 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/7a/iii/solution

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

f (r) = k r^{3}, g (θ) = - 1 .

With

r (0) = 1

and

θ (0) = 0

r (t) = \frac{1}{1 - 2 k t}, θ (t) = - t,

again on the maximal interval for which the denominator is positive.

 New to topics? Read the docs here!