Scalar Integrals With Vanishing Speed

Why Vanishing Speed Needs Attention

A scalar line integral accumulates the values of a scalar field along a curve, weighted by distance travelled. In a parametrization $\mathbf{r}(t)$, the distance element is $\|\mathbf{r}'(t)\|\,dt$. If the velocity vanishes at a parameter value, the curve is momentarily stationary in that parametrization. This does not automatically make the line integral invalid, nor does it mean that the curve stops being a piecewise smooth curve.

The key is to separate three questions: where the speed vanishes, whether the parameter interval can be divided into regular pieces, and whether another parameter describes the same geometric path more conveniently. The norm in the integral matters throughout: speed is never a signed component of velocity.

Regularity and the Scalar Line Integral

Definition: A parametrization $\mathbf{r}:[a,b]\to\mathbb{R}^n$ is regular on an interval if it is continuously differentiable there and $\mathbf{r}'(t)\ne\mathbf{0}$ at every interior parameter value. A curve is piecewise regular if its parameter interval can be divided into finitely many subintervals on which the parametrization is regular in the interior, with the pieces joining continuously. A zero derivative at a division point is allowed.

For a continuous scalar field $f$ defined along a piecewise continuously differentiable parametrized curve, its scalar line integral is computed by integrating on the smooth pieces and adding the results:

\int_C f\,ds = \sum_{j=1}^{m}\int_{t_{j-1}}^{t_j} f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt.

Here $a=t_0<t_1<\cdots<t_m=b$ is a finite partition, and $ds=\|\mathbf{r}'(t)\|\,dt$. If $\mathbf{r}'(t)=\mathbf{0}$ at an isolated point, the integrand’s distance factor is zero at that point. A single point does not change the value of an ordinary definite integral. The point may still be useful as a place to divide the curve into regular pieces.

If the velocity vanishes throughout a parameter interval, the parametrization is stationary there: it traces no additional distance on that interval, so that portion contributes zero. Such an interval is not a regular piece under the definition above. It may be removed from the parameter description if the resulting parametrization still describes the intended curve. More generally, a zero set that cannot be handled by a finite subdivision does not establish piecewise regularity by itself; the geometric curve may nevertheless admit a different piecewise regular parametrization.

Finding and Interpreting Stationary Values

To locate vanishing speed, solve the vector equation $\mathbf{r}'(t)=\mathbf{0}$, meaning that every component of the derivative is zero at once. Finding a zero of only one component is not enough. Once the parameter values are found, check whether there are finitely many and whether splitting at them leaves pieces with nonzero velocity in their interiors.

Differentiate the parametrization.
Compute each component of $\mathbf{r}'(t)$.

Solve all component equations together.
Retain only parameter values at which every component is zero.

Check the pieces.
Divide at the stationary values and test whether the velocity is nonzero in each open piece.

Integrate with speed.
Use $\|\mathbf{r}'(t)\|$, and add the piece integrals or use a valid alternative parameter.

Why a Change of Parameter Works

Suppose $\mathbf{r}(t)=\boldsymbol{\gamma}(\phi(t))$, where $\phi$ is continuously differentiable and increasing, and its range covers the parameter interval for $\boldsymbol{\gamma}$. The chain rule gives $\mathbf{r}'(t)=\boldsymbol{\gamma}'(\phi(t))\phi'(t)$. Since $\phi'(t)\ge 0$, taking norms yields $\|\mathbf{r}'(t)\|=\|\boldsymbol{\gamma}'(\phi(t))\|\phi'(t)$. Thus the scalar line integral in the $t$-parameter becomes the ordinary change of variable $u=\phi(t)$ in the integral for $\boldsymbol{\gamma}$. The speed of $\mathbf{r}$ may vanish when $\phi'$ vanishes, without changing the accumulated integral.

Reparametrization principle: If $\phi$ is a continuously differentiable increasing change of parameter covering the same traversal, then $$ \int f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt = \int f(\boldsymbol{\gamma}(u))\,\|\boldsymbol{\gamma}'(u)\|\,du. $$ The equality follows from the chain rule and change of variables. If the parameter runs in the decreasing direction, the speed introduces $|\phi'(t)|$; the scalar line integral still measures distance and does not acquire a negative sign.

This argument explains both why a stationary point can be harmless and why one must preserve the traversal being integrated. A change of parameter can remove a pause or an artificial slowing, but it cannot silently discard a portion of the path or a repeated traversal.

Worked Examples

Worked Example: A Straight Segment With One Stationary Parameter

Let $\mathbf{r}(t)=(t^3,0)$, for $-1\le t\le1$, and let $f(x,y)=1+x$. Find the stationary value and compute the scalar line integral.

Differentiate both components:

\mathbf{r}'(t)=(3t^2,0).

This vector is zero exactly when $t=0$. On $(-1,0)$ and $(0,1)$, it is nonzero, so dividing at $0$ gives two regular pieces. The speed and field value along the curve are

\|\mathbf{r}'(t)\|=3t^2,\qquad f(\mathbf{r}(t))=1+t^3.

Therefore, integrating over the full interval, or equivalently adding the two pieces, gives

\int_C f\,ds =\int_{-1}^{1}(1+t^3)3t^2\,dt =\int_{-1}^{1}(3t^2+3t^5)\,dt =\left[t^3+\frac{t^6}{2}\right]_{-1}^{1} =\frac32-\left(-\frac12\right)=2.

