Thomas Calculus 13th [Solutions]

1018 Chapter 14 Partial Derivatives
31. (a) W (20, 25) 11° F; W (30, 10) 39° F; W (15,15) 0° F
(b) W (10, 40) 65.5° F; W (50, 40) 88° F; W (60, 30) 10.2° F;
0.16
(c) W (25, 5) 17.4088° F; W 5.72 0.0684t W (25, 5) 0.36; W 0.6215 0.4275v
V 0.84 0.84
v v V T
W (25, 5) 1.3370 L( V , T ) 17.4088 0.36( V 25) 1.337( T 5) 1.337T 0.36V
15.0938
T
(d) i) W (24, 6) L(24, 6) 15.7118 15.7° F
ii) W (27, 2) L(27, 2) 22.1398 22.1° F
iii) W (5, 10) L(5, 10) 30.2638 30.2° F This value is very different because the point (5, 10)
is not close to the point (25, 5).
5.72 0.0684
0.16
W (50, 20) 59.5298° F; W t W (50, 20) 0.2651; W 0.6215 0.4275v
V v v V T
W (50, 20) 1.4209 L( V , T ) 59.5298 0.2651( V 50) 1.4209( T 20)
T
1.4209T
0.2651V
17.8568
(a) W (49, 22) L(49, 22) 62.1065 62.1° F
32.
0.84 0.84
(b) W (53, 19) L(53, 19) 58.9042 58.9° F
(c) W (60, 30) L(60, 30) 76.3898 76.4° F
33. f (2, 1) 3, fx ( x, y) 2x 3 y fx (2, 1) 1, f y ( x, y) 3 x f y (2,1) 6 L( x, y) 3 1( x 2) 6( y 1)
1
2
7 x 6 y; fxx ( x, y) 2, f yy ( x, y) 0, fxy
( x, y) 3 M 3; thus | E( x, y)| (3) | x 2| | y 1|
2
3
2
2
(0.1 0.1) 0.06
34.
y
f (2, 2) 11, fx ( x, y) x y 3 fx (2, 2) 7, f y ( x, y) x 3 f (2, 2) 0
2 y
L( x, y) 11 7( x 2) 0( y 2) 7x 3; f ( , ) 1, ( , ) 1
xx x y f yy x y , f ( , ) 1 1;
2 xy x y M thus
1 2 1
2
E( x, y) (1) | x 2| | y 2| (0.1 0.1) 0.02
2 2
35. f (0, 0) 1, fx ( x, y) cos y fx (0, 0) 1, f y ( x, y) 1 x sin y f y (0, 0) 1
L( x, y) 1 1( x 0) 1( y 0) x y 1; fxx ( x, y) 0, f yy ( x, y) x cos y, fxy
( x, y) sin y
M 1; thus
1 2 1
2
E( x, y) (1) | x| | y| (0.2 0.2) 0.08
2 2
36.
2
f (1, 2) 6, fx ( x, y) y y sin( x 1) fx (1, 2) 4, f y ( x, y) 2xy cos( x 1) f y (1, 2) 5
L( x, y) 6 4( x 1) 5( y 2) 4x 5y 8; fxx
( x, y) y cos( x 1), f yy ( x, y) 2 x,
fxy ( x, y) 2y sin( x 1); | x 1| 0.1 0.9 x 1.1 and | y 2| 0.1 1.9 y 2.1; thus the max of
fxx ( x , y ) on R is 2.1, the max of f yy ( x, y ) on R is 2.2, and the max of fxy
( x, y ) on R is
2(2.1) sin(0.9 1) 4.3 M 4.3; thus
1
2 2
| E( x, y)| (4.3) | x 1| + | y 2| (2.15)(0.1 0.1) 0.086
2
x
x
37. f (0, 0) 1, fx ( x, y) e cos y fx (0, 0) 1, f y ( x, y) e sin y f y (0, 0) 0
L( x, y) 1 1( x 0) 0( y 0) 1 x ; fxx ( x, y) e x cos y, f yy ( x, y) e x cos y, fxy
( x, y) e x sin y;
| x| 0.1 0.1 x 0.1 and | y| 0.1 0.1 y 0.1; thus the max of | fxx
( x, y )| on R is
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