Vector-Valued Functions

A vector-valued function r(t) = ⟨x(t), y(t)⟩ gives the position of a moving point at each time t. Its derivative is the velocity vector v(t) = ⟨x′(t), y′(t)⟩. The speed is the length of velocity, |v(t)| = √(x′(t)² + y′(t)²), and the derivative of velocity is the acceleration vector a(t).

Before this lesson: Arc Length of Parametric Curves, Magnitude and Direction of a Vector and Position, Velocity and Acceleration

The idea: two motions at once

A point moving in the plane is really two motions happening together: one left and right, one up and down. A vector-valued function keeps both in one object, r(t) = ⟨x(t), y(t)⟩.

Each component is an ordinary function of t, so you differentiate and integrate one component at a time. The velocity vector points the way the object is heading, and its length tells how fast it is going.

Velocity, speed and acceleration formulas

\vec{r}(t) = \langle x(t),\, y(t) \rangle, \qquad \vec{v}(t) = \vec{r}\,'(t) = \langle x'(t),\, y'(t) \rangle, \qquad \vec{a}(t) = \vec{v}\,'(t) = \langle x''(t),\, y''(t) \rangle

\text{speed} = |\vec{v}(t)| = \sqrt{x'(t)^2 + y'(t)^2}, \qquad \text{distance traveled} = \int_a^b |\vec{v}(t)|\,dt

• Velocity is a vector (direction and size). Speed is a number, the length of the velocity vector.
• To go backward, integrate each component and add a separate constant to each. An initial position or velocity gives the constants.
• Displacement from t = a to t = b is ∫ v(t) dt, a vector. Distance traveled is ∫ |v(t)| dt, a number.
• The object is at rest when both components of velocity are 0 at the same time.

Vector-valued functions examples

Example 1: Velocity, speed and acceleration

For r(t) = ⟨t³ − 3t, t²⟩, find the velocity, speed and acceleration at t = 2.

\vec{v}(t) = \langle 3t^2 - 3,\, 2t \rangle

\vec{a}(t) = \langle 6t,\, 2 \rangle

\vec{v}(2) = \langle 3(4) - 3,\, 2(2) \rangle = \langle 9, 4 \rangle

|\vec{v}(2)| = \sqrt{9^2 + 4^2} = \sqrt{97} \approx 9.85

\vec{a}(2) = \langle 12, 2 \rangle

Answer: v(2) = ⟨9, 4⟩, speed √97 ≈ 9.85, and a(2) = ⟨12, 2⟩. At t = 2 the point is at r(2) = ⟨2, 4⟩.

Example 2: Motion around a circle

For r(t) = ⟨4 cos t, 4 sin t⟩, find the speed and the acceleration. Evaluate both at t = π/3.

\vec{v}(t) = \langle -4\sin t,\, 4\cos t \rangle

|\vec{v}(t)| = \sqrt{16\sin^2 t + 16\cos^2 t} = 4

\vec{a}(t) = \langle -4\cos t,\, -4\sin t \rangle = -\vec{r}(t)

\vec{v}\left(\tfrac{\pi}{3}\right) = \langle -2\sqrt{3},\, 2 \rangle

\vec{a}\left(\tfrac{\pi}{3}\right) = \langle -2,\, -2\sqrt{3} \rangle

Answer: the speed is always 4, but the acceleration is not zero: a(t) = −r(t) keeps turning the object toward the center. At t = π/3, v = ⟨−2√3, 2⟩ and a = ⟨−2, −2√3⟩.

Example 3: Position from velocity

A particle has velocity v(t) = ⟨3t², 4t − 1⟩ and is at ⟨2, 0⟩ when t = 1. Find r(t) and the position at t = 2.

\vec{r}(t) = \langle t^3 + C_1,\, 2t^2 - t + C_2 \rangle

\langle 1 + C_1,\, 1 + C_2 \rangle = \langle 2, 0 \rangle

C_1 = 1, \quad C_2 = -1

\vec{r}(t) = \langle t^3 + 1,\, 2t^2 - t - 1 \rangle

\vec{r}(2) = \langle 8 + 1,\, 8 - 2 - 1 \rangle = \langle 9, 5 \rangle

Answer: r(t) = ⟨t³ + 1, 2t² − t − 1⟩, and the particle is at ⟨9, 5⟩ when t = 2.

Example 4: Displacement vs. distance

A point moves with r(t) = ⟨cos πt, sin πt⟩ for 0 ≤ t ≤ 1. Find its displacement and the distance it travels.

\vec{r}(1) - \vec{r}(0) = \langle -1, 0 \rangle - \langle 1, 0 \rangle = \langle -2, 0 \rangle

\vec{v}(t) = \langle -\pi\sin \pi t,\, \pi\cos \pi t \rangle

|\vec{v}(t)| = \sqrt{\pi^2\sin^2 \pi t + \pi^2\cos^2 \pi t} = \pi

\int_0^1 \pi\,dt = \pi

Answer: the displacement is ⟨−2, 0⟩ (2 units left), but the distance traveled is π ≈ 3.14, half the unit circle.

Example 5: A thrown ball (word problem)

A ball is thrown from 2 meters above the ground with initial velocity ⟨8, 15⟩ m/s. Gravity gives a(t) = ⟨0, −9.8⟩. Find r(t), the highest point, and the speed when the ball lands.

