How is an independent variable turned into a dependent variable from an exact differential?












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I didn't know how to title this exactly.. But I'll try to explain.



The following is an exact differential of a function $ f(x,y)$



$ df = (frac {partial f}{ partial x})_y dx + (frac { partial f}{ partial y})_x dy$



Where $f$ is a function in 2 independent variables $x$ and $y$



Now this is where I don't know what's going on, I can mathematically arrive at the correct result but I don't quite understand what it means:



If we consider the level curve of the function $f(x,y) = c$



Then the total differential $df = 0$



Then we get that $(frac { partial f}{ partial x})_y dx = - (frac { partial f}{ partial y})_x dy$



Dividing by the differential $dx$ , holding $f$ constant, then:



$ (frac { partial y} { partial x})_f = frac {(frac { partial f}{ partial x})_y}{-(frac { partial f}{ partial y})_x}$



We have just constructed an equation that relates $y$ to $x$ and its derivative $frac {dy}{dx}$ , $y$ just became a dependent variable on $x$



What is happening exactly, geometrically, or however you can put it intuitively? Thanks






ordinary-differential-equations







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    1












    $begingroup$


    I didn't know how to title this exactly.. But I'll try to explain.



    The following is an exact differential of a function $ f(x,y)$



    $ df = (frac {partial f}{ partial x})_y dx + (frac { partial f}{ partial y})_x dy$



    Where $f$ is a function in 2 independent variables $x$ and $y$



    Now this is where I don't know what's going on, I can mathematically arrive at the correct result but I don't quite understand what it means:



    If we consider the level curve of the function $f(x,y) = c$



    Then the total differential $df = 0$



    Then we get that $(frac { partial f}{ partial x})_y dx = - (frac { partial f}{ partial y})_x dy$



    Dividing by the differential $dx$ , holding $f$ constant, then:



    $ (frac { partial y} { partial x})_f = frac {(frac { partial f}{ partial x})_y}{-(frac { partial f}{ partial y})_x}$



    We have just constructed an equation that relates $y$ to $x$ and its derivative $frac {dy}{dx}$ , $y$ just became a dependent variable on $x$



    What is happening exactly, geometrically, or however you can put it intuitively? Thanks






    ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I didn't know how to title this exactly.. But I'll try to explain.



      The following is an exact differential of a function $ f(x,y)$



      $ df = (frac {partial f}{ partial x})_y dx + (frac { partial f}{ partial y})_x dy$



      Where $f$ is a function in 2 independent variables $x$ and $y$



      Now this is where I don't know what's going on, I can mathematically arrive at the correct result but I don't quite understand what it means:



      If we consider the level curve of the function $f(x,y) = c$



      Then the total differential $df = 0$



      Then we get that $(frac { partial f}{ partial x})_y dx = - (frac { partial f}{ partial y})_x dy$



      Dividing by the differential $dx$ , holding $f$ constant, then:



      $ (frac { partial y} { partial x})_f = frac {(frac { partial f}{ partial x})_y}{-(frac { partial f}{ partial y})_x}$



      We have just constructed an equation that relates $y$ to $x$ and its derivative $frac {dy}{dx}$ , $y$ just became a dependent variable on $x$



      What is happening exactly, geometrically, or however you can put it intuitively? Thanks






      ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      I didn't know how to title this exactly.. But I'll try to explain.



      The following is an exact differential of a function $ f(x,y)$



      $ df = (frac {partial f}{ partial x})_y dx + (frac { partial f}{ partial y})_x dy$



      Where $f$ is a function in 2 independent variables $x$ and $y$



      Now this is where I don't know what's going on, I can mathematically arrive at the correct result but I don't quite understand what it means:



      If we consider the level curve of the function $f(x,y) = c$



      Then the total differential $df = 0$



      Then we get that $(frac { partial f}{ partial x})_y dx = - (frac { partial f}{ partial y})_x dy$



      Dividing by the differential $dx$ , holding $f$ constant, then:



      $ (frac { partial y} { partial x})_f = frac {(frac { partial f}{ partial x})_y}{-(frac { partial f}{ partial y})_x}$



      We have just constructed an equation that relates $y$ to $x$ and its derivative $frac {dy}{dx}$ , $y$ just became a dependent variable on $x$



      What is happening exactly, geometrically, or however you can put it intuitively? Thanks






      ordinary-differential-equations




      ordinary-differential-equations






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      asked Jan 28 at 12:55









      khaled014zkhaled014z

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          $begingroup$

          This is a special case of the implicit function theorem. If $frac{partial f}{partial y}$ is non-zero, you can locally solve $f(x,y) = c$ for $y$, i.e. locally there is a differentiable function $g$ such that $f(x,g(x)) = c$. The derivative of $g$ is then exactly what you derived. Intuitively it is just an equation $f(x,y) = c$ and you are trying to solve it for $y$. Then $y$ becomes a function of $x$.






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            $begingroup$

            This is a special case of the implicit function theorem. If $frac{partial f}{partial y}$ is non-zero, you can locally solve $f(x,y) = c$ for $y$, i.e. locally there is a differentiable function $g$ such that $f(x,g(x)) = c$. The derivative of $g$ is then exactly what you derived. Intuitively it is just an equation $f(x,y) = c$ and you are trying to solve it for $y$. Then $y$ becomes a function of $x$.






            share|cite|improve this answer









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              1












              $begingroup$

              This is a special case of the implicit function theorem. If $frac{partial f}{partial y}$ is non-zero, you can locally solve $f(x,y) = c$ for $y$, i.e. locally there is a differentiable function $g$ such that $f(x,g(x)) = c$. The derivative of $g$ is then exactly what you derived. Intuitively it is just an equation $f(x,y) = c$ and you are trying to solve it for $y$. Then $y$ becomes a function of $x$.






              share|cite|improve this answer









              $endgroup$
















                1












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                1





                $begingroup$

                This is a special case of the implicit function theorem. If $frac{partial f}{partial y}$ is non-zero, you can locally solve $f(x,y) = c$ for $y$, i.e. locally there is a differentiable function $g$ such that $f(x,g(x)) = c$. The derivative of $g$ is then exactly what you derived. Intuitively it is just an equation $f(x,y) = c$ and you are trying to solve it for $y$. Then $y$ becomes a function of $x$.






                share|cite|improve this answer









                $endgroup$



                This is a special case of the implicit function theorem. If $frac{partial f}{partial y}$ is non-zero, you can locally solve $f(x,y) = c$ for $y$, i.e. locally there is a differentiable function $g$ such that $f(x,g(x)) = c$. The derivative of $g$ is then exactly what you derived. Intuitively it is just an equation $f(x,y) = c$ and you are trying to solve it for $y$. Then $y$ becomes a function of $x$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 15:38









                KlausKlaus

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                2,782113






























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