Reading Motion from the Coordinates
A parametrization describes a point in the plane by giving its coordinates as functions of a parameter. As the parameter increases, the point may move horizontally, vertically, or in both directions at once. The derivatives of the coordinate functions tell us how each coordinate is changing at that moment.
This gives a direct way to describe a curve’s orientation locally: inspect the sign of the horizontal coordinate’s derivative and the sign of the vertical coordinate’s derivative. The description is about how the point moves as the parameter increases; it does not, by itself, describe the curve’s tangent or overall shape.
These descriptions are relative to increasing \(t\). If \(t\) represents time, they describe the point’s direction of motion through time. If \(t\) is another parameter, they describe the direction in which the parametrized point is traced as that parameter increases.
Why the Signs Give the Direction
At a parameter value \(t_0\), compare the coordinate at \(t_0\) with its coordinate at a nearby, slightly larger parameter \(t_0+h\), where \(h>0\). The horizontal change is \(x(t_0+h)-x(t_0)\); the vertical change is \(y(t_0+h)-y(t_0)\). By the definition of the derivative, the horizontal difference quotient approaches \(x'(t_0)\) as \(h\) approaches zero from above.
If \(x'(t_0)>0\), then for all sufficiently small positive \(h\), the quotient \(\frac{x(t_0+h)-x(t_0)}{h}\) is positive. Since \(h>0\), this means \(x(t_0+h)-x(t_0)>0\): nearby, the horizontal coordinate increases as the parameter increases. The reasoning for a negative derivative is the same, with the inequality reversed. Applying this argument to \(y\) gives the vertical direction.
This reasoning describes motion near a parameter value. If a derivative keeps the same strict sign throughout an interval, the corresponding coordinate changes consistently in that direction over the interval. For example, if \(x'(t)>0\) throughout an interval, then \(x\) increases there. A derivative’s sign at just one parameter value does not establish what happens over a whole interval.
Worked Example: Moving Left and Up
Worked Example: Moving Left and Up
A point is parametrized by \(x(t)=t^2-4t\) and \(y(t)=2t+1\). Describe its motion at \(t=1\), at \(t=2\), and at \(t=3\).
First differentiate each coordinate:
At \(t=1\), the coordinate derivatives are
The horizontal derivative is negative, so the point moves left. The vertical derivative is positive, so it moves up. Its position is \(x(1)=1-4=-3\), \(y(1)=2+1=3\), or \((-3,3)\).
At \(t=2\),
The horizontal coordinate has zero instantaneous rate of change, while the vertical coordinate is increasing. Thus, at that parameter value, there is no instantaneous horizontal motion and the point is moving up. Its position is \(x(2)=4-8=-4\), \(y(2)=5\), or \((-4,5)\).
At \(t=3\),
Both derivatives are positive, so the point moves right and up. Its position is \(x(3)=9-12=-3\), \(y(3)=7\), or \((-3,7)\).
As a check on the horizontal change around \(t=2\), \(x(1)=-3\), \(x(2)=-4\), and \(x(3)=-3\). The coordinate decreases before \(t=2\) and increases after it, agreeing with the negative and positive values of \(x'(t)\) on those sides. Meanwhile, \(y'(t)=2\) throughout, and the listed \(y\)-coordinates increase from \(3\) to \(5\) to \(7\).
What a Zero Derivative Means
A zero coordinate derivative means that the coordinate has zero instantaneous rate of change at that parameter value. It does not automatically mean that the point stops moving, that the entire curve stops, or that the coordinate reverses direction. The other coordinate may still be changing, and even the coordinate with zero derivative may continue changing in the same direction on both sides of that value.
To determine what happens around a zero, inspect the derivative’s sign on either side, or compare coordinate values near the parameter value. A change from negative to positive indicates that the coordinate switches from decreasing to increasing; a change from positive to negative indicates that it switches from increasing to decreasing. If the sign does not change, the zero does not mark a reversal in that coordinate.
Worked Example: A Vertical Direction Change
Worked Example: A Vertical Direction Change
Let \(x(t)=3t-1\) and \(y(t)=t^2-4t\). Describe the motion before, at, and after \(t=2\).
Differentiate:
Since \(x'(t)=3>0\) for every \(t\), the point always moves right as the parameter increases.
