Step in the finding of curvature of ellipse as arc length parameter.

Question

I need help on following integral. $$S =\int_0^x\sqrt{4\sin^2t + 9\cos^2t}\, dt$$

Actually i am finding the curvature of ellipse as arc length parameter and i stuck on this step. I don't know how to solve this integral. Can someone help me on this. Please

Answer 1

It is the elliptic integral of second kind: $$S=3 E(x,\sqrt{5}/3).$$

Answer 2

You dont need to compute the arc length of a curve to compute its curvature.

Frenet tells you that ${1\over R}$ is the length of ${ d \vec T(t)\over ds}= ({ ds \over dt }) ^{-1}.{ d \vec T(t)\over dt} = ({ ds \over dt }) ^{-1} ( { d ^2 \vec M\over {dt} ^2} { d t\over ds} + g(t) \vec T(t))$. For a certain function $g(t)$ that is not very important.

To compute the curvature, we take the scalar product with the normal vector, or the determinant with $\vec T(t)$ (or we read wikipedia on this subject)

Then $R =$ $ \det ({d \vec M(t)\over dt}, {d ^2 \vec M(t)\over dt ^2}) \over{ {d s ^2\over {dt}^2} ^{3\over 2} } $

Here, $\vec M(t)=(2 \cos t, 3 \sin t ), {d \vec M(t)\over dt}= (-2 \sin t, 3 \cos t), M"(t)=(-2 \cos t, -3 \sin t)$, $R={12\over (4 \sin ^2 t + 9 \cos ^2 t)^{3\over 2} }$