Grad of a generalised scalar function

  • Thread starter Phyrrus
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  • #1
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Homework Statement



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

Homework Equations



Vector identities?

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?
 
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Answers and Replies

  • #2
ehild
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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)

ehild
 
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  • #3
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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?
 
  • #4
ehild
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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].

ehild
 
  • #5
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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.
 
  • #6
ehild
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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]?

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

ehild
 
  • #7
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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?
 
  • #8
ehild
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Yes, it is the chain rule again.:smile:

ehild
 

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