Question

In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.

Solution:

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

Since the rate of increase of N is proportional to N, the differential equation can be written as –

`(dN)/dt αN`

∴ `(dN)/dt` = KN, where K is constant of proportionality

∴ `(dN)/N` = k . dt

∴ `int 1/N dN = K int 1 . dt`

∴ log N = `square` + C   ...(1)

When t = 0, N = N₀ where N₀ is initial number of bacteria.

∴ log N₀ = K × 0 + C

∴ C = log N₀

Also when t = 4, N = 2N₀

∴ log (2 N₀) = K . 4 + `square`   ...[From (1)]

∴ `log((2N_0)/N_0)` = 4K,

∴ log 2 = 4K

∴ K = `square`   ...(2)

Now N = ? when t = 12

From (1) and (2)

log N = `1/4 log 2  . (12) + log N_0`

log N – log N₀ = 3 log 2

∴ `log(N_0/N_0)` = `square`

∴ N = 8 N₀

∴ Bacteria are increased 8 times in 12 hours.

Answer 1

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

Since the rate of increase of N is proportional to N, the differential equation can be written as –

`(dN)/dt αN`

∴ `(dN)/dt` = KN, where K is constant of proportionality

∴ `(dN)/N` = k . dt

∴ `int 1/N dN = K int 1 . dt`

∴ log N = Kt + C   ...(1)

When t = 0, N = N₀ where N₀ is initial number of bacteria.

∴ log N₀ = K × 0 + C

∴ C = log N₀

Also when t = 4, N = 2N₀

∴ log (2 N₀) = K . 4 + log N₀   ...[From (1)]

∴ `log((2N_0)/N_0)` = 4K,

∴ log 2 = 4K

∴ K = `bb(1/4 log  2)`   ...(2)

Now N = ? when t = 12

From (1) and (2)

log N = `1/4 log 2  . (12) + log N_0`

log N – log N₀ = 3 log 2

∴ `log(N_0/N_0)` = log (2)³

∴ N = 8 N₀

∴ Bacteria are increased 8 times in 12 hours.