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In these lecture notes http://people.math.gatech.edu/~ghomi/LectureNotes/LectureNotes5U.pdf there is a proof of Kneser's nesting theorem. In one of the steps it is stated that

$$\int_{t_0}^{t_1} \Vert \beta'(t)\Vert\,dt = \int_{t_0}^{t_1} |r'(t)|\,dt$$

where $\beta(t) := \alpha(t) + r(t)N(t)$ is the evolute of a planar curve $\alpha \in C^4(I,\Bbb R^2)$ assumed WLOG to have unit speed $\Vert \alpha\Vert = 1$. Here, $r(t)$ is the radius of the oscuating curve, i.e. the reciprocal of the curvature $\Vert \alpha''(t)\Vert$ and $N(t)=\frac {\alpha''(t)}{r(t)}$ is the normal vector.

However, my calculations show that

$$\langle \beta',\beta'\rangle = 1 + \langle \alpha''',\alpha'''\rangle + 2\langle \alpha',\alpha'''\rangle$$

which doesn't seem to be anywhere near $|r'(t)|^2$. In fact, if we imagine a curve with an interval of very slowly changing curvature, $\Vert \beta'(t)\Vert \approx 1$, independent of $r(t)$.

Rodrigo
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1 Answers1

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Because $\alpha$ is a unit speed curve, you have the Frenet equations: Letting $\alpha'(t)=T(t)$, it is the case that $T'(t)=\kappa(t)N(t)$ and $N'(t)=-\kappa(t)T(t)$. Therefore, $$\beta'(t)=\alpha'(t)+r(t)N'(t)+r'(t)N(t)= T(t)+\frac 1{\kappa(t)}(-\kappa(t)T(t))+r'(t)N(t) = r'(t)N(t).$$ So $\|\beta'(t)\| = r'(t)$, as desired.

Ted Shifrin
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