Modelling with Exponentials

This lesson brings together everything from the topic to tackle modelling problems — a key focus of the Edexcel 9MA0 specification. You will be expected to set up exponential models from context, determine constants from data, use the model to make predictions, and critically evaluate its suitability.

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The Modelling Cycle

1. Identify the appropriate model from the context
2. Set up the equation with unknown constants
3. Determine the constants using given data
4. Use the model to make predictions
5. Evaluate the model's suitability and limitations

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Setting Up Models

Growth: y = Aeᵏᵗ (k > 0)

Used when the rate of increase is proportional to the current value.

Example: A colony of insects is growing. Initially there are 200 insects. After 5 days there are 350. Assuming exponential growth, find the model.

• At t = 0: 200 = Ae⁰, so A = 200
• At t = 5: 350 = 200e⁵ᵏ
• e⁵ᵏ = 350/200 = 7/4
• 5k = ln(7/4)
• k = ln(7/4)/5 ≈ 0.1120

Model: N = 200e^(0.112t)

Decay: y = Aeᵏᵗ (k < 0) or y = Ae⁻ᵏᵗ (k > 0)

Used when the rate of decrease is proportional to the current value.

Example: A drug concentration in the blood starts at 80 mg/L and falls to 50 mg/L after 4 hours.

• At t = 0: 80 = Ae⁰, so A = 80
• At t = 4: 50 = 80e⁴ᵏ
• e⁴ᵏ = 50/80 = 5/8
• 4k = ln(5/8)
• k = ln(5/8)/4 ≈ -0.1178

Model: C = 80e^(-0.1178t)

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Models with an Asymptote: y = A + Be⁻ᵏᵗ

When the quantity approaches a non-zero value, the model has a horizontal asymptote.

Example: A room is heated. The temperature follows T = 22 - 15e^(-0.1t), where T is in °C and t is in minutes.

• At t = 0: T = 22 - 15e⁰ = 22 - 15 = 7°C (initial temperature)
• As t → ∞: e^(-0.1t) → 0, so T → 22°C (the asymptote — room equilibrium)
• At t = 10: T = 22 - 15e⁻¹ ≈ 22 - 15(0.3679) ≈ 22 - 5.52 ≈ 16.5°C

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Determining Constants from Data

Method 1: Direct Substitution

Given the form and two data points, substitute to form two equations and solve simultaneously.

Method 2: Using Logarithmic Graphs

Plot ln(y) against t to get a straight line. Read off the gradient (k) and intercept (ln A).

Method 3: Using the Ratio Method

If y₁ = Aeᵏᵗ¹ and y₂ = Aeᵏᵗ², then y₂/y₁ = e^(k(t₂-t₁)), so k = ln(y₂/y₁)/(t₂ - t₁).

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Making Predictions

Once the model is established, substitute values to predict:

Example (continuing the insect model): Predict the population after 12 days.

N = 200e^(0.112 × 12) = 200e^(1.344) ≈ 200 × 3.835 ≈ 767 insects

Example: When will the population reach 1000?

1000 = 200e^(0.112t)

• e^(0.112t) = 5
• 0.112t = ln(5)
• t = ln(5)/0.112 ≈ 14.4 days

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Evaluating Models — Exam Technique

You will often be asked: "Comment on the reliability of your answer" or "State one limitation of this model."

Standard Evaluation Points

Concern	Explanation
Extrapolation	Predictions far beyond the data range may be unreliable
Unlimited growth	Exponential growth cannot continue indefinitely (resources are finite)
Continuous vs discrete	The model assumes continuous change but populations are discrete
Constant rate	The model assumes k is constant, but in reality it may change
Initial conditions	The model may not be accurate for very small t if there is a lag phase
Environmental factors	External changes (disease, predation, policy changes) are not accounted for

Exam Tip: When asked to "comment on reliability," make a specific point related to the context. For example: "The model predicts the population will exceed 10000, but this is unlikely because the available food supply would limit growth."

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Worked Exam-Style Problem

The value V (in £) of a car can be modelled by V = 18000e⁻⁰·²ᵗ, where t is the age of the car in years.

