第五章：多元函数微分学.pdf

,000lim)0,0(0 xfxx,000lim)0,0(0 xfyy)0,0()0,0()0,0(),(lim00yfxffyxfyxyx 2200)()(limyxyxyx ,00)(1sin)(lim)0,0(220 xxxfxx,00)(1sin)(lim)0,0(220 yyyfyy)0,0()0,0()0,0(),(lim00yfxffyxfyxyx 22222200)()()()(1sin)()(limyxyxyxyx )0,0(),(yx2222221cos21sin2),(yxyxxyxxyxfx ,01sin2lim22)0,0(),(yxxyx2222)0,0(),(1cos2limyxyxxyx )0,0()1,0()1,0()1,1()0,0()1,1()1,1(fffffff .211)1(),0()1,(yxff.1.11.1),(yxyxf ,0)1,1(,2.2)1,1(,0)1,1(fff)0,0()0,1()0,1()1,1()0,0()1,1(ffffff .101)1()0,(),1(xyff1)0,0()1,1(ff22)0,0(),(22)0,0(),(2222)0,0(),(lim),(lim)(),(limyxyxyxfyxyxyxfyxyxyx 1),(lim22)0,0(),(yxyxfyx【解解1 1】直直接接法法 0),(lim)0,0(),(yxfyx,0)0,0(f),(yxf)0,0(则则，若若在在点点连连续续，否否则则不不连连续续。故故(A A）不不正正确确。【解解法法2 2】排排除除法法,0,10,),(222222 yxyxyxyxf 02),(lim2222)0,0(),(yxyxyxyxfyx,0)0,0(f02),(lim22)0,0(),(yxyxyxfyx【解解】知知，且且则则,02)0,0(),(lim22)0,0(),(yxyxfyxfyx)(2)0,0(),(yxfyxf,1)0,0(,2)0,0(yxff【解解】验验证证法法：（A A）(C C)（D D）都都不不满满足足xxfy )0,(，故故应应选选（B B）.222 yz),(22xydyyz ,2xyyz ),()2(2xxyydyxyz.12 xyyz0),(lim2200 yxyxfyx,0)0,0(f,0),(22 yxyxf,0)0,0()0,(lim0 xfxfx,0)0,0()0,(lim0 xfxfxyyxxzddd ,)(2122Cyxz 【解解1 1】【解解2 2】)(ln)(yfxfxz )()()(yfyfxfyz )(ln)(22yfxfxz ，yxz 2)()()(yfyfxf )()()()()(2222yfyfyfyfxfyz )()(ygxfxz )()(ygxfyz )()(22ygxfxz )()(2ygxfyxz )()(22ygxfyz ，.2222222)0()0()0()0()0()0(gfgfgfyxzyzxz ).0()0()0()0(gfgf yxFFxy )(0)(0 xy2)()(yxyyyxyxyxxFFyFFFyFFxy ),(),()(00000yxFyxFxyyxx ，ByxuCyuAxu 22222,02 yxu,02222 yuxu,0,0 CAB02 BAC【解解】由由题题设设可可知知,则则 ),()1sin(1)1(),0(yyyyz ,1)1(.1)1,0(yz,1422yxyxz ,),0(2yyzx.221)1,0(2 yyxyz)eln(lnyxxz yyxxxzxze)eln(.2ln21)0,1(xz，,)(1ln11 yxyxyxyyxxzyx,)(1ln1222 yxyxyxyxyxyzyx,2ln21)1,1(xz),2ln21()1,1(yz).d)(d2ln21(d)1,1(yxz 2,1 yx.0 z,lnln)ln(yxyzx ,1)ln(xyzxzyzx ).2ln1(2 xz0,1 yx.1 z,0333333222 xydzxzdyyzdxdzzdyydxx;0333 dydzdx,3222323dzzexdyzexdxzxeduyyy .32dzdydxdu .25)33(2dydxdydxdydxdu xxf)0,(.1)0,(xfxyxyxz 2),(21)(2xyxydyyxxz ,1212 yxyxz),(2121)121(222yxxyyxdxyxyz .)