山東大學(xué)高等數(shù)學(xué)第二學(xué)期期末復(fù)習(xí)

第二學(xué)期(Ch7§5-Ch11)期末復(fù)習(xí)題及參考解答一.填空題1.設(shè)點A(1,1,4)在曲面S:zf(x,y)上.若,且f(x,y)在其定義域f(x,y),則曲面S在A點f(1,1)3x內(nèi)任意一點(x,y)處滿足方程處的切平面方程為xf(x,y)yf(x,y)xy.3xyz0解:由題設(shè)有131f(1,1)4,即f(1,1)1,所以yyΠ:3(x1)(y1)(z4)0,即3xyz0.T2.設(shè)f(x,y,z)exyz2,其中是由方程xyzxyz0確定的隱函數(shù),zz(x,y)則.f(0,1,1)1xf(0,1,1)df(x,1,1)dex1.x0解:dxdxxx0xy2z20在點(6,3,3)處的切線與Oz軸正向的夾角為.63.曲線x6x6解:曲線方程可寫為yy,x(y)0,y(y)1,z(y)y,于是y26zy263},cos3,故π.τ{0,1,3}2{0,1,22264.設(shè)n是曲面2x23y2z26在點P(1,1,1)處的外法向量,則函數(shù)6x28y2在點P處沿方向n的方向?qū)?shù).u117uzn解:令F(x,y,z)2x23y2z26,則F2x,F6y,F2z,于是xyz{2,3,1},其方向余弦為cos231.又,cos,cosn141416y14u112x6uy,18,xz26x28y2z26x28y21414PPPP26x8y214,故uzz2PPun6283111．7(14)1414141414P5.設(shè)D{(x,y)x2y21},則.5π(x22sinxy31)d4D解:(x22sinxy31)dxd2sinxdyd1d23DDDDD12πcos2d1.π45π.3d00ππ4006.交換二次積分的次序:0dxf(x,y)dy1dx11x2f(x,y)dy2dyf(x,y)dx.4x22y-y24y220011x207.將直角坐標(biāo)系下的三次積分化為球坐標(biāo)系下的三次積分:1dxdy11x2y2f(x,y,z)dz1x211x211x2y2.2d2d2cosf(rcossin,rsinsin,rcos)r2sindr0008.設(shè)有曲線段Lxacost(0t),則.:xydsπa222yasintL解:x2y2dsadsaπaπa2.LL9.設(shè)C是圓周(x1)2y2R2(R1)的正向,則xdyydxπ.4x2y2CQ,故可在C的內(nèi)部作橢圓周x解:因為當(dāng)(x,y)(0,0)時,Py24x(4x2y2)22yl:4x2y2a2,且l與C同向,則xdyydxxdyydx.(11)d2πa2aπ14xy22a2a2a2Cl4x2y2a210.設(shè)是光滑閉合曲面所圍成的空間域,其體積為V,則沿外側(cè)的積分(zx)dxdy(yx)dzdx(xz)dydz.3V解:(zx)dxdy(yx)dzdx(xz)dydz(111)dV3V.11.若冪級數(shù)a(x1)n在x1處收斂,在x3處發(fā)散,則其收nn0.D[1,3)解:因為R112,R312,故R2,從而D[1,3).12.設(shè)f(x)x1,則f(n)n!.3n(1)4x解:因為af)(1),故f(nN).(n(1)n!a(n)n!nn而f(x)x1x14x3,1(x1)n3n1x131(x1)n13n11n0n1所以f(nN).n!(n)(1)3n213..exex21(2n)!x2nchxn0解:易知R.設(shè)(2n)!x2n,x(,)S(0)1,.1S(x)n0,,.1S(x)x(,)S(0)0(2n1)!x2n1n1,ex.2從而exCC11S(x)S(x)Sx()(2n2)!x2n2n1C1,解得1,故由及得CS(0)1S(0)0CC2212CC0112exex2chx.S(x)14.