These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. BC-only material. Assumptions: angles in radians; e ≈ 2.71828; no calculator for Questions 1–3, 5 and 6; a graphing calculator is allowed for Questions 4 and 7, with decimals given to 3 decimal places. Notation: r(t) = ⟨x(t), y(t)⟩, “∫ from a to b of f(t) dt” is a definite integral, and “[F(t)] from a to b” means F(b) − F(a).
Question 1 (multiple choice · foundation)
What is ∫ from 0 to 2 of ⟨3t², e^(t/2)⟩ dt?
- (A) ⟨12, e⟩
- (B) ⟨8, ½e − ½⟩
- (C) ⟨8, 2e − 2⟩
- (D) 2e + 6
Answer and explanation
Answer: (C). Integrate each component between 0 and 2. ∫ from 0 to 2 of 3t² dt = [t³] from 0 to 2 = 8. ∫ from 0 to 2 of e^(t/2) dt = [2e^(t/2)] from 0 to 2 = 2e − 2. So the integral is the vector ⟨8, 2e − 2⟩.
- (A) evaluates the integrand at t = 2 instead of integrating: ⟨3 · 4, e¹⟩.
- (B) uses ½ e^(t/2) as the antiderivative. Its derivative is ¼ e^(t/2), so the chain-rule factor has been applied the wrong way round.
- (D) adds the two components, 8 + (2e − 2). The integral of a vector is a vector, not a single number.
Question 2 (multiple choice · core)
A particle has velocity r′(t) = ⟨cos t, −2 sin(2t)⟩, and r(0) = ⟨3, 1⟩. What is r(π/2)?
- (A) ⟨1, −2⟩
- (B) ⟨4, 3⟩
- (C) ⟨4, 2⟩
- (D) ⟨4, −1⟩
Answer and explanation
Answer: (D). Integrate: x(t) = sin t + C₁ and y(t) = cos(2t) + C₂ (check: d/dt cos(2t) = −2 sin(2t)). From r(0) = ⟨3, 1⟩: C₁ = 3 and 1 + C₂ = 1, so C₂ = 0. Then r(π/2) = ⟨sin(π/2) + 3, cos π⟩ = ⟨4, −1⟩.
- (A) is the net change ∫ from 0 to π/2 of r′(t) dt = ⟨1, −2⟩. It forgets to add the starting position ⟨3, 1⟩.
- (B) uses −cos(2t) as the antiderivative of −2 sin(2t) (a sign error). Then C₂ = 2 and y(π/2) = 1 + 2 = 3.
- (C) uses the same constant, 3, for both components, so y(t) = cos(2t) + 3 and y(π/2) = 2. Each component needs its own constant.
Question 3 (multiple choice · core)
A particle has acceleration a(t) = ⟨6t, −2⟩. At t = 0 its velocity is ⟨1, 5⟩ and its position is ⟨0, 3⟩. What is its position at t = 1?
- (A) ⟨2, 7⟩
- (B) ⟨1, 2⟩
- (C) ⟨2, 4⟩
- (D) ⟨4, 3⟩
Answer and explanation
Answer: (A). First integration: v(t) = ⟨3t² + 1, −2t + 5⟩, using v(0) = ⟨1, 5⟩. Second integration: r(t) = ⟨t³ + t, −t² + 5t + 3⟩, using r(0) = ⟨0, 3⟩. So r(1) = ⟨1 + 1, −1 + 5 + 3⟩ = ⟨2, 7⟩.
- (B) leaves out the starting velocity: v(t) = ⟨3t², −2t⟩ gives r(t) = ⟨t³, −t² + 3⟩ and r(1) = ⟨1, 2⟩.
- (C) leaves out the starting position: ⟨t³ + t, −t² + 5t⟩ at t = 1 is ⟨2, 4⟩.
- (D) is the velocity v(1) = ⟨4, 3⟩, not the position. It stops after one integration.
Question 4 (multiple choice · stretch · calculator)
A particle moves so that dx/dt = √(1 + t³). At t = 1 the particle’s x-coordinate is 2. What is its x-coordinate at t = 0?
