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 > vector functions > arc length

### Calculus Main Topics

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

free ideas to save on books - bags - supplies ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com.

Vector Functions - Arc Length and The Arc Length Parameter

For the following discussion, we will consider a parameterized curve defined by the vector function $$\displaystyle{ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle }$$ which is traversed once on the continuous interval $$a \leq t \leq b$$.

 Arc Length

The equation for finding the arc length of the curve is
$$\displaystyle{ L = \int_{a}^{b}{ \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} ~dt } = \int_{a}^{b}{ \| \vec{r}'(t) \| ~dt } }$$

 Arc Length Function

As you know from parametrics, there is more than one way to describe a parametric curve. Since vector functions are parametrics written in vector form, we can also describe the same curve with a different vector function. For example, to describe the circle with radius 1 we can write $$\vec{r}(t)=\langle \cos(2t), \sin(2t) \rangle$$; $$0\leq t\leq \pi$$. Another way of describing the same circle is $$\vec{q}(s)=\langle \cos s, \sin s \rangle$$; $$0 \leq s \leq 2\pi$$. Now, let's calculate the arc length of both of these vector functions.

For $$\vec{r}(t)$$, the derivative is $$\vec{r}'(t) = \langle -2\sin(2t), 2\cos(2t) \rangle$$ and so the arc length integral is
$$\displaystyle{ \int_{0}^{\pi}{ \| \vec{r}'(t) \| ~ dt } = \int_{0}^{\pi}{ \sqrt{ (-2\sin 2t)^2 + (2\cos 2t)^2 } ~ dt } = }$$ $$\displaystyle{ \int_{0}^{\pi}{2 ~ dt} = }$$ $$\displaystyle{ \left. 2t \right|_{0}^{\pi} = 2(\pi) - 2(0) = 2\pi }$$

Now, let's do the same calculation using the other vector function, $$\vec{q}(s)=\langle \cos s, \sin s \rangle$$; $$0 \leq s \leq 2\pi$$.
The derivative is $$\vec{q}'(s)= \langle -\sin s, \cos s \rangle$$ and so the arc length integral is
$$\displaystyle{ \int_{0}^{2\pi}{ \| \vec{q}'(s) \| ~ ds } = \int_{0}^{2\pi}{ \sqrt{ (-\sin s)^2 + (\cos s)^2 } ~ ds } = }$$ $$\displaystyle{ \int_{0}^{2\pi}{ 1~ds } = }$$ $$\displaystyle{ s|_{0}^{2\pi} = 2\pi - 0 = 2\pi }$$

As expected the answer (arc length) is the same in both cases. However, let's look closely as the intermediate steps. Notice in the case of $$\vec{r}(t)$$, that $$\| \vec{r}'(t) \| = 2$$ and for $$\vec{q}(s)$$, $$\|\vec{q}'(s)\|=1$$. This is significant for our discussion. The fact that $$\|\vec{q}'(s)\|=1$$ allows us to say that the arc length for this vector function is equal to the difference in the endpoints, i.e. $$2\pi-0$$. That's what makes the parameterization $$\vec{q}(s)$$ special. We call this parameterization, the arc length parameterization, specifically for this reason, i.e. $$\|\vec{q}'(s)\|=1$$. [ The use of the variable s is not random here. Most textbooks use s to signify that they are talking about the arc length parameter. ]

So basically, we say that if we know that some curve is described by it's arc length parameter s, we can directly say that the arc length is the difference between the endpoints, without evaluating the integral.

 Finding The Arc Length Parameterization

In many cases, it is possible to convert/translate/change a vector function into one using the arc length parameterization. The reason we may want to do this is that there are some calculations that are easier to do if we have the arc length parameterization than if we have a generic parameterization. ( see the section on curvature, for example )

The general steps are
1. calculate $$\displaystyle{s(t)=\int_{0}^{t}{ \| \vec{r}'(u) \| ~du } }$$;
2. convert t to s in the vector function and the range
Let's do an example of this by using $$\vec{r}(t)=\langle \cos(2t), \sin(2t) \rangle$$; $$0\leq t\leq \pi$$ to see how we get $$\vec{q}(s)=\langle \cos s, \sin s \rangle$$; $$0 \leq s \leq 2\pi$$.

