※ 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 = _____ : 4, 求 x^3+2y^2=3xyz在(1,1,1)此點之切平面方程式 = _____ : 5, 5x : (d/dx) ∫ (sint)^5 dt = _____ : 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 = _____ : 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 : 3, ∫∫ x^2 + y^2 dxdy =____ R: x^2+y^2 <=2y : R 一、 填充題 10, f(x,y) = x^2 - (5/2)xy + y^2 考慮內部的情形: fx = 2x - (5/2)y = 0 => x = 0 , y = 0 => f(0,0) = 0 fy = -(5/2)x + 2y = 0 考慮邊界的情形: 令 x = cosΘ , y = sinΘ , 0 ≦ Θ ≦ 2π 則 f(x,y) = g(Θ) = (cosΘ)^2 - (5/2)(cosΘ)(sinΘ) + (sinΘ)^2 = 1 - (5/4)*(sin2Θ) -1 ≦ sin2Θ ≦ 1 當 sin2Θ = -1 => 2Θ = 3π/2 , 7π/2 => Θ = 3π/4 , 7π/4 g(Θ) = 1 - (5/4)*(-1) = 1 + (5/4) = 9/4 為最大值 此時 (x,y) = (-1/√2 , 1/√2) , (1/√2 , -1/√2) 二、計算題 3. 令 x = rcosΘ , y = rsinΘ , 則 |J| = r 因為 x^2 + y^2 ≦ 2y => r^2 ≦ 2rsinΘ => 0 ≦ r ≦ 2sinΘ sinΘ ≧ 0 => 0 ≦ Θ ≦ π 則 ∫∫ x^2 + y^2 dxdy R π 2sinΘ = ∫ ∫ r^2*r drdΘ 0 0 π 2sinΘ = ∫ ∫ r^3 drdΘ 0 0 π 1 |r = 2sinΘ = ∫ ---r^4 | dΘ 0 4 |r = 0 π = ∫ 4*((sinΘ)^4) dΘ 0 π = ∫ 4*(((1/2)*(1 - cos2Θ))^2) dΘ 0 π = ∫ 1 - 2cos2Θ + (cos2Θ)^2 dΘ 0 π = ∫ 1 - 2cos2Θ + (1/2)*(1 + cos4Θ) dΘ 0 π = ∫ 3/2 - 2cos2Θ + (1/2)*(cos4Θ) dΘ 0 |π = (3/2)*Θ - sin2Θ + (1/8)*(sin4Θ) | = (3/2)*π |0 --

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