求旋转抛物面z=x^2+y^2-1 在点(2,1,4) 处的切平面方程及法线方程。
求旋转抛物面z=x²+y²-1在点(2,1,4)处的切平面和法线方程解:经检查,点(2,1,4)在抛物面上。
设F(x,y,z)=x²+y²-z-1=0;
在点(2,1,4)处,∂F/∂x=2x∣(x=2)=4;∂F/∂y=2y∣(y=1)=2;∂F/∂z=-1;
故过(2,1,4)的切平面方程为:4(x-2)+2(y-1)-(z-4)=0,即4x+2y-z-6=0为所求;
过(2,1,4)的法线方程为:(x-2)/4=(y-1)/2=(z-4)/(-1).
绛旓細鏃嬭浆鎶涚墿闈=x^2+y^2,骞抽潰z=0涓庡渾鏌遍潰x^2+y^2=ax鎵鍥寸珛鏂逛綋鐨勪綋绉 =鈭埆<D>(x^2+y^2)dxdy =鈭<-蟺/2锛屜/2>du鈭<0,acosu>r^3dr =鈭<-蟺/2锛屜/2>(1/4)(acosu)^4du =(a^4/8)鈭<0,蟺/2>(1+cos2u)^2du =(a^4/8)鈭<0,蟺/2>(1+2cos2u+cos^2u)du =...
绛旓細z=1涓巣=x^2+y^2鑱旂珛锛歺^2+y^2=1锛寊=1銆傝繖涓洸绾夸负浠ワ紙0锛0锛1锛夊渾锛屽叾涓崐寰勪负1 鎵浠ラ潰绉疭=蟺 r^2 =蟺 鎶涚墿绾鏃嬭浆180掳鎵寰楀埌鐨勯潰銆傛暟瀛︿笂鐨勬姏鐗╃嚎灏辨槸鍚屼竴骞抽潰涓婂埌瀹氱偣锛堢劍鐐癸級鐨勮窛绂讳笌鍒板畾鐩寸嚎锛堝噯绾匡級鐨勮窛绂荤浉绛夌殑鐐圭殑闆嗗悎 銆
绛旓細鎶涚墿闈=x^2+y^2涓奊澶勭殑娉曞悜閲忎负(2x,2y,-1),骞抽潰x+y-z=1鐨勬硶鍚戦噺涓(1,1,-1),鍓嶈堪鐨勪袱涓钩闈㈠钩琛,绛変环涓庤繖涓や釜骞抽潰鐨勬硶鍚戦噺骞宠,鍗虫湁锛2x/1=2y/1=-1/(-1),寰梮=1/2,y=1/2,杩涜屽緱鍒皕=x^2+y^2=1/2,鍗冲緱G鍧愭爣(1/2,1/2,1/2).G鍒板钩闈+y-z=1鐨勮窛绂讳负锛歴qrt(3...
绛旓細绔嬩綋鐨勪綋绉槸蟺/2銆倂=4鈭恉銆戔埆锛1-x^2-y^2)d蟽 =4鈭0鈫1銆慬鈭紙銆0鈫掆垰锛1-y^2)銆戯紙1-x^2-y^2)dx]dy =4鈭玔銆愶紙0鈫1銆戯級鈭0鈫掆垰锛1-y^2)銆慬x-x^3/3-xy^2)dy]=4鈭玔銆愶紙0鈫1銆戯級[鈭氾紙1-y^2)-(1-y^2)^(3/2)-y^2鈭氾紙1-y^2)]dy 璁緔=sint...
绛旓細浠(x,y,z)=x^2+y^2-z 鍒檉`x锝滐紙1,2,5锛=2x锝滐紙1,2,5锛=2 f`y锝滐紙1,2,5锛=2y锝滐紙1,2,5锛=4 f`z锝滐紙1,2,5锛=-1锝滐紙1,2,5锛=-1 鏁呰繖涓鐐圭殑娉曞悜閲忎负锛2,4,-1)鍒囧钩闈负2(x-1)+4(y-2)-(z-5)=0 ...
绛旓細鍦ㄥ摢涓钩闈㈡姇褰憋紝灏辨槸闄よ骞抽潰澶栫殑鍙傛暟鍙栧间负闆躲傚锛氬湪xy骞抽潰涓婃姇褰憋紝z鐨勫彇鍊间负0銆傚湪yz骞抽潰涓婄殑鎶曞奖锛x鍙栧间负0锛屼唬鍏ュ師鏂圭▼鍚庡緱鍒版姇褰卞尯鍩熺殑杈圭晫鏂圭▼锛屽洜涓烘姇褰辨槸涓涓尯闂达紝閭d箞鍦ㄨ竟鐣屾柟绋嬩腑鐨勫尯鍩熼兘搴旇鍙栵紝鎵浠ュ姞涓婅寖鍥撮檺鍒
绛旓細娉曞悜閲忎负锛-4x,-4y,1锛 鍗 璇ョ偣鐨勬硶鍚戦噺涓猴紙-4,8锛1锛 鎵浠 鍒囧钩闈负 -4(x-1)+8(y+2)+(z-7)=0 4(x-1)-8(y+2)-(z-7)=0 閫堿
绛旓細鏇查《鏌变綋鐨勯《闈㈡槸锛z=x^2+y^2锛屽簳闈㈠尯鍩焏鏄痻oy闈㈠唴鐢眡杞淬亂杞淬亁+y=1鎵鍥 v=鈭埆 (x^2+y^2)dxdy =鈭 [0,1]鈭 [0,1](x^2+y^2)dydx =鈭 [0,1](x^2y+(1/3)y^3)dx y鐢1锛0浠e叆鐩稿噺 =鈭 [0,1](x^2+(1/3))dx =1/3x^3+1/3x x鐢1锛0浠e叆鐩稿噺 =2/3 ...
绛旓細=4, 鎵浠鏃嬭浆鎶涚墿闈=x2+y2(0鈮鈮4)鍦▁Oy闈笂鐨勬姇褰变负x²+y²鈮4.浠=0寰梲=y², 鎵浠ユ棆杞姏鐗╅潰z=x2+y2(0鈮鈮4)鍦▂Oz闈笂鐨勬姇褰变负y²鈮鈮4.浠=0寰梲=x², 鎵浠ユ棆杞姏鐗╅潰z=x2+y2(0鈮鈮4)鍦▃Ox闈笂鐨勬姇褰变负x²鈮鈮4 ...
绛旓細V=(0-2*pi)da(0-1)pdp[(2-p^2)^1/2-p^2]V=-7/6+4*2^(1/2)/3*pi 渚嬪锛氭眰涓や釜鏇查潰鍥存垚鐨勪綋绉紝杩欎釜灏辨槸涓夐噸绉垎鐨勫簲鐢紝灏辨槸琚Н鍑芥暟涓1锛岀Н鍒嗗尯鍩熶负涓ゆ洸绾垮洿鎴愮殑鍖哄煙锛岀殑涓夐噸绉垎銆∭1dv 2寮忓甫鍏1寮 锛堟秷x^2+y^2锛夋眰鍑Z=1,甯﹀叆2寮 鏂圭▼鍗硏^2+y^2=1 ...