# Growth and Decay Problems

Let N(t) denote amount of 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, then (dN)/(dt)=kN where k is the constant of proportionality.

When k is positive, population grows, when k is negative - 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, above model still provides a good approximation to she physical laws governing such a system.

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

Integration of 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 thenN_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 presently 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 N_0 is number of people initially living in the country, then N(0)=N_0 , N(2)=2N_0 , N(3)=20000 .

So, as was shown above if N(0)=N_0 then N=N_0e^(kt) .

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

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

Since N(3)=20000 then 20000=N_0e^(ln(2)/2 3) or N_0=20000e^(-1.5ln(2))~~7071 people.

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 the amount of material to decrease by half) of the material.

Let initial amount of 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 then N(t)=N_0e^(kt) .

Since N(1)=0.9N_0 then 0.9N_0=N_0e^(k*1) or k=ln(0.9) .

Now equation has 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.