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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 February 20th, 2016, 05:19 AM #1 Newbie   Joined: Feb 2016 From: UK Posts: 2 Thanks: 0 Divergence of a vector Hi, I was wondering if you could help me. I'm trying to get the divergence of my vector: [cos(x)sin(y) + picos(y)sin(x), picos(x)sin(y) + (picos(y)(sin(x)), picos(y)sin(x) + cos(x)sin(y)] however, to take the divergence of something, it is usually (dP/dx, dM/dy, dN/dz) but since there is no z component in the bottom of the vector, does that mean that it goes to 0? if so, can I cancel the vector down to a 1x2 instead of a 1x3 if needs be? Thank you! March 1st, 2016, 03:18 PM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 647 Thanks: 94 $\displaystyle divf=\nabla \cdot f$ or $\displaystyle \nabla_{x,y} \cdot \left( \begin{array}{c} P \\ Q \\R\end{array} \right)=\nabla_{x,y,z} \cdot \left( \begin{array}{c} P \\ Q \\empty-or-0\end{array} \right)$ Since $\displaystyle R\vec{k}\cdot 0 = 0 \cdot A\vec{k}$ it shows that your way is correct Last edited by idontknow; March 1st, 2016 at 03:20 PM. March 18th, 2016, 09:08 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The "divergence" of a vector valued function is NOT a vector as you show. $\displaystyle \nabla \cdot \left< P, Q, R\right>=\frac{\partial P}{\partial x}+ \frac{\partial Q}{\partial y}+ \frac{\partial R}{\partial z}$. Here, since R does not depend on z, that last derivative will be 0. Last edited by greg1313; April 29th, 2016 at 04:13 AM. Tags divergence, vector 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 harley05 Calculus 3 June 10th, 2014 10:16 AM solrob Applied Math 2 November 10th, 2013 09:06 AM razzatazz Real Analysis 8 May 10th, 2013 11:30 PM MasterOfDisaster Calculus 2 September 26th, 2011 09:17 AM patient0 Real Analysis 5 December 11th, 2010 06:17 AM

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