Confusion about derivation of equation in Gregory

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SUMMARY

The discussion focuses on the derivation of equation 1.14 from equation 1.13 in Douglas Gregory's work on Classical Mechanics, specifically utilizing Taylor expansions for the sine and cosine functions. The user clarifies that Gregory actually derives equation 1.13 from 1.14, demonstrating the relationship through the Taylor series expansion. Key formulas related to motion along a curve, including velocity and acceleration, are also highlighted, emphasizing the significance of curvature in these equations.

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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.
 
Sure...

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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