# Chain Rule 1

## 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 $\partial(f)/\partial(t')$

## Homework Equations

Chain rule of calculus

## The Attempt at a Solution

Is

$$\frac{\partial{f}}{\partial{t'}}= \frac{\partial {f}}{\partial{x}}\frac{\partial{x}}{\partial{t'}} + \frac{\partial{f}}{\partial{t}}\frac{\partial{t}}{\partial{t'}}\qquad(1)$$
?

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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tiny-tim
Homework Helper

(have a curly d: ∂ )

yes, it is ∂f/∂t'

f is a function of both x' and t', so there's no such thing as df/dt'

Ha!

Thanks TT!!!

(have a curly d: ∂ )

yes, it is ∂f/∂t'

f is a function of both x' and t', so there's no such thing as df/dt'
Hi again TT! I have decided that I do not quite agree with this

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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tiny-tim
Homework Helper
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.

But I assumed that in your original question x and t were independent variables (it looks like a Newtonian coordinate tranformation).

Yes-ish …

df/dt = (∂f/∂x)*(∂x/∂t) + ∂f/∂t.

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

Hi tiny-tim! I am having difficulty determining if they are independent 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!

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tiny-tim
Homework Helper

(just got up :zzz: …)
I am having difficulty determining if they are independent 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).

(it looks like a Newtonian coordinate tranformation).

oops! i should have said "Galilean"!

Sorry, Galileo!

(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).

oops! i should have said "Galilean"!

Sorry, Galileo!
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?

tiny-tim
Homework Helper
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.