Prescribing Speed Along A Fixed Path

Prescribing Motion Along a Path

A geometric path specifies where an object can go, but not how quickly it moves along that path. To describe motion, we also need a parameter that changes with time. When the path is written using arc length, choosing that time-dependent parameter becomes especially direct: the rate of change of arc length is the speed.

This tutorial develops that idea. Starting with a unit-speed path and a desired nonnegative speed, we integrate the speed to find the arc-length parameter as a function of time. Composing the path with that parameter gives the object's position.

Unit-Speed Paths and Arc Length

Definition: A differentiable path $\boldsymbol{\gamma}:I\to\mathbb{R}^n$, with parameter $s$ in an interval $I$, is a unit-speed path if $\|\boldsymbol{\gamma}'(s)\|=1$ for every $s\in I$. The parameter $s$ measures distance along the path, increasing in the direction of the path's orientation.

If an object follows this path with a time-dependent parameter $s(t)$, its position is $\mathbf{x}(t)=\boldsymbol{\gamma}(s(t))$. The chain rule gives its velocity. Since the path's tangent vector has length one, the magnitude of that velocity is determined entirely by the rate at which $s$ changes.

\mathbf{x}'(t)=\boldsymbol{\gamma}'(s(t))s'(t), \qquad \|\mathbf{x}'(t)\|=|s'(t)|.

Thus, for motion that progresses forward along the oriented path, a desired speed $v(t)$ is achieved by setting $s'(t)=v(t)$. Speed is a magnitude, so the prescribed function must be nonnegative. If it is zero at a time, the object is instantaneously at rest there.

Key formula: If the object starts at arc-length parameter $s_0$ at time $t_0$, and $v(t)\geq 0$ is its continuous prescribed speed, then $$ s(t)=s_0+\int_{t_0}^{t}v(u)\,du, \qquad \mathbf{x}(t)=\boldsymbol{\gamma}(s(t)). $$ The values of $s(t)$ must remain in the domain $I$ of the path.

Why Integration Produces the Required Motion

The formula follows from the relationship between speed and the rate of change of arc length. The fundamental theorem of calculus shows that the accumulated distance has the desired rate of change; the chain rule then shows that the composed position has the desired speed.

Accumulate the prescribed speed.
Define $s(t)=s_0+\int_{t_0}^{t}v(u)\,du$. Since $v$ is continuous, the fundamental theorem of calculus gives $s'(t)=v(t)$.

Compose with the path.
Set $\mathbf{x}(t)=\boldsymbol{\gamma}(s(t))$. By the chain rule, $\mathbf{x}'(t)=\boldsymbol{\gamma}'(s(t))v(t)$.

Compute the speed.
Because $\|\boldsymbol{\gamma}'(s(t))\|=1$ and $v(t)\geq 0$, $\|\mathbf{x}'(t)\|=v(t)$.

This also checks the distance traveled. The distance from $t_0$ to $t$ is the integral of the magnitude of velocity, so it equals $\int_{t_0}^{t}v(u)\,du=s(t)-s_0$. The parameter really does record the accumulated distance along the path.

Worked Examples

Worked Example: Speeding Up Along a Unit Circle

Let $\boldsymbol{\gamma}(s)=(\cos s,\sin s)$, a unit-speed circle, and suppose an object starts at $s(0)=0$ with prescribed speed $v(t)=1+t$ for $0\leq t\leq 2$. Find its position and check the speed.

First, verify that the path is unit speed:

\boldsymbol{\gamma}'(s)=(-\sin s,\cos s), \qquad \|\boldsymbol{\gamma}'(s)\| =\sqrt{\sin^2 s+\cos^2 s}=1.

Integrate the speed from the initial time to $t$:

s(t)=0+\int_0^t(1+u)\,du =\left[u+\frac{u^2}{2}\right]_0^t =t+\frac{t^2}{2}.

Composing with the path gives

\mathbf{x}(t)= \left( \cos\left(t+\frac{t^2}{2}\right), \sin\left(t+\frac{t^2}{2}\right) \right).

Differentiate to check the result. With $s(t)=t+t^2/2$, we have $s'(t)=1+t$, so

\mathbf{x}'(t) =(1+t)(-\sin s(t),\cos s(t)), \qquad \|\mathbf{x}'(t)\| =(1+t)\sqrt{\sin^2 s(t)+\cos^2 s(t)} =1+t.

The computed speed matches the prescription. At $t=2$, for example, $s(2)=4$ and the speed is $3$.

Worked Example: Increasing Speed Along a Unit Helix

Consider the helix $\boldsymbol{\gamma}(s)=\left(\frac{1}{\sqrt{2}}\cos s,\frac{1}{\sqrt{2}}\sin s,\frac{s}{\sqrt{2}}\right)$. Starting from $s(0)=0$, prescribe speed $v(t)=2t$ for $0\leq t\leq 2$. Find the position and verify the speed.

