Calculate or approximate the length

Straight paths that approach the curve

The curve is \(y = x^2\) from \(x = 0\) to \(x = 2\). One straight segment joins its two ends, and it has length \(\sqrt{2^2 + 4^2} \approx 4.47\).

Press Two segments. A new corner sits on the curve at x = 1, and the path follows the curve more closely. Its total length is about 4.58.

With four segments the total is about 4.63. A straight segment is the shortest path between two points, so a path with more corners along the curve can only be longer.

The chalk mark on the ruler is the length of the curve, which is about 4.65. The totals rise toward it, and each total stays below it.

Exact when the root simplifies

The curve is now any smooth curve from \(x = a\) to \(x = b\). The totals rise toward the arc length \(L\), which is the integral of \(\sqrt{1 + f'(x)^2}\).

The straight line that joins the two ends is the chord, the shortest path between them. This means that \(L\) is at least the length of the chord.

Sometimes \(1 + f'(x)^2\) is the square of a simpler expression. Then the root disappears, and you can find \(L\) as an exact number.

Often the root does not disappear. Then a calculator evaluates the integral, or a total of segments approximates it.

Notation

1. Start from the integral, which needs the square of the derivative.

   \[\cy{L} = \int_a^b \sqrt{1 + \left(f'(x)\right)^2}\,dx\]

2. Expand \(1 + f'(x)^2\). If it equals \(g(x)^2\) for an expression \(g\) that is positive on \([a, b]\), the root equals \(g\).

   \[\begin{aligned} \sqrt{1 + f'(x)^2} &= g(x) \\ \text{when } 1 + f'(x)^2 &= g(x)^2 \end{aligned}\]

3. If \(1 + f'(x)^2\) is linear in \(x\), substitute \(u\) for it. Then integrate a power of \(u\).

   In both cases you can find an antiderivative, so the length is exact.

4. If neither works, enter the integral in a calculator. Give the decimal to three places.

   A decimal from a calculator is an approximation, not an exact value.

5. Test the result. The curve is at least as long as the chord that joins its ends.

   \[L \ge \sqrt{(b - a)^2 + \left(f(b) - f(a)\right)^2}\]

When it fails

For \(y = x^2\), the expression \(1 + f'(x)^2\) is \(1 + 4x^2\). It is not the square of any simple expression, so the root stays.

The rules of this course give no antiderivative for this integral. The calculator gives L ≈ 4.647.

Example 1: a root that a substitution removes

Find the length of \(y = \dfrac43 x^{3/2}\) from \(x = 0\) to \(x = 2\).

1. Find the derivative. Square it and add 1. The thick segment is about \(\sqrt{1 + 4x}\,\Delta x\) long, because the integrand times \(\Delta x\) is the length of one segment.

   \[\begin{aligned} f'(x) &= 2x^{1/2} \\ 1 + f'(x)^2 &= 1 + 4x \end{aligned}\]

2. The expression \(1 + 4x\) is linear, so substitute \(u = 1 + 4x\). Then \(du = 4\,dx\). When \(x = 0\), \(u = 1\), and when \(x = 2\), \(u = 9\).

   \[\begin{aligned} L &= \int_0^2 \sqrt{1 + 4x}\,dx \\ &= \tfrac14 \int_1^9 u^{1/2}\,du \end{aligned}\]

3. Integrate the power of \(u\), and substitute the bounds, where \(9^{3/2} = 27\).

   \[\begin{aligned} L &= \tfrac14 \cdot \tfrac23 \left(27 - 1\right) \\ &= \dfrac{13}{3} \end{aligned}\]

4. Test the result. The ends are \((0,\,0)\) and \(\left(2,\,\tfrac{8\sqrt2}{3}\right)\), so the chord is \(\tfrac{2\sqrt{41}}{3} \approx 4.27\). The length \(\tfrac{13}{3} \approx 4.33\) is larger, as it must be.

Example 2: an arch of the sine curve

Find the length of \(y = \sin x\) from \(x = 0\) to \(x = \pi\). Give the answer to three decimal places.

1. Find the derivative, and write the integral.

   \[\begin{aligned} f'(x) &= \cos x \\ L &= \int_0^{\pi} \sqrt{1 + \cos^2 x}\,dx \end{aligned}\]

2. A tempting wrong move: Remove the root by writing the root of a sum as a sum of roots.

   \[\begin{aligned} \sqrt{1 + \cos^2 x} &= 1 + \cos x \\ \int_0^{\pi} (1 + \cos x)\,dx &= \pi \end{aligned}\]

   The chord from \((0,\,0)\) to \((\pi,\,0)\) has length \(\pi\). The arch rises above it, so the curve must be longer than \(\pi\). Also, \((1 + \cos x)^2\) has an extra term \(2\cos x\), so it is not \(1 + \cos^2 x\).

3. The expression \(1 + \cos^2 x\) is not the square of a simpler expression, so the root stays. Use a calculator.

   \[L \approx 3.820\]

4. Test the result. The value 3.820 is larger than the chord length \(\pi \approx 3.142\), as it must be.

Practice

Find the length of \(y = \dfrac{x^4}{16} + \dfrac{1}{2x^2}\) from \(x = 1\) to \(x = 2\).

Square the derivative and add 1. Expand the result, and look for a perfect square.

Questions

The graph shows \(y = x^2\) from \(x = 0\) to \(x = 2\). Straight segments join corners on the curve. The ruler shows the totals: 4.47 for one segment, 4.58 for two, and 4.63 for four.

1. How does the total change as more segments are added?

   • It rises, by a smaller amount each time
   • It rises by the same amount each time
   • It falls toward the length of the curve
   • It stays at 4.47
   Answer

   It rises, by a smaller amount each time

2. A student says that the curve has length 4.5. Why is this wrong?

   • The path through the corner at \(x = 1\) is already 4.58 long, and the curve is longer than that path
   • The curve must be exactly as long as the chord, which is 4.47
   • The curve must be shorter than the path, because the path lies above the curve
   Answer

   The path through the corner at \(x = 1\) is already 4.58 long, and the curve is longer than that path

3. What is the length of \(y = \dfrac{x^3}{6} + \dfrac{1}{2x}\) from \(x = 1\) to \(x = 3\)?

   • \(\dfrac{14}{3}\)
   • 4
   • \(\dfrac{13}{3}\)
   Answer

   \(\dfrac{14}{3}\)

Summary

If \(1 + f'(x)^2\) is a perfect square or linear, you find an exact length, and otherwise you give a calculator value to three decimals.

The integral is the limit of the segment totals, which rise toward the arc length and stay below it.

Test every length against the chord that joins the two ends, because the curve is at least that long.

This is one of four skills in topic 8.13 of Which & Why: AP Calculus.