I'm working through the text De curvis triangularibus of Leonard Euler. In this paper he defines orbiforms as some kind of curves with constant height. He uses these curves to find triangular curves (figure 1) as the evolute of these orbiforms. In the part where he starts working with these evolutes he uses infinitesimals in a formula which I have never seen before, can someone deduce his reasoning?

Problem setting

Consider the orbiform in figure 6 (drawn as a circle, but it doesn't need to be a circle) In the problem setting both $Ff$ and $Mm$ are normal to the curve (in each endpoint). Through a parameterization of the orbiform he deduced $$ FP = x = \frac{dS}{dp} \qquad \qquad PM = y = \frac{pdS}{dp} - S $$ for a certain function $S$. Through further calculations he ends up with: $$ \sin \phi = \frac{1}{\sqrt{1+p^2}} \quad \& \quad \cos \phi = \frac{p}{\sqrt{1+p^2}} \qquad\Rightarrow \qquad d\phi =\frac{-dp}{1+p^2} $$ He then writes:

Quod si iam brevitatis gratia ponamus $FN=v$, notum est, centrum circuli, curvam in $M$ osculantis, fore in puncto $U$, ita ut sit $$ NU = \frac{d v\sin \phi}{d \phi}; $$

roughly translated as

Let now for ease of use $FN=v$, it is known that the center of the osculationcircle in $M$ lies in the point $U$, then it is $$ NU = \frac{d v\sin \phi}{d \phi}; $$

How did he deduce this last formula? In 1778 calculus was done using infinitesimals, and I guess $d v$ should be interpreted as a tiny increase of $v$, but how should I look at $d\phi$ and how did he derive this formula?


Triangular curves Euler figure

EDIT 16 november

I've been looking trough other Euler and L'Hopital texts on curves, evolutes and involutes, but can't find any references. I did come up with an possible explanation though. Would it be reasonable that this was Euler's reasoning as well?

Consider the picture below, of a zoomed in version of fig 6. Let $NU$ progress by $d\phi$, then since two normals which are but an infinitesimal apart have the same radius of curvature, $U$ is still the same point of the osculating circle. Now apply the law of sines to $\triangle NUU'$. Also note how $|NU'| = d v$. $$ \frac{NU'}{\sin d \phi} = \frac{NU}{\sin (\phi - d \phi)} $$

If we replace $\sin d\phi \stackrel{?}{\approx} d\phi$ and $\sin(\phi-d \phi) \stackrel{?}{\approx} \sin \phi$ it follows $$ \frac{dv}{d \phi} = \frac{NU}{\sin \phi} \qquad \Rightarrow \qquad NU = \frac{d v\sin \phi}{d \phi} $$

Far fetched?



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