 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/32b/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

ε = 0

H (x, y) = 2 x^{2} + \frac{1}{2} y^{2}

is conserved, and its orbit of amplitude

a > 0

may be written

x = a cos θ, y = - 2 a sin θ, θ = 2 t .

The energy change over one period, to first order in

ε

, is

Δ H = ε \int_{0}^{π} [k (1 - 2 x^{2}) y^{2} - (1 - 3 x^{2}) y^{4}] d t + O (ε^{2}) = ε π a^{2} [2 k - (k + 6) a^{2} + 3 a^{4}] + O (ε^{2}) = 3 ε π a^{2} (a^{2} - 2) (a^{2} - k /3) + O (ε^{2}) .

Here the stated trigonometric integrals, together with

\int_{0}^{2 π} cos^{2} θ sin^{4} θ d θ = π /8

, were used. The two candidate leading-order amplitudes are therefore

a = 2, a = k /3 .

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