BC only. Unit 9 is part of Calculus BC only; Calculus AB students do not need this page.
Use this page after Unit 9 or the Unit 9 diagnostic. 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 official scoring. A calculator is allowed for Question 5 only. Angles are in radians.
Big ideas of the unit
- One variable drives both coordinates. A parametric curve gives x and y as functions of t. A vector-valued function r(t) = ⟨x(t), y(t)⟩ is the same thing written as a position vector (Topic 9.1, Topic 9.4).
- Slope comes from the chain rule. dy/dx = (dy/dt) ÷ (dx/dt) when dx/dt ≠ 0. Horizontal tangent: dy/dt = 0 and dx/dt ≠ 0; vertical tangent: the other way round (Topic 9.1).
- Concavity needs one more division. d²y/dx² is d/dt (dy/dx) divided by dx/dt. It is not y″(t), the second component of r″(t) (Topic 9.2).
- Calculus on vectors works one component at a time. Differentiate or integrate each component, each with its own constant (Topic 9.4, Topic 9.5).
- Position = starting position + integral of velocity. The known position can be at any time, not only t = 0 (Topic 9.5).
- Motion in the plane: the signs of x′ and y′ give direction, speed is √(x′² + y′²), and the speed increases when v · a > 0. The integral of speed is distance travelled (Topic 9.6).
- Arc length is the integral of speed, when the curve is traced once. Length is always at least the straight-line distance between the ends (Topic 9.3).
- A polar curve is a parametric curve in θ: x = r cos θ, y = r sin θ. dr/dθ says whether the point moves towards or away from the pole; dy/dx gives the slope (Topic 9.7).
- Polar area is built from thin sectors: ½ ∫ r² dθ for one curve, ½ ∫ (R² − r²) dθ between two, with limits from r = 0 or where the curves meet (Topic 9.8, Topic 9.9).
Key relationships and methods
| You see or need | What to do | Topics |
|---|---|---|
| Slope of a parametric or vector curve | (dy/dt) ÷ (dx/dt) | 9.1, 9.4 |
| Horizontal / vertical tangent | dy/dt = 0, dx/dt ≠ 0 / dx/dt = 0, dy/dt ≠ 0 | 9.1 |
| Concavity | d/dt (dy/dx) ÷ (dx/dt) | 9.2 |
| Velocity and acceleration | r′(t) and r″(t), component by component | 9.4, 9.6 |
| Position from velocity | r(b) = r(a) + ∫ from a to b of v(t) dt | 9.5 |
| Speed; speed increasing? | √(x′² + y′²); sign of v · a | 9.6 |
| Distance travelled or arc length | ∫ √(x′² + y′²) dt | 9.3, 9.6 |
| Polar to parametric | x = r cos θ, y = r sin θ | 9.7 |
| Towards or away from the pole | Compare the signs of r and dr/dθ | 9.7 |
| Area of a polar region | ½ ∫ r² dθ over one tracing | 9.8 |
| Area between polar curves | ½ ∫ (R² − r²) dθ; test which is outer | 9.9 |
Question 1 (multiple choice · mixed)
A curve is given by x = eᵗ and y = t². For which values of t is dy/dx positive and the curve concave down?
- (A) 0 < t < 1
- (B) t > 1
- (C) t < 0
- (D) t > 0
Answer and explanation
Answer: (B). dy/dx = 2t/eᵗ = 2t e^(−t), which is positive for t > 0. Then d/dt (dy/dx) = 2e^(−t)(1 − t), and dividing by dx/dt = eᵗ gives d²y/dx² = 2(1 − t)e^(−2t). This is negative for t > 1, so both conditions hold for t > 1.
- (A) is where the slope is positive but the curve is concave up.
- (C) is where the slope is negative.
- (D) checks the slope only.
Topics: 9.1, 9.2.
Question 2 (multiple choice · mixed)
A particle moves with position r(t) = ⟨½e^(2t) − t, 2eᵗ⟩ for 0 ≤ t ≤ ln 3. Which gives the distance it travels and the length of its displacement vector?
