多元函數(shù)微分法及其應(yīng)用期末復(fù)習(xí)題高等數(shù)學(xué)下冊(cè)(上海電機(jī)學(xué)院)

(圓滿(mǎn)版)多元函數(shù)微分法及其應(yīng)用期末復(fù)習(xí)題高等數(shù)學(xué)下冊(cè)(上海電機(jī)學(xué)院)(圓滿(mǎn)版)多元函數(shù)微分法及其應(yīng)用期末復(fù)習(xí)題高等數(shù)學(xué)下冊(cè)(上海電機(jī)學(xué)院)(圓滿(mǎn)版)多元函數(shù)微分法及其應(yīng)用期末復(fù)習(xí)題高等數(shù)學(xué)下冊(cè)(上海電機(jī)學(xué)院)第八章偏導(dǎo)數(shù)與全微分參照答案第八章偏導(dǎo)數(shù)與全微分一、選擇題1.假定u=u(x,y)是可微函數(shù)，且u(x,y)yx21,uyx2x,那么uyx2[A]xyA.11C.-1D.12B.22.函數(shù)zx2y26x2y6[D]A.在點(diǎn)(-1,3)處取極大值B.在點(diǎn)(-1,3)處取極小值C.在點(diǎn)(3,-1)處取極大值D.在點(diǎn)(3,-1)處取極小值3.二元函數(shù)fx,y在點(diǎn)x0,y0處的兩個(gè)偏導(dǎo)數(shù)fxx0,y0,fyx0,y0存在是函數(shù)f在該點(diǎn)可微的[B]A.充分而非必需條件B.必需而非充分條件C.充分必需條件D.既非充分也非必需條件4.設(shè)u=x2+2y2+3z2+xy+3x-2y-6z在點(diǎn)O(0,0,0)指向點(diǎn)A(1,1,1)方向的導(dǎo)數(shù)u[D]lA.535353536B.C.3D.365.函數(shù)zx3y33xy[B]A.在點(diǎn)(0,0)處取極大值B.在點(diǎn)(1,1)處取極小值C.在點(diǎn)(0,0),(1,1)處都取極大值D.在點(diǎn)(0,0),(1,1)處都取極小值6.二元函數(shù)fx,y在點(diǎn)x0,y0處可微是fx,y在該點(diǎn)連續(xù)的[A]A.充分而非必需條件B.必需而非充分條件C.充分必需條件D.既非充分也非必需條件7.ysinyx0(01),那么dy=[B]dxA.1cosyB.1C.1cosyD.1cosy1cosy18.函數(shù)zxy5020xy〔x>0，y>0〕[D]A.在點(diǎn)(2,5)處取極大值B.在點(diǎn)(2,5)處取極小值C.在點(diǎn)(5,2)處取極大值D.在點(diǎn)(5,2)處取極小值9.二元函數(shù)fx,y在點(diǎn)x0,y0處連續(xù)的是fx,y在點(diǎn)x0,y0處可微的[A]A.必需而非充分條件B.充分而非必需條件-1-第八章偏導(dǎo)數(shù)與全微分參照答案C.充分必需條件D.既非充分也非必需條件10.曲線x=t,y=t2,z=t3全部切線中與平面x+2y+z=4平行的切線有[B]A.1條條條D.不存在11．設(shè)f(x,y)xy，那么f(x,y)By2x2yxA.xyB.x2y2x2y2y2x2y4x2y4x4C.x4D.x4y4y412．為使二元函數(shù)f(x,y)xy沿某一特別路徑趨勢(shì)(0,0)的極限為2，這條路線應(yīng)選擇xy為BA.xyB.xyC.xD.2xy43y3213．設(shè)函數(shù)zf(x,y)知足2z2，且f(x,1)x2，fy(x,1)x1，那么f(x,y)By2A.y2(x1)y2B.y2(x1)y2C.y2(x1)y2D.y2(x1)y214．設(shè)f(x,y)3x2y，那么f(xy,f(x,y))CA.3xy4x4yB.xyx2yC.3xy6x4yD.3xy4x6y．為使二元函數(shù)f(x,y)xy2在全平面內(nèi)連續(xù)，那么它在(0,0)處應(yīng)被增補(bǔ)定義為B15x2y2D.16．函數(shù)f(xy,xy)x2y2，那么f(x,y)f(x,y)CxyA.2x2yB.2x2yC.xyD.xy17．假定f(y)x2y2(x0)，那么f(x)BxxA.x21B.x21C.x21D.xxx2118．假定zyx，那么在點(diǎn)D處有zzyxA.(0,1)B.(e,1)C.(1,e)D.(e,e)-2-第八章偏導(dǎo)數(shù)與全微分參照答案19．設(shè)zxy2，那么以下結(jié)論正確的選項(xiàng)是A2z2z2z2zA.0B.0xyyxxyyx2z2zD.二者大小沒(méi)法確立C.y0xyx0,xy020.函數(shù)f(x,y)xsin1ysin1,xy0，那么極限limf(x,y)〔C〕.x0yxy0(A)等于1(B)等于2(C)等于0(D)不存在21.