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 `2 1/2` hours.
[Take `sqrt2 = 1.414`]

Answer 1

Let x be the number of bacteria at time t.

Then the rate of increase is `"dx"/"dt"` which is proportional to x.

∴ `"dx"/"dt" prop "x"`

∴ `"dx"/"dt"` = kx, where k is a constant

∴ `"dx"/"x" = "k  dt"`

On integrating, we get

`int "dx"/"x" = "k" int "dt" + "c"`

∴ log x = kt + c

Initially, i.e. when t = 0, x = 1000

∴ log 1000 = k × 0 + c     ∴ c = log 1000

∴ log x = ly + log 1000

∴ log x - log 1000 = kt

∴ log`("x"/1000) = "kt"`     ...(1)

Now, when t = 1, x = 2 × 1000 = 2000

∴ `log (2000/1000)` = k   ∴ k = log 2

∴ (1) becomes, log`("x"/1000) = t log 2

If t = `2 1/2 = 5/2,` then

log`("x"/1000) = 5/2 log 2 = log (2)^(5/2)`

∴ `("x"/1000) = (2)^(5/2) = 4sqrt2 = 4 xx 1.414 = 5.656`

∴ x = 5.656 × 1000 = 5656

∴ number of bacteria after `2 1/2`hours = 5656.