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Module 4 · L13 of 15 ~40 min ⚡ +95 XP available

Exponential Growth & Decay

Populations explode, substances decay, and investments compound. The differential equation $\frac{dP}{dt} = kP$ has the elegant solution $P = P_0 e^{kt}$, capturing every process where the rate of change is proportional to the current amount.

Today's hook — A cup of coffee cools from $90^\circ\text{C}$ to $60^\circ\text{C}$ in 10 minutes. Will it reach $30^\circ\text{C}$ in the next 10 minutes? The answer is no — and the reason why reveals one of maths' most powerful models.

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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

Recall — your gut answer first

+5 XP warm-up

A cup of coffee cools from $90^\circ\text{C}$ to $60^\circ\text{C}$ in 10 minutes. Will it reach $30^\circ\text{C}$ in the next 10 minutes? Predict and explain in one line — no formula needed yet.

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The two moves

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Every exponential growth/decay problem follows the same two-step sequence. Lock these in and the rest is substitution.

Step 1 — Set up, find k: Write $P = P_0 e^{kt}$, substitute a known data point, and solve for $k$.
Step 2 — Then predict: Use your value of $k$ to answer any other question about time or quantity.

$$P = P_0 e^{kt}$$

Growth: $k > 0$

Doubling time: $t_2 = \dfrac{\ln 2}{k}$

Decay: $k < 0$

Half-life: $t_{1/2} = \dfrac{\ln 2}{|k|}$

Sanity check

For decay, $k$ must be negative. If your $k$ comes out positive, check your algebra.

What you'll master

Know

Key facts

The exponential model $P = P_0 e^{kt}$
Half-life formula: $t_{1/2} = \dfrac{\ln 2}{|k|}$
Doubling time formula: $t_2 = \dfrac{\ln 2}{k}$

Understand

Concepts

Why $\frac{dP}{dt} = kP$ leads to exponential solutions
The physical meaning of the growth/decay constant $k$
Why exponential decay never reaches zero

Can do

Skills

Set up and solve growth and decay models
Calculate doubling time and half-life from data
Apply models to populations, radioactivity, and finance

Key terms

Growth constantThe constant $k > 0$ in $P = P_0 e^{kt}$ that determines the rate of exponential growth.

Decay constantThe constant $k < 0$ in $P = P_0 e^{kt}$ that determines the rate of exponential decay.

Half-lifeThe time required for a quantity to reduce to half its initial value: $t_{1/2} = \frac{\ln 2}{|k|}$.

Doubling timeThe time required for a quantity to double: $t_2 = \frac{\ln 2}{k}$ (growth only).

Initial value$P_0$, the value of the quantity at time $t = 0$.

Continuous compoundingInterest compounded at every instant: $A = Pe^{rt}$ where $r$ is the annual rate.

The exponential model

core concept

When a quantity changes at a rate proportional to itself, we write $\frac{dP}{dt} = kP$. The solution is $P = P_0 e^{kt}$. For growth ($k > 0$), the quantity increases without bound. For decay ($k < 0$), the quantity approaches zero — but never reaches it in finite time.

$$P = P_0 e^{kt} \qquad t_{1/2} = \dfrac{\ln 2}{|k|} \qquad t_2 = \dfrac{\ln 2}{k}$$

$P_0$ = initial amount · $k$ = growth/decay rate · $t$ = time

The cooling coffee connection. Newton's Law of Cooling says $\frac{dT}{dt} = k(T - T_a)$ where $T_a$ is room temperature. The solution $T = T_a + (T_0 - T_a)e^{kt}$ shows the temperature approaches room temperature exponentially — each equal time interval reduces the gap by the same fraction, not the same amount. That's why 10 minutes doesn't take the coffee from $60^\circ$ to $30^\circ$ the way it took it from $90^\circ$ to $60^\circ$.

Model: $P = P_0 e^{kt}$ — arises when rate of change is proportional to current amount; Growth: $k > 0$; Decay: $k < 0$

Pause — copy the natural exponential model $P = P_0 e^{kt}$, what each parameter means, and the sign rule ($k > 0$ growth, $k < 0$ decay) into your book.

Did you get this? True or false: if a quantity decays exponentially, it will eventually reach exactly zero.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FIND k AND PREDICT

A population of 500 bacteria grows to 1500 in 4 hours. Find $k$ and predict when it reaches 10 000.

$P = 500e^{kt}$

Set up the growth model with $P_0 = 500$.

PROBLEM 2 · HALF-LIFE

A radioactive substance has half-life 10 days. Find the decay constant and the mass remaining from 100 g after 30 days.

