Why are both r and k in vector notation?

  • #1
gtfitzpatrick
380
0

Homework Statement



if [tex]\phi[/tex] = rk/r[tex]^{3}[/tex] where r=xi + yJ + zk and r is the magnitude of r, prove that [tex]\nabla[/tex][tex]\phi[/tex] = (1/r[tex]^{}5[/tex])(r[tex]^{}2[/tex]k-3(r.k)r


Right I'm not really sure where to start here...
i know

Magnitude r = [tex]\sqrt{x^{2}+y^{2}+z^{2}}[/tex]
 

Answers and Replies

  • #2
gtfitzpatrick
380
0
[tex]\nabla[/tex][tex]\phi[/tex] (is generally) = d[tex]\phi[/tex]/dx x + d[tex]\phi[/tex]/dy y + d[tex]\phi[/tex]/dz z

but could someone please give me a pointer from here. can i just multiply across the x, y, z parts of r by k/r[tex]^{3}[/tex] and then differenciate?
 
  • #3
BrendanH
63
0
Something I am perhaps missing: why are both r and k in vector notation? This would imply a dot product; do you really mean that?

Also, in your second post, you denote the unit vectors by x,y,z. Do you mean i , j , and k? I'm not trying to be a bug! I just want to try and keep consistent, so that the problem can be clarified.
 
Last edited:
  • #4
gtfitzpatrick
380
0
yes in the question they're both in vector notation
 
  • #5
BrendanH
63
0
Alright, then that would imply that [tex]\phi[/tex]=z/r^3
 
  • #6
gtfitzpatrick
380
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sorry i meant i, j, k. I'm not sure what you mean...is z is the dot product of r.k?
 
  • #7
BrendanH
63
0
Yes, exactly. i dotted with j or k is zero. only k dot k is 1, so the coefficient of z is 1.
 
  • #8
gtfitzpatrick
380
0
sorry i don't really follow. I just don't know...
 
  • #9
BrendanH
63
0
(xi + yJ + zk) dotted with k = z

After that, it's just a whole of differentiating. Could you more clearly write the 'proof' against which we should be comparing? If need be, dictate what it says instead of trying to write it out.
 
  • #10
gtfitzpatrick
380
0
Thanks for you patience with me!
so i now have [tex]\phi[/tex] = z/[tex]\sqrt{x^{2}+y^{2}+z^{2}}[/tex] amd now i just differenciate wrt x then y then z and then tidy it all up?
Thanks a mill for all the help
 
  • #11
BrendanH
63
0
yes, except you must cube the [tex]
\sqrt{x^{2}+y^{2}+z^{2}}
[/tex] !
 
  • #12
gtfitzpatrick
380
0
Thanks had it, I'm just tidying it up now...you've put me in great humour...thanks a mill...I'm off to see dinosaur jr as soon as i get it tidied up.Thanks again
 
  • #14
gtfitzpatrick
380
0
so i differenciated wrt x then y then z and tried to tidy it all up but i got1/r[tex]^{5}[/tex](-3(r.k)r)

When i differenciated wrt x i got -3x/r[tex]^{5}[/tex] and similar for y and z was this right?

then i just put these answers into [tex]\nabla\phi[/tex] = dx/d[tex]\phi[/tex] i + dy/d[tex]\phi[/tex] j + dz/d[tex]\phi[/tex] k

which gives
-3xz/r[tex]^{5}[/tex] i - 3yz/r[tex]^{5}[/tex] j - 3zz/r[tex]^{5}[/tex] k

am i right so far? it just seemed to tidy up to 1/r[tex]^{5}[/tex](-3(r.k)r)

when it should be 1/r[tex]^{5}[/tex](r[tex]^{2}[/tex]k-3(r.k)r)
 

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