# What is the slope of the tangent line of r=theta-sin((theta)/6+(2pi)/3) at theta=(pi)/4?

Jan 13, 2017

The slope is the equation:

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

evaluated at $\theta = \frac{\pi}{4}$

#### Explanation:

From the reference Tangents with polar coordinates , we obtain the equation .

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

$r \left(\theta\right) = \theta - \sin \left(\frac{\theta}{6} + \frac{2 \pi}{3}\right)$

$r ' \left(\theta\right) = 1 - \frac{1}{6} \cos \left(\frac{\theta}{6} + \frac{2 \pi}{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \frac{1}{6} \cos \left(\frac{\theta}{6} + \frac{2 \pi}{3}\right)\right) \sin \left(\theta\right) + \left(\theta - \sin \left(\frac{\theta}{6} + \frac{2 \pi}{3}\right)\right) \cos \left(\theta\right)}{\left(1 - \frac{1}{6} \cos \left(\frac{\theta}{6} + \frac{2 \pi}{3}\right)\right) \cos \left(\theta\right) - \left(\theta - \sin \left(\frac{\theta}{6} + \frac{2 \pi}{3}\right)\right) \sin \left(\theta\right)}$

The slope, $m = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\frac{\pi}{4}}$

$m = 0.776632$