Q. 26 - Calculus

Question

The arc length of polar functions: Find the arc lengths of the following polar functions. 

$r=\sin \theta +\cos \theta \mathrm{for} \theta \in [0, \pi ]$

Step-by-Step Solution

Answer

The arc length is $2\sqrt{2}\mathrm{\pi } \mathrm{units}$

Step 1: Given information.

The curve is $r=\sin \theta +\cos \theta \mathrm{for} \theta \in [0, \pi ]$

Step 2: Arc length

The arc length of the curve $r$is given by,

$$\begin{gathered} L={\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{{\left(\sin \theta +\cos \theta \right)}^2+{\left(\cos \theta -\sin \theta \right)}^2}d\theta \left[\because \frac{dr}{d\theta }=\cos \theta -\sin \theta \right] \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{\left({\sin }^2 \theta +{\cos }^2 \theta +2\sin \theta \cos \theta \right)+\left({\sin }^2 \theta +{\cos }^2 \theta -2\sin \theta \cos \theta \right)}d\theta \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{\left(1+2\sin \theta \cos \theta \right)+\left(1-2\sin \theta \cos \theta \right)} d\theta \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{2}d\theta \\ L=\sqrt{2}{\int }_0^{2\mathrm{\pi }}d\theta \\ L=\sqrt{2}\times 2\mathrm{\pi } \\ \mathrm{L}=2\sqrt{2}\mathrm{\pi } \mathrm{units} \end{gathered}$$

Thus, the arc length is $\mathrm{L}=2\sqrt{2}\mathrm{\pi } \mathrm{units}$