Exponential Growth and Decay

This lesson covers how exponential functions are used to model real-world growth and decay processes. The Edexcel 9MA0 specification requires you to set up, use, and interpret exponential models, and to understand their limitations.

The General Model

The standard exponential growth/decay model is:

y = Aeᵏᵗ

where:

A = the initial value (when t = 0)
k = the growth/decay constant
t = time
y = the value at time t

If k > 0, the model shows exponential growth. If k < 0, the model shows exponential decay.

When t = 0: y = Ae⁰ = A, confirming that A is the initial value.

Exponential Growth Examples

Population Growth

A bacteria colony initially has 500 cells and doubles every 3 hours.

Model: N = 500 × 2^(t/3), or equivalently N = 500e^(kt) where k = ln(2)/3 ≈ 0.231

t (hours)	N (bacteria)
0	500
3	1000
6	2000
9	4000
12	8000

Compound Interest

An investment of £5000 grows at 4% per annum compounded continuously.

Model: A = 5000e^(0.04t)

After 10 years: A = 5000e^(0.4) ≈ 5000 × 1.4918 ≈ £7459

Exponential Decay Examples

Radioactive Decay

A sample of 200g of a radioactive substance has a half-life of 10 years.

Since half remains after 10 years: 100 = 200e^(10k)

e^(10k) = 1/2
10k = ln(1/2) = -ln(2)
k = -ln(2)/10 ≈ -0.0693

Model: M = 200e^(-0.0693t)

After 25 years: M = 200e^(-0.0693 × 25) = 200e^(-1.7325) ≈ 200 × 0.1768 ≈ 35.4g

Newton's Law of Cooling

The temperature T of an object cooling in surroundings at temperature Tₛ is:

T = Tₛ + (T₀ - Tₛ)e⁻ᵏᵗ

where T₀ is the initial temperature.

Example: A cup of tea at 90°C is placed in a room at 20°C. After 5 minutes it is 70°C. Find the temperature after 15 minutes.

Using T = 20 + 70e⁻ᵏᵗ:

At t = 5: 70 = 20 + 70e⁻⁵ᵏ
50 = 70e⁻⁵ᵏ
e⁻⁵ᵏ = 5/7
-5k = ln(5/7)
k = -ln(5/7)/5 ≈ 0.0673

At t = 15: T = 20 + 70e^(-0.0673 × 15) = 20 + 70e^(-1.0095) ≈ 20 + 70 × 0.3645 ≈ 45.5°C

Finding the Growth/Decay Constant k

Given two data points (t₁, y₁) and (t₂, y₂), you can find k:

From y = Aeᵏᵗ:

y₁ = Aeᵏᵗ¹ and y₂ = Aeᵏᵗ²
y₂/y₁ = e^(k(t₂-t₁))
k = ln(y₂/y₁) / (t₂ - t₁)

Example: A population is 3000 at t = 0 and 5000 at t = 4. Find k.

k = ln(5000/3000) / 4 = ln(5/3) / 4 ≈ 0.5108/4 ≈ 0.1277

Half-Life and Doubling Time

Half-Life (decay)

The half-life is the time for the value to halve. If y = Aeᵏᵗ where k < 0:

A/2 = Aeᵏᵗ
1/2 = eᵏᵗ
t = ln(1/2) / k = -ln(2) / k

Doubling Time (growth)

The doubling time is the time for the value to double. If k > 0:

2A = Aeᵏᵗ
2 = eᵏᵗ
t = ln(2) / k

Exam Tip: Half-life = ln(2)/|k| and doubling time = ln(2)/k. These formulae save time in exams.

Limitations of Exponential Models

Exponential models are simplifications. Common limitations:

Model	Limitation
Population growth	Cannot grow forever — resources are finite (logistic growth is more realistic)
Radioactive decay	Accurate for large samples but breaks down for very small numbers of atoms
Cooling	Assumes constant surrounding temperature
Compound interest	Assumes constant interest rate

In exam questions, you may be asked to "comment on the suitability" of an exponential model — always consider whether the assumptions hold in the given context.

