Question

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours?

Answer 1

Let x denote the number of bacteria at time t hours.

Given = `("d"x)/"dt"` = kx

Hence `("d"x)/x` = kdt

∴ x = C e^kt

Suppose x = x₀ at time t = 0

x₀ = C e^k(0) 

= C e° = C

∴ C = x₀

Hence x = x₀ e^kt

At time 5, x = 3x₀

∵ Number triple in 5 hrs

∴ Hence 3x₀ = x₀ e^5k

∴ e^5k = 3

when t = 10,

x = x₀ e^10k 

= x₀ (e^5k)²

= x₀ 3²

= 9x₀

∴ After 10 hours, the number of bacteria as 9 times the original number of bacteria.