CALCULUS II. Paul Dawkins

Transcription

1 CALCULUS II Paul Dawkis

2 Table of Cotets Preface... iii Outlie... v Itegratio Techiques... Itroductio... Itegratio by Parts... Itegrals Ivolvig Trig Fuctios... Trig Substitutios... Partial Fractios...4 Itegrals Ivolvig Roots...4 Itegrals Ivolvig Quadratics...44 Itegratio Strategy...5 Improper Itegrals...59 Compariso Test for Improper Itegrals...66 Approximatig Defiite Itegrals...7 Applicatios of Itegrals Itroductio...80 Arc Legth...8 Surface Area...87 Ceter of Mass...9 Hydrostatic Pressure ad Force...97 Probability...0 Parametric Equatios ad Polar Coordiates Itroductio...06 Parametric Equatios ad Curves...07 Tagets with Parametric Equatios...7 Area with Parametric Equatios...4 Arc Legth with Parametric Equatios...7 Surface Area with Parametric Equatios...4 Polar Coordiates...4 Tagets with Polar Coordiates...5 Area with Polar Coordiates...55 Arc Legth with Polar Coordiates...6 Surface Area with Polar Coordiates...64 Arc Legth ad Surface Area Revisited...65 Sequeces ad Series Itroductio...67 Sequeces...69 More o Sequeces...79 Series The Basics...85 Series Covergece/Divergece...9 Series Special Series...00 Itegral Test...08 Compariso Test / Limit Compariso Test...7 Alteratig Series Test...6 Absolute Covergece... Ratio Test...6 Root Test...4 Strategy for Series...46 Estimatig the Value of a Series...49 Power Series...60 Power Series ad Fuctios...68 Taylor Series...75 Applicatios of Series Paul Dawkis i

3 Biomial Series...90 Vectors... 9 Itroductio...9 Vectors The Basics...9 Vector Arithmetic...97 Dot Product...0 Cross Product...0 Three Dimesioal Space... 6 Itroductio...6 The -D Coordiate System...8 Equatios of Lies...4 Equatios of Plaes...0 Quadric Surfaces... Fuctios of Several Variables...9 Vector Fuctios...46 Calculus with Vector Fuctios...55 Taget, Normal ad Biormal Vectors...58 Arc Legth with Vector Fuctios...6 Curvature...65 Velocity ad Acceleratio...67 Cylidrical Coordiates...70 Spherical Coordiates Paul Dawkis ii

4 Preface Here are my olie otes for my Calculus II course that I teach here at Lamar Uiversity. Despite the fact that these are my class otes, they should be accessible to ayoe watig to lear Calculus II or eedig a refresher i some of the topics from the class. These otes do assume that the reader has a good workig kowledge of Calculus I topics icludig limits, derivatives ad basic itegratio ad itegratio by substitutio. Calculus II teds to be a very difficult course for may studets. There are may reasos for this. The first reaso is that this course does require that you have a very good workig kowledge of Calculus I. The Calculus I portio of may of the problems teds to be skipped ad left to the studet to verify or fill i the details. If you do t have good Calculus I skills, ad you are costatly gettig stuck o the Calculus I portio of the problem, you will fid this course very difficult to complete. The secod, ad probably larger, reaso may studets have difficulty with Calculus II is that you will be asked to truly thik i this class. That is ot meat to isult ayoe; it is simply a ackowledgmet that you ca t just memorize a buch of formulas ad expect to pass the course as you ca do i may math classes. There are formulas i this class that you will eed to kow, but they ted to be fairly geeral. You will eed to uderstad them, how they work, ad more importatly whether they ca be used or ot. As a example, the first topic we will look at is Itegratio by Parts. The itegratio by parts formula is very easy to remember. However, just because you ve got it memorized does t mea that you ca use it. You ll eed to be able to look at a itegral ad realize that itegratio by parts ca be used (which is t always obvious) ad the decide which portios of the itegral correspod to the parts i the formula (agai, ot always obvious). Fially, may of the problems i this course will have multiple solutio techiques ad so you ll eed to be able to idetify all the possible techiques ad the decide which will be the easiest techique to use. So, with all that out of the way let me also get a couple of warigs out of the way to my studets who may be here to get a copy of what happeed o a day that you missed.. Because I wated to make this a fairly complete set of otes for ayoe watig to lear calculus I have icluded some material that I do ot usually have time to cover i class ad because this chages from semester to semester it is ot oted here. You will eed to fid oe of your fellow class mates to see if there is somethig i these otes that was t covered i class.. I geeral I try to work problems i class that are differet from my otes. However, with Calculus II may of the problems are difficult to make up o the spur of the momet ad so i this class my class work will follow these otes fairly close as far as worked problems go. With that beig said I will, o occasio, work problems off the top of my head whe I ca to provide more examples tha just those i my otes. Also, I ofte 007 Paul Dawkis iii

5 do t have time i class to work all of the problems i the otes ad so you will fid that some sectios cotai problems that were t worked i class due to time restrictios.. Sometimes questios i class will lead dow paths that are ot covered here. I try to aticipate as may of the questios as possible i writig these up, but the reality is that I ca t aticipate all the questios. Sometimes a very good questio gets asked i class that leads to isights that I ve ot icluded here. You should always talk to someoe who was i class o the day you missed ad compare these otes to their otes ad see what the differeces are. 4. This is somewhat related to the previous three items, but is importat eough to merit its ow item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Usig these otes as a substitute for class is liable to get you i trouble. As already oted ot everythig i these otes is covered i class ad ofte material or isights ot i these otes is covered i class. 007 Paul Dawkis iv

6 Outlie Here is a listig ad brief descriptio of the material i this set of otes. Itegratio Techiques Itegratio by Parts Of all the itegratio techiques covered i this chapter this is probably the oe that studets are most likely to ru ito dow the road i other classes. Itegrals Ivolvig Trig Fuctios I this sectio we look at itegratig certai products ad quotiets of trig fuctios. Trig Substitutios Here we will look usig substitutios ivolvig trig fuctios ad how they ca be used to simplify certai itegrals. Partial Fractios We will use partial fractios to allow us to do itegrals ivolvig some ratioal fuctios. Itegrals Ivolvig Roots We will take a look at a substitutio that ca, o occasio, be used with itegrals ivolvig roots. Itegrals Ivolvig Quadratics I this sectio we are goig to look at some itegrals that ivolve quadratics. Itegratio Strategy We give a geeral set of guidelies for determiig how to evaluate a itegral. Improper Itegrals We will look at itegrals with ifiite itervals of itegratio ad itegrals with discotiuous itegrads i this sectio. Compariso Test for Improper Itegrals Here we will use the Compariso Test to determie if improper itegrals coverge or diverge. Approximatig Defiite Itegrals There are may ways to approximate the value of a defiite itegral. We will look at three of them i this sectio. Applicatios of Itegrals Arc Legth We ll determie the legth of a curve i this sectio. Surface Area I this sectio we ll determie the surface area of a solid of revolutio. Ceter of Mass Here we will determie the ceter of mass or cetroid of a thi plate. Hydrostatic Pressure ad Force We ll determie the hydrostatic pressure ad force o a vertical plate submerged i water. Probability Here we will look at probability desity fuctios ad computig the mea of a probability desity fuctio. Parametric Equatios ad Polar Coordiates Parametric Equatios ad Curves A itroductio to parametric equatios ad parametric curves (i.e. graphs of parametric equatios) Tagets with Parametric Equatios Fidig taget lies to parametric curves. Area with Parametric Equatios Fidig the area uder a parametric curve. Arc Legth with Parametric Equatios Determiig the legth of a parametric curve. 007 Paul Dawkis v

7 Surface Area with Parametric Equatios Here we will determie the surface area of a solid obtaied by rotatig a parametric curve about a axis. Polar Coordiates We ll itroduce polar coordiates i this sectio. We ll look at covertig betwee polar coordiates ad Cartesia coordiates as well as some basic graphs i polar coordiates. Tagets with Polar Coordiates Fidig taget lies of polar curves. Area with Polar Coordiates Fidig the area eclosed by a polar curve. Arc Legth with Polar Coordiates Determiig the legth of a polar curve. Surface Area with Polar Coordiates Here we will determie the surface area of a solid obtaied by rotatig a polar curve about a axis. Arc Legth ad Surface Area Revisited I this sectio we will summarize all the arc legth ad surface area formulas from the last two chapters. Sequeces ad Series Sequeces We will start the chapter off with a brief discussio of sequeces. This sectio will focus o the basic termiology ad covergece of sequeces More o Sequeces Here we will take a quick look about mootoic ad bouded sequeces. Series The Basics I this sectio we will discuss some of the basics of ifiite series. Series Covergece/Divergece Most of this chapter will be about the covergece/divergece of a series so we will give the basic ideas ad defiitios i this sectio. Series Special Series We will look at the Geometric Series, Telescopig Series, ad Harmoic Series i this sectio. Itegral Test Usig the Itegral Test to determie if a series coverges or diverges. Compariso Test/Limit Compariso Test Usig the Compariso Test ad Limit Compariso Tests to determie if a series coverges or diverges. Alteratig Series Test Usig the Alteratig Series Test to determie if a series coverges or diverges. Absolute Covergece A brief discussio o absolute covergece ad how it differs from covergece. Ratio Test Usig the Ratio Test to determie if a series coverges or diverges. Root Test Usig the Root Test to determie if a series coverges or diverges. Strategy for Series A set of geeral guidelies to use whe decidig which test to use. Estimatig the Value of a Series Here we will look at estimatig the value of a ifiite series. Power Series A itroductio to power series ad some of the basic cocepts. Power Series ad Fuctios I this sectio we will start lookig at how to fid a power series represetatio of a fuctio. Taylor Series Here we will discuss how to fid the Taylor/Maclauri Series for a fuctio. Applicatios of Series I this sectio we will take a quick look at a couple of applicatios of series. Biomial Series A brief look at biomial series. Vectors Vectors The Basics I this sectio we will itroduce some of the basic cocepts about vectors. 007 Paul Dawkis vi

8 Vector Arithmetic Here we will give the basic arithmetic operatios for vectors. Dot Product We will discuss the dot product i this sectio as well as a applicatio or two. Cross Product I this sectio we ll discuss the cross product ad see a quick applicatio. Three Dimesioal Space This is the oly chapter that exists i two places i my otes. Whe I origially wrote these otes all of these topics were covered i Calculus II however, we have sice moved several of them ito Calculus III. So, rather tha split the chapter up I have kept it i the Calculus II otes ad also put a copy i the Calculus III otes. The -D Coordiate System We will itroduce the cocepts ad otatio for the three dimesioal coordiate system i this sectio. Equatios of Lies I this sectio we will develop the various forms for the equatio of lies i three dimesioal space. Equatios of Plaes Here we will develop the equatio of a plae. Quadric Surfaces I this sectio we will be lookig at some examples of quadric surfaces. Fuctios of Several Variables A quick review of some importat topics about fuctios of several variables. Vector Fuctios We itroduce the cocept of vector fuctios i this sectio. We cocetrate primarily o curves i three dimesioal space. We will however, touch briefly o surfaces as well. Calculus with Vector Fuctios Here we will take a quick look at limits, derivatives, ad itegrals with vector fuctios. Taget, Normal ad Biormal Vectors We will defie the taget, ormal ad biormal vectors i this sectio. Arc Legth with Vector Fuctios I this sectio we will fid the arc legth of a vector fuctio. Curvature We will determie the curvature of a fuctio i this sectio. Velocity ad Acceleratio I this sectio we will revisit a stadard applicatio of derivatives. We will look at the velocity ad acceleratio of a object whose positio fuctio is give by a vector fuctio. Cylidrical Coordiates We will defie the cylidrical coordiate system i this sectio. The cylidrical coordiate system is a alterate coordiate system for the three dimesioal coordiate system. Spherical Coordiates I this sectio we will defie the spherical coordiate system. The spherical coordiate system is yet aother alterate coordiate system for the three dimesioal coordiate system. 007 Paul Dawkis vii

9 007 Paul Dawkis 0

10 Itegratio Techiques Itroductio I this chapter we are goig to be lookig at various itegratio techiques. There are a fair umber of them ad some will be easier tha others. The poit of the chapter is to teach you these ew techiques ad so this chapter assumes that you ve got a fairly good workig kowledge of basic itegratio as well as substitutios with itegrals. I fact, most itegrals ivolvig simple substitutios will ot have ay of the substitutio work show. It is goig to be assumed that you ca verify the substitutio portio of the itegratio yourself. Also, most of the itegrals doe i this chapter will be idefiite itegrals. It is also assumed that oce you ca do the idefiite itegrals you ca also do the defiite itegrals ad so to coserve space we cocetrate mostly o idefiite itegrals. There is oe exceptio to this ad that is the Trig Substitutio sectio ad i this case there are some subtleties ivolved with defiite itegrals that we re goig to have to watch out for. Outside of that however, most sectios will have at most oe defiite itegral example ad some sectios will ot have ay defiite itegral examples. Here is a list of topics that are covered i this chapter. Itegratio by Parts Of all the itegratio techiques covered i this chapter this is probably the oe that studets are most likely to ru ito dow the road i other classes. Itegrals Ivolvig Trig Fuctios I this sectio we look at itegratig certai products ad quotiets of trig fuctios. Trig Substitutios Here we will look usig substitutios ivolvig trig fuctios ad how they ca be used to simplify certai itegrals. Partial Fractios We will use partial fractios to allow us to do itegrals ivolvig some ratioal fuctios. Itegrals Ivolvig Roots We will take a look at a substitutio that ca, o occasio, be used with itegrals ivolvig roots. Itegrals Ivolvig Quadratics I this sectio we are goig to look at some itegrals that ivolve quadratics. Itegratio Strategy We give a geeral set of guidelies for determiig how to evaluate a itegral. Improper Itegrals We will look at itegrals with ifiite itervals of itegratio ad itegrals with discotiuous itegrads i this sectio. 007 Paul Dawkis

11 Compariso Test for Improper Itegrals Here we will use the Compariso Test to determie if improper itegrals coverge or diverge. Approximatig Defiite Itegrals There are may ways to approximate the value of a defiite itegral. We will look at three of them i this sectio. 007 Paul Dawkis

12 Itegratio by Parts Let s start off with this sectio with a couple of itegrals that we should already be able to do to get us started. First let s take a look at the followig. x x e dx = e + c So, that was simple eough. Now, let s take a look at, x x e dx To do this itegral we ll use the followig substitutio. u = x du = x dx x dx = du x u u x xe dx = du = + c = + c e e e Agai, simple eough to do provided you remember how to do substitutios. By the way make sure that you ca do these kids of substitutios quickly ad easily. From this poit o we are goig to be doig these kids of substitutios i our head. If you have to stop ad write these out with every problem you will fid that it will take you sigificatly loger to do these problems. Now, let s look at the itegral that we really wat to do. 6x x e dx 6x If we just had a x by itself or e by itself we could do the itegral easily eough. But, we do t have them by themselves, they are istead multiplied together. There is o substitutio that we ca use o this itegral that will allow us to do the itegral. So, at this poit we do t have the kowledge to do this itegral. To do this itegral we will eed to use itegratio by parts so let s derive the itegratio by parts formula. We ll start with the product rule. fg = f g+ fg Now, itegrate both sides of this. f g dx = f g + f g dx The left side is easy eough to itegrate ad we ll split up the right side of the itegral. fg = f g dx + f g dx Note that techically we should have had a costat of itegratio show up o the left side after doig the itegratio. We ca drop it at this poit sice other costats of itegratio will be showig up dow the road ad they would just ed up absorbig this oe. Fially, rewrite the formula as follows ad we arrive at the itegratio by parts formula. 007 Paul Dawkis

13 f g dx = fg f g dx This is ot the easiest formula to use however. So, let s do a couple of substitutios. u = f ( x) v= g( x) du = f x dx dv = g x dx Both of these are just the stadard Calc I substitutios that hopefully you are used to by ow. Do t get excited by the fact that we are usig two substitutios here. They will work the same way. Usig these substitutios gives us the formula that most people thik of as the itegratio by parts formula. u dv = uv v du To use this formula we will eed to idetify u ad dv, compute du ad v ad the use the formula. Note as well that computig v is very easy. All we eed to do is itegrate dv. v = dv So, let s take a look at the itegral above that we metioed we wated to do. Example Evaluate the followig itegral. 6x x e dx Solutio So, o some level, the problem here is the x that is i frot of the expoetial. If that was t there we could do the itegral. Notice as well that i doig itegratio by parts aythig that we choose for u will be differetiated. So, it seems that choosig u = x will be a good choice sice upo differetiatig the x will drop out. Now that we ve chose u we kow that dv will be everythig else that remais. So, here are the choices for u ad dv as well as du ad v. 6x u = x dv = e dx 6x 6x du = dx v = e dx = e 6 The itegral is the, 6x x 6x 6x x dx = e 6 e dx 6 e x 6x 6x = e e + c 6 6 Oce we have doe the last itegral i the problem we will add i the costat of itegratio to get our fial aswer. 007 Paul Dawkis 4

14 Next, let s take a look at itegratio by parts for defiite itegrals. The itegratio by parts formula for defiite itegrals is, Itegratio by Parts, Defiite Itegrals b u dv uv b b = v du b a a a Note that the uv i the first term is just the stadard itegral evaluatio otatio that you should a be familiar with at this poit. All we do is evaluate the term, uv i this case, at b the subtract off the evaluatio of the term at a. At some level we do t really eed a formula here because we kow that whe doig defiite itegrals all we eed to do is do the idefiite itegral ad the do the evaluatio. Let s take a quick look at a defiite itegral usig itegratio by parts. Example Evaluate the followig itegral. 6x xe dx Solutio This is the same itegral that we looked at i the first example so we ll use the same u ad dv to get, 6x x 6x 6x xe dx = e dx 6 6 e 6x 6x x = e e = e + e Sice we eed to be able to do the idefiite itegral i order to do the defiite itegral ad doig the defiite itegral amouts to othig more tha evaluatig the idefiite itegral at a couple of poits we will cocetrate o doig idefiite itegrals i the rest of this sectio. I fact, throughout most of this chapter this will be the case. We will be doig far more idefiite itegrals tha defiite itegrals. Let s take a look at some more examples. Example Evaluate the followig itegral. t ( t + 5) cos dt 4 Solutio There are two ways to proceed with this example. For may, the first thig that they try is multiplyig the cosie through the parethesis, splittig up the itegral ad the doig itegratio by parts o the first itegral. While that is a perfectly acceptable way of doig the problem it s more work tha we really eed 007 Paul Dawkis 5

15 to do. Istead of splittig the itegral up let s istead use the followig choices for u ad dv. t u = t + 5 dv = cos dt 4 t du = dt v = 4si 4 The itegral is the, t t t ( t + 5) cos dt = 4( t + 5) si si dt t t = 4( t+ 5) si + 48cos + c 4 4 Notice that we pulled ay costats out of the itegral whe we used the itegratio by parts formula. We will usually do this i order to simplify the itegral a little. Example 4 Evaluate the followig itegral. w si 0w dw Solutio For this example we ll use the followig choices for u ad dv. u = w dv = si 0w dw The itegral is the, du = wdw v = cos 0w 0 w = + w si 0w dw cos 0w wcos 0w dw 0 5 I this example, ulike the previous examples, the ew itegral will also require itegratio by parts. For this secod itegral we will use the followig choices. u = w dv = cos( 0w) dw du = dw v = si ( 0 w) 0 So, the itegral becomes, w w w si ( 0w) dw = cos( 0w) + si ( 0w) si ( 0 ) w dw w w = cos( 0w) + si ( 0w) + cos( 0w) + c w w = cos( 0w) + si ( 0w) + cos( 0w) + c Be careful with the coefficiet o the itegral for the secod applicatio of itegratio by parts. Sice the itegral is multiplied by 5 we eed to make sure that the results of actually doig the itegral are also multiplied by 5. Forgettig to do this is oe of the more commo mistakes with itegratio by parts problems. 007 Paul Dawkis 6

16 As this last example has show us, we will sometimes eed more tha oe applicatio of itegratio by parts to completely evaluate a itegral. This is somethig that will happe so do t get excited about it whe it does. I this ext example we eed to ackowledge a importat poit about itegratio techiques. Some itegrals ca be doe i usig several differet techiques. That is the case with the itegral i the ext example. Example 5 Evaluate the followig itegral x x + dx (a) Usig Itegratio by Parts. [Solutio] (b) Usig a stadard Calculus I substitutio. [Solutio] Solutio (a) Evaluate usig Itegratio by Parts. First otice that there are o trig fuctios or expoetials i this itegral. While a good may itegratio by parts itegrals will ivolve trig fuctios ad/or expoetials ot all of them will so do t get too locked ito the idea of expectig them to show up. I this case we ll use the followig choices for u ad dv. u = x dv = x + dx The itegral is the, du = dx v = x + x x + dx = x( x + ) x + dx 5 4 = x( x+ ) ( x+ ) + c 5 (b) Evaluate Usig a stadard Calculus I substitutio. [Retur to Problems] Now let s do the itegral with a substitutio. We ca use the followig substitutio. u = x + x = u du = dx Notice that we ll actually use the substitutio twice, oce for the quatity uder the square root ad oce for the x i frot of the square root. The itegral is the, x x + dx = u u du = u u du 5 = u u + c 5 5 = ( x+ ) ( x+ ) + c 5 [Retur to Problems] 007 Paul Dawkis 7

17 So, we used two differet itegratio techiques i this example ad we got two differet aswers. The obvious questio the should be : Did we do somethig wrog? Actually, we did t do aythig wrog. We eed to remember the followig fact from Calculus I. If the f x = g x f x = g x + c I other words, if two fuctios have the same derivative the they will differ by o more tha a costat. So, how does this apply to the above problem? First defie the followig, f x = g x = x x+ The we ca compute f ( x) ad g( x ) by itegratig as follows, = = f x f x dx g x g x dx We ll use itegratio by parts for the first itegral ad the substitutio for the secod itegral. The accordig to the fact f ( x ) ad g( x ) should differ by o more tha a costat. Let s verify this ad see if this is the case. We ca verify that they differ by o more tha a costat if we take a look at the differece of the two ad do a little algebraic maipulatio ad simplificatio x x+ x+ x+ x = ( x+ ) x ( x+ ) ( x+ ) ( x ) ( 0) = + = 0 So, i this case it turs out the two fuctios are exactly the same fuctio sice the differece is zero. Note that this wo t always happe. Sometimes the differece will yield a ozero costat. For a example of this check out the Costat of Itegratio sectio i my Calculus I otes. So just what have we leared? First, there will, o occasio, be more tha oe method for evaluatig a itegral. Secodly, we saw that differet methods will ofte lead to differet aswers. Last, eve though the aswers are differet it ca be show, sometimes with a lot of work, that they differ by o more tha a costat. Whe we are faced with a itegral the first thig that we ll eed to decide is if there is more tha oe way to do the itegral. If there is more tha oe way we ll the eed to determie which method we should use. The geeral rule of thumb that I use i my classes is that you should use the method that you fid easiest. This may ot be the method that others fid easiest, but that does t make it the wrog method. 007 Paul Dawkis 8

18 Oe of the more commo mistakes with itegratio by parts is for people to get too locked ito perceived patters. For istace, all of the previous examples used the basic patter of takig u to be the polyomial that sat i frot of aother fuctio ad the lettig dv be the other fuctio. This will ot always happe so we eed to be careful ad ot get locked ito ay patters that we thik we see. Let s take a look at some itegrals that do t fit ito the above patter. Example 6 Evaluate the followig itegral. l x dx Solutio So, ulike ay of the other itegral we ve doe to this poit there is oly a sigle fuctio i the itegral ad o polyomial sittig i frot of the logarithm. The first choice of may people here is to try ad fit this ito the patter from above ad make the followig choices for u ad dv. u = dv = l x dx This leads to a real problem however sice that meas v must be, v = l x dx I other words, we would eed to kow the aswer ahead of time i order to actually do the problem. So, this choice simply wo t work. Also otice that with this choice we d get that du = 0 which also causes problems ad is aother reaso why this choice will ot work. Therefore, if the logarithm does t belog i the dv it must belog istead i the u. So, let s use the followig choices istead u = l x dv = dx du = dx v = x x The itegral is the, l x dx = x l x x dx x = x l x dx = xl x x+ c Example 7 Evaluate the followig itegral. 5 x x + dx Solutio So, if we agai try to use the patter from the first few examples for this itegral our choices for u ad dv would probably be the followig. 5 u = x dv = x + dx However, as with the previous example this wo t work sice we ca t easily compute v. v = x + dx 007 Paul Dawkis 9

19 This is ot a easy itegral to do. However, otice that if we had a x i the itegral alog with the root we could very easily do the itegral with a substitutio. Also otice that we do have a lot of x s floatig aroud i the origial itegral. So istead of puttig all the x s (outside of the root) i the u let s split them up as follows. u = x dv = x x + dx du = x dx v = ( x + ) 9 We ca ow easily compute v ad after usig itegratio by parts we get, 5 x x + dx = x ( x + ) x ( x + ) dx = x ( x + ) ( x + ) + c 9 45 So, i the previous two examples we saw cases that did t quite fit ito ay perceived patter that we might have gotte from the first couple of examples. This is always somethig that we eed to be o the lookout for with itegratio by parts. Let s take a look at aother example that also illustrates aother itegratio techique that sometimes arises out of itegratio by parts problems. Example 8 Evaluate the followig itegral. θ e cosθ dθ Solutio Okay, to this poit we ve always picked u i such a way that upo differetiatig it would make that portio go away or at the very least put it the itegral ito a form that would make it easier to deal with. I this case o matter which part we make u it will ever go away i the differetiatio process. It does t much matter which we choose to be u so we ll choose i the followig way. Note however that we could choose the other way as well ad we ll get the same result i the ed. θ u = cosθ dv = e dθ θ du = siθ dθ v = e The itegral is the, θ θ θ e cosθ dθ = e cosθ + e siθ dθ So, it looks like we ll do itegratio by parts agai. Here are our choices this time. θ u = siθ dv = e dθ du = cosθ dθ v = e θ The itegral is ow, θ θ θ θ e cosθ dθ = e cosθ + e siθ e cosθ dθ 007 Paul Dawkis 0

20 Now, at this poit it looks like we re just ruig i circles. However, otice that we ow have the same itegral o both sides ad o the right side it s got a mius sig i frot of it. This meas that we ca add the itegral to both sides to get, e θ cosθ dθ = e θ cosθ + e θ siθ All we eed to do ow is divide by ad we re doe. The itegral is, θ θ θ e cosθ dθ = ( cosθ + siθ) + c e e Notice that after dividig by the two we add i the costat of itegratio at that poit. This idea of itegratig util you get the same itegral o both sides of the equal sig ad the simply solvig for the itegral is kid of ice to remember. It does t show up all that ofte, but whe it does it may be the oly way to actually do the itegral. We ve got oe more example to do. As we will see some problems could require us to do itegratio by parts umerous times ad there is a short had method that will allow us to do multiple applicatios of itegratio by parts quickly ad easily. Example 9 Evaluate the followig itegral. x 4 x e dx Solutio We start off by choosig u ad dv as we always would. However, istead of computig du ad v we put these ito the followig table. We the differetiate dow the colum correspodig to u util we hit zero. I the colum correspodig to dv we itegrate oce for each etry i the first colum. There is also a third colum which we will explai i a bit ad it always starts with a + ad the alterates sigs as show. Now, multiply alog the diagoals show i the table. I frot of each product put the sig i the third colum that correspods to the u term for that product. I this case this would give, 007 Paul Dawkis

21 x x x x x x 4 4 x e dx= ( x ) e ( 4x ) 4e + ( x ) 8e ( 4x) 6e + ( 4) e x x x x x 4 = x e 6x e + 96x e 84xe + 768e + c We ve got the itegral. This is much easier tha writig dow all the various u s ad dv s that we d have to do otherwise. So, i this sectio we ve see how to do itegratio by parts. I your later math classes this is liable to be oe of the more frequet itegratio techiques that you ll ecouter. It is importat to ot get too locked ito patters that you may thik you ve see. I most cases ay patter that you thik you ve see ca (ad will be) violated at some poit i time. Be careful! Also, do t forget the shorthad method for multiple applicatios of itegratio by parts problems. It ca save you a fair amout of work o occasio. 007 Paul Dawkis

22 Itegrals Ivolvig Trig Fuctios I this sectio we are goig to look at quite a few itegrals ivolvig trig fuctios ad some of the techiques we ca use to help us evaluate them. Let s start off with a itegral that we should already be able to do. 5 5 cos si usig the substitutio si x x dx = u du u = x si 6 6 = x+ c This itegral is easy to do with a substitutio because the presece of the cosie, however, what about the followig itegral. Example Evaluate the followig itegral. 5 si x dx Solutio This itegral o loger has the cosie i it that would allow us to use the substitutio that we used above. Therefore, that substitutio wo t work ad we are goig to have to fid aother way of doig this itegral. Let s first otice that we could write the itegral as follows, 5 4 si x dx = si xsi x dx = si x si x dx Now recall the trig idetity, cos x+ si x= si x= cos x With this idetity the itegral ca be writte as, 5 si x dx = cos x si x dx ad we ca ow use the substitutio u = cos x. Doig this gives us, 5 si x dx = u du u 4 = + u du 5 = u u + u + c 5 = cos + cos cos x x x c So, with a little rewritig o the itegrad we were able to reduce this to a fairly simple substitutio. Notice that we were able to do the rewrite that we did i the previous example because the expoet o the sie was odd. I these cases all that we eed to do is strip out oe of the sies. 007 Paul Dawkis

23 The expoet o the remaiig sies will the be eve ad we ca easily covert the remaiig sies to cosies usig the idetity, cos x+ si x= () If the expoet o the sies had bee eve this would have bee difficult to do. We could strip out a sie, but the remaiig sies would the have a odd expoet ad while we could covert them to cosies the resultig itegral would ofte be eve more difficult tha the origial itegral i most cases. Let s take a look at aother example. Example Evaluate the followig itegral. 6 si x cos x dx Solutio So, i this case we ve got both sies ad cosies i the problem ad i this case the expoet o the sie is eve while the expoet o the cosie is odd. So, we ca use a similar techique i this itegral. This time we ll strip out a cosie ad covert the rest to sies. 6 6 si x cos x dx = si x cos x cos x dx 6 ( ) = si x si x cos x dx u = si x = 6 u u du 6 8 = u u du si x si x c = + At this poit let s pause for a secod to summarize what we ve leared so far about itegratig powers of sie ad cosie. m si x cos x dx () I this itegral if the expoet o the sies () is odd we ca strip out oe sie, covert the rest to cosies usig () ad the use the substitutio u = cos x. Likewise, if the expoet o the cosies (m) is odd we ca strip out oe cosie ad covert the rest to sies ad the use the substitutio u = si x. Of course, if both expoets are odd the we ca use either method. However, i these cases it s usually easier to covert the term with the smaller expoet. The oe case we have t looked at is what happes if both of the expoets are eve? I this case the techique we used i the first couple of examples simply wo t work ad i fact there really is t ay oe set method for doig these itegrals. Each itegral is differet ad i some cases there will be more tha oe way to do the itegral. With that beig said most, if ot all, of itegrals ivolvig products of sies ad cosies i which both expoets are eve ca be doe usig oe or more of the followig formulas to rewrite the itegrad. 007 Paul Dawkis 4

24 cos x= + cos si x= cos si xcos x= si x ( ( x) ) ( ( x) ) The first two formulas are the stadard half agle formula from a trig class writte i a form that will be more coveiet for us to use. The last is the stadard double agle formula for sie, agai with a small rewrite. Let s take a look at a example. Example Evaluate the followig itegral. si x cos x dx Solutio As oted above there are ofte more tha oe way to do itegrals i which both of the expoets are eve. This itegral is a example of that. There are at least two solutio techiques for this problem. We will do both solutios startig with what is probably the harder of the two, but it s also the oe that may people see first. Solutio I this solutio we will use the two half agle formulas above ad just substitute them ito the itegral. si x cos x dx = ( cos( x) ) ( + cos( x) ) dx cos = ( x) dx 4 So, we still have a itegral that ca t be completely doe, however otice that we have maaged to reduce the itegral dow to just oe term causig problems (a cosie with a eve power) rather tha two terms causig problems. I fact to elimiate the remaiig problem term all that we eed to do is reuse the first half agle formula give above. si x cos x dx = ( + cos( 4x) ) dx 4 = cos ( 4 x) dx 4 = x si ( 4 x) + c 4 8 = x si ( 4 x) + c 8 So, this solutio required a total of three trig idetities to complete. 007 Paul Dawkis 5

25 Solutio I this solutio we will use the half agle formula to help simplify the itegral as follows. x x dx = x x dx si cos si cos = si ( x) si = x dx 4 Now, we use the double agle formula for sie to reduce to a itegral that we ca do. si x cos x dx = cos 4x dx 8 = x si ( 4 x) + c 8 This method required oly two trig idetities to complete. Notice that the differece betwee these two methods is more oe of messiess. The secod method is ot appreciably easier (other tha eedig oe less trig idetity) it is just ot as messy ad that will ofte traslate ito a easier process. I the previous example we saw two differet solutio methods that gave the same aswer. Note that this will ot always happe. I fact, more ofte tha ot we will get differet aswers. However, as we discussed i the Itegratio by Parts sectio, the two aswers will differ by o more tha a costat. I geeral whe we have products of sies ad cosies i which both expoets are eve we will eed to use a series of half agle ad/or double agle formulas to reduce the itegral ito a form that we ca itegrate. Also, the larger the expoets the more we ll eed to use these formulas ad hece the messier the problem. Sometimes i the process of reducig itegrals i which both expoets are eve we will ru across products of sie ad cosie i which the argumets are differet. These will require oe of the followig formulas to reduce the products to itegrals that we ca do. dx siαcos β = si si α β + α + β siαsi β = cos( α β) cos( α β) + cosαcos β = cos( α β) + cos( α + β) Let s take a look at a example of oe of these kids of itegrals. 007 Paul Dawkis 6

26 Example 4 Evaluate the followig itegral. cos 5x cos 4x dx Solutio This itegral requires the last formula listed above. cos( 5x) cos( 4x) dx = cos( ) cos( 9 ) x + x dx = si ( x) + si ( 9x) + c 9 Okay, at this poit we ve covered pretty much all the possible cases ivolvig products of sies ad cosies. It s ow time to look at itegrals that ivolve products of secats ad tagets. This time, let s do a little aalysis of the possibilities before we just jump ito examples. The geeral itegral will be, m sec x ta x dx () The first thig to otice is that we ca easily covert eve powers of secats to tagets ad eve powers of tagets to secats by usig a formula similar to (). I fact, the formula ca be derived from () so let s do that. + = si x cos x si x cos x + = cos x cos x cos x ta x+ = sec x (4) Now, we re goig to wat to deal with () similarly to how we dealt with (). We ll wat to evetually use oe of the followig substitutios. u = ta x du = sec x dx u = sec x du = sec x ta x dx So, if we use the substitutio u = ta x we will eed two secats left for the substitutio to work. This meas that if the expoet o the secat () is eve we ca strip two out ad the covert the remaiig secats to tagets usig (4). Next, if we wat to use the substitutio u = sec xwe will eed oe secat ad oe taget left over i order to use the substitutio. This meas that if the expoet o the taget (m) is odd ad we have at least oe secat i the itegrad we ca strip out oe of the tagets alog with oe of the secats of course. The taget will the have a eve expoet ad so we ca use (4) to covert the rest of the tagets to secats. Note that this method does require that we have at least oe secat i the itegral as well. If there are t ay secats the we ll eed to do somethig differet. If the expoet o the secat is eve ad the expoet o the taget is odd the we ca use either case. Agai, it will be easier to covert the term with the smallest expoet. 007 Paul Dawkis 7

27 Let s take a look at a couple of examples. Example 5 Evaluate the followig itegral. sec x ta 9 5 Solutio First ote that sice the expoet o the secat is t eve we ca t use the substitutio u = ta x. However, the expoet o the taget is odd ad we ve got a secat i the itegral ad so we will be able to use the substitutio u = sec x. This meas strippig out a sigle taget (alog with a secat) ad covertig the remaiig tagets to secats usig (4). x dx Here s the work for this itegral sec x ta x dx = sec x ta x ta xsec x dx 8 8 u u du 0 8 u u u du 9 = + + = sec x sec x ta xsec x dx u = sec x = = + sec x sec x sec x c 9 Example 6 Evaluate the followig itegral. 4 6 sec x ta x dx Solutio So, i this example the expoet o the taget is eve so the substitutio u = sec x wo t work. The expoet o the secat is eve ad so we ca use the substitutio u = ta x for this itegral. That meas that we eed to strip out two secats ad covert the rest to tagets. Here is the work for this itegral sec x ta x dx = sec x ta xsec x dx 6 ( u ) 8 6 u u du 6 ta ta sec ta = x + x x dx u = x = + = + u du ta x ta x c = + + Both of the previous examples fit very icely ito the patters discussed above ad so were ot all that difficult to work. However, there are a couple of exceptios to the patters above ad i these cases there is o sigle method that will work for every problem. Each itegral will be differet ad may require differet solutio methods i order to evaluate the itegral. Let s first take a look at a couple of itegrals that have odd expoets o the tagets, but o secats. I these cases we ca t use the substitutio u = sec xsice it requires there to be at least oe secat i the itegral. 007 Paul Dawkis 8

28 Example 7 Evaluate the followig itegral. ta x dx Solutio To do this itegral all we eed to do is recall the defiitio of taget i terms of sie ad cosie ad the this itegral is othig more tha a Calculus I substitutio. si x ta x dx = dx u = cos x cos x = du u = l cos x + c rl x= l x = l cos x + c l sec x + c Example 8 Evaluate the followig itegral. ta x dx Solutio The trick to this oe is do the followig maipulatio of the itegrad. ta x dx = ta x ta x dx = ta x sec x dx = x x dx x dx ta sec ta We ca ow use the substitutio u = ta x o the first itegral ad the results from the previous example o the secod itegral. r The itegral is the, x dx = x x + c ta ta l sec Note that all odd powers of taget (with the exceptio of the first power) ca be itegrated usig the same method we used i the previous example. For istace, = ( ) = 5 ta x dx ta x sec x dx ta xsec x dx ta x dx So, a quick substitutio ( u = ta x) will give us the first itegral ad the secod itegral will always be the previous odd power. Now let s take a look at a couple of examples i which the expoet o the secat is odd ad the expoet o the taget is eve. I these cases the substitutios used above wo t work. It should also be oted that both of the followig two itegrals are itegrals that we ll be seeig o occasio i later sectios of this chapter ad i later chapters. Because of this it would t be a bad idea to make a ote of these results so you ll have them ready whe you eed them later. 007 Paul Dawkis 9

29 Example 9 Evaluate the followig itegral. sec x dx Solutio This oe is t too bad oce you see what you ve got to do. By itself the itegral ca t be doe. However, if we maipulate the itegrad as follows we ca do it. sec x( sec x+ ta x) sec x dx = dx sec x+ ta x sec x+ ta xsec x = dx sec x+ ta x I this form we ca do the itegral usig the substitutio u = sec x+ ta x. Doig this gives, sec x dx = l sec x + ta x + c The idea used i the above example is a ice idea to keep i mid. Multiplyig the umerator ad deomiator of a term by the same term above ca, o occasio, put the itegral ito a form that ca be itegrated. Note that this method wo t always work ad eve whe it does it wo t always be clear what you eed to multiply the umerator ad deomiator by. However, whe it does work ad you ca figure out what term you eed it ca greatly simplify the itegral. Here s the ext example. Example 0 Evaluate the followig itegral. sec x dx Solutio This oe is differet from ay of the other itegrals that we ve doe i this sectio. The first step to doig this itegral is to perform itegratio by parts usig the followig choices for u ad dv. u = sec x dv = sec x dx du = sec x ta x dx v = ta x Note that usig itegratio by parts o this problem is ot a obvious choice, but it does work very icely here. After doig itegratio by parts we have, sec x dx = sec x ta x sec x ta x dx Now the ew itegral also has a odd expoet o the secat ad a eve expoet o the taget ad so the previous examples of products of secats ad tagets still wo t do us ay good. To do this itegral we ll first write the tagets i the itegral i terms of secats. Agai, this is ot ecessarily a obvious choice but it s what we eed to do i this case. sec x dx = sec x ta x sec x sec x dx sec ta sec sec = x x x dx + x dx Now, we ca use the results from the previous example to do the secod itegral ad otice that 007 Paul Dawkis 0

30 the first itegral is exactly the itegral we re beig asked to evaluate with a mius sig i frot. So, add it to both sides to get, sec x dx = sec x ta x + l sec x + ta x Fially divide by two ad we re doe. sec x dx = sec x ta x + l sec x + ta x + c Agai, ote that we ve agai used the idea of itegratig the right side util the origial itegral shows up ad the movig this to the left side ad dividig by its coefficiet to complete the evaluatio. We first saw this i the Itegratio by Parts sectio ad oted at the time that this was a ice techique to remember. Here is aother example of this techique. Now that we ve looked at products of secats ad tagets let s also ackowledge that because we ca relate cosecats ad cotagets by + cot x= csc x all of the work that we did for products of secats ad tagets will also work for products of cosecats ad cotagets. We ll leave it to you to verify that. There is oe fial topic to be discussed i this sectio before movig o. To this poit we ve looked oly at products of sies ad cosies ad products of secats ad tagets. However, the methods used to do these itegrals ca also be used o some quotiets ivolvig sies ad cosies ad quotiets ivolvig secats ad tagets (ad hece quotiets ivolvig cosecats ad cotagets). Let s take a quick look at a example of this. Example Evaluate the followig itegral. 7 si x 4 dx cos x Solutio If this were a product of sies ad cosies we would kow what to do. We would strip out a sie (sice the expoet o the sie is odd) ad covert the rest of the sies to cosies. The same idea will work i this case. We ll strip out a sie from the umerator ad covert the rest to cosies as follows, 7 6 si x si x dx = si x dx 4 4 cos x cos x = = ( si x) cos 4 ( cos x) cos x 4 x si x dx si x dx At this poit all we eed to do is use the substitutio u = cos xad we re doe. 007 Paul Dawkis

31 si cos 7 x dx = x ( u ) u 4 4 du = + 4 u u u du = + + u u c + u u = + + cos x cos x c cos x cos x So, uder the right circumstaces, we ca use the ideas developed to help us deal with products of trig fuctios to deal with quotiets of trig fuctios. The atural questio the, is just what are the right circumstaces? First otice that if the quotiet had bee reversed as i this itegral, 4 cos x 7 dx si x we would t have bee able to strip out a sie. 4 4 cos x cos x dx = dx 7 6 si x si x si x I this case the stripped out sie remais i the deomiator ad it wo t do us ay good for the substitutio u = cos xsice this substitutio requires a sie i the umerator of the quotiet. Also ote that, while we could covert the sies to cosies, the resultig itegral would still be a fairly difficult itegral. So, we ca use the methods we applied to products of trig fuctios to quotiets of trig fuctios provided the term that eeds parts stripped out i is the umerator of the quotiet. 007 Paul Dawkis

32 Trig Substitutios As we have doe i the last couple of sectios, let s start off with a couple of itegrals that we should already be able to do with a stadard substitutio. x x 5x 4 dx = 5x 4 + c dx = 5x 4 + c 75 5x 4 5 Both of these used the substitutio u = 5x 4 ad at this poit should be pretty easy for you to do. However, let s take a look at the followig itegral. Example Evaluate the followig itegral. 5x 4 dx x Solutio I this case the substitutio u = 5x 4 will ot work ad so we re goig to have to do somethig differet for this itegral. It would be ice if we could get rid of the square root somehow. The followig substitutio will do that for us. x = sec θ 5 Do ot worry about where this came from at this poit. As we work the problem you will see that it works ad that if we have a similar type of square root i the problem we ca always use a similar substitutio. Before we actually do the substitutio however let s verify the claim that this will allow us to get rid of the square root. 4 5x 4 = 5 sec θ 4 = 4( sec θ ) = sec θ 5 To get rid of the square root all we eed to do is recall the relatioship, ta θ + = sec θ sec θ = ta θ Usig this fact the square root becomes, 5x 4 = ta θ = taθ Note the presece of the absolute value bars there. These are importat. Recall that x = x There should always be absolute value bars at this stage. If we kew that taθ was always positive or always egative we could elimiate the absolute value bars usig, x if x 0 x = x if x < Paul Dawkis

33 Without limits we wo t be able to determie if taθ is positive or egative, however, we will eed to elimiate them i order to do the itegral. Therefore, sice we are doig a idefiite itegral we will assume that taθ will be positive ad so we ca drop the absolute value bars. This gives, = 5x 4 ta So, we were able to elimiate the square root usig this substitutio. Let s ow do the substitutio ad see what we get. I doig the substitutio do t forget that we ll also eed to substitute for the dx. This is easy eough to get from the substitutio. x = secθ dx = secθ taθ dθ 5 5 Usig this substitutio the itegral becomes, 5x 4 taθ dx = sec θ ta θ dθ x 5 secθ 5 = ta θ dθ With this substitutio we were able to reduce the give itegral to a itegral ivolvig trig fuctios ad we saw how to do these problems i the previous sectio. Let s fiish the itegral. 5x 4 dx = sec θ dθ x = taθ θ + c θ So, we ve got a aswer for the itegral. Ufortuately the aswer is t give i x s as it should be. So, we eed to write our aswer i terms of x. We ca do this with some right triagle trig. From our origial substitutio we have, 5x hypoteuse secθ = = adjacet This gives the followig right triagle. From this we ca see that, taθ = 5x 4 We ca deal with the θ i oe of ay variety of ways. From our substitutio we ca see that, 007 Paul Dawkis 4

34 5x θ = sec While this is a perfectly acceptable method of dealig with the θ we ca use ay of the possible six iverse trig fuctios ad sice sie ad cosie are the two trig fuctios most people are familiar with we will usually use the iverse sie or iverse cosie. I this case we ll use the iverse cosie. θ = cos 5x So, with all of this the itegral becomes, 5x 4 5x 4 dx = cos + c x 5x = 5x 4 cos + c 5x We ow have the aswer back i terms of x. Wow! That was a lot of work. Most of these wo t take as log to work however. This first oe eeded lot s of explaatio sice it was the first oe. The remaiig examples wo t eed quite as much explaatio ad so wo t take as log to work. However, before we move oto more problems let s first address the issue of defiite itegrals ad how the process differs i these cases. Example Evaluate the followig itegral x x Solutio The limits here wo t chage the substitutio so that will remai the same. x = sec θ 5 Usig this substitutio the square root still reduces dow to, 5x 4 = taθ However, ulike the previous example we ca t just drop the absolute value bars. I this case we ve got limits o the itegral ad so we ca use the limits as well as the substitutio to determie the rage of θ that we re i. Oce we ve got that we ca determie how to drop the absolute value bars. Here s the limits of θ. dx 007 Paul Dawkis 5

35 x = = sec θ θ = π x = = sec θ θ = So, if we are i the rage 4 π 5 x 5 the θ is i the rage of 0 θ ad i this rage of θ s taget is positive ad so we ca just drop the absolute value bars. Let s do the substitutio. Note that the work is idetical to the previous example ad so most of it is left out. We ll pick up at the fial itegral ad the do the substitutio x x 5 4 π sec θ 0 dx = ( θ θ) = ta π = Note that because of the limits we did t eed to resort to a right triagle to complete the problem. Let s take a look at a differet set of limits for this itegral. Example Evaluate the followig itegral x x 5 4 Solutio Agai, the substitutio ad square root are the same as the first two examples. sec 5 x= θ x 4 = ta θ 5 Let s ext see the limits θ for this problem. x = = sec θ θ = π π x = = secθ θ = Note that i determiig the value of θ we used the smallest positive value. Now for this rage of x s we have π θ π ad i this rage of θ taget is egative ad so i this case we ca drop the absolute value bars, but will eed to add i a mius sig upo doig so. I other words, 5x 4 = taθ So, the oly chage this will make i the itegratio process is to put a mius sig i frot of the itegral. The itegral is the, dx π 0 dθ 007 Paul Dawkis 6

