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Thread: Polar to Cartesian Problem

  1. #1
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    Polar to Cartesian Problem

    Hi
    I am having trouble with the following question:
    How would you change the following equation in r and $\displaystyle \theta $ to an equivalent equation in x and y?

    $\displaystyle r=2sin2\theta$

    Another question:

    $\displaystyle r=tan\theta$

    $\displaystyle \sqrt{x^2+y^2}=\frac{y}{x}$

    $\displaystyle x(x+y)=y$
    $\displaystyle x^2+xy=y$

    what have i done wrong

    P.S
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Here:

    Transformation from Polar to Cartesian Coordinates


    In second one:


    $\displaystyle \sqrt{x^2+y^2}=\frac{y}{x}$

    $\displaystyle x^2+y^2=\frac{y^2}{x^2}$

    $\displaystyle \frac{x^4}{x^2-1}+y^2=0$
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  3. #3
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    in the first question i am stuck on what does $\displaystyle sin2\theta$ equal to?

    i know that sin\theta = $\displaystyle \frac{y}{r}$ so what is $\displaystyle sin2\theta$?
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    sin(2t)=2sin(t)cos(t)
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  5. #5
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    oh ok, knew it was an identity
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  6. #6
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    i get for question 1: $\displaystyle (x^2+y^2)^2=4y^2x^2$ however the book's answer is $\displaystyle (x^2+y^2)^3=4y^2x^2$.
    am i wrong or the book?
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    $\displaystyle \sqrt{x^2+y^2}=4\frac{x}{\sqrt{x^2+y^2}}\frac{y}{\ sqrt{x^2+y^2}}$

    $\displaystyle \sqrt{x^2+y^2} (x^2+y^2)=4xy$

    $\displaystyle (x^2+y^2)^{\frac{3}{2}}=4xy$

    $\displaystyle (x^2+y^2)^3=(4xy)^2$

    $\displaystyle (x^2+y^2)^3=16x^2y^2$

    Hence, you and your book mistaken...
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  8. #8
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    srry my mistake i posted the wrong question, its suppose to be $\displaystyle r=sin2\theta$
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  9. #9
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    just solved it
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  10. #10
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    Quote Originally Posted by Paymemoney View Post
    srry my mistake i posted the wrong question, its suppose to be $\displaystyle r=sin2\theta$

    So the final answer of your book is true!
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  11. #11
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    Quote Originally Posted by Paymemoney View Post
    i get for question 1: $\displaystyle (x^2+y^2)^2=4y^2x^2$ however the book's answer is $\displaystyle (x^2+y^2)^3=4y^2x^2$.
    am i wrong or the book?
    $\displaystyle \displaystyle r = 2 \sin (\theta) \cos (\theta) = 2 \left( \frac{y}{r} \right) \left(\frac{x}{r}\right) = \frac{2xy}{r^2}$

    $\displaystyle \Rightarrow r^3 = 2xy \Rightarrow (x^2 + y^2)^{3/2} = 2xy$

    and then square both sides to get the book's answer.
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