《高等數(shù)學》第07章多元函數(shù)微分法及其應用習題詳解

1、第七章 多元函數(shù)微分法及其應用習題詳解第七章多元函數(shù)微分法及其應用習題7-11.判定下列平面點集中哪些是開集、閉集、區(qū)域、有界集、無界集？并指出集合的邊界(1) (x,y)x0,y0；(2) (x,y)1x2y24；，一、2(3)(x,y)yx；(4)(x,y)x2(y1)21且x2(y1)24.解(1)集合是開集，無界集；邊界為(x,y)|x0或y0.(2)集合既非開集，又非閉集，是有界集；邊界為(x,y)|x2y21J(x,y)|x2y24.(3)集合是開集，區(qū)域，無界集；邊界為(x,y)yx2.(4)集合是閉集，有界集；邊界為(x,y)x2(y1)21J(x,y)|x2(y1)242 .

2、已知函數(shù)f(u,v)uv,試求f(xy,xy).解f(xy,xy)xy(xy).3 .設f(x,y)Jx4y42xy,證明：f(tx,ty)t2f(x,y).解 f (tx, ty).tx4ty42t2xyt2_x4y42t2xyt2x4y42xyt2f(x,y).27(x0),求f(x).212y解由于fx -1 x2-,則f15.求下列各函數(shù)的定義域:(1)22xy.22'xyy(2)zln(yx)arcsin-；x(3)ln(xy)；(4)Z21x2(5)z(6)ux2y2(1)定義域為（x,y）y(2)定義域為(x,y)(3)定義域為(x,y)xy即第一、三象限（不含坐標軸）(

3、4)定義域為(x,y)2x2a2yb2(5)定義域為(x,y)0,y0,x2y.，(6)定義域為(x,y,z)0,x26.求下列各極限:(1)lim(x,y)(2,0)(2)(3)lim(x2(x,y)(0,0)(5)(x,y)m(0,1)(1解：（1）lim(x,y)(2,0)(2)(加0,0)(3)因為(x,y)(4)y2)sin1xy)x；2xxyxy(4)(6)f(2,0)1cosx2y2ln(x2y21)lum0叫,小y2)0(x,y)m(2,0)sin(xy)(x,yl)m(2,0)lim(x,y)(0,0)lim(x,y)(2,0)(x,y)limln(1u)1cosln(x2s

4、in(xy)(1ulim-2u0u且sinxysin(xy)xxy1有界，故212;2x-2y1)2xyy)e2lim(x(x,y)(0,0)'y2)sin0；xy(5)lim(1(x,y)(0,1)xy)xlim(x,y)1(0,1)(11Xyy1xy)ee；(6)當xN0,yN0時，有0/22(xy)exy(xy)2exy而(x.yjmexylimulimu2ulim按夾逼定理得lim（x,y）（/22.xy)(xy)e0.(x,y) (0,0) x(2)設 f (x,y)0,lim(x,y) (0,0)f (x, y).7.證明下列極限不存在:(1)lim0,證明(x)x kx

5、limk0x kxkx0,（1）當（x,y）沿直線ykx趨于（0,0）時極限與k有關，上述極限不存在（2）當（x, y）沿直線x和曲線2 .x趨于(0,0)有(x,y)m(0,0)2x y42x ylimx 0 y x2x xI 2xlimx 0y x(x,y)m(0,0)limx 02 y xlim4 x 2x4故函數(shù)f（x, y）在點（0,0）處二重極PM不存在.8.指出下列函數(shù)在何處間斷:(1) z ln(x2y2)；1y2 2x解（1）函數(shù)在（0,0）處無定義，故該點為函數(shù)z ln(x2 y2)的間斷點;（2）函數(shù)在拋物線y2 2x上無定義，故y2 12x上的點均為函數(shù) z -的間斷y

