Thomas Calculus 13th [Solutions]

1076 Chapter 14 Partial Derivatives
(b) z, x are independent with
2 yz
w x e and
yz 2 yz y 2 yz
2 xe (0) zx e yx e (1);
z
w 2 yz 1 2 yz 2 yz
zx e yx e x e y z
z x
2y 2 y
(c) z, y are independent with
2 yz
w x e and
yz 2 yz 2 yz
2 xe x zx e (0) yx e (1);
z
w
z
y
yz 1 2 yz 2 yz
2xe yx e 1 x y e
2x
2 2 w w x w y
z x y
w z
z x z y z z z
2 2 y y
z x y 1 0 2 y 1 ; therefore,
z z 2 y
2 2 w w x w y
z x y
w z
z x z y z z z
2 2
z x y 1 2x x 0 x 1 ; therefore,
z z 2x
102. (a) T, P are independent with U f ( P, V , T ) and PV nRT U U P U V U T
T P T V T T T
U (0) U V U (1); PV nRT P V nR V nR ; therefore,
P V T T T T P
U U nR U
T P V P T
T
(b) V, T are independent with U f ( P, V , T ) and PV nRT U U P U V U
V P V V V T V
U P U (1) U (0); PV nRT V P P ( nR ) T 0 P P ; therefore,
P V V T V V V V
U U P U
V T P V V
CHAPTER 14 ADDITIONAL AND ADVANCED EXERCISES
(0, ) (0, 0)
1. By definition, f f
(0, 0) lim x h f
xy
x
h 0
h
so we need to calculate the first partial derivatives in the
numerator. For ( x, y ) (0, 0) we calculate fx
( x, y ) by applying the differentiation rules to the formula for
2 2 2 2
2 3 x y x x y x 2 3 2 3 3
f (2 ) (2 ) 4
( x , y ): f x y y x y y x y
( , ) ( ) (0, ) h
x x y xy .
2 2
2 2
2 2 2
2 2
2 x f h h For
2
x y x y x y x y
h
( x, y ) (0, 0) we apply the definition:
f (0, 0) lim h 0
xy
1.
h 0
h
Similarly,
3 2 3 2 3
f ( h, 0) f (0,0)
f (0, 0) lim lim 0 0
x 0. Then by definition
h 0
h
h 0
h
f f y ( h, 0) f y (0,0)
yx (0, 0) lim ,
h 0
h
so for ( x , y ) (0, 0) we have
x xy 4x y
f ( , ) ( , 0) h
y x y f ;
2 2 2 y h h for ( x, y ) (0, 0) we obtain
2
x y 2 2
x y
h
(0, ) (0, 0)
f y (0, 0) lim f h f
h 0
h
lim 0 0 0. Then by definition 0
h 0
h
f (0, 0) lim h
yx
1.
h 0
h
Note that fxy
(0, 0) f yx (0, 0) in this case.
2. w x x x x
1 e cos y w x e cos y g( y); w e sin y g ( y) 2y e sin y g ( y) 2y
x
y
2
g( y) y C; w ln 2 when x ln 2 and
C 2. Thus,
x
x 2
w x e cos y g( y) x e cos y y 2.
ln 2 2
y 0 ln 2 ln 2 e cos 0 0 C 0 2 C
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