Mathematics Advanced • Year 11 • Module 4 • Lesson 13
Exponential Growth and Decay
Build procedural fluency in the model P = P₀ e^(kt) — finding k from data, predicting future amounts, and computing half-life or doubling time.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each formula for the model P = P₀ e^(kt):
Doubling time t₂ = ____________ (assumes k > 0)
Half-life t₁/₂ = ____________ (assumes k < 0)
Q1.2 State the sign of k for each scenario:
Bacterial population doubling each hour: k ____________
Radioactive substance losing half its mass every 10 days: k ____________
Q1.3 The differential equation that the model P = P₀ e^(kt) satisfies is dP/dt = ____________. (Hint: derivative of P is k times P itself.)
2. Worked example — bacteria from 500 to 1500 in 4 hours
Follow each line. Every step has a reason on the right.
Problem. A culture of 500 bacteria grows to 1500 in 4 hours. Find k and predict when the population reaches 10 000.
Step 1 — Set up the model.
P = 500 e^(kt)
Reason: P₀ = 500, and the data is consistent with exponential growth.
Step 2 — Substitute the known data.
1500 = 500 e^(4k) ⇒ e^(4k) = 3
Reason: at t = 4, P = 1500.
Step 3 — Take ln of both sides to find k.
4k = ln 3 ⇒ k = ln 3 / 4 ≈ 0.2747
Reason: ln is the inverse of e^(·).
Step 4 — Set P = 10 000 and solve for t.
10 000 = 500 e^(0.2747 t) ⇒ e^(0.2747 t) = 20 ⇒ t = ln 20 / 0.2747 ≈ 10.9 hours
Reason: same exponential solve as in Step 3.
Conclusion. k ≈ 0.275 per hour; the population reaches 10 000 at t ≈ 10.9 hours.
3. Faded example — half-life from data
A sample of 200 g of a radioactive substance decays to 25 g in 30 days. Find the half-life. 4 marks
Step 1 — Set up the decay model:
P = ________ · e^(kt), with k ____________ (sign?).
Step 2 — Substitute the data at t = 30:
25 = 200 · e^(30k) ⇒ e^(30k) = ________ (=1/8)
Step 3 — Take ln and solve for k:
30k = ln(1/8) = ________ · ln 2 ⇒ k = ____________
Step 4 — Use half-life formula t₁/₂ = ln 2 / |k|:
t₁/₂ = ln 2 / ____________ = ____________ days
Conclusion. Half-life ≈ ____________ days. (Sanity check: 200 → 100 → 50 → 25 is 3 halvings, so 30/3 ≈ ____ matches.)
4. Graduated practice — exponential growth and decay
Use P = P₀ e^(kt). Show working for Standard and Extension items.
Foundation — direct formula (4 questions)
| Q | Question | Answer |
|---|---|---|
| 4.1 1 | P₀ = 100, k = 0.5. Find P at t = 2. | |
| 4.2 1 | P₀ = 80, k = −0.1. Find P at t = 5. | |
| 4.3 1 | State the doubling time when k = 0.2 (exact, in terms of ln 2). | |
| 4.4 1 | State the half-life when k = −0.05 (exact, in terms of ln 2). |
Standard — typical HSC difficulty (6 questions)
Show the line where you take ln of both sides, and round to 3 significant figures unless asked otherwise.
4.5 A population grows from 1000 to 4000 in 6 hours. Find k and predict P at t = 10 hours. 3 marks
4.6 A substance decays from 200 g to 25 g in 30 days. Find the half-life. 2 marks
4.7 $10 000 is invested at 5% p.a. compounded continuously. Find the value after 5 years. 2 marks
4.8 Bacteria double every 3 hours. How long until a population of 500 reaches 8000? 3 marks
4.9 The half-life of carbon-14 is 5730 years. Find the percentage remaining after 10 000 years. 3 marks
4.10 $5000 grows continuously at rate k. After 8 years the balance is $8080. Find k and the doubling time. 3 marks
Extension — combine ideas (2 questions)
4.11 A radioactive sample loses 30% of its mass in 4 days. Find (a) the decay constant k, (b) the half-life, (c) the mass after 14 days as a fraction of the initial mass. 4 marks
4.12 Show that for any growth rate k > 0, the doubling time satisfies t₂ = (ln 2) / k. Hence find k if the doubling time is exactly 12 months. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Formulas
Doubling time t₂ = (ln 2) / k. Half-life t₁/₂ = (ln 2) / |k|.
