Re: [考古] 中興99年

[rygb (再生)]

※ 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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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.34.122.244
※ 編輯: rygb            來自: 114.34.122.244       (06/25 19:50)
推 john801127:請問如何得知a b需那樣假設
→ rygb:其實想蠻久的
→ rygb:先畫出本來UV區域 應該會多少有靈感
→ rygb:然後接下來是exp(x^2) 之形式通常要換序
推 boss913:謝囉~研究中~~~
→ boss913:最後一題不太會@@變數換完之後就不太懂了
→ rygb:變數變換完 是一般Jacobian 問題
→ rygb:先找出原本r的邊界條件 再去換
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)