Chapter 3 71

Chapter 3 71 - SECTIONSJO RELATEDRATES U 221 3:m3 2m =>...

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Unformatted text preview: SECTIONSJO RELATEDRATES U 221 3. :m3+2m => 2—1:zj—Z‘fi—J:=(3m2+2)(5)=5(3m2+2).Whenn=2.z—f=5(14)=70. 4.z2+y2=25 :s 2\$:—:+2y%=0 => mZ—fz—yfg e 3—:=—%%. Wheny:4.m2+42=25 2'e2i3.For%:6,%=—fg(6)=¢s 5-Z"’::vz+y2 : 22%=2w:—:+2y% : g=é<\$%+y%>.Whenx=5and3/=l2. 22:52+122 => 222169 => z=i13.For%:-=2and%=3,%:iim(5-2+12-3)=i‘11—:. 6.3;: 1+:L‘3 :> %::—:j—::%(l+w3)ilﬂ(3m2)::=2¢%Z:.Withjz—4whenm:2and y:3,wehave4:%% => 2%:2cm/s. 7. (a) Given: a plane ﬂying horizontally at an altitude of 1 mi and a speed of 500 mi / h passes directly over a radar station. If we let t be time (in hours) and as be the horizontal distance traveled by the plane (in mi). then we are given that dzc/dt : 500 mi /h. (b) Unknown: the rate at which the distance from the plane to the station is (c) x increasing when it is 2 mi from the station. If we let y be the distance from 1'7 y the plane to the station, then we want to ﬁnd dy/dt when y = 2 mi. (d) By the Pythagorean Theorem, 3/2 : x2 + 1 => 2y(dy/dt) = 222(dx/dt). @ﬁmda: d . (6 dt _ 3E : 3(500). Since y2 = m2 + 1, when y z 2, a: : Mr}, so d—i’ = @500) : 250\/§z 433 ml/h. 8. (a) Given: the rate of decrease of the surface area is 1 cm2/min. If we let (G) t be time (in minutes) and S be the surface area (in cm2), then we are given that dS/dt = —1 cm2/s. (b) Unknown: the rate of decrease of the diameter when the diameter is 10 cm. If we let a: be the diameter, then we want to ﬁnd dm/dt when a: 2 10 cm. (d) If the radius is r and the diameter .7: 2 27', then r = gen and S : 47r7‘2 : 4n(§x)2 = 7rac2 :> E _ dﬁﬂ _ 2 dx dt — dtc dt _ 7m:dt‘ dS dz dw 1 dm 1 (e) dt 7rzc dt 2 dt 27m When an 10, dt 207r So the rate of decrease . 1 . is E cm/mm. 9. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15-ft-tall pole at a rate of 5 ft/s. If we let t be time (in s) and a: be the distance from the pole to the man (in ft), then we are given that dm/dt = 5 ft/s. (b) Unknown: the rate at which the tip of his shadow is moving when he is (c) 40 ft from the pole. If we let y be the distance from the man to the tip of 15 his shadow (in ft), then we want to ﬁnd \$06 + y) when a: = 40 ft. ...
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