 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-3/4a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Parametrize the smooth path by

γ : [0, 1] \to G

. The chain rule turns the line integral of the gradient into a one-variable derivative:

\int_{γ} F \cdot d x = \int_{0}^{1} \nabla ϕ (γ (t)) \cdot γ^{'} (t) d t = \int_{0}^{1} \frac{d}{d t} ϕ (γ (t)) d t = ϕ (b) - ϕ (a) .

For a twice continuously differentiable potential of a conservative vector field, the curl of its gradient vanishes because mixed partial derivatives commute:

(\nabla \times \nabla ϕ)_{i} = ϵ_{ijk} \partial_{j} \partial_{k} ϕ = 0.

A smooth gradient field is necessarily curl-free.

 New to topics? Read the docs here!