 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/16a/b/solution

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

Both

sin (x - t)

and

cos (x - t)

have zero average over one period, so

⟨ u ⟩ = 0

a = \frac{\partial u}{\partial t} + (u \cdot \nabla) u .

Direct differentiation yields

a_{x} = ε y sin θ + ε^{2} (1 - y^{2}) sin θ cos θ,

a_{y} = - ε cos θ + ε^{2} y cos^{2} θ .

Therefore the Eulerian time average at fixed

(x, y)

⟨ a ⟩ = (0, ε^{2} y /2) .

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