数学Paper 1讨论专区

Allmaths

HELP ME ~~
∫(2sin x + 3cos x )/(3sin x + 2 cos X) dx
Minute 发表于 16-2-2011 03:17 PM

   
自己查查看有没有错。。。

Allmaths

可以给一点提示吗?
2)If pq is a chord which is a normal at point (a,2a) ,which lies on the parabola y^2=4ax,find the length of pq.find also the area of the triangle formed by the chord and the tangents at p and q.
long_sign 发表于 16-2-2011 08:34 PM

        y^2=4ax
2y(dy/dx)=4a
   (dy/dx)=2a/y

At the point P(a, 2a), dy/dx=2a/2a
                             dy/dx=1
Normal to the tangent, m1m2=-1
                                     m2=-1
Eq of normal at the point P(a , 2a), y-2a=-(x-a)
                                                     y=3a-x   ---eq 1
Sub eq 1 into y^2=4ax, (3a-x)^2=4ax
                         9a^2-6ax+x^2=4ax
                        x^2-10ax-9a^2=0
                              (x-a)(x-9a)=0
                            x=a   or   x=9a
From eq 1, y=3a-9a
                y=-6a

Coordinate of point Q(9a , -6a)
Length of PQ=√[(9a-a)^2+(-6a-2a)^2]
                  =8(√2)a

At point P(a ,2a), dy/dx=1
Eq of tangent at point P, y-2a=x-a
                                        y=x+a   ---eq 2
At point Q(9a ,-6a), dy/dx=-1/3
Eq of tangent at point Q, y+6a=(-1/3)(x-9a)
                                  3y+9a=-x
Sub eq 2,               3(x+a)+9a=-x
                                         x=-3a
From eq 2, y=a-3a
                y=-2a

Both equation of tangent intersect at the point (-3a , -2a).

Area of triangle=(1/2)|-64a^2|
                     =32a^2 unit^2

Allmaths

可以给一点提示吗?
3)The line y-2x+4a=0 intersects the parabola y^2=4ax at the point p(ap^2,2ap) and the point q(aq^2,2aq).find the value of p+q and pq. Hence ,find the coordinates of the mid-point of pq.
long_sign 发表于 16-2-2011 08:34 PM

y=2x-4a  ---eq 1
Sub eq 1 into y^2=4ax,
           (2x-4a)^2=4ax
          (x-a)(x-4a)=0
          x=a   or   x=4a
When x=a, y=-2a
When x=4a, y =4a

Midpoint of point P and Q, ((ap^2+aq^2)/2 , ap+aq) is the same as the midpoint of point (a, -2a) and (4a , 4a)
Midpoint of point (a, -2a) and (4a , 4a) is ( 5a/2 , a)

Comparing both coordinates, (ap^2+aq^2)/2=5a/2
                                                 p^2+q^2=5  ---eq 2
                                                     ap+aq=a
                                                         p+q=1
From eq 2,    p^2+q^2=5
              (p+q)^2-2pq=5
Sub p+q=1,    1^2-2pq=5
                             pq=-2
∴p+q=1 , pq=-2

Coordinate of midpoint PQ is ( 5a/2 , a)

hongji

1）the point P has coordinates (2,-5).the straight line 3x+4y=36 cuts the x-axis at R and the y-axis at S.find the area of the triangle PRS.　

peaceboy

回复 2524# hongji


   
1）the point P has coordinates (2,-5).the straight line 3x+4y=36 cuts the x-axis at R and the y-axis at S.find the area of the triangle PRS.　

3x+4y=36

R , y=0
3x=36
x=12
R (12,0)

S, x=0
4y=36
y=9
S, (0,9)

P (2,-5) R (12,0) S, (0,9)
自己用formula

yingchin

The lines y=2X and y = X intersect the curve ysquare +7xy = 18 at points A and B respectively. Where the A and B lie on the 1st quadrant.
(b) Calculate the perpendicular distance from A to OB, Where O is the Origin.


O(∩_∩)O谢谢

peaceboy

回复 2526# yingchin


    The lines y=2X and y = X intersect the curve ysquare +7xy = 18 at points A and B respectively. Where the A and B lie on the 1st quadrant.
(b) Calculate the perpendicular distance from A to OB, Where O is the Origin.

y^2 + 7xy = 18
y=2x
4x^2 + 14x^2 =18
18x^2 =18
x=1
y=2
A(1, 2)

y=x
x^2+7x^2 =18
8x^2=18
x^2 = 9/4
x= 3/2
y=3/2
B(3/2 , 3/2)

mOB = (3/2 - 0) / (3/2-0)
        =1
OB , y=x
x-y=0
A(1, 2)
perpendicular distance= l (1 - 2) / (1+1)^(1/2) l
                               = (1) / [2^(1/2) ]
                                 = [2^(1/2) ] / 2

peaceboy

help!
long_sign 发表于 8-3-2011 11:01 PM

    什mok事情？

long_sign

long_sign 发表于 8-3-2011 11:17 PM

wah !真的很难弄弄嘞.........
可以教教我如何expand .......thank !

