# What is the equation of the line that is normal to the polar curve f(theta)=theta- sin((5theta)/2-pi/2)+tan((2theta)/3-pi/2)  at theta = pi?

May 21, 2017

$y = \frac{11}{18 \pi + 6 \sqrt{3}} x + \frac{1}{18}$

#### Explanation:

From the reference Tangents with Polar Coordinates we obtain the equation:

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

The slope of the tangent line, ${m}_{t} ,$ is equation  evaluated at $\theta = \pi$:

${m}_{t} = \frac{r}{\frac{\mathrm{dr}}{d \theta}} \text{ }$

The slope of the normal line, ${m}_{n}$ is the negative reciprocal of equation :

${m}_{n} = - \frac{\frac{\mathrm{dr}}{d \theta}}{r} \text{ }$

We are given:

$r = f \left(\theta\right) = \theta - \sin \left(\frac{5 \theta}{2} - \frac{\pi}{2}\right) + \tan \left(\frac{2 \theta}{3} - \frac{\pi}{2}\right)$

Evaluated at $\theta = \pi$:

$r = \pi + \frac{\sqrt{3}}{3} = \frac{3 \pi + \sqrt{3}}{3}$

Compute $\frac{\mathrm{dr}}{d \theta}$:

$\frac{\mathrm{dr}}{d \theta} = 1 - \frac{5}{2} \cos \left(\frac{5 \theta}{2} - \frac{\pi}{2}\right) + \frac{2}{3} {\sec}^{2} \left(\frac{2 \theta}{3} - \frac{\pi}{2}\right)$

Evaluated at $\theta = \pi$:

$- \frac{11}{18}$

${m}_{n} = \frac{11}{18 \pi + 6 \sqrt{3}}$

Compute the Cartesian point from the polar point $\left(r , \theta\right) = \left(\frac{3 \pi + \sqrt{3}}{3} , \pi\right)$:

$x = r \cos \left(\theta\right)$

$x = \frac{3 \pi + \sqrt{3}}{3} \cos \left(\pi\right)$

$x = - \frac{3 \pi + \sqrt{3}}{3}$

$y = r \sin \left(\theta\right)$

$y = \frac{3 \pi + \sqrt{3}}{3} \sin \left(\pi\right)$

$y = 0$

Use the point slope form, $y = m \left(x - {x}_{1}\right) + {y}_{1}$ for the equation of a line:

$y = \frac{11}{18 \pi + 6 \sqrt{3}} \left(x - \left(- \frac{3 \pi + \sqrt{3}}{3}\right)\right) + 0$

Simplify to slope-intercept form:

$y = \frac{11}{18 \pi + 6 \sqrt{3}} x + \frac{1}{18}$