(a) State the new price of the car. When t = 0: V = 18000e⁰ = £18000

(b) Find the value after 3 years. V = 18000e^(-0.2 × 3) = 18000e⁻⁰·⁶ ≈ 18000 × 0.5488 ≈ £9879

(c) Find the age when the value drops below £2000. 2000 = 18000e^(-0.2t)

• e^(-0.2t) = 2000/18000 = 1/9
• -0.2t = ln(1/9) = -ln(9)
• t = ln(9)/0.2 ≈ 2.197/0.2 ≈ 10.99 years So the value drops below £2000 after about 11 years.

(d) Give one reason why the model may not be suitable for large t. The model predicts V → 0 as t → ∞, meaning the car eventually has zero value. In reality, classic or rare cars can increase in value over time. Also, the car's value cannot drop below its scrap metal value, so the model breaks down for very large t.

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Summary

• Exponential models of the form y = Aeᵏᵗ (growth) or y = Ae⁻ᵏᵗ (decay) are used when the rate of change is proportional to the current value.
• Models with asymptotes (y = A + Be⁻ᵏᵗ) describe quantities approaching a limiting value.
• Determine constants using substitution, log graphs, or the ratio method.
• Always consider model limitations: extrapolation, constant rate assumptions, and contextual factors.
• In the exam, link your evaluation to the specific context of the question.

A-Level Deep Dive: Modelling with Exponentials

Spec mapping

Edexcel 9MA0-01 specification section 6 — Exponentials and logarithms (modelling sub-strand 6.7) 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). This sits at the synoptic crossroads of section 6 (logs to linearise data), Year 2 Pure section 7 (further differentiation, including $\frac{d}{dt}e^{kt} = ke^{kt}$dtd​ekt=kekt), section 8 (further integration) and especially Year 2 Pure section 11 (differential equations of the form $\frac{dN}{dt} = kN$dtdN​=kN, separation of variables, and constructing models). There is also an explicit overlap with Statistics section 4 (regression), where logarithmic transformations are used to fit $y = ax^n$y=axn or $y = ab^x$y=abx to bivariate data via linear regression of $\log y$logy against $\log x$logx or $x$x. Modelling questions are AO3-heavy: typically a substantial share of marks fall under AO3, with significant AO2 reasoning marks for "evaluate" or "comment on assumptions" sub-parts. The Edexcel formula booklet provides $\frac{d}{dx}e^{kx} = ke^{kx}$dxd​ekx=kekx and the integral of $\frac{1}{x}$x1​ but not the general solution of $\frac{dN}{dt} = kN$dtdN​=kN — students must derive $N = N_0 e^{kt}$N=N0​ekt from first principles or quote it as bookwork.

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Worked example with full mark scheme

Question (8 marks): A cup of coffee at temperature $T$T °C cools in a room of constant temperature $20$20 °C. Newton's law of cooling models the temperature as

$T = 20 + A e^{-kt}$T=20+Ae−kt

where $t$t is time in minutes, and $A$A and $k$k are positive constants. At $t = 0$t=0 the coffee is at $90$90 °C. After $5$5 minutes the temperature has fallen to $70$70 °C.

(a) Find the values of $A$A and $k$k, giving $k$k to 3 significant figures. (4)

(b) Find the time, to the nearest minute, at which the coffee first reaches a drinkable $55$55 °C. (2)

(c) State one assumption underlying this model and comment on whether it is reasonable. (2)

Solution with mark scheme:

(a) Step 1 — use the initial condition.

At $t = 0$t=0, $T = 90$T=90:

$90 = 20 + A e^{0} = 20 + A \implies A = 70$90=20+Ae0=20+A⟹A=70

M1 (AO1.1a) — substituting $t = 0$t=0 into the model and solving for the constant of integration. A1 (AO1.1b) — $A = 70$A=70.

Step 2 — use the second data point.

At $t = 5$t=5, $T = 70$T=70:

$70 = 20 + 70 e^{-5k} \implies 50 = 70 e^{-5k} \implies e^{-5k} = \dfrac{5}{7}$70=20+70e−5k⟹50=70e−5k⟹e−5k=75​

Taking natural logs:

$-5k = \ln\!\left(\dfrac{5}{7}\right) \implies k = -\dfrac{1}{5}\ln\!\left(\dfrac{5}{7}\right) = \dfrac{1}{5}\ln\!\left(\dfrac{7}{5}\right) \approx 0.0673$−5k=ln(75​)⟹k=−51​ln(75​)=51​ln(57​)≈0.0673

M1 (AO1.1b) — substituting the second point and isolating the exponential. A1 (AO1.1b) — $k \approx 0.0673$k≈0.0673 to 3 s.f. The sign-flip step $\ln(5/7) = -\ln(7/5)$ln(5/7)=−ln(7/5) is examiner-rewarded for clarity but not required for the mark.

