When a Steep Endpoint Still Has Finite Length
A graph can have a vertical tangent at an endpoint: its slope becomes unbounded as the endpoint is approached. That does not by itself tell us whether the graph has finite arc length. The length depends on how quickly the slope grows and, in some cases, on more complicated oscillations.
The graph-length formula applies on intervals where the function is sufficiently smooth. When the derivative becomes unbounded at an endpoint, we apply that formula on truncated intervals and take a limit. The resulting improper integral provides a precise test for finite length.
Length as an Improper Integral
Suppose \(b>0\), \(f\) is continuous on \([0,b]\), and \(f\) is continuously differentiable on \([\delta,b]\) for every \(\delta>0\). The continuity condition includes the endpoint; the smoothness condition is only required away from it. On each truncated interval, the graph-length formula gives the length
The length of the graph on \([0,b]\) is the limit of these truncated lengths as the lower endpoint approaches zero. The limit may be finite or infinite. Because the integrand is nonnegative, \(L(\delta)\) increases or stays the same as \(\delta\) decreases, so this limit exists in the extended sense.
To see why truncating captures the whole graph, define its length as the supremum of polygonal lengths through ordered points on the graph. Each truncated graph is part of the full graph, so its length cannot exceed the full length. Conversely, take any finite partition of \([0,b]\) and replace its initial point \(0\) by \(\delta>0\), leaving the other partition points fixed. By continuity of \(f\) at zero, the first chord length tends to the chord from \(0\) to the first partition point as \(\delta\to0^+\). Thus the truncated lengths approach at least every polygonal length on the full graph. Taking the supremum gives the claimed limit, including when it is infinite.
A Useful Comparison for the Integrand
For every real number \(u\),
The lower bound follows by squaring nonnegative quantities: \(u^2\leq 1+u^2\). For the upper bound, both sides are nonnegative and \(1+u^2\leq (1+|u|)^2=1+2|u|+u^2\). Applying these inequalities to \(u=f'(x)\) gives
Consequently, if \(|f'|\) has a finite improper integral near the endpoint, then the arc-length integral is finite there: its integrand is bounded above by \(1+|f'|\). If \(|f'|\) has an infinite improper integral, the lower bound shows that the arc length is infinite. These comparisons are useful when they apply, but they do not settle every case; oscillating derivatives, for example, require care.
Worked Examples with Power-Law Singularities
Worked Example: The Square-Root Graph
Find the length of \(y=\sqrt{x}\) on \([0,1]\). Set \(f(0)=0\), so \(f\) is continuous at the endpoint. For \(x>0\), \(f'(x)=1/(2\sqrt{x})\), which is unbounded as \(x\to0^+\). On every interval \([\delta,1]\), however, \(f\) is continuously differentiable. The length is therefore the improper integral
Use \(x=t^2\), so \(dx=2t\,dt\). Since \(t>0\), the integrand becomes \(\sqrt{1+1/(4t^2)}=\sqrt{4t^2+1}/(2t)\). The truncated limits \(x=\delta\) and \(x=1\) correspond to \(t=\sqrt{\delta}\) and \(t=1\). Hence
An antiderivative is \(\frac{t}{2}\sqrt{4t^2+1}+\frac14\ln(2t+\sqrt{4t^2+1})\). To check it, the derivative of the first term is \((4t^2+\tfrac12)/\sqrt{4t^2+1}\), and the derivative of the second is \(1/(2\sqrt{4t^2+1})\). Their sum is \((4t^2+1)/\sqrt{4t^2+1}=\sqrt{4t^2+1}\), as required. Evaluating at the limits gives
This is finite. The lower-end contribution tends to zero because \(t\to0\), \(\sqrt{4t^2+1}\to1\), and \(\ln(2t+\sqrt{4t^2+1})\to\ln(1)=0\). The antiderivative check confirms the integral evaluation, and the limiting value confirms that the endpoint singularity does not make the length infinite.
Worked Example: A Different Power Near the Endpoint
Find the length of \(y=x^{2/3}\) on \([0,1]\). Set \(f(0)=0\); this extension is continuous. For \(x>0\), \(f'(x)=\frac23x^{-1/3}\), so the derivative is unbounded at zero. On \([\delta,1]\) the graph is continuously differentiable, and
Set \(x=t^3\). Then \(dx=3t^2\,dt\), and, for \(t>0\),
The limits become \(t=\delta^{1/3}\) and \(t=1\). Therefore
An antiderivative is \((9t^2+4)^{3/2}/27\), since its derivative is \((1/27)(3/2)(9t^2+4)^{1/2}(18t)=t\sqrt{9t^2+4}\). Thus
The limit is finite, and differentiating the antiderivative verifies the evaluation. This graph also has finite length despite its unbounded endpoint derivative.
