# What is the slope of the tangent line of r=12sin(theta/3)*cos(theta/2) at theta=(3pi)/2?

Feb 13, 2016

$- 3 \cdot \sqrt{2}$

#### Explanation:

First, we need to find the derivative of the equation with respect to $\theta$

$\frac{d}{d \theta} r = 12 \cdot \frac{d}{d \theta} \left[\sin \left(\frac{\theta}{3}\right) \cdot \cos \left(\frac{\theta}{2}\right)\right]$

where we have already factored out the constant. Using the chain rule we get two terms:

$\frac{d}{d \theta} r = 12 \cdot \left[\frac{d}{d \theta} \sin \left(\frac{\theta}{3}\right)\right] \cdot \cos \left(\frac{\theta}{2}\right)$

$+ 12 \cdot \sin \left(\frac{\theta}{3}\right) \cdot \left[\frac{d}{d \theta} \cos \left(\frac{\theta}{2}\right)\right]$

Each of the derivatives in the square brackets can be completed as follows:

$\frac{d}{d \theta} \sin \left(\frac{\theta}{3}\right) = \frac{1}{3} \cos \left(\frac{\theta}{3}\right)$
and
$\frac{d}{d \theta} \cos \left(\frac{\theta}{2}\right) = - \frac{1}{2} \sin \left(\frac{\theta}{2}\right)$

plugging these back in above

$\frac{d}{d \theta} r = 4 \cos t \left(\frac{\theta}{3}\right) \cos \left(\frac{\theta}{2}\right) - 6 \sin \left(\frac{\theta}{3}\right) \sin \left(\frac{\theta}{2}\right)$

next, we plug in the value of #theta for which we want the slope and simplify

$\frac{d}{d \theta} r {|}_{\theta = 3 \frac{\pi}{2}} = 4 \cos \left(\frac{\pi}{2}\right) \cos \left(\frac{3 \pi}{4}\right) - 6 \sin \left(\frac{\pi}{2}\right) \sin \left(\frac{3 \pi}{4}\right)$

$\frac{d}{d \theta} r {|}_{\theta = 3 \frac{\pi}{2}} = 0 - 6 \cdot \frac{\sqrt{2}}{2} = - 3 \cdot \sqrt{2}$