When Arc Length Requires a Numerical Estimate
For a parametrized curve, arc length is found by integrating its speed over the parameter interval. Sometimes the speed has an elementary antiderivative, but sometimes it does not. In the latter case, the definite integral still has a well-defined value, even if it cannot be expressed using the elementary functions available in an AP Calculus course.
A calculator can approximate such an integral by numerical integration. The important work is not just entering an expression: first establish the correct speed and interval, then interpret the numerical output responsibly. A decimal estimate and a guaranteed bound are different kinds of information.
The condition \(x,y\in C^1([a,b])\) is a convenient sufficient hypothesis: both coordinate functions and their first derivatives are continuous on the whole interval. Differentiability alone is not enough to guarantee a finite arc length represented by an ordinary definite integral. In the examples below, the coordinate functions are continuously differentiable.
From the Parametrization to the Calculator Entry
The speed is obtained by differentiating each coordinate with respect to the parameter and taking the magnitude of the velocity vector. The parameter bounds must be the bounds for the motion being measured; they are not necessarily the endpoint coordinates of the curve.
Find \(x'(t)\) and \(y'(t)\) on the stated parameter interval.
Use \(v(t)=\sqrt{(x'(t))^2+(y'(t))^2}\), taking the nonnegative square root.
Set \(L=\int_a^b v(t)\,dt\), then use an appropriate numerical method if no convenient antiderivative is available.
Confirm that the integrand is nonnegative, the interval is correct, the units are length, and any reported precision is justified.
A calculator’s numerical integration command usually applies an algorithm based on sampled values of the integrand. Simpson’s Rule is one common method and is also useful for understanding what numerical quadrature does.
Worked Example: A Non-Elementary Speed Integral
Worked Example: Estimate the Length of a Cubic Parametrization
Find the length of \(x(t)=t\), \(y(t)=t^3/3\) for \(0\le t\le1\), to four decimal places. The speed is $$ x'(t)=1,\qquad y'(t)=t^2,\qquad v(t)=\sqrt{1+t^4}. $$ Thus $$ L=\int_0^1\sqrt{1+t^4}\,dt. $$ This integral has no convenient elementary antiderivative. Use composite Simpson’s Rule with \(n=8\), so \(h=(1-0)/8=0.125\). The sampled values, rounded to six decimal places, are:
| \(t_i\) | \(\sqrt{1+t_i^4}\) | Weight in Simpson’s Rule |
|---|---|---|
| 0.000 | 1.000000 | 1 |
| 0.125 | 1.000122 | 4 |
| 0.250 | 1.001951 | 2 |
| 0.375 | 1.009839 | 4 |
| 0.500 | 1.030776 | 2 |
| 0.625 | 1.073586 | 4 |
| 0.750 | 1.147347 | 2 |
| 0.875 | 1.259437 | 4 |
| 1.000 | 1.414214 | 1 |
The odd-indexed interior values sum to \(1.000122+1.009839+1.073586+1.259437=4.342984\). The even-indexed interior values sum to \(1.001951+1.030776+1.147347=3.180074\). Therefore, $$ L\approx \frac{0.125}{3}\left[1.000000+1.414214+4(4.342984)+2(3.180074)\right] =\frac{26.146298}{24} \approx1.089429. $$ A calculator’s numerical integration gives approximately the same value. Rounded to four decimal places, the estimate is \(L\approx1.0894\). The integrand is positive and the interval is one parameter unit long, so a length slightly greater than \(1\) is reasonable. A separate bound for this value appears below.
Why Numerical Quadrature Works
The speed function \(v(t)\) is continuous on the closed interval, so it is integrable there. Divide the interval into short pieces and sample the speed at points in each piece. A weighted sum of those sampled speeds, multiplied by the parameter step size, approximates the accumulated area under the speed graph. As the pieces become narrower, these sums approach the definite integral.
For Simpson’s Rule, parabolas are used to approximate the integrand across pairs of subintervals. The weights \(1,4,2,4,\ldots,2,4,1\) come from integrating those parabolic approximations. This explains why the method often works well for smooth speed functions: it uses more information about the curve of the integrand than a simple rectangle sum. It remains a numerical approximation, however; smoothness and a calculator display do not, by themselves, prove a particular number of correct decimal places.
Using Inequalities to Bound the Length
Bounds can be useful even when a numerical estimate is available. For \(0\le u\le1\), the following inequalities hold: $$ 1+\frac{u}{2}-\frac{u^2}{8}\le\sqrt{1+u}\le1+\frac{u}{2}. $$ For the upper inequality, both sides are nonnegative and \((1+u/2)^2=1+u+u^2/4\ge1+u\). For the lower inequality, let \(q=1+u/2-u^2/8\), which is positive on this interval. Then $$ q^2=1+u-\frac{u^3}{8}+\frac{u^4}{64}\le1+u, $$ because \(u^4/64\le u^3/8\) for \(0\le u\le1\). Taking nonnegative square roots proves the lower inequality.
