Why These Errors Are Easy to Make
Parametric motion describes position through coordinate functions such as \(x(t)\) and \(y(t)\). The same derivatives appear in several different questions: how fast an object moves, how far it travels, how much its position changes, and how long a curve segment is. These quantities are related, but they are not interchangeable.
A reliable solution begins by identifying what the question asks for. A velocity component can be negative; speed cannot. Displacement measures the net change in position; distance adds up the motion along the way. Arc length requires the magnitude of the velocity vector and parameter bounds that trace the intended segment.
Velocity Components Are Not Speed
The derivatives \(x'(t)\) and \(y'(t)\) are the horizontal and vertical components of velocity. Each component describes change in one coordinate per unit of time. Speed is the magnitude of the velocity vector, so it combines both components and is never negative.
This distinction follows from the distance formula. Over a small time change \(\Delta t\), the changes in the coordinates are approximately \(x'(t)\Delta t\) and \(y'(t)\Delta t\). The corresponding straight-line distance is approximately
Dividing by the positive time change \(|\Delta t|\) gives the rate of distance traveled: the magnitude of velocity. This is why adding the components, or using only one of them, generally gives the wrong speed.
Worked Example: Finding Speed from Both Components
Let \(x(t)=t^2\) and \(y(t)=3t\). Find the speed at \(t=1\).
First differentiate each coordinate:
At \(t=1\), the velocity components are \(x'(1)=2\) and \(y'(1)=3\). Apply the magnitude formula:
The speed is \(\sqrt{13}\) units per unit of time. Check: squaring the speed gives \(13\), which equals \(2^2+3^2\), the sum of the squared velocity components. The answer is positive, as speed must be.
Displacement Is Not Distance Traveled
For motion from time \(a\) to time \(b\), the displacement vector is the final position minus the initial position. It records net change, including direction. Distance traveled is the total length of the path, regardless of whether the motion later reverses and returns toward its starting point.
When velocity is continuous, the distance formula accumulates a nonnegative speed over the time interval. The displacement depends only on the two endpoint positions. Consequently, an object can travel a positive distance while having zero displacement.
Worked Example: Motion That Reverses Direction
Consider motion along the horizontal line given by \(x(t)=(t-1)^2\), \(y(t)=0\), for \(0\leq t\leq2\). Find the displacement and distance traveled.
The initial and final positions are
Therefore the displacement is
For distance, differentiate: \(x'(t)=2(t-1)\) and \(y'(t)=0\). Thus the speed is
The expression changes form at \(t=1\), so split the integral there:
The object moves from \(x=1\) to \(x=0\), then back to \(x=1\): it travels \(1+1=2\) units while its displacement is zero. The integral confirms the total distance, and the endpoint subtraction confirms the displacement.
Do Not Omit the Square Root
For a parametrized curve \(x=x(t)\), \(y=y(t)\), the arc length from \(t=a\) to \(t=b\) is obtained by accumulating the magnitude of the change in position. When \(x\) and \(y\) are continuously differentiable on \([a,b]\), the formula is
The square root is essential: it turns the sum of squared coordinate rates into the magnitude of the velocity vector. Without it, the integrand is not speed and does not represent distance per unit of parameter. The integrand is also not usually \(x'(t)+y'(t)\); those signed components can cancel even when the curve is moving.
Worked Example: Detecting a Missing Square Root
Find the length of the quarter-circle parametrization \(x(t)=3\cos t\), \(y(t)=3\sin t\), for \(0\leq t\leq \frac{\pi}{2}\).
Differentiate both coordinates:
Substitute into the speed formula and simplify using \(\sin^2t+\cos^2t=1\):
Therefore the length is
Check the segment: the endpoints are \((3,0)\) and \((0,3)\), so the parameter interval traces one quarter of the circle of radius \(3\). Its length \(3\pi/2\) is one quarter of the full circumference \(6\pi\).
Bounds Must Trace the Requested Segment
The parameter bounds are part of the problem, not a detail to choose by convenience. The values \(a\) and \(b\) specify which portion of the parametrization is being measured. In the quarter-circle example, \([0,\pi/2]\) traces one quarter of the circle. Replacing those bounds with \([0,2\pi]\) would measure the entire circle, not the requested quarter.
Before integrating, evaluate the parametrization at the stated bounds and compare those endpoints with the requested segment. If a parametrization retraces a path, a longer parameter interval can count the same geometric portion more than once. The integral measures the length traced over the parameter interval, including repeated traversal.
Decide whether the question asks for a velocity component, speed, displacement, distance, or arc length.
Use endpoint differences for displacement and the magnitude of velocity for speed or accumulated distance.
Evaluate the endpoints and ensure the interval traces exactly the motion or curve segment in the question.
Confirm that speed and distance are nonnegative and that units and endpoints make sense.
A Common Wrong Calculation
Suppose a solution finds the velocity components \(x'(t)=2\) and \(y'(t)=3\) and writes the speed as \(2+3=5\). This is incorrect because the components are perpendicular coordinate changes, not consecutive distances along one line. A second common error is to write \(2^2+3^2=13\) as the speed, omitting the square root.
The correction is to take the magnitude of the velocity vector:
As a separate check, the horizontal and vertical components cannot be substituted for total distance traveled over an interval unless the motion is known to be confined to that single coordinate direction and the appropriate absolute change is used. In general, integrate speed. Likewise, do not replace a distance integral by the magnitude of the displacement: that shortcut loses any motion that reverses or curves between the endpoints.
Use Checks to Catch Errors
Several quick checks expose many mistakes before they affect the final answer. Speed and distance cannot be negative. A displacement component can be negative because it includes direction. The speed integrand must include both coordinate derivatives, squares, and a square root. The bounds must correspond to the portion of the motion being measured.
| Quantity | How to calculate it | Useful check |
|---|---|---|
| Velocity | \((x'(t),y'(t))\) | Components can have either sign. |
| Speed | \(\sqrt{(x'(t))^2+(y'(t))^2}\) | It is nonnegative. |
| Displacement | Final position minus initial position | It depends only on endpoints. |
| Distance | Integral of speed over the interval | It includes reversals and is nonnegative. |
| Arc length | Integral of speed over the curve's parameter interval | The bounds must trace the intended segment. |
Check Your Understanding
For each question, identify the quantity being requested before calculating.
- If \(x'(t)=-4\) and \(y'(t)=3\), what is the speed? Why is \(-4\) not the speed?
- An object starts and ends at the same position but moves in between. What are its displacement and distance likely to be?
- What part of the arc length integrand is missing from \(\int_a^b ((x'(t))^2+(y'(t))^2)\,dt\)?
- A parametrization describes a full circle on \(0\leq t\leq2\pi\). What could go wrong if the requested curve segment is only a quarter-circle?
- For \(x(t)=t\), \(y(t)=0\), on \(1\leq t\leq4\), calculate the displacement and distance, and explain why they are different types of quantities.
