From Velocity Components to Speed
A velocity vector describes how position changes, including the rate of change along each coordinate axis. Speed answers a different question: how quickly is the object moving, without separating that motion into coordinate directions?
For motion in a plane, velocity may have a positive rate along one axis and a negative rate along the other. Those signs describe the component rates; they do not make speed negative. To find speed, combine the component rates using the Euclidean magnitude of the velocity vector.
The component \(v_x(t)\) is the rate of change of the \(x\)-coordinate, \(v_y(t)\) is the rate of change of the \(y\)-coordinate, and, in three dimensions, \(v_z(t)\) is the rate of change of the \(z\)-coordinate. Each component can be positive, negative, or zero. Speed, being a magnitude, is always nonnegative.
Computing Velocity from Position
If position is given as a vector-valued function, differentiate each component to obtain velocity. For example, when \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), the velocity is \(\mathbf{v}(t)=\mathbf{r}'(t)=\langle x'(t),y'(t)\rangle\). Speed is then the magnitude of this derivative vector.
This order matters: first find the velocity components, then square and add them, and finally take the nonnegative square root. The individual rates are not themselves the speed unless all other components are zero.
Worked Example: Speed from a Two-Dimensional Position Function
An object has position \(\mathbf{r}(t)=\langle 3t^2-2t,\;4t+1\rangle\), measured in meters, where \(t\) is measured in seconds. Find its instantaneous speed at \(t=1\).
Differentiate the first coordinate:
Differentiate the second coordinate:
Thus the velocity vector is \(\mathbf{v}(t)=\langle 6t-2,4\rangle\) meters per second. At \(t=1\), its components are
Take the Euclidean magnitude:
Check: the square of the result is \((4\sqrt{2})^2=32\), which equals \(4^2+4^2=32\). The \(x\)- and \(y\)-component rates are each \(4\text{ m/s}\), while the speed is \(4\sqrt{2}\text{ m/s}\), approximately \(5.66\text{ m/s}\) to two decimal places.
Why the Magnitude Gives Instantaneous Speed
The velocity vector is defined by the limit of the average velocity over a short time interval. If \(\mathbf{r}(t)\) is differentiable, then
The numerator is the displacement vector over the interval, and its magnitude is the straight-line distance between the two endpoint positions. Dividing that distance by the elapsed time gives the magnitude of average velocity. Taking the limit gives the magnitude of instantaneous velocity. The key fact is that the magnitude operation is continuous: if vectors approach a vector, their magnitudes approach that vector's magnitude.
For a nonzero time change \(h\), the average velocity is \(\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}\).
The magnitude of this average velocity is \(\frac{\|\mathbf{r}(t+h)-\mathbf{r}(t)\|}{|h|}\), since dividing a vector by \(h\) scales its magnitude by \(1/|h|\).
When the derivative exists, the average velocity approaches \(\mathbf{v}(t)\). Continuity of magnitude means the magnitudes approach \(\|\mathbf{v}(t)\|\), the instantaneous speed.
This argument also explains why the sign of a component does not determine whether the object is moving quickly or slowly. A negative component changes the direction of change along that coordinate, but its square contributes a nonnegative amount to the squared speed.
Speed in Two and Three Dimensions
The same calculation works regardless of whether motion is described with two or three coordinates: take the square of each velocity component, add those squares, and take the nonnegative square root. The result has the same units as the component rates. For instance, if the components are measured in meters per second, speed is also measured in meters per second.
Worked Example: A Negative Component Does Not Make Speed Negative
Suppose an object's velocity at a particular instant is \(\mathbf{v}=\langle -1,3,3\rangle\) meters per second. Find its speed and compare it with the component rates.
Square each component, including the negative one:
Add the squares and take the square root:
Check: \((\sqrt{19})^2=19\), matching the sum of the three squared components. The \(x\)-coordinate is changing at \(-1\text{ m/s}\), while the other two coordinates are changing at \(3\text{ m/s}\) each. The speed is \(\sqrt{19}\text{ m/s}\), approximately \(4.36\text{ m/s}\) to two decimal places; it is not \(-1\text{ m/s}\), nor is it the sum of the component rates.
Worked Example: Constant Speed with Changing Components
An object has position \(\mathbf{r}(t)=\langle 5\cos t,5\sin t\rangle\) meters. Find its speed at \(t=\frac{\pi}{6}\) seconds.
Differentiate the position components:
At \(t=\frac{\pi}{6}\), use \(\sin\frac{\pi}{6}=\frac12\) and \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}2\):
Now calculate the magnitude:
Check: the squared component rates sum to \(\frac{25}{4}+\frac{75}{4}=25\), and the square of the reported speed is \(5^2=25\). Here the components are different from one another, but their combined magnitude is exactly \(5\text{ m/s}\).
Component Rates and Speed Are Different Quantities
A component rate answers a coordinate-specific question: how quickly is one coordinate changing? Speed combines all the coordinate rates into one nonnegative quantity. In two dimensions, the component rates are the legs of a right triangle and speed is the hypotenuse; the magnitude formula is the Pythagorean theorem applied to the velocity components.
| Quantity | What it describes | Can it be negative? |
|---|---|---|
| \(v_x(t)\) | Rate of change of the \(x\)-coordinate | Yes |
| \(v_y(t)\) | Rate of change of the \(y\)-coordinate | Yes |
| \(\|\mathbf{v}(t)\|\) | Instantaneous speed | No |
In particular, speed is not generally \(v_x+v_y\), nor is it the largest component. For example, with velocity \(\langle 4,4\rangle\), the component sum is \(8\), but the speed is \(\sqrt{4^2+4^2}=4\sqrt{2}\). The components are perpendicular coordinate rates, so they combine through squares rather than direct addition.
Common Pitfall: Keeping a Component's Sign in the Speed
A frequent error is to treat a negative velocity component as a negative contribution to speed. For instance, with \(\mathbf{v}=\langle -3,4\rangle\), an incorrect calculation is
This calculation is wrong because the exponent is applied before the negative sign: \(-3^2=-(3^2)=-9\). The magnitude formula requires the square of the entire component, so the correct calculation is
Check: \(5^2=25\), which equals \((-3)^2+4^2=9+16\). The \(x\)-component is negative, but speed is \(5\), not \(\sqrt{7}\) and not a negative number. Parentheses help make clear that the entire component is being squared.
When a Component Equals Zero
A zero component means there is no instantaneous change along that coordinate at the time in question. It does not necessarily mean the object is stopped. For example, if \(\mathbf{v}=\langle 0,6\rangle\), then
Only when every velocity component is zero does the magnitude equal zero. Conversely, if the speed is zero, the sum of the squared components is zero. Since each square is nonnegative, every component must then be zero.
Summary
To calculate instantaneous speed from a position function, differentiate each coordinate to find the velocity components, square those component values, add the squares, and take the nonnegative square root. The result is the Euclidean magnitude of the velocity vector, not an individual component rate.
Check Your Understanding
For each question, keep the distinction between component rates and speed clear.
- If \(\mathbf{v}=\langle 6,8\rangle\) meters per second, what is the speed?
- If \(\mathbf{v}=\langle -5,12\rangle\) meters per second, why is the speed nonnegative, and what is its value?
- For \(\mathbf{r}(t)=\langle t^2,3t\rangle\), find the velocity components and speed at \(t=2\).
- In three dimensions, what is the speed when \(\mathbf{v}=\langle 0,-2,2\rangle\) meters per second?
- What does it mean for one velocity component to be zero? Does that alone imply the object is stopped?
