)2xycosz (b)x
2
cosz (c)−x
2
ysinz
(d)4ycosz (e)4 cosz (f)0
31.f
x=2z/x,f y=z/y,f z=ln(x
2
ycosz)−ztanz
32.f
x=y
−5/2
zsec(xz/y) tan(xz/y),f y=−xy
−7/2
zsec(xz/y) tan(xz/y)−(3/2)y
−5/2
sec(xz/y),
f
z=xy
−5/2
sec(xz/y) tan(xz/y)
33.f
x=−y
2
z
3
/
J
1+x
2
y
4
z
6
a
,f
y=−2xyz
3
/
J
1+x
2
y
4
z
6
a
,f
z=−3xy
2
z
2
/
J
1+x
2
y
4
z
6
a
34.f
x=4xyzcosh
√
zsinh
J
x
2
yz
a
cosh
J
x
2
yz
a
,f y=2x
2
zcosh
√
zsinh
J
x
2
yz
a
cosh
J
x
2
yz
a
,
f
z=2x
2
ycosh
√
zsinh
J
x
2
yz
a
cosh
J
x
2
yz
a
+(1/2)z
−1/2
sinh
√
zsinh
2
J
x
2
yz
a

12. January 27, 2005 11:55 L24-ch14 Sheet number 12 Page number 614 black
614 Chapter 14
35.∂w/∂x=yze
z
cosxz,∂w/∂y=e
z
sinxz,∂w/∂z=ye
z
(sinxz+xcosxz)
36.∂w/∂x=2x/
J
y
2
+z
2
a
,∂w/∂y=−2y
J
x
2
+z
2
a
/
J
y
2
+z
2
a
2
,∂w/∂z=2z
J
y
2
−x
2
a
/
J
y
2
+z
2
a
2
37.∂w/∂x=x/
y
x
2
+y
2
+z
2
,∂w/∂y=y/
y
x
2
+y
2
+z
2
,∂w/∂z=z/
y
x
2
+y
2
+z
2
38.∂w/∂x=2y
3
e
2x+3z
,∂w/∂y=3y
2
e
2x+3z
,∂w/∂z=3y
3
e
2x+3z
39. (a)e (b)2e (c)e
40. (a)2/
√
7 (b)4/
√
7 (c)1/
√
7
41. (a) –2
0
1
–2
–1
–1
0
1
2
y
0
1
2
z
–1
2
x
(b)
–2–10
1
2
–2
–1
0
1
2
y
–6
–4
–2
0
z
x
42.
–2
–1
0
1
2
x
–2
–1
0
1
2
–1
0
1
z
y
–2
–1
0
1
2
x
–2
–1
0
1
2
0
6
z
y
43.∂z/∂x=2x+6y(∂y/∂x)=2x, ∂z/∂x

(2,1)
=4
44.∂z/∂y=6y,∂z/∂y
2
2
(2,1)=6
45.∂z/∂x=−x
J
29−x
2
−y
2
a
−1/2
,∂z/∂x]
(4,3)
=−2
46. (a)∂z/∂y=8y,∂z/∂y]
(−1,1)
=8 (b)∂z/∂x=2x,∂z/∂x]
(−1,1)
=−2
47. (a)∂V/∂r=2πrh (b)∂V/∂h=πr
2
(c)∂V/∂r]
r=6,h=4
=48π (d)∂V/∂h]
r=8,h=10
=64π
48. (a)∂V/∂s=
πsd
2
6
√
4s
2
−d
2
(b)∂V/∂d=
πd(8s
2
−3d
2
)
24
√
4s
2
−d
2
(c)∂V/∂s]
s=10,d=16
=320π/9 (d)∂V/∂d]
s=10,d=16
=16π/9
49. (a)P=10T/V,∂P/∂T=10/V,∂P/∂T]
T=80,V=50
=1/5 lb/(in
2
K)
(b)V=10T/P, ∂V/∂P=−10T/P
2
,ifV=50 andT=80 then
P=10(80)/(50) =16,∂V/∂P]
T=80,P=16
=−25/8(in
5
/lb)
50. (a)∂T/∂x=3x
2
+1,∂T/∂x]
(1,2)
=4 (b)∂T/∂y=4y,∂T/∂y]
(1,2)
=8

13. January 27, 2005 11:55 L24-ch14 Sheet number 13 Page number 615 black
Exercise Set 14.3 615
51. (a)V=lwh, ∂V/∂l=wh=6
(b)∂V/∂w=lh=15
(c)∂V/∂h=lw=10
52. (a)∂A/∂a=(1/2)bsinθ=(1/2)(10)
J√
3/2
a
=5
√
3/2
(b)∂A/∂θ=(1/2)abcosθ=(1/2)(5)(10)(1/2) =25/2
(c)b=(2Acscθ)/a,∂b/∂a=−(2Acscθ)/a
2
=−b/a=−2
53.∂V/∂r=
2
3
πrh=
2
r
(
1
3
πr
2
h)=2V/r
54. (a)∂z/∂y=x
2
,∂z/∂y]
(1,3)
=1,j+kis parallel to the tangent line sox=1,y=3+t,
z=3+t
(b)∂z/∂x=2xy,∂z/∂x]
(1,3)
=6,i+6kis parallel to the tangent line sox=1+t,y=3,
z=3+6t
55. (a)2x−2z(∂z/∂x)=0,∂z/∂x=x/z=±3/(2
√
6) =±
√
6/4
(b)z=±
y
x
2
+y
2
−1,∂z/∂x=±x/
y
x
2
+y
2
−1=±
√
6/4
56. (a)2y−2z(∂z/∂y)=0,∂z/∂y=y/z=±4/(2
√
6) =±
√
6/3
(b)z=±
y
x
2
+y
2
−1,∂z/∂y=±y/
y
x
2
+y
2
−1=±
√
6/3
57.
3
2
J
x
2
+y
2
+z
2
a
1/2

2x+2z
∂z
∂x

=0,∂z/∂x=−x/z; similarly,∂z/∂y=−y/z
58.
4x−3z
2
(∂z/∂x)
2x
2
+y−z
3
=1,
∂z
∂x
=
4x−2x
2
−y+z
3
3z
2
;
1−3z
2
(∂z/∂y)
2x
2
+y−z
3
=0,
∂z
∂y
=
1
3z
2
59.2x+z

xy
∂z
∂x
+yz

cosxyz+
∂z
∂x
sinxyz=0,
∂z
∂x
=−
2x+yz
2
cosxyz
xyzcosxyz+ sinxyz
;
z

xy
∂z
∂y
+xz

cosxyz+
∂z
∂y
sinxyz=0,
∂z
∂y
=−
xz
2
cosxyz
xyzcosxyz+ sinxyz
60.e
xy
(coshz)
∂z
∂x
+ye
xy
sinhz−z
2
−2xz
∂z
∂x
=0,
∂z
∂x
=
z
2
−ye
xy
sinhz
e
xy
coshz−2xz
;
e
xy
(coshz)
∂z
∂y
+xe
xy
sinhz−2xz
∂z
∂y
=0,
∂z
∂y
=−
xe
xy
sinhz
e
xy
coshz−2xz
61.(3/2)
J
x
2
+y
2
+z
2
+w
2
a
1/2

2x+2w
∂w
∂x

=0,∂w/∂x=−x/w; similarly,∂w/∂y=−y/w
and∂w/∂z=−z/w
62.∂w/∂x=−4x/3,∂w/∂y=−1/3,∂w/∂z=(2x
2
+y−z
3
+3z
2
+3w)/3
63.
∂w
∂x
=−
yzwcosxyz
2w+ sinxyz
,
∂w
∂y
=−
xzwcosxyz
2w+ sinxyz
,
∂w
∂z
=−
xywcosxyz
2w+ sinxyz
64.
∂w
∂x
=
ye
xy
sinhw
z
2
−e
xy
coshw
,
∂w
∂y
=
xe
xy
sinhw
z
2
−e
xy
coshw
,
∂w
∂z
=
2zw
e
xy
coshw−z
2
65.f x=e
x
2
,fy=−e
y
2
66.f x=ye
x
2
y
2
,fy=xe
x
2
y
2

14. January 27, 2005 11:55 L24-ch14 Sheet number 14 Page number 616 black
616 Chapter 14
67. (a)−
1
4x
3/2
cosy (b)−
√
xcosy (c)−
siny
2
√
x
(d)−
siny
2
√
x
68. (a)8+84x
2
y
5
(b)140x
4
y
3
(c)140x
3
y
4
(d)140x
3
y
4
69.f x=8x−8y
4
,fy=−32xy
3
+35y
4
,fxy=fyx=−32y
3
70.f x=x/
y
x
2
+y
2
,fy=y/
y
x
2
+y
2
,fxy=fyx=−xy(x
2
+y
2
)
−3/2
71.f x=e
x
cosy, fy=−e
x
siny, fxy=fyx=−e
x
siny
72.f
x=e
x−y
2
,fy=−2ye
x−y
2
,fxy=fyx=−2ye
x−y
2
73.f x=4/(4x−5y),f y=−5/(4x−5y),f xy=fyx=20/(4x−5y)
2
74.f x=2x/(x
2
+y
2
),fy=2y/(x
2
+y
2
),fxy=−4xy/(x
2
+y
2
)
2
75.f x=2y/(x+y)
2
,fy=−2x/(x+y)
2
,fxy=fyx=2(x−y)/(x+y)
3
76.f x=4xy
2
/(x
2
+y
2
)
2
,fy=−4x
2
y/(x
2
+y
2
)
2
,fxy=fyx=8xy(x
2
−y
2
)/(x
2
+y
2
)
3
77. (a)
∂
3
f
∂x
3
(b)
∂
3
f
∂y
2
∂x
(c)
∂
4
f
∂x
2
∂y
2
(d)
∂
4
f
∂y
3
∂x
78. (a)f
xyy (b)f xxxx (c)f xxyy (d)f yyyxx
79. (a)30xy
4
−4 (b)60x
2
y
3
(c)60x
3
y
2
80. (a)120(2x−y)
2
(b)−240(2x−y)
2
(c)480(2x−y)
81. (a)f
xyy(0,1) =−30 (b)f xxx(0,1) =−125 (c)f yyxx(0,1) =150
82. (a)
∂
3
w
∂y
2
∂x
=−e
y
sinx,
∂
3
w
∂y
2
∂x
2
2
2
2
(π/4,0)
=−1/
√
2
(b)
∂
3
w
∂x
2
∂y
=−e
y
cosx,
∂
3
w
∂x
2
∂y
2
2
2
2
(π/4,0)
=−1/
√
2
83. (a)f
xy=15x
2
y
4
z
7
+2y (b)f yz=35x
3
y
4
z
6
+3y
2
(c)f xz=21x
2
y
5
z
6
(d)f zz=42x
3
y
5
z
5
(e)f zyy=140x
3
y
3
z
6
+6y (f)f xxy=30xy
4
z
7
(g)f zyx=105x
2
y
4
z
6
(h)f xxyz=210xy
4
z
6
84. (a)160(4x−3y+2z)
3
(b)−1440(4x−3y+2z)
2
(c)−5760(4x−3y+2z)
85. (a)f
x=2x+2y, f xx=2,f y=−2y+2x, f yy=−2;f xx+fyy=2−2=0
(b)z
x=e
x
siny−e
y
sinx, zxx=e
x
siny−e
y
cosx, zy=e
x
cosy+e
y
cosx,
z
yy=−e
x
siny+e
y
cosx;z xx+zyy=e
x
siny−e
y
cosx−e
x
siny+e
y
cosx=0

15. January 27, 2005 11:55 L24-ch14 Sheet number 15 Page number 617 black
Exercise Set 14.3 617
(c)z x=
2x
x
2
+y
2
−2
y
x
2
1
1+(y/x)
2
=
2x−2y
x
2
+y
2
,zxx=−2
x
2
−y
2
−2xy
(x
2
+y
2
)
2
,
z
y=
2y
x
2
+y
2
+2
1
x
1
1+(y/x)
2
=
2y+2x
x
2
+y
2
,zyy=−2
y
2
−x
2
+2xy
(x
2
+y
2
)
2
;
z
xx+zyy=−2
x
2
−y
2
−2xy
(x
2
+y
2
)
2
−2
y
2
−x
2
+2xy
(x
2
+y
2
)
2
=0
86. (a)z
t=−e
−t
sin(x/c),z x=(1/c)e
−t
cos(x/c),z xx=−(1/c
2
)e
−t
sin(x/c);
z
t−c
2
zxx=−e
−t
sin(x/c)−c
2
(−(1/c
2
)e
−t
sin(x/c)) =0
(b)z
t=−e
−t
cos(x/c),z x=−(1/c)e
−t
sin(x/c),z xx=−(1/c
2
)e
−t
cos(x/c);
z
t−c
2
zxx=−e
−t
cos(x/c)−c
2
(−(1/c
2
)e
−t
cos(x/c)) =0
87.u
x=ωsincωtcosωx, u xx=−ω
2
sincωtsinωx, u t=cωcoscωtsinωx,
u
tt=−c
2
ω
2
sincωtsinωx;u xx−
1
c
2
utt=−ω
2
sincωtsinωx−
1
c
2
(−c
2
)ω
2
sincωtsinωx=0
88. (a)∂u/∂x=∂v/∂y=2x,∂u/∂y=−∂v/∂x=−2y
(b)∂u/∂x=∂v/∂y=e
x
cosy,∂u/∂y=−∂v/∂x=−e
x
siny
(c)∂u/∂x=∂v/∂y=2x/(x
2
+y
2
),∂u/∂y=−∂v/∂x=2y/(x
2
+y
2
)
89.∂u/∂x=∂v/∂yand∂u/∂y=−∂v/∂xso∂
2
u/∂x
2
=∂
2
v/∂x∂y, and∂
2
u/∂y
2
=−∂
2
v/∂y∂x,
∂
2
u/∂x
2
+∂
2
u/∂y
2
=∂
2
v/∂x∂y−∂
2
v/∂y∂x,if∂
2
v/∂x∂y=∂
2
v/∂y∂xthen
∂
2
u/∂x
2
+∂
2
u/∂y
2
=0; thususatisfies Laplace’s equation. The proof thatvsatisfies Laplace’s
equation is similar. Adding Laplace’s equations foruandvgives Laplaces’ equation foru+v.
90.∂
2
R/∂R
2
1
=−2R
2
2
/(R1+R2)
3
,∂
2
R/∂R
2
2
=−2R
2
1
/(R1+R2)
3
,
J
∂
2
R/∂R
2
1
aJ
∂
2
R/∂R
2
2
a
=4R
2
1
R
2
2
/(R1+R2)
6
=
n
4/(R 1+R2)
4
u
[R
1R2/(R1+R2)]
2
=4R
2
/(R1+R2)
4
91.∂f/∂v=8vw
3
x
4
y
5
,∂f/∂w=12v
2
w
2
x
4
y
5
,∂f/∂x=16v
2
w
3
x
3
y
5
,∂f/∂y=20v
2
w
3
x
4
y
4
92.∂w/∂r=cosst+ue
u
cosur,∂w/∂s=−rtsinst,
∂w/∂t=−rssinst,∂w/∂u=re
u
cosur+e
u
sinur
93.∂f/∂v
1=2v 1/
J
v
2
3
+v
2
4
a
,∂f/∂v
2=−2v 2/
J
v
2
3
+v
2
4
a
,∂f/∂v
3=−2v 3
J
v
2
1
−v
2
2
a
/
J
v
2
3
+v
2
4
a
2
,
∂f/∂v
4=−2v 4
J
v
2
1
−v
2
2
a
/
J
v
2
3
+v
2
4
a
2
94.
∂V
∂x
=2xe
2x−y
+e
2x−y
,
∂V
∂y
=−xe
2x−y
+w,
∂V
∂z
=w
2
e
zw
,
∂V
∂w
=wze
zw
+e
zw
+y
95. (a)0 (b)0 (c)0 (d)0
(e)2(1 +yw)e
yw
sinzcosz (f)2xw(2 +yw)e
yw
sinzcosz
96.128,−512, 32, 64/3

16. January 27, 2005 11:55 L24-ch14 Sheet number 16 Page number 618 black
618 Chapter 14
97.∂w/∂x i=−isin(x 1+2x 2+...+nx n) 98.∂w/∂x i=
1
n

n
r
k=1
xk
L
(1/n)−1
99. (a)xy-plane,f x=12x
2
y+6xy, f y=4x
3
+3x
2
,fxy=fyx=12x
2
+6x
(b)y7 =0,f
x=3x
2
/y, fy=−x
3
/y
2
,fxy=fyx=−3x
2
/y
2
100. (a)x
2
+y
2
>1, (the exterior of the circle of radius 1 about the origin);
f
x=x/
yx
2
+y
2
−1,f y=y/
y
x
2
+y
2
−1,f xy=fyx=−xy
J
x
2
+y
2
−1
a
−3/2
(b)xy-plane,f x=2xcos(x
2
+y
3
),fy=3y
2
cos(x
2
+y
3
),fxy=fyx=−6xy
2
sin
J
x
2
+y
3
a
101.f
x(2,−1) =lim
x→2
f(x,−1)−f(2,−1)
x−2
=lim x→2
2x
2
+3x+1−15
x−2
=lim x→2
(2x+ 7) =11 and
f
y(2,−1) =lim
y→−1
f(2,y)−f(2,−1)
y+1
=lim y→−1
8−6y+y
2
−15
y+1
=lim y→−1
y−7=−8
102.f
x(x, y)=
2
3
(x
2
+y
2
)
−1/3
(2x)=
4x
3(x
2
+y
2
)
1/3
,(x, y)7 =(0,0);
f
x(0,0) =
d
dx
[f(x,0)]

x=0
=
d
dx
[x
4/3
]

x=0
=
4
3
x
1/3

x=0
=0.
103. (a)f
y(0,0) =
d
dy
[f(0,y)]

y=0
=
d
dy
[y]

y=0
=1
(b)If (x, y)7 =(0,0), thenf
y(x, y)=
1
3
(x
3
+y
3
)
−2/3
(3y
2
)=
y
2
(x
3
+y
3
)
2/3
;
f
y(x, y) does not exist wheny7 =0 andy=−x
EXERCISE SET 14.4
1.f(x, y)≈f(3,4) +f x(x−3) +f y(y−4)=5+2(x−3)−(y−4) and
f(3.01,3.98)≈5 + 2(0.01)−(−0.02) =5.04
2.f(x, y)≈f(−1,2) +f
x(x+1)+f y(y−2)=2+(x+1)+3(y−2) and
f(−0.99,2.02)≈2+0.01 + 3(0.02) =2.07
3.L(x, y, z)=f(1,2,3)+(x−1)+2(y−2)+3(z−3),
f(1.01,2.02,3.03)≈4+0.01 + 2(0.02) + 3(0.03) =4.14
4.L(x, y, z)=f(2,1,−2)−(x−2)+(y−1)−2(z+2),
f(1.98,0.99,−1.97)≈0.02−0.01−2(0.03) =−0.05
5.Supposef(x, y)=cfor all (x, y). Then at (x
0,y0)wehave
f(x
0+∆x, y0)−f(x 0,y0)
∆x
=0 and
hencef
x(x0,y0) exists and is equal to 0 (Definition 14.3.1). A similar result holds forf y. From
equation (2), it follows that ∆f=0, and then by Definition 14.4.1 we see thatfis differentiable
at (x
0,y0). An analogous result holds for functionsf(x, y, z) of three variables.

