T1

证明：$E^3$中的正则曲线 r$(t)$的曲率和挠率分别是

$$\kappa (t)=\frac{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|}{|{\mathbf{r}}^{\prime }(t){|}^3}$$
$$\tau (t)=\frac{({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))}{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t){|}^2}$$

证明：设$s$是曲线 r$(t)$的弧长参数.则$s=s(t)$与$t=t(s)$互为反函数.由于空
间曲线的 Frenet 标架和曲率与(容许的)参数选取无关，故

$$\begin{gathered} \mathbf{t}(t):=\mathbf{t}(s(t))=\frac{d\mathbf{r}(t(s))}{ds}={\mathbf{r}}^{\prime }(t)\frac{dt}{ds},\frac{dt}{ds}=\frac{1}{|{\mathbf{r}}^{\prime }(t)|} \\ \dot{\mathbf{t}}(s(t))={\mathbf{r}}^{\prime \prime }(t)(\frac{dt}{ds}{)}^2+{\mathbf{r}}^{\prime }(t)\frac{d^{2}t}{d{s}^2},\mathbf{n}(s(t))=\frac{1}{\kappa (s(t))}\dot{\mathbf{t}}(s(t)), \\ \ddot{\mathbf{t}}(s(t))={\mathbf{r}}^{\prime \prime \prime }(t)(\frac{dt}{ds}{)}^3+3{\mathbf{r}}^{\prime \prime }(t)\frac{dt}{ds}\frac{d^{2}t}{d{s}^2}+{\mathbf{r}}^{\prime }(t)\frac{d^{3}t}{d{s}^3}, \\ \mathbf{b}(s(t))=\mathbf{t}(s(t))\wedge \mathbf{n}(s(t))=\frac{1}{{\kappa }^{\prime \prime }(s(t))}\mathbf{t}(s(t))\wedge \dot{\mathbf{t}}(s(t)) \\ =\frac{1}{\kappa (s(t))}{\mathbf{r}}^{\prime }(t)\frac{dt}{ds}\wedge ({\mathbf{r}}^{\prime \prime }(t)(\frac{dt}{ds}{)}^2+{\mathbf{r}}^{\prime }(t)\frac{d^{2}t}{d{s}^2}) \\ =\frac{1}{\kappa (s(t))}(\frac{dt}{ds}{)}^3{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t) \end{gathered}$$

从而，曲率
$$\kappa (t)=\kappa (s(t))=(\frac{dt}{ds}{)}^3|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|=\frac{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|}{|{\mathbf{r}}^{\prime }(t){|}^3}.$$
由于$\dot{\mathbf{t}}(s)=\kappa (s)\mathbf{n}(s)$,故
$$\ddot{\mathbf{t}}(s)=\kappa (s)\mathbf{n}(s)+\kappa (s)\dot{\mathbf{n}}(s)=\dot{\kappa }(s)\mathbf{n}(s)+\kappa (s)(-\kappa (s)\mathbf{t}(s)+\tau (s)\mathbf{b}(s))$$

$$=-\kappa (s{)}^2\mathbf{t}(s)+\dot{\kappa }(s)\mathbf{n}(s)+\kappa (s)\tau (s)\mathbf{b}(s).$$

从而，
$$\begin{array}{rl}\kappa (s(t))\tau (s(t))& =⟨\ddot{\mathbf{t}}(s(t)),\mathbf{b}(s(t))⟩\\ & =⟨{\mathbf{r}}^{\prime \prime \prime }(t)(\frac{dt}{ds}{)}^3+3{\mathbf{r}}^{\prime \prime }(t)\frac{dt}{ds}\frac{d^{2}t}{d{s}^2}+{\mathbf{r}}^{\prime }(t)\frac{d^{3}t}{d{s}^3},\frac{1}{\kappa (s(t))}\mathbf{t}(s(t))\wedge \dot{\mathbf{t}}(s(t))⟩\\ & =\frac{1}{\kappa (s(t))}(\frac{dt}{ds}{)}^6({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t)).\end{array}$$

因此，

$$\tau (t)=\tau (s(t))=\frac{1}{\kappa (t{)}^2}\frac{1}{|{\mathbf{r}}^{\prime }(t){|}^6}({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))=\frac{({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))}{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t){|}^2}.$$

注：任意参数曲线 r$(t)$的 Frenet 标架为 { r ( t ) ; t ( t ) , n ( t ) , b ( t ) } \{\mathbf{r}(t);\mathbf{t}(t),\mathbf{n}(t),\mathbf{b}(t)\} {r(t);t(t),n(t),b(t)},其中
$$\mathbf{t}(t)=\frac{{\mathbf{r}}^{\prime }(t)}{|{\mathbf{r}}^{\prime }(t)|}=(\frac{x^{\prime }}{\sqrt{x^{\prime 2}+y^{\prime 2}+z^{\prime 2}}},\frac{y^{\prime }}{\sqrt{x^{\prime 2}+y^{\prime 2}+z^{\prime 2}}},\frac{x^{\prime }}{\sqrt{x^{\prime 2}+y^{\prime 2}+z^{\prime 2}}})$$
$$\mathbf{n}(t)=\frac{{\mathbf{r}}^{\prime \prime }(t)}{|{\mathbf{r}}^{\prime \prime }(t)|}$$
$$\mathbf{b}(t)=\frac{{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)}{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|}$$

