The Cumulative Arc-Length Function

From Speed to Distance Accumulated

A parametrized curve describes a path, but its parameter does not generally measure distance along that path. Equal changes in the parameter can correspond to different amounts of travel. To record how much distance has accumulated since a chosen starting point, integrate the curve’s speed.

The resulting function is called the cumulative arc-length function. Its value depends on where the measurement starts, while its change between two parameter values gives the length of the intervening part of the curve.

Definition

Let \(\mathbf{r}: [a,b]\to\mathbb{R}^n\) be a continuously differentiable parametrized curve, and let \(t_0\in[a,b]\) be a fixed starting parameter. The speed at parameter \(u\) is the magnitude of the velocity vector, \(\|\mathbf{r}'(u)\|\). For \(t\geq t_0\), define the cumulative arc length from \(t_0\) to \(t\) by integrating that speed.

Definition: For \(t_0\leq t\leq b\), the cumulative arc-length function starting at \(t_0\) is

s(t)=\int_{t_0}^{t}\|\mathbf{r}'(u)\|\,du.

Here \(u\) is a dummy variable of integration. The value \(s(t)\) is the distance traveled along the curve from \(\mathbf{r}(t_0)\) to \(\mathbf{r}(t)\), with the parameter moving forward from \(t_0\) to \(t\).

The integrand is the speed, not the velocity vector itself. The speed is nonnegative, so the integral adds the small amounts of distance traveled along the curve. The standard arc-length formula gives the length over any interval \([t_1,t_2]\) as the integral of speed over that interval, provided the curve is continuously differentiable there.

If values are also considered for \(t<t_0\), the oriented integral \(\int_{t_0}^{t}\|\mathbf{r}'(u)\|\,du\) is negative. That signed value is not the distance traveled from \(t_0\) to \(t\). The distance in that direction is \(\int_t^{t_0}\|\mathbf{r}'(u)\|\,du\), which is nonnegative. In this tutorial, “cumulative” refers to moving forward from the chosen starting parameter.

Why the Integral Measures Accumulated Length

Over a short parameter interval from \(u\) to \(u+\Delta u\), the curve travels approximately the distance \(\|\mathbf{r}'(u)\|\Delta u\), when \(\Delta u>0\). Adding these local distances over a partition of an interval leads to the arc-length integral. In other words, the integral accumulates the instantaneous rate of distance traveled.

This also explains how to find the rate at which cumulative length changes. Since \(\mathbf{r}'\) is continuous, its magnitude is continuous. The Fundamental Theorem of Calculus therefore applies to the defining integral:

Theorem: If \(\mathbf{r}\) is continuously differentiable, then its cumulative arc-length function is differentiable and

s'(t)=\|\mathbf{r}'(t)\|.

Thus the rate of change of accumulated distance with respect to the parameter is the speed.

This identity also verifies the meaning of the function: starting from \(s(t_0)=0\), the function increases at exactly the rate at which the curve is being traversed. If the speed is zero at an isolated point, \(s'(t)\) is zero there. If speed is zero throughout an interval, the curve does not move along that interval and cumulative length remains constant there.

Differentiate the position.
Find \(\mathbf{r}'(t)\), the velocity vector.

Find the speed.
Compute the magnitude \(\|\mathbf{r}'(t)\|\).

Integrate from the chosen start.
Evaluate \(\int_{t_0}^{t}\|\mathbf{r}'(u)\|\,du\), keeping the lower limit tied to the starting parameter.

Check the result.
Confirm \(s(t_0)=0\), \(s'(t)=\|\mathbf{r}'(t)\|\), and \(s(t)\geq 0\) for \(t\geq t_0\).

Worked Examples

Worked Example: A Curve with Constant Speed

Consider the helix \(\mathbf{r}(t)=(3\cos t,3\sin t,4t)\) for \(0\leq t\leq 2\pi\), starting at \(t_0=0\). Find its cumulative arc-length function.

Differentiate each component:

\mathbf{r}'(t)=(-3\sin t,3\cos t,4).

The speed is the magnitude of this vector. Using \(\sin^2t+\cos^2t=1\),

\|\mathbf{r}'(t)\| =\sqrt{9\sin^2t+9\cos^2t+16} =\sqrt{9+16} =5.

Since the speed is constant, integrate \(5\) from the starting parameter to \(t\):

s(t)=\int_0^t 5\,du=5t.

At \(t=0\), \(s(0)=0\), as required. Differentiating gives \(s'(t)=5\), which agrees with the speed computed from the velocity. At \(t=2\pi\), the accumulated length is \(s(2\pi)=10\pi\).

Worked Example: A Curve with Variable Speed

Let \(\mathbf{r}(t)=(t,t^2/2)\) for \(0\leq t\leq 2\), with starting parameter \(t_0=0\). Find \(s(t)\). Unlike the previous example, the speed changes with \(t\).

The velocity and its magnitude are

\mathbf{r}'(t)=(1,t), \qquad \|\mathbf{r}'(t)\|=\sqrt{1+t^2}.

Therefore, the cumulative length is

s(t)=\int_0^t\sqrt{1+u^2}\,du.

An antiderivative of \(\sqrt{1+u^2}\) is \(\frac12\bigl(u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})\bigr)\). Evaluating at the bounds gives

s(t)=\frac12\left(t\sqrt{1+t^2}+\ln(t+\sqrt{1+t^2})\right) -\frac12\ln(1) =\frac12\left(t\sqrt{1+t^2}+\ln(t+\sqrt{1+t^2})\right).

