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Conservative Vector Fields

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine if the following is conservative.
    [tex]F(x, y, z) = (4xy + z^2)i + (2x^2 + 6yz)j + (2xz)k[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I'm not entirely sure I'm doing this correctly. I've taken the partial of M with respect to y and got 4x. I then took the partial of N with respect to x and got 4x. My question is how do I check P (2xz) ? Does it matter if I take the partial with respect to x or y, or will both work for the test?
     
  2. jcsd
  3. May 12, 2009 #2

    dx

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    What test exactly are you doing? To show that F is conservative, you just have to show that ∇ x F = 0.
     
  4. May 12, 2009 #3
    I'm doing the test that says something like... if all of the partials are equal then it is conservative. Is there an easier way?
     
  5. May 12, 2009 #4

    dx

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    No there's no test like that. The definition of conservative vector field is this: F is conservative is there is a scalar field φ such that F = ∇φ. This is equivalent to showing that ∇ x F = 0 (unless we're working on regions that are not simply connected, but let's not worry about that now).
     
  6. May 12, 2009 #5

    diazona

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    That is sort of like taking the curl:
    [tex]\vec{\nabla}\times\vec{F} =
    \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right)\hat{x}
    +\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\hat{y}
    +\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{z}[/tex]
    In your first post, you basically described computing the first of the three components and verifying that it's zero (because each of the two derivatives is 4x, so they're equal). You can just do the same thing for the other two components of the curl.
     
  7. May 12, 2009 #6
    Ok, this is what I'm trying to do. I'm not sure about the tests though. Can I stop as soon as one of them doesn't match? Also, does it matter if I take P with respect to x or P with respect to y?

    The test I've been doing is if My = Nx, Mz = Px, Nz = Py then the field is conservative and I can continue to find the potential function.
     
  8. May 12, 2009 #7

    dx

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    To show that F is conservative, you have to check all three components of the curl, i.e. you have to show that

    yP = ∂zN

    zM = ∂xP

    xN = ∂yM

    (I'm assuming you're using the notation M = Fx, N = Fy, P = Fz)

    Of course, if one of them doesn't match, that's enough to show it's not conservative.
     
  9. May 12, 2009 #8
    ahhh i c. Thanks
     
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