 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/12/b/solution

D = 1 + x^{2} + y^{2} > 0

f_{x} = - \frac{4 x ( 1 + y + y ^{2} )}{D ^{2}}, f_{y} = \frac{2 ( 1 + x ^{2} - y ^{2} + 2 x ^{2} y )}{D ^{2}} .

Since

1 + y + y^{2} = (y + 1/2)^{2} + 3/4 > 0

f_{x} = 0

x = 0

. The second then reduces to

1 - y^{2} = 0

. Thus the only stationary points and their field values are

(x, y) = (0, 1), f = 1; (x, y) = (0, - 1), f = - 1.

No division by a potentially vanishing factor of

y

was used, so no stationary points have been discarded.

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