36 5 4 5 x x 5 4 π π sec θ dx = ( θ θ) = ta π = π dθ π I the last two examples we saw that we have to be very careful with defiite itegrals. We eed to make sure that we determie the limits o θ ad whether or ot this will mea that we ca drop the absolute value bars or if we eed to add i a mius sig whe we drop them. Before movig o to the ext example let s get the geeral form for the substitutio that we used i the previous set of examples. a = secθ b bx a x Let s work a ew ad differet type of example. Example 4 Evaluate the followig itegral. dx 4 x 9 x Solutio Now, the square root i this problem looks to be (almost) the same as the previous oes so let s try the same type of substitutio ad see if it will work here as well. x = secθ Usig this substitutio the square root becomes, 9 x = 9 9sec θ = sec θ = ta θ So usig this substitutio we will ed up with a egative quatity (the taget squared is always positive of course) uder the square root ad this will be trouble. Usig this substitutio will give complex values ad we do t wat that. So, usig secat for the substitutio wo t work. However, the followig substitutio (ad differetial) will work. x = siθ dx = cosθ dθ With this substitutio the square root is, 9 x = si θ = cos θ = cosθ = cosθ We were able to drop the absolute value bars because we are doig a idefiite itegral ad so we ll assume that everythig is positive. The itegral is ow, 007 Paul Dawkis 7

37 x dx = 4 9 x 8si θ = dθ 4 8 si θ csc 4 = θ dθ 8 4 ( cosθ) cosθ dθ I the previous sectio we saw how to deal with itegrals i which the expoet o the secat was eve ad sice cosecats behave a awful lot like secats we should be able to do somethig similar with this. Here is the itegral. x dx = csc csc 8 θ θ dθ 9 x = ( cot ) csc cot 8 θ + θ dθ u = θ = u du 8 + cot = θ + cot θ + c 8 4 Now we eed to go back to x s usig a right triagle. Here is the right triagle for this problem ad trig fuctios for this problem. x 9 x siθ = cotθ = x The itegral is the, x 9 x 9 x dx = + + c 9 x 8 x x 4 ( x ) Here s the geeral form for this type of square root. 9 9 x = + c 4x 8x 007 Paul Dawkis 8

38 a = siθ b a bx x There is oe fial case that we eed to look at. The ext itegral will also cotai somethig that we eed to make sure we ca deal with. Example 5 Evaluate the followig itegral. 6 5 x 0 6 ( x + ) Solutio First, otice that there really is a square root i this problem eve though it is t explicitly writte out. To see the root let s rewrite thigs a little. ( 6x + ) = 6x + = 6x + This square root is ot i the form we saw i the previous examples. Here we will use the substitutio for this root. x = taθ dx = sec θ dθ 6 6 With this substitutio the deomiator becomes, dx 6x + = ta θ + = sec θ = secθ Now, because we have limits we ll eed to covert them to θ so we ca determie how to drop the absolute value bars. x = 0 0 = taθ θ = 0 6 π x = = ta θ θ = I this rage of θ secat is positive ad so we ca drop the absolute value bars. Here is the itegral, π x 7776 ta θ dx = sec θ 0 sec θ 6 0 ( 6x + ) π 5 4 ta θ = d θ secθ 0 dθ There are several ways to proceed from this poit. Normally with a odd expoet o the taget we would strip oe of them out ad covert to secats. However, that would require that we also 007 Paul Dawkis 9

39 have a secat i the umerator which we do t have. Therefore, it seems like the best way to do this oe would be to covert the itegrad to sies ad cosies. 6 5 π x 4 0 ( 6x + ) si θ dx = d θ cos θ ( cos θ ) 007 Paul Dawkis 0 π 4 = cos θ 0 siθdθ We ca ow use the substitutio u = cosθ ad we might as well covert the limits as well. θ = 0 u = cos 0 = The itegral is the, π π θ = u = cos = x 4 dx = u u du ( 6x + ) The geeral form for this fial type of square root is = + + u u u = a + = taθ b a bx x We have a couple of fial examples to work i this sectio. Not all trig substitutios will just jump right out at us. Sometimes we eed to do a little work o the itegrad first to get it ito the correct form ad that is the poit of the remaiig examples. Example 6 Evaluate the followig itegral. x dx x 4x 7 Solutio I this case the quatity uder the root does t obviously fit ito ay of the cases we looked at above ad i fact is t i the ay of the forms we saw i the previous examples. Note however that if we complete the square o the quadratic we ca make it look somewhat like the above itegrals. Remember that completig the square requires a coefficiet of oe i frot of the x. Oce we have that we take half the coefficiet of the x, square it, ad the add ad subtract it to the

40 quatity. Here is the completig the square for this problem x x = x x+ = ( x ) = ( x ) 9 So, the root becomes, x x = x This looks like a secat substitutio except we do t just have a x that is squared. That is okay, it will work the same way. x = secθ x = + secθ dx = secθ taθ dθ Usig this substitutio the root reduces to, 4 7 = 9 = 9sec 9 = ta = ta = ta x x x θ θ θ θ Note we could drop the absolute value bars sice we are doig a idefiite itegral. Here is the itegral. x + secθ dx = sec ta θ θ dθ x 4x 7 taθ = secθ + sec θ dθ = l secθ + taθ + taθ + c Ad here is the right triagle for this problem. 4 7 sec x ta x θ θ x = = The itegral is the, x ( x ) x 4x 7 x 4x 7 dx = l c x 4x Paul Dawkis

41 Example 7 Evaluate the followig itegral. 4x e + e x dx Solutio This does t look to be aythig like the other problems i this sectio. However it is. To see this we first eed to otice that, x x e = e With this we ca use the followig substitutio. e x = taθ e x dx = sec θ d θ Remember that to compute the differetial all we do is differetiate both sides ad the tack o dx or dθ oto the appropriate side. With this substitutio the square root becomes, x x ta sec sec sec + e = + e = + θ = θ = θ = θ Agai, we ca drop the absolute value bars because we are doig a idefiite itegral. Here s the itegral. 4x x x x x e + e dx = e e + e dx x ( e ) x x e ( e ) dx ta ( sec )( sec ) ( ) 4 u u du = + = θ θ θ dθ = sec θ sec θsecθtaθ dθ u = secθ = sec θ sec θ 5 5 = + Here is the right triagle for this itegral. x x e + e taθ = secθ = = + e c x The itegral is the, 5 4x x x x e + e dx = ( + ) ( + ) + c 5 e e 007 Paul Dawkis

42 So, as we ve see i the fial two examples i this sectio some itegrals that look othig like the first few examples ca i fact be tured ito a trig substitutio problem with a little work. Before leavig this sectio let s summarize all three cases i oe place. a a bx x= siθ b a bx a x= secθ b a a + bx x= taθ b 007 Paul Dawkis

43 Partial Fractios I this sectio we are goig to take a look at itegrals of ratioal expressios of polyomials ad oce agai let s start this sectio out with a itegral that we ca already do so we ca cotrast it with the itegrals that we ll be doig i this sectio. x dx = usig 6 ad du u = x x du = ( x ) dx x x 6 u = l x x 6 + c So, if the umerator is the derivative of the deomiator (or a costat multiple of the derivative of the deomiator) doig this kid of itegral is fairly simple. However, ofte the umerator is t the derivative of the deomiator (or a costat multiple). For example, cosider the followig itegral. x + dx x x 6 I this case the umerator is defiitely ot the derivative of the deomiator or is it a costat multiple of the derivative of the deomiator. Therefore, the simple substitutio that we used above wo t work. However, if we otice that the itegrad ca be broke up as follows, x + 4 = x x 6 x x+ the the itegral is actually quite simple. x + 4 dx = dx x x 6 x x+ = 4l x l x+ + c This process of takig a ratioal expressio ad decomposig it ito simpler ratioal expressios that we ca add or subtract to get the origial ratioal expressio is called partial fractio decompositio. May itegrals ivolvig ratioal expressios ca be doe if we first do partial fractios o the itegrad. So, let s do a quick review of partial fractios. We ll start with a ratioal expressio i the form, P( x) f ( x) = Q( x) where both P(x) ad Q(x) are polyomials ad the degree of P(x) is smaller tha the degree of Q(x). Recall that the degree of a polyomial is the largest expoet i the polyomial. Partial fractios ca oly be doe if the degree of the umerator is strictly less tha the degree of the deomiator. That is importat to remember. So, oce we ve determied that partial fractios ca be doe we factor the deomiator as completely as possible. The for each factor i the deomiator we ca use the followig table to determie the term(s) we pick up i the partial fractio decompositio. 007 Paul Dawkis 4

44 Factor i deomiator ax + b ( ax + b) k Term i partial fractio decompositio A ax + b A A A ax + b ax + b ax + b k, k =,,, k Ax + B ax + bx + c ax + bx + c Ax k + B Ax+ B Ax k + Bk ax + bx + c + + +, k =,,, k ax + bx + c ax + bx + c ax + bx + c Notice that the first ad third cases are really special cases of the secod ad fourth cases respectively. There are several methods for determiig the coefficiets for each term ad we will go over each of those i the followig examples. Let s start the examples by doig the itegral above. Example Evaluate the followig itegral. x + dx x x 6 Solutio The first step is to factor the deomiator as much as possible ad get the form of the partial fractio decompositio. Doig this gives, x+ A B = + x x+ x x+ The ext step is to actually add the right side back up. x + A x+ + B x = x x+ x x+ Now, we eed to choose A ad B so that the umerators of these two are equal for every x. To do this we ll eed to set the umerators equal. x+ = A x+ + B x Note that i most problems we will go straight from the geeral form of the decompositio to this step ad ot bother with actually addig the terms back up. The oly poit to addig the terms is to get the umerator ad we ca get that without actually writig dow the results of the additio. At this poit we have oe of two ways to proceed. Oe way will always work, but is ofte more work. The other, while it wo t always work, is ofte quicker whe it does work. I this case both will work ad so we ll use the quicker way for this example. We ll take a look at the other method i a later example. 007 Paul Dawkis 5

45 What we re goig to do here is to otice that the umerators must be equal for ay x that we would choose to use. I particular the umerators must be equal for x = ad x =. So, let s plug these i ad see what we get. x= 5= A 0 + B 5 B= x= 0 = A 5 + B 0 A= 4 So, by carefully pickig the x s we got the ukow costats to quickly drop out. Note that these are the values we claimed they would be above. At this poit there really is t a whole lot to do other tha the itegral. x + 4 dx = dx x x 6 x x+ 4 = dx dx x x+ = 4l x l x+ + c Recall that to do this itegral we first split it up ito two itegrals ad the used the substitutios, u = x v= x+ o the itegrals to get the fial aswer. Before movig oto the ext example a couple of quick otes are i order here. First, may of the itegrals i partial fractios problems come dow to the type of itegral see above. Make sure that you ca do those itegrals. There is also aother itegral that ofte shows up i these kids of problems so we may as well give the formula for it here sice we are already o the subject. x = ta + c + dx x a a a It will be a example or two before we use this so do t forget about it. Now, let s work some more examples. Example Evaluate the followig itegral. x + 4 dx x + 4x 4x Solutio We wo t be puttig as much detail ito this solutio as we did i the previous example. The first thig is to factor the deomiator ad get the form of the partial fractio decompositio. x + 4 A B C = + + x x+ x x x+ x The ext step is to set umerators equal. If you eed to actually add the right side together to get 007 Paul Dawkis 6

46 the umerator for that side the you should do so, however, it will defiitely make the problem quicker if you ca do the additio i your head to get, x + 4= A x + x + Bx x + Cx x + As with the previous example it looks like we ca just pick a few values of x ad fid the costats so let s do that. x= 0 4= A( ) A= x= 8= B( )( 8) B= x= = C C = = 9 6 Note that ulike the first example most of the coefficiets here are fractios. That is ot uusual so do t get excited about it whe it happes. Now, let s do the itegral. 5 x + 4 dx = + + dx x + 4x 4x x x+ x 5 = l x + l x+ + l x + c 6 Agai, as oted above, itegrals that geerate atural logarithms are very commo i these problems so make sure you ca do them. Example Evaluate the followig itegral. x 9x+ 5 dx ( x 4) ( x + ) Solutio This time the deomiator is already factored so let s just jump right to the partial fractio decompositio. x 9x + 5 A B Cx + D = + + ( x 4) ( x + ) x 4 ( x 4) x + Settig umerators gives, x 9x + 5 = A x 4 x + + B x + + Cx + D x 4 I this case we are t goig to be able to just pick values of x that will give us all the costats. Therefore, we will eed to work this the secod (ad ofte loger) way. The first step is to multiply out the right side ad collect all the like terms together. Doig this gives, x 9x+ 5 = A+ C x + 4A+ B 8C+ D x + A+ 6C 8D x A+ B+ 6D Now we eed to choose A, B, C, ad D so that these two are equal. I other words we will eed to set the coefficiets of like powers of x equal. This will give a system of equatios that ca be solved. 007 Paul Dawkis 7

47 : + = 0 : = : = 9 : = 5 x A C x A B C D x A C D 0 x A B D A=, B= 5, C =, D= Note that we used x 0 to represet the costats. Also ote that these systems ca ofte be quite large ad have a fair amout of work ivolved i solvig them. The best way to deal with these is to use some form of computer aided solvig techiques. Now, let s take a look at the itegral dx 4 x + x x x dx = + ( x 4) ( x + ) x ( x 4) 5 x = + x x + x + ( x 4) 4 5 x = l 4 + l + + ta + x 4 dx x x c I order to take care of the third term we eeded to split it up ito two separate terms. Oce we ve doe this we ca do all the itegrals i the problem. The first two use the substitutio u = x 4, the third uses the substitutio v= x + ad the fourth term uses the formula give above for iverse tagets. Example 4 Evaluate the followig itegral. x + 0x + x+ 6 dx ( x )( x + 4) Solutio Let s first get the geeral form of the partial fractio decompositio. x + 0x + x + 6 A Bx + C Dx + E = x x + 4 x x + x + 4 Now, set umerators equal, expad the right side ad collect like terms. x x x A x Bx C x x Dx E x = ( 8 4 ) Settig coefficiet equal gives the followig system. 4 = A+ B x + C B x + A+ B C+ D x + 4B+ 4C D+ E x+ 6A 4C E 007 Paul Dawkis 8

48 : + = 0 : = : = 0 =, =, =, =, = 0 : = : 6 4 = 6 4 x A B x C B x A B C D A B C D E x B C D E 0 x A C E Do t get excited if some of the coefficiets ed up beig zero. It happes o occasio. Here s the itegral. x + 0x + x+ 6 x x dx = x x + 4 x x + x + 4 x x = + x x + x + x x = l l + 4 ta + x + 4 dx x x c To this poit we ve oly looked at ratioal expressios where the degree of the umerator was strictly less that the degree of the deomiator. Of course ot all ratioal expressios will fit ito this form ad so we eed to take a look at a couple of examples where this is t the case. dx Example 5 Evaluate the followig itegral. 4 x 5x + 6x 8 dx x x Solutio So, i this case the degree of the umerator is 4 ad the degree of the deomiator is. Therefore, partial fractios ca t be doe o this ratioal expressio. To fix this up we ll eed to do log divisio o this to get it ito a form that we ca deal with. Here is the work for that. x So, from the log divisio we see that, 4 x x x 5x + 6x 8 4 ( x x ) x + x 6 8 ( x 6x ) Paul Dawkis 9

49 4 x 5x + 6x 8 8 = x x x x x ad the itegral becomes, 4 x 5x + 6x 8 8 dx = x dx x x x x 8 = x dx dx x x The first itegral we ca do easily eough ad the secod itegral is ow i a form that allows us to do partial fractios. So, let s get the geeral form of the partial fractios for the secod itegrad. 8 = A + B + C x ( x ) x x x Settig umerators equal gives us, 8 = Ax x + B x + Cx Now, there is a variatio of the method we used i the first couple of examples that will work here. There are a couple of values of x that will allow us to quickly get two of the three costats, but there is o value of x that will just had us the third. What we ll do i this example is pick x s to get the two costats that we ca easily get ad the we ll just pick aother value of x that will be easy to work with (i.e. it wo t give large/messy umbers aywhere) ad the we ll use the fact that we also kow the other two costats to fid the third. x= 0 8 = B B= 6 x= 8 = C 9 C = x= 8 = A + B + C = A+ 4 A= The itegral is the, 4 x 5x + 6x 8 6 dx = x dx + dx x x x x x 6 = x x+ l x l x + c x I the previous example there were actually two differet ways of dealig with the x i the deomiator. Oe is to treat it as a quadratic which would give the followig term i the decompositio Ax + B x ad the other is to treat it as a liear term i the followig way, x = ( x 0) which gives the followig two terms i the decompositio, 007 Paul Dawkis 40

50 A B + x x We used the secod way of thikig about it i our example. Notice however that the two will give idetical partial fractio decompositios. So, why talk about this? Simple. This will work for x, but what about x or x 4? I these cases we really will eed to use the secod way of thikig about these kids of terms. x Let s take a look at oe more example. A B C A B C D + + x x x x x x x x 4 4 Example 6 Evaluate the followig itegral. x dx x Solutio I this case the umerator ad deomiator have the same degree. As with the last example we ll eed to do log divisio to get this ito the correct form. I ll leave the details of that to you to check. x dx = + dx = dx + dx x x x So, we ll eed to partial fractio the secod itegral. Here s the decompositio. A B = + ( x )( x+ ) x x+ Settig umerator equal gives, = A x+ + B x Pickig value of x gives us the followig coefficiets. x= = B( ) B= x= = A A= The itegral is the, x dx = dx + dx x x x+ = x+ l x l x+ + c 007 Paul Dawkis 4

51 Itegrals Ivolvig Roots I this sectio we re goig to look at a itegratio techique that ca be useful for some itegrals with roots i them. We ve already see some itegrals with roots i them. Some ca be doe quickly with a simple Calculus I substitutio ad some ca be doe with trig substitutios. However, ot all itegrals with roots will allow us to use oe of these methods. Let s look at a couple of examples to see aother techique that ca be used o occasio to help with these itegrals. Example Evaluate the followig itegral. x + dx x Solutio Sometimes whe faced with a itegral that cotais a root we ca use the followig substitutio to simplify the itegral ito a form that ca be easily worked with. u = x So, istead of lettig u be the stuff uder the radical as we ofte did i Calculus I we let u be the whole radical. Now, there will be a little more work here sice we will also eed to kow what x is so we ca substitute i for that i the umerator ad so we ca compute the differetial, dx. This is easy eough to get however. Just solve the substitutio for x as follows, x = u + dx = u du Usig this substitutio the itegral is ow, ( u + ) + 4 u du = u 5 u du u = u + u + c = ( x ) + ( x ) + c 5 So, sometimes, whe a itegral cotais the root g( x ) the substitutio, u = g( x) ca be used to simplify the itegral ito a form that we ca deal with. Let s take a look at aother example real quick. Example Evaluate the followig itegral. x x+ 0 dx Solutio We ll do the same thig we did i the previous example. Here s the substitutio ad the extra work we ll eed to do to get x i terms of u. 007 Paul Dawkis 4

52 u = x + 0 x = u 0 dx = u du With this substitutio the itegral is, 4u dx = ( u) du = du x x+ 0 u 0 u u u 0 This itegral ca ow be doe with partial fractios. 4 u = A + B u 5 u+ u 5 u+ Settig umerators equal gives, 4u= Au+ + Bu 5 Pickig value of u gives the coefficiets. 8 u = 8= B( 7) B= 7 0 u = 5 0 = A( 7) A= 7 The itegral is the, dx = + du x x+ 0 u 5 u+ 0 8 = l u 5 + l u+ + c = l x l x c 7 7 So, we ve see a ice method to elimiate roots from the itegral ad put it ito a form that we ca deal with. Note however, that this wo t always work ad sometimes the ew itegral will be just as difficult to do. 007 Paul Dawkis 4

53 Itegrals Ivolvig Quadratics To this poit we ve see quite a few itegrals that ivolve quadratics. A couple of examples are, x x dx = l x ± a + c dx ta = x ± a x + a a a We also saw that itegrals ivolvig with a trig substitutio. bx a, a bx ad a + bx could be doe Notice however that all of these itegrals were missig a x term. They all cosist of a quadratic term ad a costat. Some itegrals ivolvig geeral quadratics are easy eough to do. For istace, the followig itegral ca be doe with a quick substitutio. x + dx = ( 4 4 du u = x + x du = ( x + ) dx) 4x + x 4 u l 4 = x + x + c 4 Some itegrals with quadratics ca be doe with partial fractios. For istace, 0x 6 4 dx = 4l 5 l dx = x + x + + c x + 6x+ 5 x+ 5 x+ Ufortuately, these methods wo t work o a lot of itegrals. A simple substitutio will oly work if the umerator is a costat multiple of the derivative of the deomiator ad partial fractios will oly work if the deomiator ca be factored. This sectio is how to deal with itegrals ivolvig quadratics whe the techiques that we ve looked at to this poit simply wo t work. Back i the Trig Substitutio sectio we saw how to deal with square roots that had a geeral quadratic i them. Let s take a quick look at aother oe like that sice the idea ivolved i doig that kid of itegral is exactly what we are goig to eed for the other itegrals i this sectio. Example Evaluate the followig itegral. x + 4x + 5dx Solutio Recall from the Trig Substitutio sectio that i order to do a trig substitutio here we first eeded to complete the square o the quadratic. This gives, x + 4x+ 5= x + 4x = x+ + After completig the square the itegral becomes, x + 4x + 5dx = x + + dx 007 Paul Dawkis 44

54 Upo doig this we ca idetify the trig substitutio that we eed. Here it is, x + = taθ x = taθ dx = sec θ dθ x + + = ta θ + = sec θ = secθ = secθ Recall that sice we are doig a idefiite itegral we ca drop the absolute value bars. Usig this substitutio the itegral becomes, x + 4x + 5 dx = sec θ dθ ( sec θ ta θ l sec θ ta θ ) = c We ca fiish the itegral out with the followig right triagle. 4 5 ta x + sec x + x + θ = θ = = x + 4x+ 5 ( ) x + 4x+ 5dx= x+ x + 4x+ 5+ l x+ + x + 4x+ 5 + c So, by completig the square we were able to take a itegral that had a geeral quadratic i it ad covert it ito a form that allowed us to use a kow itegratio techique. Let s do a quick review of completig the square before proceedig. Here is the geeral completig the square formula that we ll use. b b b b x + bx + c = x + bx + + c = x + + c 4 This will always take a geeral quadratic ad write it i terms of a squared term ad a costat term. Recall as well that i order to do this we must have a coefficiet of oe i frot of the x. If ot we ll eed to factor out the coefficiet before completig the square. I other words, b c ax + bx + c = a x + x + a a complete the square o this! 007 Paul Dawkis 45

55 Now, let s see how completig the square ca be used to do itegrals that we are t able to do at this poit. Example Evaluate the followig itegral. x x+ dx Solutio Okay, this does t factor so partial fractios just wo t work o this. Likewise, sice the umerator is just we ca t use the substitutio u = x x+ 8. So, let s see what happes if we complete the square o the deomiator. x x+ = x x+ 9 9 = x x x With this the itegral is, dx = x x+ x dx Now this may ot seem like all that great of a chage. However, otice that we ca ow use the followig substitutio. u = x du = dx 4 ad the itegral is ow, dx = du 7 x x+ u + We ca ow see that this is a iverse taget! So, usig the formula from above we get, 4 4u dx = ta + c x x x = ta + c 7 7 Example Evaluate the followig itegral. x dx x + 0x+ 8 Solutio This example is a little differet from the previous oe. I this case we do have a x i the umerator however the umerator still is t a multiple of the derivative of the deomiator ad so a simple Calculus I substitutio wo t work Paul Dawkis 46

56 So, let s agai complete the square o the deomiator ad see what we get, x x x x x = = Upo completig the square the itegral becomes, x x dx = dx x + 0x+ 8 x At this poit we ca use the same type of substitutio that we did i the previous example. The oly real differece is that we ll eed to make sure that we plug the substitutio back ito the umerator as well. u = x + 5 x = u 5 dx = du ( u ) x 5 dx = du x + 0x+ 8 u + u 6 = du u + u + 6 u = l u + ta c + 6 x + 5 = l ( x+ 5) + ta + c So, i geeral whe dealig with a itegral i the form, + B Ax dx ax + bx + c () Here we are goig to assume that the deomiator does t factor ad the umerator is t a costat multiple of the derivative of the deomiator. I these cases we complete the square o the deomiator ad the do a substitutio that will yield a iverse taget ad/or a logarithm depedig o the exact form of the umerator. Let s ow take a look at a couple of itegrals that are i the same geeral form as () except the deomiator will also be raised to a power. I other words, let s look at itegrals i the form, Ax + B dx () ax + bx + c Example 4 Evaluate the followig itegral. x dx ( x 6x+ ) Solutio For the most part this itegral will work the same as the previous two with oe exceptio that will 007 Paul Dawkis 47

57 occur dow the road. So, let s start by completig the square o the quadratic i the deomiator. x x x x x 6 + = = + The itegral is the, x x dx = ( x x ) ( x ) dx Now, we will use the same substitutio that we ve used to this poit i the previous two examples. u = x x = u + dx = du x dx = u+ ( x 6x+ ) ( u + ) du u = du + ( u + ) ( u + ) Now, here is where the differeces start croppig up. The first itegral ca be doe with the substitutio v= u + ad is t too difficult. The secod itegral however, ca t be doe with the substitutio used o the first itegral ad it is t a iverse taget. It turs out that a trig substitutio will work icely o the secod itegral ad it will be the same as we did whe we had square roots i the problem. u = taθ du = sec θ dθ With these two substitutios the itegrals become, x dx = sec dv + θ dθ x 6x+ v ta θ + sec θ = + ( θ + ) 4 v 8 ta ( u + ) ( sec θ ) (( x ) + ) (( x ) + ) 4 8 dθ sec θ = dθ du = + dθ sec θ cos 4 = + θ dθ 007 Paul Dawkis 48

58 Okay, at this poit we ve got two optios for the remaiig itegral. We ca either use the ideas we leared i the sectio about itegrals ivolvig trig itegrals or we could use the followig formula. cos m θ m d θ si cos m cos m d m θ θ = + m θ θ Let s use this formula to do the itegral. 4 cos θ dθ = siθcos θ + cos 4 4 θ dθ 0 0 = siθcos θ + siθcosθ + cos d cos! 4 4 θ θ θ = = siθcos θ + siθcosθ + θ Next, let s use the followig right triagle to get this back to x s. u x x taθ = = siθ = cosθ = x + x + The cosie itegral is the, 4 x x x cos θ dθ = + + ta All told the the origial itegral is, (( x ) + ) ( x ) x x x ta = + + (( x ) + ) ( x ) 007 Paul Dawkis 49

59 x dx = ( x x ) 4 ( x ) x x x ta (( x ) 8 ) ( x ) x 9 x 9 x = + + ta + (( x ) + ) ( x ) It s a log ad messy aswer, but there it is. Example 5 Evaluate the followig itegral. x 4 x x Solutio As with the other problems we ll first complete the square o the deomiator. dx 4 x x = x + x 4 = x + x+ 4 = x+ 5 = 5 x+ The itegral is, x x dx = dx ( 4 x x ) 5 ( x + ) Now, let s do the substitutio. u = x + x = u dx = du ad the itegral is ow, x u 4 dx = du 4 x x 5 u u = du 4 ( 5 u ) ( 5 u ) I the first itegral we ll use the substitutio v= 5 u ad i the secod itegral we ll use the followig trig substitutio u = 5 siθ du = 5 cosθ dθ Usig these substitutios the itegral becomes, du c 007 Paul Dawkis 50

60 dx dv dθ x 4 = ( 5 cos ) θ 4 x x v 5 5si θ 4 5 cosθ = v 5 si ( θ ) 4 5 cosθ = dθ 4 v 5 cos θ 4 5 sec = d v 5 θ θ 5 = ( sec θ ta θ + l sec θ + ta θ ) + c v 5 We ll eed the followig right triagle to fiish this itegral out. u x+ 5 x+ siθ = = secθ = taθ = dθ ( x ) ( x ) So, goig back to x s the itegral becomes, x 5 5( x ) 5 x dx + + l = c ( 4 x x ) 5 u 5 5 ( x ) + 5 ( x+ ) 5 ( x+ ) 4x 5 x+ + 5 = l + c ( x ) ( x ) Ofte the followig formula is eeded whe usig the trig substitutio that we used i the previous example. m sec θ m m m d θ ta sec sec d m θ θ = + m θ θ Note that we ll oly eed the two trig substitutios that we used here. The third trig substitutio that we used will ot be eeded here. 007 Paul Dawkis 5

61 Itegratio Strategy We ve ow see a fair umber of differet itegratio techiques ad so we should probably pause at this poit ad talk a little bit about a strategy to use for determiig the correct techique to use whe faced with a itegral. There are a couple of poits that eed to be made about this strategy. First, it is t a hard ad fast set of rules for determiig the method that should be used. It is really othig more tha a geeral set of guidelies that will help us to idetify techiques that may work. Some itegrals ca be doe i more tha oe way ad so depedig o the path you take through the strategy you may ed up with a differet techique tha somebody else who also wet through this strategy. Secod, while the strategy is preseted as a way to idetify the techique that could be used o a itegral also keep i mid that, for may itegrals, it ca also automatically exclude certai techiques as well. Whe goig through the strategy keep two lists i mid. The first list is itegratio techiques that simply wo t work ad the secod list is techiques that look like they might work. After goig through the strategy ad the secod list has oly oe etry the that is the techique to use. If, o the other had, there is more tha oe possible techique to use we will the have to decide o which is liable to be the best for us to use. Ufortuately there is o way to teach which techique is the best as that usually depeds upo the perso ad which techique they fid to be the easiest. Third, do t forget that may itegrals ca be evaluated i multiple ways ad so more tha oe techique may be used o it. This has already bee metioed i each of the previous poits, but is importat eough to warrat a separate metio. Sometimes oe techique will be sigificatly easier tha the others ad so do t just stop at the first techique that appears to work. Always idetify all possible techiques ad the go back ad determie which you feel will be the easiest for you to use. Next, it s etirely possible that you will eed to use more tha oe method to completely do a itegral. For istace a substitutio may lead to usig itegratio by parts or partial fractios itegral. Fially, i my class I will accept ay valid itegratio techique as a solutio. As already oted there is ofte more tha oe way to do a itegral ad just because I fid oe techique to be the easiest does t mea that you will as well. So, i my class, there is o oe right way of doig a itegral. You may use ay itegratio techique that I ve taught you i this class or you leared i Calculus I to evaluate itegrals i this class. I other words, always take the approach that you fid to be the easiest. Note that this fial poit is more geared towards my class ad it s completely possible that your istructor may ot agree with this ad so be careful i applyig this poit if you are t i my class. Okay, let s get o with the strategy. 007 Paul Dawkis 5

62 . Simplify the itegrad, if possible. This step is very importat i the itegratio process. May itegrals ca be take from impossible or very difficult to very easy with a little simplificatio or maipulatio. Do t forget basic trig ad algebraic idetities as these ca ofte be used to simplify the itegral. We used this idea whe we were lookig at itegrals ivolvig trig fuctios. For example cosider the followig itegral. cos x dx This itegral ca t be doe as is, however simply by recallig the idetity, x= ( + ( x) ) cos cos the itegral becomes very easy to do. Note that this example also shows that simplificatio does ot ecessarily mea that we ll write the itegrad i a simpler form. It oly meas that we ll write the itegrad ito a form that we ca deal with ad this is ofte loger ad/or messier tha the origial itegral.. See if a simple substitutio will work. Look to see if a simple substitutio ca be used istead of the ofte more complicated methods from Calculus II. For example cosider both of the followig itegrals. x dx x x dx x The first itegral ca be doe with partial fractios ad the secod could be doe with a trig substitutio. However, both could also be evaluated usig the substitutio u = x ad the work ivolved i the substitutio would be sigificatly less tha the work ivolved i either partial fractios or trig substitutio. So, always look for quick, simple substitutios before movig o to the more complicated Calculus II techiques.. Idetify the type of itegral. Note that ay itegral may fall ito more tha oe of these types. Because of this fact it s usually best to go all the way through the list ad idetify all possible types sice oe may be easier tha the other ad it s etirely possible that the easier type is listed lower i the list. a. Is the itegrad a ratioal expressio (i.e is the itegrad a polyomial divided by a polyomial)? If so, the partial fractios may work o the itegral. b. Is the itegrad a polyomial times a trig fuctio, expoetial, or logarithm? If so, the itegratio by parts may work. c. Is the itegrad a product of sies ad cosies, secat ad tagets, or cosecats ad cotagets? If so, the the topics from the secod sectio may work. Likewise, do t forget that some quotiets ivolvig these fuctios ca also be doe usig these techiques. 007 Paul Dawkis 5

63 d. Does the itegrad ivolve bx + a, bx a, or a bx? If so, the a trig substitutio might work icely. e. Does the itegrad have roots other tha those listed above i it? If so, the the substitutio u g( x) = might work. f. Does the itegrad have a quadratic i it? If so, the completig the square o the quadratic might put it ito a form that we ca deal with. 4. Ca we relate the itegral to a itegral we already kow how to do? I other words, ca we use a substitutio or maipulatio to write the itegrad ito a form that does fit ito the forms we ve looked at previously i this chapter. A typical example here is the followig itegral. cos x + si x dx This itegral does t obviously fit ito ay of the forms we looked at i this chapter. However, with the substitutio u = si x we ca reduce the itegral to the form, which is a trig substitutio problem. + u du 5. Do we eed to use multiple techiques? I this step we eed to ask ourselves if it is possible that we ll eed to use multiple techiques. The example i the previous part is a good example. Usig a substitutio did t allow us to actually do the itegral. All it did was put the itegral ad put it ito a form that we could use a differet techique o. Do t ever get locked ito the idea that a itegral will oly require oe step to completely evaluate it. May will require more tha oe step. 6. Try agai. If everythig that you ve tried to this poit does t work the go back through the process ad try agai. This time try a techique that that you did t use the first time aroud. As oted above this strategy is ot a hard ad fast set of rules. It is oly iteded to guide you through the process of best determiig how to do ay give itegral. Note as well that the oly place Calculus II actually arises is i the third step. Steps, ad 4 ivolve othig more tha maipulatio of the itegrad either through direct maipulatio of the itegrad or by usig a substitutio. The last two steps are simply ideas to thik about i goig through this strategy. May studets go through this process ad cocetrate almost exclusively o Step (after all this is Calculus II, so it s easy to see why they might do that.) to the exclusio of the other steps. Oe very large cosequece of that exclusio is that ofte a simple maipulatio or substitutio is overlooked that could make the itegral very easy to do. Before movig o to the ext sectio we should work a couple of quick problems illustratig a couple of ot so obvious simplificatios/maipulatios ad a ot so obvious substitutio. 007 Paul Dawkis 54

64 Example Evaluate the followig itegral. ta x 4 dx sec x Solutio This itegral almost falls ito the form give i c. It is a quotiet of taget ad secat ad we kow that sometimes we ca use the same methods for products of tagets ad secats o quotiets. The process from that sectio tells us that if we have eve powers of secat to strip two of them off ad covert the rest to tagets. That wo t work here. We ca split two secats off, but they would be i the deomiator ad they wo t do us ay good there. Remember that the poit of splittig them off is so they would be there for the substitutio u = ta x. That requires them to be i the umerator. So, that wo t work ad so we ll have to fid aother solutio method. There are i fact two solutio methods to this itegral depedig o how you wat to go about it. We ll take a look at both. Solutio I this solutio method we could just covert everythig to sies ad cosies ad see if that gives us a itegral we ca deal with. ta x si x 4 dx = 4 cos x dx sec x cos x = si x cos x dx u = cos x = u du cos 4 = x+ c 4 Note that just covertig to sies ad cosies wo t always work ad if it does it wo t always work this icely. Ofte there will be a lot more work that would eed to be doe to complete the itegral. Solutio This solutio method goes back to dealig with secats ad tagets. Let s otice that if we had a secat i the umerator we could just use u = sec x as a substitutio ad it would be a fairly quick ad simple substitutio to use. We do t have a secat i the umerator. However we could very easily get a secat i the umerator simply by multiplyig the umerator ad deomiator by secat. ta x ta xsec x dx = sec 4 dx u = x 5 sec x sec x = du 5 u = + c 4 4 sec x cos 4 = x+ c Paul Dawkis 55

65 I the previous example we saw two simplificatios that allowed us to do the itegral. The first was usig idetities to rewrite the itegral ito terms we could deal with ad the secod ivolved multiplyig the umerator ad the deomiator by somethig to agai put the itegral ito terms we could deal with. Usig idetities to rewrite a itegral is a importat simplificatio ad we should ot forget about it. Itegrals ca ofte be greatly simplified or at least put ito a form that ca be dealt with by usig a idetity. The secod simplificatio is ot used as ofte, but does show up o occasio so agai, it s best to ot forget about it. I fact, let s take aother look at a example i which multiplyig the umerator ad deomiator by somethig will allow us to do a itegral. Example Evaluate the followig itegral. + si x dx Solutio This is a itegral i which if we just cocetrate o the third step we wo t get aywhere. This itegral does t appear to be ay of the kids of itegrals that we worked i this chapter. We ca do the itegral however, if we do the followig, si x dx = dx + si x + si x si x si x = dx si x This does ot appear to have doe aythig for us. However, if we ow remember the first simplificatio we looked at above we will otice that we ca use a idetity to rewrite the deomiator. Oce we do that we ca further reduce the itegral ito somethig we ca deal with. si x dx = dx + si x cos x si x = dx cos x cos x cos x = sec x ta xsec x dx = ta x sec x+ c So, we ve see oce agai that multiplyig the umerator ad deomiator by somethig ca put the itegral ito a form that we ca itegrate. Notice as well that this example also showed that simplificatios do ot ecessarily put a itegral ito a simpler form. They oly put the itegral ito a form that is easier to itegrate. Let s ow take a quick look at a example of a substitutio that is ot so obvious. 007 Paul Dawkis 56

66 Example Evaluate the followig itegral. cos x dx Solutio We itroduced this example sayig that the substitutio was ot so obvious. However, this is really a itegral that falls ito the form give by e i our strategy above. However, may people miss that form ad so do t thik about it. So, let s try the followig substitutio. u = x x = u dx = u du With this substitutio the itegral becomes, cos x dx = u cosu du This is ow a itegratio by parts itegral. Remember that ofte we will eed to use more tha oe techique to completely do the itegral. This is a fairly simple itegratio by parts problem so I ll leave the remaider of the details to you to check. cos x dx = cos x + x si x + c Before leavig this sectio we should also poit out that there are itegrals out there i the world that just ca t be doe i terms of fuctios that we kow. Some examples of these are. x si ( x) x e dx cos( x ) dx dx cos dx x e That does t mea that these itegrals ca t be doe at some level. If you go to a computer algebra system such as Maple or Mathematica ad have it do these itegrals it will retur the followig. x π e dx = erf ( x) π cos( x ) dx = FreselC x π si x dx = Si x cos ( x ) x x ( e ) dx = Ci( e ) So it appears that these itegrals ca i fact be doe. However this is a little misleadig. Here are the defiitios of each of the fuctios give above. Error Fuctio The Sie Itegral erf Si = π e x ( x) 0 ( x) = 0 x t si t dt t dt 007 Paul Dawkis 57

67 The Fresel Cosie Itegral π FreselC ( x) = cos t dt 0 The Cosie Itegral x cost Ci( x) = γ + l ( x) + dt 0 t Where γ is the Euler-Mascheroi costat. Note that the first three are simply defied i terms of themselves ad so whe we say we ca itegrate them all we are really doig is reamig the itegral. The fourth oe is a little differet ad yet it is still defied i terms of a itegral that ca t be doe i practice. It will be possible to itegrate every itegral give i this class, but it is importat to ote that there are itegrals that just ca t be doe. We should also ote that after we look at Series we will be able to write dow series represetatios of each of the itegrals above. x 007 Paul Dawkis 58

68 Improper Itegrals I this sectio we eed to take a look at a couple of differet kids of itegrals. Both of these are examples of itegrals that are called Improper Itegrals. Let s start with the first kid of improper itegrals that we re goig to take a look at. Ifiite Iterval I this kid of itegral oe or both of the limits of itegratio are ifiity. I these cases the iterval of itegratio is said to be over a ifiite iterval. Let s take a look at a example that will also show us how we are goig to deal with these itegrals. Example Evaluate the followig itegral. dx x Solutio This is a iocet eough lookig itegral. However, because ifiity is ot a real umber we ca t just itegrate as ormal ad the plug i the ifiity to get a aswer. To see how we re goig to do this itegral let s thik of this as a area problem. So istead of askig what the itegral is, let s istead ask what the area uder f ( x) = o the iterval x [, ) is. We still are t able to do this, however, let s step back a little ad istead ask what the area uder, t where t > ad t is fiite. This is a problem that we ca do. f ( x ) is o the iterval [ ] At t = dx = = x x t Now, we ca get the area uder f ( x ) o [, ) simply by takig the limit of A t as t goes to ifiity. A= lim At = lim = t t t This is the how we will do the itegral itself. t dx = lim dx t x x = lim t x = lim = t t t t 007 Paul Dawkis 59

69 So, this is how we will deal with these kids of itegrals i geeral. We will replace the ifiity with a variable (usually t), do the itegral ad the take the limit of the result as t goes to ifiity. O a side ote, otice that the area uder a curve o a ifiite iterval was ot ifiity as we might have suspected it to be. I fact, it was a surprisigly small umber. Of course this wo t always be the case, but it is importat eough to poit out that ot all areas o a ifiite iterval will yield ifiite areas. Let s ow get some defiitios out of the way. We will call these itegrals coverget if the associated limit exists ad is a fiite umber (i.e. it s ot plus or mius ifiity) ad diverget if the associated limit either does t exist or is (plus or mius) ifiity. Let s ow formalize up the method for dealig with ifiite itervals. There are essetially three cases that we ll eed to look at. t. If x dx exists for every t > athe, a f a = lim t provided the limit exists ad is fiite. b. If f t f x dx f x dx t x dx exists for every t < b the, b b = lim provided the limits exists ad is fiite. c. If f ( x) dx ad f c f x dx f x dx t x dx are both coverget the, = c + f x dx f x dx f x dx Where c is ay umber. Note as well that this requires BOTH of the itegrals to be coverget i order for this itegral to also be coverget. If either of the two itegrals is diverget the so is this itegral. Let s take a look at a couple more examples. Example Determie if the followig itegral is coverget or diverget ad if it s coverget fid its value. dx x Solutio So, the first thig we do is covert the itegral to a limit. t dx = lim dx x t x Now, do the itegral ad the limit. a t c 007 Paul Dawkis 60

70 t dx = lim l ( x) t x = lim l t = So, the limit is ifiite ad so the itegral is diverget. ( ( t) l) If we go back to thikig i terms of area otice that the area uder g( x ) = x o the iterval [, ) is ifiite. This is i cotrast to the area uder f x = which was quite small. There really is t all that much differece betwee these two fuctios ad yet there is a large differece i the area uder them. We ca actually exted this out to the followig fact. x Fact If a > 0 the is coverget if p > ad diverget if p. a p x dx Oe thig to ote about this fact is that it s i essece sayig that if a itegrad goes to zero fast eough the the itegral will coverge. How fast is fast eough? If we use this fact as a guide it looks like itegrads that go to zero faster tha x goes to zero will probably coverge. Let s take a look at a couple more examples. Example Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. 0 dx x Solutio There really is t much to do with these problems oce you kow how to do them. We ll covert the itegral to a limit/itegral pair, evaluate the itegral ad the the limit. 0 0 dx = lim dx x t x = lim x t ( t ) = lim + t = + = So, the limit is ifiite ad so this itegral is diverget. t 0 t 007 Paul Dawkis 6

71 Example 4 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. x xe dx Solutio I this case we ve got ifiities i both limits ad so we ll eed to split the itegral up ito two separate itegrals. We ca split the itegral up at ay poit, so let s choose a = 0 sice this will be a coveiet poit for the evaluatio process. The itegral is the, 0 x x x = + 0 xe dx xe dx xe dx We ve ow got to look at each of the idividual limits. 0 0 x x e = lim t e t x dx x dx x = lim t e = lim + e t = So, the first itegral is coverget. Note that this does NOT mea that the secod itegral will also be coverget. So, let s take a look at that oe. t x x e = lim 0 t e 0 x dx x dx 0 t t x = lim t e 0 t = lim e + t = This itegral is coverget ad so sice they are both coverget the itegral we were actually asked to deal with is also coverget ad its value is, 0 x x x xe dx = x dx x dx 0 e + = + = e 0 Example 5 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. si x dx Solutio First covert to a limit. t 007 Paul Dawkis 6

72 si x dx = lim si x dx t = lim t ( cos x) ( cost ) = lim cos t This limit does t exist ad so the itegral is diverget. I most examples i a Calculus II class that are worked over ifiite itervals the limit either exists or is ifiite. However, there are limits that do t exist, as the previous example showed, so do t forget about those. Discotiuous Itegrad We ow eed to look at the secod type of improper itegrals that we ll be lookig at i this sectio. These are itegrals that have discotiuous itegrads. The process here is basically the same with oe subtle differece. Here are the geeral cases that we ll look at for these itegrals.. If f ( x ) is cotiuous o the iterval [, ) b t = lim a t t ab ad ot cotiuous at x f x dx f x dx t b a = b the, provided the limit exists ad is fiite. Note as well that we do eed to use a left had limit here sice the iterval of itegratio is etirely o the left side of the upper limit.. If f ( x ) is cotiuous o the iterval (, ] b b = lim a ab ad ot cotiuous at x f x dx f x dx + t a t = a the, provided the limit exists ad is fiite. I this case we eed to use a right had limit here sice the iterval of itegratio is etirely o the right side of the lower limit. c b. If f ( x ) is ot cotiuous at x= c where a< c< b ad ad f are both coverget the, = + b c b f x dx f x dx f x dx a a c a f x dx c x dx As with the ifiite iterval case this requires BOTH of the itegrals to be coverget i order for this itegral to also be coverget. If either of the two itegrals is diverget the so is this itegral. c b 4. If f ( x ) is ot cotiuous at x= aad x= bad if ad f both coverget the, = + b c b a f x dx f x dx f x dx f x dx a a c x dx are Where c is ay umber. Agai, this requires BOTH of the itegrals to be coverget i order for this itegral to also be coverget. c 007 Paul Dawkis 6

73 Note that the limits i these cases really do eed to be right or left haded limits. Sice we will be workig iside the iterval of itegratio we will eed to make sure that we stay iside that iterval. This meas that we ll use oe-sided limits to make sure we stay iside the iterval. Let s do a couple of examples of these kids of itegrals. Example 6 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. dx 0 x Solutio The problem poit is the upper limit so we are i the first case above. t dx = lim dx x t x 0 0 t ( x ) ( t ) = lim t 0 = lim = The limit exists ad is fiite ad so the itegral coverges ad the itegral s value is. t Example 7 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. dx x Solutio This itegrad is ot cotiuous at x = 0 ad so we ll eed to split the itegral up at that poit. 0 dx = dx + dx x x x 0 Now we eed to look at each of these itegrals ad see if they are coverget. 0 t dx = lim dx x t 0 x = lim t 0 x = lim + t 0 t 8 = At this poit we re doe. Oe of the itegrals is diverget that meas the itegral that we were asked to look at is diverget. We do t eve eed to bother with the secod itegral. t 007 Paul Dawkis 64

74 Before leavig this sectio let s ote that we ca also have itegrals that ivolve both of these cases. Cosider the followig itegral. Example 8 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. dx 0 x Solutio This is a itegral over a ifiite iterval that also cotais a discotiuous itegrad. To do this itegral we ll eed to split it up ito two itegrals. We ca split it up aywhere, but pick a value that will be coveiet for evaluatio purposes. dx = dx + dx x x x 0 0 I order for the itegral i the example to be coverget we will eed BOTH of these to be coverget. If oe or both are diverget the the whole itegral will also be diverget. We kow that the secod itegral is coverget by the fact give i the ifiite iterval portio above. So, all we eed to do is check the first itegral. dx = lim dx + x t 0 x 0 = lim+ t 0 x t = lim + + t 0 t = So, the first itegral is diverget ad so the whole itegral is diverget. t 007 Paul Dawkis 65

