Limits Derivatives Integrals Infinite Series Parametrics Polar Coordinates Conics
Limits
Epsilon-Delta Definition
Finite Limits
One-Sided Limits
Infinite Limits
Trig Limits
Pinching Theorem
Indeterminate Forms
L'Hopitals Rule
Limits That Do Not Exist
Continuity & Discontinuities
Intermediate Value Theorem
Derivatives
Power Rule
Product Rule
Quotient Rule
Chain Rule
Trig and Inverse Trig
Implicit Differentiation
Exponentials & Logarithms
Logarithmic Differentiation
Hyperbolic Functions
Higher Order Derivatives
Differentials
Slope, Tangent, Normal...
Linear Motion
Mean Value Theorem
Graphing
1st Deriv, Critical Points
2nd Deriv, Inflection Points
Related Rates Basics
Related Rates Areas
Related Rates Distances
Related Rates Volumes
Optimization
Integrals
Definite Integrals
Integration by Substitution
Integration By Parts
Partial Fractions
Improper Integrals
Basic Trig Integration
Sine/Cosine Integration
Secant/Tangent Integration
Trig Integration Practice
Trig Substitution
Linear Motion
Area Under/Between Curves
Volume of Revolution
Arc Length
Surface Area
Work
Moments, Center of Mass
Exponential Growth/Decay
Laplace Transforms
Describing Plane Regions
Infinite Series
Divergence (nth-Term) Test
p-Series
Geometric Series
Alternating Series
Telescoping Series
Ratio Test
Limit Comparison Test
Direct Comparison Test
Integral Test
Root Test
Absolute Convergence
Conditional Convergence
Power Series
Taylor/Maclaurin Series
Interval of Convergence
Remainder & Error Bounds
Fourier Series
Study Techniques
Choosing A Test
Sequences
Infinite Series Table
Practice Problems
Exam Preparation
Exam List
Parametrics
Parametric Curves
Parametric Surfaces
Slope & Tangent Lines
Area
Arc Length
Surface Area
Volume
Polar Coordinates
Converting
Slope & Tangent Lines
Area
Arc Length
Surface Area
Conics
Parabolas
Ellipses
Hyperbolas
Conics in Polar Form
Vectors Vector Functions Partial Derivatives/Integrals Vector Fields Laplace Transforms Tools
Vectors
Unit Vectors
Dot Product
Cross Product
Lines In 3-Space
Planes In 3-Space
Lines & Planes Applications
Angle Between Vectors
Direction Cosines/Angles
Vector Projections
Work
Triple Scalar Product
Triple Vector Product
Vector Functions
Projectile Motion
Unit Tangent Vector
Principal Unit Normal Vector
Acceleration Vector
Arc Length
Arc Length Parameter
Curvature
Vector Functions Equations
MVC Practice Exam A1
Partial Derivatives
Directional Derivatives
Lagrange Multipliers
Tangent Plane
MVC Practice Exam A2
Partial Integrals
Describing Plane Regions
Double Integrals-Rectangular
Double Integrals-Applications
Double Integrals-Polar
Triple Integrals-Rectangular
Triple Integrals-Cylindrical
Triple Integrals-Spherical
MVC Practice Exam A3
Vector Fields
Curl
Divergence
Conservative Vector Fields
Potential Functions
Parametric Curves
Line Integrals
Green's Theorem
Parametric Surfaces
Surface Integrals
Stokes' Theorem
Divergence Theorem
MVC Practice Exam A4
Laplace Transforms
Unit Step Function
Unit Impulse Function
Square Wave
Shifting Theorems
Solve Initial Value Problems
Prepare For Calculus 1
Trig Formulas
Describing Plane Regions
Parametric Curves
Linear Algebra Review
Word Problems
Mathematical Logic
Calculus Notation
Simplifying
Practice Exams
More Math Help
Tutoring
Tools and Resources
Learning/Study Techniques
Math/Science Learning
Memorize To Learn
Music and Learning
Note-Taking
Motivation
Instructor or Coach?
Books
Math Books

You CAN Ace Calculus

17calculus > integrals > arc length

### Calculus Main Topics

Integrals

Integral Applications

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

related topics on other pages

For arc length of parametric curves, see the parametrics calculus page.

For the arc length of polar curves, see the polar calculus page.

Wikipedia - Arc Length

Pauls Online Notes - Arc Length

Applied Integration - Arc Length [ Rectangular Coordinates ]

This page covers the topic of arc length of an explicitly defined smooth curve in the xy-plane in cartesian (rectangular) coordinates. [ Arc length can also be calculated in polar coordinates and for parametric curves. ]

Setting up these integrals is very straight-forward and do not require any new techniques. However, the difficulty comes in evaluating them. We will show you a few tricks to evaluating them but first let's talk about how to set them up.

There are two sets of equations, both of which accomplish the same purpose. The reason you need two of them is because sometimes the equations describing the curve are in a form that require one or the other. Additionally, you may be able to set up both integrals but often it is possible to evaluate only one of them.

