 Solution

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

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

r = a (t)

and

r = b (t)

and subtract, thereby eliminating its time-dependent constant. For

A = a \overset{a}{˙}

and

ϕ = A lo g r + C (t)

\partial_{t} ϕ (a, t) - \partial_{t} ϕ (b, t) = \dot{A} lo g \frac{a}{b},

where each partial derivative is taken at fixed

r

. Since

u_{r} = A / r

and

p (b, t) = p_{\infty}

p (a, t) - p_{\infty} = ρ [\dot{A} lo g \frac{b}{a} - \frac{A ^{2}}{2} (\frac{1}{a ^{2}} - \frac{1}{b ^{2}})], A = a \overset{a}{˙} .

Equivalently,

\dot{A} = \overset{a}{˙}^{2} + a \overset{a}{¨}

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