M
hscscience Maths Adv · Y12
0/100daily goal
0
0
0 due
0
L1 · 0 XP
KJ
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
1

Get oriented

Set up the anchor equation and key terms for growth and decay.

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
01
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
02
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.

The rate equation dN/dt = kN and its solution, with the sign of k setting growth vs 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.
03
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
04
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.
2

Derive the model

See where the exponential growth and decay model comes from.

05
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.

3

Work through examples

Follow three applications: compound interest, decay and finding k from data.

Worked examples · 3 applications

PROBLEM 1 · COMPOUND INTEREST

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

1
$\dfrac{dA}{dt} = 0.06A,\quad A(0) = 5000$
Write the DE: rate = 6% of current amount.
1 of 3
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.

1
$t_{1/2} = \dfrac{\ln 2}{|k|} \implies k = -\dfrac{\ln 2}{12} \approx -0.0578$
Rearrange half-life formula for $k$. Negative because decay.
1 of 3
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?

1
$P(t) = 10000e^{kt}$. At $t = 5$: $12000 = 10000e^{5k}$
Let $t = 0$ be 2010. Use the second data point.
1 of 3
4

Avoid the common traps

Meet the three errors that cost marks, and check yourself.

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.

5

Drill the problems

Work the activities on half-life, doubling time and decay constants.

Activities · apply what you know

1

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

2

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

3

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

4

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

5

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?

6

Revisit and consolidate

Revisit your first answer and confirm what stuck.

09
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
7

Practise growth and decay

Work the question bank and the short-answer questions.

01
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.

02
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
Beyond the syllabus. Verifying and using $Q = Ae^{kt}$ for proportional growth and decay is core, and that is what the exam requires. The next question's logistic model and the general solving of differential equations are extension for Advanced — they belong to further applications of calculus. Attempt it as optional depth.
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].