# Find arc length of r=2\cos\theta in the range 0\le\theta\le\pi?

## I know the length formula is $\setminus {\int}_{a}^{b} \setminus \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$ ... can someone check my answer?

May 23, 2018

This is an answer from a tutor but I don't get their work.

#### Explanation:

$L = \setminus \int \mathrm{ds} = \setminus {\int}_{\setminus {\theta}_{1}}^{\setminus {\theta}_{2}} r d \setminus \theta$

$s = r \setminus \theta$ so $\mathrm{ds} = r d \setminus \theta$, thus
$2 \setminus {\int}_{0}^{\setminus} \pi \setminus \cos \setminus \theta = 2 \left(\setminus \sin \setminus \theta\right) {|}_{0}^{\setminus} \pi = 2 \left(\setminus \sin \setminus \pi - \setminus \sin 0\right) = \ldots$

Considering that this is for a semicircle perimeter of a full cycle.. $\ldots 4 \left(\setminus \sin \setminus \theta\right) {|}_{0}^{\setminus \frac{\pi}{2}} = 4 \left(\setminus \sin \left(\setminus \frac{\pi}{2}\right) - \setminus \sin 0\right) = 4 \left(1 - 0\right) = 4$

May 23, 2018

$2 \pi$

#### Explanation:

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

${\mathrm{ds}}^{2} = {\mathrm{dr}}^{2} + {r}^{2} d {\theta}^{2}$

So, for $r = 2 \cos \theta$, we have

$\mathrm{dr} = - 2 \sin \theta d \theta$

and hence

${\mathrm{ds}}^{2} = {\left(- 2 \sin \theta d \theta\right)}^{2} + {\left(2 \cos \theta\right)}^{2} d {\theta}^{2} = 4 d {\theta}^{2} \implies$

$\mathrm{ds} = 2 d \theta$

Thus, the arc length is

$L = {\int}_{\theta = 0}^{\theta = \pi} \mathrm{ds} = 2 \pi$

May 23, 2018

$2 \pi$

#### Explanation:

We seek the arc length of $r = 2 \cos \theta$ for $\theta \in \left[0 , \pi\right]$

We calculate polar arc length using the formula:

$l = {\int}_{\alpha}^{\beta} \setminus \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} \setminus d \theta$

Then, given that $r = 2 \cos \theta$ then differentiating wrt $\theta$ we get:

$\frac{\mathrm{dr}}{d \theta} = - 2 \sin \theta$

So then:

$l = {\int}_{0}^{\pi} \setminus \sqrt{{\left(2 \cos \theta\right)}^{2} + {\left(- 2 \sin \theta\right)}^{2}} \setminus d \theta$

$\setminus = {\int}_{0}^{\pi} \setminus \sqrt{4 \left({\cos}^{2} \theta + {\sin}^{2} \theta\right)} \setminus d \theta$

$\setminus = {\int}_{0}^{\pi} \setminus \sqrt{4} \setminus d \theta \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\because {\cos}^{2} \theta + {\sin}^{2} \theta \equiv 1\right)$

$\setminus = {\int}_{0}^{\pi} \setminus 2 \setminus d \theta$

$\setminus = 2 {\left[\setminus \theta \setminus\right]}_{0}^{\pi}$

$\setminus = 2 \left(\pi - 0\right)$

$\setminus = 2 \pi$

Notes:

The observant reader will note that that $r = 2 \cos \theta$ represents a circle of radius $1$ centred on $\left(1 , 0\right)$, so we can calculate the arc length as the perimeter of a unit circle, thus>

$P = \left(2 \pi\right) \left(1\right) = 2 \pi$, as above