※ 引述《lccf (~悶聲色狼~)》之銘言： : 以下是這次考完長庚轉學考微積分後整理的題型 : 大家考完閒閒沒事可以當做消遣算算吧 : 一、 填充題 70% (每題7分) : 1, ∫ x / [(9-2x-x^2)^(1/2)] dx = _____ : 2, n : lim sigma (2/n)*[(5+2i/n)^10] = _____ : n-> oo i=1 3, find the area which is inside r =3sinθ, outside r =1+sinθ = _____ r = 3sinΘ => 3sinΘ = 1 + sinΘ => sinΘ = 1/2 => Θ = π/6 , 5π/6 r = 1 + sinΘ 所求面積為 5π/6 5π/6 (1/2)*∫ (3sinΘ)^2 dΘ - (1/2)*∫ (1 + sinΘ)^2 dΘ π/6 π/6 5π/6 = (1/2)*∫ 8(sinΘ)^2 - 2sinΘ - 1 dΘ π/6 5π/6 = (1/2)*∫ 4 - 4cos2Θ - 2sinΘ - 1 dΘ π/6 |5π/6 = (1/2)*(3Θ - 2sin2Θ + 2cosΘ) | = π |π/6 4, 求 x^3+2y^2=3xyz在(1,1,1)此點之切平面方程式 = _____ 令 F(x,y,z) = x^3 + 2y^2 - 3xyz = 0 為空間一曲面 δF δF δF | 所求平面的法向量 N = [ ----- , ------ , ----- ]| δx δy δz |(1,1,1) | = [3x^2 - 3yz , 4y - 3xz , -3xy]| |(1,1,1) = [0 , 1 , -3] 因此所求切平面為 0*(x-1) + 1*(y-1) -3*(z-1) = 0 => y - 3z + 2 = 0 , x 屬於 R 5, 5x (d/dx) ∫ (sint)^5 dt = _____ x^5 d 5x ---(∫ (sint)^5 dt) = 5(sin(5x))^5 - 5x^4(sin(x^5))^5 dx x^5 : 6, find the interval of convergence : ∞ : sigma (-1)^n * {[(-3x)^n] /[ (n+1)^(1/2)]} = _____ : n=0 : 7, 求 f(x)=x-x^2 與x軸 在第一象限內所圍面積 繞x=2 之體積 = _____ : 8, 求f=(x^2)*y*z 在(1,1,1)此點之 Maximum directional derivative = _____ : δf δf δf gradf(x,y,z) = [ ----- , ----- , ----- ] δx δy δz = [ 2xyz , (x^2)*z , (x^2)*y ] | gradf(x,y,z)| = [2 , 1 , 1] |(1,1,1) 所以 f(x,y,z) 在 (1,1,1) 的 Maximum directional derivative 為 | | | |gradf(x,y,z)| | = |[2 , 1 , 1]| = √(2^2 + 1^2 + 1^2) = √6 | |(1,1,1) | 9, f=x*(y^2)*(z^3) x=(3u^2)+2v y=4u - 2v^3 z=2u^2 – 3v^2 u=1, v=1 find (df/du) = _____ 10,求 f(x,y)= x^2 – (5/2)xy +y^2 在 x^2+y^2 <= 1之Maximum = _____ : 二、 計算題 30% (每題10分) : 1, if (x,y)不等於(0,0)時 f(x,y)= [(x^2)y]/[(x^4 +y^2)] : (x,y)等於(0,0)時 f(x,y)= 0 : find lim [(x^2)y]/[(x^4 +y^2)] : (x,y)->(0,0) 2, if |g(x)| <= x^2 g(x) can derivative for all x prove g’(0) = 0 證明: |g(x)|≦x^2 => -x^2 ≦ g(x) ≦ x^2 ----------(＊) x = 0 代入 (＊) 所以 0 ≦ g(0) ≦ 0 => g(0) = 0 -x^2 ≦ g(x) - g(0) ≦ x^2 -x^2 g(x) - g(0) x^2 => -------- ≦ ------------- ≦ ------- x - 0 x - 0 x - 0 -x^2 g(x) - g(0) x^2 => lim -------- ≦ lim ------------- ≦ lim -------- x→0 x - 0 x→0 x - 0 x→0 x - 0 -x^2 x^2 => lim ------ ≦ g'(0) ≦ lim ----- x→0 x x→0 x => lim -x ≦ g'(0) ≦ lim x x→0 x→0 => 0 ≦ g'(0) ≦ 0 所以 g'(0) = 0 : 3, ∫∫ x^2 + y^2 dxdy =____ R: x^2+y^2 <=2y : R --

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