 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-3/9a/a/solution

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

Since

r = (x_{j} x_{j})^{1/2}

, the chain rule gives

\frac{\partial r}{\partial x _{i}} = \frac{x _{i}}{r} .

For the radial vector field

g (r) x

\nabla \cdot (g (r) x) = \frac{\partial}{\partial x _{i}} (g (r) x_{i}) = g^{'} (r) \frac{x _{i}}{r} x_{i} + 3 g (r) = r g^{'} (r) + 3 g (r) .

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