設(shè)是x1,0x1,10x的以為周期的Fourier級數(shù)的和函數(shù),2f(x)S(x)則.12S(4),S(7)1解:,S(7)S(1)S(1)(1)11.S(4)S(0)01122215.微分方程的通解為(1y)dx(xy2y3)dy0xxyyy34C.34解:dx(ydxxdy)y2dyy3dy0,4)0,4dxd(xy)d(y)d(y33,所以通解為C.d(xxyyy)034yxxy33y434416.用待定系數(shù)法求的特解時,應(yīng)設(shè)y4y3xcos2xy.yx[(AxB)cos2x(CxD)sin2x]二.單項選擇題1.設(shè)函數(shù)在點的某鄰域內(nèi)有定義,下列結(jié)論正確的是(B)zf(x,y)M(A)若z在點M處沿任意方向l的方,則在點M處兩個向?qū)?shù)z存在zl偏導(dǎo)數(shù)存在.(B)若z在點處可微,則在點處的梯度存在.MzM(C)若z在點M處沿任意方向l的方向?qū)?shù),則在點M處連續(xù).z存在zl(D)若z在點M處連續(xù)且沿任意方向l的方向?qū)?shù)存在,則在點zlzM處可微.32.設(shè)曲面上點P處的切平面平行于平面2x2yz10,則Pz4x2y2點的坐標(biāo)為(D)(A)(1,1,2)(C)(1,1,2)解:因為點P(x,y,z)處切平面的法向量n{2x,2y,1}平行于已知平面的法向.(B)(1,1,2).(D)(1,1,2)..量n{2,2,1},故有2x2y1,于是得xy1,而z41122,因此P122.21點的坐標(biāo)為(1,1,2)5.設(shè)在域f(x,y)D{(x,y)xy2Rxx,R0}2上連續(xù),則二重積分f(x,y)dxdy(C)D(A)2Rxx2f(x,y)dy.(B)R2y2f(x,y)dx.RdyRdx0000(C)Rdyf(x,y)dx.(D)d.f(cos,sin)dπ22Rsinyπ40RR2y28.設(shè)有球面0,是S的上半部分的上側(cè),S是S的下半S:xy2z21S212部分的下側(cè),若,zdxdy,則(B)II2zdxdy1SS2(A).(B).(C).(D).II01IIIIII11212212解:2dxdy,yIzdxdy1x21S1x2y21(1x2y2)dxdy1x2y2dxdyI.I2zdxdy1S221x2y21xy29.以下四式正確的是(B)(A)ln(1x)(1).(B)1.π2n(2n)!xn(1)n1x(1)nnn1n0(C)(1)nx(x).(D)(1)n0.2nsinxπ2n1(2n1)!x(2n1)!n0n1解:ln(1x),cosπ1,π2n(1)n(2n)!xn(1)nnn1n0(1)nx2nsinx(x0),.π2n1(1)n(2n1)!sinπππ(2n1)!xn0n110.設(shè)正項級數(shù)u收斂,且limnu存在,則(A)nnn(A)limnu0n1.(B)limnu0.nnnn(C).(D)不能確定.limnu0nn4解:因為limnulimu存在且收斂故必有u,,否則發(fā)散.ulim0unn1n1nnnnnn1n1nn11.將(ab0)展為的冪級數(shù)時,所展冪級數(shù)的收斂半徑xxabxf(x)R(D)(A)a.(B)b.(C)b.a(D)a.b解:f(x)(bx)na,當(dāng)即時級數(shù),1a1bxaxabxxxbbaabxax1xan0絕對收斂,當(dāng)即時級數(shù)發(fā)散,故a.babx1xRba12.微分方程()滿足初始條件1的解y(C)yxyx0y(2)e(A)eC.(B).(C)e(D)Ce.x22x2x22x22e2解:和不是特解,不滿足初始條件,故選(C).(或直接求解.)x22x222xeCeCe2三.解下列各題1.