Deriving with multiple variables

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SUMMARY

The discussion focuses on deriving the relativistic energy equation, specifically proving that the derivative of the expression \(\frac{mu}{\sqrt{1-u^2/c^2}}\) with respect to time \(t\) equals \(\frac{m \frac{du}{dt}}{(1-u^2/c^2)^{3/2}}\). The key to solving this derivation lies in applying the chain rule correctly, recognizing that speed \(u\) is a function of time \(t\). The participant successfully identifies that the relationship \(u = u(t)\) allows for the use of the chain rule to simplify the derivation process.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and the chain rule.
  • Familiarity with relativistic physics concepts, particularly the relationship between speed and energy.
  • Knowledge of the notation and terminology used in physics, such as \(u\) for speed and \(c\) for the speed of light.
  • Basic understanding of how to manipulate algebraic expressions involving square roots and fractions.
NEXT STEPS
  • Study the application of the chain rule in calculus, focusing on derivatives of composite functions.
  • Explore the principles of relativistic mechanics, particularly the derivation of energy equations.
  • Practice problems involving derivatives of functions that depend on multiple variables.
  • Review the mathematical representation of physical concepts, such as speed and acceleration in the context of relativity.
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying relativistic mechanics, as well as educators and tutors looking to enhance their understanding of calculus applications in physics.

Helices
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Homework Statement


I'm stuck at one of the derivations for relativistic energy. I've figured out every other single step, but I just can't wrap my head around this one:

Prove that:

{\frac{d}{dt}} {\frac {mu} {\sqrt{1-u^2/c^2}}} = {\frac {m {\frac{du} {dt}}} {(1-u^2/c^2)^{3/2}}}


Homework Equations


u is speed, so:

u = dx/dt
I don't know if that's helpful.


The Attempt at a Solution


I've tried everything that's in my calculus toolbox, but I guess that's not a whole lot. I know how to derive basic functions, but I just can't seem to figure out how turn this into something I can work with. It says to derive to t, but there's not even a t in the function. I know that u and t are related in a way, but substituting just leads to more trouble.
 
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Speed depends on time, u = u(t), so you just need to use the chain rule:\frac{d f(u(t))}{dt} = \frac{df}{du} \frac{du}{dt}
 
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Alright, I've got it now. Thanks!
 

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