# How do you find the equation of the tangent lines to the polar curve r=sin(2theta) at theta=2pi ?

Sep 25, 2014

The equation of the tangent line is $y = 0$ (in red), which looks like this. Let us look at some details.

Let us find the coordinates $\left({x}_{1} , {y}_{1}\right)$ of the point when $\theta = 2 \pi$.

Since

$\left\{\begin{matrix}x \left(\theta\right) = \sin \left(2 \theta\right) \cos \theta \\ y \left(\theta\right) = \sin \left(2 \theta\right) \sin \theta\end{matrix}\right.$,

we have $\left({x}_{1} , {y}_{1}\right) = \left(x \left(2 \pi\right) , y \left(2 \pi\right)\right) = \left(0 , 0\right)$.

Let us now find the slope $m$ of the tangent line.

By differentiating with respect to $\theta$,

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

So, we have $m = \frac{y ' \left(2 \pi\right)}{x ' \left(2 \pi\right)} = \frac{0}{2} = 0$

Hence, the equation of the tangent line is $y = 0$.