Some example word problems, with detailed step-by-step solutions explaining the applications of the composition of functions, are presented. Each solution is hidden inside a collapsible dropdown so you can practice independently before reviewing the explanations.
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) The quantity \( Q \) starts at \( 500 \, \text{cm}^3 \) and increases at \( 100 \, \text{cm}^3 \) per second:
\[ Q = 500 + 100t \]b) The volume of a cylinder is given by \( Q = \pi r^2 H \). Solving for height \( H \):
\[ H = \dfrac{Q}{\pi r^2} \]c) Substituting \( Q(t) \) into \( H(Q) \):
\[ (H \circ Q)(t) = H(Q(t)) = \dfrac{500 + 100t}{\pi r^2} \]This gives the height \( H(t) \) of water as a function of time \( t \).
d) Setting \( H(t) = 50 \) with \( r = 50 \):
\[ \dfrac{500 + 100t}{\pi (50)^2} = 50 \]Solving for \( t \):
\[ t = \dfrac{50 \times (\pi \times 50^2) - 500}{100} = 3922 \text{ seconds} \approx 1 \text{ hour} \]A small stone is thrown into still water and creates a circular wave. The radius \( r \) of the water wave increases at the rate of \( 2 \, \text{cm} \) per second.
a) Since the radius increases at \( 2 \, \text{cm} \) per second:
\[ r = 2t \]b) The composition \( (A \circ r)(t) = A(r(t)) \) represents the area of the circular wave as a function of time.
c) The area of a circle is \( A = \pi r^2 \). Substituting \( r = 2t \):
\[ (A \circ r)(t) = A(r(t)) = \pi (2t)^2 = 4\pi t^2 \]When \( t = 60 \) seconds (corrected calculation):
\[ A(60) = 4\pi (60)^2 = 14400\pi \approx 45239 \, \text{cm}^2 \]Starting from an initial radius of \( 50 \) meters, the radius \( r \) of a circular oil spill increases at the rate of \( 0.5 \) meters/second.
a) With an initial radius of \( 50 \) and rate \( 0.5 \) m/s:
\[ r = 50 + 0.5t \]b) Substituting \( r(t) \) into the area formula:
\[ (A \circ r)(t) = A(r(t)) = \pi (50 + 0.5t)^2 \]This represents the area of the oil spill as a function of time.
c) Setting up the corrected inequality for an area greater than \( 10,000 \, \text{m}^2 \):
\[ \pi (50 + 0.5t)^2 > 10000 \]Solving for \( t \):
\[ (50 + 0.5t)^2 > \dfrac{10000}{\pi} \approx 3183.10 \] \[ 50 + 0.5t > \sqrt{3183.10} \approx 56.42 \] \[ 0.5t > 6.42 \implies t > 12.8 \text{ seconds} \]After approximately 12.8 seconds, the area of the oil spill will exceed \( 10,000 \, \text{m}^2 \).
A metallic rod is being heated in an oven where the temperature \( T \) varies with time \( t \) as follows: \( T = 0.2t + 100 \) (\( T \) in degrees 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 \( T \).
From the temperature equation \( T = 0.2t + 100 \), solve for \( t \) in terms of \( T \):
\[ t = \dfrac{T - 100}{0.2} = 5(T - 100) \]Substitute this expression for \( t \) into the length formula \( L = 100 + 10^{-4}t \):
\[ L = 100 + 10^{-4}(5(T - 100)) \] \[ L = 100 + 5 \times 10^{-4}(T - 100) \]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 (assumed to be a sphere) when \( r = 10 \, \text{cm} \)?
The volume \( V \) of a sphere of radius \( r \) is given by:
\[ V = \dfrac{4}{3}\pi r^3 \]Using the chain rule with respect to time \( t \):
\[ \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt} \]Differentiating volume with respect to radius:
\[ \dfrac{dV}{dr} = 4\pi r^2 \]Substitute the given rate of change of volume (\( \dfrac{dV}{dt} = -100 \, \text{cm}^3/\text{s} \)) and \( \dfrac{dV}{dr} \):
\[ 100 = 4\pi r^2 \dfrac{dr}{dt} \]Solving for \( \dfrac{dr}{dt} \) when \( r = 10 \, \text{cm} \):
\[ \dfrac{dr}{dt} = \dfrac{100}{4\pi (10)^2} = \dfrac{100}{400\pi} = \dfrac{1}{4\pi} \approx 0.08 \, \text{cm/second} \]