Some example word problems, with detailed solutions to explain the possible applications of the compositions of functions, are presented.
Example 1A cylindrical container had 500 cm^{3} of water and is being filled at the constant rate of 100 cm^{3} 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 1a) Q = 500 + 100 tb) Q = π r^{2} × H which gives H = Q / (π r^{2}) c) (H _{o} Q)(t) = H(Q(t)) = (500 + 100 t) / (π r^{2}) This gives the height H(t) of water as a function of time t. d) H(t) = (500 + 100 t) / (π r^{2}) = 50 Solve for t: t = [ 50 × (π r^{2}) - 500 ] / 100 = 3922 seconds ≈ 1 hour.
Example 2A 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 2a) r = 2 × t = 2tb) (A _{o} r)(t) = A(r(t)) is the area as a function of time c) A = π r^{2} Hence: (A _{o} r)(t) = A(r(t)) = π (2t) ^{2} = 4πt^{2} When t = 60 seconds, A(60) = 4π60^{2} = 45289 cm^{2}
Example 3Starting 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 3a) r = 50 + 0.5tb) (A _{o} r)(t) = A(r(t)) = π r^{2}(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 m^{2}.
Example 4A 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 4a) The length of the rod changes with the temperature according to the composition of functions: L(t) = (L _{o} T)(t) = 100 + 10^{-4}t 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 cm^{3} 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 5The volume V of the balloon (supposed to be a sphere) of radius r is given by: V = (4/3) π r^{3}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 r^{2} Substitute dV/dt and dV/dr into dV/dt = dV/dr× dr/dt to obtain 100 = 4 π r^{2} dr/dt The rate of change (with time) of the radius, when r = 10 cm, is given by dr/dt = 2 / 4 π r^{2} = 100 / 4 π 10^{2} ≈ 0.08 cm/second
More References and linksComposition of Functions QuestionsQuestions on Composite Functions with Solutions. |