Conversion of polar to cartesian

**Bowlbase** · April 18th 2012, 03:54 PM

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 $r=sin(2 \theta)$ into cartesian
1. I changed it to $r=2sin(\theta) cos(\theta)$
2. Then multiplied by r twice $r^3=2rsin(\theta) rcos(\theta)$
3. $x=rcos(\theta)$ and $y=rsin(\theta)$
4. so: $r^3=2xy$
5. and since $r^2=x^2+y^2$
6. finally I get: $(x^2+y^2)(\sqrt(x^2+y^2))=2xy$

The graphs do not match so I'm getting desperate to find my mistake here.

**skeeter** · April 18th 2012, 04:16 PM

$r = \pm \sqrt{x^2+y^2}$

you have to graph ...

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

and

$-(x^2+y^2)^{3/2} = 2xy$ (graphs in quads II and IV)

**Bowlbase** · April 18th 2012, 04:19 PM

Okay, I have the method correct then, I just have to remember to do both signs.

Thanks for the help.

Thread: Conversion of polar to cartesian

LinkBack

Thread Tools

Display

Conversion of polar to cartesian

Re: Conversion of polar to cartesian

Re: Conversion of polar to cartesian

Similar Math Help Forum Discussions

Cartesian to polar of conversion

conversion from cartesian to parametric form

Conversion of polar equations into cartesian equation

polar conversion

Cartesian coordinates (Conversion)

Search Tags