What is the slope of the polar curve f(theta) = sectheta - csctheta  at theta = (5pi)/4?

Sep 4, 2017

$1$

Explanation:

$r = f \left(\theta\right) = \sec \theta - \csc \theta$

We want $\frac{\mathrm{dy}}{\mathrm{dx}}$ of this curve at $\theta = \frac{5 \pi}{4}$.

To find the slope in terms of $x$ and $y$, we have to use the identities $\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right.$

So here,

$\left\{\begin{matrix}x = \left(\sec \theta - \csc \theta\right) \cos \theta = 1 - \cot \theta \\ y = \left(\sec \theta - \csc \theta\right) \sin \theta = \tan \theta - 1\end{matrix}\right.$

Note that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}}$.

$\left\{\begin{matrix}\frac{\mathrm{dx}}{d \theta} = \frac{d}{d \theta} \left(1 - \cot \theta\right) = {\csc}^{2} \theta \\ \frac{\mathrm{dy}}{d \theta} = \frac{d}{d \theta} \left(\tan \theta - 1\right) = {\sec}^{2} \theta\end{matrix}\right.$

So:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}} = {\sec}^{2} \frac{\theta}{\csc} ^ 2 \theta = {\tan}^{2} \theta$

So the slope at $\theta = \frac{5 \pi}{4}$ is:

$m = {\tan}^{2} \left(\frac{5 \pi}{4}\right) = {\left(- 1\right)}^{2} = 1$