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Question
The rate of growth of bacteria is proportional to the number present. If, initially, there were
1000 bacteria and the number doubles in one hour, find the number of bacteria after 2½
hours.
[Take `sqrt2` = 1.414]
Solution
Let ‘N’ be the number of bacteria at time’t ’
`therefore "dN"/dt ooN`
`"dN"/dt=kN`
`"dN"/N=kdt`
Integrating on both sides, we get
`int"dN"/N=kintdt`
`logN=kt+c`
`when t=0,N=1000`
`c=log1000`
`logN=kt+log1000`
`log(N/1000)=kt`
`N=1000e^(kt)` ..........(1)
when t = 1 hour, N = 2000
`e^k=2`
`N=1000xx(2)^t` ....................from (1)
when `t=2 1/2` hours, we get
`N=1000xx(2)^(5/2)`
`=1000xx4xxsqrt2=4000xx1.414`
`N=5656`
Number of bacteria present after `2 1/2` hours is 5656
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