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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... 顯示更多 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 ?
最佳解答:
(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
其他解答:CF546184287637C5
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發問: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... 顯示更多 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 ?
最佳解答:
(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
其他解答:CF546184287637C5
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