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 1–5; in Questions 6 and 7 a calculator may be used, and final decimals should be given to 3 decimal places. Contexts and data are invented. “Euler’s method with step size h” always means yₙ₊₁ = yₙ + h · (slope at (xₙ, yₙ)).
Question 1 (multiple choice · foundation)
Let y = f(x) be the solution of dy/dx = y² − x with f(1) = 2. Using one step of Euler’s method with step size 0.1, what is the approximation of f(1.1)?
- (A) 0.3
- (B) 2.29
- (C) 2.3
- (D) 5
Answer and explanation
Answer: (C). The slope at (1, 2) is 2² − 1 = 3. So f(1.1) ≈ 2 + 0.1 × 3 = 2.3.
- (A) is the change in y, 0.1 × 3. You must add it to the starting value 2.
- (B) uses the slope at x = 1.1 with the old y-value: 4 − 1.1 = 2.9, giving 2.29. The slope must be taken at the starting point (1, 2).
- (D) forgets to multiply by the step size: 2 + 3 = 5.
Question 2 (multiple choice · core)
Let y = f(x) be the solution of dy/dx = 1 + xy with f(0) = 2. Using Euler’s method with two steps of equal size, what is the approximation of f(1)?
- (A) 3
- (B) 3.5
- (C) 3.625
- (D) 4.75
Answer and explanation
Answer: (C). h = 0.5. Step 1: slope at (0, 2) is 1 + 0 = 1, so y ≈ 2 + 0.5(1) = 2.5 at x = 0.5. Step 2: slope at (0.5, 2.5) is 1 + 0.5 × 2.5 = 2.25, so y ≈ 2.5 + 0.5(2.25) = 3.625 at x = 1.
- (A) is one step of size 1 (or reusing the first slope for both steps): 2 + 1 × 1 = 3. That ignores the change in slope.
- (B) uses the new x but the old y in step 2: slope 1 + 0.5 × 2 = 2, giving 2.5 + 1 = 3.5. Both coordinates must be updated.
- (D) forgets the step size in step 2: 2.5 + 2.25 = 4.75.
Question 3 (multiple choice · core)
Let y = f(x) be the solution of dy/dx = 4 − y with f(0) = 1. Euler’s method with two steps of size 0.5 is used to approximate f(1). Which statement is true?
- (A) The approximation is 3.25, and it is an overestimate because the solution is concave down.
- (B) The approximation is 3.25, and it is an underestimate because the solution is increasing.
- (C) The approximation is 3.25, and it is an overestimate because dy/dx > 0.
- (D) The approximation is 4, and it is an overestimate because the solution is concave down.
Answer and explanation
Answer: (A). Step 1: slope 4 − 1 = 3, so y ≈ 1 + 1.5 = 2.5 at x = 0.5. Step 2: slope 4 − 2.5 = 1.5, so y ≈ 2.5 + 0.75 = 3.25 at x = 1. Then d²y/dx² = −dy/dx = −(4 − y), which is negative while y < 4, so the solutions are concave down and the tangent steps lie above the curve: an overestimate. (The exact solution is y = 4 − 3e^(−x), with f(1) ≈ 2.896.)
- (B) has the right number but the wrong conclusion. Whether y is increasing does not decide over or under; concavity does.
- (C) reaches the right conclusion for the wrong reason. dy/dx > 0 says the solution is increasing, which says nothing about whether the tangent lines lie above or below it.
- (D) uses one step of size 1 (1 + 1 × 3 = 4), not two steps of 0.5.
Question 4 (multiple choice · core)
Let y = f(x) be the solution of dy/dx = x − 3y through the point (2, 1). Using one step of Euler’s method with step size −0.2, what is the approximation of f(1.8)?
- (A) 0.2
- (B) 0.8
- (C) 1.2
- (D) 1.24
Answer and explanation
Answer: (C). The slope at (2, 1) is 2 − 3 = −1. With h = −0.2: f(1.8) ≈ 1 + (−0.2)(−1) = 1.2. Moving left along a line of slope −1 makes y go up.
