Monday, May 4, 2009

add math folio for form5 ! ..have a look it Answer !!!!

想必各位的学校已经分题目了吧?我问老师了,老师说全国的题目都一样的,所以呢,谁做了就别那么吝啬,来分享一下
哈哈,老实说啦,我不会做啊,所以等着抄答案。。。。。。。

以下是我们论坛的会员jckkau所提供的,希望大家在抄的同时,向这位会员说声谢谢哦~~

b)
Pi or π is a mathematical constant whose value is the ratio of anycircle's circumference to its diameter in Euclidean space; this is thesame value as the ratio of a circle's area to the square of its radius.It is approximately equal to 3.14159 in the usual decimal notation (seethe table for its representation in some other bases). π is one of themost important mathematical and physical constants: many formulae frommathematics, science, and engineering involve π.

Part 2
a)
d1 (cm) d2 (cm) Length of arc PQR in terms of π (cm)÷2 Length of arc PAB in terms of π (cm) ÷2 Length of arc BCR interms of π (cm) ÷2
1 9 10 π π 9 π
2 8 10 π 2 π 8 π
3 7 10 π 3 π 7 π
4 6 10 π 4 π 6 π
5 5 10 π 5 π 5 π
6 4 10 π 6 π 4 π
7 3 10 π 7 π 3 π
8 2 10 π 8 π 2 π
9 1 10 π 9 π 1 π
10 0 10 π 10 π 0

b i)
d1 (cm) d2 (cm) d3 (cm) Length of arc PQR in termsof π (cm) ÷2 Length of arc PAB in terms of π (cm) ÷2 Length of arc BCD in terms of π (cm) ÷2 Length of arc DER interms of π (cm) ÷2
1 1 8 10 π π π 8 π
1 2 7 10 π π 2 π 7 π
1 3 6 10 π π 3 π 6 π
1 4 5 10 π π 4 π 5 π
1 5 4 10 π π 5 π 4 π
1 6 3 10 π π 6 π 3 π

part 3
(a) y = 1/2πr ^2 - 1/2 πr ^2 - 1/2 π r ^2
= 1/2 π ( 5^2 - ( x/2)^2 - ( (10- x/2 ) )^2 )
= 1/2 π ( 25 - x62/4 - (100-20x +x^2)/4 )
=( 20x -2x^2 ) / 8 π m^2
=( 10x -x^2 ) / 4 π m^2


(b) y = ( 10x -x^2 ) / 4 π m^2 = 16.5
x = 3 or 7

c) y = ( 10x -x^2 ) / 4 π m^2
y = ( 5/2 x - x2 / 4 ) π m^2
÷ x
y/x = 5/2 – 1/4 x
In the form of Y = mX + C
y/x = – 1/4 x + 5/2
y – axis = y/x
x – axis = x ( in π m^2 )
please plot the graph yourself.

d) y = ( -1/4x2+ 2/5x)π
By using differentiation
dy/dx = -1/2 x+ 5/2
dy/dx = 0
-/2 x+ 5/2 = 0
1/2 x=5/2
x = 5
Therefore
Maximum area for y when x = 5
ymax = 5π/2 (5) – π (5)2 / 4
= 19.634 m2

e) 12 semicircular flower beds, it form an arithmetic progression
S12 = 10 m = 1000 cm
a = 30 cm
Sn = n/2 [ 2a + (n-1) d ]
S12 = 12/ 2 [ 2 (30) + (12 – 1) d]
1000 = 6 ( 60 + 11d)
1000 = 360 + 66d
66d = 640
d = 9.7 cm
Number of semicircular flower beds Diameter for the flower beds / cm
1 30
2 39.7
3 49.4
4 59.1
5 68.8
6 78.5
7 88.2
8 97.9
9 107.6
10 117.3
11 127
12 136.7
Total diameter = 1000 cm

