Question

The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after `5/2` hours  `("Given"  sqrt(2) = 1.414)`

Answer 1

Let ‘x’ be the number of bacteria present at time ‘t’.

∴ `("d"x)/"dt"  ∞  x`

∴ `("d"x)/"dt"` = kx,

where k is the constant of proportionality.

∴ `("d"x)/x` = kdt

Integrating on both sides, we get

`int  ("d"x)/x = "k" int "dt"`

∴ log x = kt + c    .....(i)

When t = 0, x = 1000

∴ log (1000) = k(0) + c

∴ c = log (1000)

∴ log x = kt + log (1000)   ....(ii) ....[From (i)]

When t = 1, x = 2000

∴ log (2000) = k(1) + log (1000)

∴ log (2000) − log (1000) = k

∴ k = `log(2000/1000)`

= log 2   .....(iii)

When t =`5/2`, we get

log x = `5/2 "k" + log(1000)`   .......[From (ii)]

∴ log x = `(5/2) log 2 + log(1000)` .....[From (iii)]

= `log (2^(5/2)) + log(1000)`

= `log (4sqrt(2)) + log(1000)`

= `log (4000  sqrt(2))`

= log (4000 × 1.414)

∴ log x = log (5656)

∴ x = 5656

Thus, there will be 5656 bacteria after `5/2` hours.