 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/16c/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

b_{0}

be the outer radius when

a = a_{0}

. The area relation from part (i) gives

b^{2} = b_{0}^{2} + a^{2} - a_{0}^{2} = b_{0}^{2} + 2 a_{0}^{2} ϵ + O (ϵ^{2}),

b = b_{0} (1 + \frac{a _{0}^{2}}{b _{0}^{2}} ϵ) + O (ϵ^{2}) .

Moreover,

A = a \overset{a}{˙} = a_{0}^{2} \overset{ϵ}{˙} + O (ϵ^{2}), \dot{A} = a_{0}^{2} \overset{ϵ}{¨} + O (ϵ^{2}) .

The term

A^{2}

in the pressure is second order, while the logarithm multiplying

\dot{A}

may be evaluated at equilibrium. The linearization is therefore

p (a, t) - p_{\infty} = ρ a_{0}^{2} lo g \frac{b _{0}}{a _{0}} \overset{ϵ}{¨} + O (ϵ^{2}) .

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