75 Compariso Test for Improper Itegrals Now that we ve see how to actually compute improper itegrals we eed to address oe more topic about them. Ofte we are t cocered with the actual value of these itegrals. Istead we might oly be iterested i whether the itegral is coverget or diverget. Also, there will be some itegrals that we simply wo t be able to itegrate ad yet we would still like to kow if they coverge or diverge. To deal with this we ve got a test for covergece or divergece that we ca use to help us aswer the questio of covergece for a improper itegral. We will give this test oly for a sub-case of the ifiite iterval itegral, however versios of the test exist for the other sub-cases of the ifiite iterval itegrals as well as itegrals with discotiuous itegrads. Compariso Test f x g x If 0 o the iterval [ a, ) the,. If f ( x) dx coverges the so does a g x dx. a. If g ( x ) dx diverges the so does f a x dx. Note that if you thik i terms of area the Compariso Test makes a lot of sese. If f ( x ) is larger tha g( x ) the the area uder f ( x ) must also be larger tha the area uder g( x ). So, if the area uder the larger fuctio is fiite (i.e. f the smaller fuctio must also be fiite (i.e. the smaller fuctio is ifiite (i.e. must also be ifiite (i.e. f a a x dx coverges) the the area uder a g x dx coverges). Likewise, if the area uder a g x dx diverges) the the area uder the larger fuctio a x dx diverges). Be careful ot to misuse this test. If the smaller fuctio coverges there is o reaso to believe that the larger will also coverge (after all ifiity is larger tha a fiite umber ) ad if the larger fuctio diverges there is o reaso to believe that the smaller fuctio will also diverge. Let s work a couple of examples usig the compariso test. Note that all we ll be able to do is determie the covergece of the itegral. We wo t be able to determie the value of the itegrals ad so wo t eve bother with that. 007 Paul Dawkis 66

76 Example Determie if the followig itegral is coverget or diverget. cos x dx x Solutio Let s take a secod ad thik about how the Compariso Test works. If this itegral is coverget the we ll eed to fid a larger fuctio that also coverges o the same iterval. Likewise, if this itegral is diverget the we ll eed to fid a smaller fuctio that also diverges. So, it seems like it would be ice to have some idea as to whether the itegral coverges or diverges ahead of time so we will kow whether we will eed to look for a larger (ad coverget) fuctio or a smaller (ad diverget) fuctio. To get the guess for this fuctio let s otice that the umerator is ice ad bouded ad simply wo t get too large. Therefore, it seems likely that the deomiator will determie the covergece/divergece of this itegral ad we kow that dx x coverges sice p = > by the fact i the previous sectio. So let s guess that this itegral will coverge. So we ow kow that we eed to fid a fuctio that is larger tha cos x x ad also coverges. Makig a fractio larger is actually a fairly simple process. We ca either make the umerator larger or we ca make the deomiator smaller. I this case we ca t do a lot about the deomiator. However we ca use the fact that 0 cos x to make the umerator larger (i.e. we ll replace the cosie with somethig we kow to be larger, amely ). So, cos x x x Now, as we ve already oted dx x coverges ad so by the Compariso Test we kow that cos x dx x must also coverge. Example Determie if the followig itegral is coverget or diverget. dx x x + e Solutio Let s first take a guess about the covergece of this itegral. As oted after the fact i the last sectio about 007 Paul Dawkis 67

77 a dx p x if the itegrad goes to zero faster tha x the the itegral will probably coverge. Now, we ve got a expoetial i the deomiator which is approachig ifiity much faster tha the x ad so it looks like this itegral should probably coverge. So, we eed a larger fuctio that will also coverge. I this case we ca t really make the umerator larger ad so we ll eed to make the deomiator smaller i order to make the fuctio larger as a whole. We will eed to be careful however. There are two ways to do this ad oly oe, i this case oly oe, of them will work for us. First, otice that sice the lower limit of itegratio is we ca say that x > 0 ad we kow that expoetials are always positive. So, the deomiator is the sum of two positive terms ad if we were to drop oe of them the deomiator would get smaller. This would i tur make the fuctio larger. The questio the is which oe to drop? Let s first drop the expoetial. Doig this gives, < x x+ e x This is a problem however, sice dx x diverges by the fact. We ve got a larger fuctio that is diverget. This does t say aythig about the smaller fuctio. Therefore, we chose the wrog oe to drop. Let s try it agai ad this time let s drop the x. Also, x < = e x x x + e e t x e dx = lim t x e dx t ( e e ) = lim + = e t So, x dx e is coverget. Therefore, by the Compariso test is also coverget. dx x x + e 007 Paul Dawkis 68

78 Example Determie if the followig itegral is coverget or diverget. dx x x e Solutio This is very similar to the previous example with a couple of very importat differeces. First, otice that the expoetial ow goes to zero as x icreases istead of growig larger as it did i the previous example (because of the egative i the expoet). Also ote that the expoetial is ow subtracted off the x istead of added oto it. The fact that the expoetial goes to zero meas that this time the x i the deomiator will probably domiate the term ad that meas that the itegral probably diverges. We will therefore eed to fid a smaller fuctio that also diverges. Makig fractios smaller is pretty much the same as makig fractios larger. I this case we ll eed to either make the umerator smaller or the deomiator larger. This is where the secod chage will come ito play. As before we kow that both x ad the expoetial are positive. However, this time sice we are subtractig the expoetial from the x if we were to drop the expoetial the deomiator will become larger ad so the fractio will become smaller. I other words, > x x e x ad we kow that dx x diverges ad so by the Compariso Test we kow that dx x x e must also diverge. Example 4 Determie if the followig itegral is coverget or diverget. 4 + si ( x) dx x Solutio First otice that as with the first example, the umerator i this fuctio is goig to be bouded sice the sie is ever larger tha. Therefore, sice the expoet o the deomiator is less tha we ca guess that the itegral will probably diverge. We will eed a smaller fuctio that also diverges. 0 si. I particular, this term is positive ad so if we drop it from the umerator the umerator will get smaller. This gives, 4 + si ( x) > x x ad 4 We kow that ( x) 007 Paul Dawkis 69

79 diverges so by the Compariso Test also diverges. dx x 4 si x dx + x Okay, we ve see a few examples of the Compariso Test ow. However, most of them worked pretty much the same way. All the fuctios were ratioal ad all we did for most of them was add or subtract somethig from the umerator or deomiator to get what we wat. Let s take a look at a example that works a little differetly so we do t get too locked ito these ideas. Example 5 Determie if the followig itegral is coverget or diverget. x e dx x Solutio Normally, the presece of just a x i the deomiator would lead us to guess diverget for this itegral. However, the expoetial i the umerator will approach zero so fast that istead we ll eed to guess that this itegral coverges. To get a larger fuctio we ll use the fact that we kow from the limits of itegratio that x >. This meas that if we just replace the x i the deomiator with (which is always smaller tha x) we will make the deomiator smaller ad so the fuctio will get larger. x x e e x < = e x ad we ca show that e x dx coverges. I fact, we ve already doe this for a lower limit of ad chagig that to a wo t chage the covergece of the itegral. Therefore, by the Compariso Test x e dx x also coverges. We should also really work a example that does t ivolve a ratioal fuctio sice there is o reaso to assume that we ll always be workig with ratioal fuctios. Example 6 Determie if the followig itegral is coverget or diverget. e x dx Solutio We kow that expoetials with egative expoets die dow to zero very fast so it makes sese to guess that this itegral will be coverget. We eed a larger fuctio, but this time we do t 007 Paul Dawkis 70

80 have a fractio to work with so we ll eed to do somethig differet. x We ll take advatage of the fact that e is a decreasig fuctio. This meas that x x x > x e < e I other words, plug i a larger umber ad the fuctio gets smaller. From the limits of itegratio we kow that x > ad this meas that if we square x it will get larger. Or, x > x provided x> Note that we ca oly say this sice x >. This wo t be true if x! We ca ow use the fact x that e is a decreasig fuctio to get, x x e < e x So, e is a larger fuctio tha e ad we kow that e x dx coverges so by the Compariso Test we also kow that e x dx is coverget. x The last two examples made use of the fact that x >. Let s take a look at a example to see how do we would have to go about these if the lower limit had bee smaller tha. Example 7 Determie if the followig itegral is coverget or diverget. e x dx Solutio x x First, we eed to ote that e e is oly true o the iterval [, ) as is illustrated i the graph below. 007 Paul Dawkis 7

81 So, we ca t just proceed as we did i the previous example with the Compariso Test o the,. However, this is t the problem it might at first appear to be. We ca always iterval [ ) write the itegral as follows, x x x e dx = e dx + e dx = We used Mathematica to get the value of the first itegral. Now, if the secod itegral coverges it will have a fiite value ad so the sum of two fiite values will also be fiite ad so the origial itegral will coverge. Likewise, if the secod itegral diverges it will either be ifiite or ot have a value at all ad addig a fiite umber oto this will ot all of a sudde make it fiite or exist ad so the origial itegral will diverge. Therefore, this itegral will coverge or diverge depedig oly o the covergece of the secod itegral. As we saw i Example 6 the secod itegral does coverge ad so the whole itegral must also coverge. As we saw i this example, if we eed to, we ca split the itegral up ito oe that does t ivolve ay problems ad ca be computed ad oe that may cotai a problem that we ca use the Compariso Test o to determie its covergece. e x dx 007 Paul Dawkis 7

82 Approximatig Defiite Itegrals I this chapter we ve spet quite a bit of time o computig the values of itegrals. However, ot all itegrals ca be computed. A perfect example is the followig defiite itegral. x e dx 0 We ow eed to talk a little bit about estimatig values of defiite itegrals. We will look at three differet methods, although oe should already be familiar to you from your Calculus I days. We will develop all three methods for estimatig b f x dx a by thikig of the itegral as a area problem ad usig kow shapes to estimate the area uder the curve. Let s get first develop the methods ad the we ll try to estimate the itegral show above. Midpoit Rule This is the rule that should be somewhat familiar to you. We will divide the iterval [, ] subitervals of equal width, b a x = We will deote each of the itervals as follows, [ x0, x], [ x, x],,[ x, x] where x0 = a ad x = b The for each iterval let each subiterval with a height of ( i ) ab ito * x i be the midpoit of the iterval. We the sketch i rectagles for * f x. Here is a graph showig the set up usig = 6. We ca easily fid the area for each of these rectagles ad so for a geeral we get that, b * * * f( x) dx xf( x) + xf( x) + + xf( x ) a Or, upo factorig out a x we get the geeral Midpoit Rule. 007 Paul Dawkis 7

83 b * * * ( ) f x dx x f x f x f x a Trapezoid Rule For this rule we will do the same set up as for the Midpoit Rule. We will break up the iterval [ ab, ] ito subitervals of width, b a x = The o each subiterval we will approximate the fuctio with a straight lie that is equal to the fuctio values at either edpoit of the iterval. Here is a sketch of this case for = 6. Each of these objects is a trapezoid (hece the rule s ame ) ad as we ca see some of them do a very good job of approximatig the actual area uder the curve ad others do t do such a good job. The area of the trapezoid i the iterval [ x, i xi] is give by, x Ai = ( f ( xi ) + f ( xi) ) So, if we use subitervals the itegral is approximately, b x x x f ( x) dx ( f ( x0) + f ( x) ) + f ( x) + f ( x) + + f x a + f x ( ) Upo doig a little simplificatio we arrive at the geeral Trapezoid Rule. b x f x dx f x a 0 f x f x f x f x Note that all the fuctio evaluatios, with the exceptio of the first ad last, are multiplied by. 007 Paul Dawkis 74

84 Simpso s Rule This is the fial method we re goig to take a look at ad i this case we will agai divide up the ab, ito subitervals. However ulike the previous two methods we eed to require iterval [ ] that be eve. The reaso for this will be evidet i a bit. The width of each subiterval is, b a x = I the Trapezoid Rule we approximated the curve with a straight lie. For Simpso s Rule we are goig to approximate the fuctio with a quadratic ad we re goig to require that the quadratic agree with three of the poits from our subitervals. Below is a sketch of this usig = 6. Each of the approximatios is colored differetly so we ca see how they actually work. Notice that each approximatio actually covers two of the subitervals. This is the reaso for requirig to be eve. Some of the approximatios look more like a lie tha a quadratic, but they really are quadratics. Also ote that some of the approximatios do a better job tha others. It ca be show that the area uder the approximatio o the itervals [ x, i xi] ad [ xi, x i + ] is, x Ai = ( f ( xi ) + 4 f ( xi) + f ( xi+ ) ) If we use subitervals the itegral is the approximately, b x x f ( x) dx ( f ( x0) + 4f ( x) + f ( x) ) + ( f ( x) + 4f ( x) + f ( x4) ) a x + + f x + 4 f x + f x Upo simplifyig we arrive at the geeral Simpso s Rule. ( ( ) ( ) ) b x f x dx f x f x f x f x f x f x a Paul Dawkis 75

85 I this case otice that all the fuctio evaluatios at poits with odd subscripts are multiplied by 4 ad all the fuctio evaluatios at poits with eve subscripts (except for the first ad last) are multiplied by. If you ca remember this, this is a fairly easy rule to remember. Okay, it s time to work a example ad see how these rules work. Example Usig = 4 ad all three rules to approximate the value of the followig itegral. e x dx 0 Solutio First, for referece purposes, Maple gives the followig value for this itegral. x e dx = I each case the width of the subitervals will be, 0 x = = 4 ad so the subitervals will be, [ 0, 0.5 ], [ 0.5, ], [,.5 ], [.5, ] Let s go through each of the methods. 0 Midpoit Rule ( 0.5) e x dx = e e e e Remember that we evaluate at the midpoits of each of the subitervals here! The Midpoit Rule has a error of Trapezoid Rule ( 0) e x dx = e e e e e The Trapezoid Rule has a error of Simpso s Rule ( 0) e x dx = e e e e e The Simpso s Rule has a error of Noe of the estimatios i the previous example are all that good. The best approximatio i this case is from the Simpso s Rule ad yet it still had a error of almost. To get a better estimatio we would eed to use a larger. So, for completeess sake here are the estimates for some larger value of. 007 Paul Dawkis 76

86 Midpoit Trapezoid Simpso s Approx. Error Approx. Error Approx. Error I this case we were able to determie the error for each estimate because we could get our hads o the exact value. Ofte this wo t be the case ad so we d ext like to look at error bouds for each estimate. These bouds will give the largest possible error i the estimate, but it should also be poited out that the actual error may be sigificatly smaller tha the boud. The boud is oly there so we ca say that we kow the actual error will be less tha the boud. So, suppose that f ( x) K ad ( 4 f ) ( x) M for a x b the if E M, E T, ad E S are the actual errors for the Midpoit, Trapezoid ad Simpso s Rule we have the followig bouds, 5 K( b a) K( b a) M ( b a) EM E T E S Example Determie the error bouds for the estimatios i the last example. Solutio We already kow that = 4, a = 0, ad b = so we just eed to compute K (the largest value of the secod derivative) ad M (the largest value of the fourth derivative). This meas that we ll eed the secod ad fourth derivative of f(x). x f x = e + x ( 4 ) x 4 = 4e ( ) f x x x Here is a graph of the secod derivative. Here is a graph of the fourth derivative. 007 Paul Dawkis 77

87 So, from these graphs it s clear that the largest value of both of these are at x =. So, f = K = 98 f ( 4 ) = M = 56 We rouded to make the computatios simpler. Here are the bouds for each rule. E M E T E S ( ) ( ) ( ) = = = I each case we ca see that the errors are sigificatly smaller tha the actual bouds. 007 Paul Dawkis 78

88 007 Paul Dawkis 79

89 Applicatios of Itegrals Itroductio I this sectio we re goig to take a look at some of the applicatios of itegratio. It should be oted as well that these applicatios are preseted here, as opposed to Calculus I, simply because may of the itegrals that arise from these applicatios ted to require techiques that we discussed i the previous chapter. Here is a list of applicatios that we ll be takig a look at i this chapter. Arc Legth We ll determie the legth of a curve i this sectio. Surface Area I this sectio we ll determie the surface area of a solid of revolutio. Ceter of Mass Here we will determie the ceter of mass or cetroid of a thi plate. Hydrostatic Pressure ad Force We ll determie the hydrostatic pressure ad force o a vertical plate submerged i water. Probability Here we will look at probability desity fuctios ad computig the mea of a probability desity fuctio. 007 Paul Dawkis 80

90 Arc Legth I this sectio we are goig to look at computig the arc legth of a fuctio. Because it s easy eough to derive the formulas that we ll use i this sectio we will derive oe of them ad leave the other to you to derive. We wat to determie the legth of the cotiuous fuctio y = f ( x) o the iterval [, ] We ll also eed to assume that the derivative is cotiuous o [ ab, ]. ab. Iitially we ll eed to estimate the legth of the curve. We ll do this by dividig the iterval up ito equal subitervals each of width x ad we ll deote the poit o the curve at each poit by P i. We ca the approximate the curve by a series of straight lies coectig the poits. Here is a sketch of this situatio for = 9. Now deote the legth of each of these lie segmets by Pi Pi ad the legth of the curve will the be approximately, L Pi Pi i= ad we ca get the exact legth by takig larger ad larger. I other words, the exact legth will be, L= lim Pi Pi i = Now, let s get a better grasp o the legth of each of these lie segmets. First, o each segmet let s defie yi = yi yi = f ( xi) f ( xi ). We ca the compute directly the legth of the lie segmets as follows. i i = i i + i i = + i P P x x y y x y By the Mea Value Theorem we kow that o the iterval [ x x ], i i there is a poit * x i so that, 007 Paul Dawkis 8

91 Therefore, the legth ca ow be writte as, * = ( ) f x f x f x x x i i i i i i * ( i ) y = f x x * ( i ) P P = x x + y y i i i i i i ( i ) * = x + f x x = + f x x The exact legth of the curve is the, L= lim P P i = i = i * ( i ) = lim + f x x However, usig the defiitio of the defiite itegral, this is othig more tha, b i L = + f x dx A slightly more coveiet otatio (i my opiio ayway) is the followig. b dy L = + dx dx I a similar fashio we ca also derive a formula for x= h( y) o [, ] d dx L = + h ( y) dy = + dy c dy c Agai, the secod form is probably a little more coveiet. a a d cd. This formula is, Note the differece i the derivative uder the square root! Do t get too cofused. With oe we differetiate with respect to x ad with the other we differetiate with respect to y. Oe way to keep the two straight is to otice that the differetial i the deomiator of the derivative will match up with the differetial i the itegral. This is oe of the reasos why the secod form is a little more coveiet. Before we work ay examples we eed to make a small chage i otatio. Istead of havig two formulas for the arc legth of a fuctio we are goig to reduce it, i part, to a sigle formula. From this poit o we are goig to use the followig formula for the legth of the curve. 007 Paul Dawkis 8

92 Arc Legth Formula(s) where, L = ds dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy Note that o limits were put o the itegral as the limits will deped upo the ds that we re usig. Usig the first ds will require x limits of itegratio ad usig the secod ds will require y limits of itegratio. Thikig of the arc legth formula as a sigle itegral with differet ways to defie ds will be coveiet whe we ru across arc legths i future sectios. Also, this ds otatio will be a ice otatio for the ext sectio as well. Now that we ve derived the arc legth formula let s work some examples. π Example Determie the legth of y = l ( sec x) betwee 0 x. 4 Solutio I this case we ll eed to use the first ds sice the fuctio is i the form y = f ( x). So, let s get the derivative out of the way. dy sec x ta x ta dy = = x = ta x dx sec x dx Let s also get the root out of the way sice there is ofte simplificatio that ca be doe ad there s o reaso to do that iside the itegral. dy ta sec sec sec + = + x = x = x = x dx Note that we could drop the absolute value bars here sice secat is positive i the rage give. The arc legth is the, L = π 4 0 sec x dx = l sec x+ ta x = l + π Paul Dawkis 8

93 Example Determie the legth of x= ( y ) betwee y 4. Solutio There is a very commo mistake that studets make i problems of this type. May studets see that the fuctio is i the form x= h( y) ad they immediately decide that it will be too difficult to work with it i that form so they solve for y to get the fuctio ito the form y = f ( x). While that ca be doe here it will lead to a messier itegral for us to deal with. =. I fact, if you ca work =. There really is t a differece betwee the two so do t get excited about fuctios i the form x= h y. Sometimes it s just easier to work with fuctios i the form x h( y) with fuctios i the form y = f ( x) the you ca work with fuctios i the form x h( y) Let s compute the derivative ad the root. dx dx = ( y ) + = + y = y dy dy As you ca see keepig the fuctio i the form x= h( y) is goig to lead to a very easy itegral. To see what would happe if we tried to work with the fuctio i the form y = f ( x) see the ext example. Let s get the legth. L = = = 4 4 y y dy 4 As oted i the last example we really do have a choice as to which ds we use. Provided we ca get the fuctio i the form required for a particular ds we ca use it. However, as also oted above, there will ofte be a sigificat differece i difficulty i the resultig itegrals. Let s take a quick look at what would happe i the previous example if we did put the fuctio ito y = f x. the form Example Redo the previous example usig the fuctio i the form y = f ( x) istead. Solutio I this case the fuctio ad its derivative would be, 007 Paul Dawkis 84

94 x dy y x = + = dx The root i the arc legth formula would the be. x dy + + = + = = dx x ( ) x x x All the simplificatio work above was just to put the root ito a form that will allow us to do the itegral. Now, before we write dow the itegral we ll also eed to determie the limits. This particular ds requires x limits of itegratio ad we ve got y limits. They are easy eough to get however. Sice we kow x as a fuctio of y all we eed to do is plug i the origial y limits of itegratio ad get the x limits of itegratio. Doig this gives, 0 x ( ) Not easy limits to deal with, but there they are. Let s ow write dow the itegral that will give the legth. L = 0 x x + dx That s a really upleasat lookig itegral. It ca be evaluated however usig the followig substitutio. Usig this substitutio the itegral becomes, x x u = + du = dx x= 0 u = x= u = 4 L = 4 = u 4 = So, we got the same aswer as i the previous example. Although that should t really be all that surprisig sice we were dealig with the same curve. u du Paul Dawkis 85

95 From a techical stadpoit the itegral i the previous example was ot that difficult. It was just a Calculus I substitutio. However, from a practical stadpoit the itegral was sigificatly more difficult tha the itegral we evaluated i Example. So, the moral of the story here is that we ca use either formula (provided we ca get the fuctio i the correct form of course) however oe will ofte be sigificatly easier to actually evaluate. Okay, let s work oe more example. Example 4 Determie the legth of x= y for 0 x. Assume that y is positive. Solutio We ll use the secod ds for this oe as the fuctio is already i the correct form for that oe. Also, the other ds would agai lead to a particularly difficult itegral. The derivative ad root will the be, dx dx = y + = + y dy dy Before writig dow the legth otice that we were give x limits ad we will eed y limits for this ds. With the assumptio that y is positive these are easy eough to get. All we eed to do is plug x ito our equatio ad solve for y. Doig this gives, 0 y The itegral for the arc legth is the, L = + 0 y dy This itegral will require the followig trig substitutio. y = taθ dy = sec θ dθ y = 0 0 = taθ θ = 0 y = = taθ π θ = 4 + y = + ta θ = sec θ = secθ = secθ The legth is the, L= π 4 0 sec θ dθ = sec ta + l sec + ta = ( + l ( + )) ( θ θ θ θ ) The first couple of examples eded up beig fairly simple Calculus I substitutios. However, as this last example had show we ca ed up with trig substitutios as well for these itegrals. π Paul Dawkis 86

96 Surface Area I this sectio we are goig to look oce agai at solids of revolutio. We first looked at them back i Calculus I whe we foud the volume of the solid of revolutio. I this sectio we wat to fid the surface area of this regio. So, for the purposes of the derivatio of the formula, let s look at rotatig the cotiuous fuctio y f x ab, about the x-axis. We ll also eed to assume that the derivative is = i the iterval [ ] cotiuous o [, ] ab. Below is a sketch of a fuctio ad the solid of revolutio we get by rotatig the fuctio about the x-axis. We ca derive a formula for the surface area much as we derived the formula for arc legth. We ll start by dividig the iterval ito equal subitervals of width x. O each subiterval we will approximate the fuctio with a straight lie that agrees with the fuctio at the edpoits of each iterval. Here is a sketch of that for our represetative fuctio usig = 4. Now, rotate the approximatios about the x-axis ad we get the followig solid. 007 Paul Dawkis 87

97 The approximatio o each iterval gives a distict portio of the solid ad to make this clear each portio is colored differetly. Each of these portios are called frustums ad we kow how to fid the surface area of frustums. The surface area of a frustum is give by, A = π rl where, r = ( r+ r) r = radius of right ed r = radius of left ed ad l is the legth of the slat of the frustum. For the frustum o the iterval [ x, i xi] we have, r = f ( xi ) r = f ( xi ) l = P P ( legth of the lie segmet coectig P ad P ) i i i i ad we kow from the previous sectio that, ( * ) * [ ] P P = + f x x where x is some poit i x, x i i i i i i Before writig dow the formula for the surface area we are goig to assume that f x is cotiuous we ca the assume that, ad sice ( * ) ( * i i ad i i ) f x f x f x f x So, the surface area of the frustum o the iterval [ x x ], i i is approximately, x is small 007 Paul Dawkis 88

98 + f xi f xi Ai = π Pi Pi * * i ( i ) π f x + f x x The surface area of the whole solid is the approximately, * * i i S π f x + f x x i= ad we ca get the exact surface area by takig the limit as goes to ifiity. * * i ( i ) S = lim π f x + f x x = i b = π f x + f x dx a If we wated to we could also derive a similar formula for rotatig x= h( y) o [, ] y-axis. This would give the followig formula. d S = π h( y) + h ( y) dy c cd about the These are ot the stadard formulas however. Notice that the roots i both of these formulas f x are othig more tha the two ds s we used i the previous sectio. Also, we will replace with y ad h( y ) with x. Doig this gives the followig two formulas for the surface area. Surface Area Formulas S = π y ds rotatio about x axis where, S = π x ds rotatio about y axis dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy There are a couple of thigs to ote about these formulas. First, otice that the variable i the itegral itself is always the opposite variable from the oe we re rotatig about. Secod, we are allowed to use either ds i either formula. This meas that there are, i some way, four formulas here. We will choose the ds based upo which is the most coveiet for a give fuctio ad problem. Now let s work a couple of examples. 007 Paul Dawkis 89

99 Example Determie the surface area of the solid obtaied by rotatig x about the x-axis. y = 9 x, Solutio The formula that we ll be usig here is, S = π y ds sice we are rotatig about the x-axis ad we ll use the first ds i this case because our fuctio is i the correct form for that ds ad we wo t gai aythig by solvig it for x. Let s first get the derivative ad the root take care of. dy = ( 9 x ) ( x) = dx x ( 9 x ) dy x 9 + = + = = dx 9 x 9 x 9 x Here s the itegral for the surface area, S = π y dx 9 x There is a problem however. The dx meas that we should t have ay y s i the itegral. So, before evaluatig the itegral we ll eed to substitute i for y as well. The surface area is the, S = π 9 x dx = = 4π 6π dx 9 x Previously we made the commet that we could use either ds i the surface area formulas. Let s work a example i which usig either ds wo t create itegrals that are too difficult to evaluate ad so we ca check both ds s. Example Determie the surface area of the solid obtaied by rotatig about the y-axis. Use both ds s to compute the surface area. y = x, y Solutio Note that we ve bee give the fuctio set up for the first ds ad limits that work for the secod ds. Solutio This solutio will use the first ds listed above. We ll start with the derivative ad root. 007 Paul Dawkis 90

100 dy = dx x 4 4 dy 9x + 9x + + = + = = 4 4 dx 9x 9x x We ll also eed to get ew limits. That is t too bad however. All we eed to do is plug i the give y s ito our equatio ad solve to get that the rage of x s is x 8. The itegral for the surface area is the, 8 4 9x + S = π x dx x 8 4 π = x 9x + dx Note that this time we did t eed to substitute i for the x as we did i the previous example. I this case we picked up a dx from the ds ad so we do t eed to do a substitutio for the x. I fact if we had substituted for x we would have put y s ito the itegral which would have caused problems. Usig the substitutio the itegral becomes, 4 u = 9x + du = x dx π S = π = u u du π = 45 0 = Solutio This time we ll use the secod ds. So, we ll first eed to solve the equatio for x. We ll also go ahead ad get the derivative ad root while we re at it. dx x= y = y dy The surface area is the, dx + = + 9y dy Paul Dawkis 9

101 S = π x + y 4 dy 9 We used the origial y limits this time because we picked up a dy from the ds. Also ote that the presece of the dy meas that this time, ulike the first solutio, we ll eed to substitute i for the x. Doig that gives, 4 4 S = π y + 9y dy u = + 9y π 45 = 8 u du 0 π = 45 0 = Note that after the substitutio the itegral was idetical to the first solutio ad so the work was skipped. As this example has show we ca use either ds to get the surface area. It is importat to poit out as well that with oe ds we had to do a substitutio for the x ad with the other we did t. This will always work out that way. Note as well that i the case of the last example it was just as easy to use either ds. That ofte wo t be the case. I may examples oly oe of the ds will be coveiet to work with so we ll always eed to determie which ds is liable to be the easiest to work with before startig the problem. 007 Paul Dawkis 9

102 Ceter of Mass I this sectio we are goig to fid the ceter of mass or cetroid of a thi plate with uiform desity ρ. The ceter of mass or cetroid of a regio is the poit i which the regio will be perfectly balaced horizotally if suspeded from that poit. So, let s suppose that the plate is the regio bouded by the two curves f ( x ) ad iterval [a,b]. So, we wat to fid the ceter of mass of the regio below. g x o the We ll first eed the mass of this plate. The mass is, M = ρ ( Area of plate) b a = ρ f x g x dx Next we ll eed the momets of the regio. There are two momets, deoted by M x ad M y. The momets measure the tedecy of the regio to rotate about the x ad y-axis respectively. The momets are give by, Equatios of Momets ( ) b M x = ρ f x g x dx y a b a ( ) M = ρ x f x g x dx The coordiates of the ceter of mass, ( x, y ), are the, 007 Paul Dawkis 9

103 Ceter of Mass Coordiates b ( ) M x f x g x dx y a b x = = = x b ( f ( x) g ( x) ) dx M A a f x g x dx a ( ) b f x g x dx b M a x y = = = b ( f ( x) g ( x) ) dx M f x g x dx A a a where, b a A = f x g x dx Note that the desity, ρ, of the plate cacels out ad so is t really eeded. Let s work a couple of examples. =, y = 0 o π the iterval 0,. Solutio Here is a sketch of the regio with the ceter of mass deoted with a dot. Example Determie the ceter of mass for the regio bouded by y si ( x) Let s first get the area of the regio. A = π 0 si x dx ( x) = cos = π 0 Now, the momets (without desity sice it will just drop out) are, 007 Paul Dawkis 94

104 π π 0 y 0 M = si x dx M = xsi x dx itegratig by parts... x π π π = cos 4x dx = x cos x + cos x dx π π π = x si ( 4x) = xcos x + si x π π = = The coordiates of the ceter of mass are the, π π x = = 4 π y π = = 4 Agai, ote that we did t put i the desity sice it will cacel out. π π So, the ceter of mass for this regio is, Example Determie the ceter of mass for the regio bouded by y = x ad y x =. Solutio The two curves itersect at x = 0 ad x = ad here is a sketch of the regio with the ceter of mass marked with a box. We ll first get the area of the regio. 007 Paul Dawkis 95

105 A = x x dx 0 = x x 4 5 = Now the momets, agai without desity, are Mx = x x dx 0 6 y = ( ) = x x dx 0 = x x = x x = = 5 The coordiates of the ceter of mass is the, 5 x = = y = = 5 7 The coordiates of the ceter of mass are the,, M x x x dx Paul Dawkis 96

106 Hydrostatic Pressure ad Force I this sectio we are goig to submerge a vertical plate i water ad we wat to kow the force that is exerted o the plate due to the pressure of the water. This force is ofte called the hydrostatic force. There are two basic formulas that we ll be usig here. First, if we are d meters below the surface the the hydrostatic pressure is give by, P = ρgd where, ρ is the desity of the fluid ad g is the gravitatioal acceleratio. We are goig to assume that the fluid i questio is water ad sice we are goig to be usig the metric system these quatities become, ρ = 000 kg/m g = 9.8 m/s The secod formula that we eed is the followig. Assume that a costat pressure P is actig o a surface with area A. The the hydrostatic force that acts o the area is, F = PA Note that we wo t be able to fid the hydrostatic force o a vertical plate usig this formula sice the pressure will vary with depth ad hece will ot be costat as required by this formula. We will however eed this for our work. The best way to see how these problems work is to do a example or two. Example Determie the hydrostatic force o the followig triagular plate that is submerged i water as show. Solutio The first thig to do here is set up a axis system. So, let s redo the sketch above with the followig axis system added i. 007 Paul Dawkis 97

107 So, we are goig to oriet the x-axis so that positive x is dowward, x = 0 correspods to the water surface ad x = 4 correspods to the depth of the tip of the triagle. Next we break up the triagle ito horizotal strips each of equal width x ad i each iterval [, * xi xi] choose ay poit x i. I order to make the computatios easier we are goig to make two assumptios about these strips. First, we will igore the fact that the eds are actually goig to be slated ad assume the strips are rectagular. If x is sufficietly small this will ot affect our computatios much. Secod, we will assume that x is small eough that the hydrostatic pressure o each strip is essetially costat. Below is a represetative strip. The height of this strip is x ad the width is a. We ca use similar triagles to determie a as follows, a * = a= x * 4 4 x 4 i i 007 Paul Dawkis 98

108 Now, sice we are assumig the pressure o this strip is costat, the pressure is give by, P = ρgd = x = 980x * * i i i ad the hydrostatic force o each strip is, * * * * Fi = Pi A= Pi a x = xi xi x= xi xi x 4 4 The approximate hydrostatic force o the plate is the the sum of the forces o all the strips or, * * F 960xi xi x i= 4 Takig the limit will get the exact hydrostatic force, * * F = lim 960xi xi x i = 4 Usig the defiitio of the defiite itegral this is othig more tha, 4 F 960 = x x dx 0 4 The hydrostatic force is the, 4 F 960 x x dx = 4 0 = 960 x x 4 = N 4 0 Let s take a look at aother example. Example Fid the hydrostatic force o a circular plate of radius that is submerged 6 meters i the water. Solutio First, we re goig to assume that the top of the circular plate is 6 meters uder the water. Next, we will set up the axis system so that the origi of the axis system is at the ceter of the plate. Settig the axis system up i this way will greatly simplify our work. Fially, we will agai split up the plate ito horizotal strips each of width choose a poit with the computatios. Here is a sketch of the setup. y ad we ll * y i from each strip. We ll also assume that the strips are rectagular agai to help 007 Paul Dawkis 99

109 The depth below the water surface of each strip is, * di = 8 yi ad that i tur gives us the pressure o the strip, P = ρgd = y * i i i The area of each strip is, * i A = 4 y y i The hydrostatic force o each strip is, The total force o the plate is, * * F = PA = y 4 y y i i i i i i = ( y) * * ( i ) ( i ) F = lim y 4 y y = y dy To do this itegral we ll eed to split it up ito two itegrals. F = y dy 960 y 4 y dy 007 Paul Dawkis 00

110 The first itegral requires the trig substitutio substitutio v= 4 y. After usig these substitutio we get, y = siθ ad the secod itegral eeds the π 0 F = cos θ dθ v dv π 0 π π ( θ) = 90 + cos dθ + 0 = 90 θ + si = 90π ( θ) Note that after the substitutio we kow the secod itegral will be zero because the upper ad lower limit is the same. π π 007 Paul Dawkis 0

111 Probability I this last applicatio of itegrals that we ll be lookig at we re goig to look at probability. Before actually gettig ito the applicatios we eed to get a couple of defiitios out of the way. Suppose that we wated to look at the age of a perso, the height of a perso, the amout of time spet waitig i lie, or maybe the lifetime of a battery. Each of these quatities have values that will rage over a iterval of itegers. Because of this these are called cotiuous radom variables. Cotiuous radom variables are ofte represeted by X. Every cotiuous radom variable, X, has a probability desity fuctio, f desity fuctios satisfy the followig coditios.. f ( x) 0 for all x.. f ( x) dx = x. Probability Probability desity fuctios ca be used to determie the probability that a cotiuous radom variable lies betwee two values, say a ad b. This probability is deoted by P( a X b) ad is give by, Let s take a look at a example of this. b P a X b = f x dx x Example Let f ( x) = ( 0 x) for 0 x 0 ad f ( x ) = 0 for all other values of x Aswer each of the followig questios about this fuctio. f x is a probability desity fuctio. [Solutio] Solutio (a) Show that (b) Fid P( X 4) (c) Fid P( x 6) [Solutio] (a) Show that f [Solutio] x is a probability desity fuctio. First ote that i the rage 0 x 0 is clearly positive ad outside of this rage we ve defied it to be zero. So, to show this is a probability desity fuctio we ll eed to show that a f x dx =. 007 Paul Dawkis 0

112 0 x f ( x) dx = ( 0 x) dx x x = = 0 0 Note the chage i limits o the itegral. The fuctio is oly o-zero i these rages ad so the itegral ca be reduced dow to oly the iterval where the fuctio is ot zero. [Retur to Problems] (b) Fid P( X 4) I this case we eed to evaluate the followig itegral. 4 x P( X 4) = ( 0 x) dx x x = = So the probability of X beig betwee ad 4 is 8.658%. 4 [Retur to Problems] (c) Fid P( x 6) sice 0 is the largest value that X ca be. So the probability that X is greater tha or equal to 6 is, 0 x P ( X 6) = ( 0 x) dx 5000 Note that i this case P( x 6) is equivalet to P( 6 X 0) This probability is the 66.04% x x = = [Retur to Problems] Probability desity fuctios ca also be used to determie the mea of a cotiuous radom variable. The mea is give by, Let s work oe more example. µ = xf x dx 007 Paul Dawkis 0

113 Example It has bee determied that the probability desity fuctio for the wait i lie at a couter is give by, 0 if t < 0 f ( t) = t 0 0.e if t 0 where t is the umber of miutes spet waitig i lie. Aswer each of the followig questios about this probability desity fuctio. (a) Verify that this is i fact a probability desity fuctio. [Solutio] (b) Determie the probability that a perso will wait i lie for at least 6 miutes.\ [Solutio] (c) Determie the mea wait i lie. [Solutio] Solutio (a) Verify that this is i fact a probability desity fuctio. This fuctio is clearly positive or zero ad so there s ot much to do here other tha compute the itegral. So it is a probability desity fuctio. 0 = 0.e f t dt 0 u = lim 0.e u 0 t 0 t 0 = lim e u 0 u 0 = lim e = u (b) Determie the probability that a perso will wait i lie for at least 6 miutes. P x 6. The probability that we re lookig for here is 0 ( 6) = 0.e P X 6 u = lim 0.e u 6 t dt t 0 t dt u dt [Retur to Problems] t 0 = lim e u 6 6 u = lim e e = e = u So the probability that a perso will wait i lie for more tha 6 miutes is 54.88%. [Retur to Problems] u dt 007 Paul Dawkis 04

114 (c) Determie the mea wait i lie. Here s the mea wait time. µ = = = 0 t f t dt 0.t e t 0 dt t u 0 lim 0.t dt itegratig by parts... u e 0 = lim ( t + 0) e u t 0 = + = u 0 lim 0 ( u 0) e 0 u u 0 So, it looks like the average wait time is 0 miutes. [Retur to Problems] 007 Paul Dawkis 05

115 Parametric Equatios ad Polar Coordiates Itroductio I this sectio we will be lookig at parametric equatios ad polar coordiates. While the two subjects do t appear to have that much i commo o the surface we will see that several of the topics i polar coordiates ca be doe i terms of parametric equatios ad so i that sese they make a good match i this chapter We will also be lookig at how to do may of the stadard calculus topics such as tagets ad area i terms of parametric equatios ad polar coordiates. Here is a list of topics that we ll be coverig i this chapter. Parametric Equatios ad Curves A itroductio to parametric equatios ad parametric curves (i.e. graphs of parametric equatios) Tagets with Parametric Equatios Fidig taget lies to parametric curves. Area with Parametric Equatios Fidig the area uder a parametric curve. Arc Legth with Parametric Equatios Determiig the legth of a parametric curve. Surface Area with Parametric Equatios Here we will determie the surface area of a solid obtaied by rotatig a parametric curve about a axis. Polar Coordiates We ll itroduce polar coordiates i this sectio. We ll look at covertig betwee polar coordiates ad Cartesia coordiates as well as some basic graphs i polar coordiates. Tagets with Polar Coordiates Fidig taget lies of polar curves. Area with Polar Coordiates Fidig the area eclosed by a polar curve. Arc Legth with Polar Coordiates Determiig the legth of a polar curve. Surface Area with Polar Coordiates Here we will determie the surface area of a solid obtaied by rotatig a polar curve about a axis. Arc Legth ad Surface Area Revisited I this sectio we will summarize all the arc legth ad surface area formulas from the last two chapters. 007 Paul Dawkis 06

116 Parametric Equatios ad Curves To this poit (i both Calculus I ad Calculus II) we ve looked almost exclusively at fuctios i the form y = f ( x) or x= h( y) ad almost all of the formulas that we ve developed require that fuctios be i oe of these two forms. The problem is that ot all curves or equatios that we d like to look at fall easily ito this form. Take, for example, a circle. It is easy eough to write dow the equatio of a circle cetered at the origi with radius r. x + y = r However, we will ever be able to write the equatio of a circle dow as a sigle equatio i either of the forms above. Sure we ca solve for x or y as the followig two formulas show y =± r x x=± r y but there are i fact two fuctios i each of these. Each formula gives a portio of the circle. y = r x top x= r y right side ( bottom) ( left side) y = r x x= r y Ufortuately we usually are workig o the whole circle, or simply ca t say that we re goig to be workig oly o oe portio of it. Eve if we ca arrow thigs dow to oly oe of these portios the fuctio is still ofte fairly upleasat to work with. There are also a great may curves out there that we ca t eve write dow as a sigle equatio i terms of oly x ad y. So, to deal with some of these problems we itroduce parametric equatios. Istead of defiig y i terms of x ( y = f ( x) ) or x i terms of y ( x= h( y) ) we defie both x ad y i terms of a third variable called a parameter as follows, x= f t y = g t This third variable is usually deoted by t (as we did here) but does t have to be of course. Sometimes we will restrict the values of t that we ll use ad at other times we wo t. This will ofte be depedet o the problem ad just what we are attemptig to do. Each value of t defies a poit ( xy, ) f( t), gt = that we ca plot. The collectio of poits that we get by lettig t be all possible values is the graph of the parametric equatios ad is called the parametric curve. To help visualize just what a parametric curve is preted that we have a big tak of water that is i costat motio ad we drop a pig pog ball ito the tak. The poit ( xy, ) = ( f( t), gt ) will the represet the locatio of the pig pog ball i the tak at time t ad the parametric curve will be a trace of all the locatios of the pig pog ball. Note that this is ot always a correct aalogy but it is useful iitially to help visualize just what a parametric curve is. Sketchig a parametric curve is ot always a easy thig to do. Let s take a look at a example to see oe way of sketchig a parametric curve. This example will also illustrate why this method is usually ot the best. 007 Paul Dawkis 07

117 Example Sketch the parametric curve for the followig set of parametric equatios. x= t + t y = t Solutio At this poit our oly optio for sketchig a parametric curve is to pick values of t, plug them ito the parametric equatios ad the plot the poits. So, let s plug i some t s. t x y The first questio that should be asked at this poit is, how did we kow to use the values of t that we did, especially the third choice? Ufortuately, there is o real aswer to this questio at this poit. We simply pick t s util we are fairly cofidet that we ve got a good idea of what the curve looks like. It is this problem with pickig good values of t that make this method of sketchig parametric curves oe of the poorer choices. Sometimes we have o choice, but if we do have a choice we should avoid it. We ll discuss a alterate graphig method i later examples that will help to explai how these values of t were chose. We have oe more idea to discuss before we actually sketch the curve. Parametric curves have a directio of motio. The directio of motio is give by icreasig t. So, whe plottig parametric curves, we also iclude arrows that show the directio of motio. We will ofte give the value of t that gave specific poits o the graph as well to make it clear the value of t that gave that particular poit. Here is the sketch of this parametric curve. So, it looks like we have a parabola that opes to the right. Before we ed this example there is a somewhat importat ad subtle poit that we eed to 007 Paul Dawkis 08

118 discuss first. Notice that we made sure to iclude a portio of the sketch to the right of the poits correspodig to t = ad t = to idicate that there are portios of the sketch there. Had we simply stopped the sketch at those poits we are idicatig that there was o portio of the curve to the right of those poits ad there clearly will be. We just did t compute ay of those poits. This may seem like a uimportat poit, but as we ll see i the ext example it s more importat tha we might thik. Before addressig a much easier way to sketch this graph let s first address the issue of limits o the parameter. I the previous example we did t have ay limits o the parameter. Without limits o the parameter the graph will cotiue i both directios as show i the sketch above. We will ofte have limits o the parameter however ad this will affect the sketch of the parametric equatios. To see this effect let s look a slight variatio of the previous example. Example Sketch the parametric curve for the followig set of parametric equatios. x= t + t y = t t Solutio Note that the oly differece here is the presece of the limits o t. All these limits do is tell us that we ca t take ay value of t outside of this rage. Therefore, the parametric curve will oly be a portio of the curve above. Here is the parametric curve for this example. Notice that with this sketch we started ad stopped the sketch right o the poits origiatig from the ed poits of the rage of t s. Cotrast this with the sketch i the previous example where we had a portio of the sketch to the right of the start ad ed poits that we computed. I this case the curve starts at t = ad eds at t =, whereas i the previous example the curve did t really start at the right most poits that we computed. We eed to be clear i our sketches if the curve starts/eds right at a poit, or if that poit was simply the first/last oe that we computed. It is ow time to take a look at a easier method of sketchig this parametric curve. This method uses the fact that i may, but ot all, cases we ca actually elimiate the parameter from the parametric equatios ad get a fuctio ivolvig oly x ad y. We will sometimes call this the algebraic equatio to differetiate it from the origial parametric equatios. There will be two 007 Paul Dawkis 09

119 small problems with this method, but it will be easy to address those problems. It is importat to ote however that we wo t always be able to do this. Just how we elimiate the parameter will deped upo the parametric equatios that we ve got. Let s see how to elimiate the parameter for the set of parametric equatios that we ve bee workig with to this poit. Example Elimiate the parameter from the followig set of parametric equatios. x= t + t y = t Solutio Oe of the easiest ways to elimiate the parameter is to simply solve oe of the equatios for the parameter (t, i this case) ad substitute that ito the other equatio. Note that while this may be the easiest to elimiate the parameter, it s usually ot the best way as we ll see soo eough. I this case we ca easily solve y for t. t = + ( y ) Pluggig this ito the equatio for x gives the followig algebraic equatio, x= ( y+ ) + ( y+ ) = y + y+ 4 4 Sure eough from our Algebra kowledge we ca see that this is a parabola that opes to the right ad will have a vertex at (, 4 ). We wo t bother with a sketch for this oe as we ve already sketched this oce ad the poit here was more to elimiate the parameter ayway. Before we leave this example let s address oe quick issue. I the first example we just, seemigly radomly, picked values of t to use i our table, especially the third value. There really was o apparet reaso for choosig t =. It is however probably the most importat choice of t as it is the oe that gives the vertex. The reality is that whe writig this material up we actually did this problem first the wet back ad did the first problem. Plottig poits is geerally the way most people first lear how to costruct graphs ad it does illustrate some importat cocepts, such as directio, so it made sese to do that first i the otes. I practice however, this example is ofte doe first. So, how did we get those values of t? Well let s start off with the vertex as that is probably the most importat poit o the graph. We have the x ad y coordiates of the vertex ad we also have x ad y parametric equatios for those coordiates. So, plug i the coordiates for the vertex ito the parametric equatios ad solve for t. Doig this gives, 4 = t + t = t = t t = ( double root) 007 Paul Dawkis 0

