R3中所有密切平面通过定点P0的C3曲线x(s)(s∈(α β)为弧长)必为平面曲线．请帮忙给出正确

正确答案：证法1设所有密切平面通过定点P₀不妨设定点P₀为原点即P₀=0则有x.V₃=0．两边对s求导得到0=(x.V₃)^'=x^'.V₃^'+x.V₃一V₁.V₃+x(一τV₂)=一τx．V₂．(反证)反设τ(s₀)≠0s₀∈(αβ)．使得r(s)≠0V s∈(s₀一δs₀+δ)则x(s).V₂(s)=0V s∈(x₀一δx₀+δ)．再对上式两边关于s求导有0=(x.V₂)^'=V₁.V₂+x.V₂^'=x.(一kV₁+τV₃)=一kτx.V₁．因为K>0故x.V₁=0从而x.V₁=x.V₂=x.V₃=0．类似于习题1．3．6的证明推得x=x(s)=0V s∈(x₀一δx₀+δ)这与x(s)为正则曲线相矛盾．由此得到τ(s)≡0V s∈(αβ)．根据定理1．2．2曲线x(s)(α1(s)+μV₂(s)．由于它过定点P₀所以P₀=x(s)+λ(s)V₁(s)+μ(s)V₂(s)P₀—x(s)=λ(s)V₁(s)+μ(s)V₂(s)[P₀一x(s).V₃(s)=0．两边对s求导得到0={[P₀—x(s).V₃(s)^'=一x^'(s).V₃(s)+[P₀—x(s).V₃^'(s)=一V₁(s).V₃(s)+[P₀—x(s).[一τ(s)V₂(s)=一τ(s)[P₀—x(s).V₂(s)(反证)假设r(s₀)≠0．由τ的连续性知使得τ(s)≠0V s∈(x₀一δx₀+δ)有[P₀—x(s).V₂(s)=0 Vs∈(x₀一δx₀+δ)．再对上式两边关于s求导有0={[P₀—x(s).V₂(s))^'=一V₁(s).V₂(s)+[P₀—x(s).[一x(s)V₁(s)+τ(s)V₃(s)=k(s)[P₀—x(s).V₁(s)．因为K>0故[P₀—x(s).V₁(s)=0从而 [P₀—x(s).V₁(s)=[P₀—x(s).V₂(s)=[P₀一x(s).V₃(s)=0．类似于习题1．3．6的证明推得P₀—x(s)=0 x(s)=P₀ V s∈(x₀一δx₀+δ)．这与x(s)为正则曲线相矛盾．由此推得τ(s)≡0Vs∈(αβ)．根据定理1．2．2曲线x(s)(α＜s＜β)必为平面曲线．
证法1设所有密切平面通过定点P0,不妨设定点P0为原点,即P0=0,则有x.V3=0．两边对s求导,得到0=(x.V3)'=x'.V3'+x.V3一V1.V3+x(一τV2)=一τx．V2．(反证)反设τ(s0)≠0,s0∈(α,β)．使得r(s)≠0,Vs∈(s0一δ,s0+δ),则x(s).V2(s)=0,Vs∈(x0一δ,x0+δ)．再对上式两边关于s求导,有0=(x.V2)'=V1.V2+x.V2'=x.(一kV1+τV3)=一kτx.V1．因为K>0,故x.V1=0,从而x.V1=x.V2=x.V3=0．类似于习题1．3．6的证明推得x=x(s)=0,Vs∈(x0一δ,x0+δ),这与x(s)为正则曲线相矛盾．由此得到τ(s)≡0,Vs∈(α,β)．根据定理1．2．2,曲线x(s)(α1(s)+μV2(s)．由于它过定点P0,所以P0=x(s)+λ(s)V1(s)+μ(s)V2(s),P0—x(s)=λ(s)V1(s)+μ(s)V2(s),[P0一x(s).V3(s)=0．两边对s求导,得到0={[P0—x(s).V3(s)'=一x'(s).V3(s)+[P0—x(s).V3'(s)=一V1(s).V3(s)+[P0—x(s).[一τ(s)V2(s)=一τ(s)[P0—x(s).V2(s)(反证)假设r(s0)≠0．由τ的连续性知,,使得τ(s)≠0,Vs∈(x0一δ,x0+δ),有[P0—x(s).V2(s)=0,Vs∈(x0一δ,x0+δ)．再对上式两边关于s求导,有0={[P0—x(s).V2(s))'=一V1(s).V2(s)+[P0—x(s).[一x(s)V1(s)+τ(s)V3(s)=k(s)[P0—x(s).V1(s)．因为K>0,故[P0—x(s).V1(s)=0,从而[P0—x(s).V1(s)=[P0—x(s).V2(s)=[P0一x(s).V3(s)=0．类似于习题1．3．6的证明,推得P0—x(s)=0,x(s)=P0,Vs∈(x0一δ,x0+δ)．这与x(s)为正则曲线相矛盾．由此推得τ(s)≡0,Vs∈(α,β)．根据定理1．2．2,曲线x(s)(α＜s＜β)必为平面曲线．