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:{x=t2+2t+3y=t4−3t3⇒{dx/dt=2t+2dy/dt=4t3−9t2⇒dy/dx=4t3−9t22t+2其中{x=6y=−2⇒t=1⇒dy/dx=4−92+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)2△BPQ/△ABC=¯BP⋅¯BQ¯AB⋅¯BC=x(x+1)2△CQR/△ABC=¯CQ⋅¯CR¯AC⋅¯BC=x(x+1)2⇒△APR=△BQP=△CQR=x(x+1)2△ABC=6x(x+1)2⇒f(x)=△PQR面積=6−18x(x+1)2⇒f′(x)=18(x−1)(x+1)2=0⇒x=1⇒f(1)=6−184=32,因此{critical point at:x=1minimum of f(x)=3/2
解答:令∞∑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|,因此limn→∞|an+1an|<1⇒2e|x|<1⇒|x|<12e⇒收斂半徑為12e
解答:∫π/20cosx(lncosx2+lnsinx2)dx=∫π/20cosxln(cosx2sinx2)dx=∫π/20cosxln(12sinx)dx=ln12∫π/20cosxdx+∫π/20cosxlnsinxdx=ln12+[sinx(ln(sinx)−1)]|π/20=ln12−1=−1−ln2
解答:h(x,y,z)=g(√x2+y2+z2)⇒{∂h∂x=x√x2+y2+z2g′(√x2+y2+z2)∂h∂y=y√x2+y2+z2g′(√x2+y2+z2)∂h∂z=z√x2+y2+z2g′(√x2+y2+z2)⇒{∂2h∂x2=(1√x2+y2+z2−x2(x2+y2+z2)3/2)g′(√x2+y2+z2)+x2x2+y2+z2g″(√x2+y2+z2)∂2h∂y2=(1√x2+y2+z2−y2(x2+y2+z2)3/2)g′(√x2+y2+z2)+y2x2+y2+z2g″(√x2+y2+z2)∂2h∂z2=(1√x2+y2+z2−z2(x2+y2+z2)3/2)g′(√x2+y2+z2)+z2x2+y2+z2g″(√x2+y2+z2)⇒{∂2h∂x2(P)=59g′(1)+49g″(1)=−19∂2h∂y2(P)=59g′(1)+49g″(1)=−19∂2h∂z2(P)=89g′(1)+19g″(1)=209⇒∂2h∂x2(P)+∂2h∂y2(P)+∂2h∂z2(P)=2
解答:假設w=xcos(xy)+z2y4−7xz⇒∇w=(cos(xy)−xysin(xy)−7z,−x2sin(xy)+4z2y3,2zy4−7x)⇒∇w|P=(−6,4,2)⇒過P(0,1,1)且法向量為(−6,4,2)的切平面方程式:−6x+4(y−1)+2(z−1)=0⇒3x−2y−z+3=0;過P且方向向量為(−6,4,2)的直線方程式:x−6=y−14=z−12,即{x=−3ty=2t+1z=t+1,t∈R
解答:{x=rcosθy=rsinθ⇒R={(x,y)∣1≤x2+y2≤4,x≥0}⇒S={(r,θ)∣1≤r≤2,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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最後一題,應該是只要算上半圓
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