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Thread: Gradient vector help

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

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

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

    b) $\displaystyle \triangledown\\f(r)=$$\displaystyle f'(r)\triangledown\\r=$$\displaystyle \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 $\displaystyle f(r)$ 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 $\displaystyle r$ and then evaluate that as my function for $\displaystyle r$?

    Any suggestions?

    Thanks,

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

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

    $\displaystyle = f'(r) \nabla r $
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