# How do you graph r^2= - cos theta?

Sep 1, 2016

See the combined graph that includes an elongated circle depicting this equation, and read the related important note on ${r}^{2}$.

#### Explanation:

Here, I introduce the meaningful interpretations for the presence of

$k r , {r}^{m} \mathmr{and} m \theta$, in polar equations. This is important to

understand equations that have some family characteristics.

Examples:

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

${r}^{2} = - \cos 2 \theta$, and so on.

On par with scaling x and y in Cartesian frame,

power scaling of r, ${r}^{2}$ increases/reduces lengths and just

multiplication of $\theta$ by scalars, like $2 \theta , \frac{\theta}{2}$ produce

See the combined graph for $r = - \cos \theta , {r}^{2} = - \cos \theta ,$

$r = - \cos 2 \theta \mathmr{and} {r}^{2} = - \cos 2 \theta$.

It would be interesting to know, which is which.

The graph of ${r}^{2} = - \cos \theta$ is the elongated circle. Compared

to the circle $r = - \cos \theta$, the point $\left(\frac{1}{\sqrt{2}} , 3 \frac{\pi}{4}\right)$ moves to

$\left(\frac{1}{\sqrt{\sqrt{2}}} , 3 \frac{\pi}{4}\right)$, in that direction. See the graph, for this r-

scaling effect.

graph{(x^2+y^2+x)((x^2+y^2)^1.5+x)((x^2+y^2)^1.5+x^2-y^2)((x^2+y^2)^2+x^2-y^2)(y+x)((x+0.6)^2+(y-.6)^2 - .0015)((x+0.5)^2+(y-0.5)^2-0.001)=0[-2 2 -1.1 1.1]}