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Arc Length and Curve Length · Tutorial 419 of 998

Numerical Approximation of Arc-Length Integrals

You will translate curve-length formulas into numerical estimates using midpoint and trapezoidal sums, then check the arithmetic and report the result clearly.

Intermediate 10 min read

What You'll Learn

  • Write graph and parametric arc length as integrals of speed.
  • Choose nodes and an interval width for a numerical estimate.
  • Apply composite midpoint and trapezoidal rules to a length integrand.
  • Distinguish numerical integration of speed from adding sampled chord lengths.
  • Check numerical arithmetic and report units and rounding precision.

Why Approximate Arc Length?

Some arc-length integrals have antiderivatives that are difficult to express using elementary functions. In those cases, the length formula is still useful: it turns the geometric problem into a definite integral that can be estimated numerically.

The numerical method does not change the curve or its length. It estimates the integral of the curve’s speed over a chosen interval. The result is meaningful only if the original length integral is set up correctly, the partition is stated, and the final units and rounding are reported.

From a Curve to a Numerical Integral

For a graph \(y=f(x)\), where \(f\) has a continuous derivative on \([a,b]\), the arc length is the integral of the graph’s speed with respect to \(x\). For a parametrized plane curve \(x=x(t)\), \(y=y(t)\), with continuous first derivatives on \([\alpha,\beta]\), the speed is the magnitude of its velocity vector. These standard arc-length formulas give the integrands to be approximated.

$$ L_{\text{graph}}=\int_a^b \sqrt{1+\bigl(f'(x)\bigr)^2}\,dx $$
$$ L_{\text{parametric}}=\int_\alpha^\beta \sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2}\,dt $$

In either case, the numerical task has the form \(\int_a^b g(u)\,du\), where \(g\) is the relevant speed function. For a graph, \(g(x)=\sqrt{1+(f'(x))^2}\). For a parametrized curve, \(g(t)=\sqrt{(x'(t))^2+(y'(t))^2}\). A continuous speed function on a closed interval can be evaluated at the partition points and used in the basic numerical rules below.

Definition: A partition divides an interval \([a,b]\) into smaller subintervals. For an equal partition into \(n\) pieces, the width is \(h=(b-a)/n\), and the nodes are \(x_i=a+ih\), for \(i=0,1,\dots,n\).

Composite Trapezoidal and Midpoint Rules

The composite trapezoidal rule approximates the area under \(g\) on each subinterval by the area of a trapezoid. The composite midpoint rule approximates it by the area of a rectangle whose height is the value of \(g\) at the subinterval’s midpoint. Both methods can be used to estimate arc length because they approximate the definite integral of speed.

Rule: For \(g\) on \([a,b]\), divided into \(n\) equal subintervals of width \(h\), the composite trapezoidal estimate is $$ T_n=\frac{h}{2}\left[g(x_0)+2\sum_{i=1}^{n-1}g(x_i)+g(x_n)\right]. $$ The composite midpoint estimate is $$ M_n=h\sum_{i=0}^{n-1}g\left(\frac{x_i+x_{i+1}}{2}\right). $$

The endpoint values in \(T_n\) occur once, while each interior node occurs twice: it is the right endpoint of one trapezoid and the left endpoint of the next. In \(M_n\), each subinterval contributes one midpoint value. These weights are part of the rules, not optional adjustments.

Why the Trapezoidal Weights Have This Form

On a single subinterval \([x_i,x_{i+1}]\) of width \(h\), the trapezoid with vertical heights \(g(x_i)\) and \(g(x_{i+1})\) has area equal to its width times the average of those heights:

$$ \frac{h}{2}\bigl(g(x_i)+g(x_{i+1})\bigr). $$

Adding this expression over all \(n\) subintervals gives the composite rule. The first and last function values appear in only one term. Each interior value appears in the terms for the subinterval immediately to its left and the one immediately to its right. Therefore, the total is

$$ \begin{aligned} T_n &=\frac{h}{2}\bigl[(g(x_0)+g(x_1))+(g(x_1)+g(x_2))+\cdots+(g(x_{n-1})+g(x_n))\bigr]\\ &=\frac{h}{2}\left[g(x_0)+2\sum_{i=1}^{n-1}g(x_i)+g(x_n)\right]. \end{aligned} $$

This explains both the factor \(h/2\) and the doubled interior weights. The midpoint formula follows from adding the areas \(h\,g((x_i+x_{i+1})/2)\) of the midpoint rectangles.

