AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/9/9-6)
Unit 9 · Topic 9.6
BC only9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions
BC only. For a particle moving in the plane, the velocity is ⟨x′(t), y′(t)⟩, the speed is that vector's length √((x′(t))² + (y′(t))²), and the integral of speed gives the total distance traveled. Integrating velocity and adding a starting point gives the new position.
Key terms
- speed
- velocity vector
- magnitude
- displacement
- total distance traveled
The toolkit
This whole unit is BC only. Planar motion questions combine everything in this unit. Here's what each quantity means and how to get it.
| Quantity | Formula | Type |
|---|---|---|
| position | ⟨x(t), y(t)⟩ | vector (a point) |
| velocity | ⟨x′(t), y′(t)⟩ | vector |
| acceleration | ⟨x″(t), y″(t)⟩ | vector |
| speed | √((x′(t))² + (y′(t))²) | number |
| slope of the path | y′(t)/x′(t) | number |
| displacement from a to b | ⟨∫ₐᵇ x′(t) dt, ∫ₐᵇ y′(t) dt⟩ | vector |
| total distance from a to b | ∫ₐᵇ √((x′(t))² + (y′(t))²) dt | number |
Displacement vs. distance
Displacement is a change in position: how far and in what direction the particle ended up from where it started. Total distance traveled is the length of the path it actually covered. A particle that goes once around a circle has displacement ⟨0, 0⟩ but distance equal to the circumference.
The distance between the starting and ending points is the length of the displacement vector. The total distance traveled is at least that long, and usually longer.
Distance comes from integrating speed, which is never negative. You don't need to split the interval where the particle changes direction, as you did on a line, because speed is already non-negative.
Typical questions and how to answer them
- “Find the speed at t = 2”: evaluate √((x′(2))² + (y′(2))²).
- “Find the position at t = 3”: add the starting coordinates to the integrals of x′ and y′ from the starting time to 3.
- “Find the total distance traveled on 0 ≤ t ≤ 3”: integrate speed from 0 to 3.
- “When is the particle moving up?”: where y′(t) > 0.
- “Find the highest point on the path”: y′ changes from positive to negative there; compare y-values at that time and at the endpoints.
- “Is the speed increasing at t = 2?”: check the sign of the derivative of the speed at t = 2 (a calculator can compute it).
Units and precision
Keep your calculator in radian mode. If position is in meters and time in seconds, velocity components and speed are in meters per second, and distance is in meters. Acceleration components are in meters per second squared. On calculator questions, store intermediate values and round only the final answers to three decimal places.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Calculator: speed, distance and position
A particle moves in the plane with x′(t) = 2 + sin(t²) and y′(t) = 3 cos(√t) for t ≥ 0. At t = 0 it's at (1, 5). (a) Find the speed at t = 2. (b) Find the total distance traveled for 0 ≤ t ≤ 3. (c) Find the position at t = 3.
Show the solutionHide the solution
- Step 1: (a) x′(2) = 2 + sin 4 ≈ 1.243 and y′(2) = 3 cos(√2) ≈ 0.468. Speed = √(1.243² + 0.468²) ≈ 1.328.
- Step 2: (b) Distance = ∫₀³ √((2 + sin(t²))² + (3 cos(√t))²) dt ≈ 7.979.
- Step 3: (c) x(3) = 1 + ∫₀³ (2 + sin(t²)) dt ≈ 1 + 6.774 = 7.774.
- Step 4: y(3) = 5 + ∫₀³ 3 cos(√t) dt ≈ 5 + 3.294 = 8.294.
- Step 5: Position ≈ (7.774, 8.294).
Answer: (a) ≈ 1.328 (b) ≈ 7.979 (c) ≈ (7.774, 8.294)
- Example 2
Trap: displacement is not distance
A particle moves with r(t) = ⟨cos(πt), sin(πt)⟩ for 0 ≤ t ≤ 2. Find its displacement and the total distance traveled.
Show the solutionHide the solution
- Step 1: Displacement = r(2) − r(0) = ⟨cos 2π, sin 2π⟩ − ⟨1, 0⟩ = ⟨0, 0⟩.
- Step 2: Velocity: ⟨−π sin(πt), π cos(πt)⟩. Speed = √(π² sin²(πt) + π² cos²(πt)) = π.
- Step 3: Distance = ∫₀² π dt = 2π.
- Step 4: The particle went once around the unit circle, so it ended where it started even though it traveled 2π units.
Answer: Displacement ⟨0, 0⟩; distance 2π ≈ 6.283
Common mistakes
- Integrating x′(t) + y′(t) for distance. Distance needs the square root of the sum of squares.
- Reporting the displacement integral as the position. Add the starting point.
- Giving speed as a vector or as x′(t) + y′(t).
- Using degree mode on the calculator for trig functions of t.
On the exam
- This is the main free-response type for this unit, and it's often calculator-active. Typical parts: speed at a time, position at a time, total distance, and when the particle moves in a given direction.
- Write each setup before its value, such as “distance = ∫₀³ √((x′(t))² + (y′(t))²) dt ≈ 7.979.”
Connected topics
Videos
Check yourself
5 questions on 9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions. Pick an answer to see if you got it, and why.
A particle moves in the xy-plane with position (x(t), y(t)), where x and y are differentiable functions. Which of the following gives the distance between the particle's position at t = 0 and its position at t = 4?
| t | x(t) | y(t) | x′(t) | y′(t) |
|---|---|---|---|---|
| 0 | 1 | 2 | 3 | 4 |
| 1 | 5 | 7 | 6 | 8 |
| 2 | 6 | 17 | −5 | 12 |
| 3 | −1 | 26 | −8 | 6 |
Selected values for a particle moving in the xy-plane
What is the speed of the particle at t = 2?
Using a left Riemann sum with the three subintervals given by the table, what is the approximate total distance traveled by the particle from t = 0 to t = 3?
A particle moves in the xy-plane so that its velocity vector at time t ≥ 0 is v(t) = ⟨cos(t²), e^(sin t)⟩. At time t = 0, the particle is at the point (2, −1).
Described particle motion
What is the speed of the particle at t = 1.5?
What is the total distance traveled by the particle from t = 0 to t = 2?
0 of 5 answered