Show Derivative of Function Takes Same Form as

  • #1
3,003
3

Homework Statement



Screenshot2011-01-23at53228PM.png


Homework Equations



Chain Rule

The Attempt at a Solution



So we have that [itex]f = f(x,t)[/itex] as well as the transformations [itex]x = x' + Vt'[/itex] and [itex] t = t'[/itex]

By the chain rule:

[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'}}
[/tex]

[tex]\Rightarrow
\frac{\partial{f}}{\partial{t'}} =
\frac{\partial{f}}{\partial{x}} *
\left [ \frac{\partial{x}}{\partial{t'}}+V+t'\frac{\partial{V}}{\partial{t'}}\right ]
+ \frac{\partial{f}}{\partial{t}}\frac{\partial{t'}}{\partial{t'}}
[/tex]

[tex]\Rightarrow
\frac{\partial{f}}{\partial{t'}} =
\frac{\partial{f}}{\partial{x}} *
\left [ \frac{\partial{x}}{\partial{t'}}+V+t'\frac{\partial{V}}{\partial{t'}}\right ]
+ \frac{\partial{f}}{\partial{t}}
[/tex]

Not really sure what the next move is? Is the above equation in its simplest form? Or can I do something more with it?

Also, I don't really see how I go about transforming [itex]\rho'[/itex] and [itex]v'[/itex] into the x-t coordinates?
 
Last edited:

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
Note that there are only partial derivatives.
So then isn't
[tex]\frac{\partial x}{\partial t'} = V[/tex]?
 
  • #3
hunt_mat
Homework Helper
1,742
26
You made a mistake:
[tex]
\frac{\partial f}{\partial t'}=V\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}
[/tex]
as:
[tex]
\frac{\partial x}{\partial t'}=V\quad\frac{\partial t}{\partial t'}=1
[/tex]
 
Last edited:
  • #4
Char. Limit
Gold Member
1,204
14
You made a mistake:
[tex]
\frac{\partial f}{\partial t'}=V\frac{\partial f}{x}+\frac{\partial f}{\partial t}
[/tex]
as:
[tex]
\frac{\partial x}{\partial t'}=V\quad\frac{\partial t}{\partial t'}=1
[/tex]

Shouldn't that be [tex]\frac{\partial f}{\partial x}[/tex]?
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi Saladsamurai! :smile:

(have a rho: ρ and a curly d: ∂ :wink:)
Not really sure what the next move is? Is the above equation in its simplest form? Or can I do something more with it?

Now you put in ∂x'/∂t etc, which you can read off the Galilean transformation. :smile:
Also, I don't really see how I go about transforming [itex]\rho'[/itex] and [itex]v'[/itex] into the x-t coordinates?

But the question tells you that … ρ' = ρ, v' = v - V. :confused:
 

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