For \(t<2\), \(t-2<0\), so \(y'(t)=2(t-2)<0\). The point moves down while moving right. At \(t=2\), \(y'(2)=2(2)-4=0\): its vertical coordinate has zero instantaneous rate of change. For \(t>2\), \(t-2>0\), so \(y'(t)>0\); the point moves up while moving right.
At the transition value, the position is
We can check the vertical behavior with nearby exact values. At \(t=1\), \(y(1)=1-4=-3\); at \(t=2\), \(y(2)=-4\); at \(t=3\), \(y(3)=9-12=-3\). The vertical coordinate decreases from \(-3\) to \(-4\), then increases from \(-4\) to \(-3\), consistent with the derivative signs. The horizontal coordinates are \(x(1)=2\), \(x(2)=5\), and \(x(3)=8\), consistent with continuous rightward motion.
Worked Example: A Zero Without a Direction Reversal
Worked Example: A Zero Without a Direction Reversal
Consider \(x(t)=t^3\) and \(y(t)=2t\). Describe the motion at \(t=0\) and compare it with the motion on either side.
The coordinate derivatives are
At \(t=0\),
The point has zero instantaneous horizontal rate of change and moves up, because the vertical derivative is positive. For \(t<0\) and \(t>0\), \(3t^2>0\), so \(x'(t)>0\) on both sides of zero. The horizontal coordinate is increasing on both sides: its derivative is zero at the single parameter value \(0\), but it does not reverse direction there.
Check with values: \(x(-1)=-1\), \(x(0)=0\), and \(x(1)=1\), so the horizontal coordinate increases across these values. Also, \(y(-1)=-2\), \(y(0)=0\), and \(y(1)=2\), confirming upward movement. In particular, a zero in one coordinate derivative does not imply that the point as a whole has stopped moving.
Reading Both Coordinates Together
The two derivative signs combine to give a concise description of the point’s instantaneous direction. One coordinate can increase while the other decreases. If one derivative is zero, describe that coordinate as having zero instantaneous rate of change rather than forcing it into a left/right or up/down category.
| Sign information | Motion as \(t\) increases |
|---|---|
| \(x'(t)>0\), \(y'(t)>0\) | Right and up |
| \(x'(t)>0\), \(y'(t)<0\) | Right and down |
| \(x'(t)<0\), \(y'(t)>0\) | Left and up |
| \(x'(t)<0\), \(y'(t)<0\) | Left and down |
| Exactly one coordinate derivative is zero | That coordinate has zero instantaneous rate; use the other derivative to describe its coordinate motion. |
| Both coordinate derivatives are zero | Both coordinates have zero instantaneous rates at that value; this fact alone does not describe the motion at nearby parameter values. |
When analyzing a parametrization, calculate both derivatives before describing the motion. Then evaluate each derivative at the requested parameter value, interpret its sign separately, and combine the two descriptions. If a derivative is zero, check nearby values before claiming that a coordinate changes direction.
Common Mistake: Treating Zero as a Reversal
A common error is to see \(x'(t_0)=0\) and conclude that the point must turn around horizontally. That conclusion is not justified by the zero alone. In the example \(x(t)=t^3\), a mistaken argument might say, “\(x'(0)=0\), so the point reverses horizontal direction at \(t=0\).” But for every \(t\ne 0\), \(x'(t)=3t^2>0\). The coordinate increases on both sides, as the values \(-1,0,1\) at \(t=-1,0,1\) confirm.
The correction is to use the derivative sign on each side of the zero, or to check nearby coordinate values. A zero signals zero instantaneous rate for that coordinate at that instant. Whether the coordinate reverses, continues in the same direction, or has some other local behavior requires more information.
Check Your Understanding
For each question, focus on coordinate motion as the parameter increases.
- If \(x'(t_0)<0\) and \(y'(t_0)>0\), in which directions does the point move?
- If \(x'(t_0)=0\) and \(y'(t_0)<0\), what can you conclude about each coordinate’s instantaneous motion?
- For \(x(t)=t^2\), calculate \(x'(0)\). Does that value alone determine whether \(x\) reverses direction at zero? Explain what the derivative signs on either side show.
- For \(x(t)=2t+1\) and \(y(t)=5-t\), find both coordinate derivatives and describe the motion as \(t\) increases.
- Why is the sign of a derivative at a single parameter value not enough, by itself, to describe coordinate motion throughout an interval?
