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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/13/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 13 / a / Solution by

Codex  0 2026-10-18

For

G

, the product rule for the curl gives

\nabla \times G = \nabla (a \cdot r) \times r + (a \cdot r) (\nabla \times r) = a \times r .

Since

a \neq = 0

, this is not identically zero on

R^{3}

. A gradient field has zero curl, so no scalar potential exists for $G$ on the whole space.

For the radial vector field

K

, observe that

\nabla [\frac{1}{2} lo g (1 + x^{2} + y^{2} + z^{2})] = \frac{( x , y , z ) ^{T}}{1 + x ^{2} + y ^{2} + z ^{2}} = K .

Thus all vector-field potentials are

ϕ (r) = \frac{1}{2} lo g (1 + ∣ r ∣^{2}) + C .

These are smooth everywhere, including the origin. Any two differ by a constant because their gradient difference vanishes on the connected whole space.

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