- (A) 0.586
- (B) 0.889
- (C) 1.111
- (D) 3.111
Answer and explanation
Answer: (B). Work back in time: x(0) = x(1) − ∫ from 0 to 1 of √(1 + t³) dt. A calculator gives ∫ from 0 to 1 of √(1 + t³) dt ≈ 1.1114, so x(0) ≈ 2 − 1.1114 = 0.889. The rate is positive, so x was increasing, and x(0) must be less than 2. ✓
- (A) uses a tangent-line estimate, 2 − x′(1) · 1 = 2 − √2 ≈ 0.586, instead of the exact accumulated change.
- (C) is the value of the integral alone. It is the change in x, not the x-coordinate.
- (D) adds the change instead of subtracting it: 2 + 1.111. Going back in time, the accumulated change must be taken away, because x(1) = x(0) + ∫ from 0 to 1 of x′(t) dt.
Question 5 (calculation · core)
A particle has velocity vector r′(t) = ⟨2/(2t + 1), 6t√(t² + 1)⟩ for t ≥ 0, and r(0) = ⟨1, −2⟩.
(a) Find r(t). (b) Find the exact position of the particle at t = √3. (c) Verify your answer to (a).
Worked solution and suggested Marlbridge rubric
(a) x-component: ∫ 2/(2t + 1) dt = ln(2t + 1) + C₁ (substitute u = 2t + 1, du = 2 dt). Since 2t + 1 > 0 for t ≥ 0, no absolute value is needed. x(0) = ln 1 + C₁ = 1, so C₁ = 1.
y-component: with u = t² + 1, du = 2t dt, ∫ 6t√(t² + 1) dt = ∫ 3√u du = 2u^(3/2) = 2(t² + 1)^(3/2) + C₂. y(0) = 2 + C₂ = −2, so C₂ = −4.
r(t) = ⟨ln(2t + 1) + 1, 2(t² + 1)^(3/2) − 4⟩
(b) At t = √3: t² + 1 = 4, and 4^(3/2) = 8. So r(√3) = ⟨1 + ln(1 + 2√3), 12⟩ (about ⟨2.496, 12⟩).
(c) d/dt[ln(2t + 1) + 1] = 2/(2t + 1) ✓. d/dt[2(t² + 1)^(3/2) − 4] = 2 · (3/2)(t² + 1)^(1/2) · 2t = 6t√(t² + 1) ✓. r(0) = ⟨0 + 1, 2 − 4⟩ = ⟨1, −2⟩ ✓.
| Point | What earns it |
|---|---|
| 1 | Correct antiderivative of the x-component, ln(2t + 1) |
| 1 | Correct antiderivative of the y-component, 2(t² + 1)^(3/2) |
| 1 | Two separate constants found from r(0): C₁ = 1 and C₂ = −4 |
| 1 | Exact position ⟨1 + ln(1 + 2√3), 12⟩ |
| 1 | Verification: both derivatives and the initial value checked |
Total: 5 points. Acceptable alternative: the accumulation form r(√3) = ⟨1, −2⟩ + ∫ from 0 to √3 of r′(t) dt, evaluated exactly, earns the first four points.
Question 6 (constructed response · core)
A particle moves in the plane with acceleration a(t) = ⟨−4 sin(2t), 2⟩ for t ≥ 0. At t = 0 its velocity is ⟨2, −3⟩ and its position is (1, 0).
(a) Find the velocity vector v(t). (b) Find the position vector r(t). (c) Find the time t > 0 at which the particle is on the x-axis, and give its exact x-coordinate at that time. (d) A student writes v(t) = ⟨2 cos(2t) + 2, 2t − 3⟩. Explain the error.
