 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/13/a/ii/solution

On the curved path,

r (t) = (t, t, t^{2})^{T}

and

r^{'} = (1, 1, 2 t)^{T}

. Direct substitution gives

F (r (t)) \cdot r^{'} (t) = 2 t + (3 a - 6) t^{2} + 4 t^{3},

G (r (t)) \cdot r^{'} (t) = 2 t cos (t^{2}) + 3 c t^{2} .

Thus

\int_{C_{2}} F \cdot d r = 1 + (a - 2) + 1 = a, \int_{C_{2}} G \cdot d r = sin 1 + c .

Both values agree with part (i), for every

a, b, c

. Agreement on these two selected paths alone does not establish a conservative vector field.

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