Q. 30 - Calculus

Question

The arc length of polar functions: Find the arc lengths of the following polar functions.     $r=csc \theta \mathrm{for} \theta \in \left[\frac{-\pi }{6} ,\frac{\pi }{6} \right]$

Step-by-Step Solution

Answer

The arc length is $2 \mathrm{units}$

Step 1: Given information

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

Step 2: Arc length

The arc length of the curve is given by,

$$\begin{gathered} L={\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta \\ L={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}\sqrt{{\left(csc \theta \right)}^2+{\left(-csc \theta cot \theta \right)}^2}d\theta \left[\because \frac{dr}{d\theta }=-csc \theta cot \theta \right] \\ L={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}\sqrt{cs{c}^2\theta -cs{c}^2\theta co{t}^2\theta }d\theta \\ L={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}\sqrt{cs{c}^2\theta \left(1-co{t}^2\theta \right)} d\theta \\ L={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}\sqrt{cs{c}^2\theta \left(cs{c}^2\theta \right)}d\theta \left[1-co{t}^2\theta =cs{c}^2\theta \right] \\ L={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}cs{c}^2\theta d\theta \\ L={\left[-cot\theta \right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}} \\ L=\left[-cot\frac{3\mathrm{\pi }}{4}+cot\left(\frac{\mathrm{\pi }}{4}\right)\right] \\ \mathrm{L}=\left[1+1\right] \\ \mathrm{L}=\left[2\right] \\ L=2 \mathrm{units} \end{gathered}$$

The arc length is $2 \mathrm{units}$