Help with understanding the linear wave equation

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jwxie
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Homework Statement



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

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

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

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

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 [tex]\[\frac{d_{y}}{d_{x}}\][/tex]

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

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!
 
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jwxie said:
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.
jwxie said:
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.
 
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

[tex]\frac{dy}{dx} \,\[/tex]

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 [tex]\[dx\dot{i} + dy\dot{j} \][/tex]. The tangent of the angle wuth respect to the x-axis for this displacement is [tex]\[\frac{\mathrm{dy} }{\mathrm{d} x}\][/tex]."

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 :)
 
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? :wink: Try this: just draw a straight line on a graph, and draw a triangle to figure out the slope as [itex]\Delta x/\Delta y[/itex]. Then using the same triangle, find the angle between the line and the x-axis.