Q. 29 - Calculus

Question

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

$r=sec \theta \mathrm{for} \theta \in \left[\frac{-\pi }{6} ,\frac{\pi }{6} \right]$

Step-by-Step Solution

Answer

The arc length is $\frac{2}{\sqrt{3}} \mathrm{units}$

Step 1: Given information.

The given curve is $r=sec \theta \mathrm{for} \theta \in \left[\frac{-\pi }{6} ,\frac{\pi }{6} \right]$

Step 2: Arc length

The arc length of the curve is given by,

$$\begin{gathered} L={\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta \\ L={\int }_{\frac{-\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{6}}\sqrt{{\left(sec\theta \right)}^2+{\left(sec\theta \tan \theta \right)}^2}d\theta \left[\because \frac{dr}{d\theta }=sec\theta \tan \theta \right] \\ L={\int }_{\frac{-\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{6}}\sqrt{se{c}^2\theta +se{c}^2\theta {\tan }^2\theta }d\theta \\ L={\int }_{\frac{-\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{6}}\sqrt{se{c}^2\theta \left(1+{\tan }^2\theta \right)} d\theta \left[{\sin }^2 \theta +{\cos }^2 \theta =1\right] \\ L={\int }_{\frac{-\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{6}}\sqrt{se{c}^2\theta \left(se{c}^2\theta \right)}d\theta \left[1+{\tan }^2\theta =se{c}^2\theta \right] \\ L={\int }_{\frac{-\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{6}}se{c}^2\theta d\theta \\ L={\left[\tan \theta \right]}_{\frac{-\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{6}} \\ L=\left[\tan \frac{\mathrm{\pi }}{6}-\tan \left(-\frac{\mathrm{\pi }}{6}\right)\right] \\ \mathrm{L}=\left[\tan \frac{\mathrm{\pi }}{6}-\left(\mathrm{\pi }-\tan \left(\frac{\mathrm{\pi }}{6}\right)\right)\right] \\ \mathrm{L}=\left[2\tan \frac{\mathrm{\pi }}{6}-\mathrm{\pi }\right] \\ L=\frac{2}{\sqrt{3}} \mathrm{units} \end{gathered}$$

The arc length is $L=\frac{2}{\sqrt{3}} \mathrm{units}$