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Grad, div, and curl?

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    If scalar s=x^3 + 2xy + yz^2 and vector v = (xy^3, 2y + z, z^2) find:
    (a) grad (s)
    (b) div v
    (c) curl v


    2. Relevant equations



    3. The attempt at a solution
    I'm entirely lost at how to do this. I think that grad s is the derivative of the scalar. I think that div is the grad dotted with the vector. I think that the curl is the grad crossed with the vector. I can't do the derivative because I don't know how to handle the three variables. I can do dot product and cross product with much simpler numbers, but I can't even try without being able to find the grad first. I'm lost, and this is my first graduate level course in oceanography, please help any way you can...
     
  2. jcsd
  3. Sep 13, 2011 #2

    Dick

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    Your descriptions of what these things are pretty approximate. Why don't you look up the real definitions and try to apply them. They'll involve partial derivatives, not just derivatives. Do you know what a partial derivative is?
     
  4. Sep 14, 2011 #3
    I've heard of partial derivatives, but I've never taken a course in which I had to do one. My undergraduate math department taught sections of calc I, II, and III that I've heard everywhere else are equivalent to calc I and II. So, because I only went through calc II, I think I may have missed out on crucial information I need, like partial derivatives.

    I'm not bad at learning from media, though, is there any source that maybe you prefer, or think I should focus on?
     
  5. Sep 14, 2011 #4

    Hootenanny

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  6. Sep 14, 2011 #5
    Okay, thanks so much Hootenanny for the great overview of partial derivatives. If my math is correct, would the partial derivatives for the scalar in the problem be:

    with respect to x; 3x^2 + 2y

    with respect to y; 2x + z^2

    with respect to z; 2yz

    I'm a bit confised about the notation though... would the gradient be all of these answers separated by commas?
     
  7. Sep 14, 2011 #6

    Dick

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    Good so far. The gradient is a VECTOR. That's what the 'separated by commas' thing means. The x component of the vector is 3x^2 + 2y, the y component is 2x+x^2 and the z component is 2yz.
     
  8. Sep 14, 2011 #7
    Okay, great!

    So I think then for divergence I would do the dot product between the scalar and the vector.

    (SxVx)i + (SyVy)j + (SzVz)k

    And the curl would be the cross-product, which I would solve using a matrix, I think?
     
  9. Sep 14, 2011 #8

    Dick

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    div(v) has nothing to do with the scalar s. It's just the sum of the partial derivatives of the components of v.
     
  10. Sep 14, 2011 #9
    Ok, but isn't the divergence the dot product of the gradient and the vector?

    Oh I see I wrote dot product between scalar and vector, my mistake
     
    Last edited: Sep 14, 2011
  11. Sep 14, 2011 #10

    vela

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    In Cartesian coordinates, you have[tex]\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial x}, \frac{\partial}{\partial x}\right)[/tex]so[tex]\mathrm{div}~\vec{V} = \nabla \cdot \vec{V} = \frac{\partial V_x}{\partial x}+\frac{\partial V_y}{\partial y}+\frac{\partial V_z}{\partial z}[/tex]It's not equal to [itex](\nabla s)\cdot \vec{V}[/itex], if that's what you meant.
     
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