[情報] 99年台聯大微積分詳解

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看到考試將近，雖然我也有考...未來會是對手，但還是要積陰德^^ 分享一下台聯大99年微積分考題詳解!! 以下詳解都是我在網路上看到的，所以別問我會不會= = 純屬分享，偷偷賺P幣@ @ 人因宿命而誕生;因命運而生活;因使命而燃燒;因夢想而偉大, 以此勉勵轉學戰友... ............................................................ 甲、選擇題:共8題，每題6分，共48分。 (D)1.Use the fact that l i m(sinθ/θ)=1 to find the value of l i m(2x/tan7x). θ→0 X→0 <sol>: 1 令θ=7x 則x= ﹣θ 且 x→0 θ→0 7 2 2 2 1 1 2 l i m(2x/tan7x)=l i m(﹣θ)/(tanθ)= ﹣l i m(1/sinθ/θ)(1/cosθ)= ﹣﹣﹣= ﹣ X→0 θ→0 7 7 θ→0 7 1 1 7 答案:(D) (A)For the values of a,m,and b does the function 3, x=0 f(x)=-x^2+3x+a, 0<x<1 mx+b, 1≦x≦2 (A)2.satisdy the hypotheses of the Mean Value Theorem on the interval [0,2]? (A)a=3 m=1 b=4 (B)a=3 m=2 b=3 (C)a=3 m=4 b=1 (D)a=2 m=1 b=3 (E)None of the above. <sol>: f(x)在[0,2]必須連續且在(0,2)必需可微分，即 (Ⅰ)l i mf(x)=f(0)→l i m(-x^2+3x+a)=3→a=3 x→0+ x→0+ (Ⅱ)l i m f(x)=l i m f(x)→l i m(-x^2+3x+a)=l i m(mx+b)→5=m+b,得b=5-m x→1- x→1+ x→1- x→1+ (Ⅲ) 3 ,x=0 -2x+3 ,0<x<1 f(x)= -x^2+3x+3 ,0<x<1 → f'(x)= 1 ,x=1 mx+b ,1≦x≦2 m ,1<x<2 f'(1)負的 =f'(1)正的→-2+3=m→m=1 ∴a=3,m=1,b=4 答案:(A) (D)3.Define 1-cosx .Find f'(0) ╴╴╴, if x≠0 x f(x)= 0 , if x=0 1 (A)Does not exist (B)1 (C)2 (D)﹣(E)0 2 <sol>: f(x)-f(0) (1-cosx)/x-0 1-cosx f'(0)=l i m ╴╴╴╴ =l i m ╴╴╴╴╴╴ =l i m ╴╴╴ x→0 x-0 x→0 x x→0 x^2 1-cos^2x sin^2x sinx 1 1 1 =l i m ╴╴╴╴╴╴ =l i m ╴╴╴╴╴╴ =l i m(╴╴)^2(╴╴╴╴)=1^2(﹣)= ﹣ x^2(1+cosx) x^2(1+cosx) x 1+cosx 2 2 答案:(D) (E)4.Assume x=2tan(t),y=sec^2(t)-1.Find all equation for the kine tangent to the curve at the point where t=π/4. (A)y= -x-1 (B)y=-x+1 (C)y=x (D)y=x+1 (E)y=x-1 <sol>: dy dy/dt d/dt(sec^2t-1) (2sec^t)(tan(t)) dy π ╴ = ╴╴╴ = ╴╴╴╴╴╴╴ = ╴╴╴╴╴╴╴╴ =tant ,得切線斜率m=╴｜(t=﹣) dx dx/dt d/dt(2tan(t)) 2sec^t dx 4 =1 π π π 又當t=╴時，x=2tan(╴)=2,y=sec^2(╴)-1=(√2)^2-1=1,即切點(x,y)=(2,1) 4 4 4 