My Math Forum Divergence of unit vector in two dimensions

 Calculus Calculus Math Forum

March 14th, 2019, 01:23 AM   #1
Member

Joined: Apr 2017
From: India

Posts: 73
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.
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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,256
Thanks: 926

Math Focus: Wibbly wobbly timey-wimey stuff.
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: 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
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,529
Thanks: 1389

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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top