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x2y y2z

抛物面z=x^2+y^2被平面x+y+z=1截成一椭圆,求原点到这椭圆的最长与最短...2015-02-08 抛物面z=x2+y2被平面x+y+z=1截成一...

x2y-y2z+z2x-x2z+y2x+z2y-2xyz=(y-z)x2+(z2+y2-2yz)x+z2y-y2z=(y-z)x2+(y-z)2x-yz(y-z)=(y-z)[x2+(y-z)x-yz]=(y-z)(x+y)(x-z).故选A.

你做错了,不能那么转换。 解:原式=∫dθ∫rdr∫r^2dz (作柱面坐标变换) =2π∫r^3(2-r^2/2)dr =2π∫(2r^3-r^5/2)dr =2π(2^4/2-2^6/12) =2π(8/3) =16π/3。

曲面∑在点(0,0,0)处的法向量为n={2(x2y+y2z+z2x)(2xy+z2)+1,2(x2y+y2z+z2x)(2yz+x2)-1,-1}|(0,0,0)={1,-1,1}∴切平面方程为(x-0)-(y-0)+(z-0)=0,即x-y+z=0假设所求点的坐标P(x,y,z),则d2=F(x,y,z)=(x-2)2+(y-1)2...

z=ln(x²+y²) ∂z/∂x=2x/(x²+y²) ∂²z/∂x²=[2(x²+y²)-2x·2x]/(x²+y²)²=2(y²-x²)/(x²+y²)² ∂³z/∂x²∂y=2[2...

x=-100:0.1:100; y=-100:0.1:100; [X,Y]=meshgrid(x,y); Z=X^2+Y^2; mesh(X,Y,Z)

如图所示、

(1)原式=16x2右i1÷8x2右2+8xi右21÷8x2右2=2右1+x1;(2)原式=2f×10-12÷(9×10-f)=i×10-0.

令F(x,y,z)= x^2+y^2+z^2-14 Fx=2x,Fy=2y,Fz=2z 所以 n=(3,2,1) 从而 切平面方程为3(x-3)+2(y-2)+(z-1)=0 即 3x+2y+z=14. 法线方程为:(x-3)/3=(y-2)/2=(z-1)/1

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