 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/10b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

a_{k} x_{k} = a \cdot x

and differentiating gives

d_{ij} = - \frac{a _{i} x _{j}}{r ^{3}} + \frac{a _{j} x _{i}}{r ^{3}} + \frac{( a \cdot x ) δ _{ij}}{r ^{3}} - \frac{3 ( a \cdot x ) x _{i} x _{j}}{r ^{5}} .

d_{(ij)} = \frac{( a \cdot x ) δ _{ij}}{r ^{3}} - \frac{3 ( a \cdot x ) x _{i} x _{j}}{r ^{5}}, d_{[ij]} = \frac{a _{j} x _{i} - a _{i} x _{j}}{r ^{3}} .

Taking the trace makes the two symmetric terms cancel, so the divergence is

\partial_{i} u_{i} = 0 .

Contracting the antisymmetric part with the Levi-Civita symbol gives the curl

\nabla \times u = \frac{2 a \times x}{r ^{3}} .

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