微積分下-第二版-課后習題答案-同濟大學

xyxy第1頁共47頁xyxy第1頁共47頁習題1—1解答設(shè)設(shè)f(X,y)=xy，求f(_x,_y),f(1,-),f(xy,-),yf(x,y)Xf(_x,-y)=xyXf(_x,-y)=xy+—;yyxy2+Xy X+—;f(xy,—)xy X y

221=X+y; f(x,y)設(shè)f(X,y)=lnxlny,證明:f(xy,uv)=f(x,u)+f(x,v)+f(y,u)f(y,v)f(xy,uv)=1n(xy)‘In(uv)=(Inx+lny)(lnu+lnv)=InXJnu+lnXInv+lny”lnu+Iny”lnv=f(x,u)+f(x,v)+f(y,u)+f(y,v)3.求下列函數(shù)的定義域，并畫出定義域的圖形:(1)f(x,y)=P1—x?+Jy2_[;(2)f(x,y)J4X—y2—22ln(1—X-y)f(x,y)-I-T(4)Jx+Jy+Vzf(x,y,z)= 322-y-z解（1）={(x,y)x<1,=fx,y)0v(1)f(x,y)=P1—x?+Jy2_[;(2)f(x,y)J4X—y2—22ln(1—X-y)f(x,y)-I-T(4)Jx+Jy+Vzf(x,y,z)= 322-y-z解（1）={(x,y)x<1,=fx,y)0vx(3)(x,y)篤a(4)=tx,y,z)y>1},22^1+y<1,y<4x；>0,y>0,z>0,x2Z-172C4.求下列各極限:(1)lim—xy2XTX+yyT1—0 =10+1JOlimXTyTy)ln(X+e2Jx2+y(3)limXTyTxysin(xy)(4)lim=limXTyXTyTyyT2+4 ” (2-Jxy+4)(2+Jxy+4)lim XTyTxy(2+Jx"^)sin(xy)=2第第2頁共47頁第第2頁共47頁5.證明下列極限不存在:(1)(2)lim(1)xzfx—y2+(1)(2)lim(1)xzfx—y2+(x-y)證明如果動點P(x,y)沿y=2x趨向(0,0)則limy2x_0X,=lim亠一3;X_y XTX_2x如果動點P(x,y)沿X=2y趨向(0,0),則lim-X弍TX-y=lim空=3yTy所以極限不存在。(2)證明:如果動點P(x,y)沿y=x趨向(0,0)則lim22(2)證明:如果動點P(x,y)沿y=x趨向(0,0)則lim22芻Xy+(X—y)4

XX-0X4如果動點P(x,y)沿y=2x趨向(0,0)，則limX_J0yz2xt044x222Xy+(X-y)4x4+x12=0所以極限不存在。6.指出下列函數(shù)的間斷點:(1)2y+2xf(x,y)=y—2x(1)為使函數(shù)表達式有意義,y—2xH0，所以在y—2x=0處，函數(shù)間斷。習題(2(1)2y+2xf(x,y)=y—2x(1)為使函數(shù)表達式有意義,y—2xH0，所以在y—2x=0處，函數(shù)間斷。習題(2)為使函數(shù)表達式有意義,XHy，所以在X=y處，函數(shù)間斷。1.(1)zczcz 1ex⑵jyexcos(xy)-2ycos(xy)sin(xy)=y[cos(xy)-sin(2xy)]cz—=xcos(xy)-2xcos(xy)sin(xy)=x[cos(xy)-sin(2xy)]■~1cz y1 2 y1(3)一=y(1+xy)y=y(1+xy)lnz=yln1+xylnz=yln1+xycz——=z[In1(+xy)+xy^(1+xy)y[ln1(+xy)+1+xyXy];1+xy22第 22第 3頁共47頁22第 22第 3頁共47頁dz2yx(x-2yczdy+y)cuIncyx,Inxz(x—y)cuz(x—y)Cu(X—y)zln(X—y)1+(X—y)2zcy1+(X—y)Cz1+(x—y)2z2.(1)zx=y,zy=X,zxx=0,zxy=1,Zyy5.(1)zx=asin2(ax+by),zzXXbsin2(ax+by),2=2acos2(ax+by),Zxy=2abcos2(ax+by).zyy2=2bcos2(ax+by).xxZx2z=y2+2xz,f=2xy(0,0,1) =2,fxz(1,0,2)=-2sin2(x—L),ZtttZxxt—ln(X2zxy1+(丄)=yzxdu-2cos2(x一Zy=yzx22+z,fz=2yz+x,fxx=2,fyz(0,—1,0) =0.=sin2(x—丄)，zxt,zx=zxyzln=2乙=2cosxz=2x,ft2(x——)，Ztt2yz2z,=-cos2(x—丄)yz丄dx+zxyz6.設(shè)對角線為乙則當X=6,y=8,總X7.設(shè)兩腰分別為)+2cos2(x一,dzexZylnxdyy,zx,zy0.dxdy;,dzdx+dy;y1+(丄)yxyx,dzyzyz-ydx+xdylnlnxdzZy=0.05,Ay=-0.1時，Az dzX、y,斜邊為乙則dzxdx+ydyVX6X0.05+8X(0.1)To=-0.05(m).第第4頁共 47 頁第第4頁共 47 頁zx=Zyxdx+ydy,dz=y、z的絕對誤差分別為&、6y、右z,當X=7,y=24，蟲X<6x=0.1,Ay<6y=0.1時，z=』72+242=25Az<dz<7xO.1+24xO.1+242=0.124,z的絕對誤差6z=0.124z的相對誤差8.設(shè)內(nèi)半徑為當r=4,hAVs：dV習題1—31.duAzaz0.12425r,內(nèi)高為h，=0.496%.容積為V，則=jir2h,Vr=2兀巾,Vh=Jir22,dV=2；irhdr+?！竏h,dx=20,ir=0.1,ih=0.1時,=2咒3.14X4X20XO.1+3.14x4X0.1=55.264(cmaxaexy2z&dx cydx czdxxy21+(丄)zxy21+(丄)z心2z■2a(ax+1)y[z+axz-2axy(ax+1)](ax+1)eax(14(ax+1)2 2ax+xeC應(yīng)廳成2.—=-x-+arcsin22-X-y3.(1)4x"arcsinp1-x?-y4 4Xln(X+y)■~1