Worked solution and suggested Marlbridge rubric
(a) ∫ −4 sin(2t) dt = 2 cos(2t) + C₁. At t = 0: 2 + C₁ = 2, so C₁ = 0. ∫ 2 dt = 2t + C₂, and C₂ = −3. v(t) = ⟨2 cos(2t), 2t − 3⟩
(b) ∫ 2 cos(2t) dt = sin(2t) + D₁; at t = 0, D₁ = 1. ∫ (2t − 3) dt = t² − 3t + D₂; at t = 0, D₂ = 0. r(t) = ⟨sin(2t) + 1, t² − 3t⟩
(c) On the x-axis, y(t) = 0: t² − 3t = t(t − 3) = 0, so t = 0 or t = 3. For t > 0, t = 3, and x(3) = 1 + sin 6 (about 0.721).
(d) The student set the constant equal to the starting velocity, 2, without checking. But 2 cos(2t) already equals 2 at t = 0, so the student’s function gives v(0) = ⟨4, −3⟩, not ⟨2, −3⟩. The constant must be chosen so that the whole component equals 2 at t = 0, which gives C₁ = 0.
| Point | What earns it |
|---|---|
| 1 | Correct v(t), with both constants found from v(0) |
| 1 | Correct r(t), with both constants found from r(0) |
| 1 | Sets y(t) = 0 and identifies t = 3 |
| 1 | x(3) = 1 + sin 6, using the position function from (b) |
| 1 | Explains that the student’s v(0) would be ⟨4, −3⟩, so the constant is wrong |
Total: 5 points. A correct (c) that follows from an incorrect r(t) in (b) still earns the (c) points if the method is right.
Question 7 (constructed response · stretch · calculator)
A small remote-controlled rover crosses a fictional sports field. Its position is measured in metres, with x east and y north, and t is in seconds. Its velocity components are
dx/dt = 3 sin(√t) and dy/dt = ln(1 + t²) − 1, for 0 ≤ t ≤ 4.
At t = 1 the rover is at (2, 5).
(a) Find the rover’s position at t = 4. (b) Find the rover’s y-coordinate at t = 0. Explain why your answer is greater than 5. (c) Explain why the rover’s x-coordinate increases throughout 1 ≤ t ≤ 4.
Worked solution and suggested Marlbridge rubric
(a) x(4) = 2 + ∫ from 1 to 4 of 3 sin(√t) dt ≈ 2 + 8.6425 = 10.643. y(4) = 5 + ∫ from 1 to 4 of (ln(1 + t²) − 1) dt ≈ 5 + 2.7205 = 7.721. Position at t = 4: about (10.643, 7.721) m.
(b) y(0) = y(1) − ∫ from 0 to 1 of (ln(1 + t²) − 1) dt ≈ 5 − (−0.7361) = 5.736 m. For 0 ≤ t < 1, t² < 1, so ln(1 + t²) < ln 2 < 1 and dy/dt < 0. The y-coordinate was decreasing from t = 0 to t = 1, so it must have been larger than 5 at t = 0.
(c) For 1 ≤ t ≤ 4, 1 ≤ √t ≤ 2. Since 0 < 2 < π, sin(√t) > 0 on this interval, so dx/dt = 3 sin(√t) > 0. A positive rate of change means x(t) is increasing.
| Point | What earns it |
|---|---|
| 1 | Sets up x(4) = 2 + ∫ from 1 to 4 of dx/dt dt (starting value and limits) |
| 1 | x(4) ≈ 10.643 and y(4) ≈ 7.721 |
| 1 | y(0) = 5 − ∫ from 0 to 1 of dy/dt dt, with the subtraction |
| 1 | y(0) ≈ 5.736, explained by dy/dt < 0 on [0, 1) |
| 1 | Justifies dx/dt > 0 using 1 ≤ √t ≤ 2 < π, and links a positive derivative to increasing |
Total: 5 points. Units (metres) are expected in (a) and (b). An answer of 4.264 for (b) adds the integral instead of subtracting it and does not earn the fourth point.
How did you do?
- Q1 or Q5 wrong: check each component separately, including chain-rule factors; see “Integrate one component at a time” in the study guide.
- Q2 wrong: revisit Worked example 1, especially separate constants and the net-change trap.
- Q3 or Q6 wrong: work through Worked example 2 (acceleration to position, two initial conditions).
- Q4 or Q7 wrong: compare with Worked example 3 and the note on going backwards in time.
Then tick off the topic checklist.