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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 work-any 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
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 Originally Posted by Jilan 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 work-any insights into how to do this would be greatly appreciated!) This discussion has been going on for over a week now!
The first expression is not correct. When you take the derivative of $\displaystyle U^2$ it becomes $\displaystyle 2U~dU_x$ not $\displaystyle (dU)^2$. The second equation is correct.

-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
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Quote:
 Originally Posted by Jilan 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 work-any insights into how to do this would be greatly appreciated!) This discussion has been going on for over a week now!
Yes. I made a boo-boo. Now I'm saying that the first equation is correct and the second is incorrect.

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
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 Originally Posted by Jilan 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.)
It could be that. The confusion could also be that the x component of the velocity $\displaystyle U_x$ is also the same symbol for the partial derivative of U in the x direction $\displaystyle U_x$.

-Dan

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