# Solve for dy/dx?

## $r = 6 \setminus \cos \setminus \theta$, $\setminus \theta = \frac{3 \setminus \pi}{2}$ I know to use the formula $\frac{r ' \setminus \sin \setminus \theta + r \setminus \cos \setminus \theta}{r ' \setminus \cos \setminus \theta - r \setminus \sin \setminus \theta}$

May 25, 2018

Answer: $\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{3 \pi}{2}}$ is undefined

#### Explanation:

Consider that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$ where $r ' = \frac{\mathrm{dr}}{d \theta}$

Since we are given that $r = 6 \cos \left(\theta\right)$, we can take the derivative of $r$ with respect to $\theta$ to find $r '$:
$r ' = - 6 \sin \left(\theta\right)$

Plugging the equation for $r$ and $r '$ into the equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- 6 \sin \theta\right) \left(\sin \theta\right) + \left(6 \cos \theta\right) \left(\cos \theta\right)}{\left(- 6 \sin \theta\right) \left(\cos \theta\right) - \left(6 \cos \theta\right) \left(\sin \theta\right)}$
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If we are solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$ in general, we can continue to simply this expression:
dy/dx=(6(cos^2theta-sin^2theta))/(6(-2sinthetacostheta)

Consider the double-angle formulas: $\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$ and $\cos \left(2 \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta$

Applying these formulas we have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{2 \theta}{\sin} \left(2 \theta\right) = - \cot \left(2 \theta\right)$
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However, since we are given an angle ($\theta = \frac{3 \pi}{2}$), if we are actually asked to find $\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{3 \pi}{2}}$, then we can simply use the original plugged in expression:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- 6 \sin \theta\right) \left(\sin \theta\right) + \left(6 \cos \theta\right) \left(\cos \theta\right)}{\left(- 6 \sin \theta\right) \left(\cos \theta\right) - \left(6 \cos \theta\right) \left(\sin \theta\right)}$

Noting that $\cos \left(\frac{3 \pi}{2}\right) = 0$, we can eliminate all terms that have a $\cos \theta$:

$\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{3 \pi}{2}} = \frac{\left(- 6 \sin \theta\right) \left(\sin \theta\right) + 0}{0}$

Notice that we have division by $0$ when we do this substitution, this indicates that $\frac{\mathrm{dx}}{d \theta} = 0$, so we have a vertical tangent and $\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{3 \pi}{2}}$ is undefined.