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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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    Yes but I could not come to anything helpful. Care to give me a hand here? I'd appreciate very much.2012-06-24

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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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    @LeonidKovalev Duh! Of course! I'm too tired for this. Thanks again for all the help! :)2012-07-02