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Math Help - 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 \theta to an equivalent equation in x and y?

    r=2sin2\theta

    Another question:

    r=tan\theta

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

    x(x+y)=y
    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:


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

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

    \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 sin2\theta equal to?

    i know that sin\theta = \frac{y}{r} so what is 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: (x^2+y^2)^2=4y^2x^2 however the book's answer is (x^2+y^2)^3=4y^2x^2.
    am i wrong or the book?
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  7. #7
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    \sqrt{x^2+y^2}=4\frac{x}{\sqrt{x^2+y^2}}\frac{y}{\  sqrt{x^2+y^2}}

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

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

    (x^2+y^2)^3=(4xy)^2

    (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 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 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: (x^2+y^2)^2=4y^2x^2 however the book's answer is (x^2+y^2)^3=4y^2x^2.
    am i wrong or the book?
    \displaystyle r = 2 \sin (\theta) \cos (\theta) = 2 \left( \frac{y}{r} \right) \left(\frac{x}{r}\right) = \frac{2xy}{r^2}

    \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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