Reading Motion from Derivative Information
A parametric description gives an object's position through two coordinate functions, \(x(t)\) and \(y(t)\), where \(t\) is time. Sometimes a problem provides formulas for these functions. In other problems, it gives only a table of coordinate derivatives or a sign chart. Either way, \(x'(t)\) and \(y'(t)\) tell us how the object's horizontal and vertical coordinates are changing.
This information is enough to describe the direction of motion in each coordinate and to calculate velocity and speed at specified times. It does not always reveal the object's exact position or the path's shape. A careful description should say what the derivative information establishes—and avoid claiming more than it does.
What the Signs Tell You
A positive derivative means the corresponding coordinate is increasing; a negative derivative means it is decreasing. Thus, \(x'(t)>0\) means motion to the right, and \(x'(t)<0\) means motion to the left. Similarly, \(y'(t)>0\) means motion upward, and \(y'(t)<0\) means motion downward.
| Sign information | Coordinate behavior |
|---|---|
| \(x'(t)>0\) | Moving to the right |
| \(x'(t)<0\) | Moving to the left |
| \(y'(t)>0\) | Moving upward |
| \(y'(t)<0\) | Moving downward |
| \(x'(t)=0\) | No instantaneous horizontal motion |
| \(y'(t)=0\) | No instantaneous vertical motion |
Combining the signs gives the direction of motion in the plane. For instance, \(x'(t)>0\) and \(y'(t)<0\) mean the object moves right and down. If one component is zero while the other is nonzero, the instantaneous velocity is vertical or horizontal. If both components are zero, the velocity vector is zero and the object is instantaneously at rest.
A zero derivative at one time does not by itself tell us what happens immediately before or after that time. To determine whether a coordinate changes from increasing to decreasing, or the reverse, inspect the signs on the intervals on either side. This is why sign charts are useful: they organize the derivative information across an interval of time.
Why a Derivative Sign Gives a Direction
The derivative \(x'(t)\) is the instantaneous rate of change of the horizontal coordinate, and \(y'(t)\) is the instantaneous rate of change of the vertical coordinate. For a small positive time change \(\Delta t\), the coordinate changes are approximately
Because \(\Delta t>0\), the sign of each approximate coordinate change is the sign of the corresponding derivative. A positive horizontal change is rightward; a negative horizontal change is leftward. The same reasoning applies vertically. The approximation becomes exact as a statement about instantaneous rates in the limit that \(\Delta t\) approaches zero.
Speed uses both components because the magnitude of a vector is determined by its perpendicular coordinate components. Squaring removes their signs, so speed is never negative, even when the object moves left, down, or in both directions.
Worked Example: Describing Motion from a Sign Chart
Worked Example: Describing Motion from a Sign Chart
Suppose an object moves for \(0\le t\le 6\), and a sign chart gives the following signs on the open intervals between critical times:
| Time interval | \(x'(t)\) | \(y'(t)\) |
|---|---|---|
| \(0<t<2\) | Positive | Positive |
| \(2<t<4\) | Negative | Positive |
| \(4<t<6\) | Negative | Negative |
Describe the direction of motion on each interval and explain what the sign changes indicate.
For \(0<t<2\), \(x'(t)>0\), so the object moves right. Also, \(y'(t)>0\), so it moves up. It therefore moves right and up.
For \(2<t<4\), \(x'(t)<0\) and \(y'(t)>0\). The object moves left and up.
For \(4<t<6\), both derivatives are negative. The object moves left and down.
At \(t=2\), the sign of \(x'(t)\) changes from positive to negative, so the horizontal coordinate switches from increasing to decreasing. The chart does not specify \(y'(2)\), so it does not establish whether the object is moving vertically at that instant. At \(t=4\), the sign of \(y'(t)\) changes from positive to negative, but the chart likewise does not by itself give the derivative values exactly at \(t=4\).
The conclusions for the open intervals follow directly from the listed signs. The chart identifies changes in coordinate behavior across \(t=2\) and \(t=4\), but a claim about the exact velocity at either boundary requires derivative values there.
Tables Give Values at Times, Not Automatically Between Them
A table may list \(x'(t)\) and \(y'(t)\) at only a few individual times. These entries let you determine velocity and speed at those listed times. By themselves, they do not prove the derivative signs throughout the intervals between them.
