# What is the slope of the tangent line of r=theta-sin((10theta)/3-(pi)/8) at theta=(pi)/4?

Slope of the tangent line of
$r = \theta - \sin \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$
at $\theta = \frac{\pi}{4}$ is
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0.995$

#### Explanation:

Given:
$r = \theta - \sin \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$
$\frac{\mathrm{dr}}{d \left(\theta\right)} = 1 - \left(\frac{10}{3}\right) \cos \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$

wkt
$x = r \cos \theta$
$\frac{\mathrm{dx}}{d \left(\theta\right)} = r \left(- \sin \theta\right) + \cos \theta \left(\frac{\mathrm{dr}}{d \left(\theta\right)}\right)$
at $\theta = \frac{\pi}{4}$
$r = \frac{\pi}{4} - \sin \left(\frac{10 \frac{\pi}{4}}{3} - \frac{\pi}{8}\right)$
$= - 0.008$
$\frac{\mathrm{dr}}{d \left(\theta\right)} = 1 - \left(\frac{10}{3}\right) \cos \left(\frac{10 \frac{\pi}{4}}{3} - \frac{\pi}{8}\right)$
$= 3.029$
$\frac{\mathrm{dx}}{d \left(\theta\right)} = - 0.008 \left(- \sin \frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right) \left(3.029\right)$
$= 0.008 \left(0.707\right) + 0.707 \left(3.029\right)$
$\frac{\mathrm{dx}}{d \left(\theta\right)} = 2.147$

$y = r \sin \theta$
$\frac{\mathrm{dy}}{d \left(\theta\right)} = r \left(\cos \theta\right) + \sin \theta \left(\frac{\mathrm{dr}}{d \left(\theta\right)}\right)$
At $\theta = \frac{\pi}{4}$
$\frac{\mathrm{dy}}{d \left(\theta\right)} = - 0.008 \left(\cos \frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4}\right) \left(3.029\right)$
$\frac{\mathrm{dy}}{d \left(\theta\right)} = 2.136$
Slope of the tangent line of
$r = \theta - \sin \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$
at $\theta = \frac{\pi}{4}$ is
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \left(\theta\right)}}{\frac{\mathrm{dx}}{d \left(\theta\right)}} = \frac{2.136}{2.147}$
Thus,
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0.995$