Regularity, Stops, And Piecewise Arc Length

Why Stops and Formula Changes Need Care

A parametrized curve can slow to a stop, change its formula, or do both at the same parameter value. These features matter when describing the curve and its tangent behavior. They do not, by themselves, prevent us from calculating arc length.

The reliable method is to identify where the parametrization behaves differently, divide the parameter interval at those points, and calculate the length on each smooth piece. Adding the piece lengths gives the total distance traced. A zero derivative at an isolated parameter value is not a reason to discard that point or to assume the curve has zero length.

Regularity and Stops

Definition: A parametrization $\mathbf{r}:[a,b]\to\mathbb{R}^2$ or $\mathbb{R}^3$ is regular on an interval if it is continuously differentiable there and $\mathbf{r}'(t)\ne\mathbf{0}$ at every parameter value in that interval. A parameter value $t_0$ at which $\mathbf{r}'(t_0)=\mathbf{0}$ is called a stop. If the derivative is zero only at some points, the parametrization is not regular on an interval containing those points.

Here $\mathbf{r}'(t)$ is the vector of derivatives of the component functions. For a planar curve $\mathbf{r}(t)=(x(t),y(t))$, the derivative is $\mathbf{r}'(t)=(x'(t),y'(t))$. Its magnitude is $\|\mathbf{r}'(t)\|=\sqrt{(x'(t))^2+(y'(t))^2}$. The arc-length formula on a continuously differentiable parameter interval is the integral of this magnitude.

Key point: Regularity requires a nonzero derivative, but the arc-length integral does not. A continuously differentiable curve can have a stop and still have a well-defined length, calculated by integrating $\|\mathbf{r}'(t)\|$.

A stop can be isolated, as when the derivative vanishes at one parameter value, or it can occur throughout an interval. On a constant interval, $\mathbf{r}'(t)=\mathbf{0}$, and that interval contributes zero length. If the curve moves before and after an isolated stop, the lengths of those portions still contribute normally.

Piecewise Parametrizations

A curve may be specified by different formulas on adjacent parameter intervals. We call the parametrization piecewise $C^1$ if the interval can be divided into a finite number of pieces on each of which the formula is continuously differentiable, with finite one-sided limits at the division points. The derivative need not agree from the left and right at a join.

For adjoining formulas to describe a continuous curve through a join $c$, their position values must agree there: the left-hand and right-hand values of $\mathbf{r}(c)$ are the same. The two one-sided derivatives may differ, producing a corner. The arc length is calculated separately on each piece; a corner does not require an extra length term.

Arc length for finitely many pieces: If a curve is piecewise $C^1$ on $[a,b]$, with division points $a=t_0<t_1<\cdots<t_n=b$, then its total traced length is $$ L=\sum_{j=1}^{n}\int_{t_{j-1}}^{t_j}\|\mathbf{r}'(t)\|\,dt, $$ where on each interval the derivative is taken from that interval's formula. Stops may also be used as division points. At a finite number of endpoints or joins, the value assigned to the integrand does not change the integral.

Why Splitting Works

On a smooth piece, the arc-length formula measures the accumulated length there. Splitting at a stop or a formula change simply applies that formula to each piece separately. The lengths add because adjoining pieces cover the entire parameter interval without overlapping interiors.

Locate the special parameters.
Find zeros of $\mathbf{r}'(t)$ and every point where the formula changes.

Check the joins.
Compare the position values on either side of each formula change. Matching values give a continuous join; different values indicate a jump.

Calculate each piece length.
Use the applicable formula for $\mathbf{r}'(t)$ on that piece, and integrate its magnitude over the piece's parameter interval.

Add the results.
The sum is the total length traced across the pieces. A jump is not part of the parametrized path and must not be silently counted as a connecting segment.

For a continuous piecewise curve, additivity follows directly by splitting the integral at the division points. For example, if $a<c<b$, then

\int_a^b \|\mathbf{r}'(t)\|\,dt = \int_a^c \|\mathbf{r}'(t)\|\,dt + \int_c^b \|\mathbf{r}'(t)\|\,dt.

This equality is the integral additivity property. If the derivative has different one-sided values at $c$, the value at that single point does not affect either integral. Thus a corner, or an isolated stop at the split, needs no correction term.

Worked Examples

Worked Example: An Isolated Stop on a Polynomial Curve

Find the length of $\mathbf{r}(t)=(t^2,t^3)$ for $-1\le t\le1$, and identify any stop.

Differentiate each component:

\mathbf{r}'(t)=(2t,3t^2).

Both components vanish at $t=0$, so there is a stop there. Split the length integral at zero:

\begin{aligned} L &=\int_{-1}^{0}\sqrt{(2t)^2+(3t^2)^2}\,dt +\int_{0}^{1}\sqrt{(2t)^2+(3t^2)^2}\,dt\\ &=\int_{-1}^{0}|t|\sqrt{4+9t^2}\,dt +\int_{0}^{1}t\sqrt{4+9t^2}\,dt. \end{aligned}

The integrand is even, so these integrals are equal. On the interval from $0$ to $1$, use $u=4+9t^2$, so $du=18t\,dt$. The limits become $u=4$ and $u=13$:

\begin{aligned} L &=2\int_0^1 t\sqrt{4+9t^2}\,dt\\ &=2\left(\frac{1}{18}\int_4^{13}u^{1/2}\,du\right)\\ &=2\left(\frac{1}{18}\cdot\frac{2}{3} \left[ u^{3/2}\right]_4^{13}\right)\\ &=\frac{2}{27}\left(13\sqrt{13}-8\right) =\frac{26\sqrt{13}-16}{27}. \end{aligned}

Check the antiderivative used on the positive half: $\frac{d}{dt}\left((4+9t^2)^{3/2}/27\right)=t\sqrt{4+9t^2}$. The value at $1$ minus the value at $0$ is $(13\sqrt{13}-8)/27$; doubling accounts for the matching negative half. The isolated stop does not remove either part of the curve from the length calculation.