6、 2x點.9.用二重極限定義證明:證Jx2 y2 |OP|,于2 x2yxy lim0.(x,y) (0,0)、,2、,2xy日2 i 1-2P(x, y),其中0,0;當0時,xy_22x y成立，由二重極限定義知lim(x,y) (0,0)xy22x y0.10.設 f (x, y)2sinx ,證明f(x, y)是R上的連續(xù)函數(shù).2證設 Po(xo,yo) R .0 ,由于sinx在X0處連續(xù)，故時,|sinxsinx01以上述作P0的鄰域U(P0,)，則當P(x,y)U(P0,)時，顯然|xXo|(P,Po)從而R2|f(x,y)f(x°,y°)|sinxsinx0

7、|即f(x,y)sinx在點P0(x0,y0)連續(xù).由P0的任意性知，sinx作為x、y的二元函數(shù)在上連續(xù).習題721.設zf(x,y)在(xo,y°)處的偏導數(shù)分別為fx(x0,y°)A,fy(x0,y°)B,問下列極限是什么?(1)f(%h,yO)f(%,yo)(3)limh0hf(xo,yo2h)f(xo,yo)f(xO,yo)f(x°,y°h)hf(x°h,y°)f(xo,y。)(4)limh0hf(xoh,y°)f(xoh,y°)(2)4(x°,y°)A；f(x0,y0)f(

8、%,y0h)f(x0,y0h)f(x,y。)Zy(x0,y。)B；(3)limh0f(xo,y02h)f(xo,y°)lim2h0f(Xo,yo2h)f(Xo,y°)2h2B;（4）iim）f(%h,y°)f(Xoh,y°)limh 0f(xoh,yo)f(xo,yo)f(xo,y°)f(xoh,y°)hf(%h,y°)f(x0,y°)f(x°h,y°)“飛)。)hf(%h,y°)f(x0,y°)limf(%h,y°)f(x0,y°)AA2A2.求下列函數(shù)

9、的一階偏導數(shù):(1)xy(2)xz ln tan ； y(3)exy;(4)2y； xy(5)22x ln( x2、y )；(6)z ,Jn(xy)；(9)sec(xy)；(8)z (1xy)y ；arctan(xy)z(10)解（1）(2)tan2xsec一y1,x2x一cotsec一(3)(4)tanexy2x2xsec一yxyyye，xyxexx-2cot-I"；2x.sec一；y/2xy(x2xy2、y)y2x*2y(x222xy2、y)y222,22、z2yxy(xy)x2xy(xy)x222yxyxy(5) 2xln(x2xy2)2x .2-22x 2xln(xx yy2

10、)2x32 2x y-22xy.22，xy(6)12 Jn(xy)1-y xy1z二/=,一2x ln(xy)y111x :2 .ln(xy) xy 2y ln(xy)tan(xy)sec(xy)yytan(xy)sec(xy)，xztan(xy)sec(xy)xxtan(xy)sec(xy),y(8)_zy(1xy)y1yy2(1xy)y1,xeyln(1xy)y(1xy)ln(1xy)xy-yy1xy-rrz(xy)z1;z,x1(xy)1(xy)11 (x y)2zz(xy)z1 ( 1)z(x y)z11 (x y)2z11 (x y)2z(xy)z ln(x y)(x y)zln(x

11、 y)1 (x y)2z(10)z 1x1zyyz 1ux一 z yyln3.設 f (x, y)1nx "fy(1,0)由于f(1,y)ln1,所以fy(1,y)解法二fx(x,y)1y2xv行，fy(x,y)2x1y2x12xfx(1,0)1-02,11，fy(1,0)0124.設f(x,y)(y1)arcsinJ,求yfx(x,1).解法一由于f(x,1)x(11)arcsinx,fx(x,1)(x)1.5.設f(x,y)1，fx(x,1)ydt，求fx(x,y)解法二fx(x,y)fy(x,y).解fx(x,y)ex2*4,f(x,y)_y6.設zxyxex證明zxxzy一y

