D,t = D,y - D,x = (-8) - (-6) = -2.
second.
RELATED RATES
Thus, the length of the shadow is decreasing at the rate of 2 feet per
14.5
A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is
s = 400( — 16t
2 . How fast is the distance changing from the rocket to an observer on the ground 1800 feet away
from the launching site, when the rocket is still rising and is 2400 feet above the ground?
Fig. 14-3
I Let M be the distance from the rocket to the observer, as shown in Fig. 14-3. By the Pythagorean theorem,
w
2 = s
2 + (1800)
2
. Hence, 2u • D,u = 2s • D,s, u-D,u = s-D,s. When s = 2400, u
2 = (100)
2 • (900), u =
100-30 = 3000.
Since
s = 400/-16/
2 ,
when
s = 2400,
2400 = 400f - I6t
2 ,
t
2 - 251 + 150 = 0,
(t - W)(t - 15) = 0. So, on the way up, the rocket is at 2400 feet when / = 10. But, D,s = 400 - 32;. So,
when t=W, D,s = 400 - 32 • 10 = 80. Substituting in u-D,u = s-D t s, we obtain 3000 • D,u = 2400 • 80,
D,u = 64. So the distance from the rocket to the observer is increasing at the rate of 64 feet per second when
14.6
14.7
A small funnel in the shape of a cone is being emptied of fluid at the rate of 12 cubic centimeters per second. The
height of the funnel is 20 centimeters and the radius of the top is 4 centimeters. How fast is the fluid level
dropping when the level stands 5 centimeters above the vertex of the cone? (Remember that the volume of a
cone is \irr
2 h.)
Fig. 14-4
I The fluid in the funnel forms a cone with radius r, height h, and volume V. By similar triangles, r/4 = /i/20,
r=h/5. So, V=^irr
2 h=$Tr(h/5)
2 h= Jsirh
3 . Hence, D,V= ^trh
2 • D,h. We are given that D,V=
— 12, since the fluid is leaving at the rate of 12 cubic centimeters per second. Hence, -12 = ^irh
2 • D,h,
-300 = irh
2 • D,h. When h = 5, -300= -rr -25 • D,h, D l h = ~l2/Tr, or approximately -3.82 centimeters
per second. Hence, the fluid level is dropping at the rate of about 3.82 centimeters per second.
A balloon is being inflated by pumped air at the rate of 2 cubic inches per second,
balloon increasing when the radius is { inch?
How fast is the diameter of the
89
second.
RELATED RATES
Thus, the length of the shadow is decreasing at the rate of 2 feet per
14.5
A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is
s = 400( — 16t
2 . How fast is the distance changing from the rocket to an observer on the ground 1800 feet away
from the launching site, when the rocket is still rising and is 2400 feet above the ground?
Fig. 14-3
I Let M be the distance from the rocket to the observer, as shown in Fig. 14-3. By the Pythagorean theorem,
w
2 = s
2 + (1800)
2
. Hence, 2u • D,u = 2s • D,s, u-D,u = s-D,s. When s = 2400, u
2 = (100)
2 • (900), u =
100-30 = 3000.
Since
s = 400/-16/
2 ,
when
s = 2400,
2400 = 400f - I6t
2 ,
t
2 - 251 + 150 = 0,
(t - W)(t - 15) = 0. So, on the way up, the rocket is at 2400 feet when / = 10. But, D,s = 400 - 32;. So,
when t=W, D,s = 400 - 32 • 10 = 80. Substituting in u-D,u = s-D t s, we obtain 3000 • D,u = 2400 • 80,
D,u = 64. So the distance from the rocket to the observer is increasing at the rate of 64 feet per second when
14.6
14.7
A small funnel in the shape of a cone is being emptied of fluid at the rate of 12 cubic centimeters per second. The
height of the funnel is 20 centimeters and the radius of the top is 4 centimeters. How fast is the fluid level
dropping when the level stands 5 centimeters above the vertex of the cone? (Remember that the volume of a
cone is \irr
2 h.)
Fig. 14-4
I The fluid in the funnel forms a cone with radius r, height h, and volume V. By similar triangles, r/4 = /i/20,
r=h/5. So, V=^irr
2 h=$Tr(h/5)
2 h= Jsirh
3 . Hence, D,V= ^trh
2 • D,h. We are given that D,V=
— 12, since the fluid is leaving at the rate of 12 cubic centimeters per second. Hence, -12 = ^irh
2 • D,h,
-300 = irh
2 • D,h. When h = 5, -300= -rr -25 • D,h, D l h = ~l2/Tr, or approximately -3.82 centimeters
per second. Hence, the fluid level is dropping at the rate of about 3.82 centimeters per second.
A balloon is being inflated by pumped air at the rate of 2 cubic inches per second,
balloon increasing when the radius is { inch?
How fast is the diameter of the
89