Speed from Velocity Components

Introduction

A moving object in space has both a velocity and a speed at each time when its position is differentiable. These quantities are closely related, but they are not the same: velocity records direction as well as rate, while speed is a single nonnegative number.

To find speed from a position function, first differentiate the position to obtain the velocity components. Then take the magnitude of that vector. The resulting calculation combines the coordinate motions rather than adding their signed rates.

Velocity and Speed

Let an object's position at time $t$ be represented by a differentiable vector-valued function $\mathbf{r}(t)$. Its velocity is the derivative $\mathbf{v}(t)=\mathbf{r}'(t)$. In three dimensions, if $\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle$, then the velocity components are the derivatives of the corresponding position components.

\mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t),y'(t),z'(t)\rangle.

The magnitude of a vector $\langle a,b,c\rangle$ is $\sqrt{a^2+b^2+c^2}$. Applying this definition to velocity gives the speed.

Definition: If $\mathbf{r}$ is differentiable at time $t$, its instantaneous velocity is $\mathbf{v}(t)=\mathbf{r}'(t)$, and its instantaneous speed is the scalar $s(t)=\|\mathbf{v}(t)\|$. For $\mathbf{v}(t)=\langle v_x(t),v_y(t),v_z(t)\rangle$, $$ s(t)=\sqrt{v_x(t)^2+v_y(t)^2+v_z(t)^2}. $$ In two dimensions, omit the $v_z(t)^2$ term.

Speed is never negative because it is a magnitude. It may be zero, which means the velocity vector is zero at that instant. Velocity, in contrast, is a vector: its components indicate how position changes along the coordinate directions.

Why the Magnitude Gives the Instantaneous Rate of Displacement

The formula for speed can be justified directly from the definition of velocity. Suppose $\mathbf{r}$ is differentiable at $t$. For a nonzero time increment $h$, the displacement from time $t$ to $t+h$ is $\mathbf{r}(t+h)-\mathbf{r}(t)$. Its magnitude divided by $|h|$ is the average magnitude of displacement per unit time over that interval.

Write the displacement quotient.
The quotient is $\left\|\mathbf{r}(t+h)-\mathbf{r}(t)\right\|/|h|$.

Factor out the time increment.
For $h\ne0$, this equals $\left\|(\mathbf{r}(t+h)-\mathbf{r}(t))/h\right\|$, because $|h\mathbf{w}|=|h|\|\mathbf{w}\|$.

Take the limit.
Differentiability gives $(\mathbf{r}(t+h)-\mathbf{r}(t))/h\to\mathbf{r}'(t)$. The norm is continuous, so the magnitudes converge to $\|\mathbf{r}'(t)\|$.

\lim_{h\to0} \frac{\|\mathbf{r}(t+h)-\mathbf{r}(t)\|}{|h|} = \|\mathbf{r}'(t)\|.

This argument identifies speed as the limiting rate of the magnitude of displacement. It does not, by itself, establish that the total distance travelled along the path has that derivative: differentiability at one time alone does not rule out irregular behavior nearby. A sufficient condition is that $\mathbf{r}$ be continuously differentiable on an interval around $t$. In that case the arc-length function from a fixed time $a$ is $L(u)=\int_a^u\|\mathbf{r}'(\tau)\|\,d\tau$, and the fundamental theorem of calculus gives $L'(t)=\|\mathbf{r}'(t)\|$, since the integrand is continuous.

Computing Speed from Components

The practical method is consistent for motion in any number of coordinates: differentiate each position component, square the resulting velocity components, add the squares, and take the nonnegative square root. Keep the component signs while differentiating; the signs disappear only when each component is squared in the magnitude formula.

Worked Example: Polynomial Motion in Three Dimensions

Suppose the position, in metres, is $\mathbf{r}(t)=\langle t^2,\,t^3+1,\,2t\rangle$, where $t$ is measured in seconds. Find the velocity and speed at $t=1$ second.

Differentiate each position component with respect to time:

\mathbf{v}(t)=\mathbf{r}'(t) =\langle 2t,\,3t^2,\,2\rangle\ \text{m/s}.

At $t=1$, the velocity is

\mathbf{v}(1)=\langle 2(1),\,3(1)^2,\,2\rangle =\langle 2,3,2\rangle\ \text{m/s}.

Its magnitude is therefore

s(1)=\sqrt{2^2+3^2+2^2} =\sqrt{4+9+4} =\sqrt{17}\ \text{m/s}.