Here is a video clip deriving the integral in terms of x. This will help you understand where these equations come from.

 MIP4U - Arc Length - Part 1 of 2
 To determine the arc length of a smooth curve defined by $$y = f(x)$$ between the points $$x=a$$ and $$x=b$$, we can use the integral When the curve is given by an equation in the form $$x=g(y)$$, we need to determine the endpoints, $$y=c$$ and $$y=d$$ on the y-axis and use the integral $$\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2}dx}}$$ $$\displaystyle{s = \int_{c}^{d}{\sqrt{1 + [g'(y)]^2}dy}}$$ where $$\displaystyle{ f'(x) = \frac{dy}{dx} }$$ where $$\displaystyle{ g'(y) = \frac{dx}{dy} }$$

Okay, let's watch a quick video clip going through these equations again.

 PatrickJMT - Arc Length

Notes:
1. Notice that we use a small s to represent the arc length. Later, we will use a capital S to represent surface area. Many, if not most, mathematicians follow this standard.
2. Sometimes these equations are represented a little differently. These can be written as shown below. When you study surface area, you will see how these are used.

 $$ds = \sqrt{1 + [f'(x)]^2}dx$$ $$ds = \sqrt{1 + [g'(y)]^2}dy$$

Tricks To Evaluate Integrals

Here are some unusual situation you may encounter when evaluating integrals that have not been covered elsewhere. We will show some specific examples but we will not evaluate the integrals completely. We will just get them into a form that can be evaluated with other techniques that you are already know, like substitution or parts.

Perfect Square

Okay, so I know that you already know about perfect squares but they show up a lot in these problems, so I thought I would remind you. Notice in the integral, we have $$\sqrt{1+[f'(x)]^2}$$.   After expanding the squared derivative and adding one, the easiest problems end up with a perfect square under the square root. So, even with complicated expressions, it is usually a good use of your time to check for a perfect square first, even if it isn't obvious. Here is an example. After expanding the squared term and adding one, we have $$\displaystyle{\sqrt{ x^2 + \frac{1}{2} + \frac{1}{16x^2}}}$$

We can try to set up a perfect square using the square root of the first and last term and see if we get the middle term under the square root when we expand. So we investigate $$\displaystyle{ x + \frac{1}{4x} }$$. When we square this, sure enough we get $$1/2$$ as the middle term. So we can simplify the equation as follows
$$\displaystyle{\sqrt{ x^2 + \frac{1}{2} + \frac{1}{16x^2} } = }$$ $$\displaystyle{ \sqrt{ \left[ x + \frac{1}{4x} \right]^2 } = }$$ $$\displaystyle{ x + \frac{1}{4x} }$$
This last form is easily integrable with the techniques you know.

Getting A Common Denominator

Okay, so you have determined that you do not have a perfect square under the square root. Your next step might be to combine all terms into one fraction and then take the square root of the numerator and denominator separately. This will often allow you to move some terms outside of the square root and make the square root easier and may lead to integration by substitution or trig substitution. Here is an example.

Let's say we have an integral with the integrand $$\displaystyle{ \sqrt{ 1+\frac{y^2-2y+1}{4y} } }$$. In order to simplify this so that we can integrate it, we need to combine the one with the fraction. Doing this we get $$\displaystyle{ \sqrt{ \frac{y^2+2y+1}{4y} } }$$. Now, the numerator is a perfect square, so we factor it and simplify to get $$\displaystyle{ \frac{ \sqrt{ (y+1)^2 } }{\sqrt{4y} } =\frac{y+1}{\sqrt{4y}}}$$. So, now we have an integrand that can be integrated with basic techniques.

### Search 17Calculus

Practice Problems

Instructions - - Unless otherwise instructed, find the arc length. Give all answers in exact form.
Note: Although some of these problems are very similar, they are worked by different instructors. So watching each solution will give you a broader point of view and help you better understand how to work these problems.

 Level A - Basic

Practice A01

$$\displaystyle{y=\frac{1}{3}(x^2+2)^{3/2}, 0\leq x \leq1}$$

solution

Practice A02

$$\displaystyle{y=\frac{x^2}{2}-\frac{\ln(x)}{4}; 2\leq x\leq4}$$

solution

Practice A03

$$\displaystyle{y=x^{3/2}; 0\leq x\leq4}$$

solution

Practice A04

$$\displaystyle{y=\frac{2}{3}(x^2+1)^{3/2}; 0\leq x\leq2}$$

solution

Practice A05

$$\displaystyle{f(x)=\frac{1}{3}x^{3/2}; [0,4]}$$

solution

Practice A06

$$\displaystyle{f(x)=\frac{x^2}{8}-\ln(x); [1,e]}$$

solution

 Level B - Intermediate

Practice B01

$$\displaystyle{x=\frac{1}{3}\sqrt{y}(y-3), 1\leq y\leq9}$$