These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. BC-only material. Assumptions: no calculator for Questions 1–6; in Question 7 a calculator may be used, with final decimals to 3 decimal places. All contexts and data are invented. In every model, k and the carrying capacity are positive constants.
Question 1 (multiple choice · foundation)
A quantity y satisfies dy/dt = 0.002y(350 − y), with y(0) = 40. What is lim y(t) as t → ∞?
- (A) 0.7
- (B) 40
- (C) 175
- (D) 350
Answer and explanation
Answer: (D). The equation has the form ky(a − y) with a = 350. Since y(0) = 40 > 0, y increases towards the carrying capacity, so the limit is 350.
- (A) is k × a = 0.002 × 350 = 0.7, the relative growth rate when y is small. It is not a value of y.
- (B) is the starting value. The quantity does not stay there, because dy/dt = 0.002 × 40 × 310 > 0.
- (C) is a/2, where y grows fastest, not where it ends up.
Question 2 (multiple choice · core)
A population P satisfies dP/dt = 0.24P − 0.0004P², with P(0) = 60. For what value of P is the population growing fastest?
- (A) 60
- (B) 300
- (C) 600
- (D) 36
Answer and explanation
Answer: (B). Factor: 0.24P − 0.0004P² = 0.0004P(600 − P). So a = 600 and growth is fastest at P = a/2 = 300. Since 60 < 300, the population does pass through 300.
- (A) assumes growth is fastest at the start. At P = 60 the curve is still concave up, so the rate is still increasing.
- (C) is the carrying capacity, where the rate is 0.
- (D) is the greatest rate, 0.0004 × 300 × 300 = 36 per unit time, not the value of P.
Question 3 (multiple choice · core)
A quantity y satisfies dy/dt = 0.01y(60 − y), with y(0) = 90. Which statement describes y for t ≥ 0?
- (A) y increases without bound.
- (B) y decreases and approaches 60; its graph is concave up.
- (C) y decreases and approaches 60; its graph is concave down.
- (D) y decreases and approaches 0.
Answer and explanation
Answer: (B). At y = 90, dy/dt = 0.01 × 90 × (−30) = −27 < 0, so y decreases. It approaches the carrying capacity 60 and never crosses it, because y = 60 is a constant solution. d²y/dt² = 0.01(60 − 2y) · dy/dt; for y > 60 both factors are negative, so d²y/dt² > 0 (at y = 90 it is 32.4): concave up.
- (A) would need dy/dt > 0; above the carrying capacity the rate is negative.
- (C) gets the direction right but the concavity wrong. The fall slows down as y nears 60, which is concave up.
- (D) ignores the constant solution y = 60. The quantity cannot pass below 60.
Question 4 (multiple choice · core)
A plant disease spreads through a field of 2000 plants. The rate at which plants become infected is jointly proportional to the number I already infected and the number not yet infected. Which differential equation models this, for a positive constant k?
- (A) dI/dt = k(2000 − I)
- (B) dI/dt = kI(2000 − I)
- (C) dI/dt = kI + k(2000 − I)
- (D) dI/dt = kI/(2000 − I)
Answer and explanation
Answer: (B). “Jointly proportional” to I and to 2000 − I means proportional to their product, kI(2000 − I). This is the logistic form with carrying capacity 2000.
- (A) is proportional only to the number not infected. Its solutions approach 2000 but grow fastest at the start, not at 1000.
- (C) adds the two quantities. It simplifies to 2000k, a constant rate.
- (D) divides instead of multiplying. It makes the rate grow without bound as I nears 2000.
Question 5 (constructed response · core)
The area A, in mm², of a bacterial colony in a dish satisfies dA/dt = 0.005A(200 − A), where t is in hours, and A(0) = 20.
(a) Find lim A(t) as t → ∞. (b) Find the area of the colony when it is growing fastest, and the rate of growth at that moment. Include units. (c) Find d²A/dt² in terms of A. Is the graph of A concave up or concave down when A = 150? Justify. (d) A second dish starts with A(0) = 250. Describe how A changes over time.
Worked solution and suggested Marlbridge rubric
(a) a = 200 and A(0) = 20 > 0, so lim A(t) = 200 mm².
(b) Fastest growth at A = 200/2 = 100 mm². The rate then is 0.005 × 100 × 100 = 50 mm² per hour.
(c) dA/dt = 0.005(200A − A²), so d²A/dt² = 0.005(200 − 2A) · dA/dt = 0.005(200 − 2A) · 0.005A(200 − A). At A = 150: dA/dt = 0.005 × 150 × 50 = 37.5 > 0, and 200 − 2(150) = −100 < 0, so d²A/dt² = 0.005 × (−100) × 37.5 = −18.75 < 0. Concave down. (A is above 100, so growth is slowing.)
