 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/15b/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Put

u = tanh (λ t) = e^{ϕ / ϕ_{0}}

. The hyperbolic cotangent double-angle identity gives

coth (2 λ t) = \frac{1 + u ^{2}}{2 u} = cosh (\frac{ϕ}{ϕ _{0}}),

H (ϕ) = λ \frac{ϕ _{0}^{2}}{M _{Pl}^{2}} cosh (\frac{ϕ}{ϕ _{0}}) .

Also

csch^{2} (2 λ t) = cosh^{2} (\frac{ϕ}{ϕ _{0}}) - 1.

c^{2} V = 3 M_{Pl}^{2} H^{2} - \frac{1}{2} \dot{ϕ}^{2},

therefore gives

V (ϕ) = \frac{2 λ ^{2} ϕ _{0}^{2}}{c ^{2}} [(\frac{3 ϕ _{0}^{2}}{2 M _{Pl}^{2}} - 1) cosh^{2} (\frac{ϕ}{ϕ _{0}}) + 1] .

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