17. January 27, 2005 11:55 L24-ch14 Sheet number 17 Page number 619 black
Exercise Set 14.4 619
6.Letf(x, y)=ax+by+c. ThenL(x, y)=f(x 0,y0)+f x(x0,y0)(x−x 0)+f y(x0,y0)(y−y 0)=
ax
0+by0+c+a(x−x 0)+b(y−y 0)=ax+by+c,soL=fand thusEis zero. For three variables
the proof is analogous.
7.f
x=2x, fy=2y, fz=2zsoL(x, y, z)=0,E=f−L=x
2
+y
2
+z
2
, and
lim
(x,y,z)→(0,0,0)
E(x, y, z)
y
x
2
+y
2
+z
2
=lim
(x,y,z)→(0,0,0)
y
x
2
+y
2
+z
2
=0, sofis differentiable at (0,0,0).
8.f
x=2xr(x
2
+y
2
+z
2
)
r−1
,fy=2yr(x
2
+y
2
+z
2
)
r−1
,fz=2zr(x
2
+y
2
+z
2
)
r−1
, so the partials
offexist only ifr≥1. If so thenL(x, y, z)=0,E(x, y, z)=f(x, y, z) and
E(x, y, z)
y
x
2
+y
2
+z
2
=(x
2
+y
2
+z
2
)
r−1/2
,sofis differentiable at (0,0,0) if and only ifr>1/2.
9.dz=7dx−2dy 10.dz=ye
xy
dx+xe
xy
dy 11.dz=3x
2
y
2
dx+2x
3
ydy
12.dz=(10xy
5
−2)dx+ (25x
2
y
4
+4)dy
13.dz=
4
y/
J
1+x
2
y
2
a0
dx+
4
x/
J
1+x
2
y
2
a0
dy
14.dz=2 sec
2
(x−3y) tan(x−3y)dx−6 sec
2
(x−3y) tan(x−3y)dy
15.dw=8dx−3dy+4dz 16.dw=yze
xyz
dx+xze
xyz
dy+xye
xyz
dz
17.dw=3x
2
y
2
zdx+2x
3
yzdy+x
3
y
2
dz
18.dw=
J
8xy
3
z
7
−3y
a
dx+
J
12x
2
y
2
z
7
−3x
a
dy+
J
28x
2
y
3
z
6
+1
a
dz
19.dw=
yz
1+x
2
y
2
z
2
dx+
xz
1+x
2
y
2
z
2
dy+
xy
1+x
2
y
2
z
2
dz
20.dw=
1
2
√
x
dx+
1
2
√
y
dy+
1
2
√
z
dz
21.df=(2x+2y−4)dx+2xdy;x=1,y=2,dx=0.01,dy=0.04 so
df=0.10 and ∆f=0.1009
22.df=(1/3)x
−2/3
y
1/2
dx+(1/2)x
1/3
y
−1/2
dy;x=8,y=9,dx=−0.22,dy=0.03 sodf=−0.045
and ∆f≈−0.045613
23.df=−x
−2
dx−y
−2
dy;x=−1,y=−2,dx=−0.02,dy=−0.04 so
df=0.03 and ∆f≈0.029412
24.df=
y
2(1 +xy)
dx+
x
2(1 +xy)
dy;x=0,y=2,dx=−0.09,dy=−0.02 so
df=−0.09 and ∆f≈−0.098129
25.df=2y
2
z
3
dx+4xyz
3
dy+6xy
2
z
2
dz, x=1,y=−1,z=2,dx=−0.01,dy=−0.02,dz=0.02 so
df=0.96 and ∆f≈0.97929
26.df=
yz(y+z)
(x+y+z)
2
dx+
xz(x+z)
(x+y+z)
2
dy+
xy(x+y)
(x+y+z)
2
dz, x=−1,y=−2,z=4,dx=−0.04,
dy=0.02,dz=−0.03 sodf=0.58 and ∆f≈0.60529

18. January 27, 2005 11:55 L24-ch14 Sheet number 18 Page number 620 black
620 Chapter 14
27.Label the four smaller rectanglesA, B, C, Dstarting with the lower left and going clockwise. Then
the increase in the area of the rectangle is represented byB,CandD; and the portionsBandD
represent the approximation of the increase in area given by the total differential.
28.V+∆V=(π/3)4.05
2
(19.95)≈109.0766250π, V=320π/3,∆V≈2.40996π;
dV=(2/3)πrhdr+(1/3)πr
2
dh;r=4,h=20,dr=0.05,dh=−0.05 sodV=2.4π, and
∆V/dV≈1.00415.
29. (a)f(P)=1/5,f
x(P)=−x/(x
2
+y
2
)
−3/2

(x,y)=(4,3)
=−4/125,
f
y(P)=−y/(x
2
+y
2
)
−3/2

(x,y)=(4,3)
=−3/125,L(x, y)=
1
5
−
4
125
(x−4)−
3
125
(y−3)
(b)L(Q)−f(Q)=
1
5
−
4
125
(−0.08)−
3
125
(0.01)−0.2023342382≈−0.0000142382,
|PQ|=
y
0.08
2
+0.01
2
≈0.0008062257748,|L(Q)−f(Q)|/|PQ|≈0.000176603
30. (a)f(P)=1,f
x(P)=0.5,f y(P)=0.3,L(x, y)=1+0.5(x−1)+0.3(y−1)
(b)L(Q)−f(Q)=1+0.5(0.05)+0.3(−0.03)−1.05
0.5
0.97
0.3
≈0.00063,
|PQ|=
√
0.05
2
+0.03
2
≈0.05831,|L(Q)−f(Q)|/|PQ|≈0.0107
31. (a)f(P)=0,f
x(P)=0,f y(P)=0,L(x, y)=0
(b)L(Q)−f(Q)=−0.003 sin(0.004)≈−0.000012,|PQ|=
√
0.003
2
+0.004
2
=0.005,
|L(Q)−f(Q)|/|PQ|≈0.0024
32. (a)f(P) =ln 2,f
x(P)=1,f y(P)=1/2,L(x, y)=ln2+(x−1) +
1
2
(y−2)
(b)L(Q)−f(Q)=ln2+0.01+(1/2)(0.02)−ln 2.0402≈0.0000993383,
|PQ|=
√
0.01
2
+0.02
2
≈0.02236067978,|L(Q)−f(Q)|/|PQ|≈0.0044425
33. (a)f(P)=6,f
x(P)=6,f y(P)=3,f z(P)=2,L(x, y)=6+6(x−1)+3(y−2)+2(z−3)
(b)L(Q)−f(Q) =6 + 6(0.001) + 3(0.002) + 2(0.003)−6.018018006 =−.000018006,
|PQ|=
√
0.001
2
+0.002
2
+0.003
2
≈.0003741657387;|L(Q)−f(Q)|/|PQ|≈−0.000481
34. (a)f(P)=0,f
x(P)=1/2,f y(P)=1/2,f z(P)=0,L(x, y)=
1
2
(x+1)+
1
2
(y−1)
(b)L(Q)−f(Q)=0,|L(Q)−f(Q)|/|PQ|=0
35. (a)f(P)=e, f
x(P)=e, f y(P)=−e, f z(P)=−e, L(x, y)=e+e(x−1)−e(y+1)−e(z+1)
(b)L(Q)−f(Q)=e−0.01e+0.01e−0.01e−0.99e
0.9999
=0.99(e−e
0.9999
),
|PQ|=
√
0.01
2
+0.01
2
+0.01
2
≈0.01732,|L(Q)−f(Q)|/|PQ|≈0.01554
36. (a)f(P)=0,f
x(P)=1,f y(P)=−1,f z(P)=1,L(x, y, z)=(x−2)−(y−1)+(z+1)
(b)L(Q)−f(Q)=0.02+0.03−0.01−ln 1.0403≈0.00049086691,
|PQ|=
√
0.02
2
+0.03
2
+0.01
2
≈0.03742,|L(Q)−f(Q)|/|PQ|≈0.01312
37. (a)Letf(x, y)=e
x
siny;f(0,0) =0,f x(0,0) =0,f y(0,0) =1, soe
x
siny≈y
(b)Letf(x, y)=
2x+1
y+1
;f(0,0) =1,f
x(0,0) =2,f y(0,0) =−1, so
2x+1
y+1
≈1+2x−y

19. January 27, 2005 11:55 L24-ch14 Sheet number 19 Page number 621 black
Exercise Set 14.4 621
38.f(1,1) =1,f x(x, y)=αx
α−1
y
β
,fx(1,1) =α, f y(x, y)=βx
α
y
β−1
,fy(1,1) =β,so
x
α
y
β
≈1+α(x−1) +β(y−1)
39. (a)Letf(x, y, z)=xyz+ 2, thenf
x=fy=fz=1atx=y=z=1, and
L(x, y, z)=f(1,1,1) +f
x(x−1) +f y(y−1) +f z(z−1) =3 +x−1+y−1+z−1=x+y+z
(b)Letf(x, y, z)=
4x
y+z
, thenf
x=2,f y=−1,f z=−1atx=y=z=1, and
L(x, y, z)=f(1,1,1) +f
x(x−1) +f y(y−1) +f z(z−1)
=2+2(x−1)−(y−1)−(z−1) =2x−y−z+2
40.Letf(x, y, z)=x
α
y
β
z
γ
, thenf x=α, fy=β,fz=γatx=y=z=1, and
f(x, y, z)≈f(1,1,1) +f
x(x−1) +f y(y−1) +f z(z−1) =1 +α(x−1) +β(y−1) +γ(z−1)
41.L(x, y)=f(1,1) +f
x(1,1)(x−1) +f y(1,1)(y−1) and
L(1.1,0.9) =3.15 =3 + 2(0.1) +f
y(1,1)(−0.1) sof y(1,1) =−0.05/(−0.1) =0.5
42.L(x, y)=3+f
x(0,−1)x−2(y+1),3.3=3+f x(0,−1)(0.1)−2(−0.1), sof x(0,−1) =0.1/0.1=1
43.x−y+2z−2=L(x, y, z)=f(3,2,1) +f
x(3,2,1)(x−3) +f y(3,2,1)(y−2) +f z(3,2,1)(z−1), so
f
x(3,2,1) =1,f y(3,2,1) =−1,f z(3,2,1) =2 andf(3,2,1) =L(3,2,1) =1
44.L(x, y, z)=x+2y+3z+4=(x−0)+2(y+1)+3(z+2)−4,
f(0,−1,−2) =−4,f
x(0,−1,−2) =1,f y(0,−1,−2) =2,f z(0,−1,−2) =3
45.L(x, y)=f(x
0,y0)+f x(x0,y0)(x−x 0)+f y(x0,y0)(y−y 0),
2y−2x−2=x
2
0
+y
2
0
+2x 0(x−x 0)+2y 0(y−y 0), from which it follows thatx 0=−1,y 0=1.
46.f(x, y)=x
2
y,sof x(x0,y0)=2x 0y0,fy(x0,y0)=x
2
0
, and
L(x, y)=f(x
0,y0)+2x 0y0(x−x 0)+x
2
0
(y−y 0). ButL(x, y)=8−4x+4y, hence
−4=2x
0y0,4=x
2
0
and 8 =f(x 0,y0)−2x
2
0
y0−x
2
0
y0=−2x
2
0
y0. Thus eitherx 0=−2,y 0=1
from which it follows that 8 =−8, a contradiction, orx
0=2,y 0=−1, which is a solution since
then 8 =−2x
2
0
y0=8 is true.
47.L(x, y, z)=f(x
0,y0,z0)+f x(x0,y0,z0)(x−x 0)+f y(x0,y0,z0)(y−y 0)+f z(x0,y0,z0)(z−z 0),
y+2z−1=x
0y0+z
2
0
+y0(x−x 0)+x 0(y−y 0)+2z 0(z−z 0), so thatx 0=1,y 0=0,z 0=1.
48.L(x, y, z)=f(x
0,y0,z0)+f x(x0,y0,z0)(x−x 0)+f y(x0,y0,z0)(y−y 0)+f z(x0,y0,z0)(z−z 0).
Thenx−y−z−2=x
0y0z0+y0z0(x−x 0)+x 0z0(y−y 0)+x 0y0(z−z 0), hence
y
0z0=1,x 0z0=−1,x 0y0=−1, and−2=x 0y0z0−3x 0y0z0,orx 0y0z0=1. Since now
x
0=−y 0=−z 0, we must have|x 0|=|y 0|=|z 0|=1 or else|x 0y0z0|7 =1, impossible. Thus
x
0=1,y 0=z0=−1 (note that (−1,1,1) is not a solution).
49.A=xy,dA=ydx+xdy,dA/A=dx/x+dy/y,|dx/x|≤0.03 and|dy/y|≤0.05,
|dA/A|≤|dx/x|+|dy/y|≤0.08 =8%
50.V=(1/3)πr
2
h,dV=(2/3)πrhdr+(1/3)πr
2
dh,dV/V=2(dr/r)+dh/h,|dr/r|≤0.01 and
|dh/h|≤0.04,|dV/V|≤2|dr/r|+|dh/h|≤0.06 =6%.

20. January 27, 2005 11:55 L24-ch14 Sheet number 20 Page number 622 black
622 Chapter 14
51.z=
y
x
2
+y
2
,dz=
x
y
x
2
+y
2
dx+
y
y
x
2
+y
2
dy,
dz
z
=
x
x
2
+y
2
dx+
y
x
2
+y
2
dy=
x
2
x
2
+y
2

dx
x

+
y
2
x
2
+y
2

dy
y

,
2
2
2
2
dz
z
2
2
2
2
≤
x
2
x
2
+y
2
2
2
2
2
dx
x
2
2
2
2
+
y
2
x
2
+y
2
2
2
2
2
dy
y
2
2
2
2
,if
2
2
2
2
dx
x
2
2
2
2
≤r/100 and
2
2
2
2
dy
y
2
2
2
2
≤r/100 then
2
2
2
2
dz
z
2
2
2
2
≤
x
2
x
2
+y
2
(r/100) +
y
2
x
2
+y
2
(r/100) =
r
100
so the percentage error inzis at most aboutr%.
52. (a)z=
y
x
2
+y
2
,dz=x
J
x
2
+y
2
a
−1/2
dx+y
J
x
2
+y
2
a
−1/2
dy,
|dz|≤x
J
x
2
+y
2
a
−1/2
|dx|+y
J
x
2
+y
2
a
−1/2
|dy|;ifx=3,y=4,|dx|≤0.05, and
|dy|≤0.05 then|dz|≤(3/5)(0.05)+(4/5)(0.05) =0.07 cm
(b)A=(1/2)xy,dA=(1/2)ydx+(1/2)xdy,
|dA|≤(1/2)y|dx|+(1/2)x|dy|≤2(0.05)+(3/2)(0.05) =0.175 cm
2
.
53.dT=
π
g
y
L/g
dL−
πL
g
2
y
L/g
dg,
dT
T
=
1
2
dL
L
−
1
2
dg
g
;|dL/L|≤0.005 and|dg/g|≤0.001 so
|dT/T|≤(1/2)(0.005) + (1/2)(0.001) =0.003 =0.3%
54.dP=(k/V)dT−(kT/V
2
)dV,dP/P=dT/T−dV/V;ifdT/T=0.03 anddV/V=0.05 then
dP/P=−0.02 so there is about a 2% decrease in pressure.
55. (a)
2
2
2
2
d(xy)
xy
2
2
2
2
=
2
2
2
2
ydx+xdy
xy
2
2
2
2
=
2
2
2
2
dx
x
+
dy
y
2
2
2
2
≤
2
2
2
2
dx
x
2
2
2
2
+
2
2
2
2
dy
y
2
2
2
2
≤
r
100
+
s
100
;(r+s)%
(b)
2
2
2
2
d(x/y)
x/y
2
2
2
2
=
2
2
2
2
ydx−xdy
xy
2
2
2
2
=
2
2
2
2
dx
x
−
dy
y
2
2
2
2
≤
2
2
2
2
dx
x
2
2
2
2
+
2
2
2
2
dy
y
2
2
2
2
≤
r
100
+
s
100
;(r+s)%
(c)
2
2
2
2
d(x
2
y
3
)
x
2
y
3
2
2
2
2
=
2
2
2
2
2xy
3
dx+3x
2
y
2
dy
x
2
y
3
2
2
2
2
=
2
2
2
2
2
dx
x
+3
dy
y
2
2
2
2
≤2
2
2
2
2
dx
x
2
2
2
2
+3
2
2
2
2
dy
y
2
2
2
2
≤2
r
100
+3
s
100
;(2r+3s)%
(d)
2
2
2
2
d(x
3
y
1/2
)
x
3
y
1/2
2
2
2
2
=
2
2
2
2
3x
2
y
1/2
dx+(1/2)x
3
y
−1/2
dy
x
3
y
1/2
2
2
2
2
=
2
2
2
2
3
dx
x
+
1
2
dy
y
2
2
2
2
≤3
2
2
2
2
dx
x
2
2
2
2
+
1
2
2
2
2
2
dy
y
2
2
2
2
≤3
r
100
+
1
2
s
100
;(3r+
1
2
s)%
56.R=1/(1/R
1+1/R 2+1/R 3),∂R/∂R 1=
1
R
2
1
(1/R1+1/R 2+1/R 3)
2
=R
2
/R
2
1
, similarly
∂R/∂R
2=R
2
/R
2
2
and∂R/∂R 3=R
2
/R
2
3
so
dR
R
=(R/R
1)
dR
1
R1
+(R/R 2)
dR
2
R2
+(R/R 3)
dR
3
R3
,
2
2
2
2
dR
R
2
2
2
2
≤(R/R
1)
2
2
2
2
dR
1
R1
2
2
2
2
+(R/R
2)
2
2
2
2
dR
2
R2
2
2
2
2
+(R/R
3)
2
2
2
2
dR
3
R3
2
2
2
2
≤(R/R
1)(0.10)+(R/R 2)(0.10)+(R/R 3)(0.10)
=R(1/R
1+1/R 2+1/R 3)(0.10) =(1)(0.10) =0.10 =10%

