Derivative of the magnitude of a function

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Discussion Overview

The discussion revolves around the derivative of the magnitude of a vector function, specifically focusing on the position vector equation r(t). Participants explore the implications of the derivative expression and its relationship to the vector itself.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant presents the derivative of the magnitude of the vector function as d/dx|r(t)|=1/|r(t)| X r(t) * r'(t), questioning its meaning in relation to the derivative of r(t).
  • Another participant suggests that the expression ##\frac{r(t)}{|r(t)|}## represents the sign of r(t), indicating that if r(t) is positive, the derivative simplifies to the regular derivative, while if r(t) is negative, it adjusts for the sign correctly.
  • A participant reminds others that the magnitude can be expressed as ## |r(t)| = \sqrt{r(t) \cdot r(t)} ## and encourages differentiation of this expression.
  • Another participant reiterates the same expression for magnitude and proposes starting with ## |r(t)|^2 = r(t) \cdot r(t) ## as a simpler approach for differentiation.

Areas of Agreement / Disagreement

Participants present various interpretations and approaches to the derivative of the magnitude, but there is no consensus on a single method or understanding of the implications.

Contextual Notes

Some assumptions about the positivity or negativity of r(t) are not fully explored, and the discussion does not resolve the mathematical steps involved in differentiation.

better361
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I just learned that d/dx|r(t)|=1/|r(t) |X r(t)*r'(t), where * is the dot product and X is mutiply. What is the meaning of this statement, especially in relation to d/x r(t)=r'(t)?(Lets say r(t) is the position vector equation.)
 
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##\frac{r(t)}{|r(t)|}## is just the sign of r(t). If r(t) is positive, it is +1 and you simply get the regular derivative. If r(t) is negative, |r(t)|=-r(t) and this fraction gives the correct minus sign.
 
Recall that ## |r(t)| = \sqrt {r(t) \cdot r(t)} ##. Differentiate this.
 
voko said:
Recall that ## |r(t)| = \sqrt {r(t) \cdot r(t)} ##. Differentiate this.
A simpler starting point might be to write:

## |r(t)|^2 = r(t) \cdot r(t) ##

Chet
 

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