From Speed to Accumulated Length
For a parametrized curve, the coordinates describe where the moving point is, while its speed describes how quickly it moves along the curve. Accumulated arc length combines that speed over a parameter interval to measure the total distance traveled along the path.
This is different from measuring the straight-line distance between two positions. A point can travel along a curved path, or reverse direction, and cover more distance than its displacement suggests. The accumulated-length function keeps track of the distance traveled from a chosen starting parameter.
Definition and Conditions
Let a curve be parametrized by \(x=x(t)\) and \(y=y(t)\) on an interval containing a starting parameter \(a\). Its speed is \(v(t)=\sqrt{[x'(t)]^2+[y'(t)]^2}\). We will assume throughout that \(x\) and \(y\) are continuously differentiable on the parameter interval under consideration. Then the speed is continuous, so its integral over every closed subinterval is defined and finite.
The variable \(u\) is a dummy variable of integration; \(t\) is the parameter at which the accumulated length is evaluated. The function begins at zero because there has been no travel at the starting parameter: \(s(a)=\int_a^a v(u)\,du=0\).
The same integral can be defined for \(t<a\), but \(\int_a^t v(u)\,du\) is then nonpositive because of the orientation of a definite integral (strictly negative if the speed is positive somewhere between \(t\) and \(a\)). To represent distance accumulated while moving from \(a\) toward a smaller parameter, one can instead define \(\int_t^a v(u)\,du\). In the examples and formulas below, the starting point comes first, so \(t\ge a\).
The Fundamental Theorem and the Rate of Accumulation
Since the speed is continuous, the Fundamental Theorem of Calculus applies to the defining integral. It says that differentiating the accumulated integral returns the integrand:
Thus the derivative of accumulated arc length is the instantaneous speed. The function \(s(t)\) records total distance accumulated since \(a\), while \(s'(t)\) records the rate at which that total is increasing at the current parameter value.
Here is why the derivative formula holds. For a nonzero increment \(h\) small enough that \(t+h\) is in the interval, subtract the two accumulated lengths. The common part of the integrals cancels:
Subtract \(v(t)\) from both sides. By the integral inequality, the absolute value of the difference is at most the largest value of \(|v(u)-v(t)|\) between \(t\) and \(t+h\):
Continuity of \(v\) at \(t\) means the right-hand side approaches zero as \(h\) approaches zero. Therefore the difference quotient approaches \(v(t)\), proving \(s'(t)=v(t)\). This argument also works for negative \(h\), because the orientation of the integral and the sign of \(h\) are both included in the quotient.
Worked Example: A Straight-Line Parametrization
Worked Example: Accumulated Length at Constant Speed
A point follows \(x(t)=3t\), \(y(t)=4t\) for \(0\le t\le2\). Find its accumulated arc length \(s(t)\) from \(a=0\), and find the length traveled from \(t=\frac12\) to \(t=\frac32\).
First calculate the coordinate derivatives: \(x'(t)=3\) and \(y'(t)=4\). Therefore the speed is
Integrate the speed from the starting parameter:
The interval length is the difference of accumulated lengths:
Check the accumulated-length formula: \(s(0)=0\), as required, and \(s'(t)=5=v(t)\). The length over the requested interval is also speed times elapsed parameter: \(5\left(\frac32-\frac12\right)=5\).
Length Between Two Parameter Values
The accumulated function is based at one fixed parameter, but it also gives the length between any two later parameter values. If \(a\le c\le d\), then
Because speed is nonnegative, this difference is the distance traveled along the curve as the parameter runs from \(c\) to \(d\). This subtraction rule is useful when a problem asks for length over a subinterval rather than from the original starting point.
Worked Example: A Path That Reverses Direction
Worked Example: Accumulated Length on Both Sides of a Turn
Consider \(x(t)=t^2\), \(y(t)=0\) for \(-1\le t\le2\), with starting parameter \(a=-1\). Find \(s(t)\) and the total length traveled by \(t=2\).