(b) Step 1 — solve for $t$t when $T = 55$T=55.

$55 = 20 + 70 e^{-kt} \implies e^{-kt} = \dfrac{35}{70} = \dfrac{1}{2}$55=20+70e−kt⟹e−kt=7035​=21​

$-kt = \ln(1/2) = -\ln 2 \implies t = \dfrac{\ln 2}{k} = \dfrac{\ln 2}{0.0673} \approx 10.30$−kt=ln(1/2)=−ln2⟹t=kln2​=0.0673ln2​≈10.30

M1 (AO1.1b) — applying the model with $T = 55$T=55 and isolating $t$t via natural logs. A1 (AO1.1b) — $t \approx 10$t≈10 minutes.

(c) One assumption (any one of):

• The room temperature is constant at $20$20 °C.
• The rate of cooling is proportional to the temperature difference $(T - 20)$(T−20).
• There is no other heat source or sink (no stirring, lid, draught).

Comment: the constant-room-temperature assumption is reasonable in a closed indoor environment over short timescales ($<$< 30 minutes) but breaks down outdoors or when sunlight, draughts or radiator cycling change ambient temperature. The rate-proportional assumption is empirically well-supported for small-to-moderate temperature differences but underestimates cooling at very large differences where radiative heat loss (proportional to $T^4$T4 via Stefan–Boltzmann) becomes non-negligible.

B1 (AO3.5a) — stating an assumption. B1 (AO3.5b) — sensible critique tied to the physical context. Generic comments like "it might be wrong" earn nothing.

Total: 8 marks (M2 A4 B2, split AO1 = 5, AO3 = 3).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The mass $m$m grams of a radioactive sample is modelled by $m = m_0 e^{-\lambda t}$m=m0​e−λt, where $t$t is in years and $\lambda > 0$λ>0.

(a) Show that the half-life $T_{1/2}$T1/2​ — the time for the sample to halve in mass — is given by $T_{1/2} = \dfrac{\ln 2}{\lambda}$T1/2​=λln2​ and is independent of the starting mass $m_0$m0​. (3)

(b) Carbon-14 has $\lambda = 1.21 \times 10^{-4}$λ=1.21×10−4 per year. A sample's mass is three-tenths of its original value. Estimate its age and comment on one limitation of using this model for archaeological dating. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — setting $m = m_0/2$m=m0​/2 in the model: $m_0/2 = m_0 e^{-\lambda T_{1/2}}$m0​/2=m0​e−λT1/2​.
• M1 (AO2.1) — cancelling $m_0$m0​ (justifying $m_0 \neq 0$m0​=0) and taking logs: $-\lambda T_{1/2} = \ln(1/2) = -\ln 2$−λT1/2​=ln(1/2)=−ln2.
• A1 (AO2.4) — concluding $T_{1/2} = \ln 2 / \lambda$T1/2​=ln2/λ with the explicit observation that $m_0$m0​ cancelled, hence the half-life is independent of starting mass.

(b)

• M1 (AO3.1a) — setting $0.3 m_0 = m_0 e^{-\lambda t}$0.3m0​=m0​e−λt and reducing to $t = -\ln(0.3)/\lambda$t=−ln(0.3)/λ.
• A1 (AO1.1b) — $t \approx 9{,}950$t≈9,950 years (or any answer in the range 9,900–10,000).
• B1 (AO3.5b) — sensible limitation: e.g. assumes the initial $^{14}\text{C}$14C atmospheric concentration matched today's value (which calibration curves now correct for); assumes no contamination; assumes a pure first-order decay rate (well-supported physically); breaks down for samples beyond the detection range where remaining $^{14}\text{C}$14C falls below measurement thresholds.

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. This is a heavy AO3 distribution — characteristic of modelling sub-strand questions where reasoning and limitation-evaluation marks dominate.

