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Finally both solved.
As I said, its too long calculation, here is the pic :
8a)
8b)
8c)
Therefore at any instant we can say dx/dt = ky where k is some constant.
But we are also told that x + y = 100
Thus y = 100 - x
So dx/dt = k(100 - x)
We are also told that when x = 5, dx/dt = 1.9
1.9 = k(100 - 5)
1.9 = k95
k = 1.9 / 95
k = 0.02
dx/dt = 0.02(100 - x)
For part (ii) integrate this.
dx/dt = 0.02(100 - x)
Separate variables
dx / (100 - x) = 0.02 dt
Integrating gives
- ln (100 - x) = 0.02t + c (Note the appearance of the first minus sign)
When t = 0, x = 5
- ln (95) = c
So we now have
- ln (100 - x) = 0.02t - ln 95
ln (100 - x) = ln 95 - 0.02t
ln (100 - x) - ln 95 = -0.02t
ln [(100 - x) / 95] = -0.02t
(100 - x) / 95 = e^(-0.02t)
100 - x = 95e^(-0.02t)
100 - 95e^(-0.02t) = x
x = 100 - 95e^(-0.02t)
For part (iii) when t becomes large e^(-0.02t) becomes very small.
Thus x approaches 100
Winter 2008 :Thx for ur help!! Appreciate it very much^^
Hmmm...on/9709/2008/03 question 8 and mj/9709/2008/03 quest8 too..(
As I said, its too long calculation, here is the pic :
8a)
8b)
8c)
For part (i) we are told that the rate of formation of X is proportional to the mass of Y.please help with question 7 (ii)
I can put it into the form -In(100-x) = 0.02t-In(k)
but I can not get K to equal 95 like they appear to do in the mark scheme.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_3.pdf
Therefore at any instant we can say dx/dt = ky where k is some constant.
But we are also told that x + y = 100
Thus y = 100 - x
So dx/dt = k(100 - x)
We are also told that when x = 5, dx/dt = 1.9
1.9 = k(100 - 5)
1.9 = k95
k = 1.9 / 95
k = 0.02
dx/dt = 0.02(100 - x)
For part (ii) integrate this.
dx/dt = 0.02(100 - x)
Separate variables
dx / (100 - x) = 0.02 dt
Integrating gives
- ln (100 - x) = 0.02t + c (Note the appearance of the first minus sign)
When t = 0, x = 5
- ln (95) = c
So we now have
- ln (100 - x) = 0.02t - ln 95
ln (100 - x) = ln 95 - 0.02t
ln (100 - x) - ln 95 = -0.02t
ln [(100 - x) / 95] = -0.02t
(100 - x) / 95 = e^(-0.02t)
100 - x = 95e^(-0.02t)
100 - 95e^(-0.02t) = x
x = 100 - 95e^(-0.02t)
For part (iii) when t becomes large e^(-0.02t) becomes very small.
Thus x approaches 100