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標題: A.Maths. Questions [打印本頁]

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作者: 細薯    時間: 27/8/2005 07:10 PM
標題: A.Maths. Questions
(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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作者: 抹茶可樂    時間: 28/8/2005 12:09 AM
(a) -1/16
(b) y=16x - 63

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作者: 法皇拿破崙    時間: 28/8/2005 12:40 AM
(a) Detailed steps( you better try it yourself)
Diff. both sides w.r.t.x.
y^2+ x^1/2y=3
2y(dy/dx)+ x^1/2(dy/dx)+ (1/2x^-1/2)y=0
                             ( x^1/2+2y)dy/dx=(-1/2x^-1/2)y
                                              dy/dx=(-1/2x^-1/2)y/x^1/2+2y
                                       dy/dx|P(4,1)=-1/2(4)^-1/2(1)/(4)^1/2+2(1)
                                                       =-1/2(1/2)(1)/2+2
                                                       =-1/16

(b) Slope of the normal=16
      Equation of the normal to the curve at P: y-1= 16(x-4)
                                                                   y=16x-63

Note that the answer of your book for part (a) is WRONG. It should be -1/16. I have checked with the solution guide of the HKCEE Additonal Maths.

[ Last edited by 法皇拿破崙 on 2005-8-28 at 12:56 AM ]

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作者: 細薯    時間: 28/8/2005 12:19 PM
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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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作者: oscar    時間: 28/8/2005 12:46 PM

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...

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作者: 細薯    時間: 29/8/2005 04:42 PM
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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作者: 法皇拿破崙    時間: 29/8/2005 10:00 PM
As (p-1)^2≧0, 5-4p≧0
                          p ≦ 5/4

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作者: 細薯    時間: 1/9/2005 05:47 PM

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 ?

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作者: 抹茶可樂    時間: 1/9/2005 07:37 PM

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

咁就可以唔理 (p-1)^2 ?

無錯。

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作者: 細薯    時間: 9/9/2005 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 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 ]

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作者: 抹茶可樂    時間: 10/9/2005 12:13 AM

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

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作者: 細薯    時間: 10/9/2005 09:51 AM

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]

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