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December 7th, 2014, 09:32 AM   #1
Joined: Dec 2014
From: Glasgow

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Divergence of Functions.

Studying functions, in particular divergence and curl of functions for upcoming calculus exam.

Struggling to understand why (grad(r^n).(cxr) = 0 in the question:

Let r = (x, y, z) denote a position vector with length r =(x^2+y^2+z^2) and c is a constant
(vector) (c1,c2,c3). Prove div(r^n(c × r)) = 0

Ok, so using nabla identites ive got it down to div(r^n(cxr)) = r^ndiv(cxr) + grad(r^n).(cxr)

r^ndiv(cxr) = 0 is obvious after calculating cxr.

grad(r^n).(cxr) = (nr^(n-2) r).(cxr).

Struggling to understand why this is equal to 0.

Note: Notation of vectors here is horrible, i lost track.

Please feel free to diverge (oh the pun) onto general functions/identities/proofs.

Last edited by SeanRobinson; December 7th, 2014 at 09:52 AM.
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