# What is the shape of the graph r^2= - cos theta?

Aug 31, 2016

Correcting an error.

#### Explanation:

This graph is for $\frac{\pi}{2} \le \theta \le \frac{3 \pi}{2}$

${r}^{2} = - \cos \theta$

with

$r = \pm \sqrt{- \cos \theta}$

See the attached plot. Aug 31, 2016

See explanation for the loop.

#### Explanation:

As $r = f \left(\cos \theta\right) = f \left(\cos \left(- \theta\right)\right)$, the shape is symmetrical about

the initial line

${r}^{2} = - \cos \theta \ge 0 , \theta \in \left(\frac{1}{2} \pi , \frac{3}{2} \pi\right)$, wherein $\cos \theta \le 0$.

I strictly stick to the definition of (length) r as non-negative.

The Table for plotting the graph is

$\left(r , \theta\right) :$

$\left(0 , \frac{\pi}{2}\right) \left(\frac{1}{\sqrt{2}} , \frac{2}{3} \pi\right) \left(\frac{1}{\sqrt{\sqrt{2}}} , \frac{3}{4} \pi\right) \left(\sqrt{\frac{\sqrt{3}}{2}} , \frac{5}{6} \pi\right) \left(1 , \pi\right)$

Symmetry about the axis $\theta = \pi$ is used to draw the other

half of the loop.

Note that I have considered only one loop from

$r = \sqrt{- \cos \theta} \ge 0$

and did not consider the non-positive

$r = - \sqrt{- \cos \theta} \le 0$,

for the opposite loop, for the same

$\theta \in \left(\frac{1}{2} \pi , \frac{3}{2} \pi\right)$.

My r is a single-valued function of $\theta$.