 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/1/i/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Away from the origin,

\hat{r} = \nabla r

. Hence the curl of a gradient gives

\nabla \times \hat{r} = 0

. Its divergence is the Laplacian of the radial coordinate, which in spherical coordinates is

\nabla \cdot \hat{r} = \nabla^{2} r = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2}) = \frac{2}{r}, r > 0.

Equivalently, differentiating

x_{i} / r

3/ r - r^{2} / r^{3} = 2/ r

 New to topics? Read the docs here!