Online Idea Repository

October 1, 2005

Just a question I helped my friend solve

Filed under: Maths

\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*}
\begin{eqnarray*}<br /> \text{Mean value } &=& \frac{1}{\frac{\pi}{2}-1-0}\int_0^\frac{\pi}{2}{y\cdot\frac{dx}{d\theta}\,d\theta}\\<br /> &=& \frac{2}{\pi-2}\int_0^\frac{\pi}{2}{\left(1-\cos\theta\right)^2\,d\theta}\\<br /> &=& \frac{2}{\pi-2}\int_0^\frac{\pi}{2}{\left(1-2\cos\theta+\cos^2\theta\right)\,d\theta}\\<br /> &=& \frac{2}{\pi-2}\int_0^\frac{\pi}{2}{\left(1-2\cos\theta+\frac{\cos2\theta+1}{2}\right)\,d\theta}\\<br /> &=& \frac{1}{\pi-2}\int_0^\frac{\pi}{2}{\left(2-4\cos\theta+\cos2\theta+1\right)\,d\theta}\\<br /> &=& \frac{1}{\pi-2}\int_0^\frac{\pi}{2}{\left(3-4\cos\theta+\cos2\theta\right)\,d\theta}\\<br /> &=& \frac{1}{\pi-2}\left[3\theta-4\sin\theta+\frac{\sin2\theta}{2}\right]_0^\frac{\pi}{2}\\<br /> &=& \frac{1}{\pi-2}\left[3\cdot\frac{\pi}{2}-4\right]\\<br /> &=& \frac{3\pi-8}{2\left(\pi-2\right)}<br /> \end{eqnarray*}

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