所以切線方程式為:y-1=(1)(x-2)→y=x-1 答案:(E) 1 x (E)5.Find the limit: l i m ╴╴∫ln(t)dt x→∞ xlnx 1 (A)∞ (B)-∞ (C)-1 (D)0 (E)1 <sol>: x ∫lntdt 1 x 1 lnx (lnx)' l i m ╴╴∫lntdt=l i m ╴╴╴=l i m ╴╴╴╴╴(羅畢達)=l i m ╴╴╴╴ x→∞ xlnx 1 x→∞ xlnx x→∞ lnx+x(1/x) x→∞ (lnx+1)' 1/x =l i m ╴╴=1 x→∞ 1/x 答案:(E) xy^2 (C)6.Find the limit: l i m ╴╴╴╴. (x,y)→(0,0)x^2+y^2 1 (A)1 (B)﹣ (C)0 (D)2 (E)Does not exist 2 <sol>: 令x=rcosθ,y=rsinθ,x^2+y^2=r^2且(x,y)→(0,0) →r=0 xy^2 (rcosθ)(rsinθ)^2 l i m ╴╴╴ =l i m╴╴╴╴╴╴╴╴╴ =l i m(rcosθsin^2θ)=0 (x,y)→(0,0)x^2+y^2 r→0 r^2 r→0 答案:(C) (A)7.Find the line inretral of F=(2xyz,x^2z,x^2y)over any path from (0,0,0) to (1,2,3)? (A)6 (B)9 (C)12 (D)18 (E)None of the above <sol>: (1,2,3) → → (1,2,3) (1,2,3) (1,2,3) ∫ F˙dr=∫ 2xyzdx+x^zdy+x^2ydz=∫ d(x^2yz)=(x^yz)︳ (0,0,0) (0,0,0) (0,0,0) (0,0,0) =(1)^2(2)(3)=6 答案:(A) (B)8.Find the surface area of the portion S of the cone z^2=x^2+y^2,where z≧0 ,contained within the cylinder y^2+z^2≦1. (A)0 (B)π (C)-π (D)2π (E)-2π <sol>: 椎面通式:z^2=x^2+y^2 → x^2=z^2-y^2 → x=±√z^2-y^2, 偏微x ±y 偏微x ±z 得 ╴╴╴ = ╴╴╴╴╴,╴╴╴ = ╴╴╴╴╴ 偏微y √z^2-y^2 偏微z √z^2-y^2 所求面積S=2∫∫√1+(偏微x/偏微y)^2+(偏微x/偏微z)^2dydz R y^2 z^2 =2∫∫√1+╴╴╴╴+╴╴╴╴dydz R z^2-y^2 z^2-y^2 2z^2 =2∫∫√╴╴╴╴dydz,其中R:z=√1-y^2與y=z,y=-z所圍區域 R z^2-y^2 利用極座標轉換: π 3π 令y=rcosθ,z=rsinθ,則R={(r,θ)︳0<r<1,﹣≦θ≦╴ } 4 4 3π ╴ 4 1 2r^2sin^2θ S=2∫ ∫√╴╴╴╴╴╴╴╴╴╴╴rdrdθ π 0 r^2sin^2θ-r^2cos^2θ ﹣ 4 3π ╴ 4 1 √2 sinθ =2∫ ∫√╴╴╴╴╴╴╴╴╴╴rdrdθ π 0 √sin^2θ-cos^2θ ﹣ 4 3π ╴ 4 1 √2 sinθ 1 -1 1 1 1 =2∫ ∫╴╴╴╴╴╴╴╴dθ(∫rdr)=2∫ ╴╴╴╴(-du)(﹣r^2)︳ π 0 √1-2cos^2θ 0 1 √1-u^2 2 0 ﹣ 4 1 =2(sin^-1(1)-sin^-1(-1))(﹣) 2 π π 1 =2(﹣-(-﹣))(﹣)=π 2 2 2 答案:(B) 乙、填充題:共4題，每題8分，共32分。 