Your weak spots

Insights load after your first practice round.

Module 6 · L13 of 15 ~35 min ⚡ +95 XP available

Growth & Decay Models

Why do epidemiologists speak of 'flattening the curve'? Why do banks advertise compound interest as exponential growth? All these questions revolve around a single differential equation: $\frac{dN}{dt} = kN$. Its solution, $N(t) = N_0 e^{kt}$, describes everything from bacterial colonies to radioactive decay.

Today's hook — In early 2020, every country on Earth was racing to "flatten the curve." The curve is an exponential — and the mathematics behind pandemic modelling, compound interest, radioactive dating, and drug metabolism is identical. One equation rules them all.

0/5QUESTS

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Think First — your gut answer

+5 XP warm-up

A bank account earns 5% interest compounded continuously. If $A(t)$ is the amount after $t$ years, the differential equation is $\frac{dA}{dt} = 0.05A$. If you start with \$1000, how much will you have after 10 years? Write your gut answer first — we'll revisit it at the end.

auto-saved

The anchor equation

+5 XP to read

One equation produces four formulas — memorise the anchor and the rest follows.

The fundamental assumption: rate of change is proportional to current amount. That single idea, when solved, gives exponential functions. The sign of $k$ tells you everything: positive means growth, negative means decay.

$$\frac{dN}{dt} = kN \implies N(t) = N_0 e^{kt}$$

Half-life (decay)

$t_{1/2} = \dfrac{\ln 2}{|k|}$ — time for the amount to halve. Depends only on $k$, not on initial amount.

Doubling time (growth)

$t_{double} = \dfrac{\ln 2}{k}$ — time for the amount to double. Same formula structure as half-life.

Solving for k

Given two data points: substitute both into $N = N_0 e^{kt}$, divide the equations, then take $\ln$ of both sides.

What you'll master

Know

Key facts

$\frac{dN}{dt} = kN$ has solution $N = N_0 e^{kt}$
Half-life: $\frac{\ln 2}{|k|}$; doubling time: $\frac{\ln 2}{k}$
$k > 0$: growth; $k < 0$: decay

Understand

Concepts

Why proportional change produces exponential functions
The meaning of half-life and doubling time
Limitations of the simple exponential model

Can do

Skills

Set up and solve growth/decay DEs
Calculate half-lives and doubling times
Apply models to real-world scenarios

Key terms

Exponential growthQuantity increases by a fixed percentage per unit time; $k > 0$.

Exponential decayQuantity decreases by a fixed percentage per unit time; $k < 0$.

Half-lifeTime for a decaying quantity to halve: $t_{1/2} = \frac{\ln 2}{|k|}$.

Doubling timeTime for a growing quantity to double: $t_{double} = \frac{\ln 2}{k}$.

Initial conditionThe known value $N_0 = N(0)$ that pins the specific solution.

Decay constantThe value $k$ in $\frac{dN}{dt} = kN$; negative for decay processes.

Deriving the exponential model

core concept

The fundamental assumption is deceptively simple: the rate of change of $N$ is proportional to $N$ itself. Write this as $\frac{dN}{dt} = kN$, then solve by separation of variables.

$$\frac{dN}{N} = k\,dt \implies \ln|N| = kt + C \implies N = N_0 e^{kt}$$

where $N_0 = N(0) = e^C$ is the initial value. The key insight is that the same equation governs both growth and decay — only the sign of $k$ changes.

Growth ($k > 0$) contexts:

Populations with unlimited resources
Continuously compounded interest
Viral spread in early pandemic stages

Decay ($k < 0$) contexts:

Radioactive decay
Drug elimination from bloodstream
Newton's Law of Cooling (after substitution)

Deriving half-life: Set $N = \frac{N_0}{2}$:

$$\frac{N_0}{2} = N_0 e^{kt_{1/2}} \implies \frac{1}{2} = e^{kt_{1/2}} \implies t_{1/2} = \frac{\ln 2}{|k|}$$

COVID-19 and the Basic Reproduction Number. In early 2020, epidemiologists modelled COVID-19 spread using $\frac{dI}{dt} = rI$ where $I$ is infections and $r$ is the growth rate. When governments imposed lockdowns, they were reducing $r$ below zero, turning exponential growth into exponential decay. "Flatten the curve" means reducing $r$ so hospitals aren't overwhelmed. The mathematics is identical to radioactive decay and compound interest — only the context and sign of $k$ change.

DE: $\frac{dN}{dt} = kN$ — rate proportional to amount; Solution: $N(t) = N_0 e^{kt}$ where $N_0 = N(0)$

Pause — copy the differential equation $\dfrac{dN}{dt} = kN$ and its solution $N(t) = N_0 e^{kt}$ into your book.

Did you get this? True or false: the half-life of a radioactive substance depends on how much of the substance you start with.

Worked examples · 3 applications

PROBLEM 1 · COMPOUND INTEREST

\$5000 is invested at 6% per annum compounded continuously. Find the value after 8 years.

$\dfrac{dA}{dt} = 0.06A,\quad A(0) = 5000$

Write the DE: rate = 6% of current amount.

PROBLEM 2 · RADIOACTIVE DECAY

A substance has half-life 12 years. Find the decay constant and the amount remaining from 200 g after 30 years.

