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A_Helal · Nov 10, 2015, 11:19 AM

my question in the photo

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ebaines · Nov 10, 2015, 02:01 PM

Looks like the question is incomplete. In this part:

"and $\frac {\partial F(x+ct)}{\partial t}$ at (t=0).."

shouldn't that partial derivative be equal to something?

A_Helal · Nov 10, 2015, 02:06 PM

no I just want to show that the function is partial derivative and it will be total when t = 0

A_Helal · Nov 10, 2015, 02:15 PM

the question in an other way
If I have a function of x and t it's derivative with respect to t in case of t=0 is equal to it's dervative with respect to x why ? Why not ?

ebaines · Nov 10, 2015, 03:00 PM

Consider the total derivates with respect to x and t:

$\frac {dF}{dt}= \frac {\partial F}{\partial t} + \frac {\partial F}{\partial x}\frac {dx}{dt}$

$\frac {dF}{dx}= \frac {\partial F}{\partial x} + \frac {\partial F}{\partial t}\frac {dt}{dx}$

If these are equal then

$\frac {\partial F}{\partial t} + \frac {\partial F}{\partial x}\frac {dx}{dt} =\frac {\partial F}{\partial x} + \frac {\partial F}{\partial t}\frac {dt}{dx}$

This could be true if $\frac {dx}{dt} = \frac {dt}{dx} = 1$