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標題: 高中理科生讀A.Maths既人入黎解決我的問題 [打印本頁]

作者: oscar 時間: 30/8/2004 08:02 PM
標題: 高中理科生讀A.Maths既人入黎解決我的問題
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.

作者: 霏 時間: 30/8/2004 10:05 PM
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 ]

作者: 抹茶可樂 時間: 30/8/2004 10:41 PM
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 ]

作者: oscar 時間: 30/8/2004 11:40 PM

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...

作者: lkw922 時間: 11/9/2004 08:18 AM
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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