数学Paper 1讨论专区

hongji

Allmaths 发表于 30-3-2011 07:29 PM

    如果不是用first principle
    而是用f'(x)的话
    答案是不一样的吗？


f(x)= [4/(squareroot(x) ]
f'(x)=4 x^(1/2)
      =-2x^(-3/2)
     =-2 / (squareroot(x^3)

Allmaths

如果不是用first principle
    而是用f'(x)的话
    答案是不一样的吗？


f(x)= [4/(squ ...
hongji 发表于 30-3-2011 10:43 PM

   
一样。。

x√x=x.x^(1/2)
     =x^(1+1/2)
     =x^(3/2)

hongji

一样。。

x√x=x.x^(1/2)
     =x^(1+1/2)
     =x^(3/2)
Allmaths 发表于 30-3-2011 11:53 PM

    oo明白了

hongji

find the equation of the tangent to the curve xy=4 at point P(4,1).
Point A is a point on the x-axis such that PA is parallel to the y-axis.Tangent to curve xy=4 at P meets the y-axis at point B.The straight line passing through B and parallel to the x-axis meets the curve at point Q.Find the coordinates of Q and show that AQ is a tangent to the curve at Q.Find the coordinates of the point of intersection of the tangents to the curve xy=4 at P and Q.

从哪里开始做起。。？equation?

jessica92

有谁懂为什么
6sin 2x cos 2x=3sin 4x

Log

回复 2545# jessica92
6sin 2x cos 2x=3sin 4x


  6 sin 2x cos 2x= 3 ( 2 sin 2x cos 2x)= 3 sin4x

double angle formula---> sin 2x = 2 sin x cos x
                                   sin 2(2x) = 2 sin (2x) cos (2x)
                                   sin 4x = 2 sin 2x cos2x

Log

回复 2537# blazex
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
   Q meet the y-axis at N. Show that the area of triangle QTN is 4 units^2.  ( have u type wrong? should be normal at Qif not,it is failed to show  that triangle QTN is 4 units^2  )


  x=2t                         y=1/t^2
dx/dt=2                       dy/dt =-2/t^3
dt/dx =1/2
gradient  function tangent,  m=dy/dx= (dy/dt) (dt/dx) =-1/t^3
gradient function normal ,   m' = -1/m = t^3

eq of tangent at P -->  y-y_1=m(x-x_1)
                                 y-1/t^2 = (-1/t^3) (x-2t)
arrange , get               yt^3+x-3t=0-------(1)

eq of normal to P-->     y-y_1=m'(x-x_1)
                                  y-1/t^2=(t^3)(x-2t)
arrange , get               yt^2- xt^5+ 2t^6 -1=0----(2)


then,
eq  of  tangent at Q(-2,1)  is                          (where its parameter t=-1   coz x=2t-->-2=2t so t=-1)
                                       
insert t=-1 into (1) , so get -y+x+3=0
at y-axis, x=0,  -y+0+3=0
                        y=3
so,                   T=(0,3)


eq of normal to Q is
insert t=-1 into (2) , get   y+x+1=0
at y-axis , x=0, y+0+1=0
                       y=-1
so                    N=(0,-1)

then,u skecth 3 points triangle QTN
since TN lie on same y-axis so assumed as base and distance of Q to line TN so considered as height
thus area of triangle QTN = 1/2 (height)(base)
                                     =1/2 (distance Q to line TN)(TN)
                                     =1/2 (2)(3-(-1))
                                      = 4 units^2

hongji

TT
differentiate 我只会expand到一半就TT

find dy/dx

1)y=[√x ] [(2x-1)^2]
let y=[(x)^1/2] [(2x-1)^2]
   =[(2x-1)^2] [(1/2)(x)^-1/2]   +   [√x][(2)(2x-1)(2)]
   =[(1/2√x)(2x-1)^2   +   (√x)(4)(2x-1)

然后该怎样？



2)y=[(3x-2)^2] / [√(1+x)^2]
   
u=(3x-2)^2                v=(1+x^2)^1/2
du/dx=2(3x-2)(3)       dv/dx=[1/2][(1+x^2)^-1/2][(2x)]

dy/dx= {6(3x-2)[(1+x^2)^1/2]    +   x[(1+x)^-1/2] [(3x-2)^2  } / (1+x^2)

然后又接不下去了TT
帮帮忙><

Log

回复 2548# hongji
1.y=[√x ] [(2x-1)^2]
let y=[(x)^1/2] [(2x-1)^2]
   =[(2x-1)^2] [(1/2)(x)^-1/2]   +   [√x][(2)(2x-1)(2)]
   =[(1/2√x)(2x-1)^2   +   (√x)(4)(2x-1)]
  then, factorise
   = 1/(2√x). (2x-1)  . [ (2x-1) + 8x ]
   =  1/(2√x). (2x-1)  .(10x-1)
   =(10x-1)(2x-1)/(2√x)
  

