When Does a Particle Stop, Reverse, or Continue?
A parametric motion is described by coordinates \(x(t)\) and \(y(t)\), where \(t\) is the parameter, often interpreted as time. Its velocity components are \(x'(t)\) and \(y'(t)\). They tell us how each coordinate changes as the parameter increases.
A particle can have zero velocity at an instant and still move both before and after that instant. The key question is what happens on either side: does it retrace the path it just traveled, or does it continue through the point? A sign chart for both velocity components is a useful first test, but the coordinates may also need to be compared to confirm what the path does.
Finding a Time of Instantaneous Rest
For a parametrized position \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), define the velocity by
The speed is the magnitude of velocity, \(\sqrt{(x'(t))^2+(y'(t))^2}\). It is zero exactly when both squared terms are zero, which happens exactly when both velocity components vanish.
After finding a candidate \(t_0\), make a sign chart using intervals immediately to its left and right. A positive \(x'(t)\) means \(x\) increases as \(t\) increases; a negative \(x'(t)\) means \(x\) decreases. Interpret \(y'(t)\) in the same way. If a component keeps the same sign, that coordinate keeps moving in the same direction through \(t_0\); if its sign changes, that coordinate changes direction.
Why the Sign Chart Describes Coordinate Motion
The connection between derivative signs and coordinate change follows from the Mean Value Theorem: if a coordinate function is continuous on an interval and differentiable inside it, then its change over the interval equals its derivative at some intermediate point times the interval length. In particular, if \(x'(t)>0\) throughout an interval, then \(x\) increases there; if \(x'(t)<0\), then \(x\) decreases. The same reasoning applies to \(y\).
For example, suppose \(x'(t)<0\) just before \(t_0\) and \(x'(t)>0\) just after. As time approaches \(t_0\), \(x\) decreases toward its value at \(t_0\); after \(t_0\), it increases. So \(x\) turns around at \(t_0\). If both coordinates turn around in this way, the particle may retrace its path, but a sign chart alone does not prove that it does: the positions before and after must lie on the same local path.
Solve \(x'(t)=0\) and \(y'(t)=0\), and keep only parameter values satisfying both equations.
Determine the signs of \(x'\) and \(y'\) in intervals immediately before and after each candidate time.
Use the signs to see which coordinates turn around and which continue in the same direction.
Compare nearby positions or show that the same local part of the path is traversed in reverse order.
Worked Example: Stopping and Retracing a Path
Worked Example: Stopping and Retracing a Path
Consider \(x(t)=3+(t-2)^2\) and \(y(t)=-1+2(t-2)^2\). Find any time when the particle is instantaneously at rest and determine what happens to the motion there.
Step 1: Differentiate both coordinates.
Step 2: Find their simultaneous zero. The first equation gives \(2(t-2)=0\), so \(t=2\). Substitution into the second gives \(4(2-2)=0\). Thus both components vanish at \(t=2\), and there are no other simultaneous zeros.
Step 3: Check signs on either side. For \(t<2\), \(t-2<0\), so \(x'(t)<0\) and \(y'(t)<0\). For \(t>2\), \(t-2>0\), so \(x'(t)>0\) and \(y'(t)>0\). Both coordinates turn around at \(t=2\).
Step 4: Check whether the path is retraced. The position at \(t=2\) is \(x(2)=3\), \(y(2)=-1\). For any small \(h>0\),
The positions at equal parameter distances before and after \(2\) are identical. The particle reaches \((3,-1)\), stops instantaneously, and retraces the same local path in reverse. As a check, substituting \(t=2\) into the derivatives gives \(x'(2)=0\) and \(y'(2)=0\).
Worked Example: Passing Through While the Velocity Is Zero
Worked Example: Passing Through While the Velocity Is Zero
Consider \(x(t)=(t-1)^3\) and \(y(t)=2(t-1)^3\). Determine what happens at \(t=1\).
Step 1: Differentiate.
Both derivatives equal zero when \(t=1\), and both are positive for every \(t\ne 1\). Thus the particle is instantaneously at rest at \(t=1\), while both coordinates increase before and after that time.
