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Homework Help: Grad of a generalised scalar function

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    r=xi+yj+zk and r =[itex]\sqrt{x^2 + y^2 + z^2}[/itex] Let f(r) be a C2 scalar function

    Prove that [itex]\nabla[/itex]f = [itex]\frac{1}{2}[/itex][itex]\frac{df}{dr}[/itex]r

    2. Relevant equations

    Vector identities?

    3. The attempt at a solution

    [itex]\nabla[/itex]f = ([itex]\frac{df}{dx}[/itex] , [itex]\frac{df}{dy}[/itex] , [itex]\frac{df}{dz}[/itex])
    = df/dr]?
    = [itex]\frac{df}{dr}[/itex][itex]\hat{r}[/itex] (unit vector of r)
    = [itex]\frac{df}{dr}[/itex]r[itex]\frac{1}{r}[/itex]?

    I'm pretty sure what I've attempted isn't mathematically correct in the slightest, though in my head it seems to make some geometric sense. Am I even close though?
    Last edited: Apr 11, 2012
  2. jcsd
  3. Apr 12, 2012 #2


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    Formally f(r) is the derivative with respect to vector r: f(r)=(df/dr)=(df/dr)˙r/r. To prove the statement in the problem expand f(r) with x, y, z, and see if it is equal to (df/dr)˙r/r, using that r=√(r2)

    Last edited: Apr 12, 2012
  4. Apr 12, 2012 #3
    Sorry, I made a mistake in the OP, it's supposed to be

    [itex]\nabla[/itex]f = [itex]\frac{1}{r}[/itex][itex]\frac{df}{dr}[/itex]r

    Thanks for your reply, but I'm not really sure where the r=√(r^2) come into it? How do you arrive at the 1/r out the front? The only thing I can think of is that it's supposed to reduced it to a magnitude of only df/dr?
  5. Apr 12, 2012 #4


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    Sorry, I also made a mistake, leaving out 1/r. df/dr is multiplied by the unit vector [itex]\hat{\vec r}[/itex], so [itex]\nabla f=\frac{df}{dr}\vec r /r. [/itex]

    You get it by applying the chain rule: [tex]\frac{df(\sqrt{(\vec r)^2})}{d\vec r}=\frac{df}{dr} \frac{d \sqrt{(\vec r)^2}}{d \vec r}=\frac{df}{dr}\left(\frac{1}{2}\frac{1}{\sqrt{( \vec r)^2} }\right)2 \vec r[/tex].

  6. Apr 14, 2012 #5
    Thanks mate, finally got it in the end. Just a little bit stuck on part b of the question though, have to prove that:

    [itex]\nabla[/itex]^2f = [itex]\frac{2}{r}[/itex][itex]\frac{df}{dr}[/itex]+[itex]\frac{d^2f}{dr^2}[/itex]

    I've been using the vector identities so that I get:

    = [itex]\frac{1}{r}[/itex][itex]\frac{df}{dr}[/itex]([itex]\nabla[/itex][itex]\bullet[/itex]r) +(r[itex]\bullet[/itex][itex]\nabla[/itex])[itex]\frac{1}{r}[/itex][itex]\frac{df}{dr}[/itex]

    from here I get down to [itex]\frac{3}{r}[/itex][itex]\frac{df}{dr}[/itex]+r[itex]\bullet[/itex]([itex]\frac{1}{r}[/itex][itex]\nabla[/itex][itex]\frac{df}{dr}[/itex]+[itex]\frac{df}{dr}[/itex][itex]\nabla[/itex][itex]\frac{1}{r}[/itex])

    This is where I am confused, what is [itex]\nabla\frac{df}{dr}[/itex] is it just [itex]\frac{d^2f}{dr^2}[/itex]?

    And I keep getting [itex]\frac{df}{dr}[/itex][itex]\nabla[/itex][itex]\frac{1}{r}[/itex]=[itex]\nabla[/itex]f, which doesnt seem right.
  7. Apr 14, 2012 #6


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    No, it is [itex]\frac{d^2f}{dr^2}\frac{\vec r}{r}[/itex]

  8. Apr 14, 2012 #7
    Ahhhh yes, got it, it all works out now. Thanks a lot mate.
    But just how did you get that? I presume it's quite similar to the first question, chain rule again?
  9. Apr 14, 2012 #8


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    Yes, it is the chain rule again.:smile:

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