Build the arc-length integrand
Straight pieces along a curve
The curve is \(y = x^2/4\) for \(x\) from 0 to 4, with orange segments laid along it, corner to corner.
Take the thick segment from \((1,\,0.25)\) to \((2,\,1)\). Between its corners, the curve moves 1 unit to the right and 0.75 units up.
The two green lines show these moves, and they are the legs of a right triangle whose hypotenuse is the segment.
By the Pythagorean theorem, the segment has length \(\sqrt{1^2 + 0.75^2} = 1.25\).
Move the slider to add segments. The ruler under the graph lays them end to end. They follow the curve more closely, so their total rises toward the chalk mark at the length of the curve.
The same triangle on any curve
The curve is now any smooth curve \(y = f(x)\): it has a slope at every point, and the slope changes without jumps.
Over a short run \(\Delta x\), the curve rises by \(\Delta y\). The faint line is the tangent at the left corner of the thick segment. The dashed line is its rise over the same run, the slope \(f'(x)\) times \(\Delta x\).
The dashed line is close to the green leg \(\Delta y\), so \(\Delta y \approx f'(x)\,\Delta x\).
The segment is the hypotenuse, so its length is about \(\sqrt{\Delta x^2 + (f'(x)\,\Delta x)^2}\). Both terms under the root contain \(\Delta x^2\), so \(\Delta x\) comes out of the root.
Move the thick segment along the curve. It is about \(\sqrt{1 + f'(x)^2}\,\Delta x\) long. The slope alone sets its length for each unit of \(x\), so it is longer where the curve is steeper.
Notation
A short piece of the curve is close to a straight segment of length \(\cy{\Delta s}\), which the Pythagorean theorem gives from the two legs.
\[\cy{\Delta s} \approx \sqrt{\cg{\Delta x}^2 + \cg{\Delta y}^2}\]The rise is close to the slope times the run, so replace \(\cg{\Delta y}\) with \(\cg{f'(x)\,\Delta x}\).
\[\cg{\Delta y} \approx f'(x)\,\cg{\Delta x}\]Take \(\Delta x\) out of the root, because both terms contain \(\Delta x^2\) and \(\Delta x\) is positive.
\[\cy{\Delta s} \approx \sqrt{1 + f'(x)^2}\;\cg{\Delta x}\]Add the pieces, and let every \(\Delta x\) shrink to 0. The sum becomes an integral, the arc length \(\cy{L}\) from \(x = a\) to \(x = b\).
\[\cy{L} = \int_a^b \sqrt{1 + \big(f'(x)\big)^2}\,dx\]The integrand is the square root of 1 plus the square of the derivative, so square the derivative before you add 1.
The derivative \(f'\) must exist and be continuous on \([a, b]\), which means that the slope has no jumps.
When it fails
The curve \(y = |x - 3| + 0.6\) has a corner at \(x = 3\). The slope is \(-1\) on the left and 1 on the right, so \(f'(3)\) does not exist and the integrand has no value there.
Split the curve at the corner and use the formula on each side, where the slope exists at every point.
Example 1: a cubic curve
Write the integral for the length of \(y = \dfrac{x^3}{3}\) from \(x = 0\) to \(x = 2\).
Find the derivative. It is the slope, so the rise of a short segment is about \(x^2\,\Delta x\).
\[f'(x) = x^2\]Square the derivative. Then add 1. The root of this is the integrand, so the thick segment is about \(\sqrt{1 + x^4}\,\Delta x\) long.
\[1 + \left(x^2\right)^2 = 1 + x^4\]The bounds are the x-values of the ends of the curve.
\[L = \int_0^2 \sqrt{1 + x^4}\,dx\]This integrand has no simple antiderivative. The question asks only for the setup, so the integral is the answer.
Example 2: an exponential curve
Write the integral for the length of \(y = e^{x/2}\) from \(x = 0\) to \(x = 2\).
Find the derivative. At \(x = 0\) it is \(\tfrac12\), so the green triangle on the tangent line has legs 1 and \(\tfrac12\).
\[f'(x) = \tfrac12 e^{x/2}\]A tempting wrong move: Add 1 to the derivative. Then take the root.
\[\sqrt{1 + \tfrac12 e^{x/2}}\]For a run of 1, the integrand is the length of the hypotenuse. At \(x = 0\) the legs are 1 and \(\tfrac12\), so the hypotenuse is \(\sqrt{1 + \tfrac14} \approx 1.118\). The wrong integrand gives \(\sqrt{1.5} \approx 1.225\), drawn in red, which is longer.
The slope is a leg of the triangle, so square it before you add 1.
\[\left(\tfrac12 e^{x/2}\right)^2 = \tfrac14 e^{x}\]Add 1, take the root, and write the bounds.
\[L = \int_0^2 \sqrt{1 + \tfrac14 e^{x}}\,dx\]
Practice
Write the integral for the length of \(y = 2\sqrt{x}\) from \(x = 1\) to \(x = 4\).
The thick segment is close to the hypotenuse of a triangle with legs \(\Delta x\) and \(f'(x)\,\Delta x\), so square the slope before you add 1.
Solution
Find the derivative of \(2\sqrt{x}\). It is the slope, so the rise on the tangent, the dashed line, is \(f'(x)\,\Delta x\).
\[f'(x) = \dfrac{1}{\sqrt{x}}\]Square it. Then add 1.
\[1 + \left(\dfrac{1}{\sqrt{x}}\right)^2 = 1 + \dfrac1x\]Take the root, and write the bounds.
\[L = \int_1^4 \sqrt{1 + \dfrac1x}\,dx\]
Questions
The graph shows \(y = x^2/4\). The second orange segment is a chord from \((2,\,1)\) to \((4,\,4)\), and the small green triangle sits on the tangent line at \(x = 3\).
What is the length of the chord from \((2,\,1)\) to \((4,\,4)\)?
- \(\sqrt{13}\)
- 5
- 3
- \(\sqrt{5}\)
Answer
\(\sqrt{13}\)
The small triangle shows the slope of the tangent at \(x = 3\). What is the value of \(\sqrt{1 + f'(x)^2}\) at \(x = 3\)?
- \(\dfrac{\sqrt{13}}{2}\)
- \(\dfrac32\)
- \(\sqrt{\dfrac52}\)
- \(\dfrac{13}{4}\)
Answer
\(\dfrac{\sqrt{13}}{2}\)
Which integral gives the length of \(y = 2x^2\) from \(x = 0\) to \(x = 1\)?
- \(\displaystyle\int_0^1 \sqrt{1 + 16x^2}\,dx\)
- \(\displaystyle\int_0^1 \sqrt{1 + 4x^2}\,dx\)
- \(\displaystyle\int_0^1 \sqrt{1 + 4x^4}\,dx\)
- \(\displaystyle\int_0^1 \sqrt{1 + 4x}\,dx\)
Answer
\(\displaystyle\int_0^1 \sqrt{1 + 16x^2}\,dx\)
Summary
To write the arc-length integrand, you find the derivative, square it, add 1, and take the root.
This works because a short piece of the curve is close to the hypotenuse of a right triangle with legs \(\Delta x\) and \(f'(x)\,\Delta x\).
Use it for the length of a smooth curve \(y = f(x)\) between two values of \(x\).
This is one of four skills in topic 8.13 of Which & Why: AP Calculus.