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

Oct 24, 2016

The slope, $m \approx 2.7$

#### Explanation:

From the reference we have the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr} \left(\theta\right)}{d \theta} \sin \left(\theta\right) + r \left(\theta\right) \cos \left(\theta\right)}{\frac{\mathrm{dr} \left(\theta\right)}{d \theta} \cos \left(\theta\right) - r \left(\theta\right) \sin \left(\theta\right)}$

We are given $r \left(\theta\right)$:

$r \left(\theta\right) = {\theta}^{2} - \theta - {\cos}^{3} \left(\theta\right) + {\tan}^{2} \left(\theta\right)$

Use WolframAlpha to compute $\frac{\mathrm{dr} \left(\theta\right)}{d \theta}$:

(dr(theta))/(d theta) = 2 θ+3 sin(θ) cos^2(θ)+2 tan(θ) sec^2(θ)-1

Use WolframAlpha to evaluate $\frac{\mathrm{dr} \left(\frac{\pi}{3}\right)}{d \theta} \approx 15.6$

Evaluate $r \left(\frac{\pi}{3}\right) \approx 2.9$

When we evaluate the remaining parts of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\frac{\pi}{3}$, we obtain the slope, m, of the tangent line.

$m = \frac{15.6 \sin \left(\frac{\pi}{3}\right) + 2.9 \cos \left(\frac{\pi}{3}\right)}{15.6 \cos \left(\frac{\pi}{3}\right) - 2.9 \sin \left(\frac{\pi}{3}\right)}$

$m \approx 2.7$