微積分下模擬試卷

#/13以求絲二名二—c工y SySx3y解：由x/z=ln(z/y)得x=z(lnz-lny)；即有F(x,y,z)=z(lnz-kiy)-x=O故5z/Sx=-(aF/5X)/(aF/5z>l/[(lnz-lny)+z(l/z)]=l/[ln(z/y>4-l]5z/5y=-(5F/6y)/(5F/5z)=(z/y)/[ln(z/y)-F1]32z/5x5y=[(z/y2)ln(z/y)]/[ln(z/y)+l]2=[z(lnz-lny)]/[y(hiz-lny+l)]224,求函數(shù)f(x、y)=XJ-yj-3j;-3y1-9x的極值。arax=3xA2+6x-9=0 x2+2x-3=(x-1)(x+3)=0 x1=1,x2=-3卸Hy=-3y2+6尸0 y2-2y=y(y-2) y1=0y2=2A=717axL6x+6c=a2w=-6y+6C_ A,一C得到f(x,y)的四個穩(wěn)定點:P1(1,0)F2(1,2),P3(30),P4(32)⑴對P1點：A=12>0,B=0,C=6>0a=AC-B2=72>0f(1,0)=-5為極小值；(2)對P2,P3點不是極值點；⑶對P4^:A=-12<0,B=0,C=-6<0,A=AC-B2=72>0,f(-3J2)=31J^極大值.因此f(x,y)=x2y3+3x2+3y2-9x的極值點為：(1,0)對應(yīng)極小值:f(x5y)=f(11O)=5(32)對應(yīng)極大值:f(x,y)=f(32)=31

21、將一⑴..1展開成K-1的累級數(shù).Z--4z-3f(x)=l/(xA2+4x—3)=1(x—l)(x—3)=V?x-l>+-2X(x-l>+4)換元仁x-1=(lZ2)*[l/(t-2)-l/(t-4)]l"t—2A口Z2V工fn=O一皿)f-位yn,te仁2②f(x)=(l/2)*[l/(t-2)-l/(t-4)]=(l/2)*[(l/2)*IZ(11=0^) -(1/4)*￡(ii=0產(chǎn))(-t^/n]=(2/8)*E(n=0嚴(yán))(-1/2/* -(1/8)*12(11=0^)(-1/4Xn*Ku=(1/8)*E(口=0嚴(yán))(-1/11*(2氣1-口>4，11)*Kn=(1/8)*E(口=0嚴(yán))(-1/11*(271-口>4，11)*(x-l/n,x^(-13)22、設(shè)函數(shù)"=in(/_/_/),求出口解：ux=2x/(xA2+yA2+zA2)uy^2y/(xA2+yA2+zA2)uz^2z/(xA2+yA2+zA2)故du=uxdx+uydy+uzdz=2x/(xA2+yA2+zA2)dx+2y/(xA2+yA24-zA2)dy^2z/(xA2+yA2+zA2)dz23、計算二重積分IT(工- 仃，D是由了=毛了=5三工=1所圍成的區(qū)域°DX型：ffD(x+6y)da=f(Of1)dxf(xf5k)(k+6y)dy=J(Of1)[/(xf5x)(x+6y)dy]dx=f(0^1)[xy+Sy2|(y=x^5x)]dx=J(0。1)[x(5x)+3(5x)2]-[x(x)+3(xy]改=f(0^1)76x2dx=76x3/3|(x=0-1)=76/3Y型:fJD(x+6y)do=/(Ofl)dy/(y/5^y)(x+6y)dx+J(lf5)dyf(y/51)(x+6y)dx=r(0^1)|[I(y/5^y)(X+6y)dx]dy+/(If5)[1)(X+6y)dx]dy=J(0-1)|[x2/2+6xy|(x=y/5+y)]dy+I(1^5)[x2/2+6xy|(x=y/5-1)]dy=I(0^1)[(y2/2+6y2)-((y/5)2/2+6y2/5)]dy+T(1^5)[(1/2+6y)-((y/5)2/2+Gy2/5)]dy=/(Of1)132產(chǎn)25dy+I(1^5)(-61//50+6y+1/2)dy=76/324、求生=i一工一/一x/的通解.

dy/dx=(l-x)+yA2(l-x),dy/dx=(l-x)(l+yA2),dy/(l+yA2)=(l-x)dx.fdy/(l+ydy/dx=(l-x)+yA2(l-x),dy/dx=(l-x)(l+yA2),dy/(l+yA2)=(l-x)dx.fdy/(l+yA2)=;(l-x)dx521、計算工"X1服=2缶-2［：*iiv=2—2arctanvI=2 ' . 0 72、8_E—FlAjU八L、口於*J?1匚八求級數(shù)的收斂半徑與收斂域z——tn3*-獷.1".J21rt H解：令4=——~上，由于極限Hmia上3.當(dāng)3UKI時該級數(shù)收斂，所以此線數(shù)的收斂半徑為叫V—1.1-二v+r_7V時，此級數(shù)為,(-1/+(^r此時的級數(shù)可寫為,-一丁一，收斂。萬二二舊，烘架的筑數(shù)可寫為X當(dāng)3 kI所以原線數(shù)的收斂域為卜內(nèi)23、求函數(shù)"3y)=-2/)的極值.I.求/(k.r)=/…C1-2.T)的極許/；=右—2，F(xiàn)+2a./T-2V.百一，/=-/—vUvL/k r 1九=/T.f+2k/r+2W，+2,「(<*-zB/f■一f=一尸?/-2片廣'-4y?/r+2產(chǎn)廠了in， 3 ”/=—一/T，2/+4V?/T-小/r+4V,/Tj\=/、=0na?+2a-2y1=Ol*P+2y：-4y=0=>x=2y=>(0.0),(T2)%~=//小6=K2yl+8y+2)(2y2+8>-4)-(2/+8y)2]—[-2(2v2+8>)8]ev=-8<代曰1.曰=>°，九nJ(7「2)為極大值qf(-4,-2)=r4+2[(-4)2-2x(-2)2]=&廠24、求(1+1卜”