Arkansas Tech University
MATH 2914: Calculus I
Dr. Marcel B. Finan
2.7
Related Rates Problems
One of the applications of mathematical modeling with calculus involves the
use of implicit differentiation. Recall that the derivative of a function is a
rate of change or simply a rate. In many of the mathematical modeling
one encounters an equation involving two or more dependent variables that
depend on only one independent variable which we will assume it is time
t. Applying the implicit differentiation process we obtain a relationship between rates. Any mathematical problem that leads to such a relationship is
called a related rates problem.
Example 2.7.1
Air is being pumped into a spherical balloon at a rate of 5 cm3 /min. Determine the rate at which the radius of the balloon is increasing when the
diameter of the balloon is 20 cm.
Solution.
Let V (t) be the volume of the balloon and r(t) be its radius at time t. Then,
V (t) = 43 πr3 . Differentiating we find
dV
4
dr
= π(3r2 ) .
dt
3
dt
This equation relates the rate of change of the volume to the rate of change
of the radius. We are given that dV
dt (t) = 5 and r(t) = 10. Thus,
4
dr
5 = π(3)(102 ) .
3
dt
Solving this equation we find
dr
dt
=
1
80π
cm/min
Example 2.7.2
A tank of water in the shape of a cone is leaking water at a constant rate
of 2 ft3 /hour. The base radius of the tank is 5 ft and the height of the tank
is 14 ft.
(a) At what rate is the depth of the water in the tank changing when the
depth of the water is 6 ft?
(b) At what rate is the radius of the top of the water in the tank changing
when the depth of the water is 6 ft?
1
Solution.
Let r(t) be the radius of the water at time t and h(t) the height of the water
in the tank at time t as shown in Figure 2.7.1.
Figure 2.7.1
The volume of the water in the tank at time t is given by
1
V (t) = πr2 h.
3
(a) We are asked to find
can write
dh
dt
given that h = 6 ft. Using similar triangles we
r
5
=
h
14
5
25
3
so that r = 14 h. Hence, V = 588 πh . Differentiating this equation we find
25
dh
dV
=
(3h2 π) .
dt
588
dt
Substituting into this equation we find
−2 =
25
dh
(3)(62 )π ×
.
588
dt
Hence,
2 × 588
dh
=−
≈ −0.1386 ft/hour.
dt
25 × 3 × 62 × π
(b) Since r =
5
14 h,
we can differentiate to obtain
5 dh
dr
=
.
dt
14 dt
Hence,
dr
5 dh
5
=
= (−0.1386) ≈ −0.04951 ft/hour
dt
14 dt
14
2
Example 2.7.3
A 15 foot ladder is resting against the wall. The bottom is initially 10 feet
away from the wall and is being pushed towards the wall at a rate of 0.25
ft/sec. How fast is the top of the ladder moving up the wall 12 seconds after
we start pushing?
Solution.
Consider Figure 2.7.2.
Figure 2.7.2
Using the Pythagorean formula we find
x2 + y 2 = (15)2 = 225.
Differentiating we find
2x
or
x
dx
dy
+ 2y
=0
dt
dt
dx
dy
+y
= 0.
dt
dt
dy
We are given dx
dt = −0.25 and we are asked to find dt (12). Initially the
bottom of the ladder was at 10 ft from the wall. After 12 seconds the
bottom moved a distance of 0.25 × 12 = 3 ft√so that x(12)
√= 7 ft. Using the
2
Pythagorean formula we find that y(12) = 225 − 7 = 176. Thus,
7(−0.25) +
√
176
dy
= 0.
dt
Solving this equation we find
dy
≈ 0.1319 ft/sec
dt
Example 2.7.4
Two people are 50 feet apart. One of them starts walking north at a rate so
that the angle shown in Figure 2.7.3 is changing at a constant rate of 0.01
3
rad/min. At what rate is distance between the two people changing when
θ = 0.5 radians?
Figure 2.7.3
Solution.
We are asked to find
dx
dt (0.5).
From the figure we have
sec θ =
x
.
50
Differentiating we obtain
sec θ tan θ
dθ
1 dx
=
.
dt
50 dt
Thus,
dx
= 50(0.01) sec (0.5) tan (0.5) ≈ 0.3113 ft/min
dt
4