In the first example with the unit circle I don't get why x = -cos(teta) isn't the x-coordinate simply cos(teta) as the trigonometric coordinate are given by the point x =(x1,y1)=(cos(teta),sin(teta)) ? :)

By convention, we think of the unit circle as starting at the point (1,0). This corresponds to a x =cos(theta) parametrization. However, since the integral travels from the lower limit (-1,0) to the upper limit (1,0) we negate the cosine to switch direction of travel. Hope this helps

Haven't we learned about arc length in ordinary integral classes previously? Why are we studying this again?

It is a review leading to Arc length of parametric curves which leads to line integrals in scalar fields which leads to line integrals in vector fields and on it goes. Sometimes a little refresher review on a concept helps when a new topic depends on it.

First example, step 3, how to solve the integral using trigonometric substitution. This method is only valid for integrals of arc lengths right? Because if I find that integral in other contexts, I couldn´t think of the function inside the integral as a semicircle, because it is not a semicircle, although in the explanation is treated as a semicircle. So I don´t understand this method for solving these kind of integrals.

This cannot be solved by hand. This article and previous articles show how to set up the integral not how to solve this. The integrala in example 3 can only be approximated

in example 1, step 1, I don't understand where did the 2 come from in the denominator? in the d(1-x^2)/2(square root of /(1-x^2)

It comes from the derivative of sqrt(1-x^(2)). Take a simpler example. Let's say we want the derivative of sqrt(x). This is just x^(1/2). Using the power rule, we get (1/2)x^(-1/2) or 1/(2sqrt(x)). The example in the article is the same, but as the inside of the square is a function, you need the chain rule.

Main content

Multivariable calculus

Course: Multivariable calculus > Unit 4

Lesson 2: Line integrals for scalar functions (articles)

Arc length of function graphs, examples

Google Classroom

Practice finding the arc length of various function graphs.

Background

Arc length of function graphs, introduction

Example 1: Practice with a semicircle

Consider a semicircle of radius

1

, centered at the origin, as pictured on the right. From geometry, we know that the length of this curve is

π

. Let's practice our newfound method of computing arc length to rediscover the length of a semicircle.

By definition, all points

(x, y)

on the circle are a distance

1

from the origin, so we have

$x^{2} + y^{2} = 1$ ‍

Rearranging to write

y

as a function of

x

, we have

$y = \sqrt{1 - x^{2}}$ ‍

As you set up the arc length integral, it helps to imagine approximating this curve with a bunch of small lines.

Writing down the arc-length integral, ignoring the bounds for just a moment, we get:

$\begin{array}{r} \int \sqrt{(d x)^{2} + (d y)^{2}} \end{array}$ ‍

Just as before, we think of the integrand

\sqrt{(d x)^{2} + (d y)^{2}}

as representing the length of one of these little lines approximating the curve (via the Pythagorean theorem).

Now we start plugging in the definition of our particular curve into the integral.

Step 1: Write $d y$ ‍ in terms of $d x$ ‍

Use the fact that

y = \sqrt{1 - x^{2}}

to write

d y

in terms of

d x

$d y =$ ‍
$d x$ ‍

[Answer]

Step 2: Replace $d y$ ‍ in the integral

Plug this expression for

d y

into the integral to write the integrand completely in terms of

x

and

d x

$\int$ ‍
$d x$ ‍

[Answer]

Step 3: Place bounds on the integral and solve

Since the curve is defined between when

x = - 1

and

x = 1

, set these values as the bounds of your integral and solve it.

(Sorry, no entry box with a happy green checkmark. We know from geometry that the arc length is

π

, but the interesting part is to work through it to see how

π

pops out when using an arc length integral.)

[Answer]

Practice setting up arc length integrals

The actual integral you get for arc length is often difficult to compute. However, the important skill to practice is setting up that integral. So let's practice that a few times without worrying about computing the final integral (you can use a calculator or wolfram alpha once you get a concrete integral).

Example 2: Sine curve

What integral represents the arc length of the graph of

y = \sin (x)

between

x = 0

and

x = 2 π

$\begin{array}{r} \int_{a}^{b} \end{array}$ ‍
$d x$ ‍

a =

b =

[Answer]

Example 3: Up, not right

Consider the curve representing

$y = \pm \sqrt{x}$ ‍

For all values where

x \leq 4

. Find an integral expressing this curve's arc length. But this time, write everything in the integral in terms of

y

, not

x

$\begin{array}{r} \int_{a}^{b} \end{array}$ ‍
$d y$ ‍

a =

b =

[Answer]

Example 4: Full generality

Suppose you have any arbitrary function

f (x)

, with derivative

f^{'} (x)

. Which of the following represents the arc length of the graph

$y = f (x)$ ‍

between the points

x = a

and

x = b

[Answer]

Oftentimes people will start teaching arc length by presenting this formula. Personally, I think that takes all the fun out of discovering it yourself and getting a genuine feel for what it really represents.

