目录
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 S S的参数式为 r ( u , v ) = a ( u ) + v b ( u ) . (u,v)=\mathbf{a}(u)+v\mathbf{b}(u). (u,v)=a(u)+vb(u).
沿着直母线，曲面的切平面不变
⇔ \Leftrightarrow ⇔
对于任意 u u u, 及 v 1 ≠ v 2 , n ( u , v 1 ) / / n ( u , v 2 ) v_1\neq v_2, \textbf{ n}( u, v_1) / / \mathbf{n} ( u, v_2) v1=v2, n(u,v1)//n(u,v2)
⇔ \Leftrightarrow ⇔ r u ( u , v 1 ) ∧ r v ( u , v 1 ) / / r u ( u , v 2 ) ∧ r v ( u , v 2 ) \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) ru(u,v1)∧rv(u,v1)//ru(u,v2)∧rv(u,v2)
⇔ \Leftrightarrow ⇔
0 = ( r u ( u , v 1 ) ∧ r v ( u , v 1 ) ) ∧ ( r u ( u , v 2 ) ∧ r v ( u , v 2 ) ) = ( ( a ′ + v 1 b ′ ) ∧ b ) ∧ ( ( a ′ + v 2 b ′ ) ∧ b ) = ⟨ a ′ + v 1 b ′ , ( a ′ + v 2 b ′ ) ∧ b ⟩ b − ⟨ b , ( a ′ + v 2 b ′ ) ∧ b ⟩ ( a ′ + v 1 b ′ ) = ⟨ a ′ , v 2 b ′ ∧ b ⟩ + ⟨ v 1 b ′ , a ′ ∧ b ⟩ = ( v 1 − v 2 ) ( a ′ , b , b ′ ) \begin{align*} 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}'+v_1\mathbf{b}')\wedge\mathbf{b})\wedge((\mathbf{a}'+v_2\mathbf{b}')\wedge\mathbf{b}) \\&=\langle\mathbf{a}^{\prime}+v{1}\mathbf{b}^{\prime},(\mathbf{a}^{\prime}+v{2}\mathbf{b}^{\prime})\wedge\mathbf{b}\rangle\mathbf{b}-\langle\mathbf{b},(\mathbf{a}^{\prime}+v{2}\mathbf{b}^{\prime})\wedge\mathbf{b}\rangle(\mathbf{a}^{\prime}+v{1}\mathbf{b}^{\prime}) \\&=\langle\mathbf{a}^{\prime},v_{2}\mathbf{b}^{\prime}\wedge\mathbf{b}\rangle+\langle v_{1}\mathbf{b}^{\prime},\mathbf{a}^{\prime}\wedge\mathbf{b}\rangle=(v_{1}-v_{2})(\mathbf{a}^{\prime},\mathbf{b},\mathbf{b}^{\prime}) \end{align*} 0=(ru(u,v1)∧rv(u,v1))∧(ru(u,v2)∧rv(u,v2))=((a′+v1b′)∧b)∧((a′+v2b′)∧b)=⟨a′+v1b′,(a′+v2b′)∧b⟩b−⟨b,(a′+v2b′)∧b⟩(a′+v1b′)=⟨a′,v2b′∧b⟩+⟨v1b′,a′∧b⟩=(v1−v2)(a′,b,b′)
⇔ \Leftrightarrow ⇔
( a ′ , b , b ′ ) = 0 (\mathbf a',\mathbf b,\mathbf b')=0 (a′,b,b′)=0
⇔ \Leftrightarrow ⇔
曲面 S S S是可展曲面