I'm trying to prove if plane lies on a plane, then there is a fixed point such that all osculating planes goes through this point.
I've seen answers saying differentiate $\alpha(s)=a(s)T(s)+b(s)N(s)$ and apply Frenet.
then the Equation becomes:
\begin{align}
\alpha'(s)&=a'(s)T(s)+a(s)T'(s)+b'(s)N(s)+b(s)N'(s)\\
\alpha'(s)&=a'(s)T(s)+a(s)kN(s)+b'(s)N(s)+b(s)(-kT+tB)
\end{align}
but I can't figure out how to proceed from here. how to you get that tao goes to 0 from this equation?
also the question is if and only if so I need to prove the other way but all the answers only prove one way.
How should I prove if all osculating plane goes through a fixed point, then the curvature lies on a plane?
Best Answer
The osculating plane can be defined by the equation $\langle B,x\rangle=0$.
Let's assume the existence of some (constant) $p\in\mathbb R^3$ such that $\langle B,p\rangle\equiv0$. Since $p=0$ is a trivial solution for this, I'll also assume that $p\ne0$.
Taking the derivative we obtain $\langle B',p\rangle\equiv0$. Since $B'=\tau N$, we get $\tau\langle N,p\rangle\equiv0$.
Claim: $\tau\equiv0$. Suppose toward a contradiction that $\tau(s_0)\ne0$ for some $s_0$. Then $\langle N(s_0),p\rangle=0$. Taking derivatives, $$ \tau\langle N',p\rangle = \tau'\langle N,p\rangle + \tau\langle N',p\rangle = 0\quad(\textrm{at }s_0) $$ i.e., $$ \langle N',p\rangle = 0\quad(\textrm{at }s_0) $$ Substitute $N' = -kT-\tau B$ and get $$ k\langle T,p\rangle = k\langle T,p\rangle +\tau\langle B,p\rangle = 0\quad(\textrm{at }s_0) $$ Since $k\ne0$, it must be $\langle T(s_0),p\rangle$. Since $\langle N(s_0),p\rangle=0$ and $\langle B,p\rangle=0$ and $\{T, N, B\}$ is an orthonomal basis of $\mathbb R^3$, we get $p=0$; a contradiction. Then $\tau\equiv0$ as claimed.
Now we can use the result that says $\tau\equiv0\iff$im$(\alpha)$ is included in a plane. Since we are only interested here in the $\Rightarrow$ part, I'll recall how to prove it. Without loss of generality we will assume $\alpha(c)=0$ for some $c$ in the domain of $\alpha$.
First observe that $\tau\equiv0\implies B$ constant. Let $H$ be the osculating plane common to all points. As we saw above, $H$ is defined by the equation $\langle B,x\rangle=0$. Since $\langle B,T\rangle\equiv0$, the map $s\mapsto\langle B,\alpha\rangle$ is constant. And since it evaluates to $0$ at $c$, $$ \langle B,\alpha\rangle\equiv0, $$ i.e., $\alpha(s)\in H$ for all $s$.