21. January 27, 2005 11:55 L24-ch14 Sheet number 21 Page number 623 black
Exercise Set 14.5 623
57.dA=
1
2
bsinθda+
1
2
asinθdb+
1
2
abcosθdθ,
|dA|≤
1
2
bsinθ|da|+
1
2
asinθ|db|+
1
2
abcosθ|dθ|
≤
1
2
(50)(1/2)(1/2) +
1
2
(40)(1/2)(1/4) +
1
2
(40)(50)
-√
3/2
c
(π/90)
=35/4+50π
√
3/9≈39 ft
2
58.V=3wh,dV=whd3+3hdw+3wdh,|dV/V|≤|d3/3|+|dw/w|+|dh/h|≤3(r/100) =3r%
59.f
x=2xsiny, f y=x
2
cosyare both continuous everywhere, sofis differentiable everywhere.
60.f
x=ysinz,f y=xsinz,f z=xycoszare all continuous everywhere, sofis differentiable every-
where.
61.Thatfis differentiable means that lim
(x,y)→(x 0,y0)
Ef(x, y)
y
(x−x 0)
2
+(y−y 0)
2
=0, where
E
f(x, y)=f(x, y)−L f(x, y); hereL f(x, y) is the linear approximation tofat (x 0,y0).
Letf
xandf ydenotef x(x0,y0),fy(x0,y0) respectively. Theng(x, y, z)=z−f(x, y),
L
f(x, y)=f(x 0,y0)+f x(x−x 0)+f y(y−y 0),
L
g(x, y, z)=g(x 0,y0,z0)+g x(x−x 0)+g y(y−y 0)+g z(z−z 0)
=0−f
x(x−x 0)−f y(y−y 0)+(z−z 0)
,
and
E
g(x, y, z)=g(x, y, z)−L g(x, y, z)=(z−f(x, y)) +f x(x−x 0)+f y(y−y 0)−(z−z 0)
=f(x
0,y0)+f x(x0,y0)(x−x 0)+f y(x0,y0)(y−y 0)−f(x, y)=−E f(x, y)
Thus
|E
g(x, y, z)|
y
(x−x 0)
2
+(y−y 0)
2
+(z−z 0)
2
≤
|E
f(x, y)|
y
(x−x 0)
2
+(y−y 0)
2
so lim
(x,y,z)→(x 0,y0,z0)
Eg(x, y, z)
y
(x−x 0)
2
+(y−y 0)
2
+(z−z 0)
2
=0
andgis differentiable at (x
0,y0,z0).
62.The condition lim
(∆x,∆y)→(0,0)
∆f−f x(x0,y0)∆x−f y(x0,y0)∆y
y
(∆x)
2
+(∆y)
2
=0 is equivalent to
lim
(∆x,∆y)→(0,0)
4(∆x,∆y) =0 which is equivalent to4being continuous at (0,0) with4(0,0) =0.
Since4is continuous,fis differentiable.
EXERCISE SET 14.5
1.42t
13
2.
2(3 +t
−1/3
)
3(2t+t
2/3
)
3.3t
−2
sin(1/t) 4.
1−2t
4
−8t
4
lnt
2t
√
1+lnt−2t
4
lnt
5.−
10
3
t
7/3
e
1−t
10/3
6.(1 +t)e
t
cosh (te
t
/2) sinh (te
t
/2)
7.165t
32
8.
3−(4/3)t
−1/3
−24t
−7
3t−2t
2/3
+4t
−6

22. January 27, 2005 11:55 L24-ch14 Sheet number 22 Page number 624 black
624 Chapter 14
9.−2tcos
J
t
2
a
10.
1−512t
5
−2560t
5
lnt
2t
√
1+lnt−512t
5
lnt
11.3264 12.0
13.
dz
dt
=
∂z
∂x
dx
dt
+
∂z
∂y
dy
dt
=3(2t)
t=2−(3t
2
)t=2=12−12 =0
14.
dw
dt
=
∂w
∂x
dx
dt
+
∂w
∂y
dy
dt
+
∂w
∂z
dz
dt
=1+2(πcosπt)
t=1+ 3(2t) t=1=1−2π+6=7−2π
16.x
x
=e
xlnx
,
d
dx
(x
x
)=e
xlnx
d
dx
(xlnx)=e
xlnx
(lnx+x(1/x))
=e
xlnx
(lnx+1)=e
xlnx
+ (lnx)e
xlnx
=x
x
+ (lnx)x
x
17.∂z/∂u=24u
2
v
2
−16uv
3
−2v+3,∂z/∂v=16u
3
v−24u
2
v
2
−2u−3
18.∂z/∂u=2u/v
2
−u
2
vsec
2
(u/v)−2uv
2
tan(u/v)
∂z/∂v=−2u
2
/v
3
+u
3
sec
2
(u/v)−2u
2
vtan(u/v)
19.∂z/∂u=−
2 sinu
3 sinv
,∂z/∂v=−
2 cosucosv
3 sin
2
v
20.∂z/∂u=3+3v/u−4u, ∂z/∂v=2+3lnu+2lnv
21.∂z/∂u=e
u
,∂z/∂v=0
22.∂z/∂u=−sin(u−v) sin
J
u
2
+v
2
a
+2ucos(u−v) cos
J
u
2
+v
2
a
∂z/∂v=sin(u−v) sin
J
u
2
+v
2
a
+2vcos(u−v) cos
J
u
2
+v
2
a
23.∂T/∂r=3r
2
sinθcos
2
θ−4r
3
sin
3
θcosθ
∂T/∂θ=−2r
3
sin
2
θcosθ+r
4
sin
4
θ+r
3
cos
3
θ−3r
4
sin
2
θcos
2
θ
24.dR/dφ=5e
5φ
25.∂t/∂x=
J
x
2
+y
2
a
/
J
4x
2
y
3
a
,∂t/∂y=
J
y
2
−3x
2
a
/
J
4xy
4
a
26.∂w/∂u=
2v
2
4
u
2
v
2
−(u−2v)
2

[u
2
v
2
+(u−2v)
2
]
2
,∂w/∂v=
u
2
4
(u−2v)
2
−u
2
v
2

[u
2
v
2
+(u−2v)
2
]
2
27.∂z/∂r=(dz/dx)(∂x/∂r)=2rcos
2
θ/
J
r
2
cos
2
θ+1
a
,
∂z/∂θ=(dz/dx)(∂x/∂θ)=−2r
2
sinθcosθ/
J
r
2
cos
2
θ+1
a
28.∂u/∂x=(∂u/∂r)(dr/dx)+(∂u/∂t)(∂t/∂x)
=
J
s
2
lnt
a
(2x)+
J
rs
2
/t
aJ
y
3
a
=x(4y+1)
2
J
1+2lnxy
3
a
∂u/∂y=(∂u/∂s)(ds/dy)+(∂u/∂t)(∂t/∂y)
=(2rslnt)(4) +
J
rs
2
/t
aJ
3xy
2
a
=8x
2
(4y+1)lnxy
3
+3x
2
(4y+1)
2
/y

23. January 27, 2005 11:55 L24-ch14 Sheet number 23 Page number 625 black
Exercise Set 14.5 625
29.∂w/∂ρ=2ρ
J
4 sin
2
φ+ cos
2
φ
a
,∂w/∂φ=6ρ
2
sinφcosφ,∂w/∂θ=0
30.
dw
dx
=
∂w
∂x
+
∂w
∂y
dy
dx
+
∂w
∂z
dz
dx
=3y
2
z
3
+(6xyz
3
)(6x)+9xy
2
z
2
1
2
√
x−1
=3(3x
2
+2)
2
(x−1)
3/2
+36x
2
(3x
2
+ 2)(x−1)
3/2
+
9
2
x(3x
2
+2)
2
√
x−1
=
3
2
(3x
2
+ 2)(39x
3
−30x
2
+10x−4)
√
x−1
31.−π 32.351/2,−168
33.
√
3e
√
3
,
J
2−4
√
3
a
e
√
3
34.1161
35.F(x, y)=x
2
y
3
+ cosy,
dy
dx
=−
2xy
3
3x
2
y
2
−siny
36.F(x, y)=x
3
−3xy
2
+y
3
−5,
dy
dx
=−
3x
2
−3y
2
−6xy+3y
2
=
x
2
−y
2
2xy−y
2
37.F(x, y)=e
xy
+ye
y
−1,
dy
dx
=−
ye
xy
xe
xy
+ye
y
+e
y
38.F(x, y)=x−(xy)
1/2
+3y−4,
dy
dx
=−
1−(1/2)(xy)
−1/2
y
−(1/2)(xy)
−1/2
x+3
=
2
√
xy−y
x−6
√
xy
39.
∂F
∂x
+
∂F
∂z
∂z
∂x
=0so
∂z
∂x
=−
∂F/∂x
∂F/∂z
. 40.
∂F
∂y
+
∂F
∂z
∂z
∂y
=0so
∂z
∂y
=−
∂F/∂y
∂F/∂z
.
41.
∂z
∂x
=
2x+yz
6yz−xy
,
∂z
∂y
=
xz−3z
2
6yz−xy
42.ln(1 +z)+xy
2
+z−1=0;
∂z
∂x
=−
y
2
(1 +z)
2+z
,
∂z
∂y
=−
2xy(1 +z)
2+z
43.ye
x
−5 sin 3z−3z=0;
∂z
∂x
=−
ye
x
−15 cos 3z−3
=
ye
x
15 cos 3z+3
,
∂z
∂y
=
e
x
15 cos 3z+3
44.
∂z
∂x
=−
ze
yz
cosxz−ye
xy
cosyz
ye
xy
sinyz+xe
yz
cosxz+ye
yz
sinxz
,
∂z
∂y
=−
ze
xy
sinyz−xe
xy
cosyz+ze
yz
sinxz
ye
xy
sinyz+xe
yz
cosxz+ye
yz
sinxz
45.D=
J
x
2
+y
2
a
1/2
wherexandyare the distances of cars A andB, respectively, from the
intersection andDis the distance between them.
dD/dt=
n
x/
J
x
2
+y
2
a
1/2
u
(dx/dt)+
n
y/
J
x
2
+y
2
a
1/2
u
(dy/dt),dx/dt=−25 anddy/dt=−30
whenx=0.3 andy=0.4sodD/dt=(0.3/0.5)(−25)+(0.4/0.5)(−30) =−39 mph.
46.T=(1/10)PV,dT/dt=(V/10)(dP/dt)+(P/10)(dV/dt),dV/dt=4 anddP/dt=−1 when
V=200 andP=5sodT/dt=(20)(−1)+(1/2)(4) =−18 K/s.

24. January 27, 2005 11:55 L24-ch14 Sheet number 24 Page number 626 black
626 Chapter 14
47.A=
1
2
absinθbutθ=π/6 whena=4 andb=3soA=
1
2
(4)(3) sin(π/6) =3.
Solve
1
2
absinθ=3 forθto getθ=sin
−1

6
ab

,0≤θ≤π/2.
dθ
dt
=
∂θ
∂a
da
dt
+
∂θ
∂b
db
dt
=
1
h
1−
36
a
2
b
2

−
6
a
2
b

da
dt
+
1
h
1−
36
a
2
b
2

−
6
ab
2

db
dt
=−
6
√
a
2
b
2
−36

1
a
da
dt
+
1
b
db
dt

,
da
dt
=1 and
db
dt
=1
whena=4 andb=3so
dθ
dt
=−
6
√
144−36

1
4
+
1
3

=−
7
12
√
3
=−
7
36
√
3 radians/s
48.From the law of cosines,c=
√
a
2
+b
2
−2abcosθwherecis the length of the third side.
θ=π/3soc=
√
a
2
+b
2
−ab,
dc
dt
=
∂c
∂a
da
dt
+
∂c
∂b
db
dt
=
1
2
(a
2
+b
2
−ab)
−1/2
(2a−b)
da
dt
+
1
2
J
a
2
+b
2
−ab
a
−1/2
(2b−a)
db
dt
=
1
2
√
a
2
+b
2
−ab
7
(2a−b)
da
dt
+(2b−a)
db
dt

,
da
dt
=2 and
db
dt
=1 whena=5 andb=10
so
dc
dt
=
1
2
√
75
[(0)(2) + (15)(1)] =
√
3/2 cm/s. The third side is increasing.
49.V=(π/4)D
2
hwhereDis the diameter andhis the height, both measured in inches,
dV/dt=(π/2)Dh(dD/dt)+(π/4)D
2
(dh/dt),dD/dt=3 anddh/dt=24 whenD=30 and
h=240, sodV/dt=(π/2)(30)(240)(3) + (π/4)(30)
2
(24) =16,200πin
3
/year.
50.
dT
dt
=
∂T
∂x
dx
dt
+
∂T
∂y
dy
dt
=
y
2
x
dx
dt
+2ylnx
dy
dt
,dx/dt=1 anddy/dt=−4 at (3,2) so
dT/dt=(4/3)(1) + (4 ln 3)(−4) =4/3−16 ln 3
◦
C/s.
51. (a)V=3wh,
dV
dt
=
∂V
∂3
d3
dt
+
∂V
∂w
dw
dt
+
∂V
∂h
dh
dt
=wh
d3
dt
+3h
dw
dt
+3w
dh
dt
=(3)(6)(1) + (2)(6)(2) + (2)(3)(3) =60 in
3
/s
(b)D=
y
3
2
+w
2
+h
2
;dD/dt=(3/D)d3/dt+(w/D)dw/dt+(h/D)dh/dt
=(2/7)(1) + (3/7)(2) + (6/7)(3) =26/7 in/s
52.S=2(lw+wh+lh),
dS
dt
=
∂S
∂w
dw
dt
+
∂S
∂l
dl
dt
+
∂S
∂h
dh
dt
=2(l+h)
dw
dt
+2(w+h)
dl
dt
+2(w+l)
dh
dt
=80 in
2
/s
53. (a)f(tx, ty)=3t
2
x
2
+t
2
y
2
=t
2
f(x, y);n=2
(b)f(tx, ty)=
y
t
2
x
2
+t
2
y
2
=tf(x, y);n=1
(c)f(tx, ty)=t
3
x
2
y−2t
3
y
3
=t
3
f(x, y);n=3
(d)f(tx, ty)=5/
J
t
2
x
2
+2t
2
y
2
a
2
=t
−4
f(x, y);n=−4

25. January 27, 2005 11:55 L24-ch14 Sheet number 25 Page number 627 black
Exercise Set 14.5 627
54. (a)Iff(u, v)=t
n
f(x, y), then
∂f
∂u
du
dt
+
∂f
∂v
dv
dt
=nt
n−1
f(x, y),x
∂f
∂u
+y
∂f
∂v
=nt
n−1
f(x, y);
lett=1togetx
∂f
∂x
+y
∂f
∂y
=nf(x, y).
(b)Iff(x, y)=3x
2
+y
2
thenxf x+yfy=6x
2
+2y
2
=2f(x, y);
Iff(x, y)=
y
x
2
+y
2
thenxf x+yfy=x
2
/
y
x
2
+y
2
+y
2
/
y
x
2
+y
2
=
y
x
2
+y
2
=f(x, y);
Iff(x, y)=x
2
y−2y
3
thenxf x+yfy=3x
2
y−6y
3
=3f(x, y);
Iff(x, y)=
5 (x
2
+2y
2
)
2
thenxf x+yfy=x
5(−2)2x
(x
2
+2y
2
)
3
+y
5(−2)4y
(x
2
+2y
2
)
3
=−4f(x, y)
55. (a)
∂z
∂x
=
dz
du
∂u
∂x
,
∂z
∂y
=
dz
du
∂u
∂y
(b)
∂
2
z
∂x
2
=
dz
du
∂
2
u
∂x
2
+
∂
∂x

dz
du

∂u
∂x
=
dz
du
∂
2
u
∂x
2
+
d
2
z
du
2

∂u
∂x

2
;
∂
2
z
∂y∂x
=
dz
du
∂
2
u
∂y∂x
+
∂
∂y

dz
du

∂u
∂x
=
dz
du
∂
2
u
∂y∂x
+
d
2
z
du
2
∂u
∂x
∂u
∂y
∂
2
z
∂y
2
=
dz
du
∂
2
u
∂y
2
+
∂
∂y

dz
du

∂u
∂y
=
dz
du
∂
2
u
∂y
2
+
d
2
z
du
2

∂u
∂y

2
56. (a)z=f(u),u=x
2
−y
2
;∂z/∂x=(dz/du)(∂u/∂x)=2xdz/du
∂z/∂y=(dz/du)(∂u/∂y)=−2ydz/du, y∂z/∂x+x∂z/∂y=2xydz/du−2xydz/du=0
(b)z=f(u),u=xy;
∂z
∂x
=
dz
du
∂u
∂x
=y
dz
du
,
∂z
∂y
=
dz
du
∂u
∂y
=x
dz
du
,
x
∂z
∂x
−y
∂z
∂y
=xy
dz
du
−xy
dz
du
=0.
(c)yz
x+xzy=y(2xcos(x
2
−y
2
))−x(2ycos(x
2
−y
2
)) =0
(d)xz
x−yzy=xye
xy
−yxe
xy
=0
57.Letz=f(u) whereu=x+2y; then∂z/∂x=(dz/du)(∂u/∂x)=dz/du,
∂z/∂y=(dz/du)(∂u/∂y)=2dz/duso 2∂z/∂x−∂z/∂y=2dz/du−2dz/du=0
58.Letz=f(u) whereu=x
2
+y
2
; then∂z/∂x=(dz/du)(∂u/∂x)=2x dz/du,
∂z/∂y=(dz/du)(∂u/∂y)=2ydz/dusoy∂z/∂x−x∂z/∂y=2xydz/du−2xydz/du=0
59.
∂w
∂x
=
dw
du
∂u
∂x
=
dw
du
,
∂w
∂y
=
dw
du
∂u
∂y
=2
dw
du
,
∂w
∂z
=
dw
du
∂u
∂z
=3
dw
du
,so
∂w
∂x
+
∂w
∂y
+
∂w
∂z
=6
dw
du
60.∂w/∂x=(dw/dρ)(∂ρ/∂x)=(x/ρ)dw/dρ, similarly∂w/∂y=(y/ρ)dw/dρand
∂w/∂z=(z/ρ)dw/dρso (∂w/∂x)
2
+(∂w/∂y)
2
+(∂w/∂z)
2
=(dw/dρ)
2
61.z=f(u, v) whereu=x−yandv=y−x,
∂z
∂x
=
∂z
∂u
∂u
∂x
+
∂z
∂v
∂v
∂x
=
∂z
∂u
−
∂z
∂v
and
∂z
∂y
=
∂z
∂u
∂u
∂y
+
∂z
∂v
∂v
∂y
=−
∂z
∂u
+
∂z
∂v
so
∂z
∂x
+
∂z
∂y
=0
62.Letw=f(r, s, t) wherer=x−y,s=y−z,t=z−x;
∂w/∂x=(∂w/∂r)(∂r/∂x)+(∂w/∂t)(∂t/∂x)=∂w/∂r−∂w/∂t, similarly
∂w/∂y=−∂w/∂r+∂w/∂sand∂w/∂z=−∂w/∂s+∂w/∂tso∂w/∂x+∂w/∂y+∂w/∂z=0

26. January 27, 2005 11:55 L24-ch14 Sheet number 26 Page number 628 black
628 Chapter 14
63. (a)1=−rsinθ
∂θ
∂x
+ cosθ
∂r
∂x
and 0 =rcosθ
∂θ
∂x
+ sinθ
∂r
∂x
; solve for∂r/∂xand∂θ/∂x.
(b)0=−rsinθ
∂θ
∂y
+ cosθ
∂r
∂y
and 1 =rcosθ
∂θ
∂y
+ sinθ
∂r
∂y
; solve for∂r/∂yand∂θ/∂y.
(c)
∂z
∂x
=
∂z
∂r
∂r
∂x
+
∂z
∂θ
∂θ
∂x
=
∂z
∂r
cosθ−
1
r
∂z
∂θ
sinθ.
∂z
∂y
=
∂z
∂r
∂r
∂y
+
∂z
∂θ
∂θ
∂y
=
∂z
∂r
sinθ+
1
r
∂z
∂θ
cosθ.
(d)Square and add the results of Parts (a) and (b).
(e)From Part (c),
∂
2
z
∂x
2
=
∂
∂r

