If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Multivariable calculus>Unit 4

Lesson 2: Line integrals for scalar functions (articles)

# Arc length of function graphs, introduction

The length of a curve, called its "arc length", can be found using a certain integral.

## What is arc length?

We usually measure length with a straight line, but curves have length too. A familiar example is the circumference of a circle, which has length $2\pi r$ for radius $r$. In general, the length of a curve is called the arc length. But how do you find the arc length of an arbitrary curve? Let's find out.

## What we're building to

• You can find the arc length of a curve with an integral that looks something like this:
$\int \sqrt{\left(dx{\right)}^{2}+\left(dy{\right)}^{2}}$
The bounds of this integral depend on how you define the curve.
• If the curve is the graph of a function $y=f\left(x\right)$, replace the $dy$ term in the integral with ${f}^{\prime }\left(x\right)dx$, then factor out the $dx$.

## Warmup: Approximating arc length

Let's look at the parabola defined by the following equation:
$y=f\left(x\right)={x}^{2}$
Consider the portion of the curve between $x=-2$ and $x=2$.
Key question: What is the arc length of this curve?
Just so the question is clear, imagine the curve was a piece of string. You pull this string straight and measure it with a ruler.
If you had to guess, you could start by approximating this curve with some lines. Here's how that might look:
• A line from $\left(-2,4\right)$ to $\left(-1,1\right)$
• A line from $\left(-1,1\right)$ to $\left(0,0\right)$
• A line from $\left(0,0\right)$ to $\left(1,1\right)$
• A line from $\left(1,1\right)$ to $\left(2,4\right)$
It would be tedious, but you could calculate the length of each line segment using the Pythagorean theorem, then add them up.
Concept check: What is the length of the line from $\left(-2,4\right)$ to $\left(-1,1\right)$?
Enter answer in exact form, with a square root:

For an even more accurate estimate, you could approximate the curve with many tiny lines.
Computing all their lengths and adding them up would be painfully mind-numbing, but let's break down what it would actually look like. Zoom in on one of the little lines.
First look at the change in the $x$ value from the start of the line to its end. Let's call that $\mathrm{\Delta }x$. Similarly, let's say the change in the $y$-value is $\mathrm{\Delta }y$. Then, using the Pythagorean theorem, we can write the length of the line as
$\sqrt{\left(\mathrm{\Delta }x{\right)}^{2}+\left(\mathrm{\Delta }y{\right)}^{2}}$
Our approximation for the length of the curve will then be the sum of the lengths of all these little lines. When expressing an idea like this with symbols, it's common for writers to be a little loose with the notation and write something like this:
$\sum _{\text{all little lines}}\sqrt{\left(\mathrm{\Delta }x{\right)}^{2}+\left(\mathrm{\Delta }y{\right)}^{2}}$

## Bringing in integrals

Hmm, let's see... we are approximating a curve with many tiny steps, then adding up a very large number of very small things. With tinier steps and a larger sum, we will get a better approximation. Sound familiar?
Problems like this one are exactly what integrals were made for.
Most people first learn about integration in the context of computing the area under a curve. You imagine approximating that area with a bunch of thin rectangles. The width of each one is thought of as "$dx$", some tiny change in the $x$-value. The height of a rectangle at a given $x$-value is $f\left(x\right)$. Therefore, the area of each rectangle is
$\stackrel{\text{height}}{\stackrel{⏞}{f\left(x\right)}}\underset{\text{width}}{\underset{⏟}{dx}}$
The full area under the curve is then expressed with an integral:
${\int }_{a}^{b}f\left(x\right)dx$
This integral is a powerful machine, like a $\mathrm{\Sigma }$ on steroids. It does not merely sum over the values $f\left(x\right)dx$ for a particular tiny value of $dx$; it considers the limiting value of such a sum as the tiny width $dx$ tends toward $0$. In other words, as the approximation using rectangles comes closer and closer to the true area under the curve.
But this powerful machine can be used in many contexts unrelated to the area under a curve. Anytime you get that sensation of wanting to add a very large number of very small things, the integral swoops in to simultaneously make things less tedious and more accurate.
For example, we get that sensation when approximating arc length with the vague sum:
$\sum _{\text{all little lines}}\sqrt{\left(\mathrm{\Delta }x{\right)}^{2}+\left(\mathrm{\Delta }y{\right)}^{2}}$
So we turn it into an integral:
$\begin{array}{r}\int \sqrt{\left(dx{\right)}^{2}+\left(dy{\right)}^{2}}\end{array}$
One thing this notation does not communicate well is that $dy$, the change in height across one of our little approximation lines, is dependent on $dx$, the horizontal component of that line. Specifically, since the curve is defined by the relation $y={x}^{2}$, we can take the derivative of each side to see how $dy$ depends on $dx$,
$\begin{array}{rl}y& ={x}^{2}\\ d\left(y\right)& =d\left({x}^{2}\right)\\ dy& =2x\phantom{\rule{0.167em}{0ex}}dx\end{array}$
When we put this into our integral, it unfolds to look a bit more familiar.
$\begin{array}{rl}\int \sqrt{\left(dx{\right)}^{2}+\left(dy{\right)}^{2}}& =\int \sqrt{\left(dx{\right)}^{2}+\left(2x\phantom{\rule{0.167em}{0ex}}dx{\right)}^{2}}\\ & =\int \sqrt{\left(1+\left(2x{\right)}^{2}\right)\left(dx{\right)}^{2}}\\ & =\int \sqrt{1+4{x}^{2}}dx\end{array}$
I've been purposefully lazy about placing bounds on this integral, but now that everything inside the integral is in terms of $x$, with no $dy$'s mucking it up, it makes sense to define the bounds of integration in terms of the $x$ value, which in this case is from $-2$ to $2$.
$\begin{array}{r}{\int }_{-2}^{2}\sqrt{1+4{x}^{2}}dx\end{array}$
This looks like something we can compute. Well, actually, in this case, it turns out to be quite a tricky integral, but in this day and age, we can just plug integrals into a computer if we need to. The point is that the idea of approximating our curve's length with little lines, which was at first vaguely written with loose notation, has now turned into a concrete, computable integral.
For now, rather than getting bogged down with the details of this integral (there's plenty of that coming in the next article), I want to highlight some points from this example.

