推 transforman:推推推 114.36.128.200 07/16 16:41
※ 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
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 61.66.173.21