1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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)

    ehild
     
    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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].

    ehild
     
  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

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
  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

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Grad of a generalised scalar function
  1. Scalar Function Help (Replies: 1)

Loading...