Understanding Exponential Changes (3.4.1) | AQA GCSE Maths (Higher) Notes

Exponential growth and decay are fundamental concepts in mathematics, describing processes where quantities increase or decrease at rates proportional to their current value. These phenomena are observed in various real-world situations, including population dynamics, radioactive decay, and financial applications such as interest calculation and depreciation. This section aims to introduce the principles of exponential growth and decay, providing examples from depreciation and population change scenarios to enhance understanding.

Introduction to Exponential Growth and Decay

Exponential growth occurs when a quantity increases over time at a rate proportional to its current value, leading to a rapid rise as time progresses. Conversely, exponential decay describes a process where a quantity diminishes over time at a rate proportional to its current value, resulting in a rapid decrease. The general formula for exponential growth and decay is expressed as:

P(t) = P_0 \cdot e^{rt}

Where:

$P(t)$ is the quantity at time $t$ ,
$P_0$ is the initial quantity,
(r) is the rate of growth (if positive) or decay (if negative),
(e) is the base of the natural logarithm, approximately equal to 2.71828.

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Exponential Growth: Population Change Scenario

Example 1: Population Growth

Consider a population of 1,000 bacteria that doubles every hour. Calculate the population after 5 hours.

Solution:

Here, $P_0 = 1000$ , $r = \ln(2)$ because the population doubles (growth rate of 100%), and $t = 5$ .

P(t) = 1000 \cdot e^{\ln(2) \cdot 5}

P(5) = 1000 \cdot 2^5

P(5) = 1000 \cdot 32 = 32,000

Thus, the population of bacteria will be 32,000 after 5 hours.

Exponential Decay: Depreciation Scenario

Example 2: Depreciation of a Car

A car worth £20,000 depreciates at a rate of 10% per year. Calculate its value after 3 years.

Solution:

For depreciation, the rate $r$ is negative because the value decreases over time. Here, $P_0 = 20,000$ , $r = -0.10$ , and $t = 3$ .

P(t) = 20,000 \cdot e^{-0.10 \cdot 3}

P(3) = 20,000 \cdot e^{-0.30}

P(3) ≈ 20,000 \cdot 0.74082 ≈ £14,816.40

Therefore, the value of the car after 3 years will be approximately £14,816.40.

Understanding the Formula

The exponential growth and decay formula, $P(t) = P_0 \cdot e^{rt}$ , is versatile, allowing for the calculation of quantities over any time period. The key components to remember are:

Initial Quantity $(P_0)$ : The starting point from which growth or decay is measured.
Growth/Decay Rate $(r)$ : Expressed as a decimal, this rate determines the speed of the change. A positive rate indicates growth, while a negative rate signifies decay.
Time $(t)$ : The period over which the growth or decay occurs.
Natural Base $(e)$ : A constant (~2.71828) that ensures the formula accurately models exponential changes.

Application in Real Life

Exponential functions are not just theoretical concepts but have practical applications in everyday life, including:

Environmental Studies: Understanding population dynamics of species.
Finance: Calculating compound interest and depreciation.
Medicine: Modelling the spread of diseases or the decay of drug concentration in the body.

Real life application of exponential functions

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Written by:

Dr Rahil Sachak-Patwa

Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.