From Position to Velocity
A position function describes where a moving object is at each time. To describe how its position is changing at a particular instant, we take a derivative. Because position has several components, this derivative is a vector: it records the rate of change of every coordinate at once.
This gives a precise meaning to velocity along a space curve. It also explains why velocity points along the curve’s direction of motion when it is nonzero. The method is direct, but its interpretation depends on keeping track of the parameter, the vector components, and the units.
In three dimensions, the position components give coordinates, while the velocity components give the rates at which those coordinates change. If position is measured in metres and time in seconds, each component of velocity is measured in metres per second. The same definition applies in two dimensions or in any finite number of coordinates.
Why the Derivative Acts Component by Component
The derivative of a vector-valued function is defined by the same difference quotient used for a scalar function. Subtract the position at time \(t\) from the position at time \(t+h\), divide by the time increment \(h\), and take the limit as \(h\) approaches zero. Vector subtraction and division by a scalar act on each component.
When all three component limits exist, the vector limit exists and has those limits as its coordinates. Each coordinate limit is the ordinary derivative of its component. This proves the componentwise differentiation rule; it is not a separate physical assumption about motion.
For example, if \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), then \(\mathbf{v}(t)=\langle x'(t),y'(t),z'(t)\rangle\). A prime denotes differentiation with respect to \(t\). If a coordinate is constant, its derivative is zero: that coordinate contributes no change to velocity.
Worked Example: A Polynomial Trajectory
Worked Example: A Polynomial Trajectory
An object’s position, in metres, is \(\mathbf{r}(t)=\langle t^2,3t-1,2\rangle\), where \(t\) is measured in seconds. Find its velocity function and its velocity at \(t=2\) seconds. Interpret the direction and magnitude at that time.
Differentiate the first coordinate: \(\frac{d}{dt}(t^2)=2t\). Differentiate the second: \(\frac{d}{dt}(3t-1)=3\). The third coordinate is constant, so its derivative is \(0\). Therefore,
At \(t=2\) seconds, substitute \(2\) for \(t\): \(\mathbf{v}(2)=\langle 2(2),3,0\rangle=\langle 4,3,0\rangle\ \text{m/s}\). Its magnitude is \(\sqrt{4^2+3^2+0^2}=5\ \text{m/s}\), and its direction as a unit vector is \(\langle 4/5,3/5,0\rangle\).
Check the derivative using the difference quotient at \(t=2\). The position difference is \(\mathbf{r}(2+h)-\mathbf{r}(2)=\langle (2+h)^2-4,3h,0\rangle=\langle 4h+h^2,3h,0\rangle\). For \(h\ne0\), dividing by \(h\) gives \(\langle 4+h,3,0\rangle\), whose limit as \(h\to0\) is \(\langle 4,3,0\rangle\), as found above.
The direction of a nonzero velocity vector indicates the instantaneous direction in which position is changing. In this example, the third coordinate of velocity is zero, so there is no instantaneous change in the third position coordinate at \(t=2\). The magnitude is the length of the velocity vector; it describes the size of the position-change rate, while the unit vector gives its direction.
Worked Example: A Trigonometric Trajectory
Worked Example: A Trigonometric Trajectory
Suppose \(\mathbf{r}(t)=\langle 3\cos(2t),3\sin(2t),4t\rangle\) metres, with \(t\) in seconds and the angular factor \(2\) understood in inverse seconds. Find the velocity and evaluate it at \(t=\pi/4\) seconds.
Differentiate each component. By the chain rule, the derivative of \(\cos(2t)\) is \(-2\sin(2t)\), and the derivative of \(\sin(2t)\) is \(2\cos(2t)\). Thus,
At \(t=\pi/4\), the angle \(2t\) is \(\pi/2\). Since \(\sin(\pi/2)=1\) and \(\cos(\pi/2)=0\),
The velocity points in the negative first-coordinate direction and the positive third-coordinate direction, with no second-coordinate component at this instant. Its magnitude is \(\sqrt{(-6)^2+0^2+4^2}=\sqrt{52}=2\sqrt{13}\ \text{m/s}\).
