December 7th, 2014, 09:32 AM  #1 
Newbie Joined: Dec 2014 From: Glasgow Posts: 1 Thanks: 0  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^(n2) 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. 

Tags 
calculus, divergence, functions, identities, nabla 
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