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Homework Help: Div v and curl v

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the div v and curl v of v = (x2 + y2 + z2)-3/2(xi + yj + zk)
    2. Relevant equations

    div v = [itex]\nabla[/itex] [itex]\cdot[/itex] v and [itex]\nabla[/itex] [itex]\times[/itex] v

    3. The attempt at a solution
    I am just confused and drawing a blank in basic algebra

    Is it right to expand v like this

    v = x-3 + y-3 + z-3(xi + yj + zk)
    = (z-3xi + z-3yj + z-2k)

    is the right vector for v?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 19, 2011 #2
    [tex](x^2+y^2+z^2)^{-3/2}\neq (x^{-3}+y^{-3}+z^{-3})[/tex]
  4. Sep 19, 2011 #3
    There's your problem. Also, that's not really the right approach. Do you know what del represents?
  5. Sep 21, 2011 #4
    The del operator so instead I should be using ([itex]\partial[/itex]/[itex]\partial[/itex]x)i + ([itex]\partial[/itex]/[itex]\partial[/itex]y)j + ([itex]\partial[/itex]/[itex]\partial[/itex]z)k??
  6. Sep 21, 2011 #5
    Correct. Apply this operator to your vector [itex]v[/itex]. Are you having trouble understanding what this does to your vector?

    And just as a correction to your OP, [itex]div(v)=\triangledown \cdot v[/itex] only; it is not equal to the dot product and the cross product. The curl is the cross product.
  7. Sep 21, 2011 #6
    Yes I am struggling to apply this to my vector v? Is it right that I have to apply the product rule when applying the del operator to my vector?
  8. Sep 21, 2011 #7
    So if differentiate with respect to x that is my ([itex]\partial[/itex]/[itex]\partial[/itex]x)i term is this remotely correct

    u = (x2 + y2 + z2)-3/2
    u' = -3(x2 + y2 + z2)-5/2
    v = xi + yj + zk
    v' = 1

    then using the product rule

    (x2 + y2 + z2)-3/2 - 3x(x2 + y2 + z2)-5/2(x + y + z)

    Then I add in with respect to y and z or is the completely off track?
  9. Sep 21, 2011 #8


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    This is meaningless, u' is not defined for u a function of more than one variable.

    Pretty much. Do you not know how to take a partial derivative? I don't see any reference to partial derivatives!
  10. Sep 21, 2011 #9
    Sorry what I meant was using the product rule to take the partial derivative with respect to x then y then z and then putting it all together and getting [itex]\nabla[/itex][itex]\cdot[/itex] v

    That is
    the first term = (x2 + y2 + z2)-3/2
    then we take the partial derivative of the first tems with respect to x gives
    = -3x(x2 + y2 + z2)-5/2

    Now the second term is (xi + yj + zk)
    and the partial derivative with respect to x is 1 so using the product rule the first and second term come together to give

    -3x(x2 + y2 + z2)-5/2 + (x2 + y2 + z2)-3/2

    And doing the same by taking the partial derivative with respect to y and z then gives

    [itex]\nabla[/itex][itex]\cdot[/itex] v = (x + y + z)(x2 + y2 + z2)-5/2(-3x - 3y - 3z) + 3(x2 + y2 + z2)-3/2

    Does that look right?
  11. Sep 22, 2011 #10
    Not as such no... I got the divergence to be zero. Using straight vector notation here we have:

    [tex]\nabla \cdot \vec{v} = (\frac{\delta}{\delta x},\frac{\delta}{\delta y},\frac{\delta}{\delta z}) \cdot (x(x^2+y^2+z^2)^{-3/2},y(x^2+y^2+z^2)^{-3/2},z(x^2+y^2+z^2)^{-3/2})[/tex]

    Your partial derivative with respect to x was different. I got an x2 in front of the term with power (-5/2)
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