Vector-valued functions

A vector-valued function assigns each parameter value a position vector: $p(t) = \langle x(t), y(t) \rangle$, also written $x(t)\,\mathbf{i} + y(t)\,\mathbf{j}$ where i and j are the unit vectors along the axes. At any t, evaluate both coordinate functions and you have the object's location, read as the point at the tip of the position vector drawn from the origin.

For $p(t) = \langle t^2, 3t \rangle$ at t = 2: the position is ⟨4, 6⟩, the point (4, 6). Keep the components in their lanes: x(t) drives horizontal position only, y(t) vertical only. This is exactly a parametric function; the vector notation earns its keep in the next section, where positions get subtracted.

The displacement over an interval is the change in position, computed componentwise:

1. $\Delta p = p(b) - p(a) = \langle x(b) - x(a),\; y(b) - y(a) \rangle$: later position minus earlier, in that order.
2. Displacement is a vector: it has a direction (which way the object netted out) and a magnitude (how far the endpoints sit apart).
3. New position = old position + displacement: positions and displacements add componentwise like any vectors.

If p(1) = ⟨2, 5⟩ and p(4) = ⟨10, −1⟩, the displacement is ⟨8, −6⟩: eight right and six down, with magnitude √(64 + 36) = 10. Reversing the subtraction order gives the exact opposite vector, a trip in the wrong direction.

The magnitude of the displacement measures start-to-end separation as the crow flies. The distance traveled measures the actual path length, bends and backtracks included. The path can never be shorter than the crow's flight, so distance traveled ≥ |displacement|, with equality only for a straight, no-backtrack trip.

The extreme case makes the distinction vivid: an object that loops back to its starting point has displacement ⟨0, 0⟩, zero magnitude, while its distance traveled is the whole length of the loop. One number says "you ended where you began"; the other says how much ground you covered doing it. Exam questions love to ask for one and offer the other.

The direction of motion over an interval is the direction of its change vector, and each component reports its own axis: a change of ⟨−3, 4⟩ means 3 left (negative x-change) and 4 up (positive y-change). A negative y-component means downward motion; it says nothing about left or right. Misassigning a sign to the wrong axis is the most common motion-reading error.

Rates work per component too. The average rate of change of position over [a, b] is the displacement divided by the elapsed parameter: $\Delta p / (b - a)$, a vector whose components are the per-unit changes in x and y. Its magnitude is the average speed along the straight line between the endpoints: a single number. Two objects can share that speed with completely different headings: ⟨5, 0⟩ per second and ⟨3, 4⟩ per second both move at 5 units per second, one due right, one up a 4-in-3 diagonal.