For the first example, set \(u=t^4\). Since \(0\le t^4\le1\), the inequalities apply throughout the integration interval. Integrating both sides gives a rigorous, if not especially narrow, interval: $$ \int_0^1\left(1+\frac{t^4}{2}-\frac{t^8}{8}\right)dt \le L\le \int_0^1\left(1+\frac{t^4}{2}\right)dt, $$ $$ 1+\frac{1}{10}-\frac{1}{72}\le L\le1+\frac{1}{10}, \qquad 1.086111\ldots\le L\le1.100000. $$ The calculator estimate \(1.089429\) lies inside this interval. The bound confirms a range for the length, but it is too wide to establish the four-decimal rounding by itself.
Monotonicity gives another direct bounding method. If a continuous function \(f\) is increasing on \([a,b]\), then on each equal subinterval its left-endpoint value is no larger than its values inside the subinterval, while its right-endpoint value is no smaller. Consequently, the left sum is a lower bound for \(\int_a^b f(t)\,dt\), and the right sum is an upper bound. The speed \(\sqrt{1+t^4}\) is increasing on \([0,1]\), so these sums bound the first example’s length. With four subintervals, the left sum is approximately \(1.045019\) and the right sum is approximately \(1.148572\). These are valid but fairly broad bounds; increasing the number of subintervals narrows the gap.
Worked Example: Scaling a Curve and Its Length
Worked Example: Double Every Coordinate
Consider \(x(t)=2t\), \(y(t)=2t^3/3\) for \(0\le t\le1\). Differentiate to get \(x'(t)=2\) and \(y'(t)=2t^2\). The speed and length are $$ v(t)=\sqrt{2^2+(2t^2)^2}=2\sqrt{1+t^4}, \qquad L=\int_0^1 2\sqrt{1+t^4}\,dt. $$ Using the first example’s numerical estimate, $$ L\approx2(1.089429)=2.178858\approx2.1789 $$ to four decimal places. The final value is in the same length units as the coordinates. As a check, doubling both coordinate functions doubles each velocity component and therefore doubles the speed at every parameter value; hence the length must double. The computed result has exactly that relationship.
Worked Example: A Different Non-Elementary Speed
Worked Example: Use Simpson’s Rule and Check with Bounds
Find the length for \(x(t)=t\), \(y(t)=t^3/6\) on \(0\le t\le1\). The derivatives are \(x'(t)=1\) and \(y'(t)=t^2/2\), giving $$ v(t)=\sqrt{1+\frac{t^4}{4}}, \qquad L=\int_0^1\sqrt{1+\frac{t^4}{4}}\,dt. $$ Use Simpson’s Rule with \(n=4\), \(h=0.25\). At \(t=0,0.25,0.50,0.75,1\), the speed values are approximately \(1.000000,\ 1.000488,\ 1.007782,\ 1.038798,\ 1.118034\). Substitution gives $$ L\approx\frac{0.25}{3}\left[1.000000+1.118034+4(1.000488+1.038798)+2(1.007782)\right] \approx1.024229. $$ A calculator’s numerical integration gives approximately \(1.02421\), so the length rounds to \(1.0242\) to four decimal places. For a check, use the proved inequality with \(u=t^4/4\). This \(u\) lies between \(0\) and \(1\), and integration yields $$ 1+\frac{1}{40}-\frac{1}{1152}\le L\le1+\frac{1}{40}, \qquad 1.024132\ldots\le L\le1.025000. $$ The numerical value lies within the valid bounds. These bounds are not narrow enough to certify the four-decimal rounding, so that precision is a numerical estimate rather than a conclusion guaranteed by this inequality.
Common Pitfalls and a Practical Precision Check
A frequent error is to confuse a plausible calculator display with a proof of the displayed digits. For instance, reporting \(1.0894\) as a guaranteed four-decimal answer solely because a calculator displays \(1.0894\) is not justified: the display does not state a rigorous error bound. The correction is to identify the result as a numerical estimate, compare it with a refined computation, and use an error estimate or rigorous bounds if guaranteed rounding is required.
Another error is to omit the square root or to integrate a signed component instead of the speed. Arc length uses \(\sqrt{(x')^2+(y')^2}\), which is nonnegative. Also check the parameter interval carefully: changing the interval changes the portion of the curve whose length is measured. If the coordinates represent physical position, the velocity has distance-per-time units and integrating it with respect to time gives distance units.
- Keep enough calculator precision during intermediate steps; round only the final reported estimate.
- Record the method and, when using Simpson’s Rule, the number of subintervals.
- Do not call a numerical value exact unless an exact evaluation or proof supports that claim.
- For guaranteed rounding, establish bounds narrow enough that both endpoints round to the same requested value.
Check Your Understanding
Use the definitions and checks in this tutorial to decide what each setup or numerical claim establishes.
- For \(x(t)=t\), \(y(t)=t^3/3\) on \(0\le t\le1\), write the speed integral.
- Why does continuity of the speed matter when using the ordinary definite-integral arc-length formula?
- What two features of Simpson’s Rule distinguish its estimate from an exact integral value?
- For an increasing continuous speed, which equal-subinterval sum is a lower bound, and which is an upper bound?
- A calculator displays \(2.1789\). What additional information would be needed to claim that this is guaranteed to four decimal places?