(d) At A = 250, dA/dt = 0.005 × 250 × (−50) = −62.5 < 0. The area decreases towards 200 mm², falling more and more slowly (concave up), and never goes below 200.
| Point | What earns it |
|---|---|
| 1 | Limit 200 |
| 1 | A = 100 mm² for fastest growth |
| 1 | Greatest rate 50 mm² per hour, with units |
| 1 | Correct d²A/dt² (chain rule) and “concave down” at A = 150 with the sign shown |
| 1 | (d): decreases towards 200, with a reason (dA/dt < 0 when A > 200) |
Total: 5 points.
Question 6 (constructed response · core)
The number of users U, in thousands, of a new phone app in a fictional city is modelled by dU/dt = 0.5U(1 − U/60), where t is in months, and U(0) = 4.
(a) Rewrite the equation in the form dU/dt = kU(a − U). State k and a. (b) Find lim U(t) as t → ∞ and interpret it in context. (c) How many users does the app have when the number is growing fastest? Find the rate of growth then, with units. (d) A manager says: “The 0.5 in the model means the app gains 50% more users every month.” Explain why this is wrong.
Worked solution and suggested Marlbridge rubric
(a) 0.5U(1 − U/60) = (0.5/60) U(60 − U) = (1/120) U(60 − U). So k = 1/120 and a = 60.
(b) U(0) = 4 > 0, so lim U(t) = 60: in the long run the model predicts about 60 000 users, the most the city’s market can support.
(c) Fastest at U = 60/2 = 30, that is 30 000 users. The rate is 0.5 × 30 × (1 − 30/60) = 7.5 thousand users per month (7500 users per month).
(d) The relative growth rate is (dU/dt)/U = 0.5(1 − U/60), which changes as U changes. It is close to 0.5 only when U is small (at U = 4 it is 0.5 × 56/60 ≈ 0.467), and it falls towards 0 as U approaches 60 (at U = 45 it is 0.125). Also, even a constant relative rate of 0.5 is an instantaneous rate, not “50% more each month”.
| Point | What earns it |
|---|---|
| 1 | k = 1/120 and a = 60, with the rewriting shown |
| 1 | Limit 60, interpreted as 60 000 users in the long run |
| 1 | 30 thousand users when growing fastest |
| 1 | 7.5 thousand users per month, with units |
| 1 | (d): explains that the relative rate 0.5(1 − U/60) depends on U and decreases as U grows |
Total: 5 points.
Question 7 (constructed response · stretch, calculator allowed)
The number of birds B of a protected species on a fictional island satisfies dB/dt = 0.0008B(500 − B), where t is in years, and B(0) = 50.
(a) Use Euler’s method with two steps of 1 year to approximate B(2). (b) The solution is B(t) = 500/(1 + 9e^(−0.4t)). Find B(2) to 3 decimal places. Explain why your answer to (a) is lower, using d²B/dt². (c) Find the number of birds when the population is growing fastest, and the time at which this happens. (d) Find the greatest rate of growth, with units.
Worked solution and suggested Marlbridge rubric
(a) Step 1: slope at (0, 50) is 0.0008 × 50 × 450 = 18, so B(1) ≈ 68. Step 2: slope at (1, 68) is 0.0008 × 68 × 432 = 23.5008, so B(2) ≈ 68 + 23.5008 = 91.501 birds.
(b) B(2) = 500/(1 + 9e^(−0.8)) ≈ 99.128 birds. d²B/dt² = 0.0008(500 − 2B) · dB/dt. For 0 < B < 250, both factors are positive, so the solution is concave up. The Euler steps follow tangent lines, which lie below a concave-up curve, so 91.501 is an underestimate.
(c) Fastest growth at B = 500/2 = 250 birds. Solve 500/(1 + 9e^(−0.4t)) = 250: 1 + 9e^(−0.4t) = 2, so e^(−0.4t) = 1/9 and t = ln 9 / 0.4 ≈ 5.493 years.
(d) 0.0008 × 250 × 250 = 50 birds per year.
| Point | What earns it |
|---|---|
| 1 | Correct Euler steps, giving 68 and then 91.501 |
| 1 | B(2) ≈ 99.128 from the formula |
| 1 | Underestimate explained by d²B/dt² > 0 (concave up) for B < 250 |
| 1 | B = 250 and t = ln 9 / 0.4 ≈ 5.493 years |
| 1 | 50 birds per year, with units |
Total: 5 points. For (c) a calculator intersection of the graph of B with B = 250 is also accepted. Check of the given formula (not required): B(0) = 500/10 = 50 ✓.
How did you do?
- Q1 or Q5(a) wrong: reread “Reading the equation without solving it” in the study guide.
- Q2 or Q5(b) wrong: practise factoring into ky(a − y) and finding a/2 (Worked example 1).
- Q3 or Q5(d) wrong: look at Figure 2 and the concavity rule d²y/dt² = k(a − 2y) · dy/dt.
- Q4 or Q6 wrong: work through Worked example 2 (from a sentence to the model) and Worked example 1(d).
- Q7 wrong: combine this topic with Euler’s method, and read “Going further (optional)”.
Then tick off the topic checklist.