# What is the equation of the tangent line of r=2cos(theta-pi/2) - sin(2theta-pi) at theta=pi/4?

Oct 18, 2016

See below

#### Explanation:

First of all this simplifies to

$r = 2 \sin \theta + \sin 2 \theta$

the tangent vector is the velocity vector so we differentiate wrt time as a parameter

$\vec{r} = \left(2 \sin \theta + \sin 2 \theta\right) \hat{r}$

$\vec{r} ' = \vec{v}$

$= \frac{d}{\mathrm{dt}} \left(2 \sin \theta + \sin 2 \theta\right) \hat{r} + \left(2 \sin \theta + \sin 2 \theta\right) \frac{d}{\mathrm{dt}} \hat{r}$

where $\frac{d}{\mathrm{dt}} \hat{r} = \frac{d}{\mathrm{dt}} \left(\begin{matrix}\cos \theta \\ \sin \theta\end{matrix}\right) = \left(\begin{matrix}- \sin \theta \\ \cos \theta\end{matrix}\right) \dot{\theta} = \dot{\theta} \hat{\theta}$

$\implies \left(2 \cos \theta + 2 \cos 2 \theta\right) \setminus \dot{\theta} \setminus \hat{r} + \left(2 \sin \theta + \sin 2 \theta\right) \setminus \dot{\theta} \setminus \hat{\theta}$

dropping the $\dot{\theta}$ scalar...

$= \left(2 \cos \theta + 2 \cos 2 \theta\right) \left(\begin{matrix}\cos \theta \\ \sin \theta\end{matrix}\right) + \left(2 \sin \theta + \sin 2 \theta\right) \left(\begin{matrix}- \sin \theta \\ \cos \theta\end{matrix}\right)$

$= \left(\begin{matrix}2 {\cos}^{2} \theta + 2 \cos 2 \theta \cos \theta - 2 {\sin}^{2} \theta - \sin \theta \sin 2 \theta \\ 2 \cos \theta \sin \theta + 2 \cos 2 \theta \sin \theta + 2 \sin \theta \cos \theta + \sin 2 \theta \cos \theta\end{matrix}\right)$

$= \left(\begin{matrix}1 + 0 - 1 - \frac{1}{\sqrt{2}} \\ 1 + 0 + 1 + \frac{1}{\sqrt{2}}\end{matrix}\right)$

$= \left(\begin{matrix}- 1 \\ 2 \sqrt{2} + 1\end{matrix}\right)$

the slope is therefore

$m = - \left(1 + 2 \sqrt{2}\right)$

$r \left(\frac{\pi}{4}\right) = 1 + \sqrt{2}$

$x = r \cos \theta = \frac{1}{\sqrt{2}} + 1$

$y = r \sin \theta = \frac{1}{\sqrt{2}} + 1$

So we have

$\frac{y - \frac{1}{\sqrt{2}} - 1}{x - \frac{1}{\sqrt{2}} - 1} = - \left(1 + 2 \sqrt{2}\right)$

Or
$y = - \left(1 + 2 \sqrt{2}\right) \left(x - \frac{1}{\sqrt{2}} - 1\right) + 1 + \frac{1}{\sqrt{2}}$

that could be further simplified of course