Arc-Length Functions And Accumulated Distance

From Speed to Accumulated Distance

A parametrized curve describes a moving point. Its speed tells us how quickly the point is moving at each parameter value, but speed at a single instant does not tell us how far the point has traveled over an interval. To find that accumulated distance, we add the contributions of speed across the interval.

This leads to an important function: the arc-length function. Once it is defined from a fixed starting parameter, its value records the distance traveled so far. The Fundamental Theorem of Calculus then gives a direct connection between this accumulated distance and the speed at each instant.

Central idea: If a curve starts being measured at parameter $t_0$, accumulated distance at a later parameter $t$ is the integral of speed from $t_0$ to $t$.

Definition of the Arc-Length Function

Let a planar curve be given by $\mathbf{r}(t)=(x(t),y(t))$ on an interval containing $t_0$. Its speed is

v(t)=\|\mathbf{r}'(t)\| =\sqrt{(x'(t))^2+(y'(t))^2}.

Assume that the component functions have continuous derivatives on the interval under consideration, so that the speed is continuous. For $t\ge t_0$, define the arc-length function measured from $t_0$ by

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

The variable $u$ is a dummy variable of integration; $t$ is the parameter at which the accumulated distance is evaluated. The lower limit fixes the starting point of the measurement. In particular, $s(t_0)=0$.

Definition: The arc-length function from a fixed starting parameter $t_0$ is $s(t)=\int_{t_0}^{t}\|\mathbf{r}'(u)\|\,du$, for $t\ge t_0$. It measures distance traveled along the curve from parameter $t_0$ to parameter $t$.

Because speed is nonnegative, $s(t)$ cannot decrease as $t$ increases. If $a\le b$, the distance traveled between parameters $a$ and $b$ is

\int_a^b v(u)\,du=s(b)-s(a).

This difference formula works when $s$ is measured from any fixed starting parameter at or before $a$. It is often more efficient than recomputing an integral from the original starting point.

Why the Derivative of Arc Length Is Speed

The key result follows from the Fundamental Theorem of Calculus. Its hypotheses matter: the speed must be continuous on the interval. That condition holds, for example, when the curve’s component derivatives are continuous, because the speed is formed from those derivatives using squares, addition, and a square root.

Theorem: If $v$ is continuous and $s(t)=\int_{t_0}^{t}v(u)\,du$, then at every interior point $t$, $$ s'(t)=v(t)=\|\mathbf{r}'(t)\|. $$

Here is the direct justification. For a nonzero increment $h$ small enough that $t+h$ is in the interval, the difference quotient is

\frac{s(t+h)-s(t)}{h} = \frac{1}{h}\int_t^{t+h}v(u)\,du.

Subtract $v(t)$ from this expression. By linearity of the integral, the difference is the average over the short interval of $v(u)-v(t)$. Its absolute value satisfies

\left| \frac{1}{h}\int_t^{t+h}v(u)\,du-v(t) \right| \le \max_{u\text{ between }t\text{ and }t+h}|v(u)-v(t)|.

Continuity of $v$ at $t$ means the maximum on the right tends to zero as $h$ tends to zero. Therefore the difference quotient tends to $v(t)$, proving $s'(t)=v(t)$. At an endpoint, the corresponding one-sided derivative gives the same conclusion when it is defined.

This result has a useful interpretation: $s(t)$ is accumulated distance, and its derivative is the instantaneous rate at which that distance accumulates. The derivative is speed, not the magnitude of displacement from the starting point.

Worked Examples

Worked Example: A Polynomial Curve with Changing Speed

For $t\ge 0$, let $\mathbf{r}(t)=\left(t^2,\frac{2}{3}t^3\right)$, and measure distance from $t_0=0$. Find the arc-length function and the distance traveled by $t=2$.

First differentiate each component:

x'(t)=2t, \qquad y'(t)=2t^2.

Thus the speed is

v(t)=\sqrt{(2t)^2+(2t^2)^2} =\sqrt{4t^2+4t^4} =2t\sqrt{1+t^2},

where $t\ge0$ lets us write $\sqrt{4t^2}=2t$. Integrate from $0$ to $t$:

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

Use the substitution $w=1+u^2$, so $dw=2u\,du$. The integration limits change from $u=0$ and $u=t$ to $w=1$ and $w=1+t^2$. Therefore

s(t)=\int_1^{1+t^2}\sqrt{w}\,dw =\left[\frac{2}{3}w^{3/2}\right]_1^{1+t^2} =\frac{2}{3}\left((1+t^2)^{3/2}-1\right).

At $t=2$, the distance is

s(2)=\frac{2}{3}\left(5^{3/2}-1\right) =\frac{2}{3}\left(5\sqrt{5}-1\right).

Check by differentiating the proposed function:

s'(t)=\frac{2}{3}\cdot\frac{3}{2}(1+t^2)^{1/2}\cdot 2t =2t\sqrt{1+t^2} =v(t).