120 So, as we ca see, the value of t that will give both of these coordiates is t =. Note that the x parametric equatio gave a double root ad this will ofte ot happe. Ofte we would have gotte two distict roots from that equatio. I fact, it wo t be uusual to get multiple values of t from each of the equatios. However, what we ca say is that there will be a value(s) of t that occurs i both sets of solutios ad that is the t that we wat for that poit. We ll evetually see a example where this happes i a later sectio. Now, from this work we ca see that if we use t = we will get the vertex ad so we icluded that value of t i the table i Example. Oce we had that value of t we chose two iteger values of t o either side to fiish out the table. As we will see i later examples i this sectio determiig values of t that will give specific poits is somethig that we ll eed to do o a fairly regular basis. It is fairly simple however as this example has show. All we eed to be able to do is solve a (usually) fairly basic equatio which by this poit i time should t be too difficult. Gettig a sketch of the parametric curve oce we ve elimiated the parameter seems fairly simple. All we eed to do is graph the equatio that we foud by elimiatig the parameter. As oted already however, there are two small problems with this method. The first is directio of motio. The equatio ivolvig oly x ad y will NOT give the directio of motio of the parametric curve. This is geerally a easy problem to fix however. Let s take a quick look at the derivatives of the parametric equatios from the last example. They are, dx = t + dt dy = dt Now, all we eed to do is recall our Calculus I kowledge. The derivative of y with respect to t is clearly always positive. Recallig that oe of the iterpretatios of the first derivative is rate of chage we ow kow that as t icreases y must also icrease. Therefore, we must be movig up the curve from bottom to top as t icreases as that is the oly directio that will always give a icreasig y as t icreases. Note that the x derivative is t as useful for this aalysis as it will be both positive ad egative ad hece x will be both icreasig ad decreasig depedig o the value of t. That does t help with directio much as followig the curve i either directio will exhibit both icreasig ad decreasig x. I some cases, oly oe of the equatios, such as this example, will give the directio while i other cases either oe could be used. It is also possible that, i some cases, both derivatives would be eeded to determie directio. It will always be depedet o the idividual set of parametric equatios. The secod problem with elimiatig the parameter is best illustrated i a example as we ll be ruig ito this problem i the remaiig examples. 007 Paul Dawkis

121 Example 4 Sketch the parametric curve for the followig set of parametric equatios. Clearly idicate directio of motio. x= 5cost y = sit 0 t π Solutio Before we proceed with elimiatig the parameter for this problem let s first address agai why just pickig t s ad plottig poits is ot really a good idea. Give the rage of t s i the problem statemet let s use the followig set of t s. t x y π 0 π -5 0 π 0 - π 5 0 The questio that we eed to ask ow is do we have eough poits to accurately sketch the graph of this set of parametric equatios? Below are some sketches of some possible graphs of the parametric equatio based oly o these five poits. 007 Paul Dawkis

122 Give the ature of sie/cosie you might be able to elimiate the diamod ad the square but there is o deyig that they are graphs that go through the give poits. The last graph is also a little silly but it does show a graph goig through the give poits. Agai, give the ature of sie/cosie you ca probably guess that the correct graph is the ellipse. However, that is all that would be at this poit. A guess. Nothig actually says uequivocally that the parametric curve is a ellipse just from those five poits. That is the dager of sketchig parametric curves based o a hadful of poits. Uless we kow what the graph will be ahead of time we are really just makig a guess. So, i geeral, we should avoid plottig poits to sketch parametric curves. The best method, provided it ca be doe, is to elimiate the parameter. As oted just prior to startig this example there is still a potetial problem with elimiatig the parameter that we ll eed to deal with. We will evetually discuss this issue. For ow, let s just proceed with elimiatig the parameter. We ll start by elimiatig the parameter as we did i the previous sectio. We ll solve oe of the of the equatios for t ad plug this ito the other equatio. For example, we could do the followig, x x t = cos y = si cos 5 5 Ca you see the problem with doig this? This is defiitely easy to do but we have a greater chace of correctly graphig the origial parametric equatios by plottig poits tha we do graphig this! There are may ways to elimiate the parameter from the parametric equatios ad solvig for t is usually ot the best way to do it. While it is ofte easy to do we will, i most cases, ed up 007 Paul Dawkis

123 with a equatio that is almost impossible to deal with. So, how ca we elimiate the parameter here? I this case all we eed to do is recall a very ice trig idetity ad the equatio of a ellipse. Let s otice that we could do the followig here. x y 5cos t 4si t + = + = cos t+ si t = Elimiatig the middle steps gives the followig algebraic equatio, x y + = 5 4 ad so it looks like we ve got a ellipse. Before proceedig with this example it should be oted that what we did was probably ot all that obvious to most. However, oce it s bee doe it does clearly work ad so it s a ice idea that we ca use to elimiate the parameter from some parametric equatios ivolvig sies ad cosies. It wo t always work ad sometimes it will take a lot more maipulatio of thigs tha we did here. A alterate method that we could have used here was to solve the two parametric equatios for sie ad cosie as follows, x y cost = si t = 5 The, recall the trig idetity we used above ad these ew equatios we get, x y x y = t+ t = + = + cos si So, the same aswer as the other method. Which method you use will probably deped o which you fid easier to use. Both are perfectly valid ad will get the same result. Now, let s cotiue o with the example. We ve idetified that the parametric equatios describe a ellipse, but we ca t just sketch a ellipse ad be doe with it. First, just because the algebraic equatio was a ellipse does t actually mea that the parametric curve is the full ellipse. It is always possible that the parametric curve is oly a portio of the ellipse. I order to idetify just how much of the ellipse the parametric curve will cover let s go back to the parametric equatios ad see what they tell us about ay limits o x ad y. Based o our kowledge of sie ad cosie we have the followig, cost 5 5cost 5 5 x 5 si t si t y So, by startig with sie/cosie ad buildig up the equatio for x ad y usig basic algebraic maipulatios we get that the parametric equatios eforce the above limits o x ad y. I this case, these also happe to be the full limits o x ad y we get by graphig the full ellipse. 007 Paul Dawkis 4

124 This is the secod potetial issue alluded to above. The parametric curve may ot always trace out the full graph of the algebraic curve. We should always fid limits o x ad y eforced upo us by the parametric curve to determie just how much of the algebraic curve is actually sketched out by the parametric equatios. Therefore, i this case, we ow kow that we get a full ellipse from the parametric equatios. Before we proceed with the rest of the example be careful to ot always just assume we will get the full graph of the algebraic equatio. There are defiitely times whe we will ot get the full graph ad we ll eed to do a similar aalysis to determie just how much of the graph we actually get. We ll see a example of this later. Note as well that ay limits o t give i the problem statemet ca also affect how much of the graph of the algebraic equatio we get. I this case however, based o the table of values we computed at the start of the problem we ca see that we do ideed get the full ellipse i the rage 0 t π. That wo t always be the case however, so pay attetio to ay restrictios o t that might exist! Next, we eed to determie a directio of motio for the parametric curve. Recall that all parametric curves have a directio of motio ad the equatio of the ellipse simply tells us othig about the directio of motio. To get the directio of motio it is temptig to just use the table of values we computed above to get the directio of motio. I this case, we would guess (ad yes that is all it is a guess) that the curve traces out i a couter-clockwise directio. We d be correct. I this case, we d be correct! The problem is that tables of values ca be misleadig whe determiig a directio of motio as we ll see i the ext example. Therefore, it is best to ot use a table of values to determie the directio of motio. To correctly determie the directio of motio we ll use the same method of determiig the directio that we discussed after Example. I other words, we ll take the derivative of the parametric equatios ad use our kowledge of Calculus I ad trig to determie the directio of motio. The derivatives of the parametric equatios are, dx dy = 5si t = cost dt dt Now, at t = 0 we are at the poit ( 5,0 ) ad let s see what happes if we start icreasig t. Let s π icrease t from t = 0 to t =. I this rage of t s we kow that sie is always positive ad so from the derivative of the x equatio we ca see that x must be decreasig i this rage of t s. This, however, does t really help us determie a directio for the parametric curve. Startig at ( 5,0 ) o matter if we move i a clockwise or couter-clockwise directio x will have to decrease so we have t really leared aythig from the x derivative. The derivative from the y parametric equatio o the other had will help us. Agai, as we π icrease t from t = 0 to t = we kow that cosie will be positive ad so y must be icreasig i this rage. That however, ca oly happe if we are movig i a couter-clockwise directio. 007 Paul Dawkis 5

125 If we were movig i a clockwise directio from the poit ( 5,0 ) we ca see that y would have to decrease! Therefore, i the first quadrat we must be movig i a couter-clockwise directio. Let s move o to the secod quadrat. π 0, ad we will icrease t from t = to t = π. I this rage of t we kow that cosie will be egative ad sie will be positive. Therefore, from the derivatives of the parametric equatios we ca see that x is still decreasig ad y will ow be decreasig as well. So, we are ow at the poit I this quadrat the y derivative tells us othig as y simply must decrease to move from ( 0, ). However, i order for x to decrease, as we kow it does i this quadrat, the directio must still be movig a couter-clockwise rotatio. We are ow at ( 5,0) π ad we will icrease t from t = π to t =. I this rage of t we kow that cosie is egative (ad hece y will be decreasig) ad sie is also egative (ad hece x will be icreasig). Therefore, we will cotiue to move i a couter-clockwise motio. For the 4 th quadrat we will start at ( 0, ) π ad icrease t from t = to t = π. I this rage of t we kow that cosie is positive (ad hece y will be icreasig) ad sie is egative (ad hece x will be icreasig). So, as i the previous three quadrats, we cotiue to move i a couter-clockwise motio. At this poit we covered the rage of t s we were give i the problem statemet ad durig the full rage the motio was i a couter-clockwise directio. We ca ow fully sketch the parametric curve so, here is the sketch. Okay, that was a really log example. Most of these types of problems are t as log. We just had a lot to discuss i this oe so we could get a couple of importat ideas out of the way. The rest of the examples i this sectio should t take as log to go through. Now, let s take a look at aother example that will illustrate a importat idea about parametric equatios. Example 5 Sketch the parametric curve for the followig set of parametric equatios. Clearly 007 Paul Dawkis 6

126 idicate directio of motio. x= 5cos t y = si t 0 t π Solutio Note that the oly differece i betwee these parametric equatios ad those i Example 4 is that we replaced the t with t. We ca elimiate the parameter here usig either of the methods we discussed i the previous example. I this case we ll do the followig, ( t) ( t) 5cos 4si x y + = + = ( t) + ( t) = cos si So, we get the same ellipse that we did i the previous example. Also ote that we ca do the same aalysis o the parametric equatios to determie that we have exactly the same limits o x ad y. Namely, 5 x 5 y It s startig to look like chagig the t ito a t i the trig equatios will ot chage the parametric curve i ay way. That is ot correct however. The curve does chage i a small but importat way which we will be discussig shortly. Before discussig that small chage the t brigs to the curve let s discuss the directio of motio for this curve. Despite the fact that we said i the last example that pickig values of t ad pluggig i to the equatios to fid poits to plot is a bad idea let s do it ay way. Give the rage of t s from the problem statemet the followig set looks like a good choice of t s to use. t x y π 0 - π -5 0 π 0 π 5 0 So, the oly chage to this table of values/poits from the last example is all the ozero y values chaged sig. From a quick glace at the values i this table it would look like the curve, i this case, is movig i a clockwise directio. But is that correct? Recall we said that these tables of values ca be misleadig whe used to determie directio ad that s why we do t use them. Let s see if our first impressio is correct. We ca check our first impressio by doig the derivative work to get the correct directio. Let s work with just the y parametric equatio as the x will have the same issue that it had i the previous example. The derivative of the y parametric equatio is, dy 6cos dt = ( t) 007 Paul Dawkis 7

127 Now, if we start at t = 0 as we did i the previous example ad start icreasig t. At t = 0 the derivative is clearly positive ad so icreasig t (at least iitially) will force y to also be icreasig. The oly way for this to happe is if the curve is i fact tracig out i a couterclockwise directio iitially. Now, we could cotiue to look at what happes as we further icrease t, but whe dealig with a parametric curve that is a full ellipse (as this oe is) ad the argumet of the trig fuctios is of the form t for ay costat the directio will ot chage so oce we kow the iitial directio we kow that it will always move i that directio. Note that this is oly true for parametric equatios i the form that we have here. We ll see i later examples that for differet kids of parametric equatios this may o loger be true. Okay, from this aalysis we ca see that the curve must be traced out i a couter-clockwise directio. This is directly couter to our guess from the tables of values above ad so we ca see that, i this case, the table would probably have led us to the wrog directio. So, oce agai, tables are geerally ot very reliable for gettig pretty much ay real iformatio about a parametric curve other tha a few poits that must be o the curve. Outside of that the tables are rarely useful ad will geerally ot be dealt with i further examples. So, why did our table give a icorrect impressio about the directio? Well recall that we metioed earlier that the t will lead to a small but importat chage to the curve versus just a t? Let s take a look at just what that chage is as it will also aswer what wet wrog with our table of values. Let s start by look at t = 0. At 0 5,0 ad let s ask ourselves what values of t put us back at this poit. We saw i Example how to determie value(s) of t that put us at certai poits ad the same process will work here with a mior modificatio. t = we are at the poit Istead of lookig at both the x ad y equatios as we did i that example let s just look at the x equatio. The reaso for this is that we ll ote that there are two poits o the ellipse that will 5,0. If we set the y coordiate equal to zero we ll have a y coordiate of zero, ( 5,0 ) ad fid all the t s that are at both of these poits whe we oly wat the values of t that are at ( 5,0 ). So, because the x coordiate of five will oly occur at this poit we ca simply use the x parametric equatio to determie the values of t that will put us at this poit. Doig this gives the followig equatio ad solutio, ( t) 5 = 5cos t = cos = 0+ π t = π = 0, ±, ±, ±, Do t forget that whe solvig a trig equatio we eed to add o the + π where represets the umber of full revolutios i the couter-clockwise directio (positive ) ad clockwise directio (egative ) that we rotate from the first solutio to get all possible solutios to the equatio. Now, let s plug i a few values of startig at = 0. We do t eed egative i this case sice all of those would result i egative t ad those fall outside of the rage of t s we were 007 Paul Dawkis 8

128 give i the problem statemet. The first few values of t are the, = 0 : t = 0 = : t = = : t = = : t = = π We ca stop here as all further values of t will be outside the rage of t s give i this problem. So, what is this tellig us? Well back i Example 4 whe the argumet was just t the ellipse was traced out exactly oce i the rage 0 t π. However, whe we chage the argumet to t (ad recallig that the curve will always be traced out i a couter-clockwise directio for this problem) we are goig through the startig poit of ( 5,0 ) two more times tha we did i the previous example. I fact, this curve is tracig out three separate times. The first trace is completed i the rage π 0 t. The secod trace is completed i the rage π 4 π t ad the third ad fial trace is completed i the rage 4 π t π. I other words, chagig the argumet from t to t icrease the speed of the trace ad the curve will ow trace out three times i the rage 0 t π! This is why the table gives the wrog impressio. The speed of the tracig has icreased leadig to a icorrect impressio from the poits i the table. The table seems to suggest that betwee each pair of values of t a quarter of the ellipse is traced out i the clockwise directio whe i reality it is tracig out three quarters of the ellipse i the couter-clockwise directio. Here s a fial sketch of the curve ad ote that it really is t all that differet from the previous sketch. The oly differeces are the values of t ad the various poits we icluded. We did iclude a few more values of t at various poits just to illustrate where the curve is at for various values of t but i geeral these really are t eeded. π 4π 6π So, we saw i the last two examples two sets of parametric equatios that i some way gave the same graph. Yet, because they traced out the graph a differet umber of times we really do eed to thik of them as differet parametric curves at least i some maer. This may seem like a differece that we do t eed to worry about, but as we will see i later sectios this ca be a very 007 Paul Dawkis 9

129 importat differece. I some of the later sectios we are goig to eed a curve that is traced out exactly oce. Before we move o to other problems let s briefly ackowledge what happes by chagig the t to a t i these kids of parametric equatios. Whe we are dealig with parametric equatios ivolvig oly sies ad cosies ad they both have the same argumet if we chage the argumet from t to t we simply chage the speed with which the curve is traced out. If > we will icrease the speed ad if < we will decrease the speed. Let s take a look at a couple more examples. Example 6 Sketch the parametric curve for the followig set of parametric equatios. Clearly idetify the directio of motio. If the curve is traced out more tha oce give a rage of the parameter for which the curve will trace out exactly oce. x= si t y = cost Solutio We ca elimiate the parameter much as we did i the previous two examples. However, we ll eed to ote that the x already cotais a si t ad so we wo t eed to square the x. We will however, eed to square the y as we eed i the previous two examples. y y x+ = si t+ cos t = x= 4 4 I this case the algebraic equatio is a parabola that opes to the left. We will eed to be very, very careful however i sketchig this parametric curve. We will NOT get the whole parabola. A sketch of the algebraic form parabola will exist for all possible values of y. However, the parametric equatios have defied both x ad y i terms of sie ad cosie ad we kow that the rages of these are limited ad so we wo t get all possible values of x ad y here. So, first let s get limits o x ad y as we did i previous examples. Doig this gives, si t 0 si t 0 x cost cost y So, it is clear from this that we will oly get a portio of the parabola that is defied by the algebraic equatio. Below is a quick sketch of the portio of the parabola that the parametric curve will cover. 007 Paul Dawkis 0

130 To fiish the sketch of the parametric curve we also eed the directio of motio for the curve. Before we get to that however, let s jump forward ad determie the rage of t s for oe trace. To do this we ll eed to kow the t s that put us at each ed poit ad we ca follow the same procedure we used i the previous example. The oly differece is this time let s use the y parametric equatio istead of the x because the y coordiates of the two ed poits of the curve are differet whereas the x coordiates are the same. So, for the top poit we have, = cost t = cos = 0+ π= π, = 0, ±, ±, ±, For, pluggig i some values of we get that the curve will be at the top poit at, Similarly, for the bottom poit we have, t =, 4 π, π,0, π,4 π, = cost t = cos = π + π, = 0, ±, ±, ±, So, we see that we will be at the bottom poit at, t =, π, ππ,, π, So, if we start at say, t = 0, we are at the top poit ad we icrease t we have to move alog the curve dowwards util we reach t = π at which poit we are ow at the bottom poit. This meas that we will trace out the curve exactly oce i the rage 0 t π. This is ot the oly rage that will trace out the curve however. Note that if we further icrease t from t = π we will ow have to travel back up the curve util we reach t = π ad we are ow back at the top poit. Icreasig t agai util we reach t = π will take us back dow the curve 007 Paul Dawkis

131 util we reach the bottom poit agai, etc. From this aalysis we ca get two more rages of t for oe trace, π t π π t π As you ca probably see there are a ifiite umber of rages of t we could use for oe trace of the curve. Ay of them would be acceptable aswers for this problem. Note that i the process of determiig a rage of t s for oe trace we also maaged to determie the directio of motio for this curve. I the rage 0 t π we had to travel dowwards alog the curve to get from the top poit at t = 0 to the bottom poit at t = π. However, at t = π we are back at the top poit o the curve ad to get there we must travel alog the path. We ca t just jump back up to the top poit or take a differet path to get there. All travel must be doe o the path sketched out. This meas that we had to go back up the path. Further icreasig t takes us back dow the path, the up the path agai etc. I other words, this path is sketched out i both directios because we are ot puttig ay restrictios o the t s ad so we have to assume we are usig all possible values of t. If we had put restrictios o which t s to use we might really have eded up oly movig i oe directio. That however would be a result oly of the rage of t s we are usig ad ot the parametric equatios themselves. Note that we did t really eed to do the above work to determie that the curve traces out i both directios.i this case. Both the x ad y parametric equatios ivolve sie or cosie ad we kow both of those fuctios oscillate. This, i tur meas that both x ad y will oscillate as well. The oly way for that to happe o this particular this curve will be for the curve to be traced out i both directios. Be careful with the above reasoig that the oscillatory ature of sie/cosie forces the curve to be traced out i both directios. It ca oly be used i this example because the startig poit ad edig poit of the curves are i differet places. The oly way to get from oe of the ed poits o the curve to the other is to travel back alog the curve i the opposite directio. Cotrast this with the ellipse i Example 4. I that case we had sie/cosie i the parametric equatios as well. However, the curve oly traced out i oe directio, ot i both directios. I Example 4 we were graphig the full ellipse ad so o matter where we start sketchig the graph we will evetually get back to the startig poit without ever retracig ay portio of the graph. I Example 4 as we trace out the full ellipse both x ad y do i fact oscillate betwee their two edpoits but the curve itself does ot trace out i both directios for this to happe. Basically, we ca oly use the oscillatory ature of sie/cosie to determie that the curve traces out i both directios if the curve starts ad eds at differet poits. If the startig/edig poit is the same the we geerally eed to go through the full derivative argumet to determie the actual directio of motio. So, to fiish this problem out, below is a sketch of the parametric curve. Note that we put directio arrows i both directios to clearly idicate that it would be traced out i both directios. We also put i a few values of t just to help illustrate the directio of motio. 007 Paul Dawkis

132 To this poit we ve see examples that would trace out the complete graph that we got by elimiatig the parameter if we took a large eough rage of t s. However, i the previous example we ve ow see that this will ot always be the case. It is more tha possible to have a set of parametric equatios which will cotiuously trace out just a portio of the curve. We ca usually determie if this will happe by lookig for limits o x ad y that are imposed up us by the parametric equatio. We will ofte use parametric equatios to describe the path of a object or particle. Let s take a look at a example of that. Example 7 The path of a particle is give by the followig set of parametric equatios. x= cos( t) y = + cos ( t) Completely describe the path of this particle. Do this by sketchig the path, determiig limits o x ad y ad givig a rage of t s for which the path will be traced out exactly oce (provide it traces out more tha oce of course). Solutio Elimiatig the parameter this time will be a little differet. We oly have cosies this time ad we ll use that to our advatage. We ca solve the x equatio for cosie ad plug that ito the equatio for y. This gives, cos( ) x x t = y = + = + x 9 This time the algebraic equatio is a parabola that opes upward. We also have the followig limits o x ad y. cos t cos t x 0 cos t + cos t y So, agai we oly trace out a portio of the curve. Here is a quick sketch of the portio of the parabola that the parametric curve will cover. 007 Paul Dawkis

133 Now, as we discussed i the previous example because both the x ad y parametric equatios ivolve cosie we kow that both x ad y must oscillate ad because the start ad ed poits of the curve are ot the same the oly way x ad y ca oscillate is for the curve to trace out i both directios. To fiish the problem the all we eed to do is determie a rage of t s for oe trace. Because the ed poits o the curve have the same y value ad differet x values we ca use the x parametric equatio to determie these values. Here is that work. ( t) ( t) x = : = cos = cos t = 0+ π t = π = 0, ±, ±, ±, ( t) ( t) x = : = cos = cos t = π + π t = π + π = 0, ±, ±, ±, So, we will be at the right ed poit at t =, π, π,0, π, π, ad we ll be at the left ed poit at t =, π, π, π, π,. So, i this case there are a ifiite umber of rages of t s for oe trace. Here are a few of them. π t 0 0 t π π t π Here is a fial sketch of the particle s path with a few value of t o it. 007 Paul Dawkis 4

134 We should give a small warig at this poit. Because of the ideas ivolved i them we cocetrated o parametric curves that retraced portios of the curve more tha oce. Do ot, however, get too locked ito the idea that this will always happe. May, if ot most parametric curves will oly trace out oce. The first oe we looked at is a good example of this. That parametric curve will ever repeat ay portio of itself. There is oe fial topic to be discussed i this sectio before movig o. So far we ve started with parametric equatios ad elimiated the parameter to determie the parametric curve. However, there are times i which we wat to go the other way. Give a fuctio or equatio we might wat to write dow a set of parametric equatios for it. I these cases we say that we parameterize the fuctio. If we take Examples 4 ad 5 as examples we ca do this for ellipses (ad hece circles). Give the ellipse x y + = a b a set of parametric equatios for it would be, x= acost y = bsi t This set of parametric equatios will trace out the ellipse startig at the poit ( a,0) ad will trace i a couter-clockwise directio ad will trace out exactly oce i the rage 0 t π. This is a fairly importat set of parametric equatios as it used cotiually i some subjects with dealig with ellipses ad/or circles. Every curve ca be parameterized i more tha oe way. Ay of the followig will also parameterize the same ellipse. ( ω ) si ( ω ) ( ω ) cos( ω ) ( ω ) si ( ω ) x= acos t y = b t x= asi t y = b t x= acos t y = b t The presece of the ω will chage the speed that the ellipse rotates as we saw i Example 5. Note as well that the last two will trace out ellipses with a clockwise directio of motio (you might wat to verify this). Also ote that they wo t all start at the same place (if we thik of t = 0 as the startig poit that is). There are may more parameterizatios of a ellipse of course, but you get the idea. It is importat to remember that each parameterizatio will trace out the curve oce with a potetially differet rage of t s. Each parameterizatio may rotate with differet directios of motio ad may start at differet poits. You may fid that you eed a parameterizatio of a ellipse that starts at a particular place ad has a particular directio of motio ad so you ow kow that with some work you ca write dow a set of parametric equatios that will give you the behavior that you re after. Now, let s write dow a couple of other importat parameterizatios ad all the commets about directio of motio, startig poit, ad rage of t s for oe trace (if applicable) are still true. 007 Paul Dawkis 5

135 First, because a circle is othig more tha a special case of a ellipse we ca use the parameterizatio of a ellipse to get the parametric equatios for a circle cetered at the origi of radius r as well. Oe possible way to parameterize a circle is, x= rcost y = rsi t Fially, eve though there may ot seem to be ay reaso to, we ca also parameterize fuctios y f x x= h y. I these cases we parameterize them i the followig way, i the form = or x= t x= ht y = f t y = t At this poit it may ot seem all that useful to do a parameterizatio of a fuctio like this, but there are may istaces where it will actually be easier, or it may eve be required, to work with the parameterizatio istead of the fuctio itself. Ufortuately, almost all of these istaces occur i a Calculus III course. 007 Paul Dawkis 6

136 Tagets with Parametric Equatios I this sectio we wat to fid the taget lies to the parametric equatios give by, x= f t y = g t To do this let s first recall how to fid the taget lie to y F( x) lie is give by, dy y = F a + m x a m= = F a dx =, where = at x= a. Here the taget Now, otice that if we could figure out how to get the derivative dy from the parametric dx equatios we could simply reuse this formula sice we will be able to use the parametric equatios to fid the x ad y coordiates of the poit. So, just for a secod let s suppose that we were able to elimiate the parameter from the parametric form ad write the parametric equatios i the form y = F( x). Now, plug the parametric equatios i for x ad y. Yes, it seems silly to elimiate the parameter, the immediately put it back i, but it s what we eed to do i order to get our hads o the derivative. Doig this gives, g t = F f t Now, differetiate with respect to t ad otice that we ll eed to use the Chai Rule o the right had side. g t = F f t f t Let s do aother chage i otatio. We eed to be careful with our derivatives here. Derivatives of the lower case fuctio are with respect to t while derivatives of upper case fuctios are with respect to x. So, to make sure that we keep this straight let s rewrite thigs as follows. dy dx = F ( x) dt dt At this poit we should remid ourselves just what we are after. We eeded a formula for dy dx or F ( x) that is i terms of the parametric formulas. Notice however that we ca get that from the above equatio. 007 Paul Dawkis 7 x a dy dy dx = dt, provided 0 dx dx dt dt Notice as well that this will be a fuctio of t ad ot x.

137 As a aside, otice that we could also get the followig formula with a similar derivatio if we eeded to, Derivative for Parametric Equatios dx dx dy = dt, provided 0 dy dy dt dt Why would we wat to do this? Well, recall that i the arc legth sectio of the Applicatios of Itegral sectio we actually eeded this derivative o occasio. So, let s fid a taget lie. Example Fid the taget lie(s) to the parametric curve give by 5 x= t 4t y = t at (0,4). Solutio Note that there is apparetly the potetial for more tha oe taget lie here! We will look ito this more after we re doe with the example. The first thig that we should do is fid the derivative so we ca get the slope of the taget lie. dy dy t = dt = = 4 dx dx 5t t 5t t dt At this poit we ve got a small problem. The derivative is i terms of t ad all we ve got is a x-y coordiate pair. The ext step the is to determie the value(s) of t which will give this poit. We fid these by pluggig the x ad y values ito the parametric equatios ad solvig for t. 5 0 = t 4t = t t 4 t = 0, ± 4= t t =± Ay value of t which appears i both lists will give the poit. So, sice there are two values of t that give the poit we will i fact get two taget lies. That s defiitely ot somethig that happeed back i Calculus I ad we re goig to eed to look ito this a little more. However, before we do that let s actually get the taget lies. t = Sice we already kow the x ad y-coordiates of the poit all that we eed to do is fid the slope of the taget lie. dy m = = dx t = Paul Dawkis 8

138 The taget lie (at t = ) is the, y = 4 x 8 t = Agai, all we eed is the slope. The taget lie (at t = ) is the, dy = = m dx t = y = 4 + x 8 8 Now, let s take a look at just how we could possibly get two tagets lies at a poit. This was defiitely ot possible back i Calculus I where we first ra across taget lies. A quick graph of the parametric curve will explai what is goig o here. So, the parametric curve crosses itself! That explais how there ca be more tha oe taget lie. There is oe taget lie for each istace that the curve goes through the poit. The ext topic that we eed to discuss i this sectio is that of horizotal ad vertical tagets. We ca easily idetify where these will occur (or at least the t s that will give them) by lookig at the derivative formula. dy dy = dt dx dx dt Horizotal tagets will occur where the derivative is zero ad that meas that we ll get horizotal taget at values of t for which we have, 007 Paul Dawkis 9

139 Horizotal Taget for Parametric Equatios dy dx = 0, provided 0 dt dt Vertical tagets will occur where the derivative is ot defied ad so we ll get vertical tagets at values of t for which we have, Vertical Taget for Parametric Equatios dx dy = 0, provided 0 dt dt Let s take a quick look at a example of this. Example Determie the x-y coordiates of the poits where the followig parametric equatios will have horizotal or vertical tagets. x= t t y = t 9 Solutio We ll first eed the derivatives of the parametric equatios. dx dy = t = ( t ) = 6t dt dt Horizotal Tagets We ll have horizotal tagets where, 6t = 0 t = 0 Now, this is the value of t which gives the horizotal tagets ad we were asked to fid the x-y coordiates of the poit. To get these we just eed to plug t ito the parametric equatios. Therefore, the oly horizotal taget will occur at the poit (0,-9). Vertical Tagets I this case we eed to solve, t = 0 t =± The two vertical tagets will occur at the poits (,-6) ad (-,-6). For the sake of completeess ad at least partial verificatio here is the sketch of the parametric curve. 007 Paul Dawkis 0

140 The fial topic that we eed to discuss i this sectio really is t related to taget lies, but does fit i icely with the derivatio of the derivative that we eeded to get the slope of the taget lie. Before movig ito the ew topic let s first remid ourselves of the formula for the first derivative ad i the process rewrite it slightly. d dy d = ( y) = dt dx dx dx dt Writte i this way we ca see that the formula actually tells us how to differetiate a fuctio y (as a fuctio of t) with respect to x (whe x is also a fuctio of t) whe we are usig parametric equatios. Now let s move oto the fial topic of this sectio. We would also like to kow how to get the secod derivative of y with respect to x. d y dx ( y ) Gettig a formula for this is fairly simple if we remember the rewritte formula for the first derivative above. Secod Derivative for Parametric Equatios d dy d y d dy dt dx = = dx dx dx dx dt 007 Paul Dawkis

141 Note that, d y dx d y dt d x dt Let s work a quick example. Example Fid the secod derivative for the followig set of parametric equatios. 5 x= t 4t y = t Solutio This is the set of parametric equatios that we used i the first example ad so we already have the followig computatios completed. dy dx 4 dy = t = 5t t = dt dt dx 5t t We will first eed the followig, d ( 5t ) 4 0t = = dt 5t t 5t t 5t t The secod derivative is the, d y dt dx dx d dy = dx dt 4 0t = = ( 5t t) 4 5t t 4 0t 4 ( 5t t )( 5t t) 4 0t = t t t ( 5 ) So, why would we wat the secod derivative? Well, recall from your Calculus I class that with the secod derivative we ca determie where a curve is cocave up ad cocave dow. We could do the same thig with parametric equatios if we wated to. 007 Paul Dawkis

142 Example 4 Determie the values of t for which the parametric curve give by the followig set of parametric equatios is cocave up ad cocave dow. 7 5 x= t y = t + t Solutio To compute the secod derivative we ll first eed the followig. 6 4 dy 6 4 dx dy 7t + 5t 5 = 7t + 5t = t = = ( 7t + 5t ) dt dt dx t Note that we ca also use the first derivative above to get some iformatio about the icreasig/decreasig ature of the curve as well. I this case it looks like the parametric curve will be icreasig if t < 0 ad decreasig if t > 0. Now let s move o to the secod derivative. 4 ( 5 t + 5 t d y ) = = 5t + 5t dx t 4 It s clear, hopefully, that the secod derivative will oly be zero at t = 0. Usig this we ca see that the secod derivative will be egative if t < 0 ad positive if t > 0. So the parametric curve will be cocave dow for t < 0 ad cocave up for t > 0. Here is a sketch of the curve for completeess sake. 007 Paul Dawkis

143 Area with Parametric Equatios I this sectio we will fid a formula for determiig the area uder a parametric curve give by the parametric equatios, x= f ( t) y = g( t) We will also eed to further add i the assumptio that the curve is traced out exactly oce as t icreases from α to β. We will do this i much the same way that we foud the first derivative i the previous sectio. y = F x o a x b. We will first recall how to fid the area uder b A F ( x) dx = We will ow thik of the parametric equatio x f ( t) also assume that a = f ( α ) ad b f ( β ) a = as a substitutio i the itegral. We will = for the purposes of this formula. There is actually o reaso to assume that this will always be the case ad so we ll give a correspodig formula later b f α a= f β ). if it s the opposite case ( = ad So, if this is goig to be a substitutio we ll eed, dx = f t dt Pluggig this ito the area formula above ad makig sure to chage the limits to their correspodig t values gives us, β A = F ( f ( t) ) f ( t) dt Sice we do t kow what F(x) is we ll use the fact that y = F( x) = F f ( t) = g t ad we arrive at the formula that we wat. α Area Uder Parametric Curve, Formula I β A = g ( t) f ( t) dt Now, if we should happe to have b= f ( α ) ad a f ( β ) Area Uder Parametric Curve, Formula II α A = g ( t) f ( t) dt Let s work a example. α β = the formula would be, 007 Paul Dawkis 4

144 Example Determie the area uder the parametric curve give by the followig parametric equatios. x= 6 θ siθ y = 6 cosθ 0 θ π Solutio First, otice that we ve switched the parameter to θ for this problem. This is to make sure that we do t get too locked ito always havig t as the parameter. Now, we could graph this to verify that the curve is traced out exactly oce for the give rage if we wated to. We are goig to be lookig at this curve i more detail after this example so we wo t sketch its graph here. There really is t too much to this example other tha pluggig the parametric equatios ito the formula. We ll first eed the derivative of the parametric equatio for x however. dx 6( cosθ ) dθ = The area is the, ( θ) π A= 6 cos 0 π = 6 cos + cos 0 0 θ θ θ π = 6 cosθ + cos( θ) dθ = 6 θ siθ + si ( θ) 4 = 08π dθ d π 0 The parametric curve (without the limits) we used i the previous example is called a cycloid. I its geeral form the cycloid is, ( θ siθ) ( cosθ) x= r y = r The cycloid represets the followig situatio. Cosider a wheel of radius r. Let the poit where the wheel touches the groud iitially be called P. The start rollig the wheel to the right. As the wheel rolls to the right trace out the path of the poit P. The path that the poit P traces out is called a cycloid ad is give by the equatios above. I these equatios we ca thik of θ as the agle through which the poit P has rotated. Here is a cycloid sketched out with the wheel show at various places. The blue dot is the poit P o the wheel that we re usig to trace out the curve. 007 Paul Dawkis 5

145 From this sketch we ca see that oe arch of the cycloid is traced out i the rage 0 θ π. This makes sese whe you cosider that the poit P will be back o the groud after it has rotated through a agle of π. 007 Paul Dawkis 6

146 Arc Legth with Parametric Equatios I the previous two sectios we ve looked at a couple of Calculus I topics i terms of parametric equatios. We ow eed to look at a couple of Calculus II topics i terms of parametric equatios. I this sectio we will look at the arc legth of the parametric curve give by, x= f t y = g t α t β We will also be assumig that the curve is traced out exactly oce as t icreases from α to β. We will also eed to assume that the curve is traced out from left to right as t icreases. This is equivalet to sayig, dx 0 for α t β dt So, let s start out the derivatio by recallig the arc legth formula as we first derived it i the arc legth sectio of the Applicatios of Itegrals chapter. L = ds where, dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy We will use the first ds above because we have a ice formula for the derivative i terms of the parametric equatios (see the Tagets with Parametric Equatios sectio). To use this we ll also eed to kow that, dx dx = f ( t) dt = dt dt The arc legth formula the becomes, β β dy dy dx dt dx L dt = dt + dt dx = + dt dx dt dt α dt α This is a particularly upleasat formula. However, if we factor out the deomiator from the square root we arrive at, 007 Paul Dawkis 7

147 β dx dy dx L = + dt dx dt dt dt α dt Now, makig use of our assumptio that the curve is beig traced out from left to right we ca drop the absolute value bars o the derivative which will allow us to cacel the two derivatives that are outside the square root ad this gives, Arc Legth for Parametric Equatios β dx dy L = + dt dt α dt Notice that we could have used the secod formula for ds above if we had assumed istead that dy 0 for α t β dt If we had goe this route i the derivatio we would have gotte the same formula. Let s take a look at a example. Example Determie the legth of the parametric curve give by the followig parametric equatios. x= si ( t) y = cos( t) 0 t π Solutio We kow that this is a circle of radius cetered at the origi from our prior discussio about graphig parametric curves. We also kow from this discussio that it will be traced out exactly oce i this rage. So, we ca use the formula we derived above. We ll first eed the followig, dx dy = cos( t) = si ( t) dt dt The legth is the, 9cos π L = 9si t + t dt 0 π 0 π 0 = si t + cos t dt = = 6π dt Sice this is a circle we could have just used the fact that the legth of the circle is just the circumferece of the circle. This is a ice way, i this case, to verify our result. 007 Paul Dawkis 8

148 Let s take a look at oe possible cosequece if a curve is traced out more tha oce ad we try to fid the legth of the curve without takig this ito accout. Example Use the arc legth formula for the followig parametric equatios. x= si t y = cos t 0 t π Solutio Notice that this is the idetical circle that we had i the previous example ad so the legth is still 6π. However, for the rage give we kow it will trace out the curve three times istead oce as required for the formula. Despite that restrictio let s use the formula ayway ad see what happes. I this case the derivatives are, dx dy = 9cos t = 9si t dt dt ad the legth formula gives, π L = 8si t + 8cos t dt = 0 π 0 = 8π 9 dt The aswer we got form the arc legth formula i this example was times the actual legth. Recallig that we also determied that this circle would trace out three times i the rage give, the aswer should make some sese. If we had wated to determie the legth of the circle for this set of parametric equatios we would eed to determie a rage of t for which this circle is traced out exactly oce. This is, 0 t π. Usig this rage of t we get the followig for the legth. which is the correct aswer. π 0 π 0 = 6π L = t + t dt = 8si 8cos 9 dt Be careful to ot make the assumptio that this is always what will happe if the curve is traced out more tha oce. Just because the curve traces out times does ot mea that the arc legth formula will give us times the actual legth of the curve! Before movig o to the ext sectio let s otice that we ca put the arc legth formula derived i this sectio ito the same form that we had whe we first looked at arc legth. The oly differece is that we will add i a defiitio for ds whe we have parametric equatios. The arc legth formula ca be summarized as, L = ds 007 Paul Dawkis 9

149 where, dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy dx dy ds = + dt if x = f ( t), y = g ( t), α t β dt dt 007 Paul Dawkis 40

150 Surface Area with Parametric Equatios I this fial sectio of lookig at calculus applicatios with parametric equatios we will take a look at determiig the surface area of a regio obtaied by rotatig a parametric curve about the x or y-axis. We will rotate the parametric curve give by, x= f ( t) y = g( t) α t β about the x or y-axis. We are goig to assume that the curve is traced out exactly oce as t icreases from α to β. At this poit there actually is t all that much to do. We kow that the surface area ca be foud by usig oe of the followig two formulas depedig o the axis of rotatio (recall the Surface Area sectio of the Applicatios of Itegrals chapter). S = π y ds rotatio about x axis S = π x ds rotatio about y axis All that we eed is a formula for ds to use ad from the previous sectio we have, dx dy ds = + dt if x = f ( t), y = g ( t), α t β dt dt which is exactly what we eed. We will eed to be careful with the x or y that is i the origial surface area formula. Back whe we first looked at surface area we saw that sometimes we had to substitute for the variable i the itegral ad at other times we did t. This was depedet upo the ds that we used. I this case however, we will always have to substitute for the variable. The ds that we use for parametric equatios itroduces a dt ito the itegral ad that meas that everythig eeds to be i terms of t. Therefore, we will eed to substitute the appropriate parametric equatio for x or y depedig o the axis of rotatio. Let s take a quick look at a example. Example Determie the surface area of the solid obtaied by rotatig the followig parametric curve about the x-axis. π x= cos θ y = si θ 0 θ Solutio We ll first eed the derivatives of the parametric equatios. dx dy = cos θsiθ = si θcosθ dt dt Before pluggig ito the surface area formula let s get the ds out of the way. 007 Paul Dawkis 4

151 4 4 ds = 9 cos θ si θ + 9si θ cos θ dθ = cos si cos + si θ θ θ θ = cosθsiθ Notice that we could drop the absolute value bars sice both sie ad cosie are positive i this rage of θ give. Now let s get the surface area ad do t forget to also plug i for the y. S = π y ds = π si 0 cos si π 6 4 si 0 cos d u si = 6π π θ θ θ dθ = π θ θ θ = θ = 6π 5 4 u du Paul Dawkis 4

152 Polar Coordiates Up to this poit we ve dealt exclusively with the Cartesia (or Rectagular, or x-y) coordiate system. However, as we will see, this is ot always the easiest coordiate system to work i. So, i this sectio we will start lookig at the polar coordiate system. Coordiate systems are really othig more tha a way to defie a poit i space. For istace i the Cartesia coordiate system at poit is give the coordiates (x,y) ad we use this to defie the poit by startig at the origi ad the movig x uits horizotally followed by y uits vertically. This is show i the sketch below. This is ot, however, the oly way to defie a poit i two dimesioal space. Istead of movig vertically ad horizotally from the origi to get to the poit we could istead go straight out of the origi util we hit the poit ad the determie the agle this lie makes with the positive x- axis. We could the use the distace of the poit from the origi ad the amout we eeded to rotate from the positive x-axis as the coordiates of the poit. This is show i the sketch below. Coordiates i this form are called polar coordiates. The above discussio may lead oe to thik that r must be a positive umber. However, we also, π. allow r to be egative. Below is a sketch of the two poits 6, π ad Paul Dawkis 4

153 From this sketch we ca see that if r is positive the poit will be i the same quadrat as θ. O the other had if r is egative the poit will ed up i the quadrat exactly opposite θ. Notice as, π 7, π do. The well that the coordiates ( 6 ) describe the same poit as the coordiates ( 6 ) 7 coordiates, π 6 tells us to rotate a agle of 7 6 π from the positive x-axis, this would put us o the dashed lie i the sketch above, ad the move out a distace of. This leads to a importat differece betwee Cartesia coordiates ad polar coordiates. I Cartesia coordiates there is exactly oe set of coordiates for ay give poit. With polar coordiates this is t true. I polar coordiates there is literally a ifiite umber of coordiates for a give poit. For istace, the followig four poits are all coordiates for the same poit , π 5, π 5, π 5, π = = = Here is a sketch of the agles used i these four sets of coordiates. I the secod coordiate pair we rotated i a clock-wise directio to get to the poit. We should t forget about rotatig i the clock-wise directio. Sometimes it s what we have to do. 007 Paul Dawkis 44

154 The last two coordiate pairs use the fact that if we ed up i the opposite quadrat from the poit we ca use a egative r to get back to the poit ad of course there is both a couter clock-wise ad a clock-wise rotatio to get to the agle. These four poits oly represet the coordiates of the poit without rotatig aroud the system more tha oce. If we allow the agle to make as may complete rotatios about the axis system as we wat the there are a ifiite umber of coordiates for the same poit. I fact the poit r, θ ca be represeted by ay of the followig coordiate pairs. ( θ + π ) ( θ + ( + ) π) r, r,, where is ay iteger. Next we should talk about the origi of the coordiate system. I polar coordiates the origi is ofte called the pole. Because we are t actually movig away from the origi/pole we kow that r = 0. However, we ca still rotate aroud the system by ay agle we wat ad so the 0,θ. coordiates of the origi/pole are Now that we ve got a grasp o polar coordiates we eed to thik about covertig betwee the two coordiate systems. Well start out with the followig sketch remidig us how both coordiate systems work. Note that we ve got a right triagle above ad with that we ca get the followig equatios that will covert polar coordiates ito Cartesia coordiates. Polar to Cartesia Coversio Formulas x= rcosθ y = rsiθ Covertig from Cartesia is almost as easy. Let s first otice the followig. x + y = ( rcosθ) + ( rsiθ) = r cos θ + r si θ = r cos θ + si θ = r This is a very useful formula that we should remember, however we are after a equatio for r so let s take the square root of both sides. This gives, r = x + y 007 Paul Dawkis 45

155 Note that techically we should have a plus or mius i frot of the root sice we kow that r ca be either positive or egative. We will ru with the covetio of positive r here. Gettig a equatio for θ is almost as simple. We ll start with, y rsiθ = = taθ x rcosθ Takig the iverse taget of both sides gives, y θ = ta x We will eed to be careful with this because iverse tagets oly retur values i the rage π π < θ <. Recall that there is a secod possible agle ad that the secod agle is give by θ + π. Summarizig the gives the followig formulas for covertig from Cartesia coordiates to polar coordiates. Cartesia to Polar Coversio Formulas r = x + y r = x + y Let s work a quick example. θ ta y x = Example Covert each of the followig poits ito the give coordiate system. π (a) 4, ito Cartesia coordiates. [Solutio] (b) (-,-) ito polar coordiates. [Solutio] Solutio (a) Covert π 4, ito Cartesia coordiates. This coversio is easy eough. All we eed to do is plug the poits ito the formulas. π x = 4cos = 4 = π y = 4si = 4 = So, i Cartesia coordiates this poit is (, ). [Retur to Problems] 007 Paul Dawkis 46

156 (b) Covert (-,-) ito polar coordiates. Let s first get r. r = + = Now, let s get θ. π θ = ta = ta = 4 This is ot the correct agle however. This value of θ is i the first quadrat ad the poit we ve bee give is i the third quadrat. As oted above we ca get the correct agle by addig π oto this. Therefore, the actual agle is, π 5π θ = + π = 4 4, π. Note as well that we could have used the first θ that we got by usig a egative r. I this case the poit could also be writte i polar coordiates, π. 5 So, i polar coordiates the poit is ( 4 ) as ( 4 ) [Retur to Problems] We ca also use the above formulas to covert equatios from oe coordiate system to the other. Example Covert each of the followig ito a equatio i the give coordiate system. (a) Covert x 5x = + xy ito polar coordiates. [Solutio] (b) Covert r = 8cosθ ito Cartesia coordiates. [Solutio] Solutio (a) Covert x 5x = + xy ito polar coordiates. I this case there really is t much to do other tha pluggig i the formulas for x ad y (i.e. the Cartesia coordiates) i terms of r ad θ (i.e. the polar coordiates). rcosθ 5 rcosθ = + rcosθ rsiθ (b) Covert rcos 5r cos = + r cos si θ θ θ θ r = 8cosθ ito Cartesia coordiates. [Retur to Problems] This oe is a little trickier, but ot by much. First otice that we could substitute straight for the r. However, there is o straight substitutio for the cosie that will give us oly Cartesia coordiates. If we had a r o the right alog with the cosie the we could do a direct substitutio. So, if a r o the right side would be coveiet let s put oe there, just do t forget to put oe o the left side as well. r = 8rcosθ We ca ow make some substitutios that will covert this ito Cartesia coordiates. x + y = 8x [Retur to Problems] 007 Paul Dawkis 47

