Calculating the Gradient of a Vector Function with a Power Function

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SUMMARY

The discussion focuses on calculating the gradient of the vector function defined as f(x,y,z) = |r|^-n, where r = xî + yĵ + zk. The correct gradient is established as ∇f = -nr / |r|^(n+2). Participants clarify that the absolute value notation refers to the norm of the vector r, leading to confusion regarding the presence of the (i + j + k) term in the gradient calculation. The resolution emphasizes the importance of recognizing |r| as the Euclidean norm, which is calculated as √(x² + y² + z²).

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Homework Statement


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:
 
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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.
 

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