Why Endpoints Need Special Attention
A polar arc-length calculation depends on more than finding an integrand and integrating it. The parameter interval must describe the intended piece of the curve, and its endpoints may require care. At a cusp, the parametrization can have zero speed; at a pole crossing, the radius is zero. Neither fact alone implies that the arc length is infinite or undefined.
The key is to distinguish three questions: what curve is traced, how often it is traced, and whether the length integral converges at any endpoint. A zero of the radius is a geometric event, not automatically a singularity in the length calculation.
When \(r\) and \(r'\) are continuous up to an endpoint, the formula applies there directly, including when \(r=0\). If the interval contains a cusp or another point where the parametrization changes regularity, split the calculation there and add the lengths of the pieces. Splitting does not itself change the length; it makes endpoint behavior and tracing easier to inspect.
What Happens at a Cusp?
A cusp is a sharp point on a curve. In a parametrization, it often occurs where the velocity becomes zero while the curve approaches the point along a non-smooth-looking path. The speed in polar coordinates is \(\sqrt{r^2+(r')^2}\), so a vanishing speed at the cusp usually makes the integrand smaller, not larger. The endpoint still needs to be checked, especially if the formula has only been defined on an open interval.
A useful local model is \(r(\theta_0+u)\sim c u^p\) for \(u>0\), where \(c\ne 0\), \(p>0\), and the derivative has the corresponding behavior \(r'(\theta_0+u)\sim cp u^{p-1}\). Here \(\sim\) means that the ratio of the two sides tends to \(1\) as \(u\to0^+\). In this model the derivative term determines the leading size of the speed.
This comparison is a local test for the stated type of behavior, not a claim that every curve called a cusp has exactly a power-law parametrization. When a specific curve is given, use its actual speed and evaluate the endpoint limit. A cusp should not be mistaken for an infinite-length endpoint merely because the direction of the curve changes sharply.
Worked Example: Length Up to a Polar Cusp
Worked Example: Length Up to a Polar Cusp
Consider \(r=\theta^2\) on \(0\leq\theta\leq1\). At \(\theta=0\), the curve reaches the pole. Locally, \(x=r\cos\theta\sim\theta^2\) and \(y=r\sin\theta\sim\theta^3\), so the two coordinate components have the characteristic unequal orders of a cusp when the curve is continued through the parameter value.
First differentiate the radius: $$ r'(\theta)=2\theta. $$ Then the polar speed is $$ \sqrt{r^2+(r')^2} =\sqrt{\theta^4+4\theta^2} =\theta\sqrt{\theta^2+4}, $$ where \(\theta\geq0\) on this interval. Thus $$ L=\int_0^1\theta\sqrt{\theta^2+4}\,d\theta. $$ Set \(u=\theta^2+4\), so \(du=2\theta\,d\theta\). The limits are \(u=4\) and \(u=5\), giving $$ L=\frac12\int_4^5 u^{1/2}\,du =\frac13\left[u^{3/2}\right]_4^5 =\frac{5\sqrt5-8}{3}. $$ This is finite. To check the antiderivative, differentiate \(\frac13(\theta^2+4)^{3/2}\): $$ \frac{d}{d\theta}\left(\frac13(\theta^2+4)^{3/2}\right) =\theta\sqrt{\theta^2+4}, $$ the required integrand. Substitution of the bounds gives the stated length.
The endpoint contributes no separate length at the cusp: it is already included by the convergent integral. If a curve is described on both sides of the cusp, calculate each side on its own interval and add the lengths; do not assume that one side accounts for both branches.
What Happens at a Pole Crossing?
A pole crossing occurs when \(r(\theta_0)=0\). The point on the curve is then the origin. If \(r\) is differentiable at \(\theta_0\), the speed there is $$ \sqrt{r(\theta_0)^2+r'(\theta_0)^2}=|r'(\theta_0)|. $$ In particular, a simple zero of \(r\), where \(r'(\theta_0)\ne0\), gives a finite, nonzero speed. The radius changes sign as the curve passes through the pole, but the length formula uses \(r^2\) and \((r')^2\); it does not require replacing a negative radius by a positive one.
For an interval that includes a pole crossing, it is often helpful to split the integral at that parameter value. This makes it easier to check the two one-sided pieces and to see whether the interval traces a loop more than once. The lengths on the pieces are then added. If the speed is continuous and finite at the crossing, the split is not required for convergence, but can still clarify the geometry.
