如果使用的是一般参数 $t$, Frenet 方程仍形如
\[\dot{e}_i(t)=\sum_{j}\omega_{ij}e_j(t),\]
特别地, 对于 $\mathbb{E}^3$ 中的曲线, 有
\[
\begin{pmatrix}
\dot{e}_1(t)\\
\dot{e}_2(t)\\
\dot{e}_3(t)
\end{pmatrix}=|\dot{c}(t)|
\begin{pmatrix}
0 & \kappa(t) & 0\\
-\kappa(t) & 0 & \tau(t)\\
0 & -\tau(t) & 0
\end{pmatrix}
\begin{pmatrix}
e_1(t)\\
e_2(t)\\
e_3(t)
\end{pmatrix}
\]
我们一般用 $e_1(t),e_2(t),e_3(t)$ 来代替 $\vec{v}(t),\vec{n}(t),\vec{b}(t)$.
假设有这样的向量 $\xi$, 满足
\[\dot{e_1}(t)=\xi\times e_1(t),\quad\dot{e_2}(t)=\xi\times e_2(t),\quad\dot{e_3}(t)=\xi\times e_3(t),\]
则推出
\[
\begin{aligned}
\xi\times e_1(t)&=|\dot{c}(t)|\kappa(t)e_2(t),\\
\xi\times e_2(t)&=|\dot{c}(t)|\bigl(-\kappa(t)e_1(t)+\tau(t)e_3(t)\bigr),\\
\xi\times e_3(t)&=-|\dot{c}(t)|\tau(t)e_2(t).
\end{aligned}
\]
这说明 $\xi\perp e_2(t)$, 因此可设 $\xi=\alpha e_1(t)+\beta e_3(t)$, 代入上面的三式, 得
\[
\begin{aligned}
(\alpha e_1(t)+\beta e_3(t))\times e_1(t)&=|\dot{c}(t)|\kappa(t)e_2(t),\\
(\alpha e_1(t)+\beta e_3(t))\times e_2(t)&=|\dot{c}(t)|\bigl(-\kappa(t)e_1(t)+\tau(t)e_3(t)\bigr),\\
(\alpha e_1(t)+\beta e_3(t))\times e_3(t)&=-|\dot{c}(t)|\tau(t)e_2(t).
\end{aligned}
\]
化解为
\[
\begin{aligned}
\beta e_2(t)&=|\dot{c}(t)|\kappa(t)e_2(t),\\
\alpha e_3(t)-\beta e_1(t)&=|\dot{c}(t)|\bigl(-\kappa(t)e_1(t)+\tau(t)e_3(t)\bigr),\\
-\alpha e_2(t)&=-|\dot{c}(t)|\tau(t)e_2(t).
\end{aligned}
\]
解得
\[\alpha=|\dot{c}(t)|\tau(t),\qquad\beta=|\dot{c}(t)|\kappa(t).\]
因此这样的向量 $\xi$ 确实存在, 并且等于
\[|\dot{c}(t)|\bigl(\tau(t)e_1(t)+\kappa(t)e_3(t)\bigr).\]