# Determining the Slope and Tangent Lines for a Polar Curve

## Key Questions

• By converting into parametric equations,

{(x(theta)=r(theta)cos theta=cos2theta cos theta), (y(theta)=r(theta)sin theta=cos2theta sin theta):}

By Product Rule,

$x ' \left(\theta\right) = - \sin 2 \theta \cos \theta - \cos 2 \theta \sin \theta$

$x ' \left(\frac{\pi}{2}\right) = - \sin \left(\pi\right) \cos \left(\frac{\pi}{2}\right) - \cos \left(\pi\right) \sin \left(\frac{\pi}{2}\right) = 1$

$y ' \left(\theta\right) = - \sin 2 \theta \sin \theta + \cos 2 \theta \cos \theta$

$y ' \left(\frac{\pi}{2}\right) = - \sin \left(\pi\right) \sin \left(\frac{\pi}{2}\right) + \cos \left(\pi\right) \cos \left(\frac{\pi}{2}\right) = 0$

So, the slope $m$ of the curve can be found by

$m = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{\pi}{2}} = \frac{y ' \left(\frac{\pi}{2}\right)}{x ' \left(\frac{\pi}{2}\right)} = \frac{0}{1} = 0$

I hope that this was helpful.

The equation of the tangent line is

$y - \frac{7}{2} = \frac{11 \sqrt{3}}{5} \left(x - \frac{7 \sqrt{3}}{2}\right)$

#### Explanation:

In order to find the equation of a line, we need two pieces of information:

$\left\{\begin{matrix}1. \text{Point: " (x_1y_1) \\ 2. "Slope: } m\end{matrix}\right.$

Let us find $\left({x}_{1} , {y}_{1}\right)$.

Since

$\left\{\begin{matrix}x \left(\theta\right) = r \cos \theta = \left(3 + 8 \sin \theta\right) \cos \theta \\ y \left(\theta\right) = r \sin \theta = \left(3 + 8 \sin \theta\right) \sin \theta\end{matrix}\right.$,

${x}_{1} = x \left(\frac{\pi}{6}\right) = \left[3 + 8 \sin \left(\frac{\pi}{6}\right)\right] \cos \left(\frac{\pi}{6}\right) = \frac{7 \sqrt{3}}{2}$

${y}_{1} = y \left(\frac{\pi}{6}\right) = \left[3 + 8 \sin \left(\frac{\pi}{6}\right)\right] \sin \left(\frac{\pi}{6}\right) = \frac{7}{2}$

Now, let us find $m$.

By differentiating with respect to theta,

$\frac{\mathrm{dx}}{d \theta} = 8 \cos \theta \cdot \cos \theta + \left(3 + 8 \sin \theta\right) \cdot \left(- \sin \theta\right)$

$= 8 \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) - 3 \sin \theta$

$= 8 \cos \left(2 \theta\right) - 3 \sin \theta$

$\frac{\mathrm{dy}}{d \theta} = 8 \cos \theta \cdot \sin \theta + \left(3 + 8 \sin \theta\right) \cdot \cos \theta$

$= 8 \left(2 \sin \theta \cos \theta\right) + 3 \cos \theta$

$= 8 \sin \left(2 \theta\right) + 3 \cos \theta$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}} = \frac{8 \sin \left(2 \theta\right) + 3 \cos \theta}{8 \cos \left(2 \theta\right) - 3 \sin \theta}$

Now, we can find $m$.

m={dy}/{dx}|_{theta=pi/6} ={8({sqrt{3}}/2)+3({sqrt{3}}/2)}/{8(1/2)-3(1/2)}={11sqrt{3}}/5#

By Point-Slope Form: $y - {y}_{1} = m \left(x - {x}_{1}\right)$,

$y - \frac{7}{2} = \frac{11 \sqrt{3}}{5} \left(x - \frac{7 \sqrt{3}}{2}\right)$

• A polar equation of the form $r = r \left(\theta\right)$ can be converted into a pair of parametric equations

$\left\{\begin{matrix}x \left(\theta\right) = r \left(\theta\right) \cos \theta \\ y \left(\theta\right) = r \left(\theta\right) \sin \theta\end{matrix}\right.$.

The slope $m$ of the tangent line at $\theta = {\theta}_{0}$ can be expressed as

$m = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = {\theta}_{0}} = \frac{\frac{\mathrm{dy}}{d \theta} {|}_{\theta = {\theta}_{0}}}{\frac{\mathrm{dx}}{d \theta} {|}_{\theta = {\theta}_{0}}} = \frac{y ' \left({\theta}_{0}\right)}{x ' \left({\theta}_{0}\right)}$.

I hope that this was helpful.