Question

求曲面e^x-z+xy=3在点(2,1,0)处的切平面及法线方程.

Answer 1

∵e^x-z+xy=3 ==>z=e^x+xy-3

==>αz/αx│(2,1,0)=e²+1,αz/αy│(2,1,0)=2

∴在点(2,1,0)处切平面的法向量是(e²+1,2,-1)

故 所求切平面是(e²+1)(x-2)+2(y-1)-(z-0)=0,即(e²+1)x+2y-z=2(e²+2)

所求法线方程是(x-2)/(e²+1)=(y-1)/2=z/(-1)