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2022年9月10日 星期六

109年台綜大轉學考-微積分A詳解

臺灣綜合大學系統109學年度學士班轉學生聯合招生考試

科目名稱:微積分A

解答(a)limn3n+22n+1=limn3+2/n2+1/n=32(b)limx0cos(2x)1x2=limx0(cos(2x)1)(x2)=limx0sin(2x)x=limx0(sin(2x))(x)=limx02cos(2x)1=2
解答{x=t2+2t+3y=t43t3{dx/dt=2t+2dy/dt=4t39t2dy/dx=4t39t22t+2{x=6y=2t=1dy/dx=492+2=54
解答¯BP¯AP=¯CQ¯BQ=¯AR¯CR=x{¯AP=3/(x+1)¯BP=3x/(x+1)¯BQ=5/(x+1)¯CQ=5x/(x+1)¯CR=4/(x+1)¯AR=4x/(x+1){APR/ABC=¯AP¯AR¯AB¯AC=x(x+1)2BPQ/ABC=¯BP¯BQ¯AB¯BC=x(x+1)2CQR/ABC=¯CQ¯CR¯AC¯BC=x(x+1)2APR=BQP=CQR=x(x+1)2ABC=6x(x+1)2f(x)=PQR=618x(x+1)2f(x)=18(x1)(x+1)2=0x=1f(1)=6184=32{critical point at:x=1minimum of f(x)=3/2
解答n=0(2n)nn!xn=n=0an|an+1an|=|(2n+2)n+1xn+1(n+1)!n!(2n)nxn|=|(2n+2)(n+1)(2n+22n)nx|=|2(1+1n)nx|limn|an+1an|<12e|x|<1|x|<12e12e
解答π/20cosx(lncosx2+lnsinx2)dx=π/20cosxln(cosx2sinx2)dx=π/20cosxln(12sinx)dx=ln12π/20cosxdx+π/20cosxlnsinxdx=ln12+[sinx(ln(sinx)1)]|π/20=ln121=1ln2
解答h(x,y,z)=g(x2+y2+z2){hx=xx2+y2+z2g(x2+y2+z2)hy=yx2+y2+z2g(x2+y2+z2)hz=zx2+y2+z2g(x2+y2+z2){2hx2=(1x2+y2+z2x2(x2+y2+z2)3/2)g(x2+y2+z2)+x2x2+y2+z2g(x2+y2+z2)2hy2=(1x2+y2+z2y2(x2+y2+z2)3/2)g(x2+y2+z2)+y2x2+y2+z2g(x2+y2+z2)2hz2=(1x2+y2+z2z2(x2+y2+z2)3/2)g(x2+y2+z2)+z2x2+y2+z2g(x2+y2+z2){2hx2(P)=59g(1)+49g(1)=192hy2(P)=59g(1)+49g(1)=192hz2(P)=89g(1)+19g(1)=2092hx2(P)+2hy2(P)+2hz2(P)=2
解答w=xcos(xy)+z2y47xzw=(cos(xy)xysin(xy)7z,x2sin(xy)+4z2y3,2zy47x)w|P=(6,4,2)P(0,1,1)(6,4,2):6x+4(y1)+2(z1)=03x2yz+3=0;P(6,4,2):x6=y14=z12{x=3ty=2t+1z=t+1,tR
解答{x=rcosθy=rsinθR={(x,y)1x2+y24,x0}S={(r,θ)1r2,θ[π/2,π/2]}
解答\cases{x(t)= \cos(t) \\ y(t)=\sin(t)\\ z(t)=t} \Rightarrow \cases{x'(t)= -\sin(t)\\ y'(t)= \cos(t)\\ z'(t)=1 \\ \mathbf{F}= (t\cos(t), t\sin(t), \cos^3(t))} \Rightarrow 曲線C長度=\int_0^a \sqrt{x'(t)^2 +y'(t)^2 +z'(t)^2}\,dt\\ =\int_0^a \sqrt{2}\,dt = \sqrt 2a = {\sqrt 2\over 4}\pi  \Rightarrow a={\pi\over 4}\\ \\\Rightarrow \int_C \mathbf{F}\cdot dr = \int_0^a \left(-t\sin(t)\cos(t)+ t\sin(t)\cos(t) +\cos^3(t) \right)\,dt =\int_0^a \cos^3(t)\,dt \\ ={1\over 9}\left(9\sin(t)+ \sin(3t) \right)|_0^{\pi/4} ={1\over 9}(9\cdot {\sqrt 2\over 2}+ {\sqrt 2\over 2})= \bbox[red,2pt]{5\sqrt 2/9}
解答F=(3x,2y,5z) \Rightarrow \nabla\cdot F=3+2+5=10 \Rightarrow \iiint \nabla\cdot F\,dV = \iiint 10\,dV =10\times {4\over 3}\pi = \bbox[red, 2pt]{{40\over 3}\pi}

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