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Would this be a general formula for the gradient of a function r^n?

  1. Jan 27, 2017 #1

    grandpa2390

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    Had to find the general formula for the gradient of a function r^n. r is the length of the vector connecting (x,y,z) with (x',y',z')

    I took the gradient of r^n and simplified it. If I plug in any number for in in r^n and go through the process, I will get the same result as if I take this function and plug that number into n at the end.
    Does that make sense? Would this be considered the general formula.
     
  2. jcsd
  3. Jan 27, 2017 #2

    mfb

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    I moved the thread to our homework section.

    It would help to see the formulas you got.

    In general, it should not matter when you plug in values - and in the general case you don't even have to do that.
     
  4. Jan 27, 2017 #3

    grandpa2390

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    I plugged in values to check.

    let R be the vector (x-x')+(y-y')+(z-z')
    let r be the length (sqrt of (x-x')^2...
    R' be the unit vector

    I get the general formula for del(r^n) as (n*R)/(n^(2-n)) or (n*R')/(r^(1-n))
    I plugged in a number for n to check if my result would give me the same answer as if I used that number instead of n and it does. Does that mean that this is a general formula?
     
  5. Jan 27, 2017 #4

    haruspex

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    I guess you mean (n*R)/(r^(2-n)).
    In LaTeX, you have
    ##\nabla|\vec R|^n=n\vec R|\vec R|^{n-2}##, right?
     
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