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高中理科生讀A.Maths既人入黎解決我的問題

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1#
發表於 30/8/2004 08:02 PM | 只看該作者 回帖獎勵 |倒序瀏覽 |閱讀模式
我有條有關Differentiation/Integration(微積分)既問題唔係好明。
(由於我用英文學,因此以下問題將會以英文發問。)
1) Find the area enclosed by the curve y=x^3+2x^2-x-2 and the line tangent to the curve at point (-2,0).

2) Find the value of m for which the area enclosed by the parabola y=2x-x^2 and the x-axis is divided into two equal halves by the line y=mx

3) Find the value of a for which the two regions enclosed by the curve y=x(a-x) and the curve y=x^2(a-x), have equal areas. Assume that a>1.
2#
發表於 30/8/2004 10:05 PM | 只看該作者
呢D係Form5 gar數...
我都唔係好識-=-

Definate Itegral...唔易解釋-.-

[ Last edited by 幻將 on 2004-8-30 at 10:57 PM ]
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3#
發表於 30/8/2004 10:41 PM | 只看該作者
先解決第一題


此題同時涉及微分及積分
先微分以找出切線(tangent)
再判斷切線的位置
才用積分計算面積
dy/dx
=3x^2+4x-1
=(x-0.21)(x+1.54)
When x<-1.54,it is increasing
When -1.54<x<0.21,it is decreasing
When x>0.21,it is increasing

The equation of tangent=
y-0 / x-(-2) = 3
y=3x+6

The meeting point
=x^3+2x^2-x-2 - (3x+6)
=x^3+2x^2-4x-8
=(x-2)(x+2)(x+2)
x=-2 OR 2
There are only two meeting points.

On (-2,0)
3(-2)^2+4(-2)-1
=3
dy/dx is increasing on (-2,0),
y=x^3+2x^2-x-2 is below the tangent in enclosed area.

The area=
   2
∫  [(3x+6)-(x^3+2x^2-x-2)]dx
-2
                                         2
=[-(1/4 x^4) -(2/3 x^3) +(2 x^2) +(8 x)]
                                        -2
=64/3

我是升中四的,有錯出聲

[ Last edited by 落雷 on 2004-8-30 at 10:44 PM ]
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4#
 樓主| 發表於 30/8/2004 11:40 PM | 只看該作者
落雷  在 2004-8-30 22:41 發表:

先解決第一題


此題同時涉及微分及積分
先微分以找出切線(tangent)
再判斷切線的位置
才用積分計算面積
dy/dx
=3x^2+4x-1
=(x-0.21)(x+1.54)
When x<-1.54,it is increasing
When -1.54<x<0 ...


嘩...升中四都咁勁?
不過唔該曬...仲有2題...
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5#
發表於 11/9/2004 08:18 AM | 只看該作者
我無一題識。-,-


Warning: Please post some constructive things. If you do not know how to answer it, please do not post anything on that topic. You will be given a serious warning if you do this once again.


rogergeot

[ Last edited by rogergeot on 2004-9-12 at 08:33 AM ]
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