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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-2/5a/iv/solution

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Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 5A / iv / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

At time two, continuity of the second temperature gives

θ_{2} (2) = (T_{0} - T_{\infty}) e^{- 2 a} - \frac{1 - e ^{- a}}{a} .

(1)

The forcing then turns off, so Newton's law of cooling again gives exponential decay:

T_{2} (t) = T_{\infty} + (T_{0} - T_{\infty}) e^{- a t} - \frac{1 - e ^{- a}}{a} e^{- a (t - 2)}, t > 2.

(2)

Subtracting this from the first temperature and expressing both corrections with the same exponential factor gives

T_{1} (t) - T_{2} (t) = (\frac{e ^{a} - 1}{a} - 1) e^{- a (t - 1)}, t > 2.

(3)

Because

e^{a} > 1 + a

for

a > 0

, this difference is positive. Thus the cup cooled instantaneously is warmer after the addition interval: its cooling occurred earlier and has undergone more subsequent exponential attenuation.

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