高數(shù)1-d8習(xí)題課三、多元函數(shù)微分法應(yīng)用

PAGEPAGE7一、基本概念可顯顯示結(jié)1.分析復(fù)合結(jié)隱式結(jié)(畫變量關(guān)系圖自變量=變量總個(gè)數(shù)–方程總自變量與因變量由所求對(duì)象正確使用求導(dǎo)“分段用乘,分叉用加,單路全導(dǎo),叉路偏導(dǎo)注意正確使用求導(dǎo)符利用一階微分形式不在幾何中的應(yīng)求曲線在切線及法平面(關(guān)鍵抓住切向量)求曲面的切平面及法線(關(guān)鍵:抓住法向量方向?qū)?shù)和梯極值與最值問極值的必要條件與充求條件極值的方法(消元法 日乘數(shù)法求解最值問例例討論二重極限x0x解ykx,原式limkx01yx2x原式x3x所以極限不存在(x例設(shè)二元函f(x,y)x x2y2x2y2求使f(x,y)在點(diǎn)(0,0)處連續(xù)的正整數(shù)解：當(dāng)n1時(shí) (10-11,三沿直線ykx時(shí)(x,y)(0,0)x2 ykxx2 x0(k2xylimxylim(k1)xf(x,y)在點(diǎn)(0,0)處不連(x(xf(x,y)x x2y2x2y2n2ykx(xy)2lim(x(k1)2(k(x,y)(0,0)x ykxx2222lim x0(k2 k2f(x,y)(0,0)(xf(x,y)x x2y2x2y2n0|f(x,y)||xy|n[(xy)2x2 x2n nnx222(x2y2)0(xy)n0f(x,y)(0,0)x2yf(x,y)(0,0) 設(shè)函數(shù)f(x,y)可微，且對(duì)任意的x，y都f(xy)0f(x,y)0,則f(xyf(x,y1 2成立的一個(gè)充分條件是（D(A)x1x2,y1(C)x1x2,y1x1x2,y1(D)x1x2,y1解:f(x,y)0f(x,y)0,f(x 關(guān)于y則當(dāng)x1x2y1y2f(x1,y1)f(x2,y2例x2y 證明f(xy)(x2y2)3 xy x2y2在點(diǎn)(0,0)處連續(xù)且偏導(dǎo)數(shù)存在,但不解利用x2y2x2y2)2,知f(xy)1(x2y22 limf(x,y)0f(0,f(0,0)連續(xù)f(x,0f(0y0fx(0,0fy(0,0而而f(0,0)(x)2([(x)2(y)2]3當(dāng)x0，y0時(shí)f(x)2( [(x)2(y)2(x)2(0所以f在點(diǎn)(0,0)不可微 如果函數(shù)f(x,y)在(0,0)處連續(xù)，那么下列命正確的是 (A)若極限limf(xy)存在，則f(x,y(0,0x0|x||y(B)若極限limf(x,x0x2y存在，則f(xy)(0,0處可(C)若極限f(x,y(0,0處可微f(x,x0|x||y存(D若極限f(x,y)(0,0處可微，則limf(x,x0x2存ff(x,(Blxim0x2y2f(xy(0,0解:f(xy(0,0若limf(xy)x0x2f(0,0)0,f(0,0)limf(x,0)0,f(0,0)limf(0,y)xxy f(x,y)f(0,0)f(0,0)xfxlimf(x,y)x2則f(xy(0,0(A)若極限f(x,x0|x||yf(xy)(0,0f(x,y)|x||y(Cf(xy(0,0f(x,x0|x||y存(Df(xy(0,0limf(x,x0x2f(x,y)例函例函數(shù)zf(x,y)滿足limf(x,y2xy2x2(y求dz解:由題f(x,y2xy2(x2y1)2)f(x,y2xy2(x2y1)2)則f(0,1)1f(0x,1yf(0,12xy(x2y2dz|(0,1)2dx例f二階連續(xù)可微,求下列函數(shù)的二階偏x2z(1)zxfy2)xxf 2) 2yf 2 23x2y 2)2f (2)(2)zf(x ) fy 2z2yf2yf xx2y2x2 x(1x2)(3)zf(x )yx xz2yf 2z2y2y(f2x2x x2f例uf(xyz有二階連續(xù)偏導(dǎo)數(shù),zx2sintln(xy),求xu,2u解ff(2xsint2 xcost)2u (xcostx21f 1 (xcostxy)(2xsintx2costxyf32xcost1xx