(2yy 21)2,(xxxu xxxuxxu2)2,(2)2,(1211 xxxu)2,(1)2,(2)2,(21 xxuxxu0)2,(4)2,(2)2,(2)2,(22211211 xxuxxuxxuxxu0)2,(4)2,(51211 xxuxxu.34)2,(11xxxu ,11)(21221FxFyxzFFxz ,11)(21221FxFyFyzFyz .yzyxzx)(xyz ttyxfytxft),(),2(lim00000 tyxftyxfyxfytxft),(),(),(),2(lim000000000 ),(),(20000yxfyxfyx 5122 0)1(22),(lim2201 yxyxyxfyx,2)0,1(,1)0,1(yxfftftfftfttftftt)0,1()2,10,1()0,1(lim)2,1()0,1(lim00 .5)0,1(2)0,1(yxff 1010)()()()()(xyxydttfxytdttftxydttftxyz 1100)()()()(xyxyxyxydttfxydtttfdtttfdttfxy 122220)()()()()()(xyxyxdttfyxyfxyxyfxyxyfxyxyfxytfyz 10)()(xyxydttfydttfy)(2)()(222xyfyxyfyxyfyzxx )(22xyfxzyy).()(222xyfyxzzyyxx ;2231yxxffxu ;)()(222222322223322131122yxxyfyxxfyxxffxz ,1)1,1()1,1(,1()1(fff 1213d)(d)(3)(dd xxxxxxx 121212),(),()(,(,(),(,()(3 xxxfxxfxxfxfxxfxfx.51)32(3213 .ddddddxzzfxyyfxfxu 2e xyxy,0dddde xyxyxyxyxy.ddxyxy zxxttt0dsine,dd1)sin(e xzzxzxx.)sin()(e1ddzxzxxzx .)sin()(e1ddzfzxzxyfxyxfxux ,vzuzxz ,2vzauzyz ,22222222vzvuzuzxz ,4422222222vzavuzauzyz .)2(2222222vzavuzauzyxz 【解解1 1】.0)6()510(2222 vzaavuza062 aa,0510 a.3 a.2,22 avuyavaux,212yzaxzaauyyzuxxzuz vyyzvxxyzavyyxzuxxzaavuz2222222212.21)2(2)2(22222222yzayxzaaxzaa 【解解2 2】,uxy)()(ufyxz xyyzxz .)(2)()1(2uufufu Cuuuuuuf 12)1ln(11)(.2)1ln()(yxxyyxz )()(),(22yxgrgyxf rxrgyxxrgxf )()(22322)()(rxyrgrxyrgyxf 3222sincos)(sincos)(rrrgrrrg 02 yxf.0)(1)(rgrrg221)(CrCrg .)(),(2221CyxCyxf 22yxr ,drdzrxxrdrdzxz ,132222222drdzrxdrdzrdrzdrxxz ,132222222drdzrydrdzrdrzdryyz .222rzdrzd .2sincos221 rrCrCz 0)(240)(2433yxyfyxxfyx)0,0(),1,1(),1,1()1,1(),1,1(02 BAC0 A)0,0(.02 BAC;0)0,0(f.02),(4 xxxf.0)0,(24 xxxf ),2(2),(2yxyxfx .1ln2),(2 yyxyxfy ,0),(,0),(yxfyxfyx e1,0【解解】令令解解得得唯唯一一驻驻点点.由由于于,e122)2(2e1,02e1,02 yfAxx,04e1,0e1,0 xyfBxye,12e1,0e1,02 yxfCyy 0e12e222 BAC.0 A.e1e1,0 f所所以以极极小小值值为为0)1(g,)(21fxgyf yxz .)()()()(22212121112fxgfxgyfxgfxgfxyfyxz ).1,1()1,1()1,1(12111112fffyxzyx ),(),(,(),(,(121yxfyxfyxfyxfyxfxz ),(,(),(),(),(,(),(,(212212112yxfyxfyxfyxfyxfyxfyxfyxfyxz ).,(),(,(),(,(),(222211yxfyxfyxfyxfyxfyxf ,0)1,1(,0)1,1(21 ff).1,1()2,2()2,2(12211)1,1(2fffyxz .0824,08824yyyxxxzxzz zyzxzzz zx1 z78 