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Conservative vector field, potential function

  1. Jun 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A vector field is defined by F(x) = (y+z, x+y, x+z).
    Find the Jacobian and determine if the field is conservative in a finite region. If it is conservative, find the potential function.


    2. Relevant equations
    F = delta p AKA
    F = (upsidedown triangle) p


    3. The attempt at a solution
    I found the Jacobian and determined that the field is conservative in a finite region. My problem is finding the potential function. So far I got:

    dp ---> x(y+z) + C(y,z)
    dx

    dp ---> 1/2y (2x + y) + C(x,z)
    dx

    dp ---> 1/2z (2x + z) + C(x,y)
    dx

    I'm not sure on finding the potential function. From examples I've seen Icn set C(x,z) and C(x,y) to both equal zero, and then make C(y,z) a constant (using only y and z) to make dp/dx equal the other ones. But do they have to be exactly equal to each other? I'm wondering if anyone would give me some help with finding this potential function, please. Any help is greatly appreciated!!! :!!)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 26, 2007 #2

    NateTG

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    Well, it seems like one way to do it would be to pick a point to set as zero, and then integrate the work done by the force as a particle is moved to an arbitrary location. If the force is conservative, the path taken is arbitrary.
     
  4. Jun 26, 2007 #3

    Dick

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    Your first step is good. So p=x(y+z)+C(y,z). Now set dp/dy=x+y. What can you conclude about dC(y,z)/dy?
     
  5. Jul 3, 2007 #4
    Hi there guys! Sorry I didn't respond and thank you in a while; I went on a bit of a vacation!

    So thank you Dick, for the hint. I think I got it, but I want to double check with you when what I've done... Here goes.

    p(x,y,z) = x(y+z) + c1(y,z) = xy + 1/2y^2 + c2(x,z) = xz + 1/2z^2 + c3(x,y)

    p = x(y+z) + c1(y,z)

    dp/dy = x + y

    (x + y) = y + c1(y,z)

    So therefore c1(y,z) = x.

    But I'm unsure if I can do this because I thought c1(y,z) had to be a function with only y and z, no x. I'll double check some things, but in the mean time, give it some thought please!

    Thank you!! :D :D :D
     
  6. Jul 3, 2007 #5
    Wait.

    I had meant that c1(y,z) = xy.

    What should I do to confirm this?
     
  7. Jul 4, 2007 #6

    Dick

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    You aren't thinking very clearly here. p=x(y+z)+C(y,z). So dp/dx=y+z. Now I want dp/dy=x+y. From the above I compute dp/dy=x+dC(y,z)/dy. What must dC(y,z)/dy be??
     
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