Q1.2 — Sign of k
Doubling bacterial population: k > 0 (growth). Radioactive decay: k < 0.
Q1.3 — Differential equation
dP/dt = k P. (The defining property of exponential change: rate is proportional to current amount.)
Q3 — Faded example, half-life from data
Step 1: P = 200 · e^(kt), k < 0. Step 2: e^(30k) = 1/8.
Step 3: 30k = ln(1/8) = −3 · ln 2, so k = −ln 2 / 10 ≈ −0.0693.
Step 4: t₁/₂ = ln 2 / (ln 2 / 10) = 10 days. ✓ Matches 3 halvings in 30 days.
Q4.1 — P₀ = 100, k = 0.5, t = 2
P = 100 · e^(0.5 · 2) = 100 · e¹ ≈ 271.83.
Q4.2 — P₀ = 80, k = −0.1, t = 5
P = 80 · e^(−0.5) ≈ 80 · 0.6065 ≈ 48.52.
Q4.3 — Doubling time when k = 0.2
t₂ = (ln 2) / 0.2 = 5 ln 2 ≈ 3.47 units of time.
Q4.4 — Half-life when k = −0.05
t₁/₂ = (ln 2) / 0.05 = 20 ln 2 ≈ 13.86 units of time.
Q4.5 — 1000 → 4000 in 6 hours
4000 = 1000 · e^(6k) ⇒ e^(6k) = 4 ⇒ k = (ln 4)/6 ≈ 0.231 per hour.
At t = 10: P = 1000 · e^(2.31) ≈ 10 080.
Q4.6 — 200 g → 25 g in 30 days
25/200 = 1/8 = e^(30k) ⇒ k = −(ln 8)/30 = −(3 ln 2)/30 = −(ln 2)/10. t₁/₂ = ln 2 / |k| = 10 days.
Q4.7 — $10 000 at 5% continuous for 5 years
A = 10 000 · e^(0.05 · 5) = 10 000 · e^(0.25) ≈ 10 000 · 1.2840 ≈ $12 840.
Q4.8 — Doubling every 3 hours, 500 → 8000
k = (ln 2)/3 ≈ 0.231. 8000 = 500 · e^(0.231 t) ⇒ e^(0.231 t) = 16 ⇒ t = ln 16 / 0.231 = 4 ln 2 / 0.231 = 12 hours (4 doublings × 3 hours).
Q4.9 — Carbon-14 after 10 000 years (half-life 5730)
k = −(ln 2)/5730. Fraction remaining = e^(k · 10000) = e^(−10000 ln 2 / 5730) = 2^(−10000/5730) = 2^(−1.745) ≈ 0.298. ≈ 29.8 % remains.
Q4.10 — $5000 → $8080 in 8 years
8080 = 5000 · e^(8k) ⇒ e^(8k) = 1.616 ⇒ k = ln 1.616 / 8 ≈ 0.480 / 8 ≈ 0.0600 per year.
Doubling time = (ln 2)/0.06 ≈ 11.6 years.
Q4.11 — Loses 30% in 4 days
(a) After 4 days, fraction = 0.7. So 0.7 = e^(4k) ⇒ k = (ln 0.7)/4 ≈ −0.0892 per day.
(b) t₁/₂ = ln 2 / 0.0892 ≈ 7.77 days.
(c) After 14 days, fraction = e^(14 · −0.0892) ≈ e^(−1.249) ≈ 0.287, i.e. 28.7 % of the initial mass.
Q4.12 — Doubling-time derivation
Set 2 P₀ = P₀ · e^(k t₂) ⇒ e^(k t₂) = 2 ⇒ k t₂ = ln 2 ⇒ t₂ = (ln 2)/k. ✓
If t₂ = 12 months, then k = (ln 2)/12 ≈ 0.0578 per month (≈ 5.78 % per month continuously).