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Dy/dx of vector valued function F(x,y)

  1. Jul 11, 2011 #1
    F(x,y) = F1(x,y)i + F2(x,y)j, F is a vector valued function of x,y. How to find dy/dx?

    The book gives the answer as

    dy/dx = (F1(x,y))/(F2(x,y)). But I do not know how to get to this answer at all. Find the partial derivative is what I can do, but how to get to dy/dx when the main function is F(x,y)?
     
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  3. Jul 14, 2011 #2

    hunt_mat

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    I would be tempted to look at the norm of the function and then differentiate.
     
  4. Jul 14, 2011 #3
    excuse me but what does norm mean?
     
  5. Jul 14, 2011 #4

    hunt_mat

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    Think of it as dot product.
    [tex]
    \mathbf{F}(x,y)=F_{1}(x,y)\mathbf{i}+F_{2}(x,y) \mathbf{j} \Rightarrow |\mathbf{F}(x,y)|^{2}=F_{1}(x,y)^{2}+F_{2}(x,y)^{2}
    [/tex]
     
  6. Jul 14, 2011 #5

    LCKurtz

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    This appears to be just a vector function of two apparently independent variables x and y. Not a situation where you would normally even ask for dy/dx. What is the context of your question?
     
  7. Jul 15, 2011 #6
    OK so this question is from the book called "Div, Grad, Curl, and all that: an informal text on vector calculus" by h. m. Schey. This is the the question 1-6 from the book.

    The question asks to prove that the F(x,y) = F1(x,y)i + F2(x,y) is infact a solution to the differential equation dy/dx=(F2(x,y))/(F1(x,y)).

    Sorry but I just realized from the small print in the ebook that the dy/dx=(F2(x,y))/(F1(x,y)) and not upside down.
     
  8. Jul 15, 2011 #7

    HallsofIvy

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    The question, as given, makes no sense. In order to find dy/dx, y must be a function of x. But just telling us that "F(x,y)= F1(x,y)i+ F2(x,y)j" does not imply that. It might well be that x and y are completely independent parameters.

    Please check your book again and tell us what the problem really says.
     
  9. Jul 16, 2011 #8

    LCKurtz

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    It's always a challenge to figure out what the problem is when the poster doesn't understand it will enough to state it correctly. So here goes a wild guess. First I'm guessing that what the OP called F1(x,y) and F2(x,y) are really F1(x,y) and F2(x,y) and are alternate notations for the partials Fx(x,y) and Fy(x,y). Then I'm going to further guess that his question has something to do with the fact that if you differentiate F(x,y) = C implicitly with respect to x you get

    Fx(x,y) + Fy(x,y)y' = 0

    [tex] y' = -\frac{F_x(x,y)}{F_y(x,y)}[/tex]
     
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