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Thread: Conversion of polar to cartesian

  1. #1
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    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.
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  2. #2
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    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)
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  3. #3
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    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.
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