 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/16/b/solution

Differentiating the parametrization gives

d x = a cos θ d θ

and

d y = - b sin θ d θ

. Thus the arc length element is

d s = a^{2} cos^{2} θ + b^{2} sin^{2} θ d θ = a 1 - e^{2} sin^{2} θ d θ .

All four quadrants give the same contribution. Consequently the ellipse perimeter is

L = a \int_{0}^{2 π} 1 - e^{2} sin^{2} θ d θ = 4 a E (e, π /2) .

e

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