 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/40c/b/iii/solution

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

r = a

are

⟨ K ⟩ = \frac{ρ _{0}}{4} ∣ u_{r} ∣^{2} = \frac{ρ _{0} ω ^{2} ϵ ^{2}}{4},

⟨ W ⟩ = \frac{∣ p ∣ ^{2}}{4 ρ _{0} c _{0}^{2}} = \frac{ρ _{0} ω ^{2} ϵ ^{2}}{4} \frac{x ^{2}}{1 + x ^{2}} .

From part (ii),

∣ ⟨ I_{r} ⟩ ∣ = \frac{ρ _{0} ω ^{2} ϵ ^{2} c _{0}}{2} \frac{x ^{2}}{1 + x ^{2}} .

Consequently

\frac{c _{0} ⟨ K + W ⟩}{∣ ⟨ I _{r} ⟩ ∣} = \frac{1 + 2 x ^{2}}{2 x ^{2}} = 1 + \frac{c _{0}^{2}}{2 ω ^{2} a ^{2}} .

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