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

Apr 24, 2017

The slope, $m \approx - 1.58$

#### Explanation:

To find the slope, m, of the tangent line, we must compute $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $\theta$ and then evaluate it at $\theta = \frac{3 \pi}{8}$.

Here is a reference Tangents with Polar Coordinates that will give us the general equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $\theta$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} \sin \left(\theta\right) + r \cos \left(\theta\right)}{\frac{\mathrm{dr}}{d \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)} \text{ [1]}$

We are given r:

$r \left(\theta\right) = \theta + \cot \left(\theta\right) + \theta {\sin}^{3} \left(\theta\right) \text{ [2]}$

We must compute $\frac{\mathrm{dr}}{d \theta}$

(dr)/(d theta) = 1 - csc^2(θ) + sin^3(θ) + 3 θ sin^2(θ) cos(θ)" [3]"

Substitute the right sides of equations [2] and [3] into equation [1]:

dy/dx = ((1 - csc^2(θ) + sin^3(θ) + 3 θ sin^2(θ) cos(θ))sin(theta)+(theta + cot(theta)+thetasin^3(theta))cos(theta))/((1 - csc^2(θ) + sin^3(θ) + 3 θ sin^2(θ) cos(θ))cos(theta)-(theta + cot(theta)+thetasin^3(theta))sin(theta))" [4]"

When we evaluate equation [4] at $\theta = \frac{3 \pi}{8}$, the left side of the equation becomes the slope, m:

$m = \frac{\left(1 - {\csc}^{2} \left(\frac{3 \pi}{8}\right) + {\sin}^{3} \left(\frac{3 \pi}{8}\right) + 3 \frac{3 \pi}{8} {\sin}^{2} \left(\frac{3 \pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)\right) \sin \left(\frac{3 \pi}{8}\right) + \left(\frac{3 \pi}{8} + \cot \left(\frac{3 \pi}{8}\right) + \frac{3 \pi}{8} {\sin}^{3} \left(\frac{3 \pi}{8}\right)\right) \cos \left(\frac{3 \pi}{8}\right)}{\left(1 - {\csc}^{2} \left(\frac{3 \pi}{8}\right) + {\sin}^{3} \left(\frac{3 \pi}{8}\right) + 3 \frac{3 \pi}{8} {\sin}^{2} \left(\frac{3 \pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)\right) \cos \left(\frac{3 \pi}{8}\right) - \left(\frac{3 \pi}{8} + \cot \left(\frac{3 \pi}{8}\right) + \frac{3 \pi}{8} {\sin}^{3} \left(\frac{3 \pi}{8}\right)\right) \sin \left(\frac{3 \pi}{8}\right)} \text{ [5]}$

$m \approx - 1.58$