Calculus Calculus Math Forum

 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,345
Thanks: 986

Math Focus: Wibbly wobbly timey-wimey stuff.
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!
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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,345
Thanks: 986

Math Focus: Wibbly wobbly timey-wimey stuff.
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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,345
Thanks: 986

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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 Tags dimensions, velocity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post quarkz Calculus 0 April 18th, 2014 06:34 AM MattJ81 New Users 3 July 28th, 2011 12:40 AM bignick79 Algebra 1 June 24th, 2010 06:21 PM Joachim Linear Algebra 1 October 26th, 2009 11:41 AM redandtheblue Applied Math 0 February 15th, 2009 12:07 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      