12、xyz.解由于yexyxexyexyexyxex所以zy一yyexyexyyxyex(xy)xyyexxyyxexxyxy7. (1) z(2)z41xy,在點(1,1,J3)處的切線與y軸正向所成的傾角是多少？x1解(1)按偏導數(shù)的幾何意義，zx(2,4)就是曲線在點(2,4,5)處的切線對于x軸正向所成傾角的斜率，而1.，Zx(2,4),Xx21,即ktan(2)按偏導數(shù)的幾何意義,Zy(1,1)就是曲線在點(1,1,J3)處的切線對于y軸正向所成tan傾角的斜率，而zy(1,1)一=2y2J18.求下列函數(shù)的二階偏函數(shù)：(1)已知z3.3.xsinyysinx,(2)已知z(3)已知zl

13、n(xx2),lnxy2十z,求；xy2z；xy(4)+y+2zzarctan求一2xx解(1)3x2sinyx3ycosx,3x2cosy3y2cosxylnxlny1xlnylnxyxlnxlnylnxInxyInx1(1Inxlny);(3)22xxy22、xy2x2x2,x2y2(x,x21x23，2y2)y2，2z2x2x2.x22z2y2.x2(4)x12yx121工xc2一22z2滿足12xyy222x y 2y22 2x y22yX22 2，x y2222z x y 2xyxx2 y2 222yx22 2 .x y9.設 f (x, y, z)xy2yz2 zx2 ,求 fxx

14、(0,0,1),fxz(1,0,2) , fyz(0, 1,0)及fzzx(2,0,1).2斛因為fxy2xz,fxx2z,fxz2x,2fy2xyz,fyz2z,2fz2yzx,fzz2y,fzzx0,所以fxx(0,0,1)2,fxz(1,0,2)2,fyz(0,1,0)0,fzzx(2,0,1)0.10.驗證:(1)kn2ty(2)sinnx滿足t(1)kn2kn2tesinnxnekn2tcosnx,2y2xkn2tsinnxkn2t.sinnx(2)因為x2r由函數(shù)關于自變量的對稱性,2r2z2z3r2r所以x2r2y2J2z2y3r2z3r習題2J“、st11uu-22-2；St(

15、2)(x22xyy)e(4) Z,x(3)zarcsin(y0)；yyz(6) u x ./L,/222、(5)uln(xyz)；(1)2s(s22_2t)2s(st2)t224st2TV2t(s2 t2)4s2t2t(s2t2)2st2(2)(3)(4)dux22xe4st2:y2xyt2ds4s2t2dtt24stt22(tdssdt)(x22)e22xyxy22xy(x222xyy2)y22xyxy2x44xy2，xy由函數(shù)關于自變量的對稱性可得dzdzdz22xyxy2x44xy2xydx2y22xyexy2y2xyxydyxarcsin一y12x2y1一dxyxFdyyyy2ydxx

16、dydxdy(5)dudln(z2)2dz2xdx2ydy2zdz2,I,、22-2-222(xdxydyzdz);xyzxyz(6)dudxyzyzxyz1dxxyzzlnxdyxyzylnxdzxyz1yzdxxzlnxdyxylnxdz.2.求下列函數(shù)的全微分:(1)zln(1x2y2)在x2處的全微分;(2)z,xarctan21y1處的全微分.解（1）因為dzdln(1y2)11x2d(1yy2)(2xdx2ydy)y所以dz11(2dx264dy)3dx(2)因為dzdarctan-12d所以dzxy3.求函數(shù)z解因為dz所以當4.求函數(shù)z并求兩者之差.解因為dz1y2.221yx

17、122dxxdx2dzxy2xdx2xy2Tdyy,2xydx-2dy1y2xy；2dy1y2,y2xy3dx0.02,12dxdy.3x2y2dy0.02,y2xy3x3x20.01時全微分為0.080.120.2.0.01時的全微分.2,x0.01,y0.03時的全微分和全增量,xy22xy22、d(xy)xyd(xy)22xy(ydx+xdy)xy(2xdx2ydy)233一2.xyydx+x+xydy所以當x2,y1dz3y2x而當x2,全增量與全微分之差為1.設U0.01y0.03時全微分的值為32x+x+xyy(x,y)(2,1)x0.01,y0.030.01,0.250.0277

