Arc Length of a Curve

Published

October 26, 2025

The arc length of a curve defined by a function [ y = f(x) ] from [ x = a ] to [ x = b ] can be calculated using the following formula:

\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

where \(\frac{dy}{dx}\) is the derivative of the function \(f(x)\). To compute the arc length, follow these steps:

Steps to Calculate Arc Length

  1. Find the Derivative: Compute \(\frac{dy}{dx}\) for the function $f(x) $.
  2. Set Up the Integral: Substitute \(\frac{dy}{dx}\) into the arc length formula.
  3. Evaluate the Integral: Calculate the definite integral from \(a\) to \(b\). This will give you the length of the curve between the points $ (a, f(a))$ and \((b, f(b))\).

Example 1

Consider the function \(f(x) = x^2\) from \(x = 0\) to \(x = 1\):

  1. Find the Derivative: \(\frac{dy}{dx} = 2x\)
  2. Set Up the Integral: \(L = \int_{0}^{1} \sqrt{1 + (2x)^2} \, dx = \int_{0}^{1} \sqrt{1 + 4x^2} \, dx\)
  3. Evaluate the Integral: This integral can be computed using a suitable method (e.g., trigonometric substitution or numerical integration) to find the arc length. The result will give you the length of the curve \(y = x^2\) from \(x = 0\) to \(x = 1\).
# R code to compute the arc length of y = x^2 from x = 0 to x = 1
library(pracma) 
f <- function(x) { sqrt(1 + (2*x)^2) }
arc_length <- integral(f, 0, 1)
arc_length
[1] 1.478943

Via: Trigonometric Substitution

You can use trigonometric substitution to evaluate the integral:

  1. Let \(x = \frac{1}{2} \tan(\theta)\), then \(dx = \frac{1}{2} \sec^2(\theta) d\theta\).

  2. Substitute into the integral:

\[L = \int \sqrt{1 + 4\left(\frac{1}{2} \tan(\theta)\right)^2} \cdot \frac{1}{2} \sec^2(\theta) d\theta\]

  1. Simplify:

\[ L = \frac{1}{2} \int \sec(\theta) \sec^2(\theta) d\theta = \frac{1}{2} \int \sec^3(\theta) d\theta\]

The integral of \(\sec^3(\theta)\) can be computed using integration by parts as folows: \[\int \sec^3(\theta) d\theta = \int \sec(\theta)(\sec^2(\theta)) d\theta\]

  1. Integration by parts:

Let \(u = \sec(\theta)\) and \(dv = \sec^2(\theta) d\theta\), then \(du = \sec(\theta) \tan(\theta) d\theta\) and \(v = \tan(\theta)\).

Applying integration by parts: \[\int \sec^3(\theta) d\theta = \sec(\theta) \tan(\theta) - \int \sec(\theta) \tan^2(\theta) d\theta\] Using the identity \(\tan^2(\theta) = \sec^2(\theta) - 1\), we have: \[\int \sec^3(\theta) d\theta = \sec(\theta) \tan(\theta) - \int \sec^3(\theta) d\theta + \int \sec(\theta) d\theta.\] Rearranging gives: \[2 \int \sec^3(\theta) d\theta = \sec(\theta) \tan(\theta) + \int \sec(\theta) d\theta.\] The integral of \(\sec(\theta)\) is: \[\int \sec(\theta) d\theta = \ln|\sec(\theta) + \tan(\theta)| + .C\]

  1. Putting it all together, we get:

\[L = \frac{1}{2} \left[ \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| \right] + C\].

Note, the first comes from the substitution for dx.

Get back to x

To express the result back in terms of \(x\), we

  1. use the substitutions:

    • \(\sec(\theta) = \sqrt{1 + 4x^2}\)

      . (this comes from the original substitution and thinking about the right triangle where the hypotenuse is \(\sqrt{1 + 4x^2}\) and the adjacent side is 1

    • \(\tan(\theta) = 2x\).

  2. Then we substitute these back into the expression for \(L\) and evaluate it from \(x = 0\) to \(x = 1\) to find the arc length of the curve \(y = x^2\) over that interval:

\[L = \frac{1}{4} \left[ \sqrt{1 + 4x^2} \cdot 2x + \ln\left|\sqrt{1 + 4x^2} + 2x\right| \right]_{0}^{1}\]

Evaluating this expression at the limits will give you the arc length of the curve from \(x = 0\) to \(x = 1\).

# R code to compute the arc length of y = x^2 from x = 0 to x = 1 using the trigonometric substitution result
L = (1/4) * ((sqrt(1 + 4*1^2) * 2*1 + log(abs(sqrt(1 + 4*1^2) + 2*1))) - 
            (sqrt(1 + 4*0^2) * 2*0 + log(abs(sqrt(1 + 4*0^2) + 2*0))))
L
[1] 1.478943

Example 2

Find the arc length of the curve \[ y^2 = x^3\] from (1,1) to (4,8).

  1. Express y in terms of x: From the equation \(y^2 = x^3\), we have \(y = x^{3/2}\).

  2. Find the Derivative: \(\frac{dy}{dx} = \frac{3}{2} x^{1/2}\)

  3. Set Up the Integral: \[ L = \int_{1}^{4} \sqrt{1 + \left( \frac{3}{2} x^{1/2} \right)^2} \, dx = \int_{1}^{4} \sqrt{1 + \frac{9}{4} x} \, dx \]

  4. Evaluate the Integral: This integral can be computed using u-substitution) to find the arc length.

  5. The result will give you the length of the curve \(y^2 = x^3\) from \((1,1)\) to \((4,8)\).

Step-by-step solution via u-substitution

  1. Let \(u = 1 + \frac{9}{4} x\), then \(du = \frac{9}{4} dx\) or \(dx = \frac{4}{9} du\).

  2. Change the limits of integration:

    • When \(x = 1\), \(u = 1 + \frac{9}{4} \cdot 1 = \frac{13}{4}\).
    • When \(x = 4\), \(u = 1 + \frac{9}{4} \cdot 4 = 10\).

  3. Substitute into the integral: \[ L = \int_{\frac{13}{4}}^{10} \sqrt{u} \cdot \frac{4}{9} du = \frac{4}{9} \int_{\frac{13}{4}}^{10} u^{1/2} du \]

  4. Evaluate the integral: \[ L = \frac{4}{9} \left[ \frac{2}{3} u^{3/2} \right]_{\frac{13}{4}}^{10} = \frac{8}{27} \left[ 10^{3/2} - \left(\frac{13}{4}\right)^{3/2} \right] \]

# R code to compute the arc length of y^2 = x^3 from (1,1) to (4,8)
L2 = (8/27) * (10^(3/2) - (13/4)^(3/2))
L2
[1] 7.633705