Thomas Calculus 13th [Solutions]

1066 Chapter 14 Partial Derivatives
2 2 2 2
f , , 0 , f
4 4 2 x , , 0 , f
4 4 2 y , , 0 , f , , 0
4 4 2 z 4 4 2
2 2 2 2 2 2 2 2
L( x, y, z) x y ( z 0) x y z
2 2 4 2 4 2 2 2 2 2
59.
2 2 2
V r h dV 2 rh dr r dh dV | (1.5,5280) 2 (1.5)(5280) dr (1.5) dh 15,840 dr 2.25 dh.
You should be more careful with the diameter since it has a greater effect on dV.
60. df (2 x y) dx ( x 2 y) dy df | (1,2) 3 dy f is more sensitive to changes in y; in fact, near the point
(1, 2) a change in x does not change f.
61. dI 1 V | 1 24
2 (24,100) |
100 2 dV 1, dR 20 0.01 (480)(.0001) 0.038,
R
dV dR dI dV dR dI or
R
100
increases by 0.038 amps; % change in V (100) 1 4.17%; % change in 20
24
R (100) 20%;
100
I 24 0.24 estimated % change in I dI 100 0.038 100 15.83% more sensitive to voltage
100
I 0.24
change.
62. A ab dA b da a db dA| (10,16) 16 da 10 db; da 0.1 and db 0.1
dA 26 (0.1) 2.6 and A (10)(16) 160 dA 100 2.6
A 160
100 1.625%
63. (a) y uv dy v du u dv ; percentage change in u 2% | du | 0.02, and percentage change in v 3%
dy v du u dv
dy
| dv| 0.03; du dv 100 du 100 dv 100 du 100 dv 100
y uv u v y u v u v
2% 3% 5%
(b) z u v dz du dv du dv du dv (since u 0, v 0 )
z u v u v u v u v
dz 100 du dy
100 dv 100 100
z u v y
7
( 0.425)(7)
( 0.725)(7)
C C and Ch
0.425 1.725
71.84w h 71.84w h
71.84w
h
dC 2.975 dw 5.075 dh ; thus when w 70 and h 180 we have
1.425 0.725 0.425 1.725
71.84w h 71.84w h
dC| (70,180) (0.00000225) dw (0.00000149) dh 1 kg error in weight has more effect
64.
0.425 0.725 w
1.425 0.725
65. fx
( x, y) 2x y 2 0 and f y ( x, y) x 2y 2 0 x 2 and y 2 ( 2, 2) is the critical
point;
2
fxx ( 2, 2) 2, f yy ( 2, 2) 2, fxy ( 2, 2) 1 fxx f yy fxy
3 0 and f xx 0 local minimum
value of f ( 2, 2) 8
66. fx
( x, y) 10x 4y 4 0 and f y ( x, y) 4x 4y 4 0 x 0 and y 1 (0, 1) is the critical point;
2
fxx (0, 1) 10, f yy (0, 1) 4, fxy (0, 1) 4 fxx f yy f xy 56 0 saddle point with f (0, 1) 2
67.
2
fx ( x , y ) 6 x 3 y 0 and
2 2
f y ( x , y ) 3 x 6 y 0 y 2 x and
4 3
3x 6 4x 0 x 1 8x
0
x 0 and y 0, or x 1 and y 1 the critical points are (0, 0) and 1 , 1 . For (0, 0):
2
2
2 2
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