peaceboy

(1/3)^(3r-2) + (1/3)^(3r-1) = (1/3)^(3r) / (1/3)^(2)  +  (1/3)^(3r)/(1/3)
                                          = (1/27^r) / (1/9) + (1/27^r) / (1/3)
                                         = 9/27^r + 3/27^r
                                         =12/27^r
sum 12/27^r , r=1 to n
a= 12/27
r=1/27
using formula ,
(12/27) (1- 1/27^n) / (26/27) = (8/13)(1- 1/27^n)

sum to infinity 自己用formula找

ultikiller

如何转换？

ultikiller

还有第二题。

peaceboy

回复 2531# ultikiller


    partial fraction 啊

let 1/(3+t)(3-t) = a/(3+t) + b/(3-t)
                      = [a(3-t) + b(3+t)] / (3+t)(3-t)
by comparison
[a(3-t) + b(3+t)]  = 1
when t = 3 , 6b = 1
                   b= 1/6
when t=-3 , 6a =1
                   a= 1/6
therefore,
let 1/(3+t)(3-t) = 1/6(3+t) + 1/6(3-t)
                       =(1/6)[1/(3+t) + 1/(3-t)]

peaceboy

回复 2532# ultikiller


    let u = 3 + 2 sin x
x=0 , u=3
x= pi/6 , u=4
du/dx = 2 cos x
dx = du/2cos x

int cos x / (3+2sin x)^2 dx =int cos x / 2u^2  (du/2cos x) _u from 3 to 4
                                      = int 1/2u^2 du _ u from 3 to 4
                                       = [- 1/2u]_u from 3 to 4
                                      = -1/2(4) - -1/2(3)
                                       = -1/8 + 1/6
                                      = 1/24

blazex

Each of the surfaces of a cube is expanding at the rate of 10cm^2 per second. find the rate of change of the volume of the cube when the surface area is 150cm^2.

whyyie

回复 2535# blazex

Let x be the side length of the cube
surface area of one side of the cube, A = x^2
dA/dt = (dA/dx) x (dx/dt)
A=x^2
dA / dx = 2x
dA/dt = 2x (dx/dt)
10 = 2x (dx/dt)
dx/dt = (5/x) cm/s

Let V be the volume of the cube
V = x^3
dV/dt = 3x^2 (dx/dt) = (3x^2) (5/x) = 15x

Total surface area = 6x^2
150 = 6x^2
x = 5 cm

At x = 5,
dV/dt = 15(5) = 75 cm^3 / s

blazex

1)A right circular cone has base radius,r and height,h. As r and h vary, its curved  
   area is kept constant. Show that its volume is a maximum when r=rh(2^1/2).
2)A curve has parametric equation x=2t,y=1/t^2,where t is a real non-zero
   parameter.
   Find the equations of the tangent and the normal to the curve at the point  
   P(2t,1/t^2). The tangent at point Q(-2,1) meets the y-axis at T.The normal at
   P meet the y-axis at N. Show that the area of triangle QTN is 4 units^2.

peaceboy

回复 2537# blazex

很可怜，米有人要solve
我很懒惰

    1)A right circular cone has base radius,r and height,h. As r and h vary, its curved  
   area is kept constant. Show that its volume is a maximum when r=rh(2^1/2).

its curved area is kept constant , A= pi r S
s=(r^2 + h^2)^(1/2)
A= pi r (r^2 + h^2)^(1/2)
r^2 + h^2 = [A/ (pi r)]^2
h^2 = [A^2 - (pi)^2 r^4] / (pi r)^2
h=[A^2 - (pi)^2 r^4]^(1/2) / (pi r)
pi r h = [A^2 - (pi)^2 r^4]^(1/2) ---@

V= (1/3) pi r^2 h
  =(1/3) (r) [A^2 - (pi)^2 r^4]^(1/2)
dv/dr= (r/3) [1/[A^2 - (pi)^2 r^4]^(1/2)] (-4 (pi)^2 r^3) + (1/3) [A^2 - (pi)^2 r^4]^(1/2)
sub @
dv/dr = (r/3)(1/2) [1/pi r h] (-4pi^2 r^3) + (1/3) pi r h
        = (pi r h) /3 - 2 (pi)^2 r^3 / (3 pi  h)
         = (pi^2 r h^2 - 2 (pi)^2 r^3) / 3pi h
dv/dr =0
(pi^2 r h^2 - 2 (pi)^2 r^3) / 3pi h=0
pi^2 r h^2 - 2 (pi)^2 r^3=0
rpi^2[h^2 -2  r^2]=0
r pi^2 =0 ,  h^2 -2r^2=0
h^2 -2r^2=0

2r^2=  h^2
r=  h / (2)^(1/2)

第二等我的空才作

hongji

y=[4/square root(x)] use by first principle ,solve

let f(x) =4/square root(x)                      8=delta
  f(x+8x)=4/[square root(x+8x)]
  f'(x)=lim    {4/[square root (x+8x)  -  4/[square root(x)]  }  / 8x
           8-0
       =lim  (4/[square root(x)]   - 4/[square root(x)]  )  (1/8x)
         8-0
      =lim  {4(squareroot (x)  -  4(square root(x+8x) }  /  (squareroot (x+8x)(squareroots(x)(8x)

then?
TT please can teach me ,tq
       8-0

Allmaths

y=[4/square root(x)] use by first principle ,solve

let f(x) =4/square root(x)                     ...
hongji 发表于 29-3-2011 09:18 PM