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

Sep 23, 2017

1) Differentiate both sides
2) Examine $\frac{d}{d \theta} r \left(\frac{- \pi}{3}\right)$
3) Get that the slope is $3$

#### Explanation:

The slope of the tangent line of r at $\theta = \frac{- \pi}{3}$, is the same as the derivative of the function for that exact x-value. Therefore, we need to differentiate both sides of the function, so we get an expression for $\frac{d}{d \theta}$ r.

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

From this point, we need to know how to differentiate $\sin \theta$ and $\cos \theta$:

$\frac{d}{d \theta} \sin \theta = \cos \theta$

$\frac{d}{d \theta} \cos \theta = - \sin \theta$

We need to know how the chain rule works as well. In this case, we have to substitute $3 \theta$ with $u$ and $\frac{\theta}{2}$ with $v$. When we then differentiate, we have to also differentiate our substitute. (If you want to have a further explanation, notify me).

We then have:
$\frac{d}{d \theta} \left[r\right] = \frac{d}{d \theta} \left[- 2 \sin \left(u\right)\right] - \frac{d}{d \theta} \left[12 \cos \left(v\right)\right]$
$\frac{d}{d \theta} \left[r\right] = - 2 \cos \left(u\right) \cdot \frac{d}{d \theta} \left[u\right] - \left(- 12 \sin \left(v\right) \cdot \frac{d}{d \theta} \left[v\right]\right)$

Let's now substitute $u$ and $v$ back to our original functions.

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

$\frac{d}{d \theta} \left[r\right] = - 2 \cos \left(3 \theta\right) \cdot 3 + 12 \sin \left(\frac{\theta}{2}\right) \cdot \frac{1}{2} = - 6 \cos \left(3 \theta\right) + 6 \sin \left(\frac{\theta}{2}\right)$

Now we have an expression for $\frac{d}{d \theta} r \left(\theta\right)$, where we can put in any values we want, and get the slope for whatever $\theta$-value we want. So let's put in $\theta = \frac{- \pi}{3}$:

$\frac{d}{d \theta} \left[r \left(\frac{- \pi}{3}\right)\right]$ = $\frac{d}{d \theta} \left[r\right] = - 6 \cos \left(3 \frac{- \pi}{3}\right) + 6 \sin \left(\frac{- \pi}{3 \cdot 2}\right)$
$= - 6 \cos \left(- \pi\right) + 6 \sin \left(- \frac{\pi}{6}\right)$

$\cos \left(- \pi\right) = - 1$ and $\sin \left(\frac{- \pi}{6}\right) = - \frac{1}{2}$

This gives us that the slope, for $\theta = \left(- \frac{\pi}{3}\right)$, is:
$- 6 \left(- 1\right) + 6 \cdot \left(- \frac{1}{2}\right) = 6 - 3 = 3$