函數(shù)zxy在點(diǎn)(0,0)(D).(A)有極大值(B)有極小值(C)不是駐點(diǎn)(D)無(wú)極值22.二元函數(shù)zx2y2在原點(diǎn)(0,0)處〔A〕.(A)連續(xù)，但偏導(dǎo)不存在(B)可微(C)偏導(dǎo)存在，但不連續(xù)(D)偏導(dǎo)存在，但不能夠微23．設(shè)uf(r)，而rx2y2z2，f(r)擁有二階連續(xù)導(dǎo)數(shù)，那么2u2u2ux2y2z2B〕.(A)f''(r)1f'(r)(B)f''(r)2f'(r)rr(C)1f''(r)1f'(r)(D)1f''(r)2f'(r)r2rr2r24．函數(shù)zf(x,y)在點(diǎn)(x0,y0)處連續(xù)是它在該點(diǎn)偏導(dǎo)存在的〔D〕.(A)必需而非充分條件(B)充分而非必需條件(C)充分必需條件(D)既非充分又非必需條件25．函數(shù)z1x2y2的極大值點(diǎn)是〔D〕.(A)(1,1)(B)(1,0)(C)(0,1)(D)(0,0)26．設(shè)f(x,y)y，那么fx(2,1)〔B〕.arcsinx1(B)11(D)1(A)44(C)2227．極限limx2y2〔B〕.x4yx0y0-3-第八章偏導(dǎo)數(shù)與全微分參照答案(A)等于0(B)不存在(C)等于21(D)存在且不等于0及1228．zf(x,y)假定在點(diǎn)P0(x0,y0)處的兩個(gè)一階偏導(dǎo)數(shù)存在，那么〔B〕.(A)f(x,y)在點(diǎn)P0連續(xù)(B)zf(x,y0)在點(diǎn)x0連續(xù)(C)dzz|Pdxz|Pdy(D)A,B,C都不對(duì)x0y029.設(shè)函數(shù)zxy，那么dz=〔A〕．(A)．yxy1dxxylnxdy(B)．yxy1dxxydy(C)．xydxxylnxdy(D)．yxy1dxxylnydyzu2lnv,ux,vxy,那么z30.yy〔C〕2x2lnxyx22x2x2333lnxy3〔A〕yy〔B〕yy2x2lnxyx22xlnxyxy3y322〔C〕〔D〕yy31．函數(shù)z=1x2y2的定義域是〔D〕〔A.〕D={(x,y)|x2+y2=1}〔B.〕D={(x,y)|x2+y21}〔C.〕D={(x,y)|x2+y2<1}〔D.〕D={(x,y)|x2+y21}32．設(shè)f(x,y)xy，那么以下式中正確的選項(xiàng)是〔C〕；x2y2(A)fx,yf(x,y)；(B)f(xy,xy)f(x,y)；x(C)f(y,x)f(x,y)；(D)f(x,y)f(x,y)33．設(shè)zexcosy，那么2z〔D〕；xy(A)exsiny；(B)exexsiny；(C)excosy；(D)exsiny34．f(xy,xy)x2y2，那么ff〔C〕；xy-4-第八章偏導(dǎo)數(shù)與全微分參照答案(A)2x2y；(B)xy；(C)2x2y(D)xy．z35.設(shè)z2x23xyy2，那么xy〔B〕〔A〕6〔B〕3〔C〕-2〔D〕2.zfx,y,那么z36.設(shè)xx0,y0〔B〕limfx0x,y0yfx0,y0limfx0x,yfx0,y0xx〔A〕x0〔B〕x0limfx0x,y0fx0,y0fx0x,y0xlimx〔C〕x0〔D〕x0設(shè)由方程ezxyz0確立的隱函數(shù)zfx,y,那么z37.x〔B〕zzyy〔A〕1z〔B〕xz1〔C〕x1z〔D〕x1z38.二次函數(shù)zln(4x2y2)x21的定義域是〔D〕y21A.1＜x2y2≤4；B.–1≤x2y2＜4；C.–1≤x2y2≤4；D.1＜x2y2＜4。39.f(x,y)在點(diǎn)(x,y)處的偏導(dǎo)數(shù)fx(x,y)和fy(x,y)連續(xù)是f(x,y)可微分的〔B〕充分必需條件；B.充分非必需條件；C.必需非充分條件；D.非充分又非必需條件。40.拋物面zx2y2上點(diǎn)P處的切平面平行于平面2xyz30，那么點(diǎn)P的坐標(biāo)是〔C〕A.(1,1,0)；B.(1,1,0)；C.(1,1,5)；D.(1,1,5)22242441.設(shè)zexyyx2，那么z︱(1,2)〔B〕yA.e1；B.e21；C.2e1；D.2e1。42.設(shè)二元函數(shù)zx3y33x23y29x的極小值點(diǎn)是〔A〕A.〔1，0〕；B.〔1，2〕；C.〔-3，0〕；D.〔-3，2〕-5-第八章偏導(dǎo)數(shù)與全微分參照答案uxy,那么u43.設(shè)x1,1〔B〕1〔A〕0〔B〕2〔C〕-1〔D〕144.設(shè)zfx,y是由方程xyzsin(xyz)決定的隱函數(shù)，那么

