Finding the major and minor axis of ellipse

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Discussion Overview

The discussion revolves around the mathematical approach to finding the major and minor axes of an ellipse, specifically examining the use of derivatives in the context of the equation \(\rho(t)^2 = x^2(t) + y^2(t)\). The focus is on the reasoning behind choosing to differentiate \(\rho^2(t)\) rather than \(\rho(t)\) itself.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant notes that the ellipse is represented by \(\rho(t)^2 = x^2(t) + y^2(t)\) and questions the choice of using \(\frac{d(\rho^2(t))}{d t}=0\) instead of \(\frac{d\rho(t)}{d t}=0\).
  • Another participant suggests that differentiating \(\rho^2(t)\) is easier and yields the same results as long as \(\rho\) is not zero.
  • A further reply confirms that the choice is primarily to avoid the complexity introduced by the square root in the expression.
  • One participant expresses concern about potentially missing a critical aspect of the reasoning.

Areas of Agreement / Disagreement

Participants appear to agree on the reasoning that differentiating \(\rho^2(t)\) simplifies the process, but there is no consensus on whether there are additional considerations that might warrant using \(\frac{d\rho(t)}{d t}=0\).

Contextual Notes

There is an implicit assumption that \(\rho\) is never zero, which may affect the validity of the reasoning presented.

yungman
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An ellipse is represented by \rho(t)^2=x^2(t) + y^2(t) where \rho(t) is the distance from origin to the ellipse at a given time.

The way the article used to find the major and minor axis is the take the derivative \frac{d(\rho^2(t))}{d t}=0 to find the maximum and minimum.

My question is why it use \frac{d(\rho^2(t))}{d t}=0, not \frac{d\rho(t)}{d t}=0?
 
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yungman said:
My question is why it use \frac{d(\rho^2(t))}{d t}=0, not \frac{d\rho(t)}{d t}=0?
Because it's easier and yields the same answers so long as ρ is never 0.
 
D H said:
Because it's easier and yields the same answers so long as ρ is never 0.

Thanks, so all it is, is to avoid dealing with the square root x^2 + y^2?
 
That's all it is. Why bother with the added complexity?
 
Thanks, I thought I missed something.
 

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