 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-2/15d/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

P_{r} / c^{2} = ρ_{r} /3

, so

d lo g ρ_{r} = - 4 d lo g a

. Normalization at

a = 1

gives $ρ_{r} = ρ_{r 0} a^{- 4}$ .

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