zx〔D〕xsinyzcosyzz〔A〕z〔B〕yz〔C〕yz〔D〕xzexyyx2,那么z45.設(shè)y1,2(B)〔A〕e1〔B〕e21〔C〕2e1〔D〕2e1二、填空題1.xlim(1x)ye22yy2.函數(shù)u=ln(x2y2z2)在點(diǎn)M(1,2,-2)2的梯度gradu={1,2,-2}93.limsin(xy)2x2yy04.zf(xy)是可微函數(shù)，那么dzyf'(xy)dxxf'(xy)dy5.limxy=4(x,y)(0,0)xy42．設(shè)rx2y2z2，那么2rrrgradr＝2xi2yj2zk67．曲線z1x2y2在點(diǎn)(1,1,3)處的切線與Y軸的正向夾角是3x18．設(shè)rln(x2y222xr2yr2zrz)，那么gradrx2y2z2i2y2z2jx2y2z2kx9．函數(shù)zxy的中斷點(diǎn)是xy0x3y310．函數(shù)uxyz在點(diǎn)(1,1,1)沿方向(2,1,3)的方導(dǎo)游數(shù)是0-6-第八章偏導(dǎo)數(shù)與全微分參照答案函數(shù)ulnxyz的定義域是(x,y,z)x0,y0,z0或x0,y0,z0或x0,y0,z0或x0,y0,z012.二元函數(shù)zln4arcsin1的定義域是1x2y24x2y2x2y213．函數(shù)u3x2y22y4x6z在原點(diǎn)沿方向l{2,3,1}的方導(dǎo)游數(shù)為

