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Homework Help: Another mathamatician neede to check this! CURL-GRAD-DIV

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data

    I have repeated this problem over and over again so either the question is wrong or my likely its my method thats at fault. Please can someone check this for me.

    Prove this is true:
    [tex]\nabla\times(\nabla \times F)=-\nabla^{2}F+\nabla(\nabla.F)[/tex]

    for [tex]F=x^{2}z^{3}i[/tex]

    3. The attempt at a solution

    Im getting for the LHS:

    [tex](\nabla \times F)= 3x^{2}z^{3} j[/tex]

    [tex]\nabla\times(\nabla \times F)= -6x^{2}z i + 6xz^{2}k[/tex]

    and for the right:

    [tex]-\nabla^{2}F=-2z^{3} i[/tex]


    [tex]\nabla(\nabla.F)=2z^{3}i + 6xz^2k[/tex]

    [tex]-\nabla^{2}F+\nabla(\nabla.F)=6xz^{2} k[/tex]

    [tex]\Rightarrow \nabla\times(\nabla \times F)\neq-\nabla^{2}F+\nabla(\nabla.F)[/tex]
    for [tex]F=x^{2}z^{3}i[/tex]
  2. jcsd
  3. May 16, 2009 #2
    isn't [tex]\nabla[/tex].[tex]\nabla[/tex] = [tex]\nabla[/tex]^2= d^2/dx^2+d^2/dy^2+etc.

    ugh the text never comes out right why is line 1 not equal to negative line 3?
    Last edited: May 16, 2009
  4. May 16, 2009 #3
    This is not correct.
  5. May 16, 2009 #4


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    Homework Helper

    It seems you've taken the scalar Laplacian and then made a vector out of it. Make sure you use the vector Laplacian and you will get the extra term you need to proof the equality.
  6. May 16, 2009 #5


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    This question doesn't make any sense. Using the vector laplacian is trivial since it's defined by using the other two terms in your equations.
  7. May 16, 2009 #6


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    Staff Emeritus
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    Presumably, given that we are in Cartesian coordinates, you use the fact that [tex]\nabla^2\vec{F}=\nabla^2F_x\mathbf{i}+\nabla^2F_y\mathbf{j}+\nabla^2F_z\mathbf{k}[/tex]
  8. May 17, 2009 #7


    [tex]A_{x}=x^{2}z^{3} [/tex]

    [tex] A_{y}=0 [/tex]

    [tex] A_{z}=0[/tex]




    is this not right?
    Last edited by a moderator: Apr 24, 2017
  9. May 17, 2009 #8


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    Staff Emeritus
    Science Advisor

    This isn't correct. [tex]\nabla^2 F_x=\nabla^2x^2z^3=2z^3+6zx^2[/tex]
    Last edited by a moderator: Apr 24, 2017
  10. May 17, 2009 #9


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    Homework Helper

    The normal Laplacian is given by [tex]\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}[/tex]. You already stated that [itex]A_x=x^2 z^2[/itex]. Now all you have to do is apply the normal Laplacian to A_x.
  11. May 17, 2009 #10
    Thanks guys, sorted it.
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