Speed from Parametric Velocity

From Velocity Components to Speed

A particle moving in the plane has a position that can be described by two coordinate functions: \(x(t)\) and \(y(t)\). Each coordinate changes with time, so each has its own rate of change. These rates tell us how the particle moves horizontally and vertically, but neither one alone gives its total speed.

The particle may move in both coordinate directions at once. Its overall speed accounts for both contributions, just as the length of a diagonal depends on both perpendicular sides of a right triangle. This tutorial develops that calculation and shows how to evaluate it carefully at a specified time.

Velocity and Speed

Suppose the particle's position at time \(t\) is \(\mathbf{r}(t)=(x(t),y(t))\), where \(x\) and \(y\) are differentiable functions. The velocity vector records the rate of change of position. Its horizontal and vertical components are the derivatives of the corresponding coordinates.

The component \(x'(t)\) describes the rate of change of horizontal position, and \(y'(t)\) describes the rate of change of vertical position. These component values can be positive, negative, or zero: their signs indicate the direction of motion along each coordinate axis.

Speed, by contrast, is a scalar: it gives the nonnegative rate at which the particle moves, without specifying a direction. In the plane, it is the magnitude, or Euclidean length, of the velocity vector.

The squares matter: a negative velocity component contributes positively to the speed. The square root makes the result a length per unit time, and therefore a nonnegative rate.

Why the Magnitude Gives Speed

The formula follows from how velocity is defined. Over a short time interval, the change in position is a vector with horizontal change \(\Delta x\) and vertical change \(\Delta y\). The straight-line distance between the two positions is the length of that vector, so the Pythagorean theorem gives \(\sqrt{(\Delta x)^2+(\Delta y)^2}\).

To obtain the instantaneous rate of change of position, divide the displacement vector by elapsed time and take a limit. Differentiability of the coordinate functions means that this vector limit is the velocity vector. The length function is continuous, so taking the length commutes with this limit. In detail, for a nonzero time increment \(h\), the distance between the two positions divided by the elapsed time is

As \(h\) approaches zero, the vector inside the norm approaches \(\mathbf{r}'(t)=\mathbf{v}(t)\). Continuity of the norm then gives the magnitude of velocity. This establishes the instantaneous speed formula.

This argument uses the distance between nearby positions to describe instantaneous motion. It does not require finding the total distance traveled over a longer time interval.

A Reliable Calculation Process

When the coordinates are given as functions of time and the question asks for speed at a particular time, keep differentiation and substitution in that order. First find both component functions of velocity; then evaluate them at the requested time; finally take their magnitude.

Worked Examples

Units and Interpretation

If \(x(t)\) and \(y(t)\) are measured in meters and \(t\) in seconds, both coordinate derivatives are measured in meters per second. Squaring them gives units of square meters per square second; adding these quantities preserves those units, and taking the square root returns meters per second. The final speed therefore has the correct units.

The same reasoning works with any consistent units. For example, coordinates measured in feet and time measured in minutes produce speed in feet per minute. If the two coordinate functions use the same position units and time variable, their derivative units agree, so the magnitude is well-defined.

Speed also has a useful interpretation when one or both velocity components are zero. If \(y'(t)=0\), for example, then the particle has no instantaneous vertical motion at that time, and the formula reduces to \(\sqrt{(x'(t))^2}=|x'(t)|\). The absolute value is important: speed is not negative even if the horizontal component is negative. If both components are zero, the magnitude formula gives speed \(0\) at that time.

A Common Mistake

A frequent error is to add the velocity components directly and call the result speed. For instance, if a particle has velocity components \(3\) m/s and \(4\) m/s, the calculation \(3+4=7\) m/s is not the speed. It combines two perpendicular directions as though they lay along the same line.

The correction is to use the magnitude formula:

Another related error is to report \(3^2+4^2=25\) as the speed. That value is the squared magnitude, with squared speed units; the square root is necessary to return to speed units. A quick check is to square your proposed speed and see whether it equals the sum of the squared components.

What to Keep in Mind

The velocity vector contains directional information in its signed components. Speed removes the directional information by taking the vector's nonnegative magnitude. For parametric motion in the plane, the calculation is always based on the two coordinate derivatives:

These steps answer a question about how fast the particle is moving at one instant. They do not by themselves determine whether the particle is speeding up or slowing down, nor do they give the total distance traveled over an interval; those are different questions.