Undergrad Confusion about derivation of equation in Gregory

[terahertz]

TL;DR

Classical Mechanics, Douglas Gregory

Can anyone tell me how Gregory gets equation 1.14 from equation 1.13?

 

[berkeman]

Can you upload a pic of that page? Use the "Attach files" link below the Edit window. Thanks.

 

[terahertz]

Sure...

 

[wrobel]

Looks like Taylor expansions for $\cos$ and $\sin$

 

[PeroK]

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.

 

[wrobel]

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.