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Div and curl

  1. May 14, 2008 #1
    1. The problem statement, all variables and given/known data
    F = [tex]\frac{r}{r}[/tex]

    Find divF
    and curl F

    2. Relevant equations
    r = x[tex]\widehat{i}[/tex] + y[tex]\widehat{j}[/tex] + z[tex]\widehat{k}[/tex]

    r = [tex]\sqrt{(x^{2} + y^{2} + z^{2})}[/tex]

    3. The attempt at a solution


    F = [tex]\frac{x}{(\sqrt{x^{2} + y^{2} + z^{2}})}[/tex][tex]\widehat{i}[/tex] + [tex]\frac{y}{(\sqrt{x^{2} + y^{2} + z^{2}})}[/tex][tex]\widehat{j}[/tex] + [tex]\frac{z}{(\sqrt{x^{2} + y^{2} + z^{2}})}[/tex][tex]\widehat{k}[/tex]


    div F = [tex]\frac{\partial}{\partial x}[/tex] [tex](x(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}}[/tex] + [tex]\frac{\partial}{\partial y}[/tex][tex](y(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}}[/tex] + [tex]\frac{\partial}{\partial z}[/tex][tex](z(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}}[/tex]

    Take the partial derivative of x first.

    let u = [tex]x^{2} + y^{2} + z^{2}[/tex]
    and a = [tex] xu^{2}[/tex]

    [tex]\frac{\partial u}{\partial x}[/tex] = 2x
    [tex]\frac{\partial a}{\partial u}[/tex] = [tex]\frac{-1}{2}[/tex]x[tex]u^{\frac{-3}{2}}[/tex]

    [tex]\frac{\partial a}{\partial x}[/tex] = -[tex]x^{2}[/tex][tex]u^{\frac{-3}{2}}[/tex]
    = -[tex]x^{2}[/tex]([tex]\frac{1}{\sqrt{(x^{2} + y^{2} + z^{2})}}[/tex][tex])^{3}[/tex]

    The other derivatives would give similar answers, and the final answer would be

    -[tex]\frac{x^{2}}{r^{3}}[/tex]-[tex]\frac{y^{2}}{r^{3}}[/tex]-[tex]\frac{z^{2}}{r^{3}}[/tex]

    This is apparently the incorrect answer, can anybody help?
     
  2. jcsd
  3. May 14, 2008 #2
    The three parts that didnt come up are partial derivatives; du/dx da/du and da/dx

    And the r as the numerator in the first equation is meant to be in bold to denote that it is a vector.
     
  4. May 14, 2008 #3

    tiny-tim

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    Hi ineedmunchies! :smile:

    (here's √ and ² and ∂ for you to copy-and-paste
    oh … and tags like B don't work in LaTeX
    hmm … nice LaTeX apart from that, though! :smile:)

    ooooh … this is horrible … can't look … need air:cry:

    Just use r² = x² + y² + z², so 2r∂r/∂x = … ? :smile:
     
  5. May 14, 2008 #4
    hmmm sorry where are you getting 2r∂r/∂x from?
     
  6. May 14, 2008 #5

    tiny-tim

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    Differentiating r² with respect to x. :smile:

    (Chain rule: ∂r²/∂x = dr²/dr ∂r/∂x = 2r ∂r/∂x.)

    (Can you read r² on your computer? if not, it's supposed to be r^2. :redface:)
     
  7. May 14, 2008 #6
    ah actually never mind, i've worked it out. I had left out part of the product rule in my differentiation. I would post the full worked solution but its quite long and the forums seem to be running slow for me today.
     
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