高数1-d8习题课三、多元函数微分法应用

PAGEPAGE7一、基本概念可显显示结1.分析复合结隐式结(画变量关系图自变量=变量总个数–方程总自变量与因变量由所求对象正确使用求导“分段用乘,分叉用加,单路全导,叉路偏导注意正确使用求导符利用一阶微分形式不在几何中的应求曲线在切线及法平面(关键抓住切向量)求曲面的切平面及法线(关键:抓住法向量方向导数和梯极值与最值问极值的必要条件与充求条件极值的方法(消元法 日乘数法求解最值问例例讨论二重极限x0x解ykx,原式limkx01yx2x原式x3x所以极限不存在(x例设二元函f(x,y)x x2y2x2y2求使f(x,y)在点(0,0)处连续的正整数解：当n1时 (10-11,三沿直线ykx时(x,y)(0,0)x2 ykxx2 x0(k2xylimxylim(k1)xf(x,y)在点(0,0)处不连(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) 设函数f(x,y)可微，且对任意的x，y都f(xy)0f(x,y)0,则f(xyf(x,y1 2成立的一个充分条件是（D(A)x1x2,y1(C)x1x2,y1x1x2,y1(D)x1x2,y1解:f(x,y)0f(x,y)0,f(x 关于y则当x1x2y1y2f(x1,y1)f(x2,y2例x2y 证明f(xy)(x2y2)3 xy x2y2在点(0,0)处连续且偏导数存在,但不解利用x2y2x2y2)2,知f(xy)1(x2y22 limf(x,y)0f(0,f(0,0)连续f(x,0f(0y0fx(0,0fy(0,0而而f(0,0)(x)2([(x)2(y)2]3当x0，y0时f(x)2( [(x)2(y)2(x)2(0所以f在点(0,0)不可微 如果函数f(x,y)在(0,0)处连续，那么下列命正确的是 (A)若极限limf(xy)存在，则f(x,y(0,0x0|x||y(B)若极限limf(x,x0x2y存在，则f(xy)(0,0处可(C)若极限f(x,y(0,0处可微f(x,x0|x||y存(D若极限f(x,y)(0,0处可微，则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则f(xy(0,0(A)若极限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)例函例函数zf(x,y)满足limf(x,y2xy2x2(y求dz解:由题f(x,y2xy2(x2y1)2)f(x,y2xy2(x2y1)2)则f(0,1)1f(0x,1yf(0,12xy(x2y2dz|(0,1)2dx例f二阶连续可微,求下列函数的二阶偏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有二阶连续偏导数,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与F分别具有一阶导数或偏导数,求dz.d解:x求导dzfxf(1d)d dxfdydzfxfdxdFFdyFdz12d 3dFdyFdzF2d 3ddzdxfxF1fxF2ff xfF3xffxf 例例zf(lnx1其中f(u)可微，求xzy2y 解:z1f,z1f 则xzy2z 例在曲面zxy上求一点使该点处的法线垂直于平面x3yz90,并写出该法线方程.解:设所求点为(x0y0z0则法线方程为xx0yy0z利 y0x0 z0x0得x03,y01,z0则法线方程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在约束条件(xy0下的一个极值点,下列选项正确的是(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极解f(1x2xy2xx x和fyxy2y yx 在(1,0)A2e2B0Ce xfxyy(1x2)eACB20A0,(1,0)极小值f(1,0exfx(y2 在(1,0)A2eB0C ACB0,A2(1,0)为极大值点极大值f(1,0e例在第一卦限作椭球 1的切平面 使其在三坐标轴上的截距的平方和最小,并求切点 z解:设F(x,y,z) 切点为M(xyz 00n(Fx,Fy,FzM2x0,2y0,2z0 c202x(xx)02y(yy)02z(zz)000即 x0xy0yz0za2a2b2切平面在三坐标轴上的截距x ,y,问题归结为求sa2b2c222xyz在条件x 1下的条件极值问题 y2z设 日函F a2b2 c2 x y2zxy z (x0,y0,z令Fx x a2x唯一驻F2bb2y xyyy bFz c2zyzz x2y2z21z由实际意义可知Maaabbccabcab ab,b,abc为所求切点例在例在第一卦限内作椭球 x2y2z21的切平使与三坐标面围成的四面体体积最小,并求此体积解:设切点为(x0y0z0则切平面为x0xy0yz0z 所指四面体体 V 6x0y0V最小等价于f(x,y,z)=xyz最大,故 日函Fxyz(x2y2z2 日乘数法可求出(x0,y0,z0)1 x2y2V Fxyz(6x0y0 zFyz2 唯一驻 3 x Fyxzb2 y3Fzxyc2 z z3 a2 xyz 由实际意义可知所求切平面 ab所求体V32 设zxf(xy),F(x,y,z)0,其中f与F分别具有一阶导数或偏导数,求dz.(1999考研)d解:方程两边对x求导,dzfxf(1d)d dxfdydzfxfdxdFFdyFdz2d 3dFdyFdzF2d 3ddzd21xF1fxF2ffxfF3 求旋转抛物面 求旋转抛物面zx2y2与平面xy2z之间的最短距离解P(xyz)zx2y2上任一点，xy2z20d1xy2z6目标函数:(xy2z2)2(约束条件:x2y2zF(x,y,z)(xy2z2)2(zx2y2Fx2(xy2z2)2xFy2(xy2z2)2y令Fz2(xy2z2)(2)zx2解此方程组得唯一驻点x1,y1,z. ,d11112644 4F(x,y,z)(xy2z2)2(zx2y2 3在在第一卦限内作椭球 x2y2z21的切平使与三坐标面围成的四面体体积最小,并求此体积提示:设切点