 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-332/2/b/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

ζ = 0

- 1 = - \frac{θ _{\infty}}{Ω} (e^{r Ω H} - 1) .

Therefore

\frac{Vh}{κ} = H = \frac{1}{r Ω} lo g (1 + \frac{Ω}{θ _{\infty}}), Ω = 1 + \frac{S}{r C} .

The thickness is proportional to thermal diffusivity divided by pulling speed. Latent-heat release modifies the exponential decay rate through

Ω

, while the far-liquid temperature fixes how far the profile can remain below the liquidus.

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