18、77,0.03時的全增量為zdzsint,0.028252xy-2x0.028252,V(x,y)(2,1)0.010.030.0277770.000475.解duudxxdtudyydt2.設zarccos(u解dzzduudx3.設zzdvvdx4x222x(4x2uv3)24.設z習題74,3tdut，求證.2y.cost(uxcosy,v一2uvvx2ex2y3t23x,求.dxv)2xsin2.，、3xsinycosy(cosysiny)2uv12x2cosy33x(sin2sin2ycosy3cosu3xsint2t3exsinyy2cos2,_2、(cost6t).(Iv)232

19、uvsiny2u2uvxcosyysiny).5.6(3x2y)ln6.設uf(u,x,y)ln(u222ulnv3v-(3xx2y)2ulnv2ysinx),12uysinxy14(3x2y)ln-(3xxy-ycosxuysinx2y)2.7.設z2x2e-2Te2e2xy./2sin(x(rs(rs(rs,xarctan,yycosxysinxsinxysinx),stststtr)(str)(r12uysinx1-sinxysinxt)t)tr)(rs)uv,y2xrsstutr,zrst,求2y(st)c/22zstcos(xz2)2.2rstcos2x2y(r22rstcos2x2

20、y(s22rstcos(rt)(rr)(rt)2(rs22zrtcos(xst)2(rs22zrscos(xst)2(rsstststtr)2z2)tr)2z2)tr)2(rst)2,(rst)2,(rst)28.設zf(x,y,t)(u2(uv)2v)(uv)2xsint,cost,解dzzdxxdtdyydt2xcost2y(sint)9.求下列函數(shù)的一階偏導數(shù)（其中f具有一階連續(xù)偏導數(shù)）(1)z-22f(xy)；(2)u(3)uf(x,xy,xyz)(4)解（1）z2xfx(x2(2)2yf(x2、y)；(3)(4)10.設zf1f2f2f1yzf3xf2xzf32xf1xyyef22y

21、f1xyxF(u),而x2ydzdt2sinf(x2v2.v2t1.xyx,e,lnx).f1xyf3；*yrxef2.F（u）為可導函數(shù)，證明:yzxy.yxF (u) yxyxyF(u)xF(u)xyxxyF(u)-F(u)xyxF(u)xyxF(u)xyzxy.11.設zycos(xy),試證：.xyy、丁z證一xysin(xy)cos(xy)sin(xy)12.設ucos(xy)-y且函數(shù)F-,具有一階連續(xù)偏導數(shù),xx試證:u y-yku .uxx、丁u證xkxk1FFiz2xF2k-xF21F1k-xF1uxxuzzkkkxFx1zF11yF11yF11zF1ku.13.設zsiny

22、f(sinsiny),試證:zsecxxzdsecy-1.yfcosx，ycosy(cosy)f,zzsecxsecyxysecxcosxfsecycosysecy(cosy)f1.14.求下列函數(shù)的二階偏導數(shù)222-z,-2（其中f具有二階連續(xù)偏導數(shù)）xxyy(1)zf(xy,y)；,22zf(xy)；第七章多元函數(shù)微分法及其應用習題詳解i6(3)zf(x2y,xy2)；(4)zf(sinx,cosy,exy).解(1)令sxy,ty,則zf(xy,y),s和t是中間變量fizsdtyf1，f1f2xf1yydyf2.因為f(s,t)是s和t的函數(shù)，所以f1和f2也是s和t的函數(shù),從而f1和

