# What is the arclength of r=2sin(theta) -3theta on theta in [(-3pi)/8,(7pi)/8]?

Apr 14, 2017

I used WolframAlpha $\text{Length} \approx 12.6829$

#### Explanation:

From the reference Arc Length with Polar Coordinates

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

Given: $r = 2 \sin \left(\theta\right) - 3 \theta , \alpha = \frac{- 3 \pi}{8} , \mathmr{and} \beta = \frac{7 \pi}{8}$

Square the function:

${r}^{2} = 4 {\sin}^{2} \left(\theta\right) - 12 \theta \sin \left(\theta\right) + 9 {\theta}^{2}$

Compute the derivative of the function:

$\frac{\mathrm{dr}}{d \theta} = 2 \cos \left(\theta\right) - 3$

Square the derivative:

${\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} = 4 {\cos}^{2} \left(\theta\right) - 12 \cos \left(\theta\right) + 9$

Compute the argument under the radical:

${r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} = 4 {\sin}^{2} \left(\theta\right) - 12 \theta \sin \left(\theta\right) + 9 {\theta}^{2} + 4 {\cos}^{2} \left(\theta\right) - 12 \cos \left(\theta\right) + 9$

Use the identity ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$:

${r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} = 4 \left(1\right) - 12 \theta \sin \left(\theta\right) + 9 {\theta}^{2} - 12 \cos \left(\theta\right) + 9$

Combine like terms:

${r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} = - 12 \theta \sin \left(\theta\right) + 9 {\theta}^{2} - 12 \cos \left(\theta\right) + 13$

Substitute into the integral:

$L = {\int}_{- 3 \frac{\pi}{8}}^{7 \frac{\pi}{8}} \sqrt{- 12 \theta \sin \left(\theta\right) + 9 {\theta}^{2} - 12 \cos \left(\theta\right) + 13} d \theta$

I used WolframAlpha to evaluate the integral:

$L = {\int}_{- 3 \frac{\pi}{8}}^{7 \frac{\pi}{8}} \sqrt{- 12 \theta \sin \left(\theta\right) + 9 {\theta}^{2} - 12 \cos \left(\theta\right) + 13} d \theta \approx 12.6829$