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Math Help - y as an implicit function of x in a 2-dimensional level curve

  1. #1
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    y as an implicit function of x in a 2-dimensional level curve

    Hi,

    Given F(x,y) = 0, show that

    \frac{\delta^2y}{\delta x^2} = -\frac{F_{xx}F_{y}^2 - 2F_{xy}F_{x}F_{y} + F_{yy}F_{x}^2}{F_{y}^3}.

    So what I think is going on is that when we write \frac{\delta^2y}{\delta x^2} we are assuming we can have y as a function of x, ie y = \phi(x)? So, we can also have F(x,\phi(x)) which is subject to F(x, \phi(x)) = 0. To be honest, I am not sure how to start this problem or how the relationship of all those partials has anything to do with \frac{\delta^2y}{\delta x^2} .

    Any help much appreciated, thanks!!
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  2. #2
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    Quote Originally Posted by matt.qmar View Post
    Hi,

    Given F(x,y) = 0, show that

    \frac{\delta^2y}{\delta x^2} = -\frac{F_{xx}F_{y}^2 - 2F_{xy}F_{x}F_{y} + F_{yy}F_{x}^2}{F_{y}^3}.

    So what I think is going on is that when we write \frac{\delta^2y}{\delta x^2} we are assuming we can have y as a function of x, ie y = \phi(x)? So, we can also have F(x,\phi(x)) which is subject to F(x, \phi(x)) = 0. To be honest, I am not sure how to start this problem or how the relationship of all those partials has anything to do with \frac{\delta^2y}{\delta x^2} .

    Any help much appreciated, thanks!!
    Your Idea is sound.

    So starting with \displaystyle F(x,y(x))=0 and using the Chain rule gives

    \displaystyle 0=\frac{\partial F}{\partial x}\frac{\partial x}{d \partial x}+\frac{\partial F}{\partial }\frac{\partial y}{\partial x}=F_x+F_y\frac{dy}{dx}

    This gives

    \displaystyle \frac{dy}{dx}=\frac{-F_x(x,y)}{F_y(x,y)}

    Now using the above again we can take another derivative

    \displaystyle \frac{d}{dx}0=\frac{d}{dx}\left(F_x+F_y\frac{dy}{d  x} \right)

    \displaystyle 0=F_{xx}\frac{dx}{dx}+F_{xy}\frac{dy}{dx}+\frac{dy  }{dx}\left( F_{yx} \frac{dx}{dx}+F_{yy}\frac{dy}{dx}\right)+F_y\frac{  d^2y}{dx^2}

    \displaystyle \frac{dy}{dx}=\frac{-F_x(x,y)}{F_y(x,y)}
    for the derivative and solve for the 2nd.

    Now just plug in
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  3. #3
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    Aha, the chain rule. Thank you!
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