# 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