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Thomas Calculus 13th [Solutions]

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Section 3.10 Related Rates 189<br />

27.<br />

2<br />

2<br />

3 2<br />

V 1 r h, h 3 (2 r)<br />

3r r 4h V 1 4h h 16 h dV 16 h dh<br />

3 8 4 3 3 3 27 dt 9 dt<br />

(a) dh<br />

9 (10) 90 0.1119 m/sec 11.19 cm/sec<br />

dt 2<br />

h 4 16 4 256<br />

(b) r 4h<br />

dr 4 dh 4 90 15 0.1492 m/sec 14.92 cm/sec<br />

3 dt 3 dt 3 256 32<br />

2<br />

2<br />

3<br />

28. (a) V 1 r h and r 15h<br />

V 1 15h<br />

h 75 h<br />

3<br />

2 3 2 4<br />

8 0.0113 m/min 1.13 cm/min<br />

225<br />

(b) r 15h<br />

dr 15 dh dr 15 8 4<br />

2 dt 2 dt dt h 5 2 225 15<br />

dV 225<br />

2<br />

h dh dh 4( 50)<br />

dt 4 dt dt h 5<br />

2<br />

225 (5)<br />

0.0849 m/sec 8.49 cm/sec<br />

2<br />

29. (a) V y (3 R y) dV [2 (3 )<br />

3<br />

dt 3 y R y 2 dy dy<br />

( 1)]<br />

2<br />

1<br />

y<br />

dt dt 3 (6 Ry 3 y ) dV at R 13 and y 8<br />

dt<br />

dy<br />

we have 1 ( 6) 1 m/min<br />

dt 144 24<br />

(b) The hemisphere is one the circle 2 (13 ) 2<br />

2<br />

r y 169 r 26 y y m<br />

2 1/2<br />

(c) r (26 y y ) dr 1 2 1/2<br />

dt 2 (26 y y ) (26 2 ) dy dr 13 y dy<br />

y<br />

dt dt<br />

2<br />

26 y y<br />

dt<br />

dr<br />

dt<br />

y 8<br />

13 8 1<br />

26 8 64 24<br />

5<br />

288 m/min<br />

30.<br />

4 3 2<br />

If V r , S 4 , and dV<br />

2<br />

r<br />

kS 4 k r , then dV<br />

3<br />

dt<br />

dt<br />

Therefore, the radius is increasing at a constant rate.<br />

2 2<br />

4 r dr 4k r<br />

dt<br />

2<br />

4 r dr<br />

dt<br />

dr<br />

dt<br />

k , a constant.<br />

31.<br />

4 3<br />

If V r , r 5, and dV<br />

3<br />

100 ft /min, then dV 2<br />

4 r dr dr 1 ft/min. Then S<br />

3<br />

dt<br />

dt dt dt<br />

dS<br />

2<br />

8 r dr 8 (5)(1) 40 ft /min, the rate at which the surface area is increasing.<br />

dt dt<br />

2<br />

4 r<br />

32. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.<br />

(a) We have s<br />

2 x<br />

2 36 dx s ds s ds . Therefore, the boat is approaching the dock at<br />

dt x dt<br />

s<br />

2 36<br />

dt<br />

dx 10 ( 2) 2.5 ft/sec.<br />

dt<br />

2<br />

s 10 10 36<br />

(b) cos 6 sin d 6 dr d 6 dr . Thus, r 10, x 8, and sin 8<br />

r dt 2 dt dt 2<br />

r<br />

r sin dt<br />

10<br />

d 6 ( 2) 3 rad/sec<br />

dt 2 8<br />

10<br />

20<br />

10<br />

33. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal<br />

distance between the balloon and the bicycle. The relationship between the variables is s 2 h 2 x<br />

2<br />

ds 1 h dh x dx ds 1 [68(1) 51(17)] 11 ft/sec.<br />

dt s dt dt dt 85<br />

34. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is<br />

(b)<br />

V 9 h dV 9 dh the rate the coffee is rising is dh 1 dV 10 in/min.<br />

dt dt<br />

dt 9 dt 9<br />

2<br />

3<br />

Let h the height of the coffee in the pot. From the figure, the radius of the filter r h V 1 r h h<br />

2 3 12 ,<br />

the volume of the filter. The rate the coffee is falling is dh 4 dV 4 ( 10) 8 in/min.<br />

dt 2<br />

h dt 25 5<br />

Copyright<br />

2014 Pearson Education, Inc.

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