Question

Show that the total arc length of the ellipse x=acost, y=bsint, 0<= t <= 2pi for a>b>0 is given by
4a times int_{pi/2}^{0} sqrt{1-k ^{2} times cos^2 t} dt
where k=sqrt(a^2 -b^2) /a

Answer 1

Arc length is given by
\[ ds = \sqrt{1 + \left ( {dy \over dx} \right ) ^2
}dx \\
\int ds = \int \sqrt{1 + \left ( {dy \over dx} \right ) ^2} dx = 4\int_0^a \sqrt{1 + \left ( {dy \over dx} \right ) ^2} dx \\
= 4 \int_0^{\pi/2} \sqrt{ 1 + {(b \cos \theta)^2 \over (a \sin \theta)^2}}(-b \sin \theta) d\theta \\ =
4 \int_{\pi/2}^0 \sqrt{ a^2 \sin^2 \theta + b ^2\cos^2 \theta }(b /a) d\theta
\]

Answer 2

\[ = 4 \int_{\pi/2}^0\sqrt{a^2 + (b^2 - a^2)\cos ^2 \theta } (b/a) d\theta \\
4b \int_{\pi/2}^0\sqrt{1 + {(b^2 - a^2) \over a^2}\cos ^2 \theta } d\theta= \]

Answer 3

looks like i didn't get it right!!

Answer 4

-b sin theta = a sin theta

Answer 5

it's elliptical integral of second kind
http://en.wikipedia.org/wiki/Elliptic_integral
i guess