x 2sint(xy)cost(xuxyxx例u例uf(xyz,yy(xzz(x)exyxyex xzsind0t d解dyyexyyyd xexy x 1dsin(xz)dsin(xex(xsin(xxdud fyx1e(xz)sin(xx例zxf(xy)F(xyz0f與F分別具有一階導(dǎo)數(shù)或偏導(dǎo)數(shù),求dz.d解:x求導(dǎo)dzfxf(1d)d dxfdydzfxfdxdFFdyFdz12d 3dFdyFdzF2d 3ddzdxfxF1fxF2ff xfF3xffxf 例例zf(lnx1其中f(u)可微，求xzy2y 解:z1f,z1f 則xzy2z 例在曲面zxy上求一點(diǎn)使該點(diǎn)處的法線垂直于平面x3yz90,并寫出該法線方程.解:設(shè)所求點(diǎn)為(x0y0z0則法線方程為xx0yy0z利 y0x0 z0x0得x03,y01,z0則法線方程x y z 例求例求grad(xyyz解:grad(xyzy(y,xz,1)y2yf(xy(xy均可微y(xy0(x0y0f(x在約束條件(xy0下的一個(gè)極值點(diǎn),下列選項(xiàng)正確的是(Dfx(x0y00,fy(x0y0若fx(x0y00,fy(x0y0若fx(x0y00,fy(x0y0若fx(x0y00,fy(x0y0提示:Ff(xyxFxfx(x,y)x(x,y) y(x0y00,f(xy,代入()f(x,y)f(x,y(x,yx0例f例f(xyx2極解f(1x2xy2xx x和fyxy2y yx 在(1,0)A2e2B0Ce xfxyy(1x2)eACB20A0,(1,0)極小值f(1,0exfx(y2 在(1,0)A2eB0C ACB0,A2(1,0)為極大值點(diǎn)極大值f(1,0e例在第一卦限作橢球 1的切平面 使其在三坐標(biāo)軸上的截距的平方和最小,并求切點(diǎn) z解:設(shè)F(x,y,z) 切點(diǎn)為M(xyz 00n(Fx,Fy,FzM2x0,2y0,2z0 c202x(xx)02y(yy)02z(zz)000即 x0xy0yz0za2a2b2切平面在三坐標(biāo)軸上的截距x ,y,問題歸結(jié)為求sa2b2c222xyz在條件x 1下的條件極值問題 y2z設(shè) 日函F a2b2 c2 x y2zxy z (x0,y0,z令Fx x a2x唯一駐F2bb2y xyyy bFz c2zyzz x2y2z21z由實(shí)際意義可知Maaabbccabcab ab,b,abc為所求切點(diǎn)例在例在第一卦限內(nèi)作橢球 x2y2z21的切平使與三坐標(biāo)面圍成的四面體體積最小,并求此體積解:設(shè)切點(diǎn)為(x0y0z0則切平面為x0xy0yz0z 所指四面體體 V 6x0y0V最小等價(jià)于f(x,y,z)=xyz最大,故 日函Fxyz(x2y2z2 日乘數(shù)法可求出(x0,y0,z0)1 x2y2V Fxyz(6x0y0 zFyz2 唯一駐 3 x Fyxzb2 y3Fzxyc2 z z3 a2 xyz 由實(shí)際意義可知所求切平面 ab所求體V32 設(shè)zxf(xy),F(x,y,z)0,其中f與F分別具有一階導(dǎo)數(shù)或偏導(dǎo)數(shù),求dz.(1999考研)d解:方程兩邊對(duì)x求導(dǎo),dzfxf(1d)d dxfdydzfxfdxdFFdyFdz2d 3dFdyFdzF2d 3ddzd21xF1fxF2ffxfF3 求旋轉(zhuǎn)拋物面 求旋轉(zhuǎn)拋物面zx2y2與平面xy2z之間的最短距離解P(xyz)zx2y2上任一點(diǎn)，xy2z20d1xy2z6目標(biāo)函數(shù):(xy2z2)2(約束條件:x2y2zF(x,y,z)(xy2z2)2(zx2y2Fx2(xy2z2)2xFy2(xy2z2)2y令Fz2(xy2z2)(2)zx2解此方程組得唯一駐點(diǎn)x1,y1,z. ,d11112644 4F(x,y,z)(xy2z2)2(zx2y2 3在在第一卦限內(nèi)作橢球 x2y2z21的切平使與三坐標(biāo)面圍成的四面體體積最小,并求此體積提示:設(shè)切點(diǎn)