\vec{v}(t) = \langle 8,\, 15 - 9.8t \rangle

\vec{r}(t) = \langle 8t,\, 2 + 15t - 4.9t^2 \rangle

15 - 9.8t = 0

t = \dfrac{15}{9.8} \approx 1.531

y \approx 2 + 15(1.531) - 4.9(1.531)^2 \approx 13.48

4.9t^2 - 15t - 2 = 0

t = \dfrac{15 + \sqrt{225 + 39.2}}{9.8} \approx 3.189

\vec{v}(3.189) \approx \langle 8,\, -16.25 \rangle

|\vec{v}| \approx \sqrt{64 + 264.2} \approx 18.12

Answer: r(t) = ⟨8t, 2 + 15t − 4.9t²⟩. The ball peaks at about 13.5 m after 1.53 s, lands after about 3.19 s (25.5 m away), and hits the ground at about 18.1 m/s.

Common mistakes

Related methods

Parametric equations

A vector-valued function is a pair of parametric equations written as one vector. The slope of the path is still dy/dx = y′(t)/x′(t), from derivatives of parametric equations.

Distance traveled

The distance formula here is the same integral as in arc length of parametric curves: the integral of speed.

Vectors and projectile motion

The components work like the vectors in vector word problems, and Example 5 is the calculus version of projectile motion with parametric equations. Calculus 3 extends all of this to curves in space.

Practice problems

1. For r(t) = ⟨t², t³⟩, find v(1) and a(1).
2. Find the speed in problem 1 at t = 1.
3. For r(t) = ⟨e²ᵗ, e⁻ᵗ⟩, find v(0) and the speed at t = 0.
4. Show that r(t) = ⟨3 sin t, 3 cos t⟩ has constant speed.
5. Find r(t) if v(t) = ⟨2t, 6t²⟩ and r(0) = ⟨1, −2⟩.
6. Find v(t) if a(t) = ⟨2, −6t⟩ and v(0) = ⟨1, 4⟩.
7. A particle has v(t) = ⟨t² − 4, 2t − 6⟩ for t ≥ 0. When is it moving straight up or down? Is it ever at rest?
8. Find the distance traveled when v(t) = ⟨1 − cos t, sin t⟩ for 0 ≤ t ≤ 2π.
9. For v(t) = ⟨2t, 3t²⟩ and r(0) = ⟨1, 1⟩, find the displacement from t = 0 to t = 2 and the position at t = 2.
10. A particle has v(t) = ⟨3t² − 12, 2t⟩ for t ≥ 0 and r(0) = ⟨0, 5⟩. Where is it, and how fast is it going, when it moves straight up?

\vec{v}(t) = \langle 2t,\, 3t^2 \rangle, \quad \vec{a}(t) = \langle 2,\, 6t \rangle

\vec{v}(1) = \langle 2, 3 \rangle, \quad \vec{a}(1) = \langle 2, 6 \rangle

|\vec{v}(1)| = \sqrt{2^2 + 3^2} = \sqrt{13}

\vec{v}(t) = \langle 2e^{2t},\, -e^{-t} \rangle

\vec{v}(0) = \langle 2, -1 \rangle

|\vec{v}(0)| = \sqrt{4 + 1} = \sqrt{5}

\vec{v}(t) = \langle 3\cos t,\, -3\sin t \rangle

|\vec{v}(t)| = \sqrt{9\cos^2 t + 9\sin^2 t} = 3

\vec{r}(t) = \langle t^2 + C_1,\, 2t^3 + C_2 \rangle

C_1 = 1, \quad C_2 = -2

\vec{v}(t) = \langle 2t + C_1,\, -3t^2 + C_2 \rangle

C_1 = 1, \quad C_2 = 4

t^2 - 4 = 0 \Rightarrow t = 2

2(2) - 6 = -2

2t - 6 = 0 \Rightarrow t = 3

|\vec{v}(t)| = \sqrt{2 - 2\cos t} = 2\sin\dfrac{t}{2}

\int_0^{2\pi} 2\sin\dfrac{t}{2}\,dt = 8

\int_0^2 \langle 2t,\, 3t^2 \rangle\,dt = \langle 4, 8 \rangle

\langle 1, 1 \rangle + \langle 4, 8 \rangle = \langle 5, 9 \rangle

3t^2 - 12 = 0 \Rightarrow t = 2

\vec{r}(t) = \langle t^3 - 12t,\, t^2 + 5 \rangle

\vec{r}(2) = \langle 8 - 24,\, 4 + 5 \rangle = \langle -16, 9 \rangle

|\vec{v}(2)| = \sqrt{0^2 + 4^2} = 4

Vector-valued functions FAQ

What is a vector-valued function?

It is a function whose output is a vector, such as r(t) = ⟨x(t), y(t)⟩. As t changes, the tip of the vector traces a curve, so it describes motion along a path.

How do you find speed from a velocity vector?

Take the length of the velocity vector: speed = √(x′(t)² + y′(t)²). Speed is never negative.

What is the difference between velocity and speed?

Velocity is a vector that gives both direction and size. Speed is only the size, the length of the velocity vector.

How do you find position from velocity?

Integrate each component of velocity and add a separate constant to each. Then use a known position to find the constants.

Summary

• Position r(t) = ⟨x(t), y(t)⟩; velocity v(t) = r′(t); acceleration a(t) = v′(t).
• Differentiate and integrate one component at a time.
• Speed = |v(t)| = √(x′² + y′²), a number.
• Displacement = ∫ v dt (a vector); distance = ∫ |v| dt (a number).
• Each component gets its own constant of integration.
• Constant speed does not mean zero acceleration when the path curves.

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