假设 $c:I\rightarrow\mathbb{R}^3$ 是一以弧长为参数的空间曲线. 设 $s_0\in I$. 则有
\[
\begin{split}
c(s)-c(s_0)&=\biggl((s-s_0)-\frac{(s-s_0)^3}{6}\kappa^2(s_0)\biggr)e_1(s_0)\\
&\quad+\biggl(\frac{(s-s_0)^2}{2}\kappa(s_0)+\frac{(s-s_0)^3}{6}\dot{\kappa}(s_0)\biggr)e_2(s_0)\\
&\qquad+\biggl(\frac{(s-s_0)^3}{6}\kappa(s_0)\tau(s_0)\biggr)e_3(s_0)+o((s-s_0)^3).
\end{split}
\]
动画演示
1
利用向量值函数的 Taylor 展式, 得
\[c(s)=c(s_0)+\frac{\dot{c}(s_0)}{1!}(s-s_0)+\frac{\ddot{c}(s_0)}{2!}(s-s_0)^2+\frac{\dddot{c}(s_0)}{3!}(s-s_0)^3+o((s-s_0)^3).\tag{$*$}\]
回忆问题724的回答中对 $\dddot{c}(s)$ 的计算:
\[e_1(s)=\dot{c}(s),\quad e_2(s)=\frac{\ddot{c}(s)}{|\ddot{c}(s)|},\quad e_3(s)=\frac{\dot{c}(s)\times\ddot{c}(s)}{|\ddot{c}(s)|},\quad\kappa(s)=|\ddot{c}(s)|\]
\[
\begin{split}
\dddot{c}(s)&=\dot{\kappa}(s)e_2(s)+\kappa(s)\dot{e}_2(s)\\
&=\dot{\kappa}(s)e_2(s)+\kappa(s)[-\kappa(s)e_1(s)+\tau(s)e_3(s)]\\
&=\dot{\kappa}(s)e_2(s)-\kappa^2(s)e_1(s)+\kappa(s)\tau(s)e_3(s)
\end{split}
\]
\[
\begin{split}
\dddot{c}(s)&=\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|}\frac{\ddot{c}(s)}{|\ddot{c}(s)|}-|\ddot{c}(s)|^2\dot{c}(s)+|\ddot{c}(s)|\tau(s)\frac{\dot{c}(s)\times\ddot{c}(s)}{|\ddot{c}(s)|}\\
&=\frac{\langle\dddot{c}(s),\ddot{c}(s)\rangle}{|\ddot{c}(s)|^2}\ddot{c}(s)-|\ddot{c}(s)|^2\dot{c}(s)+(\dot{c}(s)\times\ddot{c}(s))\tau(s),
\end{split}
\]
我们这里只需要 $\dot{c}(s_0)=e_1(s_0)$, $\ddot{c}(s_0)=|\ddot{c}(s_0)|e_2(s_0)$, 和
\[\dddot{c}(s_0)=\dot{\kappa}(s_0)e_2(s_0)-\kappa^2(s_0)e_1(s_0)+\kappa(s_0)\tau(s_0)e_3(s_0)\]
代入 $(*)$ 式即可.