 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/17c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

\frac{d r}{r d θ} = \frac{u _{r}}{u _{θ}} = - cot (γ θ) .

Integration gives

lo g r = - \frac{1}{γ} lo g sin (γ θ) + C .

The radius is smallest on the wedge bisector

θ = α /2

, where

sin (γ θ) = 1

. Labelling that radius by

r_{m i n}

therefore gives

r (θ) = r_{m i n} (sin \frac{π θ}{α})^{- α / π} .

Every streamline arrives from infinity along one wall, reaches its closest point on the bisector, and returns to infinity along the other wall:

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