As a check, the geometric segment is the $x$-axis from $-1$ to $1$, so $ds=dx$. Directly, $\int_{-1}^{1}(1+x)\,dx=[x+x^2/2]_{-1}^{1}=2$, agreeing with the parametrized calculation.

Worked Example: A Cusp That Is Piecewise Regular

Consider $\mathbf{r}(t)=(t^2,t^3)$ for $-1\le t\le1$, and compute the length of the curve, which is the scalar line integral with $f=1$.

The velocity is $\mathbf{r}'(t)=(2t,3t^2)$, so it vanishes only at $t=0$. On each of $(-1,0)$ and $(0,1)$, the velocity is nonzero. The curve is therefore piecewise regular when divided at the cusp. Its speed is

\|\mathbf{r}'(t)\| =\sqrt{4t^2+9t^4} =|t|\sqrt{4+9t^2}.

The absolute value matters because speed cannot be negative. The integrand is symmetric, so the two halves have equal length:

\int_C 1\,ds =2\int_0^1 t\sqrt{4+9t^2}\,dt.

Since $\frac{d}{dt}(4+9t^2)^{3/2}=27t\sqrt{4+9t^2}$, an antiderivative is $(4+9t^2)^{3/2}/27$. Hence

\int_C 1\,ds =2\left[\frac{(4+9t^2)^{3/2}}{27}\right]_0^1 =\frac{2(13\sqrt{13}-8)}{27} =\frac{26\sqrt{13}-16}{27}.

As a check, each branch has length $(13\sqrt{13}-8)/27$, and adding the two branch lengths gives the displayed result. The zero velocity at the joining point does not remove either branch from the curve.

Worked Example: Removing a Zero by an Alternative Parameter

Let $\mathbf{r}(t)=(\cos(t^3),\sin(t^3))$ for $-1\le t\le1$, and let $f(x,y)=x$. Locate the stationary value and evaluate the integral.

Differentiating gives $\mathbf{r}'(t)=(-3t^2\sin(t^3),3t^2\cos(t^3))$, so $\|\mathbf{r}'(t)\|=3t^2$. The speed is zero at $t=0$. Along the curve, $f(\mathbf{r}(t))=\cos(t^3)$, and therefore

\int_C f\,ds =\int_{-1}^{1}\cos(t^3)3t^2\,dt.

Set $u=t^3$, so $du=3t^2\,dt$, with bounds $u=-1$ and $u=1$. Then

\int_C f\,ds =\int_{-1}^{1}\cos u\,du =[\sin u]_{-1}^{1} =2\sin 1.

To check, use the regular unit-circle parameter $\boldsymbol{\gamma}(u)=(\cos u,\sin u)$ on $[-1,1]$. Its speed is $1$, and the field restricted to it is $\cos u$, giving the same value $2\sin1$. This alternative parameter removes the artificial slowing at $t=0$.

A Common Mistake: Using Signed Velocity as Speed

For $\mathbf{r}(t)=(t^2,0)$, $-1\le t\le1$, the velocity is $(2t,0)$. The curve travels from $x=1$ to $x=0$, then back to $x=1$. For $f=1$, an incorrect calculation might use the first component $2t$ as the speed:

\int_{-1}^{1}2t\,dt=0.

This result wrongly cancels distance travelled in opposite directions. Speed is the norm, $\|\mathbf{r}'(t)\|=2|t|$, so the correct calculation is

\int_C 1\,ds =\int_{-1}^{1}2|t|\,dt =2\left(\int_{-1}^{0}(-t)\,dt+\int_0^1t\,dt\right) =2\left(\frac12+\frac12\right)=2.

The check is geometric: the segment from $x=0$ to $x=1$ is traversed twice, giving total length $1+1=2$. A zero velocity at the turnaround point does not cancel the distance from either traversal.

When to Split and When to Reparametrize

Subdivision is usually the simplest choice when there are finitely many stationary values and the velocity is nonzero between them. It makes each regular portion explicit and allows the integral to be computed piece by piece. An alternative parameter is useful when the zero results from an avoidable change in parameter, as in the circular-arc example.

Whichever method is used, retain the same path and traversal, and calculate the norm of the entire velocity vector. If a parameter interval contains a stationary portion, that portion contributes no distance; if the path is traversed more than once, each traversal contributes its own integral. These checks keep the geometry and the calculation consistent.

Check Your Understanding

For each question, identify the role of the velocity and the distance element.

For $\mathbf{r}(t)=(t^5,t^2)$, find every parameter value in $[-1,1]$ where the velocity vanishes.
Why can an isolated zero of $\mathbf{r}'(t)$ be used as a subdivision point without necessarily making the curve non-piecewise-regular?
For $\mathbf{r}(t)=(t^2,0)$ and $f=1$ on $[-1,1]$, explain why integrating $2t$ does not give the scalar line integral.
In a change of parameter $u=\phi(t)$, what factor appears in the speed when $\phi'(t)$ can be negative?
For $\mathbf{r}(t)=(\cos(t^3),\sin(t^3))$, what regular parameter removes the zero speed at $t=0$?