CalcII_ArcLength

Calculus II

Preface Here are my online notes for my Calculus II course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn Calculus II or needing a refresher in some of the topics from the class. These notes do assume that the reader has a good working knowledge of Calculus I topics including limits, derivatives and basic integration and integration by substitution. Calculus II tends to be a very difficult course for many students. There are many reasons for this. The first reason is that this course does require that you have a very good working knowledge of Calculus I. The Calculus I portion of many of the problems tends to be skipped and left to the student to verify or fill in the details. If you don’t have good Calculus I skills, and you are constantly getting stuck on the Calculus I portion of the problem, you will find this course very difficult to complete. The second, and probably larger, reason many students have difficulty with Calculus II is that you will be asked to truly think in this class. That is not meant to insult anyone; it is simply an acknowledgment that you can’t just memorize a bunch of formulas and expect to pass the course as you can do in many math classes. There are formulas in this class that you will need to know, but they tend to be fairly general. You will need to understand them, how they work, and more importantly whether they can be used or not. As an example, the first topic we will look at is Integration by Parts. The integration by parts formula is very easy to remember. However, just because you’ve got it memorized doesn’t mean that you can use it. You’ll need to be able to look at an integral and realize that integration by parts can be used (which isn’t always obvious) and then decide which portions of the integral correspond to the parts in the formula (again, not always obvious). Finally, many of the problems in this course will have multiple solution techniques and so you’ll need to be able to identify all the possible techniques and then decide which will be the easiest technique to use. So, with all that out of the way let me also get a couple of warnings out of the way to my students who may be here to get a copy of what happened on a day that you missed.

1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn’t covered in class.

2. In general I try to work problems in class that are different from my notes. However, with Calculus II many of the problems are difficult to make up on the spur of the moment and so in this class my class work will follow these notes fairly close as far as worked problems go. With that being said I will, on occasion, work problems off the top of my head when I can to provide more examples than just those in my notes. Also, I often

© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

Calculus II

don’t have time in class to work all of the problems in the notes and so you will find that some sections contain problems that weren’t worked in class due to time restrictions.

3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible in writing these up, but the reality is that I can’t anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are.

4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class.

© 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx

Calculus II

Arc Length In this section we are going to look at computing the arc length of a function. Because it’s easy enough to derive the formulas that we’ll use in this section we will derive one of them and leave the other to you to derive. We want to determine the length of the continuous function ( )y f x= on the interval [ ],a b . Initially we’ll need to estimate the length of the curve. We’ll do this by dividing the interval up into n equal subintervals each of width x∆ and we’ll denote the point on the curve at each point by Pi. We can then approximate the curve by a series of straight lines connecting the points. Here is a sketch of this situation for 9n = .

Now denote the length of each of these line segments by 1i iP P− and the length of the curve will then be approximately,

11

n

i ii

L P P−=

≈∑

and we can get the exact length by taking n larger and larger. In other words, the exact length will be,

11

limn

i in iL P P−→∞

=

= ∑

Now, let’s get a better grasp on the length of each of these line segments. First, on each segment let’s define ( ) ( )1 1i i i i iy y y f x f x− −∆ = − = − . We can then compute directly the length of the line segments as follows.

( ) ( )2 2 2 21 1 1i i i i i i iP P x x y y x y− − −= − + − = ∆ + ∆

By the Mean Value Theorem we know that on the interval [ ]1,i ix x− there is a point *

ix so that,

( ) ( ) ( )( )

( )

*1 1

*

i i i i i

i i

f x f x f x x x

y f x x

− −′− = −

′∆ = ∆

© 2007 Paul Dawkins 3 http://tutorial.math.lamar.edu/terms.aspx

Calculus II

Therefore, the length can now be written as,

( ) ( )

( )

( )

2 21 1 1

22 * 2

2*1

i i i i i i

i

i

P P x x y y

x f x x

f x x

− − −= − + −

′= ∆ + ∆

′= + ∆

The exact length of the curve is then,

( )

11

2*

1

lim

lim 1

n

i in in

in i

L P P

f x x

−→∞=

→∞=

=

′= + ∆

∑

∑

However, using the definition of the definite integral, this is nothing more than,

( ) 21

b

aL f x dx′= + ⌠

⌡

A slightly more convenient notation (in my opinion anyway) is the following.

2

1b

a

dyL dxdx

= +

⌠⌡

In a similar fashion we can also derive a formula for ( )x h y= on [ ],c d . This formula is,

( )2

21 1

dd

cc

dxL h y dy dydy

′= + = +

⌠⌠ ⌡

⌡

Again, the second form is probably a little more convenient. Note the difference in the derivative under the square root! Don’t get too confused. With one we differentiate with respect to x and with the other we differentiate with respect to y. One way to keep the two straight is to notice that the differential in the “denominator” of the derivative will match up with the differential in the integral. This is one of the reasons why the second form is a little more convenient. Before we work any examples we need to make a small change in notation. Instead of having two formulas for the arc length of a function we are going to reduce it, in part, to a single formula. From this point on we are going to use the following formula for the length of the curve. Arc Length Formula(s)


Calculus II

L ds= ∫ where,

( )

( )

2

2

1 if ,

1 if ,

dyds dx y f x a x bdx

dxds dy x h y c y ddy

= + = ≤ ≤

= + = ≤ ≤

Note that no limits were put on the integral as the limits will depend upon the ds that we’re using. Using the first ds will require x limits of integration and using the second ds will require y limits of integration. Thinking of the arc length formula as a single integral with different ways to define ds will be convenient when we run across arc lengths in future sections. Also, this ds notation will be a nice notation for the next section as well. Now that we’ve derived the arc length formula let’s work some examples.

