# What is the slope of the tangent line of r=2theta^2-3thetacos(2theta-(pi)/3) at theta=(-5pi)/3?

##### 1 Answer
Apr 9, 2018

Derivative with Polar Coordinates is  (\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)

#### Explanation:

$\frac{\setminus \partial r}{\setminus \partial \theta} = 4 \theta - 3 \cdot \cos \left(2 \theta - \frac{\pi}{3}\right) - 3 \theta \cdot \left(- \sin \left(2 \theta - \frac{\pi}{3}\right)\right) \cdot 2 =$
$= 4 \theta - 3 \cdot \cos \left(2 \theta - \frac{\pi}{3}\right) + 6 \theta \cdot \left(\sin \left(2 \theta - \frac{\pi}{3}\right)\right)$
The numerator:
$\frac{\setminus \partial r}{\setminus \partial \theta} \cdot \sin \left(\theta\right) =$
=((4 theta-3*cos(2 theta-(pi)/(3))+6 theta*(sin(2 theta - pi/3)))*sin(theta)
$r \cdot \cos \left(\theta\right) = \left(2 {\theta}^{2} - 3 \theta \cos \left(2 \theta - \frac{\pi}{3}\right)\right) \cdot \cos \left(\theta\right)$

The denominator:
$\frac{\setminus \partial r}{\setminus \partial \theta} \cdot \cos \left(\theta\right) =$
$4 \theta - 3 \cdot \cos \left(2 \theta - \frac{\pi}{3}\right) + 6 \theta \cdot \left(\sin \left(2 \theta - \frac{\pi}{3}\right)\right) \cdot \cos \left(\theta\right)$
$r \cdot \sin \left(\theta\right) = \left(2 {\theta}^{2} - 3 \theta \cos \left(2 \theta - \frac{\pi}{3}\right)\right) \cdot \sin \left(\theta\right)$
You are calculating the slope at $\theta = - \frac{5 \pi}{3}$, insert $\theta = - \frac{5 \pi}{3}$ in the formulae above:

The numerator:
$\frac{\setminus \partial r}{\setminus \partial \theta} \cdot \sin \left(\theta\right) = - 42.999$
$r \cdot \cos \left(\theta\right) = 31.343$

The denominator:
$\frac{\setminus \partial r}{\setminus \partial \theta} \cdot \cos \left(\theta\right) = - 24.825$
$r \cdot \sin \left(\theta\right) = 54.287$

$\frac{\setminus \partial x}{\setminus \partial y} \left(\theta = - \frac{5 \pi}{3}\right) =$
$= \frac{- 42.999 + 31.343}{- 24.825 - 54.287} = 0.15$

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