Velocity Vectors, Speed, And Direction Of Motion

Position and Motion

A parametrized path describes where an object is at each value of a parameter, often time. Its position at time $t$ can be represented by a vector $\mathbf r(t)$. In two dimensions, for example, $\mathbf r(t)=\langle x(t),y(t)\rangle$; in three dimensions, $\mathbf r(t)=\langle x(t),y(t),z(t)\rangle$.

Position alone tells us where the object is, not how it is moving. Differentiating the position components tells us how each coordinate changes per unit time. Together, those coordinate rates form the velocity vector. The magnitude of that vector is speed, while its orientation indicates the direction of motion.

Definition: If $\mathbf r(t)$ is differentiable, its velocity vector is $\mathbf v(t)=\mathbf r'(t)$. If $\mathbf r(t)=\langle x(t),y(t),z(t)\rangle$, then $\mathbf v(t)=\langle x'(t),y'(t),z'(t)\rangle$. The speed is the magnitude of velocity: $$ \|\mathbf v(t)\|=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}. $$ In two dimensions, omit the $z$-component.

Velocity is a vector, so it contains both a magnitude and a direction. Speed is a scalar: it gives the nonnegative rate of motion but not the direction. When $\mathbf v(t)\ne\mathbf 0$, the unit vector in the forward direction of motion is $\mathbf T(t)=\mathbf v(t)/\|\mathbf v(t)\|$. This vector has magnitude $1$. It points in the direction the object moves as time increases.

Why the Derivative Gives the Direction of Motion

The derivative of a position vector is defined through changes in position. For a small time increment $h$, the displacement from time $t$ to time $t+h$ is $\mathbf r(t+h)-\mathbf r(t)$. Differentiability means that the displacement per unit time approaches velocity:

\lim_{h\to 0}\frac{\mathbf r(t+h)-\mathbf r(t)}{h} =\mathbf r'(t)=\mathbf v(t).

Thus, when $h$ is small and positive, displacement is approximately $h\mathbf v(t)$. If velocity is nonzero, this makes the forward displacement point approximately in the direction of $\mathbf v(t)$. More precisely, normalizing the displacement gives the limiting direction:

\lim_{h\to 0^+} \frac{\mathbf r(t+h)-\mathbf r(t)} {\|\mathbf r(t+h)-\mathbf r(t)\|} = \frac{\mathbf v(t)}{\|\mathbf v(t)\|}.

To see why, for $h>0$ write the expression on the left as \[ \frac{(\mathbf r(t+h)-\mathbf r(t))/h} {\|(\mathbf r(t+h)-\mathbf r(t))/h\|}. \] The numerator approaches $\mathbf v(t)$, and the denominator approaches $\|\mathbf v(t)\|$. If $\mathbf v(t)\ne\mathbf 0$, division is valid and the limit is $\mathbf T(t)$. This establishes that the velocity direction is the forward direction of motion at that instant.

The same reasoning distinguishes a tangent line from a direction of travel. A tangent line is an unoriented geometric line; velocity points along that line in the direction corresponding to increasing time. If $h<0$, the displacement points oppositely, toward the earlier position. At a self-intersection, the same position may also occur at different parameter values with different velocities, so position alone need not determine direction of motion.

Speed and Accumulated Distance

Velocity’s magnitude combines the coordinate rates using the Euclidean norm. The components should not generally be added: motion in perpendicular coordinate directions contributes to the magnitude by the Pythagorean theorem. For a continuously differentiable path, the rate of change of accumulated distance traveled is speed:

\frac{ds}{dt}=\|\mathbf r'(t)\| =\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2},

where $s$ denotes distance accumulated along the path. This describes how quickly distance is being covered, regardless of which direction the object is traveling. A change in direction can occur while speed remains positive.

Worked Example: Finding Velocity, Speed, and Direction

Suppose an object’s position, in meters, at time $t$ seconds is $\mathbf r(t)=\langle 2t,t^2\rangle$. Find its position, velocity, speed, and forward direction at $t=1$.

First evaluate position: $$ \mathbf r(1)=\langle 2(1),1^2\rangle=\langle 2,1\rangle\text{ m}. $$ Differentiate each component: $$ \mathbf v(t)=\mathbf r'(t)=\langle 2,2t\rangle\text{ m/s}. $$ Therefore, $$ \mathbf v(1)=\langle 2,2\rangle\text{ m/s}. $$ Its speed is the magnitude: $$ \|\mathbf v(1)\|=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\text{ m/s}. $$ Dividing velocity by its magnitude gives the forward unit direction: $$ \mathbf T(1)=\frac{\langle 2,2\rangle}{2\sqrt{2}} =\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle. $$

Check the result: the direction has magnitude $\sqrt{(1/\sqrt{2})^2+(1/\sqrt{2})^2}=1$, and multiplying it by the speed recovers velocity: $(2\sqrt{2})\mathbf T(1)=\langle 2,2\rangle\text{ m/s}$.

Velocity Along a Curved Path

For a curved path, velocity usually changes direction as the object moves. Its magnitude and direction must therefore be found from the velocity vector at the particular parameter value of interest. The following example illustrates this for an ellipse.

Worked Example: Motion on an Ellipse

Let $\mathbf r(t)=\langle 3\cos t,2\sin t\rangle$ meters, with $t$ measured in seconds. Find the velocity, speed, and forward direction at $t=\pi/3$.

