Confusion about derivation of equation in Gregory

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Discussion Overview

The discussion revolves around the derivation of equation 1.14 from equation 1.13 in Douglas Gregory's work on Classical Mechanics. Participants explore the use of Taylor expansions for trigonometric functions in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests clarification on how Gregory derives equation 1.14 from 1.13.
  • Another participant suggests that Gregory actually derives equation 1.13 from 1.14 using Taylor expansions for sine and cosine functions.
  • Specific Taylor expansions are presented, showing the expansion of ##\sin(\kappa s)## and ##1 - \cos(\kappa s)##.
  • A participant notes a tendency among physicists to work backwards in derivations, expressing personal discomfort with this approach.
  • Additional formulas related to motion along a curve and the concept of curvature are provided, but their relevance to the main question is not explicitly stated.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the derivation process, with differing views on the direction of the derivation and the use of Taylor expansions.

Contextual Notes

There are assumptions about the familiarity with Taylor expansions and the specific equations referenced, which may not be clear to all participants. The discussion includes unresolved mathematical steps related to the derivation.

terahertz
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TL;DR
Classical Mechanics, Douglas Gregory
Can anyone tell me how Gregory gets equation 1.14 from equation 1.13?
 
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Can you upload a pic of that page? Use the "Attach files" link below the Edit window. Thanks.
 
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20251121_194240.webp
 
Looks like Taylor expansions for ##\cos## and ##\sin##
 
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terahertz said:
TL;DR: Classical Mechanics, Douglas Gregory

Can anyone tell me how Gregory gets equation 1.14 from equation 1.13?
The way I would look at it, he actually gets equation 1.13 from 1.14 (using Taylor expansions). I.e.:
$$\kappa^{-1} \sin(\kappa s) = \kappa^{-1}(\kappa s - \frac{\kappa^3 s^3}{3!} + \dots) = s - \frac{\kappa^2}{3!}s^3 + \dots$$And:
$$\kappa^{-1}(1 - \cos(\kappa s)) = \kappa^{-1}(\frac{\kappa^2 s^2}{2!} - \frac{\kappa^4 s^4}{4!} \dots) = \frac 1 2 \kappa s^2 - \frac{\kappa^3 s^4}{4!} + \dots$$In my experience, physicists have this somewhat curious tendency to work backwards without saying so. For me, personally, it seems odd. But, you have to get used to it.
 
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The following formulas are useful. Try to get them by yourself. Suppose we have a curve ##\boldsymbol r=\boldsymbol r(s).## And let a a point slides along this curve in accordance with the law of motion ##s=s(t).## Then the velocity of the point is
$$\boldsymbol v=\dot s\boldsymbol t(s(t));$$
the acceleration of the point is
$$\boldsymbol a=\ddot s\boldsymbol t(s(t))+\dot s^2 \kappa(s(t))\boldsymbol n(s(t)).$$
The last formula explains why ##1/\kappa## is called the radius of curvature as well.
Another useful exercise is to find the angular velocity of the Frenet frame in such a motion.
 
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