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.

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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.
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.

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.
F = z^4-3xz+2x+y^2
F'(x) = -3z+2 = -1
F'(y) = 2y = 2
F'(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)