T2

求 下 列 曲 线 的 曲 率 和 挠 率 :

$(1)\text{ r}(t)=(a\cosh t,a\sinh t,bt)(a>0);$
$(2)\text{ r}(t)=(3t-t^2,3t^2,3t+t^2);$
$(3)\text{ r}(t)=(a(1-\sin t),a(1-\cos t),bt)(a>0);$
$(4)\text{ r}(t)=(at,\sqrt{2}a\log t,\frac{a}{t})(a>0).$

(1)直接计算，得
$${\mathbf{r}}^{\prime }(t)=(a\sinh t,a\cosh t,b)$$

$${\mathbf{r}}^{\prime \prime }(t)=(a\cosh t,a\sinh t,0)$$

$${\mathbf{r}}^{\prime \prime \prime }(t)=(a\sinh t,a\cosh t,0)$$

从而，

$$|{\mathbf{r}}^{\prime }(t)|=\sqrt{a^2\cosh 2t+b^2},$$

$${\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)=(-ab\sinh t,ab\cosh t,-a^2),$$

$$|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|=a\sqrt{b^2\cosh 2t+a^2},$$

$$({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime \prime }(t))=⟨{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime \prime }(t)⟩=a^{2}b.$$

故曲率

$$\kappa (t)=\frac{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|}{|{\mathbf{r}}^{\prime }(t){|}^3}=\frac{a\sqrt{b^2\cosh 2t+a^2}}{(a^2\cosh 2t+b^2{)}^{\frac{3}{2}}},$$

挠率

$$\tau (t)=\frac{({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))}{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t){|}^2}=\frac{b}{b^2\cosh 2t+a^2}.$$

T3

求 下 列 曲 线 的 曲 率 和 挠 率 :

$(1)\text{ r}(t)=(a\cosh t,a\sinh t,bt)(a>0);$
$(2)\text{ r}(t)=(3t-t^2,3t^2,3t+t^2);$
$(3)\text{ r}(t)=(a(1-\sin t),a(1-\cos t),bt)(a>0);$
$(4)\text{ r}(t)=(at,\sqrt{2}a\log t,\frac{a}{t})(a>0).$

(2)直接计算，得

$${\mathbf{r}}^{\prime }(t)=(3-2t,6t,3+2t),$$
$${\mathbf{r}}^{\prime \prime }(t)=(-2,6,2),$$
$${\mathbf{r}}^{\prime \prime \prime }(t)=0$$
从而，
$$|{\mathbf{r}}^{\prime }(t)|=\sqrt{2(4t^2+18t+9)}$$
$${\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)=(-18,-12,18)$$
$$|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|=6\sqrt{22}$$
$$({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))=0$$

故曲率

$$\kappa (t)=\frac{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|}{|{\mathbf{r}}^{\prime }(t){|}^3}=\frac{3\sqrt{11}}{(4t^2+18t+9{)}^{\frac{3}{2}}}$$

挠率

$$\tau (t)=\frac{({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))}{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t){|}^2}=0.$$

T4

求 下 列 曲 线 的 曲 率 和 挠 率 :

$(1)\text{ r}(t)=(a\cosh t,a\sinh t,bt)(a>0);$
$(2)\text{ r}(t)=(3t-t^2,3t^2,3t+t^2);$
$(3)\text{ r}(t)=(a(1-\sin t),a(1-\cos t),bt)(a>0);$
$(4)\text{ r}(t)=(at,\sqrt{2}a\log t,\frac{a}{t})(a>0).$

(3)直接计算，得

$$\begin{gathered} {\mathbf{r}}^{\prime }(t)=(-a\cos t,a\sin t,b), \\ {\mathbf{r}}^{\prime \prime }(t)=(a\sin t,a\cos t,0), \\ {\mathbf{r}}^{\prime \prime \prime }(t)=(a\cos t,-a\sin t,0). \end{gathered}$$

从而，

$$|{\mathbf{r}}^{\prime }(t)|=\sqrt{a^2+b^2},$$

$${\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)=(-ab\cos t,ab\sin t,-a^2)$$
$$|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|=a\sqrt{a^2+b^2},$$
$$({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))=⟨{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime \prime }(t)⟩=-a^{2}b.$$

故曲率

$$\kappa (t)=\frac{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|}{|{\mathbf{r}}^{\prime }(t){|}^3}=\frac{a\sqrt{a^2+b^2}}{(a^2+b^2{)}^{\frac{3}{2}}}=\frac{a}{a^2+b^2},$$

挠率

$$\tau (t)=\frac{({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))}{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t){|}^2}=\frac{-a^{2}b}{a^2(a^2+b^2)}=-\frac{b}{a^2+b^2}.$$

T5

求 下 列 曲 线 的 曲 率 和 挠 率 :

