Question

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

Answer 1

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

`kr, r^m 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 2theta`,

`r^2 = - cos 2theta`, 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 `2theta, theta/2` produce

slower/faster rotations, about the pole.

See the combined graph for `r = - cos theta, r^2 = - cos theta,`

`r = - cos 2 theta and r^2 = - cos 2theta`.

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 `(1/sqrt2, 3pi/4)` moves to

`(1/sqrt(sqrt2), 3pi/4)`, 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]}