z)1,0,2().78,0,716()78,0,716(取取极极小小值值,取取极极大大值值.)1,0,2()1()1(),(22yxoyxyxf 0)0,1(f.1)0,1()0,1(yxff ,222121yfxefgxfyefgxyyxyx .0)0,0(,0)0,0(yxgg2222121121122)2()2(2fxxfyefyefyexfyefgxyxyxyxyx xyfxefexyefyeyfxefgxyxyxyxyxyxy2)2()()2(222111211 2222121121122)2()2(2fyyfxefxefxeyfxefgxyxyxyxyy 1)0,1()0,0(,2)0,1(2)0,0(122 fgBfgAxyx，2)0,1(2)0,0(22 fgCy032 BAC0 A0)0,1()0,0(fg，,024,02222yxyfxyxfyx.2)1,2(),1,2(f)22(0:1 xyL,)0,()(2xxfxg )0(4:222 yyxL).22(85)4,()(242 xxxxxfxh.8)2,0()0(fh.4723,2525 fh【解解】)254(212),(2222 yxyxyxyxF 0254)2(02412)1(0812222yxFyyxFxyxFyx 令令由由(1 1)和和(2 2)式式知知：0)2(606)41(yxyx 且且有有非非零零解解.,026641 417,221 则则 解解得得 21 .50),3,2(),3,2(21 zPP时时,驻驻点点4172 .4425),4,23(),4,23(43 zPP时时,驻驻点点0412,0122 yxzyxzyx得得驻驻点点0)0,0(),0,0(z)10(2),(222 zyxyzxyzyxF 01002202202222zyxFzyFyzxFxyFzyx 【解解】令令 1 1）当当时时，0 ,02,0 zxy);2,5,1(1P);2,5,1(2 P);2,5,1(3 P);2,5,1(4 P2 2）当当时时，0 ,5,222yxzx );2,0,22(5 P);2,0,22(6 P;55)()(41 PuPu;55)()(32 PuPu;0)()(65 PuPu).4()(),(22222 zyxzyxzyxzyxF ,04,0,02,022,02222zyxFzyxFzFyyFxxFzyx 【解解】令令 ),2,1,1(),(111 zyx).8,2,2(),(222 zyx解解方方程程组组，得得 故故所所求求的的最最大大值值为为7 72 2，最最小小值值为为6 6.)(xXxyyY yxyxxy33)(yxYyxX31,3100 )81(21xyS )0,0(yx)1323(),(22 yxyxxyyxF 22yx ,81 yx41 S )53()2(),(2222 zyxzyxzzyxL ,53,02,0342,02,02222zyxzyxzzLyLxLzyx yx ,532,02222zxzx ,5,5,5zyx .1,1,1zyx)()23(),(222222zyxzyxzyxzyxL 261311d 613111 d613112 d 0123022020322222zyxLzyxLzzLyyLxxLzyx xy)0(kkqypxqp【证证】求求函函数数在在条条件件下下的的最最大大值值。)(),(kqypxxyyxLqp 01 pxxyL 01 qyyxL kqypxqp 令 qkykxp1,1 qxpxkkkkxyqpqpqp 1111（I I）10000)0,0(,6,220 CyyCxxC.2642010000),(22yyxxyxC （I II I）,500,2)50()50(642010000)50,()(22 xxxxxxxCxf03623)(xxf.24 x（I II II I）32)220(26242624 yxyxxxC yxyxCyxRyxL2),(),(),(3642321422 yxyxyx【解解】1 1）令令 0823202214yxyLyxxL3,4 yx40)3,4(max L )62(),(),(yxyxLyxF 06202823202214yxFyxFyxFyx 2 2）令令2,2 yx28)2,2(max L)(2),(),(22yxyxfyxg 122 yx,1),(yxg1)0,0(g),(yxg),(00yx0),(),(0000 yyxgxyxg0004),(xxyxf 0004),(yyyxf .16)(16),(),(2020200200 yxyyxfxyxf【证证】考考查查函函数数在在单单位位圆圆周周上上，.而而.可可在在单单位位圆圆内内，从从而而有有取取到到它它的的最最小小值值.设设该该点点是是