求曲面x2-y2-3z=0上点（2，-1,1）处的切平面方程及～法线方程 求曲面x2+2y2+3z2=21在点(1,-2,2)处的切平...
\u6c42\u66f2\u9762x2+2y2+3z2=6\u5728\u70b9\uff081\uff0c-1\uff0c1\uff09\u5904\u7684\u5207\u5e73\u9762\u53ca\u6cd5\u7ebf\u65b9\u7a0b\u7531\u9898\u610f\uff0c\u8bbeF\uff08x\uff0cy\uff0cz\uff09=x2+2y2+3z2-6\uff0c\u5219\u66f2\u9762x2+2y2+3z2=6\u5728\u70b9\uff081\uff0c-1\uff0c1\uff09\u5904\u7684\u6cd5\u5411\u91cf\u5e73\u884c\u4e8en=(Fx\uff0cFy\uff0cFz)|(1\uff0c-1\uff0c1)=2\uff081\uff0c-2\uff0c3\uff09\u53d6\u6cd5\u5411\u91cf\u4e3a\uff081\uff0c-2\uff0c3\uff09\uff0c\u5219\u6240\u6c42\u5207\u5e73\u9762\u65b9\u7a0b\u4e3a\uff1a\uff08x-1\uff09-2\uff08y+1\uff09+3\uff08z-1\uff09=0\u5373x-2y+3z=6\u6240\u6c42\u6cd5\u7ebf\u65b9\u7a0b\u4e3a\uff1ax-11=y+2-2=z-13
F(x\u3001y\u3001z)=x²+2y²+3z²-21
n=\uff08Fx\u3001Fy\u3001Fz\uff09=\uff082x\u30014y\u30016z\uff09
n|(1\u3001-2\u30012)=\uff082\u3001-8\u300112\uff09
\u5207\u5e73\u9762\u65b9\u7a0b\uff1a
2\uff08x-1\uff09-8\uff08y+2\uff09+12\uff08z-2\uff09=0
2x-8y+12z-42=0
\u6cd5\u7ebf\u65b9\u7a0b\uff1a
\uff08x-1\uff09/2=\uff08y+2\uff09/\uff08-8\uff09=\uff08z-2\uff09/12
则Fx(x,y,z)=2x,Fy(x,y,z)=-2y,Fz(x,y,z)=-3
Fx(2,-1,1)=4,Fy(2,-1,1)=2,Fxz(2,-1,1)=-3
曲面x2-y2-3z=0上点（2，-1,1）处的切平面方程为4（x-2)+2(y+2)-3(z-1)=0
法线方程是（x-2)/4=(y+2)/2=(z-1)/(-3)
绛旓細鏇查潰x2-y2-3z=0涓婄偣锛2,-1,1锛夊鐨勫垏骞抽潰鏂圭▼涓4锛坸-2)+2(y+2)-3(z-1)=0 娉曠嚎鏂圭▼鏄紙x-2)/4=(y+2)/2=(z-1)/(-3)
绛旓細鏇查潰x2-y2-3z=0涓婄偣锛2锛-1,1锛夊鐨勫垏骞抽潰鏂圭▼涓4锛坸-2)+2(y+2)-3(z-1)=0 娉曠嚎鏂圭▼鏄紙x-2)/4=(y+2)/2=(z-1)/(-3)
绛旓細娉曞悜閲忎负锛(2x0,-2y0,-3)鎵浠 鍒囧钩闈负 2x0(x-x0)-2y0(y-y0)-3(z-z0)=0 2x0x-2y0y-3z -2x0²+2y0²+3z0=0 鍥犱负鍜岀嚎x锛2=y锛1=z锛2骞宠 鎵浠 鍚戦噺鍨傜洿,鍗 4x0-2y0-6=0 y0=2x0-3 鏈夊钩闈㈣繃鐐癸紙0,0,-1锛夋墍浠 3-2x0²+2y0²+3...
绛旓細(2x/3,4y/3,-1)=(4/3,4/3,-1)(x-2)/(4/3)=(y-1)/(4/3)=(z-2)/(-1)3(x-2)/4=3(y-1)/4=2-z
绛旓細璇佹槑鏇查潰f(x-2z,y-3z)=0涓婁换鎰忎竴鐐瑰鐨勫垏骞抽潰閮戒笌鐩寸嚎x/2=y/3=z/1骞宠 璇佹槑鏇查潰f(x-2z,y-3z)=0涓婁换鎰忎竴鐐瑰鐨勫垏骞抽潰閮戒笌鐩寸嚎x/2=y/3=z/1骞宠... 璇佹槑鏇查潰f(x-2z,y-3z)=0涓婁换鎰忎竴鐐瑰鐨勫垏骞抽潰閮戒笌鐩寸嚎x/2=y/3=z/1骞宠 灞曞紑 鎴戞潵绛 ...
绛旓細Fz=-3 璁惧垏鐐逛负锛x0,y0,z0锛夋硶鍚戦噺涓猴細(2x0,-2y0,-3)鎵浠 鍒囧钩闈负 2x0(x-x0)-2y0(y-y0)-3(z-z0)=0 2x0x-2y0y-3z -2x0²+2y0²+3z0=0 鍥犱负鍜岀嚎x锛2=y锛1=z锛2骞宠 鎵浠 鍚戦噺鍨傜洿,鍗 4x0-2y0-6=0 y0=2x0-3 鏈夊钩闈㈣繃鐐癸紙0,0,-1锛夋墍浠 ...
绛旓細瑙 :
绛旓細瀵鏇查潰鏂圭▼姹傚亸瀵煎嵆涓鸿鐐瑰鐨勬硶鍚戦噺 璁綟(x,y,z)=tan(x²+2y²+3z³)鍒橣x=2x/cos(x²+2y²+3z³)Fy=4y/cos(x²+2y²+3z³)Fz=9z/cos(x²+2y²+3z³)鎵浠ョ偣(1,-1,-1)澶勭殑娉曞悜閲忎负(2,-4,-9)鍒欐硶绾挎柟绋嬩负...
绛旓細F=x^2-y^2-z,G=x^2+2y^2+3z^2-3,鍒 F'<x>=2x,F'<y>=-2y,F'<z>=-1;G'<x>=2x,G'<y>=4y,G'<z>=6z,鍦ㄧ偣(1,1,0),鏇查潰F,G鐨勬硶鍚戦噺鍒嗗埆涓 n1={2锛-2锛-1}锛宯2={1锛2锛0}锛屽垯鍒囩嚎鍚戦噺鏄 n1脳n2={2锛-1,6}锛屽垏绾挎柟绋嬩负 (x-1)2=(y-1)/(-1)=z/6...
绛旓細璁 f(x锛y锛寊) = x^2 + 2y + 3z - 1 锛屽垯 fx ' = 2x锛宖y ' = 2锛宖z ' = 3锛屽皢锛1锛0锛0锛夊潗鏍囦唬鍏ュ緱娉曞悜閲忎负锛2锛2锛3锛夈
