Growth Decay and Iterative Processes

Exponential growth and decay

Many real world phenomena involve quantities that grow or shrink at a rate proportional to their current size. In these growth and decay problems, the quantity is multiplied by the same number in each specific time period. This is the foundation of exponential change.

If we let $q$ represent the initial size of a population at time $t = 0$, and the population is multiplied by a factor $x$ every hour, then the size of the population after $n$ hours is given by the formula $qx^n$. The nature of the change depends entirely on the value of the factor $x$:

1. If $x > 1$, the population is in a state of growth. An example is a bacteria colony that trebles in size every hour, meaning $x = 3$.
2. If $0 < x < 1$, the population is in a state of decay. An example is a radioactive substance losing half of its radioactivity every 6 hours, where $x = 1/2$.
3. If $x = 1$, the population remains static.

Compound interest

Compound interest is a specific application of exponential growth. When money is invested with compound interest, the interest earned in one period is added to the total, and the interest for the next period is calculated on this new, larger amount.

If an initial sum, called the principal $P$, is placed in an account and the rate of compound interest is $r\%$ per annum (which means per year), then after $n$ years the total amount of money $A$ in the account is calculated using the formula:

$A = P\left(1 + \frac{r}{100}\right)^n$

Iterative processes

An iterative process is a procedure where a basic set of instructions or mathematical rules are applied repeatedly, usually over time. Each output from the rule becomes the input for the next step. Compound interest is an example of an iterative process because the rule (add $r\%$ of the current balance) is applied again and again at the end of every year.

Worked Example: Growth and decay in an epidemic

Consider an epidemic where the number of patients in a town trebles every day. At the end of day 1, there are 50 patients.

Part a: How many patients are there at the end of day 4?

To move from the end of day 1 to the end of day 4, three days (or three time intervals) must pass. Since the population trebles each day, the multiplier is 3. We apply the multiplier three times to the initial measurement of 50:

$50 \times 3^3 = 50 \times 27 = 1350$

There are 1350 patients at the end of day 4.

Part b: When the number of patients reaches 4000, a cure is found. The number of patients then decreases exponentially. After 4 days of using the cure, there are 500 patients. How many patients will there be after 6 days?

First, we must find the rate of decay, $r$. Following the logic that 4 days represents 3 intervals of decay from the starting count of 4000:

$4000r^3 = 500$

$r^3 = \frac{500}{4000} = \frac{1}{8}$

$r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

Now, to find the number of patients after 6 days of the cure, we calculate the population after 5 intervals from the start:

$4000r^5 = 4000 \times (\frac{1}{2})^5 = 4000 \times \frac{1}{32} = 125$

There will be 125 patients remaining after 6 days.

Worked Example: Compound interest calculation

£1000 is invested in a savings account at 4% per annum compound interest for 5 years. How much interest has been received after 5 years? Give your answer to the nearest penny.

First, calculate the total amount in the account after 5 years using $P = 1000$, $r = 4$, and $n = 5$:

$A = 1000\left(1 + \frac{4}{100}\right)^5 = 1000 \times 1.04^5$

$A = 1216.6529...$

To the nearest penny, the total amount is £1216.65. To find the interest earned, subtract the original principal from this total:

Interest $= 1216.65 - 1000 = 216.65$

The total interest received is £216.65.