Population Growth and Growth of Bacteria

There are many real situations where the relation describing the RATE of change of a quantity is known directly from the physics or biology of the situation, and this rate description is itself a differential equation that can be solved to recover the quantity as a function of time.

It is an observed fact that the number of bacteria in a culture — or, more generally, the size of a population — increases with time, and that the RATE of that increase is proportional to the population present at that instant. If P(t) is the population at time t, this statement translates directly to dP/dt ∝ P, i.e. dP/dt = k·P for a constant k > 0. Separating variables, dP/P = k·dt, and integrating both sides gives log P = kt + c₁, i.e. P = c·e^(kt), where c = e^(c₁) is a new constant (equal to the population at t=0). This exponential formula gives the population at any time t, once the two constants c and k are pinned down by whatever numerical data the problem supplies (typically the population at t=0, and the population at one later instant).

Two solved examples work this out fully:

Ex.1 A town's population increases from 40,000 to 60,000 over 40 years; find the population 20 years after that (i.e. at 60 years). With P = a·e^(kt) and P=40,000 at t=0, a = 40,000. Using P=60,000 at t=40: e^(40k) = 60,000/40,000 = 3/2. At t = 40+20 = 60 years: P = 40,000·e^(60k) = 40,000·(e^(40k))^(3/2) = 40,000·(3/2)^(3/2); using the given value √(3/2) = 1.2247, (3/2)^(3/2) = (3/2)·1.2247 ≈ 1.837, giving P ≈ 40,000 × 1.837 ≈ 73,482. So the required population is 73,482. …