Mathematics Advanced • Year 12 • Module 6 • Lesson 13
Growth and Decay Models
Apply exponential growth and decay to epidemiology, finance, archaeology, ecology and pharmacology.
Problem 1 — Early-pandemic exponential growth (R0)
In the early phase of an outbreak, infections obey
dI/dt = r I, r = (R0 − 1)/D
where R0 is the basic reproduction number and D is the infectious period (days). Take R0 = 2.5 and D = 5 days. Initial infections I(0) = 100.
Set up: What are we solving for?
(i) Compute r and state its units. Write I(t). 2 marks
(ii) Find I(14) and the doubling time. 3 marks
(iii) After lockdown, R0 drops to 0.6 (so r < 0). What does "flattening the curve" mean mathematically? In one sentence, state the new long-run behaviour of I(t). 2 marks
Stuck? Revisit lesson § The Model — COVID callout.Problem 2 — Continuous inflation (price of a takeaway coffee)
A Sydney coffee costs $5 today and prices grow continuously at 3.2% per year.
Set up: What are we solving for?
(i) Write P(t) and find P(20) to the nearest cent. 2 marks
(ii) Find the doubling time (years), giving both an exact form and a decimal to 1 d.p. 2 marks
(iii) In what year (from today) will a coffee first cost $10? Give an exact and decimal answer. 2 marks
Problem 3 — Dating Indigenous rock art (carbon-14)
Charcoal embedded in pigment from rock art in the Kimberley region contains 41% of the carbon-14 ratio of a living organism. Carbon-14 decay: dN/dt = −k N, with half-life 5730 years.
Set up: What are we solving for?
(i) Find k (exact form) and state its units. 2 marks
(ii) Solve 0.41 = e−kt for t. Give an answer to the nearest 100 years. 3 marks
(iii) Carbon-14 dating is reliable up to roughly 50,000 years. Briefly explain (1-2 sentences) why beyond that the method fails. 2 marks
Problem 4 — Endangered koala population
A NSW koala population is in decline. In 2010 the population was 36,000; by 2020 it had fallen to 24,000. Assume exponential decline P(t) = P0 ekt with t in years since 2010.
Set up: What are we solving for?
(i) Find k (exact form using ln(2/3) and decimal to 4 d.p.). Is this growth or decay? 2 marks
(ii) If the trend continues, predict the population in 2040. 2 marks
(iii) "Functionally extinct" is sometimes defined as population below 1000 in a region. In what year would this threshold be crossed (under the same model)? 3 marks
Stuck? Set P(t) = 1000 and solve for t (years since 2010); add 2010 for the calendar year.Problem 5 — Dosing schedule (ibuprofen)
Ibuprofen has a plasma half-life of about 2 hours. A patient takes a 400 mg dose at t = 0.
Set up: What are we solving for?
(i) Write C(t) (mg in bloodstream) and evaluate C(6) — both directly via the formula and via the half-life shortcut. They should agree. 3 marks
(ii) Pharmacists consider a second dose safe once C(t) ≤ 80 mg. How many hours after the first dose can the second be taken? Give an exact and decimal answer. 3 marks
(iii) Why does the simple exponential model break down for repeated dosing every 6 hours? (One sentence.) 1 mark
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — COVID growth
Set up. Compute the continuous growth rate r from R0 and D, write I(t), evaluate and find doubling time, then interpret lockdown.
(i) r = (2.5 − 1)/5 = 0.3 / day. I(t) = 100 e0.3 t.
(ii) I(14) = 100 e4.2 ≈ 100 × 66.69 ≈ 6669 infections. Doubling time = (ln 2)/0.3 ≈ 2.31 days.
(iii) "Flatten the curve" means reducing r (so that R0 < 1) to turn exponential growth into exponential decay. With r < 0, I(t) → 0 as t → ∞ — the outbreak dies out.
Problem 2 — Coffee inflation
Set up. Standard exponential growth in price.
(i) P(t) = 5 e0.032 t. P(20) = 5 e0.64 ≈ 5 × 1.8965 ≈ $9.48.
(ii) tdouble = (ln 2)/0.032 ≈ 21.7 years.
(iii) 10 = 5 e0.032 t ⇒ e0.032 t = 2 ⇒ t = (ln 2)/0.032 ≈ 21.7 years (same as doubling time — makes sense, since $5 → $10 is exactly doubling).
Problem 3 — Carbon-14 rock art
Set up. Use the half-life to find k, then solve 0.41 = e−kt.
(i) k = (ln 2)/5730 ≈ 1.2097 × 10−4 / year.
(ii) 0.41 = e−kt ⇒ −kt = ln(0.41) ⇒ t = −ln(0.41)/k = ln(1/0.41)/k ≈ 0.8916 / (1.2097 × 10−4) ≈ 7370 years (to nearest 100: 7400 years).
(iii) After roughly 8–10 half-lives (≈ 50,000 years), so little carbon-14 remains that detection error swamps the measurement; the signal-to-noise ratio becomes too small for reliable dating.
Problem 4 — Koala decline
Set up. Exponential decay model with negative k.
(i) 24000 = 36000 e10k ⇒ e10k = 24/36 = 2/3 ⇒ k = (ln(2/3))/10 ≈ −0.0405 /year. Decay (k < 0).
(ii) 2040 corresponds to t = 30. P(30) = 36000 e30k = 36000 (2/3)³ = 36000 × 8/27 ≈ 10,667.
(iii) 1000 = 36000 ekt ⇒ ekt = 1/36 ⇒ t = ln(1/36)/k = −ln(36)/(−0.0405) ≈ 3.5835 / 0.0405 ≈ 88.5 years from 2010 ⇒ year 2098–2099.
Problem 5 — Ibuprofen dosing
Set up. Half-life 2 hours ⇒ k = (ln 2)/2 ≈ 0.3466.
(i) C(t) = 400 e−0.3466 t. C(6) = 400 e−3 ln 2 = 400 × 2−3 = 50 mg. Shortcut: 6 h = 3 half-lives ⇒ 400 × (1/2)³ = 50 mg ✓.
(ii) 80 = 400 e−kt ⇒ e−kt = 0.2 ⇒ t = (ln 5)/k = 2 ln 5 / ln 2 ≈ 4.64 hours.
(iii) With repeated doses, drug accumulates: each new dose is added on top of what's still in the system, so the simple single-dose decay model underestimates the true plasma level — a multi-dose model with superposition is needed.