設(shè)是由方程確定的隱函數(shù),求2z.zz(x,y)xyzezx2解:將方程兩邊對求導(dǎo)得1zezz,解得x1;zxxxez1兩邊再對求導(dǎo)得ez(z)ez2z,解得22zxxx2x2ez(z)2x1ez2zez.(1ez)3x22.設(shè)是由方程確定的隱函數(shù),且可微,F(u,v)zz(x,y)F(yx2,x2z2)02試計算xz.yzyx解:,,1,故F2yFF2zF22xF2xF1Fx2yzzy2x[FF]2yFxy.1yzxxx2zF12y2zFz223.求在閉域上的最大值與最小值.z2xyxyD:xy12222x40解:令zxy,得惟一駐點(0,0),且.z(0,0)02yx0zycosxt(0t2)上,函數(shù)化為在邊界,即xy212ysintzz(t)2costsintsintcost225(0t2).1cos2t1sin2t3222令得,t23.8z(t)sin2tcos2t0t81832,211113z()2222238332.22)1111z(2222經(jīng)比較得32,z0.zmax2min4.將分成三個正數(shù)之和,使得函數(shù)達到最小值.44x,y,zux22y23z2解:此問題為:求在下的最小值.ux2y23z2xyz4402作,().L(x,y,z,)x2y3z(xyz44)x,y,z0,02222x0Lxy04令L,解得惟一駐點(24,12,8).由于u存在最小值(無y6z0L0zxyz44L最大值),故當(dāng)x24,y12,z8時u達到最小.5.在橢球面的第一卦限部分上求一點,使得橢球面在該點處x2y2z124的切平面在三個軸上的截距的平方和最小.解:設(shè)所求點為.曲面在點的法向量z,切平面Π的方n{2x,2y,}2P(x,y,z)P程為化為2)0,即.xXyYzZ(x2y2zxXyYzZ14441,立即可得Π在三個坐標(biāo)軸上的截距為.于是114,,X1xY1yZ4xyzz問題歸結(jié)為:求u1(x0,y0,z0)在條件116x2y2z2下的最小值.x2y2z124作21)().x,y,z0,01116(x2y2zz2L(x,y,z,)4x2y26L22x0xx3x12L22y0yx,得惟一解1.令yy3z22xy2z2Lz32z0x2y2z212zx342Lx2y2z104由問題的實際意義知,點11即為所求的點.(,,2)226.設(shè)三角形的周長為2p,問三角形的三邊各為多少時,才能使它繞自己的的旋轉(zhuǎn)體體積最大?一邊旋轉(zhuǎn)所得B解:設(shè)ABC三邊長分別為x,y,z,則xyz2p且繞其邊AC旋轉(zhuǎn)(見圖)..xzh若記AC上的高為h,則ABC的面積yAC4p(px)(py)(pz);2yhp(px)(py)(pz),從而hS12y24πp(px)(py)(pz),故旋轉(zhuǎn)體體積V13yπh23y其中xyz2p(x0,y0,z0).為簡化計算,我們求函數(shù)uln(px)ln(py)ln(pz)lnyxyz2p下的駐點.為此作輔助函數(shù)在條件.L(x,y,z,)ln(px)ln(py)ln(pz)lny(xyz2p)1px1Lx0py10得惟一駐點解方程組L.(3p,1p,3p)yy4241pzxyz2p0Lz0L由問題的實際意義知,34是V在xyz2p下的最大值點,(p,1p,3p)24即當(dāng)ABBC,pAC12p,且繞AC旋轉(zhuǎn)時,所得體積最大,Vπp3.1234max7.用重積分表示并計算出由曲面z2x2y2與zxy2所圍立體的2表面積.解:記S:z2xy2,S:zx2y2,S.在平面上的投影S,SxOySS1212212域均為D:x2y21.dS,.2dxdydS14x24y2dxdy12故S(214x24y)dxdy2D7dxdy2πd12142d00D1.2π2π121(142)2π(551)π63208.一質(zhì)點在平面力場F1y2i(11)yj的作用下,沿曲線xy22y1x3x2由點A(1,0)運動到點B(4,1),求力場所作的功.1y2dx(11)ydy.