My Math Forum Divergence of a vector

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 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.

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