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March 14th, 2019, 01:23 AM   #1
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Divergence of unit vector in two dimensions

I am confused in opting the correct option of this question. Please help me with the correct answer. This is a part of my assignment.  Attached Images 20190314_144026.jpg (79.5 KB, 23 views) March 14th, 2019, 04:09 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics Have you tried computing the divergence? Thanks from topsquark March 14th, 2019, 06:12 PM   #3
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 Originally Posted by shashank dwivedi I am confused in opting the correct option of this question. Please help me with the correct answer. This is a part of my assignment.  Probably the most direct way is to use xy coordinates. You have a vector in xy components as $\displaystyle \left ( \begin{matrix} a_x \\ a_y \end{matrix} \right )$ so

$\displaystyle \vec{ \nabla } \cdot \left ( \begin{matrix} a_x \\ a_y \end{matrix} \right ) = \dfrac{d}{dx} (a_x) + \dfrac{d}{dy} (a_y)$

$\displaystyle a_x = cos( \theta )$ and $\displaystyle a_y = sin( \theta )$

How do you take the derivatives?

The other way is to convert $\displaystyle cos( \theta ) \hat{i} + sin( \theta ) \hat{j}$ to polar coordinates and then take the divergence. As the problem was given in rectangular coordinates I suspect they want the first way I suggested.

-Dan March 15th, 2019, 09:43 AM #4 Member   Joined: Apr 2017 From: India Posts: 73 Thanks: 0 When I tried solving it, I got the answer as 1/r however, that is not there in the option. That is the reason why I posted the question for help. I would love to see the approach opted by the mathematicians over here. Thank you. March 15th, 2019, 09:59 AM   #5
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Quote:
 Originally Posted by shashank dwivedi When I tried solving it, I got the answer as 1/r however, that is not there in the option. That is the reason why I posted the question for help. I would love to see the approach opted by the mathematicians over here. Thank you.
I get $\dfrac 1 r$ as well

One has to assume that $r,~\theta$ are the usual polar coordinates.

$r = \left(\dfrac{x}{\sqrt{x^2+y^2}},\dfrac{y}{\sqrt{x^ 2+y^2}}\right)$

$\nabla\cdot r = \dfrac{y^2}{\left(x^2+y^2\right)^{3/2}} + \dfrac{x^2}{\left(x^2+y^2\right)^{3/2}} = \dfrac{1}{\sqrt{x^2 + y^2}} =\dfrac 1 r$ Tags dimensions, divergence, unit, 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 mojojojo17 Differential Equations 2 March 18th, 2016 09:08 AM harley05 Calculus 3 June 10th, 2014 10:16 AM solrob Applied Math 2 November 10th, 2013 09:06 AM furor celtica Algebra 8 January 26th, 2012 10:09 AM MasterOfDisaster Calculus 2 September 26th, 2011 09:17 AM

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