这两道题目怎么做?
2个回答
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3e^(xy) + x + y = 0, x = 0 时, 3+y = 0, y = -3
3(y+xy')e^(xy) + 1 + y' = 0, y' = -[1+3ye^(xy)]/[1+3xe^(xy)]
x = 0, y = -3 时, y' = -(1-9)/1 = 8
15. 记 F = e^z+xy-z-3, 则 Fx = y, Fy = x, Fz = e^z-1
在点 (2, 1, 0)处, Fx = 1, Fy = 2, Fz= 0
切平面方程 1(x-2) + 2(y-1) + 0(z-0) = 0, 即 x+2y = 4
法线方程 (x-2)/1 = (y-1)/2 = z/0
3(y+xy')e^(xy) + 1 + y' = 0, y' = -[1+3ye^(xy)]/[1+3xe^(xy)]
x = 0, y = -3 时, y' = -(1-9)/1 = 8
15. 记 F = e^z+xy-z-3, 则 Fx = y, Fy = x, Fz = e^z-1
在点 (2, 1, 0)处, Fx = 1, Fy = 2, Fz= 0
切平面方程 1(x-2) + 2(y-1) + 0(z-0) = 0, 即 x+2y = 4
法线方程 (x-2)/1 = (y-1)/2 = z/0
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