Position Records Where the Path Is
A parametrized curve describes a moving point by assigning a position to each value of a parameter. When the parameter is time, this records where an object is at each instant. The positions at the two endpoints tell us the net change in location, but they do not tell us how much path the object covered between them.
This distinction is important whenever a path bends, loops, or reverses direction. The endpoint difference is a vector: it records the overall change in position. Distance traveled is a nonnegative scalar: it measures the length of the route taken.
Position Vectors and Endpoint Displacement
For a curve in the plane, write its coordinates as \(x(t)\) and \(y(t)\). Its position vector is \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). At parameter value \(t\), this vector points from the origin to the point on the curve. In three dimensions, the same notation applies with a third coordinate.
Subtracting vectors means subtracting their corresponding coordinates. The result points from the starting position to the ending position. It does not, by itself, describe the route between those positions.
Distance Along the Path
Assume throughout that \(\mathbf{r}\) is continuously differentiable on \([a,b]\). Its derivative \(\mathbf{r}'(t)\) describes the rate and direction of change of position. The speed at \(t\) is the magnitude \(\|\mathbf{r}'(t)\|\); speed is nonnegative, even when the point is moving back along part of the path.
For a plane curve \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), the magnitude in this formula is \(\sqrt{(x'(t))^2+(y'(t))^2}\). Thus, calculate the two coordinate derivatives, combine them using the Euclidean norm, and integrate over the parameter interval. This formula counts all the path covered, including portions retraced later.
The parameter interval matters: it specifies which portion of the path is traversed. If the point stops momentarily, then \(\mathbf{r}'(t)=\mathbf{0}\) at that parameter value, so the instantaneous speed is zero there. A stop does not make distance negative or cancel distance accumulated elsewhere.
Worked Example: A Straight Path at Constant Speed
Worked Example: A Straight Path at Constant Speed
Let \(\mathbf{r}(t)=\langle 3t,4t\rangle\) for \(0\le t\le2\). Find the endpoint displacement, the straight-line distance between the endpoints, and the distance traveled along the path.
The starting and ending positions are $$ \mathbf{r}(0)=\langle 0,0\rangle,\qquad \mathbf{r}(2)=\langle 6,8\rangle. $$ Therefore the displacement is $$ \mathbf{r}(2)-\mathbf{r}(0) =\langle 6,8\rangle-\langle 0,0\rangle =\langle 6,8\rangle. $$ Its magnitude is $$ \sqrt{6^2+8^2}=\sqrt{36+64}=10. $$
For the path length, first differentiate: $$ \mathbf{r}'(t)=\langle 3,4\rangle. $$ The speed is constant: $$ \|\mathbf{r}'(t)\|=\sqrt{3^2+4^2}=\sqrt{25}=5. $$ Thus the distance traveled is $$ D=\int_0^2 5\,dt=5(2-0)=10. $$ The traveled distance agrees with the magnitude of displacement because the path is a straight segment traversed in one direction. As a check, the endpoint difference has length \(10\), and differentiating the position vector gives the constant speed \(5\), which over two parameter units gives \(10\).
Why Displacement Cannot Exceed Distance
The magnitude of displacement is always less than or equal to the distance traveled. The straight segment between two points is no longer than a route connecting them. The integral formula makes this relationship precise.
To prove this, apply the Fundamental Theorem of Calculus to each coordinate of \(\mathbf{r}\). It gives $$ \mathbf{r}(b)-\mathbf{r}(a)=\int_a^b \mathbf{r}'(t)\,dt. $$ If the displacement is nonzero, let \(\mathbf{u}\) be the unit vector in its direction: \(\mathbf{u}=(\mathbf{r}(b)-\mathbf{r}(a))/\|\mathbf{r}(b)-\mathbf{r}(a)\|\). Then $$ \|\mathbf{r}(b)-\mathbf{r}(a)\| =\mathbf{u}\mathbin{\cdot}\int_a^b\mathbf{r}'(t)\,dt =\int_a^b\mathbf{u}\mathbin{\cdot}\mathbf{r}'(t)\,dt \le\int_a^b\|\mathbf{r}'(t)\|\,dt. $$ The last inequality follows from the Cauchy–Schwarz inequality: since \(\|\mathbf{u}\|=1\), we have \(\mathbf{u}\mathbin{\cdot}\mathbf{r}'(t)\le\|\mathbf{r}'(t)\|\). If the displacement is zero, the claimed inequality follows because the integral of speed is nonnegative. This proves the result in both cases.
