# Conversion of polar to cartesian

• Apr 18th 2012, 03:54 PM
Bowlbase
Conversion of polar to cartesian
I'm getting pretty frustrate with polar coordinates and I can't find any videos online that are very helpful. So I have this and a few other questions I'm posting.

The first problem, I thought, would be simple but my graphs dont match:
Convert$\displaystyle r=sin(2 \theta)$ into cartesian
1. I changed it to $\displaystyle r=2sin(\theta) cos(\theta)$
2. Then multiplied by r twice $\displaystyle r^3=2rsin(\theta) rcos(\theta)$
3. $\displaystyle x=rcos(\theta)$ and $\displaystyle y=rsin(\theta)$
4. so: $\displaystyle r^3=2xy$
5. and since $\displaystyle r^2=x^2+y^2$
6. finally I get: $\displaystyle (x^2+y^2)(\sqrt(x^2+y^2))=2xy$

The graphs do not match so I'm getting desperate to find my mistake here.
• Apr 18th 2012, 04:16 PM
skeeter
Re: Conversion of polar to cartesian
$\displaystyle r = \pm \sqrt{x^2+y^2}$

you have to graph ...

$\displaystyle (x^2+y^2)^{3/2} = 2xy$ (note that this equation graphs in quads I and III)

and

$\displaystyle -(x^2+y^2)^{3/2} = 2xy$ (graphs in quads II and IV)
• Apr 18th 2012, 04:19 PM
Bowlbase
Re: Conversion of polar to cartesian
Okay, I have the method correct then, I just have to remember to do both signs.

Thanks for the help.