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

Jan 9, 2017

The slope, $m \approx = - 5.12$

#### Explanation:

Given: $r \left(\theta\right) = \sec \left(\theta\right) - \cos \left(\theta\right) - \theta \text{ [1]}$

The reference Tangents to a polar curve gives us the following equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{r ' \left(\theta\right) \sin \left(\theta\right) + r \left(\theta\right) \cos \left(\theta\right)}{r ' \left(\theta\right) \cos \left(\theta\right) - r \left(\theta\right) \sin \left(\theta\right)} \text{ [2]}$

Compute the first derivative of equation [1]:

$r ' \left(\theta\right) = \sin \left(\theta\right) + \tan \left(\theta\right) \sec \left(\theta\right) - 1 \text{ [3]}$

Substitute equation [1] and equation [3] into equation [2]:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\sin \left(\theta\right) + \tan \left(\theta\right) \sec \left(\theta\right) - 1\right) \sin \left(\theta\right) + \left(\sec \left(\theta\right) - \cos \left(\theta\right) - \theta\right) \cos \left(\theta\right)}{\left(\sin \left(\theta\right) + \tan \left(\theta\right) \sec \left(\theta\right) - 1\right) \cos \left(\theta\right) - \left(\sec \left(\theta\right) - \cos \left(\theta\right) - \theta\right) \sin \left(\theta\right)} \text{ [4]}$

Evaluate equation [4] at $\theta = \frac{5 \pi}{4}$:

$m \approx = - 5.12$
Here is the equation of the tangent line:

$y = - 5.12 \left(x - 3.2768\right) + 3.2768$

Here is a graph of the curve, the point of tangency, and the tangent line: