# How would you graph this

1. Sep 22, 2009

### Bachelier

In polar equations, how can I graph r^2= sin $$\theta$$ on TI-89?

The r is squared. I tried to graph r = sqrt(sin $$\theta$$ ) and r = -sqrt(sin $$\theta$$ ) separately on [0, $$\pi$$ ], however I am not sure this is the correct answer?

thx

2. Sep 22, 2009

### tiny-tim

Hi Bachelier!

(have a theta: θ and a pi: π and a square-root: √ and try using the X2 tag just above the Reply box )

(I don't know anything about the TI-89, but … )

If this r and θ are the usual polar coordinates, then r can't be negative, and so you don't need r = -√(sinθ).

3. Sep 22, 2009

### Bachelier

the formula in polar coordinates is that of a Lemniscate:

r squared equals sin (theta)

I am used to graphing R, not R squared.