Thomas Calculus 13th [Solutions]

998 Chapter 14 Partial Derivatives
83. w cos ( x ct) 2sin (2x 2 ct),
w c cos ( x ct) 2c sin (2 x ct);
w sin ( x ct) 4cos (2x 2 ct),
x
t
2
x
2 w 2 2
2
2 2
2
c sin ( x ct) 4c cos (2x 2 ct) w c sin ( x ct) 4cos (2x 2 ct)
c w
2 2 2
t t x
2
84. w 1 ,
x x ct
w
t
c ;
x ct
2
w
x
1 ,
( x ct)
2 2
2 2 2 2
w c w 2 2
c 1 c w
2 2 2 2 2
t ( x ct) t ( x ct)
x
85.
w 2
2sec (2x
2 ct),
w
2
2csec (2x 2 ct);
w 2
8sec (2x 2 ct) tan (2x 2 ct),
x
t
2
x
2 w 2 2
2
2 2 2
2
8c sec (2x 2 ct) tan (2x 2 ct) ux w c 8sec (2x 2 ct) tan (2x 2 ct)
c w
2 2 2
t t x
2
86. w
x ct
15sin (3x 3 ct) e , w
x ct
15c sin (3x 3 ct) ce ; w
x ct
45cos (3x 3 ct) e ,
x
t
2
x
2 2 2
w 2 2 x ct
2 x ct 2
45c cos (3x 3 ct) c e w c 45cos (3x 3 ct)
e c w
2 2 2
t t x
2
87.
2
2 2
f f f 2 2 f
ac ac ac a c
2 2 2
w u ( ) w ( ) ( ) ;
t u t u t u u
2 2 2 2
2
2 f 2 2 f 2 2 f 2
a w a c c a c w
2 2 2 2 2
u t u u x
2
2
f f f
2 2
w u a w a a
x u x u x u
88. If the first partial derivatives are continuous throughout an open region R, then by Theorem 3 in this section of
the text, f ( x, y) f ( x0, y0) fx
( x0, y0) x f y ( x0 , y0 ) y 1 x 2 y , where 1 , 2 0 as x, y 0.
Then as ( x, y) ( x0 , y0), x 0 and
every point ( x0 , y 0)
in R.
y 0 lim f ( x, y) f ( x0 , y0
) f is continuous at
( x, y) ( x , y )
89. Yes, since fxx, f yy , fxy , and f yx are all continuous on R, use the same reasoning as in Exercise 88 with
fx ( x, y) fx ( x0 , y0) fxx ( x0 , y0) x fxy
( x0 , y0)
y 1 x 2 y and
f y ( x, y) f y ( x0, y0) f yx ( x0, y0 ) x f yy ( x0, y0 ) y 1 x 2
y . Then
lim fx
( x, y) fx
( x0 , y 0)
and lim f y ( x, y) f y ( x0, y0).
( x, y) ( x , y )
( x, y) ( x , y )
0 0
0 0
0 0
90. To find and so that ut uxx ut
sin( x)
e
t
and ux
cos( x)
e
t
uxx
2 sin( x ) e
t ; then ut
uxx
sin( x) e
t 2
sin( x) e
t
, thus ut u xx only if
2
91.
h 0
2
h
2
0
4
0
f (0 h, 0) f (0, 0)
f (0, 0) lim lim lim 0
x 0;
h 0
h
h 0
h
h 0
h
0 h
2
0
2
h
4
f (0, 0 h) f (0, 0)
f y (0, 0) lim lim
h 0
h
h 0
h
2 2
ky y 4
ky k k
2
2
4
2 4 4 2 2
lim 0 0; lim f ( x, y ) lim lim lim
h 0
h
( x, y) (0, 0) y 0 ky y y 0 k y y y 0 k 1 k 1
different limits for
2
along x ky
different values of k lim f ( x, y ) does not exist f ( x, y ) is not continuous at (0, 0) by Theorem
( x, y) (0, 0)
4, f ( x, y ) is not differentiable at (0, 0).
0
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