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Math Help - inverse transformation?

  1. #1
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    inverse transformation?

    Hi

    I'm working on some questions and I'm just unsure how one of the solutions work.

    I'm given

    x=1/2(u^2-v^2)
    y=uv
    z=z

    and in the solution it is transformed into

    u=[x+(x^2+y^2)^1/2]^1/2
    v=[-x+(x^2+y^2)^1/2]^1/2
    z=z

    I'm unsure how it was transformed. The question says to use "inverse transformation" but I'm not sure what that is.

    Thanks!
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  2. #2
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    Quote Originally Posted by romtik View Post
    Hi

    I'm working on some questions and I'm just unsure how one of the solutions work.

    I'm given

    x=1/2(u^2-v^2)
    y=uv
    z=z

    and in the solution it is transformed into

    u=[x+(x^2+y^2)^1/2]^1/2
    v=[-x+(x^2+y^2)^1/2]^1/2
    z=z

    I'm unsure how it was transformed. The question says to use "inverse transformation" but I'm not sure what that is.

    Thanks!
    u^2 - v^2 = 2x .... (1)

    uv = y \Rightarrow u = \frac{y}{v} .... (2)

    Substitute (2) into (1):

    \left( \frac{y}{v} \right)^2 - v^2 = 2x

    \Rightarrow y^2 - v^4 = 2x v^2

    \Rightarrow v^4 + 2x v^2 - y^2 = 0

    whch is a quadratic in v^2. From the quadratic formula:

    \Rightarrow v^2 = \frac{-2x \pm \sqrt{4x^2 + 4y^2}}{2} = -x \pm \sqrt{x^2 + y^2}

    Discard the negative root solution since v^2 > 0:

    \Rightarrow v^2 = -x + \sqrt{x^2 + y^2}

    etc.
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