- (A) Distance 9/2 + ln 3; displacement length √((4 − ln 3)² + 16)
- (B) Distance √((4 − ln 3)² + 16); displacement length 4 + ln 3
- (C) Distance 8 − ln 3; displacement length √((4 − ln 3)² + 16)
- (D) Distance 4 + ln 3; displacement length √((4 − ln 3)² + 16)
Answer and explanation
Answer: (D). x′ = e^(2t) − 1 and y′ = 2eᵗ. Squaring and adding gives e^(4t) + 2e^(2t) + 1 = (e^(2t) + 1)², so the speed is e^(2t) + 1. The distance is [½e^(2t) + t] from 0 to ln 3 = (9/2 + ln 3) − ½ = 4 + ln 3, about 5.099. The particle starts at (½, 2) and ends at (9/2 − ln 3, 6), so the displacement is ⟨4 − ln 3, 4⟩, with length about 4.941.
- (A) forgets to subtract the value at the lower limit.
- (B) swaps the two answers. The distance can never be less than the length of the displacement.
- (C) integrates x′ + y′ instead of the speed, which just adds the changes in x and y.
Topics: 9.3, 9.4, 9.6.
Question 3 (multiple choice · mixed)
A polar curve r = f(θ) has f(π/3) = 2 and f′(π/3) = −1. Let A(β) = ½ ∫ from 0 to β of (f(θ))² dθ. Which describes dA/dβ at β = π/3, and the motion of the point (f(θ), θ) as θ increases through π/3?
- (A) dA/dβ = 2; moving away from the pole
- (B) dA/dβ = −1; moving towards the pole
- (C) dA/dβ = 2; moving towards the pole
- (D) dA/dβ = 4; moving towards the pole
Answer and explanation
Answer: (C). By the Fundamental Theorem of Calculus, dA/dβ = ½(f(β))² = ½(4) = 2: the swept area grows at 2 square units per radian. r = 2 > 0 and dr/dθ = −1 < 0, so the distance from the pole is decreasing.
- (A) reverses the meaning of dr/dθ < 0 when r > 0.
- (B) uses f′ instead of ½f².
- (D) leaves out the ½.
Topics: 9.7, 9.8.
Question 4 (constructed response · mixed)
A particle moves in the plane for 0 ≤ t ≤ 4 with velocity v(t) = ⟨2t − 2, t² − 4⟩. At t = 0 it is at (1, 2).
(a) Find r(t) and the position at t = 3. (b) Find the equation of the tangent line to the path at t = 3. (c) Find d²y/dx² at t = 3. Is the path concave up or concave down there? (d) Find the speed at t = 3. Is the speed increasing or decreasing? Justify. (e) For which t is the particle moving left and down?
Worked solution and suggested Marlbridge rubric
(a) x(t) = 1 + ∫ from 0 to t of (2s − 2) ds = t² − 2t + 1 = (t − 1)². y(t) = 2 + ∫ from 0 to t of (s² − 4) ds = t³/3 − 4t + 2. At t = 3: (4, −1).
(b) dy/dx = (t² − 4)/(2t − 2) = 5/4 at t = 3. Tangent line: y + 1 = (5/4)(x − 4).
(c) By the quotient rule, d/dt (dy/dx) = (t² − 2t + 4)/(2(t − 1)²), which is 14/16 = 7/8 at t = 3. Divide by dx/dt = 4: d²y/dx² = 7/32 > 0, concave up.
(d) v(3) = ⟨4, 5⟩, so the speed is √41. a(t) = ⟨2, 2t⟩, so a(3) = ⟨2, 6⟩ and v · a = 8 + 30 = 38 > 0. The speed is increasing.
(e) x′ < 0 for t < 1 and y′ < 0 for t < 2. Both hold for 0 ≤ t < 1.
| Point | What earns it |
|---|---|
| 1 | x(t) and y(t), each with its constant from (1, 2) |
| 1 | Position (4, −1) |
| 1 | Slope 5/4 and the tangent line |
| 1 | d/dt (dy/dx) divided by dx/dt |
| 1 | 7/32 and concave up |
| 1 | √41, and increasing justified by v · a > 0 |
| 1 | 0 ≤ t < 1, from the signs of both components |
Total: 7 points. Topics: 9.1, 9.2, 9.5, 9.6.
