F.4 POLYNOMIALS

a. Let r be the base radius of the cylinder.
Using Pythagora's theorem, r^2 + (x/2)^2 = 10^2
r^2 = 100 - x^2/4

y = (pi)r^2x
= (pi)(100 - x^2/4)x
= 100pi x - pi x^3/4

b. 198pi = 100pi x - pi x^3/4
x^3 - 400x + 792 = 0
(x - 2)(x^2 + 2x - 396) = 0
x = 2 or 18.9(rej.) or -20.9(rej.)
So the height is 2cm.

3a. Let V be capacity of the pool.
For pipe B, flow rate = V/x
For pipe A, flow rate = V/(x + 5)
For pipe C, flow rate = V/(x + 2)
Hence 4(V/x + V/(x + 5) + V/(x + 2)) = V
4(x + 2)(x + 5) + 4x(x + 2) + 4x(x + 5) = x(x + 2)(x + 5)
x^3 - 5x^2 - 46x - 40 = 0

b. Using long division, x^3 - 5x^2 - 46x - 40 = (x - 10)(x^2 + 5x + 4)
= (x - 10)(x + 1)(x + 4) = 0
x = 10 or -1(rej.) or -4(rej.)

2010-12-30 20:28:36 補充：
flow rate x time taken = capacity of the pool

2010-12-30 20:54:06 補充：
因為立體上下,左右都對稱,所以圓柱既中心點同球心係同一點。
從球心畫斜線至圓柱底既圓周上任何一點,當中有3樣野可以知道:
1. 條線向下/向上移左 x/2 (再次重申,圓柱既中心點同球心係同一點)
2. 條線向左/向右移左 r
3. 條線實際長度即球體半徑,10

(a) The radius r = √[100-(x/2)^2]

y=πr^2h=π[100-(x/2)^2]x=100πx-πx^3/4

(b) 198π=100πx-πx^3/4
792π=400πx-πx^3
x^3-400x+792=0
x=2 (There is no reason and the answer is obtained by trial and error)
So, the height is 2 cm


3(a) 4[1/(x+5)+1/x+1/(x+2)]=1
4[x(x+2)+(x+2)(x+5)+x(x+5)]=x(x+2)(x+5)
4[x^2+2x+x^2+7x+10+x^2+5x]=x^3+7x^2+10x
4(3x^2+14x+10)=x^3+7x^2+10x
x^3-5x^2-46x-40=0

(b) 10 is a root. (Again there is no reason) So x^3-5x^2-46x-40=0 becomes
(x-10)(x^2+5x+4)=0
(x-10)(x+1)(x+4)=0
x=10 or -1 or -4
So, B alone takes 10 hours to fill up the pool.



2010-12-30 20:57:20 補充：
因為圓柱體是內嵌於球體。所以理論上你可以將個圓柱體氹氹轉仍然不會有實質上的影響。換句話說圚柱體的軸應該是在球體的正中間。