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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/2020/paper-1/1/a/iv/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 / 2020 / Paper 1 / 1 / a / iv / Solution by

Codex  0 2026-10-18

For a finite separation use the Cartesian distance, rather than substituting finite coordinate differences into the local line element. Expanding the two horizontal components gives

(ρ cos ϕ - ρ^{'} cos ϕ^{'})^{2} + (ρ sin ϕ - ρ^{'} sin ϕ^{'})^{2} = ρ^{2} + ρ^{'2} - 2 ρ ρ^{'} cos (ϕ - ϕ^{'}) .

(1)

The vertical contribution is

(z - z^{'})^{2}

, independently. Thus

h = ρ^{2} + ρ^{'2} - 2 ρ ρ^{'} cos (ϕ - ϕ^{'}), j = (z - z^{'})^{2},

(2)

and

R = ρ^{2} + ρ^{'2} - 2 ρ ρ^{'} cos (ϕ - ϕ^{'}) + (z - z^{'})^{2} .

(3)

Equivalently, the horizontal contribution is

(ρ - ρ^{'})^{2} + 4 ρ ρ^{'} sin^{2} [(ϕ - ϕ^{'}) /2]

, which also makes its nonnegativity explicit.

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