23、f2是以s和t為中間變量的x和y的函數(shù).故2 z2 x一yfixyfii一 yfi yfifiifi2dtdyfixyfiiyfi22z2 yxfiyf2xfiifi2dtdyf2if2*x2fii 2xfi2f22(2)令 s x2 y2,則 z一一22.f(xy)是以s為中間變量的x和y的函數(shù).z's-'zsf2xf,f-y fi 2xyf2 xyfxxyy因為f(s)是s的函數(shù)，所以f也是s的函數(shù)，從而f是以s中間變量的x和y的函數(shù).故22f 2xf 2x 2f 4x2 fzz2xfxxxx2zz2xf2xf2y4xyf,xyyxy2-zz2yf2f2yf2y2f4y2f

24、yyyy(3)令sxy2tx2y,貝Ust2zfi-f2yfi2xyf2,xxyst2fif2-2xyfixf2.yy2zz""2"xxx第七章多元函數(shù)微分法及其應用習題詳解I7(4)2z2x、,2fsftyfiifi2xx2yf22yfi2yfi2yfi2xf12xf12xf1dufiTdx2yf22xyf21f222c，2r22rfii2xyfi22yf22xyy£?143,yfii4xyfi2,224xyf22,y2fi2xyf22xyf22fii-y2xy3f123-2xf22xyfii一2xyf1ys2xyfii一y2xy2xyfiix22-4

25、xyfiisinxf1sinxf1exy2xf2yfi22xf225xy2fI2f22xyf2i2xy2xyf2132xyf22,f122f124x3yfi2cosy,cosxf1cosxficosxcosxsinxfif212xyf2iXf22.exyf3，exyx2f22dvdysf22乂旺22f3xysinyf2ef3.dufiiVdxfi3xdu3i展cosxf11cos2xfiiyfl3exyexycosxf31exyf33cosxfiex2exycosxfi3f33，第七章 多元函數(shù)微分法及其應用習題詳解y f dvf32 dy_dv_wcosxf12f13dyy77cosxXyx

26、ysinyf12ef13ef3xyxyesinyf32ef33exyf3cosxsinyf12eycosxf13exysinyf322z2ysinyf2exycosyf2sinyf22dyf23exyexycosyf2sinysinyf22xyef23exyf3cosyf2sin2yf222exysinyf23f32dyf33sinyf32xyref33習題751.設xcos y ex2y 0,求曳 dx解設F(x, y)x 2cosy e x ydydxFxFex 2xysin y2.設 xylnIndydx解設F(x, y)xyln y當x 1時，由xyex 2xysin ydydxFyl

27、n y ln x 1 知 y1,所以dy dx2xy3.設lnjxy2arctan"，求dy.xdx解設F(x,y)ln7x2y2arctan,則xdydxFxFyx4.設cos21xx2cos2cos解設F(x,y,z)2cosx2cos2ycos5.設方程F(xx2yy2yy2yx2y6.設由方程y(x,z),z證因為所以y2cosxsin2coszsiny乙xyFxFzFiFiyzF(x,y,z)0z(x,y)證明:FyFxsin2xsin2zFyFz2cosysiny2coszsinzsin2ysin2zzx)0確定了函數(shù)z(yz)F2(yx)F2FyFzz(x,y),其中F

28、存在偏導函數(shù),F1(xz)F2F1(yx)F2分別可確定具有連續(xù)偏導數(shù)的函數(shù)xx(y,z),FzFy7.設(U,V)具有連續(xù)偏導數(shù)f(x,y)滿足abc.xyFzFyFxFz證明由方程(cxaz,cybz)0所確定的函數(shù)證令ucxaz,vcybz,8.設cv，xyzc.解設F(x,y,z)Fx9.設2z-2.xxyz,則Fxyz,Fzxy.yzyzz盤、y(exy)xyzzexy2zexyyzt2yexy22z32yze2xyzxyz(x,y)是由方程解設F(x,y,z)zexzxz0所確定的隱函數(shù),(0,1)Fxx,Fy2y.日_zZExFxFzFFz2yezzze2yzex2yezx2yz

