Partial derivatives with Wave Function

AI Thread Summary
The discussion focuses on finding the partial derivatives of the wave function y(x,t) = Acos(kx-ωt). The solutions provided include dy/dt = ωAsin(kx-ωt), dy/dx = -kAsin(kx-ωt), d^2y/dt^2 = -(ω^2)Acos(kx-ωt), and d^2y/dx^2 = -(k^2)Acos(kx-ωt). Questions raised include the reasoning behind treating kx and ωt as constants during differentiation and the significance of the derivatives, particularly dy/dx as a slope and d^2y/dx^2. The discussion concludes with a reference to a relationship involving the second derivatives with respect to time and space.
abelanger
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Homework Statement


Knowing: y(x,t) = Acos(kx-ωt)
Find the partial derivatives of:
1) dy/dt
2) dy/dx
3) d^2y/dt^2
4) d^2y/dx^2

Homework Equations




The Attempt at a Solution


These are the answers the actual answers:
1) dy/dt = ωAsin(kx-ωt) = v(x,t) of a particle
2) dy/dx = -kAsin(kx-ωt)
3) d^2y/dt^2 = -(ω^2)Acos(kx-ωt) = a(x,t) of a particle
4) d^2y/dx^2 = -(k^2)Acos(ks-wt)

now here are my questions:
1) how come when I do the partial derivative of y with respect to t, kx becomes the constant and vice versa with dy/dx, how come ωt becomes the constant? is it because of implicit differentiation?
2) What does it give me to find: dy/dx? the slope? also what about d^2y/dx^2?

Thanks for the Help!
 
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abelanger said:
also what about d^2y/dx^2?

Thanks for the Help!

<br /> \dfrac{\partial ^{2}y}{\partial x^{2}}=\dfrac {1}{c^2}\dfrac{\partial ^{2}y}{\partial t^{2}}<br />
 
klondike said:
<br /> \dfrac{\partial ^{2}y}{\partial x^{2}}=\dfrac {1}{c^2}\dfrac{\partial ^{2}y}{\partial t^{2}}<br />

Ooooh.. Interesting!

Thanks
 
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