# What is the arc length of the polar curve f(theta) = 5sintheta-4theta  over theta in [pi/8, pi/3] ?

Jul 30, 2017

L $\approx$ 0.47235219096 $\approx$ 0.472 (3 decimal places)

#### Explanation:

The derivation of the arc length of a polar curve over the interval $\left[a , b\right]$ could be found in another one of my posts, the link to which is here: https://socratic.org/questions/what-is-the-arclength-of-the-polar-curve-f-theta-3sin-3theta-2cot4theta-over-the#455941

Applying to this question, we must first find the derivative of the function with respect to $\theta$:

$f ' \left(\theta\right) = \frac{d}{d \theta} \left(5 \sin \left(\theta\right) - 4 \theta\right)$

Since $\frac{d}{d \theta} \left(\sin \left(\theta\right)\right) = \cos \left(\theta\right)$,

$f ' \left(\theta\right) = 5 \cos \left(\theta\right) - 4$

The formula for the arc length of a polar curve over the interval $\left[a , b\right]$ is given by:

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

where $r = f \left(\theta\right)$ and therefore it follows that $\frac{\mathrm{dr}}{d \theta} = f ' \left(\theta\right)$

Therefore, the arc length for the given function over the given interval would be:

$L = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} \sqrt{{\left(5 \sin \left(\theta\right) - 4 \theta\right)}^{2} + {\left(5 \cos \left(\theta\right) - 4\right)}^{2}} d \theta$

Using a graphing utility (for the sake of simplicity), the integral equals about $0.47235219096$, or to three decimal places, it is equal to $0.472$