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Synoptic links

Connects to:

1. Year 2 Pure section 11 — Differential equations: the exponential model $N = N_0 e^{kt}$N=N0​ekt is the solution of $\frac{dN}{dt} = kN$dtdN​=kN, derived by separating variables: $\int \frac{1}{N}\,dN = \int k\,dt$∫N1​dN=∫kdt, giving $\ln|N| = kt + C$ln∣N∣=kt+C, hence $N = A e^{kt}$N=Aekt. Every exponential modelling question implicitly invokes a first-order ODE.

2. Year 2 Pure section 8 — Integration: integrating $e^{kt}$ekt to recover total quantity (e.g. total radiation emitted, total bacteria-hours) uses $\int e^{kt}\,dt = \frac{1}{k}e^{kt} + C$∫ektdt=k1​ekt+C. Modelling questions often combine an instantaneous-rate model with an integral-quantity follow-up.

3. Statistics section 4 — Regression and correlation: when fitting an exponential model $y = ab^x$y=abx to data, the standard method is to take logs ($\log y = \log a + x \log b$logy=loga+xlogb), perform linear regression on $(\,x, \log y)\,$(x,logy), and back-transform. This directly tests the synoptic link between log-laws and statistical fitting.

4. Mechanics — variable-acceleration motion and resistance: a body falling through air with resistance proportional to velocity satisfies $\frac{dv}{dt} = g - kv$dtdv​=g−kv, whose solution $v = \frac{g}{k}(1 - e^{-kt})$v=kg​(1−e−kt) is a "saturating exponential" identical in form to capacitor-charging in physics. Recognising the same model across mechanics and physics earns AO2/AO3 cross-context marks.

5. Biology — limited-growth (logistic) populations: a population growing in a finite-resource environment satisfies $\frac{dN}{dt} = rN(1 - N/K)$dtdN​=rN(1−N/K) — derived in the additional A* section below. Questions in this strand explicitly ask candidates to choose between an unbounded exponential model and a logistic model based on context and data.

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Mark-scheme literacy

Modelling questions on 9MA0 split AO marks heavily toward AO3:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	smaller share	Substituting initial conditions, taking logs to solve for $t$t, applying the formula correctly, computing numerical answers
AO2 (reasoning / interpretation)	moderate share	Choosing the correct model from context, justifying parameter signs, interpreting the meaning of $k$k or $\lambda$λ in physical terms
AO3 (problem-solving / modelling)	dominant share	Evaluating model assumptions, commenting on limitations, identifying the validity range, suggesting model refinements (e.g. switching to logistic, adding a constant offset)

Examiner-rewarded phrasing for "evaluate the model" questions: name the assumption explicitly (e.g. "the model assumes the rate of cooling is proportional to $(T - T_{\text{room}})$(T−Troom​)"), then say whether it is reasonable in this specific context and why. Examiner-rewarded phrasing for "comment on assumptions" questions: identify the quantitative validity range (e.g. "the model is reasonable for short times but breaks down for longer periods because…") rather than vague "it might not always work". Phrases that lose marks: "the model is just an approximation" (true but content-free); "it could be wrong" (no engagement with context); ignoring the assumption-evaluation sub-part entirely (forfeits AO3 marks even if the calculation is perfect).

A specific Edexcel pattern: when a question gives a model and asks for predictions outside the data range used to fit it, the AO3 mark is for stating that extrapolation beyond the fitted range is unreliable. This is almost always a single mark and is almost always missed.

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Grade-band model answers

3-mark question

Question: A bacterial colony's population $N$N thousand at time $t$t hours is modelled by $N = 5 e^{0.4t}$N=5e0.4t. State one assumption of this model and explain whether it is reasonable in the long term.

Grade C response (~210 words):

The model assumes the population grows at a constant rate proportional to its current size. This means the bigger the colony gets, the faster it grows. In the long term this is not reasonable because the bacteria will eventually run out of food, space or nutrients. A real colony would slow down and stop growing, but the exponential model just keeps going up forever. So while the model might work for a short time near the start, it will overestimate the population once resources start to run out.

Examiner commentary: Full marks (3/3). The candidate correctly identifies the proportionality assumption (B1), recognises the unbounded growth flaw (B1), and gives a context-specific reason — finite resources — for why the assumption breaks down (B1). The phrase "constant rate proportional to its current size" is the right level of precision for a 3-mark question. A weaker answer would say "it grows fast" or "it gets too big", which gestures at the issue without naming it. The "in the long term" phrasing of the question is a signal that examiners want the limitation framed in terms of validity range, and this answer delivers that.