CHAPTER 14
8 feet in diameter at the top. At the time the water is 12 feet deep, the water level is observed to be rising 4
inches per minute. How fast is the water leaking out?
I Let h be the depth of the water, and let r be the radius of the water surface (Fig. 14-11). The water's volume
V=^trr
2 h. By similar triangles, r/4 = fc/16, r=\h, V= ^Tr(h/4)
2 h = ±rrh
3 . So D,V= ^-rrh
2 • D,h.
We are told that when h = 12, D,h = j . Hence, at that moment, D,V = T feir(l44)( j ) = 3ir. Since the rate
at which the water is pouring in is 10, the rate of leakage is (10 - ITT) ft
3 /min.
14.30 An airplane is ascending at a speed of 400 kilometers per hour along a line making an angle of 60° with the ground.
How fast is the altitude of the plane changing?
I Let h be the altitude of the plane, and let u be the distance of the plane from the ground along its flight path
(Fig. 14-12). Then hlu = sin60° = V3/2, 2/i = V3u, 2 • D,h = V3D,« = V5 • 400. Hence, D,/j = 200V5
kilometers per hour.
Fig. 14-12
Fig. 14-13
14.31
How fast is the shadow cast on level ground by a pole 50 feet tall lengthening when the angle a of elevation of the
sun is 45° and is decreasing by \ radian per hour? (See Fig. 14.13.)
I Let x be the length of the shadow. tana=50/jr. By the chain rule, sec" a • D,a = (-50/*
2 )- D,x. When
a =45°, tana = l, sec
2 a = 1 + tan
2 a = 2, A: = 50. So, 2(-$) = -&• D,x. Hence, D,* = 25ft/h.
14.32
A revolving beacon is situated 3600 feet off a straight shore. If the beacon turns at 477 radians per minute, how
fast does its beam sweep along the shore at its nearest point A1
14.33
f Let x be the distance from A to the point on the shore hit by the beacon, and let a be the angle between the line
from the lighthouse 5 to A, and the beacon (Fig. 14-14). Then tan a = je/3600, so sec
2 a • D,a = 3555 • D,x.
We are told that D,a=4?r. When the beacon hits point A, a=0, seca = l, so 4n=jsooD,x,
D,x = 14,40077- ft/min = 2407T ft/s.
Two sides of a triangle are 15 and 20 feet long, respectively. How fast is the third side increasing when the angle
a between the given sides is 60° and is increasing at the rate of 2° per second?
I Let x be the third side. By the law of cosines, x
2 = (15)
2 + (20)
2 -2(15)(20)-cos a. Hence, 2x • D,x =
600sino-D,a. x- D,x = 300sin a • D,a. We are told that D,a =2- (Tr/180) = 7r/90rad/s. When o =
60°, sina=V3/2,
cosa = |,
x
2 = 225 + 400-600- \ = 325,
x = 5VT5.
Hence,
5VT3-D,x =
300-(V3/2)-(7r/90), D,x = (7T/V39) ft/s.
94
Fig. 14-14
8 feet in diameter at the top. At the time the water is 12 feet deep, the water level is observed to be rising 4
inches per minute. How fast is the water leaking out?
I Let h be the depth of the water, and let r be the radius of the water surface (Fig. 14-11). The water's volume
V=^trr
2 h. By similar triangles, r/4 = fc/16, r=\h, V= ^Tr(h/4)
2 h = ±rrh
3 . So D,V= ^-rrh
2 • D,h.
We are told that when h = 12, D,h = j . Hence, at that moment, D,V = T feir(l44)( j ) = 3ir. Since the rate
at which the water is pouring in is 10, the rate of leakage is (10 - ITT) ft
3 /min.
14.30 An airplane is ascending at a speed of 400 kilometers per hour along a line making an angle of 60° with the ground.
How fast is the altitude of the plane changing?
I Let h be the altitude of the plane, and let u be the distance of the plane from the ground along its flight path
(Fig. 14-12). Then hlu = sin60° = V3/2, 2/i = V3u, 2 • D,h = V3D,« = V5 • 400. Hence, D,/j = 200V5
kilometers per hour.
Fig. 14-12
Fig. 14-13
14.31
How fast is the shadow cast on level ground by a pole 50 feet tall lengthening when the angle a of elevation of the
sun is 45° and is decreasing by \ radian per hour? (See Fig. 14.13.)
I Let x be the length of the shadow. tana=50/jr. By the chain rule, sec" a • D,a = (-50/*
2 )- D,x. When
a =45°, tana = l, sec
2 a = 1 + tan
2 a = 2, A: = 50. So, 2(-$) = -&• D,x. Hence, D,* = 25ft/h.
14.32
A revolving beacon is situated 3600 feet off a straight shore. If the beacon turns at 477 radians per minute, how
fast does its beam sweep along the shore at its nearest point A1
14.33
f Let x be the distance from A to the point on the shore hit by the beacon, and let a be the angle between the line
from the lighthouse 5 to A, and the beacon (Fig. 14-14). Then tan a = je/3600, so sec
2 a • D,a = 3555 • D,x.
We are told that D,a=4?r. When the beacon hits point A, a=0, seca = l, so 4n=jsooD,x,
D,x = 14,40077- ft/min = 2407T ft/s.
Two sides of a triangle are 15 and 20 feet long, respectively. How fast is the third side increasing when the angle
a between the given sides is 60° and is increasing at the rate of 2° per second?
I Let x be the third side. By the law of cosines, x
2 = (15)
2 + (20)
2 -2(15)(20)-cos a. Hence, 2x • D,x =
600sino-D,a. x- D,x = 300sin a • D,a. We are told that D,a =2- (Tr/180) = 7r/90rad/s. When o =
60°, sina=V3/2,
cosa = |,
x
2 = 225 + 400-600- \ = 325,
x = 5VT5.
Hence,
5VT3-D,x =
300-(V3/2)-(7r/90), D,x = (7T/V39) ft/s.
94
Fig. 14-14