Growth and Decay Problems

Let `N(t)` denote the amount of a substance (or population) that is either growing or decaying. If we assume that `(dN)/(dt)`, the time rate of change of this amount of substance, is proportional to the amount of substance present, we have that `(dN)/(dt)=kN`, where `k` is the constant of proportionality.

When `k` is positive, the population grows, and on the contrary, when `k` is negative, the population decays.

We are assuming that `N(t)` is a differentiable, hence continuous, function of time. For population problems, where `N(t)` is actually discrete and integer-valued, this assumption is incorrect. Nonetheless, the above model still provides a good approximation to the physical laws governing such a system.

This differential equation is separable. Rewriting it, we obtain `(dN)/N=kdt`.

Integrating both sides gives: `ln(N)=kt+c_1`, or `N=Ce^(kt)`, where `C=e^(c_1)`.

If we are given that `N(0)=N_0`, it can be stated that `N_0=Ce^(k*0)`, or `C=n_0`.

So, `N=N_0e^(kt)`.

Example 1. The population of a certain country is known to increase at a rate proportional to the number of people currently living in the country. If after two years the population has doubled and after three years the population is 20000, estimate the number of people initially living in the country.

Let's assume that `N_0` is the number of people initially living in the country; then, `N(0)=N_0`, `N(2)=2N_0`, and `N(3)=20000`.

So, as shown above, if `N(0)=N_0`, we obtain that `N=N_0e^(kt)`.

Next, since `N(2)=2N_0`, we have that `2N_0=N_0e^(k*2)`, or `2k=ln(2)`. This gives that `k=ln(2)/2`.

Now, the equation can be rewritten as `N=N_0e^(ln(2)/2t)`.

Since `N(3)=20000`, we have that `20000=N_0e^(ln(2)/2 3)`, or `N_0=20000e^(-1.5ln(2))~~7071` people.

Let's work another useful example.

Example 2. A certain radioactive material is known to decay at a rate proportional to the amount present. If after one hour it is observed that 10 percent of the material has decayed, find the half-life (period of time it takes for an amount of material to decrease by half) of the material.

Let's assume that the initial amount of the radioactive material is `N_0`; then, `N(0)=N_0`, `N(1)=(1-0.1)N_0=0.9N_0`.

We need to find such `t` that `N(t)=(N_0)/2`.

So, as was shown above, if `N(0)=N_0`, we have that `N(t)=N_0e^(kt)`.

Since `N(1)=0.9N_0`, we obtain that `0.9N_0=N_0e^(k*1)` or `k=ln(0.9)`.

Now, the equation has the form `N(t)=N_0e^(ln(0.9)t)`.

Now, find `t` such that `N(t)=0.5 N_0`: `0.5 N_0=N_0e^(ln(0.9)t)`, or `t=(ln(0.5))/(ln(0.9))~~6.58` hours.