∂z
∂r
cosθ−
1
r
∂z
∂θ
sinθ

∂r
∂x
+
∂
∂θ

∂z
∂r
cosθ−
1
r
∂z
∂θ
sinθ

∂θ
∂x
=

∂
2
z
∂r
2
cosθ+
1
r
2
∂z
∂θ
sinθ−
1
r
∂
2
z
∂r∂θ
sinθ

cosθ
+

∂
2
z
∂θ∂r
cosθ−
∂z
∂r
sinθ−
1
r
∂
2
z
∂θ
2
sinθ−
1
r
∂z
∂θ
cosθ

−
sinθ
r

=
∂
2
z
∂r
2
cos
2
θ+
2
r
2
∂z
∂θ
sinθcosθ−
2
r
∂
2
z
∂θ∂r
sinθcosθ+
1
r
2
∂
2
z
∂θ
2
sin
2
θ+
1
r
∂z
∂r
sin
2
θ.
Similarly, from Part (c),
∂
2
z
∂y
2
=
∂
2
z
∂r
2
sin
2
θ−
2
r
2
∂z
∂θ
sinθcosθ+
2
r
∂
2
z
∂θ∂r
sinθcosθ+
1
r
2
∂
2
z
∂θ
2
cos
2
θ+
1
r
∂z
∂r
cos
2
θ.
Add to get
∂
2
z
∂x
2
+
∂
2
z
∂y
2
=
∂
2
z
∂r
2
+
1
r
2
∂
2
z
∂θ
2
+
1
r
∂z
∂r
.
64.z
x=
−2y
x
2
+y
2
,zxx=
4xy
(x
2
+y
2
)
2
,zy=
2x
x
2
+y
2
,zyy=−
4xy
(x
2
+y
2
)
2
,zxx+zyy=0;
z=tan
−1
2r
2
cosθsinθ
r
2
(cos
2
θ−sin
2
θ)
=tan
−1
tan 2θ=2θ+kπfor some fixedk;z r=0,z θθ=0
65. (a)By the chain rule,
∂u
∂r
=
∂u
∂x
cosθ+
∂u
∂y
sinθand
∂v
∂θ
=−
∂v
∂x
rsinθ+
∂v
∂y
rcosθ, use the
Cauchy-Riemann conditions
∂u
∂x
=
∂v
∂y
and
∂u
∂y
=−
∂v
∂x
in the equation for
∂u
∂r
to get
∂u
∂r
=
∂v
∂y
cosθ−
∂v
∂x
sinθand compare to
∂v
∂θ
to see that
∂u
∂r
=
1
r
∂v
∂θ
. The result
∂v
∂r
=−
1
r
∂u
∂θ
can be obtained by considering
∂v
∂r
and
∂u
∂θ
.
(b)u
x=
2x
x
2
+y
2
,vy=2
1
x
1
1+(y/x)
2
=
2x
x
2
+y
2
=ux;
u
y=
2y
x
2
+y
2
,vx=−2
y
x
2
1
1+(y/x)
2
=−
2y
x
2
+y
2
=−u y;
u=lnr
2
,v=2θ, u r=2/r, vθ=2, sou r=
1
r
v
θ,uθ=0,v r=0, sov r=−
1
r
u
θ

27. January 27, 2005 11:55 L24-ch14 Sheet number 27 Page number 629 black
Exercise Set 14.5 629
66. (a)u x=f
y
(x+ct),u xx=f
yy
(x+ct),u t=cf
y
(x+ct),u tt=c
2
f
yy
(x+ct);u tt=c
2
uxx
(b)Substitutegforfand−cforcin Part (a).
(c)Since the sum of derivatives equals the derivative of the sum, the result follows from Parts
(a) and (b).
(d)sintsinx=
1
2
(−cos(x+t) + cos(x−t))
67.∂w/∂ρ=(sinφcosθ)∂w/∂x+ (sinφsinθ)∂w/∂y+ (cosφ)∂w/∂z
∂w/∂φ=(ρcosφcosθ)∂w/∂x+(ρcosφsinθ)∂w/∂y−(ρsinφ)∂w/∂z
∂w/∂θ=−(ρsinφsinθ)∂w/∂x+(ρsinφcosθ)∂w/∂y
68. (a)
∂w
∂x
=
∂f
∂x
+
∂f
∂z
∂z
∂x
(b)
∂w
∂y
=
∂f
∂y
+
∂f
∂z
∂z
∂y
69.w
r=e
r
/(e
r
+e
s
+e
t
+e
u
),wrs=−e
r
e
s
/(e
r
+e
s
+e
t
+e
u
)
2
,
w
rst=2e
r
e
s
e
t
/(e
r
+e
s
+e
t
+e
u
)
3
,
w
rstu=−6e
r
e
s
e
t
e
u
/(e
r
+e
s
+e
t
+e
u
)
4
=−6e
r+s+t+u
/e
4w
=−6e
r+s+t+u−4w
70.∂w/∂y 1=a1∂w/∂x1+a2∂w/∂x2+a3∂w/∂x3,
∂w/∂y
2=b1∂w/∂x1+b2∂w/∂x2+b3∂w/∂x3
71. (a)dw/dt=
4
r
i=1
(∂w/∂xi)(dxi/dt)
(b)∂w/∂v
j=
4
r
i=1
(∂w/∂xi)(∂xi/∂vj) forj=1,2,3
72.Letu=x
2
1
+x
2
2
+...+x
2
n
; thenw=u
k
,∂w/∂xi=ku
k−1
(2xi)=2kx iu
k−1
,
∂
2
w/∂x
2
i
=2k(k−1)x iu
k−2
(2xi)+2ku
k−1
=4k(k−1)x
2
i
u
k−2
+2ku
k−1
fori=1,2,...,n
so
n
r
i=1
∂
2
w/∂x
2
i
=4k(k−1)u
k−2
n
r
i=1
x
2
i
+2kn u
k−1
=4k(k−1)u
k−2
u+2kn u
k−1
=2ku
k−1
[2(k−1) +n]
which is 0 ifk=0orif2(k−1) +n=0,k=1−n/2.
73.dF/dx=(∂F/∂u)(du/dx)+(∂F/∂v)(dv/dx)
=f(u)g
y
(x)−f(v)h
y
(x)=f(g(x))g
y
(x)−f(h(x))h
y
(x)
74.Represent the line segmentCthat joinsAandBby
x=x
0+(x 1−x0)t,y=y 0+(y1−y0)tfor 0≤t≤1. Let
F(t)=f(x
0+(x 1−x0)t, y0+(y1−y0)t) for 0≤t≤1; then
f(x
1,y1)−f(x 0,y0)=F(1)−F(0).
Apply the Mean Value Theorem toF(t) on the interval [0,1] to get
[F(1)−F(0)]/(1−0) =F
y
(t
∗
),F(1)−F(0) =F
y
(t
∗
) for somet
∗
in (0,1) so
f(x
1,y1)−f(x 0,y0)=F
y
(t
∗
). By the chain rule,
F
y
(t)=f x(x, y)(dx/dt)+f y(x, y)(dy/dt)=f x(x, y)(x 1−x0)+f y(x, y)(y 1−y0).
Let (x
∗
,y
∗
) be the point onCfort=t
∗
then
f(x
1,y1)−f(x 0,y0)=F
y
(t
∗
)=f x(x
∗
,y
∗
)(x1−x0)+f y(x
∗
,y
∗
)(y1−y0).

28. January 27, 2005 11:55 L24-ch14 Sheet number 28 Page number 630 black
630 Chapter 14
75.Let (a, b) be any point in the region, if (x, y) is in the region then by the result of Exercise 74
f(x, y)−f(a, b)=f
x(x
∗
,y
∗
)(x−a)+f y(x
∗
,y
∗
)(y−b) where (x
∗
,y
∗
) is on the line segment joining
(a, b) and (x, y). Iff
x(x, y)=f y(x, y) =0 throughout the region then
f(x, y)−f(a, b) =(0)(x−a) + (0)(y−b)=0,f(x, y)=f(a, b)sof(x, y) is constant on the region.
EXERCISE SET 14.6
1.∇f(x, y)=(3y/2)(1 +xy)
1/2
i+(3x/2)(1 +xy)
1/2
j,∇f(3,1) =3i+9j,
D
uf=∇f·u=12/
√
2=6
√
2
2.∇f(x, y)=2ye
2xy
i+2xe
2xy
j,∇f(4,0) =8j,D uf=∇f·u=32/5
3.∇f(x, y)=
4
2x/
J
1+x
2
+y
a0
i+
4
1/
J
1+x
2
+y
a0
j,∇f(0,0) =j,D uf=−3/
√
10
4.∇f(x, y)=−
4
(c+d)y/(x−y)
2

i+
4
(c+d)x/(x−y)
2

j,
∇f(3,4) =−4(c+d)i+3(c+d)j,D
uf=−(7/5)(c+d)
5.∇f(x, y, z)=20x
4
y
2
z
3
i+8x
5
yz
3
j+12x
5
y
2
z
2
k,∇f(2,−1,1) =320i−256j+ 384k,D uf=−320
6.∇f(x, y, z)=yze
xz
i+e
xz
j+(xye
xz
+2z)k,∇f(0,2,3) =6i+j+6k,D uf=45/7
7.∇f(x, y, z)=
2x
x
2
+2y
2
+3z
2
i+
4y
x
2
+2y
2
+3z
2
j+
6z
x
2
+2y
2
+3z
2
k,
∇f(−1,2,4) =(−2/57)i+(8/57)j+ (24/57)k,D
uf=−314/741
8.∇f(x, y, z)=yzcosxyzi+xzcosxyzj+xycosxyzk,
∇f(1/2,1/3,π)=(π
√
3/6)i+(π
√
3/4)j+(
√
3/12)k,D uf=(1−π)/12
9.∇f(x, y)=12x
2
y
2
i+8x
3
yj,∇f(2,1) =48i+64j,u=(4/5)i−(3/5)j,D uf=∇f·u=0
10.∇f(x, y)=(2x−3y)i+
J
−3x+12y
2
a
j,∇f(−2,0) =−4i+6j,u=(i+2j)/
√
5,D uf=8/
√
5
11.∇f(x, y)=
J
y
2
/x
a
i+2ylnxj,∇f(1,4) =16i,u=(−i+j)/
√
2,D uf=−8
√
2
12.∇f(x, y)=e
x
cosyi−e
x
sinyj,∇f(0,π/4) =(i−j)/
√
2,u=(5i−2j)/
√
29,D uf=7/
√
58
13.∇f(x, y)=−
4
y/
J
x
2
+y
2
a0
i+
4
x/
J
x
2
+y
2
a0
j,
∇f(−2,2) =−(i+j)/4,u=−(i+j)/
√
2,D uf=
√
2/4
14.∇f(x, y)=(e
y
−ye
x
)i+(xe
y
−e
x
)j,∇f(0,0) =i−j,u=(5i−2j)/
√
29,D uf=7/
√
29
15.∇f(x, y, z)=
J
3x
2
z−2xy
a
i−x
2
j+
J
x
3
+2z
a
k,∇f(2,−1,1) =16i−4j+10k,
u=(3i−j+2k)/
√
14,D uf=72/
√
14
16.∇f(x, y, z)=−x
J
x
2
+z
2
a
−1/2
i+j−z
J
x
2
+z
2
a
−1/2
k,∇f(−3,1,4) =(3/5)i+j−(4/5)k,
u=(2i−2j−k)/3,D
uf=0
17.∇f(x, y, z)=−
1
z+y
i−
z−x
(z+y)
2
j+
y+x
(z+y)
2
k,∇f(1,0,−3) =(1/3)i+(4/9)j+(1/9)k,
u=(−6i+3j−2k)/7,D
uf=−8/63

29. January 27, 2005 11:55 L24-ch14 Sheet number 29 Page number 631 black
Exercise Set 14.6 631
18.∇f(x, y, z)=e
x+y+3z
(i+j+3k),∇f(−2,2,−1) =e
−3
(i+j+3k),u=(20i−4j+5k)/21,
D
uf=(31/21)e
−3
19.∇f(x, y)=(y/2)(xy)
−1/2
i+(x/2)(xy)
−1/2
j,∇f(1,4) =i+(1/4)j,
u=cosθi+ sinθj=(1/2)i+
J√
3/2
a
j,D uf=1/2+
√
3/8
20.∇f(x, y)=[2y/(x+y)
2
]i−[2x/(x+y)
2
]j,∇f(−1,−2) =−(4/9)i+(2/9)j,u=j,D uf=2/9
21.∇f(x, y) =2 sec
2
(2x+y)i+ sec
2
(2x+y)j,∇f(π/6,π/3) =8i+4j,u=(i−j)/
√
2,D uf=2
√
2
22.∇f(x, y) =coshxcoshyi+ sinhxsinhyj,∇f(0,0) =i,u=−i,D
uf=−1
23.∇f(x, y)=y(x+y)
−2
i−x(x+y)
−2
j,∇f(1,0) =−j,
−→
PQ=−2i−j,u=(−2i−j)/
√
5,
D
uf=1/
√
5
24.∇f(x, y)=−e
−x
secyi+e
−x
secytanyj,
∇f(0,π/4) =
√
2(−i+j),
−→
PO=−(π/4)j,u=−j,D uf=−
√
2
25.∇f(x, y)=
ye
y
2
√
xy
i+

√
xye
y
+
xe
y
2
√
xy

j,∇f(1,1) =(e/2)(i+3j),u=−j,D
uf=−3e/2
26.∇f(x, y)=−y(x+y)
−2
i+x(x+y)
−2
j,∇f(2,3) =(−3i+2j)/25, ifD uf=0 thenuand∇fare
orthogonal, by inspection 2i+3jis orthogonal to∇f(2,3) sou=±(2i+3j)/
√
13.
27.∇f(2,1,−1) =−i+j−k.
−→
PQ=−3i+j+k,u=(−3i+j+k)/
√
11,D uf=3/
√
11
28.∇f(−1,−2,1) =13i+5j−20k,u=−k,D
uf=20
29.Solve the system (3/5)f
x(1,2)−(4/5)f y(1,2) =−5, (4/5)f x(1,2)+(3/5)f y(1,2) =10 for
(a)f
x(1,2) =5 (b)f y(1,2) =10
(c)∇f(1,2) =5i+10j,u=(−i−2j)/
√
5,D uf=−5
√
5.
30.∇f(−5,1) =−3i+2j,
−→
PQ=i+2j,u=(i+2j)/
√
5,D uf=1/
√
5
31.fincreases the most in the direction of III.
32.The contour lines are closer atP, so the function is increasing more rapidly there, hence∇fis
larger atP.
33.∇z=4i−8j 34. ∇z=−4e
−3y
sin 4xi−3e
−3y
cos 4xj
35.∇w=
x
x
2
+y
2
+z
2
i+
y
x
2
+y
2
+z
2
j+
z
x
2
+y
2
+z
2
k
36.∇w=e
−5x
sec(x
2
yz)
4J
2xyztan(x
2
yz)−5
a
i+x
2
ztan(x
2
yz)j+x
2
ytan(x
2
yz)k

37.∇f(x, y) =3(2x+y)
J
x
2
+xy
a
2
i+3x
J
x
2
+xy
a
2
j,∇f(−1,−1) =−36i−12j
38.∇f(x, y)=−x
J
x
2
+y
2
a
−3/2
i−y
J
x
2
+y
2
a
−3/2
j,∇f(3,4) =−(3/125)i−(4/125)j

30. January 27, 2005 11:55 L24-ch14 Sheet number 30 Page number 632 black
632 Chapter 14
39.∇f(x, y, z)=[y/(x+y+z)]i+[y/(x+y+z) + ln(x+y+z)]j+[y/(x+y+z)]k,
∇f(−3,4,0) =4i+4j+4k
40.∇f(x, y, z)=3y
2
ztan
2
xsec
2
xi+2yztan
3
xj+y
2
tan
3
xk,∇f(π/4,−3) =54i−6j+9k
41.f(1,2) =3,
level curve 4x−2y+3=3,
2x−y=0;
∇f(x, y)=4i−2j
∇f(1,2) =4i−2j(1, 2)
4i – 2j
x
y
42.f(−2,2) =1/2,
level curvey/x
2
=1/2,
y=x
2
/2 forx7 =0.
∇f(x, y)=−
J
2y/x
3
a
i+
J
1/x
2
a
j
∇f(−2,2) =(1/2)i+(1/4)j
1
2
1 4
i + j
(−2, 2)
x
y
43.f(−2,0) =4,
level curvex
2
+4y
2
=4,
x
2
/4+y
2
=1.
∇f(x, y)=2xi+8yj
∇f(−2,0) =−4i
–4i
1
2
x
y
44.f(2,−1) =3,
level curvex
2
−y
2
=3.
∇f(x, y)=2xi−2yj
∇f(2,−1) =4i+2j
(2, −1)
4i + 2j
x
y
45.∇f(x, y)=8xyi+4x
2
j,∇f(1,−2) =−16i+4jis normal to the level curve throughPso
u=±(−4i+j)/
√
17.
46.∇f(x, y)=(6xy−y)i+
J
3x
2
−x
a
j,∇f(2,−3) =−33i+10jis normal to the level curve through
Psou=±(−33i+10j)/
√
1189.
47.∇f(x, y)=12x
2
y
2
i+8x
3
yj,∇f(−1,1) =12i−8j,u=(3i−2j)/
√
13,∇f(−1,1)=4
√
13
48.∇f(x, y)=3i−(1/y)j,∇f(2,4) =3i−(1/4)j,u=(12i−j)/
√
145,∇f(2,4)=
√
145/4
49.∇f(x, y)=x
J
x
2
+y
2
a
−1/2
i+y
J
x
2
+y
2
a
−1/2
j,
∇f(4,−3) =(4i−3j)/5,u=(4i−3j)/5,∇f(4,−3)=1
50.∇f(x, y)=y(x+y)
−2
i−x(x+y)
−2
j,∇f(0,2) =(1/2)i,u=i,∇f(0,2)=1/2

31. January 27, 2005 11:55 L24-ch14 Sheet number 31 Page number 633 black
Exercise Set 14.6 633
51.∇f(1,1,−1) =3i−3j,u=(i−j)/
√
2,∇f(1,1,−1)=3
√
2
52.∇f(0,−3,0) =(i−3j+4k)/6,u=(i−3j+4k)/
√
26,∇f(0,−3,0)=
√
26/6
53.∇f(1,2,−2) =(−i+j)/2,u=(−i+j)/
√
2,∇f(1,2,−2)=1/
√
2
54.∇f(4,2,2) =(i−j−k)/8,u=(i−j−k)/
√
3,∇f(4,2,2)=
√
3/8
55.∇f(x, y)=−2xi−2yj,∇f(−1,−3) =2i+6j,u=−(i+3j)/
√
10,−∇f(−1,−3)=−2
√
10
56.∇f(x, y)=ye
xy
i+xe
xy
j;∇f(2,3) =e
6
(3i+2j),u=−(3i+2j)/
√
13,−∇f(2,3)=−
√
13e
6
57.∇f(x, y)=−3 sin(3x−y)i+ sin(3x−y)j,
∇f(π/6,π/4) =(−3i+j)/
√
2,u=(3i−j)/
√
10,−∇f(π/6,π/4)=−
√
5
58.∇f(x, y)=
y
(x+y)
2
h
x+y
x−y
i−
x
(x+y)
2
h
x+y
x−y
j,∇f(3,1) =(
√
2/16)(i−3j),
u=−(i−3j)/
√
10,−∇f(3,1)=−
√
5/8
59.∇f(5,7,6) =−i+11j−12k,u=(i−11j+12k)/
√
266,−∇f(5,7,6)=−
√
266
60.∇f(0,1,π/4) =2
√
2(i−k),u=−(i−k)/
√
2,−∇f(0,1,π/4)=−4
61.∇f(4,−5) =2i−j,u=(5i+2j)/
√
29,D uf=8/
√
29
62.Letu=u
1i+u 2jwhereu
2
1
+u
2
2
=1, butD uf=∇f·u=u 1−2u 2=−2sou 1=2u 2−2,
(2u
2−2)
2
+u
2
2
=1, 5u
2
2
−8u 2+3=0,u 2=1oru 2=3/5thusu 1=0oru 1=−4/5;u=jor
u=−
4
5
i+
3
5
j.
63. (a)At (1,2) the steepest ascent seems to be in the directioni+jand the slope in that direction
seems to be 0.5/(
√
2/2) =1/
√
2, so∇f≈
1
2
i+
1
2
j, which has the required direction and
magnitude.
(b)The direction of−∇f(4,4) appears to be
−i−jand its magnitude appears to be
1/0.8=5/4.
5
5
x
y
−∇f(4, 4)
64. (a)
500
P
0 ft
100
200
300
400
Depart from each contour line in a direction orthogonal to that contour line, as an approxi-
mation to the optimal path.

