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Thread: 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 $\displaystyle F(x,y) = 0$, show that

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

    So what I think is going on is that when we write $\displaystyle \frac{\delta^2y}{\delta x^2} $ we are assuming we can have y as a function of x, ie $\displaystyle y = \phi(x)$? So, we can also have $\displaystyle F(x,\phi(x))$ which is subject to $\displaystyle F(x, \phi(x)) = 0$. 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 $\displaystyle \frac{\delta^2y}{\delta x^2} $.

    Any help much appreciated, thanks!!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by matt.qmar View Post
    Hi,

    Given $\displaystyle F(x,y) = 0$, show that

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

    So what I think is going on is that when we write $\displaystyle \frac{\delta^2y}{\delta x^2} $ we are assuming we can have y as a function of x, ie $\displaystyle y = \phi(x)$? So, we can also have $\displaystyle F(x,\phi(x))$ which is subject to $\displaystyle F(x, \phi(x)) = 0$. 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 $\displaystyle \frac{\delta^2y}{\delta x^2} $.

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

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

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

    This gives

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

    Now using the above again we can take another derivative

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

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

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