The proof also highlights the different kinds of quantities involved. Integrating the derivative gives the net vector change, because changes in opposite directions can cancel. Integrating its magnitude gives distance, because the magnitude is never negative and therefore does not cancel motion in another direction.
Worked Example: Returning to the Starting Point
Worked Example: Returning to the Starting Point
Consider the path \(\mathbf{r}(t)=\langle t^2-1,0\rangle\) for \(-1\le t\le1\). Find its displacement and distance traveled.
At both endpoints, the position is the origin: $$ \mathbf{r}(-1)=\langle (-1)^2-1,0\rangle=\langle 0,0\rangle, \qquad \mathbf{r}(1)=\langle 1^2-1,0\rangle=\langle 0,0\rangle. $$ So the displacement is $$ \mathbf{r}(1)-\mathbf{r}(-1)=\langle 0,0\rangle. $$
Differentiate the position: $$ \mathbf{r}'(t)=\langle 2t,0\rangle, \qquad \|\mathbf{r}'(t)\|=\sqrt{(2t)^2}=2|t|. $$ The absolute value changes form at \(t=0\), so split the distance integral there: $$ D=\int_{-1}^0(-2t)\,dt+\int_0^1 2t\,dt =\big[-t^2\big]_{-1}^0+\big[t^2\big]_0^1 =1+1=2. $$ The point moves from \(x=0\) to \(x=-1\), then returns to \(x=0\). The total distance is \(2\), although the displacement is the zero vector. As a check, the position at each endpoint is identical, while the speed integral is positive and gives the two one-unit portions of the path.
Worked Example: A Portion of a Circle
Worked Example: A Portion of a Circle
Let \(\mathbf{r}(t)=\langle\cos t,\sin t\rangle\) for \(0\le t\le 3\pi/2\). Find the displacement and distance traveled.
The endpoint positions are $$ \mathbf{r}(0)=\langle 1,0\rangle,\qquad \mathbf{r}(3\pi/2)=\langle 0,-1\rangle. $$ Therefore $$ \mathbf{r}(3\pi/2)-\mathbf{r}(0) =\langle -1,-1\rangle, \qquad \|\langle -1,-1\rangle\|=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}. $$
The derivative and speed are $$ \mathbf{r}'(t)=\langle-\sin t,\cos t\rangle, \qquad \|\mathbf{r}'(t)\| =\sqrt{\sin^2t+\cos^2t}=1. $$ Consequently, $$ D=\int_0^{3\pi/2}1\,dt=\frac{3\pi}{2}. $$ As a check, the derivative has unit magnitude at every parameter value, so the distance equals the interval length \(3\pi/2\). This is greater than the endpoint separation \(\sqrt{2}\), as the theorem requires: the path follows a circular arc rather than the straight segment between its endpoints.
Common Mistake: Treating the Endpoint Difference as the Distance
A frequent error is to calculate the displacement vector and then report its magnitude as the distance traveled. That works for some paths, such as a straight segment traversed without reversing, but it does not work in general.
A useful check is to ask whether the computed quantity is a vector or a scalar. Displacement is a vector, while distance is a nonnegative scalar. If the route bends or retraces itself, compare the accumulated distance with the magnitude of the endpoint displacement rather than assuming they are equal.
When the Distinction Is Useful
Displacement answers, “Where did the point end up relative to where it started?” Distance answers, “How much path did it cover?” These questions have different uses. Endpoint displacement summarizes net change; distance measures the full route and cannot be reduced by a return trip or loop.
For any continuously differentiable path, distance is at least the magnitude of displacement. Equality holds for a straight path traversed without backtracking; curved routes and reversals generally make the traveled distance larger. In the examples, the straight path had equal values, while the returning path and circular arc had strictly greater traveled distances.
Check Your Understanding
For each question, distinguish the endpoint vector change from the accumulated path length.
- State the formula for the displacement of \(\mathbf{r}(t)\) on \([a,b]\).
- State the formula for distance traveled, and identify the quantity integrated.
- If a path starts and ends at the same point, what must its displacement be? Does that determine its distance traveled?
- Why is the magnitude of displacement no greater than the distance traveled?
- For \(\mathbf{r}(t)=\langle 2t,0\rangle\) on \(0\le t\le3\), calculate both the displacement and distance.