高数。求切平面方程。请问这个1怎么处理? 高数题。空间曲面的切平面方程 。怎么算都是1
\u9ad8\u6570\u7ebf\u6027\u4ee3\u6570\u3002\u5207\u5e73\u9762\u65b9\u7a0b\u3002\u8bf7\u95ee\u8fd9\u4e2a1\u600e\u4e48\u5904\u7406\uff1f\u8bbe\u5207\u5e73\u9762\u65b9\u7a0b\u4e3ax+2y+z+D=0.\u5219\u5207\u5e73\u9762\u6cd5\u5411\u91cf=\uff081,2,1\uff09,\u7531\u4e8e\u6cd5\u5411\u91cf\u53ef\u6b63\u53ef\u8d1f\uff0c\u6240\u4ee5\u5207\u5e73\u9762\u6cd5\u5411\u91cf\u8fd8\u7b49\u4e8e\uff08-1\uff0c-2\uff0c-1\uff09. \u518d\u4ee4F(x,y,z)=2x²+y²-z,\u5219\u66f2\u9762\u7684\u6cd5\u5411\u91cf=\uff08Fx,Fy,Fz)=(4x,2y,-1)=\uff08-1,-2,-1\uff09\uff0c\u89e3\u5f97\uff0cx=-1/4,y=-1\uff0c\u4e0b\u6765\u4f60
x²/a²+y²/b²+z²/c²=1,上点(x0,y0,z0)处的切平面方程为:
x0x/a²+y0y/b²+z0z/c²=1
推导过程:
令F(x,y,z)=x²/a²+y²/b²+z²/c²-1
Fx=2x/a²,Fy=2y/b²,Fz=2z/c²
n=(x0/a²,y0/b²,z0/c²)
切平面方程为
x0/a²(x-x0)+y0/b²(y-y0)+z0/c²(z-z0)=0
化简即得。
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