Re: [考古] 中興99年
※ 引述《boss913 (伯斯)》之銘言：
: http://recruit.nchu.edu.tw/college-exam/transfer/transfer-index2009.htm
: 第二題是硬算嗎= =?
Use Leibniz rule and Chain rule
1
f'(x) = ----------------- * g'(x)
[1+g(x)^3]^1/2
g'(x) = [1+sin(cos(x)^2) ] *(-sinx)
f'(1/2π) = 1 * g'(x) = (1+sin0) * (-sin(1/2π) = 0
cos(1/2π)=0 so g(x) = 0
: 5.(a),6,7,9不會@@
: 希望有高手解答~3Q
5.(a)
Use 分式積分
5 4
x + x = x(x + 1)
3 2
1 A Bx + Cx + Dx +E
∫ -------- dx = ∫[------ + ------------------ ] dx
5 4
x + x x x + 1
Solve A,B,C,D,E=? get five equation A + B = 0 ,C=D=E=0 , A= 1
so 3
1 1 -x 4
∫------ dx = ∫[---- + ----- ] dx = ln|x| - ln | x + 1| +c
5 4
x + x x x + 1
6.
Area of surface by rotating respect to x-axis : A = ∫2πyds
2 2 2
where ds = dx + dy dx = -6costcostsintdt dy = 6sintsintcostdt
ds = 6sintcost dt
3
A = ∫2π2sin (t) 6sintcostdt
24π
= 24π / 5 (1-0) = -----
5
9.
set a = v - u v = a + b + 1
b = u - 1 u = 0 + b + 1 |J| = 1
2 2
(a+1) (a+1) 1
∫∫ e dbda = ∫(a+1) e d (a+1) = ----- (e - 1)
2
b = 0 , b = 1 , a + 1 =b , a = 0, is new R'
--
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◆ From: 114.34.122.244
※ 編輯: rygb 來自: 114.34.122.244 (06/25 19:50)
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7.
x = 2 cos t - cos 2t
y = 2 sin t - sin 2t
Def:
vertical tangent : dx = 0 and dy ≠ 0
horizontal tangent : dy = 0 and dx ≠ 0
So first we use the condition to find vertical tangent:
dx = - 2 sin t + 2 sin 2t = 0
so sin t = 2 cos t sin t gives that sin t ( 1 - 2 cos t) = 0
sin t = 0 or cos t = 1/ 2 --> t = 0 , π , ±1/3π
Same , find dy = 0,
2 2 2
dy = 2 cos t - 2 cos 2t = 0 cost = cos t - sin t =2 cos t - 1
1 ± 3
cos t = -------------- = -1/2 or 1 gives t = 0 , ±2/3π
2*2
by def t = 0 , dx=0 dy=0 , so delete it.
and by the Domain [0,3/2π] we could get the answer
vertical tangent (-3, 0) , ( 3/2 , √3/2)
horizontal tangent (-1/2, 3√3/2)
This is a "cardioid" you can plot it by:
"plot x = 2cost - cos2t and y = 2sint -sin2t"
enter the word above in this site:
http://www.wolframalpha.com/
※ 編輯: rygb 來自: 114.34.122.244 (06/26 00:50)
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