Deriving the Arc-Length Formula for Graphs

From Chords to a Length Integral

The arc length of a graph can be understood by joining nearby points on it with straight chords. As the points become closer together, the sum of the chord lengths approaches the curve’s length. The earlier material on arc length as a limit of chord sums establishes this geometric starting point; here we derive the integral that results for a graph.

Let \(f\) be a function on \([a,b]\), where \(a<b\). Divide the interval into subintervals with endpoints \(a=x_0<x_1<\cdots<x_n=b\). The graph points above \(x_{i-1}\) and \(x_i\) have horizontal separation \(\Delta x_i=x_i-x_{i-1}\) and vertical separation \(\Delta y_i=f(x_i)-f(x_{i-1})\). By the distance formula, the chord joining them has length

c_i=\sqrt{(\Delta x_i)^2+(\Delta y_i)^2}.

The key step is to write the vertical change in terms of the slope of the secant line: \(\Delta y_i/\Delta x_i\). This lets us connect each chord to the derivative, which describes the graph’s local slope.

Definition: The mesh of a partition is the length of its widest subinterval: \(\lVert P\rVert=\max_i\Delta x_i\). For a graph whose chord sums have a limit as the mesh tends to zero, that limit is its arc length on \([a,b]\).

The Smoothness Condition

A convenient sufficient assumption is that \(f\) is continuously differentiable on \([a,b]\), written \(f\in C^1([a,b])\). This means that \(f\) is continuous on the closed interval, differentiable in its interior, and its derivative is continuous there and extends continuously to the endpoints. In particular, \(f'\) is bounded and continuous on the interval.

These conditions are useful for two reasons. First, the Mean Value Theorem applies on every subinterval. Second, the function built from the derivative, \(\sqrt{1+[f'(x)]^2}\), is continuous, so its Riemann sums converge to its integral. Continuous differentiability is a sufficient condition for the derivation; it is not a claim that no less restrictive conditions can ever give a valid length formula.

Why the condition matters: Differentiability at every point alone does not guarantee that the derivative behaves well enough for this Riemann-sum argument. Continuity of \(f'\) ensures the derivative values vary in a controlled way and makes the integrand continuous.

Deriving the Formula

We now prove the formula under the stated \(C^1\) assumption. On each subinterval, \(\Delta x_i>0\), so we can factor it out of the chord length:

\begin{aligned} c_i &=\sqrt{(\Delta x_i)^2+(\Delta y_i)^2}\\ &=\Delta x_i\sqrt{1+\left(\frac{\Delta y_i}{\Delta x_i}\right)^2}. \end{aligned}

The Mean Value Theorem says that for each \(i\), there is a point \(\xi_i\in(x_{i-1},x_i)\) for which \(\Delta y_i/\Delta x_i=f'(\xi_i)\). Substituting that equality into the chord length gives an exact expression, not an approximation:

c_i=\Delta x_i\sqrt{1+[f'(\xi_i)]^2}.

Adding the chords produces a sum of the form used to define a Riemann integral. The function \(g(x)=\sqrt{1+[f'(x)]^2}\) is continuous because \(f'\) is continuous and the square-root function is continuous on positive inputs. Each \(\xi_i\) lies in its own subinterval, so these are valid sample points for Riemann sums. As the mesh tends to zero, the sums converge to the integral of \(g\).

Write the chord using its secant slope.
Factor the positive horizontal change \(\Delta x_i\) out of the distance formula.

Apply the Mean Value Theorem.
On each subinterval, the secant slope equals \(f'(\xi_i)\) at some interior point.

Recognize a Riemann sum.
The chord sum has terms \(g(\xi_i)\Delta x_i\), where \(g(x)=\sqrt{1+[f'(x)]^2}\).

Take the limit.
Continuity of \(g\) gives the integral of \(g\) from \(a\) to \(b\).

Theorem: If \(f\in C^1([a,b])\), the arc length \(L\) of the graph \(y=f(x)\) from \(x=a\) to \(x=b\) is

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

The argument also explains what the integrand means. Over a small horizontal step \(dx\), the tangent’s vertical change is approximately \(f'(x)\,dx\). The corresponding small distance is therefore approximately \(\sqrt{(dx)^2+(f'(x)\,dx)^2}=\sqrt{1+[f'(x)]^2}\,dx\). The proof makes this local picture rigorous by using exact chords and a limit.

Worked Examples

Worked Example: A Straight-Line Graph

Find the arc length of \(y=3x-2\) on \(0\le x\le2\), using both chord sums and the formula.

Take any partition \(0=x_0<x_1<\cdots<x_n=2\). On the \(i\)th subinterval, \(\Delta y_i=3x_i-2-(3x_{i-1}-2)=3\Delta x_i\). Thus

c_i=\sqrt{(\Delta x_i)^2+(3\Delta x_i)^2} =\sqrt{10}\,\Delta x_i.

Summing gives \(\sum_{i=1}^n c_i=\sqrt{10}\sum_{i=1}^n\Delta x_i =\sqrt{10}(2-0)=2\sqrt{10}\). This is exact for every partition, so it is also the limit as the mesh tends to zero.

For the integral formula, \(f'(x)=3\), so

L=\int_0^2\sqrt{1+3^2}\,dx =\int_0^2\sqrt{10}\,dx =2\sqrt{10}.

Check: the endpoint heights are \(f(0)=-2\) and \(f(2)=4\). The endpoint chord has length \(\sqrt{(2-0)^2+(4-(-2))^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}\), agreeing with both calculations.