Also, $s(0)=\frac{2}{3}(1-1)=0$, as required by the chosen starting parameter.

Worked Example: Distance over a Later Parameter Interval

Let $\mathbf{r}(t)=(3t+1,4t-2)$, and define $s$ using the starting parameter $t_0=1$. Find the distance traveled from $t=1$ to $t=4$, and the rate at which accumulated distance changes.

The component derivatives are $x'(t)=3$ and $y'(t)=4$, so

v(t)=\sqrt{3^2+4^2}=\sqrt{9+16}=5.

Consequently, for $t\ge1$,

s(t)=\int_1^t 5\,du=5(t-1).

The distance from $1$ to $4$ is

$$ s(4)-s(1)=5(4-1)-5(1-1)=15-0=15. $$

As a check, the position at $t=1$ is $(4,2)$, while the position at $t=4$ is $(13,14)$. The displacement is $(9,12)$, whose magnitude is $\sqrt{9^2+12^2}=\sqrt{225}=15$, agreeing with the distance. Differentiating the arc-length function gives $s'(t)=5=v(t)$, as expected.

Worked Example: A Complete Circular Trip

Let $\mathbf{r}(t)=(2\cos(3t),2\sin(3t))$, for $0\le t\le \frac{2\pi}{3}$, with starting parameter $t_0=0$. Find the arc-length function, the total distance traveled, and the displacement over the full interval.

Differentiate the components:

x'(t)=-6\sin(3t), \qquad y'(t)=6\cos(3t).

The speed is constant:

v(t)=\sqrt{36\sin^2(3t)+36\cos^2(3t)} =6\sqrt{\sin^2(3t)+\cos^2(3t)} =6.

Integrating from the starting parameter gives

s(t)=\int_0^t 6\,du=6t.

At the end of the interval, the total distance is

s\left(\frac{2\pi}{3}\right) =6\cdot\frac{2\pi}{3} =4\pi.

The endpoint positions are both $(2,0)$: at $t=0$ the position is $(2,0)$, and at $t=\frac{2\pi}{3}$ the angle $3t$ is $2\pi$, so the position is again $(2,0)$. The displacement is therefore $(0,0)$, even though the distance traveled is $4\pi$. Finally, differentiating $s(t)=6t$ gives $s'(t)=6=v(t)$.

Distance Is Not Displacement

Displacement compares the positions at two endpoints. Its magnitude is the length of the straight line between them. Distance traveled measures the length of the path actually followed. These quantities agree for some paths, such as the straight-line example, but need not agree in general.

The full circular trip illustrates the difference sharply: the point returns to its starting position, so its displacement is zero, but its accumulated distance is $4\pi$. More generally, if $a\le b$, then the path distance is $\int_a^b v(u)\,du$, whereas the displacement vector is $\mathbf{r}(b)-\mathbf{r}(a)$. Taking the magnitude of that vector does not usually give the path distance.

A Common Mistake

A tempting but incorrect shortcut is to calculate the magnitude of the change in position and call it the distance traveled. For the circular example, this mistake gives

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

That calculation correctly finds the magnitude of the displacement, but not the distance along the curve. The correction is to integrate speed over the parameter interval:

\int_0^{2\pi/3}6\,dt =6\left(\frac{2\pi}{3}\right) =4\pi.

A second useful check is to differentiate any proposed arc-length function. If it is correct and the speed is continuous, its derivative must equal the speed, and its value at the starting parameter must be zero.

Using the Arc-Length Function

The arc-length function packages distance information for every parameter value after the chosen start. It is useful when a question asks for the distance accumulated by a particular time, the distance between two later parameter values, or the instantaneous rate of change of accumulated distance.

To find distance from the starting parameter $t_0$ to $t$, evaluate $s(t)$.
To find distance from $a$ to $b$, with $t_0\le a\le b$, calculate $s(b)-s(a)$.
To find the rate of change of accumulated distance, differentiate: $s'(t)=v(t)$, provided speed is continuous.

The integral definition also makes the role of the parameter explicit: the limits specify which portion of the motion is being measured. A parameter interval may describe a path that bends, loops, or returns to a previous position; integrating speed still adds the distance traveled along that path.

Check Your Understanding

Use the definition of accumulated distance and the relationship between its derivative and speed to answer these questions.

For a curve with continuous speed $v(t)$, write the arc-length function measured from $t_0$, and state its value at $t_0$.
If $s(t)=\int_2^t v(u)\,du$ and $v$ is continuous, what is $s'(t)$?
If $s(1)=4$ and $s(5)=13$, what distance is traveled from parameter $1$ to parameter $5$?
A moving point returns to its starting position after traveling around a loop. Must its distance traveled be zero? Explain the distinction involved.
What two checks can confirm a proposed formula for an arc-length function?