# How do you graph the lemniscate r^2=36cos2theta?

Feb 24, 2015

Hello,

• You have to graph two curves of polar equations
$r = 6 \sqrt{\cos \left(2 \theta\right)}$ and $r = - 6 \sqrt{\cos \left(2 \theta\right)}$

• It's enough to graph the first one : the second curve is the O-symmetric of the first one. The range is
$\left\{\theta \in \mathbb{R} | \cos \left(2 \theta\right) \ge 0\right\} = {\bigcup}_{k \in \mathbb{Z}} \left[- \frac{\pi}{4} + k \pi , \frac{\pi}{4} + k \pi\right]$

• Reduce the range of $r$ :
1) $r \left(\theta + \pi\right) = r \left(\theta\right)$ : you can study $r$ only on $\left[- \frac{\pi}{4} , \frac{\pi}{4}\right]$.
2) $r \left(- \theta\right) = r \left(\theta\right)$ : you can study $r$ on $\left[0 , \frac{\pi}{4}\right]$ but you have to complete the curve with $O x$-axis.

• Calculate derivative of $r$ :
$r ' \left(\theta\right) = 6 \frac{- 2 \sin \left(2 \theta\right)}{2 \sqrt{\cos \left(2 \theta\right)}} = - 6 \sin \frac{2 \theta}{\sqrt{\cos \left(2 \theta\right)}}$

• Study the sign of $r '$ on $\left[0 , \frac{\pi}{2}\right]$ : because $2 \theta \in \left[0 , \pi\right]$, $\sin \left(2 \theta\right) \ge 0$, so $r ' \left(\theta\right) \le 0$ and $r$ is a decreasing function on $\left[0 , \frac{\pi}{2}\right]$.

• Find particular tangents. Remember that :
1) if $M \left(\theta\right) \setminus \ne O$, the tangent at $M \left(\theta\right)$ is directed by $r ' \left(\theta\right) {\vec{u}}_{\theta} + r \left(\theta\right) {\vec{v}}_{\theta}$, where $\left({\vec{u}}_{\theta} , {\vec{v}}_{\theta}\right)$ is the polar basis.
2) if $M \left(\theta\right) = O$, the tangent at $O$ is directed by ${\vec{u}}_{\theta}$.

Here, you have :
1) tangent at $M \left(0\right) = \left(6 , 0\right)$ : directed by $0 {\vec{u}}_{0} + 6 {\vec{v}}_{0}$, so directed by ${\vec{v}}_{0} = \left(0 , 1\right)$.
2) tangent at $M \left(\frac{\pi}{4}\right) = O$ : directed by ${\vec{u}}_{\frac{\pi}{4}}$.
3) there exists a horizontal tangent : $y = r \sin \left(\theta\right)$, so $y ' = r ' \sin \left(\theta\right) + r \cos \left(\theta\right)$ and so
$y ' = 0 \iff \frac{r '}{r} = - \frac{1}{\tan} \left(\theta\right) \iff \tan \left(2 \theta\right) \tan \left(\theta\right) = 1$.
But, $\tan \left(2 \theta\right) = \frac{2 \tan \left(\theta\right)}{1 - {\tan}^{2} \left(\theta\right)}$, so you have to solve
$2 {\tan}^{2} \left(\theta\right) = 1 - {\tan}^{2} \left(\theta\right) \iff \tan \left(\theta\right) = \frac{1}{\sqrt{3}}$
Finally, $\theta = \arctan \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.

Graph.

1) On $\left[0 , \frac{\pi}{4}\right]$ :

2) On $\left[- \frac{\pi}{4} , \frac{\pi}{4}\right]$ (with $O x$-symmetry) :

3) Finally the all curve with $O$-symmetry :