# Applications of Composition of Functions in Mathematics

Some example word problems, with detailed solutions to explain the possible applications of the compositions of functions, are presented.

## Examples with Solutions

### Example 1

A cylindrical container had 500 cm3 of water and is being filled at the constant rate of 100 cm3 per second. The radius of the container is 50 cm.
a) Write a formula for the quantity Q of water in the container after t seconds.
b) Write a formula for the height H of water in the container in terms of Q.
c) Find an expression for the composition (H
o Q)(t) and its meaning.
b) How long does it take the height H of the water in the container to reach 50cm?

Solution to Example 1
a) Q = 500 + 100 t
b) Q = π r2 × H which gives H = Q / (π r2)
c) (H o Q)(t) = H(Q(t)) = (500 + 100 t) / (π r2)
This gives the height H(t) of water as a function of time t.
d) H(t) = (500 + 100 t) / (π r2) = 50
Solve for t: t = [ 50 × (π r2) - 500 ] / 100 = 3922 seconds ≈ 1 hour.

### Example 2

A small stone is thrown into still water and create a circular wave. The radius r of the water wave increases at the rate of 2 cm per second.
a) Find an expression for the radius r in terms of time t (in seconds) after the stone was thrown.
b) If A is the area of the water wave, what is the meaning of the composition (A
o r)(t)?
c) Find the area A of the water wave after 60 seconds.

Solution to Example 2
a) r = 2 × t = 2t
b) (A o r)(t) = A(r(t)) is the area as a function of time
c) A = π r2
Hence: (A o r)(t) = A(r(t)) = π (2t) 2 = 4πt2
When t = 60 seconds, A(60) = 4π602 = 45289 cm2

### Example 3

Starting from 50 meters, the radius r of a circular oil spill increases at the rate of 0.5 meters/second.
a) Express the radius r as a function of time.
b) The area A of a circular shape is given by A = π r
2. Find the composite function (A o r)(t) and explain its meaning.
c) How long will it take the area to be larger 10,000 m
2?

Solution to Example 3
a) r = 50 + 0.5t
b) (A o r)(t) = A(r(t)) = π r2(t) = π (50 + 0.5t)2
It is the area of the oil spill as a function of time
c) π (50 + 0.5t) > 10,000
Solve the above inequality to obtain
t > 79 seconds
After approximately 79 seconds, the area of the oil spill will be larger than 10,000 m2.

### Example 4

A metallic rod is being heated in a oven where the temperature T varies with the time t as follows: T = 0.2 t + 100 (T in degree Celsius and t in seconds). The length L of the rod varies with temperature and therefore with time according to the formula: L = 100 + 10-4t (L in cm). Find L as a function of the temperature T.

Solution to Example 4
a) The length of the rod changes with the temperature according to the composition of functions:
L(t) = (L o T)(t) = 100 + 10-4t
Since T = 0.2 t + 100, we can write t = (T - 100) / 0.2
We now substitute t by (T - 100) / 0.2 in L(t) to obtain
L = 100 + 10-4 (T - 100) / 0.2)
L = 100 + 5×10-4(T - 100)

### Example 5 ( calculus skills are needed )

Air escapes from a balloon at the constant rate of 100 cm3 per second. What is the rate of change of the radius of the balloon (supposed to be a sphere) when r = 10 cm?

Solution to Example 5
The volume V of the balloon (supposed to be a sphere) of radius r is given by: V = (4/3) π r3
V and r are related by the composition of functions
V(t) = (V o t)(t) = V(r(t))
Using the chain rule:
dV/dt = dV/dr× dr/dt
where dV/dt is the rate of change (with time) of the volume and dr/dt is the rate of change (with time) of the radius.
calculate derivative: dV/dr = 4 π r2
Substitute dV/dt and dV/dr into dV/dt = dV/dr× dr/dt to obtain
100 = 4 π r2 dr/dt
The rate of change (with time) of the radius, when r = 10 cm, is given by dr/dt = 2 / 4 π r2 = 100 / 4 π 102 ≈ 0.08 cm/second