- (A) is the change in y (+0.2), not the new value.
- (B) uses h = +0.2 by mistake: 1 + 0.2(−1) = 0.8. That would approximate f(2.2), not f(1.8).
- (D) uses the slope at x = 1.8 (1.8 − 3 = −1.2) instead of at the starting point.
Question 5 (constructed response · core, no calculator)
Let y = f(x) be the particular solution of dy/dx = 2y − x with f(0) = 1.
(a) Write the tangent line to the graph of f at x = 0 and use it to approximate f(0.4). (b) Use Euler’s method with two steps of equal size, starting at x = 0, to approximate f(0.4). Show your steps. (c) Find d²y/dx² in terms of x and y. (d) Is your answer to (b) an overestimate or an underestimate of f(0.4)? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) Slope at (0, 1) is 2(1) − 0 = 2. Tangent line: y = 1 + 2x. So f(0.4) ≈ 1 + 0.8 = 1.8.
(b) h = 0.2.
| Step | (xₙ, yₙ) | Slope 2yₙ − xₙ | 0.2 × slope | New point |
|---|---|---|---|---|
| 1 | (0, 1) | 2 | 0.4 | (0.2, 1.4) |
| 2 | (0.2, 1.4) | 2.8 − 0.2 = 2.6 | 0.52 | (0.4, 1.92) |
f(0.4) ≈ 1.92.
(c) d²y/dx² = 2 dy/dx − 1 = 2(2y − x) − 1 = 4y − 2x − 1.
(d) For 0 ≤ x ≤ 0.4 and y ≥ 1 (true along the steps and for the solution, which starts at 1 with positive slope and keeps increasing), 4y − 2x − 1 ≥ 4 − 0.8 − 1 = 2.2 > 0. At the starting point it equals 3. The solutions are concave up in this region, so the Euler segments lie below the curve: 1.92 is an underestimate.
| Point | What earns it |
|---|---|
| 1 | Tangent line y = 1 + 2x and the estimate 1.8 |
| 1 | Correct first Euler step to (0.2, 1.4) |
| 1 | Correct second step, slope 2.6 at (0.2, 1.4), giving 1.92 |
| 1 | d²y/dx² = 4y − 2x − 1, with the chain rule applied to y |
| 1 | “Underestimate”, justified by d²y/dx² > 0 (concave up) in the relevant region |
Total: 5 points. The (d) point needs the sign of d²y/dx², not just “y is increasing”. For reference (not needed): the exact value is f(0.4) = 0.45 + 0.75e^0.8 ≈ 2.119, so both estimates are low, and Euler (1.92) beats the single tangent line (1.8).
Question 6 (constructed response · core, calculator allowed)
During steady rain, water flows into a rain barrel from a roof at 8 litres per hour and leaks out of a small hole at 0.6√W litres per hour, where W is the volume of water in litres. So dW/dt = 8 − 0.6√W, with t in hours. At t = 0 the barrel holds 100 litres.
(a) Find the rate of change of W at t = 0. Include units. (b) Use Euler’s method with three steps of 1 hour to approximate W(3). (c) Find d²W/dt² in terms of W. Use it to decide whether your answer to (b) is an overestimate or an underestimate. (d) A classmate repeats (b) with six steps of 0.5 hours. Should the new estimate be larger or smaller than yours? Explain.
Worked solution and suggested Marlbridge rubric
(a) dW/dt = 8 − 0.6√100 = 8 − 6 = 2 litres per hour.
(b)
| Step | (tₙ, Wₙ) | Slope 8 − 0.6√Wₙ (L/h) | New W (L) |
|---|---|---|---|
| 1 | (0, 100) | 2 | 102 |
| 2 | (1, 102) | ≈ 1.940297 | ≈ 103.940297 |
| 3 | (2, 103.940297) | ≈ 1.882933 | ≈ 105.823230 |
W(3) ≈ 105.823 litres.
