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I am stucked on problem 1.7.2.b of Differential Geometry of Curves and Surfaces by Manfredo do Carmo. The problem is similar as this topic, but here the exercise defines the osculator circle, ie, this circle of exercise is whose we call osculator circle, and is in $\mathbb{R}^3$. My difficult is another of that topic.

(Adaptated exercise) Let $\alpha(s) \colon I \to \mathbb{R}^3$ be the parametrization by arc length of a curve. Consider a circle that passes through the three points $\alpha(0)$, $\alpha(h_1)$, and $\alpha(h_2)$. Prove that as $(h_1, h_2) \to (0, 0)$, the limiting position of this circle has center at the line that contains $n(0)$ and radius $1/k(0)$.

We can consider the canonical form of the curve at $0$.

As some similar exercises of the book, I beginned witring $P: ax+by+cz=0$ as the plane containing $\alpha(0),\alpha(h_1),\alpha(h_2)$ (by item a) of the question we know that its limiting position is the osculating plane). The center $C$ of this circle $\Pi$ is given by the intersection of mediatrixs at $P$, with directions

$$v_1=(a,b,c)\wedge \alpha(h_1)\quad,\quad v_2=(a,b,c)\wedge\alpha(h_2)$$

Am I correct...?

Well, considering $|(a,b,c)|=1$, we obtain:

$$v_1\cdot v_2=\begin{vmatrix}|(a,b,c)|^2&0\\ 0&\alpha(h_1)\cdot \alpha(h_2) \end{vmatrix}=\alpha(h_1)\cdot \alpha(h_2)$$

We ask for $k_1,k_2$ s.t.

$$\alpha(h_1)+v_1k_1=\alpha(h_2)+v_2k_2$$

I wrote some calculus from here but they are confused and I could not finish...

Many thanks in advance.

Quiet_waters
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3 Answers3

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If $\alpha(0)$, $\alpha(h_1)$, and $\alpha(h_2)$ are three distinct non-aligned points, then there exist a unique circle passing through them. If $h_1,h_2$ are distinct and close enough to $0$, this is granted if $\kappa(0) \neq 0$.

This circle lies in the plane $\alpha(0) + span\{\alpha(h_1)-\alpha(0), \alpha(h_2)-\alpha(h_0)\}$. Call $c$ the center of the circle and set: \begin{align} C:=&c-\alpha(0)\\ U:=&\alpha(h_1)-\alpha(0)\\ V:=&\alpha(h_2)-\alpha(0)\\ W:=&V-\dfrac{\langle V, U \rangle}{\langle U,U \rangle}U \end{align} Notice $W\neq 0$, since the three points $\alpha(0), \alpha(h_1), \alpha(h_2)$ are not aligned. The fact $c$ is equidistant from $\alpha(0), \alpha(h_1), \alpha(h_2)$ corresponds to the following to the equations: \begin{equation} \Vert C \Vert^2 = \Vert C-U \Vert^2 \end{equation} \begin{equation} \Vert C \Vert^2 = \Vert C-V \Vert^2 \end{equation} Setting $C=aU+bW$ you can solve these equations to find $a,b$ and, therefore, the center of the circle (use the fact $U\perp W$). All you have to do now is taking the limit as $h_1,h_2\to 0$ with $h_1\neq h_2$. The Frenet formulas should be helpful. After long calculation you should be able to show what is claimed.

FormulaWriter
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Taking $F(s)=(x(s)-a)^2+(y(s)-b)^2+(z(s)-c)^2-r^2$ where $(a,b,c)$ is the center of a circle crossing $\alpha(0),\alpha(h_1),\alpha(h_2)$. Consider $F'(s)=2(x'(x-a)+y'(y-b)+z'(z-c))$ using local canonical form of $\alpha$ and taking the limit as $s$ approaches $0$ gives $F'(0)=-2a$. Similarly for $F''(s)=2(x''(x-a)+x'^2)+2(y''(y-b)+y'^2)+2(z''(z-c)+z'^2)$ we get $F''(0)=2-kc$. Using the mean value theorem twice and noticing $F(0)=F(h_1)=F(h_2)=0$ (and consequently points $F'(k_1)=F'(k_2)=0$ and $F''(j)=0$). If $h_1,h_2 \rightarrow 0$ then $a\rightarrow 0$ and $b\rightarrow \frac{1}{k}$ we know from part a $c\rightarrow 0$ so we know the center is $\frac{n(0)}{k}$ as desired

Mars
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  • I think this method can be modified to solve the next problem by setting $G(s)=F\circ\pi(s)$ and noticing the mean value theorem implies there exist points such that $G'(k_1)=\frac{-\kappa\tau (h_1)^2}{6}$ and $G'(k_2)=\frac{-\kappa\tau (h_2)^2}{6}$ and $G''(j)=\frac{-\kappa\tau (h_1+h_2)}{6}$ which approach zero as $h_1,h_2\rightarrow 0$ so the conclusion $a\rightarrow 0$ and $b\rightarrow \frac{1}{k}$ holds for $G(s)$. However a mimic arguement using the circle defined by $\alpha\circ\pi(h_1)$ and $\alpha\circ\pi(h_1)$ gives $a\rightarrow 0$ and $b\rightarrow \frac{1}{k_{\pi}}$ – Mars Sep 10 '21 at 17:39
  • It seems to me that your argument involves at least uniform convergence of $F_{h_1,h_2}$ to a function $F$, in order to conclude (for example) that $F'_{h_1,h_2}(k(h_1)) \to F'(0)$, so I was wondering how to prove that? – Lorenzo Vanni Dec 28 '23 at 20:23
  • @LorenzoVanni Perhaps the differential curve is sufficiently nice so that the hypothesis of Dini's theorem apply and give uniform convergence. – Mars Jan 11 '24 at 07:07
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I think I get the answer by taking circle

$$C:(x-a)^2+(y-b)^2+(z-c)^2=a^2+b^2+c^2=r^2$$

instead of plane $P$,

defining

$$F(s)=(x-a)^2+(y-b)^2+(z-c)^2-r^2$$

and doing some derivatives and calculus of these kind ( Derivatives of first/second order going to zero).

Many thanks for attention on topic and I am open to suggestions and corrections and still offer the bounty.

Quiet_waters
  • 1,475
  • 1
    I calculate $F'(0)=-2a$ and $F''(0)=2-2\kappa b$ which using the mean value theorem like in part a gives us $a=0$ and $b=\frac{1}{\kappa}$. As we know the limit is in the oscilating plane from part a we get $c=0$ so the result follows. – Mars Sep 09 '21 at 02:26