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March 14th, 2019, 01:23 AM  #1 
Member Joined: Apr 2017 From: India Posts: 49 Thanks: 0  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. 
March 14th, 2019, 04:09 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics 
Have you tried computing the divergence?

March 14th, 2019, 06:12 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,079 Thanks: 845 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \vec{ \nabla } \cdot \left ( \begin{matrix} a_x \\ a_y \end{matrix} \right ) = \dfrac{d}{dx} (a_x) + \dfrac{d}{dy} (a_y)$ $\displaystyle a_x = cos( \theta )$ 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: 49 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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276  Quote:
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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dimensions, divergence, unit, vector 
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