# What is the equation of the tangent line of r=cot^2theta - sintheta at theta=pi/4?

Dec 13, 2016

$y \approx - 4.707 x + 4.707 \frac{\pi}{4} + 0.293$

#### Explanation:

$\frac{\mathrm{dr}}{d \theta} = - 2 \cot \left(\theta\right) {\csc}^{2} \left(\theta\right) - \cos \left(\theta\right) = r ' \left(\theta\right)$

$r ' \left(\frac{\pi}{4}\right) = - \frac{2}{\tan} \left(\frac{\pi}{4}\right) \cdot \frac{1}{\sin} ^ 2 \left(\frac{\pi}{4}\right) - \cos \left(\frac{\pi}{4}\right)$

$= - 4 - \frac{\sqrt{2}}{2} \approx - 4.707$

This is our instantaneous rate of change (slope of the tangent) when $\theta = \frac{\pi}{4}$

Then, we can just a point-slope formula to figure out the equation of that tangent line by plugging in values into the original function:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$r \left(\frac{\pi}{4}\right) = {\cot}^{2} \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{4}\right) = 1 - \frac{\sqrt{2}}{2} \approx 0.293$

Thus:

$y - 0.293 = 3.293 \left(x - \frac{\pi}{4}\right)$

$y \approx - 4.707 x + 4.707 \frac{\pi}{4} + 0.293$