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Let $r$ be a unit-speed bi-regular curve. (It passes the point $s_0$) Let $distP(q)$ be the distance between the plane $P$ and the point $q$.

Question.

The plane is equal to the osculating plane of $r$ at $s_0$ if and only if $P$ contains $r(s_0)$,$$ \lim_{s \to s_0} \frac{distP(r(s))}{(s-s_0)} = 0$$ $$\lim_{s \to s_0} \frac{distP(r(s))}{(s-s_0)^2} =0$$

Math-Nerd
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    Any attempt at all? – Display Name Mar 23 '14 at 02:06
  • First limit is easy; since P contains the tangent line of r. You can just use the equation a•(X-P)/|a| to get the distP(r(s)) and find out that limit is really zero. – Math-Nerd Mar 23 '14 at 02:26
  • You should always put an attempt into the question. The members of Stack exchange are likely to down-vote questions not showing any attempt. – Display Name Mar 23 '14 at 02:32
  • Okay, it might look dirty. By using the distance formula, distP(r(s))/(s-s0)^2=|r'Xr"|•(r(s)-r(s0))/|r'Xr"|(s-s0)^2. Here I can't advance more. Stucked. – Math-Nerd Mar 23 '14 at 02:49
  • I think "unit speed curve" can be an important key, but I don't know where to use it. – Math-Nerd Mar 23 '14 at 02:55

1 Answers1

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I found the answer. Using the L'Hospital's law, the equation means that the binormal vector of r(s) perpendicular to the normal vector. And it is true because it is bi-regular curve.

Math-Nerd
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