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).