Aim for FiveCalculus BC

Unit 9 · Topic 9.4

BC only

9.4 Defining and Differentiating Vector-Valued Functions

BC only. A vector-valued function r(t) = ⟨x(t), y(t)⟩ gives a position vector for each t. You differentiate it one component at a time: r′(t) is the velocity vector and r″(t) is the acceleration vector.

Key terms

  • vector-valued function
  • component
  • position vector
  • velocity vector
  • acceleration vector

Vector-valued functions

This whole unit is BC only. A vector-valued function packages a pair of parametric equations into one object: r(t) = ⟨x(t), y(t)⟩. The angle brackets hold the components. You may also see it written x(t)i + y(t)j, where i and j are unit vectors along the axes. Both mean the same thing.

Picture r(t) as an arrow from the origin to the moving point (x(t), y(t)). As t changes, the tip of the arrow traces the curve.

Units follow the usual pattern. If position is in feet and t is in seconds, the velocity components are in feet per second and the acceleration components are in feet per second per second.

Derivatives component by component

Every derivative rule you know applies to each component separately:

r′(t) = ⟨x′(t), y′(t)⟩ and r″(t) = ⟨x″(t), y″(t)⟩.

If r(t) is position, then v(t) = r′(t) is the velocity vector and a(t) = r″(t) is the acceleration vector. The velocity vector points in the direction of motion and is tangent to the path.

What the velocity vector tells you

  • Horizontal direction: x′(t) > 0 means moving right; x′(t) < 0 means moving left.
  • Vertical direction: y′(t) > 0 means moving up; y′(t) < 0 means moving down.
  • Slope of the path: dy/dx = y′(t)/x′(t), when x′(t) ≠ 0.
  • Speed: the length (magnitude) of the velocity vector, |v(t)| = √((x′(t))² + (y′(t))²).
  • At rest: the particle is at rest at time t only when both components of velocity are 0.

Second derivatives and the path

The acceleration vector ⟨x″(t), y″(t)⟩ is not the same thing as d²y/dx². Acceleration describes how the velocity changes over time. d²y/dx² describes how the path bends, and you still compute it as in 9.2: differentiate dy/dx = y′(t)/x′(t) with respect to t, then divide by x′(t).

Likewise, the slope of the path, y′(t)/x′(t), and the speed, √((x′(t))² + (y′(t))²), are both built from the velocity vector but answer different questions. One is a direction, the other a rate.

Why this matters

Vector notation is mostly a way of writing parametric motion compactly. Exam questions switch freely between “x′(t) and y′(t)” and “the velocity vector ⟨x′(t), y′(t)⟩.” Getting comfortable with both forms lets you read the question and answer in the form it asks for. Velocity is a vector; speed is a number.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    Velocity, acceleration and speed

    A particle's position is r(t) = ⟨3t², ln(t + 1)⟩ for t ≥ 0. Find the velocity vector, the acceleration vector and the speed at t = 1.

    Show the solution
    1. Step 1: Velocity: v(t) = ⟨6t, 1/(t + 1)⟩. At t = 1: ⟨6, 1/2⟩.
    2. Step 2: Acceleration: a(t) = ⟨6, −1/(t + 1)²⟩. At t = 1: ⟨6, −1/4⟩.
    3. Step 3: Speed at t = 1: √(6² + (1/2)²) = √(36.25) = √145/2 ≈ 6.021.

    Answer: v(1) = ⟨6, 1/2⟩, a(1) = ⟨6, −1/4⟩, speed = √145/2 ≈ 6.021

  2. Example 2

    Trap: at rest means both components are zero

    A particle moves with r(t) = ⟨t² − 4t, t³ − 3t²⟩. (a) Describe its motion at t = 1. (b) At what times is it at rest?

    Show the solution
    1. Step 1: v(t) = ⟨2t − 4, 3t² − 6t⟩.
    2. Step 2: (a) v(1) = ⟨−2, −3⟩: moving left (x′ < 0) and down (y′ < 0). The slope of the path is (−3)/(−2) = 3/2.
    3. Step 3: (b) x′(t) = 0 at t = 2. y′(t) = 3t(t − 2) = 0 at t = 0 and t = 2.
    4. Step 4: Both are 0 only at t = 2. At t = 0, x′(0) = −4 ≠ 0, so the particle is still moving (horizontally).
    5. Step 5: Common wrong answer: t = 0 and t = 2.

    Answer: (a) Moving left and down, along a path with slope 3/2. (b) At rest only at t = 2.

Common mistakes

  • Giving speed as a vector, or velocity as a single number.
  • Saying the particle is at rest when only one component of velocity is zero.
  • Forgetting the chain rule on a component like ln(t + 1) or sin(3t).
  • Mixing up position and velocity components when reading a table.

On the exam

  • BC free-response motion questions usually give r(t) or its derivative in component form and ask for velocity, acceleration or speed at a time. Answer in the form asked: vectors in brackets, speed as a number.
  • Describe direction with the sign of each component, like “moving to the left and upward.”

Connected topics

Videos

  • Calculus BC – 9.4 Defining and Differentiating Vector-Valued Functions

    The AlgebrosWatch on YouTube (opens in a new tab)

  • Second derivatives (vector-valued functions) | Advanced derivatives | AP Calculus BC | Khan Academy

    Khan AcademyWatch on YouTube (opens in a new tab)

  • Derivative of the vector function (KristaKingMath)

    Krista KingWatch on YouTube (opens in a new tab)

  • The Derivative of a Vector Valued Function

    Mathispower4uWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 9.4 Defining and Differentiating Vector-Valued Functions. Pick an answer to see if you got it, and why.

Question 1 of 4

The position of a particle moving in the xy-plane is r(t) = ⟨e²ᵗ, cos(3t)⟩. What is the acceleration vector of the particle at t = 0?

Question 2 of 4

A particle moves in the xy-plane with position vector r(t) = ⟨t² − 4t, t³ − 12t⟩. At time t = 3, in which direction is the particle moving?

Question 3 of 4

A particle moves along a path with position vector r(t) = ⟨√t, t² − 3t⟩ for t > 0. What is the slope of the line tangent to the path at t = 4?

Question 4 of 4

A particle moves in the xy-plane with position vector r(t) = ⟨t³ − 3t, t² + 1⟩. At what time t > 0 is the particle moving straight upward?

0 of 4 answered