Worked Example: A Simple Pole Crossing
Worked Example: A Simple Pole Crossing
Let \(r=2\cos\theta\) on \(0\leq\theta\leq\pi\). The radius is zero at \(\theta=\pi/2\), and it changes from positive to negative there. Differentiate: $$ r'(\theta)=-2\sin\theta. $$ The speed is $$ \sqrt{(2\cos\theta)^2+(-2\sin\theta)^2} =\sqrt{4\cos^2\theta+4\sin^2\theta} =2. $$ Therefore the full length on the interval is $$ L=\int_0^\pi 2\,d\theta=2\pi. $$ To check that this interval traces the circle once, write its Cartesian parametrization: $$ x=r\cos\theta=2\cos^2\theta=1+\cos(2\theta), \qquad y=r\sin\theta=2\sin\theta\cos\theta=\sin(2\theta). $$ As \(\theta\) runs from \(0\) to \(\pi\), \(2\theta\) runs from \(0\) to \(2\pi\), so the circle centered at \((1,0)\) with radius \(1\) is traced once. Splitting at the pole crossing gives $$ \int_0^{\pi/2}2\,d\theta=\pi, \qquad \int_{\pi/2}^{\pi}2\,d\theta=\pi. $$ The two pieces meet at the origin and together have length \(2\pi\), as required.
Choosing an Interval Without Repeated Tracing
Polar coordinates can describe the same Cartesian point using different pairs \((r,\theta)\). Consequently, an interval that appears to cover a full angular turn need not trace the curve just once. A useful check is to convert to Cartesian coordinates or to identify a period of the Cartesian parametrization. The polar length integrand alone does not reveal whether the interval repeats the curve.
Find where \(r=0\), where a cusp occurs, and where the formula or its derivative changes behavior.
Use the curve’s geometry or Cartesian parametrization to determine when the desired arc or loop has been traced once.
Evaluate separate pieces when this makes one-sided limits or the traced branches clearer.
Use the speed formula, taking endpoint limits if necessary. Add lengths only for distinct pieces of the intended arc.
Worked Example: Avoiding a Double Count
Worked Example: Avoiding a Double Count
For the same curve \(r=2\cos\theta\), consider the longer interval \(0\leq\theta\leq2\pi\). The speed remains \(2\), so direct integration gives $$ \int_0^{2\pi}2\,d\theta=4\pi. $$ That value is not the circumference of the circle: the interval traces the circle twice. Indeed, the Cartesian coordinates are \(x=1+\cos(2\theta)\) and \(y=\sin(2\theta)\). As \(\theta\) goes from \(0\) to \(2\pi\), the angle \(2\theta\) goes from \(0\) to \(4\pi\), making two full turns.
Thus the correct interval for the circumference is \(0\leq\theta\leq\pi\), with length $$ \int_0^\pi 2\,d\theta=2\pi. $$ The check is both geometric and computational: the Cartesian parametrization completes one turn on \([0,\pi]\), and the resulting length is the circumference of a circle of radius \(1\). Using \([0,2\pi]\) correctly measures the length of the parametrized traversal, but not the length of the circle counted just once.
Common Pitfalls and Practical Checks
Pitfall: treating a zero radius as an infinite-length singularity. For \(r=2\cos\theta\) at \(\theta=\pi/2\), the radius is zero, but the speed is exactly \(2\). There is no divergent endpoint integral. Check \(\sqrt{r^2+(r')^2}\), rather than deciding convergence from \(r=0\) alone.
Pitfall: using a signed speed. A speed is nonnegative. For example, when simplifying a square root near a cusp, use \(\sqrt{u^2}=|u|\), not \(u\) on an interval where \(u\) might be negative. Choose the sign only after the parameter interval is specified.
Pitfall: integrating over an interval that traces the same arc again. The calculation \(\int_0^{2\pi}2\,d\theta=4\pi\) for \(r=2\cos\theta\) is valid for that parametrized interval, but it counts the circle twice. Inspect the Cartesian parametrization or determine the period of the traced curve before interpreting the result as the length of one loop.
Near a cusp or pole crossing, a reliable calculation therefore has two separate checks. First, use limits and the speed to establish that the endpoint length is finite. Second, verify that the parameter interval covers precisely the desired arc, without omissions or repeated tracing. These checks address different issues and neither can replace the other.
Check Your Understanding
Use the speed and the curve’s tracing behavior to answer each question.
- If \(r(\theta_0)=0\) and \(r'(\theta_0)\ne0\), what is the polar speed at the pole crossing?
- For a local model \(r(\theta_0+u)\sim cu^p\) with \(p>0\) and the corresponding derivative asymptotic, why is the endpoint length finite?
- For \(r=2\cos\theta\), why does the interval \([0,\pi]\) trace the circle once while \([0,2\pi]\) traces it twice?
- When an interval is split at a pole crossing, how should the lengths of its two pieces be combined?
- Why is \(r=0\) by itself not enough to conclude that a polar arc-length integral diverges?