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: angles in radians; no calculator is needed. Notation: ⟨a, b⟩ is the vector a i + b j, and r′(t) means dr/dt.
Question 1 (multiple choice · foundation)
Let r(t) = ⟨e^(2t), t sin t⟩. Which of the following is r′(t)?
- (A) ⟨2e^(2t), sin t + t cos t⟩
- (B) ⟨e^(2t), t cos t⟩
- (C) ⟨2e^(2t), cos t⟩
- (D) 2e^(2t) + sin t + t cos t
Answer and explanation
Answer: (A). Differentiate each component. d/dt [e^(2t)] = 2e^(2t) (chain rule). d/dt [t sin t] = 1 · sin t + t cos t (product rule). So r′(t) = ⟨2e^(2t), sin t + t cos t⟩.
- (B) drops the chain-rule factor 2 in the first component and differentiates only the sin t part of the product in the second.
- (C) multiplies the derivatives of t and sin t (1 · cos t). The derivative of a product is not the product of the derivatives.
- (D) has the right pieces but adds them into a single number. The derivative of a vector-valued function is a vector.
Question 2 (multiple choice · foundation)
What is the domain of r(t) = ⟨√(6 − t), ln(t − 1)⟩?
- (A) t ≤ 6
- (B) t > 1
- (C) 1 < t ≤ 6
- (D) 1 ≤ t < 6
Answer and explanation
Answer: (C). √(6 − t) needs 6 − t ≥ 0, so t ≤ 6. ln(t − 1) needs t − 1 > 0, so t > 1. Both components must be defined, so the domain is 1 < t ≤ 6.
- (A) uses only the first component.
- (B) uses only the second component.
- (D) swaps the inequalities at the ends. ln 0 is undefined, so t = 1 is excluded; √0 = 0 is fine, so t = 6 is included.
Question 3 (multiple choice · core)
A point moves so that its position is r(t) = ⟨t² − 6t, 5 − t³/3⟩. At t = 2, which statement describes its direction of motion and the slope of the tangent line?
- (A) Moving left and down; slope 2
- (B) Moving right and up; slope 2
- (C) Moving left and down; slope ½
- (D) Moving left and up; slope −2
Answer and explanation
Answer: (A). r′(t) = ⟨2t − 6, −t²⟩, so r′(2) = ⟨−2, −4⟩. x′ < 0 means moving left; y′ < 0 means moving down. The slope is y′/x′ = (−4)/(−2) = 2.
- (B) reads the positive slope as “right and up”. A positive slope only says the tangent line rises to the right; the point can travel along it in either direction. The signs of x′ and y′ decide.
- (C) computes x′/y′ = ½, run over rise.
- (D) loses the minus sign on y′, giving r′(2) = ⟨−2, 4⟩.
Question 4 (multiple choice · stretch)
Let r(t) = ⟨t² + 1, t³ + 2t⟩. Which gives r″(1) and the value of d²y/dx² at t = 1?
- (A) r″(1) = ⟨2, 6⟩ and d²y/dx² = ¼
- (B) r″(1) = ⟨2, 6⟩ and d²y/dx² = 3
- (C) r″(1) = ⟨2, 5⟩ and d²y/dx² = 5/2
- (D) r″(1) = ⟨2, 6⟩ and d²y/dx² = 6
Answer and explanation
Answer: (A). r′(t) = ⟨2t, 3t² + 2⟩ and r″(t) = ⟨2, 6t⟩, so r″(1) = ⟨2, 6⟩. For d²y/dx², use Topic 9.2: dy/dx = (3t² + 2)/(2t) = (3/2)t + 1/t. Then d/dt [dy/dx] = 3/2 − 1/t², and d²y/dx² = (3/2 − 1/t²) ÷ (2t). At t = 1: (3/2 − 1) ÷ 2 = ¼.
- (B) divides y″ by x″: 6 ÷ 2 = 3. That is not how d²y/dx² is found.
- (C) gives r′(1) = ⟨2, 5⟩ instead of r″(1), and then uses its ratio 5/2, which is dy/dx at t = 1.
- (D) takes the y-component of r″(1) as d²y/dx². The vector r″ and the scalar d²y/dx² are different quantities.
Question 5 (calculation · core)
Let r(t) = ⟨t² e^(−t), cos³ t⟩ for 0 < t < π.
(a) Find r′(t). (b) Find the value of t at which the curve has a vertical tangent, and the value at which it has a horizontal tangent. Justify each.
Worked solution
(a) First component (product rule): d/dt [t² e^(−t)] = 2t e^(−t) − t² e^(−t) = t(2 − t) e^(−t). Second component (chain rule): d/dt [(cos t)³] = 3 cos² t · (−sin t) = −3 sin t cos² t. r′(t) = ⟨t(2 − t) e^(−t), −3 sin t cos² t⟩
(b) Vertical tangent: x′(t) = 0 when t = 0 or t = 2. Only t = 2 is in 0 < t < π. There y′(2) = −3 sin 2 cos² 2 ≠ 0, because sin 2 ≠ 0 and cos 2 ≠ 0 (2 is not a multiple of π/2). So r′(2) = ⟨0, nonzero⟩: vertical tangent. Horizontal tangent: y′(t) = 0 when sin t = 0 or cos t = 0. In 0 < t < π, only cos t = 0 works, at t = π/2. There x′(π/2) = (π/2)(2 − π/2) e^(−π/2) ≠ 0, because 0 < π/2 < 2. So r′(π/2) = ⟨nonzero, 0⟩: horizontal tangent.
