Handout 7, Summer 2014
Math 1823-171
03 June 2014
Ex. 1. Air is being pumped into a spherical balloon so that its volume
increases at a rate of 100 cm3 /s. How fast is the radius of the
balloon increasing when the diameter is 50 cm?
Ex. 2. A ladder 10 ft. long rests against a vertical wall. If the bottom
of the ladder slides away from the wall at a rate of 1 ft/s, how
fast is the top of the ladder sliding down the wall when the
bottom of the ladder is 6 ft from the wall?
Ex. 3. A water tank has the shape of an inverted circular cone with
base radius 2 m and height 4 m. If water is being pumped into
the tank at a rate of 2 m3 /min, find the rate at which the water
level is rising when the water is 3 m deep.
Ex. 4. A man walks along a straight path at a speed of 4 ft/s. A
searchlight is located on the ground 20 ft from the path and
is kept focused on the man. At what rate is the searchlight
rotating when the man is 15 ft from the point on the path
closest to the searchlight?
1. Each side of a square is increasing at a rate of 6 cm/s. At what
rate is the area of the square increasing when the area of the
square is 16 cm2 ?
Solution. Let s(t) be the length of any side of the square. We are then
told that ds
= 6. Let A(t) be the area of the square. We wish to find
dt
dA .
dt A(t)=16
Recall that for a square, the area is given by A(t) = s(t)2 . Therefore,
differentiating we have:
dA
ds
= 2s(t)
dt
dt
dA We know ds
,
and
we
are
looking
for
, which means we must
dt
dt A(t)=16
find s(t) when A(t) = 16. Well, if A(t) = 16, we must have that
s(t) = 4, so that:
dA = 2 · 4 · 6 = 48
dt A(t)=16
Therefore, we have dA
= 48 cm2 /s.
dt A(t)=16
1
2
2. The radius of a sphere is increasing at a rate of 4 mm/s. How
fast is the volume increasing when the diameter is 80 mm?
Solution. Let r(t) be the radius of the sphere and V (t)
its volume.
dr
dV Then we are told that dt = 4, and we are looking for dt r(t)=40 .
Recall that the volume of a sphere is given by V (t) = 43 πr(t)3 . Differentiating both sides with respect to t, we have:
dV
4
dr
2 dr
= π 3r(t)
= 4πr(t)2
dt
3
dt
dt
Plugging in the known information, we have:
dV = 4π(40)2 · 4 = 25600π
dt r(t)=40
= 25600π mm3 /s. This gives the result.
That is, dV
dt r(t)=40
3. A cylindrical tank with radius 5 m is being filled with water
at a rate of 3 m3 /min. How fast is the height of the water
increasing?
Solution. Let h(t) be the height of water in the tank, and V (t) be its
volume. We are told that the radius r of the tank is 5 m, and that
dV
= 3. We are asked to find dh
.
dt
dt
Since the shape of the water inside the tank is itself a cylinder, we
have that its volume is given by V (t) = πr2 h(t) = 25πh(t), where we
plug in 5 for the radius of the water in the tank, as this is an unchanging
constant. Differentiating both sides with respect to t, we have
dV
dh
= 25π
dt
dt
Plugging in all the known information, we have:
dh
3 = 25π
dt
dh
3
So that dt = 25π m/min. This gives the result.