Worked Example: A Graph with Two Subintervals

Worked Example: A Graph with Two Subintervals

Consider the graph \(y=x^2\) on \(0\leq x\leq1\), with coordinates measured in consistent length units. Its derivative is \(f'(x)=2x\), so the speed integrand is \(g(x)=\sqrt{1+4x^2}\). Use two equal subintervals, so \(h=(1-0)/2=0.5\), and calculate both the trapezoidal and midpoint estimates.

The trapezoidal nodes are \(x_0=0\), \(x_1=0.5\), and \(x_2=1\). The corresponding values are

$$ g(0)=1,\qquad g(0.5)=\sqrt{2}\approx1.414214,\qquad g(1)=\sqrt{5}\approx2.236068. $$

Substitute these values, retaining the doubled interior value:

$$ \begin{aligned} T_2 &=\frac{0.5}{2}\bigl(1+2\sqrt{2}+\sqrt{5}\bigr)\\ &=0.25\bigl(1+2(1.414213562)+2.236067978\bigr)\\ &=0.25(6.064495102)\\ &\approx1.516124. \end{aligned} $$

For the midpoint estimate, the two midpoints are \(0.25\) and \(0.75\). Their speed values are

$$ g(0.25)=\sqrt{1.25}\approx1.118034,\qquad g(0.75)=\sqrt{3.25}\approx1.802776. $$

Thus

$$ M_2=0.5(1.118034+1.802776)\approx1.460405. $$

Both estimates are lengths in the chosen coordinate units. As an arithmetic check, \(1.118034+1.802776=2.920810\), and half of that sum is \(1.460405\). The trapezoidal calculation likewise uses \(0.25\) times \(6.064495102\), which is \(1.516123776\) before rounding to six decimal places.

Worked Example: A Parametrized Curve

Worked Example: A Parametrized Curve

Let \(x(t)=t^2\) metres and \(y(t)=t^3\) metres for \(0\leq t\leq1\), where \(t\) is dimensionless. Estimate the length using the composite trapezoidal rule with two equal subintervals. The derivatives are \(x'(t)=2t\) metres and \(y'(t)=3t^2\) metres, so the speed is

$$ g(t)=\sqrt{4t^2+9t^4}\ \text{metres per unit of }t. $$

Here \(h=(1-0)/2=0.5\), with nodes \(0\), \(0.5\), and \(1\). Evaluate the speed at each node:

$$ \begin{aligned} g(0)&=0,\\ g(0.5)&=\sqrt{4(0.5)^2+9(0.5)^4} =\sqrt{1+0.5625} =1.25,\\ g(1)&=\sqrt{4+9} =\sqrt{13} \approx3.605551. \end{aligned} $$

Apply the composite trapezoidal formula:

$$ \begin{aligned} T_2 &=\frac{0.5}{2}\bigl(g(0)+2g(0.5)+g(1)\bigr)\\ &=0.25\bigl(0+2(1.25)+3.605551275\bigr)\\ &=0.25(6.105551275)\\ &\approx1.526388\ \text{m}. \end{aligned} $$

To check the arithmetic, \(2(1.25)=2.5\), and \(2.5+3.605551275=6.105551275\). Multiplying by \(0.25\) gives \(1.526387819\), which rounds to \(1.526388\) metres to six decimal places. The parameter is dimensionless, so integrating speed over \(t\) leaves the length unit as metres.

Worked Example: Choosing a Finer Partition

Worked Example: Choosing a Finer Partition

For the graph \(y=x^3\) on \(0\leq x\leq1\), let both coordinates be measured in the same length unit. Use the composite trapezoidal rule with four equal subintervals. Since \(f'(x)=3x^2\), the integrand is \(g(x)=\sqrt{1+9x^4}\), and \(h=0.25\).

