填空题 曲面z-(xy) z +2xy=3在点(1,2,0)处的切平面方程为 1
【正确答案】
【答案解析】4x+2y+(1-ln2)z=8 [解析] 令F(x,y,z)=z-(xy) 2 +2xy-3
则 F x =2y-z(xy) z-1 y,F x (1,2,0)=4
F y =2x-z(xy) z-1 x,F y (1,2,0)=2
F z =1-(xy) z ln(xy),F z (1,2,0)=1-ln2
则所求切平面方程为
4(x-1)+2(y-2)+(1-ln2)z=0
即 4x+2y+(1-ln2)z=8.