Location and Motion Are Different Kinds of Information
A location tells where a particle is. A velocity tells how its location is changing at a particular instant. A velocity field connects these ideas by assigning a velocity vector to each location in a region.
This lets us interpret motion locally without first knowing the particle’s entire path. At a chosen point, the field gives the particle’s instantaneous rate of change of position if the particle is there. Its components give the coordinate rates of motion; its magnitude gives the speed.
The distinction between a location and a velocity matters. A point such as \((2,1)\) is a location, while a vector such as \(\langle 1,3\rangle\) can describe the velocity at that location. The field’s arrow is attached to the location, but its components describe motion, not a second location.
In three dimensions, the same definition uses a vector with three components: \(\mathbf{v}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle\). A velocity field can be displayed by a formula, a table of sampled values, or arrows attached to locations.
From Components to Local Speed and Direction
For a two-dimensional velocity vector \(\langle a,b\rangle\), the first component is the rate of change of the \(x\)-coordinate and the second is the rate of change of the \(y\)-coordinate. A positive component points in the positive coordinate direction; a negative component points in the negative coordinate direction. If a component is zero, there is no instantaneous motion in that coordinate direction.
The speed is the nonnegative magnitude of the velocity. When the velocity vector is nonzero, its direction can be represented by a unit vector: divide the velocity by its magnitude. The unit vector preserves direction while having length one. If the velocity is zero, its magnitude is zero, but it has no direction, so a unit direction vector cannot be formed.
In three dimensions, the magnitude is \(\sqrt{P^2+Q^2+R^2}\). These formulas give the velocity and speed at a particular location after substituting that location into the component functions.
Worked Example: Evaluating a Velocity Field at a Location
Consider the field \(\mathbf{v}(x,y)=\langle x-y,x+y\rangle\), with coordinates recorded in meters and velocity components in meters per second. Find the velocity, speed, and direction at \((2,1)\).
Step 1: Substitute the location into each component.
The particle at this location has instantaneous velocity \(1\) meter per second in the positive \(x\)-direction and \(3\) meters per second in the positive \(y\)-direction.
Step 2: Compute the speed.
Step 3: Find the unit direction vector. Since \(\sqrt{10}\ne 0\), the direction is defined:
Check: The magnitude of this direction vector is \(\sqrt{(1/\sqrt{10})^2+(3/\sqrt{10})^2}=\sqrt{10/10}=1\). Multiplying it by the speed recovers the velocity: \(\sqrt{10}\langle 1/\sqrt{10},3/\sqrt{10}\rangle=\langle 1,3\rangle\) meters per second.
Reading a Velocity Arrow Diagram
A vector-field diagram places an arrow at each sampled location. The arrow’s orientation indicates the direction of the assigned velocity, and—when a common length scale is provided—its length indicates speed. A diagram is therefore a visual encoding of the field’s values, not a picture of one particle’s whole route.
To read an arrow quantitatively, first identify the diagram’s scale. Resolve the arrow into horizontal and vertical components, using signs to record left, right, down, or up. Then use the component values to calculate the speed. Comparing arrow lengths is meaningful only if the same scale is used throughout the diagram.
Worked Example: Reading Components from a Scaled Diagram
Suppose a diagram uses a scale in which one grid unit of arrow length represents \(2\) meters per second. At location \(A\), the arrow points two grid units right and one grid unit up. At location \(B\), it points one grid unit left and two grid units down. Find both velocities, speeds, and directions.
At \(A\): The rightward and upward components are positive. Multiplying the grid components by the scale gives
Thus the speed is \(\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}\) meters per second. The unit direction is \(\langle 4,2\rangle/(2\sqrt{5})=\langle 2/\sqrt{5},1/\sqrt{5}\rangle\).
At \(B\): Leftward and downward components are negative, so
The speed is \(\sqrt{(-2)^2+(-4)^2}=\sqrt{20}=2\sqrt{5}\) meters per second, and the unit direction is \(\langle -1/\sqrt{5},-2/\sqrt{5}\rangle\).
Check: Both arrows have the same length in grid units: each has length \(\sqrt{2^2+1^2}=\sqrt{5}\) grid units. Applying the scale gives \(\sqrt{5}(2)=2\sqrt{5}\) meters per second for each. Their speeds match, but their directions differ because the component signs differ.
