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Why are both r and k in vector notation?

  1. May 12, 2008 #1
    1. The problem statement, all variables and given/known data

    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]
  2. jcsd
  3. May 12, 2008 #2
    [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?
  4. May 12, 2008 #3
    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: May 12, 2008
  5. May 12, 2008 #4
    yes in the question they're both in vector notation
  6. May 12, 2008 #5
    Alright, then that would imply that [tex]\phi[/tex]=z/r^3
  7. May 12, 2008 #6
    sorry i meant i, j, k. I'm not sure what you mean...is z is the dot product of r.k?
  8. May 12, 2008 #7
    Yes, exactly. i dotted with j or k is zero. only k dot k is 1, so the coefficient of z is 1.
  9. May 12, 2008 #8
    sorry i dont really follow. I just dont know...
  10. May 12, 2008 #9
    (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.
  11. May 12, 2008 #10
    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
  12. May 12, 2008 #11
    yes, except you must cube the [tex]
    [/tex] !!!!!
  13. May 12, 2008 #12
    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. May 12, 2008 #13
  15. May 13, 2008 #14
    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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