Help with understanding the linear wave equation

[jwxie]

Homework Statement

Suppose an element of a string, called \[\triangle x\] with T being the tension.
The net force acting on the element in the vertical direction is

\[\sum F_{y} = Tsin(\theta _{B}) - Tsin(\theta _{A}) = T(sin\theta _{B} - sin\theta _{A})\]<br />

1. Since the angels are small, we can use the small-angle approximation \[sin\theta \approx tan\theta \], and rewrite

\[\sum F_{y} = T(tan\theta _{B} - tan\theta _{A})\]

I know what small-approximation is, but I suspect there is a definitive reason to why we choose sin ~= tan and not sin ~= delta. But y/x is arctan.. if we are talking about that.. So what is it?

If we extended the displacement outward and gives infinitesimal x and y components, then the tangent of the angle with respect to the x-axis for this displacement is \[\frac{d_{y}}{d_{x}}\]

2. Because we evaluate this tangent at a particular instant of time, we must express it in partial form as \[\frac{\partial y }{\partial x}\]

To be more clear, the reason we use partial is because the function contains two variables, x and t, right?
Any help is appreciated! Thank you!

 

[diazona]

I know what small-approximation is, but I suspect there is a definitive reason to why we choose sin ~= tan and not sin ~= delta. But y/x is arctan.. if we are talking about that.. So what is it?

Well, for small-angle approximations you can set sin(θ) ≈ tan(θ) or sin(θ) ≈ θ, depending on which one is more useful for the particular calculation you're doing. In this case it appears that they want to use the derivative dy/dx, which is equal to the tangent of the angle, so it's more useful to choose tangent.

To be more clear, the reason we use partial is because the function contains two variables, x and t, right?

Right, and because you want to take the derivative of y with respect to only x, leaving t constant.

 

[jwxie]

LOL I am so stupid. tan = sin/cos, and I always thought x/y. It was opposite / adjacent, which makes dy/dx.

After reading a bit on derivative on Wiki,

In Leibniz's notation, such an infinitesimal change in x is denoted by dx, and the derivative of y with respect to x is written

\frac{dy}{dx} \,\

According to the book

"Imagine undergoing an infinitesimal displacement outward from the end of the rope element along the blue line representation the force T. This displacement has infinitesimal x and y components and can be represented by vector \[dx\dot{i} + dy\dot{j} \]. The tangent of the angle wuth respect to the x-axis for this displacement is \[\frac{\mathrm{dy} }{\mathrm{d} x}\]."

1. So why do the physicists imagine this "infinitesimal displacement"?2. So in essence, the rate of change, dy/dx gives the rate of change. If we interpret it in dy/dx form, we have the slope of a tangent line. I see the relationship between dy/dx and tan, but how do I see the relationship between the slope of the tangent line and tan?

Thank you! I hope I don't sound dumb :)

 

[diazona]

1. As opposed to a finite displacement or something? I'm not sure I understand what you're confused about here. Generally speaking, that's just one way to think about a derivative, you move an infinitesimal amount in the x direction and see how much your function changes in the y direction. (If it were a finite displacement, you might have different slopes at different points within the interval.)

2. Well why do you think they call it the tangent function? Try this: just draw a straight line on a graph, and draw a triangle to figure out the slope as \Delta x/\Delta y. Then using the same triangle, find the angle between the line and the x-axis.

 

[jwxie]

Thank you!