Models of exponential growth and decay

Summary

This lesson applies the exponential and logarithmic functions to the description of quantities that change by the same proportion over equal periods. It studies discrete growth at a constant rate and the continuous model $P(t) = P_0 e^{kt}$ , proves that in the latter the doubling time and the half-life do not depend on the instant at which they are measured, and applies the result to radioactive decay, to carbon-14 dating and to Newton’s law of cooling. It then shows how to determine the parameters of a model from two data points, proves that exponential data line up on a semi-log plot, and ends with the logistic model, whose growth is bounded and whose graph is symmetric about its midpoint.

Learning objectives

Recognize the models of discrete and continuous growth and decay, and relate the rate of change per period to the constant of the model.
Compute and use the doubling time and the half-life, and apply them to radioactive decay, to carbon-14 dating and to the cooling of a body.
Determine the parameters of an exponential model from two data points and recognize exponential data on a semi-log plot.
Study the logistic model: its boundedness, its monotonicity and the symmetry of its graph about the point at which it reaches half its bound.

Growth at a constant rate

A quantity can increase in two elementary ways. If the same amount is added to it in each period, its values form an arithmetic progression and its graph is a line; this is how a capital grows with simple interest, receiving the same interest every year. If, instead, it is multiplied by the same factor in each period, its values form a geometric progression; this is how a capital grows with compound interest, whose interest each year is computed on the accumulated capital, and this is how a population of bacteria grows in which each cell divides after a fixed time. Both progressions were studied in the lesson on summations, progressions and induction; this lesson deals with the second one and with its continuous version.

Let $P_0 > 0$ be the initial value of a quantity and $P_n$ its value after $n$ periods. The quantity is said to change at a constant rate $r$ per period, with $r > -1$ , when each value is obtained from the one before it by multiplying it by $1 + r$ , that is, when the quotient of two consecutive values is always the same:

\frac{P_{n + 1}}{P_n} = 1 + r

The number $r$ is usually expressed as a percentage: a population that increases by $3$ percent per year has $r = \frac{3}{100}$ , and a substance that loses a fifth of its mass every hour has $r = -\frac{1}{5}$ . If $r > 0$ we speak of growth; if $-1 < r < 0$ , of decay.

LemmaGrowth at a constant rate

If $P_{n + 1} = (1 + r) P_n$ for every $n$ , with $P_0 > 0$ and $r > -1$ , then $P_n = P_0 (1 + r)^n$ for every natural number $n$ . All the values are positive; the sequence is strictly increasing if $r > 0$ and strictly decreasing if $-1 < r < 0$ .

The formula is the general term of a geometric progression with common ratio $1 + r$ , and it is proved by induction. The figure shows the first values with $P_0 = 1$ , first with $r = \frac{1}{2}$ and then with $r = -\frac{1}{2}$ .

Proof

$P_1 = (1 + r) P_0 = P_0 (1 + r)^{\resaltar{1}}$
For $n = 1$ , the formula is the hypothesis applied to $P_0$ .
$P_{n + 1} = (1 + r) \resaltar{P_n} = (1 + r) \resaltar{P_0 (1 + r)^n} = \dato{f}{P_0 (1 + r)^{n + 1}}$
If the formula holds for $n$ , the hypothesis gives it for $n + 1$ ; by the principle of induction, it holds for every natural number.
$P_0 > 0, \quad 1 + r > 0 \rightarrow \dato{p}{P_n > 0}$
Since $P_0$ and $1 + r$ are positive, so are all their powers and products: all the values are positive.
$P_{n + 1} - P_n = \resaltar{r P_n}$
The difference between two consecutive values is $r P_n$ , which has the sign of $r$ : the sequence increases if $r > 0$ and decreases if $-1 < r < 0$ .

For example, a capital of $1000$ at $5$ percent per year, compounded annually, becomes after three years

C_3 = 1000 \left( 1 + \frac{5}{100} \right)^3 = \frac{9261}{8}

that is, 1157.625, whereas with simple interest it would become $1150$ . The difference, small at first, grows without bound with the number of years; solved problem 2 proves that, from the second year on, compound interest always exceeds simple interest, and the systematic study of capitals and interest rates belongs to the course Financial Mathematics.