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Thomas Calculus 13th [Solutions]

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190 Chapter 3 Derivatives<br />

35.<br />

1 dy 1 dQ 2<br />

y QD D QD dD 1 (0) 233 ( 2) 466 L/min increasing about 0.2772 L/min<br />

dt dt dt 41 2<br />

(41) 1681<br />

2<br />

36. Let P( x, y ) represent a point on the curve y x and the angle of inclination of a line containing P and the<br />

y<br />

2<br />

origin. Consequently, tan tan x<br />

2<br />

x sec d dx d 2<br />

cos dx . Since dx 10 m/sec and<br />

x<br />

x<br />

dt dt dt<br />

dt dt<br />

2<br />

2 2<br />

cos<br />

x 3 1 , we have d 1 rad/sec.<br />

x 3 2 2 2 2<br />

y x 9 3 10 dt x 3<br />

2 2<br />

37. The distance from the origin is s x y and we wish to find<br />

ds 1 2 2 1/2<br />

(5)( 1) (12)( 5)<br />

2 ( ) 2 2 dy<br />

x y x dx y<br />

dt 5m/sec<br />

(5,12) dt dt<br />

(5,12) 25 144<br />

38. Let s distance of the car from the foot of perpendicular in the textbook diagram tan s<br />

132<br />

sec 2 d 1 ds d cos 2 ds<br />

dt 132 dt dt 132 dt<br />

; ds 264 and 0 d 2 rad/sec. A half second later the car has<br />

dt<br />

dt<br />

2<br />

traveled 132 ft right of the perpendicular | | , cos 1 , and ds 264 (since s increases)<br />

4<br />

2 dt<br />

1<br />

d<br />

( )<br />

2<br />

(264) 1 rad/sec.<br />

dt 132<br />

2<br />

39. Let s 16t represent the distance the ball has<br />

fallen, h the distance between the ball and the<br />

ground, and I the distance between the shadow and<br />

the point directly beneath the ball. Accordingly,<br />

s h 50 and since the triangle LOQ and triangle<br />

2<br />

PRQ are similar we have I 30h<br />

h 50 16t<br />

50 h<br />

2<br />

30(50 16 )<br />

and I t 1500 30 dI 1500<br />

2<br />

50 (50 16 t ) 16t<br />

2 dt 8t<br />

3<br />

dI 1500 ft/sec.<br />

dt t<br />

1<br />

2<br />

40. When x represents the length of the shadow, then tan<br />

We are given that d 0.27 3 rad/ min. At<br />

dt<br />

2000<br />

3 ft/min 0.589 ft/min 7.1in./min.<br />

16<br />

80 2<br />

sec d 80 dx dx<br />

x dt 2<br />

x dt dt<br />

x 60, cos 3 dx<br />

5 dt<br />

x<br />

2 2<br />

sec<br />

80<br />

d<br />

x<br />

2 2<br />

sec<br />

80<br />

d<br />

dt<br />

d 3 5<br />

dt and sec<br />

dt 2000 3<br />

3 3<br />

41. The volume of the ice is 4 4 4 dV 2<br />

V r<br />

4<br />

3 3 dt<br />

r dr dr<br />

dt dt r 6<br />

5 in./min when dV 3<br />

10 in /min,<br />

72<br />

dt<br />

the thickness of the ice is decreasing at 5<br />

2<br />

in/min. The surface area is S 4 r ds 8 r dr dS<br />

72<br />

dt dt dt r 6<br />

48 5 10 2<br />

72 3 in /min, the outer surface area of the ice is decreasing at 10 2<br />

in /min.<br />

3<br />

42. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between<br />

2 2<br />

the car and plane 9 s r ds r dr ds 5 ( 160) 200 mph speed of plane speed<br />

dt 2<br />

r 9<br />

dt dt r 5 16<br />

of car 200 mph the speed of the car is 80 mph.<br />

.<br />

Copyright<br />

2014 Pearson Education, Inc.

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