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

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.

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, \( A(60) = 4\pi 60^2 = 45289 \) cm^{2}

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 \( 10,000 \text{m}^2 \).

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

\( L = 100 + 10^{-4} \left(\dfrac{T - 100}{0.2}\right) \)

\( L = 100 + 5 \times 10^{-4}(T - 100) \)

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

Composition of Functions Questions

Questions on Composite Functions with Solutions.