求下列曲面在指定点的切平面与法线方程:
(1)z=y+In(x/y),在点(1,1,1)处;
(2)z2=x2+y2,在点(3,4,5)处;
(3)x3+y3+z3+xyz-6=0,在点(1,2,-1)处;
(4)ez-z+xy=3,在点(2,1,0)处.
(1)令F(x,y ,z)=z-y-In(x/y),则: Fx (1,1,1)=1,Fy(1,1,1)=0,FZ(1,1,1)=1 ∴切平面方程为:(x-1)•1+(y-1)•0+(z-1)•(-1)=0 即:x-z=0 法线为:(x-1)/1=(y-1)/0=(z-1)/-1 (2)令F(戈,y,z)=z2-y2-y2,则: Fx(3,4,5)=-6,Fy(3,4,5)=-8,Fz(3,4,5)=10 ∴切平面方程为(-6)•(x-3)+(-8)(y-4)+(z-5)•10=0, 即:3x+4y-5z=0 法线为:(x-3)/3=(y-4)/4=(z-5)/-5 (3)令F(x,y,z) =x3+y3+z3+xyz-6,则: Fx,(1,2,-1)=1,Fy(1,2,-1)=11,(1,2,-1)=5 ∴切平面方程为:(x-1)•1+(y+2)•11+(z+1)•5=0,即:x+11y+5z-18=0 法线为(x-1)/1=(y-2)/11=(z+1)/5 (4)令F(x,y,z)=ez-z+xy-3,则: Fx(2,1,0)=1,Fy(2,1,0)=2,Fz(2,1,0)=0 ∴切平面方程为:(x-2)•1+(y-1)•2+(z-0)•0=0, 即:x+2y-4=0 法线为:(x-2)/1=(y-1)/2=z/0