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Divergence Problem

  1. Jun 24, 2006 #1
    This problem has me stumped:


    If [tex]r = (x^2 + y^2)^{1/2}[/tex], show that

    [tex] div \left( \frac{h(r)}{r^2}(x \vec{i} + y \vec{j}) \right) = \frac{h'(r)}{r}[/tex]


    My trouble is with mixing the polar coordinates with the position vector. If I write the above as

    [tex] div \left( \frac{h((x^2 + y^2)^{1/2})}{(x^2 + y^2)}(x \vec{i} + y \vec{j}) \right) = \frac{h'((x^2 + y^2)^{1/2})}{(x^2 + y^2)^{1/2}}[/tex]

    I can try to get from the left side to the right side by computing the partial derivatives, but when I started this it involved so much messy differentiation (double chain rule, quotient rule) that it just seemed like it wasn't the right approach, so I didn't even try it.

    But even if I did, what would h' be if h is a function of two variables? I've never seent the notation f '(x,y), just directional derivatives and partials...

    What I'm saying is, even if I did do all the messy differentiation and kept all my stuff in order, I don't see how I would be able to write an h'(r) in the end.

    Where am I going off track?

    Thanks.
     
    Last edited: Jun 24, 2006
  2. jcsd
  3. Jun 24, 2006 #2

    0rthodontist

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    h is a function of one variable, in this case r. You could write h(10) for example. Remember when you differentiate it with respect to x or y, that for example d(h(r))/dx = d(h(r))/dr * dr/dx = h'(r) dr/dx

    It doesn't get too complicated. The more you use the chain rule or product rule, the less trouble it seems like. Just work on one part of the expression at a time to save rewriting, and use symmetry between x and y to reduce work.


    By the way, you probably shouldn't expand out r.
     
    Last edited: Jun 25, 2006
  4. Jun 25, 2006 #3
    This is exactly what I needed. I've seen that trick a handful of times, but I always forget it when I need it! :biggrin:

    Thanks Orthodontist.

    BTW, this reminds me of a side question. I call the above a "trick" because I think of it as multiplying the derivative by 1=dr/dr... But is that a mathematically precise way to think about it? (I'm guessing that it isn't.) If not, what is the mathematically correct way to think of this "trick"?

    In other words, how would you get from d(h(r))/dx to d(h(r))/dr*dr/dx without thinking "hmmm... I'll multiply it by dr/dr"?? Because what is dr/dr anyway?


    Thanks again.
     
  5. Jun 25, 2006 #4

    arildno

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    Well, the easy answer is dr/dr=1..:biggrin:

    A slightly more expanded answer is the following:
    Let R(x,y) be the radial function, that is, the variable r satisfies
    [tex]r=R(x,y)=\sqrt{x^{2}+y^{2}}[/tex]
    whenever r has the interpretation as a radius, and x and y are the associated Cartesian variables.

    Let h(r) be some function of the variable r.
    We may therefore define a function H(x,y):
    [tex]H(x,y)=h(R(x,y))[/tex]
    If we want to calculate, say, [itex]\frac{\partial{H}}{\partial{x}}[/itex],
    we simply invoke the chain rule:
    [tex]\frac{\partial{H}}{\partial{x}}=\frac{dh}{dr}\mid_{(r=R(x,y))}\frac{\partial{R}}{\partial{x}}[/tex]

    Many would regard the use of notations like H and R as unnecessarily pedantic, however I think it is rather enlightening to use, on occasion, a (somewhat) cumbersome but unambiguous notation.
     
    Last edited: Jun 25, 2006
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