32. January 27, 2005 11:55 L24-ch14 Sheet number 32 Page number 634 black
634 Chapter 14
(b)
500
P
0 ft
100
200
300400
At the top there is no contour line, so head for the nearest contour line. From then on depart
from each contour line in a direction orthogonal to that contour line, as in Part (a).
65.∇z=6xi−2yj,∇z=
y
36x
2
+4y
2
=6if36x
2
+4y
2
=36; all points on the ellipse 9x
2
+y
2
=9.
66.∇z=3i+2yj,∇z=
y
9+4y
2
,so∇∇z=
4y
y
9+4y
2
j, and∇∇z
2
2
2
2
(x,y)=(5,2)
=
8
5
j
67. r=ti−t
2
j,dr/dt=i−2tj=i−4jat the point (2,−4),u=(i−4j)/
√
17;
∇z=2xi+2yj=4i−8jat (2,−4), hencedz/ds=D
uz=∇z·u=36/
√
17.
68. (a)∇T(x, y)=
y
J
1−x
2
+y
2
a (1 +x
2
+y
2
)
2
i+
x
J
1+x
2
−y
2
a
(1 +x
2
+y
2
)
2
j,∇T(1,1) =(i+j)/9,u=(2i−j)/
√
5,
D
uT=1/
J
9
√
5
a
(b) u=−(i+j)/
√
2, opposite to∇T(1,1)
69. (a)∇V(x, y)=−2e
−2x
cos 2yi−2e
−2x
sin 2yj,E=−∇V(π/4,0) =2e
−π/2
i
(b)V(x, y) decreases most rapidly in the direction of−∇V(x, y) which isE.
70.∇z=−0.04xi−0.08yj,ifx=−20 andy=5 then∇z=0.8i−0.4j.
(a) u=−ipoints due west,D
uz=−0.8, the climber will descend becausezis decreasing.
(b) u=(i+j)/
√
2 points northeast,D uz=0.2
√
2, the climber will ascend at the rate of 0.2
√
2
mpermoftravelinthexy−plane.
(c)The climber will travel a level path in a direction perpendicular to∇z=0.8i−0.4j,by
inspection±(i+2j)/
√
5 are unit vectors in these directions; (i+2j)/
√
5 makes an angle of
tan
−1
(1/2)≈27
◦
with the positivey-axis so−(i+2j)/
√
5 makes the same angle with the
negativey-axis. The compass direction should be N 27
◦
EorS27
◦
W.
71.Letube the unit vector in the direction ofa, then
D
uf(3,−2,1) =∇f(3,−2,1)·u=∇f(3,−2,1)cosθ=5 cosθ=−5, cosθ=−1,θ=πso
∇f(3,−2,1) is oppositely directed tou;∇f(3,−2,1) =−5u=−10/3i+5/3j+10/3k.
72. (a)∇T(1,1,1) =(i+j+k)/8,u=−(i+j+k)/
√
3,D uT=−
√
3/8
(b)(i+j+k)/
√
3 (c)
√
3/8
73. (a)∇r=
x
y
x
2
+y
2
i+
y
y
x
2
+y
2
j=r/r
(b)∇f(r)=
∂f(r)
∂x
i+
∂f(r)
∂y
j=f
y
(r)
∂r
∂x
i+f
y
(r)
∂r
∂y
j=f
y
(r)∇r

33. January 27, 2005 11:55 L24-ch14 Sheet number 33 Page number 635 black
Exercise Set 14.6 635
74. (a)∇
J
re
−3r
a
=
(1−3r)
r
e
−3r
r
(b)3r
2
r=
f
y
(r) r
rsof
y
(r)=3r
3
,f(r)=
3
4
r
4
+C,f(2) =12 +C=1,C=−11;f(r)=
3
4
r
4
−11
75. u
r=cosθi+ sinθj,u θ=−sinθi+ cosθj,
∇z=
∂z
∂x
i+
∂z
∂y
j=

∂z
∂r
cosθ−
1
r
∂z
∂θ
sinθ

i+

∂z
∂r
sinθ+
1
r
∂z
∂θ
cosθ

j
=
∂z
∂r
(cosθi+ sinθj)+
1
r
∂z
∂θ
(−sinθi+ cosθj)=
∂z
∂r
u
r+
1
r
∂z
∂θ
u
θ
76. (a)∇(f+g)=(f x+gx)i+(f y+gy)j=(f xi+f yj)+(g xi+g yj)=∇f+∇g
(b)∇(cf)=(cf
x)i+(cf y)j=c(f xi+f yj)=c∇f
(c)∇(fg)=(fg
x+gfx)i+(fg y+gfy)j=f(g xi+g yj)+g(f xi+f yj)=f∇g+g∇f
(d)∇(f/g)=
gf
x−fgx
g
2
i+
gf
y−fgy
g
2
j=
g(f
xi+f yj)−f(g xi+g yj)
g
2
=
g∇f−f∇g
g
2
(e)∇(f
n
)=
J
nf
n−1
fx
a
i+
J
nf
n−1
fy
a
j=nf
n−1
(fxi+f yj)=nf
n−1
∇f
77. r
y
(t)=v(t)=k(x, y)∇T=−8k(x, y)xi−2k(x, y)yj;
dx
dt
=−8kx,
dy
dt
=−2ky. Divide and solve
to gety
4
=256x; one parametrization isx(t)=e
−8t
,y(t)=4e
−2t
.
78. r
y
(t)=v(t)=k∇T=−2k(x, y)xi−4k(x, y)yj. Divide and solve to gety=
3
25
x
2
; one parametriza-
tion isx(t)=5e
−2t
,y(t)=3e
−4t
.
79.
–33
–5
5
x
y
C = 0
C = –5
C = –10
C = –15
80. 4
–4
–66
T = 80
T = 95
T = 97
(5, 3)
T = 90
81. (a)
z
x
y
(c)∇f=[2x−2x(x
2
+3y
2
)]e
−(x
2
+y
2
)
i
+[6y−2y(x
2
+3y
2
)]e
−(x
2
+y
2
)
j
(d)∇f=0ifx=y=0orx=0,y=±1orx=±1,y=0.

34. January 27, 2005 11:55 L24-ch14 Sheet number 34 Page number 636 black
636 Chapter 14
82.dz/dt=(∂z/∂x)(dx/dt)+(∂z/∂y)(dy/dt)
=(∂z/∂xi+∂z/∂yj)·(dx/dti+dy/dtj)=∇z·r
y
(t)
83.∇f(x, y)=f
x(x, y)i+f y(x, y)j,if∇f(x, y) =0 throughout the region then
f
x(x, y)=f y(x, y) =0 throughout the region, the result follows from Exercise 71, Section 14.5.
84.Letu
1andu 2be nonparallel unit vectors for which the directional derivative is zero. Letube
any other unit vector, thenu=c
1u1+c2u2for some choice of scalarsc 1andc 2,
D
uf(x, y)=∇f(x, y)·u=c 1∇f(x, y)·u 1+c2∇f(x, y)·u 2
=c1Du1f(x, y)+c 2Du2f(x, y)=0.
85.∇f(u, v, w)=
∂f
∂x
i+
∂f
∂y
j+
∂f
∂z
k
=

∂f
∂u
∂u
∂x
+
∂f
∂v
∂v
∂x
+
∂f
∂w
∂w
∂x

i+

∂f
∂u
∂u
∂y
+
∂f
∂v
∂v
∂y
+
∂f
∂w
∂w
∂y

j
+

∂f
∂u
∂u
∂z
+
∂f
∂v
∂v
∂z
+
∂f
∂w
∂w
∂z

k=
∂f
∂u
∇u+
∂f
∂v
∇v+
∂f
∂w
∇w
86. (a)The distance between (x
0+su1,y0+su2) and (x 0,y0)is|s|
y
u
2
1
+u
2
2
=|s|, so the condition
lim
s→0
E(s)
|s|
=0 is exactly the condition of Definition 14.4.1, with the local linear approximation
offgiven byL(s)=f(x
0,y0)+f x(x0,y0)su1+fy(x0,y0)su2, which in turn says that
g
y
(0) =f x(x0,y0)+f y(x0,y0).
(b)The functionE(s) of Part (a) has the same values as the functionE(x, y) whenx=x
0+
su
1,y=y 0+su2, and the distance between (x, y) and (x 0,y0)is|s|, so the limit in Part (a)
is equivalent to the limit (5) of Definition 14.4.2.
(c)Letf(x, y) be differentiable at (x
0,y0) and letu=u 1i+u 2jbe a unit vector. Then by Parts
(a) and (b) the directional derivativeD
u
d
ds
[f(x
0+su1,y0+su2)]s=0exists and is given by
f
x(x0,y0)u1+fy(x0,y0)u2.
87. (a)
d
ds
f(x
0+su1,y0+su2)ats=0 is by definition equal to lim
s→0
f(x0+su1,y0+su2)−f(x 0,y0)
s
,
and from Exercise 86(a) this value is equal tof
x(x0,y0)u1+fy(x0,y0)u2.
(b)For any number4>0anumberδ>0 exists such that whenever 0<|s|<δthen
2
2
2
2
f(x
0+su1,y0+su2)−f(x 0,y0)−f x(x0,y0)su1−fy(x0,y0)su2
s
2
2
2
2
<4.
(c)For any number4>0 there exists a numberδ>0 such that
|E(x, y)|
y
(x−x 0)
2
+(y−y 0)
2
<4
whenever 0<
y
(x−x 0)
2
+(y−y 0)
2
<δ.
(d)For any number4>0 there exists a numberδ>0 such that
2
2
2
2
f(x
0+su1,y0+su2)−f(x 0,y0)−f x(x0,y0)su1−fy(x0,y0)su2|s
2
2
2
2
<4when 0<|s|<δ.
(e)Sincefis differentiable at (x
0,y0), by Part (c) the Equation (5) of Definition 14.2.1 holds.
By Part (d), for any4>0 there existsδ>0 such that
2
2
2
2
f(x
0+su1,y0+su2)−f(x 0,y0)−f x(x0,y0)su1−fy(x0,y0)su2
s
2
2
2
2
<4when 0<|s|<δ.

35. January 27, 2005 11:55 L24-ch14 Sheet number 35 Page number 637 black
Exercise Set 14.7 637
By Part (a) it follows that the limit in Part (a) holds, and thus that
d
ds
f(x
0+su1,y0+su2)

s=0
=fx(x0,y0)u1+fy(x0,y0)u2,
which proves Equation (4) of Theorem 14.6.3.
EXERCISE SET 14.7
1.AtP,∂z/∂x=48 and∂z/∂y=−14, tangent plane 48x−14y−z=64, normal linex=1+48t,
y=−2−14t,z=12−t.
2.AtP,∂z/∂x=14 and∂z/∂y=−2, tangent plane 14x−2y−z=16, normal linex=2+14t,
y=4−2t,z=4−t.
3.AtP,∂z/∂x=1 and∂z/∂y=−1, tangent planex−y−z=0, normal linex=1+t,y=−t,
z=1−t.
4.AtP,∂z/∂x=−1 and∂z/∂y=0, tangent planex+z=−1, normal linex=−1−t,y=0,
z=−t.
5.AtP,∂z/∂x=0 and∂z/∂y=3, tangent plane 3y−z=−1, normal linex=π/6,y=3t,
z=1−t.
6.AtP,∂z/∂x=1/4 and∂z/∂y=1/6, tangent plane 3x+2y−12z=−30, normal linex=4+t/4,
y=9+t/6,z=5−t.
7.By implicit differentiation∂z/∂x=−x/z,∂z/∂y=−y/zso atP,∂z/∂x=3/4 and
∂z/∂y=0, tangent plane 3x−4z=−25, normal linex=−3+3t/4,y=0,z=4−t.
8.By implicit differentiation∂z/∂x=(xy)/(4z),∂z/∂y=x
2
/(8z)soatP,∂z/∂x=3/8 and
∂z/∂y=−9/16, tangent plane 6x−9y−16z=5, normal linex=−3+3t/8,y=1−9t/16,
z=−2−t.
9.The tangent plane is horizontal if the normal∂z/∂xi+∂z/∂yj−kis parallel tokwhich occurs
when∂z/∂x=∂z/∂y=0.
(a)∂z/∂x=3x
2
y
2
,∂z/∂y=2x
3
y;3x
2
y
2
=0 and 2x
3
y=0 for all (x, y)onthex-axis ory-axis,
andz=0 for these points, the tangent plane is horizontal at all points on thex-axis or
y-axis.
(b)∂z/∂x=2x−y−2,∂z/∂y=−x+2y+ 4; solve the system 2x−y−2=0,−x+2y+4=0,
to getx=0,y=−2.z=−4at(0,−2), the tangent plane is horizontal at (0,−2,−4).
10.∂z/∂x=6x,∂z/∂y=−2y,so6x
0i−2y 0j−kis normal to the surface at a point (x 0,y0,z0)on
the surface. 6i+4j−kis normal to the given plane. The tangent plane and the given plane are
parallel if their normals are parallel so 6x
0=6,x 0=1 and−2y 0=4,y 0=−2.z=−1at(1,−2),
the point on the surface is (1,−2,−1).
11.∂z/∂x=−6x,∂z/∂y=−4yso−6x
0i−4y 0j−kis normal to the surface at a point (x 0,y0,z0)on
the surface. This normal must be parallel to the given line and hence to the vector
−3i+8j−kwhich is parallel to the line so−6x
0=−3,x 0=1/2 and−4y 0=8,y 0=−2.
z=−3/4at(1/2,−2). The point on the surface is (1/2,−2,−3/4).
12.(3,4,5) is a point of intersection because it satisfies both equations. Both surfaces have
(3/5)i+(4/5)j−kas a normal so they have a common tangent plane at (3,4,5).

36. January 27, 2005 11:55 L24-ch14 Sheet number 36 Page number 638 black
638 Chapter 14
13. (a)2t+7=(−1+t)
2
+(2+t)
2
,t
2
=1,t=±1 so the points of intersection are (−2,1,5) and
(0,3,9).
(b)∂z/∂x=2x,∂z/∂y=2yso at (−2,1,5) the vectorn=−4i+2j−kis normal to the surface.
v=i+j+2kis parallel to the line;n·v=−4 so the cosine of the acute angle is
[n·(−v)]/(n−v)=4/
J√
21
√
6
a
=4/
J
3
√
14
a
. Similarly, at (0,3,9) the vector
n=6j−kis normal to the surface,n·v=4 so the cosine of the acute angle is
4/
J√
37
√
6
a
=4/
√
222.
14.z=xf(u) whereu=x/y,∂z/∂x=xf
y
(u)∂u/∂x+f(u)=(x/y)f
y
(u)+f(u)=uf
y
(u)+f(u),
∂z/∂y=xf
y
(u)∂u/∂y=−(x
2
/y
2
)f
y
(u)=−u
2
f
y
(u). If (x 0,y0,z0) is on the surface then, with
u
0=x0/y0,[u0f
y
(u0)+f(u 0)]i−u
2
0
f
y
(u0)j−kis normal to the surface so the tangent plane is
[u
0f
y
(u0)+f(u 0)]x−u
2
0
f
y
(u0)y−z=[u 0f
y
(u0)+f(u 0)]x0−u
2
0
f
y
(u0)y0−z0
=
7
x
0
y0
f
y
(u0)+f(u 0)

x0−
x
2
0
y
2
0
f
y
(u0)y0−z0
=x0f(u0)−z 0=0
so all tangent planes pass through the origin.
15. (a)f(x, y, z)=x
2
+y
2
+4z
2
,∇f=2xi+2yj+8zk,∇f(2,2,1) =4i+4j+8k,
n=i+j+2k,x+y+2z=6
(b) r(t)=2i+2j+k+t(i+j+2k),x(t)=2+t, y(t)=2+t, z(t)=1+2t
(c)cosθ=
n·k
n
=
√
2
√
3
,θ≈35.26
◦
16. (a)f(x, y, z)=xz−yz
3
+yz
2
,n=∇f(2,−1,1) =i+3k; tangent planex+3z=5
(b)normal linex=2+t,y=−1,z=1+3t
(c)cosθ=
n·k
n
=
3
√
10
,θ≈18.43
◦
17.Setf(x, y)=z+x−z
4
(y−1), thenf(x, y, z)=0,n=±∇f(3,5,1) =±(i−j−19k),
unit vectors±
1
√
363
(i−j−19k)
18.f(x, y, z) =sinxz−4 cosyz,∇f(π, π,1) =−i−πk; unit vectors±
1
√
1+π
2
(i+πk)
19.f(x, y, z)=x
2
+y
2
+z
2
,if(x 0,y0,z0) is on the sphere then∇f(x 0,y0,z0)=2(x 0i+y 0j+z 0k)
is normal to the sphere at (x
0,y0,z0), the normal line isx=x 0+x0t,y=y 0+y0t,z=z 0+z0t
which passes through the origin whent=−1.
20.f(x, y, z)=2x
2
+3y
2
+4z
2
,if(x 0,y0,z0) is on the ellipsoid then
∇f(x
0,y0,z0)=2(2x 0i+3y 0j+4z 0k) is normal there and hence so isn 1=2x 0i+3y 0j+4z 0k;
n
1must be parallel ton 2=i−2j+3kwhich is normal to the given plane son 1=cn 2
for some constantc. Equate corresponding components to getx 0=c/2,y 0=−2c/3, and
z
0=3c/4; substitute into the equation of the ellipsoid yields 2
J
c
2
/4
a
+3
J
4c
2
/9
a
+4
J
9c
2
/16
a
=9,
c
2
=108/49,c=±6
√
3/7. The points on the ellipsoid are
J
3
√
3/7,−4
√
3/7,9
√
3/14
a
and
J
−3
√
3/7,4
√
3/7,−9
√
3/14
a
.