Differentiate the position components: $$ \mathbf v(t)=\langle -3\sin t,2\cos t\rangle\text{ m/s}. $$ At $t=\pi/3$, use $\sin(\pi/3)=\sqrt{3}/2$ and $\cos(\pi/3)=1/2$: $$ \mathbf v(\pi/3) =\left\langle-\frac{3\sqrt{3}}{2},1\right\rangle\text{ m/s}. $$ The speed is $$ \|\mathbf v(\pi/3)\| =\sqrt{\left(-\frac{3\sqrt{3}}{2}\right)^2+1^2} =\sqrt{\frac{27}{4}+\frac{4}{4}} =\frac{\sqrt{31}}{2}\text{ m/s}. $$ Thus the forward unit direction is $$ \mathbf T(\pi/3) =\frac{\left\langle-3\sqrt{3}/2,1\right\rangle}{\sqrt{31}/2} =\left\langle-\frac{3\sqrt{3}}{\sqrt{31}},\frac{2}{\sqrt{31}}\right\rangle. $$

For a check, the squared magnitude of this direction is \[ \frac{27}{31}+\frac{4}{31}=1. \] Multiplying it by the speed gives \[ \frac{\sqrt{31}}{2} \left\langle-\frac{3\sqrt{3}}{\sqrt{31}},\frac{2}{\sqrt{31}}\right\rangle = \left\langle-\frac{3\sqrt{3}}{2},1\right\rangle\text{ m/s}, \] which is the velocity found by differentiation.

When Velocity Is Zero

If $\mathbf v(t)=\mathbf 0$, the object is instantaneously at rest in the sense that its instantaneous rate of change of position is zero. Its speed is zero as well. But the formula for the unit direction requires division by $\|\mathbf v(t)\|$, so it does not define a direction at that instant. Nearby directions can sometimes give more information about how the path approaches or leaves the point, but they are not a direction supplied by the velocity at the stop.

Worked Example: A Stop with Different Nearby Directions

Consider $\mathbf r(t)=\langle t^2,t^3\rangle$, and examine the motion at $t=0$. Differentiate: $$ \mathbf v(t)=\langle 2t,3t^2\rangle. $$ At the specified time, $$ \mathbf r(0)=\langle 0,0\rangle,\qquad \mathbf v(0)=\langle 0,0\rangle. $$ The speed at $t=0$ is therefore $$ \|\mathbf v(0)\|=\sqrt{0^2+0^2}=0. $$ There is no unit direction $\mathbf v(0)/\|\mathbf v(0)\|$, because that would divide by zero.

For $t\ne 0$, the speed is $$ \|\mathbf v(t)\| =\sqrt{4t^2+9t^4} =|t|\sqrt{4+9t^2}. $$ For positive $t$, the unit velocity is $$ \frac{\mathbf v(t)}{\|\mathbf v(t)\|} =\left\langle\frac{2}{\sqrt{4+9t^2}},\frac{3t}{\sqrt{4+9t^2}}\right\rangle, $$ which approaches $\langle 1,0\rangle$ as $t\to0^+$. For negative $t$, the first component of the unit velocity is $-2/\sqrt{4+9t^2}$, while the second is $3|t|/\sqrt{4+9t^2}$; the unit velocity approaches $\langle-1,0\rangle$ as $t\to0^-$. Thus the two nearby forward-motion directions have opposite limits, although the velocity at the stop itself is zero.

Check the calculation by substituting $t=0$ into the differentiated components: $2(0)=0$ and $3(0)^2=0$. The zero velocity correctly gives zero speed, but it does not provide an instantaneous unit direction.

A Common Mistake: Adding Velocity Components

A frequent error is to treat the speed as the sum of the component rates. For example, for $\mathbf r(t)=\langle 2t,2t\rangle$, the velocity is $\langle 2,2\rangle$. Adding its components would incorrectly give speed $2+2=4$. The correct calculation uses the magnitude: $$ \|\langle 2,2\rangle\| =\sqrt{2^2+2^2} =\sqrt{8} =2\sqrt{2}. $$ The correction matters because the two components describe perpendicular coordinate directions. The velocity points diagonally, and its magnitude is not the sum of the lengths of those perpendicular component rates.

Also distinguish speed from signed coordinate rates. A component of velocity can be negative, indicating motion toward decreasing values of that coordinate. Speed cannot be negative: it is the magnitude of the full velocity vector.

Why These Quantities Matter

Velocity, speed, and direction answer different questions about a parametrized path. Velocity gives the complete instantaneous motion in a single vector. Speed measures how rapidly distance is accumulated. The normalized velocity gives the direction of forward travel when the velocity is nonzero. Keeping these interpretations separate helps prevent errors when analyzing paths that curve, change coordinate direction, or contain stops.

Key takeaway: Differentiate the position vector to find velocity; take its magnitude to find speed; and, only when velocity is nonzero, divide by its magnitude to find the forward unit direction.

Check Your Understanding

Use the definitions and reasoning in this tutorial to answer the following questions.

If $\mathbf r(t)=\langle x(t),y(t)\rangle$, what are the component formulas for velocity and speed?
For $\mathbf v=\langle 5,12\rangle$, find its speed and its unit direction.
Why does the velocity vector give the forward direction of motion when it is nonzero?
Can speed be negative if one component of velocity is negative? Explain.
What prevents the velocity vector from defining a unit direction at a time when it equals $\mathbf 0$?