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3.11 Perimeter of the ellipse E

Suppose $a\geq b$ , and let $\kappa=\sqrt{1-\frac{b^{2}}{a^{2}}}$ . Then the perimeter of the ellipse is

a\int_{0}^{2\pi}\sqrt{1-\kappa^{2}\sin^{2}t}\,dt=4a\int_{0}^{\frac{\pi}{2}}% \sqrt{1-\kappa^{2}\sin^{2}t}\,dt.

Proof. We parametrize the ellipse by setting $x=a\sin t$ , $y=b\cos t$ . Then $\frac{dx}{dt}=a\cos t$ and $\frac{dy}{dt}=-b\sin t$ , hence

\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}={a^{2}\cos^{2}t+% b^{2}\sin^{2}t=}\,{a^{2}+(b^{2}-a^{2})\sin^{2}t}

={a^{2}\left(1+\left(\frac{b^{2}}{a^{2}}-1\right)\sin^{2}t\right)=}\,{a^{2}(1-% \kappa^{2}\sin^{2}t).}

Now we apply the arc length formula from 3.5.