ä¸ãéæ©é¢ï¼1ãBï¼ 2ãCï¼ 3ãBï¼ 4ãDï¼ 5ãAï¼ 6ãAï¼
äºã1ãxââlimxsin(1/x)=xââlim[sin(1/x)/(1/x)=xââlim(1/x)/(1/x)=1ï¼æ
y=1æ¯å
¶æ°´å¹³æ¸è¿çº¿ï¼æ
åç´æ¸è¿çº¿ï¼
2ãé¢ç§¯=Ïa²ï¼
3ã
4ãto=Ï/2ï¼x'=1-costï¼y'=sintï¼z'=2cos(t/2)ï¼æ
x'o=1ï¼y'o=1ï¼z'o=â2ï¼xo=Ï/2-1ï¼yo=1ï¼zo=2â2ï¼æ
å线æ¹ç¨ä¸ºï¼(x-Ï/2+1)/1=(y-1)/1=(z-2â2)/â2ï¼
æ³çº¿æ¹ç¨ä¸ºï¼ï¼x-Ï/2+`)+(y-1)+(â2)(z-2â2)=0
5ã交æ¢ç§¯å次åºå¾ã0ï¼1ãâ«dxãx²ï¼xãâ«f(xï¼y)dy
6ãAB=(1ï¼-6ï¼3)ï¼æ
å¹³é¢æ¹ç¨ä¸º(x-2)-6(y+1)+3(z-2)=0ï¼å³x-6y+3z-14=0为ææ±ã
7ãä¸ä¸ªç¹è§£ä¸ºy*=(1/2)(e^x)cos2x.
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1ãxââlim[ln(1-1/x)/(arctanx-Ï/2)]=xââlim[1/x(x-1)]/[1/(1+x²)]=xââlim[(1+x²)/(x²-x)]
=xââlim[[(1/x²+1)/(1-1/x)]=1
2ãxâ0ï¼yâ0lim{(x²+y²)/[â(x²+y²+1)-1]}=xâ0ï¼yâ0lim{(x²+y²)[â(x²+y²+1)+1]/(x²+y²)}
=xâ0ï¼yâ0lim[â(x²+y²+1)+1]=2ï¼
3ãã0ï¼ln2ãâ«â(e^x-1)dxï¼ä»¤e^x-1=u²ï¼å(e^x)dx=2uduï¼dx=2udu/(1+u²)ï¼
x=0æ¶u=0ï¼x=ln2æ¶u=1
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ã0ï¼ln2ãâ«â(e^x-1)dx=ã0ï¼1ã2â«u²du/(1+u²)=ã0ï¼1ã2â«[1-1/(1+u²)]du
=2[u-arctanu]ã0ï¼1ã=2(1-Ï/4)2-Ï/2.
4ãã-âï¼+âãâ«[1/(9+x²)]dx=(1/3)arctan(x/3)ã-âï¼+âã=(1/3)(Ï/2+Ï/2)=Ï/3
5ï¼z=f(x²-y²ï¼e^(xy))ï¼ä»¤u=x²-y²ï¼v=e^(xy)ï¼
å∂z/∂x=(∂f/∂u)(∂u/∂x)+(∂f/∂v)(∂v/∂x)=2x(∂f/∂u)+ye^(xy)(∂f/∂v)
∂z/∂y=(∂f/∂u)(∂u/∂y)+(∂f/∂v)(∂v/∂y)=-2y(∂f/∂u)+xe^(xy)(∂f/∂v)
6ãx=te^tï¼y=t²e^tï¼æ
dy/dx=y'=(dy/dt)/(dx/dt)=(2te^t+t²e^t)/(e^t+te^t)=(2t+t²)/(1+t)
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d²y/dx²=(dy'/dt)/(dx/dt)={[(1+t)(2+2t)-(2t+t²)]/(1+t)²}/(e^t+te^t)=(t²+2t+2)/[(1+t)³e^t]
åãæ±æç©çº¿y=-x²+4x-3åå
¶ç¹(0ï¼-3)å(3ï¼0)çå线æå´å¾å½¢çé¢ç§¯ã
解ï¼y'=-2x+4ï¼y'(0)=4ï¼y'(3)=-2
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è¿M(0ï¼-3)çå线æ¹ç¨ä¸ºy=4x-3ï¼è¿N(3ï¼0)çå线æ¹ç¨ä¸ºy=-2(x-3)=-2x+6ï¼
令4x-3=-2x+6ï¼å¾6x=9ï¼æ
x=9/6=3/2ï¼y=-3+6=3ï¼å³äºå线交ç¹Qçåæ 为(3/2ï¼3)ï¼
设å线MQä¸xè½´ç交ç¹ä¸ºPï¼åPç¹çåæ 为(3/4ï¼0)ï¼
æå´å¾å½¢çé¢ç§¯S=â³PNQçé¢ç§¯S₁-æç©çº¿ä¸xè½´æå´é¢ç§¯S₂+æç©çº¿ä¸å线MQåä¸xè½´æå´
é¢ç§¯S₃.