1 1.Evaluate the indefinite integral ∫╴╴╴dx. 1+e^x <sol>: 1 (1+e^x)-e^x e^x 1 ∫╴╴╴dx=∫╴╴╴╴╴╴dx=∫(1-╴╴╴)dx=x-∫╴╴╴d(1+e^x)=x-ln︱1+e^x︳+C 1+e^x 1+e^x 1+e^x 1+e^x 答案:x-ln︱1+e^x︳+C 2.Find the points on the graph of z=3x^2-4y^2 at which the vector n=<3,2,2> is normal to the tangent plane. <sol>: 令ψ(x,y,z)=3x^2-4y^2-z,則梯度▽ψ=<6x,-8y,-1> → ∵曲面z=3x^2-4y^2在點(a,b,c)處的梯度▽ψ=<6a,-8b,-1>與n=<3,2,2>平行 6a=3t 1 1 1 即<6a,-8b,-1>=t<3,2,2> → -8b=2t,得t=-﹣,a=-﹣,b=﹣ -1=2t 2 4 8 3 1 1 又c=3a^2-4b^2= ╴ - ╴ = ﹣ 16 16 8 1 1 1 → ∴曲面 z=3x^2-4y^2在點(-﹣,﹣,﹣)處的切平面法向量為n=<3,2,2> 4 8 8 1 1 1 答案:(-﹣,﹣,﹣) 4 8 8 4 2 3.What is the value of the double integral ∫∫ √x^3+1dxdy? 0 √y <sol>: ∵積分範圍R={(x,y)︳√y≦x≦2,0≦y≦4}={(x,y)︳0≦y≦x^2,0≦x≦2} 4 2 2 x^2 2 x^2 2 ∴∫∫ √x^3+1dxdy=∫∫ √x^3+1dydx=∫[y√x^3+1]︳ dx=∫x^2√x^3+1dx 0 √y 0 0 0 0 0 2 1 1 1 2 3 2 2 52 =∫(x^3+1)^﹣﹣d(x^3+1)=(﹣﹣(x^3+1)^﹣)︳= ﹣(27-1)=╴ 0 2 3 3 3 2 0 9 9 52 答案:╴ 9 4.What is the largest value that the directional derivative of f(x,y,z)=xyz can have at the point (1,1,1)? <sol>: ∵f(x,y,z)=xyz在點(1,1,1)的梯度▽f︳ = <yz,xz,xy>︳ = <1,1,1> (1,1,1) (1,1,1) ∴f(x,y,z)=xyz在點(1,1,1)沿著<1,1,1>的方向可得最大方向導數︱▽f︳︱=√1^2+1^2+1^2=√3 (1,1,1) 答案:√3 丙、計算、證明題:共2大題，每題10分，共20分。 ∞ n 1 1.(a)Determine if the series Σ(-1) cos﹣converges or diverges? (5分) n=1 n 1 dx (b)Determine if the improper intergral ∫╴╴╴ converges or diverges? (5分) 0x-sinx <sol>: n 1 (a)當n→∞且n為偶數時，(-1) cos(﹣)將趨近於1 n n 1 當n→∞且n為奇數時，(-1) cos(﹣)將趨近於-1 n n 1 ∞ n 1 即l i m(-1) cos(﹣)不存在，故由終值檢驗法知Σ(-1) cos﹣發散 n→∞ n n=1 n 1 1 11 11 (b)∵﹣< ╴╴╴,0<x<1 ;又∫﹣dx=l i m∫﹣dx x x-sinx 0x t→0+ tx 1 =l i m(ln︱x︳)︳=l i m(-ln︱t︳)=+∞發散 t→0+ t t→0+ 1 1 ∴∫╴╴╴dx發散 0x-sinx ............................................................................ 太多了..今天先打完選擇題8題和4題填充.明天再把最後的計算題打完= = 修改文章好像不能多P幣...但未來還是會打完(版主居然M了= =) 希望大家也幫我集氣^^ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.116.115.213 ※ 編輯: c487o6k9 來自: 140.116.115.213 (06/27 23:35)

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