Partial Derivatives and the Linear Wave Equation

4 replies · 2K views · last reply
Taulant Sholla
Messages
96
Reaction score
5

Homework Statement


I'm reading through the derivations of the linear wave equation. I'm following everything, except the passage I highlighted in yellow in the below attachment:
Textbook page 16.6 The Linear Wave Equation showing equations 16.22 and 16.23 for net transverse force on a string element, with Figure 16.19 of a string element under tension T at angles theta A and theta B.


Homework Equations


I'm not understanding why partials must be used because "we evaluate this tangent at a particular instant of time."

The Attempt at a Solution


It seems if x, y, and t were all changing then there'd be a need for a partial. As stated in the highlighted text, I'm not getting it. Any help is appreciated.
 
The displacement y of the string from its equilibrium position (y=0) is a function of both x and t - it depends on where along the string you look, and on the time at which you look at it. Figure 16.19 shows the position of the string at a given time t. So when you find the slope of the tangent at this time t, you are holding the time constant. That means that you must take the partial derivative of y with respect to x - the definition of a partial derivative requires you to hold fixed all independent variables other than the one by which you differentiate.

On your point 3, it is because you are holding t fixed that you need to use the partial derivative. Note that x and t are your independent variables, and that y is the dependent variable. were you to allow both x and t to change, you would need a different kind of derivative that you will probably not have studied yet, and its physical interpretation would be quite different.
 
Very helpful - thank you so much! What kind of derivative is needed if x and t were both changing?
 
It is called the directional derivative, or the covariant derivative.