 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/13/c/solution

Parametrize the diagonal by

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

for

0 \leq t \leq 1

. Then

F \cdot x^{'} = 2 t^{2} cosh t + t^{3} sinh t, G \cdot x^{'} = 3 t^{2} cosh t + t^{3} sinh t .

The second expression is

(t^{3} cosh t)^{'}

, immediately giving

\int G \cdot d x = cosh 1

. The first differs from it by

t^{2} cosh t

. Two applications of integration by parts give

\int_{0}^{1} t^{2} cosh t d t = [t^{2} sinh t - 2 t cosh t + 2 sinh t]_{0}^{1} = 3 sinh 1 - 2 cosh 1.

Hence

\int_{C} F \cdot d x = 3 (cosh 1 - sinh 1) = \frac{3}{e}, \int_{C} G \cdot d x = cosh 1 .

The changed value for

F

explicitly demonstrates path dependence, while

G

gives the same endpoint-dependent result.

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