Conversion of polar to cartesian

Jan 2011
52
0
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.
 

skeeter

MHF Helper
Jun 2008
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6,765
North Texas
\(\displaystyle r = \pm \sqrt{x^2+y^2}\)

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)
 
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Jan 2011
52
0
Okay, I have the method correct then, I just have to remember to do both signs.

Thanks for the help.