How the Field Relates to a Particle’s Motion
A particle’s position at time \(t\) can be represented by a differentiable function \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). Its instantaneous velocity is \(\mathbf{r}'(t)=\langle x'(t),y'(t)\rangle\). If the particle moves according to a velocity field \(\mathbf{v}\), the defining relation is
This equation says that at every time, the particle’s velocity equals the field vector evaluated at its current location. It does not say that the particle’s position equals the field vector. A field gives the local motion rule; the particle’s changing positions describe its actual motion over time.
Justification. Differentiability and the velocity relation give
For \(h>0\), dividing the displacement by \(h\) changes its length but not its direction. Because the limiting vector is nonzero, the vectors \((\mathbf{r}(t+h)-\mathbf{r}(t))/h\) are nonzero for sufficiently small positive \(h\), and normalizing them gives unit vectors that approach \(\mathbf{v}(\mathbf{r}(t))/\|\mathbf{v}(\mathbf{r}(t))\|\). Thus the displacement direction approaches the field direction. The argument relies on the field vector being nonzero: a zero vector cannot be normalized to a direction.
Worked Example: A Particle at a Zero-Velocity Location
Consider the path \(\mathbf{r}(t)=\langle t^3,0\rangle\) and the field \(\mathbf{v}(x,y)=\langle 3|x|^{2/3},0\rangle\). Verify that the path follows the field, and examine what happens at \(t=0\).
Step 1: Differentiate the path.
Step 2: Evaluate the field along the path. Since \(x=t^3\),
Here \(|t^3|^{2/3}=t^2\), since \(|t^3|=|t|^3\) and \((|t|^3)^{2/3}=|t|^2=t^2\). Therefore \(\mathbf{r}'(t)=\mathbf{v}(\mathbf{r}(t))\) for every \(t\).
Step 3: Inspect \(t=0\). At this time, \(\mathbf{r}(0)=\langle 0,0\rangle\) and \(\mathbf{v}(\mathbf{r}(0))=\langle 0,0\rangle\). The field vector has no direction. Yet for \(h>0\), the displacement is
which points right. This does not contradict the theorem: its conclusion about approaching the field direction requires a nonzero field vector. At this point the instantaneous velocity is zero, while the particle still has a nonzero displacement over a positive time interval.
Check: The displacement divided by \(h\) is \(\langle h^2,0\rangle\), which tends to \(\langle 0,0\rangle=\mathbf{r}'(0)\) as \(h\to0^+\). This verifies the instantaneous-velocity definition directly.
A Common Misreading
A frequent mistake is to treat the location coordinates as though they were the velocity components. For the field in the first example, a mistaken calculation might say, “At \((2,1)\), the velocity is \(\langle 2,1\rangle\), so the speed is \(\sqrt{5}\).” That simply copies the location into a velocity vector and ignores the field formula.
The correction is to evaluate the component functions at the location: \(\mathbf{v}(2,1)=\langle 2-1,2+1\rangle=\langle 1,3\rangle\). The speed is therefore \(\sqrt{1^2+3^2}=\sqrt{10}\), not \(\sqrt{5}\). Always use the field rule first; the location is the input, not automatically the output.
A related error is to interpret each arrow as the particle’s whole path. An arrow reports the instantaneous velocity at its base point. The particle’s position may change, so the field vector evaluated at its new position may point in a different direction. A field diagram shows these local assignments across space; it does not, by itself, specify which particle is moving or where that particle began.
Why Velocity Fields Are Useful
Velocity fields provide a compact way to describe motion throughout a region. Instead of recording a separate velocity for each possible particle, one component formula assigns the appropriate instantaneous velocity at every location. Evaluating the field answers local questions: which way motion is directed at a point, whether the speed there is large or small, and whether the instantaneous velocity vanishes.
Arrow diagrams make these patterns visible, while component formulas allow exact calculations. The two representations work together: the formula supplies precise values, and the diagram helps reveal how direction and speed vary from place to place. In either representation, keep the key interpretation in view: the vector at a location is the instantaneous velocity assigned there, not the location itself and not a full trajectory.
Check Your Understanding
Use the definitions and methods in this tutorial to answer the following questions.
- For \(\mathbf{v}(x,y)=\langle 2x,-y\rangle\), what is the instantaneous velocity at \((1,3)\), and what is its speed?
- What information do the two components of a planar velocity vector provide?
- A diagram’s arrows all use the same length scale. What does an arrow’s orientation represent, and what does its length represent?
- Why does the direction of a sufficiently short displacement approach the field direction when the field vector is nonzero?
- Can a zero velocity vector determine a direction? Explain why a particle could nevertheless have a nonzero displacement over a finite time interval.