Worked Example: Chords of a Parabola

Show directly how equal-interval chord sums for \(y=x^2\) on \([0,1]\) lead to the graph-length integral.

Divide \([0,1]\) into \(n\) equal subintervals, so \(x_i=i/n\) and \(\Delta x_i=1/n\). The vertical change on the \(i\)th interval is

\Delta y_i =\left(\frac{i+1}{n}\right)^2-\left(\frac{i}{n}\right)^2 =\frac{(i+1)^2-i^2}{n^2} =\frac{2i+1}{n^2}.

Consequently, the secant slope is \(\Delta y_i/\Delta x_i=(2i+1)/n\), and the chord length is

c_i=\frac{1}{n}\sqrt{1+\left(\frac{2i+1}{n}\right)^2}.

The midpoint of the interval is \(m_i=(i+\tfrac12)/n\), so \(2m_i=(2i+1)/n\). The complete chord sum is therefore

\sum_{i=0}^{n-1}c_i =\frac1n\sum_{i=0}^{n-1}\sqrt{1+4m_i^2}.

These are midpoint Riemann sums for the continuous function \(\sqrt{1+4x^2}\) on \([0,1]\). Their limit is \(\int_0^1\sqrt{1+4x^2}\,dx\). Since \(f'(x)=2x\), the theorem gives exactly this integral. As a check on the finite calculation, when \(n=1\) there is one chord from \((0,0)\) to \((1,1)\); the sum is \(\sqrt{1+1}=\sqrt2\), as the displayed chord formula gives.

Worked Example: A Cubic Graph

For \(y=x^3\) on \([0,1]\), derive the limit of the equal-partition chord sums and compare it with the formula.

Set \(x_i=i/n\), giving \(\Delta x_i=1/n\). Factor the difference of cubes to obtain

\Delta y_i =\left(\frac{i+1}{n}\right)^3-\left(\frac{i}{n}\right)^3 =\frac{(i+1)^3-i^3}{n^3} =\frac{3i^2+3i+1}{n^3}.

The secant slope is \((3i^2+3i+1)/n^2\). With midpoint \(m_i=(i+\tfrac12)/n\), we have \(3m_i^2+1/(4n^2)=(3i^2+3i+1)/n^2\). Hence the chord sum is

\sum_{i=0}^{n-1}c_i =\frac1n\sum_{i=0}^{n-1} \sqrt{1+\left(3m_i^2+\frac{1}{4n^2}\right)^2}.

As \(n\) grows, \(1/(4n^2)\) tends to zero uniformly across the intervals. The function \(u\mapsto\sqrt{1+u^2}\) is continuous, so these sums have the same limit as the midpoint sums with terms \(\frac1n\sqrt{1+9m_i^4}\). Those sums converge to \(\int_0^1\sqrt{1+9x^4}\,dx\). Since \(f'(x)=3x^2\), this is the graph-length formula. For \(n=1\), the chord is from \((0,0)\) to \((1,1)\), with length \(\sqrt2\); the finite-sum expression also gives \(\sqrt{1+(3/4+1/4)^2}=\sqrt2\).

A Common Mistake: Using an Endpoint Derivative

The Mean Value Theorem does not say that the secant slope equals the derivative at the left endpoint, the right endpoint, or the midpoint. It says that it equals the derivative at some point inside the interval. Replacing that point by an endpoint can give an incorrect chord length.

Worked Example: Correcting an Endpoint-Slope Error

Consider \(y=x^2\) on \([0,1]\). A tempting but incorrect calculation uses \(f'(0)=0\) as the slope for the whole interval. It would give the alleged chord length \(\sqrt{1+0^2}=1\). But the endpoints are \((0,0)\) and \((1,1)\), so their actual distance is \(\sqrt{(1-0)^2+(1-0)^2}=\sqrt2\), not \(1\).

The correct secant slope is \((f(1)-f(0))/(1-0)=(1-0)/1=1\). The Mean Value Theorem supplies the interior point \(\xi=\tfrac12\), where \(f'(\xi)=2(\tfrac12)=1\). Thus the chord length is \(\sqrt{1+f'(\tfrac12)^2}=\sqrt2\), matching the endpoint distance. The endpoint derivative was not a valid substitute for the secant slope.

Why the Derivation Is Useful

The formula turns a geometric measurement into an integral whose integrand is determined by the graph’s slope. A horizontal graph has \(f'(x)=0\), so its length contribution per unit of horizontal change is \(1\). A steeper graph has a larger factor \(\sqrt{1+[f'(x)]^2}\), reflecting the extra distance traveled vertically as well as horizontally.

The assumptions should remain attached to the result: for \(f\in C^1([a,b])\), the chord-sum argument above proves the formula and the integrand is continuous. When a graph does not meet these assumptions, one should not automatically apply this proof unchanged. The behavior of the curve and the convergence of the relevant length sums must be justified separately.

Check Your Understanding

Use the derivation, not just the final formula, to answer these questions.

How can a chord length be rewritten using its horizontal change and secant slope?
What does the Mean Value Theorem provide on each subinterval of a partition?
Why does continuity of \(f'\) ensure that the resulting sums converge to a Riemann integral?
For \(f(x)=x^2\) on \([0,1]\), what is the secant slope on the single interval, and at what interior point does it equal \(f'(x)\)?
Why is using \(f'(x_{i-1})\) as the secant slope generally an error?