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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 15A / iii / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2024 ib Paper 4 15A iii
Created 2026-09-24 Updated 2026-09-25  0 By others on same topic  0 Discussions Create my own version
Since A′/A=−iℏB=−2i​tan(ϕ−it) and
dtd​logcos(ϕ−it)=itan(ϕ−it),
(1)
integration gives
A(t)=A0​[cos(ϕ−it)]−1/2​,
(2)
with the branch chosen continuously from the initial value. The constant A0​ fixes normalization.

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