臺灣綜合大學系統108學年度學士班轉學生聯合招生考試
科目名稱:微積分B
解答:V=43πr3⇒dVdt4πr2drdt⇒5=4π⋅22⋅drdt⇒drdt=516π⇒速度為516πcm/sec
解答:(x2+y2)′=((2x2+2y2−x)2)′⇒2x+2yy′=2(2x2+2y2−x)(4x+4yy′−1)將y(0)=12代入上式⇒y′(0)=2⋅12(2y′(0)−1)⇒y′(0)=1⇒過(0,12)之切線:y=x+12
解答:f(x)=∫√x1ettdt⇒f′(x)=e√x2x>0,∀x∈[1,∞)⇒f(x)在區間[1,∞)為嚴格遞增函數⇒最小值=f(1)=0
解答:底面為單位圓:x2+y2=1⇒x2=1−y2,只考慮第一象限與z軸所圍體積:dV=2xx′=2(1−y2)dy⇒∫dV=∫102(1−y2)dy=43⇒全部體積=4×43=163
解答:{u=tan−1x⇒du=dx1+x2dv=xdx(1+x2)2⇒v=−12(1+x2)⇒∫∞0xtan−1x(1+x2)2dx=−tan−1x2(1+x2)|∞0+12∫1(1+x2)2dx=−tan−1x2(1+x2)+14(x1+x2+tan−1x)|∞0=π8註:v=tan−1x⇒{dv=dx1+x21+x2=1+tan2u=sec2v⇒∫1(1+x2)2dx=∫dvsec2v=∫cos2vdv=12∫cos2v+1dv=12(12sin2v+v)=12(sinvcosv+v)=12(x1+x2+tan−1x)
解答:limn→∞|an+1an|=limn→∞|(−2)n+1⋅xn+1(n+1)2⋅n2(−2)nxn|=2|x|<1⇒|x|<12⇒{收斂半徑=1/2收斂區間=(−1/2,1/2)
解答:{fx(0,0)=limh→0f(h,0)−f(0,0)h=limh→0f(h,0)h=limh→00/h2h=0fy(0,0)=limh→0f(0,h)−f(0,0)h=limh→0f(0,h)h=limh→0h/2h=12⇒{fx(0,0)=0fy(0,0)=12
解答:{f(x,y,z)=xyzg(x,y,z)=xy+2xz+2yz−12⇒{fx=λgxfy=λgyfz=λgzg=0⇒{yz=λ(y+2z)⋯(1)xz=λ(x+2z)⋯(2)xy=λ(2x+2y)⋯(3)xy+2xz+2yz=12⋯(4)因此{(1)(2)⇒yx=y+2zx+2z⇒x=y(2)(3)⇒zy=x+2z2x+2y⇒y=2z(1)(3)⇒zx=y+2z2x+2y⇒x=2z代入(4)⇒4z2+4z2+4z2=12⇒z2=1⇒z=±1⇒{A=(2,2,1)B=(−2,−2,−1)⇒{f(A)=4f(B)=−4⇒極值為±4
解答:令{x=rcosθy=rsinθ⇒∫a−a∫√a2−x2−√a2−x2cos(x2+y2)dydx=∫2π0∫a0cos(r2)rdrdθ=∫2π0[12sin(r2)]|a0dθ=∫2π012sin(a2)dθ=πsin(a2)
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