The derivative of the path is

\boldsymbol{\gamma}'(s) =\left(-\frac{1}{\sqrt{2}}\sin s,\frac{1}{\sqrt{2}}\cos s,\frac{1}{\sqrt{2}}\right), \qquad \|\boldsymbol{\gamma}'(s)\|^2 =\frac{1}{2}\sin^2s+\frac{1}{2}\cos^2s+\frac{1}{2}=1.

Therefore the helix is unit speed. Integrating the prescribed speed gives

s(t)=\int_0^t 2u\,du =\left[u^2\right]_0^t=t^2.

The position is the composition

\mathbf{x}(t) =\left(\frac{1}{\sqrt{2}}\cos(t^2), \frac{1}{\sqrt{2}}\sin(t^2), \frac{t^2}{\sqrt{2}}\right).

Differentiate each coordinate:

\mathbf{x}'(t) =\left(-\sqrt{2}t\sin(t^2), \sqrt{2}t\cos(t^2), \sqrt{2}t\right).

For $0\leq t\leq 2$, $t\geq 0$, and hence

\|\mathbf{x}'(t)\|^2 =2t^2\sin^2(t^2)+2t^2\cos^2(t^2)+2t^2 =4t^2, \qquad \|\mathbf{x}'(t)\|=2t.

This is the prescribed speed, including the initial instant $t=0$, when the object starts at rest.

Worked Example: A Motion That Comes to a Stop

Use the unit-speed circle $\boldsymbol{\gamma}(s)=(2\cos(s/2),2\sin(s/2))$. An object starts at $s(0)=0$. Its speed is $v(t)=2$ for $0\leq t\leq 1$, and $v(t)=2(2-t)$ for $1\leq t\leq 2$. Find the parameter and position on both intervals, and check the stop at $t=2$.

The circle is unit speed because $\boldsymbol{\gamma}'(s)=(-\sin(s/2),\cos(s/2))$, whose magnitude is $1$. On the first interval, integration gives

s(t)=\int_0^t 2\,du=2t, \qquad 0\leq t\leq 1.

At $t=1$, the parameter is $s(1)=2$. For $1\leq t\leq 2$, add the distance accumulated during the second interval to that initial value:

s(t)=2+\int_1^t 2(2-u)\,du =2+\left[4u-u^2\right]_1^t =2+(4t-t^2-3) =4t-t^2-1.

Consequently, the position is

\mathbf{x}(t)= \begin{cases} (2\cos t,2\sin t),&0\leq t\leq 1,\\ (2\cos((4t-t^2-1)/2),2\sin((4t-t^2-1)/2)),&1\leq t\leq 2. \end{cases}

At the joining time, both formulas give $(2\cos 1,2\sin 1)$, so position is continuous. On the first interval, $s'(t)=2=v(t)$. On the second, $s'(t)=4-2t=2(2-t)=v(t)$. Since the path has unit speed, the object's speed is $v(t)$ on each interval. At $t=2$, $s'(2)=0$, so the object is instantaneously at rest; the total distance traveled is $s(2)-s(0)=3$.

A Common Mistake: Using Speed as the Parameter

The speed $v(t)$ is a rate of change, not the accumulated arc length. For example, if $v(t)=1+t$ and $s(0)=0$, it is incorrect to set $s(t)=1+t$. That gives $s'(t)=1$, so a unit-speed path would be followed at speed $1$, not at speed $1+t$.

The correction is to integrate the speed:

s(t)=\int_0^t(1+u)\,du=t+\frac{t^2}{2}, \qquad s'(t)=1+t.

Always distinguish the speed $v(t)$, with units of distance per time, from the arc-length parameter $s(t)$, with units of distance. Integration converts the former into the latter.

Conditions to Check

Before composing a prescribed speed with a path, check the assumptions that make the construction valid:

The path is differentiable and unit speed on the parameter values that will be used.
The prescribed speed is nonnegative. If it is continuous, the integral defines a differentiable parameter function with $s'(t)=v(t)$.
The initial value $s_0$ identifies the intended starting point on the path.
The accumulated parameter remains in the domain of the path for the entire time interval.

A negative value cannot represent speed. If a model calls for traveling backward along an oriented path, the parameter must decrease, and its signed rate must be chosen separately; the speed itself remains the magnitude of the velocity.

Why the Method Is Useful

This method separates two decisions: the geometry of the route is encoded in $\boldsymbol{\gamma}(s)$, while the timing is encoded in $s(t)$. Once a route has been expressed as a unit-speed path, changing the desired speed changes the timing without changing the path. The integral also makes stops and changes in pace visible: intervals with $v(t)=0$ leave $s(t)$ unchanged, while larger speeds make it increase more rapidly.

Check Your Understanding

Use the definitions and chain-rule argument above to answer these questions.

Why does a unit-speed path imply that the object's speed is $|s'(t)|$?
If $s(0)=3$ and $v(t)=2t$, what is $s(t)$ for $t\geq 0$?
For forward motion, why must the prescribed speed be nonnegative?
If the speed is zero at one instant, what does that say about the parameter's instantaneous rate of change?
What condition must the values of $s(t)$ satisfy in relation to the domain of $\boldsymbol{\gamma}$?