由jckkau提供。。。



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PART 1
http://en.wikipedia.org/wiki/Pi

PART 2
(a) arc PQR = arc PAB + arc BCR (From the table)

(b) (i) Tabulate (the same way as (a))

arc PQR = arc PAB + arc BCD + arc DER

(ii) length of the arc of the outer semicircle = total length of arcs of n inner semicircles

(c) let diameter of the outer semicircle PQR = D (where D is any real number greater than 0)

there are n inner semicircles where n = 2, 3, 4...
dn = diameter of nth inner semicircle
D = d1 + d2 + ... + dn

arc PQR = (pi)(D)
arc PQR = (pi)(d1 + d2 + ... + dn)
arc PQR = (pi)(d1) + (pi)(d2) + ... + (pi)(dn)
arc PQR = arc of 1st inner semicircle + arc of 2nd inner semicircle + ... + arc of nth inner semicircle

therefore, for any value of diameter of outer semicircle,

length of the arc of the outer semicircle = total length of arcs of n inner semicircles

PART 3
(a) y = (pi/2)(10/2)^2 - (pi/2)(x/2)^2 - (pi/2)((10 - x)/2)^2
y = (pi/8)(10^2 - x^2 - (10 - x)^2)
y = (pi/8)(10^2 - x^2 - (10^2 - 20x + x^2))
y = (pi/8)(10^2 - x^2 - 10^2 + 20x - x^2)
y = (pi/8)(10^2 - x^2 - 10^2 + 20x - x^2)
y = (pi/8)(20x - 2x^2)

y = (pi/4)(10x - x^2)

(b) substitute y = 16.5,
16.5 = (pi/4)(10x - x^2)
16.5(4/pi) = 10x - x^2
16.5(4)/(22/7) = 10x - x^2
21 = 10x - x^2
x^2 - 10x + 21 = 0
(x -3)(x - 7) = 0
x = 3 or 7


therefore, the diameters of the two ponds are 3m and 7m.

(c) y = (pi/4)(10x - x^2)
(4/pi)y = 25 - 25 + 10x - x^2
(4/pi)y = 25 - (x - 5)^2

let Y = (4/pi)y, X = (x - 5)^2
Y = 25 - X (Draw a straight line graph Y versus X)
x = 4.5, X = 0.25 (find Y from the graph using X = 0.25)
Y = 24.7 or 24.8 (exact value = 24.75, but we are not able to see it accurately from the graph)
(4/pi)y = 24.7 or 24.8
y = 19.4 or 19.5

The area of the flower plot = 19.4 or 19.5 m^2 when the diameter of one of the fish pond is 4.5m.

(d) method I
y = (pi/4)(10x - x^2)
dy/dx = (pi/4)(10 - 2x)

when dy/dx = 0, (pi/4)(10 - 2x)=0
x = 5
d^2y/dx^2 = (pi/4)(-2) <>

maximum value of y = (pi/4)(10(5) - 5^2) = 19.64
therefore, at x = 5, y = 19.64 is maximum and the cost of construct the garden is minimum.

the area of the flower plot = 19.64 m^2


method II
the construct of the garden is minimum when y is maximum.
Y = (4/pi)y is also maximum.
0 <>
-5 <>
| x - 5 | <>
0 <= (x - 5)^2 <>
0 <= X <>
from graph Y versus X, maximum value of Y = 25 when X = 0.
maximum value of y = (pi/4)(25) = 19.64

the area of the flower plot = 19.64 m^2

(e) arithmetic progression (a = 30, d = common difference)
sum of 12 diameters = 1000 cm
n/2(2a + (n - 1)d) = Sn
12/2(60 + 11d) = 1000
360 + 66d = 1000
66d = 640
d = 320/33

diameter of 2nd flower bed = (30 + 320/33) cm

diameter of 3rd flower bed = (30 + 2(320/33)) cm

diameter of 4th flower bed = (30 + 3(320/33)) cm

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