A.Maths. Questions

細薯

(暫時一條..)
(To be cintinued.......)
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New Trend Additional Mathematics Volume 2 Ex 16A Q.27
Application of Differentiation
P(4,1) is a point on the curve y^2 + y x^1/2 = 3 , where x>0.
(a) Find the value of dy/dx at P.
(b) Find the equation of the normal to the curve ar P.
                                                                                (HKCEE 1995)
my answer :
(a) -1/8
(b) 8x-y-31=0

my book's answer :
(a) -1/6
(b) y=16x - 63

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細薯

Originally posted by 落雷 at 2005-9-10 12:13 AM:

v=2t-4
s=∫(2t-4)dt
=1/2．t^2 - 4t + C

1/2...?
why...[E-FeB]

抹茶可樂

Originally posted by smallpotato226 at 2005-9-9 08:46 PM:
A question again

New Trend Additional Mathematics Volume 2 Ex 17B Q.14
Indefinite Integral and Its Applications

In the question,v,t represent the velocity and time of a particle moving in a  ...

v=2t-4
s=∫(2t-4)dt
=1/2．t^2 - 4t + C

細薯

A question again

New Trend Additional Mathematics Volume 2 Ex 17B Q.14
Indefinite Integral and Its Applications

In the question,v,t represent the velocity and time of a particle moving in a straight line.

If v=2t-4 , find the distance that the particle moves
(a) from t=0 to t=2.
(b) from t=0 to t=5.

I don't know how to do part b.
(red part)

v=2t-4
s=∫(2t-4)dt
=t^2 - 4t + C
When t=0,s=0^2 - 4(0) + C = C

(a)
When t=2 , s = (2)^2 - 4(2) + C = -4 + C
∴Distance = C -(-4 + C) = 4

(b)
When t>2,v>0
When t=5,s=(5)^2 - 4(5) + C = 5 + C
∴Distance = 2(4) + 5 + C - C =13

[ Last edited by smallpotato226 on 2005-9-9 at 09:38 PM ]

抹茶可樂

Originally posted by smallpotato226 at 2005-9-1 05:47 PM:

咁就可以唔理 (p-1)^2 ?

無錯。

細薯

Originally posted by 法皇拿破崙 at 2005-8-29 10:00 PM:
As (p-1)^2≧0, 5-4p≧0
                          p ≦ 5/4

咁就可以唔理 (p-1)^2 ?

法皇拿破崙

As (p-1)^2≧0, 5-4p≧0
                          p ≦ 5/4

細薯

又有唔識．．．

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(節錄)
New Trend Additional Mathematics
Ch.1 Quadratic Equation and Quadratic Functions
P.38

Q.21 (c)
q^2 - 3(p-1)q + (p-1)^2 (p+1) = 0 ........(**)

Find the range of values of p such that the quadratic equation (**) in q has real roots.
                                                         (HKCEE 1989)

Answer p ≦ 5/4

我的steps：
q^2 - 3(p-1)q + (p-1)^2 (p+1) = 0

D ≧ 0

[- 3(p-1) ]^2 - 4(p-1)^2 (p+1) ≧ 0

9(p-1)^2 - 4(p-1)^2 (p+1) ≧ 0

(p-1)^2 [ 9 - 4(p+1)] ≧ 0

(p-1)^2 (5-4p) ≧ 0

∴
p ≧ 1　　　　　　　or　p ≦ 1
p ≦ 5/4　　　　　　　　p ≧ 5/4
1 ≦ p ≦ 5/4　　　　　no solution

1 ≦ p ≦ 5/4

......a little bit different......
am i wrong ?

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oscar

Originally posted by smallpotato226 at 2005-8-28 12:19:
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Oh......=.="
我岩岩睇返我D steps.....



無奈爆><


thx~
我明啦
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=.=
你將4乘4當4加4咁做...
越高年級既學生越唔識得計加減乘除數...

細薯

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Oh......=.="
我岩岩睇返我D steps.....

...
..
.
dy/dx =(-1/2 x^-1/2 y )/ (2y + x^1/2)
dy/dx |(4,1) = (-1/4) / (2+2)
=-1/8

無奈爆><


thx~
我明啦
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