Mathematics Advanced • Year 11 • Module 4 • Lesson 13
Exponential Growth and Decay
Apply P = P₀ e^(kt) to population, finance, cooling, archaeology and pharmacology contexts.
Problem 1 — Town population
A coastal town's population P(t) (in thousands) is modelled by P = 12 · e^(kt), where t is years since 2020. In 2025 the population was 14.4 thousand.
Set up: What are we solving for?
(i) Use the 2025 data to find k correct to 3 decimal places. 2 marks
(ii) Predict the population in 2030 (t = 10) to the nearest hundred people. 2 marks
(iii) Find the doubling time of the population to the nearest year, and state in which year the population first reaches 24 thousand. 3 marks
Stuck? Revisit lesson § Worked example 1.Problem 2 — Carbon-14 dating
Carbon-14 has a half-life of 5730 years. A wooden artefact contains 38% of the carbon-14 expected in a fresh sample.
Set up: What are we solving for?
(i) Find the decay constant k (exact form using ln 2). 1 mark
(ii) Set up the equation 0.38 = e^(kt) and solve for t. Round to the nearest hundred years. 3 marks
(iii) Estimate, in one sentence, how the answer would change if the laboratory's 38% reading had an uncertainty of ±2%. (Give a rough range, no need for full calculation.) 2 marks
Problem 3 — Continuous vs annual compounding
$5000 is invested at a stated rate of 8% per annum for 10 years.
Set up: What are we solving for?
(i) Find the final balance under annual compounding: A = 5000 · (1.08)^10. Round to the nearest dollar. 1 mark
(ii) Find the final balance under continuous compounding: A = 5000 · e^(0.08 · 10). Round to the nearest dollar. 2 marks
(iii) Compute the difference (continuous − annual) and express it as a percentage of the annual balance. Comment in one sentence on whether the gap grows with time. 3 marks
Stuck? Revisit lesson § SAQ 3 — compare compounding.Problem 4 — Drug clearance
After a dose, the amount of drug D (in mg) in the bloodstream of a patient follows D = 200 · e^(−0.25 t), where t is hours since the dose.
Set up: What are we solving for?
(i) Find D after 6 hours, to 1 decimal place. 1 mark
(ii) Find the half-life of the drug (in this patient) to the nearest 0.1 hour. 2 marks
(iii) A safe re-dose can only be administered when the level drops below 25 mg. How long after the dose must the patient wait? Round up to the nearest 0.5 hour. 3 marks
Problem 5 — Newton's law of cooling
A cup of coffee starts at 90 °C in a 20 °C room. After 10 minutes the temperature is 60 °C. Newton's law of cooling gives
T(t) = 20 + (90 − 20) · e^(kt) = 20 + 70 e^(kt)
Set up: What are we solving for?
(i) Use T(10) = 60 to find k correct to 3 decimal places. 3 marks
(ii) Predict the temperature after 20 minutes. Compare with a "guess" of 30 °C and explain the discrepancy in one sentence. 3 marks
(iii) Find when the temperature drops to 25 °C. Round to the nearest minute. 2 marks
Stuck on (ii)? The model halves the gap to room temperature, not the temperature itself.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Town population
Set up. We find the growth constant from one data point, then predict and find a doubling time.
(i) 14.4 = 12 · e^(5k) ⇒ e^(5k) = 1.2 ⇒ k = (ln 1.2)/5 ≈ 0.036 per year.
(ii) P(10) = 12 · e^(0.36) ≈ 12 · 1.4333 ≈ 17.2 thousand, i.e. 17 200 people.
(iii) Doubling time = (ln 2)/0.036 ≈ 19.0 years, so ≈ 19 years. Population reaches 24 thousand when 12 · e^(0.036 t) = 24 ⇒ t = (ln 2)/0.036 ≈ 19.0 years, i.e. in year 2020 + 19 = 2039.
Problem 2 — Carbon-14 dating
Set up. We use the half-life to find k, then solve for the age from the surviving fraction.
(i) k = −(ln 2)/5730 ≈ −1.21 × 10⁻⁴ per year.
(ii) 0.38 = e^(kt) ⇒ t = (ln 0.38)/k = (ln 0.38) / (−ln 2 / 5730) = 5730 · ln(1/0.38) / ln 2 = 5730 · (0.9676 / 0.6931) ≈ 5730 · 1.396 ≈ 7997 years ≈ 8000 years.
(iii) A ±2% change in the fraction (i.e. 36% or 40%) shifts ln(1/fraction) by about ±5%, so the age changes by about ±400 years; the laboratory should report the age as roughly 8000 ± 400 years.
Problem 3 — Compounding
Set up. We compare two compounding regimes and quantify the gap.
(i) A_annual = 5000 · (1.08)^10 ≈ 5000 · 2.1589 ≈ $10 795.
(ii) A_continuous = 5000 · e^(0.8) ≈ 5000 · 2.2255 ≈ $11 128.
(iii) Difference = 11 128 − 10 795 = $333, i.e. 333 / 10 795 ≈ 3.08% of the annual balance. The gap grows in absolute terms (both balances grow exponentially) and the percentage gap converges toward (e^r − (1 + r))/(1 + r)^t ≈ constant ratio — for r = 0.08, the gap stays close to 3% per re-period.
Problem 4 — Drug clearance
Set up. We use the given decay model to compute amounts, half-life and safe re-dose time.
(i) D(6) = 200 · e^(−1.5) ≈ 200 · 0.2231 ≈ 44.6 mg.
(ii) t₁/₂ = (ln 2) / 0.25 ≈ 2.7726 ≈ 2.8 hours.
(iii) 25 = 200 · e^(−0.25 t) ⇒ e^(−0.25 t) = 0.125 = 1/8 ⇒ −0.25 t = ln(1/8) = −3 ln 2 ⇒ t = 12 ln 2 ≈ 8.318 hours. Rounded up to the nearest 0.5 hour: wait at least 8.5 hours.
Problem 5 — Newton's law of cooling
Set up. We fit k from the 10-minute data, predict another time, then find when the cup reaches a target temperature.
(i) 60 = 20 + 70 · e^(10k) ⇒ e^(10k) = 40/70 = 4/7 ≈ 0.5714. So k = (ln(4/7))/10 ≈ −0.056 per minute.
(ii) T(20) = 20 + 70 · e^(−1.12) ≈ 20 + 70 · 0.3266 ≈ 20 + 22.86 ≈ 42.9 °C. The naive guess of 30 °C is wrong because Newton's law halves the gap-to-room (70 → 40 → ~23), not the temperature itself.
(iii) 25 = 20 + 70 · e^(kt) ⇒ e^(kt) = 5/70 = 1/14 ⇒ t = ln(1/14)/k = −2.639 / −0.056 ≈ 47.1 minutes ≈ 47 minutes.