Proving d<r>/dt = <dr/dt>: A Case Study

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SUMMARY

The discussion centers on proving the equation d/dt = using the running average formula (t)/dt = 1/2Pi * Integrate[r(t)*dt,{r,t-Pi,t+Pi}]. The user seeks assistance in demonstrating that the order of differentiation and time-averaging does not affect the outcome. The application of Leibniz's rule and the symmetry of the integration interval are suggested as potential methods for solving the problem.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and integration.
  • Familiarity with Leibniz's rule for differentiation under the integral sign.
  • Knowledge of sinusoidal functions and their properties.
  • Basic concepts of running averages in mathematical analysis.
NEXT STEPS
  • Study the application of Leibniz's rule in calculus.
  • Research the properties of sinusoidal functions and their integrals.
  • Explore the concept of running averages in mathematical contexts.
  • Practice problems involving differentiation and integration of oscillatory functions.
USEFUL FOR

Students studying calculus, particularly those tackling problems involving differentiation and integration of oscillatory functions, as well as anyone preparing for advanced mathematics coursework.

Silva_physics
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Let <r> (t)/dt= 1/2Pi * Integrate[r(t)*dt,{r,t-Pi,t+Pi}] denote the running average of r over one cycle of the sinosoidal oscillation.

I have to show that d<r>/dt = <dr/dt>, it does not matter whether we differentiate or time-average first.


Should I work with Leibniz rule? Can I use some symmetry of interval, I don't really know, something I tried to do but not successfully.

Can anybody, please, show solution?
 
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