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Math Help - Orthogonal trajectories

  1. #1
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    Orthogonal trajectories

    Show that the orthogonal trajectories for the family of curves y = ae^{-x} (a constant) are a family of parabolas.

    Attempt:

    y = ae^{-x}

    \frac{dy}{dx} = -ae^{-x}



    \frac{dy}{dx} = \frac{1}{ae^{-x}}

    \int dy = \int \frac{1}{ae^{-x}}dx

    y = \frac{1}{a}\int e^{x}dx

    y = \frac{1}{a}e^{x} + C

    but the correct answer was:

    \frac{1}{2}y^2 = x +C

    I cant seem to see where I went wrong though.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Welll, maybe you could just change this part:

        \dfrac{dy}{dx} = \dfrac{1}{ae^{-x}}

        \dfrac{dy}{dx} = \dfrac{1}{y}

    \int y \ dy = \int \ dx

    \frac12 y^2 = x + C

    From what you did though.. maybe you can do this:

        y = \frac{1}{a}e^{x} + C

        y = \frac{1}{ae^{-x}} + C

        y = \frac{1}{y} + C

        y^2 = 1 + yC

        y^2 = 1 + ae^{-x}C

    But then, it gets complicated for me... the constant becomes another constant...
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