![\begin{eqnarray*}<br />
x &=& \theta - \sin\theta\\<br />
y &=& 1 - \cos\theta\\<br />
\frac{dx}{d\theta} &=& 1 - \cos\theta\\<br />
\frac{dy}{d\theta} &=& \sin\theta\\<br />
\text{Arc length} &=& \int_0^{2\pi}{\left\{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2\right\}^\frac{1}{2}\,d\theta}\\<br />
&=& \int_0^{2\pi}{\left\{\left(1 - \cos\theta\right)^2 + \left(\sin\theta\right)^2\right\}^\frac{1}{2}\,d\theta}\\<br />
&=& \int_0^{2\pi}{\left\{1-2\cos\theta+\cos^2\theta+\sin^2\theta\right\}^\frac{1}{2}\,d\theta}\\<br />
&=& \int_0^{2\pi}{\left\{2(1-\cos\theta)\right\}^\frac{1}{2}\,d\theta}\\<br />
&=& \sqrt{2}\int_0^{2\pi}{\left\{1-\left(1-2\sin^2\frac{\theta}{2}\right)\right\}^\frac{1}{2}\,d\theta}\\<br />
&=& \sqrt{2}\int_0^{2\pi}{\left\{2\sin\frac{\theta}{2}\right\}^\frac{1}{2}\,d\theta}\\<br />
&=& 2\left[\frac{\cos\frac{\theta}{2}}{\frac{1}{2}}\right]_0^{2\pi}\\<br />
&=& 4\left|-1-1\right|\\<br />
&=& 8 \text{ units}<br />
\end{eqnarray*}](http://img346.imageshack.us/img346/5995/math7ty.gif)
October 20, 2005
October 1, 2005
Times
There seems to be a problem with the times. I can’t make them Singapore time (UTC +8 I think). So I’m changing the times manually. If your comments are of the wrong time, please post another comment with the correct time. Thanks. Sorry for any inconveniences caused.
Please post all comments with a timestamp. If it does not reflect Singapore time, please amend. Thanks.