## Takeaways

• The central expression to remember is $\sqrt{d{x}^{2}+d{y}^{2}}$, which represents a tiny unit of arc length in terms of $x$ and $y$.
• The arc length integral you set up starts its life looking like this:
$\int \sqrt{\left(dx{\right)}^{2}+\left(dy{\right)}^{2}}$
• Before computing the integral, we had to write the differential $dy$ in terms of the differential $dx$. To do this, we took the derivative of the function defining the curve.
• In general, an integral can only be computed with respect to a single differential, and finding relations between differentials can be done using the derivative.
• Maybe the most important lesson to take away from this is that integrals can be used to do things other than finding the area under a curve.

## Practice

To solidify your understanding, you can practice more arc length problems in the next article.

## Want to join the conversation?

• Why can we take out (dx)^2 from the square root as simply dx, isn't that supposed to be plus or minus dx?
• Since dx is a tiny change in x direction how do you imagine it being negative?
• has anyone tried to create a way to get the most accurate answer.
• I'm sorry, I don't quite understand your question... do you mean "is there a more precice way of approximating the arclengh of a graph then the method used above"? If that is the question, combining infinetly small units infenantly many times is so far our best and most accurate way of doing that. We don't explore /more precice/ options because it is pretty exact and it will just become negligable, as the equation for a derivative prooves to us (review your early calculus if this sounds wierd).
• At my school we do it a different way, on the graphing calculator we we take the number and divide by 360 and multiply by 2π. it would look like this #/360 times 2π. IS this another way to do0 it or am I confused with a different type of problem
• Yes, the arc length you're talking about is changing a circles degree measurement into a measure. This is because the circumference of a circle is C = pi * d

We also know that a circle has 2pi radians in it (convertible as 2pi = 360 degrees, thus 1 degree = 2pi / 360 radians) This is where you're getting that, you taking the degree, i.e. 45 degrees, and multiplying it by 2pi / 360 to get the radian measure which would be pi/4 radians in this case.

This is only for the unit circle however. Imagine we had a much larger circle, with diameter 30. Then the circumference is 30pi units. This logically means an arc with angle measure 180 degrees would have a length of 15pi units (it is half of the circle). Using our above formula won't simply work: 180 * 2pi / 360 = pi. But our circumference should be 15pi, not just pi.

The unit circle is based on a circle with radius of 1, therefore it's diameter it's full circumference is 2pi (it has a diameter of 2, i.e. C = d * pi = 2 * pi). In our case we changed this value to C = 30 pi. This also means we changed the definition for our radians, since 1 degree is now actually equal in length to: Degree = 1/360 * 30pi radians). This conversion would give the accurrate answers: 180 degrees * 30 pi / 360 = 15 pi.

This arc length problem involves non circular arcs, it is talking about "curve length", i.e. not a straight line (though equally aplicable if you wished).

It does not extend because these arcs don't have circular properties. Imagine the center of a square, then trying to take it's "arc length" as you would a circle, it doesn't work, because a square is a bit different in it's construction and it's properties are different.
• After you calculate the integral for arc length - such as: the integral of ((1 + (-2x)^2))^(1/2) dx from 0 to 3 and get an answer for the length of the curve: y = 9 - x^2 from 0 to 3 which equals approximately 9.7 - what is the unit you would associate with that answer? I drew it to scale in inches and the answer approximately equaled the length of my piece of yarn in inches? Is that always the case?