急 ! F5 變分計法 7q03

16.
(a)
C ∝ 1/n
Then C = k/n
where k is constant.

When n = 3, C = 100 000 :
100 000 = k/3
k = 300 000

Hence, C = 300 000/n

(b)
When n = 5 (in 2010) :
C = 300 000/5
C = 60 000

Value of the car in the year 2010 = $60 000

(c)
50 000 = 300 000/n
n = 6
2008 - 3 + 6 = 2011

It is in the year of 2011.


18.
(a)
V ∝ r²h
Then V = kr²h
when k is constant.

When r = 1 and h = 3, V = π :
π = k(1)²(3)
k = π/3

V = (π/3)r²h

(b)
V ≥ 3π
(π/3)(1.5)²h ≥ 3π
h ≥ 4

Minimum height = 4 cm


20.
F ∝ Mn/d²
Then, F = kMn/d²
where k is constant.

When M = Mo, m = mo and d = do, F = Fo:
Fo = kMomo/do² ...... [1]

When M = Mo(1 + 21%), m = mo(1 - 19%) aand F = Fo:
Fo = kMo(1 + 21%)mo(1 - 19%)/d² ...... [2]

[1] = [2] :
kMomo/do² = Fo = kMo(1 + 21%)mo(1 - 19%)/d²
1/do² = (1.21)(0.81)/d²
d² = 0.9801do²

Since d > 0 and do > 0, then
d = 0.99do

% change in distance
= [(0.99do - do)/do] x 100%
= -1%

16a)
C = k/n

16b)
100 000 = k/3
k = 300 000

C = 300 000/2 = 150 000

16c)
50 000 = 300 000/n
n = 6 .Therefore , the value of the car will be $50000 in 2014

18a)
V = k(r^2)h
pi = k(1^2)(3) => k = pi/3

V = (pi)(r^2)h/3

18b)
3(pi) >= (pi)(1.5^2)h/3
h >= 4 , the min. of height is 4cm

20)
F = k*mn/d^2

((Since the force unchange , the percentage equals to 0)

k*(1.21)m(0.81)n/(xd)^2 - k*mn/d^2)/(k*mn/d^2) = 0

(k*mn/d^2)*(1.21*0.81/x^2 - 1)/(k*mn/d^2) = 0
1.21*0.81/x^2 = 1
x = 0.99 or -0.99
Since the mass is decreased (1.21*0.81 <1) , then F =k*mn/d^2 .If the force is unchanged and the mass decrease , the distance should be increase, Therefore , the distance should be increased by 99%


最後果題唔肯定,希望冇錯

2012-08-28 01:36:49 補充：
睇番002 我知錯咩野啦 應該係 1%

2012-08-28 01:38:41 補充：
002 我想問下 你2邊開方之後 咪會出正負既,點先可以rejected走-0.99 果個??