Displacement and Distance from Parametric Coordinates

Net Change and Total Travel

A particle can travel for a while, stop, reverse direction, and return to where it began. Its final position then matches its initial position, so its net displacement is zero. But it has still traveled a positive distance.

To describe motion precisely, keep two quantities separate. Displacement records the change from the initial position to the final position. Total distance traveled accumulates the motion along the way. For parametric motion, endpoint coordinates give displacement, while integrating speed gives total distance.

Displacement from Parametric Coordinates

Suppose a particle moves in the plane according to \(x=x(t)\) and \(y=y(t)\) on the time interval \([a,b]\). Its position is \((x(t),y(t))\). The componentwise displacement is the final coordinate minus the initial coordinate in each component:

Together these components form the displacement vector \(\langle \Delta x,\Delta y\rangle\). Its magnitude, \(\sqrt{(\Delta x)^2+(\Delta y)^2}\), is the straight-line distance between the endpoints. It is not necessarily the total distance the particle traveled.

The Fundamental Theorem of Calculus connects endpoint displacement to velocity. Since the velocity components are \(x'(t)\) and \(y'(t)\), provided the coordinate functions are differentiable with continuous derivatives on \([a,b]\),

These integrals are signed: motion in a coordinate’s positive direction contributes positively, and motion in its negative direction contributes negatively. If the particle reverses and later returns to its starting coordinate, the positive and negative contributions cancel in the net displacement.

Total Distance from Speed

Velocity is the rate of change of position. Its magnitude is the speed, which is the rate at which distance accumulates. For a differentiable parametric path, the speed is

Therefore, on \([a,b]\), total distance traveled is found by accumulating speed over time:

This integral adds nonnegative contributions throughout the interval. Unlike the signed component integrals for displacement, a change in direction does not cancel previously traveled distance. If the velocity becomes zero and the particle reverses, the speed is still nonnegative on both sides of that time.

Why the Displacement Formula Works

The coordinate formula follows directly from the Fundamental Theorem of Calculus. For the horizontal coordinate, an antiderivative of \(x'(t)\) is \(x(t)\). Thus

The same reasoning applied to \(y(t)\) gives \(\int_a^b y'(t)\,dt=y(b)-y(a)\). These equations show why integrating velocity components gives net change rather than total travel: the integral records the signed accumulation of each component. By contrast, the speed integral accumulates the nonnegative rate of travel.

Worked Examples

A Common Mistake: Integrating Signed Velocity for Distance

For the motion \(x(t)=t^2-4t\) on \([0,4]\), a tempting but incorrect calculation is to call the integral of velocity the total distance:

The calculation itself is correct, but it gives the net horizontal displacement, not the total distance. The negative velocity before \(t=2\) cancels the positive velocity after \(t=2\). For distance, use speed and account for its sign change:

In two dimensions, use the magnitude of the velocity vector, not a signed component by itself. A particle can also change direction without reversing along a single coordinate axis, so the full speed is \(\sqrt{(x')^2+(y')^2}\).

How to Choose the Right Quantity

Keep units consistent: if coordinates are in meters and time is in seconds, velocity and speed are in meters per second, while displacement and distance are in meters. A useful final check is that total distance cannot be less than the magnitude of displacement, because the direct endpoint separation cannot exceed the length of the route traveled.