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Divergence of a function

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    The question is to draw the function:
    [tex]V=\frac {\boldsymbol{\hat {r}}}{r^2}[/tex]
    And to compute it's divergence: [itex]\nabla \cdot V[/itex]
    2. Relevant equations
    [tex]\nabla\cdot V=\left ( \frac {\partial}{\partial x} \boldsymbol{\hat {x}}+\frac {\partial}{\partial y} \boldsymbol{\hat {y}}+\frac {\partial}{\partial z} \boldsymbol{\hat {z}}\right )\left ( v_x \boldsymbol{\hat {x}}+v_y \boldsymbol{\hat {y}}+v_z \boldsymbol{\hat {z}}\right )=\frac {\partial v_x}{\partial x}+\frac {\partial v_y}{\partial y}+\frac {\partial v_z}{\partial z}[/tex]
    3. The attempt at a solution
    [tex]V=\frac {x}{x^2+y^2+z^2}\boldsymbol{\hat {x}}+\frac {y}{x^2+y^2+z^2}\boldsymbol{\hat {y}}+\frac {x}{x^2+y^2+z^2}\boldsymbol{\hat {x}}[/tex]
    The x component:
    [tex]\frac {\partial}{\partial x}\left( \frac {x}{x^2+y^2+z^2}\right)=\frac {-2x^2+y^2+z^2}{(x^2+y^2+z^2)^2}[/tex]
    And the other components are similar
    I don't think this is the answer, since i was told to expect something special, maybe a number?
    The vector field is, to my opinion, as in the picture attached

    Attached Files:

  2. jcsd
  3. Feb 18, 2012 #2


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    This derivative is incorrect. And even if it were it wouldn't be "the answer" because you haven't calculated the other two partial derivatives.

  4. Feb 18, 2012 #3
    Why isn't the derivative correct?
    The derivative of a fraction:
    [tex]\frac {u}{v}=\frac {u'v-v'u}{v^2}[/tex]
    And, in our case:
    [tex]\left ( \frac {x}{x^2+y^2+z^2} \right )'=\frac {1 \cdot (x^2+y^2+z^2)-2x \cdot x}{(x^2+y^2+z^2)^2}=\frac {-x^2+y^2+z^2}{(x^2+y^2+z^2)^2}[/tex]
  5. Feb 18, 2012 #4


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    Science Advisor

    Yes, that is correct. And not what you had before.

    Now calculate the other two partial derivatives and add them.
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