157 Before movig o to the ext subject let s do a little more work o the secod part of the previous example. The equatio give i the secod part is actually a fairly well kow graph; it just is t i a form that most people will quickly recogize. To idetify it let s take the Cartesia coordiate equatio ad do a little rearragig. x + 8x+ y = 0 Now, complete the square o the x portio of the equatio. x + 8x+ 6 + y = 6 ( x ) So, this was a circle of radius 4 ad ceter (-4,0). This leads us ito the fial topic of this sectio y = 6 Commo Polar Coordiate Graphs Let s idetify a few of the more commo graphs i polar coordiates. We ll also take a look at a couple of special polar graphs. Lies Some lies have fairly simple equatios i polar coordiates.. θ = β. We ca see that this is a lie by covertig to Cartesia coordiates as follows θ = β y ta = β x y = ta β x y = ( ta β ) x This is a lie that goes through the origi ad makes a agle of β with the positive x- axis. Or, i other words it is a lie through the origi with slope of ta β.. rcosθ = a This is easy eough to covert to Cartesia coordiates to x= a. So, this is a vertical lie.. rsiθ = b Likewise, this coverts to y = b ad so is a horizotal lie. 007 Paul Dawkis 48

158 π Example Graph θ =, r cosθ = 4 ad r siθ = o the same axis system. 4 Solutio There really is t too much to this oe other tha doig the graph so here it is. Circles Let s take a look at the equatios of circles i polar coordiates.. r = a. This equatio is sayig that o matter what agle we ve got the distace from the origi must be a. If you thik about it that is exactly the defiitio of a circle of radius a cetered at the origi. So, this is a circle of radius a cetered at the origi. This is also oe of the reasos why we might wat to work i polar coordiates. The equatio of a circle cetered at the origi has a very ice equatio, ulike the correspodig equatio i Cartesia coordiates.. r = acosθ. We looked at a specific example of oe of these whe we were covertig equatios to Cartesia coordiates. This is a circle of radius a ad ceter ( a,0). Note that a might be egative (as it was i our example above) ad so the absolute value bars are required o the radius. They should ot be used however o the ceter.. r = bsiθ. 0,b. This is similar to the previous oe. It is a circle of radius b ad ceter 4. r = acosθ + bsiθ. This is a combiatio of the previous two ad by completig the square twice it ca be show that this is a circle of radius a + b ad ceter ( ab, ). I other words, this is the geeral equatio of a circle that is t cetered at the origi. 007 Paul Dawkis 49

159 Example 4 Graph r = 7, r = 4cosθ, ad r = 7siθ o the same axis system. Solutio The first oe is a circle of radius 7 cetered at the origi. The secod is a circle of radius cetered at (,0). The third is a circle of radius 7 cetered at 7 0,. Here is the graph of the three equatios. Note that it takes a rage of 0 θ π for a complete graph of r = a ad it oly takes a rage of 0 θ π to graph the other circles give here. Cardioids ad Limacos These ca be broke up ito the followig three cases.. Cardioids : r = a± acosθ ad r = a± asiθ. These have a graph that is vaguely heart shaped ad always cotai the origi.. Limacos with a ier loop : r = a± bcosθ ad r = a± bsiθ with a< b. These will have a ier loop ad will always cotai the origi.. Limacos without a ier loop : r = a± bcosθ ad r = a± bsiθ with a > b. These do ot have a ier loop ad do ot cotai the origi. Example 5 Graph r = 5 5siθ, r = 7 6cosθ, ad r = + 4cosθ. Solutio These will all graph out oce i the rage 0 θ π. Here is a table of values for each followed by graphs of each. 007 Paul Dawkis 50

160 θ r = 5 5siθ r = 7 6cosθ r = + 4cosθ π 0 7 π 5 - π 0 7 π Paul Dawkis 5

161 There is oe fial thig that we eed to do i this sectio. I the third graph i the previous example we had a ier loop. We will, o occasio, eed to kow the value of θ for which the graph will pass through the origi. To fid these all we eed to do is set the equatio equal to zero ad solve as follows, π 4π 0 = + 4cosθ cos θ = θ =, 007 Paul Dawkis 5

162 Tagets with Polar Coordiates We ow eed to discuss some calculus topics i terms of polar coordiates. We will start with fidig taget lies to polar curves. I this case we are goig to assume that the equatio is i the form r = f ( θ ). With the equatio i this form we ca actually use the equatio for the derivative dy we derived whe we looked at taget lies with parametric dx equatios. To do this however requires us to come up with a set of parametric equatios to represet the curve. This is actually pretty easy to do. From our work i the previous sectio we have the followig set of coversio equatios for goig from polar coordiates to Cartesia coordiates. x= rcosθ y = rsiθ =. Now, we ll use the fact that we re assumig that the equatio is i the form r f ( θ ) Substitutig this ito these equatios gives the followig set of parametric equatios (with θ as the parameter) for the curve. cos x= f θ θ y = f θ siθ Now, we will eed the followig derivatives. dx dy = f ( θ) cosθ f ( θ) siθ = f ( θ) siθ + f ( θ) cosθ dθ dθ dr dr = cosθ rsiθ = siθ + rcosθ dθ dθ The derivative dy dx is the, Derivative with Polar Coordiates dy = dx dr siθ + r cosθ dθ dr cosθ r siθ dθ Note that rather tha tryig to remember this formula it would probably be easier to remember how we derived it ad just remember the formula for parametric equatios. Let s work a quick example with this. 007 Paul Dawkis 5

163 Example Determie the equatio of the taget lie to r = + 8siθ at Solutio We ll first eed the followig derivative. The formula for the derivative dy dx becomes, dy dx dr 8cosθ dθ = π θ =. 6 8cosθ siθ + + 8siθ cosθ 6 cosθ siθ + cosθ = = 8cos θ + 8siθ siθ 8cos θ siθ 8si θ The slope of the taget lie is, 4 + dy m = = = dx π θ = Now, at θ = π 6 we have r = 7. We ll eed to get the correspodig x-y coordiates so we ca get the taget lie. π 7 π 7 x= 7 cos = y = 7si = 6 6 The taget lie is the, 7 7 y = + x 5 For the sake of completeess here is a graph of the curve ad the taget lie. 007 Paul Dawkis 54

164 Area with Polar Coordiates I this sectio we are goig to look at areas eclosed by polar curves. Note as well that we said eclosed by istead of uder as we typically have i these problems. These problems work a little differetly i polar coordiates. Here is a sketch of what the area that we ll be fidig i this sectio looks like. We ll be lookig for the shaded area i the sketch above. The formula for fidig this area is, A= β α r d Notice that we use r i the itegral istead of f ( θ ) so make sure ad substitute accordigly whe doig the itegral. Let s take a look at a example. Example Determie the area of the ier loop of r = + 4cosθ. Solutio We graphed this fuctio back whe we first started lookig at polar coordiates. For this problem we ll also eed to kow the values of θ where the curve goes through the origi. We ca get these by settig the equatio equal to zero ad solvig. 0 = + 4cosθ π 4π cos θ = θ =, Here is the sketch of this curve with the ier loop shaded i. θ 007 Paul Dawkis 55

165 Ca you see why we eeded to kow the values of θ where the curve goes through the origi? These poits defie where the ier loop starts ad eds ad hece are also the limits of itegratio i the formula. So, the area is the, 4π A= + 4cos π 4π π 4π π 4π π ( θ) = ( 4 + 6cos θ + 6cos θ) dθ ( ) = + 8cosθ cos θ dθ = 6 + 8cosθ + 4cos θ dθ ( 6θ 8siθ si ( θ) ) = + + = 4π 6 =.74 dθ 4π π You did follow the work doe i this itegral did t you? You ll ru ito quite a few itegrals of trig fuctios i this sectio so if you eed to you should go back to the Itegrals Ivolvig Trig Fuctios sectios ad do a quick review. So, that s how we determie areas that are eclosed by a sigle curve, but what about situatios like the followig sketch were we wat to fid the area betwee two curves. 007 Paul Dawkis 56

166 I this case we ca use the above formula to fid the area eclosed by both ad the the actual area is the differece betwee the two. The formula for this is, Let s take a look at a example of this. β A= ( r r ) dθ o i α Example Determie the area that lies iside r = + siθ ad outside r =. Solutio Here is a sketch of the regio that we are after. To determie this area we ll eed to kow the values of θ for which the two curves itersect. We ca determie these poits by settig the two equatios ad solvig. 007 Paul Dawkis 57

167 + siθ = 7π π si θ = θ =, 6 6 Here is a sketch of the figure with these agles added. Note as well here that we also ackowledged that aother represetatio for the agle 6π is π. This is importat for this problem. I order to use the formula above the area must be 6 eclosed as we icrease from the smaller to larger agle. So, if we use 7 π 6 to π 6 we will ot eclose the shaded area, istead we will eclose the bottom most of the three regios. However, if we use the agles π 6 to 7 π 6 we will eclose the area that we re after. So, the area is the, 7π 6 (( θ) ) A= + si π 6 7π 6 = ( 5 + si θ + 4si θ) dθ π 6 7π 6 = ( 7 + si θ cos ( θ) ) dθ π 6 = 7 cos si ( θ θ ( θ) ) 4π = + = 4.87 dθ 7 6 π π 6 Let s work a slight modificatio of the previous example. 007 Paul Dawkis 58

168 Example Determie the area of the regio outside r = + siθ ad iside r =. Solutio This time we re lookig for the followig regio. So, this is the regio that we get by usig the limits 7 π to π. The area for this regio is, 6 6 π 6 7π 6 π 7π 6 π 6 7π 6 ( ( θ) ) A= + si 6 = ( 5 siθ 4si θ) dθ = ( 7 siθ + cos( θ) ) dθ = + + ( 7θ cosθ si ( θ) ) 7π = =.96 dθ Notice that for this area the outer ad ier fuctio were opposite! Let s do oe fial modificatio of this example. Example 4 Determie the area that is iside both r = + siθ ad r =. Solutio Here is the sketch for this example. π π 007 Paul Dawkis 59

169 We are ot goig to be able to do this problem i the same fashio that we did the previous two. There is o set of limits that will allow us to eclose this area as we icrease from oe to the other. Remember that as we icrease θ the area we re after must be eclosed. However, the oly two rages for θ that we ca work with eclose the area from the previous two examples ad ot this regio. I this case however, that is ot a major problem. There are two ways to do get the area i this problem. We ll take a look at both of them. Solutio I this case let s otice that the circle is divided up ito two portios ad we re after the upper portio. Also otice that we foud the area of the lower portio i Example. Therefore, the area is, Area = Area of Circle Area from Example = π.96 = 0.70 Solutio I this case we do pretty much the same thig except this time we ll thik of the area as the other portio of the limaco tha the portio that we were dealig with i Example. We ll also eed to actually compute the area of the limaco i this case. So, the area usig this approach is the, 007 Paul Dawkis 60

170 Area = Area of Limaco Area from Example π ( si θ) dθ = + π = ( 9 + si θ + 4si θ) dθ π = ( + si θ cos ( θ) ) dθ π = ( θ cos ( θ) si ( θ) ) 4.87 = π 4.87 = 0.70 A slightly loger approach, but sometimes we are forced to take this loger approach. As this last example has show we will ot be able to get all areas i polar coordiates straight from a itegral Paul Dawkis 6

171 Arc Legth with Polar Coordiates We ow eed to move ito the Calculus II applicatios of itegrals ad how we do them i terms of polar coordiates. I this sectio we ll look at the arc legth of the curve give by, r = f ( θ) α θ β where we also assume that the curve is traced out exactly oce. Just as we did with the taget lies i polar coordiates we ll first write the curve i terms of a set of parametric equatios, x= rcosθ y = rsiθ = f ( θ) cosθ = f ( θ) siθ ad we ca ow use the parametric formula for fidig the arc legth. We ll eed the followig derivatives for these computatios. dx dy = f ( θ) cosθ f ( θ) siθ = f ( θ) siθ + f ( θ) cosθ dθ dθ dr dr = cosθ rsiθ = siθ + rcosθ dθ dθ We ll eed the followig for our ds. dx dy dr dr + = cosθ rsiθ + siθ + rcosθ dθ dθ dθ dθ dr dr = cos θ r cosθsiθ + r si θ dθ dθ dr dr + si θ + r cosθsiθ + r cos θ dθ dθ dr = ( cos θ + si θ) + r ( cos θ + si θ) dθ dr = r + dθ The arc legth formula for polar coordiates is the, where, L = ds dr ds = r + dθ dθ Let s work a quick example of this. 007 Paul Dawkis 6

172 Example Determie the legth of r = θ 0 θ. Solutio Okay, let s just jump straight ito the formula sice this is a fairly simple fuctio. L= θ + dθ We ll eed to use a trig substitutio here. θ = ta x dθ = sec x dx θ = 0 0 = ta x x= 0 θ = = ta x π x= 4 θ + = ta x+ = sec x = sec x = sec x 0 The arc legth is the, L = θ + 0 = π 4 0 sec dθ x dx = sec ta + l sec + ta ( x x x x) ( l ( )) = + + π 4 0 Just as a aside before we leave this chapter. The polar equatio r = θ is the equatio of a spiral. Here is a quick sketch of r = θ for 0 θ 4π. 007 Paul Dawkis 6

173 Surface Area with Polar Coordiates We will be lookig at surface area i polar coordiates i this sectio. Note however that all we re goig to do is give the formulas for the surface area sice most of these itegrals ted to be fairly difficult. We wat to fid the surface area of the regio foud by rotatig, r = f ( θ) α θ β about the x or y-axis. As we did i the taget ad arc legth sectios we ll write the curve i terms of a set of parametric equatios. x= rcosθ y = rsiθ = f θ cosθ = f θ siθ If we ow use the parametric formula for fidig the surface area we ll get, where, S = π y ds rotatio about x axis S = π x ds rotatio about y axis dr ds r d r f = + θ = θ, α θ β dθ Note that because we will pick up a dθ from the ds we ll eed to substitute oe of the parametric equatios i for x or y depedig o the axis of rotatio. This will ofte mea that the itegrals will be somewhat upleasat. 007 Paul Dawkis 64

174 Arc Legth ad Surface Area Revisited We wo t be workig ay examples i this sectio. This sectio is here solely for the purpose of summarizig up all the arc legth ad surface area problems. Over the course of the last two chapters the topic of arc legth ad surface area has arise may times ad each time we got a ew formula out of the mix. Studets ofte get a little overwhelmed with all the formulas. However, there really are t as may formulas as it might seem at first glace. There is exactly oe arc legth formula ad exactly two surface area formulas. These are, L = ds S = π y ds rotatio about x axis S = π x ds rotatio about y axis The problems arise because we have quite a few ds s that we ca use. Agai studets ofte have trouble decidig which oe to use. The examples/problems usually suggest the correct oe to use however. Here is a complete listig of all the ds s that we ve see ad whe they are used. dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy dx dy ds = + dt if x = f ( t), y = g ( t), α t β dt dt dr ds r d r f = + θ if = θ, α θ β dθ Depedig o the form of the fuctio we ca quickly tell which ds to use. There is oly oe other thig to worry about i terms of the surface area formula. The ds will itroduce a ew differetial to the itegral. Before itegratig make sure all the variables are i terms of this ew differetial. For example if we have parametric equatios we ll use the third ds ad the we ll eed to make sure ad substitute for the x or y depedig o which axis we rotate about to get everythig i terms of t. Likewise, if we have a fuctio i the form x h( y) = the we ll use the secod ds ad if the rotatio is about the y-axis we ll eed to substitute for the x i the itegral. O the other had if we rotate about the x-axis we wo t eed to do a substitutio for the y. 007 Paul Dawkis 65

175 Keep these rules i mid ad you ll always be able to determie which formula to use ad how to correctly do the itegral. 007 Paul Dawkis 66

176 Sequeces ad Series Itroductio I this chapter we ll be takig a look at sequeces ad (ifiite) series. Actually, this chapter will deal almost exclusively with series. However, we also eed to uderstad some of the basics of sequeces i order to properly deal with series. We will therefore, sped a little time o sequeces as well. Series is oe of those topics that may studets do t fid all that useful. To be hoest, may studets will ever see series outside of their calculus class. However, series do play a importat role i the field of ordiary differetial equatios ad without series large portios of the field of partial differetial equatios would ot be possible. I other words, series is a importat topic eve if you wo t ever see ay of the applicatios. Most of the applicatios are beyod the scope of most Calculus courses ad ted to occur i classes that may studets do t take. So, as you go through this material keep i mid that these do have applicatios eve if we wo t really be coverig may of them i this class. Here is a list of topics i this chapter. Sequeces We will start the chapter off with a brief discussio of sequeces. This sectio will focus o the basic termiology ad covergece of sequeces More o Sequeces Here we will take a quick look about mootoic ad bouded sequeces. Series The Basics I this sectio we will discuss some of the basics of ifiite series. Series Covergece/Divergece Most of this chapter will be about the covergece/divergece of a series so we will give the basic ideas ad defiitios i this sectio. Series Special Series We will look at the Geometric Series, Telescopig Series, ad Harmoic Series i this sectio. Itegral Test Usig the Itegral Test to determie if a series coverges or diverges. Compariso Test/Limit Compariso Test Usig the Compariso Test ad Limit Compariso Tests to determie if a series coverges or diverges. Alteratig Series Test Usig the Alteratig Series Test to determie if a series coverges or diverges. Absolute Covergece A brief discussio o absolute covergece ad how it differs from covergece. 007 Paul Dawkis 67

177 Ratio Test Usig the Ratio Test to determie if a series coverges or diverges. Root Test Usig the Root Test to determie if a series coverges or diverges. Strategy for Series A set of geeral guidelies to use whe decidig which test to use. Estimatig the Value of a Series Here we will look at estimatig the value of a ifiite series. Power Series A itroductio to power series ad some of the basic cocepts. Power Series ad Fuctios I this sectio we will start lookig at how to fid a power series represetatio of a fuctio. Taylor Series Here we will discuss how to fid the Taylor/Maclauri Series for a fuctio. Applicatios of Series I this sectio we will take a quick look at a couple of applicatios of series. Biomial Series A brief look at biomial series. 007 Paul Dawkis 68

178 Sequeces Let s start off this sectio with a discussio of just what a sequece is. A sequece is othig more tha a list of umbers writte i a specific order. The list may or may ot have a ifiite umber of terms i them although we will be dealig exclusively with ifiite sequeces i this class. Geeral sequece terms are deoted as follows, a first term a secod term th a term a + ( ) st + term Because we will be dealig with ifiite sequeces each term i the sequece will be followed by aother term as oted above. I the otatio above we eed to be very careful with the subscripts. The subscript of + deotes the ext term i the sequece ad NOT oe plus the th term! I other words, a a + + so be very careful whe writig subscripts to make sure that the + does t migrate out of the subscript! This is a easy mistake to make whe you first start dealig with this kid of thig. There is a variety of ways of deotig a sequece. Each of the followig are equivalet ways of deotig a sequece. {,,,,, } { } { } a a a a a a + = I the secod ad third otatios above a is usually give by a formula. A couple of otes are ow i order about these otatios. First, ote the differece betwee the secod ad third otatios above. If the startig poit is ot importat or is implied i some way by the problem it is ofte ot writte dow as we did i the third otatio. Next, we used a startig poit of = i the third otatio oly so we could write oe dow. There is absolutely o reaso to believe that a sequece will start at =. A sequece will start where ever it eeds to start. Let s take a look at a couple of sequeces. 007 Paul Dawkis 69

179 Example Write dow the first few terms of each of the followig sequeces. + (a) (b) = + ( ) b = (c) { } = 0 [Solutio] [Solutio] th, where b digit of π = [Solutio] Solutio + (a) = To get the first few sequece terms here all we eed to do is plug i values of ito the formula give ad we ll get the sequece terms =,,,,, = = = = = 4 = 5 Note the iclusio of the at the ed! This is a importat piece of otatio as it is the oly thig that tells us that the sequece cotiues o ad does t termiate at the last term. [Retur to Problems] + (b) = 0 This oe is similar to the first oe. The mai differece is that this sequece does t start at =. +,,,,, = = 0 Note that the terms i this sequece alterate i sigs. Sequeces of this kid are sometimes called alteratig sequeces. [Retur to Problems] b = (c) { } th, where b = digit of π This sequece is differet from the first two i the sese that it does t have a specific formula for each term. However, it does tell us what each term should be. Each term should be the th digit of π. So we kow that π = The sequece is the, {,,4,,5,9,,6,5,,5, } [Retur to Problems] 007 Paul Dawkis 70

180 I the first two parts of the previous example ote that we were really treatig the formulas as fuctios that ca oly have itegers plugged ito them. Or, + + ( ) f = g = This is a importat idea i the study of sequeces (ad series). Treatig the sequece terms as fuctio evaluatios will allow us to do may thigs with sequeces that could t do otherwise. Before delvig further ito this idea however we eed to get a couple more ideas out of the way. First we wat to thik about graphig a sequece. To graph the sequece { a } we plot the poits ( a, ) as rages over all possible values o a graph. For istace, let s graph the + sequece =. The first few poits o the graph are, 4 5 6,,,,,, 4,, 5,, The graph, for the first 0 terms of the sequece, is the, This graph leads us to a importat idea about sequeces. Notice that as icreases the sequece terms i our sequece, i this case, get closer ad closer to zero. We the say that zero is the limit (or sometimes the limitig value) of the sequece ad write, + lim a = lim = 0 This otatio should look familiar to you. It is the same otatio we used whe we talked about the limit of a fuctio. I fact, if you recall, we said earlier that we could thik of sequeces as fuctios i some way ad so this otatio should t be too surprisig. Usig the ideas that we developed for limits of fuctios we ca write dow the followig workig defiitio for limits of sequeces. 007 Paul Dawkis 7

181 Workig Defiitio of Limit. We say that lim a = L if we ca make a as close to L as we wat for all sufficietly large. I other words, the value of the a s approach L as approaches ifiity.. We say that lim a = if we ca make a as large as we wat for all sufficietly large. Agai, i other words, the value of the a s get larger ad larger without boud as approaches ifiity.. We say that lim a = if we ca make a as large ad egative as we wat for all sufficietly large. Agai, i other words, the value of the a s are egative ad get larger ad larger without boud as approaches ifiity. The workig defiitios of the various sequece limits are ice i that they help us to visualize what the limit actually is. Just like with limits of fuctios however, there is also a precise defiitio for each of these limits. Let s give those before proceedig Precise Defiitio of Limit. We say that lim a = L if for every umber ε > 0 there is a iteger N such that. We say that lim a a L < ε wheever > N. We say that lim a = if for every umber M > 0 there is a iteger N such that a > M wheever > N = if for every umber M < 0 there is a iteger N such that a < M wheever > N We wo t be usig the precise defiitio ofte, but it will show up occasioally. Note that both defiitios tell us that i order for a limit to exist ad have a fiite value all the sequece terms must be gettig closer ad closer to that fiite value as icreases. Now that we have the defiitios of the limit of sequeces out of the way we have a bit of termiology that we eed to look at. If lim a exists ad is fiite we say that the sequece is coverget. If lim a does t exist or is ifiite we say the sequece diverges. Note that sometimes we will say the sequece diverges to if lim a = ad if lim a = we will sometimes say that the sequece diverges to. 007 Paul Dawkis 7

182 Get used to the terms coverget ad diverget as we ll be seeig them quite a bit throughout this chapter. So just how do we fid the limits of sequeces? Most limits of most sequeces ca be foud usig oe of the followig theorems. Theorem Give the sequece { a } if we have a fuctio f ( x ) such that f = a ad lim the lim a = L x f x = L This theorem is basically tellig us that we take the limits of sequeces much like we take the limit of fuctios. I fact, i most cases we ll ot eve really use this theorem by explicitly writig dow a fuctio. We will more ofte just treat the limit as if it were a limit of a fuctio ad take the limit as we always did back i Calculus I whe we were takig the limits of fuctios. So, ow that we kow that takig the limit of a sequece is early idetical to takig the limit of a fuctio we also kow that all the properties from the limits of fuctios will also hold. Properties lim a ± b = lim a ± lim b.. lim ca = c lim a. lim ( ab) = ( lim a)( lim b) 4. a lim a lim =, provided lim b 0 b lim b p 5. lim a = lim a provided a 0 p These properties ca be proved usig Theorem above ad the fuctio limit properties we saw i Calculus I or we ca prove them directly usig the precise defiitio of a limit usig early idetical proofs of the fuctio limit properties. Next, just as we had a Squeeze Theorem for fuctio limits we also have oe for sequeces ad it is pretty much idetical to the fuctio limit versio. Squeeze Theorem for Sequeces a c b for all > N for some N ad lim a = lim b = L the lim c = L. If 007 Paul Dawkis 7

183 Note that i this theorem the for all > N for some N is really just tellig us that we eed to have a c b for all sufficietly large, but if it is t true for the first few that wo t ivalidate the theorem. As we ll see ot all sequeces ca be writte as fuctios that we ca actually take the limit of. This will be especially true for sequeces that alterate i sigs. While we ca always write these sequece terms as a fuctio we simply do t kow how to take the limit of a fuctio like that. The followig theorem will help with some of these sequeces. Theorem If lim a = 0 the lim a = 0. Note that i order for this theorem to hold the limit MUST be zero ad it wo t work for a sequece whose limit is ot zero. This theorem is easy eough to prove so let s do that. Proof of Theorem The mai thig to this proof is to ote that, a a a The ote that, ( a ) lim = lim a = 0 We the have ( a ) lim = lim a = 0 ad so by the Squeeze Theorem we must also have, lim a = 0 The ext theorem is a useful theorem givig the covergece/divergece ad value (for whe it s coverget) of a sequece that arises o occasio. Theorem The sequece { } = 0 r coverges if < r ad diverges for all other values of r. Also, lim r 0 if < r < = if r = Here is a quick (well ot so quick, but defiitely simple) partial proof of this theorem. Partial Proof of Theorem We ll do this by a series of cases although the last case will ot be completely prove. Case : r > We kow from Calculus I that lim r x x = if r > ad so by Theorem above we also kow 007 Paul Dawkis 74

184 that lim r = ad so the sequece diverges if r >. Case : r = I this case we have, lim r = lim = lim = So, the sequece coverges for r = ad i this case its limit is. Case : 0< r < x We kow from Calculus I that lim r = 0 if 0< r < ad so by Theorem above we also kow x that lim r = 0 ad so the sequece coverges if 0< r < ad i this case its limit is zero. Case 4 : r = 0 I this case we have, lim r = lim 0 = lim 0 = 0 So, the sequece coverges for r = 0 ad i this case its limit is zero. Case 5 : < r < 0 First let s ote that if < r < 0 the 0< r < the by Case above we have, lim r = lim r = 0 Theorem above ow tells us that we must also have, lim r = 0 ad so if < r < 0 the sequece coverges ad has a limit of 0. Case 6 : r = I this case the sequece is, { r } 007 Paul Dawkis 75 { } {,,,,,,,, } = 0 = = = 0 = 0 lim ad hopefully it is clear that does t exist. Recall that i order of this limit to exist the terms must be approachig a sigle value as icreases. I this case however the terms just alterate betwee ad - ad so the limit does ot exist. So, the sequece diverges for r =. Case 7 : r < I this case we re ot goig to go through a complete proof. Let s just see what happes if we let r = for istace. If we do that the sequece becomes, { r } { } {,, 4, 8,6,, } = 0 = = = 0 = 0 So, if r = we get a sequece of terms whose values alterate i sig ad get larger ad larger

185 ad so lim does t exist. It does ot settle dow to a sigle value as icreases or do the terms ALL approach ifiity. So, the sequece diverges for r =. We could do somethig similar for ay value of r such that r < ad so the sequece diverges for r <. Let s take a look at a couple of examples of limits of sequeces. Example Determie if the followig sequeces coverge or diverge. If the sequece coverges determie its limit. (a) [Solutio] 0+ 5 = e (b) [Solutio] (c) ( ) (d) = = { } = 0 [Solutio] [Solutio] Solutio (a) 0+ 5 = I this case all we eed to do is recall the method that was developed i Calculus I to deal with the limits of ratioal fuctios. See the Limits At Ifiity, Part I sectio of my Calculus I otes for a review of this if you eed to. To do a limit i this form all we eed to do is factor from the umerator ad deomiator the largest power of, cacel ad the take the limit. lim = lim = lim = So the sequece coverges ad its limit is 5. [Retur to Problems] 007 Paul Dawkis 76

186 (b) e = We will eed to be careful with this oe. We will eed to use L Hospital s Rule o this sequece. The problem is that L Hospital s Rule oly works o fuctios ad ot o sequeces. Normally this would be a problem, but we ve got Theorem from above to help us out. Let s defie x f ( x) = e x ad ote that, f = e Theorem says that all we eed to do is take the limit of the fuctio. x x e e e lim = lim = lim = x x x So, the sequece i this part diverges (to ). More ofte tha ot we just do L Hospital s Rule o the sequece terms without first covertig to x s sice the work will be idetical regardless of whether we use x or. However, we really should remember that techically we ca t do the derivatives while dealig with sequece terms. [Retur to Problems] (c) ( ) = We will also eed to be careful with this sequece. We might be tempted to just say that the limit of the sequece terms is zero (ad we d be correct). However, techically we ca t take the limit of sequeces whose terms alterate i sig, because we do t kow how to do limits of fuctios that exhibit that same behavior. Also, we wat to be very careful to ot rely too much o ituitio with these problems. As we will see i the ext sectio, ad i later sectios, our ituitio ca lead us astray i these problem if we are t careful. So, let s work this oe by the book. We will eed to use Theorem o this problem. To this we ll first eed to compute, ( ) lim = lim = 0 Therefore, sice the limit of the sequece terms with absolute value bars o them goes to zero we kow by Theorem that, ( ) lim = 0 which also meas that the sequece coverges to a value of zero. [Retur to Problems] 007 Paul Dawkis 77

187 { } = 0 (d) ( ) For this theorem ote that all we eed to do is realize that this is the sequece i Theorem above usig r =. So, by Theorem this sequece diverges. [Retur to Problems] We ow eed to give a warig about misusig Theorem. Theorem oly works if the limit is zero. If the limit of the absolute value of the sequece terms is ot zero the the theorem will ot hold. The last part of the previous example is a good example of this (ad i fact this warig is the whole reaso that part is there). Notice that lim = lim = ad yet, lim does t eve exist let aloe equal. So, be careful usig this Theorem. You must always remember that it oly works if the limit is zero. Before movig oto the ext sectio we eed to give oe more theorem that we ll eed for a proof dow the road. Theorem 4 For the sequece { } lim a = L. a if both lim a = L ad lim a + = L the { a } is coverget ad Proof of Theorem 4 Let ε > 0. The sice lim a Likewise, because lim a + = L there is a N > 0 such that if > N we kow that, a L < ε = L there is a N > 0 such that if > N we kow that, a + L < ε Now, let N = max{ N, N + } ad let > N. The either a = ak for some k > N or a = a k + for some k > N ad so i either case we have that, a L < ε Therefore, lim a = L ad so { a } is coverget. 007 Paul Dawkis 78

188 More o Sequeces I the previous sectio we itroduced the cocept of a sequece ad talked about limits of sequeces ad the idea of covergece ad divergece for a sequece. I this sectio we wat to take a quick look at some ideas ivolvig sequeces. Let s start off with some termiology ad defiitios. Give ay sequece { a } we have the followig.. We call the sequece icreasig if a < a + for every.. We call the sequece decreasig if a > a + for every.. If { a } is a icreasig sequece or { } a is a decreasig sequece we call it mootoic. 4. If there exists a umber m such that m a for every we say the sequece is bouded below. The umber m is sometimes called a lower boud for the sequece. 5. If there exists a umber M such that a M for every we say the sequece is bouded above. The umber M is sometimes called a upper boud for the sequece. 6. If the sequece is both bouded below ad bouded above we call the sequece bouded. Note that i order for a sequece to be icreasig or decreasig it must be icreasig/decreasig for every. I other words, a sequece that icreases for three terms ad the decreases for the rest of the terms is NOT a decreasig sequece! Also ote that a mootoic sequece must always icrease or it must always decrease. Before movig o we should make a quick poit about the bouds for a sequece that is bouded above ad/or below. We ll make the poit about lower bouds, but we could just as easily make it about upper bouds. A sequece is bouded below if we ca fid ay umber m such that m a for every. Note however that if we fid oe umber m to use for a lower boud the ay umber smaller tha m will also be a lower boud. Also, just because we fid oe lower boud that does t mea there wo t be a better lower boud for the sequece tha the oe we foud. I other words, there are a ifiite umber of lower bouds for a sequece that is bouded below, some will be better tha others. I my class all that I m after will be a lower boud. I do t ecessarily eed the best lower boud, just a umber that will be a lower boud for the sequece. Let s take a look at a couple of examples. 007 Paul Dawkis 79

189 Example Determie if the followig sequeces are mootoic ad/or bouded. Solutio (a) { } [Solutio] = 0 (a) { } + { } (b) (c) = 0 [Solutio] = = 5 [Solutio] This sequece is a decreasig sequece (ad hece mootoic) because, > ( + ) for every. Also, sice the sequece terms will be either zero or egative this sequece is bouded above. We ca use ay positive umber or zero as the boud, M, however, it s stadard to choose the smallest possible boud if we ca ad it s a ice umber. So, we ll choose M = 0 sice, 0 for every This sequece is ot bouded below however sice we ca always get below ay potetial boud by takig large eough. Therefore, while the sequece is bouded above it is ot bouded. As a side ote we ca also ote that this sequece diverges (to if we wat to be specific). [Retur to Problems] + { } (b) ( ) = The sequece terms i this sequece alterate betwee ad - ad so the sequece is either a icreasig sequece or a decreasig sequece. Sice the sequece is either a icreasig or decreasig sequece it is ot a mootoic sequece. The sequece is bouded however sice it is bouded above by ad bouded below by -. Agai, we ca ote that this sequece is also diverget. [Retur to Problems] (c) = 5 This sequece is a decreasig sequece (ad hece mootoic) sice, > + The terms i this sequece are all positive ad so it is bouded below by zero. Also, sice the 007 Paul Dawkis 80

190 sequece is a decreasig sequece the first sequece term will be the largest ad so we ca see that the sequece will also be bouded above by 5. Therefore, this sequece is bouded. We ca also take a quick limit ad ote that this sequece coverges ad its limit is zero. [Retur to Problems] Now, let s work a couple more examples that are desiged to make sure that we do t get too used to relyig o our ituitio with these problems. As we oted i the previous sectio our ituitio ca ofte lead us astray with some of the cocepts we ll be lookig at i this chapter. Example Determie if the followig sequeces are mootoic ad/or bouded. (a) [Solutio] + Solutio (a) + (b) = 4 = = 0 [Solutio] We ll start with the bouded part of this example first ad the come back ad deal with the icreasig/decreasig questio sice that is where studets ofte make mistakes with this type of sequece. First, is positive ad so the sequece terms are all positive. The sequece is therefore bouded below by zero. Likewise each sequece term is the quotiet of a umber divided by a larger umber ad so is guarateed to be less tha oe. The sequece is the bouded above by oe. So, this sequece is bouded. Now let s thik about the mootoic questio. First, studets will ofte make the mistake of assumig that because the deomiator is larger the quotiet must be decreasig. This will ot always be the case ad i this case we would be wrog. This sequece is icreasig as we ll see. To determie the icreasig/decreasig ature of this sequece we will eed to resort to Calculus I techiques. First cosider the followig fuctio ad its derivative. x f ( x) = f ( x) = x + x + We ca see that the first derivative is always positive ad so from Calculus I we kow that the fuctio must the be a icreasig fuctio. So, how does this help us? Notice that, f = = a + Therefore because < + ad f x is icreasig we ca also say that, 007 Paul Dawkis 8

191 + a = = f < f + = = a a < a I other words, the sequece must be icreasig. Note that ow that we kow the sequece is a icreasig sequece we ca get a better lower boud for the sequece. Sice the sequece is icreasig the first term i the sequece must be the smallest term ad so sice we are startig at = we could also use a lower boud of for this sequece. It is importat to remember that ay umber that is always less tha or equal to all the sequece terms ca be a lower boud. Some are better tha others however. A quick limit will also tell us that this sequece coverges with a limit of. Before movig o to the ext part there is a atural questio that may studets will have at this poit. Why did we use Calculus to determie the icreasig/decreasig ature of the sequece whe we could have just plugged i a couple of s ad quickly determied the same thig? The aswer to this questio is the ext part of this example! [Retur to Problems] (b) = 0 This is a messy lookig sequece, but it eeds to be i order to make the poit of this part. First, otice that, as with the previous part, the sequece terms are all positive ad will all be less tha oe (sice the umerator is guarateed to be less tha the deomiator) ad so the sequece is bouded. Now, let s move o to the icreasig/decreasig questio. As with the last problem, may studets will look at the expoets i the umerator ad deomiator ad determie based o that that sequece terms must decrease. This however, is t a decreasig sequece. Let s take a look at the first few terms to see this. a = a = a = a4 = a5 = a6 = a7 = a8 = a9 = a0 = = The first 0 terms of this sequece are all icreasig ad so clearly the sequece ca t be a 007 Paul Dawkis 8

192 decreasig sequece. Recall that a sequece ca oly be decreasig if ALL the terms are decreasig. Now, we ca t make aother commo mistake ad assume that because the first few terms icrease the whole sequece must also icrease. If we did that we would also be mistake as this is also ot a icreasig sequece. This sequece is either decreasig or icreasig. The oly sure way to see this is to do the Calculus I approach to icreasig/decreasig fuctios. I this case we ll eed the followig fuctio ad its derivative. 4 x x x 0000 f ( x) = f 4 ( x) = 4 x x This fuctio will have the followig three critical poits, 4 4 x= 0, x= , x= Why critical poits? Remember these are the oly places where the fuctio may chage sig! Our sequece starts at = 0 ad so we ca igore the third oe sice it lies outside the values of that we re cosiderig. By pluggig i some test values of x we ca quickly determie that the 4 derivative is positive for 0 < x < ad so the fuctio is icreasig i this rage. Likewise, we ca see that the derivative is egative for x > ad so the fuctio will be decreasig i this rage. So, our sequece will be icreasig for 0 ad decreasig for. Therefore the fuctio is ot mootoic. Fially, ote that this sequece will also coverge ad has a limit of zero. [Retur to Problems] So, as the last example has show we eed to be careful i makig assumptios about sequeces. Our ituitio will ofte ot be sufficiet to get the correct aswer ad we ca NEVER make assumptios about a sequece based o the value of the first few terms. As the last part has show there are sequeces which will icrease or decrease for a few terms ad the chage directio after that. Note as well that we said first few terms here, but it is completely possible for a sequece to decrease for the first 0,000 terms ad the start icreasig for the remaiig terms. I other words, there is o magical value of for which all we have to do is check up to that poit ad the we ll kow what the whole sequece will do. The oly time that we ll be able to avoid usig Calculus I techiques to determie the icreasig/decreasig ature of a sequece is i sequeces like part (c) of Example. I this case icreasig oly chaged (i fact icreased) the deomiator ad so we were able to determie the behavior of the sequece based o that. 007 Paul Dawkis 8

193 I Example however, icreasig icreased both the deomiator ad the umerator. I cases like this there is o way to determie which icrease will wi out ad cause the sequece terms to icrease or decrease ad so we eed to resort to Calculus I techiques to aswer the questio. We ll close out this sectio with a ice theorem that we ll use i some of the proofs later i this chapter. Theorem If { a } is bouded ad mootoic the { } a is coverget. Be careful to ot misuse this theorem. It does ot say that if a sequece is ot bouded ad/or ot mootoic that it is diverget. Example b is a good case i poit. The sequece i that example was ot mootoic but it does coverge. Note as well that we ca make several variats of this theorem. If { a } is bouded above ad icreasig the it coverges ad likewise if { a } is bouded below ad decreasig the it coverges. 007 Paul Dawkis 84

194 Series The Basics I this sectio we will itroduce the topic that we will be discussig for the rest of this chapter. That topic is ifiite series. So just what is a ifiite series? Well, let s start with a sequece { a} = (ote the = is for coveiece, it ca be aythig) ad defie the followig, s = a s = a+ a s = a+ a + a s4 = a+ a + a + a4 s = a + a + a + a + + a = a 4 i i= s are called partial sums ad otice that they will form a sequece, { } The. Also recall = that the Σ is used to represet this summatio ad called a variety of ames. The most commo ames are : series otatio, summatio otatio, ad sigma otatio. s You should have see this otatio, at least briefly, back whe you saw the defiitio of a defiite itegral i Calculus I. If you eed a quick refresher o summatio otatio see the review of summatio otatio i my Calculus I otes. s. = Now back to series. We wat to take a look at the limit of the sequece of partial sums, { } Notatioally we ll defie, We will call i= a i where our origial sequece, { } lim s = lim a = a i i i = i = a ifiite series ad ote that the series starts at i = because that is a =, started. Had our origial sequece started at the our ifiite series would also have started at. The ifiite series will start at the same value that the sequece of terms (as opposed to the sequece of partial sums) starts. s If the sequece of partial sums, { } = ifiite series, i= series is also called diverget. a i, is coverget ad its limit is fiite the we also call the coverget ad if the sequece of partial sums is diverget the the ifiite Note that sometimes it is coveiet to write the ifiite series as, ai = a+ a + a+ + a + i= 007 Paul Dawkis 85

195 We do have to be careful with this however. This implies that a ifiite series is just a ifiite sum of terms ad as we ll see i the ext sectio this is ot really true. I the ext sectio we re goig to be discussig i greater detail the value of a ifiite series, provided it has oe of course as well as the ideas of covergece ad divergece. This sectio is goig to be devoted mostly to otatioal issues as well as makig sure we ca do some basic maipulatios with ifiite series so we are ready for them whe we eed to be able to deal with them i later sectios. First, we should ote that i most of this chapter we will refer to ifiite series as simply series. If we ever eed to work with both ifiite ad fiite series we ll be more careful with termiology, but i most sectios we ll be dealig exclusively with ifiite series ad so we ll just call them series. Now, i i= a i the i is called the idex of summatio or just idex for short ad ote that the letter we use to represet the idex does ot matter. So for example the followig series are all the same. The oly differece is the letter we ve used for the idex. = = i + k + + i= 0 k= 0 = 0 etc. It is importat to agai ote that the idex will start at whatever value the sequece of series terms starts at ad this ca literally be aythig. So far we ve used = 0 ad = but the idex could have started aywhere. I fact, we will usually use a to represet a ifiite series i which the startig poit for the idex is ot importat. Whe we drop the iitial value of the idex we ll also drop the ifiity from the top so do t forget that it is still techically there. We will be droppig the iitial value of the idex i quite a few facts ad theorems that we ll be seeig throughout this chapter. I these facts/theorems the startig poit of the series will ot affect the result ad so to simplify the otatio ad to avoid givig the impressio that the startig poit is importat we will drop the idex from the otatio. Do ot forget however, that there is a startig poit ad that this will be a ifiite series. Note however, that if we do put a iitial value of the idex o a series i a fact/theorem it is there because it really does eed to be there. Now that some of the otatioal issues are out of the way we eed to start thikig about various ways that we ca maipulate series. We ll start this off with basic arithmetic with ifiite series as we ll eed to be able to do that o occasio. We have the followig properties. 007 Paul Dawkis 86

196 Properties If a ad b are both coverget series the, 6. ca, where c is ay umber, is also coverget ad = ca c a 7. a ± b = k = k is also coverget ad, a ± b = ( a ± b). = k = k = k The first property is simply tellig us that we ca always factor a multiplicative costat out of a ifiite series ad agai recall that if we do t put i a iitial value of the idex that the series ca start at ay value. Also recall that i these cases we wo t put a ifiity at the top either. The secod property says that if we add/subtract series all we really eed to do is add/subtract the series terms. Note as well that i order to add/subtract series we eed to make sure that both have the same iitial value of the idex ad the ew series will also start at this value. Before we move o to a differet topic let s discuss multiplicatio of series briefly. We ll start both series at = 0 for a later formula ad the ote that, a b ( ab ) = 0 = 0 = 0 To covice yourself that this is t true cosider the followig product of two fiite sums. ( + x)( 5x+ x ) = 6 7x x + x Yeah, it was just the multiplicatio of two polyomials. Each is a fiite sum ad so it makes the poit. I doig the multiplicatio we did t just multiply the costat terms, the the x terms, etc. Istead we had to distribute the through the secod polyomial, the distribute the x through the secod polyomial ad fially combie like terms. Multiplyig ifiite series (eve though we said we ca t thik of a ifiite series as a ifiite sum) eeds to be doe i the same maer. With multiplicatio we re really askig us to do the followig, a b = ( a0 + a + a + a + )( b0 + b + b + b + ) = 0 = 0 To do this multiplicatio we would have to distribute the a 0 through the secod term, distribute the a through, etc the combie like terms. This is pretty much impossible sice both series have a ifiite set of terms i them, however the followig formula ca be used to determie the product of two series. a b = c where c= ab i i = 0 = 0 = 0 i= Paul Dawkis 87

197 We also ca t say a lot about the covergece of the product. Eve if both of the origial series are coverget it is possible for the product to be diverget. The reality is that multiplicatio of series is a somewhat difficult process ad i geeral is avoided if possible. We will take a brief look at it towards the ed of the chapter whe we ve got more work uder our belt ad we ru across a situatio where it might actually be what we wat to do. Util the, do t worry about multiplyig series. The ext topic that we eed to discuss i this sectio is that of idex shift. To be hoest this is ot a topic that we ll see all that ofte i this course. I fact, we ll use it oce i the ext sectio ad the ot use it agai i all likelihood. Despite the fact that we wo t use it much i this course does t mea however that it is t used ofte i other classes where you might ru across series. So, we will cover it briefly here so that you ca say you ve see it. The basic idea behid idex shifts is to start a series at a differet value for whatever the reaso (ad yes, there are legitimate reasos for doig that). Cosider the followig series, = + 5 Suppose that for some reaso we wated to start this series at = 0, but we did t wat to chage the value of the series. This meas that we ca t just chage the = to = 0 as this would add i two ew terms to the series ad thus chage its value. Performig a idex shift is a fairly simple process to do. We ll start by defiig a ew idex, say i, as follows, i = Now, whe =, we will get i = 0. Notice as well that if = the i = =, so oly the lower limit will chage here. Next, we ca solve this for to get, = i+ We ca ow completely rewrite the series i terms of the idex i istead of the idex simply by pluggig i our equatio for i terms of i. + 5 ( i + ) + 5 i+ 7 = = i+ i+ = i= 0 i= 0 To fiish the problem out we ll recall that the letter we used for the idex does t matter ad so we ll chage the fial i back ito a to get, = + = = 0 To covice yourselves that these really are the same summatio let s write out the first couple of terms for each of them, 007 Paul Dawkis 88

198 = = = = So, sure eough the two series do have exactly the same terms. There is actually a easier way to do a idex shift. The method give above is the techically correct way of doig a idex shift. However, otice i the above example we decreased the iitial value of the idex by ad all the s i the series terms icreased by as well. This will always work i this maer. If we decrease the iitial value of the idex by a set amout the all the other s i the series term will icrease by the same amout. Likewise, if we icrease the iitial value of the idex by a set amout, the all the s i the series term will decrease by the same amout. Let s do a couple of examples usig this shorthad method for doig idex shifts. Example Perform the followig idex shifts. (a) Write ar as a series that starts at = 0. = (b) Write as a series that starts at =. + = Solutio (a) I this case we eed to decrease the iitial value by ad so the s (okay the sigle ) i the term must icrease by as well. ( + ) ar = ar = ar = = 0 = 0 (b) For this problem we wat to icrease the iitial value by ad so all the s i the series term must decrease by. ( ) ( ) = = + ( ) + = = = The fial topic i this sectio is agai a topic that we ll ot be seeig all that ofte i this class, although we will be seeig it more ofte tha the idex shifts. This fial topic is really more about alterate ways to write series whe the situatio requires it. Let s start with the followig series ad ote that the = startig poit is oly for coveiece sice we eed to start the series somewhere. = a = a + a + a + a + a Paul Dawkis 89

