Mathematics Advanced • Year 12 • Module 6 • Lesson 13

Growth and Decay Models

Build fluency in the exponential model N(t) = N₀ e^kt, half-lives, doubling times and rate-constant calculations.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the exponential model:

dN/dt = k N ⇒ N(t) = ____________________ (in terms of N₀, k, t)

Q1.2 Write the formulas (use |k| where needed):

Half-life t_1/2 = ____________ Doubling time t_double = ____________

Q1.3 Classify each as growth or decay: k = 0.08 → ____________; k = −0.05 → ____________; k = 0 → ____________ (special case).

Stuck? Revisit lesson § The Exponential Model.

2. Worked example — radioactive decay with half-life 12 years

Problem. A substance with half-life 12 years starts at 200 g. Find the decay constant k and the mass remaining after 30 years.

Step 1 — Decay constant from half-life.

t_1/2 = (ln 2)/|k| ⇒ |k| = (ln 2)/12 ≈ 0.0578 / year

Since this is decay, k = −0.0578 / year.

Step 2 — Write N(t).

N(t) = 200 e^{−0.0578 t}

Step 3 — Evaluate at t = 30.

N(30) = 200 e^−1.733 ≈ 200 × 0.1767 ≈ 35.4 g

Sanity check. 30 years = 2.5 half-lives. Expect 200 × (1/2)^2.5 = 200 × 0.1768 ≈ 35.4 ✓.

3. Faded example — bacterial growth, doubling time 3 hours

Starting with 100 bacteria. Find k and the population after 10 hours. 4 marks

Step 1 — Rate constant from doubling time.

t_double = (ln 2)/k ⇒ k = ____________ ≈ ____________ / hour

Step 2 — Write N(t).

N(t) = ________ e^{____________ t}

Step 3 — Evaluate at t = 10.

N(10) = ____________

Sanity check. 10 hours = ________ doubling times; (2)^10/3 × 100 ≈ ____________.

Stuck? Revisit lesson § Applications — Example 3 (bacterial growth).

4. Graduated practice

Foundation — half-life and doubling time (4 questions)

Q	Given	Find
4.1 1	k = −0.05 / year	Half-life (years)
4.2 1	k = 0.10 / year	Doubling time (years)
4.3 1	Half-life = 25 years	k (per year)
4.4 1	Doubling time = 7 days	k (per day)

Standard — apply the exponential model (6 questions)

4.5 An investment of $2000 grows continuously at 4.5% p.a. Find the value after 15 years. 2 marks

4.6 A substance with decay constant k = −0.05 /year starts at 80 g. Find the mass after 20 years. 2 marks

4.7 A bacteria culture triples every 5 hours. How long until it grows to 10 times its original size? 3 marks

4.8 A radioactive isotope has half-life 8 days. Starting with 50 g, find the mass after 20 days. 3 marks

4.9 A city's population grows from 50,000 to 65,000 in 10 years (exponential model). Find k. 2 marks

4.10 Continuing 4.9, when will the population reach 100,000? 2 marks

Extension (2 questions)

4.11 Compare $1000 invested at 5% for 10 years compounded (a) annually, (b) continuously. Find the dollar difference. 3 marks

4.12 A drug has half-life 4 hours. A patient takes 100 mg. How much remains after 12 hours? Solve two ways: (a) using the formula N(t) = 100 e^kt; (b) using the half-life shortcut. Confirm the two agree. 3 marks

Stuck on 4.12(b)? 12 hours = 3 half-lives, so the remaining fraction is (1/2)³.

5. Self-check the easy 3

Tick once you have verified your work on the first three Standard problems.

For 4.5 I used A(t) = 2000 e^{0.045 t} and obtained A(15) ≈ $3928.

For 4.6 I used N(t) = 80 e^{−0.05 t} and obtained N(20) ≈ 29.4 g.

For 4.7 I solved 3 = e^5k to get k = (ln 3)/5, then solved 10 = e^kt to get t = (ln 10)/k ≈ 10.5 hours.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — General solution

N(t) = N₀ e^kt.

Q1.2 — Half-life and doubling

t_1/2 = (ln 2)/|k|. t_double = (ln 2)/k (for growth, k > 0).

Q1.3 — Sign of k

k = 0.08 → growth; k = −0.05 → decay; k = 0 → constant (no growth, no decay).

Q3 — Faded bacterial-growth example

Step 1: k = (ln 2)/3 ≈ 0.231 /hour.
Step 2: N(t) = 100 e^{0.231 t}.
Step 3: N(10) = 100 e^2.31 ≈ 1006 bacteria.
Sanity check: 10/3 ≈ 3.33 doubling times; (2)^3.33 × 100 ≈ 1006 ✓.

Q4.1 — 4.4 Half-life / doubling

4.1 t_1/2 = (ln 2)/0.05 ≈ 13.9 years. 4.2 t_double = (ln 2)/0.10 ≈ 6.9 years. 4.3 k = −(ln 2)/25 ≈ −0.0277 /year. 4.4 k = (ln 2)/7 ≈ 0.099 /day.

Q4.5 — $2000 at 4.5% for 15 years

A(15) = 2000 e^{0.045 × 15} = 2000 e^0.675 ≈ 2000 × 1.9640 ≈ $3928.

Q4.6 — 80 g, k = −0.05, t = 20

N(20) = 80 e^−1.0 ≈ 80 × 0.3679 ≈ 29.4 g.

Q4.7 — Tripling every 5 h, reach 10×

3 = e^5k ⇒ k = (ln 3)/5 ≈ 0.2197 /hour. 10 = e^kt ⇒ t = (ln 10)/k = 5 ln 10 / ln 3 ≈ 10.48 hours.

Q4.8 — Half-life 8 days, 50 g, t = 20

k = −(ln 2)/8 ≈ −0.0866 /day. N(20) = 50 e^−1.733 ≈ 50 × 0.1768 ≈ 8.84 g. (Sanity check: 20/8 = 2.5 half-lives; 50 × (1/2)^2.5 = 50 × 0.1768 ≈ 8.84 ✓.)

Q4.9 — Find k from 50,000 → 65,000 in 10 years

65000 = 50000 e^10k ⇒ e^10k = 1.3 ⇒ k = (ln 1.3)/10 ≈ 0.0262 /year.

Q4.10 — Reach 100,000

100000 = 50000 e^kt ⇒ e^kt = 2 ⇒ t = (ln 2)/k = (ln 2)/0.0262 ≈ 26.4 years (from start).

Q4.11 — Annual vs continuous compounding

Annual: 1000 × (1.05)¹⁰ ≈ $1628.89. Continuous: 1000 e^0.5 ≈ $1648.72. Difference ≈ $19.83 in favour of continuous.

Q4.12 — Drug half-life 4 h, 100 mg after 12 h

(a) k = −(ln 2)/4. N(12) = 100 e^{−3 ln 2} = 100 × 2⁻³ = 12.5 mg.
(b) 12 h = 3 half-lives, so remaining = 100 × (1/2)³ = 12.5 mg ✓. The formulas agree because the half-life shortcut is the exponential formula evaluated at integer multiples of t_1/2.