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Course: Multivariable calculus > Unit 4

Lesson 2: Line integrals for scalar functions (articles)

Arc length of parametric curves

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

Background:

What we're building to

To find the arc length of a curve, set up an integral of the form
$\begin{array}{r} \int \sqrt{(d x)^{2} + (d y)^{2}} \end{array}$ ‍
We now care about the case when the curve is defined parametrically, meaning $x$ ‍ and $y$ ‍ are defined as functions of some new variable $t$ ‍. To apply the arc length integral, first take the derivative of both these functions to get $d x$ ‍ and $d y$ ‍ in terms of $d t$ ‍.
$d x = \frac{d x}{d t} d t$ ‍
$d y = \frac{d y}{d t} d t$ ‍
Plug these expressions into the integral and factor the $d t^{2}$ ‍ term out of the radical.

The length of a parametric curve

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

$x (t) = t^{3} - t$ ‍

$y (t) = 2 e^{- t^{2}}$ ‍

If we let

t

range from

- 1.5

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?

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See video transcript

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:

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

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,
$\sqrt{d x^{2} + d y^{2}}$ ‍
$d x$ ‍ and $d y$ ‍ represent the tiny change in $x$ ‍ and $y$ ‍ values from the start to the end of the line.

This same integral can apply to parametric curves as well as function graphs. This time, since

x

and

y

are given as functions of

t

, we write

d x

and

d y

in terms of

d t

by taking the derivative of these two functions.

For example, differentiating the function defining

x

, we get

$\begin{aligned} x & = t^{3} - t \\ d (x) & = d (t^{3} - t) \\ d x & = (3 t^{2} - 1) d t \end{aligned}$ ‍

And similarly with

y

$\begin{aligned} y & = 2 e^{- t^{2}} \\ d (y) & = d (2 e^{- t^{2}}) \\ d y & = (2 (- 2 t) e^{- t^{2}}) d t \\ d y & = - 4 t e^{- t^{2}} d t \end{aligned}$ ‍

You can think of these expressions as answering the question "when you take some value

t

, and increase it slightly by some tiny amount

d t

, how much does it change

x

and

y

?" The answer is expressed in terms of

t

d t

Putting these into the integral, we get

$\begin{aligned} \int \sqrt{(d x)^{2} + {(d y)}^{2}} & = \int \sqrt{((3 t^{2} - 1) d t)^{2} + ((- 4 t e^{- t^{2}}) d t)^{2}} \\ = \int \sqrt{({(3 t^{2} - 1)}^{2} + {(- 4 t e^{- t^{2}})}^{2}) d t^{2}} \\ = \int \sqrt{9 t^{4} - 6 t^{2} + 1 + 16 t^{2} e^{- 2 t^{2}}} d t \end{aligned}$ ‍

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

1.5

, so we have

$\begin{array}{r} \int_{- 1.5}^{1.5} \sqrt{9 t^{4} - 6 t^{2} + 1 + 16 t^{2} e^{- 2 t^{2}}} d t \end{array}$ ‍

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

$x (t) = t^{3} - 3 t$ ‍

$y (t) = 3 t^{2}$ ‍

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

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

To get this integral in terms of

t

, we must write

d x

and

d y

each as some expression of

t

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

What is

d x

in terms of

t

$d x =$ ‍
$d t$ ‍

What is

d y

in terms of

t

$d y =$ ‍
$d t$ ‍

Step 2: Put these expressions in the integral

What does our integral look like after we plug in these expressions for

d x

and

d y

? Simplify it down to the point where there is no radical.

$\int$ ‍
$d t$ ‍

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

The problem states that the curve runs from

- 2

2

. Solve the integral with these bounds.

$\begin{aligned} \int_{- 2}^{2} (3 t^{2} + 3) d t & = {[t^{3} + 3 t]}_{- 2}^{2} \\ = (2^{3} + 3 (2)) - ((- 2)^{3} + 3 (- 2)) \\ = 14 - (- 14) \\ = 28 \end{aligned}$ ‍

So evidently the length of this curve is

28

. Looking at the picture, this seems about right. The curve starts from about

y \approx 12

, goes down to the

x

-axis and back, which takes at least 24 units of length. Since it has some curvature, wandering left and right as it goes down and up, the true length is a bit more than

24

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
$\begin{array}{r} \int \sqrt{(d x)^{2} + (d y)^{2}} \end{array}$ ‍
When the curve is defined parametrically, with $x$ ‍ and $y$ ‍ given as functions of $t$ ‍, take the derivative of both these functions to get $d x$ ‍ and $d y$ ‍ in terms of $d t$ ‍.
$d x = \frac{d x}{d t} d t$ ‍
$d y = \frac{d y}{d t} d t$ ‍
plug these expressions into the integral and factor the $d t^{2}$ ‍ term out of the radical.

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Maria Elvina
Posted 8 years ago. Direct link to Maria Elvina's post “The link to the backgroun...”
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
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MustardManExtremeTurboEdition
Posted 2 months ago. Direct link to MustardManExtremeTurboEdition's post “damn there are almost no ...”
damn there are almost no comments here
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(1 vote)
Answer

Multivariable calculus

Course: Multivariable calculus > Unit 4

Background:

What we're building to

The length of a parametric curve

Practice a parametric arc length integral

Step 1: Write dx‍ and dy‍ in terms of t‍

Step 2: Put these expressions in the integral

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

What's next?

Summary

Want to join the conversation?

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