IB Maths AI HL Modelling with Functions Paper 1 & 2 Growth, decay, half-life, cooling ~9 min read

Exponential Models

Exponential models — y = Aa^t or y = Ae^kt (sometimes with a constant added on) — describe quantities that change by a constant percentage per unit time: population growth, radioactive decay, half-lives, depreciation, cooling, and saturation.

📘 What you need to know

Growth/decay (base form): y = Aa^t; growth if a > 1, decay if 0 < a < 1.
Continuous form: y = Ae^kt; growth if k > 0, decay if k < 0.
Initial value: A = y(0). The exponential factor multiplies it as t increases.
Percent change: a = 1 + r for growth rate r; a = 1 − r for decay (depreciation).
Half-life form: M(t) = M₀(½)^t/T where T is the half-life.
Bounded models: y = Ae^−kt + c (cooling toward asymptote c) or y = L(1 − e^−kt) (saturation toward L).

The three modelling shapes

Three patterns cover almost every exponential application. Pure growth (population, compound interest, virus spread) uses Aa^t with a > 1 — the curve starts at A and climbs without bound. Pure decay (depreciation, radioactive isotope mass) uses the same form with 0 < a < 1 — the curve falls toward zero. Bounded models add a constant: cooling falls from a hot start toward room temperature (y → c), and saturation rises from zero up to a ceiling (y → L). The asymptote in each case carries physical meaning.

Building a model from data

Two common setups. Given the initial value and a rate (“$24,000, depreciating 12% per year”), write V(t) = 24000(0.88)^t directly — the base is 1 − 0.12. Given two data points (t₁, y₁) and (t₂, y₂), divide one equation by the other to eliminate A and solve for the base or growth constant, then back-substitute for A. For continuous form Ae^kt, the divide trick gives e^kΔt = y₂/y₁, and k = ln(y₂/y₁)/Δt.

Pure growth runs away to infinity; bounded models tend toward an asymptote that carries physical meaning (room temperature, market saturation).

Exponential model at a glance y = Aa^t or y = Ae^kt half-life form M = M₀(½)^t/T · cooling y = Ae^−kt + c · saturation y = L(1 − e^−kt)

Solving for time: logs or GDC

“When does the population reach …?” and “How long until the coffee is …?” are solve-for-t problems. Isolate the exponential first, then take a logarithm: ln on both sides for an e-form, or log₁₀/log on both sides for an a^t form. Equivalently, type both sides into the GDC and use the intersect feature. Always check that your answer is positive (negative time is nonsensical) and inside the model’s domain.

Log shortcut: for Aa^t = B, taking logs gives t = log(B/A) / log(a). Any base works as long as you use the same base on top and bottom.

🧭 Recipe — exponential modelling

Identify the model type: growth, decay, half-life, cooling, or saturation.
Write the form: Aa^t, Ae^kt, plus a constant if the model has an asymptote.
Use the given data to find the parameters: initial value → A; rate or 2nd point → a or k.
Check the model by substituting back into the data points.
Apply: predict by substituting t; find time by isolating the exponential and using logs (or GDC intersect).

Worked examples

WE 1

Population growth from data

A town has population 5000 in 2010 and 8000 in 2018. Assume the population grows as P(t) = P₀·a^t where t = years since 2010. (a) Find a (4 d.p.). (b) Predict the population in 2030.

(a) initial: P₀ = 5000 P(8) = 5000a⁸ = 8000 a⁸ = 8000/5000 = 1.6 a = 1.6⁷ֿ⁸ ≈ 1.0605 P(t) ≈ 5000(1.0605)ᵗ (b) t = 20 (year 2030) P(20) = 5000(1.0605)²⁰ = 5000(1.6)²·⁵ ≈ 5000 × 3.2382 P(2030) ≈ 16,191 about 16,000 people — round sensibly in context.

WE 2

Radioactive decay (half-life)

An 80 g sample of an isotope has a half-life of 30 years. (a) Write M(t) for the mass after t years. (b) Find the mass after 100 years (2 d.p.). (c) After how long is only 5 g left?

(a) half-life form M(t) = 80(½)ᵗֿ³⁰ (b) M(100) M = 80(½)¹⁰⁰ֿ³⁰ = 80(½)³·³³ ≈ 80 × 0.0992 ≈ 7.94 M(100) ≈ 7.94 g (c) solve M(t) = 5 5 = 80(½)ᵗֿ³⁰ (½)ᵗֿ³⁰ = 5/80 = 1/16 = (½)⁴ t/30 = 4 ⇒ t = 120 t = 120 years 1/16 is exactly 4 half-lives — clean answer, no GDC needed.

