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maths ..

chung

2007-04-04 7:17 am

In a certain culture of bacteria the rate of change increase is proportional to the number present.

Let x (t) be the bacteria at time t hours.

(a) If it is found that the number doubles in 4 hours and x (0) = 500 , how many may be expected at the end of 12 hours ?

(b) If there are 800 at the end of 3 hours and 3200 at the end of 5 hours, how many were there in the beginning ?

回答 (1)

myisland8132

2007-04-04 7:42 am

✔ 最佳答案

(a)
We have
dx/dt=λx (where λ is the rate of increase of bacteria)
(1/x)dx=λdt
integrate both sides (from 0 to T) , we have
ln x(T)- ln x(0)=λT
x(T)=x(0)[e^(λT)]
Since x (0) = 500
x(T)=500[e^(λT)]
or x(t)=500[e^(λt)] (it's the same)
x(4)=2x(0)=1000
500[e^(4λ)]=1000
λ=0.1733
So x(t)=500[e^(0.1733t)]
The bacteria expected at the end of 12 hours
=x(12)
= 500[e^(0.1733*12)]
=4000
(b)
Now
x(3)=x (0) [e^(3λ)]=800...(1)
x(5)=x (0) [e^(5λ)]=3200...(2)
(2)/(1):
e^(2λ)=4
λ=0.6931
So x(3)=x (0) [e^(0.6913*3)]=800
x(0)=100
There were 100 bacteria in the beginning

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