81414.函數(shù)zln(xlny)的定義域是{(x,y)|x0,y1或x0,0y1}15.曲面exxyz3在點(diǎn)(0,1,2)處的法線方程為xy1z220116．極限lim2xy41xy4x0y017．假定f(x,y)3x2y，那么f[xy,f(x,y)]6x4y3xy18．設(shè)有函數(shù)u(x,y,z)xyz，那么du|(1,2,2)4dxdz19.函數(shù)z1x2y2的極大值點(diǎn)是(0,0)xy2r2,0,2},那么方導(dǎo)游數(shù)u20．設(shè)函數(shù)uz3,l{222l1,1,121．設(shè)函數(shù)zfxy,x2y2可微,那么zxf12yf2y22．曲面z2x2y2上一點(diǎn)〔1，-1，3〕處的切平面方程為4x2yz3023.4zx2y2在點(diǎn)P〔0,1,3〕處的切平面方程2y+z=5，法線方程xy1z302124、設(shè)zex22xy，那么全微分dz=2ex22xyxydxxdy25、設(shè)z=11n(x2y2),那么2z=2xy2xy(x2y2)226、f(xy,xy)x2y2,f(x,y)f(x,y)2x2yxy-7-第八章偏導(dǎo)數(shù)與全微分參照答案27.2xy4=1limxy4x0y0lnzxzzzyz，那么xxzx28.29.zsinxy,那么dzdzycosxydxxcosxydy三、計(jì)算與證明1.設(shè)z=f(x+y,xy)的二階偏導(dǎo)數(shù)連續(xù),求2zxy解：z=f1'f2'yx2zf12''(xy)xyf22f2=f11'''''xy2.求平面xyz1和柱面x2y21的交線上與xoy平面距離最短的點(diǎn)3410解：設(shè)(x,y,z)是交線上任一點(diǎn)，由，距離函數(shù)f(x,y,z)=z又設(shè)L(x,y,z,,)z(xyz1)(x2y21)3410Lx32x0(1)Ly42y0(2)令：Lz2z100(3)Lxyz10(4)3410Lx2y210(5)(1)與(2)比較，得：y3x，443代入(5),得：x；相應(yīng)的有：y55進(jìn)而得交線上的兩點(diǎn)：(4,3,35)，(4,3,85)556556此中：點(diǎn)(4,3,35)到xoy平面的距離是355566點(diǎn)(4,3,85)到xoy平面的距離是855566比較得：所求點(diǎn)是(4,3,35)556-8-第八章偏導(dǎo)數(shù)與全微分參照答案3.證明極限limxy2不存在x2y4x0y0證明：當(dāng)(x,y)沿著曲線y2=x趨于(0,0)時(shí)，limxy2＝limy4124y4y42y0x0當(dāng)(x,y)沿著曲線2y2=x趨于(0,0)時(shí)，limxy2＝lim2y422444x0xyy04yy5y0所以，極限limxy2不存在2y4x0y0x4.設(shè)z=xf(xy,ey),求2zxy解：z=ff1'xyx2zeyf2x2yf11''xyeyf12=2xf1''''xy5.求曲線x=t-sint,y=1-cost,z=4sint,在點(diǎn)M(21,1,22)處的切線及法平面方程2解：因?yàn)閤'=1-cost,yt'=sint,z'=2costtt2而點(diǎn)M(21,1,22)所對(duì)應(yīng)的參數(shù)為t=2點(diǎn)M的切向量T={1,1,2}x1y1z22故點(diǎn)M處的切線方程為2112點(diǎn)M處法平面方程為:x+y+2z=426.求曲面ezzxy3在點(diǎn)(2,1,0)處的切平面方程及法線方程解：令F(x,y,z)=ezzxy3那么Fx'y,Fy'x,Fz'ex1-9-第八章偏導(dǎo)數(shù)與全微分參照答案故Fx'(2,1,0)1,F'(2,1,0)2,F'(2,1,0)0yz所以:點(diǎn)(2,1,0)處的切平面方程為x-2+2(y-1)=0,即：x+2y-4=0點(diǎn)(2,1,0)x2y1處的法線方程為120z=ysin(x+y),求全微分dz及梯度gradz解：zycos(xy)，zsin(xy)ycos(xy)xy故：dz=[ycos(x+y)]dx+[sin(x+y)+ycos(x+y)]dygradz=(ycos(x+y),sin(x+y)+ycos(x+y))xyb0與曲面zx2y28.