← Back to course

Vector-Valued Functions & Motion

You’ll be able to

Motion in the plane as a vector

A particle moving in the plane has a position vector ⟨x(t), y(t)⟩. Differentiating component-by-component gives the velocity vector ⟨x′(t), y′(t)⟩, and differentiating again gives the acceleration vector ⟨x″(t), y″(t)⟩. Each component is handled with ordinary single-variable calculus; the vector just bundles the two independent motions together. This is exactly the parametric setup, now read as physical motion.

Speed as the magnitude of velocity
speed = |v(t)| = √( (dx/dt)² + (dy/dt)² )
Speed is a scalar: the length of the velocity vector. Velocity carries direction; speed is just how fast.

Speed is the magnitude of velocity

The velocity vector ⟨x′, y′⟩ points in the direction of motion, but speed is its scalar length √((x′)² + (y′)²). This is the same Pythagorean combination that appears in arc length — no accident, since speed is the rate at which arc length accumulates. To find total distance traveled over [a, b], integrate the speed: ∫ₐᵇ √((x′)² + (y′)²) dt. Speed is never negative, while a velocity component can be.

Recovering position from velocity

Given the velocity vector and a starting position, you recover position by integrating each component and using the initial condition to fix each constant. x(t) = x(t₀) + ∫_{t₀}^t x′(s) ds, and likewise for y. On the calculator section this is often done as a definite integral added to the known starting coordinate — the Fundamental Theorem of Calculus applied one component at a time.

Worked example

A particle has velocity vector v(t) = ⟨2t, 3⟩. Find its speed at t = 2 and its acceleration vector.

  1. 1.Speed is the magnitude of velocity: |v| = √((2t)² + 3²) = √(4t² + 9).
  2. 2.At t = 2: √(4·4 + 9) = √(16 + 9) = √25 = 5.
  3. 3.Acceleration is the derivative of velocity, component by component: a(t) = ⟨d/dt[2t], d/dt[3]⟩.
  4. 4.This gives a(t) = ⟨2, 0⟩, constant for all t.
Answer: Speed at t = 2 is 5, and the acceleration vector is ⟨2, 0⟩ (constant).
Watch out

Speed is a single nonnegative number — the magnitude √((x′)² + (y′)²) — not a vector. Reporting speed as ⟨x′, y′⟩ confuses it with velocity. Whenever a problem asks "how fast," it wants the scalar magnitude.

Checkpoint

A particle has velocity vector ⟨3, 4⟩ at a certain instant. What is its speed at that instant?

Checkpoint

If a particle’s position is ⟨t², sin(t)⟩, what is its acceleration vector?

On the exam

For planar motion, keep three vectors straight: position ⟨x, y⟩, velocity ⟨x′, y′⟩, acceleration ⟨x″, y″⟩. "Speed" and "total distance" both use the magnitude √((x′)² + (y′)²) — speed at an instant, distance as its integral over time.

Answer the 2 checkpoints as you read.

Sign in to save your progress