29、ezzex由ezxzy20,知z(0,1)0,得一z-l2.xy(0,1)10.求由方程xyz西z2J2所確定的函數(shù)zz(x,y)在點(1,0,1)處的全微分dz.解設F(x,y,z)xyzJx2yz./FxxFzxy小xzzFy2:y2z2J2,則x222.'222xyzyz.xyzxzxy；x2y2z2z222xyzy222222xyzxzxyzyyFzxy-=zxy.222xyz222zzyzJxyzxdzdxdy.xyxy,x2y2z2:dz(1,0,1)dx2dy.11.求由卜列方程組所確定的函數(shù)的導數(shù)或偏導數(shù)：，222xyzz222(xzxyzy-dx-Y=-dy,'

30、;222乙xyxyzz22(1)設zxy,才222x2y3z20,xuyv0,u(2)設求，-yuxv1,xu、廳xeusinv,u(3)設口求一uyeuucosv,x解(1)分別在兩個方程兩端對dzdx2x稱項，得%dxdxuvv一，一，yxy，,?yxyx求導，得2x2y%dxdydz4y6z0.dxdxdydz2y2x，dxdxcdycdz2y3zxdxdx在解方程組得2y 12y 3z6yz 2y0的條件下,2x1dyx3z6xzxx(6z1)dxD6yz2y2y(3z1)2y2xdz2yx2xyxdxD6yz2y3z1(2)此方程組確定兩個二元隱函數(shù)uu(x,y),vv(x,y),將

31、所給方程的兩邊對x求導并移項，得uvxyu,xxuvyxv.xxxy22c_、在Jxy0的條件下，yxuyuvxxuyv22，xxyxyyxxuvyvyuxv1-22,xxyxyyx將所給方程的兩邊對y求導，用同樣方法在Jx2y20的條件下可得uxvyuvxuyv22,22yxyyxy(3)此方程組確定兩個二元隱函數(shù)uu(x,y),vv(x,y)是已知函數(shù)的反函數(shù)，令u1(F,G)11ucosvsinvxJ(x,v)J0usinvue(sinvcosv)1u1(F,G)10ucosvcosvyJ(y,v)J1usinveu(sinvcosv)1v1(F,G)1u.一esinv1ucosvexJ

32、(u,x)Juecosv0ueu(sinvcosv)1v1(F,G)1u.一esinv0.一usinvexJ(u,x)Juecosv1ueu(sinvcosv)1習題76組得FxGxF(x,y,u,v) x1, Fy0, Gy(F,G)(u,v)u sin v, G(x,y,u,v) yue u cosv.0,u eu eFuGusin vcosvsin v , Fvu cosv,cosvucosvusin v,Gv u sin v.ueu (sin v cosv)u 0的條件下，解方程1.求下列曲線在指定點處的切線方程和法平面方程:(1)(2)一3.t在(1,0,1)處;2.zt在t1的對應

33、點處;(3)sint ,cost , z 4sin -在點21,1,2 衣處;2(4)2 x2 y2 y2 z10100, 在點(1,1,3)處.0,(1)因為xt2t, yt1 ， Zt(1,0,1)所對應的參數(shù)t 1,所以(2,1,3).于是，切線方程為法平面方程為2(x1)y3(z1)0,2xy3z50.(2)因為xt12)(1t)ytt(it)t2zt2t,t1對應著點11,2,1,所以24,1,2.于是，切線方程為1x21法平面方程為2x8y16z10.(3)因為x所以2切線方程為法平面方程為(4)將2x2y由此得2y2zcost10100,0,ytsint,zt2cos-2占八、1