czX4+y44y3arcsinJ1-x-y+y4J(1-X2-y2)(x2+y2)-y+arcsin2-y34yX4+y4yin(X4+y4)J(1-x2-y2、，2)(x2+y2)du——=2xfcu+yexycucyf2,cu-2yf1+xexyf2.&=iyfyzf3,f2zduLfczcu =xf2+xzf3,cucz=xyf3.第第5頁共47 頁第第5頁共47 頁.(1)-cxcXcycz次22xf1+yf2+f3yf1—=xf1+色=2yf!+xf2+f3,cyf2,2=y(f11y}=yJ,cucz=^3.r Ff:(yf1Uyj1+y(f1「X+f122iXyfiyf12,￡= (xf1+f2)=x2 扭y仁+2xyf2,——=2xyf2f1+2xyf2)=y=y2(f11=2yf2+c2z dddy cy=2yf1=2yf1cz戲1 Cf2 + =x(f11 -x+f仁)+f21”X+f22di &魚+2yf2+2xyf'/+f12Cxy)+2yf2+2xy(f214 3 2 2yf11+4xy弟+4xyf22(y2fi+2xyf2)=2yfi2y(f11'2xy+f122=Xf"+2xf12+f222■y+f222xy)°f1+2xf2+2xycy2X)+2xf2+2xy(f213 3 2 2+2xf2+2xyf1^2xyf22+5xyC 2=一pxyj+xf2)=2x4+2xy內(nèi) cy=2xft+2xy(f11'2xy+f122?X)+x2(f21f12cy*2xy+=2xf!+4x2y2fn+4x3yfi2x4f22”2xy+f22x)f22?X2)L OU5打一csf c recu ex cucy= + cuexrecs cycs2excuctcuex +<xct cycucyctcu,1cu + 2 ex 2cyA2cscu)2=1(4ex+v3cuEuexcy+3(皀：24cy),(Cu、2)2=3(竺)Ct 4exV3cu點u,1 十—re2 excy 4(竺2cycu二(―)GS和2+(—)

Ctdu2=(—)ex2J).6(1)設(shè)(X,y,z)=X+y,FyFz=1+e4十七）0 F0 F1第第頁共47頁bn=一[f(x)sinlbxdxxsinn；!xdx+J(2_X)sinxdx[xd(cosn；!n；!xcos——x)+2fsinxdxxsinn；!xdxcosn；!xdxcosn；!xd(cosx)2n兀4n兀1--——cos+2-Sin——xnj!2(nji)2022n；!02222n；!xcoscosxdxn；!cosn兀nj!cosLn（n兀）(n；!)ii：bna。cossinsinf（X）；=sin-0cosn兀cosn；!cosn；!cosn；!n；!(")2sinnJI-0□cn-1sinn;isinx,0<x<2用余弦級數(shù)逼近：作偶延拓，由系數(shù)公式:=0,(n=0,1,2，…)Ln(x)cosxdxxcosxdx(2—x)cosxdx[f(x)dxxsinxdx2-xdx=1sinxdx+sinxd(sinsin(n兀)2(n兀)f(x)coscosx)cosxsinnJ!xdxxcosxdxsinxdxn；!2（n兀）cossinn；!□csinsinnJ!sinn；