NBT Mathematical Modelling Flashcards

A dam loses water through evaporation at a rate proportional to its current volume V. If the dam starts at 800 000 litres and drops to 600 000 litres after 10 days, which expression correctly models V(t) after t days?

Answer: V(t) = 800 000 · (3/4)^(t/10)

When a quantity decreases at a rate proportional to its current value, the model is exponential decay: V(t) = V₀ · rᵗ. After 10 days, V(10) = 800 000r¹⁰ = 600 000, so r¹⁰ = 3/4, giving r = (3/4)^(1/10). Therefore V(t) = 800 000 · (3/4)^(t/10). Option B is a linear model (constant absolute loss), not proportional. Option C uses a different decay constant (e^−0.025 ≈ 0.975 per day, not matching the data). Option D is a power function, not exponential decay.

A company's profit P (in rands) is modelled by P(x) = −2x² + 120x − 1 000, where x is the number of units sold. A rival offers to buy the company for R800. At what minimum number of units must the company sell per period to make selling to the rival financially irrational (i.e. to exceed R800 profit)?

Answer: The profit never exceeds R800

The maximum of P(x) = −2x² + 120x − 1000 occurs at x = −b/(2a) = −120/(−4) = 30. P(30) = −2(900) + 120(30) − 1000 = −1800 + 3600 − 1000 = 800. This is a maximum, not a range — the parabola opens downward and peaks at exactly R800. The profit never exceeds R800, so the rival's offer of R800 matches the best possible outcome. Selling to the rival is never strictly irrational.

Water fills a conical tank (apex down) at a constant rate of 2 m³/min. The cone has height 6 m and base radius 3 m. When the water depth is 4 m, the rate at which the water depth is increasing is:

Answer: 1/(2π) m/min

For a cone with H=6 m and R=3 m, r/h = 1/2, so r = h/2. Volume V = (πh³)/12. Differentiating with respect to time: dV/dt = (πh²/4)·(dh/dt). At h = 4 m and dV/dt = 2 m³/min: 2 = (π·16/4)·(dh/dt) = 4π·(dh/dt). Therefore dh/dt = 2/(4π) = 1/(2π) m/min. A common error is using H and R directly without finding the radius at the current water level first.

A population of bacteria is modelled by N(t) = N₀ · 2^(t/3), where t is in hours. A researcher observes the population at t = 0 and t = 9. She calculates the average rate of change over [0, 9] and compares it to the instantaneous rate of change at t = 3. Which statement is correct?

Answer: The average rate over [0, 9] is greater than the instantaneous rate at t = 3

Average rate = [N(9)−N(0)]/(9−0) = N₀(2³−1)/9 = 7N₀/9. Instantaneous rate: N′(t) = N₀·2^(t/3)·(ln2/3). At t=3: N′(3) = N₀·2·(ln2/3) = 2N₀·ln2/3 ≈ 2N₀·0.6931/3 ≈ 0.4621N₀. Average rate = 7N₀/9 ≈ 0.7778N₀. Since 0.7778 > 0.4621, the average rate over [0,9] is greater than the instantaneous rate at t=3. The MVT guarantees a point where they are equal, but that point is not necessarily t=3. For a convex exponential function, the average rate over an interval exceeds the instantaneous rate at the midpoint.

A farmer encloses three identical rectangular pens in a row using a total of 240 m of fencing. The pens share interior dividing walls (running parallel to the shorter side). If x is the length of the longer side of the entire enclosure, which function correctly models the total area A(x) of all three pens combined?

Answer: A(x) = x(240 − 2x)/4

Let x = total length, w = width of the combined enclosure. Fencing: 2 long sides + 4 short sides (2 outer + 2 interior dividers sharing walls) → 2x + 4w = 240, so w = (240−2x)/4. Area A = x·w = x(240−2x)/4. To find maximum: dA/dx = (240−4x)/4 = 0 → x = 60 m, giving maximum area. Option C incorrectly uses (240−x) instead of (240−2x) — a common error from forgetting both long sides require fencing.

A model for the spread of a disease in a town of 5 000 people is given by I(t) = 5000/(1 + 4999e^(−0.8t)), where I is the number infected and t is in days. On which day does the infection rate (number of new cases per day) reach its maximum?

Answer: Day 11

For a logistic model I(t) = K/(1+Ae^(−rt)), the rate of new infections dI/dt is maximised at the inflection point where I = K/2. Setting I = 2500: 5000/(1+4999e^(−0.8t)) = 2500 → 4999e^(−0.8t) = 1 → t = ln(4999)/0.8 ≈ 8.517/0.8 ≈ 10.6 days, so the maximum rate occurs on day 11. A frequent error is solving for when I is maximum (which is never, as logistic growth asymptotes to K) rather than when dI/dt is maximum.