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Potential function of conservative vector field

  1. Dec 5, 2004 #1
    Hey ya'll,

    How do I find the potential function of this conservative vector field (It is conservative isn't it?? I did check, but i might've messed that up too!).

    [itex] \int (2x-3y-1)dx - (3x+y-5)dy [/itex]

    I know to break the function:

    [itex] F(x,y)= (2x-3y-1)i - (3x+y-5)j [/itex]

    apart and integrate each part WRT x or y like:

    [itex] f(x,y)= \int (2x-3y-1)dx [/itex]

    [itex] f(x,y)= \int (3x+y-5)dy [/itex]

    To get:

    [itex] x^2-3xy-x+g(y)+K[/itex]


    [itex] -3xy + y^2/2 - 5y +h(x) + K[/itex]

    Respectivlly. K being the constant of integration, but then i don't know how to combine/cancle/manipulate thoes to get one function....

    I thought (and my book seems to show) that you have to find what g(y) and h(x) are but I don't know how to do that, and even if I did I would again be stuck and put them together.

  2. jcsd
  3. Dec 6, 2004 #2


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    First notice that

    [tex]\frac {\partial F_x}{\partial y} - \frac {\partial F_y}{\partial x} = 0[/tex]

    which establishes that the field is conservative.

    Therefore, [itex]\vec F(x, y) = - \nabla \Phi[/itex] and you can determine the potential by finding [itex]\Phi[/itex] such that

    [tex]\frac {\partial \Phi}{\partial x} = -F_x[/tex]


    [tex]\frac {\partial \Phi}{\partial y} = -F_y[/itex]

    which is what you have done. To find g and f simply set your two expressions equal to each other and choose g and f to make the statement true. E.g. h(x) must be

    [tex]h(x) = 3x^2 - x[/tex]
  4. Dec 6, 2004 #3


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    You are looking for a function F(x,y) such that Fx= 2x-3y-1 and Fy= -(3x+ y- 5)= -3x- y+ 5.

    Since Fx= 2x- 3y- 1 you must have F= x2- 3xy- x+ g(y).
    (NOTICE: since this is partial differentiation, the "constant" of integration may be a function of y!)

    Differentiating that with respect to y, Fy= -3x+ g'(y) and that must be equal to -3x- y+ 5. That is: -3x+ g'= -3x- y+ 5. Notice that the "-3x" terms cancel! That has to happen since g(y) is a function of y only so g'(y) must depend on y only- if the field had NOT been conservative, if Tide's
    [tex]\frac {\partial F_x}{\partial y} - \frac {\partial F_y}{\partial x} = 0[/tex]
    check had not worked, that wouldn't happen. Since this is "conservative" (that's really a physics term. Mathematically, we would say that this is an "exact differential".) we have g'(y)= -y+ 5. Integrate that to find g(y) and substitute back into F= x2- 3xy- x+ g(y).
  5. Dec 6, 2004 #4
    Thanks guys,

    Very helpful, I'll have to read these a few times trough to get it down but you guys acctually explain it much better than my texts.

    Thanks again,
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