Since $4+9+4=17$, this is approximately $4.12\ \text{m/s}$ to two decimal places. Checking the component calculation, the derivatives of $t^2$, $t^3+1$, and $2t$ are respectively $2t$, $3t^2$, and $2$; substituting $t=1$ gives the stated velocity and speed.

For polynomial positions, the velocity components are usually polynomials as well. The expression for speed may remain a square root of a polynomial expression; it need not simplify to a polynomial. At a particular time, exact squares or familiar square roots can make the final value straightforward to evaluate.

Worked Example: Trigonometric Motion with Constant Speed

Consider $\mathbf{r}(t)=\langle 3\cos(2t),\,3\sin(2t),\,4t\rangle$, with position measured in metres and time in seconds. Find the speed as a function of time.

Differentiate each component, applying the chain rule to the sine and cosine terms:

\mathbf{v}(t) = \langle -6\sin(2t),\,6\cos(2t),\,4\rangle\ \text{m/s}.

Now square and add the components:

\begin{aligned} s(t) &=\sqrt{(-6\sin(2t))^2+(6\cos(2t))^2+4^2}\\ &=\sqrt{36\sin^2(2t)+36\cos^2(2t)+16}\\ &=\sqrt{36\bigl(\sin^2(2t)+\cos^2(2t)\bigr)+16}\\ &=\sqrt{36+16}\\ &=\sqrt{52}=2\sqrt{13}\ \text{m/s}. \end{aligned}

The identity $\sin^2(\theta)+\cos^2(\theta)=1$ shows that the speed is constant, even though the velocity vector changes direction as $t$ changes. Substituting any time into the velocity and taking its magnitude produces the same result, because the trigonometric terms always have squared sum $1$.

Speed Is Not the Velocity Vector

A speed value gives the magnitude of motion at an instant but not its direction. Distinct velocity vectors can have the same speed. For example, $\langle 3,4\rangle$ and $\langle -3,4\rangle$ both have magnitude $5$, although their first components point in opposite coordinate directions. A speed cannot distinguish between these velocities.

Likewise, a negative component does not mean that the speed is negative. It means that position is decreasing along that coordinate direction. The component is squared when calculating the magnitude, so its sign does not make the total speed negative.

Worked Example: Signed Components and a Nonnegative Speed

At a particular instant, suppose an object's velocity is $\mathbf{v}=\langle 3,-4,12\rangle\ \text{m/s}$. Find its speed and explain what the negative component means.

Use the three-dimensional magnitude formula:

\begin{aligned} s &=\|\mathbf{v}\|\\ &=\sqrt{3^2+(-4)^2+12^2}\\ &=\sqrt{9+16+144}\\ &=\sqrt{169}\\ &=13\ \text{m/s}. \end{aligned}

The speed is $13\ \text{m/s}$, not a signed quantity. The component $-4\ \text{m/s}$ indicates motion in the negative direction of the second coordinate axis at that instant; it does not make the speed negative. Squaring and adding all three components gives $169$, and its nonnegative square root is $13$.

A Common Mistake

A frequent error is to add the velocity components instead of taking their Euclidean magnitude. For the velocity $\langle 3,-4,12\rangle$, one might incorrectly write $s=3+(-4)+12=11\ \text{m/s}$. This is wrong because a sum of components depends on their signs and is not the vector's magnitude. The correct calculation squares each component, adds the squares, and takes the nonnegative square root:

s=\sqrt{3^2+(-4)^2+12^2} =\sqrt{169} =13\ \text{m/s}.

Another useful check is dimensional: if each velocity component is measured in metres per second, its magnitude is also measured in metres per second. A result with units of metres or seconds alone cannot be a speed.

Key takeaway: Differentiate position to obtain the velocity vector, then take its magnitude to obtain speed. The velocity contains directional information; speed retains only the nonnegative magnitude.

Check Your Understanding

Answer each question by applying the component formula and keeping the distinction between a vector and a scalar clear.

If $\mathbf{v}=\langle -5,12\rangle\ \text{m/s}$, what is the speed, and what does the negative first component indicate?
For $\mathbf{r}(t)=\langle 2t,\,t^2\rangle$, find the velocity and speed as functions of $t$.
Why can two different velocity vectors describe motion with the same speed?
For a continuously differentiable position function, what is the derivative of the arc-length function?
What is the error in computing speed by adding the signed velocity components?

What You'll Learn