The nodes and speed values, rounded to six decimal places for display, are

\(x_i\)Calculation of \(g(x_i)\)Approximate value
0\(\sqrt{1+9(0)^4}\)1.000000
0.25\(\sqrt{1+9(0.25)^4}\)1.017426
0.50\(\sqrt{1+9(0.50)^4}\)1.250000
0.75\(\sqrt{1+9(0.75)^4}\)1.961544
1\(\sqrt{10}\)3.162278

Using the unrounded values in the calculation,

$$ \begin{aligned} T_4 &=\frac{0.25}{2}\bigl[g(0)+2(g(0.25)+g(0.50)+g(0.75))+g(1)\bigr]\\ &=0.125\bigl[1+2(1.01742629+1.25+1.96154435)+3.16227766\bigr]\\ &\approx0.125(12.62021894)\\ &\approx1.577527. \end{aligned} $$

As a check, the three interior values sum to approximately \(4.22897064\); doubling gives \(8.45794128\). Adding the endpoints gives \(1+8.45794128+3.16227766=12.62021894\). Multiplication by \(0.125\) gives \(1.57752737\), or \(1.577527\) length units to six decimal places. The displayed table values are rounded, so using only those displayed values may change the last digit.

Choosing a Partition and Reporting the Estimate

The partition should reflect the interval and the behavior of the speed function. Equal subintervals are convenient because the formulas above apply directly. A larger \(n\) means a smaller width \(h\), and hence more evaluations of the integrand. If the speed changes rapidly in some part of the interval, a uniform partition may need many points to represent it adequately. When using unequal subintervals, apply the single-interval trapezoid formula separately on each piece, using that piece’s actual width.

Before computing, identify the variable of integration and the full interval. For a graph this is the \(x\)-interval; for a parametrized curve it is the parameter interval. Then evaluate the speed—not the original function or its derivative alone—at the required nodes or midpoints. If the curve is described in pieces or the speed formula changes at a point, split the integral at that point and estimate each piece separately.

Report the method, the number of subintervals, and the units with the result. Keep additional digits during the arithmetic and round only the final estimate to a stated precision, such as six decimal places or three significant figures. The number of digits shown describes rounding; by itself it does not guarantee that the estimate is accurate to that many digits.

A Common Weighting Mistake

A frequent error is to evaluate \(g\) at the correct trapezoidal nodes but add all values with equal weight. For the first example, that incorrect calculation would be

$$ T_{\mathrm{wrong}} =\frac{0.5}{2}(1+\sqrt{2}+\sqrt{5}) =0.25(4.650281540) \approx1.162570. $$

The interior value must be counted twice because it contributes to both adjacent trapezoids. The correctly weighted calculation is

$$ T_2 =\frac{0.5}{2}(1+2\sqrt{2}+\sqrt{5}) =0.25(6.064495102) \approx1.516124. $$

The difference is not a rounding issue; it comes from using the wrong weights. A useful check is to write the area of each individual trapezoid first, then add them. That procedure makes it clear which values occur once and which occur twice.

Numerical Integration Is Not a Chord Sum

There is another distinction worth keeping clear. The trapezoidal and midpoint rules approximate the integral of the speed function. They do not directly add the lengths of straight segments joining sampled points on the curve. A sum of such chord lengths is a separate geometric approximation. Both approaches use sampled information, but the formulas and interpretations differ. In this tutorial, the stated estimates come from numerical integration of the arc-length integrand.

Numerical estimates are useful when an exact antiderivative is unavailable or inconvenient. Their reliability depends on a correct length setup, appropriate sampling of the speed, and transparent reporting. The methods here provide practical estimates; selecting an error guarantee or a systematic refinement procedure requires additional analysis.

Check Your Understanding

Use the definitions and rules above to answer these questions. Include the partition and units where appropriate.

  1. For a graph \(y=f(x)\), what speed function must be integrated to estimate its length on \([a,b]\)?
  2. For \(n=4\) equal subintervals on \([2,6]\), what is the width \(h\)?
  3. In the composite trapezoidal rule, how many times is each interior node value counted, and why?
  4. What is the difference between a midpoint estimate of the speed integral and a sum of sampled chord lengths?
  5. Why should intermediate values generally retain more digits than the final reported estimate?

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