The derivatives are \(x'(t)=2t\) and \(y'(t)=0\), so the speed is
The absolute value changes form at \(t=0\), so compute the integral separately on each side. For \(-1\le t\le0\), \(|u|=-u\):
For \(0\le t\le2\), split the accumulation at zero:
The total length at \(t=2\) is \(s(2)=1+2^2=5\). Check both pieces: they agree at zero, with \(s(0)=1\); their derivatives are \(-2t=2|t|\) for \(-1<t<0\) and \(2t=2|t|\) for \(0<t<2\). At the start, \(s(-1)=1-(-1)^2=0\). The accumulated length increases even while the position moves back toward the origin.
Worked Example: A Curved Path
Worked Example: Length on a Subinterval of a Parametrized Curve
For \(0\le t\le\pi\), let \(x(t)=t-\sin t\) and \(y(t)=1-\cos t\). Find the accumulated length from \(0\), and then find the length traveled from \(t=\frac{\pi}{3}\) to \(t=\pi\).
Differentiate each coordinate: \(x'(t)=1-\cos t\) and \(y'(t)=\sin t\). The speed simplifies as follows:
Using \(2-2\cos t=4\sin^2(t/2)\), and \(\sin(t/2)\ge0\) for \(0\le t\le\pi\), this becomes \(v(t)=2\sin(t/2)\). Therefore
Evaluate at the two parameter values: \(s(\pi)=4(1-\cos(\pi/2))=4\), and \(s(\pi/3)=4(1-\cos(\pi/6))=4(1-\frac{\sqrt3}{2})=4-2\sqrt3\). The requested length is
Check by differentiating: \(s'(t)=4\cdot\frac12\sin(t/2)=2\sin(t/2)=v(t)\), and \(s(0)=4(1-1)=0\). The subinterval length is positive, as it must be.
Distance Traveled Is Not Displacement
Displacement compares the final and initial positions directly. Distance traveled measures the length of the path between them. In the reversal example, the point starts at \(x(-1)=1\) and ends at \(x(2)=4\), so its displacement is \(4-1=3\). But it travels from \(1\) to \(0\), then from \(0\) to \(4\), for a total distance of \(1+4=5\). The accumulated-length function gives that total as \(s(2)=5\).
A common error is to use the magnitude of displacement as the traveled length. For this same example, the incorrect calculation would be
That is the straight-line distance between the endpoints, not the distance along the path. The correction is to integrate speed, or equivalently to use the accumulated function: \(s(2)-s(-1)=5-0=5\).
Why the Accumulated Function Is Useful
Once \(s(t)\) is known, it answers several questions consistently. Its value gives distance traveled from the chosen starting parameter, its derivative gives the instantaneous speed, and differences of its values give lengths over subintervals. In particular, \(s\) is nondecreasing because \(s'(t)=v(t)\ge0\) wherever the derivative formula applies.
The regularity condition matters. Continuous differentiability of the coordinates ensures that speed is continuous and the accumulation integral is finite on a closed interval. Differentiability alone does not guarantee that the speed has a finite integral. If the coordinates have weaker regularity, additional hypotheses are needed before interpreting an integral as the curve's geometric length or applying the derivative conclusion at a point.
Check Your Understanding
Use the accumulated-length definition and the Fundamental Theorem of Calculus to answer these questions.
- If \(s(t)=\int_a^t v(u)\,du\) and \(v\) is continuous, what is \(s'(t)\)?
- For \(x(t)=t\), \(y(t)=2t\), find the speed and the length traveled from \(t=0\) to \(t=3\).
- Why is \(s(d)-s(c)\) the length traveled from \(c\) to \(d\), when \(a\le c\le d\)?
- For \(x(t)=t^2\), \(y(t)=0\), explain why the length from \(t=-1\) to \(t=2\) is not the magnitude of the displacement.
- What regularity assumption in this tutorial ensures the speed integral is defined and finite on a closed parameter interval?
