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Standard test

  1. Apr 10, 2008 #1
    Hiya, I've just done a line integral question, and the final part of the question is, "by applying a standard test, determine whether the value of the line integral depends on the path followed between the given initial and final points." The only standard test, I know, is dq/dx, and dp/dy to find if its a perfect differential. I somehow doubt thats the correct standard test :-p. Does it mean, to try calculating it again, but with different forumla?
     
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  3. Apr 10, 2008 #2

    tiny-tim

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    … contour integral …

    Hi Davio! :smile:

    Join two different paths together, in opposite directions, to make a circuit (a contour). Then the value of the line integral is independent of the path iff the contour integral is always zero.

    Do you know a test for the contour integral being zero? :smile:
     
  4. Apr 10, 2008 #3

    HallsofIvy

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    Why do you doubt that is the correct test? It sounds perfectly reasonable to me.
     
  5. Apr 12, 2008 #4
    I doubt its the correct test, because, well it doesn't mean anything to me, I've only ever used it to determine if its a perfect differential, and then solved it! Surely it means you can just express it in a certain form?
    @Tiny tim, nope I don't know it? Unless its the perfect differential thing I mentioned .... in which case I'm being silly and need to read up on what perfect differentials mean!
     
  6. Apr 12, 2008 #5

    tiny-tim

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    Hi Davio! :smile:

    If you haven't done it yet in class, then just follow HallsofIvy's good advice! :smile:
     
  7. Apr 12, 2008 #6
    :-D ok will do! Like I said, guess i need to read up on perfect differential's!
     
  8. Apr 12, 2008 #7

    HallsofIvy

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    Of you have used it to determine "if it is a perfect differential" (I would say exact differential) but don't know how that would help you whether the integral depends only on the endpoints, then yes, you need to read up on what perfect differentials mean!
    It doesn't help to determine whether a differential is "perfect" (exact) if you don't know what it means!
     
  9. Apr 17, 2008 #8
    Yup, my notes say both exact and perfect, I don't know how it helps either, I have 2 equations, F (small x) = 2x-y and F(small y) 1-x, denoting the 2 seperate components, I have the line integral, by going along x, then y, I'm guessing its asking if it makes a different if I just go from (-1,-1) to (1,1)? If I differentiate, Fx and Fy to y and x respectively, I get -1 in both cases, does this help me?
     
  10. Apr 17, 2008 #9

    HallsofIvy

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    Then learn what a "perfect" differential is!


    If F(x,y) is a function of the two variables, x and y, and we evaluate F along a path in the plane with parametric equation x(t), y(t), by the chain rule,
    [tex]\frac{dF}{dt}= \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt}[/tex]
    or, in "differential notation"
    [tex]\dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy[/tex]
    which does NOT depend on the particular parameter t!

    In other words, f(x,y)dx+ g(x,y)dy is a "perfect" differential if there really does exist a function F(x,y) such that dF= f(x,y)dx+ g(x,y)dy. That is the same as saying that
    [tex]f(x,y)= \frac{\partial F}{\partial x}[/tex]
    and
    [tex]g(x,y)= \frac{\partial F}{\partial y}[/itex]

    If that is true, then for any curve, C, between point P and Q
    [tex]\int_C f(x,y)dx+ g(x,y)dy= F(x,y)\right|^Q_P[/tex]
    which depends only on the endpoints P and Q, not on the curve itself.

    I can think of no reason for learning how to test whether or not a differential is a "perfect" differential without knowing what a "perfect" differential is!
     
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