Suggested mark points (4): 1 for each correct component of r′(t); 1 for t = 2 with the check y′(2) ≠ 0; 1 for t = π/2 with the check x′(π/2) ≠ 0.
Common error: missing the check that the other component is nonzero. If both were 0, no conclusion about the tangent would be possible.
Question 6 (constructed response · core)
Let r(t) = ⟨3 cos t, 2 sin(2t)⟩.
(a) Find r′(t) and r″(t). (b) Find r(π/6) and r′(π/6). (c) Find an equation of the tangent line to the curve at t = π/6. (d) Describe the direction of motion at t = π/6.
Worked solution and suggested Marlbridge rubric
(a) r′(t) = ⟨−3 sin t, 4 cos(2t)⟩. r″(t) = ⟨−3 cos t, −8 sin(2t)⟩.
(b) r(π/6) = ⟨3 · (√3/2), 2 sin(π/3)⟩ = ⟨3√3/2, √3⟩. r′(π/6) = ⟨−3 · ½, 4 cos(π/3)⟩ = ⟨−3/2, 2⟩.
(c) Slope = y′/x′ = 2 ÷ (−3/2) = −4/3. Tangent line: y − √3 = −(4/3)(x − 3√3/2), which simplifies to y = −(4/3)x + 3√3.
(d) x′ = −3/2 < 0 and y′ = 2 > 0, so the point is moving left and up.
| Point | What earns it |
|---|---|
| 1 | (a) Correct r′(t), including the chain-rule factor 2 in the second component |
| 1 | (a) Correct r″(t) |
| 1 | (b) Correct r(π/6) and r′(π/6) |
| 1 | (c) Slope −4/3 from y′/x′ and a correct line through (3√3/2, √3) |
| 1 | (d) Left and up, justified by the signs of x′ and y′ |
Total: 5 points. Any correct form of the line earns the (c) point. Extra: r″(π/6) = ⟨−3√3/2, −4√3⟩, not needed for the marks.
Question 7 (constructed response · stretch)
A fictional remote-controlled boat moves on a lake. Its position is r(t) = ⟨x(t), y(t)⟩ metres, where x is measured east and y north of a jetty, and t is in seconds, 0 ≤ t ≤ 4. A student records these values:
| t (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| x(t) (m) | 0 | 7 | 12 | 15 | 16 |
| y(t) (m) | 5 | 2.25 | 1 | 2.75 | 9 |
(a) Use the table to estimate r′(2). Show the method and give units. (b) The values fit the model x(t) = 8t − t², y(t) = ¼t³ − 3t + 5. Use the model to find r′(2) exactly. In which direction is the boat moving at t = 2? (c) Find r″(2) from the model and explain what its first component tells you. (d) Show that at t = 4 the boat is moving due north.
Worked solution and suggested Marlbridge rubric
(a) Use the symmetric difference quotient on [1, 3]: x′(2) ≈ (15 − 7)/(3 − 1) = 4 and y′(2) ≈ (2.75 − 2.25)/(3 − 1) = 0.25. r′(2) ≈ ⟨4, 0.25⟩ m/s.
(b) r′(t) = ⟨8 − 2t, ¾t² − 3⟩. r′(2) = ⟨4, 0⟩ m/s. The boat is moving due east (x′ > 0, y′ = 0). The estimate in (a) was close; its small y-component comes from averaging over [1, 3].
(c) r″(t) = ⟨−2, (3/2)t⟩, so r″(2) = ⟨−2, 3⟩ m/s². The first component, −2 m/s², means the eastward component of the velocity is decreasing by 2 m/s every second at t = 2.
(d) r′(4) = ⟨8 − 8, ¾ · 16 − 3⟩ = ⟨0, 9⟩. The east component is 0 and the north component is positive, so the boat is moving due north (at 9 m/s).
| Point | What earns it |
|---|---|
| 1 | (a) Difference quotients using values on both sides of t = 2, giving ⟨4, 0.25⟩ with units m/s |
| 1 | (b) Correct r′(t) and r′(2) = ⟨4, 0⟩ |
| 1 | (b) Direction due east, justified by the signs of the components |
| 1 | (c) r″(2) = ⟨−2, 3⟩ with the meaning of −2 (eastward velocity component decreasing) |
| 1 | (d) r′(4) = ⟨0, 9⟩ and the conclusion “due north” from x′ = 0, y′ > 0 |
Total: 5 points. Acceptable alternative for (a): a one-sided quotient such as (x(3) − x(2))/1 = 3 and (y(3) − y(2))/1 = 1.75, if the method is shown; the symmetric quotient is usually more accurate. Units: metres per second for r′, metres per second per second for r″.
How did you do?
- Q1 or Q5(a) wrong: practise the product and chain rules inside components; see Worked example 1 in the study guide.
- Q2 wrong: reread the domain paragraph in “What is a vector-valued function?”.
- Q3, Q5(b) or Q6 wrong: reread “What r′(t) tells you about the curve”.
- Q4 wrong: work through “r″(t) is not d²y/dx²”.
- Q7 wrong: compare with Worked example 2 and check your units.
Then tick off the topic checklist.