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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.
人气:405 ℃ 时间:2020-02-06 01:41:11
解答
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)...
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