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Math Help - Gradient vector help

  1. #1
    Senior Member Danneedshelp's Avatar
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    Gradient vector help

    Q: let r=\sqrt{x^2+y^2}. Show that...

    a) \triangledown\\r=\frac{\vec{r}}{r}, where \vec{r}=<x,y>

    b) \triangledown\\f(r)= f'(r)\triangledown\\r= \big({\frac{f'(r)}{r}})\vec{r}

    I got the first part just fine, but I am stuck on part b. I'm not sure what f(r) really is. If I let the derivative equal 1, then everything falls into place and I get equivalent answers, but I doubt thats how I am supposed to do it. Do I solve for r and then evaluate that as my function for r?

    Any suggestions?

    Thanks,

    Oh, and how do I make my parenthesis proportional to what’s inside them?! Its angering me....
    Last edited by Danneedshelp; July 6th 2009 at 10:09 PM.
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  2. #2
    Super Member Random Variable's Avatar
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    chain rule

     \nabla f(r) = \nabla f(\sqrt{x^{2}+y^{2}}) = \Big (f'(\sqrt{x^{2}+y^{2}} * \frac {x}{\sqrt{x^{2}+y^{2}}} , f'(\sqrt{x^{2}+y^{2}}) * \frac{y}{ \sqrt{x^{2}+y^{2}}} \Big)

     = f'(r) \nabla r
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