Calculating Partial Derivative of f with Respect to t'”

In summary, if f is the velocity of the fluid at each particular place and time, then in the first case it's speed = f(x,t), and in the second case it's speed = f(x0,t).
  • #1
Saladsamurai
3,020
7

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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  • #2
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
Ha!

Thanks TT! :smile:
 
  • #4
tiny-tim said:
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:

Hi again TT! I have decided that I do not quite agree with this :redface:

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)

See http://en.wikipedia.org/wiki/Total_derivative#The_total_derivative_via_differentials" maybe I am missing something?
 
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  • #5
Saladsamurai said:
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
tiny-tim said:
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 no one will notice :wink:
 
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  • #7
Hi Saladsamurai! :smile:

(just got up :zzz: …)
Saladsamurai said:
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:
tiny-tim said:
(it looks like a Newtonian coordinate tranformation). :wink:

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

Sorry, Galileo! :biggrin:
 
  • #8
tiny-tim said:
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
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.
 

What is a partial derivative?

A partial derivative is a mathematical concept that calculates the rate of change of a function with respect to one of its variables while holding all other variables constant.

Why do we need to calculate partial derivatives?

Calculating partial derivatives allows us to understand how a function changes in relation to one specific variable, which can be useful in areas such as physics, economics, and engineering.

How do you calculate a partial derivative?

To calculate a partial derivative, you first need to determine which variable you are taking the derivative with respect to. Then, you simply treat all other variables as constants and use the rules of differentiation to find the derivative of the function.

What does it mean to take the partial derivative of a function with respect to time?

Taking the partial derivative of a function with respect to time means that we are looking at how the function changes over time, while keeping all other variables constant. This is useful in areas such as physics, where time is a crucial variable.

Can partial derivatives be negative?

Yes, partial derivatives can be negative. This indicates that the function is decreasing in value with respect to the variable being considered. It is important to keep in mind the context of the function and variable when interpreting a negative partial derivative.

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