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

Oct 14, 2017

$\frac{\pi - 4}{\pi + 4}$

#### Explanation:

First, note that we can simplify the equation:

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

The slope of the equation is given by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

To do this, we need to find the use the relationship between $r , \theta$ and $x , y$. Recall that $x = r \cos \theta$ and $y = r \sin \theta$.

Thus, we can do:

dy/dx=(dy//d theta)/(dx//d theta)=(d/(d theta)rsintheta)/(d/(d theta)rcostheta

Since we have $r = {\theta}^{-} 1$, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{d}{d \theta} {\theta}^{-} 1 \sin \theta}{\frac{d}{d \theta} {\theta}^{-} 1 \cos \theta}$

Using the product rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {\theta}^{-} 2 \sin \theta + {\theta}^{-} 1 \cos \theta}{- {\theta}^{-} 2 \cos \theta - {\theta}^{-} 1 \sin \theta}$

Multiply through by ${\theta}^{2} / {\theta}^{2}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sin \theta + \theta \cos \theta}{- \cos \theta - \theta \sin \theta}$

Recalling that $\sin \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, we see that the slope at $\theta = \frac{\pi}{4}$ the slope is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{1}{\sqrt{2}} + \frac{\pi}{4 \sqrt{2}}}{- \frac{1}{\sqrt{2}} - \frac{\pi}{4 \sqrt{2}}}$

Multiply through by $\frac{4 \sqrt{2}}{4 \sqrt{2}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 + \pi}{- 4 - \pi} = \frac{\pi - 4}{\pi + 4}$