Mathematics Advanced • Year 12 • Module 6 • Lesson 13

Growth and Decay Models

Practise HSC-style writing on exponential growth and decay, including the carrying-capacity limitation.

Master · Past-Paper Style

1. Short-answer questions

1.1 A radioactive isotope has half-life 8 days. Starting with 50 g, find the mass remaining after 20 days. Show all working. 3 marks Band 4

1.2 A city's population grows from 50,000 to 65,000 in 10 years. Assuming exponential growth, (a) find the growth constant k, (b) predict the population in 25 years. 3 marks Band 4-5

1.3 Compare $1000 invested at 5% for 10 years under (a) annual compounding and (b) continuous compounding. State both values to the nearest cent and the dollar difference. 3 marks Band 4

Stuck on 1.2(b)? Use the k from (a) and evaluate at t = 25.

2. Extended response

2.1 The simple exponential model dP/dt = k P assumes unlimited resources. In reality populations face a carrying capacity K.
(a) Explain the limitation of the simple model in one sentence.
(b) Write down the logistic differential equation, defining each symbol.
(c) For small P (P much less than K), show that the logistic model reduces to the simple exponential model. Hence justify why exponential growth is still a good approximation for the early stages of growth.
(d) Sketch (on the same axes) the exponential solution and the logistic solution starting at the same small P₀, with the carrying capacity K labelled. Describe in one sentence what happens to each curve as t → ∞. 8 marks Band 5-6

Explicit marking criteria

Part (a) — 1 mark

• 1 mark — states that unlimited resources are unrealistic, leading to unbounded growth that cannot occur physically (food, space or other constraints).

Part (b) — 2 marks

• 1 mark — writes dP/dt = k P (1 − P/K).

• 1 mark — defines K as carrying capacity and k as intrinsic growth rate.

Part (c) — 2 marks

• 1 mark — observes that for P ≪ K, the factor (1 − P/K) ≈ 1.

• 1 mark — concludes dP/dt ≈ k P, the simple exponential model, justifying its use early on.

Part (d) — 3 marks

• 1 mark — sketches the exponential curve (concave up, unbounded growth).

• 1 mark — sketches the logistic curve (S-shape) with horizontal asymptote P = K labelled.

• 1 mark — describes long-run behaviour: exponential → ∞, logistic → K.

Your response:

Stuck on (c)? When P/K is very small, 1 − P/K is very close to 1, so the factor can be ignored.

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Answers — sample responses + marking notes

1.1 — Isotope half-life 8 days, 50 g for 20 days (3 marks)

Sample response. k = −(ln 2)/8 ≈ −0.0866 /day. N(t) = 50 e^kt. N(20) = 50 e^{−0.0866 × 20} = 50 e^−1.733 ≈ 8.84 g.

Marking notes. 1 mark — correct k (with sign indicating decay). 1 mark — correct N(t) and substitution at t = 20. 1 mark — correct numerical mass. A common error is omitting the negative sign on k; this gives 50 e^1.733 ≈ 283 g, which fails the sanity check that mass should decrease.

1.2 — Population growth, 50,000 → 65,000 in 10 years (3 marks)

Sample response. (a) 65000 = 50000 e^10k ⇒ e^10k = 1.3 ⇒ k = (ln 1.3)/10 ≈ 0.0262 /year.
(b) P(25) = 50000 e^25k = 50000 × 1.3^2.5 ≈ 50000 × 1.928 ≈ 96,400.

Marking notes. (a) 1 mark — correct k. (b) 1 mark — correct setup using k from (a); 1 mark — correct numerical value (accept 96,000–97,000 depending on rounding). Students who use a wrong t (e.g. 15 instead of 25) lose part (b).

1.3 — Compounding comparison (3 marks)

Sample response. Annual: A_a = 1000 × (1.05)¹⁰ ≈ $1628.89. Continuous: A_c = 1000 e^0.5 ≈ $1648.72. Difference: $19.83 (continuous is larger).

Marking notes. 1 mark each for the two values; 1 mark for the difference and stating that continuous beats annual.

2.1 — Extended response (8 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) The simple exponential model dP/dt = k P predicts unlimited growth, but real populations are constrained by finite food, space and other resources, so the model becomes unrealistic at large P. [1 mark — clear limitation.]

(b) The logistic differential equation is

dP/dt = k P (1 − P/K). [1 mark — equation.]

Here k > 0 is the intrinsic per-capita growth rate (units 1/time) and K > 0 is the carrying capacity (the equilibrium population, units the same as P). [1 mark — both symbols defined.]

(c) When P ≪ K, the ratio P/K is close to 0, so 1 − P/K ≈ 1. [1 mark — approximation justified.]
Substituting: dP/dt ≈ k P (1) = k P. This is the simple exponential model. Hence exponential growth is a good approximation in the early stages, when the population is small compared to the carrying capacity — which is why epidemiologists, ecologists and economists begin with the simple model before adding capacity effects. [1 mark — reduces to exponential model with interpretation.]

(d) Sketch (described): both curves start at the same small (0, P₀).
• Exponential P(t) = P₀ e^kt: concave up, no asymptote, P → ∞ as t → ∞. [1 mark — exponential curve.]
• Logistic P(t): coincides with the exponential at small t, then bends over and approaches the horizontal asymptote P = K (labelled). S-shape (sigmoid). [1 mark — logistic S-curve with K asymptote.]
Long-run behaviour: exponential → ∞ (unrealistic); logistic → K (realistic equilibrium). [1 mark — limits stated.] ▮

Total: 8/8.

Band descriptors for marker.

Band 3: Vague answer to (a) ("not accurate"); writes the logistic equation but does not define symbols; shows little or no work for (c) and (d). ≈ 2-3 marks.

Band 4: (a)-(b) complete; (c) states the approximation but does not justify why P ≪ K matters; sketch missing asymptote label. ≈ 4-5 marks.

Band 5: All four parts answered with mostly correct reasoning; one of the sketch elements missing (e.g. K label) or the long-run interpretation incomplete. ≈ 5-6 marks.

Band 6: Full response: definition of K, reduction explained step-by-step, sketch with both curves and asymptote, and long-run limits stated explicitly. 7-8 marks.