高数切平面方程问题 高数--切平面方程和法平面方程
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1\u3001\u5207\u5e73\u9762\u65b9\u7a0b\u662fF'x(x0\uff0cy0\uff0cz0)(x-x0)+F'y(x0\uff0cy0\uff0cz0)(y-y0)+F'z(x0\uff0cy0\uff0cz0)(z-z0)=0\u3002
2\u3001\u6cd5\u5e73\u9762\u65b9\u7a0b\u662f0(x-1)+1(y-1)+2(z-1)=0\u3002
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再令F(x,y,z)=2x²+y²-z,则曲面的法向量=(Fx,Fy,Fz)=(4x,2y,-1)=(-1,-2,-1),解得,x=-1/4,y=-1,下来你自己就能做出来了。
3、设f(x,y,z)=√x+√y+√z-2,
f 对 x、y、z 的偏导数分别为
fx'=1/2√x,fy'=1/2√y,fz'=1/2√z,
设(x0,y0,z0)是曲面上任一点,
则切平面方程为 1/2√x0(x-x0)
+1/2√y0(y-y0)+1/2√z0(z-z0)=0,
注意到 √x0+√y0+√z0=2,切平面方程可化为 x/(2√x0)+y/(2√y0)+z/(2√z0)=1,
所以三个截距之和为
2√x0+2√y0+2√z0=2*2=4 。
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