For a check, the first component’s difference quotient at \(t=\pi/4\) is \(3[\cos(\pi/2+2h)-\cos(\pi/2)]/h=-3\sin(2h)/h\), which tends to \(-6\). The second component’s quotient is \(3[\sin(\pi/2+2h)-\sin(\pi/2)]/h=3[\cos(2h)-1]/h\), which tends to \(0\). The third component’s quotient is \(4h/h=4\). These limits reproduce \(\langle -6,0,4\rangle\).
Direction, Magnitude, and a Zero Velocity
A vector’s direction is meaningful when the vector is nonzero. At such a time, the velocity points in the direction of the curve’s instantaneous motion. Dividing velocity by its magnitude produces a unit vector in that direction. If velocity is the zero vector, this division is not defined, and the velocity itself does not specify an instantaneous direction.
Worked Example: Zero Velocity at an Instant
Let \(\mathbf{r}(t)=\langle t^2,t^3,\cos t\rangle\) metres. Find the velocity at \(t=0\) and state what can and cannot be concluded from it.
Differentiate the components: \(\frac{d}{dt}(t^2)=2t\), \(\frac{d}{dt}(t^3)=3t^2\), and \(\frac{d}{dt}(\cos t)=-\sin t\). Thus \(\mathbf{v}(t)=\langle 2t,3t^2,-\sin t\rangle\ \text{m/s}\), so \(\mathbf{v}(0)=\langle 0,0,0\rangle\ \text{m/s}\).
Check by the difference quotient at zero. We have \(\mathbf{r}(h)-\mathbf{r}(0)=\langle h^2,h^3,\cos h-1\rangle\). Dividing by \(h\ne0\) gives \(\langle h,h^2,(\cos h-1)/h\rangle\). As \(h\to0\), the first two components tend to zero, and the third tends to zero because \(\cos h-1\) is of order \(h^2\). The limit is \(\langle 0,0,0\rangle\), confirming the result.
At \(t=0\), the velocity has magnitude zero and no direction unit vector. This conclusion concerns the instantaneous velocity; it does not by itself describe the object’s motion at other times.
A Common Differentiation Mistake
A frequent error is to differentiate a vector as if the whole vector were one scalar expression, or to omit the chain-rule factor inside a trigonometric component. For instance, for the first coordinate \(3\cos(2t)\), the incorrect calculation \( \frac{d}{dt}[3\cos(2t)]=-3\sin(2t)\) forgets that the inside function \(2t\) has derivative \(2\).
The correction is to differentiate each component using the ordinary differentiation rules, including the chain rule wherever a component is composed with another function. Only after all components have been differentiated should they be assembled into the velocity vector.
A Reliable Procedure
Identify each coordinate function and the parameter \(t\).
Apply the relevant scalar derivative rules, including the chain rule.
Keep the coordinate order unchanged and attach units of position per unit time.
Substitute the specified time; if velocity is nonzero, its direction and magnitude can be read from the resulting vector.
This procedure matters because a space curve can change in several coordinate directions at once. Velocity combines those coordinate rates into one vector, so it gives a compact description of the curve’s instantaneous motion. Careful componentwise differentiation also makes it possible to check each coordinate independently and catch sign or chain-rule errors before interpreting the result.
Check Your Understanding
For each question, identify the derivative rule you need and keep the velocity components in the same order as the position coordinates.
- If \(\mathbf{r}(t)=\langle 2t^3,t-4,7\rangle\), what is \(\mathbf{v}(t)\)?
- For \(\mathbf{r}(t)=\langle \sin(3t),\cos(3t),t^2\rangle\), what is the velocity at \(t=0\)?
- What does the direction of a nonzero velocity vector represent at a given instant?
- When is the unit direction vector \(\mathbf{v}(t)/\|\mathbf{v}(t)\|\) defined?
- Why must differentiating \(\cos(2t)\) include a factor of \(2\)?