Question 5 (constructed response · mixed)
Calculator allowed. A fictional robotic cart moves across a flat courtyard. Its position is in metres, with x east and y north, and t is in seconds. For 0 ≤ t ≤ 6,
dx/dt = 2 + cos(t²/4) and dy/dt = 3 cos(0.5t) − 0.2t.
At t = 0 the cart is at (5, 2).
(a) Explain why the cart never moves west. (b) Find the cart’s greatest y-coordinate for 0 ≤ t ≤ 6. Justify your answer. (c) Find the cart’s position at t = 6. (d) Find the speed at t = 4. Is the speed increasing or decreasing at t = 4? Justify. (e) Find the total distance the cart travels for 0 ≤ t ≤ 6.
Worked solution and suggested Marlbridge rubric
(a) cos(t²/4) ≥ −1, so dx/dt ≥ 1 > 0. The x-coordinate always increases, so the cart never moves west.
(b) dy/dt = 0 at t ≈ 2.770 (the only zero in the interval). dy/dt > 0 before it (y′(1) ≈ 2.433) and < 0 after it (y′(4) ≈ −2.048), so y has its maximum there. y(2.770) = 2 + ∫ from 0 to 2.770 of y′(t) dt ≈ 7.129 m. (Ends: y(0) = 2, y(6) ≈ −0.753.)
(c) x(6) = 5 + ∫ from 0 to 6 of x′(t) dt ≈ 18.406 and y(6) = 2 + ∫ from 0 to 6 of y′(t) dt ≈ −0.753. Position (18.406, −0.753).
(d) v(4) ≈ ⟨1.346, −2.048⟩, so the speed is ≈ 2.451 m/s. Differentiating each component, a(4) ≈ ⟨1.514, −1.564⟩. v · a ≈ 2.038 + 3.204 ≈ 5.242 > 0, so the speed is increasing.
(e) Distance = ∫ from 0 to 6 of √((x′)² + (y′)²) dt ≈ 19.323 m.
| Point | What earns it |
|---|---|
| 1 | (a): dx/dt ≥ 1 > 0 |
| 1 | t ≈ 2.770 with the sign change of dy/dt |
| 1 | y ≈ 7.129 using the starting value 2 |
| 1 | (18.406, −0.753) with both integrals shown |
| 1 | Speed ≈ 2.451 |
| 1 | Increasing, from v · a > 0 |
| 1 | Distance ≈ 19.323 from the integral of speed |
Total: 7 points. Topics: 9.3, 9.4, 9.5, 9.6.
Question 6 (constructed response · mixed)
The circle r = 3 cos θ and the limaçon r = 2 − cos θ are drawn on the same polar axes. Both are symmetric about the x-axis.
(a) For r = 2 − cos θ, find dr/dθ at θ = π/3. Is the point moving towards or away from the pole? Give a reason. (b) Find the slope of r = 3 cos θ at θ = π/3. (c) Find where the curves meet, for −π/2 ≤ θ ≤ π/2. For 0 < θ < π/3 and π/3 < θ < π/2, state which curve is nearer the pole, with a test value in each. (d) Find the exact area of the region inside both curves. (e) Find the exact area inside r = 3 cos θ but outside r = 2 − cos θ.
Worked solution and suggested Marlbridge rubric
(a) dr/dθ = sin θ = √3/2 at π/3. r = 3/2 > 0 and dr/dθ > 0, so the point is moving away from the pole.
(b) On r = 3 cos θ: x = 3 cos²θ and y = 3 sin θ cos θ = (3/2) sin 2θ. dx/dθ = −3 sin 2θ = −3√3/2 and dy/dθ = 3 cos 2θ = −3/2 at π/3. Slope = 1/√3.
