Mathematics Advanced • Year 11 • Module 4 • Lesson 13
Exponential Growth and Decay
Practise HSC-style writing on exponential models, including a structured cooling-law problem with explicit marking criteria.
1. Short-answer questions
1.1 A bacterial population of 800 doubles every 5 hours. Write down the model P = P₀ e^(kt), find k exactly, and predict the population after 15 hours. 3 marks Band 3
1.2 A radioactive sample has half-life 8 days. Starting with 100 g, how much remains after 20 days? Give your answer to the nearest 0.1 g and show ln working. 3 marks Band 3-4
1.3 A population P satisfies dP/dt = 0.04 P, with P = 250 at t = 0.
(a) Show that P(t) = 250 e^(0.04 t) satisfies both the differential equation and the initial condition.
(b) Find the time at which the population reaches 1000, to the nearest year. 4 marks Band 4
2. Extended response
2.1 A forensic scientist arrives at a crime scene at 10:00 pm. The victim's body has temperature 30 °C, and the room is held at a constant 20 °C. Newton's law of cooling models the body temperature as
T(t) = 20 + (T₀ − 20) e^(kt), where T₀ is the temperature at time t = 0 (time of death) and k < 0.
At 11:00 pm the body has cooled to 28 °C.
(a) Taking time of death t = 0, use the two readings to derive a single equation that the body's initial temperature T₀ and k must satisfy. State both readings as equations T(t₁) = … and T(t₂) = … with appropriate t-values.
(b) The pathologist assumes T₀ = 37 °C (normal body temperature). Using this and the 10 pm reading T = 30, find k correct to 3 decimal places.
(c) Use the same model to estimate t₁ = (the hour at which the 10:00 pm reading was taken, measured from the time of death). Hence determine the time of death, to the nearest 15 minutes.
(d) Comment in one sentence on a source of uncertainty in this estimate. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — writes 30 = 20 + (T₀ − 20) e^(k · t₁) AND 28 = 20 + (T₀ − 20) e^(k · (t₁ + 1)), labelling t₁ as time from death to 10:00 pm.
Part (b) — 2 marks
• 1 mark — substitutes T₀ = 37 into the 10 pm equation to get 30 = 20 + 17 e^(k · t₁).
• 1 mark — uses both equations to eliminate t₁; solves e^k = 8/10 ⇒ k = ln(0.8) ≈ −0.223. (Award for any algebra leading to k = ln(8/10) once both equations are used.)
Part (c) — 3 marks
• 1 mark — solves 30 = 20 + 17 e^(k · t₁) ⇒ e^(k · t₁) = 10/17 ⇒ t₁ = (ln(10/17))/k.
• 1 mark — substitutes k ≈ −0.223: t₁ ≈ ln(0.5882) / −0.223 ≈ −0.5306 / −0.223 ≈ 2.38 hours ≈ 2 h 23 min.
• 1 mark — concludes time of death ≈ 10:00 pm − 2 h 23 min ≈ 7:37 pm, rounded to the nearest 15 minutes ≈ 7:30 pm.
Part (d) — 1 mark
• 1 mark — names a credible uncertainty (e.g. T₀ may not have been exactly 37 °C; room temperature may have changed; ambient draught speeds up cooling; clothing changes k).
Your response:
Stuck on (b)? Divide the 11 pm equation by the 10 pm equation to cancel (T₀ − 20).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Bacteria doubling every 5 h (3 marks)
Sample response. Model: P = 800 e^(kt). Doubling every 5 hours ⇒ 2 = e^(5k) ⇒ k = (ln 2)/5. After 15 hours: P = 800 · e^((ln 2)/5 · 15) = 800 · e^(3 ln 2) = 800 · 2³ = 6400 bacteria.
Marking notes. 1 mark — sets up P = 800 e^(kt) and uses 2 = e^(5k). 1 mark — solves k = (ln 2)/5 (exact form). 1 mark — clean answer 6400, recognising 15 h = 3 doublings.
1.2 — Radioactive sample, half-life 8 days (3 marks)
Sample response. k = −(ln 2)/8 ≈ −0.0866 per day. After 20 days: P = 100 · e^(20 · −0.0866) = 100 · e^(−1.733) ≈ 100 · 0.1768 ≈ 17.7 g.
Marking notes. 1 mark — correct k from half-life formula. 1 mark — substitution and exponential evaluation shown. 1 mark — final answer 17.7 g to 1 dp (accept 17.6 or 17.7 g).
1.3 — Differential equation (4 marks)
(a) Sample response. Differentiate: dP/dt = 250 · 0.04 · e^(0.04 t) = 0.04 · (250 e^(0.04 t)) = 0.04 P. ✓ Initial condition: P(0) = 250 · e⁰ = 250. ✓ Both checks pass, so the proposed P(t) is a solution.
(b) Sample response. 1000 = 250 · e^(0.04 t) ⇒ e^(0.04 t) = 4 ⇒ 0.04 t = ln 4 ⇒ t = (ln 4)/0.04 = 25 · ln 4 ≈ 25 · 1.3863 ≈ 34.66 years ≈ 35 years.
Marking notes. (a) 1 mark — differentiates and recognises 0.04 · P. 1 mark — checks initial condition. (b) 1 mark — sets up e^(0.04 t) = 4. 1 mark — rounds correctly to 35 years.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Let t₁ be the number of hours between the time of death (t = 0) and the 10:00 pm reading. Then T(t₁) = 30 and T(t₁ + 1) = 28, giving
30 = 20 + (T₀ − 20) e^(k · t₁) …(1)
28 = 20 + (T₀ − 20) e^(k · (t₁ + 1)) …(2). [1 mark — both equations stated with labelled t-values]
Part (b). Substituting T₀ = 37 into (1): 30 = 20 + 17 e^(k · t₁) ⇒ e^(k · t₁) = 10/17. [1 mark]
Dividing (2) by (1): (28 − 20)/(30 − 20) = e^(k · (t₁ + 1)) / e^(k · t₁) = e^k. So e^k = 8/10 = 0.8 ⇒ k = ln(0.8) ≈ −0.223 per hour. [1 mark]
Part (c). From (1): e^(k · t₁) = 10/17 ≈ 0.5882. [1 mark]
So t₁ = ln(10/17) / k = (−0.5306) / (−0.2231) ≈ 2.38 hours = 2 h 23 min. [1 mark]
Time of death ≈ 10:00 pm − 2 h 23 min ≈ 7:37 pm. Rounded to the nearest 15 minutes: ≈ 7:30 pm. [1 mark]
Part (d). The biggest uncertainty is the assumption T₀ = 37 °C — the victim could have been febrile (higher) or hypothermic (lower), which would shift the calculated time of death by tens of minutes. (Also acceptable: room temperature changes; clothing or wind altering the cooling constant k; thermometer accuracy.) [1 mark]
Total: 7/7.
Band descriptors for marker.
Band 3: Writes the model and one of the two equations; substitutes T₀ = 37 but does not eliminate t₁; no time-of-death estimate; ≈ 2-3 marks.
Band 4: Both equations stated; k found by dividing equations; gets t₁ but rounds carelessly or fails to convert to a clock time; ≈ 4-5 marks.
Band 5: All algebra correct, time of death stated; uncertainty mentioned but vague; ≈ 5-6 marks.
Band 6: All four parts complete, uncertainty named specifically (T₀, room temp, ambient draught), times rounded to the requested 15-minute precision. 7/7.