 Solution

ID: past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2021/paper-ne1/2/f/i/solution

Differentiating the supplied relation gives

\frac{d ρ}{d ω} = Λ + \frac{β λ}{( ω + λ ) ^{2}}, \frac{d ρ}{d ω}_{0} = Λ + \frac{β}{λ} .

ρ ≃ ω (Λ + β / λ)

and

T_{tangent} = \frac{Λ + β / λ}{ρ} = 70.5 s .

It is about

16.4%

longer than

60.58 s

and therefore predicts slower growth. For this positive step it is not conservative as a growth-rate estimate. The local expansion needs

∣ ω ∣ ≪ λ

; here the actual ratio is approximately

0.165

, so its error is appreciable.

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