求曲面z=xlny—ylnx在点（1,1,0）处的切平面方程与法线方程 求曲面z=xlny-ylnx在点（1,1,0）处的切平面方程...
\u6c42\u66f2\u9762ez-z+xy=3\u5728\u70b9\uff082\uff0c1\uff0c0\uff09\u5904\u7684\u5207\u5e73\u9762\u53ca\u6cd5\u7ebf\u65b9\u7a0b\u5207\u5e73\u9762\u65b9\u7a0b\u4e3a\uff1ax+2y-4=0\uff0c\u6cd5\u7ebf\u65b9\u7a0b\u4e3a\uff1ax−2/1\uff1dy−1/2
\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u7531\u9898\u610f\uff0c\u8bbeF\uff08x\uff0cy\uff0cz\uff09=ez-z+xy-3
\u5219\u66f2\u9762\u5728\u70b9\uff082\uff0c1\uff0c0\uff09\u5904\u7684\u6cd5\u5411\u91cf\u4e3a\uff1a
n\uff1d(Fx\uff0cFy\uff0cFz)|(2\uff0c1\uff0c0)=\uff08y\uff0cx\uff0cez-1\uff09|\uff082\uff0c1\uff0c0\uff09=\uff081\uff0c2\uff0c0\uff09
\u2234\u6240\u6c42\u5207\u5e73\u9762\u65b9\u7a0b\u4e3a\uff1a
\uff08x-2\uff09+2\uff08y-1\uff09=0
\u5373 x+2y-4=0
\u6240\u6c42\u6cd5\u7ebf\u65b9\u7a0b\u4e3a\uff1ax−2/1\uff1dy−1/2\uff0cz\uff1d0
\u2234x\uff1d2+t\uff0cy\uff1d1+2t\uff0cz\uff1d0
\u6269\u5c55\u8d44\u6599\u6c42\u66f2\u9762\u5207\u5e73\u9762\u53ca\u6cd5\u7ebf\u65b9\u7a0b\u65b9\u6cd5\uff1a
\u6cd5\u7ebf\u659c\u7387\u4e0e\u5207\u7ebf\u659c\u7387\u4e58\u79ef\u4e3a-1\uff0c\u5373\u82e5\u6cd5\u7ebf\u659c\u7387\u548c\u5207\u7ebf\u659c\u7387\u5206\u522b\u7528\u03b1\u3001\u03b2\u8868\u793a\uff0c\u5219\u5fc5\u6709\u03b1*\u03b2=-1\u3002\u6cd5\u7ebf\u53ef\u4ee5\u7528\u4e00\u5143\u4e00\u6b21\u65b9\u7a0b\u6765\u8868\u793a\uff0c\u5373\u6cd5\u7ebf\u65b9\u7a0b\u3002\u4e0e\u5bfc\u6570\u6709\u76f4\u63a5\u7684\u8f6c\u6362\u5173\u7cfb\u3002
\u66f2\u7ebf\u5728\u70b9\uff08x0\uff0cy0\uff09\u7684\u6cd5\u7ebf\u65b9\u7a0b\u516c\u5f0f\uff1a
z=xlny-ylnx
\u4ee4F(x,y,z)=xlny-ylnx-z
\u90a3\u4e48dF/dx=lny-y/x dF/dy=x/y-lnx dF/dz=-1
\u6240\u4ee5\u5207\u5e73\u9762\u5c31\u662f
dF/dx | (1,1,0) (x-1)+dF/dy | (1,1,0) (y-1)+dF/dz | (1,1,0) (z-0)=0
\u90a3\u4e48\u5c31\u662fx-y+z=0
\u6cd5\u7ebf\u65b9\u7a0b\u5f0f\uff1a
(x-1)/dF/dx | (1,1,0)=(y-1)/dF/dy | (1,1,0)=(z-0)/dF/dz | (1,1,0)
\u4e5f\u5c31\u662f(x-1)/-1=(y-1)/1=z/-1
对x求偏导得 -lny+y/x
对y求偏导得 -x/y+lnx
对z求偏导得 1
将点（1,1,0）代入以上表达式得切平面法向量（1,-1,1）
则切平面方程(点法式） x-y+z=0
法线方程（点向式） (x-1)/1=(y-1)/-1=z/1
绛旓細z=xlny-ylnx 浠(x,y,z)=xlny-ylnx-z 閭ｄ箞dF/dx=lny-y/x dF/dy=x/y-lnx dF/dz=-1 鎵浠ュ垏骞抽潰灏辨槸 dF/dx | (1,1,0) (x-1)+dF/dy | (1,1,0) (y-1)+dF/dz | (1,1,0) (z-0)=0 閭ｄ箞灏辨槸x-y+z=0 娉曠嚎鏂圭▼寮忥細(x-1)/dF/dx | (1,1,0)=(y-1)...