Summary

The model y = Aeᵏᵗ describes growth (k > 0) or decay (k < 0), with A as the initial value.
Find k from two data points using k = ln(y₂/y₁)/(t₂ - t₁).
Half-life = ln(2)/|k|; doubling time = ln(2)/k.
Common applications: population, radioactive decay, cooling, investment.
Always consider the limitations and assumptions of the model.

A-Level Deep Dive: Exponential Growth and Decay

Spec mapping

Edexcel 9MA0-01 specification section 6 — Exponentials and logarithms, modelling sub-strand covers exponential growth and decay in modelling, giving consideration to limitations and refinements of exponential models (refer to the official specification document for exact wording). The canonical model is $N = A e^{kt}$ , where $A$ is the initial value (at $t = 0$ ), $k$ controls the rate, and the sign of $k$ determines growth ( $k > 0$ ) or decay ( $k < 0$ ). This sub-strand sits at the synoptic crossroads of Pure Mathematics: it draws on section 6 (logarithms, used to invert the exponential and solve for $t$ or $k$ ), Year 2 section 11 (differential equations of the form $\frac{dN}{dt} = kN$ , whose solution is this model), section 8 (integration via separation of variables, which derives the model from the differential equation) and Statistics section 4 (regression on log-transformed data, used to estimate $k$ from real-world measurements). The Edexcel formula booklet does not list the exponential model — it must be recalled and justified from the differential equation it solves.

Worked example with full mark scheme

Question (8 marks):

A scientist models the mass $m$ grams of a radioactive isotope at time $t$ years by $m = A e^{kt}$ , where $A$ and $k$ are constants. At $t = 0$ the mass is 80 grams, and at $t = 10$ years the mass has decayed to 50 grams.

(a) Find the value of $A$ and the value of $k$ , giving $k$ to 4 decimal places. (4)

(b) Find the mass remaining at $t = 25$ years. (2)

Solution with mark scheme:

(a) Step 1 — use the initial condition.

At $t = 0$ : $m = A e^{0} = A$ . The given initial mass is 80 grams, so $A = 80$ .

M1 (AO1.1a) — substituting $t = 0$ and recognising $e^{0} = 1$ .

A1 (AO1.1b) — $A = 80$ .

Step 2 — use the second data point to find $k$ .

At $t = 10$ , $m = 50$ :

$50 = 80 e^{10k}$ $e^{10k} = \dfrac{50}{80} = \dfrac{5}{8}$

Take the natural logarithm of both sides:

$10k = \ln\left(\dfrac{5}{8}\right)$ $k = \dfrac{1}{10}\ln\left(\dfrac{5}{8}\right) = \dfrac{1}{10}(-0.4700...) \approx -0.0470$

M1 (AO1.1b) — applying $\ln$ to both sides to invert the exponential.

A1 (AO1.1b) — $k = -0.0470$ (4 d.p.). Note that $k < 0$ , consistent with decay.

(b) Step 1 — substitute $t = 25$ into the model.

$m = 80 e^{25 \times (-0.0470)} = 80 e^{-1.175} = 80 \times 0.3088... \approx 24.7 \text{ grams}$

M1 (AO1.1b) — correct substitution into $m = A e^{kt}$ using the values from (a).

A1 (AO1.1b) — $m \approx 24.7$ grams (3 s.f.).

$80 e^{kt} < 10$ , so $e^{kt} < \dfrac{1}{8}$ . Take $\ln$ :

$kt < \ln\left(\dfrac{1}{8}\right) = -\ln 8$

Since $k < 0$ , dividing flips the inequality:

$t > \dfrac{-\ln 8}{k} = \dfrac{-\ln 8}{-0.0470} = \dfrac{2.0794...}{0.0470} \approx 44.2$

M1 (AO1.1b) — applying $\ln$ and remembering to flip the inequality on dividing by negative $k$ .