(c) d²W/dt² = −0.6 · (1/(2√W)) · dW/dt = −(0.3/√W)(8 − 0.6√W). For W < (8/0.6)² ≈ 177.8 L, 8 − 0.6√W > 0, so d²W/dt² < 0. The water stays between 100 L and 177.8 L here, so the solution is concave down and the Euler steps lie above it: 105.823 L is an overestimate.
(d) Smaller. Halving the step size usually brings the estimate closer to the true value. Since the true value is below 105.823 L, the new estimate should be lower. (With h = 0.5 the estimate is about 105.781 L; a very fine step gives about 105.740 L.)
| Point | What earns it |
|---|---|
| 1 | 2 litres per hour, with units |
| 1 | Correct first two steps (102 and ≈ 103.940) |
| 1 | W(3) ≈ 105.823 L, with full values carried between steps |
| 1 | Correct d²W/dt² and “overestimate” justified by its negative sign |
| 1 | “Smaller”, with the reason that a smaller step usually moves the estimate towards the true value, which lies below it |
Total: 5 points. For the third point, accept answers within 0.002 of 105.823 when working is shown. Rounding at each step (for example to 1 decimal place, which gives 105.8) does not earn the third point.
Question 7 (constructed response · stretch, calculator allowed)
Let y = f(x) be the solution of dy/dx = y(2 − x) with f(0) = 1.
(a) Show that f(x) = e^(2x − x²/2) satisfies the differential equation and the initial condition. (b) Without a calculator, use Euler’s method with two steps of 0.5 to approximate f(1). (c) Use Euler’s method with four steps of 0.25 to approximate f(1). (d) Find the error (exact value − estimate) in (b) and in (c). Comment on how the step size affected the error.
Worked solution and suggested Marlbridge rubric
(a) f′(x) = e^(2x − x²/2) · (2 − x) = f(x)(2 − x). ✓ And f(0) = e⁰ = 1. ✓
(b) Step 1: slope at (0, 1) is 1 × 2 = 2, so y ≈ 1 + 1 = 2 at x = 0.5. Step 2: slope at (0.5, 2) is 2 × 1.5 = 3, so y ≈ 2 + 1.5 = 3.5 at x = 1.
(c)
| Step | (xₙ, yₙ) | Slope yₙ(2 − xₙ) | New y |
|---|---|---|---|
| 1 | (0, 1) | 2 | 1.5 |
| 2 | (0.25, 1.5) | 2.625 | 2.15625 |
| 3 | (0.5, 2.15625) | 3.234375 | ≈ 2.964844 |
| 4 | (0.75, 2.964844) | ≈ 3.706055 | ≈ 3.891357 |
f(1) ≈ 3.891.
(d) Exact: f(1) = e^1.5 ≈ 4.482. Error with h = 0.5: e^1.5 − 3.5 ≈ 0.982. Error with h = 0.25: e^1.5 − 3.891357 ≈ 0.590 (keep the unrounded values; 4.482 − 3.891 would give 0.591). Halving the step size made the error smaller (by a factor of about 1.7 here), but did not remove it. Both estimates are too low, which matches d²y/dx² = y[(2 − x)² − 1] > 0 for 0 ≤ x < 1: the solution is concave up there.
| Point | What earns it |
|---|---|
| 1 | Derivative by the chain rule shown equal to y(2 − x), and f(0) = 1 checked |
| 1 | 3.5 from two correct steps |
| 1 | 3.891 from four correct steps, with full values carried |
| 1 | Both errors, about 0.982 and 0.590 (0.591 from rounded values is also fine) |
| 1 | Comment: smaller step gives smaller error, but the error is not zero |
Total: 5 points. The concavity remark in (d) is good practice but is not required for the last point.
How did you do?
- Q1 or Q4 wrong: reread “The procedure” and “Stepping backwards” in the study guide.
- Q2 or Q7 wrong: practise the table layout from Worked example 1, updating both x and y every step.
- Q3 or Q5(d) wrong: work through “Overestimate or underestimate? Use concavity”.
- Q6 wrong: compare with Worked example 2 (Euler’s method in context) and check your units.
- Q7(d) wrong: revisit “How step size affects accuracy”.
Then tick off the topic checklist.