Question

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

Answer 1

Correcting an error.

Explanation:

This graph is for `pi/2 le theta le (3pi)/2`

`r^2=-costheta`

with

`r = pm sqrt(-costheta)`

See the attached plot.

Answer 2

See explanation for the loop.

Explanation:

As `r = f(cos theta)=f(cos(-theta))`, the shape is symmetrical about

the initial line

`r^2=-cos theta >=0, theta in (1/2pi, 3/2pi)`, wherein `cos theta <=0`.

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

The Table for plotting the graph is

`(r, theta):`

`(0, pi/2) (1/sqrt 2, 2/3pi) (1/sqrt(sqrt2), 3/4pi) (sqrt(sqrt 3/2), 5/6pi) (1, pi)`

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)>=0`

and did not consider the non-positive

`r = - sqrt(- cos theta)<=0`,

for the opposite loop, for the same

`theta in (1/2pi, 3/2pi)`.

My r is a single-valued function of `theta`.