# What is the slope of the tangent line of r=theta/3+sin((3theta)/8-(5pi)/3) at theta=(7pi)/6?

Jan 24, 2018

$- 1.847$

#### Explanation:

First, let's go ahead and find the values of $r$ and $\frac{\mathrm{dr}}{d \theta}$ at $\theta = \frac{7 \pi}{6}$, just to make things easier later.

$r = \frac{\theta}{3} + \sin \left(\frac{3 \theta}{8} - \frac{5 \pi}{3}\right)$

$= \frac{7 \pi}{18} + \sin \left(\frac{21 \pi}{48} - \frac{5 \pi}{3}\right)$

$\approx 1.881$

$\frac{\mathrm{dr}}{d \theta} = \frac{1}{3} + \frac{3}{8} \cos \left(\frac{3 \theta}{8} - \frac{5 \pi}{3}\right)$

$= \frac{1}{3} + \frac{3}{8} \cos \left(\frac{21 \pi}{48} - \frac{5 \pi}{3}\right)$

$\approx 0.0514$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, the slope of the tangent line at any point is $\frac{\mathrm{dy}}{\mathrm{dx}}$, but the problem is that we don't have $y$ and $x$, we have $r$ and $\theta$.

Luckily, we can apply a version of the chain rule which states that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\mathrm{dy} \text{/"d theta)/(dx"/} d \theta\right)$

We will also have to use the rectangular --> polar coordinate formulas:

$x = r \cos \theta$
$y = r \sin \theta$

Since we have expressions for $x$ and $y$, we can use the product rule to derive them with respect to $\theta$.

$\frac{\mathrm{dy}}{d \theta} = \frac{d}{d \theta} r \sin \theta = \frac{\mathrm{dr}}{d \theta} \sin \theta + r \cos \theta$

$\frac{\mathrm{dx}}{d \theta} = \frac{d}{d \theta} r \cos \theta = \frac{\mathrm{dr}}{d \theta} \cos \theta - r \sin \theta$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

With these formulas, we can find the slope of the line at $\theta = \frac{7 \pi}{6}$.

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

We already know that at $\theta = \frac{7 \pi}{6}$, $r = 1.881$ and $\frac{\mathrm{dr}}{d \theta} = 0.0514$. All we need to do now is find the values of $\sin \theta$ and $\cos \theta$, and then plug everything in.

$\sin \left(\frac{7 \pi}{6}\right) = - 0.5$
$\cos \left(\frac{7 \pi}{6}\right) = - 0.866$

Now we can find our slope:

$\frac{\frac{\mathrm{dr}}{d \theta} \sin \theta + r \cos \theta}{\frac{\mathrm{dr}}{d \theta} \cos \theta - r \sin \theta} = \frac{0.0514 \cdot \left(- 0.5\right) + 1.881 \cdot \left(- 0.866\right)}{0.0514 \cdot \left(- 0.866\right) - 1.881 \cdot \left(- 0.5\right)}$

$= - 1.847$