IB Maths AI SL Topic 3 — Modelling with Functions Paper 1 & 2 Growth & decay ~7 min read

Exponential Models

An exponential model describes anything that grows or decays at a constant percentage rate: compound interest, radioactive decay, bacteria multiplying, hot drinks cooling, a car losing value. The general form is f(x) = k·a^x + c (or f(x) = ke^rx + c). Each parameter has direct real-world meaning: c is the long-term boundary (asymptote), k + c is the initial value, and the base a (or rate r) controls how fast the change happens.

📘 What you need to know

Standard forms: f(x) = k·a^x + c (with a > 0, a ≠ 1) or f(x) = ke^rx + c using the natural base e ≈ 2.718.
Initial value = k + c, NOT just c: at x = 0 the exponential factor equals 1, so f(0) = k(1) + c = k + c.
Boundary y = c: the horizontal asymptote — the long-term limit the quantity approaches but never reaches. Room temperature, salvage value, background level, carrying capacity.
Growth vs decay: growth if a > 1 with k > 0 (or r > 0); decay if 0 < a < 1 with k > 0 (or r < 0). Negative k flips the curve below the asymptote.
Percentage interpretation: a = 1.04 means +4% per period; a = 0.85 means −15% per period (loss of 15% of the depreciable amount).
When to use exponential: the data changes by a roughly constant percentage per unit time, not by a constant amount (which would be linear). Real giveaway: doubling time or half-life is roughly steady.

Reading the parameters in context

c = boundary (asymptote). For a decay model (something cooling toward room temperature, a car losing value toward its scrap value, a drug clearing from the bloodstream), c is the long-term limit the quantity approaches but never reaches.

k + c = initial value. The value of f(0). Often the question gives this directly, allowing you to read off k once c is known.

Sign of k tells you which side of the boundary the curve sits. Positive k: curve approaches c from above (decay), or rises away from c (growth with no top limit). Negative k: curve approaches c from below (e.g. a pool heating up toward 30°C).

a (or r) sets the speed. Closer to 1 (or smaller |r|) means slower change; further from 1 (larger |r|) means faster.

Exponential model f(x) = k·a^x + c or f(x) = ke^rx + c

initial value f(0) = k + c · asymptote (boundary) y = c

The dashed red line is the boundary c = 20 (room temperature). The curve starts at k + c = 80 (initial coffee temperature) and decays toward the asymptote, halving its excess every hour.

Spotting an exponential pattern

The clue is constant percentage change. If a quantity changes by the same percentage each period — not the same amount — the model is exponential. Examples:

Decay: cooling toward room temperature, radioactive decay (constant fraction per half-life), car depreciation (X% loss per year), drug elimination from blood, capacitor discharge.

Growth: compound interest (X% gain per year), bacterial colonies (constant doubling time), early phase of viral spread, certain investments.

Linear vs exponential check: take differences between successive values. Constant differences ⇒ linear (add the same each step). Constant ratios ⇒ exponential (multiply by the same each step).

The initial-value trap

In a quadratic or cubic, the constant term IS the initial value. In an exponential f(x) = k·a^x + c, the constant term c is the boundary, not the initial value. The initial value is f(0) = k(1) + c = k + c. Forgetting this is the most common exponential-model error.

Limitations

An exponential growth model has no upper bound — it would predict infinite bacteria, infinite money, infinite tumour size. Real systems eventually run out of resources and the growth tapers off (use a different model for that phase). Likewise, exponential decay never reaches zero in finite time, but in practice “below detection threshold” or “negligible” is good enough.

🧭 Recipe — using an exponential model

Read off c as the boundary: the long-term value the quantity approaches (room temp, salvage value, baseline level). Then the asymptote is y = c.
Initial value = k + c: not just c. Substitute x = 0 to confirm.
Evaluate at a given input: substitute the value of x (or t) into the model. Round answers sensibly for the context (money to cents, populations to whole numbers).
For “when does the quantity reach N“: plot the model and the horizontal line y = N on the GDC, then use the intersect tool. No need for logarithms at AI SL.
Building a model from features: use c = asymptote, k = initial − asymptote, then one more data point to find a (or r).

Worked examples

WE 1

Coffee cooling — decay toward room temperature

A cup of coffee cools according to T(t) = 60(0.5)^t + 20, where T is in °C and t is in hours. (a) State the initial temperature. (b) Find the temperature after 3 hours. (c) State the long-term limit and interpret it.

(a) initial = k + c T(0) = 60(1) + 20 = 80 °C (b) substitute t = 3 T(3) = 60(0.5)³ + 20 = 60(0.125) + 20 = 7.5 + 20 = 27.5 °C (c) long-term limit as t → ∞, (0.5)ᵗ → 0 so T → 0 + 20 = 20 °C interpretation: 20 °C is room temperature (a) 80 °C · (b) 27.5 °C · (c) T → 20 °C (room temp) notice the “halving” pattern: T − 20 starts at 60, halves to 30 (at t=1), to 15 (at t=2), to 7.5 (at t=3). The base 0.5 IS the halving each hour.

WE 2

Bacterial growth

A bacterial colony’s size at time t hours is modelled by N(t) = 200 × 3^t. (a) State the initial population. (b) Find N(4). (c) Find the time at which the population reaches 5400.

(a) initial: at t = 0, 3⁰ = 1 N(0) = 200 × 1 = 200 (b) N(4) N(4) = 200 × 3⁴ = 200 × 81 = 16 200 (c) solve 200 × 3ᵗ = 5400 3ᵗ = 5400/200 = 27 3ᵗ = 3³ ⇒ t = 3 hours (or use GDC intersect tool to confirm) (a) 200 bacteria · (b) 16 200 after 4 h · (c) reaches 5400 at t = 3 h base 3 means the colony TRIPLES every hour: 200 → 600 → 1800 → 5400 → 16 200 → … Notice no +c term here, so the asymptote is just y = 0 (no bacteria below zero).

