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Characterizing Unit-Speed Curves

See why unit speed makes a parameter measure distance traveled, and how the cumulative arc-length formula proves the converse.

Intermediate 10 min read

What You'll Learn

  • Define unit speed and the cumulative arc-length coordinate for a parametrized curve.
  • Prove that unit speed is equivalent to the arc-length coordinate differing from the parameter by a constant.
  • Distinguish distance traveled along a curve from straight-line distance between its points.
  • Check whether a circle or helix is parametrized by arc length.
  • Identify and correct a common speed-calculation error.

Why Unit Speed Characterizes Arc Length

A parametrized curve describes a path together with a rule for how its position changes as the parameter varies. The parameter may increase at a different rate from the distance traveled. Unit speed is the precise condition that removes this discrepancy: each unit of parameter change corresponds to one unit of distance along the path.

The key fact is an equivalence. If a differentiable curve has speed one throughout an interval, its cumulative arc-length coordinate is the parameter plus a constant. Conversely, if its cumulative arc-length coordinate differs from the parameter by a constant, its speed must be one.

Definition: Let \(\mathbf r:I\to\mathbb R^n\) be a continuously differentiable parametrized curve, where \(I\) is an interval. Its speed at \(t\) is \(v(t)=\|\mathbf r'(t)\|\). The curve has unit speed on \(I\) if \(v(t)=1\) for every \(t\in I\).

For a fixed reference parameter \(t_0\in I\), define the cumulative arc-length coordinate by

$$ S(t)=C+\int_{t_0}^{t}\|\mathbf r'(u)\|\,du, $$

where \(C\) is any constant. The integral is signed when \(t<t_0\), as is standard for a definite integral. The actual distance traveled from \(t_0\) to \(t\) is nonnegative and equals \(|S(t)-S(t_0)|\). This distinction lets one use a single increasing arc-length coordinate across the whole parameter interval.

The Characterization Theorem

Theorem: Suppose \(\mathbf r:I\to\mathbb R^n\) is continuously differentiable, and let \(S\) be its cumulative arc-length coordinate. The curve has unit speed on \(I\) if and only if there is a constant \(K\) such that \(S(t)=t+K\) for every \(t\in I\).

The constant accounts for where the arc-length coordinate is set to zero. In particular, if we choose \(S(t_0)=0\), then unit speed gives \(S(t)=t-t_0\). The distance traveled from \(t_0\) to \(t\) is therefore \(|t-t_0|\), not necessarily \(t-t_0\).

1
Differentiate the arc-length coordinate.
By the Fundamental Theorem of Calculus, \(S'(t)=\|\mathbf r'(t)\|\).
2
Assume unit speed.
Then \(S'(t)=1\) throughout the interval, so \(S(t)-t\) is constant. Thus \(S(t)=t+K\).
3
Assume the parameter differs from arc length by a constant.
If \(S(t)=t+K\), then \(S'(t)=1\), and hence \(\|\mathbf r'(t)\|=1\). The curve has unit speed.

This argument requires differentiability of the curve and the usual Fundamental Theorem of Calculus for the continuous speed function. The equivalence is a statement over the whole interval: an equality at just one parameter value, or a comparison of total lengths only at the endpoints, does not establish unit speed everywhere.

Worked Example: A Circle Traversed at Unit Speed

Worked Example: A Circle Traversed at Unit Speed

Consider \(\mathbf r(t)=(3\cos(t/3),3\sin(t/3))\). Determine whether \(t\) measures distance along this curve, up to an additive constant.

Differentiate each component:

$$\mathbf r'(t)=(-\sin(t/3),\cos(t/3)).$$

Its speed is

$$\|\mathbf r'(t)\|=\sqrt{(-\sin(t/3))^2+(\cos(t/3))^2}=\sqrt{\sin^2(t/3)+\cos^2(t/3)}=1.$$

Thus the curve has unit speed. For example, take \(t_0=1\) and \(t=4\). The distance traveled is

$$\int_1^4\|\mathbf r'(u)\|\,du=\int_1^4 1\,du=4-1=3.$$

The parameter change is also \(4-1=3\). More generally, with \(S(t_0)=0\), the cumulative coordinate is \(S(t)=\int_{t_0}^t1\,du=t-t_0\), so the parameter measures distance up to the constant \(-t_0\).

Check: The derivative has norm \(1\) at every \(t\), and the computed distance from \(1\) to \(4\) equals the parameter change.

Worked Example: A Circle Traversed at Twice Unit Speed

Worked Example: A Circle Traversed at Twice Unit Speed

Consider \(\mathbf r(t)=(2\cos t,2\sin t)\), for \(0\leq t\leq \pi/3\). Determine whether \(t\) is an arc-length coordinate.