$t_{1/2} = \dfrac{\ln 2}{|k|} \implies k = -\dfrac{\ln 2}{12} \approx -0.0578$

Rearrange half-life formula for $k$. Negative because decay.

PROBLEM 3 · FIND k FROM DATA

A town's population was 10,000 in 2010 and 12,000 in 2015. Assuming exponential growth, when will it reach 20,000?

$P(t) = 10000e^{kt}$. At $t = 5$: $12000 = 10000e^{5k}$

Let $t = 0$ be 2010. Use the second data point.

Quick check: A substance has decay constant $k = -0.05$ per year. What is its half-life?

Common errors · 3 traps that cost marks

Trap 01

Using half-life formula for growth

$t_{1/2} = \frac{\ln 2}{|k|}$ is for decay only. For growth problems you want doubling time: $t_{double} = \frac{\ln 2}{k}$. The formulas look similar, so always check the context.

Trap 02

Forgetting the initial condition

The general solution is $N = Ae^{kt}$ with arbitrary $A$. You need $N(0) = N_0$ to fix $A = N_0$. Students who skip this step get the wrong specific solution.

Trap 03

Getting $k$ positive for decay

When the half-life formula gives $k = \frac{\ln 2}{t_{1/2}}$ (positive), remember to attach the negative sign: $k = -\frac{\ln 2}{t_{1/2}}$. A positive $k$ in a decay context will make $N$ grow, not shrink.

Fill in the blank: A bacteria colony doubles every 3 hours, so its doubling time is 3 h and its growth constant is $k = \dfrac{\ln 2}{3}$. Starting from 100 bacteria, after 9 hours there will be ___ bacteria.

Activities · apply what you know

Find the half-life of a substance with decay constant $k = -0.05$ per year.

An investment of \$2000 grows at 4.5% compounded continuously. Find its value after 15 years.

A bacteria culture triples every 5 hours. How long until it reaches 10 times its original size?

Why does the simple exponential model eventually fail for population growth?

Compare \$1000 at 5% annual compounding versus continuous compounding over 10 years. What is the difference?

Odd one out: Three of these are examples of exponential decay; one is exponential growth. Which is the odd one out?

Revisit your thinking

Earlier: \$1000 at 5% continuously compounded. The solution is $A(t) = 1000e^{0.05t}$. After 10 years: $A(10) = 1000e^{0.5} \approx 1000 \times 1.6487 = \$1648.72$. This is slightly more than annual compounding ($1000 \times 1.05^{10} \approx \$1628.89$) because interest earns interest more frequently. The limit as compounding frequency approaches infinity gives the continuous formula — and this limit is exactly what the DE $\frac{dA}{dt} = rA$ describes.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next.

Short answer

ApplyBand 43 marks

Q1. A radioactive isotope has half-life 8 days. Starting with 50 g, how much remains after 20 days? Show all working. (3 marks)

auto-saved

ApplyBand 54 marks

Q2. A city's population grows from 50,000 to 65,000 in 10 years. Assuming exponential growth: (a) Find the growth constant $k$. (b) Predict the population in 25 years. (c) When will the population reach 100,000? (4 marks)

auto-saved

AnalyseBand 54 marks

Q3. The simple exponential model $\frac{dP}{dt} = kP$ assumes unlimited resources. Explain how this limitation changes the model, what the modified differential equation (logistic growth) looks like, and why the exponential model is still useful in the early stages of growth. (4 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity 1:

1. $t_{1/2} = \frac{\ln 2}{0.05} \approx 13.9$ years.

2. $A(15) = 2000e^{0.045 \times 15} = 2000e^{0.675} \approx \$3927.80$.

3. $3 = e^{5k}$, so $k = \frac{\ln 3}{5}$. Then $10 = e^{kt}$, so $t = \frac{\ln 10}{k} = \frac{5\ln 10}{\ln 3} \approx 10.5$ hours.

Activity 2:

4. Unlimited resources don't exist. The logistic model $\frac{dP}{dt} = kP(1 - \frac{P}{K})$ includes carrying capacity $K$.

5. Annual: $1000 \times 1.05^{10} = \$1628.89$. Continuous: $1000e^{0.5} = \$1648.72$. Difference: \$19.83.

Q1 (3 marks): $k = -\frac{\ln 2}{8} \approx -0.0866$ per day [1]. $N(20) = 50e^{-0.0866 \times 20} = 50e^{-1.733} \approx 8.84$ g [2].

Q2 (4 marks): (a) $65000 = 50000e^{10k}$, $k = \frac{\ln 1.3}{10} \approx 0.0263$ [1]. (b) $P(25) = 50000e^{0.657} \approx 92950$ [1.5]. (c) $t = \frac{\ln 2}{0.0263} \approx 26.4$ years [1.5].

Q3 (4 marks): Simple model predicts unbounded growth, unrealistic [1]. Logistic: $\frac{dP}{dt} = kP(1 - \frac{P}{K})$ where $K$ is carrying capacity [1]. When $P \ll K$, $(1-P/K) \approx 1$, so logistic $\approx$ exponential [1]. Exponential models work well in early pandemic/colonisation stages [1].

Boss battle · The Epidemiologist

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

Enter the arena

Science Jump · platform challenge

Climb platforms by solving exponential growth and decay problems, calculating half-lives and doubling times.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 12 · Separable Equations Lesson 14 · Motion Applications →

Module overview · Maths Advanced · Checkpoint 2