(c) 3 cos θ = 2 − cos θ gives cos θ = ½, so θ = ±π/3 (both r = 3/2). The limaçon has r ≥ 1, so the pole is not shared. At θ = π/6: 3 cos θ ≈ 2.598 and 2 − cos θ ≈ 1.134, so the limaçon is nearer on (0, π/3). At θ = 5π/12: 0.776 and 1.741, so the circle is nearer on (π/3, π/2).
(d) The region inside both is bounded by the nearer curve. Double the upper half: Area = 2[½ ∫ from 0 to π/3 of (2 − cos θ)² dθ + ½ ∫ from π/3 to π/2 of 9 cos²θ dθ] = (3π/2 − 15√3/8) + (3π/4 − 9√3/8) = 9π/4 − 3√3 (about 1.872).
(e) The circle r = 3 cos θ has radius 3/2, so its area is 9π/4. Subtract (d): 9π/4 − (9π/4 − 3√3) = 3√3 (about 5.196).
| Point | What earns it |
|---|---|
| 1 | dr/dθ = √3/2 and “away”, using r > 0 |
| 1 | dx/dθ and dy/dθ from x = r cos θ, y = r sin θ |
| 1 | Slope 1/√3 |
| 1 | θ = ±π/3 and the nearer curve on each interval, with test values |
| 1 | Correct integrals with limits 0, π/3, π/2, doubled |
| 1 | 9π/4 − 3√3 |
| 1 | 3√3 |
Total: 7 points. Topics: 9.7, 9.8, 9.9.
Question 7 (constructed response · mixed)
A curve is given by x = 4t − ½t² and y = (8/3)t^(3/2), for 0 ≤ t ≤ 6.
(a) Find dx/dt and dy/dt, and hence dy/dx. (b) Find the point where the curve has a vertical tangent. Explain why it has no horizontal tangent for 0 < t ≤ 6. (c) Find d²y/dx² in terms of t, and the interval of t on which the curve is concave up. (d) Show that (dx/dt)² + (dy/dt)² = (4 + t)², and find the exact length of the curve.
Worked solution and suggested Marlbridge rubric
(a) dx/dt = 4 − t and dy/dt = 4√t. So dy/dx = 4√t/(4 − t), for t ≠ 4.
(b) Vertical: dx/dt = 0 at t = 4, where dy/dt = 8 ≠ 0. Point (8, 64/3). Horizontal would need dy/dt = 0, but 4√t > 0 for t > 0.
(c) By the quotient rule, d/dt (dy/dx) = [(2/√t)(4 − t) + 4√t] ÷ (4 − t)² = 2(4 + t)/(√t (4 − t)²). Divide by dx/dt = 4 − t: d²y/dx² = 2(4 + t)/(√t (4 − t)³). The sign is the sign of 4 − t: concave up for 0 < t < 4.
(d) (4 − t)² + 16t = 16 + 8t + t² = (4 + t)², so the speed is 4 + t > 0 and the curve is traced once. Length = ∫ from 0 to 6 of (4 + t) dt = 24 + 18 = 42.
| Point | What earns it |
|---|---|
| 1 | Both derivatives, and dy/dx = 4√t/(4 − t) |
| 1 | Vertical tangent at (8, 64/3), checking dy/dt ≠ 0 |
| 1 | No horizontal tangent, because dy/dt > 0 for t > 0 |
| 1 | d²y/dx² = 2(4 + t)/(√t (4 − t)³), dividing by dx/dt |
| 1 | Concave up for 0 < t < 4 |
| 1 | Integrand simplified to 4 + t, and length 42 |
Total: 6 points. Topics: 9.1, 9.2, 9.3.
How did you do?
Add up your points from Questions 4–7 (27 in total) and your correct answers to Questions 1–3. The total is only a guide, not a predicted exam score. More useful: note which topics your lost points came from (each answer lists them), then tick off those topic checklists:
9.1 · 9.2 · 9.3 · 9.4 · 9.5 · 9.6 · 9.7 · 9.8 · 9.9
If many topics need work, go back to the Unit 9 diagnostic and use its “Your next step” table to choose where to start.