Why a Zero-Speed Point Deserves a Closer Look
A parametric curve is described by coordinate functions \(x(t)\) and \(y(t)\). Its velocity is \(\bigl(x'(t),y'(t)\bigr)\), and its speed is the magnitude of that velocity. Usually, the speed tells us how quickly the point moves along the curve. But at a parameter value where both coordinate derivatives vanish, the speed is zero: the point is momentarily stationary.
That fact alone does not tell us whether the curve has a cusp, whether the motion reverses, or whether the arc length integral is difficult. Those are separate questions. A zero can be a harmless pause along a smooth path, a cusp, or a point where the parametrization retraces part of the curve. Careful diagnosis keeps these possibilities distinct.
For a continuously differentiable parametrization on \([a,b]\), the arc length is $$ L=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\,dt. $$ The integrand is the speed. If the speed is continuous, it remains defined and finite when it equals zero. Thus an isolated zero-speed point does not, by itself, invalidate the arc length formula.
First Diagnostic: Check the Motion on Both Sides
Finding \(x'(t_0)=y'(t_0)=0\) identifies an instant of rest, but not what happens immediately before and after it. To investigate, examine the velocity for parameter values on each side of \(t_0\). When the speed is positive, the unit velocity vector $$ \frac{\bigl(x'(t),y'(t)\bigr)}{v(t)} $$ gives the direction of motion. Its one-sided limits, when they exist, can reveal whether the motion passes through in a consistent direction or whether its direction changes abruptly.
A curve can have a cusp even though the arc length integrand is perfectly integrable there. Conversely, a parametrization can stop at a point on a smooth line and continue in the same direction. A third possibility is that the parametrization turns around and retraces part of its path. The geometric shape and the motion encoded by the parameter should both be considered.
Worked Example: A Harmless Stop on a Straight Path
Consider \(x(t)=t^3\), \(y(t)=t^3\) for \(-1\le t\le1\). Find the zero-speed parameter, determine the motion through it, and calculate the arc length.
First differentiate the coordinates: $$ x'(t)=3t^2,\qquad y'(t)=3t^2. $$ Both derivatives are zero at \(t=0\), so the speed is zero there. For every \(t\ne0\), both derivatives are positive, so the point moves in the direction \((1,1)\) on both sides of zero. It pauses at the origin, then continues in the same direction along the line \(y=x\).
The speed is $$ v(t)=\sqrt{(3t^2)^2+(3t^2)^2} =\sqrt{18t^4} =3\sqrt{2}\,t^2. $$ Therefore, $$ L=\int_{-1}^{1}3\sqrt{2}\,t^2\,dt =3\sqrt{2}\left[\frac{t^3}{3}\right]_{-1}^{1} =\sqrt{2}\bigl(1-(-1)\bigr) =2\sqrt{2}. $$ The endpoints are \((-1,-1)\) and \((1,1)\); the straight-line distance between them is $$ \sqrt{(1-(-1))^2+(1-(-1))^2} =\sqrt{8} =2\sqrt{2}. $$ This agrees with the integral. The zero speed is a pause, not a cusp or a reversal.
A Zero Can Occur at a Cusp
A cusp is a feature of the curve’s geometry, not just a name for any point where the velocity vanishes. One useful diagnostic is to compare the limiting directions of motion from the two sides. If those directions approach opposite vectors, the curve may meet itself at a sharp point rather than pass smoothly through it. This diagnostic must be combined with the actual coordinate behavior; a zero derivative alone is not enough to classify the point.
Worked Example: Arc Length Through a Cusp
Consider \(x(t)=t^2\), \(y(t)=t^3\) on \([-1,1]\). Locate the zero-speed point, examine the two sides, and find the arc length.
The derivatives are $$ x'(t)=2t,\qquad y'(t)=3t^2. $$ Both vanish at \(t=0\). For \(t<0\), \(x'(t)<0\) and \(y'(t)>0\); for \(t>0\), both derivatives are positive. More precisely, the velocity direction has the one-sided limits $$ \lim_{t\to0^-}\frac{(2t,3t^2)}{\sqrt{4t^2+9t^4}}=(-1,0), \qquad \lim_{t\to0^+}\frac{(2t,3t^2)}{\sqrt{4t^2+9t^4}}=(1,0). $$ The point approaches the origin from the left-hand branch and leaves along the right-hand branch with the opposite limiting direction. The curve has a cusp at the origin.
The speed is $$ v(t)=\sqrt{4t^2+9t^4}=|t|\sqrt{4+9t^2}. $$ Because of the absolute value, it is convenient to split the integral at \(0\). The two contributions are equal, so $$ L=2\int_0^1 t\sqrt{4+9t^2}\,dt. $$ Set \(u=4+9t^2\), so \(du=18t\,dt\). When \(t=0\), \(u=4\); when \(t=1\), \(u=13\). Then $$ L=2\cdot\frac{1}{18}\int_4^{13}u^{1/2}\,du =\frac{1}{9}\left[\frac{2}{3}u^{3/2}\right]_4^{13} =\frac{2}{27}\left(13^{3/2}-4^{3/2}\right) =\frac{2}{27}\left(13\sqrt{13}-8\right). $$ Differentiating the substitution antiderivative confirms the calculation: $$ \frac{d}{dt}\left(\frac{(4+9t^2)^{3/2}}{27}\right) =t\sqrt{4+9t^2}. $$ The integrand is continuous at \(0\), where its value is zero. The cusp does not make the length infinite or prevent the integral from being evaluated.
