Question

Bacteria increase at the rate proportional to the number of bacteria present. If the original number N doubles in 3 hours, find in how many hours the number of bacteria will be 4N?

Answer 1

Let x be the number of bacteria at time t.

Since the rate of increase of x is proporational x,

The differential equation can be written as:

`dx/dt` = kx

where k is constant of proportionality.

Solving this differential equation we have

x = c₁·e^kt, where c₁ = e^c  ...(1)

Given that x = N when t = 0

∴ From equation (1) we get

N = c₁·1

∴ c₁ = N

∴ x = N·e^kt  ...(2)

Again given that x = 2N when t = 3

∴ From equation (2), we have

2N = N·e^3k   ...(3)

e^3k = 2

Now we have to find t, when x = 4N

∴ From equation (2), we have

4N = N·e^kt

i.e. 4 = e^kt = `(e^(3k))^(t/3)`

∴ 2² = `2^(t/3)`   ...By equation (3)

∴ `t/3` = 2

∴ t = 6

Therefore, the number of bacteria will be 4N in 6 hours.