34、,1,2.2對應在的參數(shù)1,1,2.的兩邊對x求導并移項，得2x,dydxx一,ydz0,dzdx0.2y2y2x2y02y2z4xy4yzdydx(1,1,3)dzdx(1,1,3)從而故所求切線方程為法平面方程為1T1,1-.3x1y1z333-V3x3yz30.2.在曲線xt,yt22斛(1) F (x, y, z) 3x y z 27 ,ztn (Fx, Fy,Fz) (6x,2y, 2z),n(3,i,i)(18,2, 2).所以在點(3,1,1)處的切平面方程為9(x 3) (y 1) (z 1) 0, 即9x y z 27 0.法線方程為(2) F(x,y, z) ln(1 x2

35、 2y2) z,上求一點，使此點的切線平行于平面x2yz4.解因為xt1,yt2t,zt3t2,設所求點對應的參數(shù)為t0,于是曲線在該點處的切向量可取為T(1,2to,3t2).已知平面的法向量為n(1,2,1),由切線與平面平行，得2-，1Tn0,即14t03to0,解得to1和.于是所求點為(1,1,1)或3111.3'9'273.求下列曲面在指定點處的切平面和法線方程：(1) 3x2y2z227在點(3,1,1)處；22.(2) zln(1x2y)在點(1,1,ln4)處；(3) zarctany在點1,1-處.x4n(Fx,Fy,Fz)2x22c21x2y-4y-,1,

36、1x22y2所以在點(1,1,ln4)處的切平面方程為法線方程為y(3)F(x,y,z)arctan上x所以在點1,1-4法線方程為4.求曲面11(1,1,ln4)2,i,2y2z3z,n(Fx，F(xiàn)y,Fz)處的切平面方程為2y23z21,1l44ln20.F(x,y,z)n(Fx,Fy,Fz)行，得代入曲面方程得解得所以切點為所求切平面方程為2ln22zyx2,2yx0.-,1，y21上平行于平面2y23z221,x4y6z0的切平面方程.則曲面在點(x,y,z)處的一個法向量(2x,4y,6z).已知平面的法向量為(1,4,6),由已知平面與所求切平面平2x11-z,2yz.2,2z23z

37、221.1,2,2.4y6z21xy5 .證明：曲面F(xaz,ybz)0上任息點處的切平面與直線z平行(a,bab為常數(shù)，函數(shù)F(u,v)可微).證曲面F(xaz,ybz)0的法向量為n(RE,aFibF?),而直線的方向向量s(a,b,1),由ns0知ns,即曲面F0上任意點的切平面與已知直線Jxyz平行.ab6 .求旋轉橢球面3x2y2z216上點(1,2,3)處的切平面與xOy面的夾角的余弦.解令F(x,y,z)3x2y2z216,曲面的法向量為n(Fx,Fy,Fz)(6x,2y,2z),曲面在點(1,2,3)處的法向量為n1n,一q、(6,4,6),xOy面的法向量(1,2,3)n2

38、(0,0,1),記n1與山的夾角為，則所求的余弦值為n1n263COSr;jccc.n11ln2|4621V22按題意，方向l2.求函數(shù)z解依題意,3.求函數(shù)z向的方向導數(shù).(1,2)(1,73),ez2y,y2,(1,2)ln(x2ei(1,2)2244,（1,1）解先求切線斜率:法線斜率為內法線方向1,32,-2123.2y2）在點（1,1）處沿與x軸正向夾角為60“的方向的方向導數(shù).12,22x2y(1,2)112yb2dydx(b,（1,1）在點2yb22y2y處沿曲線1兩端分別對x求導,2x2ab2x2ay2ydyb2dxdydxab=,-=22ab'a),ei,a2b2，.a22-y21在這點的內法線方ba_一b2ab2/24.求函數(shù)解因為所以5.求函數(shù).2ab2,2ab了中逆b1,2(a2b2).abxyz2在點(1,0,1)處沿該點到(3,1,3)方向的方向導數(shù).2x2,(1,0,1)2,(1,0,1)u2z,z1,(1,0,1)2.1,2(1,0,1)ei2z在曲線yt2,.3zt上點(1,1,1)處，沿曲線在該點的切線正方向(對應于t增大的方向)的方向導數(shù)解先求曲線在