求曲面z=arctany/x在点（1.1.π/4）的切平面与法线方程 求曲面z=arctany/x在点（1.1.π/4）的切平面与...
\u66f2\u9762z=arctany/x\u5728\u70b9(1,1,PI/4)\u5904\u7684\u5207\u5e73\u9762\u65b9\u7a0bz=arctan(y/x),
dz=d(y/x)/[1+(y/x)^2]=(xdy-ydx)/(x^2+y^2),
\u5728\u70b9(1,1,\u03c0/4)\u5904\uff0c\u5207\u5e73\u9762\u7684\u6cd5\u5411\u91cf\u4e3a(1/2,-1/2,-1),
\u2234\u5207\u5e73\u9762\u65b9\u7a0b\u662f(1/2)(x-1)-(1/2)(y-1)-(z-\u03c0/4)=0,
\u5373x-y-2z+\u03c0/2=0.
\u6cd5\u7ebf\u65b9\u7a0b\u662fx-1=(y-1)/(-1)=(z-\u03c0/4)/(-2).
z=arctan(y/x),
dz=d(y/x)/[1+(y/x)^2]=(xdy-ydx)/(x^2+y^2),
\u5728\u70b9(1,1,\u03c0/4)\u5904\uff0c\u5207\u5e73\u9762\u7684\u6cd5\u5411\u91cf\u4e3a(1/2,-1/2,-1),
\u2234\u5207\u5e73\u9762\u65b9\u7a0b\u662f(1/2)(x-1)-(1/2)(y-1)-(z-\u03c0/4)=0,
\u5373x-y-2z+\u03c0/2=0.
\u6cd5\u7ebf\u65b9\u7a0b\u662fx-1=(y-1)/(-1)=(z-\u03c0/4)/(-2).
dz=d(y/x)/[1+(y/x)^2]=(xdy-ydx)/(x^2+y^2),
在点(1,1,π/4)处，切平面的法向量为(1/2,-1/2,-1),
∴切平面方程是(1/2)(x-1)-(1/2)(y-1)-(z-π/4)=0,
即x-y-2z+π/2=0.
法线方程是x-1=(y-1)/(-1)=(z-π/4)/(-2).
绛旓細z=arctan(y/x),dz=d(y/x)/[1+(y/x)^2]=(xdy-ydx)/(x^2+y^2),鍦ㄧ偣(1,1,蟺/4)澶勶紝鍒囧钩闈㈢殑娉曞悜閲忎负(1/2,-1/2,-1),鈭村垏骞抽潰鏂圭▼鏄(1/2)(x-1)-(1/2)(y-1)-(z-蟺/4)=0,鍗硏-y-2z+蟺/2=0.娉曠嚎鏂圭▼鏄痻-1=(y-1)/(-1)=(z-蟺/4)/(-2).
绛旓細z=arctan(y/x),dz=d(y/x)/[1+(y/x)^2]=(xdy-ydx)/(x^2+y^2),鍦ㄧ偣(1,1,蟺/4)澶勶紝鍒囧钩闈㈢殑娉曞悜閲忎负(1/2,-1/2,-1),鈭村垏骞抽潰鏂圭▼鏄(1/2)(x-1)-(1/2)(y-1)-(z-蟺/4)=0,鍗硏-y-2z+蟺/2=0.娉曠嚎鏂圭▼鏄痻-1=(y-1)/(-1)=(z-蟺/4)/(-2).
绛旓細zx=1/(1+y^2/x^2)*(-y/x^2)=-y/(x^2+y^2)zy=x/(x^2+y^2)zx(1,1,PI/4)=-1/2zy(1,1,PI/4)=1/2n=(1/2,-1/2,1)=1/2(1,-1,2)鍒囧钩闈㈡柟绋嬩负(x-1)-(y-1)+2(z-蟺/4)=0x-y+2z-蟺/2=0
绛旓細zy(1,1,PI/4)=1/2 n=(1/2,-1/2,1)=1/2(1,-1,2)鍒囧钩闈㈡柟绋嬩负 (x-1)-(y-1)+2(z-蟺/4)=0 x-y+2z-蟺/2=0