$t_{1/2} = \dfrac{\ln 2}{|k|} = 10$, so $|k| = \dfrac{\ln 2}{10} \approx 0.0693$

Use half-life formula. Since it is decay, $k = -0.0693$.

PROBLEM 3 · CONTINUOUS COMPOUNDING

$5000 is invested at 6% p.a. compounded continuously. Find the value after 8 years and the doubling time.

$A = 5000e^{0.06t}$

Continuous compounding formula with $k = 0.06$.

Quick check: A substance has half-life 5 years. After 15 years, what fraction of the original remains?

Common errors · the 3 traps that cost marks

Trap 01

Using the wrong sign for $k$

For growth, $k > 0$. For decay, $k < 0$. Using a positive $k$ for decay makes the quantity grow instead of shrink. Always check that your answer makes physical sense before writing it down.

Trap 02

Confusing half-life with time to reach zero

Half-life is the time to reach half the initial amount. Exponential decay never reaches zero in finite time. After $n$ half-lives, the amount is $P_0\left(\frac{1}{2}\right)^n$, which approaches but never equals zero.

Trap 03

Using $\frac{\ln 2}{k}$ instead of $\frac{\ln 2}{|k|}$ for half-life

If $k$ is negative (decay), $\frac{\ln 2}{k}$ gives a negative time. Use $\frac{\ln 2}{|k|}$ or $-\frac{\ln 2}{k}$ to get a positive half-life. The absolute value avoids sign errors.

Fill in the gap: The half-life of an exponentially decaying substance with decay constant $k$ (where $k < 0$) is $t_{1/2} =$ (write as a fraction using ln2 and |k|).

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

A population grows from 1000 to 4000 in 6 hours. Find $k$ and predict the population after 10 hours.

A substance decays from 200 g to 25 g in 30 days. Find the half-life.

$10\,000 is invested at 5% p.a. compounded continuously. Find the value after 5 years.

Bacteria double every 3 hours. How long until a population of 500 reaches 8000?

The half-life of carbon-14 is 5730 years. Find the percentage remaining after 10 000 years.

Odd one out: Three of these describe exponential growth. Which is the odd one out?

Revisit your thinking

Earlier you predicted whether coffee at $60^\circ\text{C}$ would reach $30^\circ\text{C}$ in another 10 minutes. It will not — Newton's Law of Cooling is exponential, so each 10-minute interval reduces the gap to room temperature by the same fraction, not the temperature itself. The model $T = T_a + (T_0 - T_a)e^{kt}$ captures this asymptotic approach: the temperature never quite reaches room temperature in finite time.

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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. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

Q1. A population grows from 2000 to 4500 in 5 hours. Model with $P = P_0 e^{kt}$ and find $k$ to 3 decimal places. (3 marks)

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ApplyBand 43 marks

Q2. A radioactive substance has half-life 8 days. Starting with 100 g, how much remains after 20 days? (3 marks)

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ApplyBand 54 marks

Q3. $5000 is invested at 8% p.a. Find the value after 10 years for: (i) annual compounding, (ii) continuous compounding. Find the difference. (4 marks)

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Comprehensive answers (click to reveal)

Drill 1: $k = \frac{\ln 4}{6} \approx 0.231$; $P(10) = 1000e^{2.31} \approx 10\,079$

Drill 2: $200 \to 25$ is 3 halvings, so half-life $= 30/3 = 10$ days

Drill 3: $10000e^{0.25} \approx \$12\,840$

Drill 4: $k = \frac{\ln 2}{3}$; $8000 = 500e^{kt}$, so $t = \frac{3\ln 16}{\ln 2} = 12$ hours

Drill 5: $100e^{-(\ln 2/5730)\times 10000} \approx 29.9\%$

Q1 (3 marks): $4500 = 2000e^{5k}$, so $e^{5k} = 2.25$ [0.5]; $k = \frac{\ln 2.25}{5} \approx 0.162$ [2]; $k \approx 0.162$ per hour [0.5]

Q2 (3 marks): $k = -\frac{\ln 2}{8} \approx -0.0866$ [0.5]; $P = 100e^{-0.0866 \times 20} = 100e^{-1.733} \approx 17.7$ g [2.5]

Q3 (4 marks): (i) $A = 5000(1.08)^{10} \approx \$10\,795$ [1.5]; (ii) $A = 5000e^{0.8} \approx \$11\,128$ [1.5]; Difference $\approx \$333$ [1]

Boss battle · The Alchemist

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 answering exponential growth and decay questions.

Mark lesson as complete

Tick when you've finished the practice and review.

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Module overview · Maths Advanced