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

May 14, 2016

Choosing $r = \frac{1}{\sqrt{\theta}}$, with theta in radian (1 radian = 0.3183) ,
the graph is a spiral through $\left\{\left(r , \theta\right)\right\} = \left\{\ldots \left(4 , \frac{1}{16}\right) \left(3 , \frac{1}{9}\right) \left(2 , \frac{1}{4}\right) \left(1. 1\right) , \left(\frac{1}{2} , 4\right) \left(\frac{1}{3} , 9\right) \left(\frac{1}{4} , 16\right) \ldots \left(0 , \infty\right)\right\}$.

#### Explanation:

${r}^{2} = \frac{1}{\theta} > 0$. So, $\theta > 0$, As $\theta$ increases, r decreases.

If negative r is allowed, $r = \pm \frac{1}{\sqrt{\theta}}$.

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

Choosing $r = \frac{1}{\sqrt{\theta}}$, with theta in radian (1 radian = 0.3183) ,

the graph is a spiral through $\left\{\left(r , \theta\right)\right\} = \left\{\ldots \left(4 , \frac{1}{16}\right) \left(3 , \frac{1}{9}\right) \left(2 , \frac{1}{4}\right) \left(1. 1\right) , \left(\frac{1}{2} , 4\right) \left(\frac{1}{3} , 9\right) \left(\frac{1}{4} , 16\right) \ldots \left(0 , \infty\right)\right\}$.