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 Calculus Calculus Math Forum

 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^(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. Tags calculus, divergence, functions, identities, nabla Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post vysero Calculus 0 December 5th, 2014 09:21 PM razzatazz Real Analysis 8 May 10th, 2013 11:30 PM walter r Real Analysis 4 May 4th, 2013 07:34 PM math221 Calculus 3 April 8th, 2013 06:39 AM patient0 Real Analysis 5 December 11th, 2010 06:17 AM

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