# What is the slope of the tangent line of r=(sin^2theta)/(-thetacos^2theta) at theta=(pi)/4?

Oct 11, 2016

The slope is $m = \frac{4 - 5 \pi}{4 - 3 \pi}$

#### Explanation:

Here is a reference to Tangents with polar coordinates

From the reference, we obtain the following equation:

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

We need to compute $\frac{\mathrm{dr}}{d \theta}$ but please observe that $r \left(\theta\right)$ can be simplified by using the identity $\sin \frac{x}{\cos} \left(x\right) = \tan \left(x\right)$:

$r = - {\tan}^{2} \frac{\theta}{\theta}$

$\frac{\mathrm{dr}}{d \theta} = \left(g \frac{\theta}{h \left(\theta\right)}\right) ' = \frac{g ' \left(\theta\right) h \left(\theta\right) - h ' \left(\theta\right) g \left(\theta\right)}{h \left(\theta\right)} ^ 2$

$g \left(\theta\right) = - {\tan}^{2} \left(\theta\right)$
$g ' \left(\theta\right) = - 2 \tan \left(\theta\right) {\sec}^{2} \left(\theta\right)$

$h \left(\theta\right) = \theta$
$h ' \left(\theta\right) = 1$

$\frac{\mathrm{dr}}{d \theta} = \frac{- 2 \theta \tan \left(\theta\right) {\sec}^{2} \left(\theta\right) + {\tan}^{2} \left(\theta\right)}{\theta} ^ 2$

Let's evaluate the above at $\frac{\pi}{4}$

${\sec}^{2} \left(\frac{\pi}{4}\right) = 2$
$\tan \left(\frac{\pi}{4}\right) = 1$

$r ' \left(\frac{\pi}{4}\right) = \frac{- 2 \left(\frac{\pi}{4}\right) \left(1\right) \left(2\right) + 1}{\frac{\pi}{4}} ^ 2$

$r ' \left(\frac{\pi}{4}\right) = \left(- 2 \left(\frac{\pi}{4}\right) \left(1\right) \left(2\right) + 1\right) \left(\frac{16}{{\pi}^{2}}\right)$

$r ' \left(\frac{\pi}{4}\right) = \frac{16 - 16 \pi}{{\pi}^{2}}$

Evaluate r at $\frac{\pi}{4}$:

$r \left(\frac{\pi}{4}\right) = - \frac{4}{\pi} = - \frac{4 \pi}{\pi} ^ 2$

Note: I made the above denominator ${\pi}^{2}$ so that it was common with the denominator of $r '$ and would, therefore, cancel when we put them in the following equation:

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

At $\frac{\pi}{4}$ the sines and cosines are equal, therefore, they will cancel.

We are ready to write an equation for the slope, m:

$m = \frac{16 - 16 \pi + - 4 \pi}{16 - 16 \pi - - 4 \pi}$

$m = \frac{4 - 5 \pi}{4 - 3 \pi}$