Mathematics Advanced • Year 11 • Module 4 • Lesson 3
The Number e and Natural Exponentials
Apply ex and A = Pert to investment, population, cooling and decay scenarios.
Problem 1 — Continuous investment
$8 000 is invested at 7% per annum compounded continuously: A = 8000 e0.07 t.
Set up: What are we solving for?
(i) Find A(5) to the nearest dollar. 2 marks
(ii) By trial, estimate the smallest whole number of years until the investment doubles in value. 2 marks
(iii) Solve algebraically for the exact doubling time (give your answer in terms of ln 2, then to 1 dp in years). 2 marks
Stuck on (iii)? 2P = Pe0.07 t ⇒ 2 = e0.07 t ⇒ take ln.Problem 2 — Three ways of compounding
$10 000 is invested at 8% per annum for 5 years.
Set up: What are we solving for?
(i) Annual compounding: A = 10 000 × (1.08)5. Compute to the nearest dollar. 1 mark
(ii) Monthly compounding: A = 10 000 × (1 + 0.08/12)60. Compute to the nearest dollar. 2 marks
(iii) Continuous compounding: A = 10 000 × e0.4. Compute to the nearest dollar, then rank the three methods from lowest to highest balance. Comment in one sentence on what (1 + r/n)n tends to as n → ∞. 2 marks
Problem 3 — Radioactive decay (continuous)
A 100 g sample of a radioactive isotope decays continuously at 5% per year: M(t) = 100 e−0.05 t.
Set up: What are we solving for?
(i) Find M(10), M(20) and M(50) to 2 dp. 3 marks
(ii) Algebraically find the half-life (time for M to halve), to 1 dp in years. Hence comment whether the half-life is constant or depends on the starting mass. 3 marks
(iii) A safety inspector wants to know when the sample first drops below 10 g. Find this time, to 1 dp in years. 2 marks
Problem 4 — Cooling coffee (Newton's law, e-form)
A cup of coffee cools according to T(t) = 20 + 70 e−0.1 t, with t in minutes and T in °C.
Set up: What are we solving for?
(i) State the room temperature (the asymptote) and the initial coffee temperature. 2 marks
(ii) Find T(15) to 1 dp. 1 mark
(iii) Find the time (to the nearest minute) at which the coffee reaches a comfortable drinking temperature of 60°C. 3 marks
Stuck on (iii)? Solve 60 = 20 + 70 e−0.1 t: subtract 20, divide by 70, take ln.Problem 5 — Comparing decay rates
Two isotopes have continuous decay constants kA = 0.02 yr−1 and kB = 0.10 yr−1. Both start with 1 g.
Set up: What are we solving for?
(i) Write decay models MA(t) and MB(t). 1 mark
(ii) Find the half-life of each isotope. 2 marks
(iii) Without using a calculator, decide which sample has more mass remaining at t = 10 years, and justify in one line. Then check with a calculation. 2 marks
Stuck on (iii)? Larger k means faster decay (more rapid mass loss).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Continuous investment at 7%
Set up. Use A = Pert for compounding, then logs for doubling time.
(i) A(5) = 8000 · e0.35 ≈ 8000 × 1.4191 ≈ $11 353.
(ii) Trial: A(9) = 8000 · e0.63 ≈ 8000 × 1.878 ≈ $15 023 (< $16 000). A(10) = 8000 · e0.7 ≈ 8000 × 2.014 ≈ $16 110 (> $16 000). Smallest whole-year doubling: t = 10 years.
(iii) 2 = e0.07 t ⇒ 0.07 t = ln 2 ⇒ t = (ln 2)/0.07 ≈ 0.6931/0.07 ≈ 9.9 years. (Confirms 10 yr is the first whole year crossing.)
Problem 2 — Three compounding methods
Set up. Compute the balance under three formulas and compare.
(i) Annual: 10 000 × (1.08)5 = 10 000 × 1.46933 ≈ $14 693.
(ii) Monthly: 10 000 × (1 + 0.08/12)60 = 10 000 × (1.006667)60 ≈ 10 000 × 1.48985 ≈ $14 898.
(iii) Continuous: 10 000 × e0.4 ≈ 10 000 × 1.4918 ≈ $14 918. Ranking lowest → highest: annual ($14 693) < monthly ($14 898) < continuous ($14 918). As n → ∞, (1 + r/n)n → er, so continuous compounding is the limiting case of ever-more-frequent compounding.
Problem 3 — Radioactive decay
Set up. Continuous-decay model M(t) = 100 e−0.05 t; use logs for half-life and threshold time.
(i) M(10) = 100 · e−0.5 ≈ 60.65 g. M(20) = 100 · e−1 ≈ 36.79 g. M(50) = 100 · e−2.5 ≈ 8.21 g.
(ii) 50 = 100 e−0.05 t ⇒ 0.5 = e−0.05 t ⇒ −0.05 t = ln 0.5 ⇒ t = ln 2 / 0.05 ≈ 13.9 years. The half-life depends only on the decay constant k = 0.05, not on the starting mass — it would be the same for a 1 g or 1 kg sample.
(iii) 10 = 100 e−0.05 t ⇒ 0.1 = e−0.05 t ⇒ −0.05 t = ln 0.1 ⇒ t = ln 10 / 0.05 ≈ 2.303 / 0.05 ≈ 46.1 years.
Problem 4 — Cooling coffee
Set up. T = ambient + (initial − ambient) e−kt; read off room and initial values, then solve for time at a target temperature.
(i) Room (asymptote) = 20°C. Initial T(0) = 20 + 70 · 1 = 90°C.
(ii) T(15) = 20 + 70 e−1.5 ≈ 20 + 70 × 0.2231 ≈ 20 + 15.6 = 35.6°C.
(iii) 60 = 20 + 70 e−0.1 t ⇒ 40 = 70 e−0.1 t ⇒ e−0.1 t = 4/7 ⇒ −0.1 t = ln(4/7) ⇒ t = −ln(4/7)/0.1 = ln(7/4)/0.1 ≈ 0.5596 / 0.1 ≈ 6 minutes (5.6 to 1 dp).
Problem 5 — Two isotopes
Set up. Model both samples, compute half-lives, compare at t = 10.
(i) MA(t) = e−0.02 t. MB(t) = e−0.10 t.
(ii) Half-life of A: ln 2 / 0.02 ≈ 34.7 yr. Half-life of B: ln 2 / 0.10 ≈ 6.9 yr. (Larger k → shorter half-life.)
(iii) Larger k = faster decay, so isotope A (smaller k = 0.02) retains more mass. Check: MA(10) = e−0.2 ≈ 0.819 g; MB(10) = e−1 ≈ 0.368 g. ✓