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Chain Rule 1

  • #1
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Homework Statement



I have some function f(x,t) and I know that both x and t undergo a coordinate transformation according to x = x' + Vt' and t = t'. I am asked to find [itex]\partial(f)/\partial(t')[/itex]




Homework Equations



Chain rule of calculus

The Attempt at a Solution



Is

[tex]\frac{\partial{f}}{\partial{t'}}=
\frac{\partial {f}}{\partial{x}}\frac{\partial{x}}{\partial{t'}}
+
\frac{\partial{f}}{\partial{t}}\frac{\partial{t}}{\partial{t'}}\qquad(1)[/tex]
?

I feel like it is, but I also feel like the equation given by (1) is the total derivative wrt t'. That is, I think (1) is the same as df/dt'.

Can someone help clarify this?
 
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Answers and Replies

  • #2
tiny-tim
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Hi Saladsamurai! :smile:

(have a curly d: ∂ :wink:)

yes, it is ∂f/∂t'

f is a function of both x' and t', so there's no such thing as df/dt' :wink:
 
  • #3
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2
Ha!

Thanks TT!!! :smile:
 
  • #4
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Last edited by a moderator:
  • #5
tiny-tim
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If f = f(x,t) and x = x(t), then the total derivative of f wrt t does indeed exist yes?:

df/dt = (∂f/∂x)*(∂x/∂t)
Yes-ish …

df/dt = (∂f/∂x)*(∂x/∂t) + ∂f/∂t. :smile:

But I assumed that in your original question x and t were independent variables (it looks like a Newtonian coordinate tranformation). :wink:
 
  • #6
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Yes-ish …

df/dt = (∂f/∂x)*(∂x/∂t) + ∂f/∂t. :smile:

But I assumed that in your original question x and t were independent variables (it looks like a Newtonian coordinate tranformation). :wink:
(Edit: ah yes + ∂f/∂t )

Hi tiny-tim! I am having difficulty determining if they are independent :redface: I do not think that they are. It is from a fluid dynamics course, so I believe that x is position x(t) though they don't say it explicitly. I am going to break the rules here and cross-post :shudder:. This is the full problem statement and the progress I have made on it. Perhaps you can respond to that thread if you have any thoughts you would like to share. Thanks again! :smile:

https://www.physicsforums.com/showthread.php?t=466146" ...maybe noone will notice :wink:
 
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  • #7
tiny-tim
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Hi Saladsamurai! :smile:

(just got up :zzz: …)
I am having difficulty determining if they are independent :redface: I do not think that they are. It is from a fluid dynamics course, so I believe that x is position x(t) though they don't say it explicitly.
No.

You're confusing a function with a coordinate, just because it's convenient to call them by the same name.

btw, the function is even more complicated than you're suggesting … the position x of a particle is is a function of t and of x0, the position of that particle at t = 0 … x = x(x0,t). :wink:

You're just going to have to think about this! :smile:
(it looks like a Newtonian coordinate tranformation). :wink:

oops! i should have said "Galilean"! :redface:

Sorry, Galileo! :biggrin:
 
  • #8
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2
Hi Saladsamurai! :smile:

(just got up :zzz: …)


No.

You're confusing a function with a coordinate, just because it's convenient to call them by the same name.

btw, the function is even more complicated than you're suggesting … the position x of a particle is is a function of t and of x0, the position of that particle at t = 0 … x = x(x0,t). :wink:

You're just going to have to think about this! :smile:


oops! i should have said "Galilean"! :redface:

Sorry, Galileo! :biggrin:
I am sorry, but now I am getting more confused. What have we concluded then?
Is f = f(x(t),t) or not? Or as you like to put it f = f(x(x0,t),t) ? Or is it just plain old f = f(x,t) where x and t are completely independent of each other?
 
  • #9
tiny-tim
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If f is, say, the velocity of the fluid at each particular place and time, the you can express f as a function of either

i] that place and time (ie x and t)

or

ii] the original position (at t = 0) of the particle which is now at that place, and time (ie x0 and t)

In the first case, f = f(x,t), in the second case, f = f(x0,t).

But the two fs are (obviously) different functions.

If you choose to follow a particular particle (x0 = 5 say), and find the acceleration,

then in the second case it's speed = f(5,t), so acceleration = ∂f/∂t(5,t),

but in the first case it's speed = f(x(5,t),t), which is more complicated to write and to differentiate.
 

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