Summary

We use the words arc length to describe how long a curve is. If you imagine the curve as a string, this is the length of that string once you pull it taut.
You can find the arc length of a curve with an integral of the form
$\begin{array}{r} \int \sqrt{(d x)^{2} + (d y)^{2}} \end{array}$ ‍
If the curve is the graph of a function $y = f (x)$ ‍, replace the $d y$ ‍ term in the integral with $f^{'} (x) d x$ ‍, then factor out the $d x$ ‍. The boundary values of the integral will be the leftmost and rightmost $x$ ‍-values of the curve.
When setting up an arc length integral, it helps to think about how you would actually choose to walk along the curve.

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Aryan Jamali
Posted 8 years ago. Direct link to Aryan Jamali's post “In the first example with...”
In the first example with the unit circle I don't get why x = -cos(teta) isn't the x-coordinate simply cos(teta) as the trigonometric coordinate are given by the point x =(x1,y1)=(cos(teta),sin(teta)) ? :)
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(11 votes)
Answer
- jjakow
  Posted 6 years ago. Direct link to jjakow's post “By convention, we think o...”
  By convention, we think of the unit circle as starting at the point (1,0). This corresponds to a
  x =cos(theta) parametrization. However, since the integral travels from the lower limit (-1,0) to the upper limit (1,0) we negate the cosine to switch direction of travel. Hope this helps
  Button navigates to signup page
  (5 votes)
MichaelSuitang
Posted 7 years ago. Direct link to MichaelSuitang's post “Haven't we learned about ...”
Haven't we learned about arc length in ordinary integral classes previously? Why are we studying this again?
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(3 votes)
Answer
- Stefen
  Posted 7 years ago. Direct link to Stefen's post “It is a review leading to...”
  It is a review leading to Arc length of parametric curves which leads to line integrals in scalar fields which leads to line integrals in vector fields and on it goes.
  Sometimes a little refresher review on a concept helps when a new topic depends on it.
  Button navigates to signup page
  (13 votes)
jp2338
Posted a year ago. Direct link to jp2338's post “For the part of x = -cos(...”
For the part of x = -cos(theta), I think a better wway to describe theta is the angle between the radius and the positive side of x-axis ( in the picture it is the angle between the radius and negative side of x-axis). And then the angle between the radius and negative side of x-axis (i.e the theta in current picture) would be pi-theta, then x = cos(pi-theta) = -cos(theta). Changing the definition of theta would be easier for student to know, otherwise it seems x=-cos(theta) comes from nowhere.
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(2 votes)
Answer
Victor Gutierrez
Posted 6 years ago. Direct link to Victor Gutierrez's post “First example, step 3, ho...”
First example, step 3, how to solve the integral using trigonometric substitution. This method is only valid for integrals of arc lengths right? Because if I find that integral in other contexts, I couldn´t think of the function inside the integral as a semicircle, because it is not a semicircle, although in the explanation is treated as a semicircle. So I don´t understand this method for solving these kind of integrals.
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(1 vote)
Answer
- jjakow
  Posted 6 years ago. Direct link to jjakow's post “This cannot be solved by ...”
  This cannot be solved by hand. This article and previous articles show how to set up the integral not how to solve this. The integrala in example 3 can only be approximated
  Button navigates to signup page
  (2 votes)
bluemoon29yume
Posted 7 months ago. Direct link to bluemoon29yume's post “in example 1, step 1, I d...”
in example 1, step 1, I don't understand where did the 2 come from in the denominator? in the d(1-x^2)/2(square root of /(1-x^2)
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(1 vote)
Answer
- Venkata
  Posted 7 months ago. Direct link to Venkata's post “It comes from the derivat...”
  It comes from the derivative of sqrt(1-x^(2)). Take a simpler example. Let's say we want the derivative of sqrt(x). This is just x^(1/2). Using the power rule, we get (1/2)x^(-1/2) or 1/(2sqrt(x)). The example in the article is the same, but as the inside of the square is a function, you need the chain rule.
  Button navigates to signup page
  (2 votes)
jimlkport
Posted 4 years ago. Direct link to jimlkport's post “I put in 1+4y^2 instead o...”
I put in 1+4y^2 instead of 4y^2+1 and got wrong answer! Nice going........
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(0 votes)
Answer

Multivariable calculus

Course: Multivariable calculus > Unit 4

Background

Example 1: Practice with a semicircle

Step 1: Write dy‍ in terms of dx‍

Step 2: Replace dy‍ in the integral

Step 3: Place bounds on the integral and solve

Practice setting up arc length integrals

Example 2: Sine curve

Example 3: Up, not right

Example​ 4: Full generality

Summary

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Step 1: Write $d y$ ‍ in terms of $d x$ ‍

Step 2: Replace $d y$ ‍ in the integral

Example 4: Full generality