I've been staring at this problem for a while, and I would really appreciate any insight into finding the solution. The problem in its original state is as follows:
Show that the arc length of the graph of $y = \sin x$ on the interval $[0, 2\pi]$ is equal to the circumference of the ellipse $x^2 + 2y^2 = 2$.
I have gotten far enough to see that the arc length of $sinx$ on the given interval is $4\sqrt2E\left(\frac{1}{2}\right)$ which, according to WolframAlpha "is the complete elliptic integral of the second kind with parameter $m=k^2$."
My question is first of all, where do I go from here? Will I have to use elliptic integrals of any kind in order to arrive at a coherent solution? And how does arc length relate to elliptic circumference?
Thank you!
Notes: This is for an Honors Calculus 2 homework assignment.
I'm not sure if this is within the scope of 'Calculus 2':
The length of a smooth curve $\gamma:[0,2 \pi] \to \mathbb{R}^2$ is give by $l(\gamma) = \int_0^{2 \pi} \|\gamma'(t)\| dt$.
The first curve is $\gamma_1(t) = (t, \sin t)$, the second can be written as $\gamma_2(t) = (\sqrt{2} \sin t, \cos t)$ (I choose this parameterisation, as opposed to $t \mapsto (\sqrt{2} \cos t, \sin t)$, since the equality is immediate).
You will find that $\|\gamma_1'(t) \| = \|\gamma_2'(t) \| $, hence the lengths are equal.