From Position to Instantaneous Motion
A particle moving in the plane can be described by two position functions: \(x(t)\), its horizontal coordinate, and \(y(t)\), its vertical coordinate. Together they specify where the particle is at time \(t\). To describe how it is moving at a particular instant, differentiate each coordinate with respect to time.
The results, \(x'(t)\) and \(y'(t)\), are the particle’s velocity components. Their signs and values give an oriented description of its instantaneous motion: right or left, up or down, and how the horizontal and vertical changes compare.
This tutorial assumes that \(x\) and \(y\) are differentiable at the time being considered. At an interior time of their domain, the ordinary derivatives give the instantaneous components. The focus here is interpreting those components, not finding the length of the velocity vector.
Why Differentiating the Coordinates Gives Velocity
Over a short time interval from \(t\) to \(t+h\), the horizontal change is \(x(t+h)-x(t)\), and the vertical change is \(y(t+h)-y(t)\). Dividing each change by the elapsed time \(h\) gives the average rate of change of each coordinate. The instantaneous rates are the limits of those averages as the time interval shrinks to zero.
The position can also be written as the vector \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). Its difference quotient has one coordinate for each of the two scalar difference quotients. Taking the limit therefore gives both derivatives in the corresponding coordinates.
This establishes why the coordinate derivatives are the components of velocity: differentiating the position vector differentiates each coordinate. In practice, find \(x'(t)\) and \(y'(t)\) separately, then evaluate both at the specified time.
Reading Direction from the Signs
The first component describes the instantaneous horizontal change; the second describes the instantaneous vertical change. A positive horizontal component means the particle is moving to the right at that instant, while a negative one means it is moving to the left. A positive vertical component means it is moving upward, and a negative one means it is moving downward.
| Horizontal component | Vertical component | Instantaneous direction |
|---|---|---|
| Positive | Positive | Right and up |
| Positive | Negative | Right and down |
| Negative | Positive | Left and up |
| Negative | Negative | Left and down |
If one component is zero and the other is nonzero, the instantaneous direction is along one coordinate axis. If both components are zero, the velocity vector is the zero vector, so its components do not specify a direction at that instant. Avoid assigning a compass direction to a zero vector.
The values matter as well as the signs. For instance, if the components are \(\langle 4,2\rangle\), the particle is moving right and up, with horizontal component \(4\) and vertical component \(2\) in the stated units. These are rates along the two coordinate axes; they should not be replaced by a single slope.
Worked Examples
Worked Example: Right and Up at a Specified Time
A particle’s position, in meters, is \(x(t)=t^2+1\) and \(y(t)=2t-3\), where \(t\) is in seconds. Find its velocity components and describe its instantaneous direction at \(t=2\).
Differentiate each coordinate with respect to \(t\):
Evaluate at \(t=2\):
Thus \(\mathbf{v}(2)=\langle 4,2\rangle\) meters per second. Both components are positive, so the particle is moving to the right and upward at that instant.
Check the horizontal derivative directly with a difference quotient at \(t=2\):
which approaches \(4\) as \(h\to 0\). For the vertical coordinate, \(y(2+h)=1+2h\) and \(y(2)=1\), so \(\frac{y(2+h)-y(2)}{h}=2\). Both component values are confirmed.
Worked Example: Motion Along a Coordinate Axis
Let \(x(t)=3\cos t\) and \(y(t)=3\sin t\), with position measured in meters and \(t\) in seconds. Determine the instantaneous direction at \(t=\frac{\pi}{2}\).
Differentiate the coordinate functions:
Substitute \(t=\frac{\pi}{2}\), using \(\sin(\frac{\pi}{2})=1\) and \(\cos(\frac{\pi}{2})=0\):
The velocity vector is \(\left\langle -3,0\right\rangle\) meters per second. The horizontal component is negative and the vertical component is zero, so the particle’s instantaneous direction is horizontally to the left.
Check by differentiating the original coordinate formulas: the derivative of \(3\cos t\) is \(-3\sin t\), and the derivative of \(3\sin t\) is \(3\cos t\). Evaluating those expressions gives exactly \(-3\) and \(0\).
Worked Example: Left and Up with Unequal Components
A particle has position \(x(t)=4-t^2\) and \(y(t)=t^3-3t\), in meters, with time in seconds. Find its velocity components and direction at \(t=2\).
Differentiate both coordinates:
Evaluate at \(t=2\):
Therefore, \(\mathbf{v}(2)=\langle -4,9\rangle\) meters per second. The negative horizontal component means left, and the positive vertical component means up. The particle is moving left and upward; the component values are \(-4\) and \(9\) meters per second, respectively.
As a check, \(x(2)=4-4=0\). For a small change \(h\), \(x(2+h)=4-(2+h)^2=-4h-h^2\), so \(\frac{x(2+h)-x(2)}{h}=-4-h\), which approaches \(-4\). Also, \(y(2)=2\) and \(y(2+h)=(2+h)^3-3(2+h)=2+9h+6h^2+h^3\). Thus \(\frac{y(2+h)-y(2)}{h}=9+6h+h^2\), which approaches \(9\).
Direction Is Not the Same as Slope
When \(x'(t)\ne 0\), the slope of the parametric curve at the point corresponding to \(t\) is \(\frac{dy}{dx}=\frac{y'(t)}{x'(t)}\). That ratio describes the slope of the tangent line, but it is not the pair of velocity components. The velocity components preserve the separate horizontal and vertical rates and their signs.
For example, the components \(\langle 4,2\rangle\) give slope \(\frac{2}{4}=\frac12\), but the velocity vector is still \(\langle 4,2\rangle\), not \(\frac12\). The slope alone cannot tell you whether the particle is moving right and up or left and down: \(\langle 4,2\rangle\) and \(\langle -4,-2\rangle\) have the same ratio of vertical to horizontal component but point in opposite directions.
A Common Mistake
Suppose \(x(t)=t^2\) and \(y(t)=3t\), with position measured in meters and time in seconds. At \(t=2\), a mistaken calculation might be:
The quotient \(\frac34\) is the slope of the tangent line, not the velocity vector. To find velocity, keep the derivatives as separate components:
Check by differentiating the position functions: \(\frac{d}{dt}(t^2)=2t\), which is \(4\) at \(t=2\), and \(\frac{d}{dt}(3t)=3\). The corrected components say the particle moves right and up. Their ratio, \(3/4\), can describe tangent slope here, but it cannot stand in for the two rates.
Check Your Understanding
For each question, use the coordinate derivatives to identify both the component values and the direction they indicate.
- If \(x(t)=t^2-1\) and \(y(t)=5-t\), what are the velocity components at \(t=1\), and in which directions is the particle moving horizontally and vertically?
- A particle has velocity components \(\langle -2,-7\rangle\) at an instant. Describe its direction of motion.
- What does it mean for \(x'(t)=0\) while \(y'(t)>0\) at a particular instant?
- Why does \(\frac{y'(t)}{x'(t)}\), when defined, not give both velocity components?
- If both coordinate derivatives equal zero at an instant, what can you conclude about the direction from those components alone?
