Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 4 / c

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- Solution c

Solution

The fundamental theorem of calculus and part (a) give

I_{θ} = E (θ, k) 1 - k^{2} sin^{2} θ = E (θ, k) E_{θ} (θ, k) = \partial_{θ} (\frac{1}{2} E (θ, k)^{2}) .

The difference is constant in

θ

. Since

I (0, k) = E (0, k) = 0

, the constant is zero, proving

I (θ, k) = \frac{1}{2} E (θ, k)^{2} .

This is simply the chain rule integrated, with

E

itself as the antiderivative of the square-root factor.

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