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?

Answer 1

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.