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Vector field

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    confirm that the given function is apotential for the given vector field

    [itex]ln(x^{2} + y^{2}) for \frac{2x}{\sqrt{x^{2}+y^{2}}} \vec{i} + \frac{2y}{\sqrt{x^{2}+y^{2}}} \vec{j}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    the first thing i did was let my equation = [itex]P\vec{i}+Q \vec{j}[/itex]

    then if they are conservative[itex]\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} [/itex]

    [itex]\frac{\partial P}{\partial y} = \frac{-2xy}{\sqrt{x^{2}+y^{2}}} [/itex]

    and

    [itex]\frac{\partial Q}{\partial x} = \frac{-2xy}{\sqrt{x^{2}+y^{2}}} [/itex]

    so the vector field is conservative.

    then

    f(x,y) = [itex]\int P(x,y) dx[/itex] and f(x,y) = [itex]\int Q(x,y) dy[/itex]


    from tables i get f(x,y) = 2([itex]\sqrt{x^2 + y^2}[/itex]

    what am i doing wrong here? am i getting my integration wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 24, 2012 #2

    sharks

    User Avatar
    Gold Member

    The conservative vector field, [itex]\vec F[/itex], is:
    [tex]\nabla \phi = P\hat i + Q \hat j[/tex]
    But,
    [tex]\nabla \phi = \frac{\partial \phi}{\partial x}\hat i + \frac{\partial \phi}{\partial y} \hat j[/tex]
    So,
    [tex]\frac{\partial \phi}{\partial x} = P \, ... equation \, (1)[/tex]
    and
    [tex]\frac{\partial \phi}{\partial y} = Q \, ... equation \, (2)[/tex]

    Actually, i got [tex]\phi (x,y)=2\sqrt{x^2+y^2}+C[/tex]
    By definition, the potential of the vector field is [itex]- \phi(x,y)[/itex].
     
    Last edited: Apr 24, 2012
  4. Apr 24, 2012 #3
    thats what i got, but i left out the constant of integration.
    So do you think the question is wrong...
     
  5. Apr 24, 2012 #4

    sharks

    User Avatar
    Gold Member

    The potential of the vector field [itex]\vec F[/itex] should be:
    [tex]-\phi (x,y)=-2\sqrt{x^2+y^2}-C[/tex]Unless there is a nice way to convert the above expression into:
    [tex]\ln (x^2 + y^2)[/tex] then, i don't see how.
     
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