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Let $\alpha$ be a curve (in $\mathbb{R^{3}}$) with natural (arc length) parametrization which all osculating planes have exactly one point in common. Show that $\alpha$ is a spherical curve (lies on a sphere).

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May I ask "...intersect (?) in one point." :) –  B. S. Jun 24 '12 at 16:16
    
@BabakSorouh: As I understand it: intersection of exactly two planes is a line, intersection of more than 2 planes is a point. Am I wrong? –  MasterM Jun 24 '12 at 16:19
    
The intersection of more than 2 planes does not necessarily make a point. For example, in linear algebra, we can have a system of 3 equations without even a finite set of solutions. I think this problem needs more information. :) –  B. S. Jun 24 '12 at 16:53
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Is the curve in 3-dimensional space, or in $n$ dimensions? Also, are you sure you mean tangent plane, and not normal plane? The tangent space to a curve is a line, not a plane. –  user31373 Jun 24 '12 at 16:59
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@MasterM I would begin like this: Let's say that the common point is $(0,0,0)$. Let $\vec r$ be the position vector of the curve. Your assumption says that $\vec r\cdot \vec B\equiv 0$. And what you want to prove is $\vec r \cdot \vec T\equiv 0$. Have you tried playing with TNB equations? –  user31373 Jun 24 '12 at 17:16
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Taking the derivative of $\vec r\cdot \vec B\equiv 0$ (using the product rule) we get $\vec T\cdot \vec B+\vec r\cdot \vec B'\equiv 0$. But $\vec T\cdot \vec B=0$ always, so we have $\vec r\cdot \vec B'\equiv 0$. If $\vec B$ is nonconstant, then there is an open interval $I$ of parameter $s$ in which $\vec B'\ne 0$. Since $\vec B'=-\tau \vec N$, we conclude that $\vec r\cdot \vec N=0$ on $I$. It follows that $\vec r$ is collinear to $\vec T$ on $I$. In other words, the curve moves along the line through the origin.

... Now you get to decide whether such a line is a counterexample to your statement. Perhaps not, because we assume that the osculating plane is defined, which requires $\vec N$ to be defined. Proceeding on this assumption, we conclude that $\vec B$ is constant, which makes the curve planar. The rest is easy.

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I can see how differential geometry comes and solve the problem. Nice answer. –  B. S. Jun 25 '12 at 5:57
    
So if I understand correctly I then need to prove that $\kappa$ is constant, so the curve is in fact a circle and concluding from that it indeed lies on a sphere? –  MasterM Jul 2 '12 at 22:36
    
@MasterM I wrote "the rest is easy" to mean "I did not think about what's next". If the curve lies in a plane P, then its osculating plane is just P itself, so the condition is satisfied but the curve does not have to lie on a sphere (it could be an ellipse or something else). Is this a counterexample? –  user31373 Jul 2 '12 at 22:43
    
@LeonidKovalev Clearly the only planar curve that can lie on a sphere is a circle (or an arc). I now vaguely recall from the lectures that we have indeed arrived to the conclusion that $\vec{r}$ is planar and a circle so your reasoning may be sound but incomplete. –  MasterM Jul 2 '12 at 22:50
    
@MasterM But what is to stop $\vec r$ from being an ellipse, or any smooth planar curve whatsoever? As far as I can tell, every planar curve (with nonvanishing curvature, so that the osculating plane is well-defined) satisfies the assumption of the problem. –  user31373 Jul 2 '12 at 22:58
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