# What is the equation of the tangent line of r=tan^2theta - costheta at theta=pi/4?

Nov 6, 2016

slope= $\frac{5}{3 + \frac{2}{\sqrt{2}}}$ = $\frac{5}{3 + \sqrt{2}}$

#### Explanation:

It is known that Cartesian to Polar is x= r$\cos \theta$ and y= r $\sin \theta$

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

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

Therefore slope dy/dx =((dr)/(d theta) sin theta + r cos theta)/((dr)/(d theta) cos theta - r sin theta

When $\theta = \frac{\pi}{4}$ , slope would be =$\frac{\frac{\mathrm{dr}}{d \theta} + r}{\frac{\mathrm{dr}}{d \theta} - r}$ , because for $\theta = \frac{\pi}{4}$, $\sin \theta = \cos \theta$

Now given r= ${\tan}^{2} \theta - \cos \theta$, $\frac{\mathrm{dr}}{d \theta} = 2 \tan \theta {\sec}^{2} \theta + \sin \theta$. For $\theta = \frac{\pi}{4}$, $\frac{\mathrm{dr}}{d \theta} = 2 \cdot 1 \cdot 2 + \frac{1}{\sqrt{2}}$ = $4 + \frac{1}{\sqrt{2}}$ and r= 1- $\frac{1}{\sqrt{2}}$

Hence slope= $\frac{5}{3 + \frac{2}{\sqrt{2}}}$ = $\frac{5}{3 + \sqrt{2}}$