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November 14th, 2015, 08:47 AM  #1 
Newbie Joined: Nov 2015 From: Morpeth.UK Posts: 3 Thanks: 0  Help with velocity in two dimensions
There is an impasse at the physics forum. We have a velocity: U^2 = U_x^2 + U_y^2 We have two expressions that seem to be incompatible. dU^2 = dU_x^2 + dU_y^2 which looks believable when you draw a graph... And ... dU=(U_x dU_x +U_y dU_y)/U derived from the partial derivatives. The second one seems to be undefined when U=0. Can anyone shed any light on this and if the two expressions can be reconciled? (Sorry I cannot the the latex to workany insights into how to do this would be greatly appreciated!) This discussion has been going on for over a week now! 
November 14th, 2015, 10:06 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 986 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
November 14th, 2015, 01:01 PM  #3 
Newbie Joined: Nov 2015 From: Morpeth.UK Posts: 3 Thanks: 0 
Dan, thanks for that. But if your consider U=0, it breaks down. Also if you draw the graphs only the first equation makes any sense.

November 14th, 2015, 03:49 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 986 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Let's do this a bit more detail. $\displaystyle U_x$ and $\displaystyle U_y$ are velocity components in the x and y directions respectively. So we indeed have $\displaystyle [d(U)]^2 = [d(U_x)]^2 + [d(U_y)]^2$ as you stated. Now look at $\displaystyle U^2 = U_x^2 + U_y^2$. The LHS: $\displaystyle d(U^2) = \frac{\partial (U^2)}{\partial x} ~ dx + \frac{\partial (U^2)}{\partial y}~dy$ So $\displaystyle d(U^2) = 2 U \frac{\partial U}{\partial x} ~ dx + 2 U \frac{\partial U}{\partial y}~dy$ The RHS: $\displaystyle d(U_x^2 + U_y^2) = 2 U_x \frac{\partial (U_x)}{\partial x}~dx + 2 U_x \frac{\partial (U_x)}{\partial y}~dy + 2U_y \frac{\partial (U_y)}{\partial x}~dx + 2U_y \frac{\partial (U_y)}{\partial y}~dy$ I see no way to make LHS = RHS look anything like your second equation. Dan Last edited by topsquark; November 14th, 2015 at 03:52 PM.  
November 14th, 2015, 11:31 PM  #5 
Newbie Joined: Nov 2015 From: Morpeth.UK Posts: 3 Thanks: 0 
Thanks Dan. Perhaps the confusion has been in the placing of the brackets? (Differentiating U^2 is clearly not the same thing as differentiating U then squaring it.)

November 15th, 2015, 10:41 AM  #6  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 986 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  

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