证明直纹面是可展曲面的当且仅当沿着直母线，曲面的切平面不变

直纹面是可展曲面当且仅当沿着直母线，曲面的切平面不变.

证明：设直纹面$S$的参数式为 r$(u,v)=\mathbf{a}(u)+v\mathbf{b}(u).$

沿着直母线，曲面的切平面不变

$\iff$

对于任意 $u$, 及 $v_1\ne v_2,\text{ n}(u,v_1)//\mathbf{n}(u,v_2)$

$\iff$$${\mathbf{r}}_u(u,v_1)\wedge {\mathbf{r}}_v(u,v_1)//{\mathbf{r}}_u(u,v_2)\wedge {\mathbf{r}}_v(u,v_2)$$

$\iff$

$$\begin{array}{rl}0& =({\mathbf{r}}_u(u,v_1)\wedge {\mathbf{r}}_v(u,v_1))\wedge ({\mathbf{r}}_u(u,v_2)\wedge {\mathbf{r}}_v(u,v_2))\\ & =(({\mathbf{a}}^{\prime }+v_1{\mathbf{b}}^{\prime })\wedge \mathbf{b})\wedge (({\mathbf{a}}^{\prime }+v_2{\mathbf{b}}^{\prime })\wedge \mathbf{b})\\ & =⟨{\mathbf{a}}^{\prime }+v_1{\mathbf{b}}^{\prime },({\mathbf{a}}^{\prime }+v_2{\mathbf{b}}^{\prime })\wedge \mathbf{b}⟩\mathbf{b}-⟨\mathbf{b},({\mathbf{a}}^{\prime }+v_2{\mathbf{b}}^{\prime })\wedge \mathbf{b}⟩({\mathbf{a}}^{\prime }+v_1{\mathbf{b}}^{\prime })\\ & =⟨{\mathbf{a}}^{\prime },v_2{\mathbf{b}}^{\prime }\wedge \mathbf{b}⟩+⟨v_1{\mathbf{b}}^{\prime },{\mathbf{a}}^{\prime }\wedge \mathbf{b}⟩=(v_1-v_2)({\mathbf{a}}^{\prime },\mathbf{b},{\mathbf{b}}^{\prime })\end{array}$$

$\iff$

$$({\mathbf{a}}^{\prime },\mathbf{b},{\mathbf{b}}^{\prime })=0$$

$\iff$

曲面$S$是可展曲面