WE 3

Car depreciation — with a salvage boundary

A car’s value (in dollars) at time t years after purchase is modelled by V(t) = 24000(0.8)^t + 4000. (a) Find the original purchase price. (b) State the long-term value of the car and explain. (c) Use your GDC to find the year when the car is first worth less than $12 000 (give your answer to 3 s.f.).

(a) original price = V(0) = k + c V(0) = 24000(1) + 4000 = $28 000 (b) long-term: as t → ∞, (0.8)ᵗ → 0 V → 0 + 4000 = $4 000 interpretation: $4 000 is the SALVAGE/scrap value the car loses value but stays worth at least $4 000 (c) solve V(t) = 12000 on GDC plot Y₁ = 24000(0.8)ᵗ + 4000 and Y₂ = 12000 intersect tool gives t ≈ 4.92 years (a) $28 000 · (b) $4 000 (salvage) · (c) after 4.92 years CAREFUL: the initial value is k + c = $28 000, NOT just k = $24 000. Forgetting to add the boundary c is the classic exponential-model error.

WE 4

Endangered species — decay toward stable level

The population of an endangered species in a reserve at time t years is modelled by P(t) = 600(0.5)^t/10 + 200. (a) State the current population. (b) Find P(10) and P(20). (c) State the long-term population and explain its meaning.

(a) P(0) = k + c P(0) = 600(1) + 200 = 800 (b) substitute t = 10 and t = 20 P(10) = 600(0.5)¹ + 200 = 300 + 200 = 500 P(20) = 600(0.5)² + 200 = 150 + 200 = 350 (c) long-term: as t → ∞, (0.5)ᵗᐟ¹⁰ → 0 P → 0 + 200 = 200 interpretation: 200 is the stable long-term population (a) 800 · (b) P(10) = 500, P(20) = 350 · (c) P → 200 (stable level) the model says the population’s EXCESS over 200 halves every 10 years. Starting at 600 above the floor, it drops to 300 above (at t=10), then 150 above (at t=20), etc. The reserve never quite reaches 200, but gets arbitrarily close.

WE 5

Build a model from features

A scientist measures the amount (mg) of a substance over time, and proposes the model m(t) = k·a^t + c. The initial amount is 80 mg, the long-term level is 5 mg, and after 1 hour the amount is 42.5 mg. Find k, a and c.

Step 1 — long-term gives c c = 5 Step 2 — initial gives k k + c = 80 ⇒ k = 80 − 5 = 75 Step 3 — use m(1) = 42.5 to find a 75·a¹ + 5 = 42.5 75a = 37.5 a = 0.5 k = 75, a = 0.5, c = 5 ⇒ m(t) = 75(0.5)ᵗ + 5 three unknowns, three pieces of info. The “long-term” or asymptote always pins c immediately; the initial value pins k; then any extra data point fixes a. Quick three-step setup, no simultaneous equations needed.

WE 6

Heater warming a pool — negative k

A swimming pool’s water temperature (°C) at time t minutes after the heater is switched on is modelled by T(t) = −25(0.97)^t + 30. (a) State the initial temperature. (b) Find the temperature after 60 minutes (to 3 s.f.). (c) State the maximum temperature the pool can reach and explain why this is a limit.

(a) initial T(0) = k + c T(0) = −25(1) + 30 = 5 °C (b) substitute t = 60 T(60) = −25(0.97)⁶⁰ + 30 = −25(0.1614…) + 30 ≈ −4.03 + 30 ≈ 26.0 °C (c) long-term: as t → ∞, (0.97)ᵗ → 0 T → −25(0) + 30 = 30 °C k = −25 < 0, so the curve sits BELOW the asymptote T rises toward 30 °C but never reaches it (a) 5 °C · (b) 26.0 °C · (c) max 30 °C (heater limit, never reached) negative k flips the standard decay shape: the curve approaches the asymptote from BELOW instead of above. This is the model for things being WARMED toward a target (heating, charging a battery, filling a container with a max level).

💡 Top tips

Always check if k + c matches the stated initial value: this catches sign errors in k and confirms you’ve used the right form.
The base IS the multiplier per period: base 1.05 means +5% each period; base 0.9 means −10% each period. Read it straight off.
For “when does the quantity reach N“: plot the curve and y = N on the GDC, use the intersect tool. AI SL doesn’t expect log algebra here.
State the meaning of c in context: room temperature, salvage value, baseline level, carrying capacity. Numbers without interpretation lose marks.
Half-life / doubling-time shortcut: if the base is 0.5 (or 2), the period for the model is the half-life (or doubling time). For other bases, find the time when the model halves/doubles using GDC.

⚠ Common mistakes

Calling c the initial value: c is the BOUNDARY (asymptote). The initial value is k + c. This is the most common exponential-model error.
Forgetting that exponentials never reach the asymptote: a cooling object gets arbitrarily close to room temperature but never quite touches it. Don’t write “T = room temp at t = …” — it never happens exactly.
Mixing up “growth rate” and “base”: a growth rate of 4% means base 1.04 (not 0.04). The base is one PLUS the rate.
Predicting growth forever: an exponential growth model has no upper limit, but real populations and resources do. Outside the validated range, the model is just maths.
Sign error with negative k: if k < 0 the curve sits BELOW the asymptote (rising toward it). Don’t sketch it above.

Up next: Direct & Inverse Variation. Two variables vary directly when their ratio is constant (y = kxⁿ) and inversely when their product is constant (y = k/xⁿ). These describe physical laws — how time to complete a job depends on the number of workers, how light intensity depends on distance from the source.

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