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).
Show that curve lies on a sphere
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0May I ask "...intersect (?) in one point." :) – 2012-06-24
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0@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? – 2012-06-24
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0The 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. :) – 2012-06-24
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0Hmm.... in this case we have infinitely many planes and the assumption is that they intersect in a point. I don't really have any more information. – 2012-06-24
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1Is 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. – 2012-06-24
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0@LeonidKovalev: I meant osculating planes and the curve is indeed in 3-dimensional space. Edited my post for clarity. – 2012-06-24
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0@LeonidKovalev: Do you think if we change tangent plane to the normal plane, his problem would make sense? – 2012-06-24
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0@BabakSorouh The problem makes perfect sense to me now. It would be better to say: all osculating planes have a point in common. Or even better (since we can choose coordinates as we wish): all osculating planes pass through $(0,0,0)$. – 2012-06-24
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0Dear @Babak Sorouh: I don't see any point of confusion. If the tangent plane is replaced with normal plane, intuitively, the common point of intersection would be the possible choice for the center of the sphere. However, in this case, I think the assumption implies that normal planes are intersecting in a point. – 2012-06-24
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0@LeonidKovalev: Edited my post accordingly. Thank you. – 2012-06-24
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0@LeonidKovalev: I v' got the point. Now it is clear to me. Thanks for comment. – 2012-06-24
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0@ehsanmo: Thanks – 2012-06-24
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2@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](http://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas)? – 2012-06-24
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0Yes but I could not come to anything helpful. Care to give me a hand here? I'd appreciate very much. – 2012-06-24
1 Answers
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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0I can see how differential geometry comes and solve the problem. Nice answer. – 2012-06-25
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0So 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? – 2012-07-02
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0@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? – 2012-07-02
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0@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. – 2012-07-02
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0@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. – 2012-07-02
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0@LeonidKovalev That's what bugs me too. I could have sworn the teacher proved that to us but clearly an ellipse satisfies all the conditions and is a counterexample. I'm going to accept your solution, thank you for all your help! :) – 2012-07-02
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0@LeonidKovalev Hmm... on the other hand what if I change the problem to state that all osculating planes have *exactly one* point in common? I'm not sure my translation (from Polish to English) is 100% correct so that may be the case. – 2012-07-02
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0@MasterM Then there are no such curves at all, because we already saw that the curve must be planar. – 2012-07-02
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0@LeonidKovalev Duh! Of course! I'm too tired for this. Thanks again for all the help! :) – 2012-07-02