S₁=(1/2)Ã(3-3/4)Ã3=27/8
S₂=ã1ï¼3ãâ«(-x²+4x-3)dx=[-x³/3+2x²-3x]ã1ï¼3ã=(-9+18-9)-(-1/3+2-3)=4/3
S₃=ã0ï¼1ãâ£â«(-x²+4x-3)dxâ£-(1/2)Ã(3/4)Ã3=â£-x³/3+2x²-3xâ£ã0ï¼1ã-9/8=4/3-9/8=5/24
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é¢ç§¯S=27/8-4/3ï¼5/24=(81-32+5)/24=54/24=9/4
äºãæ±å½æ°f(x)=1/xå¨ç¹xo=2å¤çæ³°åå±å¼å¼ï¼å¹¶æ±å
¶æ¶æåã
解ï¼f(2)=1/2ï¼f'(x)=-1/x²ï¼f'(2)=-1!/2²ï¼f''(x)=2!/x³ï¼f(2)=2!/2³ï¼
f'''(x)=-3!/xⁿï¼f'''(2)=-3!/2⁴ï¼.......ï¼f⁽ⁿ⁾(x)=(-1)ⁿn!/xⁿ⁺¹ï¼f⁽ⁿ⁾(2)=(-1)ⁿn!/2ⁿ⁺¹ï¼
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å±å¼å¼ä¸º1/x=1/2-(1/2²)(x-2)+(1/2³)(x-2)²-(1/2⁴)(x-2)³+......+[(-1)ⁿ/2ⁿ⁺¹](x-2)ⁿ+........
nââlimâ£R‹n›(x)â£=nââlimâ£(x-2)â£ⁿ⁺¹/2ⁿ⁺²<nââlim[â£x-2â£/2]ⁿ⁺²
ç±â£x-2â£/2<1ï¼å¾-2<x-2<2ï¼æ
å¾0<x<4ï¼å³æ¶æå为(0ï¼4).
å
ã计ç®ãcãâ®[(e^x)cosy+2y]dx-[(e^x)siny]dyï¼å
¶ä¸c为è´ååå¨x²+y²=a².
解ï¼â®[(e^x)cosy+2y]dx-[(e^x)siny]dy=ãDã-â«â«[(-e^x)siny+(e^x)siny-2]dxdy
=ãDã4â«â«dxdy=ã-aï¼aã4â«dyã-â(a²-y²)ï¼â(a²-y²)ãâ«dx
=ã-aï¼aã8â«â(a²-y²)dy=[(y/2)â(a²-y²)+(a²/2)arcsin(y/a)]ã-aï¼aã
=(a²/2)arcsin1-(a²/2)arcsin(-1)=Ïa²/2
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1ããDãâ«â«arctan(y/x)dxdyï¼å
¶ä¸Dï¼1â¦x²+y²â¦4ï¼xâ§0ï¼yâ§0.
解ï¼ç¨æåæ ï¼1â¦r²â¦4ï¼æ
1â¦râ¦2ï¼0â¦Î¸â¦Ï/2
ãDãâ«â«arctan(y/x)dxdy=ã0ï¼Ï/2ãâ«dθã1ï¼2ãâ«arctan(sinθ/cosθ)rdr
=ã0ï¼Ï/2ãâ«dθã1ï¼2ãâ«arctan(tanθ)rdr=ã0ï¼Ï/2ãâ«Î¸dθã1ï¼2ãâ«rdr
=ã0ï¼Ï/2ãâ«Î¸dθ(r²/2)ã1ï¼2ã=ã0ï¼Ï/2ã(3/2)â«Î¸dθ=(3/2)(θ²/2)ã0ï¼Ï/2ã
=(3/2)(ϲ/8)=3ϲ/16
2ãæ±çé¢x²+y²+z²=a²è¢«å¹³é¢z=a/2ï¼åz=a/4æ夹é¨åæ²é¢çé¢ç§¯
解ï¼æ夹é¨åæ²é¢çé¢ç§¯=2Ïa(a/2-a/4)=Ïa²/2
å
«ãè¿æ¯è¦è¯æè¬å¸é¢ç§¯S=ÏR²+2ÏRH+2ÏRâ(R²+h²)=ÏR[R+H+2â(R²+h²)]å¨æ»¡è¶³å¸è¬å®¹ç§¯V=ÏR²H+(1/3)ÏR²h=(H+h/3)ÏR²=kçæ¡ä»¶ä¸è·å¾æå°å¼æ¶å¿
须满足æ¡ä»¶R=(â5)Hï¼h=2H.
è¯æï¼S=ÏR[R+H+2â(R²+h²)]ï¼éå æ¡ä»¶Ï(Rï¼Hï¼h)=(H+h/3)ÏR²-k=0
ä½å½æ°F(Rï¼Hï¼h)=ÏR²+2ÏRH+2ÏRâ(R²+h²)+λ[(H+h/3)ÏR²-k]
∂F/∂R=2ÏR+2ÏH+2Ïâ(R²+h²)+2ÏR²/â(R²+h²)+2ÏRλ[(H+h/3)=0
å³æR+H+â(R²+h²)+R²/â(R²+h²)+Rλ(H+h/3)=0.............(1)
∂F/∂H=2ÏR+ÏR²Î»=0ï¼å³æ2+Rλ=0ï¼æ
λ=-2/R............(2)
∂F/∂h=h/â(R²+h²)+Rh/â(R²+h²)+(1/3)Rλ=0
å³æ3h+3Rh+Rλâ(R²+h²)=0...............(3)
(H+h/3)ÏR²-k=0.................(4)
å个æ¹ç¨èç«æ±è§£ï¼å³å¯å¾è¯ããæççµèåºé®é¢äºï¼å¾ä¸å¥½ç¨ï¼ä½ èªå·±æ±è§£å§ã
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