show arc lengths are equal

[galactus]

Here's an interesting problem I seen somewhere.

"Show that the arc length of \(\displaystyle sin(x)\) over the interval \(\displaystyle [0,2\pi]\) is equal to the circumference of the ellipse \(\displaystyle x^{2}+2y^{2}=2\)".

It's interesting because they are equal.

The arc length formula for sin(x) works out to be \(\displaystyle \L\\\int_{0}^{2\pi}\sqrt{1+cos^{2}(x)}dx\)

The arc length integral for the circumference of the ellipse is:

\(\displaystyle \L\\2\int_{-\sqrt{2}}^{\sqrt{2}}\sqrt{\frac{1}{2}-\frac{1}{x^{2}-2}}dx\)

The thing is to show they're equal.

These two arc lengths integrals are not easily done by elementary means, so their is an easier method to show they're equal.

 

[pka]

Write the ellipse in parametric form: \(\displaystyle \L x(t) = \sqrt 2 \cos (t),\quad y(t) = \sin (t),\quad 0 \le t \le 2\pi\).

The circumference of the ellipse is:
\(\displaystyle \L
\begin{array}{rcl}
\int\limits_0^{2\pi } {\sqrt {\left[ {x'(t)} \right]^2 + \left[ {y'(t)} \right]^2 } } dt & = & \int\limits_0^{2\pi } {\sqrt {2\sin ^2 (t) + \cos ^2 (t)} dt} \\
& = & \int\limits_0^{2\pi } {\sqrt {1 + \sin ^2 (t)} dt} \\
& = & \int\limits_0^{2\pi } {\sqrt {1 + \cos ^2 (t)} dt} \\
\end{array}\).

 

[galactus]

I knew you'd get it, pka. Believe it or not, that's how I approached it.

Though, admittedy, I scratched my head for a while before it dawned on me.