37. January 27, 2005 11:55 L24-ch14 Sheet number 37 Page number 639 black
Exercise Set 14.7 639
21.f(x, y, z)=x
2
+y
2
−z
2
,if(x 0,y0,z0) is on the surface then∇f(x 0,y0,z0)=2(x 0i+y 0j−z 0k)
is normal there and hence so isn
1=x0i+y 0j−z 0k;n1must be parallel to
−→
PQ=3i+2j−2kso
n
1=c
−→
PQfor some constantc. Equate components to getx 0=3c,y 0=2candz 0=2cwhich
when substituted into the equation of the surface yields 9c
2
+4c
2
−4c
2
=1,c
2
=1/9,c=±1/3
so the points are (1,2/3,2/3) and (−1,−2/3,−2/3).
22.f
1(x, y, z)=2x
2
+3y
2
+z
2
,f2(x, y, z)=x
2
+y
2
+z
2
−6x−8y−8z+ 24,
n
1=∇f 1(1,1,2) =4i+6j+4k,n 2=∇f 2(1,1,2) =−4i−6j−4k,n 1=−n 2son1andn 2are
parallel.
23. n
1=2i−2j−k,n 2=2i−8j+4k,n 1×n2=−16i−10j−12kis tangent to the line, so
x(t)=1+8t, y(t)=−1+5t, z(t)=2+6t
24.f(x, y, z)=
y
x
2
+y
2
−z,n 1=∇f(4,3,5) =
4
5
i+
3
5
j−k,n
2=i+2j+2k,n 1×n2=(16i−13j+5k)/5
is tangent to the line,x(t)=4+16t, y(t)=3−13t, z(t)=5+5t
25.f(x, y, z)=x
2
+z
2
−25,g(x, y, z)=y
2
+z
2
−25,n 1=∇f(3,−3,4) =6i+8k,
n
2=∇g(3,−3,4) =−6j+8k,n 1×n2=48i−48j−36kis tangent to the line,
x(t)=3+4t, y(t)=−3−4t, z(t)=4−3t
26. (a)f(x, y, z)=z−8+x
2
+y
2
,g(x, y, z)=4x+2y−z,n 1=4j+k,n 2=4i+2j−k,
n
1×n2=−6i+4j−16kis tangent to the line,x(t)=3t, y(t)=2−2t, z(t)=4+8t
27.Use implicit differentiation to get∂z/∂x=−c
2
x/
J
a
2
z
a
,∂z/∂y=−c
2
y/
J
b
2
z
a
.At(x 0,y0,z0),
z
07=0, a normal to the surface is−
4
c
2
x0/
J
a
2
z0
a0
i−
4
c
2
y0/
J
b
2
z0
a0
j−kso the tangent plane is
−
c
2
x0
a
2
z0
x−
c
2
y0
b
2
z0
y−z=−
c
2
x
2
0
a
2
z0
−
c
2
y
2
0
b
2
z0
−z0,
x
0x
a
2
+
y
0y
b
2
+
z
0z
c
2
=
x
2
0
a
2
+
y
2
0
b
2
+
z
2
0
c
2
=1
28.∂z/∂x=2x/a
2
,∂z/∂y=2y/b
2
.At(x 0,y0,z0) the vector
J
2x 0/a
2
a
i+
J
2y
0/b
2
a
j−kis normal
to the surface so the tangent plane is
J
2x
0/a
2
a
x+
J
2y
0/b
2
a
y−z=2x
2
0
/a
2
+2y
2
0
/b
2
−z0, but
z
0=x
2
0
/a
2
+y
2
0
/b
2
so
J
2x 0/a
2
a
x+
J
2y
0/b
2
a
y−z=2z
0−z0=z0,2x0x/a
2
+2y 0y/b
2
=z+z 0
29. n1=fx(x0,y0)i+f y(x0,y0)j−kandn 2=gx(x0,y0)i+g y(x0,y0)j−kare normal, respectively,
toz=f(x, y) andz=g(x, y)atP;n
1andn 2are perpendicular if and only ifn 1·n2=0,
f
x(x0,y0)gx(x0,y0)+f y(x0,y0)gy(x0,y0)+1=0,
f
x(x0,y0)gx(x0,y0)+f y(x0,y0)gy(x0,y0)=−1.
30. n
1=fxi+f yj−k=
x
0
y
x
2
0
+y
2
0
i+
y
0y
x
2
0
+y
2
0
j−k; similarlyn 2=−
x
0y
x
2
0
+y
2
0
i−
y
0y
x
2
0
+y
2
0
j−k;
since a normal to the sphere isN=x
0i+y 0j+z 0k, andn 1·N=
y
x
2
0
+y
2
0
−z0=0,
n
2·N=−
y
x
2
0
+y
2
0
−z0=0, the result follows.
31.∇f=f
xi+f yj+f zkand∇g=g xi+g yj+g zkevaluated at (x 0,y0,z0) are normal, respectively,
to the surfacesf(x, y, z) =0 andg(x, y, z)=0at(x
0,y0,z0). The surfaces are orthogonal at
(x
0,y0,z0) if and only if∇f·∇g=0sof xgx+fygy+fzgz=0.

38. January 27, 2005 11:55 L24-ch14 Sheet number 38 Page number 640 black
640 Chapter 14
32.f(x, y, z)=x
2
+y
2
+z
2
−a
2
=0,g(x, y, z)=z
2
−x
2
−y
2
=0,
f
xgx+fygy+fzgz=−4x
2
−4y
2
+4z
2
=4g(x, y, z)=0
33.z=
kxy
; at a point

a, b,
k
ab

on the surface,
S
−
k
a
2
b
,−
k
ab
2
,−1
e
and hence
t
bk, ak, a
2
b
2
m
is
normal to the surface so the tangent plane isbkx+aky+a
2
b
2
z=3abk. The plane cuts thex,
y, andz-axes at the points 3a, 3b, and
3k
ab
, respectively, so the volume of the tetrahedron that is
formed isV=
1
3

3k
ab
17
1
2
(3a)(3b)

=
9
2
k, which does not depend on a andb.
EXERCISE SET 14.8
1. (a)minimum at (2,−1), no maxima (b)maximum at (0,0), no minima
(c)no maxima or minima
2. (a)maximum at (−1,5), no minima (b)no maxima or minima
(c)no maxima or minima
3.f(x, y)=(x−3)
2
+(y+2)
2
, minimum at (3,−2), no maxima
4.f(x, y)=−(x+1)
2
−2(y−1)
2
+ 4, maximum at (−1,1), no minima
5.f
x=6x+2y=0,f y=2x+2y=0; critical point (0,0);D=8>0 andf xx=6>0 at (0,0),
relative minimum.
6.f
x=3x
2
−3y=0,f y=−3x−3y
2
=0; critical points (0,0) and (−1,1);D=−9<0 at (0,0),
saddle point;D=27>0 andf
xx=−6<0at(−1,1), relative maximum.
7.f
x=2x−2xy=0,f y=4y−x
2
=0; critical points (0,0) and (±2,1);D=8>0 andf xx=2>0
at (0,0), relative minimum;D=−16<0at(±2,1), saddle points.
8.f
x=3x
2
−3=0,f y=3y
2
−3 =0; critical points (−1,±1) and (1,±1);D=−36<0at(−1,1)
and (1,−1), saddle points;D=36>0 andf
xx=6>0 at (1,1), relative minimum;D=36>0
andf
xx=−36<0at(−1,−1), relative maximum.
9.f
x=y+2=0,f y=2y+x+ 3 =0; critical point (1,−2);D=−1<0at(1,−2), saddle point.
10.f
x=2x+y−2=0,f y=x−2 =0; critical point (2,−2);D=−1<0at(2,−2), saddle point.
11.f
x=2x+y−3=0,f y=x+2y=0; critical point (2,−1);D=3>0 andf xx=2>0at(2,−1),
relative minimum.
12.f
x=y−3x
2
=0,f y=x−2y=0; critical points (0,0) and (1/6,1/12);D=−1<0 at (0,0),
saddle point;D=1>0 andf
xx=−1<0at(1/6,1/12), relative maximum.
13.f
x=2x−2/
J
x
2
y
a
=0,f y=2y−2/
J
xy
2
a
=0; critical points (−1,−1) and (1,1);D=32>0
andf
xx=6>0at(−1,−1) and (1,1), relative minima.
14.f
x=e
y
=0 is impossible, no critical points.
15.f
x=2x=0,f y=1−e
y
=0; critical point (0,0);D=−2<0at(0,0), saddle point.

39. January 27, 2005 11:55 L24-ch14 Sheet number 39 Page number 641 black
Exercise Set 14.8 641
16.f x=y−2/x
2
=0,f y=x−4/y
2
=0; critical point (1,2);D=3>0 andf xx=4>0at(1,2),
relative minimum.
17.f
x=e
x
siny=0,f y=e
x
cosy=0, siny=cosy=0 is impossible, no critical points.
18.f
x=ycosx=0,f y=sinx=0; sinx=0ifx=nπforn=0,±1,±2,...and cosx7 =0 for these
values ofxsoy=0; critical points (nπ,0) forn=0,±1,±2,...;D=−1<0at(nπ,0), saddle
points.
19.f
x=−2(x+1)e
−(x
2
+y
2
+2x)
=0,f
y=−2ye
−(x
2
+y
2
+2x)
=0; critical point (−1,0);D=4e 2
>0
andf
xx=−2e<0at(−1,0), relative maximum.
20.f
x=y−a
3
/x
2
=0,f y=x−b
3
/y
2
=0; critical point
J
a
2
/b, b
2
/a
a
;ifab >0 thenD=3>0 and
f
xx=2b
3
/a
3
>0at
J
a
2
/b, b
2
/a
a
, relative minimum; ifab <0 thenD=3>0 and
f
xx=2b
3
/a
3
<0at
J
a
2
/b, b
2
/a
a
, relative maximum.
21.
–2 –101 2
–2
–1
0
1
2
∇f=(4x−4y)i−(4x−4y
3
)j=0whenx=y, x=y
3
,sox=y=0orx=y=±1. At
(0,0),D=−16, a saddle point; at (1,1) and (−1,−1),D=32>0,f
xx=4, a relative minimum.
22.
–10 –5 0 5 10
–10
–5
0
5
10
∇f=(2y
2
−2xy+4y)i+(4xy−x
2
+4x)j=0when 2y
2
−2xy+4y=0,4xy−x
2
+4x=0, with
solutions (0,0),(0,−2),(4,0),(4/3,−2/3). At (0,0),D=−16, a saddle point. At (0,−2),
D=−16, a saddle point. At (4,0),D=−16, a saddle point. At (4/3,−2/3),D=16/3,
f
xx=4/3>0, a relative minimum.
23. (a)critical point (0,0);D=0
(b)f(0,0) =0,x
4
+y
4
≥0sof(x, y)≥f(0,0), relative minimum.
24. (a)critical point (0,0);D=0
(b)The trace of the surface on the planex=0 has equationz=−y
4
, which has a maximum
at (0,0,0); the trace of the surface on the planey=0 has equationz=x
4
, which has a
minimum at (0,0,0).

40. January 27, 2005 11:55 L24-ch14 Sheet number 40 Page number 642 black
642 Chapter 14
25. (a)f x=3e
y
−3x
2
=3
J
e
y
−x
2
a
=0,f
y=3xe
y
−3e
3y
=3e
y
J
x−e
2y
a
=0,e
y
=x
2
and
e
2y
=x,x
4
=x,x
J
x
3
−1
a
=0sox=0,1; critical point (1,0);D=27>0 andf xx=−6<0
at (1,0), relative maximum.
(b)lim
x→−∞
f(x,0) =lim
x→−∞
J
3x−x
3
−1
a
=+∞so no absolute maximum.
26.f
x=8xe
y
−8x
3
=8x(e
y
−x
2
)=0,f y=4x
2
e
y
−4e
4y
=4e
y
(x
2
−e
3y
)=0,x
2
=e
y
and
x
2
=e
3y
,e
3y
=e
y
,e
2y
=1, soy=0 andx=±1; critical points (1,0) and (−1,0).D=128>0
andf
xx=−16<0 at both points so a relative maximum occurs at each one.
27.f
x=y−1=0,f y=x−3 =0; critical point (3,1).
Alongy=0:u(x)=−x; no critical points,
alongx=0:v(y)=−3y; no critical points,
alongy=−
4
5
x+4:w(x)=−
4
5
x
2
+
27
5
x−12; critical point (27/8,13/10).
(x, y)(3,1)(0,0)(5,0)(0,4)(27/8,13/10)
f(x, y)−3 0 −5−12 −231/80
Absolute maximum value is 0, absolute minimum value is−12.
28.f
x=y−2=0,f y=x=0; critical point (0,2), but (0,2) is not in the interior ofR.
Alongy=0:u(x)=−2x; no critical points,
alongx=0:v(y) =0; no critical points,
alongy=4−x:w(x)=2x−x
2
; critical point (1,3).
(x, y)(0,0)(0,4)(4,0)(1,3)
f(x, y)0 0 −8 1
Absolute maximum value is 1, absolute minimum value is−8.
29.f
x=2x−2=0,f y=−6y+ 6 =0; critical point (1,1).
Alongy=0:u
1(x)=x
2
−2x; critical point (1,0),
alongy=2:u
2(x)=x
2
−2x; critical point (1,2)
alongx=0:v
1(y)=−3y
2
+6y; critical point (0,1),
alongx=2:v
2(y)=−3y
2
+6y; critical point (2,1)
(x, y)(1,1)(1,0)(1,2)(0,1)(2,1)(0,0)(0,2)(2,0)(2,2)
f(x, y)2 −1−1 3 3 0 0 0 0
Absolute maximum value is 3, absolute minimum value is−1.
30.f
x=e
y
−2x=0,f y=xe
y
−e
y
=e
y
(x−1) =0; critical point (1,ln 2).
Alongy=0:u
1(x)=x−x
2
−1; critical point (1/2,0),
alongy=1:u
2(x)=ex−x
2
−e; critical point (e/2,1),
alongx=0:v
1(y)=−e
y
; no critical points,
alongx=2:v
2(y)=e
y
−4; no critical points.
(x, y)(0,0)(0,1)(2,1)(2,0)(1,ln 2)(1/2,0) (e/2,1)
f(x, y)−1−ee−4−3 −1 −3/4e(e−4)/4≈−0.87
Absolute maximum value is−3/4, absolute minimum value is−3.

41. January 27, 2005 11:55 L24-ch14 Sheet number 41 Page number 643 black
Exercise Set 14.8 643
31.f x=2x−1=0,f y=4y=0; critical point (1/2,0).
Alongx
2
+y
2
=4:y
2
=4−x
2
,u(x)=8−x−x
2
for−2≤x≤2; critical points (−1/2,±
√
15/2).
(x, y)(1/2,0)
J
−1/2,
√
15/2
aJ
−1/2,−
√
15/2
a
(−2,0)(2,0)
f(x, y)−1/4 33/4 33/4 6 2
Absolute maximum value is 33/4, absolute minimum value is−1/4.
32.f
x=y
2
=0,f y=2xy=0; no critical points in the interior ofR.
Alongy=0:u(x) =0; no critical points,
alongx=0:v(y) =0; no critical points
alongx
2
+y
2
=1:w(x)=x−x
3
for 0≤x≤1; critical point
-
1/
√
3,
y
2/3
c
.
(x, y)(0,0)(0,1)(1,0)
-
1/
√
3,
y
2/3
c
f(x, y)0 0 0 2
√
3/9
Absolute maximum value is
2
9
√
3, absolute minimum value is 0.
33.MaximizeP=xyzsubject tox+y+z=48,x>0,y>0,z>0.z=48−x−yso
P=xy(48−x−y)=48xy−x
2
y−xy
2
,Px=48y−2xy−y
2
=0,P y=48x−x
2
−2xy=0. But
x7 =0 andy7 =0so48−2x−y=0 and 48−x−2y=0; critical point (16,16).P
xxPyy−P
2
xy
>0
andP
xx<0at(16,16), relative maximum.z=16 whenx=y=16, the product is maximum for
the numbers 16,16,16.
34.MinimizeS=x
2
+y
2
+z
2
subject tox+y+z=27,x>0,y>0,z>0.z=27−x−yso
S=x
2
+y
2
+ (27−x−y)
2
,Sx=4x+2y−54 =0,S y=2x+4y−54 =0; critical point (9,9);
S
xxSyy−S
2
xy
=12>0 andS xx=4>0at(9,9), relative minimum.z=9 whenx=y=9, the
sum of the squares is minimum for the numbers 9,9,9.
35.Maximizew=xy
2
z
2
subject tox+y+z=5,x>0,y>0,z>0.x=5−y−zso
w=(5−y−z)y
2
z
2
=5y
2
z
2
−y
3
z
2
−y
2
z
3
,wy=10yz
2
−3y
2
z
2
−2yz
3
=yz
2
(10−3y−2z)=0,
w
z=10y
2
z−2y
3
z−3y
2
z
2
=y
2
z(10−2y−3z) =0, 10−3y−2z=0 and 10−2y−3z=0; critical
point wheny=z=2;w
yywzz−w
2
yz
=320>0 andw yy=−24<0 wheny=z=2, relative
maximum.x=1 wheny=z=2,xy
2
z
2
is maximum at (1,2,2).
36.Minimizew=D
2
=x
2
+y
2
+z
2
subject tox
2
−yz=5.x
2
=5+yzsow=5+yz+y
2
+z
2
,
w
y=z+2y=0,w z=y+2z=0; critical point wheny=z=0;w yywzz−w
2
yz
=3>0 and
w
yy=2>0 wheny=z=0, relative minimum.x
2
=5,x=±
√
5 wheny=z=0. The points
J
±
√
5,0,0
a
are closest to the origin.
37.The diagonal of the box must equal the diameter of the sphere, thus we maximizeV=xyzor, for
convenience,w=V
2
=x
2
y
2
z
2
subject tox
2
+y
2
+z
2
=4a
2
,x>0,y>0,z>0;z
2
=4a
2
−x
2
−y
2
hencew=4a
2
x
2
y
2
−x
4
y
2
−x
2
y
4
,wx=2xy
2
(4a
2
−2x
2
−y
2
)=0,w y=2x
2
y
J
4a
2
−x
2
−2y
2
a
=0,
4a
2
−2x
2
−y
2
=0 and 4a
2
−x
2
−2y
2
=0; critical point
J
2a/
√
3,2a/
√
3
a
;
w
xxwyy−w
2
xy
=
4096
27
a
8
>0 andw xx=−
128
9
a
4
<0at
J
2a/
√
3,2a/
√
3
a
, relative maximum.
z=2a/
√
3 whenx=y=2a/
√
3, the dimensions of the box of maximum volume are
2a/
√
3,2a/
√
3,2a/
√
3.
38.MaximizeV=xyzsubject tox+y+z=1,x>0,y>0,z>0.z=1−x−ysoV=xy−x
2
y−xy
2
,
V
x=y(1−2x−y)=0,V y=x(1−x−2y) =0, 1−2x−y=0 and 1−x−2y=0; critical point
(1/3,1/3);V
xxVyy−V
2
xy
=1/3>0 andV xx=−2/3<0at(1/3,1/3), relative maximum. The
maximum volume isV=(1/3)(1/3)(1/3) =1/27.