設(shè)直線l:z3在平面上，而平面相切于點(diǎn)xay0M(1,-2,5),求a,b之值解：點(diǎn)M處曲面的法向量n={2x,2y,-1}M={2,-4,-1}點(diǎn)M處切平面方程為2(x-1)-4(y+2)-(z-5)=0即:2x-4y-z-5=0,此即平面之方程由直線l可得y=-x-b,z=x-a(x+b)-3代入得:(5+a)x+4b+ab-2=0解得:a=-5,b=-29.設(shè)函數(shù)z=f(u,v),那么u,v擁有二階連續(xù)偏導(dǎo)數(shù)，此中x2zu=3x+2y,v=，求yxy解：z=3f1'1f2'xy2z=6f11''xf22''23x)f12''1f2'xyy3(y2y2y10.limx6y6能否存在？假如存在，等于多少？假如不存在，說(shuō)明原因。(x2y4)5(x,y)(0,0)解：不存在。limx6y60。limx6y6limx9。24524525x02x0y)x0,yx(xy)(2x)y0(x11.求u對(duì)于x,y,z的一階偏導(dǎo)數(shù)：uxyz解：uyzxyz1。uzyz1xyzlnxuxyzyzlnxlnyxyz-10-第八章偏導(dǎo)數(shù)與全微分參照答案12、說(shuō)明函數(shù)在何時(shí)獲得極值，并求出該極值：z(xy1)2解：函數(shù)定義域R2。因?yàn)閦0，故xy10時(shí)極小；無(wú)極大。z2(xy1)0x解方程組，可知函數(shù)駐點(diǎn)散布在直線xy10上。z2(xy1)0y對(duì)于此直線上的點(diǎn)都有z0?？墒莦0恒建立。所以函數(shù)在直線xy10上的各點(diǎn)獲得極小值z(mì)0。13.lim(x2y2)x2y2(x,y)(0,0)解：lim(x2y2)x2y2=limex2y2ln(x2y2)(x,y)(0,0)(x,y)(0,0)而x2y2ln(x2y2)1x2y22ln(x2y2)4lim1(x2y2)2ln(x2y2)0，。故原式=e01x,y(0,0)414.求u的一階全微分：uzx2y2解：du2z2(xdxydy)dz(x22)x2y2yxxt15、求函數(shù)u在點(diǎn)M〔1，2，-2〕沿曲線y2t2在此點(diǎn)的切線方向上的x2y2z2z2t4方導(dǎo)游數(shù)。解：uy2z2，uxy3，x3y(x2y2z2)2(x2y2z2)2uxz。z3(x2y2z2)2在點(diǎn)〔1，2，-2〕它們的值分別是8,2,2272727曲線在該點(diǎn)切線方向余弦為1,4,8。999方導(dǎo)游數(shù)為u81242816g()gg)l2799243M27279-11-第八章偏導(dǎo)數(shù)與全微分參照答案sin(xy)lim(x,y)(0,a)x解：limsin(xy)=sin(xy)gxlimxyy=a(x,y)(0,a)(x,y)(0,a)17.求由下式?jīng)Q定的隱函數(shù)z對(duì)于x和y的一階偏導(dǎo)數(shù)：xyze(xyz)。解：等式兩頭對(duì)x求偏導(dǎo)數(shù)，得1ze(xyz)(1z)xx故z1。利用對(duì)稱(chēng)性可得z1xy18.用拉格朗日法求條件極值：zx2y2,xy1(a0,b0)ab解：設(shè)F(x,y)x2y2(xy1)，解方程組abF2x10xaF2y10ybxy1ab2a2b2ab2a2b可得a2b2,xa2b2,ya2b2。因?yàn)楫?dāng)x或y時(shí)都有z。故函數(shù)只幸好有限處獲得極小值〔最小〕值：當(dāng)xab2,ya2b2時(shí)，函數(shù)獲得極小〔最小〕值z(mì)a2b2a2b2a2b22ab19.求極限lim1x2y132sin(xy).x0xyy0解：原式1x2y1sin(xy)limx2yxyx0lim00lim0012

