AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/4/4-9)
Unit 4 · Topic 4.9
4.9 Vector-Valued Functions
A vector-valued function p(t) = ⟨x(t), y(t)⟩ gives a particle's position as a vector at each time t. A velocity vector shows which way the particle is moving and how fast. The magnitude of the position vector is the distance from the origin; the magnitude of the velocity vector is the speed.
Key terms
- vector-valued function
- position vector
- velocity vector
- parameter
Position vectors
If a particle's position is given by the parametric function (x(t), y(t)), you can write it as a vector-valued function: p(t) = ⟨x(t), y(t)⟩ = x(t)i + y(t)j.
At each time t, p(t) is the vector from the origin to the particle. Its magnitude, ‖p(t)‖ = √(x(t)² + y(t)²), is the particle's distance from the origin at that time.
Velocity vectors
A velocity vector v(t) = ⟨v₁(t), v₂(t)⟩ gives the particle's horizontal rate of motion, v₁(t), and vertical rate of motion, v₂(t), at time t.
- If v₁(t) > 0 the particle is moving right; if v₁(t) < 0 it's moving left.
- If v₂(t) > 0 the particle is moving up; if v₂(t) < 0 it's moving down.
- The magnitude ‖v(t)‖ = √(v₁(t)² + v₂(t)²) is the speed.
Paths, circles and straight lines
As t changes, the head of the position vector p(t) traces the particle's path, which is the same curve as the parametric function (x(t), y(t)).
If ‖p(t)‖ stays constant, the particle stays the same distance from the origin. For p(t) = ⟨3 cos t, 3 sin t⟩, ‖p(t)‖ = √(9 cos² t + 9 sin² t) = 3 for every t, so the particle moves around a circle of radius 3.
If the velocity vector is constant, such as v = ⟨2, −1⟩ meters per second, the particle moves in a straight line: 2 meters right and 1 meter down every second, at a speed of √5 ≈ 2.236 meters per second.
Average velocity
Over a time interval [t₁, t₂], the change in position is the displacement vector p(t₂) − p(t₁). Dividing by the elapsed time gives the average velocity: (p(t₂) − p(t₁))/(t₂ − t₁).
Each component is just the average rate of change of x(t) or y(t) from topic 4.3. The direction of the average velocity matches the direction of the secant line from the first position to the second.
In AP Calculus, velocity at an instant comes from shrinking this interval, using derivatives of x(t) and y(t).
Distance, displacement and speed
These three ideas sound alike but answer different questions. ‖p(t)‖ is how far the particle is from the origin right now. The displacement p(t₂) − p(t₁) is the straight-line change in position between two times. Speed, ‖v(t)‖, is how fast it's moving.
A particle can be far from the origin and moving slowly, or close to the origin and moving quickly.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Distance from the origin
A particle's position is p(t) = ⟨t² − 3, 2t + 1⟩. How far is it from the origin at t = 2?
Show the solutionHide the solution
- Step 1: p(2) = ⟨4 − 3, 4 + 1⟩ = ⟨1, 5⟩.
- Step 2: ‖p(2)‖ = √(1 + 25) = √26 ≈ 5.099.
Answer: √26 ≈ 5.099 units from the origin.
- Example 2
Reading a velocity vector
A particle has velocity v(t) = ⟨2t − 4, −3⟩. Describe its motion and speed at t = 1 and t = 3.
Show the solutionHide the solution
- Step 1: At t = 1: v(1) = ⟨−2, −3⟩. Both components are negative: moving left and down. Speed √(4 + 9) = √13 ≈ 3.606.
- Step 2: At t = 3: v(3) = ⟨2, −3⟩. Moving right and down. Speed √13 ≈ 3.606 again.
Answer: At t = 1: left and down; at t = 3: right and down; speed √13 ≈ 3.606 at both times.
- Example 3
Trap: distance from the origin is not speed
For p(t) = ⟨t² − 3, 2t + 1⟩, find the average velocity on [0, 2]. A student says the particle's speed at t = 2 is √26 because ‖p(2)‖ = √26. What's wrong?
Show the solutionHide the solution
- Step 1: p(0) = ⟨−3, 1⟩ and p(2) = ⟨1, 5⟩. Displacement: ⟨4, 4⟩. Average velocity: ⟨4, 4⟩/2 = ⟨2, 2⟩, which points right and up.
- Step 2: ‖p(2)‖ = √26 is the distance from the origin, a fact about where the particle is.
- Step 3: Speed is the magnitude of the velocity vector, a fact about how fast it's moving. The position vector can't tell you that.
Answer: Average velocity ⟨2, 2⟩. The student found distance from the origin, not speed; speed needs the velocity vector.
Common mistakes
- Treating the magnitude of the position vector as the speed.
- Reading direction from the position vector's signs. Direction of motion comes from the velocity's signs.
- Forgetting to divide the displacement by the elapsed time when finding average velocity.
On the exam
- Unit 4 is not on the AP Precalculus Exam, so you'll see this topic on class tests rather than in May.
- Expect to interpret position and velocity vectors: distance from the origin, direction of motion, and speed.
Connected topics
Videos
Check yourself
3 questions on 4.9 Vector-Valued Functions. Pick an answer to see if you got it, and why.
A particle moves in the plane, and its velocity vector at time t is v(t) = ⟨2t − 4, 3⟩. Which of the following describes the particle's motion at time t = 1?
A particle's position at time t is p(t) = ⟨2t + 1, t²⟩. What is the average rate of change of p, the average velocity vector, over the interval 1 ≤ t ≤ 3?
The position of a particle at time t is given by the vector-valued function p(t) = ⟨t², 3t − 1⟩. What is the distance between the particle's positions at t = 1 and t = 3?
0 of 3 answered