 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1/16c/a/ii/solution

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

x

δ x \times δy

- \frac{\partial p}{\partial x} δ x δy + o (δ x δy) .

τ_{x y} = μ u_{y}

, so the net viscous force is

μ u_{yy} δ x δy + o (δ x δy) .

There is no convective acceleration. Dividing the momentum balance by the area and taking the limit gives

ρ u_{t} = - \frac{d p}{d x} + μ u_{yy} .

The

y

p_{y} = 0

, so the pressure depends only on

x

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