Question

find the exact length of the curve. use a graph to determine the parameter interval.

$r=cos ^{2}(\frac{\theta}{2})$

Explanation:

Step1: Find the derivative of \(r\) with respect to \(\theta\)

We have \(r = \cos^{2}(\frac{\theta}{2})\). Using the chain - rule, if \(u=\frac{\theta}{2}\), then \(r=\cos^{2}(u)\). The derivative of \(r\) with respect to \(u\) is \(2\cos(u)(-\sin(u))\), and the derivative of \(u\) with respect to \(\theta\) is \(\frac{1}{2}\). So \(\frac{dr}{d\theta}=2\cos(\frac{\theta}{2})(-\sin(\frac{\theta}{2}))\times\frac{1}{2}=-\cos(\frac{\theta}{2})\sin(\frac{\theta}{2})\)

Step2: Use the formula for the length of a polar curve \(L=\int_{\alpha}^{\beta}\sqrt{r^{2}+(\frac{dr}{d\theta})^{2}}d\theta\)

First, find \(r^{2}+(\frac{dr}{d\theta})^{2}\):

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Since \(r = \cos^{2}(\frac{\theta}{2})\geq0\) when \(-\pi\leq\theta\leq\pi\) (by graphing \(y = \cos^{2}(x/2)\), we know the curve is traced out once in the interval \([-\pi,\pi]\)). Then \(L=\int_{-\pi}^{\pi}\sqrt{\cos^{2}(\frac{\theta}{2})}d\theta=\int_{-\pi}^{\pi}\vert\cos(\frac{\theta}{2})\vert d\theta\)
Because \(\cos(\frac{\theta}{2})\geq0\) for \(-\pi\leq\theta\leq\pi\), so \(L = 2\int_{0}^{\pi}\cos(\frac{\theta}{2})d\theta\)

Step3: Evaluate the integral

Let \(u=\frac{\theta}{2}\), then \(d\theta = 2du\). When \(\theta = 0\), \(u = 0\); when \(\theta=\pi\), \(u=\frac{\pi}{2}\)

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Answer:

\(4\)