 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/16/a/solution

\frac{d F}{d t} = (3 x^{2} yz + y) (- 3 sin t) + (x^{3} z + x) (3 cos t) + (x^{3} y + 1) 2.

Substituting the three parametrizing functions also writes the complete derivative as

\frac{d F}{d t} = 162 cos^{3} t sin t + 162 t (cos^{4} t - 3 cos^{2} t sin^{2} t) + 9 cos 2 t + 2.

t = π /2

, the point is

(0, 3, π)

, its velocity is

(- 3, 0, 2)

, and

\nabla F = (3, 0, 1)

. Hence

\frac{d F}{d t}_{t = π /2} = \nabla F \cdot r^{'} = - 9 + 2 = - 7 .

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