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Question
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Solution
Let the number of bacteria at some time t be y.
Given: `dy/dt prop y`
or `dy/dt = ky`
and `dy/y = k dt`
On integration
`int dy/y = int k dt + C`
⇒ log y = kt + C
When t = 0, y = y0
∴ log y0 = 0 + C
⇒ log y = kt + log y0
or log y - log y0 = kt
`=> log y/(y_0) = kt` ....(i)
The number of bacteria increases by 10% in 2 hours.
t = 2,
`y = y_0 + 10/100 y_0 = 110/100 y_0`
Putting t = 2, y = `11/10` y0 in equation (i),
`therefore log (11/10 y_0)/y_0 = k . 2`
or `log 11/10` = 2k
`therefore k = 1/2 log 11/10`
Putting the value of k in (i),
`log y/y_0 = (1/2 log 11/10)` t .....(ii)
Let the bacteria increase from 1,00,000 to 2,00,000 in time t.
`therefore y_1/y_0 = (2,00,000)/(1,00,000) = 2`
Putting the value of `y_1/y_0` in equation (ii),
log 2 = `(1/2 log 11/10)` t
`therefore t = (2 log 2)/(log 11/10)`
Finally, in ` (2 log 2)/(log 11/10)` hours, the bacteria will increase from 1,00,000 to 2,00,000.
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