解:WFdsx3x2L(A,B)L(A,B),因為P2yQ故積分與路徑無關(guān),于是yxx3(4,1)1y2dx(11)ydy41W1dxx31(11)ydy16x3x2(1,0)0.151713232161y2f(xy)dxx[y2f(xy)1]dy,其中L是經(jīng)過O(0,0),9.計算曲線積分yy2L和B(1,2)三點的圓周上從A點到B點的一段劣弧,f為可微函數(shù).23A(3,)f(xy)1Q,故積分與路徑無關(guān),于是選擇沿解:因為Pyy2f(xy)xy3xy2yxx2(x從3變到1)積分得BC:L(A,B)yxAC14x[4x2f(2)f(2)1]原式1dx213x2Ox224x3xx2.1xdx4310.設(shè)S是曲面x2y2z21(z0)的下側(cè),求x3dydzy3dzdxz3dxdy.S解:記,S,所圍域記為.:x2y21原式3(x0d2y2z2)dV內(nèi)上x2y2132πd6π.5π2d1r2r2sindr000811.計算1dxdy,其中是由曲面Ix3dydzf(ey)y3dzdxf(ey)z3SzzzSzxy2,z1xy和z4xy所圍立體的表面外側(cè),具22222f有連續(xù)導(dǎo)函數(shù).解:記所圍立體為,由Gauss公式得SI3(xyz)dV32π4ddrrπ4sindr22222001.935(22)π12.計算Ixzdydzyzdzdxzdxdy,其中是球面介于平面xyz24222和之間的部分,并取法向量與軸正向成銳角的那一側(cè).z3z2Oz解:記S:z3;,S在xOy平面的投影域D:x;所圍域記為,2y21Sxy則I(zz2z)dV(3)2dD上S下S下xy.4dd142zdz3ππ3π4π2π00315.判斷級數(shù)(e)nn!（0）的斂散性.2nnn1(e)2n1(n1)!解:因為liman1nn(e)2limna(n1)n1nn!nnn(11)nn,elim故當(dāng)1,即時,級數(shù)收斂,當(dāng)e時級數(shù)顯然收斂;當(dāng)1,即eeee時,級數(shù)發(fā)散.16.求冪級數(shù)1xn的收斂域及和函數(shù).n22nn!n0(n1)21解:因為lima,所以.2nn!limn0(,)n1aDn1(n1)!n21n2n1(x)n!2n2(x)n21nnS(x)xnn!22nn!n0n0n0x2n2x2netnet(n1)!tn1n!n1[n1]][xtnxtn1etett22(n1)!(n1)!n1n19,t[tet]e2(t2t)et(1xx)e2xx(,).xe2x22417.將函數(shù)展為的冪級數(shù).(x1)1f(x)x23x2解:f(x)11(x1)(x2)1x2x11x23x212(x1)3(x1)11121x131x11123(x1)n(x1)n,3n1213(1)n(1)n2nn0n0,x(1,3).)(x1)n11(1)n(2n13n1n018.將(0x2)展為以為周期的Fourier級數(shù),并求級數(shù)πx(x)2π2的和函數(shù).cosnxn2n1解:(1)(x)在上滿足Dirichlet條件,且πx2[0,2π),1π2πdx0x1π22π(x)dxaaπ002π0πxcosnxdx21π12π12πxcosnxdx2π2ππcosnxdxn002πsinnxdx]001[(n1,2,),xsinnxn2π1n02π002ππxsinnxdx2ππsinnxdx2πxsinnxdx1π12π12πb2n0001[]102πcosnxdx(n1,2,);xcosnx2π12πnnn00πx,(0,2π).2故sinnx(x)~nx00,n1(2)設(shè)f(x)cosnx,x[0,2π),其中f(0);π261n2n2n1n1π,x(0,2π),所以因為sinnxπxx22f(x)n2n1f(0)xf(x)dx2π23x26πx,x[0,2π).