Differentiate and calculate the speed:

$$\mathbf r'(t)=(-2\sin t,2\cos t),\qquad \|\mathbf r'(t)\|=\sqrt{4\sin^2t+4\cos^2t}=2.$$

The speed is \(2\), not \(1\). The distance traveled from \(0\) to \(\pi/3\) is

$$\int_0^{\pi/3}2\,du=2\left(\frac{\pi}{3}-0\right)=\frac{2\pi}{3}.$$

But the parameter change is only \(\pi/3\). The cumulative arc-length coordinate with reference value \(S(0)=0\) is \(S(t)=\int_0^t2\,du=2t\), not \(t+K\). Therefore this parametrization is not unit speed and its parameter does not measure distance traveled up to an additive constant.

Check: Differentiating \(S(t)=2t\) gives \(S'(t)=2\), which agrees with the speed calculated directly.

Worked Example: A Unit-Speed Helix and an Additive Constant

Worked Example: A Unit-Speed Helix and an Additive Constant

Consider the space curve

$$\mathbf r(t)=\left(3\cos(t/5),3\sin(t/5),4t/5\right).$$

Differentiate its three components:

$$\mathbf r'(t)=\left(-\frac35\sin(t/5),\frac35\cos(t/5),\frac45\right).$$

Then

$$\|\mathbf r'(t)\|=\sqrt{\frac9{25}\sin^2(t/5)+\frac9{25}\cos^2(t/5)+\frac{16}{25}}=\sqrt{\frac9{25}+\frac{16}{25}}=1.$$

So the helix has unit speed. Choose \(t_0=-1\) and set \(S(-1)=7\). Then \(C=7\) in the definition of \(S\), and

$$S(t)=7+\int_{-1}^t1\,du=7+(t+1)=t+8.$$

For instance, from \(t=-1\) to \(t=2\), the distance traveled is \(\int_{-1}^{2}1\,du=3\). The arc-length coordinate changes from \(7\) to \(10\), also a change of \(3\). The constant \(8\) shifts the coordinate’s origin without changing the distance it measures.

Check: Differentiating \(S(t)=t+8\) gives \(S'(t)=1\), equal to the speed found from \(\mathbf r'(t)\).

Distance Along the Curve Is Not Straight-Line Distance

Arc length measures how far the parametrized motion travels along its path. It is not, in general, the straight-line distance between the starting and ending positions. A curve can bend or even return to a point it has visited before while continuing to accumulate arc length.

For example, the unit-speed circle in the first example returns to its starting position when the parameter increases by \(6\pi\). Its endpoints then have straight-line distance \(0\), but the distance traveled along the circle is \(6\pi\). The parameter records accumulated travel, not the separation of the endpoint positions.

This also explains why a unit-speed curve need not be one-to-one. Repeated positions do not invalidate the characterization: the speed and arc length count the path as it is traversed.

A Common Speed-Calculation Pitfall

For \(\mathbf r(t)=(2\cos t,2\sin t)\), a common error is to drop the factor \(2\) while differentiating and then conclude that the speed is one:

$$ \text{Incorrect: }\quad \mathbf r'(t)=(-\sin t,\cos t), \qquad \|\mathbf r'(t)\|=1. $$

The derivatives of \(2\cos t\) and \(2\sin t\) retain that factor. The correct calculation is

$$ \mathbf r'(t)=(-2\sin t,2\cos t), \qquad \|\mathbf r'(t)\|=\sqrt{4\sin^2t+4\cos^2t}=2. $$

Thus its cumulative arc length from \(0\) is \(S(t)=2t\), not \(t+K\). Always differentiate each component before computing the norm; confusing a curve’s position with its velocity can lead to an incorrect unit-speed conclusion.

Why the Characterization Is Useful

The equivalence provides two ways to recognize an arc-length parameter. If the velocity is easy to compute, checking \(\|\mathbf r'(t)\|=1\) immediately identifies the parameter as arc length up to a constant. If the cumulative length is already known, differentiating its relation to the parameter gives the speed.

It also clarifies the role of scaling. A constant speed \(c>0\) gives \(S(t)=ct+K\), not \(S(t)=t+K\), unless \(c=1\). In that case the parameter is proportional to distance rather than equal to distance up to an additive constant. Unit speed is the exact condition for the latter.

Takeaway: For a continuously differentiable curve, the cumulative arc-length coordinate satisfies \(S'(t)=\|\mathbf r'(t)\|\). Therefore \(S(t)=t+K\) on an interval exactly when the curve has unit speed throughout that interval.

Check Your Understanding

Use the speed and cumulative arc-length relationship to answer these questions.

  1. State the definition of unit speed for a differentiable parametrized curve.
  2. If \(S(t)=t-4\) throughout an interval, what is the curve’s speed there?
  3. For \(\mathbf r(t)=(5\cos t,5\sin t)\), is \(t\) an arc-length parameter? Find the speed.
  4. Why can a curve return to a previously visited point while its cumulative arc length continues to increase?
  5. If the speed is the constant \(3\), what is the form of the cumulative arc-length coordinate?

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