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绛旓細2014-07-03 宸茬煡xlny+ylnx=0,姹倅鐨勫鏁 2016-10-29 姹備笅鍒楁柟绋嬬殑闅愬嚱鏁y=y(x)鍦ㄦ寚瀹氱偣鐨勫鏁 (1)璁緔=x... 2015-12-01 y鏄痻鐨勯殣鍑芥暟,xlny+ylnx=1,姹俤y/dx 3 2014-11-12 x鐨剏鏂=y鐨剎鏂,姹倄瀵箉鐨勫鏁?涓ょ鏂规硶绛旀涓嶅悓,鍖栦负l... 2017-01-24 xlny+ylnx>(x+y)ln(x+y)/2,x>...
绛旓細xlny-ylnx=1 d/dx(xlny-ylnx)=0 (x/y).y' + lny - lnx.y' - (y/x) =0 x^2.y'+ xylny- xylnx.y' - y^2 =0 (x^2-xylnx).y'= y^2-xylny y'= (y^2-xylny)/(x^2-xylnx)y'|(x,y)=(1,e)=(e^2-e)/(e^2-e)=1 ...
绛旓細鍏跺疄X^y=Y^x锛屼袱杈瑰彇瀵规暟灏卞彲浠ュ寲涓xlny=ylnx 鍗硏lny-ylnx=0 鐜板湪鍙互姹備簡鍚绗簩涓F(x,y,z)=z-f(x,y)(鎴戞妸浣犵殑F閮芥崲鎴愪簡f)鍒橣x=-fx,Fy=-fy,Fz=1 鍦ㄧ偣P澶刦x宸茬煡锛宖(x,y)=z,浠ｅ叆f(x,y)婊¤冻鐨勯偅涓柟绋 鍙互姹傚嚭fy 鍦ㄧ偣P澶勭殑Fx,Fy,Fz宸叉眰鍑猴紝涔嬪悗灏辨槸鐩存帴浠ｅ叕寮忎簡...
绛旓細xlny=ylnx y^x=x^y 4^2=2^4 x鈮爕涔熸垚绔 涓嶈兘纭畾y=x
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绛旓細鍙栧鏁ylnx=xlny锛岃繖鏍锋眰瀵硷細y'lnx+y/x=lny+(x/y)y'姹傚嚭y'锛岀殑纭鍒╃敤閭ｄ釜缁撹锛屼笉杩囨垜杩欎箞姹傚氨鐩存帴鍖栫畝浜嗐備篃灏辨槸姹傛硶鍚戦噺鍚э紝锛團x锛孎y锛岋紞1锛塅x锛3,鍐嶅埄鐢▁*Fx(x,y)+y*Fy(x,y)=F(x,y),1脳3锛嬶紙锛1锛塅y锛2锛岃В鍑篎y 銆愭敞鎰忚繃锛1锛岋紞1锛2锛夛紝鎵浠z锛1锛岋紞1锛夛紳2锛併...
绛旓細杩欐牱鍛
绛旓細ylnx=xlny 鑻=1,鍒檡=1锛涘嵆杩囷紙1锛1锛塴ny/y=lnx/x;鑰屽嚱鏁發nz/z鍦e澶勫彇鏈鍊硷紝鍙兘(x,y)涔熻繃锛坋锛宔锛墄锛寉瀹屽叏瀵圭瓑锛屽嵆 y=f锛坸锛夛紱x=f(y)锛涘攭锛屽嚱鏁颁笌鍙嶅嚱鏁扮浉绛夛紝杩欐槸涓涓嚜韬叧浜巠=x瀵圭О銆佷笖杩囷紙1锛1锛夊拰锛坋锛宔锛夌殑鍑芥暟 涓嶈繃鎮ㄥソ鍍忔病鏈夐檺瀹氬畾涔夊煙锛涘湪0<x<1鏃跺彧鑳統=x锛...
