 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-3/11a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

[F \times (\nabla \times G)]_{i} = ϵ_{ijk} F_{j} ϵ_{k ℓ m} \partial_{ℓ} G_{m} = F_{j} \partial_{i} G_{j} - F_{j} \partial_{j} G_{i} .

(F \cdot \nabla) G_{i} = F_{j} \partial_{j} G_{i}

cancels the second term. Interchanging

F, G

gives another pair whose sum is

G_{j} \partial_{i} F_{j}

. The total right-hand side is therefore

F_{j} \partial_{i} G_{j} + G_{j} \partial_{i} F_{j} = \partial_{i} (F_{j} G_{j}),

the

i

th component of the gradient of the dot product. Since this holds for each component, the dot-product gradient identity is proved:

\nabla (F \cdot G) = (F \cdot \nabla) G + (G \cdot \nabla) F + F \times (\nabla \times G) + G \times (\nabla \times F) .

It holds for any continuously differentiable vector fields; no irrotationality assumption is used.

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