# Unused Cycles

## May 26, 2008

### Circumference of a Circle

Filed under: Mathematics — Kevin @ 2:06 am

So I was thinking about my previous area problem and the thought occurred to me that the same method could also be used to derive the circumference of the circle.

Using the same setup as before, the perimeter of the polygon is just $Nb$. So again if we let $N\rightarrow\infty$ we should get the circumference of a circle.

Recall that I found the formula for $b$ in terms of the number of sides and the radius: it was

$\displaystyle b=\frac{r\sin(2\pi/N)}{\sin(\pi/2-\pi/N)}$

The circumference of a circle is then

$\begin{array}{rcl}\displaystyle C&\displaystyle=&\displaystyle \lim_{N\rightarrow \infty}N\frac{r\sin(2\pi/N)}{\sin(\pi/2-\pi/N)}\vspace{0.3 cm}\\&\displaystyle=&\displaystyle \lim_{N\rightarrow \infty}\frac{r\sin(2\pi/N)}{N^{-1}\sin(\pi/2-\pi/N)}\vspace{0.3 cm}\\&\displaystyle=&\displaystyle \lim_{N\rightarrow\infty}\frac{r\cos(2\pi/N)(-2\pi N^{-2})}{-N^{-2}\sin(\pi/2-\pi/N)+\pi N^{-3}\cos(\pi/2-\pi/N)}\vspace{0.3 cm}\\&\displaystyle=&\displaystyle \lim_{N\rightarrow\infty}\frac{r\cos(2\pi/N)(-2\pi)}{-\sin(\pi/2-\pi/N)+\pi N^{-1}\cos(\pi/2-\pi/N)}\vspace{0.3 cm}\\&\displaystyle=&\displaystyle 2\pi r,\end{array}$

which is again exactly as expected.

And now with that out of the way, I can get some sleep. Good night!