数学Paper 1讨论专区

Furudesuya

回复 1559# 强


    哇！！“强”已跑到这里来宣传了~~~

peaceboy

1 A right circular cone of height h is inscribed in a sphere of radius R. Show that the volume V of  ...
芭樂 发表于 20-8-2010 09:16 PM

   1. draw a triangle from a circle , with center 0 , take r as hypotenuse , l h-r l = one of the side ,

therefore , another side =[ r^2 - （h-r）^2 ]^(1/2)= [r^2 - h^2 +2hr - r^2]^(1/2)
                                                           = [2hr- h^2]^(1/2)<<< the r of the cone

so cone formula =  (pie/3)r^2h
                       =  (pie/3)[2hr- h^2]h
                       = (pie/3)(2rh^2-h^3)


2 A right circular cone of base radius r and height h has a totalsurface area S and volume V.Show that 9V^2 = r^2(S^2-2pier^2S). Henceor otherwise, show that for a fixed surface area S, the maximum volumeof the cone occurs when its   semi-vertical angle A is given that bytan A = 8^-.5

S= pi *r*(h^2+r^2)^(1/2)  + pi * r^2

S- pi * r^2 =  pi *r*(h^2+r^2)^(1/2)
S^2 + pi^2 * r^4 -2Spi * r^2=  pi^2 *r^2 *(h^2+r^2)
                                         =pi^2 *r^2 *h^2 + pi^2 *r^4

S^2-2S*pi * r^2 = pi^2 *r^2 *h^2 ---1


V = 1/3 pi r^2h
V^2 = 1/9 (pi r^2h)^2
9V^2 = (pi r^2h)^2
        = r^2(pi^2 r^2h^2)
sub back eqn 1
        =r^2(S^2-2S*pi * r^2 )


find dv/dr = 0 (S is constant)
然后angel A不懂在那里...

3 A right pyramid has a square base of side x m and a total surfacearea (base and four sides) 72m^2. Show that the volume, Vcm^3, is givenby  v^2 = 144x^2 - 4x^4. If x varies, find the value of x for which Vis a maximum and obtain the maximum values of V.


xm 是square的side的长？

芭樂

回复 1562# peaceboy


1）h-r 那一part不同
3）对

peaceboy

回复 1563# 芭樂

1. 什么不同？

    3 A right pyramid has a square base of side x m and a total surfacearea(base and four sides) 72m^2. Show that the volume, Vcm^3, isgivenby  v^2 = 144x^2 - 4x^4. If x varies, find the value of x forwhich Vis a maximum and obtain the maximum values of V.

72m^2= (xm)^2 + 4(xm/2)(h^2+(xm/2)^2)^(1/2)
72m^2  - (xm)^2    =2xm(h^2+(xm/2)^2)^(1/2)
36m/x - xm/2 = (h^2+(xm/2)^2)^(1/2)
(36m/x - xm/2)^2 = h^2+(xm/2)^2
1296m^2/x^2 - 36m^2 + (xm/2)^2 = h^2+(xm/2)^2
h^2=1296m^2/x^2 - 36m^2

V= (1/3)(xm)^2 (h)
V^2 = (1/9)(xm)^4 (h)^2
       =(1/9)(xm)^4(1296m^2/x^2 - 36m^2 )
       = 144x^2m^6 - 4x^4 m^6
       = (m^6)(144x^2 - 4x^4)

最多show到这边

芭樂

回复 1564# peaceboy


    写错，是不明白

peaceboy

回复  peaceboy


    写错，是不明白
芭樂 发表于 21-8-2010 02:02 PM

   

画到很白痴
因为要找cone,所以我们要先找cone的radius
所以我们知道cone在sphere 里面
所以蓝色的部分就是cone的r
深青的是sphere的R
如果我们蓝线的长度，就拿[（深青）^2 - （粉红）^2 ]^(1/2)
深青=R
不过粉红我们不懂
所以我分了2个可能性
第一，h<R
所以粉红就等于R-h

第二h > R
所以粉红就等于
h-R

不过为了方便，所以把两个情况结合成 l h-R l , 因为square掉它后答案都一样的....

whyyie

Razor_1130

SHOW THAT THE EQUATION OF THE TANGENT TO THE PARABOLA
Y^2 = 4ax at P(at^2,2at) is ty=x + at^2

find the coordinates of the midpoint of PT in terms of a and t

show that for all value of t , this midpoint lies on the curve y^2(2x+a) = a(3x+a)^2

后面的show不会。。拜托各位下



the roots of the equation x^3 -3x^2 -3x -7 = 0 are p,q,r
find the value of p^2 + q^2 + r^2


先说谢谢各位。。拜托下

Allmaths

the roots of the equation x^3 -3x^2 -3x -7 = 0 are p,q,r
find the value of p^2 + q^2 + r^2
Razor_1130 发表于 23-8-2010 03:18 PM

Allmaths

SHOW THAT THE EQUATION OF THE TANGENT TO THE PARABOLA
Y^2 = 4ax at P(at^2,2at) is ty=x + at^2

s ...
Razor_1130 发表于 23-8-2010 03:18 PM

你的coordinate geometry 的题目应该是不完整吧...

Razor_1130

你的coordinate geometry 的题目应该是不完整吧...
Allmaths 发表于 23-8-2010 04:49 PM

    少抄了一句。。补上去了

Razor_1130

Allmaths 发表于 23-8-2010 04:15 PM

    完全明白！谢谢指导哦

Allmaths

完全明白！谢谢指导哦
Razor_1130 发表于 23-8-2010 06:22 PM

好像还少了些info...
是STPM past year 1982的吗?

kelfaru

whyyie 发表于 23-8-2010 02:06 PM

有 final answer 吗???我做到的答案很奇怪...==

Allmaths

SHOW THAT THE EQUATION OF THE TANGENT TO THE PARABOLA
Y^2 = 4ax at P(at^2,2at) is ty=x + at^2

f ...
Razor_1130 发表于 23-8-2010 03:18 PM

whyyie

回复 1574# kelfaru

我没有答案..

kelfaru

回复 1576# whyyie


请问你用什么方法???

peaceboy

whyyie 发表于 23-8-2010 02:06 PM

    确定下，问题有没有写错？

Lov瑜瑜4ever

(m^2-3m-4)x^2+(m^2+2)x+12=0
A=alpha ; B=Beta
A<-1<B
A+1<0<B+1
so, A+1<0 and B+1>0

(A+1)(B+1)<0
AB+(A+B)+1<0
[12/(m^2-3m-4)]-[(m^2+2)/(m^2-3m-4)]+[(m^2-3m-4)/(m^2-3m-4)]<0
(m^2-3m-4)(12-m^2-2+m^2-3m-4)<0
(m-4)(m+1)(-3m+6)<0
(m-4)(m+1)(m-2)>0
so, m>4 ; -1<m<2

这是我的做法啦
不懂对不对><

peaceboy

(m^2-3m-4)x^2+(m^2+2)x+12=0
A=alpha ; B=Beta
A
Lov瑜瑜4ever 发表于 24-8-2010 03:04 PM

    厉害