$(1)\text{ r}(t)=(a\cosh t,a\sinh t,bt)(a>0);$
$(2)\text{ r}(t)=(3t-t^2,3t^2,3t+t^2);$
$(3)\text{ r}(t)=(a(1-\sin t),a(1-\cos t),bt)(a>0);$
$(4)\text{ r}(t)=(at,\sqrt{2}a\log t,\frac{a}{t})(a>0).$

(4)注意到 $t\in (0,+\infty ).$ 直接计算，得

$${\mathbf{r}}^{\prime }(t)=(a,\frac{\sqrt{2}a}{t},-\frac{a}{t^2}),$$

$${\mathbf{r}}^{\prime \prime }(t)=(0,-\frac{\sqrt{2}a}{t^2},\frac{2a}{t^3}),$$

$${\mathbf{r}}^{\prime \prime \prime }(t)=(0,\frac{2\sqrt{2}a}{t^3},-\frac{6a}{t^4}).$$

从而，

$$|{\mathbf{r}}^{\prime }(t)|=\frac{a(t^2+1)}{t^2},$$

$${\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)=(\frac{\sqrt{2}{a}^2}{t^4},-\frac{2a^2}{t^3},-\frac{\sqrt{2}{a}^2}{t^2}),$$

$$|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|=\frac{\sqrt{2}{a}^2(t^2+1)}{t^4},$$

$$({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))=⟨{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime \prime }(t)⟩=\frac{2\sqrt{2}{a}^3}{t^6}.$$

故曲率

$$\kappa (t)=\frac{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t)|}{|{\mathbf{r}}^{\prime }(t){|}^3}=\frac{\sqrt{2}{t}^2}{a(t^2+1{)}^2},$$

挠率

$$\tau (t)=\frac{({\mathbf{r}}^{\prime }(t),{\mathbf{r}}^{\prime \prime }(t),{\mathbf{r}}^{\prime \prime \prime }(t))}{|{\mathbf{r}}^{\prime }(t)\wedge {\mathbf{r}}^{\prime \prime }(t){|}^2}=\frac{\sqrt{2}{t}^2}{a(t^2+1{)}^2}.$$

T6

证明：曲线

$$\mathbf{r}(s)=(\frac{(1+s{)}^{\frac{3}{2}}}{3},\frac{(1-s{)}^{\frac{3}{2}}}{3},\frac{s}{\sqrt{2}})(-1<s<1)$$
以$s$为弧长参数，并求它的曲率、挠率和 Frenet 标架。

证明：由于

$${\mathbf{r}}^{\prime }(s)=(\frac{(1+s{)}^{\frac{1}{s}}}{2},-\frac{(1-s{)}^{\frac{1}{s}}}{2},\frac{1}{\sqrt{2}})(-1<s<1),$$
$$|{\mathbf{r}}^{\prime }(s)|=\sqrt{\frac{1+s}{4}+\frac{1-s}{4}+\frac{1}{2}}=1.$$

故$s$是弧长参数.从而，
$$\mathbf{t}(s)={\mathbf{r}}^{\prime }(s),$$
$$\dot{\mathbf{t}}(s)=(\frac{(1+s{)}^{-\frac{1}{2}}}{4},-\frac{(1-s{)}^{-\frac{1}{2}}}{4},0).$$
故曲率
$$\kappa (s)=|\dot{\mathbf{t}}(s)|=\sqrt{\frac{1}{8(1-s^2)}}=\frac{\sqrt{2}(1-s^2{)}^{-\frac{1}{2}}}{4}.$$
从而
$$\mathbf{n}(s)=\frac{1}{\kappa }\dot{\mathbf{t}}(s)=(\frac{\sqrt{2}(1-s{)}^{\frac{1}{2}}}{2},\frac{\sqrt{2}(1+s{)}^{\frac{1}{2}}}{2},0),$$
$$\dot{\mathbf{n}}(s)=(-\frac{\sqrt{2}(1-s{)}^{-\frac{1}{2}}}{4},\frac{\sqrt{2}(1+s{)}^{-\frac{1}{2}}}{4},0),$$
$$\mathbf{b}(s)=\mathbf{t}(s)\wedge \mathbf{n}(s)=(-\frac{(1+s{)}^{\frac{1}{2}}}{2},\frac{(1-s{)}^{\frac{1}{2}}}{2},\frac{\sqrt{2}}{2}).$$
挠率
$$\tau (s)=⟨\dot{\mathbf{n}}(s),\mathbf{b}(s)⟩(=-⟨\dot{\mathbf{b}}(s),\mathbf{n}(s)⟩)=\frac{\sqrt{2}(1-s^2{)}^{-\frac{1}{2}}}{4}.$$
最后，曲线$\mathbf{r}(s)$的 Frenet 标架为 { r ( s ) ; t ( s ) , n ( s ) , b ( s ) } \{\mathbf{r}(s);\mathbf{t}(s),\mathbf{n}(s),\mathbf{b}(s)\} {r(s);t(s),n(s),b(s)},其中$\mathbf{t},\mathbf{n},\mathbf{b}$如上