# What is the slope of the polar curve f(theta) = theta^2 - sec^3theta+tantheta  at theta = (3pi)/4?

Jun 21, 2016

$f ' \left(\frac{3 \pi}{4}\right) = - 1.77$

#### Explanation:

Anytime a problem asks for the slope of a curve at a given point, it is equivalent to asking what the value of the derivative is at that point.

Step 1: Find the derivative of $f \left(\theta\right)$ with respect to $\theta$
$f ' \left(\theta\right) = \frac{d}{d \left(\theta\right)} f \left(\theta\right)$
$f ' \left(\theta\right) = \frac{d}{d \left(\theta\right)} \left({\theta}^{2} - {\sec}^{3} \theta + \tan \theta\right)$
$f ' \left(\theta\right) = 2 \theta - 3 \cdot \tan \theta \sec \theta \cdot {\sec}^{2} \theta + {\sec}^{2} \theta$
$f ' \left(\theta\right) = 2 \theta - 3 \cdot \tan \theta {\sec}^{3} \theta + {\sec}^{2} \theta$

Step 2: Plug $\theta = \frac{3 \pi}{4}$ into $f ' \left(\theta\right)$

$f ' \left(\frac{3 \pi}{4}\right) = 2 \cdot \left(\frac{3 \pi}{4}\right) - 3 \cdot \tan \left(\frac{3 \pi}{4}\right) {\sec}^{3} \left(\frac{3 \pi}{4}\right) + {\sec}^{2} \left(\frac{3 \pi}{4}\right)$

Finally,

$f ' \left(\frac{3 \pi}{4}\right) = - 1.77$