A1 (AO1.1b) — $t = 45$ years (the first integer year at which the mass is below 10 grams; $t = 44$ gives $m \approx 10.1$ grams, still above the threshold).

Total: 8 marks (M4 A4). A common loss-point is in (c): students forget to flip the inequality when dividing by a negative $k$ , producing $t < 44.2$ and quoting $t = 44$ — both wrong.

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The temperature $\theta$ °C of a cup of coffee at time $t$ minutes after being poured is modelled by

$\theta = 20 + 70 e^{-0.05 t}$

(a) State, with a reason, the room temperature predicted by this model. (2)

(b) Find the time at which the coffee has cooled to 50°C, giving your answer to the nearest minute. (3)

Mark scheme decomposition by AO:

(a)

B1 (AO3.4) — recognising that as $t \to \infty$ , $e^{-0.05 t} \to 0$ , so $\theta \to 20$ .
B1 (AO2.4) — stating "the room temperature is 20°C" with the reason "the asymptote of the cooling curve".

(b)

M1 (AO1.1b) — setting $50 = 20 + 70 e^{-0.05 t}$ and rearranging to $e^{-0.05 t} = \dfrac{30}{70} = \dfrac{3}{7}$ .
M1 (AO1.1b) — applying $\ln$ : $-0.05 t = \ln(3/7)$ .
A1 (AO1.1b) — $t = \dfrac{-\ln(3/7)}{0.05} \approx 16.94$ , so $t = 17$ minutes (nearest minute).

(c)

B1 (AO3.5b) — any reasonable refinement: e.g. "the model assumes a constant room temperature, but in reality the room cools overnight"; or "the model predicts asymptotic approach to 20°C and never actually reaches it, whereas in practice the coffee equilibrates with the room".

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. Modelling questions weight AO3 (problem-solving / model evaluation) more heavily than typical AO1-dominated procedural questions — the "state, with a reason" and "explain one limitation" command words are the AO3 hooks.

Synoptic links

Connects to:

Year 2 section 11 — Differential equations: the model $N = A e^{kt}$ is the general solution of the first-order ODE $\dfrac{dN}{dt} = kN$ . The defining property of exponential growth/decay is that the rate of change is proportional to the current quantity. Recognising and writing down this differential equation is itself a marked step in modelling questions: "the rate of decay is proportional to the mass present, so $\dfrac{dm}{dt} = km$ ".
Section 8 — Integration (separation of variables): the derivation $\dfrac{dN}{dt} = kN \Rightarrow \int \dfrac{1}{N}\,dN = \int k\,dt \Rightarrow \ln|N| = kt + c \Rightarrow N = A e^{kt}$ is a Year 2 separation-of-variables exercise. Knowing where the model comes from earns marks even when the question only asks you to use it.
Section 6 — Logarithms: every "solve for $t$ " step requires $\ln$ to invert $e^{kt}$ . The key fact $\ln(e^x) = x$ is what makes the natural logarithm — rather than $\log_{10}$ — the right tool here.
Modelling and refinement (AO3 strand): the spec explicitly demands "consideration to limitations and refinements". Pure exponential growth predicts unbounded population — the logistic refinement $\dfrac{dN}{dt} = rN(1 - N/K)$ caps growth at carrying capacity $K$ , and Newton's law of cooling $\dfrac{d\theta}{dt} = -k(\theta - \theta_{\text{room}})$ gives the shifted exponential $\theta = \theta_{\text{room}} + A e^{-kt}$ .
Statistics section 4 — Regression: in the lab, $k$ is estimated by plotting $\ln N$ against $t$ . Since $\ln N = \ln A + kt$ , the gradient of the linear regression is $k$ and the intercept is $\ln A$ . This synoptic move from non-linear to linear via log-transformation appears in both Pure section 6 and Statistics section 4.

