1. Limited time only! Sign up for a free 30min personal 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!

Gradient of |xi + yj + zk|^-n

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f(x,y,z)= |r|-n where r = x[tex]\hat{i}[/tex] + y[tex]\hat{j}[/tex] + z[tex]\hat{k}[/tex]

    Show that

    [tex]\nabla[/tex] f = -nr / |r|n+2

    2. The attempt at a solution
    Ok, I don't care about the absolute value (yet at least).

    I take partial derivatives of (xi + yj + zk)^-n and get

    [tex]\nabla[/tex] f = i(-n)(xi + yj + zk)^(-n-1) + j(-n)(xi + yj + zk)^(-n-1) + k(-n)(xi + yj + zk)^(-n-1)

    = -n(i + j + k)*(xi + yj + zk)^(-n-1)

    But according to problem statement what I should get is:
    -nr / |r|n+2 = -n (i x + j y + k z)^(-1 - n)

    I don't understand where the (i + j + k) term goes! :eek:
     
  2. jcsd
  3. Mar 21, 2009 #2
    The | | does not refer to the absolute value, but the norm in this case. In fact xi+yj+zk is a vector and it does not make sense to compute the power of a vector.

    So what you have to use is that [tex]|r|=\sqrt{x^2+y^2+z^2}[/tex], then compute the partials.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gradient of |xi + yj + zk|^-n
Loading...