Use this page after studying Unit 7, Differential Equations, or after the Unit 7 diagnostic. The unit is shared by Calculus AB and BC. Question 7 is BC only (Euler’s method and logistic growth); AB students skip it. These are original Marlbridge practice questions, not past exam questions; data are invented. The rubrics are a suggested Marlbridge rubric, not official scoring. No calculator.
Big ideas of the unit
- A differential equation is a statement about rates. Words become a rate equation; the sign shows increase or decrease (Topic 7.1).
- Checking a solution is differentiation. Both sides must agree on an interval (Topic 7.2).
- A slope field draws the equation. Each segment has the slope the equation gives at that point (Topic 7.3).
- You can reason without solving. The signs of dy/dx and d²y/dx² give direction and bending; a constant solution cannot be crossed (Topic 7.4).
- Euler’s method follows the field in steps (BC only). Each step moves along a tangent line, so concavity says whether the estimate is too high or too low (Topic 7.5).
- Separation of variables gives two antiderivatives and one constant (Topic 7.6).
- An initial condition picks one curve, with a domain. Find C at once, choose the sign of any root, state where it is valid (Topic 7.7).
- “Rate proportional to amount” means exponential. dy/dt = ky gives y = y₀e^(kt); other rate laws give other shapes (Topic 7.8).
- Logistic growth levels off (BC only). With y₀ > 0, y approaches a, and grows fastest at y = a/2 (Topic 7.9).
Key relationships and methods
| You see or need | What it means or what to do | Topics |
|---|---|---|
| A sentence about a rate | Rate = k × (stated quantity); add a minus sign for a decrease | 7.1 |
| “Is this a solution?” | Differentiate, substitute into both sides, check the initial value | 7.2 |
| Horizontal segments | Set dy/dx = 0; a horizontal line here is a constant solution | 7.3, 7.4 |
| Estimate near a point | Tangent line, or Euler steps (BC only); concavity gives over or under | 7.4, 7.5 |
| dy/dx = f(x)g(y) | Separate, antidifferentiate, one constant | 7.6 |
| Initial condition given | Find C at once; choose the sign; state the open interval | 7.7 |
| No elementary antiderivative | y = y₀ + ∫ (a to x) f(t) dt | 7.7 |
| dy/dt = ky | y = y₀e^(kt); doubling time ln 2/k | 7.8 |
| dy/dt = ky(a − y) | Limit a; fastest at a/2 (BC only) | 7.9 |
Question 1 (multiple choice · mixed)
A bacterial colony grows at a rate proportional to the number of cells N present, with t in hours. When N = 400, the colony is growing at 60 cells per hour. How long does it take the colony to double in size?
- (A) (20/3) ln 2 hours
- (B) 0.15 ln 2 hours
- (C) 20/3 hours
- (D) (ln 2)/60 hours
Answer and explanation
Answer: (A). dN/dt = kN with 60 = k(400) gives k = 0.15 per hour. N = N₀e^(0.15t) doubles when e^(0.15t) = 2, so t = (ln 2)/0.15 = (20/3) ln 2 hours (about 4.6).
- (B) multiplies by k instead of dividing.
- (C) assumes a steady 60 cells per hour, but the rate rises as N grows.
- (D) uses 60 as k. k is the rate per cell, 60/400.
Topics: 7.1, 7.8.
Question 2 (multiple choice · mixed)
Let y = f(x) be the particular solution of dy/dx = 9x/y with f(0) = −3. Which statement is true?
- (A) f(x) = −3√(x² + 1), and f has a relative minimum at x = 0.
- (B) f(x) = −3√(x² + 1), and f has a relative maximum at x = 0.
- (C) f(x) = 3√(x² + 1), and f has a relative minimum at x = 0.
- (D) f(x) = −√(9x² + 9/2), and f has a relative maximum at x = 0.
