 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/9e/c/solution

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

For the natural inflation potential, write

x = ϕ / f

. Then

V^{'} = - \frac{V _{0}}{f} sin x

and the half-angle identity gives

\frac{V ^{'}}{V} = \frac{1}{f} \frac{s i n x}{1 + c o s x} = \frac{1}{f} tan \frac{x}{2} .

Therefore, on the stated rolling branch,

ϵ (ϕ) = \frac{M _{Pl}}{2 f} tan \frac{ϕ}{2 f} .

Substitution into part (b) gives

N = \frac{f}{M _{Pl}^{2}} \int_{ϕ_{f}}^{ϕ_{i}} cot \frac{ϕ}{2 f} d ϕ .

Since

\int cot \frac{ϕ}{2 f} d ϕ = 2 f lo g sin \frac{ϕ}{2 f},

N = \frac{2 f ^{2}}{M _{Pl}^{2}} [lo g (sin \frac{ϕ _{i}}{2 f}) - lo g (sin \frac{ϕ _{f}}{2 f})] .

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