Show that |x(t)|d/dt*|x(t)| = x(t)x`(t)

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SUMMARY

The discussion focuses on proving the equation |x(t)| d/dt |x(t)| = x(t) x`(t). The user attempts to demonstrate this using the function x(t) = 2t, resulting in the derivative x`(t) = 2. They calculate |x(t)| and |x`(t)|, yielding consistent results. The conversation emphasizes the importance of considering different cases for x(t) based on its sign and suggests using the derivative of the absolute value function with the chain rule for a comprehensive proof applicable to any path x(t).

PREREQUISITES
  • Understanding of calculus, specifically differentiation and the chain rule.
  • Familiarity with absolute value functions and their properties.
  • Knowledge of piecewise functions to handle cases where x(t) changes sign.
  • Basic experience with mathematical proofs and problem-solving techniques.
NEXT STEPS
  • Study the chain rule in depth, particularly its application to absolute value functions.
  • Explore piecewise function definitions and their implications in calculus.
  • Practice deriving and manipulating functions to solidify understanding of derivatives.
  • Investigate the implications of differentiating functions at critical points, such as where x(t) = 0.
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Students studying calculus, particularly those focusing on differentiation and mathematical proofs, as well as educators looking for examples of teaching absolute value derivatives.

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Homework Statement



[itex]Show that |x(t)| d/dt |x(t)| = x(t) \cdot x`(t)[/itex]

The Attempt at a Solution



I don't know what they want to see here but my go at this is to take x(t) = 2t. Differentiating this gives x`(t) = 2.

Taking the normal of both |x(t)| and |x`(t)| gives 2t and 2, respectively.

So 2t * 2 = 4t.

Then just do the same with x(t) and x`(t) to get 4t again?

I don't know if this will suffice as a proper answer
 
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I assume they expect you to show the result for any path x(t), not just x(t)=2t .

I think working with the cases (for the absolute value) ; when is x(t)>0 , when is it <0 ,=0 ,

etc. will be helpful.
 
Hint: Use the fact that [itex]\frac{d}{dt} |t|=\frac{t}{|t|}[/itex] with the chain rule to compute your derivative in the two cases [itex]x(t) > 0[/itex] and [itex]x(t) < 0[/itex] . You have to consider the case [itex]x(t)=0[/itex] carefully.
 

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