Answer and explanation
Answer: (B). Separate: y dy = 9x dx, so y²/2 = 9x²/2 + C. At (0, −3), C = 9/2, so y² = 9x² + 9. f(0) < 0, so take the negative root: f(x) = −3√(x² + 1). Since y < 0, dy/dx = 9x/y is positive for x < 0 and negative for x > 0, so f has a relative maximum at x = 0.
- (A) has the right formula but pictures 3√(x² + 1), not its negative.
- (C) takes the positive root, which does not pass through (0, −3).
- (D) forgets to double C when clearing the halves. Then f(0) = −√(9/2), not −3.
Topics: 7.4, 7.6, 7.7.
Question 3 (multiple choice · mixed)
Let y = F(x) be the particular solution of dy/dx = e^(−x²) with F(1) = 3. Which statement is true?
- (A) F(0) = 3 + ∫ (0 to 1) e^(−t²) dt, so 3 < F(0) < 4.
- (B) F(0) = ∫ (1 to 0) e^(−t²) dt, so F(0) < 0.
- (C) F(0) = 3 − 1/e exactly.
- (D) F(0) = 3 − ∫ (0 to 1) e^(−t²) dt, so 2 < F(0) < 3.
Answer and explanation
Answer: (D). There is no elementary antiderivative, so use the integral form: F(x) = 3 + ∫ (1 to x) e^(−t²) dt. Then F(0) = 3 − ∫ (0 to 1) e^(−t²) dt. On [0, 1], 0 < e^(−t²) ≤ 1, so the integral is between 0 and 1, and 2 < F(0) < 3. (Every slope is positive, so F(0) < F(1) fits.)
- (A) forgets that the integral from 1 to 0 is negative.
- (B) drops the initial value 3.
- (C) is the tangent line estimate, not the exact value. (F(0) is about 2.25.)
Topics: 7.4, 7.7.
Question 4 (constructed response · mixed)
In an invented model, the charge Q (per cent) of a phone battery increases at a rate proportional to the difference between 100 and Q, with t in minutes. At t = 0, Q = 20 and the charge is rising at 4 per cent per minute.
(a) Write a differential equation for Q and find the constant of proportionality, with units. (b) Find d²Q/dt² in terms of Q. Explain what its sign says about the charging. (c) Use the tangent line at t = 0 to estimate Q(2). Is this an overestimate or an underestimate? Justify. (d) Use separation of variables to find Q(t), and find when the battery reaches 60 per cent.
Worked solution and suggested Marlbridge rubric
(a) dQ/dt = k(100 − Q), and 4 = k(80), so k = 1/20 = 0.05 per minute.
(b) d²Q/dt² = −0.05 · dQ/dt = −0.0025(100 − Q). This is negative for Q < 100: the battery charges more slowly as it fills.
(c) Q(2) ≈ 20 + 4(2) = 28 per cent. d²Q/dt² < 0, so Q is concave down and the tangent line lies above it: 28 is an overestimate (the model gives about 27.6).
(d) For Q < 100, separate: dQ/(100 − Q) = 0.05 dt. Antidifferentiate: −ln(100 − Q) = 0.05t + C. At t = 0, −ln 80 = C. So ln(100 − Q) = ln 80 − 0.05t, giving 100 − Q = 80e^(−0.05t) and Q(t) = 100 − 80e^(−t/20). Then 80e^(−t/20) = 40 gives t = 20 ln 2 minutes (about 13.9).
| Point | What earns it |
|---|---|
| 1 | dQ/dt = k(100 − Q) |
| 1 | k = 0.05 per minute, with units |
| 1 | d²Q/dt² = −0.0025(100 − Q) and the meaning of its sign |
| 1 | Estimate 28 and “overestimate” justified by concavity |
| 1 | Correct separation and antiderivatives with a constant |
| 1 | Q(t) = 100 − 80e^(−t/20) and t = 20 ln 2 minutes |
Total: 6 points. Topics: 7.1, 7.4, 7.6, 7.7.
Question 5 (constructed response · mixed)
Consider the differential equation dy/dx = (2 − y)/(x² + 1).
