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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.
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 March 14th, 2019, 04:09 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 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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Quote:
 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: 64 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.

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

$\nabla\cdot r = \dfrac{y^2}{\left(x^2+y^2\right)^{3/2}} + \dfrac{x^2}{\left(x^2+y^2\right)^{3/2}} = \dfrac{1}{\sqrt{x^2 + y^2}} =\dfrac 1 r$

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