# What is the equation of the tangent line to the polar curve  r(theta)=theta^2costheta-theta+sin(theta/3)  at theta = pi?

Apr 30, 2017

The equation of the tangent line is:

$y = \frac{- {\pi}^{2} - \pi + \frac{\sqrt{3}}{2}}{- 2 \pi - \frac{7}{6}} \left(x - {\pi}^{2} - \pi + \frac{\sqrt{3}}{2}\right)$

#### Explanation:

Given: $r \left(\theta\right) = {\theta}^{2} \cos \left(\theta\right) - \theta + \sin \left(\frac{\theta}{3}\right) , \theta = \pi$

$r \left(\pi\right) = {\pi}^{2} \cos \left(\pi\right) - \pi + \sin \left(\frac{\pi}{3}\right)$

$r \left(\pi\right) = - {\pi}^{2} - \pi + \frac{\sqrt{3}}{2}$

The polar point is $\left(- {\pi}^{2} - \pi + \frac{\sqrt{3}}{2} , \pi\right)$

To convert to rectangular coordinates, use:

$x = r \cos \left(\theta\right)$ and y = $r \sin \left(\theta\right)$

x = ${\pi}^{2} + \pi - \frac{\sqrt{3}}{2}$

$y = 0$

Using the point-slope form of the equation of a line:

$y = m \left(x - {\pi}^{2} - \pi + \frac{\sqrt{3}}{2}\right) + 0$

To find the value of the slope, m, we must write $\frac{\mathrm{dy}}{\mathrm{dx}}$ as a function of $\theta$ and then evaluate it at $\theta = \pi$.

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} \sin \left(\theta\right) + r \cos \left(\theta\right)}{\frac{\mathrm{dr}}{d \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)} \text{ [1]}$

Because $\theta = \pi , \sin \left(\pi\right) = 0 , \mathmr{and} \cos \left(\pi\right) = - 1$, this simplifies to:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{r}{\frac{\mathrm{dr}}{d \theta}} \text{ [2]}$

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

$\frac{d \left({\theta}^{2} \cos \left(\theta\right) - \theta + \sin \left(\frac{\theta}{3}\right)\right)}{d \theta} = 2 \theta \cos \left(\theta\right) - {\theta}^{2} \sin \left(\theta\right) - 1 - \frac{1}{3} \cos \left(\frac{\theta}{3}\right)$

Evaluate $2 \theta \cos \left(\theta\right) - {\theta}^{2} \sin \left(\theta\right) - 1 + \frac{1}{3} \cos \left(\frac{\theta}{3}\right)$ at $\theta = \pi$:

$- 2 \pi - 1 - \frac{1}{6} = - 2 \pi - \frac{7}{6}$

$m = \frac{- {\pi}^{2} - \pi + \frac{\sqrt{3}}{2}}{- 2 \pi - \frac{7}{6}}$

The equation of the tangent line is:

$y = \frac{- {\pi}^{2} - \pi + \frac{\sqrt{3}}{2}}{- 2 \pi - \frac{7}{6}} \left(x - {\pi}^{2} - \pi + \frac{\sqrt{3}}{2}\right)$