(a) Show that y = 2 is a solution. Where in the plane are the slopes positive? (b) Along the row y = 0, find the slopes at x = 0, x = 1 and x = 2. Describe how the segments change as you move away from the y-axis. (c) Let y = f(x) be the particular solution through (0, 0). Find d²y/dx² at (0, 0) and say whether the graph of f is concave up or concave down there. (d) Find f(x). (e) A student says, “From the slope field, f(x) increases towards 2 as x → ∞.” Find the limit of f(x) as x → ∞ and comment.
Worked solution and suggested Marlbridge rubric
(a) If y = 2, the left side is 0 and the right side is (2 − 2)/(x² + 1) = 0 for every x, so y = 2 is a solution. x² + 1 > 0, so slopes are positive wherever y < 2.
(b) The slopes are 2/1 = 2, 2/2 = 1 and 2/5 = 0.4. The segments get flatter away from the y-axis, on both sides.
(c) Using the quotient rule and dy/dx: d²y/dx² = [−(dy/dx)(x² + 1) − (2 − y)(2x)]/(x² + 1)². At (0, 0), dy/dx = 2, so d²y/dx² = (−2 − 0)/1 = −2 < 0: concave down.
(d) For y < 2, separate: dy/(2 − y) = dx/(x² + 1). Antidifferentiate: −ln(2 − y) = arctan x + C. At (0, 0), −ln 2 = C. So ln(2 − y) = ln 2 − arctan x and 2 − y = 2e^(−arctan x). f(x) = 2 − 2e^(−arctan x).
(e) As x → ∞, arctan x → π/2, so f(x) → 2 − 2e^(−π/2) (about 1.58). f increases, but the slopes shrink so fast that it levels off below 2. A slope field shows directions, not limits.
| Point | What earns it |
|---|---|
| 1 | Both sides shown equal to 0 for y = 2, and slopes positive for y < 2 |
| 1 | Slopes 2, 1, 0.4, with “flatter away from the y-axis” |
| 1 | Correct expression for d²y/dx² (implicit differentiation) |
| 1 | −2 at (0, 0), concave down |
| 1 | Correct separation with ln(2 − y) and arctan x |
| 1 | f(x) = 2 − 2e^(−arctan x) |
| 1 | Limit 2 − 2e^(−π/2), with a comment that the curve does not approach 2 |
Total: 7 points. Topics: 7.2, 7.3, 7.4, 7.6, 7.7.
Question 6 (constructed response · mixed)
A boat cuts its engine at t = 0 seconds, with velocity 4 m/s, at x = 0 m. Two invented models describe how it slows:
- Model 1: the boat’s velocity v decreases at a rate proportional to the square of v.
- Model 2: v decreases at a rate proportional to v.
In both, the acceleration at t = 0 is −1 m/s².
(a) Write a differential equation for each model and find each constant of proportionality. (b) Solve Model 1 to find v(t), and verify your answer. (c) Give v(t) for Model 2 and find the position x(t) for each model. (d) A buoy is 20 m from the boat’s starting point. According to each model, does the boat reach it? If so, when?
Worked solution and suggested Marlbridge rubric
(a) Model 1: dv/dt = −kv². At t = 0: −1 = −k(16), so k = 1/16. Model 2: dv/dt = −kv. At t = 0: −1 = −k(4), so k = 1/4.
(b) Separate: v^(−2) dv = −(1/16) dt. Antidifferentiate: −1/v = −t/16 + C. At t = 0, −1/4 = C. So 1/v = t/16 + 1/4 = (t + 4)/16, and v(t) = 16/(t + 4). Verify: dv/dt = −16/(t + 4)² = −(1/16)v², and v(0) = 4.
(c) Model 2 is exponential decay: v(t) = 4e^(−t/4). Position is accumulated velocity from x(0) = 0. Model 1: x(t) = ∫ (0 to t) 16/(s + 4) ds = 16 ln((t + 4)/4). Model 2: x(t) = ∫ (0 to t) 4e^(−s/4) ds = 16 − 16e^(−t/4).
