Determining Arc Length Using Integrals

Introduction

Besides calculating area and volume of solids of revolution, integrals have another important application: determining arc length of a curve. This application further demonstrates the importance and versatility of integral calculus.

Prerequisites

To understand this material, you should master:

• Distance between two points
• Riemann sums
• Derivatives of algebraic functions
• Definite integrals of algebraic and trigonometric functions

Note: The theory is relatively simple, but the integration calculations can be challenging.

Conceptual Foundation

Idea Behind Arc Length Formula

To calculate arc length, we approximate the curve with many small straight line segments (red lines in the diagram).

The length of each small segment from point \(C(x_{k-1}, y_{k-1})\) to \(D(x_k, y_k)\) is:

\[ \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2} \]

The total approximate length is the sum of all segments:

\[ \text{Approximate length} = \sum_{k=1}^n \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2} \]

As we take more and smaller segments (\(n \to \infty\)), this becomes an integral:

\[ \text{Arc length} = \lim_{n \to \infty} \sum_{k=1}^n \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2} \]

Arc Length Formulas

Standard Arc Length Formulas

For curve \(y = f(x)\) from point \(A(a, c)\) to \(B(b, d)\):

Using X-boundaries (integrate with respect to x):

\[ \text{Arc length} = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

Using Y-boundaries (integrate with respect to y):

\[ \text{Arc length} = \int_c^d \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} \, dy \]

Note: Choose the formula that gives the easier integration.

Example Problems

Example 1: Curve \(9y^2 = 4x^3\)

Find the arc length of curve \(9y^2 = 4x^3\) from \(A(0, 0)\) to \(B(3, 2\sqrt{3})\).

Solution (using X-boundaries):

1. Rewrite function: \(9y^2 = 4x^3 \Rightarrow y = \frac{2}{3}x^{3/2}\)

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

3. Set up integral (x from 0 to 3):

Arc length = \(\int_0^3 \sqrt{1 + (x^{1/2})^2} \, dx = \int_0^3 \sqrt{1 + x} \, dx\)

4. Calculate: \(= \int_0^3 (1 + x)^{1/2} \, dx = \left[ \frac{2}{3}(1 + x)^{3/2} \right]_0^3\)

5. Evaluate: \(= \frac{2}{3}(4^{3/2}) - \frac{2}{3}(1^{3/2}) = \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}\)

Answer: Arc length = \(\frac{14}{3}\) units.

Note: Using Y-boundaries would be more difficult for this problem.

Example 2: Straight Line \(y = 3x\)

Find the arc length of line \(y = 3x\) from \(A(0, 0)\) to \(B(2, 6)\) using both methods.

Solution Method 1 (X-boundaries):

1. Derivative: \(\frac{dy}{dx} = 3\)

2. Arc length: \(\int_0^2 \sqrt{1 + 3^2} \, dx = \int_0^2 \sqrt{10} \, dx\)

3. Calculate: \(= \left[ \sqrt{10}x \right]_0^2 = 2\sqrt{10}\)

Solution Method 2 (Y-boundaries):

1. Rewrite: \(y = 3x \Rightarrow x = \frac{1}{3}y\)

2. Derivative: \(\frac{dx}{dy} = \frac{1}{3}\)

3. Arc length: \(\int_0^6 \sqrt{\left(\frac{1}{3}\right)^2 + 1} \, dy = \int_0^6 \sqrt{\frac{1}{9} + 1} \, dy = \int_0^6 \sqrt{\frac{10}{9}} \, dy\)

4. Calculate: \(= \int_0^6 \frac{\sqrt{10}}{3} \, dy = \left[ \frac{\sqrt{10}}{3}y \right]_0^6 = 2\sqrt{10}\)

Answer: Arc length = \(2\sqrt{10}\) units (both methods give same result).

Parametric Form Arc Length

Arc Length for Parametric Equations

For parametric equations \(x = f(t)\), \(y = g(t)\) with \(a \leq t \leq b\):

\[ \text{Arc length} = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Derivation: Starting from \(\int_a^b \sqrt{(dx)^2 + (dy)^2}\), multiply by \(\frac{dt}{dt}\):

\[ = \int_a^b \frac{\sqrt{(dx)^2 + (dy)^2}}{dt} \, dt = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Example 3: Parametric Path of a Particle

A particle moves along a path described by \(x = 3t\) and \(y = \frac{8}{3}t^{3/2}\), where \(t\) is in minutes. Find the distance traveled in the first minute.

Solution:

1. Find derivatives:

\(\frac{dx}{dt} = 3\)

\(\frac{dy}{dt} = \frac{8}{3} \cdot \frac{3}{2}t^{1/2} = 4t^{1/2}\)

2. Set up integral (t from 0 to 1):

Distance = \(\int_0^1 \sqrt{3^2 + (4t^{1/2})^2} \, dt = \int_0^1 \sqrt{9 + 16t} \, dt\)

3. Calculate using substitution \(u = 9 + 16t\), \(du = 16 dt\):

\(= \frac{1}{16} \int \sqrt{u} \, du = \frac{1}{16} \cdot \frac{2}{3} u^{3/2} = \frac{1}{24} u^{3/2}\)

4. Evaluate from t=0 to t=1:

\(= \left[ \frac{1}{24}(9 + 16t)^{3/2} \right]_0^1 = \frac{1}{24}(25^{3/2}) - \frac{1}{24}(9^{3/2})\)

5. Simplify: \(= \frac{1}{24}(125) - \frac{1}{24}(27) = \frac{1}{24}(98) = \frac{49}{12} = 4\frac{1}{12}\)

Answer: Distance traveled = \(4\frac{1}{12}\) units.

Key Points and Tips

Choosing the Right Formula

1. Standard form \(y = f(x)\): Use \(\int \sqrt{1 + (dy/dx)^2} \, dx\)

2. Standard form \(x = f(y)\): Use \(\int \sqrt{(dx/dy)^2 + 1} \, dy\)

3. Parametric form: Use \(\int \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt\)

Tip: Choose the form that gives the simplest derivative and integral.

Common Integration Techniques

Arc length problems often require:

• Substitution (u-substitution)
• Trigonometric substitution
• Integration of square roots
• Recognizing perfect squares under the radical

Verification Strategy

For straight lines, you can verify your answer using the distance formula:

\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

For Example 2: \(\sqrt{(2-0)^2 + (6-0)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\) ✓

Summary of Formulas

Form	Arc Length Formula
\(y = f(x)\)	\(\displaystyle L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)
\(x = f(y)\)	\(\displaystyle L = \int_c^d \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} \, dy\)
Parametric	\(\displaystyle L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\)