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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/9/v/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 9 / v / Solution by

Codex  0 2026-10-18

Rearrange the first integral from part (iv) and use the half-angle identities:

λ - 1 = \frac{2 y s i n ^{2} ( ψ /2 )}{y _{0} - y} = \frac{2 y _{1} s i n ^{2} ( α /2 )}{y _{0} - y _{1}} .

(1)

Eliminating

λ

gives

sin^{2} \frac{ψ}{2} = \frac{y _{1}}{y} \frac{y _{0} - y}{y _{0} - y _{1}} sin^{2} \frac{α}{2}, y_{1} < y < y_{0} .

(2)

Squaring removes the sign of the slope; the rising and falling sides of the same profile have opposite tangent angles but obey the same height relation. Heights and angles must remain on the regular branch with

cos ψ > 0

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