199 Notice that if we igore the first term the remaiig terms will also be a series that will start at = istead of = So, we ca rewrite the origial series as follows, a = a + a = = I this example we say that we ve stripped out the first term. We could have stripped out more terms if we wated to. I the followig series we ve stripped out the first two terms ad the first four terms respectively. a = a + a + a = = a = a + a + a + a + a 4 = = 5 Beig able to strip out terms will, o occasio, simplify our work or allow us to reuse a prior result so it s a importat idea to remember. Notice that i the secod example above we could have also deoted the four terms that we stripped out as a fiite series as follows, 4 a = a + a + a + a + a = a + a 4 = = 5 = = 5 This is a coveiet otatio whe we are strippig out a large umber of terms or if we eed to strip out a udetermied umber of terms. I geeral, we ca write a series as follows, N a = a + a = = = N+ We ll leave this sectio with a importat warig about termiology. Do t get sequeces ad series cofused! A sequece is a list of umbers writte i a specific order while a ifiite series is a limit of a sequece of fiite series ad hece, if it exists will be a sigle value. So, oce agai, a sequece is a list of umbers while a series is a sigle umber, provided it makes sese to eve compute the series. Studets will ofte cofuse the two ad try to use facts pertaiig to oe o the other. However, sice they are differet beasts this just wo t work. There will be problems where we are usig both sequeces ad series so we ll always have to remember that they are differet. 007 Paul Dawkis 90

200 Series Covergece/Divergece I the previous sectio we spet some time gettig familiar with series ad we briefly defied covergece ad divergece. Before worryig about covergece ad divergece of a series we wated to make sure that we ve started to get comfortable with the otatio ivolved i series ad some of the various maipulatios of series that we will, o occasio, eed to be able to do. As oted i the previous sectio most of what we were doig there wo t be doe much i this chapter. So, it is ow time to start talkig about the covergece ad divergece of a series as this will be a topic that we ll be dealig with to oe extet or aother i almost all of the remaiig sectios of this chapter. So, let s recap just what a ifiite series is ad what it meas for a series to be coverget or diverget. We ll start with a sequece { a} = ad agai ote that we re startig the sequece at = oly for the sake of coveiece ad it ca, i fact, be aythig. Next we defie the partial sums of the series as, s = a s = a+ a s = a+ a + a s = a + a + a + a ad these form a ew sequece, { } 4 4 s = a + a + a + a + + a = a 4 i i= s. = A ifiite series, or just series here sice almost every series that we ll be lookig at will be a ifiite series, is the the limit of the partial sums. Or, i= a = lim s i If the sequece of partial sums is a coverget sequece (i.e. its limit exists ad is fiite) the the series is also called coverget ad i this case if lim s = s the, i= a i = s. Likewise, if the sequece of partial sums is a diverget sequece (i.e. its limit does t exist or is plus or mius ifiity) the the series is also called diverget. Let s take a look at some series ad see if we ca determie if they are coverget or diverget ad see if we ca determie the value of ay coverget series we fid. Example Determie if the followig series is coverget or diverget. If it coverges 007 Paul Dawkis 9

201 determie its value. Solutio To determie if the series is coverget we first eed to get our hads o a formula for the geeral term i the sequece of partial sums. s = = i i= This is a kow series ad its value ca be show to be, + s = i = i= Do t worry if you did t kow this formula (I d be surprised if ayoe kew it ) as you wo t be required to kow it i my course. So, to determie if the series is coverget we will first eed to see if the sequece of partial sums, ( + ) = is coverget or diverget. That s ot terribly difficult i this case. The limit of the sequece terms is, ( + ) lim = Therefore, the sequece of partial sums diverges to ad so the series also diverges. So, as we saw i this example we had to kow a fairly obscure formula i order to determie the covergece of this series. I geeral fidig a formula for the geeral term i the sequece of partial sums is a very difficult process. I fact after the ext sectio we ll ot be doig much with the partial sums of series due to the extreme difficulty faced i fidig the geeral formula. This also meas that we ll ot be doig much work with the value of series sice i order to get the value we ll also eed to kow the geeral formula for the partial sums. We will cotiue with a few more examples however, sice this is techically how we determie covergece ad the value of a series. Also, the remaiig examples we ll be lookig at i this sectio will lead us to a very importat fact about the covergece of series. So, let s take a look at a couple more examples. Example Determie if the followig series coverges or diverges. If it coverges determie its sum. 007 Paul Dawkis 9

202 = Solutio This is actually oe of the few series i which we are able to determie a formula for the geeral term i the sequece of partial fractios. However, i this sectio we are more iterested i the geeral idea of covergece ad divergece ad so we ll put off discussig the process for fidig the formula util the ext sectio. The geeral formula for the partial sums is, s = = i= i 4 ( + ) ad i this case we have, lim s = lim = 4 ( ) + 4 The sequece of partial sums coverges ad so the series coverges also ad its value is, = 4 = Example Determie if the followig series coverges or diverges. If it coverges determie its sum. = 0 ( ) Solutio I this case we really do t eed a geeral formula for the partial sums to determie the covergece of this series. Let s just write dow the first few partial sums. s0 = s = = 0 s = + = s = + = 0 etc. So, it looks like the sequece of partial sums is, { s } = {, 0,, 0,, 0,, 0,, } = 0 ad this sequece diverges sice lim s does t exist. Therefore, the series also diverges. Example 4 Determie if the followig series coverges or diverges. If it coverges determie its sum. 007 Paul Dawkis 9

203 = Solutio Here is the geeral formula for the partial sums for this series. s = = i i= Agai, do ot worry about kowig this formula. This is ot somethig that you ll ever be asked to kow i my class. I this case the limit of the sequece of partial sums is, lim s = lim = The sequece of partial sums is coverget ad so the series will also be coverget. The value of the series is, = = As we already oted, do ot get excited about determiig the geeral formula for the sequece of partial sums. There is oly goig to be oe type of series where you will eed to determie this formula ad the process i that case is t too bad. I fact, you already kow how to do most of the work i the process as you ll see i the ext sectio. So, we ve determied the covergece of four series ow. Two of the series coverged ad two diverged. Let s go back ad examie the series terms for each of these. For each of the series let s take the limit as goes to ifiity of the series terms (ot the partial sums!!). lim = ( ) 007 Paul Dawkis 94 this series diverged lim = 0 this series coverged lim does't exist this series diverged lim = 0 this series coverged Notice that for the two series that coverged the series term itself was zero i the limit. This will always be true for coverget series ad leads to the followig theorem. Theorem If a coverges the lim a = 0. Proof First let s suppose that the series starts at =. If it does t the we ca modify thigs

204 as appropriate below. The the partial sums are, s = a = a + a + a + a + + a s = a = a + a + a + a + + a + a i 4 i 4 i= i= Next, we ca use these two partial sums to write, a = s s Now because we kow that is also = coverget ad that lim s = sfor some fiite value s. However, sice as we also have lim s = s. a is coverget we also kow that the sequece { } s We ow have, lim a = lim s s = lim s lim s = s s = 0 Be careful to ot misuse this theorem! This theorem gives us a requiremet for covergece but ot a guaratee of covergece. I other words, the coverse is NOT true. If lim a = 0 the series may actually diverge! Cosider the followig two series. = = I both cases the series terms are zero i the limit as goes to ifiity, yet oly the secod series coverges. The first series diverges. It will be a couple of sectios before we ca prove this, so at this poit please believe this ad kow that you ll be able to prove the covergece of these two series i a couple of sectios. Agai, as oted above, all this theorem does is give us a requiremet for a series to coverge. I order for a series to coverge the series terms must go to zero i the limit. If the series terms do ot go to zero i the limit the there is o way the series ca coverge sice this would violate the theorem. This leads us to the first of may tests for the covergece/divergece of a series that we ll be seeig i this chapter. Divergece Test If lim a 0 the a will diverge. Agai, do NOT misuse this test. This test oly says that a series is guarateed to diverge if the series terms do t go to zero i the limit. If the series terms do happe to go to zero the series may or may ot coverge! Agai, recall the followig two series, 007 Paul Dawkis 95

205 = = diverges coverges Oe of the more commo mistakes that studets make whe they first get ito series is to assume that if lim a = 0 the a will coverge. There is just o way to guaratee this so be careful! Let s take a quick look at a example of how this test ca be used. Example 5 Determie if the followig series is coverget or diverget. 4 = Solutio With almost every series we ll be lookig at i this chapter the first thig that we should do is take a look at the series terms ad see if they go to zero or ot. If it s clear that the terms do t go to zero use the Divergece Test ad be doe with the problem. That s what we ll do here lim = 0 The limit of the series terms is t zero ad so by the Divergece Test the series diverges. The divergece test is the first test of may tests that we will be lookig at over the course of the ext several sectios. You will eed to keep track of all these tests, the coditios uder which they ca be used ad their coclusios all i oe place so you ca quickly refer back to them as you eed to. Next we should briefly revisit arithmetic of series ad covergece/divergece. As we saw i the previous sectio if a ad b are both coverget series the so are ca ad = k ( a ± b ). Furthermore, these series will have the followig sums or values. = ( ± ) = ± ca c a a b a b = k = k = k We ll see a example of this i the ext sectio after we get a few more examples uder our belt. At this poit just remember that a sum of coverget series is coverget ad multiplyig a coverget series by a umber will ot chage its covergece. We eed to be a little careful with these facts whe it comes to diverget series. I the first case if a is diverget the ca will also be diverget (provided c is t zero of course) sice multiplyig a series that is ifiite i value or does t have a value by a fiite value (i.e. c) wo t 007 Paul Dawkis 96

206 chage the fact that the series has a ifiite or o value. However, it is possible to have both a ad b be diverget series ad yet have ( a ± b ) be a coverget series. = k Now, sice the mai topic of this sectio is the covergece of a series we should metio a stroger type of covergece. A series a is said to coverge absolutely if a also coverges. Absolute covergece is stroger tha covergece i the sese that a series that is absolutely coverget will also be coverget, but a series that is coverget may or may ot be absolutely coverget. I fact if a coverges ad a diverges the series a is called coditioally coverget. At this poit we do t really have the tools at had to properly ivestigate this topic i detail or do we have the tools i had to determie if a series is absolutely coverget or ot. So we ll ot say aythig more about this subject for a while. Whe we fially have the tools i had to discuss this topic i more detail we will revisit it. Util the do t worry about it. The idea is metioed here oly because we were already discussig covergece i this sectio ad it ties ito the last topic that we wat to discuss i this sectio. I the previous sectio after we d itroduced the idea of a ifiite series we commeted o the fact that we should t thik of a ifiite series as a ifiite sum despite the fact that the otatio we use for ifiite series seems to imply that it is a ifiite sum. It s ow time to briefly discuss this. First, we eed to itroduce the idea of a rearragemet. A rearragemet of a series is exactly what it might soud like, it is the same series with the terms rearraged ito a differet order. For example, cosider the followig ifiite series. = A rearragemet of this series is, = a = a + a + a + a + a + a + a a = a + a + a + a + a + a + a The issue we eed to discuss here is that for some series each of these arragemets of terms ca have differet values despite the fact that they are usig exactly the same terms. Here is a example of this. It ca be show that, + ( ) = = l () = Sice this series coverges we kow that if we multiply it by a costat c its value will also be multiplied by c. So, let s multiply this by to get, 007 Paul Dawkis 97

207 = l () Now, let s add i a zero betwee each term as follows = l () Note that this wo t chage the value of the series because the partial sums for this series will be the partial sums for the () except that each term will be repeated. Repeatig terms i a series will ot affect its limit however ad so both () ad () will be the same. We kow that if two series coverge we ca add them by addig term by term ad so add () ad () to get, = l (4) Now, otice that the terms of (4) are simply the terms of () rearraged so that each egative term comes after two positive terms. The values however are defiitely differet despite the fact that the terms are the same. Note as well that this is ot oe of those tricks that you see occasioally where you get a cotradictory result because of a hard to spot math/logic error. This is a very real result ad we ve ot made ay logic mistakes/errors. Here is a ice set of facts that gover this idea of whe a rearragemet will lead to a differet value of a series. Facts Give the series a,. If a is absolutely coverget ad its value is s the ay rearragemet of a will also have a value of s.. If a is coditioally coverget ad r is ay real umber the there is a rearragemet of a whose value will be r. Agai, we do ot have the tools i had yet to determie if a series is absolutely coverget ad so do t worry about this at this poit. This is here just to make sure that you uderstad that we have to be very careful i thikig of a ifiite series as a ifiite sum. There are times whe we ca (i.e. the series is absolutely coverget) ad there are times whe we ca t (i.e. the series is coditioally coverget). As a fial ote, the fact above tells us that the series, = Paul Dawkis 98

208 must be coditioally coverget sice two rearragemets gave two separate values of this series. Evetually it will be very simple to show that this series is coditioally coverget. 007 Paul Dawkis 99

209 Series Special Series I this sectio we are goig to take a brief look at three special series. Actually, special may ot be the correct term. All three have bee amed which makes them special i some way, however the mai reaso that we re goig to look at two of them i this sectio is that they are the oly types of series that we ll be lookig at for which we will be able to get actual values for the series. The third type is diverget ad so wo t have a value to worry about. I geeral, determiig the value of a series is very difficult ad outside of these two kids of series that we ll look at i this sectio we will ot be determiig the value of series i this chapter. So, let s get started. Geometric Series A geometric series is ay series that ca be writte i the form, = ar or, with a idex shift the geometric series will ofte be writte as, = 0 These are idetical series ad will have idetical values, provided they coverge of course. If we start with the first form it ca be show that the partial sums are, a( r ) a ar s = = r r r The series will coverge provided the partial sums form a coverget sequece, so let s take the limit of the partial sums. a ar lim s = lim r r a ar = lim lim r r a a = lim r r r Now, from Theorem from the Sequeces sectio we kow that the limit above will exist ad be fiite provided < r. However, ote that we ca t let r = sice this will give divisio by zero. Therefore, this will exist ad be fiite provided < r < ad i this case the limit is zero ad so we get, a lim s = r ar 007 Paul Dawkis 00

210 Therefore, a geometric series will coverge if < r <, which is usually writte r <, its value is, a = = r ar ar = = 0 Note that i usig this formula we ll eed to make sure that we are i the correct form. I other words, if the series starts at = 0 the the expoet o the r must be. Likewise if the series starts at = the the expoet o the r must be. Example Determie if the followig series coverge or diverge. If they coverge give the value of the series. + + (a) 9 4 [Solutio] Solutio (a) = (b) 9 4 = 4 [Solutio] = This series does t really look like a geometric series. However, otice that both parts of the series term are umbers raised to a power. This meas that it ca be put ito the form of a geometric series. We will just eed to decide which form is the correct form. Sice the series starts at = we will wat the expoets o the umbers to be. It will be fairly easy to get this ito the correct form. Let s first rewrite thigs slightly. Oe of the s i the expoet has a egative i frot of it ad that ca t be there i the geometric form. So, let s first get rid of that ( ) = 9 4 = 9 = = = Now let s get the correct expoet o each of the umbers. This ca be doe usig simple expoet properties = = = = = Now, rewrite the term a little = 6( 9) = 44 = = 9 = 9 4 So, this is a geometric series with a = 44 ad r = 9 <. Therefore, sice r < we kow the series will coverge ad its value will be, 007 Paul Dawkis 0

211 = = = = [Retur to Problems] (b) = 0 ( 4) 5 Agai, this does t look like a geometric series, but it ca be put ito the correct form. I this case the series starts at = 0 so we ll eed the expoets to be o the terms. Note that this meas we re goig to eed to rewrite the expoet o the umerator a little ( 4) = = 5 = = 0 = 0 = 0 = 0 64 So, we ve got it ito the correct form ad we ca see that a = 5 ad r = 5. Also ote that r ad so this series diverges. [Retur to Problems] Back i the Series Basics sectio we talked about strippig out terms from a series, but did t really provide ay examples of how this idea could be used i practice. We ca ow do some examples. Example Use the results from the previous example to determie the value of the followig series. + + (a) 9 4 [Solutio] Solutio (a) = 0 (b) 9 4 = [Solutio] = + + I this case we could just ackowledge that this is a geometric series that starts at = 0 ad so we could put it ito the correct form ad be doe with it. However, this does provide us with a ice example of how to use the idea of strippig out terms to our advatage. Let s otice that if we strip out the first term from this series we arrive at, = = = 0 = = From the previous example we kow the value of the ew series that arises here ad so the value of the series i this example is, 007 Paul Dawkis 0

212 = = + = 5 5 [Retur to Problems] (b) = I this case we ca t strip out terms from the give series to arrive at the series used i the previous example. However, we ca start with the series used i the previous example ad strip terms out of it to get the series i this example. So, let s do that. We will strip out the first two terms from the series we looked at i the previous example = = = = = We ca ow use the value of the series from the previous example to get the value of this series = = 08 = = = 5 5 [Retur to Problems] Notice that we did t discuss the covergece of either of the series i the above example. Here s why. Cosider the followig series writte i two separate ways (i.e. we stripped out a couple of terms from it). Let s suppose that we kow = a a = a + a + a + a 0 = 0 = is a coverget series. This meas that it s got a fiite value ad addig three fiite terms oto this will ot chage that fact. So the value of fiite ad so is coverget. Likewise, suppose that = 0 a this value we will remai fiite ad arrive at the value of so this series will also be coverget. = 0 a is also is coverget. I this case if we subtract three fiite values from = a. This is ow a fiite value ad I other words, if we have two series ad they differ oly by the presece, or absece, of a fiite umber of fiite terms they will either both be coverget or they will both be diverget. The differece of a few terms oe way or the other will ot chage the covergece of a series. This is a importat idea ad we will use it several times i the followig sectios to simplify some of the tests that we ll be lookig at. 007 Paul Dawkis 0

213 Telescopig Series It s ow time to look at the secod of the three series i this sectio. I this portio we are goig to look at a series that is called a telescopig series. The ame i this case comes from what happes with the partial sums ad is best show i a example. Example Determie if the followig series coverges or diverges. If it coverges fid its value. = Solutio We first eed the partial sums for this series. s = i + i+ Now, let s otice that we ca use partial fractios o the series term to get, = = i + i+ i+ i+ i+ i+ i= 0 I ll leave the details of the partial fractios to you. By ow you should be fairly adept at this sice we spet a fair amout of time doig partial fractios back i the Itegratio Techiques chapter. If you eed a refresher you should go back ad review that sectio. So, what does this do for us? Well, let s start writig out the terms of the geeral partial sum for this series usig the partial fractio form. s = i= 0 i+ i+ = = + Notice that every term except the first ad last term caceled out. This is the origi of the ame telescopig series. This also meas that we ca determie the covergece of this series by takig the limit of the partial sums. lim s = lim = + The sequece of partial sums is coverget ad so the series is coverget ad has a value of = = I telescopig series be careful to ot assume that successive terms will be the oes that cacel. Cosider the followig example. 007 Paul Dawkis 04

214 Example 4 Determie if the followig series coverges or diverges. If it coverges fid its value. = + 4+ Solutio As with the last example we ll leave the partial fractios details to you to verify. The partial sums are, s = i i i = = + + i= i+ i+ = = I this case istead of successive terms cacelig a term will cacel with a term that is farther dow the list. The ed result this time is two iitial ad two fial terms are left. Notice as well that i order to help with the work a little we factored the out of the series. The limit of the partial sums is, 5 5 lim s = lim = So, this series is coverget (because the partial sums form a coverget sequece) ad its value is, 5 = = + 4+ Note that it s ot always obvious if a series is telescopig or ot util you try to get the partial sums ad the see if they are i fact telescopig. There is o test that will tell us that we ve got a telescopig series right off the bat. Also ote that just because you ca do partial fractios o a series term does ot mea that the series will be a telescopig series. The followig series, for example, is ot a telescopig series despite the fact that we ca partial fractio the series terms. + = = = I order for a series to be a telescopig series we must get terms to cacel ad all of these terms are positive ad so oe will cacel. Next, we eed to go back ad address a issue that was first raised i the previous sectio. I that sectio we stated that the sum or differece of coverget series was also coverget ad that the presece of a multiplicative costat would ot affect the covergece of a series. Now that we have a few more series i had let s work a quick example showig that. 007 Paul Dawkis 05

215 Example 5 Determie the value of the followig series = + 4+ Solutio To get the value of this series all we eed to do is rewrite it ad the use the previous results = = + 4+ = + 4+ = + + = = = 5 96 = = 5 We did t discuss the covergece of this series because it was the sum of two coverget series ad that guarateed that the origial series would also be coverget. Harmoic Series This is the third ad fial series that we re goig to look at i this sectio. Here is the harmoic series. = The harmoic series is diverget ad we ll eed to wait util the ext sectio to show that. This series is here because it s got a ame ad so I wated to put it here with the other two amed series that we looked at i this sectio. We re also goig to use the harmoic series to illustrate a couple of ideas about diverget series that we ve already discussed for coverget series. We ll do that with the followig example. Example 6 Show that each of the followig series are diverget. 5 (a) = (b) = 4 Solutio 5 (a) = To see that this series is diverget all we eed to do is use the fact that we ca factor a costat out of a series as follows, 5 = 5 = = 007 Paul Dawkis 06

216 Now, is diverget ad so five times this will still ot be a fiite umber ad so the series = has to be diverget. I other words, if we multiply a diverget series by a costat it will still be diverget. (b) = 4 I this case we ll start with the harmoic series ad strip out the first three terms. = = 6 = = 4 = 4 = I this case we are subtractig a fiite umber from a diverget series. This subtractio will ot chage the divergece of the series. We will either have ifiity mius a fiite umber, which is still ifiity, or a series with o value mius a fiite umber, which will still have o value. Therefore, this series is diverget. Just like with coverget series, addig/subtractig a fiite umber from a diverget series is ot goig to chage the divergece of the series. So, some geeral rules about the covergece/divergece of a series are ow i order. Multiplyig a series by a costat will ot chage the covergece/divergece of the series ad addig or subtractig a costat from a series will ot chage the covergece/divergece of the series. These are ice ideas to keep i mid. 007 Paul Dawkis 07

217 Itegral Test The last topic that we discussed i the previous sectio was the harmoic series. I that discussio we stated that the harmoic series was a diverget series. It is ow time to prove that statemet. This proof will also get us started o the way to our ext test for covergece that we ll be lookig at. So, we will be tryig to prove that the harmoic series, = diverges. We ll start this off by lookig at a apparetly urelated problem. Let s start off by askig what the area uder f ( x) = o the iterval [, ). From the sectio o Improper Itegrals we x kow that this is, dx = x ad so we called this itegral diverget (yes, that s the same term we re usig here with series.). So, just how does that help us to prove that the harmoic series diverges? Well, recall that we ca always estimate the area by breakig up the iterval ito segmets ad the sketchig i rectagles ad usig the sum of the area all of the rectagles as a estimate of the actual area. Let s do that for this problem as well ad see what we get. We will break up the iterval ito subitervals of width ad we ll take the fuctio value at the left edpoit as the height of the rectagle. The image below shows the first few rectagles for this area. So, the area uder the curve is approximately, 007 Paul Dawkis 08

218 A = = = Now ote a couple of thigs about this approximatio. First, each of the rectagles overestimates the actual area ad secodly the formula for the area is exactly the harmoic series! Puttig these two facts together gives the followig, A > dx = x Notice that this tells us that we must have, > = = = = Sice we ca t really be larger tha ifiity the harmoic series must also be ifiite i value. I other words, the harmoic series is i fact diverget. So, we ve maaged to relate a series to a improper itegral that we could compute ad it turs out that the improper itegral ad the series have exactly the same covergece. Let s see if this will also be true for a series that coverges. Whe discussig the Divergece Test we made the claim that = coverges. Let s see if we ca do somethig similar to the above process to prove this. We will try to relate this to the area uder f ( x) = is o the iterval [ ) Improper Itegral sectio we kow that, ad so this itegral coverges. x dx = x,. Agai, from the We will oce agai try to estimate the area uder this curve. We will do this i a almost idetical maer as the previous part with the exceptio that istead of usig the left ed poits for the height of our rectagles we will use the right ed poits. Here is a sketch of this case, 007 Paul Dawkis 09

219 I this case the area estimatio is, A = This time, ulike the first case, the area will be a uderestimatio of the actual area ad the estimatio is ot quite the series that we are workig with. Notice however that the oly differece is that we re missig the first term. This meas we ca do the followig, = < + dx = + = = 4 5 x Area Estimatio Or, puttig all this together we see that, = < With the harmoic series this was all that we eeded to say that the series was diverget. With this series however, this is t quite eough. For istace < ad if the series did have a value of the it would be diverget (whe we wat coverget). So, let s do a little more work. First, let s otice that all the series terms are positive (that s importat) ad that the partial sums are, s = i Because the terms are all positive we kow that the partial sums must be a icreasig sequece. I other words, + s = < s = + i i i= i= i= 007 Paul Dawkis 0

220 I s + we are addig a sigle positive term oto s ad so must get larger. Therefore, the partial sums form a icreasig (ad hece mootoic) sequece. Also ote that, sice the terms are all positive, we ca say, s = < < s < i= i = ad so the sequece of partial sums is a bouded sequece. I the secod sectio o Sequeces we gave a theorem that stated that a bouded ad mootoic sequece was guarateed to be coverget. This meas that the sequece of partial sums is a coverget sequece. So, who cares right? Well recall that this meas that the series must the also be coverget! So, oce agai we were able to relate a series to a improper itegral (that we could compute) ad the series ad the itegral had the same covergece. We wet through a fair amout of work i both of these examples to determie the covergece of the two series. Luckily for us we do t eed to do all this work every time. The ideas i these two examples ca be summarized i the followig test. Itegral Test Suppose that f ( x ) is a cotiuous, positive ad decreasig fuctio o the iterval [, ) f that = a the, x dx is coverget so is a.. If f. If f k k = k x dx is diverget so is a. = k A formal proof of this test ca be foud at the ed of this sectio. k ad There are a couple of thigs to ote about the itegral test. First, the lower limit o the improper itegral must be the same value that starts the series. Secod, the fuctio does ot actually eed to be decreasig ad positive everywhere i the iterval. All that s really required is that evetually the fuctio is decreasig ad positive. I other words, it is okay if the fuctio (ad hece series terms) icreases or is egative for a while, but evetually the fuctio (series terms) must decrease ad be positive for all terms. To see why this is true let s suppose that the series terms icrease ad or are egative i the rage k N ad the decrease ad are positive for N +. I this case the series ca be writte as, N a = a + a = k = k = N+ Now, the first series is othig more tha a fiite sum (o matter how large N is) of fiite terms ad so will be fiite. So the origial series will be coverget/diverget oly if the secod ifiite series o the right is coverget/diverget ad the test ca be doe o the secod series as it satisfies the coditios of the test. 007 Paul Dawkis

221 A similar argumet ca be made usig the improper itegral as well. The requiremet i the test that the fuctio/series be decreasig ad positive everywhere i the rage is required for the proof. I practice however, we oly eed to make sure that the fuctio/series is evetually a decreasig ad positive fuctio/series. Also ote that whe computig the itegral i the test we do t actually eed to strip out the icreasig/egative portio sice the presece of a small rage o which the fuctio is icreasig/egative will ot chage the itegral from coverget to diverget or from diverget to coverget. There is oe more very importat poit that must be made about this test. This test does NOT give the value of a series. It will oly give the covergece/divergece of the series. That s it. No value. We ca use the above series as a perfect example of this. All that the test gave us was that, < = So, we got a upper boud o the value of the series, but ot a actual value for the series. I fact, from this poit o we will ot be askig for the value of a series we will oly be askig whether a series coverges or diverges. I a later sectio we look at estimatig values of series, but eve i that sectio still wo t actually be gettig values of series. Just for the sake of completeess the value of this series is kow. π = = < 6 Let s work a couple of examples. = Example Determie if the followig series is coverget or diverget. = l Solutio I this case the fuctio we ll use is, f ( x) = xl x This fuctio is clearly positive ad if we make x larger the deomiator will get larger ad so the fuctio is also decreasig. Therefore, all we eed to do is determie the covergece of the followig itegral. t dx = lim dx u = l x xl x t xl x t ( ( x) ) ( ( t) ) = lim l l t = lim l l l l t = The itegral is diverget ad so the series is also diverget by the Itegral Test. 007 Paul Dawkis

222 Example Determie if the followig series is coverget or diverget. e = 0 Solutio The fuctio that we ll use i this example is, f ( x) = xe x This fuctio is always positive o the iterval that we re lookig at. Now we eed to check that the fuctio is decreasig. It is ot clear that this fuctio will always be decreasig o the iterval give. We ca use our Calculus I kowledge to help us however. The derivative of this fuctio is, x f x = e x This fuctio has two critical poits (which will tell us where the derivative chages sig) at x = ±. Sice we are startig at = 0 we ca igore the egative critical poit. Pickig a couple of test poits we ca see that the fuctio is icreasig o the iterval 0, ad it is decreasig o, ). Therefore, evetually the fuctio will be decreasig ad that s all that s required for us to use the Itegral Test. t x x e lim 0 t e 0 x dx = x dx u = x x = lim t e 0 t = lim e = t The itegral is coverget ad so the series must also be coverget by the Itegral Test. We ca use the Itegral Test to get the followig fact/test for some series. Fact ( The p series Test) If k > 0 the coverges if p > ad diverges if p k p. = Sometimes the series i this fact are called p-series ad so this fact is sometimes called the p- series test. This fact follows directly from the Itegral Test ad a similar fact we saw i the Improper Itegral sectio. This fact says that the itegral, dx p k x coverges if p > ad diverges if p. t 007 Paul Dawkis

223 Usig the p-series test makes it very easy to determie the covergece of some series. Example Determie if the followig series are coverget or diverget. (a) 7 = 4 (b) = Solutio (a) I this case p = 7> ad so by this fact the series is coverget. (b) For this series p = ad so the series is diverget by the fact. The last thig that we ll do i this sectio is give a quick proof of the Itegral Test. We ve essetially doe the proof already at the begiig of the sectio whe we were itroducig the Itegral Test, but let s go through it formally for a geeral fuctio. Proof of Itegral Test First, for the sake of the proof we ll be workig with the series = a. The origial test statemet was for a series that started at a geeral = k ad while the proof ca be doe for that it will be easier if we assume that the series starts at =. Aother way of dealig with the = k is we could do a idex shift ad start the series at = ad the do the Itegral Test. Either way provig the test for = will be sufficiet. Let s start off ad estimate the area uder the curve o the iterval [, ] ad we ll uderestimate the area by takig rectagles of width oe ad whose height is the right edpoit. This gives the followig figure. Now, ote that, f = a f ( ) = a f = a 007 Paul Dawkis 4

224 The approximate area is the, A f + f ( ) + + f = a + a+ a ad we kow that this uderestimates the actual area so, i= i a = a + a + a < f x dx Now, let s suppose that f ( x) dx is coverget ad so f Also, because f ( x ) is positive we kow that, f ( x ) dx < f ( x ) dx x dx must have a fiite value. This i tur meas that, i < < i= a f x dx f x dx Our series starts at = so this is t quite what we eed. However, that s easy eough to deal with. i i i= i= a = a + a < a + f x dx = M So, just what has this told us? Well we ow kow that the sequece of partial sums, are bouded above by M. s = ai i= Next, because the terms are positive we also kow that, ad so the sequece { } of partial sums { } + s s + a = a + a = a = s s s + i + i + + i= i= s = s coverges ad hece our series = a is coverget. = is also a icreasig sequece. So, we ow kow that the sequece So, the first part of the test is prove. The secod part is somewhat easier. This time let s overestimate the area uder the curve by usig the left edpoits of iterval for the height of the rectagles as show below. 007 Paul Dawkis 5

225 I this case the area is approximately, A f + f + + f = a + a + a Sice we kow this overestimates the area we also the kow that, i i= s = a = a + a + a > f x dx Now, suppose that f ( x) dx is diverget. I this case this meas that f because f ( x) 0. However, because as we also kow that f ( x) dx. Therefore, sice x dx as s > f x dx we kow that as we must have. This i tur tells us that s as. s So, we ow kow that the sequece of partial sums, { } = = a is a diverget series. s, is a diverget sequece ad so It is importat to ote before leavig this sectio that i order to use the Itegral Test the series terms MUST evetually be decreasig ad positive. If they are ot the the test does t work. Also remember that the test oly determies the covergece of a series ad does NOT give the value of the series. 007 Paul Dawkis 6

226 Compariso Test / Limit Compariso Test I the previous sectio we saw how to relate a series to a improper itegral to determie the covergece of a series. While the itegral test is a ice test, it does force us to do improper itegrals which are t always easy ad i some cases may be impossible to determie the covergece of. For istace cosider the followig series. 0 = + I order to use the Itegral Test we would have to itegrate dx 0 x + x ad I m ot eve sure if it s possible to do this itegral. Nicely eough for us there is aother test that we ca use o this series that will be much easier to use. First, let s ote that the series terms are positive. As with the Itegral Test that will be importat i this sectio. Next let s ote that we must have x > 0 sice we are itegratig o the iterval 0 x <. Likewise, regardless of the value of x we will always have x > 0. So, if we drop the x from the deomiator the deomiator will get smaller ad hece the whole fractio will get larger. So, + < Now, is a geometric series ad we kow that sice will be, 0 = r = = = = 0 < the series will coverge ad its value Now, if we go back to our origial series ad write dow the partial sums we get, s = i + i i= 0 Sice all the terms are positive addig a ew term will oly make the umber larger ad so the sequece of partial sums must be a icreasig sequece. + s = < i = s i + + i + i The sice, i= 0 i= Paul Dawkis 7

227 + < ad because the terms i these two sequeces are positive we ca also say that, s = < i < i = s < + i i= 0 i= 0 i= 0 Therefore, the sequece of partial sums is also a bouded sequece. The from the secod sectio o sequeces we kow that a mootoic ad bouded sequece is also coverget. So, the sequece of partial sums of our series is a coverget sequece. This meas that the series itself, 0 = + is also coverget. So, what did we do here? We foud a series whose terms were always larger tha the origial series terms ad this ew series was also coverget. The sice the origial series terms were positive (very importat) this meat that the origial series was also coverget. To show that a series (with oly positive terms) was diverget we could go through a similar argumet ad fid a ew diverget series whose terms are always smaller tha the origial series. I this case the origial series would have to take a value larger tha the ew series. However, sice the ew series is diverget its value will be ifiite. This meas that the origial series must also be ifiite ad hece diverget. We ca summarize all this i the followig test. Compariso Test Suppose that we have two series a ad b with a, b 0 for all ad a b for all. The,. If b is coverget the so is a.. If a is diverget the so is b. I other words, we have two series of positive terms ad the terms of oe of the series is always larger tha the terms of the other series. The if the larger series is coverget the smaller series must also be coverget. Likewise, if the smaller series is diverget the the larger series must also be diverget. Note as well that i order to apply this test we eed both series to start at the same place. A formal proof of this test is at the ed of this sectio. Do ot misuse this test. Just because the smaller of the two series coverges does ot say aythig about the larger series. The larger series may still diverge. Likewise, just because we kow that the larger of two series diverges we ca t say that the smaller series will also diverge! Be very careful i usig this test 007 Paul Dawkis 8

228 Recall that we had a similar test for improper itegrals back whe we were lookig at itegratio techiques. So, if you could use the compariso test for improper itegrals you ca use the compariso test for series as they are pretty much the same idea. Note as well that the requiremet that a, b 0 ad a b really oly eed to be true evetually. I other words, if a couple of the first terms are egative or a b for a couple of the first few terms we re okay. As log as we evetually reach a poit where a, b 0 ad a b for all sufficietly large the test will work. To see why this is true let s suppose that the series start at = k ad that the coditios of the test are oly true for for N + ad for k N at least oe of the coditios is ot true. If we the look at a (the same thig could be doe for b ) we get, N a = a + a = k = k = N+ The first series is othig more tha a fiite sum (o matter how large N is) of fiite terms ad so will be fiite. So the origial series will be coverget/diverget oly if the secod ifiite series o the right is coverget/diverget ad the test ca be doe o the secod series as it satisfies the coditios of the test. Let s take a look at some examples. Example Determie if the followig series is coverget or diverget. = cos Solutio Sice the cosie term i the deomiator does t get too large we ca assume that the series terms will behave like, = which, as a series, will diverge. So, from this we ca guess that the series will probably diverge ad so we ll eed to fid a smaller series that will also diverge. Recall that from the compariso test with improper itegrals that we determied that we ca make a fractio smaller by either makig the umerator smaller or the deomiator larger. I this case the two terms i the deomiator are both positive. So, if we drop the cosie term we will i fact be makig the deomiator larger sice we will o loger be subtractig off a positive quatity. Therefore, The, sice > = cos 007 Paul Dawkis 9

229 = diverges (it s harmoic or the p-series test) by the Compariso Test our origial series must also diverge. Example Determie if the followig series coverges or diverges. + 4 = + 5 Solutio I this case the + ad the +5 do t really add aythig to the series ad so the series terms should behave pretty much like = 4 which will coverge as a series. Therefore, we ca guess that the origial series will coverge ad we will eed to fid a larger series which also coverges. This meas that we ll either have to make the umerator larger or the deomiator smaller. We ca make the deomiator smaller by droppig the +5. Doig this gives, + + < At this poit, otice that we ca t drop the + from the umerator sice this would make the term smaller ad that s ot what we wat. However, this is actually the furthest that we eed to go. Let s take a look at the followig series. + 4 = = = = = + 4 = = As show, we ca write the series as a sum of two series ad both of these series are coverget by the p-series test. Therefore, sice each of these series are coverget we kow that the sum, + 4 = is also a coverget series. Recall that the sum of two coverget series will also be coverget. Now, sice the terms of this series are larger tha the terms of the origial series we kow that the origial series must also be coverget by the Compariso Test. The compariso test is a ice test that allows us to do problems that either we could t have doe with the itegral test or at the best would have bee very difficult to do with the itegral test. That does t mea that it does t have problems of its ow. Cosider the followig series. 0 = 007 Paul Dawkis 0

230 This is ot much differet from the first series that we looked at. The origial series coverged because the gets very large very fast ad will be sigificatly larger tha the. Therefore, the does t really affect the covergece of the series i that case. The fact that we are ow subtractig the off istead of addig the o really should t chage the covergece. We ca say this because the gets very large very fast ad the fact that we re subtractig off wo t really chage the size of this term for all sufficietly large values of. So, we would expect this series to coverge. However, the compariso test wo t work with this series. To use the compariso test o this series we would eed to fid a larger series that we could easily determie the covergece of. I this case we ca t do what we did with the origial series. If we drop the we will make the deomiator larger (sice the was subtracted off) ad so the fractio will get smaller ad just like whe we looked at the compariso test for improper itegrals kowig that the smaller of two series coverges does ot mea that the larger of the two will also coverge. So, we will eed somethig else to do help us determie the covergece of this series. The followig variat of the compariso test will allow us to determie the covergece of this series. Limit Compariso Test Suppose that we have two series a ad b with a 0, b > 0 for all. Defie, a c = lim b If c is positive (i.e. c > 0 ) ad is fiite (i.e. c <) the either both series coverge or both series diverge. The proof of this test is at the ed of this sectio. Note that it does t really matter which series term is i the umerator for this test, we could just have easily defied c as, b c = lim a ad we would get the same results. To see why this is, cosider the followig two defiitios. a b c= lim c = lim b a Start with the first defiitio ad rewrite it as follows, the take the limit. a c = lim = lim = = b b b lim c a a I other words, if c is positive ad fiite the so is c ad if c is positive ad fiite the so is c. Likewise if c = 0 the c = ad if c = the c = 0. Both defiitios will give the same results from the test so do t worry about which series terms should be i the umerator ad which should be i the deomiator. Choose this to make the limit easy to compute. 007 Paul Dawkis

231 Also, this really is a compariso test i some ways. If c is positive ad fiite this is sayig that both of the series terms will behave i geerally the same fashio ad so we ca expect the series themselves to also behave i a similar fashio. If c = 0 or c = we ca t say this ad so the test fails to give ay iformatio. The limit i this test will ofte be writte as, c= lim a b sice ofte both terms will be fractios ad this will make the limit easier to deal with. Let s see how this test works. Example Determie if the followig series coverges or diverges. 0 = Solutio To use the limit compariso test we eed to fid a secod series that we ca determie the covergece of easily ad has what we assume is the same covergece as the give series. O top of that we will eed to choose the ew series i such a way as to give us a easy limit to compute for c. We ve already guessed that this series coverges ad sice it s vaguely geometric let s use 0 = as the secod series. We kow that this series coverges ad there is a chace that sice both series have the i it the limit wo t be too bad. Here s the limit. c = lim = lim Now, we ll eed to use L Hospital s Rule o the secod term i order to actually evaluate this limit. c = lim l = So, c is positive ad fiite so by the Compariso Test both series must coverge sice coverges. 0 = 007 Paul Dawkis

232 Example 4 Determie if the followig series coverges or diverges = + Solutio Fractios ivolvig oly polyomials or polyomials uder radicals will behave i the same way as the largest power of will behave i the limit. So, the terms i this series should behave as, = = 7 7 ad as a series this will diverge by the p-series test. I fact, this would make a ice choice for our secod series i the limit compariso test so let s use it lim = lim = lim = = 4 = c So, c is positive ad fiite ad so both limits will diverge sice diverges. = Fially, to see why we eed c to be positive ad fiite (i.e. c 0 ad c ) cosider the followig two series. = = The first diverges ad the secod coverges. Now compute each of the followig limits. lim = lim = lim = lim = 0 I the first case the limit from the limit compariso test yields c = ad i the secod case the limit yields c = 0. Clearly, both series do ot have the same covergece. Note however, that just because we get 0 c = or c = does t mea that the series will have the opposite covergece. To see this cosider the series, 007 Paul Dawkis

233 = = Both of these series coverge ad here are the two possible limits that the limit compariso test uses. lim = lim = 0 lim = lim = So, eve though both series had the same covergece we got both c = 0 ad c =. The poit of all of this is to remid us that if we get c = 0 or c = from the limit compariso test we will kow that we have chose the secod series icorrectly ad we ll eed to fid a differet choice i order to get ay iformatio about the covergece of the series. We ll close out this sectio with proofs of the two tests. Proof of Compariso Test The test statemet did ot specify where each series should start. We oly eed to require that they start at the same place so to help with the proof we ll assume that the series start at =. If the series do t start at = the proof ca be redoe i exactly the same maer or you could use a idex shift to start the series at = ad the this proof will apply. We ll start off with the partial sums of each series. s = a t = b i i i= i= Let s otice a couple of ice facts about these two partial sums. First, because a, b 0 we kow that, + s s + a = a + a = a = s s s + i + i + + i= i= + t t + b = b + b = b = t t t + i + i + + i= i= So, both partial sums form icreasig sequeces. Also, because a b for all we kow that we must have s t for all. With these prelimiary facts out of the way we ca proceed with the proof of the test itself. Let s start out by assumig that = b is a coverget series. Sice b 0 we kow that, t = b b i i i= i= 007 Paul Dawkis 4

234 However, we also have established that s t for all ad so for all we also have, s bi Fially sice b is a coverget series it must have a fiite value ad so the partial sums, s = are bouded above. Therefore, from the secod sectio o sequeces we kow that a mootoic i= s ad bouded sequece is also coverget ad so { } is coverget. Next, let s assume that a is a coverget sequece ad so a = is diverget. Because a 0 we the kow that we must have = s as. However, we also kow that for all we have s kow that t as. t ad therefore we also = t So, { } is a diverget sequece ad so b = = is diverget. Proof of Limit Compariso Test Because 0 < c < we ca fid two positive ad fiite umbers, m ad M, such that m< c< M. a a Now, because lim c = we kow that for large eough the quotiet must be close to c b b ad so there must be a positive iteger N such that if > N we also have, a m< < M b Multiplyig through by b gives, mb < a < Mb provided > N. Now, if b diverges the so does mb ad so sice mb < a for all sufficietly large by the Compariso Test also diverges. a Likewise, if b coverges the so does Mb ad sice a < Mb for all sufficietly large by the Compariso Test a also coverges. 007 Paul Dawkis 5

235 Alteratig Series Test The last two tests that we looked at for series covergece have required that all the terms i the series be positive. Of course there are may series out there that have egative terms i them ad so we ow eed to start lookig at tests for these kids of series. The test that we are goig to look ito i this sectio will be a test for alteratig series. A alteratig series is ay series, a, for which the series terms ca be writte i oe of the followig two forms. a = b b 0 + a = b b 0 There are may other ways to deal with the alteratig sig, but they ca all be writte as oe of the two forms above. For istace, + = = ( ) = ( ) ( ) = ( ) + + There are of course may others, but they all follow the same basic patter of reducig to oe of the first two forms give. If you should happe to ru ito a differet form tha the first two, do t worry about covertig it to oe of those forms, just be aware that it ca be ad so the test from this sectio ca be used. Alteratig Series Test Suppose that we have a series for all. The if, a ad either a = ( ) b or + a = b where b 0. lim b = 0 ad,. { b } is a decreasig sequece the series a is coverget. A proof of this test is at the ed of the sectio. There are a couple of thigs to ote about this test. First, ulike the Itegral Test ad the Compariso/Limit Compariso Test, this test will oly tell us whe a series coverges ad ot if a series will diverge. Secodly, i the secod coditio all that we eed to require is that the series terms, b will be evetually decreasig. It is possible for the first few terms of a series to icrease ad still have the test be valid. All that is required is that evetually we will have b b + for all after some poit. To see why this is cosider the followig series, 007 Paul Dawkis 6

236 = ( ) b Let s suppose that for N b is ot decreasig ad that for N decreasig. The series ca the be writte as, { } N ( ) b = ( ) b + ( ) b = = = N+ b is + { } The first series is a fiite sum (o matter how large N is) of fiite terms ad so we ca compute its value ad it will be fiite. The covergece of the series will deped solely o the covergece of the secod (ifiite) series. If the secod series has a fiite value the the sum of two fiite values is also fiite ad so the origial series will coverge to a fiite value. O the other had if the secod series is diverget either because its value is ifiite or it does t have a value the addig a fiite umber oto this will ot chage that fact ad so the origial series will be diverget. The poit of all this is that we do t eed to require that the series terms be decreasig for all. We oly eed to require that the series terms will evetually be decreasig sice we ca always strip out the first few terms that are t actually decreasig ad look oly at the terms that are actually decreasig. Note that, i practice, we do t actually strip out the terms that are t decreasig. All we do is check that evetually the series terms are decreasig ad the apply the test. Let s work a couple of examples. Example Determie if the followig series is coverget or diverget. + = Solutio First, idetify the b for the test. + ( ) + = ( ) b = = = Now, all that we eed to do is ru through the two coditios i the test. lim b = lim = 0 b = > = b+ + Both coditios are met ad so by the Alteratig Series Test the series must coverge. The series from the previous example is sometimes called the Alteratig Harmoic Series. Also, the ( ) + could be ( ) or ay other form of alteratig sig ad we d still call it a Alteratig Harmoic Series. 007 Paul Dawkis 7

237 I the previous example it was easy to see that the series terms decreased sice icreasig oly icreased the deomiator for the term ad hece made the term smaller. I geeral however we will eed to resort to Calculus I techiques to prove the series terms decrease. We ll see a example of this i a bit. Example Determie if the followig series is coverget or diverget. ( ) = + 5 Solutio First, idetify the b for the test. ( ) = ( ) b = = = Let s check the coditios. lim b = lim = So, the first coditio is t met ad so there is o reaso to check the secod. Sice this coditio is t met we ll eed to use aother test to check covergece. I these cases where the first coditio is t met it is usually best to use the divergece test. So, the divergece test requires us to compute the followig limit. ( ) lim + 5 This limit ca be somewhat tricky to evaluate. For a secod let s cosider the followig, ( ) lim = ( lim ( ) ) lim Splittig this limit like this ca t be doe because this operatio requires that both limits exist ad while the secod oe does the first clearly does ot. However, it does show us how we ca at least covice ourselves that the overall limit does ot exist (eve if it wo t be a direct proof of that fact). So, let s start with, lim = lim ( ) Now, the secod part of this clearly is goig to as while the first part just alterates betwee ad -. So, as the terms are alteratig betwee positive ad egative values that are gettig closer ad closer to ad - respectively. I order for limits to exist we kow that the terms eed to settle dow to a sigle umber ad sice these clearly do t this limit does t exist ad so by the Divergece Test this series 007 Paul Dawkis 8

