Choosing a Method for Arc Length
Arc-length problems do not all end in the same kind of answer. Some speed functions have manageable antiderivatives, so an exact expression is practical. Others lead to integrals that are best represented by a special function or approximated with numerical quadrature. The important first step is to set up the length correctly; the next is to choose a method suited to the resulting integral.
In each case, a geometric check provides an independent way to assess whether the answer is plausible. A curve joining two points must be at least as long as the straight segment between them. Bounds based on horizontal and vertical change can also give a useful upper estimate.
The expression under the integral is the speed of the parametrized curve, or the graph’s length factor with respect to \(x\). Once this integrand is identified, ask whether it has a useful elementary antiderivative. If so, exact evaluation is usually the clearest route. If not, a definite integral can still define the length exactly, but numerical quadrature may be the practical way to obtain a decimal estimate.
Worked Example: A Quadratic Graph
Worked Example: Exact Length of \(y=x^2\)
Find the length of \(y=x^2\) for \(-1\le x\le 1\). Differentiate the graph function: \(y'=2x\). The graph formula gives
Here an exact antiderivative is available. To derive it, set \(u=2x\), so \(dx=du/2\). Then use integration by parts on \(\int\sqrt{1+u^2}\,du\), taking the factors \(\sqrt{1+u^2}\) and \(du\):
Since \(u^2=(1+u^2)-1\), the remaining integral is \(\int\sqrt{1+u^2}\,du-\int(1+u^2)^{-1/2}\,du\). Solving for the original integral and using \(\int(1+u^2)^{-1/2}\,du=\operatorname{asinh}(u)\) gives
Returning to \(x\), an antiderivative is
Therefore, using the endpoint values,
Check by differentiation: differentiating the first term of \(F\) gives \(\frac12\sqrt{1+4x^2}+\frac{2x^2}{\sqrt{1+4x^2}}\). Differentiating the second gives \(\frac{1}{2\sqrt{1+4x^2}}\). Combining these terms yields \(\frac{1+4x^2}{\sqrt{1+4x^2}}=\sqrt{1+4x^2}\), as required.
Geometric check: the endpoints are \((-1,1)\) and \((1,1)\), so their straight-line distance is \(2\). The curve’s total horizontal change is \(2\), and its total vertical variation is \(2\), because it falls from height \(1\) to \(0\) and rises back to \(1\). Since length is at least the endpoint distance and at most the sum of these total changes, \(2\le L\le4\). The result \(2.958\) satisfies these bounds.
Worked Example: One Cycloid Arch
Worked Example: Exact Length of a Cycloid Arch
Consider the parametrized cycloid \(x=a(t-\sin t)\), \(y=a(1-\cos t)\), where \(a>0\) and \(0\le t\le2\pi\). This parameter interval traces one arch, from the cusp at \((0,0)\) to the next cusp at \((2\pi a,0)\). Differentiate both coordinates:
The speed simplifies as follows:
The last equality uses \(2-2\cos t=4\sin^2(t/2)\) and the fact that \(\sin(t/2)\ge0\) for \(0\le t\le2\pi\). Thus the length is
Check: differentiating \(-4a\cos(t/2)\) gives \(2a\sin(t/2)\), the speed. The arch’s endpoints are \(2\pi a\) apart, so its length must be at least \(2\pi a\). Also, the total horizontal change is \(2\pi a\), and the total vertical variation is \(2a\): the curve rises from \(0\) to \(2a\) and returns to \(0\). Hence \(2\pi a\le L\le2\pi a+2a\). Since \(2\pi a\le8a\le2\pi a+2a\), the exact result passes the check.
Worked Example: Numerical Length of an Ellipse
Worked Example: Approximating the Circumference of an Ellipse
Take the ellipse parametrized by \(x=5\cos t\), \(y=3\sin t\), for \(0\le t\le2\pi\). Its speed is \(\sqrt{25\sin^2t+9\cos^2t}\). Symmetry makes the full circumference four times the first-quadrant length, so
This integral is an elliptic integral; a convenient elementary antiderivative is not available. We can retain the definite integral as an exact description, or estimate it by numerical quadrature. For illustration, apply Simpson’s rule with four equal subintervals, so \(h=\pi/8\). Set \(f(t)=\sqrt{9+16\sin^2t}\). The five function values are
Using the exact trigonometric values at these points, Simpson’s approximation for the quarter circumference is
Thus \(C\approx4Q\approx25.529\), reported to three decimal places. The definite integral, rather than this rounded estimate, is the exact circumference. A more accurate decimal estimate can be obtained by refining the quadrature and checking its error.
Geometric check: the ellipse passes through the four axis endpoints. The four straight segments between consecutive endpoints each have length \(\sqrt{5^2+3^2}=\sqrt{34}\), so the circumference is at least \(4\sqrt{34}\approx23.324\). For an upper bound, each quadrant’s arc length is no greater than the sum of its horizontal and vertical changes, \(5+3=8\). Therefore \(4\sqrt{34}\le C\le32\), and the estimate \(25.529\) lies within this interval.
How to Choose: Exact Expression or Quadrature?
The examples illustrate a practical distinction. For the quadratic graph, the speed integral has an elementary antiderivative, so evaluating it exactly is straightforward. For the cycloid arch, simplification makes the speed elementary and its definite integral exact. For the ellipse, the integral is still an exact mathematical description, but its antiderivative belongs to the theory of elliptic integrals; numerical quadrature is a useful way to obtain a decimal circumference.
Geometric checks do not replace an antiderivative check or an error analysis. They can, however, expose a wrong setup, an impossible sign, or an implausible numerical result. For example, any proposed length smaller than the distance between the endpoints is certainly incorrect.
A Common Sign Mistake
When simplifying a square root, remember that \(\sqrt{u^2}=|u|\), not \(u\) in general. In the cycloid example, writing \(\sqrt{2-2\cos t}=2\sin(t/2)\) is valid on the one-arch interval \(0\le t\le2\pi\), because \(\sin(t/2)\ge0\) there. It is not valid across multiple arches without considering the sign.
For example, on \(0\le t\le4\pi\), the speed is \(2a|\sin(t/2)|\). The incorrect calculation that drops the absolute value gives
which cannot be a curve length. The correction is to retain the absolute value, or split the integral at \(t=2\pi\). Each arch has length \(8a\), so the two-arch length is \(16a\), which is positive and consistent with the geometry.
Check Your Understanding
Use the formulas and checks from this tutorial to answer the following questions.
- For a graph \(y=f(x)\), what condition on \(f\) is used in the stated arc-length formula?
- Why is \(2a\sin(t/2)\) a valid expression for the cycloid speed on one arch, but not on two arches without an absolute value?
- What is the difference between an exact definite-integral description and a decimal estimate obtained by quadrature?
- For the ellipse with semiaxes \(5\) and \(3\), why must its circumference be at least \(4\sqrt{34}\)?
- What basic geometric inequality can immediately reveal that a proposed curve length is wrong?