(1x2y1)(1x2yx2y(1x2y1)1sin(xy)1x2y1xy(2分).

1)sin(xy)(2分)xy(1分)-12-第八章偏導(dǎo)數(shù)與全微分參照答案20.設(shè)zf(x2y2,xy)，求2z.xy解：zf1'2xf2'y2xf1'yf2',(2分)x2z2x[f11''(2y)f12''x]f2'y[f21''(2y)f22''x]xy4xyf11''2x2f12''f2'2y2f21''xyf22''(3分).21.求拋物面zx2y2到平面xyz10的近來(lái)距離。解：設(shè)M(x,y,z)在zx2y2上，M到xyz10的距離為d，那么|xyz1|(1分),d3d2(xyz1)2.3記L(x,y,z,)(xyz1)2(x2y2z)，Lx2(xyz1)2x0Ly2(xyz1)2y0令2(xyz1)0(2分)LzLx2y2z0解得：xy1,z122

(2分).所以d1|1111|1(2分).32222322.求曲面zx2y2上與平面2x4yz0平行的切平面方程。解：曲面zx2y2的切平面的法向量為n1{2x,2y,1}(2分)，平面2x4yz0的法向量為n2{2,4,1}.要使zx2y2切平面與平面2x4yz0平行，必有n1//n2，即-13-第八章偏導(dǎo)數(shù)與全微分參照答案2x2y1分).24(21解之得，x1,y2,進(jìn)而z5(2分).所以為2(x1)4(y2)(z5)0,y23．函數(shù)zarctan,求dz|(1,1).z1yy1分),解：因?yàn)閥2(2)x2y2(2x(1,1)x(1,1)212x(1,1)z11x1分),y(1,1)y2x2y2(21x(1,1)2x2(1,1)所以dz|1dx1dy(1分).(1,1)22xf(y)確立，求z。24．設(shè)函數(shù)zz(x,y)由方程x2y2z2xx解：(方法一)令F(x,y,z)x2y2z2xf(y).x那么Fx2xf(y)yf'(y),Fy2yf'(y),Fz2z(2分)，xxxx所以zFxf(y)yf'(y)2xxxx(3分).xFz2z(方法二)方程x2y2z2xf(y)兩邊對(duì)x求導(dǎo)，并注意z是x,y的函數(shù)，得x2x2zzyyyyyyxf( )xf'( )(x2)f()f'(),xxxxx解得zf(y)yf'(y)2xxxx.x2z25a分紅兩個(gè)正數(shù)x,yxpqpq．怎樣將正數(shù)之和，使得為最大，此中、是的正-14-第八章偏導(dǎo)數(shù)與全微分參照答案數(shù)。解：由拉格朗日乘數(shù)法，令L(x,y,)xpyq(xya)(2分).Lxpxp1yq0由Lyqxpyq10(2分)Lxya0解得駐點(diǎn)(ap,aq)(2分).pqpq又由題意當(dāng)點(diǎn)(x,y)趨于界限x0或y0時(shí)，目標(biāo)函數(shù)f趨于零，所以連續(xù)函數(shù)f在駐點(diǎn)取最大值。所以當(dāng)xapq,yaq時(shí)，xpyq的值最大ppq26．設(shè)zf(x,y)g(u,v),ux3,vxy，此中f,g擁有一階連續(xù)偏導(dǎo)數(shù)，求z.解：zuvxfx'gu'gv'(2分)xxxfx'3x2gu'yxy1gv'(3分).27．求曲線x2t2,ycos(t),z2lnt在對(duì)應(yīng)于t2點(diǎn)處的切線及法平面方程。解：當(dāng)t2時(shí)，對(duì)應(yīng)點(diǎn)的坐標(biāo)為(8,1,2ln2)；又參數(shù)方程的切線方向向量為：n|t2{4t,sin(t),2}|t2{8,0,1}(2分)，t故切線方程為x8y1z2ln2(2分)，801x88(z2ln2)或10.y而法平面方程為8(x8)(z2ln2)0(2分).28.求函數(shù)uxy2z3在點(diǎn)M0(1,1,1)處方導(dǎo)游數(shù)的最大值和最小值。解：u在點(diǎn)M0(1,1,1)處沿方向l的方導(dǎo)游數(shù)為：u(y2z3cos2xyz3cos3xy2z2cos)|M0lM0cos2cos3cos(2分).令l0{cos,cos,cos},g{1,2,3},-15-第八章偏導(dǎo)數(shù)與全微分參照答案u000那么gl|g||l|cos,為g與l的夾角。要使

