Thomas Calculus 13th [Solutions]

206 Chapter 3 Derivatives
90.
2 t 2
1 t 2 d d
dt dt
0 d (2 t
dt
1) 2 d
2 1/3
2 1 ; r
dt t
( 7)
dr 1 2 2/3
( 7) (2 ) 2 2 2/3 2
( 7) ; now 0 and
d 3 3 t t 1 1 so that d dt t 0, 1
and dr 2 2/3
d 1 3 (1 7) 1 dr dr dr 1 ( 1) 1
6 dt t 0 d t 0 d t 0 6 6
2
1
1
1
91.
3 2 dy
y y 2cos x 3y
dx
dy
dx
dy 2
2sin x (3y 1) 2sin
dx
dy
dx
x
2
2sin x dy
3y
1 dx
(0,1)
2sin(0)
3 1
0;
2
d y
dx
2
(3y 1)( 2cos x) ( 2sin x) 6 y
2 2 2
(3y
1)
dy
dx
2
d y
2
dx (0,1)
(3 1)( 2cos 0)( 2sin 0)(6 0) 1
2
(3 1)
2
92.
1/3 1/3
x y
4
1 2/3 1 2/3 dy
x y 0
3 3 dx
dy
dx
2/3 2 1/3 dy 2/3 2 1/3
2 x y y x
2
3 dx
3 d y
2
2/3
2 2
d y
dx
x
dx
y
x
(8, 8)
2/3
2/3
2/3
dy
1; dy y
dx
2/3
(8, 8)
dx x
8
2
8 ( 1) 8
2
8
2/3 1/3 2/3 1/3
3 3
4/3
8
1 1 2
3 3 3
2/3
8
1
4 6
93. f ( t) 1 and f ( t h)
2t
1
2 f ( t)
(2t 2h 1)(2t
1)
1 1
1 f ( t h) f ( t)
2( t h) 1 2t 1 2t 1 (2t 2h
1)
2( t h) 1 h
h (2t 2h 1)(2t 1) h
( ) ( )
lim f t h f t lim 2 2
h 0
h
2
h 0 (2 t 2 h 1)(2 t 1) (2t
1)
2h
(2t 2h 1)(2t 1) h
94.
2
g( x) 2x 1 and g( x h)
2
4xh
2h
h
4x 2 h g ( x)
2 2 2 g( x h) g( x)
2( x h) 1 2x 4xh 2h
1
h
( ) ( )
lim g x h g x lim (4x 2 h) 4x
h 0
h
h 0
2 2 2
(2x 4xh 2h 1) (2x
1)
h
95. (a)
(b)
(c)
2
2
lim f ( x) lim x 0 and lim f ( x) lim x 0 lim f ( x ) 0. Since lim f ( x ) 0 f (0) it
x 0 x 0
x 0 x 0
x 0
x 0
follows that f is continuous at x 0.
lim f ( x) lim (2 x)
0 and lim f ( x)
lim ( 2 x)
0 lim f ( x ) 0. Since this limit exists, it
x 0 x 0
x 0 x 0
x 0
follows that f is differentiable at x 0.
96. (a)
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