238 diverges. Example Determie if the followig series is coverget or diverget. ( ) = Solutio Notice that i this case the expoet o the - is t or +. That wo t chage how the test works however so we wo t worry about that. I this case we have, so let s check the coditios. b = + 4 The first is easy eough to check. lim b = lim = The secod coditio requires some work however. It is ot immediately clear that these terms will decrease. Icreasig to + will icrease both the umerator ad the deomiator. Icreasig the umerator says the term should also icrease while icreasig the deomiator says that the term should decrease. Sice it s ot clear which of these will wi out we will eed to resort to Calculus I techiques to show that the terms decrease. Let s start with the followig fuctio ad its derivative. x 4 x f ( x) = f ( x) = x + 4 x x+ 4 Now, there are three critical poits for this fuctio, x = 4, x = 0, ad x = 4. The first is outside the boud of our series so we wo t eed to worry about that oe. Usig the test poits, f = f ( 5) = ad so we ca see that the fuctio i icreasig o 0 x 4 ad decreasig o x 4. Therefore, sice f = b we kow as well that the b are also icreasig o 0 4 ad decreasig o 4. The b are the evetually decreasig ad so the secod coditio is met. Both coditios are met ad so by the Alteratig Series Test the series must be covergig. As the previous example has show, we sometimes eed to do a fair amout of work to show that the terms are decreasig. Do ot just make the assumptio that the terms will be decreasig ad let it go at that. Let s do oe more example just to make a poit. Example 4 Determie if the followig series is coverget or diverget. 007 Paul Dawkis 9

239 = cos ( π ) Solutio The poit of this problem is really just to ackowledge that it is i fact a alteratig series. To see this we eed to ackowledge that, cos( π ) = ( ) ad so the series is really, cos( π ) ( ) = b = = = Checkig the two coditio gives, lim b = lim = 0 b = > = b + + The two coditios of the test are met ad so by the Alteratig Series Test the series is coverget. It should be poited out that the rewrite we did i previous example oly works because is a iteger ad because of the presece of the π. Without the π we could t do this ad if was t guarateed to be a iteger we could t do this. Let s close this sectio out with a proof of the Alteratig Series Test. Proof of Alteratig Series Test Without loss of geerality we ca assume that the series starts at =. If ot we could modify the proof below to meet the ew startig place or we could do a idex shift to get the series to start at =. First, otice that because the terms of the sequece are decreasig for ay two successive terms we ca say, b b + 0 Now, let s take a look at the eve partial sums. s = b b 0 s4 = b b + b b4 = s + b b4 s because b b4 0 s6 = s4 + b5 b6 s4 because b5 b6 0 s = s + b b s because b b Paul Dawkis 0

240 So, { s } is a icreasig sequece. Next, we ca also write the geeral term as, s = b b + b b4 + b5 + b + b b = b b b b b + b b b 4 5 Each of the quatities i parethesis are positive ad by assumptio we kow that b is also positive. So, this tells us that s b for all. We ow kow that { s } is a icreasig sequece that is bouded above ad so we kow that it must also coverge. So, let s assume that its limit is s or, lim s = s Next, we ca quickly determie the limit of the sequece of odd partial sums, { s + } lim s = lim ( s + b ) = lim s + lim b = s+ 0 = s + + +, as follows, So, we ow kow that both { s } ad { s + } are coverget sequeces ad they both have the same limit ad so we also kow that { s } is a coverget sequece with a limit of s. This i tur tells us that a is coverget. 007 Paul Dawkis

241 Absolute Covergece Whe we first talked about series covergece we briefly metioed a stroger type of covergece but did t do aythig with it because we did t have ay tools at our disposal that we could use to work problems ivolvig it. We ow have some of those tools so it s ow time to talk about absolute covergece i detail. First, let s go back over the defiitio of absolute covergece. Defiitio A series a is called absolutely coverget if a is coverget. If a is coverget ad a is diverget we call the series coditioally coverget. We also have the followig fact about absolute covergece. Fact If a is absolutely coverget the it is also coverget. Proof First otice that say, a is either a or it is a depedig o its sig. This meas that we ca the 0 a + a a Now, sice we are assumig that a is coverget the a is also coverget sice we ca just factor the out of the series ad times a fiite value will still be fiite. This however allows us to use the Compariso Test to say that a + a is also a coverget series. Fially, we ca write, ad so a a = a + a a is the differece of two coverget series ad so is also coverget. This fact is oe of the ways i which absolute covergece is a stroger type of covergece. Series that are absolutely coverget are guarateed to be coverget. However, series that are coverget may or may ot be absolutely coverget. Let s take a quick look at a couple of examples of absolute covergece. Example Determie if each of the followig series are absolute coverget, coditioally coverget or diverget. 007 Paul Dawkis

242 (a) (b) Solutio (a) = = ( ) [Solutio] ( ) + [Solutio] = si (c) [Solutio] = This is the alteratig harmoic series ad we saw i the last sectio that it is a coverget series so we do t eed to check that here. So, let s see if it is a absolutely coverget series. To do this we ll eed to check the covergece of. ( ) = = = This is the harmoic series ad we kow from the itegral test sectio that it is diverget. Therefore, this series is ot absolutely coverget. It is however coditioally coverget sice the series itself does coverge. [Retur to Problems] (b) = ( ) + I this case let s just check absolute covergece first sice if it s absolutely coverget we wo t eed to bother checkig covergece as we will get that for free. + ( ) = = = This series is coverget by the p-series test ad so the series is absolute coverget. Note that this does say as well that it s a coverget series. [Retur to Problems] si (c) = I this part we eed to be a little careful. First, this is NOT a alteratig series ad so we ca t use ay tools from that sectio. What we ll do here is check for absolute covergece first agai sice that will also give covergece. This meas that we eed to check the covergece of the followig series. 007 Paul Dawkis

243 si = si = = To do this we ll eed to ote that si si ad so we have, si Now we kow that = coverges by the p-series test ad so by the Compariso Test we also kow that si coverges. = Therefore the origial series is absolutely coverget (ad hece coverget). [Retur to Problems] Let s close this sectio off by recappig a topic we saw earlier. Whe we first discussed the covergece of series i detail we oted that we ca t thik of series as a ifiite sum because some series ca have differet sums if we rearrage their terms. I fact, we gave two rearragemets of a Alteratig Harmoic series that gave two differet values. We closed that sectio off with the followig fact, Facts Give the series a,. If a is absolutely coverget ad its value is s the ay rearragemet of a will also have a value of s. 4. If a is coditioally coverget ad r is ay real umber the there is a rearragemet of a whose value will be r. Now that we ve got the tools uder our belt to determie absolute ad coditioal covergece we ca make a few more commets about this. First, as we showed above i Example a a Alteratig Harmoic is coditioally coverget ad so o matter what value we chose there is some rearragemet of terms that will give that value. Note as well that this fact does ot tell us what that rearragemet must be oly that it does exist. Next, we showed i Example b that, 007 Paul Dawkis 4

244 ( ) + = is absolutely coverget ad so o matter how we rearrage the terms of this series we ll always get the same value. I fact, it ca be show that the value of this series is, = ( ) + π = 007 Paul Dawkis 5

245 Ratio Test I this sectio we are goig to take a look at a test that we ca use to see if a series is absolutely coverget or ot. Recall that if a series is absolutely coverget the we will also kow that it s coverget ad so we will ofte use it to simply determie the covergece of a series. Before proceedig with the test let s do a quick remider of factorials. This test will be particularly useful for series that cotai factorials (ad we will see some i the applicatios) so let s make sure we ca deal with them before we ru ito them i a example. If is a iteger such that 0 the factorial is defied as,! = if 0! = by defiitio Let s compute a couple real quick.! =! = =! = = 6 4! = 4 = 4 5! = 5 4 = 0 I the last computatio above, otice that we could rewrite the factorial i a couple of differet ways. For istace, 5! = 5( 4)( ) = 5 4! 4! 5! = 54 = 54! I geeral we ca always strip out terms from a factorial as follows.! = k k+! ( ) ( ( ))! (! ) = k k+ = k k We will eed to do this o occasio so do t forget about it. Also, whe dealig with factorials we eed to be very careful with parethesis. For istace,!! as we ca see if we write each of the followig factorials out. ( )! = ( )( ) ( )! = ( )( ) ( ) 007 Paul Dawkis 6

246 Agai, we will ru across factorials with parethesis so do t drop them. This is ofte oe of the more commo mistakes that studets make whe they first ru across factorials. Okay, we are ow ready for the test. Ratio Test Suppose we have the series a. Defie, a lim + L = a The,. if L < the series is absolutely coverget (ad hece coverget).. if L > the series is diverget.. if L = the series may be diverget, coditioally coverget, or absolutely coverget. A proof of this test is at the ed of the sectio. Notice that i the case of L = the ratio test is pretty much worthless ad we would eed to resort to a differet test to determie the covergece of the series. Also, the absolute value bars i the defiitio of L are absolutely required. If they are ot there it will be impossible for us to get the icorrect aswer. Let s take a look at some examples. Example Determie if the followig series is coverget or diverget. ( 0) = Solutio With this first example let s be a little careful ad make sure that we have everythig dow correctly. Here are the series terms a. ( 0) a = Recall that to compute a + all that we eed to do is substitute + for all the s i a. + + ( 0) ( 0) a+ = = ( + ) ( ) Now, to defie L we will use, L= lim a + a sice this will be a little easier whe dealig with fractios as we ve got here. So, 007 Paul Dawkis 7

247 + ( ) ( ) ( + ) lim ( ) L = lim = = lim = < 6 So, L < ad so by the Ratio Test the series coverges absolutely ad hece will coverge. As see i the previous example there is usually a lot of cacelig that will happe i these. Make sure that you do this cacelig. If you do t do this kid of cacelig it ca make the limit fairly difficult. Example Determie if the followig series is coverget or diverget.! 0 5 = Solutio Now that we ve worked oe i detail we wo t go ito quite the detail with the rest of these. Here is the limit. ( +! ) 5 ( +! ) L = lim = lim + 5! 5! I order to do this limit we will eed to elimiate the factorials. We simply ca t do the limit with the factorials i it. To elimiate the factorials we will recall from our discussio o factorials above that we ca always strip out terms from a factorial. If we do that with the umerator (i this case because it s the larger of the two) we get, ( + )! L = lim 5! at which poit we ca cacel the! for the umerator a deomiator to get, ( + ) L = lim => 5 So, by the Ratio Test this series diverges. Example Determie if the followig series is coverget or diverget.! = Solutio I this case be careful i dealig with the factorials. 007 Paul Dawkis 8

248 ( + ) ( ) ( ( + ) )! L = lim! ( + ) ( ) ( + )! = lim! ( + ) ( ) ( + )( )! = lim! ( + ) = lim = 0< ( + )( ) So, by the Ratio Test this series coverges absolutely ad so coverges. Example 4 Determie if the followig series is coverget or diverget. 9 + = Solutio Do ot mistake this for a geometric series. The i the deomiator meas that this is t a geometric series. So, let s compute the limit ( ) L = lim = lim ( )( + ) 9 = lim + 9 = > Therefore, by the Ratio Test this series is diverget. I the previous example the absolute value bars were required to get the correct aswer. If we 9 had t used them we would have gotte L = < which would have implied a coverget series! Now, let s take a look at a couple of examples to see what happes whe we get L =. Recall that the ratio test will ot tell us aythig about the covergece of these series. I both of these examples we will first verify that we get L = ad the use other tests to determie the covergece. 007 Paul Dawkis 9

249 Example 5 Determie if the followig series is coverget or diverget. Solutio Let s first get L. = 0 ( ) + + ( ) + + ( ) ( ) L = lim = lim = So, as implied earlier we get L = which meas the ratio test is o good for determiig the covergece of this series. We will eed to resort to aother test for this series. This series is a alteratig series ad so let s check the two coditios from that test. lim b = lim = 0 + b = > = b The two coditios are met ad so by the Alteratig Series Test this series is coverget. We ll leave it to you to verify this series is also absolutely coverget. Example 6 Determie if the followig series is coverget or diverget. + = Solutio Here s the limit ( + )( + 7) L = lim = lim = Agai, the ratio test tells us othig here. We ca however, quickly use the divergece test o this. I fact that probably should have bee our first choice o this oe ayway. + lim = By the Divergece Test this series is diverget. So, as we saw i the previous two examples if we get L = from the ratio test the series ca be either coverget or diverget. There is oe more thig that we should ote about the ratio test before we move oto the ext sectio. The last series was a polyomial divided by a polyomial ad we saw that we got L = from the ratio test. This will always happe with ratioal expressio ivolvig oly polyomials or polyomials uder radicals. So, i the future it is t eve worth it to try the ratio test o these kids of problems sice we ow kow that we will get L =. 007 Paul Dawkis 40

250 Also, i the secod to last example we saw a example of a alteratig series i which the positive term was a ratioal expressio ivolvig polyomials ad agai we will always get L = i these cases. Let s close the sectio out with a proof of the Ratio Test. Proof of Ratio Test First ote that we ca assume without loss of geerality that the series will start at = as we ve doe for all our series test proofs. Let s start off the proof here by assumig that L < ad we ll eed to show that absolutely coverget. To do this let s first ote that because that L< r <. a is L < there is some umber r such Now, recall that, a lim + L = a ad because we also have chose r such that L< r there is some N such that if N we will have, a+ < r a+ < ra a Next, cosider the followig, So, for k =,,, we have look at the followig series. a N+ < ra a < ra < r a N N+ N+ N a < ra < r a N+ N+ N a < ra < r a k N+ k N+ k N k an+ k< r an. Just why is this importat? Well we ca ow k = 0 a N r k This is a geometric series ad because 0< r < we i fact kow that it is a coverget series. k Also because a + < r a by the Compariso test the series N k N is coverget. However sice, a = a N+ k = N+ k= N a = a + a = = = N+ 007 Paul Dawkis 4

251 we kow that = terms ad hece fiite. Therefore a is also coverget sice the first term o the right is a fiite sum of fiite = a is absolutely coverget. Next, we eed to assume that L > ad we ll eed to show that that, a lim + L = a ad because L > we kow that there must be some N such that if a+ > a+ > a a However, if a+ > a for all N the we kow that, lim a 0 a is diverget. Recallig N we will have, because the terms are gettig larger ad guarateed to ot be egative. This i tur meas that, lim a 0 Therefore, by the Divergece Test a is diverget. Fially, we eed to assume that L = ad show that we could get a series that has ay of the three possibilities. To do this we just eed a series for each case. We ll leave the details of checkig to you but all three of the followig series have L = ad each oe exhibits oe of the possibilities. absolutely coverget = = = ( ) coditioally coverget diverget 007 Paul Dawkis 4

252 Root Test This is the last test for series covergece that we re goig to be lookig at. As with the Ratio Test this test will also tell whether a series is absolutely coverget or ot rather tha simple covergece. Root Test Suppose that we have the series a. Defie, L= lim a = lim a The, 4. if L < the series is absolutely coverget (ad hece coverget). 5. if L > the series is diverget. 6. if L = the series may be diverget, coditioally coverget, or absolutely coverget. A proof of this test is at the ed of the sectio. As with the ratio test, if we get L = the root test will tell us othig ad we ll eed to use aother test to determie the covergece of the series. Also ote that if L = i the Ratio Test the the Root Test will also give L =. We will also eed the followig fact i some of these problems. Fact lim = Let s take a look at a couple of examples. Example Determie if the followig series is coverget or diverget. + = Solutio There really is t much to these problems other tha computig the limit ad the usig the root test. Here is the limit for this problem. So, by the Root Test this series is diverget. L + + = lim = lim = => Example Determie if the followig series is coverget or diverget. 5 = Solutio 007 Paul Dawkis 4

253 Agai, there is t too much to this series. 5 5 L = lim lim = = = < Therefore, by the Root Test this series coverges absolutely ad hece coverges. Note that we had to keep the absolute value bars o the fractio util we d take the limit to get the sig correct. Example Determie if the followig series is coverget or diverget. ( ) = Solutio Here s the limit for this series. ( ) L = lim = lim = = > After usig the fact from above we ca see that the Root Test tells us that this series is diverget. Proof of Root Test First ote that we ca assume without loss of geerality that the series will start at = as we ve doe for all our series test proofs. Also ote that this proof is very similiar to the proof of the Ratio Test. Let s start off the proof here by assumig that L < ad we ll eed to show that absolutely coverget. To do this let s first ote that because that L< r <. Now, recall that, L= lim a = lim a a is L < there is some umber r such ad because we also have chose r such that L< r there is some N such that if N we will have, Now the series a < r a < r = 0 is a geometric series ad because 0< r < we i fact kow that it is a coverget series. Also because a < r N by the Compariso test the series r 007 Paul Dawkis 44

254 is coverget. However sice, we kow that = terms ad hece fiite. Therefore a = N a N a = a + a = = = N is also coverget sice the first term o the right is a fiite sum of fiite = a is absolutely coverget. Next, we eed to assume that L > ad we ll eed to show that that, L= lim a = lim a ad because L > we kow that there must be some N such that if a > a > = However, if a > for all N the we kow that, lim a 0 a is diverget. Recallig N we will have, This i tur meas that, lim a 0 Therefore, by the Divergece Test a is diverget. Fially, we eed to assume that L = ad show that we could get a series that has ay of the three possibilities. To do this we just eed a series for each case. We ll leave the details of checkig to you but all three of the followig series have L = ad each oe exhibits oe of the possibilities. absolutely coverget = = = ( ) coditioally coverget diverget 007 Paul Dawkis 45

255 Strategy for Series Now that we ve got all of our tests out of the way it s time to thik about orgaizig all of them ito a geeral set of guidelies to help us determie the covergece of a series. Note that these are a geeral set of guidelies ad because some series ca have more tha oe test applied to them we will get a differet result depedig o the path that we take through this set of guidelies. I fact, because more tha oe test may apply, you should always go completely through the guidelies ad idetify all possible tests that ca be used o a give series. Oce this has bee doe you ca idetify the test that you feel will be the easiest for you to use. With that said here is the set of guidelies for determiig the covergece of a series.. With a quick glace does it look like the series terms do t coverge to zero i the limit, i.e. does lim a 0? If so, use the Divergece Test. Note that you should oly do the Divergece Test if a quick glace suggests that the series terms may ot coverge to zero i the limit.. Is the series a p-series ( p ) or a geometric series ( ar or ar )? If so use = 0 = the fact that p-series will oly coverge if p > ad a geometric series will oly coverge if r <. Remember as well that ofte some algebraic maipulatio is required to get a geometric series ito the correct form.. Is the series similar to a p-series or a geometric series? If so, try the Compariso Test. 4. Is the series a ratioal expressio ivolvig oly polyomials or polyomials uder radicals (i.e. a fractio ivolvig oly polyomials or polyomials uder radicals)? If so, try the Compariso Test ad/or the Limit Compariso Test. Remember however, that i order to use the Compariso Test ad the Limit Compariso Test the series terms all eed to be positive. 5. Does the series cotai factorials or costats raised to powers ivolvig? If so, the the Ratio Test may work. Note that if the series term cotais a factorial the the oly test that we ve got that will work is the Ratio Test. 6. Ca the series terms be writte i the form a = ( ) b or + a the Alteratig Series Test may work. 7. Ca the series terms be writte i the form 8. work. a = b? If so, the = b? If so, the the Root Test may If a = f for some positive, decreasig fuctio ad f the the Itegral Test may work. x dx is easy to evaluate a 007 Paul Dawkis 46

256 Agai, remember that these are oly a set of guidelies ad ot a set of hard ad fast rules to use whe tryig to determie the best test to use o a series. If more tha oe test ca be used try to use the test that will be the easiest for you to use ad remember that what is easy for someoe else may ot be easy for you! Also just so we ca put all the tests ito oe place here is a quick listig of all the tests that we ve got. Divergece Test If lim a 0 the a will diverge Itegral Test Suppose that f ( x) is a positive, decreasig fuctio o the iterval [ k, ) ad that f = a the, x dx is coverget so is a.. If f. If f k k = k x dx is diverget so is a. Compariso Test Suppose that we have two series The,. If b. If a = k a ad b is coverget the so is a. is diverget the so is b. with a, 0 b for all ad a b for all. Limit Compariso Test Suppose that we have two series a ad b with a, b 0 for all. Defie, a c = lim b If c is positive (i.e. c > 0 ) ad is fiite (i.e. c <) the either both series coverge or both series diverge. Alteratig Series Test Suppose that we have a series for all. The if, a ad either a = ( ) b or + a = b where b 0. lim b = 0 ad,. { b } is evetually a decreasig sequece the series a is coverget 007 Paul Dawkis 47

257 Ratio Test Suppose we have the series a. Defie, a + L = lim The,. if L < the series is absolutely coverget (ad hece coverget).. if L > the series is diverget.. if L = the series may be diverget, coditioally coverget, or absolutely coverget. Root Test Suppose that we have the series a. Defie, a L= lim a = lim a The,. if L < the series is absolutely coverget (ad hece coverget).. if L > the series is diverget.. if L = the series may be diverget, coditioally coverget, or absolutely coverget. 007 Paul Dawkis 48

258 Estimatig the Value of a Series We have ow spet quite a few sectios determiig the covergece of a series, however, with the exceptio of geometric ad telescopig series, we have ot talked about fidig the value of a series. This is usually a very difficult thig to do ad we still are t goig to talk about how to fid the value of a series. What we will do is talk about how to estimate the value of a series. Ofte that is all that you eed to kow. Before we get ito how to estimate the value of a series let s remid ourselves how series covergece works. It does t make ay sese to talk about the value of a series that does t coverge ad so we will be assumig that the series we re workig with coverges. Also, as we ll see the mai method of estimatig the value of series will come out of this discussio. So, let s start with the series = a (the startig poit is ot importat, but we eed a startig poit to do the work) ad let s suppose that the series coverges to s. Recall that this meas that if we get the partial sums, s = ai the they will form a coverget sequece ad its limit is s. I other words, lim s = s Now, just what does this mea for us? Well, sice this limit coverges it meas that we ca make the partial sums, s, as close to s as we wat simply by takig large eough. I other words, if we take large eough the we ca say that, i= s s This is oe method of estimatig the value of a series. We ca just take a partial sum ad use that as a estimatio of the value of the series. There are ow two questios that we should ask about this. First, how good is the estimatio? If we do t have a idea of how good the estimatio is the it really does t do all that much for us as a estimatio. Secodly, is there ay way to make the estimate better? Sometimes we ca use this as a startig poit ad make the estimatio better. We wo t always be able to do this, but if we ca that will be ice. So, let s start with a geeral discussio about the determiig how good the estimatio is. Let s first start with the full series ad strip out the first terms. a = a + a () i i i i= i= i= + Note that we coverted over to a idex of i i order to make the otatio cosistet with prior otatio. Recall that we ca use ay letter for the idex ad it wo t chage the value. 007 Paul Dawkis 49

259 Now, otice that the first series (the terms that we ve stripped out) is othig more tha the partial sum s. The secod series o the right (the oe startig at i = + ) is called the remaider ad deoted by R. Fially let s ackowledge that we also kow the value of the series sice we are assumig it s coverget. Takig this otatio ito accout we ca rewrite () as, s = s + R We ca solve this for the remaider to get, R = s s So, the remaider tells us the differece, or error, betwee the exact value of the series ad the value of the partial sum that we are usig as the estimatio of the value of the series. Of course we ca t get our hads o the actual value of the remaider because we do t have the actual value of the series. However, we ca use some of the tests that we ve got for covergece to get a pretty good estimate of the remaider provided we make some assumptios about the series. Oce we ve got a estimate o the value of the remaider we ll also have a idea o just how good a job the partial sum does of estimatig the actual value of the series. There are several tests that will allow us to get estimates of the remaider. We ll go through each oe separately. Itegral Test Recall that i this case we will eed to assume that the series terms are all positive ad will evetually be decreasig. We derived the itegral test by usig the fact that the series could be thought of as a estimatio of the area uder the curve of f ( x ) where f = a. We ca do somethig similar with the remaider. As we ll soo see if we ca get a upper ad lower boud o the value of the remaider we ca use these bouds to help us get upper ad lower bouds o the value of the series. We ca i tur use the upper ad lower bouds o the series value to actually estimate the value of the series. So, let s first recall that the remaider is, R = a = a + a + a + a + i i= + Now, if we start at x= +, take rectagles of width ad use the left edpoit as the height of the rectagle we ca estimate the area uder f ( x ) o the iterval [ +, ) as show i the sketch below. 007 Paul Dawkis 50

260 We ca see that the remaider, R, is the area estimatio ad it will overestimate the exact area. So, we have the followig iequality. + R f x dx () Next, we could also estimate the area by startig at x=, takig rectagles of width agai ad the usig the right edpoit as the height of the rectagle. This will give a estimatio of the,. This is show i the followig sketch. area uder f ( x ) o the iterval [ ) Agai, we ca see that the remaider, R, is agai this estimatio ad i this case it will uderestimate the area. This leads to the followig iequality, Combiig () ad () gives, R f x dx () 007 Paul Dawkis 5

261 + f x dx R f x dx So, provided we ca do these itegrals we ca get both a upper ad lower boud o the remaider. This will i tur give us a upper boud ad a lower boud o just how good the partial sum, s, is as a estimatio of the actual value of the series. I this case we ca also use these results to get a better estimate for the actual value of the series as well. First, we ll start with the fact that Now, if we use () we get, Likewise if we use () we get, Puttig these two together gives us, s = s + R = s s R s f x dx = + + s s R s f x dx s f x dx s s + + f x dx + (4) This gives a upper ad a lower boud o the actual value of the series. We could the use as a estimate of the actual value of the series the average of the upper ad lower boud. Let s work a example with this. Example Usig = 5 to estimate the value of =. Solutio First, for compariso purposes, we ll ote that the actual value of this series is kow to be, π = = = Usig = 5 let s first get the partial sum. 5 s5 = = i i= Note that this is close to the actual value i some sese, but is t really all that close either. Now, let s compute the itegrals. These are fairly simple itegrals so we ll leave it to you to verify the values. 007 Paul Dawkis 5

262 dx = dx = x 5 x Pluggig these ito (4) gives us, s s Both the upper ad lower boud are ow very close to the actual value ad if we take the average of the two we get the followig estimate of the actual value. That is pretty dar close to the actual value. s So, that is how we ca use the Itegral Test to estimate the value of a series. Let s move o to the ext test. Compariso Test I this case, ulike with the itegral test, we may or may ot be able to get a idea of how good a particular partial sum will be as a estimate of the exact value of the series. Much of this will deped o how the compariso test is used. First, let s remid ourselves o how the compariso test actually works. Give a series a let s assume that we ve used the compariso test to show that it s coverget. Therefore, we foud a secod series b that coverged ad a b for all. What we wat to do is determie how good of a job the partial sum, s = ai i= will do i estimatig the actual value of the series a. Agai, we will use the remaider to do this. Let s actually write dow the remaider for both series. R = a T = b i i i= + i= + Now, sice a b we also kow that R T Whe usig the compariso test it is ofte the case that the b are fairly ice terms ad that we might actually be able to get a idea o the size of T. For istace, if our secod series is a p- series we ca use the results from above to get a upper boud o T as follows, where = R T g x dx g b 007 Paul Dawkis 5

263 Also, if the secod series is a geometric series the we will be able to compute T exactly. If we are uable to get a idea of the size of T the usig the compariso test to help with estimates wo t do us much good. Let s take a look at a example. Example Usig = 5 to estimate the value of. = Solutio To do this we ll first eed to go through the compariso test so we ca get the secod series. So, = ad = 0 is a geometric series ad coverges because r = <. Now that we ve gotte our secod series let s get the estimate. 5 s5 = = = 0 So, how good is it? Well we kow that, R5 T5 = = 6 will be a upper boud for the error betwee the actual value ad the estimate. Sice our secod series is a geometric series we ca compute this directly as follows. 5 = + = 0 = 0 = 6 The series o the left is i the stadard form ad so we ca compute that directly. The first series o the right has a fiite umber of terms ad so ca be computed exactly ad the secod series o the right is the oe that we d like to have the value for. Doig the work gives, 5 = = 6 = 0 = 0 = ( ) = So, accordig to this if we use s as a estimate of the actual value we will be off from the exact value by o more tha ad that s ot too bad. 007 Paul Dawkis 54

264 I this case it ca be show that = = ad so we ca see that the actual error i our estimatio is, Error = Actual Estimate = = Note that i this case the estimate of the error is actually fairly close (ad i fact exactly the same) as the actual error. This will ot always happe ad so we should t expect that to happe i all cases. The error estimate above is simply the upper boud o the error ad the actual error will ofte be less tha this value. Before movig o to the fial part of this sectio let s agai ote that we will oly be able to determie how good the estimate is usig the compariso test if we ca easily get our hads o the remaider of the secod term. The reality is that we wo t always be able to do this. Alteratig Series Test Both of the methods that we ve looked at so far have required the series to cotai oly positive terms. If we allow series to have egative terms i it the process is usually more difficult. However, with that said there is oe case where it is t too bad. That is the case of a alteratig series. Oce agai we will start off with a coverget series a = ( ) b which i this case happes to be a alteratig series that satisfies the coditios of the alteratig series test, so we kow that b 0 for all. Also ote that we could have ay power o the - we just used for the sake of coveiece. We wat to kow how good of a estimatio of the actual series value will the partial sum, s, be. As with the prior cases we kow that the remaider, R, will be the error i the estimatio ad so if we ca get a hadle o that we ll kow approximately how good the estimatio is. From the proof of the Alteratig Series Test we ca see that s will lie betwee s ad s + for ay ad so, s s s s = b Therefore, + + R = s s b + We eeded absolute value bars because we wo t kow ahead of time if the estimatio is larger or smaller tha the actual value ad we kow that the b s are positive. Let s take a look at a example. Example Usig = 5 to estimate the value of ( ). = 007 Paul Dawkis 55

265 Solutio This is a alteratig series ad it does coverge. I this case the exact value is kow ad so for compariso purposes, = ( ) π = = Now, the estimatio is, s 5 ( ) = = = From the fact above we kow that R5 = s s5 b6 = = So, our estimatio will have a error of o more tha I this case the exact value is kow ad so the actual error is, R5 = s s5 = I the previous example the estimatio had oly half the estimated error. It will ofte be the case that the actual error will be less tha the estimated error. Remember that this is oly a upper boud for the actual error. Ratio Test This will be the fial case that we re goig to look at for estimatig series values ad we are goig to have to put a couple of fairly striget restrictios o the series terms i order to do the work. Oe of the mai restrictios we re goig to make is to assume that the series terms are positive. We ll also be addig o aother restrictio i a bit. I this case we ve used the ratio test to show that a a lim + L = a ad foud that L <. is coverget. To do this we computed As with the previous cases we are goig to use the remaider, R, to determie how good of a estimatio of the actual value the partial sum, s, is. To get a estimate of the remaider let s first defie the followig sequece, a+ r = a We ow have two possible cases.. If { } r is a decreasig sequece ad < the, r Paul Dawkis 56

266 . If { r } is a icreasig sequece the, R R a r + + a+ L Proof Both parts will eed the followig work so we ll do it first. We ll start with the remaider. R = a = a + a + a + a + i i= + a a a = a a+ a+ a+ Next we eed to do a little work o a couple of these terms. a a a a 4 a a R = a a+ a+ a+ a+ a+ a+ a a a a a a = a a+ a+ a+ a+ a+ a+ Now use the defiitio of r to write this as, R = a + r + r r + r r r + Okay ow let s do the proof For the first part we are assumig that { r } is decreasig ad so we ca estimate the remaider as, ( ) R = a + r + r r + r r r ( ) a + r + r + r k + r+ k = 0 = a Fially, the series here is a geometric series ad because r + < we kow that it coverges ad we ca compute its value. So, a+ R r Paul Dawkis 57

267 For the secod part we are assumig that { r } is icreasig ad we kow that, ad so we kow that r a lim r lim + = = a < L for all. The remaider ca the be estimated as, ( ) R = a + r + r r + r r r ( ) a + L+ L + L + = a L + k = 0 k This is a geometric series ad sice we are assumig that our origial series coverges we also kow that L < ad so the geometric series above coverges ad we ca compute its value. So, a+ R L L Note that there are some restrictios o the sequece { r } ad at least oe of its terms i order to use these formulas. If the restrictios are t met the the formulas ca t be used. Let s take a look at a example of this. Example 4 Usig = 5 to estimate the value of. 0 = Solutio First, let s use the ratio test to verify that this is a coverget series. + + L = lim = lim = < + So, it is coverget. Now let s get the estimate. 5 s = = = To determie a estimate o the remaider, ad hece the error, let s first get the sequece { r }. + + r = = = + + The last rewrite was just to simplify some of the computatios a little. Now, otice that, f ( x) = + f ( x) = < 0 x x f = r ad so, this sequece is decreasig. Sice this fuctio is always decreasig ad 007 Paul Dawkis 58

268 Also ote that r = + <. Therefore we ca use the first case from the fact above to get, 6 6 R a = = r6 + 6 So, it looks like our estimate is probably quite good. I this case the exact value is kow. = = 0 4 ad so we ca compute the actual error. R5 = s s5 = This is less tha the upper boud, but ulike i the previous example this actual error is quite close to the upper boud. I the last two examples we ve see that the upper boud computatios o the error ca sometimes be quite close to the actual error ad at other times they ca be off by quite a bit. There is usually o way of kowig ahead of time which it will be ad without the exact value i had there will ever be a way of determiig which it will be. Notice that this method did require the series terms to be positive, but that does t mea that we ca t deal with ratio test series if they have egative terms. Ofte series that we used ratio test o are also alteratig series ad so if that is the case we ca always resort to the previous material to get a upper boud o the error i the estimatio, eve if we did t use the alteratig series test to show covergece. Note however that if the series does have egative terms, but does t happe to be a alteratig series the we ca t use ay of the methods discussed i this sectio to get a upper boud o the error. 007 Paul Dawkis 59

269 Power Series We ve spet quite a bit of time talkig about series ow ad with oly a couple of exceptios we ve spet most of that time talkig about how to determie if a series will coverge or ot. It s ow time to start lookig at some specific kids of series ad we ll evetually reach the poit where we ca talk about a couple of applicatios of series. I this sectio we are goig to start talkig about power series. A power series about a, or just power series, is ay series that ca be writte i the form, = 0 ( ) c x a where a ad c are umbers. The c s are ofte called the coefficiets of the series. The first thig to otice about a power series is that it is a fuctio of x. That is differet from ay other kid of series that we ve looked at to this poit. I all the prior sectios we ve oly allowed umbers i the series ad ow we are allowig variables to be i the series as well. This will ot chage how thigs work however. Everythig that we kow about series still holds. I the discussio of power series covergece is still a major questio that we ll be dealig with. The differece is that the covergece of the series will ow deped upo the values of x that we put ito the series. A power series may coverge for some values of x ad ot for other values of x. Before we get too far ito power series there is some termiology that we eed to get out of the way. First, as we will see i our examples, we will be able to show that there is a umber R so that the power series will coverge for, x a < Rad will diverge for x a > R. This umber is called the radius of covergece for the series. Note that the series may or may ot coverge if x a = R. What happes at these poits will ot chage the radius of covergece. Secodly, the iterval of all x s, icludig the edpoits if eed be, for which the power series coverges is called the iterval of covergece of the series. These two cocepts are fairly closely tied together. If we kow that the radius of covergece of a power series is R the we have the followig. a R< x< a+ R x< a R ad x> a+ R power series coverges power series diverges The iterval of covergece must the cotai the iterval a R< x< a+ R sice we kow that the power series will coverge for these values. We also kow that the iterval of covergece ca t cotai x s i the rages x< a R ad x> a+ R sice we kow the power series diverges for these value of x. Therefore, to completely idetify the iterval of covergece all that we have to do is determie if the power series will coverge for x= a R or x= a+ R. If the power series coverges for oe or both of these values the we ll eed to iclude those i the iterval of covergece. 007 Paul Dawkis 60

270 Before gettig ito some examples let s take a quick look at the covergece of a power series for the case of x= a. I this case the power series becomes, 0 c a a = c 0 = c 0 + c 0 = c + 0= c + 0= c = 0 = 0 = = ad so the power series coverges. Note that we had to strip out the first term sice it was the oly o-zero term i the series. It is importat to ote that o matter what else is happeig i the power series we are guarateed to get covergece for x= a. The series may ot coverge for ay other value of x, but it will always coverge for x= a. Let s work some examples. We ll put quite a bit of detail ito the first example ad the ot put quite as much detail i the remaiig examples. Example Determie the radius of covergece ad iterval of covergece for the followig power series. ( ) ( x + ) = 4 Solutio Okay, we kow that this power series will coverge for x =, but that s it at this poit. To determie the remaider of the x s for which we ll get covergece we ca use ay of the tests that we ve discussed to this poit. After applicatio of the test that we choose to work with we will arrive at coditio(s) o x that we ca use to determie which values of x for which the power series will coverge ad which values of x for which the power series will diverge. From this we ca get the radius of covergece ad most of the iterval of covergece (with the possible exceptio of the edpoits). With all that said, the best tests to use here are almost always the ratio or root test. Most of the power series that we ll be lookig at are set up for oe or the other. I this case we ll use the ratio test. L = lim = lim ( ) ( + )( x+ ) ( )( x ) ( ) ( x+ ) + 4 Before goig ay farther with the limit let s otice that sice x is ot depedet o the limit it ca be factored out of the limit. Notice as well that i doig this we ll eed to keep the absolute value bars o it sice we eed to make sure everythig stays positive ad x could well be a value that will make thigs egative. The limit is the, + L= x+ lim 4 = x + 4 So, the ratio test tells us that if L < the series will coverge, if L > the series will diverge, 007 Paul Dawkis 6

271 ad if L = we do t kow what will happe. So, we have, x+ < 4 x+ < 4 series coverges x+ > 4 x+ > 4 series diverges We ll deal with the L = case i a bit. Notice that we ow have the radius of covergece for this power series. These are exactly the coditios required for the radius of covergece. The radius of covergece for this power series is R = 4. Now, let s get the iterval of covergece. We ll get most (if ot all) of the iterval by solvig the first iequality from above. 4< x + < 4 7< x < So, most of the iterval of validity is give by 7< x <. All we eed to do is determie if the power series will coverge or diverge at the edpoits of this iterval. Note that these values of x will correspod to the value of x that will give L =. The way to determie covergece at these poits is to simply plug them ito the origial power series ad see if the series coverges or diverges usig ay test ecessary. x = 7 : I this case the series is, ( ) ( ) ( 4) = = = = = = = = = This series is diverget by the Divergece Test sice lim = 0. x = : I this case the series is, ( ) 4 ( 4) = ( ) = = This series is also diverget by the Divergece Test sice lim does t exist. So, i this case the power series will ot coverge for either edpoit. The iterval of covergece is the, 7< x < 007 Paul Dawkis 6

272 I the previous example the power series did t coverge for either edpoit of the iterval. Sometimes that will happe, but do t always expect that to happe. The power series could coverge at either both of the edpoits or oly oe of the edpoits. Example Determie the radius of covergece ad iterval of covergece for the followig power series. ( 4x 8) = Solutio Let s jump right ito the ratio test. + + ( 4x 8) L = lim + 4x 8 ( x ) 4 8 = lim + = 4x 8 lim + = 4x 8 So we will get the followig covergece/divergece iformatio from this. 4x 8 < series coverges 4x 8 > series diverges We eed to be careful here i determiig the iterval of covergece. The iterval of covergece requires x a < Rad x a > R. I other words, we eed to factor a 4 out of the absolute value bars i order to get the correct radius of covergece. Doig this gives, 8 x < x < 8 series coverges 8 x > x > 8 series diverges So, the radius of covergece for this power series is R =. 8 Now, let s fid the iterval of covergece. Agai, we ll first solve the iequality that gives covergece above. < x < < x < 8 8 Now check the edpoits. 007 Paul Dawkis 6

273 5 x = : 8 The series here is, 5 8 = = = = = = = ( ) ( ) This is the alteratig harmoic series ad we kow that it coverges. 7 x = : 8 The series here is, 7 8 = = = = = = = This is the harmoic series ad we kow that it diverges. So, the power series coverges for oe of the edpoits, but ot the other. This will ofte happe so do t get excited about it whe it does. The iterval of covergece for this power series is the, 5 7 x < 8 8 We ow eed to take a look at a couple of special cases with radius ad itervals of covergece. Example Determie the radius of covergece ad iterval of covergece for the followig power series. = 0 ( x+ )! Solutio We ll start this example with the ratio test as we have for the previous oes. 007 Paul Dawkis 64

274 L = lim = lim + ( +! ) ( x+ )! ( x+ ) ( + )! ( x+ )! ( ) = x+ lim + At this poit we eed to be careful. The limit is ifiite, but there is that term with the x s i frot of the limit. We ll have L => provided x. So, this power series will oly coverge if x =. If you thik about it we actually already kew that however. From our iitial discussio we kow that every power series will coverge for x= a ad i this case a =. Remember that we get a from ( x a), ad otice the coefficiet of the x must be a oe! I this case we say the radius of covergece is R = 0 ad the iterval of covergece is x =, ad yes we really did mea iterval of covergece eve though it s oly a poit. Example 4 Determie the radius of covergece ad iterval of covergece for the followig power series. ( x 6) = Solutio I this example the root test seems more appropriate. So, L = lim ( x 6) x 6 = lim = x 6 lim = 0 So, sice L = 0< regardless of the value of x this power series will coverge for every x. I these cases we say that the radius of covergece is R = ad iterval of covergece is < x <. So, let s summarize the last two examples. If the power series oly coverges for x= a the the radius of covergece is R = 0 ad the iterval of covergece is x= a. Likewise if the power series coverges for every x the radius of covergece is R = ad iterval of covergece is < x <. 007 Paul Dawkis 65

275 Let s work oe more example. Example 5 Determie the radius of covergece ad iterval of covergece for the followig power series. x = Solutio First otice that a = 0 i this problem. That s ot really importat to the problem, but it s worth poitig out so people do t get excited about it. The importat differece i this problem is the expoet o the x. I this case it is rather tha the stadard. As we will see some power series will have expoets other tha a ad so we still eed to be able to deal with these kids of problems. This oe seems set up for the root test agai so let s use that. x L = lim ( ) x = lim x = So, we will get covergece if x < x < The radius of covergece is NOT however. The radius of covergece requires a expoet of o the x. Therefore, x < x Be careful with the absolute value bars! I this case it looks like the radius of covergece is R =. Notice that we did t bother to put dow the iequality for divergece this time. The iequality for divergece is just the iterval for covergece that the test gives with the iequality switched ad geerally is t eeded. We will usually skip that part. Now let s get the iterval of covergece. First from the iequality we get, < x < Now check the edpoits. x = : Here the power series is, < 007 Paul Dawkis 66

276 ( ) ( ) Calculus II = = = = (( ) ) = = ( ) ( ) ( ) ( ) = This series is diverget by the Divergece Test sice lim does t exist. x = : Because we re squarig the x this series will be the same as the previous step. which is diverget. ( ) ( ) = = = The iterval of covergece is the, < x < 007 Paul Dawkis 67

277 Power Series ad Fuctios We opeed the last sectio by sayig that we were goig to start thikig about applicatios of series ad the promptly spet the sectio talkig about covergece agai. It s ow time to actually start with the applicatios of series. With this sectio we will start talkig about how to represet fuctios with power series. The atural questio of why we might wat to do this will be aswered i a couple of sectios oce we actually lear how to do this. Let s start off with oe that we already kow how to do, although whe we first ra across this series we did t thik of it as a power series or did we ackowledge that it represeted a fuctio. Recall that the geometric series is a ar = provided r < = 0 r Do t forget as well that if r the series diverges. Now, if we take a = ad r = x this becomes, x = x provided x < () = 0 Turig this aroud we ca see that we ca represet the fuctio f ( x) = x () with the power series x provided x < () = 0 This provisio is importat. We ca clearly plug ay umber other tha x = ito the fuctio, however, we will oly get a coverget power series if x <. This meas the equality i () will oly hold if x <. For ay other value of x the equality wo t hold. Note as well that we ca also use this to ackowledge that the radius of covergece of this power series is R = ad the iterval of covergece is x <. This idea of covergece is importat here. We will be represetig may fuctios as power series ad it will be importat to recogize that the represetatios will ofte oly be valid for a rage of x s ad that there may be values of x that we ca plug ito the fuctio that we ca t plug ito the power series represetatio. I this sectio we are goig to cocetrate o represetig fuctios with power series where the fuctios ca be related back to (). 007 Paul Dawkis 68

278 I this way we will hopefully become familiar with some of the kids of maipulatios that we will sometimes eed to do whe workig with power series. So, let s jump ito a couple of examples. Example Fid a power series represetatio for the followig fuctio ad determie its iterval of covergece. g( x) = + x Solutio What we eed to do here is to relate this fuctio back to (). This is actually easier tha it might look. Recall that the x i () is simply a variable ad ca represet aythig. So, a quick rewrite of g( x ) gives, g x = x ad so the x i g( x ) holds the same place as the x i (). Therefore, all we eed to do is replace the x i () ad we ve got a power series represetatio for g( x ). = 0 g x = x provided x < Notice that we replaced both the x i the power series ad i the iterval of covergece. All we eed to do ow is a little simplificatio. = 0 provided g x = x x < x < So, i this case the iterval of covergece is the same as the origial power series. This usually wo t happe. More ofte tha ot the ew iterval of covergece will be differet from the origial iterval of covergece. Example Fid a power series represetatio for the followig fuctio ad determie its iterval of covergece. x h( x) = + x Solutio This fuctio is similar to the previous fuctio. The differece is the umerator ad at first glace that looks to be a importat differece. Sice () does t have a x i the umerator it appears that we ca t relate this fuctio back to that. However, ow that we ve worked the first example this oe is actually very simple sice we ca use the result of the aswer from that example. To see how to do this let s first rewrite the fuctio a little. h( x) = x + x 007 Paul Dawkis 69

279 Now, from the first example we ve already got a power series for the secod term so let s use that to write the fuctio as, provided h x = x x x < = 0 Notice that the presece of x s outside of the series will NOT affect its covergece ad so the iterval of covergece remais the same. The last step is to brig the coefficiet ito the series ad we ll be doe. Whe we do this make sure ad combie the x s as well. We typically oly wat a sigle x i a power series. + h x = x provided x < = 0 As we saw i the previous example we ca ofte use previous results to help us out. This is a importat idea to remember as it ca ofte greatly simplify our work. Example Fid a power series represetatio for the followig fuctio ad determie its iterval of covergece. x f ( x) = 5 x Solutio So, agai, we ve got a x i the umerator. So, as with the last example let s factor that out ad see what we ve got left. f ( x) = x 5 x If we had a power series represetatio for g( x) = 5 x f x. we could get a power series represetatio for So, let s fid oe. We ll first otice that i order to use (4) we ll eed the umber i the deomiator to be a oe. That s easy eough to get. g( x) = 5 x 5 Now all we eed to do to get a power series represetatio is to replace the x i () with x 5. Doig this gives, x x g( x) = provided < 5 = Now let s do a little simplificatio o the series. 007 Paul Dawkis 70

280 g x x = 5 5 The iterval of covergece for this series is, x < x < 5 5 x < 5 = = 0 Okay, this was the work for the power series represetatio for g( x ) let s ow fid a power series represetatio for the origial fuctio. All we eed to do for this is to multiply the power series represetatio for g( x ) by x ad we ll have it. = 0 x 5 + f ( x) = x 5 x x = x 5 x = = 0 + = 0 The iterval of covergece does t chage ad so it will be x < 5. So, hopefully we ow have a idea o how to fid the power series represetatio for some fuctios. Admittedly all of the fuctios could be related back to () but it s a start. We ow eed to look at some further maipulatio of power series that we will eed to do o occasio. We eed to discuss differetiatio ad itegratio of power series. Let s start with differetiatio of the power series, = 0 0 f x = c x a = c + c x a + c x a + c x a + Now, we kow that if we differetiate a fiite sum of terms all we eed to do is differetiate each of the terms ad the add them back up. With ifiite sums there are some subtleties ivolved that we eed to be careful with, but are somewhat beyod the scope of this course. Nicely eough for us however, it is kow that if the power series represetatio of f ( x ) has a radius of covergece of R > 0 the the term by term differetiatio of the power series will also have a radius of covergece of R ad (more importatly) will i fact be the power series f x provided we stay withi the radius of covergece. represetatio of Agai, we should make the poit that if we are t dealig with a power series the we may or may ot be able to differetiate each term of the series to get the derivative of the series. 007 Paul Dawkis 7

