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Worked solution

Solve and assess an exponential rod-expansion model

When a metal rod is heated, its length LL increases at a rate proportional to LL.

(i) Write a differential equation for dL/dtdL/dt in terms of LL.

(ii) Show that its general solution is L=AektL=Ae^{kt}.

(iii) The initial length is 5050 cm. Sketch LL against tt.

(iv) Explain why this differential equation does not appropriately describe the expansion of the heated rod.

[2006/RJC/Promo/11]

CJC · Discussion · Question 7

Answer

dL/dt=kLdL/dt=kL, L=AektL=Ae^{kt}; initially L=50ektL=50e^{kt}, whose unlimited growth exposes the model limitation.

(i) Translate proportionality into an equation

dL/dt=kLdL/dt=kL, with k>0k>0 because the length is increasing.

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(ii) Separate using the positive physical length

∫dL/L=∫k dt\int dL/L=\int k\,dt gives ln⁡L=kt+C\ln L=kt+C since L>0L>0. Thus L=eCekt=AektL=e^Ce^{kt}=Ae^{kt}, with A=eC>0A=e^C>0.

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(iii) Apply the initial length and sketch

L(0)=50L(0)=50 gives A=50A=50, so L=50ektL=50e^{kt} for t≥0t\ge0. The graph starts at (0,50)(0,50), is increasing and curves upwards.

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(iv) Explain the model limitation

For constant k>0k>0, the model predicts unbounded, ever-faster growth as tt increases. The printed solution states that expansion cannot be non-terminating; this unlimited exponential prediction does not appropriately describe a heated metal rod over indefinitely long times.

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Increasing, upward-curving exponential length graph L=50 exp(kt), k positive, for nonnegative time. It starts at length 50 cm and grows without bound. The horizontal axis is time t with no specified numerical scale.
Increasing, upward-curving exponential length graph L=50 exp(kt), k positive, for nonnegative time. It starts at length 50 cm and grows without bound. The horizontal axis is time t with no specified numerical scale.
Increasing, upward-curving exponential length graph L=50 exp(kt), k positive, for nonnegative time. It starts at length 50 cm and grows without bound. The horizontal axis is time t with no specified numerical scale.

Common mistakes

Solving a model correctly does not establish that its long-term physical prediction is appropriate.

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