20 C – Related Rates
In a related rates problem, we are given the rate of change of one quantity and we are asked to
find the rate of change of a related quantity. To do this, we find an equation that relates the two
quantities and use the Chain Rule to differentiate both sides of the equation with respect to time.
Many practical situations involve related rates of change. For example, suppose the side of a
square piece of metal increases at a rate of 0.1cm per second when it is heated. As a result, the
area of the square surface of the metal also increases – but at what rate?
Let the square piece of metal have the sides of x cm. Then, its area is given by A = x2.
We are given that
We want
𝑑𝐴
𝑑𝑡
𝑑𝑥
𝑑𝑡
, the rate of increase of the length of a side with respect to time, is 0.1 cm s-1.
, the rate of increase of the area with respect to time.
𝑑𝐴
Now A = x2, therefore 𝑑𝑥 = 2𝑥. We are given that
𝑑𝐴
𝑑𝑡
𝑑𝐴 𝑑𝑥
= 𝑑𝑥 𝑑𝑡 = 2𝑥 ∗ 0.1
𝑑𝐴
𝑑𝑡
𝑑𝑥
𝑑𝑡
= 0.1, so
= 0.2𝑥
The rate of increase of the area is 0.2x cm2s-1.
Example 1. A spherical snowball is melting in such a way that its volume is decreasing at a rate of
1cm3/min. At what rate is the radius decreasing when the radius is 5 cm?
4
V   r3
3
dV
dr
 1cm3 / min Find
when r  5.
dt
dt
dV dV dr
dr

 4 r 2
dt
dr dt
dt
dr
1 dV

dt 4 r 2 dt
Now we put r = 5 and
dV
 1 into the equation and we get
dt
dr
1
1

(1)  
2
dt 4 (5)
100
The radius of the snowball is decreasing at a rate of
1
 0.003 cm / min
100
Example 2. A water tank is built in the shape of a circular cone with height 5 m and diameter 6 m at
the top. Water is being pumped into the tank at a rate of 1.6 m 3/min. Find the rate at which the
water level is rising when the water is 2 m deep.
Let V be the volume of the water and let r and h be the radius of the surface and the height at time
t, where t is measured in minutes.
We are given the rate of increase of V, that is
𝑑𝑉
𝑑𝑡
= 1.6 𝑚3/min
and we are asked to find
𝑑ℎ
𝑑𝑡
when h = 2 m.
The quantities V and h are related by the equation
1
V   r 2h
3
but we have to express V as a function of h alone. To eliminate r we look for a relationship
between r and h. We use the similar triangles in the figure to write:
𝑟 3
=
ℎ 5
3
Thus r = ℎ and we have
5
1 3 
3 3
V    h h 
h
3 5 
25
2
Differentiating both sides with respect to t, we have
dV 3
dh 9 2 dh

3h 2 

h

dt
25
dt 25
dt
dh 25 1 dV

dt 9 h 2 dt
When h = 2 and
dV
 1.6 we have
dt
dh 25 1
10

(1.6) 
2
dt 9 2
9
The water level is rising at a rate of
10
 0.4 m / min .
9