Mark-scheme literacy

Modelling questions on 9MA0 split AO marks more evenly than procedural pure-maths questions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	45–55%	Substituting initial conditions, applying $\ln$ to both sides, evaluating $e^{kt}$ at given times
AO2 (reasoning / interpretation)	20–30%	Recognising $k < 0$ for decay, identifying $A$ as the initial value, interpreting asymptotes and limits
AO3 (problem-solving / modelling)	25–35%	Setting up the model from first principles, evaluating fit, stating and justifying limitations, proposing refinements

Examiner-rewarded phrasing: "the rate of change is proportional to the current population, so $\dfrac{dN}{dt} = kN$ "; "since the population is decaying, $k < 0$ "; "as $t \to \infty$ , $e^{kt} \to 0$ , so $N \to 0$ — the model predicts eventual extinction"; "the model assumes constant per-capita growth rate, which is unrealistic when resources are limited". Phrases that lose marks: "exponential decay" used as a label without the equation; " $A$ is a constant" without identifying it as the initial value; quoting a numerical answer without a unit; failing to state whether $k$ is positive or negative when the question asks you to find it.

A specific Edexcel pattern to watch: questions ending "evaluate the model" or "comment on the suitability of the model" carry an explicit AO3 mark that is not awarded for re-quoting the equation. The mark requires a sentence about a limitation (e.g. "the model assumes the environment is unchanged") and its consequence ("so the prediction is unreliable beyond the data range").

Grade-band model answers

3-mark question

Question: A bacterial population is modelled by $N = 200 e^{0.05 t}$ , where $t$ is measured in hours. State the initial population and the value of $N$ when $t = 10$ .

Grade C response (~210 words):

The initial population is when $t = 0$ . Substituting:

$N = 200 e^{0 \times 0.05} = 200 e^{0} = 200 \times 1 = 200$ .

So the initial population is 200.

When $t = 10$ :

$N = 200 e^{0.05 \times 10} = 200 e^{0.5} = 200 \times 1.6487... = 329.74...$ .

So $N \approx 330$ when $t = 10$ .

Examiner commentary: Full marks (3/3). The candidate correctly identifies that "initial" means $t = 0$ , applies $e^0 = 1$ explicitly, and evaluates the second part to a sensible number of significant figures. The rounding to 330 is appropriate for a population (which must be a whole number). A weaker candidate would either miss that $e^0 = 1$ (and evaluate $200 e^{0.05 \times 0}$ on a calculator without seeing the simplification) or quote $329.74$ as a population — neither catastrophic, but the rounded 330 is the physically meaningful answer. Working is brief but every step is justified, which is exactly the right balance for a 3-mark question.

Grade A response (~250 words):*

For the initial population, set $t = 0$ :

$N(0) = 200 e^{0.05 \times 0} = 200 e^{0} = 200 \times 1 = 200$

So the initial population is 200 — note that this is precisely the coefficient $A$ in the model $N = A e^{kt}$ , which is the standard interpretation: $A$ is always the value at $t = 0$ .

For $t = 10$ :

$N(10) = 200 e^{0.05 \times 10} = 200 e^{0.5}$

Using $e^{0.5} \approx 1.6487$ :

$N(10) \approx 200 \times 1.6487 = 329.74$

Since $N$ represents a count of organisms, we round to the nearest whole number: $N(10) \approx 330$ .

Exponential Growth and Decay

Exponential Growth and Decay

The General Model

Exponential Growth Examples

Population Growth

Compound Interest

Exponential Decay Examples

Radioactive Decay

Newton's Law of Cooling

Finding the Growth/Decay Constant k

Half-Life and Doubling Time

Half-Life (decay)

Doubling Time (growth)

Limitations of Exponential Models

Summary

A-Level Deep Dive: Exponential Growth and Decay

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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