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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

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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 lineno 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

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Pick your answer, then rate your confidencethat 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