 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/13d/b/solution

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

For

L = \frac{1 + y ^{'2}}{x},

one has

L_{y} = 0

and

L_{y^{'}} = \frac{y ^{'}}{x 1 + y ^{'2}} .

R

centred at

(0, c)

on the

y

-axis satisfies

x^{2} + (y - c)^{2} = R^{2}, y^{'} = - \frac{x}{y - c} .

On any arc that does not cross

y = c

1 + y^{'2} = \frac{R}{∣ y - c ∣}, L_{y^{'}} = - \frac{sgn ( y - c )}{R},

which is constant. Hence

d L_{y^{'}} / d x = 0 = L_{y}

δ I = 0

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