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GRAD DIV CURL question

  1. Dec 26, 2011 #1
    1. The problem statement, all variables and given/known data
    If [itex]\phi[/itex]= xy[itex]^{2}[/itex]
    A=xzi-z[itex]^{2}[/itex]j+xy[itex]^{2}[/itex]k
    B=zi+xj+yk

    Verify that
    [itex]\nabla[/itex].([itex]\phi[/itex]A)=A.[itex]\nabla[/itex][itex]\phi[/itex]+[itex]\phi[/itex].[itex]\nabla[/itex]A

    2. Relevant equations



    3. The attempt at a solution
    I have worked out the first two parts of the question:
    [itex]\phi[/itex]A = (x[itex]^{2}[/itex]y[itex]^{2}[/itex]z, -xy[itex]^{2}[/itex]z[itex]^{2}[/itex],x[itex]^{2}[/itex]y[itex]^{4}[/itex])
    div([itex]\phi[/itex]A) = 2xy[itex]^{2}[/itex]z-2xyz[itex]^{2}[/itex]

    A.grad([itex]\phi[/itex]) = (xy[itex]^{2}[/itex]z-2xyz[itex]^{2}[/itex])

    I'm struggling to work out the last part:
    [itex]\phi[/itex].[itex]\nabla[/itex]A

    I tried working out [itex]\phi[/itex].grad(A)? but the answer sheet has
    div(A) = z
    [itex]\phi[/itex]div(A) = xy[itex]^{2}[/itex]z

    why?
    Any help appreciated.

    Merry Christmas and Happy new year
     
  2. jcsd
  3. Dec 26, 2011 #2
    Ok, so to start compute [itex]\nabla A[/itex] which will just be [itex] (\frac{\partial}{\partial x}\vec{A},\frac{\partial}{\partial y}\vec{A},\frac{\partial}{\partial z}\vec{A})[/itex]
    You will end up with a scalar, which you can multiply by your scalar [itex]\phi[/itex] and you should end up with [itex]xy^2z[/itex].
     
    Last edited: Dec 26, 2011
  4. Dec 26, 2011 #3

    vela

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    The last term of the expression doesn't make sense. You can't dot a scalar into anything. It should be
    $$\nabla\cdot(\phi \vec{A}) = \vec{A}\cdot\nabla \phi + \phi\nabla\cdot\vec{A}$$
     
  5. Dec 27, 2011 #4
    Thank you.
     
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