 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2/section-b/6/c/solution

Parametrize the straight line by

x (t) = (t, t, t)

0 \leq t \leq 1

\int F \cdot d x = \int_{0}^{1} [(2 t^{2} + t) + (t^{2} + 2 t) + t] d t = \int_{0}^{1} (3 t^{2} + 4 t) d t = 3 .

The change in scalar potential is

ψ (1, 1, 1) - ψ (0, 0, 0) = - 3

, so the answer is

- Δ ψ

F = - \nabla ψ

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