 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-354/2/b/ii/solution

L = z^{- 2} 1 + z^{'2}

has no explicit

x

dependence, so its Beltrami identity is conserved:

L - z^{'} \frac{\partial L}{\partial z ^{'}} = \frac{1}{z ^{2} 1 + z ^{'2}} = constant .

z = z_{m a x}

and

z^{'} = 0

, the constant is

z_{m a x}^{- 2}

. Rearrangement gives

1 + z^{'2} = \frac{z _{m a x}^{4}}{z ^{4}}, \frac{d z}{d x} = \pm \frac{z _{m a x}^{4}}{z ^{4}} - 1 .

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