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Find a formula for dy/dx

  1. Jul 6, 2014 #1
    Let y be a function of x satisfying [tex]F(x,y, x+y)=0[/tex] where [tex]F(x,y,z)[/tex] is a given function. find a formula for [tex]\frac{dy}{dx}[/tex]



    I know

    [tex]\frac{dy}{dx} = \frac{-f_{x}}{{f_y}}[/tex]

    but how does this change now that I have 3 variables
     
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  3. Jul 6, 2014 #2

    Fredrik

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    You should probably put a little more effort into your explanations of what you're thinking. It took me a while to decode what you're saying here. You appear to have done the following, and then solved for dy/dx:
    $$0=\frac{d}{dx}f(x,y)=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}.$$ What do you get when you try the same approach with F?
    $$0=\frac{d}{dx}F(x,y,x+y)=?$$
     
  4. Jul 6, 2014 #3
    [tex] \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}[/tex]

    [tex]\frac{-\frac{\partial f}{\partial x}(1+1)}{\frac{\partial f}{\partial y}(1+1)} = \frac{dy}{dx}[/tex]

    ??
     
  5. Jul 6, 2014 #4

    Fredrik

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    What you're dealing with is F(x,y,z) where z=x+y, so the last term should include ##\partial F/\partial z##, not ##\partial F/\partial x##.
     
  6. Jul 6, 2014 #5
    [tex]\frac{-\frac{\partial f}{\partial x}\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}} = \frac{dy}{dx}[/tex]

    ??
     
  7. Jul 6, 2014 #6

    Fredrik

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    That's not what I get. It's hard to tell where you went wrong, since you didn't post the calculation. The key to this problem is to apply the chain rule correctly at the start of the calculation, so I'm guessing that you did that wrong.
     
  8. Jul 6, 2014 #7

    verty

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    Could F be the zero function here? If so, I don't how there is enough information to answer the question.
     
  9. Jul 6, 2014 #8

    HallsofIvy

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    You said that F(x, y, x+ y)= 0. While F could be F(x,y,z)= 0 for all x, y, and z, it does NOT have to be. For example, F(x, y, z)= x+ y- z satisfies F(x, y, x+ y)= x+ y- (x+ y)= 0 for all x and y.

    But certainly the best way to handle this problem is to treat F(x, y, z) with the provision that is z= x+ y then F(x, y, x+ y)= 0. Letting f(x,y)= F(x, y, x+ y), so that f is identically 0, then
    [tex]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy= 0[/tex]
    [tex]df= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}(dx+ dy)[/tex]
    [tex]= \left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial z}\right)dx+ \left(\frac{\partial F}{\partial y}+ \frac{\partial F}{\partial z}\right)dy= 0[/tex]
     
  10. Jul 6, 2014 #9


    [tex] \left(\frac{\partial F}{\partial y}+ \frac{\partial F}{\partial z}\right)dy= - \left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial z}\right)dx[/tex]

    [tex] \frac{dy}{dx}= \frac{- \left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial z}\right)}{\left(\frac{\partial F}{\partial y}+ \frac{\partial F}{\partial z}\right)}[/tex]
     
  11. Jul 6, 2014 #10

    Fredrik

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    The end result happens to be correct, but your way of finding it is not. You can't just treat dy and dx as numbers and solve for dy/dx. If you use HallsofIvy's approach, you have to understand the significance of a result like the one he got.

    I would recommend that you stick to the original approach and use it to find this result. Then we can discuss HallsofIvy's approach as an alternative if you want. His approach may be a part of something that you haven't yet studied.
     
    Last edited: Jul 6, 2014
  12. Jul 6, 2014 #11
    I used the implicit function theorem

    2-variable case

    suppose [tex]f_{y}(x_{0},y_{0}) \neq 0[/tex]

    so that y can be thought as a function of x (locally)

    [tex]f(x,y)=K; f: \mathbb{R}^2 \rightarrow \mathbb{R}[/tex]

    [tex]\vec{D}f = [ f_{x}\,\,\,f_{y}][/tex]

    and

    [tex]\vec{i}: \mathbb{R} \rightarrow \mathbb{R}^2[/tex]

    [tex]\vec{i}(x) = \left[\begin{array}{cc}x\\ y(x)\end{array}\right][/tex]

    [tex]\vec{D}f(\vec{i}(x))D\vec{i}(x) = [f_{x}(x,y(x))\,\,\,f_{y}(x,y(x))] \left[\begin{array}{cc}1\\\frac{dy}{dx}\end{array}\right]=0[/tex]

    [tex]f_{x}(x,y(x))+f{y}(x,y(x))\frac{dy}{dx}=0[/tex]

    [tex]\frac{dy}{dx} = \frac{-f_{x}(x,y(x))}{f_{y}(x,y(x))}[/tex]
     
  13. Jul 6, 2014 #12

    Fredrik

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    Unless the problem statement says that the relationship between x and y is given by ##f(x,y)=0##, I see no need to invoke the implicit function theorem to justify why y is a function of x. We can just use the chain rule:
    $$0=\frac{d}{dx}f(x,y) =\frac{\partial f}{\partial x} +\frac{\partial f}{\partial y}\frac{dy}{dx} ~\Rightarrow~ \frac{dy}{dx} =\frac{-\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}.$$ It looks like you've been given that y is a function of x in your problem, so you can use essentially the same method. You just need to apply the chain rule correctly when you evaluate ##\frac{d}{dx}F(x,y,x+y)##.

    Regarding the alternative method that HallsofIvy suggested, I said earlier that you can't treat dx and dy as numbers and solve for dy/dx, but after some thought I see that you can. Specifically, you can view dx as an arbitrary number and dy as defined by dy=(dy/dx)dx. Then you can solve for dy/dx. But I wouldn't recommend this approach unless you have studied a definition of df in your book, and understand it well.
     
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