Retracing Is Different from Passing Through
A parametrization can also stop and reverse direction, tracing a portion of the same geometric path twice. Then the integral of speed measures the distance traveled with multiplicity: each traversal contributes its length. This is why the integral need not equal the length of the set of distinct points on the curve.
Worked Example: Stopping at the Origin and Retracing a Line
Consider \(x(t)=t^2\), \(y(t)=t^2\) on \([-1,1]\). Locate the zero-speed point and compute the length traveled. In this example, the parametrization travels toward the origin and then retraces the same line away from it.
The derivatives are $$ x'(t)=2t,\qquad y'(t)=2t, $$ so both vanish at \(t=0\). When \(t<0\), both derivatives are negative, and the point moves toward \((0,0)\). When \(t>0\), both are positive, and it moves away from the origin. The endpoints are both \((1,1)\).
The speed and length are $$ v(t)=\sqrt{(2t)^2+(2t)^2}=2\sqrt{2}\,|t|, $$ $$ L=\int_{-1}^{1}2\sqrt{2}\,|t|\,dt =2\sqrt{2}\left(\int_{-1}^{0}(-t)\,dt+\int_0^1 t\,dt\right). $$ The first integral is $$ \int_{-1}^{0}(-t)\,dt =\left[-\frac{t^2}{2}\right]_{-1}^{0} =\frac{1}{2}, $$ and the second is $$ \int_0^1 t\,dt =\left[\frac{t^2}{2}\right]_0^1 =\frac{1}{2}. $$ Thus $$ L=2\sqrt{2}\left(\frac{1}{2}+\frac{1}{2}\right)=2\sqrt{2}. $$ The line segment from the origin to \((1,1)\) has length \(\sqrt{2}\), and it is traversed twice. The computed length \(2\sqrt{2}\) accounts for both traversals.
Why Splitting at a Zero Works
When the coordinate functions have continuous derivatives, the speed is a continuous, nonnegative function. If \(t_0\) is inside the interval, additivity of definite integrals gives $$ \int_a^b v(t)\,dt = \int_a^{t_0}v(t)\,dt+\int_{t_0}^b v(t)\,dt. $$ This equality does not require \(v(t_0)>0\). If \(v(t_0)=0\), the value at that single point does not create a gap or a divergence in the integral. Splitting is useful because the speed may have different formulas on the two sides—for example, \(v(t)=|t|\)—or because the motion’s direction changes there.
Solve \(x'(t)=0\) and \(y'(t)=0\) together; a zero of only one coordinate derivative is not a zero-speed point unless the other is also zero.
Check the signs of the derivatives and, when useful, the one-sided limits of the unit velocity vector.
Use \(v(t)=\sqrt{(x'(t))^2+(y'(t))^2}\). Simplify carefully, retaining absolute values when required.
Split at a zero if it clarifies a sign change, cusp, reversal, or piecewise formula; then add all contributions.
A Short Derivation: Why the Speed Integrand Cannot Be Negative
Over a small parameter change \(\Delta t\), the displacement is approximately $$ \bigl(x'(t)\Delta t,\ y'(t)\Delta t\bigr). $$ Its magnitude is approximately $$ \sqrt{(x'(t)\Delta t)^2+(y'(t)\Delta t)^2} = \sqrt{(x'(t))^2+(y'(t))^2}\,|\Delta t|. $$ The absolute value matters: elapsed parameter change has a nonnegative size whether \(t\) increases or decreases. Dividing by \(|\Delta t|\) gives the magnitude of velocity, the speed. Adding these small distance contributions along the interval leads to the arc length integral of speed. This also explains why the integrand is a square root of squares and why a zero value means an instant with no motion, not a negative contribution.
Common Mistake: Treating Every Zero as a Cusp
A frequent error is to conclude “there is a cusp” as soon as both derivatives vanish. For \(x(t)=t^3\), \(y(t)=t^3\), both derivatives vanish at \(0\), but the point continues in the same direction along a straight line. The zero marks a pause, not a cusp. Another error is to omit the absolute value when simplifying a speed: for \(x=t^2,\ y=t^2\), writing \(v(t)=2\sqrt{2}\,t\) is wrong on \([-1,0)\), since that expression is negative there. The correct calculation is $$ v(t)=\sqrt{8t^2}=2\sqrt{2}\,|t|. $$ Speed cannot be negative. Use the coordinate derivatives and nearby motion—not the zero alone—to distinguish a pause, cusp, or reversal.
Why These Diagnostics Matter
The arc length integral records how far the parametrized point travels, including repeated portions. A zero-speed point can be geometrically harmless while still requiring an absolute value in the integrand, as in the retraced-line example. A cusp can have a finite, straightforward length integral. The parameter behavior determines whether the path is traversed once or more than once; the speed integral measures the total travel accordingly.
For AP problems, the practical checks are direct: solve for simultaneous zeros, identify the speed without losing absolute values, and determine whether splitting the integral helps. If the derivatives or parametrization are only defined piecewise, apply the formula on each interval where the required derivatives exist and add the resulting lengths. Do not infer a singularity, cusp, or infinite length solely from zero speed.
Check Your Understanding
For each question, distinguish what the zero speed establishes from what must be checked separately.
- For \(x(t)=t^3,\ y(t)=t^3\), at what parameter is the speed zero, and does the motion reverse there?
- For \(x(t)=t^2,\ y(t)=t^3\), what are the one-sided limits of the unit velocity direction at \(t=0\)?
- Why must the speed for \(x(t)=t^2,\ y(t)=t^2\) be written using an absolute value?
- What total length is counted when a parametrization traverses the same segment twice?
- Under what regularity condition does an isolated zero of the speed remain an ordinary, finite point in the arc length integral?
