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...this speed will be less than the prior $c$, and the decrease in speed will be $= \frac{(Mv - Mm)c}{Mv}$. To find the value of this, let $MO$ be the radius of osculation The radius of the circle that most closely touches and matches the curve at point M; essentially, a measure of how sharply the path is curving. of the curve at $M = r$, and let the element An "element" refers to an infinitely small segment of the path. $Mm = ds$; it will then be, because angle $O =$ angle $m M v$,
$MO : Mm = Mm : mv$, from which it follows that $mv = \frac{ds^2}{r}$, and $Mv = \sqrt{ds^2 + \frac{ds^4}{r^2}} = \frac{ds}{r} \sqrt{r^2 + ds^2} = ds + \frac{ds^2}{2r^2}$. From this, the decrease in speed will now be obtained while the body traverses the element of the curve $ds = \frac{cds^2}{2r^2}$, the integral In calculus, the "integral" is the sum of all these tiny changes over a finite distance. of which will give the decrease in speed while the body traverses a finite portion of the curve $A M$. But the expression $\frac{cds^2}{2rr}$ is equivalent to a differential of the second degree A mathematical quantity so small that it is the "square" of an already infinitely small value.; therefore, its integral will be a differential of the first degree. For this reason, the decrease in speed after the body has traversed any arc of a given curve will be infinitely small, and the body will be carried with uniform motion Motion at a constant, unchanging speed. through the entire curve $A M$, provided that the radius of osculation $r$ is nowhere infinitely small. original: "Q. E. D."; abbreviation for Quod Erat Demonstrandum, meaning "Which was to be demonstrated."
13. Therefore, in every curve in which the radius of osculation is nowhere infinitely small, a body will move uniformly, provided that it is prompted by no forces and suffers no friction.
14. If the radius of osculation is infinitely small This refers to a "sharp" corner or a cusp in the path., then $\frac{cds^2}{2r^2}$ is either a finite quantity or a differential of the first degree.