# What is the slope of r=tantheta^2-theta at theta=(3pi)/8?

May 11, 2017

Slope is given by $\frac{\mathrm{dy}}{\mathrm{dx}}$. However, this is not explicitly defined when we have a polar function. We have to use the identities for $x$ and $y$:

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

And to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we need to see that:

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

Using the product rule, we can say that:

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

Since $r = {\tan}^{2} \theta - \theta$, we see that:

$\frac{\mathrm{dr}}{d \theta} = 2 \tan \theta \left({\sec}^{2} \theta\right) - 1$

So:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 \tan \theta {\sec}^{2} \theta - 1\right) \sin \theta + \left({\tan}^{2} \theta - \theta\right) \cos \theta}{\left(2 \tan \theta {\sec}^{2} \theta - 1\right) \cos \theta - \left({\tan}^{2} \theta - \theta\right) \sin \theta}$

Evaluate this at $\theta = \frac{3 \pi}{8}$ for the slope.