Mathematics Advanced • Year 11 • Module 4 • Lesson 4
Exponential Modelling and Equations
Apply Aekt modelling and log-based solving to populations, cooling, depreciation, half-life and savings.
Problem 1 — Bacteria growth
A bacterial colony grows from P(0) = 500 to P(3) = 800 cells over 3 hours and is assumed to follow P(t) = A ekt.
Set up: What are we solving for?
(i) Identify A from the initial condition. 1 mark
(ii) Find k to 3 sf. 2 marks
(iii) Predict the time (to 1 dp in hours) at which the colony reaches 2000 cells, and state the predicted population at t = 24 hours (one day later). Comment in one sentence whether this prediction is realistic. 3 marks
Stuck on (iii)? Set P(t) = 2000 and take ln.Problem 2 — Coffee cooling (Newton's law)
A cup of coffee in a 20°C room is modelled by T(t) = 85 e−0.1 t + 20.
Set up: What are we solving for?
(i) Find the initial temperature and the temperature at t = 10 minutes. 2 marks
(ii) Solve for the time at which T = 50°C (drinkable). Give exact and to 1 dp. 3 marks
(iii) Explain in one sentence why T(t) never quite reaches 20°C, even for large t. 1 mark
Problem 3 — Radioactive half-life
A radioactive isotope has mass M(t) = 100 e−0.05 t grams.
Set up: What are we solving for?
(i) Find the half-life (time to reach 50 g) exactly and to 1 dp. 2 marks
(ii) Find the time at which M(t) = 10 g (90% decayed). 2 marks
(iii) Without further calculation, state the time at which M(t) = 25 g, using the half-life from (i). Explain in one sentence why you can answer without re-solving. 2 marks
Stuck on (iii)? 25 g is half of 50 g, which is half of 100 g — two half-lives.Problem 4 — Car depreciation (e-form)
A car worth $30 000 depreciates exponentially. After 2 years it is worth $20 000.
Set up: What are we solving for?
(i) Fit V(t) = 30 000 ekt using the two data points. Find k to 4 sf. 2 marks
(ii) Predict V(5) to the nearest dollar. 2 marks
(iii) When does the car first fall below $10 000? Give to the nearest year. 2 marks
Problem 5 — Doubling-time rule of thumb
Bankers use the "Rule of 70": at r% continuous interest, the doubling time is roughly 70/r years.
Set up: What are we solving for?
(i) Show algebraically that the exact doubling time at continuous rate r (decimal) is t = ln 2 / r. 2 marks
(ii) Compare the rule "70/r" with the exact answer for r = 5%, 7%, 10% (give percentages, not decimals, and use ln 2 ≈ 0.6931). State the percentage error of the rule at r = 10%. 3 marks
(iii) Suggest a more accurate single-number rule (e.g. "Rule of 69" or "Rule of 72") and explain in one line why it could be better. 1 mark
Stuck on (i)? Set 2P = Pert and take ln.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bacteria growth
Set up. Fit P(t) = Aekt, then use it to predict time and population.
(i) A = P(0) = 500.
(ii) 800 = 500 e3k ⇒ e3k = 1.6 ⇒ 3k = ln 1.6 ⇒ k = ln 1.6 / 3 ≈ 0.4700 / 3 ≈ 0.157.
(iii) 2000 = 500 e0.157 t ⇒ e0.157 t = 4 ⇒ t = ln 4 / 0.157 ≈ 1.3863 / 0.157 ≈ 8.8 hours. P(24) = 500 e0.157 × 24 = 500 e3.768 ≈ 500 × 43.3 ≈ 21 660 cells. This figure may be unrealistic in practice — bacterial growth typically plateaus once nutrients run low, so the exponential model overpredicts at large t.
Problem 2 — Coffee cooling
Set up. Use T(t) = 85 e−0.1 t + 20 to find temperatures and the time when T = 50.
(i) T(0) = 85 + 20 = 105°C. T(10) = 85 e−1 + 20 ≈ 85 × 0.3679 + 20 ≈ 31.3 + 20 = 51.3°C.
(ii) 50 = 85 e−0.1 t + 20 ⇒ 30 = 85 e−0.1 t ⇒ e−0.1 t = 30/85 = 6/17. Take ln: −0.1 t = ln(6/17) ⇒ t = −ln(6/17) / 0.1 = ln(17/6) / 0.1 ≈ 1.0414 / 0.1 ≈ 10.4 minutes.
(iii) The exponential factor 85 e−0.1 t > 0 for all finite t, so T = 20 + (positive) > 20 always; T approaches 20°C as t → ∞ but never equals it (horizontal asymptote).
Problem 3 — Radioactive half-life
Set up. Use logs to solve for time at given mass thresholds; use the constancy of half-life.
(i) 50 = 100 e−0.05 t ⇒ e−0.05 t = 0.5 ⇒ t = ln 2 / 0.05 (exact) ≈ 13.9 years.
(ii) 10 = 100 e−0.05 t ⇒ e−0.05 t = 0.1 ⇒ t = ln 10 / 0.05 ≈ 2.303 / 0.05 ≈ 46.1 years.
(iii) 25 g is half of 50 g, which is half of 100 g — i.e. two half-lives. So t = 2 × 13.9 = 27.7 years. The half-life is constant (depends only on the decay constant), so two halvings always take exactly two half-lives.
Problem 4 — Depreciation
Set up. Fit V = 30 000 ekt using V(2) = 20 000; project to year 5 and find when V < $10 000.
(i) 20 000 = 30 000 e2k ⇒ e2k = 2/3 ⇒ 2k = ln(2/3) ⇒ k = ½ · ln(2/3) ≈ ½ · (−0.4055) ≈ −0.2027.
(ii) V(5) = 30 000 e5 × −0.2027 = 30 000 e−1.0135 ≈ 30 000 × 0.3629 ≈ $10 887.
(iii) 10 000 = 30 000 e−0.2027 t ⇒ e−0.2027 t = 1/3 ⇒ −0.2027 t = ln(1/3) = −ln 3 ⇒ t = ln 3 / 0.2027 ≈ 1.0986 / 0.2027 ≈ 5.42 yr → first whole year V < $10 000 is t = 6 years.
Problem 5 — Rule of 70
Set up. Derive and compare doubling-time formulas.
(i) 2P = Pert ⇒ 2 = ert ⇒ rt = ln 2 ⇒ t = ln 2 / r.
(ii) Express r as a decimal; "70/r%" interprets r as a percent so it's equivalent to (70/100)/rdec = 0.70 / rdec, vs the exact 0.6931 / rdec. At r = 5%: exact t = 0.6931 / 0.05 = 13.86 yr; rule = 70/5 = 14.0 yr. At r = 7%: exact = 9.90 yr; rule = 10.0 yr. At r = 10%: exact = 6.931 yr; rule = 7.0 yr. Percentage error at r = 10% is (7.0 − 6.931) / 6.931 × 100 ≈ 1.0%.
(iii) The "Rule of 69" (or 69.3, since ln 2 ≈ 0.6931) is closer to the exact factor. The Rule of 72 trades a small accuracy loss at low rates for easy mental division (72 has many integer divisors).