Example 1 Determine the length of ( )ln secy x= between 04

x π≤ ≤ .

Solution In this case we’ll need to use the first ds since the function is in the form ( )y f x= . So, let’s get the derivative out of the way.

2

2sec tan tan tansec

dy x x dyx xdx x dx

= = =

Let’s also get the root out of the way since there is often simplification that can be done and there’s no reason to do that inside the integral.

2

2 21 1 tan sec sec secdy x x x xdx

+ = + = = =

Note that we could drop the absolute value bars here since secant is positive in the range given. The arc length is then,

( )

40

40

sec

ln sec tan

ln 2 1

L x dx

x x

π

π

=

= +

= +

∫

Example 2 Determine the length of ( )32

2 13

x y= − between 1 4y≤ ≤ .

Solution


Calculus II

There is a very common mistake that students make in problems of this type. Many students see that the function is in the form ( )x h y= and they immediately decide that it will be too difficult to work with it

in that form so they solve for y to get the function into the form ( )y f x= . While that can be done here it will lead to a messier integral for us to deal with. Sometimes it’s just easier to work with functions in the form ( )x h y= . In fact, if you can work with

functions in the form ( )y f x= then you can work with functions in the form ( )x h y= . There really

isn’t a difference between the two so don’t get excited about functions in the form ( )x h y= . Let’s compute the derivative and the root.

( )21

21 1 1 1dx dxy y ydy dy

= − ⇒ + = + − =

As you can see keeping the function in the form ( )x h y= is going to lead to a very easy integral. To see

what would happen if we tried to work with the function in the form ( )y f x= see the next example. Let’s get the length.

4

1

432

1

23

143

L y dy

y

=

=

=

∫

As noted in the last example we really do have a choice as to which ds we use. Provided we can get the function in the form required for a particular ds we can use it. However, as also noted above, there will often be a significant difference in difficulty in the resulting integrals. Let’s take a quick look at what would happen in the previous example if we did put the function into the form ( )y f x= . Example 3 Redo the previous example using the function in the form ( )y f x= instead. Solution In this case the function and its derivative would be,

2 13 33 31

2 2x dy xy

dx

− = + =

The root in the arc length formula would then be.


Calculus II

( )

( )( )

( )

( )

222 33 33 22

2 2 13 3 33 3 32 2 2

1111 1xx

x x x

dydx

++ + = + = =

All the simplification work above was just to put the root into a form that will allow us to do the integral. Now, before we write down the integral we’ll also need to determine the limits. This particular ds requires x limits of integration and we’ve got y limits. They are easy enough to get however. Since we know x as a function of y all we need to do is plug in the original y limits of integration and get the x limits of integration. Doing this gives,

( )32

20 33

x≤ ≤

Not easy limits to deal with, but there they are. Let’s now write down the integral that will give the length.

( )

( )

( )32

2 3 233 32

13 320

1x

xL dx

+=⌠⌡

That’s a really unpleasant looking integral. It can be evaluated however using the following substitution.

2 13 33 31

2 2x xu du dx

− = + =

( )

32

0 12 3 43

x u

x u

= ⇒ =

= ⇒ =

Using this substitution the integral becomes,

4

1

432

1

23

143

L u du

u

=

=

=

∫

So, we got the same answer as in the previous example. Although that shouldn’t really be all that surprising since we were dealing with the same curve. From a technical standpoint the integral in the previous example was not that difficult. It was just a Calculus I substitution. However, from a practical standpoint the integral was significantly more difficult than the integral we evaluated in Example 2. So, the moral of the story here is that we can use either formula (provided we can get the function in the correct form of course) however one will often be significantly easier to actually evaluate. Okay, let’s work one more example.


Calculus II

Example 4 Determine the length of 212

x y= for 102

x≤ ≤ . Assume that y is positive.

Solution We’ll use the second ds for this one as the function is already in the correct form for that one. Also, the other ds would again lead to a particularly difficult integral. The derivative and root will then be,

2

21 1dx dxy ydy dy

= ⇒ + = +

Before writing down the length notice that we were given x limits and we will need y limits for this ds. With the assumption that y is positive these are easy enough to get. All we need to do is plug x into our equation and solve for y. Doing this gives, 0 1y≤ ≤ The integral for the arc length is then,

1 2

01L y dy= +∫

This integral will require the following trig substitution. 2tan secy dy dθ θ θ= =

0 0 tan 0

1 1 tan4

y

y

θ θπθ θ

= ⇒ = ⇒ =

= ⇒ = ⇒ =

2 2 21 1 tan sec sec secy θ θ θ θ+ = + = = = The length is then,

( )

( )( )

340

4

0

sec

1 sec tan ln sec tan21 2 ln 1 22

L dπ

π

θ θ

θ θ θ θ

=

= + +

= + +

∫

The first couple of examples ended up being fairly simple Calculus I substitutions. However, as this last example had shown we can end up with trig substitutions as well for these integrals.


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CalcII_ArcLength