Worked Example: A Stop That Retraces a Segment

Find the length traced by $\mathbf{r}(t)=((t-1)^2,0)$ for $0\le t\le2$. Determine where it stops.

The derivative is

\mathbf{r}'(t)=(2(t-1),0).

It vanishes at $t=1$, so split there. On the first interval, $t-1\le0$, and on the second, $t-1\ge0$. Therefore,

\begin{aligned} L &=\int_0^1 2(1-t)\,dt+\int_1^2 2(t-1)\,dt\\ &=\left[2t-t^2\right]_0^1+\left[(t-1)^2\right]_1^2\\ &=(2-1-0)+\bigl(1-0\bigr)\\ &=2. \end{aligned}

Check: the first segment has length $1$, and the second also has length $1$, for a total of $2$. Although the curve's image is just the interval from $(0,0)$ to $(1,0)$, the parametrization traces that segment once in each direction. The integral counts both traversals.

Worked Example: A Continuous Join With a Corner

Let

\mathbf{r}(t)= \begin{cases} (t,t),&0\le t\le1,\\ (2-t,t),&1<t\le2. \end{cases}

Find the total length. At the join $t=1$, the first formula gives $(1,1)$. The right-hand formula also tends to $(1,1)$, so the curve is continuous there.

On the first piece, $\mathbf{r}'(t)=(1,1)$, whose magnitude is $\sqrt{2}$. On the second piece, $\mathbf{r}'(t)=(-1,1)$, also of magnitude $\sqrt{2}$. Thus

\begin{aligned} L &=\int_0^1\sqrt{1^2+1^2}\,dt +\int_1^2\sqrt{(-1)^2+1^2}\,dt\\ &=\int_0^1\sqrt{2}\,dt+\int_1^2\sqrt{2}\,dt\\ &=\sqrt{2}(1-0)+\sqrt{2}(2-1)\\ &=2\sqrt{2}. \end{aligned}

Check: each straight segment has displacement components with absolute values $1$ and $1$, hence length $\sqrt{1^2+1^2}=\sqrt{2}$. The two lengths sum to $2\sqrt{2}$. The derivative changes at the join, but the lengths remain additive.

Common Mistake: Treating a Stop as Zero Length

A common error is to see $\mathbf{r}'(t_0)=\mathbf{0}$ and conclude that the curve has zero length, or that the length integral must fail. A zero derivative at one parameter value says only that the instantaneous rate of change is zero there. It does not say that the curve is constant on the full interval.

For example, for $\mathbf{r}(t)=((t-1)^2,0)$, the incorrect calculation $L=0$ because $\mathbf{r}'(1)=\mathbf{0}$ ignores all parameter values other than $1$. The correct calculation is

L=\int_0^2\|\mathbf{r}'(t)\|\,dt =\int_0^2 2|t-1|\,dt =1+1=2.

The correction is to check where the derivative vanishes, split there when useful, and integrate the magnitude over the entire interval.

When a Formula Change Is a Jump

If adjacent formulas do not agree at a join, they do not describe a continuous curve through that point. The arc-length integral on the pieces measures the length of the portions actually traced; it does not include the straight-line gap between their endpoints. If the intended path includes that gap, it must be specified as an additional segment and its length calculated separately.

This distinction prevents a different kind of error: assuming that every pair of formulas written in one piecewise definition automatically meets continuously. Always compare the endpoint positions before treating the pieces as one connected path.

Summary

Regularity is a condition on the derivative: a regular parametrization is continuously differentiable with a nonzero derivative. Stops are parameter values where the derivative is zero. Neither a stop nor a corner automatically prevents arc-length calculation. For a piecewise $C^1$ curve, calculate the integral of the derivative's magnitude on each smooth piece and add the results. Check that the position formulas agree at joins when the curve is meant to be continuous.

Check Your Understanding

Use the definitions and splitting method to answer the following questions.

What condition on $\mathbf{r}'(t)$ makes a continuously differentiable parametrization regular?
For $\mathbf{r}(t)=(t^3,t^2)$, find any stop in $[-1,1]$. Does its presence prevent the arc-length integral from being defined?
Why can a corner at a continuous piecewise join be handled by adding the lengths on the two sides?
For $\mathbf{r}(t)=((t-2)^2,0)$ on $[1,3]$, identify the stop and set up the two integrals for total length.
If the endpoint positions of two adjacent formulas do not match, does the sum of their arc lengths include the gap between them? Explain.

Regularity, Stops, And Piecewise Arc Length

What You'll Learn

Why Stops and Formula Changes Need Care

Regularity and Stops

Piecewise Parametrizations

Why Splitting Works

Worked Examples

Worked Example: An Isolated Stop on a Polynomial Curve

Worked Example: A Stop That Retraces a Segment

Worked Example: A Continuous Join With a Corner

Common Mistake: Treating a Stop as Zero Length

When a Formula Change Is a Jump

Summary

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Regularity, Stops, And Piecewise Arc Length

What You'll Learn

Why Stops and Formula Changes Need Care

Regularity and Stops

Piecewise Parametrizations

Why Splitting Works

Worked Examples

Worked Example: An Isolated Stop on a Polynomial Curve

Worked Example: A Stop That Retraces a Segment

Worked Example: A Continuous Join With a Corner

Common Mistake: Treating a Stop as Zero Length

When a Formula Change Is a Jump

Summary

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Previous

Next