WE 3

Newton’s cooling toward an asymptote

A cup of coffee in a room cools according to T(t) = 75e^−0.05t + 20, where T is in °C and t in minutes. (a) Initial temperature. (b) Temperature after 15 minutes (2 d.p.). (c) Time when T = 35°C (2 d.p.). (d) Long-run temperature.

(a) T(0) = 75e⁰ + 20 = 75 + 20 = 95 95 °C (b) T(15) = 75e⁻⁰·⁷⁵ + 20 e⁻⁰·⁷⁵ ≈ 0.4724 T ≈ 75(0.4724) + 20 ≈ 55.43 T(15) ≈ 55.43 °C (c) solve 35 = 75e⁻⁰·⁰⁵ᵗ + 20 15 = 75e⁻⁰·⁰⁵ᵗ e⁻⁰·⁰⁵ᵗ = 0.2 −0.05t = ln(0.2) ⇒ t = −ln(0.2)/0.05 t ≈ 32.19 min (d) as t → ∞, e⁻⁰·⁰⁵ᵗ → 0 T → 20 °C (room temperature)

WE 4

Compound depreciation

A car bought new for $24,000 loses 12% of its value each year. (a) Write a model V(t) for its value after t years. (b) Find its value after 5 years. (c) After how long is it worth $10,000 (2 d.p.)?

(a) base = 1 − 0.12 = 0.88 V(t) = 24000(0.88)ᵗ (b) V(5) = 24000(0.88)⁵ (0.88)⁵ ≈ 0.5277 V ≈ 24000 × 0.5277 V(5) ≈ $12,665.57 (c) solve 10000 = 24000(0.88)ᵗ (0.88)ᵗ = 10000/24000 = 5/12 t · ln(0.88) = ln(5/12) t = ln(5/12)/ln(0.88) ≈ (−0.8755)/(−0.1278) t ≈ 6.85 years

WE 5

Saturation model

The number of people in a town of 5000 who have heard a rumour t days after it starts is modelled by N(t) = 5000(1 − e^−0.3t). (a) N(0). (b) N(7) (nearest whole). (c) When does N = 4000? (d) Long-term value.

(a) N(0) = 5000(1 − 1) = 0 0 (no one has heard yet) (b) N(7) = 5000(1 − e⁻²·¹) e⁻²·¹ ≈ 0.1225 N ≈ 5000(0.8775) ≈ 4387.72 N(7) ≈ 4388 people (c) 4000 = 5000(1 − e⁻⁰·³ᵗ) 0.8 = 1 − e⁻⁰·³ᵗ e⁻⁰·³ᵗ = 0.2 ⇒ t = −ln(0.2)/0.3 t ≈ 5.36 days (d) as t → ∞, e⁻⁰·³ᵗ → 0 N → 5000 (everyone hears)

WE 6

Continuous model from two data points

A population (in millions) is modelled by P(t) = Ae^kt, where t = years since 2000. In 2005 P = 12; in 2015 P = 18. (a) Find k (4 d.p.) and A (2 d.p.). (b) Predict P in 2025.

(a) form two equations Ae⁵ᵏ = 12 & Ae¹⁵ᵏ = 18 divide to eliminate A e¹⁰ᵏ = 18/12 = 1.5 10k = ln(1.5) ⇒ k = ln(1.5)/10 k ≈ 0.0405 find A: e⁵ᵏ = √1.5 A = 12/√1.5 = 4√6 A ≈ 9.80 million (b) P(25) = A · e²⁵ᵏ = A · (e⁵ᵏ)⁵ = (12/√1.5) · (√1.5)⁵ = 12 · (√1.5)⁴ = 12 · 1.5² = 12 × 2.25 = 27 P(2025) = 27 million (exact) algebra collapses to a clean answer — check by GDC.

💡 Top tips

Read the wording for the model type: “doubles”, “half-life”, “depreciates by X%”, “approaches” all hint at the right form.
Percent ↔ base: 7% growth ⇒ base 1.07; 12% loss ⇒ base 0.88.
Divide two equations to eliminate A when you’re fitting Ae^kt to two points.
Use ln for e-forms, log/log for base-a forms — or just let the GDC’s intersect tool handle it.
State the asymptote for cooling/saturation models — it’s a natural “what happens long-term?” answer.

⚠ Common mistakes

Using 0.12 instead of 0.88 for 12% depreciation — the base is what remains, not what’s lost.
Adding the rate to the initial value (linear thinking) instead of multiplying by a base each period.
Forgetting to take logs when solving for t; you can’t simply divide an exponent away.
Mixing ln and log₁₀ — pick one base and use it consistently on both sides.
Reporting a negative time from a log calculation as if it were valid — it usually means a sign or domain error.

Next up: Logarithmic Models — y = a + b ln(x), used for sound intensity (decibels), pH, and growth that flattens out.

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