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Potential Function

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that ln(x^2+y^2) is a potential function for the following vector field

    [itex] \displaystyle \frac{2x \vec i}{(x^2+y^2)^{1/2}} +\frac{2y \vec j}{(x^2+y^2)^{1/2}}[/itex]

    I calculate [itex]\nabla \phi [/itex] as

    [itex] \displaystyle \frac{2y^2}{(x^2+y^2)^{3/2}} \vec i +\frac{2x^2}{(x^2+y^2)^{3/2}} \vec j[/itex]

    I dont know how this connects to the log function...?

  2. jcsd
  3. Jan 13, 2012 #2


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    It doesn't because you have calculated it incorrectly. ##\nabla f = \langle f_x, f_y\rangle##. Is that what you did?
  4. Jan 13, 2012 #3
    I just took the gradient

    [itex] \displaystyle \nabla \phi =\frac{\partial}{\partial x} \frac{2x \vec i}{\sqrt {x^2+y^2}}+\frac{\partial}{\partial y} \frac{2y \vec j}{\sqrt{ x^2+y^2}}[/itex]..?
  5. Jan 13, 2012 #4


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    You were supposed to show that ##\phi(x,y) = \ln{(x^2+y^2)}## is a potential function for your vector field. ##\phi## is what you should be taking the gradient of.
  6. Jan 13, 2012 #5
    To find the potential function, set [tex] f_{x}=\frac{2x \vec i}{(x^2+y^2)^{1/2}}[/tex]Then take the integral w.r.t. x. which then rewrite f(x,y)=...+g(y). Then take the derivative of this w.r.t. y and equate it to the second one. Then f(x,y), the potential function satisfies the vector field.
  7. Jan 14, 2012 #6
    Thanks, noted.

    Then I calculate this, I dont know where the sqrt comes into it unless its a typo in the question?

    [itex]\displaystyle \frac{2y}{x^2+y^2} \vec j +\frac{2x}{x^2+y^2} \vec i[/itex]
  8. Jan 14, 2012 #7


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    This is what I get also.

    It looks like the correct potential for that vector field is [itex]\displaystyle \phi(x,\ y)=-2\sqrt{x^2+y^2}+C[/itex]
  9. Jan 14, 2012 #8
    V. good.

    Thank you SammyS.
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