2.)y=[(3x-2)^2] / [√(1+x)^2]
   
u=(3x-2)^2                v=(1+x^2)^1/2
du/dx=2(3x-2)(3)       dv/dx=[1/2][(1+x^2)^-1/2][(2x)]

dy/dx= 【6(3x-2)[(1+x^2)^1/2]    +   x[(1+x)^-1/2] [(3x-2)^2] 】/ 【1+x^2】
then, 分子及分母各乘 (1+x&#178; )^1/2
get     = 【6(3x-2)(1+x&#178; ) - x(3x-2)&#178;】/ 【(1+x&#178; )^3/2】
          =  【(3x-2)[6(1+x&#178; ) - x(3x-2)]】/【(1+x&#178; )^3/2】
          =   【(3x-2)(3x&#178;+2x+6)】/【(1+x&#178; )^3/2】

Log

回复 2544# hongji
skecth the graph and all the points so u can see their position clearly
xy=4
y=4/x
m= dy/dx=-4/x^2
find dy/dx then u noe can find its gradient of tangent at P
at P(4,1), m= dy/dx = -4/4^2 =-1/4
so eq of tangent at P is
y-1=m(x-4)
.....
∴ 4y+x=8------>(1)
then find coordinate B
at y-axis, x=0, 4y+0=8
                        y=2
              so     B=(0,2)
since straight line passing through B and parallel to the x-axis meets the curve at point Q that means y-coordinate of B= y-coordinate of  Q
Q=(x,2)
Q lies on the curve xy=4
x(2)=4
x=2
∴Q=(2,2)

as A is a foot of line PA on x-axis, so A=(4,0)

then,  find gradient line AQ ( use  m=y1-y2 / x1-x2) and its eq  and find gradient tangent at Q using  dy/dx as well its eq too
if their eq same, so AQ is tangent to curve at Q
eq : y+x=4----(2)
point of intersection:  solve   (1) and (2) simultaneously

hongji

回复 2549# Log


    thank you 看到你的作法我应该会expand下去了

A curve has parametric equations x=2cos 2 t and y=5 sin t.
find dy/dx when t=TT / 2

whyyie

回复 2551# hongji

dx/dt = - 4 sin 2t;   dy/dt = 5 cos t
dy/dx = 5 cos t/ - 4 sin 2t
         = 5 cos t/ - 4 (2 sin t cos t)
         = -5 / 8 sin t
At t = pi/2,
dy/dx = - 5/8

hongji

我有几题又expand到一半就？？

y=(1/√x  - √x )^2
let y=([x^-1/2] - [x^1/2])^2  是这样吗<<还是 把它分开(1/√x - √x)(1/√x - √x)
  dy/dx= 2([x^-1/2]  - [x^1/2]) ([-1/2x^3/2] -[1/2 x^-1/2])
      然后不会接下去TT




还有另一题和答案不一样
y=[√3x^2-2] /[x-3]

u= (3x^2 -2)^1/2                                   v=x-3
du/dx=(1/2)[(3x^2 -2)^-1/2](6x)                 dv/dx=1

dy/dx=(x-3)(1/2)[(3x^2-2)^-1/2] (6x) - [(3x^2-2)^1/2] / [(x-3)^2]
then 分子分母*(3x^2-2)^1/2
=[(x-3)(1/2)(6x)] - [(3x^2-2)^1/2]   / [(x-3)^2][√(3x^2-2)^1/2]
=-9x+2 /  [(x-3)^2][√(3x^2-2)^1/2]
但是答案是[(3x^4) -(9x^3) -(5x^2) +6x + 2 ]/  [(x-3)^2][√(3x^2-2)^1/2]
能告诉我错在哪里吗
谢谢。。

hongji

[√3+x ] [[√(1+x)^3] find dy/dx

long_sign

对不起，我有问题要问....differentiation 的.........
1)A solidcircular cylinder has a given volume .show that its total surface area will beleast when its height is equal to diameter of the base.
2)

Lov瑜瑜4ever

a) let P=pi
V=(P)(r^2)h , so h=V/[(P)(r^2)]
A=2(P)(r^2)+2(P)rh
  =2(P)(r^2)+2(P)r[V/[(P)(r^2)]]
  =2Pr^2+2V/r
  =2Pr^2+2Vr^-1
dA/dr=4Pr-2V/r^2
d^2A/dr^2=4P+4V/r^3
when dA/dr=0,
        4Pr=2V/r^2
         r^3=V/2P
           r=(V/2P)^1/3
when r=(V/2P)^1/3, d^2A/dr^2=12P>0
so A is minimum when r^3=V/2P
r^3=V/2P
r^3=P(r^2)h/2P , V=P(r^2)h
   r=h/2
   h=2r
   h=d (shown)

Lov瑜瑜4ever

对不起，我有问题要问....differentiation 的.........
1)A solidcircular cylinder has a given volume . ...
long_sign 发表于 8-4-2011 11:43 PM

我想问下。。。
y=4+36/x
这个对吗？

long_sign

回复 2557# Lov瑜瑜4ever


答安是这个.但4少了个x.
你是如何做的?我的答安跑出

x^-2

麻烦show 一下,我想compare my answer  

Lov瑜瑜4ever

回复  Lov瑜瑜4ever


答安是这个.但4少了个x.
你是如何做的?我的答安跑出
x^-2
麻烦show 一下,我想 ...
long_sign 发表于 9-4-2011 10:32 PM

那么我的答案对吗？
还是是xy=4x+36哦？

long_sign

对是这个xy=4x+36