Arc Length as a Function
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 2.4: “Arc Length of a Curve and Surface Area” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/2-4-arc-length-of-a-curve-and-surface-area |
| Textbook used in class | Stewart, Calculus, Section 8.1: “Arc Length” |
Opening Scenario
The arc length formula $L = \int_a^b \sqrt{1+[f'(x)]^2}\,dx$ gives a single number: the total length of the curve over $[a,b]$. But sometimes the question is not “how long is the whole curve?” but “how far along the curve have I traveled from $a$ to position $x$?”
Treating the upper limit as a variable $x$ turns the total length into a running distance function $s(x)$. This function is the foundation for the idea of parameterizing a curve by arc length, which appears in physics, differential geometry, and computer graphics.
Quick Reference
Arc length function. If $f'$ is continuous on $[a,b]$, define: $$s(x) = \int_a^x \sqrt{1 + [f'(t)]^2}\,dt, \quad a \leq x \leq b.$$
$s(x)$ is the distance along the curve $y = f(t)$ from $(a, f(a))$ to $(x, f(x))$.
By the Fundamental Theorem of Calculus (Part 1): $$s'(x) = \frac{ds}{dx} = \sqrt{1 + [f'(x)]^2}.$$
The quantity $ds = \sqrt{1 + [f'(x)]^2}\,dx$ is called the arc length element.
Key Concepts
1. Building s(x) from the Arc Length Integral
The arc length formula replaces the fixed upper limit $b$ with the variable $x$: $$s(x) = \int_a^x \sqrt{1 + [f'(t)]^2}\,dt.$$
Immediate consequences:
- $s(a) = 0$: zero distance at the starting point.
- $s(b) = L$: the total arc length from $a$ to $b$.
- $s'(x) = \sqrt{1+(f'(x))^2} \geq 1 > 0$: the arc length function is strictly increasing.
The derivative $s'(x)$ is the speed at which arc length accumulates as $x$ increases.
2. The Arc Length Element
Think of the arc length element geometrically. An infinitesimal step from $x$ to $x + dx$:
- Moves $dx$ horizontally.
- Moves $dy = f'(x)\,dx$ vertically.
- The diagonal length is $ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1+(f')^2}\,dx$.
This element $ds$ appears directly in the surface area formula $S = 2\pi \int y\,ds$.
3. Parameterization by Arc Length
A curve can be described using any parameter. If the parameter is $s$ (distance along the curve), the curve is said to be parameterized by arc length or in unit speed form. Under arc length parameterization, the velocity vector has magnitude 1 at every point.
Because $s'(x) > 0$, the function $s$ is strictly increasing and therefore has an inverse $s^{-1}$. The substitution $x = s^{-1}(u)$ converts from the $x$-parameterization to the arc length parameterization.
Worked Example
Find $s'(x)$ and the arc length element $ds$ for $y = \ln(\sec x)$ on $[0, \pi/4]$. Then compute the total arc length $L$.
$f'(x) = \dfrac{\sec x \tan x}{\sec x} = \tan x$.
$1 + [f'(x)]^2 = 1 + \tan^2 x = \sec^2 x$.
$s'(x) = \sqrt{\sec^2 x} = \sec x$ (since $\sec x > 0$ on $[0, \pi/4]$).
$ds = \sec x\,dx$.
Total arc length: $$L = \int_0^{\pi/4} \sec x\,dx = \left[\ln|\sec x + \tan x|\right]_0^{\pi/4} = \ln(\sqrt{2}+1).$$
Boxed answers: $s'(x) = \sec x$, $ds = \sec x\,dx$, $L = \ln(\sqrt{2}+1) \approx 0.881$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Confusing $s'(x)$ with $f'(x)$ | Writing $s'(x) = \tan x$ for $f(x) = \ln(\sec x)$ | $s'(x) = \sqrt{1+(f')^2} = \sec x$, not $f'(x) = \tan x$ |
| Forgetting $s(a) = 0$ | Using $s(a) = f(a)$ | The arc length accumulates from zero at $x = a$, not from the $y$-value |
| Dropping the “1 +” inside the radical | Writing $ds = |f'|\,dx$ | $ds = \sqrt{1+(f')^2}\,dx$; the 1 is essential |
Common Misconceptions
the rate of arc length accumulation $s'(x)$ equals the slope $f'(x)$ of the curve.
This is the height-vs-slope error applied to arc length. The arc length function $s(x)$ accumulates distance along the curve, not height. By the Fundamental Theorem, $s'(x) = \sqrt{1+[f'(x)]^2}$, which includes both the horizontal and vertical components of each infinitesimal step. For $f(x) = \ln(\sec x)$, the slope is $f'(x) = \tan x$, but $s'(x) = \sec x$, a larger quantity. Setting $s'(x) = f'(x)$ ignores the horizontal displacement $dx$ and undercounts the true arc length at every point.
the arc length element is $ds = f(x)\,dx$, using the function value rather than the formula involving the derivative.
This is the concept-image-conflicts-definition error. The function value $f(x)$ gives the height of the curve at $x$, which is unrelated to how much arc the curve traces in moving from $x$ to $x + dx$. The arc length element $ds = \sqrt{1+[f'(x)]^2}\,dx$ comes from the Pythagorean theorem applied to the infinitesimal right triangle with legs $dx$ and $dy = f'(x)\,dx$. Using $f(x)\,dx$ in place of $ds$ would give the area under the curve, not the curve’s length.
Leveled Practice
Level 1 -- Computing s’(x)
Problem 1. For $y = x^2$, write $s'(x)$.
Show answer
$f'(x) = 2x$.
$s'(x) = \sqrt{1 + 4x^2}$.
Problem 2. For $y = \cosh x$, show that $s'(x) = \cosh x$ and find $s(x) - s(0)$.
Show answer
$f'(x) = \sinh x$. $\;1 + \sinh^2 x = \cosh^2 x$.
$s'(x) = \cosh x$.
$s(x) - s(0) = \displaystyle\int_0^x \cosh t\,dt = \sinh x$.
So the arc length from $0$ to $x$ is $\sinh x$ -- a special property of the catenary $y = \cosh x$.
Level 2 -- Conceptual
Problem 3. Without computing anything, explain why $s(x) \geq x - a$ for all $x \in [a,b]$.
Show answer
$s'(x) = \sqrt{1+(f')^2} \geq 1$ for all $x$.
A function with derivative at least 1 grows at least as fast as the identity. Therefore $s(x) - s(a) \geq x - a$, i.e., $s(x) \geq x - a$ (since $s(a) = 0$).
Geometrically: the arc length from $a$ to $x$ is always at least the horizontal distance $x - a$.
Mastery Checklist
Mental Model
The arc length function is an odometer on the curve. The odometer starts at zero at $x = a$ and ticks at rate $s'(x) = \sqrt{1+(f')^2}$. This rate is always at least 1 -- you accumulate at least as much curve-distance as horizontal distance. On a steep section where $|f'|$ is large, the odometer ticks faster because each horizontal step covers more diagonal distance.
Connections
Looking back
- Arc length formula (Section 8.1): $s(x)$ is the arc length integral with variable upper limit.
- FTC Part 1 (Section 4.3): Differentiating $s(x)$ applies FTC directly.
Looking ahead
- Parametric arc length (Section 8.1): The same idea extends to $s(t) = \int_{t_0}^t \sqrt{(x')^2+(y')^2}\,du$.
- Surface area (Section 8.2): The arc length element $ds = \sqrt{1+(f')^2}\,dx$ appears in $S = 2\pi\int y\,ds$.