Applications of Composition of Functions in Mathematics

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

Examples with Solutions

Example 1

A cylindrical container had \( 500 \, \text{cm}^3 \) of water and is being filled at the constant rate of \( 100 \, \text{cm}^3 \) per second. The radius of the container is \( 50 \, \text{cm} \).
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 \( (H \circ Q)(t) \) and its meaning.
b) How long does it take the height \( H \) of the water in the container to reach \( 50 \text{cm} \)?

Solution to Example 1
a) \( Q = 500 + 100 t \)
b) \( Q = \pi r^2 \times H \) which gives \( H = \dfrac{Q}{\pi r^2} \)
c) \( (H \circ Q)(t) = H(Q(t)) = \dfrac{500 + 100 t}{\pi r^2} \)
This gives the height H(t) of water as a function of time \( t \).
d) \( H(t) = \dfrac{500 + 100 t}{\pi r^2} = 50 \)
Solve for t: \( t = \dfrac{50 \times (\pi r^2) - 500}{100} = 3922 \) 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 \( 2 \text{cm} \) 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 \( (A \circ r)(t) \)?
c) Find the area \( A\) of the water wave after 60 seconds.

Solution to Example 2
a) \( r = 2t \)
b) \( (A \circ r)(t) = A(r(t)) \) is the area as a function of time
c) \( A = \pi r^2 \)
Hence: \( (A \circ r)(t) = A(r(t)) = \pi (2t)^2 = 4\pi t^2 \)
When \( t = 60 \) seconds, \( A(60) = 4\pi 60^2 = 45289 \) 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 \( A = \pi r^2 \). Find the composite function \( (A \circ r)(t) \) and explain its meaning.
c) How long will it take the area to be larger \( 10,000 \text{m}^2 \)?

Solution to Example 3
a) \( r = 50 + 0.5t \)
b) \( (A \circ r)(t) = A(r(t)) = \pi r^2(t) = \pi (50 + 0.5t)^2 \)
It is the area of the oil spill as a function of time
c) \( \pi (50 + 0.5t) > 10,000 \)
Solve the above inequality to obtain
\( t > 79 \) seconds
After approximately 79 seconds, the area of the oil spill will be larger than \( 10,000 \text{m}^2 \).

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: \( L = 100 + 10^{-4}t \) (\( L \) 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:
\( L(t) = (L \circ T)(t) = 100 + 10^{-4}t \)
Since \( T = 0.2 t + 100 \), we can write \( t = \dfrac{T - 100}{0.2} \)
We now substitute t by \( \dfrac{T - 100 }{0.2} \) in \( L(t) \) to obtain
\( L = 100 + 10^{-4} \left(\dfrac{T - 100}{0.2}\right) \)
\( L = 100 + 5 \times 10^{-4}(T - 100) \)

Example 5 ( calculus skills are needed )

Air escapes from a balloon at the constant rate of \( 100 \; \text{cm}^3 \) per second. What is the rate of change of the radius of the balloon (supposed to be a sphere) when \( r = 10 \; \text{cm}\)?

Solution to Example 5
The volume \( V \) of the balloon (supposed to be a sphere) of radius \( r \) is given by: \( V = \dfrac{4}{3} \pi r^3 \)
\( V \) and \( r \) are related by the composition of functions
\( V(t) = (V \circ t)(t) = V(r(t)) \)
Using the chain rule:
\( \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt} \)
where \( \dfrac{dV}{dt} \) is the rate of change (with time) of the volume and \( \dfrac{dr}{dt} \) is the rate of change (with time) of the radius.
calculate derivative: \( \dfrac{dV}{dr} = 4 \pi r^2 \)
Substitute \( \dfrac{dV}{dt} \) and \( \dfrac{dV}{dr} \) into \( \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt} \) to obtain
\( 100 = 4 \pi r^2 \dfrac{dr}{dt} \)
The rate of change (with time) of the radius, when \( r = 10 \text{cm} \), is given by \( \dfrac{dr}{dt} = \dfrac{2}{4 \pi r^2} = \dfrac{100}{4 \pi 10^2} \approx 0.08 \) cm/second