3
2.
Gas is being pumped into a spherical balloon at a rate of 5 ft min . If the pressure is
constant, find the rate at which the radius is changing when the diameter is 18 in . (Note here
that 18 in must be converted to feet.)
dV
dr
D = 18" ⇒ r = 9" = 3 4 ft ,
= 5ft 3 / min , find
dt
dt
V=
.
r= 3 4
4 3
πr
3
dV
dr
dr
1 dV
= 4π r 2
⇒
=
dt
dt
dt 4π r 2 dt
dr
1
5
20
=
=
2 •5 =
9 π 9π
dt r= 3
4π 3 4
4
4
( )
When the diameter is 18”, the radius is increasing at a rate of 0.707ft/min or 8.488in/min.
3.
A light is at the top of a 16 ft pole. A boy 5 ft tall walks away from the pole at a rate of
ft
4 sec . At what rate is the tip of his shadow moving when he is 18 ft from
the pole? At what rate is the length of his shadow increasing?
16 ft
dx
= 4ft/sec . (A) Find the rate at which the tip of the shadow is
dt
moving.
Since he is moving at 4ft/sec, the tip of the shadow also moves at 4ft/sec.
Given:
(B) Find the rate at which the length of the shadow increasing?
Let x be the distance from the base of the pole to the tip of the shadow.
Let s be the length of the boy’s shadow.
ds
s
x
Using similar triangles, = . We’re to find
.
dt x=18
5 16
s
x
=
5 16
1 ds 1 dx
=
5 dt 16 dt
ds 5 dx
=
dt 16 dt
ds
5•4 5
=
= 4 ft/sec
dt x=18
16
When the boy is 18ft from the pole, the shadow is increasing at a rate of 1.25ft/sec.
5 ft
Shadow
A girl starts at a point A and runs east at a rate of 10 ft sec . One minute later, another girl
starts at A and runs north at a rate of 8 ft sec . At what rate is
the distance between them changing 1 minute after the
z
second girl starts?
y
dy
dx
dz
= 8ft/sec;
= 10ft/sec. Find
dt
dt
dt t =1min
The east-bound girl has a 1-minute head start during which
x+600
she runs 600ft.
2
y 2 + ( x + 600 ) = z 2
4.
dx
dy
dz
One minute after the second
+ 2 ( x + 600 ) = 2z
dt
dt
dt
runner starts,
dx
dy
x = 10 • 60 = 600ft
dz y dt + ( x + 600 ) dt
=
y = 8 • 60 = 480ft, and
z
dt
480 ( 8 ) + ( 600 + 600 ) (10 ) z = 480 2 + ( 600 + 600 )2
dz
=
dt t =1min
240 29
z = 240 29ft
dz
66 29
=
≈ 12.256ft/sec
29
dt t =1min
2y
One minute after the second runner starts, the distance between the girls is increasing at 12.256ft
/sec.
5.
A boy flying a kite holds the string 5 ft above ground level and pays out string at a rate of
ft
2 sec as the kite moves horizontally at an altitude of 105 ft . Assuming there is no sag in the
string, find the rate at which the kite is moving when 125 ft of string has been paid out.
Given:
dz
dx
= 2ft/sec , find
dt
dt
.
z
z=125
100
x 2 + 100 2 = z 2
dx
dz
= 2z
When z = 125,
dt
dt
x 2 + 100 2 = 125 2
dx z dz
=
dt x dt
x = 75
dx
125
=
( 2 ) = 10 3 ft/sec
dt z=125 75
2x
x
When 125 feet of string have been paid out, the kite is moving horizontally at 10/3ft/sec.
6.
Suppose a spherical snowball is melting and the radius is decreasing at a constant rate,
changing from 12in to 8in in 45 minutes. How fast was the volume changing when the radius
is10in?
Given:
dr 8 − 12 −4
dV
=
=
in/min , find
45
45
dt
dt
.
r =10
V = 4 3π r 3
dV
dr
= 4π r 2
dt
dt
−4 −320π
dV
2
= 4π (10 ) •
=
= −111.701in 3 / min
45
9
dt r =10
When the radius is 10in, the volume is decreasing at -111.701in3/min.
9.
A softball diamond has the shape of a square with sides 60ft long. If a player is running
from second base to third base at a speed of 24ft/sec, at what rate is her distance from home plate
changing when she is 20ft from third?
Given:
dx
dz
= −24ft/sec , find
dt
dt
.
x 2 + 60 2 = z 2
dx
dz
2x
= 2z
dt
dt
dz x dx
=
dt z dt
20 ( −24 )
dz
=
dt x = 20
20 10
dz
dt
=
x = 20
x
x = 20
z
When x = 20,
60
20 2 + 60 2 = z 2
z = 20 10
−12 10
≈ −7.589ft/sec
5
When the girl is 20ft from third base, her distance from home plate is decreasing at 7.589ft/sec.
10. A ladder 20ft long leans against a vertical building. If the bottom of
the ladder slides away from the building horizontally at a rate of 2ft/sec, at
what rate is the angle between the ladder and the ground changing when
the top of the ladder is 12ft above the ground?
y
Given:
dx
dθ
= 2ft/sec , find
dt
dt
.
y=12
x
20
1 dx
dθ
− sin θ
=
dt 20 dt
When y = 12,
−1 dx
dθ
=
12 3
dt 20 sin θ dt
=
sin θ =
20 5
−1( 2 )
dθ
=
dt y=12 20 3
5
dθ
= −1 6 ≈ −0.167rad/sec
dt y=12
cosθ =
20
θ
x
( )
When the top of the ladder is 12ft from the ground, the angle between the ladder and the ground
is decreasing at 0.167rad/sec.