42. January 27, 2005 11:55 L24-ch14 Sheet number 42 Page number 644 black
644 Chapter 14
39.Letx,y, andzbe, respectively, the length, width, and height of the box. Minimize
C=10(2xy) + 5(2xz+2yz) =10(2xy+xz+yz) subject toxyz=16.z=16/(xy)
soC=20(xy+8/y+8/x),C
x=20(y−8/x
2
)=0,C y=20(x−8/y
2
)=0;
critical point (2,2);C
xxCyy−C
2
xy
=1200>0
andC
xx=40>0 at (2,2), relative minimum.z=4 whenx=y=2. The cost of materials is
minimum if the length and width are 2 ft and the height is 4 ft.
40.Maximize the profitP=500(y−x)(x−40) + [45,000 + 500(x−2y)](y−60)
=500(−x
2
−2y
2
+2xy−20x+ 170y−5400).
P
x=1000(−x+y−10) =0,P y=1000(−2y+x+ 85) =0; critical point (65,75);
P
xxPyy−P
2
xy
=1,000,000>0 andP xx=−1000<0 at (65,75), relative maximum. The profit
will be maximum whenx=65 andy=75.
41. (a)x=0:f(0,y)=−3y
2
, minimum−3, maximum 0;
x=1,f(1,y)=4−3y
2
+2y,
∂f
∂y
(1,y)=−6y+2=0aty=1/3, minimum 3,
maximum 13/3;
y=0,f(x,0) =4x
2
, minimum 0, maximum 4;
y=1,f(x,1) =4x
2
+2x−3,
∂f
∂x
(x,1) =8x+27 =0 for 0<x<1, minimum−3, maximum 3
(b)f(x, x)=3x
2
, minimum 0, maximum 3;f(x,1−x)=−x
2
+8x−3,
d
dx
f(x,1−x)=−2x+87 =0
for 0<x<1, maximum 4, minimum−3
(c)f
x(x, y)=8x+2y=0,f y(x, y)=−6y+2x=0, solution is (0,0), which is not an interior
point of the square, so check the sides: minimum−3, maximum 13/3.
42.MaximizeA=absinαsubject to 2a+2b=3,a>0,b>0, 0<α<π .b=(3−2a)/2so
A=(1/2)(3a−2a
2
) sinα,A a=(1/2)(3−4a) sinα,A α=(a/2)(3−2a) cosα; sinα7 =0 so from
A
a=0wegeta=3/4 and then fromA α=0 we get cosα=0,α=π/2.A aaAαα−A
2
aα
=3
2
/8>0
andA
aa=−2<0 whena=3/4 andα=π/2, the area is maximum.
43.MinimizeS=xy+2xz+2yzsubject toxyz=V,x>0,y>0,z>0 wherex,y, andzare,
respectively, the length, width, and height of the box.z=V/(xy)soS=xy+2V/y+2V/x,
S
x=y−2V/x
2
=0,S y=x−2V/y
2
=0; critical point (
3
√
2V,
3
√
2V);S xxSyy−S
2
xy
=3>0 and
S
xx=2>0 at this point so there is a relative minimum there. The length and width are each
3
√
2V, the height isz=
3
√
2V/2.
44.The altitude of the trapezoid isxsinφand the lengths of the lower and upper bases are, respectively,
27−2xand 27−2x+2xcosφso we want to maximize
A=(1/2)(xsinφ)[(27−2x) + (27−2x+2xcosφ)] =27xsinφ−2x
2
sinφ+x
2
sinφcosφ.
A
x=sinφ(27−4x+2xcosφ),
A
φ=x(27 cosφ−2xcosφ−xsin
2
φ+xcos
2
φ)=x(27 cosφ−2xcosφ+2xcos
2
φ−x).
sinφ7 =0 so fromA
x=0 we get cosφ=(4x−27)/(2x),x7 =0 so fromA φ=0weget
(27−2x+2xcosφ) cosφ−x=0 which, for cosφ=(4x−27)/(2x), yields 4x−27−x=0,
x=9. Ifx=9 then cosφ=1/2,φ=π/3. The critical point occurs whenx=9 andφ=π/3;
A
xxAφφ−A
2
xφ
=729/2>0 andA xx=−3
√
3/2<0 there, the area is maximum whenx=9 and
φ=π/3.

43. January 27, 2005 11:55 L24-ch14 Sheet number 43 Page number 645 black
Exercise Set 14.8 645
45. (a)
∂g
∂m
=
n
r
i=1
2(mx i+b−y i)xi=2

m
n
r
i=1
x
2
i
+b
n
r
i=1
xi−
n
r
i=1
xiyi
L
=0if

n
r
i=1
x
2
i
L
m+

n
r
i=1
xi
L
b=
n
r
i=1
xiyi,
∂g
∂b
=
n
r
i=1
2(mx i+b−y i)=2

m
n
r
i=1
xi+bn−
n
r
i=1
yi
L
=0if

n
r
i=1
xi
L
m+nb=
n
r
i=1
yi
(b)
n
r
i=1
(xi−¯x)
2
=
n
r
i=1
J
x
2
i
−2¯xx i+¯x
2
a
=
n
r
i=1
x
2
i
−2¯x
n
r
i=1
xi+n¯x
2
=
n
r
i=1
x
2
i
−
2
n

n
r
i=1
xi
L
2
+
1
n

n
r
i=1
xi
L
2
=
n
r
i=1
x
2
i
−
1
n

n
r
i=1
xi
L
2
≥0son
n
r
i=1
x
2
i
−

n
r
i=1
xi
L
2
≥0
This is an equality if and only if
n
r
i=1
(xi−¯x)
2
=0, which meansx i=¯xfor eachi.
(c)The system of equationsAm+Bb=C, Dm+Eb=Fin the unknownsmandbhas a unique
solution providedAE7 =BD, and if so the solution ism=
CE−BF
AE−BD
,b=
F−Dm
E
, which
after the appropriate substitution yields the desired result.
46. (a)g
mm=2
n
r
i=1
x
2
i
,gbb=2n, gmb=2
n
r
i=1
xi,
D=g
mmgbb−g
2
mb
=4

n
n
r
i=1
x
2
i
−

n
r
i=1
xi
L
2

>0 andg
mm>0
(b)g(m, b) is of the second-degree inmandbso the graph ofz=g(m, b) is a quadric surface.
(c)The functionz=g(m, b), as a function ofmandb, has only one critical point, found in
Exercise 47, and tends to +∞as either|m|or|b|tends to infinity, sinceg
mmandg bbare
both positive. Thus the only critical point must be a minimum.
47.n=3,
3
r
i=1
xi=3,
3
r
i=1
yi=7,
3
r
i=1
xiyi=13,
3
r
i=1
x
2
i
=11,y=
3
4
x+
19
12
48.n=4,
4
r
i=1
xi=7,
4
r
i=1
yi=4,
4
r
i=1
x
2
i
=21,
4
r
i=1
xiyi=−2,y=−
36
35
x+
14
5
49.
4
r
i=1
xi=10,
4
r
i=1
yi=8.2,
4
r
i=1
x
2
i
=30,
4
r
i=1
xiyi=23,n=4;m=0.5,b=0.8,y=0.5x+0.8.

44. January 27, 2005 11:55 L24-ch14 Sheet number 44 Page number 646 black
646 Chapter 14
50.
5
r
i=1
xi=15,
5
r
i=1
yi=15.1,
5
r
i=1
x
2
i
=55,
5
r
i=1
xiyi=39.8,n=5;m=−0.55,b=4.67,y=4.67−0.55x
51. (a)y=
8843
140
+
57
200
t≈63.1643 + 0.285t
(b)
600(1930)
60
80 (c)y=
2909
35
≈83.1143
52. (a)y≈119.84−1.13x
(b) 90
60
35 50
(c)about 52 units
53. (a)P=
2798
21
+
171
350
T≈133.2381 + 0.4886T
(b)
1200
130
190 (c)T≈−
139,900
513
≈−272.7096
◦
C
54. (a)for example,z=y
(b)For example, on 0≤x≤1,0≤y≤1 letz=
3
yif 0<x<1,0<y<1
1/2ifx=0,1ory=0,1
55.f(x
0,y0)≥f(x, y) for all (x, y) inside a circle centered at (x 0,y0) by virtue of Definition 14.8.1.
Ifris the radius of the circle, then in particularf(x
0,y0)≥f(x, y 0) for allxsatisfying
|x−x
0|<rsof(x, y 0) has a relative maximum atx 0. The proof is similar for the function
f(x
0,y).

45. January 27, 2005 11:55 L24-ch14 Sheet number 45 Page number 647 black
Exercise Set 14.9 647
EXERCISE SET 14.9
1. (a)xy=4 is tangent to the line, so the maximum value offis 4.
(b)xy=2 intersects the curve and so gives a smaller value off.
(c)Maximizef(x, y)=xysubject to the constraintg(x, y)=x+y−4=0,∇f=λ∇g,
yi+xj=λ(i+j), so solve the equationsy=λ, x=λwith solutionx=y=λ, butx+y=4,
sox=y=2, and the maximum value offisf=xy=4.
2. (a)x
2
+y
2
=25 is tangent to the line at (3,4), so the minimum value offis 25.
(b)A larger value offyields a circle of a larger radius, and hence intersects the line.
(c)Minimizef(x, y)=x
2
+y
2
subject to the constraintg(x, y)=3x+4y−25 =0,∇f=λ∇g,
2xi+2yj=3λi+4λj, so solve 2x=3λ,2y=4λand 3x+4y−25 =0; solution isx=3,y=4,
minimum =25.
3. (a)
31.5–31.5
–27
15 (b)one extremum at (0,5) and one at
approximately (±5,0), so minimum
value−5, maximum value≈25
(c)Find the minimum and maximum values off(x, y)=x
2
−ysubject to the constraint
g(x, y)=x
2
+y
2
−25 =0,∇f=λ∇g,2xi−j=2λxi+2λyj, so solve
2x=2λx,−1=2λy, x
2
+y
2
−25 =0. Ifx=0 theny=±5,f=∓5, and ifx7 =0 then
λ=1,y=−1/2,x
2
=25−1/4=99/4,f=99/4+1/2 =101/4, so the maximum value off
is 101/4at(±3
√
11/2,−1/2) and the minimum value offis−5at(0,5).
4. (a)
0123456
0
1
2
3
4
5
6
(b)f≈15
(d)Setf(x, y)=x
3
+y
3
−3xy, g(x, y)=(x−4)
2
+(y−4)
2
−4; minimizefsubject to the
constraintg=0:∇f=λg,(3x
2
−3y)i+(3y
2
−3x)j=2λ(x−4)i+2λ(y−4)j, so solve (use
a CAS) 3x
2
−3y=2λ(x−4),3y
2
−3x=2λ(y−4) and (x−4)
2
+(y−4)
2
−4 =0; minimum
valuef=14.52 at (2.5858,2.5858)
5.y=8xλ,x=16yλ;y/(8x)=x/(16y),x
2
=2y
2
so 4
J
2y
2
a
+8y
2
=16,y
2
=1,y=±1. Test
J
±
√
2,−1
a
and (±
√
2,1).f
J
−
√
2,−1
a
=f
J√
2,1
a
=
√
2,f
J
−
√
2,1
a
=f
J√
2,−1
a
=−
√
2.
Maximum
√
2at
J
−
√
2,−1
a
and
J√
2,1
a
, minimum−
√
2at
J
−
√
2,1
a
and
J√
2,−1
a
.

46. January 27, 2005 11:55 L24-ch14 Sheet number 46 Page number 648 black
648 Chapter 14
6.2x=2xλ,−2y=2yλ, x
2
+y
2
=25. Ifx7 =0 thenλ=1 andy=0sox
2
+0
2
=25,x=±5.
Ifx=0 then 0
2
+y
2
=25,y=±5. Test (±5,0) and (0,±5):f(±5,0) =25,f(0,±5) =−25,
maximum 25 at (±5,0), minimum−25 at (0,±5).
7.12x
2
=4xλ,2y=2yλ.Ify7 =0 thenλ=1 and 12x
2
=4x,12x(x−1/3) =0,x=0orx=1/3
so from 2x
2
+y
2
=1 we find thaty=±1 whenx=0,y=±
√
7/3 whenx=1/3. Ify=0
then 2x
2
+ (0)
2
=1,x=±1/
√
2. Test (0,±1),
J
1/3,±
√
7/3
a
, and
J
±1/
√
2,0
a
.f(0,±1) =1,
f
J
1/3,±
√
7/3
a
=25/27,f
J
1/
√
2,0
a
=
√
2,f
J
−1/
√
2,0
a
=−
√
2. Maximum
√
2at
J
1/
√
2,0
a
,
minimum−
√
2at
J
−1/
√
2,0
a
.
8.1=2xλ,−3=6yλ;1/(2x)=−1/(2y),y=−xsox
2
+3(−x)
2
=16,x=±2. Test (−2,2) and
(2,−2).f(−2,2) =−9,f(2,−2) =7. Maximum 7 at (2,−2), minimum−9at(−2,2).
9.2=2xλ,1=2yλ,−2=2zλ;1/x=1/(2y)=−1/zthusx=2y,z=−2yso
(2y)
2
+y
2
+(−2y)
2
=4,y
2
=4/9,y=±2/3. Test (−4/3,−2/3,4/3) and (4/3,2/3,−4/3).
f(−4/3,−2/3,4/3) =−6,f(4/3,2/3,−4/3) =6. Maximum 6 at (4/3,2/3,−4/3), minimum−6
at (−4/3,−2/3,4/3).
10.3=4xλ,6=8yλ,2=2zλ;3/(4x)=3/(4y)=1/zthusy=x,z=4x/3, so
2x
2
+4x
2
+(4x/3)
2
=70,x
2
=9,x=±3. Test (−3,−3,−4) and (3,3,4).
f(−3,−3,−4) =−35,f(3,3,4) =35. Maximum 35 at (3, 3, 4), minimum−35 at (−3,−3,−4).
11.yz=2xλ,xz=2yλ,xy=2zλ;yz/(2x)=xz/(2y)=xy/(2z)thusy
2
=x
2
,z
2
=x
2
so
x
2
+x
2
+x
2
=1,x=±1/
√
3. Test the eight possibilities withx=±1/
√
3,y=±1/
√
3, and
z=±1/
√
3 to find the maximum is 1/
J
3
√
3
a
at
J
1/
√
3,1/
√
3,1/
√
3
a
,
J
1/
√
3,−1/
√
3,−1/
√
3
a
,
J
−1/
√
3,1/
√
3,−1/
√
3
a
, and
J
−1/
√
3,−1/
√
3,1/
√
3
a
; the minimum is−1/
J
3
√
3
a
at
J
1/
√
3,1/
√
3,−1/
√
3
a
,
J
1/
√
3,−1/
√
3,1/
√
3
a
,
J
−1/
√
3,1/
√
3,1/
√
3
a
, and
J
−1/
√
3,−1/
√
3,−1/
√
3
a
.
12.4x
3
=2λx,4y
3
=2λy,4z
3
=2λz;ifx(oryorz)7 =0 thenλ=2x
2
(or 2y
2
or 2z
2
).
Assume for the moment that|x|≤|y|≤|z|. Then:
Case I:x, y, z7 =0soλ=2x
2
=2y
2
=2z
2
,x=±y=±z,3x
2
=1,x=±1/
√
3,
f(x, y, z)=3/9=1/3
Case II:x=0,y,z7 =0; theny=±z,2y
2
=1,y=±z=±1/
√
2,f(x, y, z)=2/4=1/2
Case III:x=y=0,z7 =0; thenz
2
=1,z=±1,f(x, y, z)=1
Thusfhas a maximum value of 1 at (0,0,±1),(0,±1,0), and (±1,0,0) and a minimum value of
1/3 at (±1/
√
3,±1/
√
3,±1/
√
3).
13.f(x, y)=x
2
+y
2
;2x=2λ,2y=−4λ;y=−2xso 2x−4(−2x)=3,x=3/10. The point is
(3/10,−3/5).
14.f(x, y)=(x−4)
2
+(y−2)
2
,g(x, y)=y−2x−3; 2(x−4) =−2λ,2(y−2) =λ;x−4=−2(y−2),
x=−2y+8soy=2(−2y+8)+3,y=19/5. The point is (2/5,19/5).
15.f(x, y, z)=x
2
+y
2
+z
2
;2x=λ,2y=2λ,2z=λ;y=2x,z=xsox+ 2(2x)+x=1,x=1/6.
The point is (1/6,1/3,1/6).
16.f(x, y, z)=(x−1)
2
+(y+1)
2
+(z−1)
2
;2(x−1) =4λ,2(y+1)=3λ,2(z−1) =λ;x=4z−3,
y=3z−4so4(4z−3) + 3(3z−4) +z=2,z=1. The point is (1,−1,1).
17.f(x, y)=(x−1)
2
+(y−2)
2
;2(x−1) =2xλ,2(y−2) =2yλ;(x−1)/x=(y−2)/y,y=2x
sox
2
+(2x)
2
=45,x=±3.f(−3,−6) =80 andf(3,6) =20 so (3,6) is closest and (−3,−6) is
farthest.

