问题及解答

1. $\vec{r}=e^t(\sin t,\cos t,1)$,
2. $\vec{r}=a(\cosh t,\sinh t,t)$.

回顾曲率和挠率的计算公式(问题724)

\[\kappa(t)=\frac{|\dot{c}(t)\times\ddot{c}(t)|}{|\dot{c}(t)|^3},\quad\tau(t)=\frac{\det(\dot{c}(t),\ddot{c}(t),\dddot{c}(t))}{|\dot{c}(t)\times\ddot{c}(t)|^2}.\]

这里曲线 $c(t)$ 以一般参数表示. 我们这里简记曲线 $\vec{r}(t)$ 为 $r(t)$. 根据公式, 我们分别计算 $\dot{r}(t)$, $\ddot{r}(t)$, $\dddot{r}(t)$, $\dot{r}(t)\times\ddot{r}(t)$ 及它们的模长.

\[
\begin{split}
\dot{r}(t)&=(e^t)\'(\sin t,\cos t,1)+e^t(\cos t,-\sin t,0)\\
&=e^t(\sin t+\cos t,\cos t-\sin t,1),\\
\ddot{r}(t)&=(e^t)\'(\sin t+\cos t,\cos t-\sin t,1)+e^t(\cos t-\sin t,-\sin t-\cos t,0)\\
&=e^t(2\cos t,-2\sin t,1),\\
\dddot{r}(t)&=(e^t)\'(2\cos t,-2\sin t,1)+e^t(-2\sin t,-2\cos t,0)\\
&=e^t(2\cos t-2\sin t,-2\sin t-2\cos t,1),
\end{split}
\]

所以

\[
\begin{split}
|\dot{r}(t)|&=|e^t(\sin t+\cos t,\cos t-\sin t,1)|\\
&=e^t\sqrt{(\sin t+\cos t)^2+(\cos t-\sin t)^2+1^2}\\
&=\sqrt{3}e^t,
\end{split}
\]

\[
\begin{split}
\dot{r}(t)\times\ddot{r}(t)&=e^t(\sin t+\cos t,\cos t-\sin t,1)\times e^t(2\cos t,-2\sin t,1)\\
&=e^{2t}
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
\sin t+\cos t & \cos t-\sin t & 1\\
2\cos t & -2\sin t & 1
\end{vmatrix}\\
&=e^{2t}(\cos t+\sin t,\cos t-\sin t, -2),
\end{split}
\]

故

\[
\begin{split}
|\dot{r}(t)\times\ddot{r}(t)|&=e^{2t}\sqrt{(\cos t+\sin t)^2+(\cos t-\sin t)^2+(-2)^2}\\
&=\sqrt{6}e^{2t}.
\end{split}
\]

\[
\begin{split}
&\det(\dot{r}(t),\ddot{r}(t),\dddot{r}(t))\\
=&\bigl\langle\dot{r}(t)\times\ddot{r}(t),\dddot{r}(t)\bigr\rangle\\
=&\bigl\langle e^{2t}(\cos t+\sin t,\cos t-\sin t, -2),e^t(2\cos t-2\sin t,-2\sin t-2\cos t,1)\bigr\rangle\\
=&e^{3t}\bigl[2(\cos^2 t-\sin^2 t)-2(\cos^2 t-\sin^2 t)-2\bigr]\\
=&-2e^{3t}.
\end{split}
\]

因此, 曲线的曲率和挠率分别为

\[
\begin{split}
\kappa(t)&=\frac{|\dot{r}(t)\times\ddot{r}(t)|}{|\dot{r}(t)|^3}\\
&=\frac{\sqrt{6}e^{2t}}{(\sqrt{3}e^t)^3}\\
&=\frac{\sqrt{2}}{3}e^{-t},
\end{split}
\]

\[
\begin{split}
\tau(t)&=\frac{\det(\dot{r}(t),\ddot{r}(t),\dddot{r}(t))}{|\dot{r}(t)\times\ddot{r}(t)|^2}\\
&=\frac{-2e^{3t}}{(\sqrt{6}e^{2t})^2}\\
&=-\frac{1}{3}e^{-t}.
\end{split}
\]

2. $\vec{r}=a(\cosh t, \sinh t, t)$,  这里不妨设 $a > 0$.

\[
\begin{aligned}
\dot{r}(t)&=a(\sinh t, \cosh t, 1)\\
\ddot{r}(t)&=a(\cosh t, \sinh t, 0)\\
\dddot{r}(t)&=a(\sinh t, \cosh t, 0)\\
\end{aligned}
\]

\[
|\dot{r}(t)|=a\sqrt{\sinh^2 t+\cosh^2 t+1}=\sqrt{2}a\cosh t,
\]

\[
\dot{r}(t)\times\ddot{r}(t)=a^2
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
\sinh t & \cosh t & 1\\
\cosh t & \sin h t & 0
\end{vmatrix}
=a^2(-\sinh t, \cosh t, -1)
\]

因此

\[
|\dot{r}(t)\times\ddot{r}(t)|=\sqrt{2}a^2\cosh t
\]

\[
\begin{split}
\det(\dot{r}(t),\ddot{r}(t),\dddot{r}(t))&=\langle\dot{r}(t)\times\ddot{r}(t),\dddot{r}(t)\rangle\\
&\langle a^2(-\sinh t,\cosh t, -1),\ a(\sinh t, \cosh t, 0)\rangle\\
&=a^3(-\sinh^2 t+\cosh^2 t+0)\\
&=a^3
\end{split}
\]

因此, 曲率和挠率分别为

\[
\kappa(t)=\frac{|\dot{r}(t)\times\ddot{r}(t)|}{|\dot{r}(t)|^3}=\frac{1}{2a\cosh^2 t},
\]

\[
\tau(t)=\frac{\det(\dot{r}(t),\ddot{r}(t),\dddot{r}(t))}{|\dot{r}(t)\times\ddot{r}(t)|^2}=\frac{1}{2a\cosh^2 t}.
\]

值得注意的是, 曲率和挠率算出的数值是一样的.