Partial derivatives with Wave Function

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SUMMARY

The discussion focuses on calculating partial derivatives of the wave function y(x,t) = Acos(kx-ωt). The specific derivatives computed include dy/dt = ωAsin(kx-ωt), dy/dx = -kAsin(kx-ωt), d²y/dt² = -(ω²)Acos(kx-ωt), and d²y/dx² = -(k²)Acos(kx-ωt). Participants clarify the role of constants during differentiation, emphasizing that kx is treated as a constant when differentiating with respect to t, and ωt is treated as a constant when differentiating with respect to x. The discussion also touches on the significance of these derivatives in understanding wave behavior, such as velocity and acceleration.

PREREQUISITES
  • Understanding of wave functions and their mathematical representation
  • Knowledge of partial differentiation techniques
  • Familiarity with trigonometric functions and their derivatives
  • Basic concepts of physics related to wave motion
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  • Study the application of the chain rule in partial differentiation
  • Explore the physical interpretation of wave functions in classical mechanics
  • Learn about the wave equation and its solutions
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Students of physics, mathematics, and engineering who are studying wave mechanics and need to understand the application of partial derivatives in wave functions.

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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