Growth Decay and Iterative Processes for the TMUA

Exponential Growth and Decay

Problems involving growth and decay generally require a specific rate where a quantity is multiplied by the same factor in every time period. This is the foundation of exponential modelling.

If the initial size of a population at time $t = 0$ is $q$, and this population is multiplied by a factor $x$ every hour, then after $n$ hours, the population size is given by the formula $qx^n$.

The value of the multiplier $x$ determines the type of change:

1. If $x > 1$, the quantity is undergoing growth (for example, a bacterial colony doubling or trebling).
2. If $0 < x < 1$, the quantity is undergoing decay (for example, a radioactive substance losing half its mass).
3. If $x = 1$, the population remains static.

Worked Example: Epidemic Growth and Decay

Example: In a certain town, the number of patients in an epidemic trebles every day. At the end of day 1, there are 50 patients.

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 daily intervals pass. Since the population trebles each day, the multiplier $x$ is 3. The calculation is: $50 \times 3^3 = 50 \times 27 = 1350$ patients.

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 of using the cure?

Let the rate of decay be $r$. Following the iterative logic where the start of the cure is the first data point, we find that after 4 days, the population is $4000r^3$. Setting up the equation: $4000r^3 = 500$ $r^3 = \frac{500}{4000} = \frac{1}{8}$ $r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

To find the number of patients after 6 days, we use the fifth power (as 5 intervals have passed since the start of the cure treatment): $4000r^5 = 4000 \times (\frac{1}{2})^5 = 4000 \times \frac{1}{32} = 125$ patients.

Compound Interest

Compound interest is a specific application of exponential growth used in finance. Unlike simple interest, where the interest is calculated only on the original sum, compound interest adds the interest back into the account so that the interest itself earns interest in the next period.

If an initial sum $P$ (the principal) is invested in an account and the rate of compound interest is $r\%$ per annum (per year), then after $n$ years, the total amount $A$ in the account is given by: $A = P(1 + \frac{r}{100})^n$

Worked Example: Savings Investment

Example: £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: $A = 1000 \times (1 + \frac{4}{100})^5$ $A = 1000 \times 1.04^5$ $A = 1216.6529...$ To the nearest penny, the amount is £1216.65.

To find the interest earned, subtract the original principal from the total amount: $Interest = 1216.65 - 1000 = 216.65$ The interest received is £216.65.

Iterative Processes

An iterative process is defined as one where a basic set of instructions or rules are applied again and again, usually over a duration of time.

Compound interest serves as a primary example of an iterative process because the same rule (adding a fixed percentage of the current balance) is repeated every year. In broader mathematics, iteration involves taking the output of one step and using it as the input for the next step. This process can model a wide variety of real world phenomena beyond finance, such as population dynamics or the cooling of an object.