Q. 27 - Calculus

Question

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

$r=1-\sin \theta \mathrm{for} \theta \in [0, 2\pi ]$

Step-by-Step Solution

Answer

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

Step 1: Given information.

The curve is $r=1-\sin \theta \mathrm{for} \theta \in [0, 2\pi ]$

Step 2: Arc length.

The length of an arc is given by,

$$\begin{gathered} L={\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{{\left(1-\sin \theta \right)}^2+{\left(-\cos \theta \right)}^2}d\theta \left[\because \frac{dr}{d\theta }=-\cos \theta \right] \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{\left(1+{\sin }^2 \theta -2\sin \theta \right)+\left({\cos }^2 \theta \right)}d\theta \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{\left(2-2\sin \theta \right)} d\theta \left[{\sin }^2 \theta +{\cos }^2 \theta =1\right] \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{2\left(1-\sin \theta \right)}d\theta \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{2\left(1-\cos \left(\frac{\mathrm{\pi }}{2}- \theta \right)\right)}d\theta \\ L={\int }_0^{2\mathrm{\pi }}\sqrt{2\left(2{\sin }^2\left(\frac{\mathrm{\pi }}{4}- \frac{\mathrm{\theta }}{2}\right)\right)}d\theta \\ L=2{\int }_0^{2\mathrm{\pi }}\sin \left(\frac{\mathrm{\pi }}{4}- \frac{\mathrm{\theta }}{2}\right)d\theta \\ L=2{\left[\frac{1}{2}\cos \left(\frac{\mathrm{\pi }}{4}- \frac{\mathrm{\theta }}{2}\right)\right]}_0^{2\mathrm{\pi }} \\ L=\left[\cos \left(\frac{\mathrm{\pi }}{4}- \frac{2\mathrm{\pi }}{2}\right)-\cos \left(\frac{\mathrm{\pi }}{4}- \frac{0}{2}\right)\right] \\ L=\left[\cos \left(\frac{-3\mathrm{\pi }}{4}\right)-\cos \left(\frac{\mathrm{\pi }}{4}\right)\right] \\ L=\left[-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right] \\ L=\left|-\frac{2}{\sqrt{2}}\right| \\ \mathrm{L}=\sqrt{2} \mathrm{units} \end{gathered}$$

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