Question

How do you graph `r^2 theta=1`?

Answer 1

Choosing `r=1/sqrt theta`, with theta in radian (1 radian = 0.3183) ,
the graph is a spiral through `{(r, theta)}={...(4, 1/16) (3, 1/9) (2, 1/4) (1. 1), (1/2, 4) (1/3, 9) (1/4, 16)...(0, oo)}`.

Explanation:

`r^2=1/theta>0`. So, `theta>0`, As `theta` increases, r decreases.

If negative r is allowed, `r=+-1/sqrt theta`.

The each graph is the mirror image of the other with respect to the pole r = 0.

Choosing `r=1/sqrt theta`, with theta in radian (1 radian = 0.3183) ,

the graph is a spiral through `{(r, theta)}={...(4, 1/16) (3, 1/9) (2, 1/4) (1. 1), (1/2, 4) (1/3, 9) (1/4, 16)...(0, oo)}`.