Show Derivative of Function Takes Same Form as

Saladsamurai
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Homework Statement



Screenshot2011-01-23at53228PM.png


Homework Equations



Chain Rule

The Attempt at a Solution



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

By the chain rule:

\frac{\partial{f}}{\partial{t&#039;}} = <br /> \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t&#039;}} +<br /> \frac{\partial{f}}{\partial{t}}\frac{\partial{t}}{\partial{t&#039;}}<br />

\Rightarrow<br /> \frac{\partial{f}}{\partial{t&#039;}} = <br /> \frac{\partial{f}}{\partial{x}} * <br /> \left [ \frac{\partial{x}}{\partial{t&#039;}}+V+t&#039;\frac{\partial{V}}{\partial{t&#039;}}\right ]<br /> + \frac{\partial{f}}{\partial{t}}\frac{\partial{t&#039;}}{\partial{t&#039;}}<br />

\Rightarrow<br /> \frac{\partial{f}}{\partial{t&#039;}} = <br /> \frac{\partial{f}}{\partial{x}} * <br /> \left [ \frac{\partial{x}}{\partial{t&#039;}}+V+t&#039;\frac{\partial{V}}{\partial{t&#039;}}\right ]<br /> + \frac{\partial{f}}{\partial{t}}<br />

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 \rho&#039; and v&#039; into the x-t coordinates?
 
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Note that there are only partial derivatives.
So then isn't
\frac{\partial x}{\partial t&#039;} = V?
 
You made a mistake:
<br /> \frac{\partial f}{\partial t&#039;}=V\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}<br />
as:
<br /> \frac{\partial x}{\partial t&#039;}=V\quad\frac{\partial t}{\partial t&#039;}=1<br />
 
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hunt_mat said:
You made a mistake:
<br /> \frac{\partial f}{\partial t&#039;}=V\frac{\partial f}{x}+\frac{\partial f}{\partial t}<br />
as:
<br /> \frac{\partial x}{\partial t&#039;}=V\quad\frac{\partial t}{\partial t&#039;}=1<br />

Shouldn't that be \frac{\partial f}{\partial x}?
 
Hi Saladsamurai! :smile:

(have a rho: ρ and a curly d: ∂ :wink:)
Saladsamurai said:
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 \rho&#039; and v&#039; into the x-t coordinates?

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