To check the formula, differentiate it. The derivative of \(t\sqrt{1+t^2}\) is \(\sqrt{1+t^2}+t^2/\sqrt{1+t^2}\), and the derivative of \(\ln(t+\sqrt{1+t^2})\) is \(1/\sqrt{1+t^2}\). Hence

s'(t) =\frac12\left(\frac{1+2t^2}{\sqrt{1+t^2}}+ \frac{1}{\sqrt{1+t^2}}\right) =\frac{1+t^2}{\sqrt{1+t^2}} =\sqrt{1+t^2}.

This is exactly the speed. Also, substituting \(t=0\) into the formula gives \(s(0)=\frac12\ln(1)=0\). At \(t=2\), the accumulated length is \(s(2)=\sqrt{5}+\frac12\ln(2+\sqrt{5})\).

Worked Example: Changing the Starting Parameter

Take \(\mathbf{r}(t)=(t^2/2,t^2/2,0)\) on \(0\leq t\leq 3\), but now start measuring at \(t_0=1\). Find the cumulative arc length for \(1\leq t\leq 3\).

Differentiating and taking the magnitude gives

\mathbf{r}'(t)=(t,t,0), \qquad \|\mathbf{r}'(t)\|=\sqrt{t^2+t^2}=\sqrt{2}\,t,

where \(t\geq 1\), so \(\sqrt{t^2}=t\). Integrate from \(1\), not from \(0\):

s(t)=\int_1^t \sqrt{2}\,u\,du =\frac{\sqrt{2}}{2}(t^2-1).

At the starting parameter, \(s(1)=\frac{\sqrt{2}}{2}(1-1)=0\). Differentiation gives \(s'(t)=\sqrt{2}\,t\), the speed. At \(t=3\), \(s(3)=\frac{\sqrt{2}}{2}(9-1)=4\sqrt{2}\). If the measurement had instead started at \(0\), the value at \(t=3\) would be \(\frac{9\sqrt{2}}{2}\). The difference comes only from the choice of starting point.

How Values of the Function Relate to Curve Length

For any two parameters \(t_1\) and \(t_2\) with \(t_0\leq t_1\leq t_2\), subtracting the two defining integrals gives

s(t_2)-s(t_1) =\int_{t_0}^{t_2}\|\mathbf{r}'(u)\|\,du -\int_{t_0}^{t_1}\|\mathbf{r}'(u)\|\,du =\int_{t_1}^{t_2}\|\mathbf{r}'(u)\|\,du.

The final integral is the arc length of the curve segment between those parameter values. Thus \(s(t_2)-s(t_1)\) is the distance traveled over that segment. This difference formula is often useful when a problem asks for length over a later part of the motion, rather than from the original starting point.

Because speed is nonnegative, \(s\) is nondecreasing as \(t\) increases. If the speed is positive throughout an interval, then \(s\) is strictly increasing on that interval. If the speed vanishes over some part of the parameter interval, the function may be constant over that part. Neither conclusion requires the curve to move at constant speed.

A Common Mistake: Using Displacement Instead of Distance

Displacement is the change in position vector. It records the net change from the beginning to the end, not the total distance traveled along the curve. The two quantities can differ because a curve may bend or loop.

For the helix in the first example, the displacement from \(t=0\) to \(t=\pi/2\) is

\mathbf{r}(\pi/2)-\mathbf{r}(0)=(-3,3,2\pi).

Its magnitude is \(\sqrt{18+4\pi^2}\), whereas the distance traveled along the helix is \(s(\pi/2)=5\pi/2\). These are not the same quantity. For example, the square of the displacement magnitude is \(18+4\pi^2\), while the square of the arc length is \(25\pi^2/4\); their difference is \(9\pi^2/4-18>0\), since \(\pi^2>8\).

A related error is to integrate the velocity vector and treat the result as the arc length. The integral \(\int_{t_0}^{t}\mathbf{r}'(u)\,du\) is the displacement vector \(\mathbf{r}(t)-\mathbf{r}(t_0)\). Arc length instead requires integrating the scalar speed \(\|\mathbf{r}'(u)\|\). The magnitude must be taken before integration.

Why the Starting Point Matters

The curve and its speed determine how quickly \(s(t)\) changes, but they do not determine where the function equals zero. The chosen starting parameter supplies that information: \(s(t_0)=0\). Changing the lower limit changes the accumulated values, although the difference \(s(t_2)-s(t_1)\) over a fixed segment remains its arc length.

For constant speed \(c\geq 0\), the integral simplifies to \(s(t)=c(t-t_0)\). For variable speed, the integral may have an elementary antiderivative, as in the parabola example, or may require numerical estimation. In either case, the definition is the same: accumulate speed from the specified starting parameter.

Check Your Understanding

Use the definition of cumulative arc length and the distinction between speed and displacement to answer these questions.

For \(\mathbf{r}(t)=(2\cos t,2\sin t)\) starting at \(t_0=0\), find the speed and write \(s(t)\) for \(t\geq 0\).
State the condition that lets the Fundamental Theorem of Calculus give \(s'(t)=\|\mathbf{r}'(t)\|\).
If \(s(t)=\int_2^t\|\mathbf{r}'(u)\|\,du\), what is \(s(2)\), and what does \(s(5)-s(3)\) represent?
Explain why integrating the velocity vector does not generally give arc length.
For \(\mathbf{r}(t)=(t,t^2/2)\), identify the speed and explain why its cumulative arc-length function is not linear in \(t\).