设φ(u v)=常数 φ(u v)=常数为曲面M上的两族正则曲线．证明：两族曲线正交证明：Eu=2x

正确答案：因为E=x_u^'.x_u^'F=x_u^'.x_v^'G=x_v^'.x_v^'所以E_u^'=(x_u^'.x_u^')_u^'=2x_u^'.x_uu^''E_v^'=(x_u^'.x_u^')_v^'=2x_u^'.x_uvG_v^'=(x_v^'.x_v^')_v^'=2x_v^'.x_vv^'' G_u^'=(x_v^'.x_v^')_u^'=2x_v^'.x_uv^''F_u^'=(x_u^'.x_v^')_u^'=x_uu^''.x_v^'+x_u^'.x_uv^''F_v^'=(x_u^'.x_v^')_v^'=x_vv^''.x_v^'+x_u^'.x_vv^''2F_u^'一E_v^'=2(x_uu^''.x_v^'+x_u^'.x_uv^'')一2x_u^'.x_uv^''=2x_u^'.x_uu^''2F_v^'一G_u^'=2(x_uv^''.x_v^'+x_u^'.x_vv^'')一2x_v^'.x_uv^''=2x_u^'.x_vv^''．
因为E=xu'.xu',F=xu'.xv',G=xv'.xv',所以Eu'=(xu'.xu')u'=2xu'.xuu'',Ev'=(xu'.xu')v'=2xu'.xuv,Gv'=(xv'.xv')v'=2xv'.xvv'',Gu'=(xv'.xv')u'=2xv'.xuv'',Fu'=(xu'.xv')u'=xuu''.xv'+xu'.xuv'',Fv'=(xu'.xv')v'=xvv''.xv'+xu'.xvv'',2Fu'一Ev'=2(xuu''.xv'+xu'.xuv'')一2xu'.xuv''=2xu'.xuu'',2Fv'一Gu'=2(xuv''.xv'+xu'.xvv'')一2xv'.xuv''=2xu'.xvv''．