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Unformatted text preview: Chapter 3 Review
d g(x dx d dx 125 (f) f(x)) f(x))(1 g (x f(x)) (x f(x)) (d) Since cos t
4 2 0, the particle first reaches the origin at g (x At x g (0
dw ds dw dr dr ds f (x)) f(0))[1 1 f (0)] . The velocity is given by 10 sin t
4 2 0, the derivative is g (0 1)[1 ( 2)] (1)( 1) v(t) t
4 , so the velocity at 10, and the is 10 sin
4 68. cos ( r At s
dw ds 2) d [sin ( dr 1 2 r r 2)] d 8 sin s ds 6 6 speed at t
6 is 10
4 10. The acceleration is given , so the acceleration at by a(t) t
4 10 cos t 10 cos
2 8 cos s 8 sin 0
1 is 0. 64 32
ds dt 0, we have r cos ( 4 2)
6 4 and so 71. (a)
6 ds dt d 2s dt 2 d (64t dt d (64 dt 16t 2) 32t) 32t 8 cos 0 cos 0 4 8 cos 2 4 1 8 3 4 2 3 (b) The maximum height is reached when 69. Solving t
1
2 0, which 2 t
2 1 for t, we have
1 occurs at t (c) When t (d) Since
2 1 2 sec. 0,
ds dt , and we may write:
dr dt dt d dr d ( dt d dr ( 2 dt 2 4( 2 3(
2 64, so the velocity is 64 ft/sec. 2.6t 2) 64 5.2t, the
64 5.2 dr d d 2 ( d 1 2 ( 3 dr dt ds dt d (64t dt 7)1/3
2/3 maximum height is reached at t The maximum height is s
64 5.2 12.3 sec. )
2 393.8 ft.
4 sec. 7 7) (2 )
2/3 2 3 ) 72. (a) Solving 160 490t 2, it takes
4 7 2 ( 2 3( 2 7)
3 ) 7) 2) 2/3 The average velocity is 160 1,
ds dt 280 cm/sec. At t and so 0, we may solve
dr dt 2(1)4(12 3(1 7) 2) t
2/3 1 to obtain
2(8)
2/3 (b) Since v(t) (980) 980t, the velocity is
dv dt 3 1 . 6 4 = 560 cm/sec. Since a(t) 7 980, the 70. (a) One possible answer: x(t) (b) s(0) 10 cos t 10 cos
4 acceleration is 980 cm/sec 2. , y(t) 2 1 73.
dV dx d dx 10 x 2) 3 x 2 x 3 d dx 10x 2 1 3 x 3 4 5 (20x 74. (a) r(x) (c) Farthest left: When cos t Farthest right: When cos t
4 4 x 2 x 40 9x 3 2 x 20 1 3 x 1600 1, we have s(t) 10. (b) The marginal revenue is r (x) 9
3 x 10 3 2 x 1600 1, we have s(t) 10. 3 (x 2 1600 3 (x 1600 160x 40)(x 4800) 120), which is zero when x 40 or x 120. Since the bus holds only 60 people, we require 0 x 60. The marginal revenue is 0 when there are 40 people, and the 40 2 corresponding fare is p(40) 3 $4.00.
40 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.
 Fall '08
 GERMAN
 Derivative

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