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, complete the following activity to find the number of times the bacteria are increased in 12 hours.

Answer 1

Let N be the number of bacteria present at time t.

∵ The rate of increase is proportional to the number present.

`(dN)/(dt) ∝ N`

∴ `(dN)/(dt)` = k. N

Where k is the constant of proportionality.

∴ `(dN)/N` = k. dt

Integrating both sides,

`int 1/N. dN = k int dt`

log N = k. t + C  ......(1)

Now at, t = 0, N = N₀

∴ From (1),

log N₀ = C  ......(2)

At t = 4, N = 2N₀

∴ From (1),

log 2N₀ = 4K + log N₀  ......[∵ C = log N₀]

⇒ log 2N₀ – log N₀ = 4k

⇒ `log((2N_0)/N_0)` = 4k

⇒ log 2 = 4k

⇒ k = `1/4` log 2

When t = 12,

log N = `1/4 log 2 xx 12 + log N_0`

⇒ `log (N/N_0)` = 3 log 2 = log2³

⇒ `log(N/N_0)` = log 8

⇒ N = 8N₀.