ππ22x2xf(x)641202,x1.試求在內(nèi)的(x)0,x1(,)19.設(shè)有微分方程,其中y2y(x)連續(xù)函數(shù)yy(x),使之在(,1)和(1,)內(nèi)都滿足所給方程,且滿足條件y(0)0.10解:(1)當(dāng)時x1,方程為,通解為,由得,所22yCeC111y(0)0yy2x1以.ye12x(2)當(dāng)時,方程為,通解為.要使在處20e()1yyyyxyC2xxx1連續(xù)2,必須2,即,于是得2,所以2limCe2xlim(e2x1)e21C1eCe22xy(1e2)e2x1.x1(3)補充定義21,則得在(,)內(nèi)的連續(xù)函數(shù)y(1)e滿足所給方程及初始條件e2x1,x1yy(x)(1e2)e2x,x1.20.一質(zhì)量為的汽艇以速度v行駛,在時刻關(guān)閉動力繼續(xù)行駛.假定0阻力與行駛速度v的n次方成正比(n為常數(shù)),求與關(guān)閉動vt0m水對汽艇的力后行駛距離之間的函數(shù)關(guān)系.解:設(shè)經(jīng)過t時間后,行駛的距離為x(t)(t0),阻力Rkvn(k0),由Newton第二定律得,由題設(shè)有x(0)0,x(0)v.x(t)k[x(t)]n0m0因為,所以dv,于是方程dxm化為0,分離變量得dvkvn1x(t)vdx.1ndvkmxt()vdxv(1)當(dāng)時,通解為,由得v2n,v2kxCmn2nv(0)x(0)vC02n2n0所以(2n)x,即m);2v2nknm(n2)kx(vnv2nv200(2)當(dāng)時,通解為,由得,故x,k(0)(0)vkmn2evCmv0vve0xvxC0即.vmxln0kv21.設(shè)f(x)具有二階連續(xù)導(dǎo)數(shù),且積分fx[ex2()f(x)]ydxf(x)dyL(為常數(shù))與路徑無關(guān),試求f(x).積分與路徑無關(guān),故有PQ,即解:因為f(x),整理得2f(x)f(x)exyx.,所以.()2()f(x)exr22rfxfx10rr11Ye(CCx)x212當(dāng)時,設(shè),代入原方程定出,即,(1)21(1)2ex1yyAexA此時通解為;exye(CCx)x(1)212當(dāng)時,設(shè)x,代入原方程定出1,即,此時2xy2ex1AyAx2e211通解為2).e(CCxxyx222.設(shè)高為厘米(為時間,單位為小時)的雪堆,在融化過程中其側(cè)面方t12h(t)程為zh(t)2(x2y2),其體積減少的速率與側(cè)面積成正比(比例系數(shù)h(t)為0.9),問高度為130厘米的雪堆全部融化需要多少小時?解:設(shè)雪堆的體積為V,側(cè)面積為S,則Vht(d)zdxdy,[h2(t)h(t)z]dzh3(t)π)π4ht(20021x2y[h(t2)h(t)z]2116(x2y2)h2(t)S1z2z2dxdydxdyxy1x2y212h2(t)x2y22h2(t)13πh2(t).122πh(t)h2(t)162dh(t)20由題設(shè)知dV0.9S,即3πh2(t)dh(t)913πh2(t),于是dt4dt1012dh(t)13,故h(t),由得C130,即h(0)13013tCdt1010h(t)13t130.10令h(t)0得t100(小時),即高度為130厘米的雪堆全部融化需要100小時.xyx23.求微分方程d(2x2y4)dy的通解.dx2解：方程可化為：dyyxyx1，這是Bernoulli方程.3dz4z2y3,令zx1(1)x2,則方程化為線性方程dyydyy4(C2lny),zeC2y3e4dyy4dyy故通解為y4(C2lny).x224.求冪級數(shù)x的和函數(shù).3n(3n)!n0解：易知收斂半徑為，設(shè)顯然x3ns(x),x(,),s(0)1.