Power-Law Behavior: A Practical Criterion
Suppose that for some \(p>0\) there are constants \(c,C>0\) and \(\eta>0\) such that for \(0<x<\eta\),
This says that the derivative’s magnitude is bounded above and below by constant multiples of the same power near the endpoint. The comparison inequalities show that the arc-length integrand is at least \(c x^{-p}\) and at most \(1+C x^{-p}\). The power integral \(\int_0^\eta x^{-p}\,dx\) is finite for \(p<1\) and infinite for \(p\geq1\). It follows that the graph has finite length near zero when \(p<1\), and infinite length when \(p\geq1\).
For \(f(x)=x^\alpha\), where \(\alpha>0\), the derivative is \(\alpha x^{\alpha-1}\), with magnitude a constant times \(x^{-(1-\alpha)}\) when \(0<\alpha<1\). Here \(p=1-\alpha<1\), so the length is finite. Both worked examples fit this pattern. The criterion requires the two-sided power bounds; it should not be applied just because a derivative is unbounded.
Unbounded Derivative Does Not Guarantee Finite Length
Worked Example: An Oscillating Graph with Infinite Length
Define \(f(0)=0\) and \(f(x)=x^2\sin(1/x^2)\) for \(0<x\leq1\). Since \(|f(x)|\leq x^2\), \(f(x)\to0\) as \(x\to0^+\), so \(f\) is continuous at zero. It is continuously differentiable on every \([\delta,1]\), and
For each \(\delta>0\), the length integrand is at least \(|f'(x)|\). The inequality \(|a+b|\geq |a|-|b|\) therefore gives
In the first integral, substitute \(u=1/x^2\), so \(dx/x=-du/(2u)\). It follows that
This last integral diverges as \(\delta\to0^+\). Indeed, on each interval \([2\pi k,2\pi k+\pi/3]\), with positive integer \(k\), \(|\cos u|\geq1/2\). The contribution from that interval is at least \(\frac{\pi/6}{2\pi k+\pi/3}\). The sum of these lower bounds diverges by comparison with the harmonic series. Meanwhile, \(2\int_\delta^1 x\,dx=1-\delta^2\leq1\). Thus the lower bound for the truncated arc length tends to infinity, so the graph has infinite length. Its derivative oscillates as well as becoming large; the two-sided power-law criterion does not apply.
A Common Miscalculation
A frequent error is to treat every unbounded derivative as though it forced the length integral to diverge. For \(f(x)=\sqrt{x}\), one might incorrectly claim that \(\sqrt{1+(f'(x))^2}=\sqrt{1+1/(4x)}\) behaves like \(1/x\), and hence that the length is infinite. That is not the correct leading behavior: the dominant term inside the square root is \(1/(4x)\), so its square root has size \(1/(2\sqrt{x})\), not \(1/x\).
The comparison also gives a direct finite bound on \((0,1]\):
Since the actual integrand is bounded above by an integrable function, its improper integral is finite. The exact calculation above gives its value. The correction is to analyze the square root of the squared derivative, rather than confusing \(|f'|\) with \((f')^2\).
How to Test an Endpoint
Confirm that the graph is continuous at the endpoint and continuously differentiable on every truncated interval.
Integrate \(\sqrt{1+(f')^2}\) from a positive cutoff to the other endpoint.
Evaluate the integral or compare its local behavior with a known convergent or divergent improper integral.
A finite limit means finite graph length; a limit of infinity means infinite graph length.
The distinction matters geometrically. A curve can become arbitrarily steep while adding only a finite amount of length, as in the square-root example. Conversely, an oscillating graph can keep accumulating length near an endpoint even while its height tends to zero. The improper-integral test measures that accumulation rather than relying on the appearance of the graph or on the size of its derivative at a single point.
Check Your Understanding
For each question, pay attention to the endpoint limit, not just the value of the derivative at the endpoint.
- What regularity conditions allow the graph-length formula to be applied on every truncated interval \([\delta,b]\)?
- Why does \(|f'(x)|\leq\sqrt{1+(f'(x))^2}\leq1+|f'(x)|\) help test length?
- If \(|f'(x)|\) is bounded above and below by positive constant multiples of \(x^{-p}\), for which values of \(p\) is the length finite?
- Why is an unbounded derivative at an endpoint not enough to conclude that the graph has infinite length?
- For \(f(x)=x^2\sin(1/x^2)\), what integral comparison establishes that the length is infinite?