Step 2: Compare positions around the time. At \(t=1\), the position is \((0,0)\). For any \(h>0\),
Before \(t=1\), the particle is on the portion of the line where both coordinates are negative; after \(t=1\), it is on the portion where both are positive. It continues through the origin rather than returning along the part of the path it came from. As a check, \(x'(1)=3(0)^2=0\) and \(y'(1)=6(0)^2=0\); on both sides the derivatives are positive.
Worked Example: One Coordinate Turns, the Other Does Not
Worked Example: One Coordinate Turns, the Other Does Not
Consider \(x(t)=(t-1)^2\) and \(y(t)=(t-1)^3\). Classify the motion at \(t=1\).
Step 1: Differentiate and find the candidate.
Both components vanish at \(t=1\), so the particle is instantaneously at rest there. For \(t<1\), \(x'(t)<0\) and \(y'(t)>0\). For \(t>1\), \(x'(t)>0\) and \(y'(t)>0\). The \(x\)-coordinate turns around, while the \(y\)-coordinate continues to increase.
Step 2: Check the nearby positions. At \(t=1\), the position is \((0,0)\). For \(h>0\),
The two positions have the same \(x\)-coordinate but opposite \(y\)-coordinates, so they are not the same point. The particle does not retrace its local path: it approaches the origin along one branch and leaves along another. It is accurate to say that \(x\) reverses while \(y\) continues to increase; it is not accurate to label the entire motion a simple reversal along the same path. The derivative check gives \(x'(1)=0\) and \(y'(1)=0\), as required.
Interpreting Sign Patterns Carefully
A sign chart gives coordinate-by-coordinate information. The following patterns are useful, provided the signs are recorded on both sides of a time when both derivatives vanish.
| Behavior of a component | Sign before and after \(t_0\) | Coordinate interpretation |
|---|---|---|
| \(x'\) changes from negative to positive | \(-\) then \(+\) | \(x\) decreases, then increases; \(x\) turns around |
| \(x'\) changes from positive to negative | \(+\) then \(-\) | \(x\) increases, then decreases; \(x\) turns around |
| \(x'\) keeps a positive sign | \(+\) then \(+\) | \(x\) continues to increase |
| \(x'\) keeps a negative sign | \(-\) then \(-\) | \(x\) continues to decrease |
Apply the same interpretations to \(y'\). If both coordinates turn around, that suggests a reversal, but confirmation requires checking whether positions on either side match or otherwise lie on the same local path. If both coordinates continue in their respective directions, the signs support continued motion through the point; comparing positions can confirm that the particle moves from one side of the point to another. If the two coordinates behave differently, describe their behavior separately instead of forcing the event into a simple reversal/continuation label.
Common Pitfall: Treating One Zero as a Full Stop
A common error is to call a time a stop as soon as one coordinate derivative is zero. For example, if \(x'(t_0)=0\) but \(y'(t_0)\ne 0\), then the velocity is \(\langle 0,y'(t_0)\rangle\), which is not the zero vector. The \(x\)-coordinate may momentarily stop changing, but the particle itself is still moving.
Another error is to decide “reversal” solely because both signs switch. The signs establish that each coordinate turns around; they do not, by themselves, establish that the particle retraces the same path. The first worked example confirms retracing by showing identical positions at \(2-h\) and \(2+h\). The third example shows why that extra check matters: \(x\) turns, but the positions on the two sides are different.
Why the Distinction Matters
Velocity describes motion, not just the shape of a curve. The same point on a parametric curve can be approached, left, retraced, or passed through in different ways. Checking the simultaneous zeros and the signs on either side distinguishes these possibilities and prevents a zero velocity from being mistaken for a complete description of the motion.
For a reliable conclusion, state three things: whether both velocity components vanish, how each component’s sign behaves on either side, and whether nearby positions show a retracing or a passage through the point. This gives a precise description even when the two coordinates behave differently.
Check Your Understanding
For each question, distinguish what the signs establish from what would require comparing positions.
- What two equations must hold for a particle to be instantaneously at rest?
- If \(x'\) changes from negative to positive but \(y'\) stays positive, what happens to each coordinate?
- Why does a zero velocity at an interior time not, by itself, imply a reversal?
- What position comparison confirms that a particle retraces its path around a candidate time \(t_0\)?
- Why is a two-sided sign check generally unavailable at a parameter endpoint?