(3n)!n0s(x)(3n1)!,s(0)0;s(x)x3n1(3n2)!,s(0)0;xn32n1n112s(x)(3n3)!x3n3xn3s(x);(3n)!n1n0s(x)s(x)0,或()()()ex,sxsxsx故得s(0)1s,(0)s0,(0)0s(0).1,s(0)0.s(x)s(x)0,的求解過程如下：（1）s(0)1,s(0)0,s(0)0r310,r1,r13i,s(x)CexeCcos3xCsin3x.x2222212,31233CcosCsin3x,又33Cs(x)Cexe1Cx1x2222222123233Ccos3x3CCsin3x,s(x)Cexe1C1x222222212323C1,CC1,3112于是由初始條件得解得所以和函數(shù)CC3C0,C2,12232123C0.3C1C3C0,22123s(x)1ex2ecos3x,x(,).x2332s(x)s(x)s(x)ex,的求解過程如下：（2）s(0)1,s(0)0r2r10,r13i,S(x)eCcos3xCsin3x.x222221,212因為不是特征根，故設(shè)，代入原方程，得13i1sAex,A即s于是1ex,3(x)eCcos3xCsin3x1ex.x2s(x)S(x)s22312s(x)1ex+ecos3xCsin3x.由s(0)1得，即2323xC2322123Csin3，s(x)1ex+e3Ccos3xxx112232323222由s(0)0得，所以和函數(shù)C02s(x)1ex2ecos3x,x(,).x233213四.證明題2.若limna0,且收斂,試證收斂.[(n1)ana]annn1nnn1n1證:記,,,則收斂.nnsa[(k1)aka]n1,2,{}nknkk1nk1k1(2aa)(3a2a)(4a3a)[(n1)ana]1223342snan1n,n2a2a2a2ana123nn1nn1.因為收斂,所以要證收斂,只需證收斂{na}.s11na{}{s}22nnn1nnn1事實上,由limna0知lim(n1)a0,于是nn1nn.nlimnalim[n1(n1)a]0n1n1nn,即收斂且.(].)n1因此{(lán)s}收斂12aa[(n1)anannnnn1n1n13.若冪級數(shù)axn的收斂半徑,則冪級數(shù)axn的收斂半徑R.R10nn!nn0n0證:因為,故對于x(0,1),級數(shù)絕對收斂,從而{axn}有界,設(shè)R1axn000n0nn0axnM.于是x(,),有a.而由比值法xnaxn0n!xnMxnxnnnn!n0n!xn00知,x(,),,收斂從而絕對收斂,故其收斂半aMxnxnnn!n!xnn0n00徑R.4.設(shè)(nN),試證:a2,a11(a1)2an1nn)收斂.(1)lima存在;(2)(a1nannn1n1a21n2a證:(1)(nN),即{a}有下界;又1a2,a1(a1)2a1n1nnnna1a20,即{a}單調(diào)遞減,故lima存在.an2an1nnnnn(2)因為{a}單調(diào)遞減,所以,即(1)是正項級數(shù),又因naa10naann1n1n1aaan1aa為a1,所以(nN);而lima存在,故1nnaan1nnnnn1n1收斂,于是由比較判別法知,(an1)收斂.(aa)n1ann1n1n11411115.證明:(1)(1);(2)n1.1收斂n23lnnlimlnnnnn1證:(1)因為0lnn11ln(11)1111111oo,nnnnnn2n2n22n2n2(),收斂所以.而1收斂n11lnnnn2n1n1(2)(11)lim(ln(k1)lnk)1lnnnnnkn1nk1lim[1ln21ln3ln211ln(n1)lnn]23nn,lim[1111ln(n1)]n23n()1收斂,即因為存在,故1lim[1111ln(n1)]lnnnn23nnn11