求曲面z^4－3xz+2x+y^2=1上点（1,1,1）处的切平面和法线方程.
问题描述：
求曲面z^4－3xz+2x+y^2=1上点（1,1,1）处的切平面和法线方程.
答
F = z^4－3xz+2x+y^2F'(x) = -3z+2 = -1F'(y) = 2y = 2F'(z) = 4z³-3x = 1因此在点(1,1,1)处的法向量为(-1,2,1)切平面方程为-(x-1)+2(y-1)+(z-1) = 0也就是-x+2y+z=2切线方程为-(x-1) = (y-1)/2 = (z-1)...