47. January 27, 2005 11:55 L24-ch14 Sheet number 47 Page number 649 black
Exercise Set 14.9 649
18.f(x, y, z)=x
2
+y
2
+z
2
;2x=yλ,2y=xλ,2z=−2zλ.Ifz7 =0 thenλ=−1so2x=−yand
2y=−x,x=y=0; substitute intoxy−z
2
=1togetz
2
=−1 which has no real solution. If
z=0 thenxy−(0)
2
=1,y=1/x, and also (from 2x=yλand 2y=xλ), 2x/y=2y/x,y
2
=x
2
so (1/x)
2
=x
2
,x
4
=1,x=±1. Test (1,1,0) and (−1,−1,0) to see that they are both closest to
the origin.
19.f(x, y, z)=x+y+z,x
2
+y
2
+z
2
=25 wherex,y, andzare the components of the vector;
1=2xλ,1=2yλ,1=2zλ;1/(2x)=1/(2y)=1/(2z);y=x,z=xsox
2
+x
2
+x
2
=25,
x=±5/
√
3.f
J
−5/
√
3,−5/
√
3,−5/
√
3
a
=−5
√
3 andf
J
5/
√
3,5/
√
3,5/
√
3
a
=5
√
3 so the vector
is 5(i+j+k)/
√
3.
20.x
2
+y
2
=25 is the constraint; solve 8x−4y=2xλ,−4x+2y=2yλ.Ifx=0 theny=0 and
conversely; butx
2
+y
2
=25, soxandyare nonzero. Thusλ=(4x−2y)/x=(−2x+y)/y,so
0=2x
2
+3xy−2y
2
=(2x−y)(x+2y), hencey=2xorx=−2y.Ify=2xthenx
2
+(2x)
2
=25,
x=±
√
5. Ifx=−2ythen
J
−2y
2
a
+y
2
=25,y=±
√5.T
J
−
√
5,−2
√
5
a
=T
J√
5,2
√
5
a
=0 and
T
J
2
√
5,−
√
5
a
=T
J
−2
√
5,
√
5
a
=125. The highest temperature is 125 and the lowest is 0.
21.Minimizef=x
2
+y
2
+z
2
subject tog(x, y, z)=x+y+z−27 =0.∇f=λ∇g,
2xi+2yj+2zk=λi+λj+λk, solutionx=y=z=9, minimum value 243
22.Maximizef(x, y, z)=xy
2
z
2
subject tog(x, y, z)=x+y+z−5=0,∇f=λ∇g=λ(i+j+k),
λ=y
2
z
2
=2xyz
2
=2xy
2
z, λ=0 is impossible, hencex, y, z7 =0, andz=y=2x,5x−5=0,
x=1,y=z=2, maximum value 16 at (1,2,2)
23.Minimizef=x
2
+y
2
+z
2
subject tox
2
−yz=5,∇f=λ∇g,2x=2xλ,2y=−zλ,2z=−yλ.
Ifλ7 =±2, theny=z=0,x=±
√
5,f=5; ifλ=±2 thenx=0, and since−yz=5,
y=−z=±
√
5,f=10, thus the minimum value is 5 at (±
√
5,0,0).
24.The diagonal of the box must equal the diameter of the sphere so maximizeV=xyzor, for
convenience, maximizef=V
2
=x
2
y
2
z
2
subject tog(x, y, z)=x
2
+y
2
+z
2
−4a
2
=0,∇f=λ∇g,
2xy
2
z
2
=2λx,2x
2
yz
2
=2λy,2x
2
y
2
z=2λz. SinceV7 =0 it follows thatx, y, z7 =0, hence
x=±y=±z,3x
2
=4a
2
,x=±2a/
√
3, maximum volume 8a
3
/(3
√
3).
25.Letx,y, andzbe, respectively, the length, width, and height of the box. Minimize
f(x, y, z) =10(2xy) + 5(2xz+2yz) =10(2xy+xz+yz) subject tog(x, y, z)=xyz−16 =0,
∇f=λ∇g,20y+10z=λyz,20x+10z=λxz,10x+10y=λxy. SinceV=xyz=16,x,y,z7 =0,
thusλz=20 + 10(z/y) =20 + 10(z/x), sox=y. From this and 10x+10y=λxyit follows that
20 =λx,so10z=20x, z=2x=2y, V=2x
3
=16 and thusx=y=2ft,z=4ft,f(2,2,4) =240
cents.
26. (a)Ifg(x, y)=x=0 then 8x+2y=λ,−6y+2x=0; butx=0, soy=λ=0,
f(0,0) =0 maximum,f(0,1) =−3, minimum.
Ifg(x, y)=x−1 =0 then 8x+2y=λ,−6y+2x=0; butx=1, soy=1/3,
f(1,1/3) =13/3 maximum,f(1,0) =4,f(1,1) =3 minimum.
Ifg(x, y)=y=0 then 8x+2y=0,−6y+2x=λ; buty=0sox=λ=0,
f(0,0) =0 minimum,f(1,0) =4, maximum.
Ifg(x, y)=y−1 =0 then 8x+2y=0,−6y+2x=λ; buty=1sox=−1/4, no solution,
f(0,1) =−3 minimum,f(1,1) =3 maximum.

48. January 27, 2005 11:55 L24-ch14 Sheet number 48 Page number 650 black
650 Chapter 14
(b)Ifg(x, y)=x−y=0 then 8x+2y=λ,−6y+2x=−λ; butx=yso solution
x=y=λ=0,f(0,0) =0 minimum,f(1,1) =3 maximum. Ifg(x, y)=1−x−y=0 then
8x+2y=−1,−6y+2x=−1; butx+y=1 so solution isx=−2/13,y=3/2 which is not
on diagonal,f(0,1) =−3 minimum,f(1,0) =4 maximum.
27.MaximizeA(a, b, α)=absinαsubject tog(a, b, α)=2a+2b−3=0,∇
(a,b,α) f=λ∇
(a,b,α) g,
bsinα=2λ, asinα=2λ, abcosα=0 with solutiona=b(=3/4),α=π/2 maximum value if
parallelogram is a square.
28.Minimizef(x, y, z)=xy+2xz+2yzsubject tog(x, y, z)=xyz−V=0,∇f=λ∇g,
y+2z=λyz, x+2z=λxz,2x+2y=λxy;λ=0 leads tox=y=z=0, impossible, so solve for
λ=1/z+2/x=1/z+2/y=2/y+2/x,sox=y=2z,x
3
=2V, minimum value 3(2V)
2/3
29. (a)Maximizef(α, β, γ) =cosαcosβcosγsubject tog(α, β, γ)=α+β+γ−π=0,
∇f=λ∇g,−sinαcosβcosγ=λ,−cosαsinβcosγ=λ,−cosαcosβsinγ=λwith solution
α=β=γ=π/3, maximum value 1/8
(b)for example,f(α, β) =cosαcosβcos(π−α−β)
f
J a
30.Find maxima and minimaz=x
2
+4y
2
subject to the constraintg(x, y)=x
2
+y
2
−1=0,
∇z=λ∇g,2xi+8yj=2λxi+2λyj, solve 2x=2λx,8y=2λy.Ify7 =0 thenλ=4,x=0,y
2
=1
andz=x
2
+4y
2
=4. Ify=0 thenx
2
=1 andz=1, so the maximum height is obtained for
(x, y)=(0,±1),z=4 and the minimum height isz=1at(±1,0).
REVIEW EXERCISES, CHAPTER 14
1. (a)f(lny, e
x
)=e
lny
lne
x
=xy (b)e
r+s
ln(rs)
2. (a)
x
y
y =
1
x
(b)
x
y
–11

49. January 27, 2005 11:55 L24-ch14 Sheet number 49 Page number 651 black
Review Exercises, Chapter 14 651
3.z=
y
x
2
+y
2
=cimpliesx
2
+y
2
=c
2
, which is the equation of a circle;x
2
+y
2
=cis also the
equation of a circle (forc>0).
–3 3
–3
3
x
y
z = x
2
+ y
2
–33
–3
3
x
y
z = √x
2
+ y
2
4. (b)f(x, y, z)=z−x
2
−y
2
5.x
4
−x+y−x
3
y=(x
3
−1)(x−y), limit =−1, not defined on the liney=xso not continuous at
(0,0)
6.
x
4
−y
4
x
2
+y
2
=x
2
−y
2
, limit =lim
(x,y)→(0,0)
(x
2
−y
2
) =0, continuous
7. (a)They approximate the profit per unit of any additional sales of the standard or high-resolution
monitors, respectively.
(b)The rates of change with respect to the two directionsxandy, and with respect to time.
9. (a)P=
10T
V
,
dP
dt
=
∂P
∂T
dT
dt
+
∂P
∂V
dV
dt
=
10
V
·3−
10T
V
2
·0=
30
V
=
30
2.5
=12 N/(m
2
min) =12 Pa/min
(b)
dP
dt
=
∂P
∂T
dT
dt
+
∂P
∂V
dV
dt
=
10
V
·0−
10T
V
2
·(−3) =
30T
V
2
=
30·50
(2.5)
2
=240 Pa/min
10. (a)z=1−y
2
, slope =
∂z
∂y
=−2y=4 (b)z=1−4x
2
,
∂z
∂x
=−8x=−8
11.w
x=2xsec
2
(x
2
+y
2
)+
√
y, wxy=8xysec
2
(x
2
+y
2
) tan(x
2
+y
2
)+
1
2
y
−1/2
,
w
y=2ysec
2
(x
2
+y
2
)+
1
2
xy
−1/2
,wyx=8xysec
2
(x
2
+y
2
) tan(x
2
+y
2
)+
1
2
y
−1/2
12.∂w/∂x=
1
x−y
−sin(x+y),∂
2
w/∂x
2
=−
1
(x−y)
2
−cos(x+y),
∂w/∂y=−
1
x−y
−sin(x+y),∂
2
w/∂y
2
=−
1
(x−y)
2
−cos(x+y)=∂
2
w/∂x
2

50. January 27, 2005 11:55 L24-ch14 Sheet number 50 Page number 652 black
652 Chapter 14
13.F x=−6xz, F xx=−6z,F y=−6yz, F yy=−6z,F z=6z
2
−3x
2
−3y
2
,
F
zz=12z,F xx+Fyy+Fzz=−6z−6z+12z=0
14.f
x=yz+2x, f xy=z,fxyz=1,f xyzx=0;f z=xy−(1/z),f zx=y, fzxx=0,f zxxy=0
16.∆w=(1.1)
2
(−0.1)−2(1.1)(−0.1)+(−0.1)
2
(1.1)−0=0.11,
dw=(2xy−2y+y
2
)dx+(x
2
−2x+2yx)dy=−(−0.1) =0.1
17.dV=
2
3
xhdx+
1
3
x
2
dh=
2
3
2(−0.1) +
1
3
(0.2) =−0.06667 m
3
;∆V=−0.07267 m
3
18.f x

1
3
,π

=πcos
π
3
=
π
2
,f
y

1
3
,π

=
1
3
cos
π
3
=
1
6
,so
L(x, y)=
√
3
2
+
π
2

x−
1
3

+
1
6
(y−π)
19.
dz
dt
=
∂z
∂x
dx
dt
+
∂z
∂y
dy
dt
, so whent=0, 4

−
1
2

+2
dy
dt
=2. Solve to obtain
dy
dt
2
2
2
2
t=0
=2
20. (a)
dy
dx
=−
6x−5y+ysec
2
xy
−5x+xsec
2
xy
. (b)
dy
dx
=−
lny+ cos(x−y)
x/y−cos(x−y)
21.
dy
dx
=−
f
x
fy
,
d
2
y
dx
2
=−
f
y(d/dx)f x−fx(d/dx)f y
f
2
y
=−
f
y(fxx+fxy(dy/dx))−f x(fxy+fyy(dy/dx))
f
2
y
=−
f
y(fxx+fxy(−fx/fy))−f x(fxy+fyy(−fx/fy))
f
2
y
=
−f
2
y
fxx+2f xfyfxy−f
2
x
fyy
f
3
y
22. (a)
d
dt

∂z
∂x

=
∂
∂x

∂z
∂x

dx
dt
+
∂
∂y

∂z
∂x

dy
dt
=
∂
2
z
∂x
2
dx
dt
+
∂
2
z
∂y∂x
dy
dt
by the Chain Rule, and
d
dt

∂z
∂y

=
∂
∂x

∂z
∂y

dx
dt
+
∂
∂y

∂z
∂y

dy
dt
=
∂
2
z
∂x∂y
dx
dt
+
∂
2
z
∂y
2
dy
dt
(b)
dz
dt
=
∂z
∂x
dx
dt
+
∂z
∂y
dy
dt
,
d
2
z
dt
2
=
dx
dt

∂
2
z
∂x
2
dx
dt
+
∂
2
z
∂y∂x
dy
dt

+
∂z
∂x
d
2
x
dt
2
+
dy
dt

∂
2
z
∂x∂y
dx
dt
+
∂
2
z
∂y
2
dy
dt

+
∂z
∂y
d
2
y
dt
2
25.∇f=
y
x+y
i+

ln(x+y)+
y
x+y

j, so when (x, y)=(−3,5),
∂f
∂u
=∇f·u=
7
5
2
i+

ln 2 +
5
2

j

·
7
3
5
i+
4
5
j

=
3
2
+2+
4
5
ln 2 =
7
2
+
4
5
ln 2
26. (a) uis a unit vector parallel to the gradient, sou=
2
5

2i+
3
2
j

=
4
5
i+
3
5
j. The maximum
value is∇f(0,0)·u=
8
5
+
9
10
=
5
2
(b)The unit vector to give the minimum has the opposite sense of the vector in Part(a), so
u=−
4
5
i−
3
5
jand∇f(0,0)·u=−
5
2
.

51. January 27, 2005 11:55 L24-ch14 Sheet number 51 Page number 653 black
Review Exercises, Chapter 14 653
27.Use the unit vectorsu=-
1
√
2
,
1
√
2
c,v=-0,−1c,w=-J
1
√
5
,−
2
√
5
c=−
√
2
√
5
u+
1
√
5
v, so that
D
wf=−
√
2
√
5
Duf+
1
√
5
Dvf=−
√
2
√
5
2
√
2+
1
√
5
(−3) =−
7
√
5
28. (a) n=z
xi+z yj−k=8i+8j−k, tangent plane 8x+8y−z=4 + 8 ln 2, normal line
x(t)=1+8t, y(t)=ln2+8t, z(t)=4−t
(b) n=3i+10j−14k, tangent plane 3x+10y−14z=30, normal line
x(t)=2+3t, y(t)=1+10t, z(t)=−1−14t
29.The origin is not such a point, so assume that the normal line at (x
0,y0,z0)7 =(0,0,0) passes
through the origin, thenn=z
xi+zyj−k=−y 0i−x 0j−k; the line passes through the origin and
is normal to the surface if it has the formr(t)=−y
0ti−x 0tj−tkand (x 0,y0,z0)=(x 0,y0,2−x 0y0)
lies on the line if−y
0t=x 0,−x0t=y 0,−t=2−x 0y0, with solutionsx 0=y0=−1,
x
0=y0=1,x 0=y0=0; thus the points are (0,0,2),(1,1,1),(−1,−1,1).
30. n=
2
3
x
−1/3
0
i+
2
3
y
−1/3
0
j+
2
3
z
−1/3
0
k, tangent planex
−1/3
0
x+y
−1/3
0
y+z
−1/3
0
z=x
2/3
0
+y
2/3
0
+z
2/3
0
=1;
intercepts arex=x
1/3
0
,y=y
1/3
0
,z=z
1/3
0
, sum of squares of intercepts isx
2/3
0
+y
2/3
0
+z
2/3
0
=1.
31.A tangent to the line is 6i+4j+k, a normal to the surface isn=18xi+8yj−k, so solve
18x=6k,8y=4k,−1=k;k=−1,x=−1/3,y=−1/2,z=2
32.Solve (t−1)
2
/4+16e
−2t
+(2−
√
t)
2
=1 fortto gett=1.833223,2.839844; the particle strikes the
surface at the pointsP
1(0.83322,0.639589,0.646034),P 2(1.83984,0.233739,0.314816). The velocity
vectors are given byv=
dxdt
i+
dy
dt
j+
dz
dt
k=i−4e
−t
j−1/(2
√
t)k, and a normal to the surface is
n=∇(x
2
/4+y
2
+z
2
)=x/2i+2yj+2zk. At the pointsP ithese are
v
1=i−0.639589j−0.369286k,v 2=i−0.233739j+0.296704k;
n
1=0.41661i+1.27918j+1.29207kandn 2=0.91992i+0.46748j+0.62963kso
cos
−1
[(vi·ni)/(v ini)] =112.3
◦
,61.1
◦
; the acute angles are 67.7
◦
,61.1
◦
.
33.∇f=(2x+3y−6)i+(3x+6y+3)j=0if 2x+3y=6,x+2y=−1,x=15,y=−8,D=3>0,
f
xx=2>0, sofhas a relative minimum at (15,−8).
34.∇f=(2xy−6x)i+(x
2
−12y)j=0if 2xy−6x=0,x
2
−12y=0; ifx=0 theny=0, and
ifx7 =0 theny=3,x=±6, thus the gradient vanishes at (0,0),(−6,3),(6,3);f
xx=0atall
three points,f
yy=−12<0,D=−4x
2
,so(±6,3) are saddle points, and near the origin we write
f(x, y)=(y−3)x
2
−6y
2
; sincey−3<0 when|y|<3,fhas a local maximum by inspection.
35.∇f=(3x
2
−3y)i−(3x−y)j=0ify=x
2
,3x=y,sox=y=0orx=3,y=9; at
x=y=0,D=−9, saddle point; atx=3,y=9,D=9,f
xx=18>0, relative minimum
36.∇f=(8x−12y)i+(−12x+18y)j=0ify=
2
3
x;f
xx=8,f xy=−12,f yy=18,D=0, from which
we can draw no conclusion. Upon inspection, however,f(x, y)=(2x−3y)
2
,sofhas a relative
(and an absolute) minimum of 0 at every point on the liney=
2
3
x, no relative maximum.
37. (a)y
2
=8−4x
2
, find extrema off(x)=x
2
(8−4x
2
)=−4x
4
+8x
2
defined for−
√
2≤x≤
√
2.
Thenf
y
(x)=−16x
3
+16x=0 whenx=0,±1,f
yy
(x)=−48x
2
+ 16, sofhas a relative
maximum atx=±1,y=±2 and a relative minimum atx=0,y=±2
√
2. At the endpoints
x=±
√
2,y=0 we obtain the minimumf=0 again.

52. January 27, 2005 11:55 L24-ch14 Sheet number 52 Page number 654 black
654 Chapter 14
(b)f(x, y)=x
2
y
2
,g(x, y)=4x
2
+y
2
−8=0,∇f=2xy
2
i+2x
2
yj=λ∇g=8λxi+2λyj,so
solve 2xy
2
=λ8x,2x
2
y=λ2y.Ifx=0 theny=±2
√
2, and ify=0 thenx=±
√
2. In
either casefhas a relative and absolute minimum. Assumex, y7 =0, theny
2
=4λ, x
2
=λ,
useg=0 to obtainx
2
=1,x=±1,y=±2, andf=4 is a relative and absolute maximum
at (±1,±2).
38.Let the first octant corner of the box be (x, y, z), so that (x/a)
2
+(y/b)
2
+(z/c)
2
=1. Maximize
V=8xyzsubject tog(x, y, z)=(x/a)
2
+(y/b)
2
+(z/c)
2
=1,solve∇V=λ∇g,or
8(yzi+xzj+xyk)=(2λx/a
2
)i+(2λy/b
2
)j+(2λz/c
2
)k,8a
2
yz=2λx,8b
2
xz=2λy,8c
2
xy=2λz.
For the maximum volume,x, y, z7 =0; divide the first equation by the second to obtaina
2
y
2
=b
2
x
2
;
the first by the third to obtaina
2
z
2
=c
2
x
2
, and finallyb
2
z
2
=c
2
y
2
. Fromg=1 get
3(x/a)
2
=1,x=±a/
√
3, and theny=±b/
√
3,z=±c/
√
3. The dimensions of the box are
2a
√
3
×
2b
√
3
×
2c
√
3
, and the maximum volume is 8abc/(3
√
3).
39.Denote the currentsI
1,I2,I3byx, y, zrespectively. Then minimizeF(x, y, z)=x
2
R1+y
2
R2+z
2
R3
subject tog(x, y, z)=x+y+z−I=0, so solve∇F=λ∇g,2xR 1i+2yR 2j+2zR 3k=λ(i+j+k),
λ=2xR
1=2yR 2=2zR 3, so the minimum value ofFoccurs whenI 1:I2:I3=
1
R1
:
1
R2
:
1
R3
.
40. (a)
P = 1
P = 3
P = 2
12345
1
2
3
4
5
L
K (b)
012345
0
1
2
3
4
5
41. (a)∂P/∂L=cαL
α−1
K
β
,∂P/∂K=cβL
α
K
β−1
(b)the rates of change of output with respect to labor and capital equipment, respectively
(c)K(∂P/∂K)+L(∂P/∂L)=cβL
α
K
β
+cαL
α
K
β
=(α+β)P=P
42. (a)MaximizeP=1000L
0.6
(200,000−L)
0.4
subject to 50L+100K=200,000 orL=2K=4000.
600

K
L

0.4
=λ,400

L
K

0.6
=2λ, L+2K=4000; so
2
3

L
K

=2, thusL=3K,
L=2400,K=800,P(2400,800) =1000·2400
0.6
·800
0.4
=1000·3
0.6
·800 =800,000·3
0.6
≈
$1,546,545.64
(b)The value of labor is 50L=120,000 and the value of capital is 100K=80,000.