微分幾何答案(第二章)

§1曲面的概念rucosv解rcosvsinv,sinvcosvsinv,{000000ruv,usinv00r證rvvvav,bvv000000av,bvv000ruuuu,buu000000u,buu000r=asin,asin,asin}解r=asincos,asinsin,acos}racossin,acoscos，=xacoscosasincosyacossinzasinasinsinacoscosacos0acossin0即++-a=0；xayaza。x2y21a2b2。13x2y21的參數(shù)方程為x=,y=,z=t,解橢圓柱面a2b2rasin,bcos,rtxacosybsinzt0，即x+y－ab=0asinbcos0001t。a3u,v,}r證ra3a3xy}r}z3，。3uvauu2vv2a21a93V3|u|3|v|a36||2rra,b,2},ra,b,2u},Erab4v,解2222uvuFrra2b24uv,Gr2a2b24u2,uvv(ab4v)du(ab4uv)dudv(ab4u)dv2?！郔=2222222222rucosvsinv,Frr0，解r{cosv,sinvrusinv,ucosv,}，Er12，uvuuvGr2u2b2I=du2(u2b2)dv2v142u=vI=222得解由條件2222,沿曲線u=v有，將其代入2222=22=,u=vv到v21|coshvdv|sinhvsinhv|為v。2211(ua)dv2u+v=0I=du2=0221F0Gua2u解E，，2v1F0Ga2。+v=0與u–v=0的交點(diǎn)為u=v=E，，v曲線u+v=0=,u–v=0,，1a1aEduuGdvu22=。Eu2Edu2Gdv22Gvz=x=x=y00解rx=xrxy}，000ry=yr,yyry0000xrra2xy0y=x與y=y的夾角為=xy0|r||r|00011a2x2a2y20xy0求解=++Gd=得+=0.+=0.P2++R215+dudu+2Q+dv()2dvPduuRduuQduu,,則dvv=，+=dvvPdvvPduuduuE++G=0dvvdvv-+=0.E22.證uv,(vu)(vv)(Fdv)(Gdv)2222。EuGv222EG2F2)2F2)2F2F2E22.9．設(shè)曲面的第一基本形式為I=udu2(u2a2)dv2u=av,vvo解011au2adudvuadudv222aua0ua1uauaadudv)uadu2222a0ua01623[ua)uuaauua|=22222222aa022[2)]。=a23r=asin,asin,asin}解r=，r=asin,asinsin,a}asin,a2222E=r=arr=0,G=r2=a2dadacosdacossin|aS=42222.2222022rr,t21}1.0<+v,u2=+v,u212u22,t2)td1+v,t=u,則222t21u21u21)udv)22211u2u2u211=(u222(u2u22=I.222u21u217+v,t=u21.r解r={sinhucosv,sinhusinv,1},ruvr={coshucosv,coshusinv,0},rr={-coshucosv,-coshusinv,0},Er2=2FrrGr22uuvv所以I=2u2+2udv2.n=rr1ucosv,usinv,sinhusin},=uvcosh2uEGF2coshucoshu1,.sinh1sinh122=-2+dv2。2x5x4xx2x21223125{x,x,x2xxx}解r212,212122rx2x},rx2x},r,xxx12(0,0)x12(0,0)1211r,rxx,E=F=0,G=1=5,M=2,N,xx2122dx5dx4dxdx2dxdx22,2212.2112vsinvr解,sin,0},{sin,,}，rrvvruvuvbuvrrEr21，F(xiàn)rr0，uuv18bGr2u2b2，M=,N=0-+0.vu2b21(axby)z222解r0,}rby}r}r,,,x(0,0)y(0,0)bdyadxdx222r}k.dyn2已知平面解1d21k1d21d2k1dk2=1.nIIIknk222211LMN1()EFGRIRRn22vsinvrrrv,sinv,0},r{usinv,uv,}，ruuuv(r,r,r)(r,r,r)uvuuvuuuvvvEGFEGF2219xyz2解r{x,y,xy}r0,y},rxy},r2,2,xyr},rEr214y4,Frr2xy2,Gr214x2y2.xxyy2y14xyy2xLM,N.14xyy2242242MdxdyNdy4ydxdy2xdy0,一族為即漸近線的微分方程為Ldx22,即2xyc,或xyc,c為常數(shù)..yc,c即ln22112證:rr(s)S:(s,t)r(s)t(s)，()()()(1(1ttststtt，，stst=0,nnststF2證rr0,f'},rg}.rf''},rrg'''xyxxxyyyrrrxy0MxyEGF2vsinvrrrr解,sin,0},{sin,,}，rvvuvuvbuuuv20brEr21，F(xiàn)rr0Gr2u2b2，,uuvvu2b2dv12dudvdu210bub0dvdu22u2b200u2b2vuub)cvubu)c.222212a1ay,Faxy,G1ax,LM,.解E22222221axay2222dy2dxdyaxydx21ax1axa2y2dx2))a2x2dy2,積分得由2222222得a001a2xay222ln(ax1ax)ln(ay1ay)c2222.abuv{(uv),(uv),}r222a2b2v2a2b2a2b2u2E,F,G,L解4442F2()(),積分得:a2b2u2dv2a2b2v2du2uabu)vabv)c.222222LL21LL有dnL有?ndrnn0,////0,即-·則有或nn而ndndr與正交n得,所以·n.若則L·nL，nndn=，L=0nnndr‖n‖dn‖n）而L是曲率線，所以沿L有d‖dnnn0,L若n，?n·n0n0n以沿L有dn‖dr·nnn‖，dn‖d0證vsinv解設(shè)正螺面的向量表示為rr解,sin,0},{sin,,}，rvvruvuvbuuuvEr21，F(xiàn)rr0，r，ruuvbGr2u2b2，M=,N=vu2b222a22）-（LG-2FM+EN）+2=0得=。F2M2(ua)NNN222aa,。1u2a22ua22yx22{0,0,2}r{x,y,a(xy)}r2}r{0,1,2ay}，解r，22，yyxyr{0,0,2}，r{0,0,0},r{0,0,2},xxxyyy2=2a,=2a.2aN12N證+，212)cossin22，n12)cos)sin)sincos2222222n1212))2n12ncossin2證由22得tg1n122arctgarctg1,12.又-+121221112證3232證0,2(F)2MF22K=.a2由H=12=0有==0或20.2112)cossin,若=,222112nk22220In2212若122F=M=0H=0，M+=0,因E>0>0<02L=M=N=0M2<2tg12即=212H0LG2FMNE02022u2MLdududvduuNdu,()2MN02所以L，所以，所以dvvLdvvdv24uvduududv(EduuFduvdvuGdvvdvvE)[F()G]dvvN2M[EF()G]0=dvvLL10H()0，所以，所以高斯曲率K0證法三：M21212M02L=N=0，若M=0M0H0LG2FMNE0MF=0F=0(xb)za(ba)在y=222,r,,L=M=N=2解E=,0,a(bacos)2-M2),b>a>0>-M2的3<和<-M222<<,=222或-M22(u,v)(dudv)I222ft{gt),gt)sin,(r證rft{gt),gt)sin,('2f'2g'2f'2Igt)(gdtd2)udt，v=22g2gIg2[t(u)](du2dv2),S、SSS121225S、S12證S、Sn、nS、S121212n與nn·nn·n121212dn·n+nnSdndr12