# What is the slope of the polar curve f(theta) = theta - sec^2theta+costhetasin^3theta  at theta = (5pi)/6?

Mar 23, 2017

From the reference Polar Curve Tangents :
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dr}}{d \text{theta)sin(theta)+rcos(theta))/((dr)/(d} \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)\right)$

#### Explanation:

Given: $r \left(\theta\right) = \theta - {\sec}^{2} \left(\theta\right) + \cos \left(\theta\right) {\sin}^{3} \left(\theta\right)$

Evaluate at $5 \frac{\pi}{6}$

$r \left(5 \frac{\pi}{6}\right) = 5 \frac{\pi}{6} - {\sec}^{2} \left(5 \frac{\pi}{6}\right) + \cos \left(5 \frac{\pi}{6}\right) {\sin}^{3} \left(5 \frac{\pi}{6}\right)$

$r \left(5 \frac{\pi}{6}\right) \approx 1.18$

Compute (dr)/(d""theta):

(dr)/(d""theta)=1-2tan(theta)sec^2(theta)+sin^2(theta)(2cos^2(theta)-2sin^2(theta)+1)

Evaluate at $5 \frac{\pi}{6}$

(dr(5pi/6))/(d""theta)~~3.0

The slope of the tangent line, m, is $\frac{\mathrm{dy}}{\mathrm{dx}}$ evaluated at $5 \frac{\pi}{6}$:

m = (3sin(5pi/6)+1.18cos(5pi/6))/(3cos(5pi/6)-1.18sin(5pi/6)

$m \approx - 0.15$