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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 6A / b / Solution

The stationary equations are

4 x^{3} - 6 x + 2 y = 0, 2 x + 2 y = 0.

Thus

y = - x

and

4 x (x^{2} - 2) = 0

, giving

(0, 0), (2, - 2), (- 2, 2) .

The Hessian matrix is

H = (12 x^{2} - 6 2 22) .

At either nonzero stationary point,

H = (18222), det H = 32 > 0,

so both are strict local minima.

Along a line through the origin,

g (t) = f (t cos γ, t sin γ) = t^{2} q (γ) + t^{4} cos^{4} γ,

where

q (γ) = - 3 cos^{2} γ + 2 sin γ cos γ + sin^{2} γ .

When

cos γ \neq = 0

and

r = tan γ

,

q (γ) = cos^{2} γ (r - 1) (r + 3) .

Consequently the origin is a local minimum along the line when

tan γ \leq - 3 or tan γ \geq 1,

including the equality cases because the positive quartic term then leads. It is also a minimum on the vertical line. It is a local maximum along the line when

- 3 < tan γ < 1 .

In the first case the graph is locally bowl-shaped. In the second it initially bends downward from the origin, but the positive quartic term eventually turns it upward, producing the usual double-well profile.

Solved by gpt-5.6-sol high.

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