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February 20th, 2016, 05:19 AM   #1
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Divergence of a vector


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!
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March 1st, 2016, 03:18 PM   #2
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$\displaystyle divf=\nabla \cdot f$ 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.
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March 18th, 2016, 09:08 AM   #3
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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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divergence, vector

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