作者rod24574575 (天然呆)
看板NTU-Exam
標題[試題] 99上 薛克民 微積分甲上 第八次小考 (附詳解)
時間Sat Jan 8 22:26:40 2011
課程名稱︰微積分甲上
課程性質︰必修
課程教師︰薛克民
開課學院:電資學院、工學院、管理學院
開課系所︰電機系、資工系、材料系、資管系
考試日期(年月日)︰99/12/21
考試時限(分鐘):30分鐘
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試題 :
˙每題十分
˙請詳述計算過程,無計算過程的答案不予計分
1. Find the length of the curve. y = ㏑[sec(x)] , 0 ≦ x ≦ π/4 (10%)
d ㏑[sec(x)] tan(x)sec(x)
y' = ─────── = ─────── = tan(x)
dx sec(x)
π/4 ________ π/4
∫ √1 + [tan(x)]^2 dx = ∫ sec(x) dx
0 0
π/4 sec(x) + tan(x)
= ∫ sec(x) * ─────────dx
0 sec(x) + tan(x)
┌ ┐π/4
= │ ㏑│sec(x) + tan(x)│ │
└ ┘0
= ㏑[√(2) + 1] - ㏑(1)
=
㏑[√(2) + 1]
#
2. Find the area of the resulting surface.
y = x^2 from (1,1) to (2,4) is rotated about the y-axis. (10%)
d(x^2)
y' = ──── = 2x
dx
2 ______ 2 _____
∫ 2πx √1 + (2x)^2 dx = 2π * ∫ √4x^2 + 1 dx
1 1
┌ 3/2 ┐2
= │ (π/6)(4x^2 + 1) │
└ ┘1
┌ 3/2 3/2 ┐
=
(π/6)│ 17 - 5 │
└ ┘
#
t+z
3. Solve the differential equatation. dz/dt + e = 0 (10%)
t z
dz/dt = -(e * e )
-z t
=> ∫(-e ) dz = ∫(e ) dt
=> e^(-z) = e^(t) + C
=> -z = ㏑│e^(t) + C│
│ 1 │
=>
z = ㏑│──────│
│ e^(t) + C │
#
4. Solve the differential equation. xy' = y + (x^2)*sin(x) , y(π) = 0 (10%)
y' - y/x = x*sin(x)
∫(-1/x) dx -㏑|x|
∴ I(x) = e = e = 1/x
兩邊同乘 I(x):
y' y
=> ──- ───= sin(x)
x x^2
=> (y/x)' = sin(x)
=> ∫ d(y/x) = ∫ sin(x) dx
=> y/x = -cos(x) + C
=> y = (-x)*cos(x) + C*x
y(π) = π + π*C = 0
=> C = -1
∴
y = (-x)*cos(x) - x
#
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