(d) Model 1: 16 ln((t + 4)/4) = 20 gives (t + 4)/4 = e^(1.25), so the boat reaches the buoy at t = 4e^(1.25) − 4 seconds (about 10.0). Model 2: x(t) < 16 for every t, so the boat never reaches the buoy.
| Point | What earns it |
|---|---|
| 1 | Both differential equations with minus signs |
| 1 | k = 1/16 and k = 1/4 |
| 1 | Correct separation and antiderivative for Model 1 |
| 1 | v(t) = 16/(t + 4), verified in the equation and at t = 0 |
| 1 | v(t) = 4e^(−t/4) and both position functions |
| 1 | Model 1 time 4e^(1.25) − 4 seconds |
| 1 | Model 2 never reaches 20 m, justified by x(t) < 16 |
Total: 7 points. Topics: 7.1, 7.2, 7.6, 7.7, 7.8.
Question 7 (constructed response · mixed) (BC only)
A plant spreads over a 600 m² pond. The area A (m²) it covers grows at a rate jointly proportional to A and to the uncovered area, with t in weeks. At t = 0, A = 100 and the plant is spreading at 25 m² per week.
(a) Write a differential equation for A and find the constant of proportionality. (b) Use Euler’s method with two steps of 2 weeks to approximate A(4). (c) Find d²A/dt² in terms of A. Is your answer to (b) an overestimate or an underestimate? Justify. (d) Find lim A(t) as t → ∞. Find the area covered when the plant is spreading fastest, and that greatest rate, with units. (e) A second pond has no plant at t = 0. Use the differential equation to explain what the model predicts for it.
Worked solution and suggested Marlbridge rubric
(a) dA/dt = kA(600 − A). At t = 0: 25 = k(100)(500), so k = 1/2000 = 0.0005.
(b) Step 1: the slope at A = 100 is 25, so A(2) ≈ 100 + 2(25) = 150. Step 2: the slope at A = 150 is 0.0005(150)(450) = 33.75, so A(4) ≈ 150 + 2(33.75) = 217.5 m².
(c) d²A/dt² = 0.0005(600 − 2A) · dA/dt = 0.0005²(600 − 2A) · A(600 − A). This is positive for 0 < A < 300. A rises by at most 45 m² per week (see (d)), so A(4) ≤ 280 and the solution is concave up on 0 ≤ t ≤ 4. Tangent lines lie below a concave up curve, so 217.5 is an underestimate.
(d) 0 < A(0) < 600, so A increases towards the carrying capacity: lim A(t) = 600 m². The rate is greatest at A = 300 m², half of 600, where dA/dt = 0.0005(300)(300) = 45 m² per week.
(e) With A = 0, dA/dt = 0, so A = 0 is a constant solution: the plant never appears. It can only spread from plant already present.
| Point | What earns it |
|---|---|
| 1 | dA/dt = kA(600 − A) and k = 0.0005 |
| 1 | First Euler step: A(2) ≈ 150 |
| 1 | Second step with slope 33.75 at A = 150: A(4) ≈ 217.5 |
| 1 | Correct d²A/dt² in terms of A |
| 1 | Underestimate, justified by d²A/dt² > 0 on the interval |
| 1 | Limit 600 and A = 300 |
| 1 | Greatest rate 45 m² per week, with units |
| 1 | (e): A = 0 gives dA/dt = 0, a constant solution, so no growth |
Total: 8 points. Topics: 7.1, 7.4, 7.5, 7.9.
How did you do?
Add your points from Questions 4–6 (20 in total) to your correct answers in Questions 1–3; BC students add Question 7 (8 points). This is a guide, not a predicted score. Note which topics your lost points came from (each answer lists them), then use those topic checklists:
7.1 · 7.2 · 7.3 · 7.4 · 7.5 (BC only) · 7.6 · 7.7 · 7.8 · 7.9 (BC only)
If many topics need work, use the “Your next step” table in the Unit 7 diagnostic.