ul

取最大值，那么cos＝1，即＝0，也就是g與l0同向時(shí)，uM0l

取最大值，0即：當(dāng)l01{1,2,3}時(shí)，u14l

取最大值|g|14(3分).0同理，要使

ul

取最小值，那么cos＝-1，即＝，也就是g與l0反向時(shí)，uM0l

取最小0值，即：當(dāng)l01{1,2,3}時(shí)，u取最小值|g|14(3分).14lM029.設(shè)函數(shù)zf(x2y,exy)，求z，z．xy解：設(shè)ux2y，vexy，那么u2xy，ux2，vyexy，vxexyxyxy故zfufv=2xyf+yexyfxuxvxuvzfufv=x2f+xexyfyuyvyuv30.設(shè)zzx,y是由x3y3z3xyz60所確立的隱函數(shù)，求它在點(diǎn)〔1，2，-1〕及z處的偏導(dǎo)數(shù)xy的值。z3x2yzxM03z2xyz3y2xzyM032xyz

1分5,M0=(1,2,1)(3)M011分)5M0斜邊長(zhǎng)為m的全部直角三角形中，求有最大周長(zhǎng)的直角三角形直角邊的邊長(zhǎng)．解：設(shè)兩條直角邊的邊長(zhǎng)為x，y，周長(zhǎng)為S，那么Smxy〔1分〕并知足x2y2m2．由-16-第八章偏導(dǎo)數(shù)與全微分參照答案F(x,y,)mxy(x2y2m2)〔2分〕F12x0x令F12y0〔3分〕yFx2y2m20解得xy2m2因?yàn)槿恐苯侨切蔚闹苯菢O點(diǎn)位于直徑為m的半圓周上，最小周長(zhǎng)不存在，進(jìn)而實(shí)詰問(wèn)題只有最大值，此時(shí)有最大周長(zhǎng)的直角三角形的邊長(zhǎng)均是2m。2eusinv，而uzz32.．設(shè)zxy,vxy,求x,yzzuzvxuxvx=eusinvyeucosv1=uysinvcosv〔3分〕ezzuzvyuyvy=eusinvxeucosv1=euxsinvcosv33.．設(shè)zfx2y2,且f可微，求yzxz。xyz2xf(2分)z2yf(2分)yzxz0(2分)xyxy34．求曲面ezzxy3在點(diǎn)2,1,0處的切平面與法線的方程.fx,y,zezzxy3那么f1,f2,f0〔3分〕x2,1,0y2,1,0z2,1,0切平面方程為x22y10z00即x2y40〔2分〕-17-第八章偏導(dǎo)數(shù)與全微分參照答案x2y1法線方程為12〔2分〕z035.將正數(shù)12分紅三個(gè)正數(shù)x,y,z之和，使得ux3y2z為最大.〔8分〕Fx3x2y2z0解：令F(x,y,z)x3y2z(xyz12)，那么Fy2x3yz0〔3分〕Fzx3y20xyz12解得獨(dú)一駐點(diǎn)(6,4,2)〔4分〕，故最大值為umax634