# Find arc length of r =

Find arc length of $r=2\mathrm{cos}\theta$ in the range $0\le \theta \le \pi$ ?
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Explanation:
$L=\int ds={\int }_{{\theta }_{1}}^{{\theta }_{2}}rd\theta$
$s=r\theta$ so $ds=rd\theta$ , thus
$2{\int }_{0}^{\pi }\mathrm{cos}\theta =2\left(\mathrm{sin}\theta \right){|}_{0}^{\pi }=2\left(\mathrm{sin}\pi -\mathrm{sin}0\right)=...$
Considering that this is for a semicircle perimeter of a full cycle..
$4\left(\mathrm{sin}\right]theta\right){|}_{0}^{\frac{\pi }{2}}=4\left(\mathrm{sin}\left(\frac{\pi }{2}\right)-\mathrm{sin}0\right)=4\left(1-0\right)=4$

Audrina Jackson
It is easy to see that the curve is a circle of radius 1. It's length is obviously $2\pi$
A more analytic solution would go as follows
$d{s}^{2}=d{r}^{2}+{r}^{2}d{\theta }^{2}$
So, for $r=2\mathrm{cos}\theta$ , we have
$dr=-2\mathrm{sin}\theta d\theta$
and hence
$d{s}^{2}=\left(-2\mathrm{sin}\theta d\theta {\right)}^{2}+\left(2\mathrm{cos}\theta {\right)}^{2}d{\theta }^{2}=4d{\theta }^{2}⇒$
$ds=2d\theta$
Thus, the arc length is
$L={\int }_{\theta =0}^{\theta =\pi }ds=2\pi$