First, we calculate $$\displaystyle{s(t)=\int_{0}^{t}{ \| \vec{r}'(u) \| ~du } }$$. Notice we use zero as the lower limit of integration since the range $$0\leq t\leq \pi$$ starts with zero. Notice also that the upper limit is t, so we use u as our variable of integration. Any variable of integration would work except for t since t is the upper limit.
In our example, we now have $$\displaystyle{ s(t)=\int_{0}^{t}{ \sqrt{ (-2\sin 2u)^2 + (2\cos 2u)^2 ~du } } = }$$ $$\displaystyle{ \int_{0}^{t}{ 2 ~ du} = 2u|_{0}^{t} = 2t }$$. So now we have $$s=2t$$. We solve this for t to give us $$t=s/2$$.

For the second step, we substitute $$s/2$$ for $$t$$ in our original vector function and range to get $$\vec{r}(s) = \langle \cos s, \sin s \rangle$$; $$0\leq s \leq 2\pi$$ which is equal to $$\vec{q}(s)$$.

 Intuitive Understanding of the Arc Length Parameter

Okay, so you now you know how to find the arc length parameterization of a curve, but what does it really mean? Why is it important?
The idea is very simple but it is hidden in all the equations. Let's look at the number line below and the example above.

Let's think about the unit circle for a moment. We can take the unit circle and 'unroll it' and place it on the number line. Each angle will match up perfectly from zero to $$2\pi$$. Now, let's look at the equations.
From the previous example, the two parameterizations of the unit circle are

 1 $$\vec{r}(t)=\langle \cos(2t), \sin(2t) \rangle$$ $$0\leq t\leq \pi$$ 2 $$\vec{q}(s)=\langle \cos s, \sin s \rangle$$ $$0 \leq s \leq 2\pi$$

Let's look at the first parameterization of the unit circle, $$\vec{r}(t)=\langle \cos(2t), \sin(2t) \rangle$$; $$0\leq t\leq \pi$$. Intuitively, if we were given the time $$0\leq t\leq \pi$$ to walk along the unit circle, we would have to do it at a rate of 2 steps for each t.

However, the second parameterization is on the range, $$0\leq s\leq 2\pi$$. This arc length parameterization means that if we were to think about walking on the unit circle and we were given $$s=2\pi$$ time to walk the entire circle, we could walk at a constant speed of exactly 1 to traverse the entire circle, i.e. for every step we take along the curve, we take exactly the same distance along the number line. So the interval $$0\leq s\leq 2\pi$$ matches up perfectly with the arc length.

This is true for any curve that is described by the arc length parameter, i.e. you can just unroll the curve and place it on the number line that is the same length as the interval and they will match up one-to-one.

So how does this help us? Well, if you are given an extremely complicated set of equations for a curve and you are asked to find the arc length on an interval, say $$a\leq t \leq b$$ but you know that the parameter describing the curve is the arc length parameter, you can say that the arc length is just $$b-a$$ and you do not need to do any other work. Pretty cool, eh?

Okay, time for some practice problems.

### Search 17Calculus

Practice Problems

Instructions - - Unless otherwise instructed, give your answers in exact form.

 Level A - Basic

Practice A01

Find the arc length parameterization of the vector function $$\vec{r}(t) = \langle 4\cos t, 4\sin t, 3t \rangle$$ for $$0 \leq t \leq 2\pi$$

solution

Practice A02

Find the arc length of $$\vec{r}(t)=t\sqrt{2}\vhat{i}+e^t\vhat{j}+e^{-t}\vhat{k}$$ for $$0\leq t\leq 1$$.

solution

Practice A03

Find the arc length of $$\vec{r}(t)=\langle 2\cos(t),2\sin(t),2t\rangle$$ for $$0\leq t\leq \pi$$.

solution

Practice A04

Find the arc length of $$\vec{r}(t)=\langle\cos^3(t),\sin^3(t)\rangle$$ on the interval $$[0,\pi]$$.

solution

Practice A05

Find the arc length of $$\vec{r}(t)=\langle 5\cos(4t),5\sin(4t),5t\rangle$$ for $$-5\leq t\leq 8$$.

solution

Practice A06

Find the arc length of $$\vec{r}(t)=\langle 2\ln t,2t,t^2/2\rangle$$, $$t \in [1,2]$$.

Find the arc length parameterization of the vector function $$\vec{r}(t)=\cos t\vhat{i}+\sin t\vhat{j}+t\vhat{k}$$ for $$t\geq 0$$.