# What is the slope of the polar curve f(theta) = theta + cottheta+thetasin^2theta  at theta = (3pi)/8?

Jul 5, 2016

$= 1.51$

#### Explanation:

• The slope of any curve/function at a certain point is always the function's first derivative.
Hence, the slope of $f \left(\theta\right) = f ' \left(\theta\right)$
• $f ' \left(\theta\right) = 1 - {\csc}^{2} \theta + \theta \cdot 2 \sin \theta \cdot \cos \theta + {\sin}^{2} \theta$
$= 1 - {\csc}^{2} \theta + \theta \sin 2 \theta + {\sin}^{2} \theta$
• Therefore, the slope at $\theta = \frac{3 \pi}{8}$is
$f ' \left(\frac{3 \pi}{8}\right) = 1 - {\csc}^{2} \left(\frac{3 \pi}{8}\right) + \frac{3 \pi}{8} \sin 2 \left(\frac{3 \pi}{8}\right) + {\sin}^{2} \left(\frac{3 \pi}{8}\right)$
$= 1 - 1.17 + 0.83 + 0.85$
$= 1.51$
Jul 5, 2016