# What is the arclength of the polar curve f(theta) = -3sin(3theta)-2cot4theta  over theta in [0,pi/8] ?

Jul 25, 2017

 L= ∞

#### Explanation:

Let us start with the derivation of the formula to find the arclength of a polar curve, where $r = f \left(\theta\right)$

From basic geometry and trigonometry,
$x = r \cdot \cos \left(\theta\right)$ and $y = r \cdot \sin \left(\theta\right)$

since $r = f \left(\theta\right)$,
$x = f \left(\theta\right) \cdot \cos \left(\theta\right)$ and $y = f \left(\theta\right) \cdot \sin \left(\theta\right)$

Starting with $x$ first and differentiating both sides with respect to $\theta$using the product rule,
$\frac{\mathrm{dx}}{d \theta} = \frac{d}{d \theta} \left(f \left(\theta\right) \cdot \cos \left(\theta\right)\right) = f ' \left(\theta\right) \cos \left(\theta\right) - f \left(\theta\right) \sin \left(\theta\right)$
$= \frac{\mathrm{dr}}{d \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)$

Doing the same with the $y$ equation,
$\frac{\mathrm{dy}}{d \theta} = \frac{d}{d \theta} \left(f \left(\theta\right) \cdot \sin \left(\theta\right)\right) = f ' \left(\theta\right) \sin \left(\theta\right) + f \left(\theta\right) \cos \left(\theta\right)$
$= \frac{\mathrm{dr}}{d \theta} \sin \left(\theta\right) + r \cos \left(\theta\right)$

To find $\mathrm{dl}$, a differential of the arclength, each infinitesimal piece of the arclength can be seen as the hypotenuse (or simply a straight line segment) whose length can be figured out by finding

$\mathrm{dl} = \sqrt{{\left(\frac{\mathrm{dx}}{d \theta}\right)}^{2} + {\left(\frac{\mathrm{dy}}{d \theta}\right)}^{2}} d \left(\theta\right)$

let us concentrate on the ${\left(\frac{\mathrm{dx}}{d \theta}\right)}^{2} + {\left(\frac{\mathrm{dy}}{d \theta}\right)}^{2}$ bit first.

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

=((dr)/(d theta))^2cos^2(theta)-2r(dr)/(d theta)cos(theta)sin(theta)+r^2sin^2(theta) + ((dr)/(d theta))^2sin^2(theta)+2r(dr)/(d theta)cos(theta)sin(theta)+r^2cos^2(theta)

simplifying,
$= {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} \left({\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right)\right) + {r}^{2} \left({\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right)\right)$

and since ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$ (trigonometric identity),

${\left(\frac{\mathrm{dx}}{d \theta}\right)}^{2} + {\left(\frac{\mathrm{dy}}{d \theta}\right)}^{2} = {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} + {r}^{2}$

therefore,

$\sqrt{{\left(\frac{\mathrm{dx}}{d \theta}\right)}^{2} + {\left(\frac{\mathrm{dy}}{d \theta}\right)}^{2}} = \sqrt{{\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} + {r}^{2}}$

$\mathrm{dl} = \left(\sqrt{{\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} + {r}^{2}}\right) d \left(\theta\right)$

integrating all the infinitesimal pieces of length, say from $\theta = a$ to $\theta = b$

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

note that $r = f \left(\theta\right)$
in your case, $\theta$ ranges in the interval $\left[0 , \frac{\pi}{8}\right]$
$f ' \left(\theta\right) = \frac{\mathrm{dr}}{d \theta} = \frac{d}{d \theta} \left(- 3 \sin \left(3 \theta\right) - 2 \cot \left(4 \theta\right)\right)$
$= - 9 \cos \left(3 \theta\right) + 8 {\csc}^{2} \left(4\right)$

Arc-length =
${\int}_{0}^{\frac{\pi}{8}} \left(\sqrt{{\left(- 9 \cos \left(3 \theta\right) + 8 {\csc}^{2} \left(4 \theta\right)\right)}^{2}} + {\left(- 3 \sin \left(3 \theta\right) - 2 \cot \left(4 \theta\right)\right)}^{2}\right) d \left(\theta\right)$

You could use a calculator to evaluate the integral, but by even graphing, the arc length reaches up and down towards infinity, so that there is not a closed curve to calculate its length, which can be seen graphically on a Cartesian plane:
graph{-3sin(3x)-2cot(4x) [-14.24, 14.24, -7.12, 7.12]}

There are discontinuous segments throughout its domain, which are infinite, so its arc length cannot be finite.

Therefore,  L= ∞