For example, knowing that \(x'(1)>0\) and \(x'(3)>0\) does not establish that \(x'(t)>0\) for every \(t\) between 1 and 3. The derivative could change sign in between. To describe motion over an entire interval, use a sign chart or other information that establishes the signs across that interval. When only isolated table values are supplied, limit the conclusion to those times.
Worked Example: Comparing Velocity and Speed at Selected Times
Worked Example: Comparing Velocity and Speed at Selected Times
A table gives the coordinate derivatives for a moving object. The units of each derivative are meters per second.
| Time \(t\) (seconds) | \(x'(t)\) (m/s) | \(y'(t)\) (m/s) |
|---|---|---|
| 1 | 3 | 4 |
| 3 | \(-3\) | 4 |
| 5 | \(-3\) | \(-4\) |
Find the velocity and speed at each listed time. Compare the results.
At \(t=1\), the velocity is
Its speed is
At \(t=3\), the velocity is
Its speed is
At \(t=5\), the velocity is
Its speed is
All three speeds are \(5\text{ m/s}\), but the velocity vectors are different. At \(t=1\), the object moves right and up; at \(t=3\), it moves left and up; at \(t=5\), it moves left and down. Substituting each velocity's components into the magnitude formula verifies the speed in every case.
Interpreting Zero Components Carefully
A zero component is not the same as zero speed. If \(x'(t)=0\) but \(y'(t)\ne 0\), then
The object has no instantaneous horizontal motion, but it is still moving vertically. Likewise, if \(y'(t)=0\) and \(x'(t)\ne 0\), the object moves horizontally with positive speed \(|x'(t)|\). Only when both derivatives are zero is the speed zero.
Worked Example: A Zero Component Is Not a Stop
At \(t=2\), suppose \(x'(2)=0\text{ cm/s}\) and \(y'(2)=-6\text{ cm/s}\). Determine the velocity, speed, and direction.
The velocity vector is
The horizontal component is zero, and the vertical component is negative, so the object is moving downward with no instantaneous horizontal motion. Its speed is
Checking the result, the magnitude of \(\langle 0,-6\rangle\) is \(6\), not zero. Thus the object is moving, and its speed is \(6\text{ cm/s}\).
A Common Mistake: Treating Velocity as Speed
Velocity is a vector with a direction; speed is a nonnegative number. A common error is to add the velocity components, or to report a negative component as the speed.
For example, with \(\mathbf{v}=\langle -3,4\rangle\text{ m/s}\), an incorrect calculation might be
This is wrong: adding components does not calculate the vector's magnitude. The correct calculation is
The velocity remains \(\langle -3,4\rangle\text{ m/s}\), which describes left-and-up motion. Its speed is \(5\text{ m/s}\). Keep the vector and its magnitude separate, and include units appropriate to each.
A Reliable Reading Process
Check whether the data give values at individual times or signs across entire intervals.
Use \(x'(t)\) for left-right motion and \(y'(t)\) for up-down motion.
State the two coordinate directions together, such as right and down.
Square both components, add, and take the nonnegative square root.
A zero component removes motion in one coordinate; only two zero components imply zero instantaneous speed.
Why These Descriptions Matter
A parametric motion problem may ask for more than a numerical velocity. It may ask whether the object moves left or right on an interval, whether it changes vertical direction, or whether two moments have the same speed. Derivative signs answer directional questions; derivative values give the velocity components; the magnitude formula gives speed.
These are related but distinct conclusions. Two objects—or the same object at two different times—can have equal speeds and different velocities. A sign chart can establish motion directions over intervals, but isolated table entries cannot generally establish those interval-wide claims. Keeping these distinctions clear makes descriptions accurate and prevents unsupported conclusions.
Check Your Understanding
Use derivative signs or values precisely, and distinguish interval conclusions from conclusions at a single time.
- If \(x'(t)<0\) and \(y'(t)>0\) throughout an interval, describe the direction of motion.
- If \(x'(t)=0\) and \(y'(t)=5\text{ ft/s}\) at a particular time, find the velocity, speed, and direction.
- At one time the velocity is \(\langle 6,8\rangle\text{ m/s}\). Calculate the speed, showing the intermediate arithmetic.
- Can a table showing \(x'(1)>0\) and \(x'(4)>0\) alone establish that \(x'(t)>0\) for all \(1<t<4\)? Explain.
- If \(x'(t)=0\) and \(y'(t)=0\) at a time, what can you conclude about the instantaneous velocity and speed? Does that information alone establish the motion immediately before or after that time?
