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### Course: Multivariable calculus > Unit 4

Lesson 2: Line integrals for scalar functions (articles)# Arc length of parametric curves

How to find the length of a parametric curve? This will lead to the idea of a line integral.

## What we're building to

- To find the arc length of a curve, set up an integral of the form
- We now care about the case when the curve is defined parametrically, meaning
and$x$ are defined as functions of some new variable$y$ . To apply the arc length integral, first take the derivative of both these functions to get$t$ and$dx$ in terms of$dy$ .$dt$ Plug these expressions into the integral and factor the term out of the radical.$d{t}^{2}$

## The length of a parametric curve

Consider the parametric curve defined by the following set of equations:

If we let $t$ range from $-1.5$ to $1.5$ , the resulting curve looks like this:

Key question: What is the length of this curve?

That is, imagine pulling the line straight, as if you were tightening a loose piece of string, then measuring it with a ruler. What value would you get?

In the last article, we saw how to find the arc length of

*function graphs*, not parametric curves. We started by writing down the following integral:Let's quickly recap the meaning behind this integral.

- Imagine approximating the curve with a bunch of tiny straight lines.
- The length of each such tiny line is given using the Pythagorean theorem,
and$dx$ represent the tiny change in$dy$ and$x$ values from the start to the end of the line.$y$

This same integral can apply to $x$ and $y$ are given as functions of $t$ , we write $dx$ and $dy$ in terms of $dt$ by taking the derivative of these two functions.

*parametric*curves as well as function graphs. This time, sinceFor example, differentiating the function defining $x$ , we get

And similarly with $y$ :

You can think of these expressions as answering the question "when you take some value $t$ , and increase it slightly by some tiny amount $dt$ , how much does it change $x$ and $y$ ?" The answer is expressed in terms of $t$ an $dt$ .

Putting these into the integral, we get

Now everything inside the integral is written in terms of $t$ , so the bounds we place on the integral correspond with the starting and ending values of the parameter $t$ . In this case, we are letting $t$ range from $-1.5$ to $1.5$ , so we have

This is a very nasty integral to compute. I'm not even sure that an antiderivative exists. However, we've at least reduced the arc length problem down to a state where you can plug it into a computer.

## Practice a parametric arc length integral

Let's look at the parametric curve defined by

Consider the segment of this curve between the points where $t=-2$ and $t=2$ .

What is the length of this segment?

Since our curve is expressed in terms of $x$ and $y$ , our arc length integrals begin life looking like

To get this integral in terms of $t$ , we must write $dx$ and $dy$ each as some expression of $t$

### Step 1: Write $dx$ and $dy$ in terms of $t$

What is $dx$ in terms of $t$ ?

What is $dy$ in terms of $t$ ?

### Step 2: Put these expressions in the integral

What does our integral look like after we plug in these expressions for $dx$ and $dy$ ? Simplify it down to the point where there is no radical.

### Step 3: Place the appropriate bounds on the integral and solve

The problem states that the curve runs from $-2$ to $2$ . Solve the integral with these bounds.

## What's next?

Arc length of parametric curves is a natural starting place for learning about line integrals, a central notion in multivariable calculus. To keep things from getting too messy as we do so, I first need to go over some more compact notation for these arc length integrals, which you can find in the next article.

## Summary

- To find the arc length of a curve, set up an integral of the form
- When the curve is defined parametrically, with
and$x$ given as functions of$y$ , take the derivative of both these functions to get$t$ and$dx$ in terms of$dy$ .$dt$ plug these expressions into the integral and factor the term out of the radical.$d{t}^{2}$

## Want to join the conversation?

- The link to the background topic 'Parametric curves' is broken.

I believe this is the correct one: https://www.khanacademy.org/math/multivariable-calculus/thinking-about-multivariable-function/ways-to-represent-multivariable-functions/a/parametric-functions(8 votes) - damn there are almost no comments here(1 vote)