281 So, what all this meas for us is that, d f x = c x a = c+ c x a + c x a + = c x a dx = 0 = Note the iitial value of this series. It has bee chaged from = 0 to =. This is a ackowledgemet of the fact that the derivative of the first term is zero ad hece is t i the derivative. Notice however, that sice the =0 term of the above series is also zero, we could start the series at = 0 if it was required for a particular problem. I geeral however, this wo t be doe i this class. We ca ow fid formulas for higher order derivatives as well ow. f x = c x a etc. = f x = c x a = Oce agai, otice that the iitial value of chages with each differetiatio i order to ackowledge that a term from the origial series differetiated to zero. Let s ow briefly talk about itegratio. Just as with the differetiatio, whe we ve got a ifiite series we eed to be careful about just itegratio term by term. Much like with derivatives it turs out that as log as we re workig with power series we ca just itegrate the terms of the series to get the itegral of the series itself. I other words, f ( x) dx = c ( x a) dx = 0 = 0 = c x a dx = C+ c ( x a) = Notice that we pick up a costat of itegratio, C, that is outside the series here. Let s summarize the differetiatio ad itegratio ideas before movig o to a example or two. Fact has a radius of covergece of 0 = 0 If f ( x) = c ( x a) R > the, f x = c x a f x dx = C + c + ( x a) = = 0 + ad both of these also have a radius of covergece of R. 007 Paul Dawkis 7

282 Now, let s see how we ca use these facts to geerate some more power series represetatios of fuctios. Example 4 Fid a power series represetatio for the followig fuctio ad determie its iterval of covergece. g( x) = ( x) Solutio To do this problem let s otice that d = ( x) dx x The sice we ve got a power series represetatio for x all that we ll eed to do is differetiate that power series to get a power series represetatio for g( x ). g( x) = x d = dx x d = x dx = 0 = The sice the origial power series had a radius of covergece of R = the derivative, ad hece g(x), will also have a radius of covergece of R =. Example 5 Fid a power series represetatio for the followig fuctio ad determie its iterval of covergece. h( x) = l ( 5 x) Solutio I this case we eed to otice that = x dx = l 5 5 x ad the recall that we have a power series represetatio for 5 x ( x) Remember we foud a represetatio for this i Example. So, 007 Paul Dawkis 7

283 l ( 5 x) = dx 5 x = = C = 0 = 0 x 5 + dx + x 5 ( + ) We ca fid the costat of itegratio, C, by pluggig i a value of x. A good choice is x = 0 sice that will make the series easy to evaluate. + 0 l ( 5 0) = C l 5 = C = 0 + So, the fial aswer is, ( x) l 5 = l 5 = 0 + x 5 ( + ) + Note that it is okay to have the costat sittig outside of the series like this. I fact, there is o way to brig it ito the series so do t get excited about it. 007 Paul Dawkis 74

284 Taylor Series I the previous sectio we started lookig at writig dow a power series represetatio of a fuctio. The problem with the approach i that sectio is that everythig came dow to eedig to be able to relate the fuctio i some way to x ad while there are may fuctios out there that ca be related to this fuctio there are may more that simply ca t be related to this. So, without takig aythig away from the process we looked at i the previous sectio, what we eed to do is come up with a more geeral method for writig a power series represetatio for a fuctio. So, for the time beig, let s make two assumptios. First, let s assume that the fuctio f ( x ) does i fact have a power series represetatio about x = a, = f x = c x a = c + c x a + c x a + c x a + c x a + Next, we will eed to assume that the fuctio, f ( x ), has derivatives of every order ad that we ca i fact fid them all. Now that we ve assumed that a power series represetatio exists we eed to determie what the coefficiets, c, are. This is easier tha it might at first appear to be. Let s first just evaluate everythig at x= a. This gives, = c0 f a So, all the terms except the first are zero ad we ow kow what c 0 is. Ufortuately, there is t ay other value of x that we ca plug ito the fuctio that will allow us to quickly fid ay of the other coefficiets. However, if we take the derivative of the fuctio (ad its power series) the plug i x= a we get, f ( x) = c+ c( x a) + c( x a) + 4c4( x a) + f ( a) = c ad we ow kow c. Let s cotiue with this idea ad fid the secod derivative. f ( x) = c + c( x a) + 4 c4( x a) + f ( a) = c So, it looks like, f ( a) c = Usig the third derivative gives, 007 Paul Dawkis 75

285 4 f x = c + c x a + 4 f a = c c = f ( a) Usig the fourth derivative gives, 4 f x = 4 c c x a 4 5 ( 4 ) ( a) ( 4 ) f f ( a) = 4 c4 c4 = 4 Hopefully by this time you ve see the patter here. It looks like, i geeral, we ve got the followig formula for the coefficiets. ( f ) ( a) c =! This eve works for =0 if you recall that 0! = ad defie ( 0 ) f x = f x. So, provided a power series represetatio for the fuctio f ( x ) about x= a exists the Taylor Series for f ( x ) about x= ais, Taylor Series ( f ) ( a) f ( x) = ( x a ) = 0! f ( a) f ( a) = f ( a) + f ( a)( x a) + ( x a ) + ( x a ) +!! If we use a = 0, so we are talkig about the Taylor Series about x = 0, we call the series a f x or, Maclauri Series for Maclauri Series = 0 ( ) ( 0) f f ( x) = x! ( 0) f ( 0) f f ( 0) f ( 0) x x x!! = Before workig ay examples of Taylor Series we first eed to address the assumptio that a Taylor Series will i fact exist for a give fuctio. Let s start out with some otatio ad defiitios that we ll eed. To determie a coditio that must be true i order for a Taylor series to exist for a fuctio let s f x as, first defie the th degree Taylor polyomial of 007 Paul Dawkis 76

286 i= 0 i! ( i ) ( a) f T x = x a Note that this really is a polyomial of degree at most. If we were to write out the sum without the summatio otatio this would clearly be a th degree polyomial. We ll see a ice applicatio of Taylor polyomials i the ext sectio. Notice as well that for the full Taylor Series, ( f ) ( a) ( x a ) = 0! the th degree Taylor polyomial is just the partial sum for the series. Next, the remaider is defied to be, R x = f x T x So, the remaider is really just the error betwee the fuctio f ( x ) ad the th degree Taylor polyomial for a give. With this defiitio ote that we ca the write the fuctio as, f x = T x + R x We ow have the followig Theorem. Theorem = +. The if, Suppose that f ( x) T ( x) R ( x) for x a < R the, o x a < R. I geeral showig that R ( x) ( x) lim R = 0 = 0! ( ) ( a) f f x = x a lim = 0 is a somewhat difficult process ad so we will be assumig that this ca be doe for some R i all of the examples that we ll be lookig at. Now let s look at some examples. x Example Fid the Taylor Series for f ( x ) = e about x = 0. Solutio This is actually oe of the easier Taylor Series that we ll be asked to compute. To fid the Taylor Series for a fuctio we will eed to determie a geeral formula for f a. This is oe of the i 007 Paul Dawkis 77

287 few fuctios where this is easy to do right from the start. To get a formula for ad so, ( ) ( 0) f all we eed to do is recogize that, f ( ) x = e = 0,,,, f x ( ) 0 0 = e = = 0,,,, x Therefore, the Taylor series for f ( x ) = e about x=0 is, e x x = x =!! = 0 = 0 x Example Fid the Taylor Series for f x = e about x = 0. Solutio There are two ways to do this problem. Both are fairly simple, however oe of them requires sigificatly less work. We ll work both solutios sice the loger oe has some ice ideas that we ll see i other examples. Solutio ( As with the first example we ll eed to get a formula for f ) ( 0). However, ulike the first oe we ve got a little more work to do. Let s first take some derivatives ad evaluate them at x=0. 0 x 0 f x = e f 0 = ( ) x f x = e f 0 = x f x = e f 0 = f x x = e f 0 = ( ) x f x = e f 0 = = 0,,, After a couple of computatios we were able to get geeral formulas for both ( ) ( 0) f ( ) x ad f. We ofte wo t be able to get a geeral formula for f x so do t get too excited about gettig that formula. Also, as we will see it wo t always be easy to get a geeral formula for ( f ) ( a ). So, i this case we ve got geeral formulas so all we eed to do is plug these ito the Taylor Series formula ad be doe with the problem. x ( ) x e =! = Paul Dawkis 78

288 Solutio The previous solutio was t too bad ad we ofte have to do thigs i that maer. However, i this case there is a much shorter solutio method. I the previous sectio we used series that we ve already foud to help us fid a ew series. Let s do the same thig with this oe. We x already kow a Taylor Series for e about x = 0 ad i this case the oly differece is we ve got a -x i the expoet istead of just a x. So, all we eed to do is replace the x i the Taylor Series that we foud i the first example with -x. x ( x) ( ) x e = =!! = 0 = 0 This is a much shorter method of arrivig at the same aswer so do t forget about usig previously computed series where possible (ad allowed of course). 4 x Example Fid the Taylor Series for f x = x e about x = 0. Solutio For this example we will take advatage of the fact that we already have a Taylor Series for about x = 0. I this example, ulike the previous example, doig this directly would be sigificatly loger ad more difficult. ( x ) 4 x 4 x e = x = x = 4 = 0 = 0 = 0 ( ) ( ) To this poit we ve oly looked at Taylor Series about x = 0 (also kow as Maclauri Series) so let s take a look at a Taylor Series that is t about x = 0. Also, we ll pick o the expoetial fuctio oe more time sice it makes some of the work easier. This will be the fial Taylor Series for expoetials i this sectio. x Example 4 Fid the Taylor Series for Solutio Fidig a geeral formula for The Taylor Series is the, ( ) ( 4) ( ) e f x! x! x! f is fairly simple. + 4 = e about x = 4. ( ) 4 x f x = f 4 = e x e 007 Paul Dawkis 79

289 e x = 0 ( ) 4 e = +! ( x 4) Okay, we ow eed to work some examples that do t ivolve the expoetial fuctio sice these will ted to require a little more work. Example 5 Fid the Taylor Series for f ( x) cos( x) = about x = 0. Solutio First we ll eed to take some derivatives of the fuctio ad evaluate them at x= f x = cos x f 0 = f x = si x f 0 = 0 f x = cos x f 0 = f x = si x f 0 = f x = cos x f 0 = 5 5 f x = si x f 0 = 0 6 f x = cos x 6 f 0 = I this example, ulike the previous oes, there is ot a easy formula for either the geeral derivative or the evaluatio of the derivative. However, there is a clear patter to the evaluatios. So, let s plug what we ve got ito the Taylor series ad see what we get, ( f ) ( 0 ) cos x= x = 0! ( 4 ) ( 5 f 0 ) f 0 f 0 4 f ( 0) 5 = f ( 0) + f ( 0) x+ x + x + x + x +!! 4! 5! 4 6 = + 0 x x + 0 x +! 4! 6! = 0 = = = 5 = = 4 = 6 So, we oly pick up terms with eve powers o the x s. This does t really help us to get a geeral formula for the Taylor Series. However, let s drop the zeroes ad reumber the terms as follows to see what we ca get. 4 6 cos x= x + x x +! 4! 6! = 0 = = = By reumberig the terms as we did we ca actually come up with a geeral formula for the Taylor Series ad here it is, 007 Paul Dawkis 80

290 cos x = = 0 ( ) x! This idea of reumberig the series terms as we did i the previous example is t used all that ofte, but occasioally is very useful. There is oe more series where we eed to do it so let s take a look at that so we ca get oe more example dow of reumberig series terms. Example 6 Fid the Taylor Series for f ( x) si ( x) = about x = 0. Solutio As with the last example we ll start off i the same maer. 0 0 f x = si x f 0 = 0 f x = cos x f 0 = f x = si x f 0 = 0 f x = cos x f 0 = 4 4 f x = si x f 0 = f x = cos x f 0 = 6 f x = si x 6 f 0 = 0 So, we get a similar patter for this oe. Let s plug the umbers ito the Taylor Series. ( f ) ( 0 ) si x= x = 0! 5 7 = x x + x x +!! 5! 7! I this case we oly get terms that have a odd expoet o x ad as with the last problem oce we igore the zero terms there is a clear patter ad formula. So reumberig the terms as we did i the previous example we get the followig Taylor Series. + ( ) x si x = +! = 0 We really eed to work aother example or two i which f ( x ) is t about x = 0. Example 7 Fid the Taylor Series for f ( x) l ( x) Solutio Here are the first few derivatives ad the evaluatios. = about x =. 007 Paul Dawkis 8

291 0 0 f x = l x f = l ( ) ( ) f x = f = x ( ) ( ) f x = f = x ( ) ( ) f x = f = x ( 4 ) ( 4 ) f x = f 4 = 4 x ( 5 ) ( 4) ( 5 ) 4 f x = f 5 = 5 x ( ) ( ) ( ) + + ( )!! f ( x) = f = =,,, x Note that while we got a geeral formula here it does t work for = 0. This will happe o occasio so do t worry about it whe it does. I order to plug this ito the Taylor Series formula we ll eed to strip out the = 0 term first. ( f ) l ( x) = ( x )! = 0 = = = ( ) ( x ) f = f +! + ( ) (! ) ( x ) = l +! ( ) + ( x ) = l + Notice that we simplified the factorials i this case. You should always simplify them if there are more tha oe ad it s possible to simplify them. Also, do ot get excited about the term sittig i frot of the series. Sometimes we eed to do that whe we ca t get a geeral formula that will hold for all values of. 007 Paul Dawkis 8

292 Example 8 Fid the Taylor Series for f ( x) = about x =. x Solutio Agai, here are the derivatives ad evaluatios. ( 0 ) ( 0 ) f x = f ( ) = = x ( ) ( ) 007 Paul Dawkis 8 ( ) ( ) f x = f ( ) = = x ( ) 4 ( ) 5 ( ) ( ) f ( x) = f = = 4 x ( ) 4 4 f ( x) = f = = 4 5 x ( ) ( ) + ( ) ( ) +! +! f ( x) = f = = ( +! + ) x Notice that all the egative sigs will cacel out i the evaluatio. Also, this formula will work for all, ulike the previous example. Here is the Taylor Series for this fuctio. f = x + x = 0! ( +! ) = ( x + )! = 0 = 0 ( )( x ) = + + Now, let s work oe of the easier examples i this sectio. The problem for most studets is that it may ot appear to be that easy (or maybe it will appear to be too easy) at first glace. Example 9 Fid the Taylor Series for f ( x) = x 0x + 6 about x =. Solutio Here are the derivatives for this problem. 0 0 f x = x 0x + 6 f = 57 ( ) ( 4 ) f x = x 0x f = f x = 6x 0 f = f x = 6 f = 6 f x = 0 f = 0 4

293 This Taylor series will termiate after =. This will always happe whe we are fidig the Taylor Series of a polyomial. Here is the Taylor Series for this oe. ( ) f ( ) x 0x + 6 = ( x )! = 0 ( x ) ( x ) ( x ) ( ) f ( ) f = f ( ) + f ( )( x ) + x + x + 0!! = 57 + Whe fidig the Taylor Series of a polyomial we do t do ay simplificatio of the right had side. We leave it like it is. I fact, if we were to multiply everythig out we just get back to the origial polyomial! While it s ot apparet that writig the Taylor Series for a polyomial is useful there are times where this eeds to be doe. The problem is that they are beyod the scope of this course ad so are t covered here. For example, there is oe applicatio to series i the field of Differetial Equatios where this eeds to be doe o occasio. So, we ve see quite a few examples of Taylor Series to this poit ad i all of them we were able to fid geeral formulas for the series. This wo t always be the case. To see a example of oe that does t have a geeral formula check out the last example i the ext sectio. Before leavig this sectio there are three importat Taylor Series that we ve derived i this sectio that we should summarize up i oe place. I my class I will assume that you kow these formulas from this poit o. x e = cos x = si x = = 0 = 0 = 0 x! ( ) x ( )! ( ) x ( + ) +! 007 Paul Dawkis 84

294 Applicatios of Series Now, that we kow how to represet fuctio as power series we ca ow talk about at least a couple of applicatios of series. There are i fact may applicatios of series, ufortuately most of them are beyod the scope of this course. Oe applicatio of power series (with the occasioal use of Taylor Series) is i the field of Ordiary Differetial Equatios whe fidig Series Solutios to Differetial Equatios. If you are iterested i seeig how that works you ca check out that chapter of my Differetial Equatios otes. Aother applicatio of series arises i the study of Partial Differetial Equatios. Oe of the more commoly used methods i that subject makes use of Fourier Series. May of the applicatios of series, especially those i the differetial equatios fields, rely o the fact that fuctios ca be represeted as a series. I these applicatios it is very difficult, if ot impossible, to fid the fuctio itself. However, there are methods of determiig the series represetatio for the ukow fuctio. While the differetial equatios applicatios are beyod the scope of this course there are some applicatios from a Calculus settig that we ca look at. Example Determie a Taylor Series about x = 0 for the followig itegral. si x dx x Solutio To do this we will first eed to fid a Taylor Series about x = 0 for the itegrad. This however is t terribly difficult. We already have a Taylor Series for sie about x = 0 so we ll just use that as follows, + si x ( ) x ( ) x = = x x +! +! We ca ow do the problem. si x dx = x = 0 = 0 = 0 C = + ( ) x dx ( +! ) + ( ) x = 0 ( + ) ( + )! So, while we ca t itegrate this fuctio i terms of kow fuctios we ca come up with a series represetatio for the itegral. This idea of derivig a series represetatio for a fuctio istead of tryig to fid the fuctio itself is used quite ofte i several fields. I fact, there are some fields where this is oe of the mai ideas used ad without this idea it would be very difficult to accomplish aythig i those fields. 007 Paul Dawkis 85

295 Aother applicatio of series is t really a applicatio of ifiite series. It s more a applicatio of partial sums. I fact, we ve already see this applicatio i use oce i this chapter. I the Estimatig the Value of a Series we used a partial sum to estimate the value of a series. We ca do the same thig with power series ad series represetatios of fuctios. The mai differece is that we will ow be usig the partial sum to approximate a fuctio istead of a sigle value. We will look at Taylor series for our examples, but we could just as easily use ay series represetatio here. Recall that the th degree Taylor Polyomial of f(x) is give by, ( i f ) ( a) i T ( x) = ( x a ) i Let s take a look at example of this. i= 0! Example For the fuctio f ( x) = cos( x) plot the fuctio as well as T ( x ), T4 T8 ( x ) o the same graph for the iterval [-4,4]. x, ad Solutio Here is the geeral formula for the Taylor polyomials for cosie. T ( x) = i= 0 i ( ) ( i) x! The three Taylor polyomials that we ve got are the, x T ( x) = 4 x x T4 ( x) = x x x x T8 ( x) = Here is the graph of these three Taylor polyomials as well as the graph of cosie. i 007 Paul Dawkis 86

296 As we ca see from this graph as we icrease the degree of the Taylor polyomial it starts to look more ad more like the fuctio itself. I fact by the time we get to T8 ( x ) the oly differece is right at the eds. The higher the degree of the Taylor polyomial the better it approximates the fuctio. Also the larger the iterval the higher degree Taylor polyomial we eed to get a good approximatio for the whole iterval. Before movig o let s write dow a couple more Taylor polyomials from the previous example. Notice that because the Taylor series for cosie does t cotai ay terms with odd powers o x we get the followig Taylor polyomials. T T T 5 9 ( x) ( x) ( x) x = 4 x x = x x x x = These are idetical to those used i the example. Sometimes this will happe although that was ot really the poit of this. The poit is to otice that the th degree Taylor polyomial may actually have a degree that is less tha. It will ever be more tha, but it ca be less tha. The fial example i this sectio really is t a applicatio of series ad probably beloged i the previous sectio. However, the previous sectio was gettig too log so the example is i this sectio. This is a example of how to multiply series together ad while this is t a applicatio of series it is somethig that does have to be doe o occasio i the applicatios. So, i that sese it does belog i this sectio. 007 Paul Dawkis 87

297 x Example Fid the first three o-zero terms i the Taylor Series for cos x = 0. f x = e x about Solutio Before we start let s ackowledge that the easiest way to do this problem is to simply compute the first -4 derivatives, evaluate them at x = 0, plug ito the formula ad we d be doe. However, as we oted prior to this example we wat to use this example to illustrate how we multiply series. We will make use of the fact that we ve got Taylor Series for each of these so we ca use them i this problem. x x ( ) x e cos x = = 0! = 0 ( )! We re ot goig to completely multiply out these series. We re goig to do eough of the multiplicatio to get a aswer. The problem statemet says that we wat the first three o-zero terms. That o-zero bit is importat as it is possible that some of the terms will be zero. If oe of the terms are zero this would mea that the first three o-zero terms would be the costat term, x term, ad x term. However, because some might be zero let s assume that if we get all the terms up through x 4 we ll have eough to get the aswer. If we ve assumed wrog it will be very easy to fix so do t worry about that. Now, let s write dow the first few terms of each series ad we ll stop at the x 4 term i each. 4 4 cos x x x x x x e x= + x Note that we do eed to ackowledge that these series do t stop. That s the purpose of the + at the ed of each. Just for a secod however, let s suppose that each of these did stop ad ask ourselves how we would multiply each out. If this were the case we would take every term i the secod ad multiply by every term i the first. I other words, we would first multiply every term i the secod series by, the every term i the secod series by x, the by x etc. By stoppig each series at x 4 we have ow guarateed that we ll get all terms that have a expoet of 4 or less. Do you see why? Each of the terms that we eglected to write dow have a expoet of at least 5 ad so multiplyig by or ay power of x will result i a term with a expoet that is at a miimum 5. Therefore, oe of the eglected terms will cotribute terms with a expoet of 4 or less ad so were t eeded. So, let s start the multiplicatio process. 007 Paul Dawkis 88

298 4 4 x x x x x x e cos x= + x x x x x x x x = x Secod Series Secod Series x Secod Series x 6 Secod Series x x x x x x x Secod Series x 4 Now, collect like terms igorig everythig with a expoet of 5 or more sice we wo t have all those terms ad do t wat them either. Doig this gives, x 4 e cos x= + x+ + x + + x + + x x x = + x + 6 There we go. It looks like we over guessed ad eded up with four o-zero terms, but that s okay. If we had uder guessed ad it tured out that we eeded terms with x 5 i them all we would eed to do at this poit is go back ad add i those terms to the origial series ad do a couple quick multiplicatios. I other words, there is o reaso to completely redo all the work. 007 Paul Dawkis 89

299 Biomial Series I this fial sectio of this chapter we are goig to look at aother series represetatio for a fuctio. Before we do this let s first recall the followig theorem. Biomial Theorem If is ay positive iteger the, where, i i a+ b = a b i= 0 i ( ) = a + a b + a b + + ab + b! ( )( ) ( + ) i = i =,,, i i! = 0 This is useful for expadig ( a+ b) for large whe straight forward multiplicatio would t be easy to do. Let s take a quick look at a example. Example Use the Biomial Theorem to expad ( x ) 4 Solutio There really is t much to do other tha pluggig ito the theorem i i ( x ) = ( x) ( ) i= 0 i = x + x + x + x = ( x) + 4( x) ( ) + ( x ) ( ) + 4 ( x )( ) + ( ) 4 = 6x 96x + 6x 6x Now, the Biomial Theorem required that be a positive iteger. There is a extesio to this however that allows for ay umber at all. 007 Paul Dawkis 90

300 Biomial Series If k is ay umber ad x < the, where, k k + = x = 0 k( k ) k( k )( k ) kx x x!! ( x) = ( )( ) ( + ) k k k k k = =,,,! k = 0 So, similar to the biomial theorem except that it s a ifiite series ad we must have x < i order to get covergece. Let s check out a example of this. Example Write dow the first four terms i the biomial series for 9 x Solutio So, i this case k = ad we ll eed to rewrite the term a little to put it ito the form required. x x 9 x = = The first four terms i the biomial series is the, x 9 x = + 9 x = = 0 9 x ( ) x ( )( ) x = x x x = Paul Dawkis 9

301 Vectors Itroductio This is a fairly short chapter. We will be takig a brief look at vectors ad some of their properties. We will eed some of this material i the ext chapter ad those of you headig o towards Calculus III will use a fair amout of this there as well. Here is a list of topics i this chapter. Vectors The Basics I this sectio we will itroduce some of the basic cocepts about vectors. Vector Arithmetic Here we will give the basic arithmetic operatios for vectors. Dot Product We will discuss the dot product i this sectio as well as a applicatio or two. Cross Product I this sectio we ll discuss the cross product ad see a quick applicatio. 007 Paul Dawkis 9

302 Vectors The Basics Let s start this sectio off with a quick discussio o what vectors are used for. Vectors are used to represet quatities that have both a magitude ad a directio. Good examples of quatities that ca be represeted by vectors are force ad velocity. Both of these have a directio ad a magitude. Let s cosider force for a secod. A force of say 5 Newtos that is applied i a particular directio ca be applied at ay poit i space. I other words, the poit where we apply the force does ot chage the force itself. Forces are idepedet of the poit of applicatio. To defie a force all we eed to kow is the magitude of the force ad the directio that the force is applied i. The same idea holds more geerally with vectors. Vectors oly impart magitude ad directio. They do t impart ay iformatio about where the quatity is applied. This is a importat idea to always remember i the study of vectors. I a graphical sese vectors are represeted by directed lie segmets. The legth of the lie segmet is the magitude of the vector ad the directio of the lie segmet is the directio of the vector. However, because vectors do t impart ay iformatio about where the quatity is applied ay directed lie segmet with the same legth ad directio will represet the same vector. Cosider the sketch below. Each of the directed lie segmets i the sketch represets the same vector. I each case the vector starts at a specific poit the moves uits to the left ad 5 uits up. The otatio that we ll use for this vector is, v =,5 ad each of the directed lie segmets i the sketch are called represetatios of the vector. Be careful to distiguish vector otatio,,5, from the otatio we use to represet coordiates of poits, (,5). The vector deotes a magitude ad a directio of a quatity while the poit deotes a locatio i space. So do t mix the otatios up! 007 Paul Dawkis 9

303 A represetatio of the vector v = a, a i two dimesioal space is ay directed lie segmet, AB, from the poit A= ( xy, ) to the poit B= ( x+ a, y+ a ). Likewise a represetatio of the vector v = a, a, a i three dimesioal space is ay directed lie segmet, AB, from the A xyz,, B= x+ a, y+ a, z+ a. poit = to the poit Note that there is very little differece betwee the two dimesioal ad three dimesioal formulas above. To get from the three dimesioal formula to the two dimesioal formula all we did is take out the third compoet/coordiate. Because of this most of the formulas here are give oly i their three dimesioal versio. If we eed them i their two dimesioal form we ca easily modify the three dimesioal form. There is oe represetatio of a vector that is special i some way. The represetatio of the vector v = a, a, a that starts at the poit ( 0,0,0) B= a, a, a is called the positio vector of the poit ( a, a, a ). So, whe we talk about positio vectors we are specifyig the iitial ad fial poit of the vector. A = ad eds at the poit Positio vectors are useful if we ever eed to represet a poit as a vector. As we ll see there are times i which we defiitely are goig to wat to represet poits as vectors. I fact, we re goig to ru ito topics that ca oly be doe if we represet poits as vectors. Next we eed to discuss briefly how to geerate a vector give the iitial ad fial poits of the represetatio. Give the two poits A= ( a, a, a) ad B= ( b, b, b) the vector with the represetatio AB is, v = b a, b a, b a Note that we have to be very careful with directio here. The vector above is the vector that starts at A ad eds at B. The vector that starts at B ad eds at A, i.e. with represetatio BA is, w= a b, a b, a b These two vectors are differet ad so we do eed to always pay attetio to what poit is the startig poit ad what poit is the edig poit. Whe determiig the vector betwee two poits we always subtract the iitial poit from the termial poit. Example Give the vector for each of the followig. (a) The vector from (, 7,0) to (,, 5). (b) The vector from (,, 5) to(, 7,0). (c) The positio vector for ( 90, 4) Solutio (a) Remember that to costruct this vector we subtract coordiates of the startig poit from the edig poit. 007 Paul Dawkis 94

304 (b) Same thig here., ( 7 ), 5 0 =,4, 5, 7 ( ), 0 ( 5) =, 4, 5 Notice that the oly differece betwee the first two is the sigs are all opposite. This differece is importat as it is this differece that tells us that the two vectors poit i opposite directios. (c) Not much to this oe other tha ackowledgig that the positio vector of a poit is othig more tha a vector with the poit s coordiates as its compoets. 90, 4 We ow eed to start discussig some of the basic cocepts that we will ru ito o occasio. Magitude The magitude, or legth, of the vector v = a, a, a is give by, v = a + a + a Example Determie the magitude of each of the followig vectors. (a) a =, 5,0 (b) u =, 5 5 (c) w = 0,0 (d) i =,0,0 Solutio There is t too much to these other tha plug ito the formula. (a) a = = 4 (c) w = 0+ 0 = 0 (b) 4 u = + = = 5 5 (d) i = = We also have the followig fact about the magitude. If a = 0 the a = 0 This should make sese. Because we square all the compoets the oly way we ca get zero out of the formula was for the compoets to be zero i the first place. Uit Vector Ay vector with magitude of, i.e. u =, is called a uit vector. 007 Paul Dawkis 95

305 Example Which of the vectors from Example are uit vectors? Solutio Both the secod ad fourth vectors had a legth of ad so they are the oly uit vectors from the first example. Zero Vector The vector w = 0,0 that we saw i the first example is called a zero vector sice its compoets are all zero. Zero vectors are ofte deoted by 0. Be careful to distiguish 0 (the umber) from 0 (the vector). The umber 0 deotes the origi i space, while the vector 0 deotes a vector that has o magitude or directio. Stadard Basis Vectors The fourth vector from the secod example, i =,0,0, is called a stadard basis vector. I three dimesioal space there are three stadard basis vectors, i =,0,0 j = 0,,0 k = 0,0, I two dimesioal space there are two stadard basis vectors, i =, 0 j = 0, Note that stadard basis vectors are also uit vectors. Warig We are pretty much doe with this sectio however, before proceedig to the ext sectio we should poit out that vectors are ot restricted to two dimesioal or three dimesioal space. Vectors ca exist i geeral -dimesioal space. The geeral otatio for a -dimesioal vector is, v = a, a, a,, a ad each of the a i s are called compoets of the vector. Because we will be workig almost exclusively with two ad three dimesioal vectors i this course most of the formulas will be give for the two ad/or three dimesioal cases. However, most of the cocepts/formulas will work with geeral vectors ad the formulas are easily (ad aturally) modified for geeral -dimesioal vectors. Also, because it is easier to visualize thigs i two dimesios most of the figures related to vectors will be two dimesioal figures. So, we eed to be careful to ot get too locked ito the two or three dimesioal cases from our discussios i this chapter. We will be workig i these dimesios either because it s easier to visualize the situatio or because physical restrictios of the problems will eforce a dimesio upo us. 007 Paul Dawkis 96

306 Vector Arithmetic I this sectio we eed to have a brief discussio of vector arithmetic. We ll start with additio of two vectors. So, give the vectors a = a, a, a ad b = b, b, b the additio of the two vectors is give by the followig formula. a+ b = a + b, a + b, a + b The followig figure gives the geometric iterpretatio of the additio of two vectors. This is sometimes called the parallelogram law or triagle law. Computatioally, subtractio is very similar. Give the vectors a = a, a, a b = b, b, b the differece of the two vectors is give by, a b = a b, a b, a b ad Here is the geometric iterpretatio of the differece of two vectors. 007 Paul Dawkis 97

307 It is a little harder to see this geometric iterpretatio. To help see this let s istead thik of subtractio as the additio of a ad b. First, as we ll see i a bit b is the same vector as b with opposite sigs o all the compoets. I other words, a + b. b. Here is the vector set up for b goes i the opposite directio as + As we ca see from this figure we ca move the vector represetig a ( b) we ve got i the first figure showig the differece of the two vectors. to the positio Note that we ca t add or subtract two vectors uless they have the same umber of compoets. If they do t have the same umber of compoets the additio ad subtractio ca t be doe. The ext arithmetic operatio that we wat to look at is scalar multiplicatio. Give the vector a = a, a, a ad ay umber c the scalar multiplicatio is, ca = ca, ca, ca So, we multiply all the compoets by the costat c. To see the geometric iterpretatio of scalar multiplicatio let s take a look at a example. Example For the vector a =, 4 the same axis system. compute a, a ad a. Graph all four vectors o Solutio Here are the three scalar multiplicatios. a = 6, a =, a = 4, 8 Here is the graph for each of these vectors. 007 Paul Dawkis 98

308 I the previous example we ca see that if c is positive all scalar multiplicatio will do is stretch (if c > ) or shrik (if c < ) the origial vector, but it wo t chage the directio. Likewise, if c is egative scalar multiplicatio will switch the directio so that the vector will poit i exactly the opposite directio ad it will agai stretch or shrik the magitude of the vector depedig upo the size of c. There are several ice applicatios of scalar multiplicatio that we should ow take a look at. The first is parallel vectors. This is a cocept that we will see quite a bit over the ext couple of sectios. Two vectors are parallel if they have the same directio or are i exactly opposite directios. Now, recall agai the geometric iterpretatio of scalar multiplicatio. Whe we performed scalar multiplicatio we geerated ew vectors that were parallel to the origial vectors (ad each other for that matter). So, let s suppose that a ad b are parallel vectors. If they are parallel the there must be a umber c so that, a = cb So, two vectors are parallel if oe is a scalar multiple of the other. Example Determie if the sets of vectors are parallel or ot. (a) a =, 4,, b = 6,, (b) a = 4,0, b =, 9 Solutio (a) These two vectors are parallel sice b = a (b) These two vectors are t parallel. This ca be see by oticig that ( ) 0 = 5 9. I other words we ca t make a be a scalar multiple of b. The ext applicatio is best see i a example. 4 = ad yet 007 Paul Dawkis 99

309 Example Fid a uit vector that poits i the same directio as w = 5,,. Solutio Okay, what we re askig for is a ew parallel vector (poits i the same directio) that happes to be a uit vector. We ca do this with a scalar multiplicatio sice all scalar multiplicatio does is chage the legth of the origial vector (alog with possibly flippig the directio to the opposite directio). Here s what we ll do. First let s determie the magitude of w. w = = 0 Now, let s form the followig ew vector, 5 u = w= 5,, =,, w The claim is that this is a uit vector. That s easy eough to check u = + + = = This vector also poits i the same directio as w sice it is oly a scalar multiple of w ad we used a positive multiple. So, i geeral, give a vector w w, u = will be a uit vector that poits i the same directio w as w. Stadard Basis Vectors Revisited I the previous sectio we itroduced the idea of stadard basis vectors without really discussig why they were importat. We ca ow do that. Let s start with the vector a = a, a, a We ca use the additio of vectors to break this up as follows, a = a, a, a = a,0,0 + 0, a,0 + 0,0, a Usig scalar multiplicatio we ca further rewrite the vector as, a = a,0,0 + 0, a,0 + 0,0, a = a,0,0 + a 0,,0 + a 0,0, Fially, otice that these three ew vectors are simply the three stadard basis vectors for three dimesioal space. a, a, a = ai + a j+ ak 007 Paul Dawkis 00

310 So, we ca take ay vector ad write it i terms of the stadard basis vectors. From this poit o we will use the two otatios iterchageably so make sure that you ca deal with both otatios. Example 4 If a =, 9, ad w= i + 8k compute a w. Solutio I order to do the problem we ll covert to oe otatio ad the perform the idicated operatios. a w=, 9,, 0,8 = 6, 8,,0, 4 = 9, 8, We will leave this sectio with some basic properties of vector arithmetic. Properties If v, w ad u are vectors (each with the same umber of compoets) ad a ad b are two umbers the we have the followig properties. v+ w= w+ v u+ ( v+ w) = ( u+ v) + w v+ 0= v v = v a v + w = av + aw a + b v = av + bv The proofs of these are pretty much just computatio proofs so we ll prove oe of them ad leave the others to you to prove. Proof of a ( v + w) = av + aw We ll start with the two vectors, v = v, v,, v ad w= w, w,, w ad yes we did mea for these to each have compoets. The theorem works for geeral vectors so we may as well do the proof for geeral vectors. Now, as oted above this is pretty much just a computatioal proof. What that meas is that we ll compute the left side ad the do some basic arithmetic o the result to show that we ca make the left side look like the right side. Here is the work. a v+ w = a v, v,, v + w, w,, w ( ) = a v + w, v + w,, v + w,,, = a v + w a v + w a v + w = av + aw, av + aw,, av + aw = av, av,, av + aw, aw,, aw = a v, v,, v + a w, w,, w = av + aw 007 Paul Dawkis 0

311 Dot Product The ext topic for discussio is that of the dot product. Let s jump right ito the defiitio of the dot product. Give the two vectors a = a, a, a ad b = b, b, b the dot product is, a b= ab + ab + ab () Sometimes the dot product is called the scalar product. The dot product is also a example of a ier product ad so o occasio you may hear it called a ier product. Example Compute the dot product for each of the followig. (a) v = 5i 8 j, w= i + j (b) a = 0,, 7, b =,, Solutio Not much to do with these other tha use the formula. (a) vw= 5 6 = (b) ab= = Here are some properties of the dot product. Properties u v w uv uw cv w v cw c vw vw= wv v0= 0 vv = v If vv = 0 the v= 0 ( + ) = + = = The proofs of these properties are mostly computatioal proofs ad so we re oly goig to do a couple of them ad leave the rest to you to prove. Proof of u ( v + w ) = uv + uw We ll start with the three vectors, u = u, u,, u, v = v, v,, v ad w= w, w,, w ad yes we did mea for these to each have compoets. The theorem works for geeral vectors so we may as well do the proof for geeral vectors. Now, as oted above this is pretty much just a computatioal proof. What that meas is that we ll compute the left side ad the do some basic arithmetic o the result to show that we ca make the left side look like the right side. Here is the work. 007 Paul Dawkis 0

312 u v w u u u v v v w w w ( + ) =,,, (,,, +,,, ) = u, u,, u v + w, v + w,, v + w,,, = u v + w u v + w u v + w = uv + uw, uv + u w,, uv + u w = uv, uv,, uv + uw, u w,, u w = u, u,, u v, v,, v + u, u,, u w, w,, w = uv + uw Proof of : If vv= 0 the v = 0 This is a pretty simple proof. Let s start with v = v, v,, v ad compute the dot product. v v = v, v,, v v, v,, v = 0 = v + v + + v Now, sice we kow vi 0 for all i the the oly way for this sum to be zero is to i fact have v = 0. This i tur however meas that we must have v = 0 ad so we must have had v = 0. i i There is also a ice geometric iterpretatio to the dot product. First suppose that θ is the agle betwee a ad b such that 0 θ π as show i the image below. We ca the have the followig theorem. 007 Paul Dawkis 0

313 Theorem ab = a b cosθ () Proof Let s give a modified versio of the sketch above. The three vectors above form the triagle AOB ad ote that the legth of each side is othig more tha the magitude of the vector formig that side. The Law of Cosies tells us that, a b = a + b a b cosθ Also usig the properties of dot products we ca write the left side as, a b = ( a b) ( a b) = aa ab ba + bb = a ab + b Our origial equatio is the, a b = a + b a b cosθ a ab + b = a + b a b cosθ ab = a b cosθ ab = a b cosθ 007 Paul Dawkis 04

314 The formula from this theorem is ofte used ot to compute a dot product but istead to fid the agle betwee two vectors. Note as well that while the sketch of the two vectors i the proof is for two dimesioal vectors the theorem is valid for vectors of ay dimesio (as log as they have the same dimesio of course). Let s see a example of this. Example Determie the agle betwee a =, 4, ad b = 0,5,. Solutio We will eed the dot product as well as the magitudes of each vector. ab = a = 6 b = 9 The agle is the, ab cosθ = = = a b 6 9 θ = = cos radias=4.4 degrees The dot product gives us a very ice method for determiig if two vectors are perpedicular ad it will give aother method for determiig whe two vectors are parallel. Note as well that ofte we will use the term orthogoal i place of perpedicular. Now, if two vectors are orthogoal the we kow that the agle betwee them is 90 degrees. From () this tells us that if two vectors are orthogoal the, ab= 0 Likewise, if two vectors are parallel the the agle betwee them is either 0 degrees (poitig i the same directio) or 80 degrees (poitig i the opposite directio). Oce agai usig () this would mea that oe of the followig would have to be true. ab = a b = ab = a b = ( θ 0 ) OR ( θ 80 ) Example Determie if the followig vectors are parallel, orthogoal, or either. (a) a = 6,,, b =,5, (b) u = i j, v = i + j 4 Solutio (a) First get the dot product to see if they are orthogoal. ab= 0 = 0 The two vectors are orthogoal. (b) Agai, let s get the dot product first. 007 Paul Dawkis 05

315 5 uv= = 4 4 So, they are t orthogoal. Let s get the magitudes ad see if they are parallel. u 5 5 = 5 v = = 6 4 Now, otice that, 5 5 uv 5 = = = u v 4 4 So, the two vectors are parallel. There are several ice applicatios of the dot product as well that we should look at. Projectios The best way to uderstad projectios is to see a couple of sketches. So, give two vectors a ad b we wat to determie the projectio of b oto a. The projectio is deoted by proj a b. Here are a couple of sketches illustratig the projectio. So, to get the projectio of b oto a we drop straight dow from the ed of b util we hit (ad form a right agle) with the lie that is parallel to a. The projectio is the the vector that is parallel to a, starts at the same poit both of the origial vectors started at ad eds where the dashed lie hits the lie parallel to a. There is a ice formula for fidig the projectio of b oto a. Here it is, proj a b = a ab a Note that we also eed to be very careful with otatio here. The projectio of a oto b is give by 007 Paul Dawkis 06

316 proj b a = b ab b We ca see that this will be a totally differet vector. This vector is parallel to b, while proj a b is parallel to a. So, be careful with otatio ad make sure you are fidig the correct projectio. Here s a example. Example 4 Determie the projectio of b =,, oto a =, 0,. Solutio We eed the dot product ad the magitude of a. ab = 4 a = 5 The projectio is the, proj a ab b = a a 4 =, 0, =,0, 5 5 For compariso purposes let s do it the other way aroud as well. Example 5 Determie the projectio of a =, 0, otob =,,. Solutio We eed the dot product ad the magitude of b. ab = 4 b = 6 The projectio is the, proj b a = ab b b 4 =,, 6 4 =,, As we ca see from the previous two examples the two projectios are differet so be careful. 007 Paul Dawkis 07

317 Directio Cosies This applicatio of the dot product requires that we be i three dimesioal space ulike all the other applicatios we ve looked at to this poit. Let s start with a vector, a, i three dimesioal space. This vector will form agles with the x- axis (α ), the y-axis (β ), ad the z-axis (γ ). These agles are called directio agles ad the cosies of these agles are called directio cosies. Here is a sketch of a vector ad the directio agles. The formulas for the directio cosies are, ai a aj a ak a α = = β = = γ = = a a a a a a cos cos cos where i, j ad k are the stadard basis vectors. Let s verify the first dot product above. We ll leave the rest to you to verify. ai = a, a, a,0,0 = a Here are a couple of ice facts about the directio cosies.. The vector u = cos α, cos β, cosγ is a uit vector. cos α + cos β + cos γ =.. a = a cos α,cos β,cosγ Let s do a quick example ivolvig directio cosies. 007 Paul Dawkis 08

318 Example 6 Determie the directio cosies ad directio agles for a =,, 4. Solutio We will eed the magitude of the vector. a = = The directio cosies ad agles are the, cosα = α =.9 radias = 64. degrees cos β = β =.5 radias = degrees cosγ = 4 γ =.6 radias = degrees 007 Paul Dawkis 09

319 Cross Product I this fial sectio of this chapter we will look at the cross product of two vectors. We should ote that the cross product requires both of the vectors to be three dimesioal vectors. Also, before gettig ito how to compute these we should poit out a major differece betwee dot products ad cross products. The result of a dot product is a umber ad the result of a cross product is a vector! Be careful ot to cofuse the two. So, let s start with the two vectors a = a, a, a ad b = b, b, b the the cross product is give by the formula, a b= ab ab, ab ab, ab ab This is ot a easy formula to remember. There are two ways to derive this formula. Both of them use the fact that the cross product is really the determiat of a x matrix. If you do t kow what this is that is do t worry about it. You do t eed to kow aythig about matrices or determiats to use either of the methods. The otatio for the determiat is as follows, i j k a b = a a a b b b The first row is the stadard basis vectors ad must appear i the order give here. The secod row is the compoets of a ad the third row is the compoets of b. Now, let s take a look at the differet methods for gettig the formula. The first method uses the Method of Cofactors. If you do t kow the method of cofactors that is fie, the result is all that we eed. Here is the formula. where, a a a a a a = + b b b b b b a b i j k a b ad bc c d = This formula is ot as difficult to remember as it might at first appear to be. First, the terms alterate i sig ad otice that the x is missig the colum below the stadard basis vector that multiplies it as well as the row of stadard basis vectors. The secod method is slightly easier; however, may textbooks do t cover this method as it will oly work o x determiats. This method says to take the determiat as listed above ad the copy the first two colums oto the ed as show below. 007 Paul Dawkis 0

320 i j k i j a b = a a a a a b b b b b We ow have three diagoals that move from left to right ad three diagoals that move from right to left. We multiply alog each diagoal ad add those that move from left to right ad subtract those that move from right to left. This is best see i a example. We ll also use this example to illustrate a fact about cross products. Example If a =,, ad b =, 4, compute each of the followig. (a) a b (b) b a Solutio (a) Here is the computatio for this oe. i j k i j a b = 4 4 = i + j + k 4 j i 4 k = 5i + j + k (b) Ad here is the computatio for this oe. i j k i j b a = 4 4 = i 4 + j + k j i k 4 = 5i j k Notice that switchig the order of the vectors i the cross product simply chaged all the sigs i the result. Note as well that this meas that the two cross products will poit i exactly opposite directios sice they oly differ by a sig. We ll formalize up this fact shortly whe we list several facts. There is also a geometric iterpretatio of the cross product. First we will let θ be the agle betwee the two vectors a ad b ad assume that 0 θ π, the we have the followig fact, a b = a b siθ () ad the followig figure. 007 Paul Dawkis

321 There should be a atural questio at this poit. How did we kow that the cross product poited i the directio that we ve give it here? First, as this figure, implies the cross product is orthogoal to both of the origial vectors. This will always be the case with oe exceptio that we ll get to i a secod. Secod, we kew that it poited i the upward directio (i this case) by the right had rule. This says that if we take our right had, start at a ad rotate our figers towards b our thumb will poit i the directio of the cross product. Therefore, if we d sketched i b a above we would have gotte a vector i the dowward directio. Example A plae is defied by ay three poits that are i the plae. If a plae cotais the,0,0,, R =,, fid a vector that is orthogoal to the plae. poits P =, Q = ad Solutio The oe way that we kow to get a orthogoal vector is to take a cross product. So, if we could fid two vectors that we kew were i the plae ad took the cross product of these two vectors we kow that the cross product would be orthogoal to both the vectors. However, sice both the vectors are i the plae the cross product would the also be orthogoal to the plae. So, we eed two vectors that are i the plae. This is where the poits come ito the problem. Sice all three poits lie i the plae ay vector betwee them must also be i the plae. There are may ways to get two vectors betwee these poits. We will use the followig two, PQ =, 0, 0 = 0,, PR =, 0, 0 =,, The cross product of these two vectors will be orthogoal to the plae. So, let s fid the cross product. 007 Paul Dawkis

322 PQ PR = i j k i j 0 0 = 4i + j k So, the vector 4i + j k will be orthogoal to the plae cotaiig the three poits. Now, let s address the oe time where the cross product will ot be orthogoal to the origial vectors. If the two vectors, a ad b, are parallel the the agle betwee them is either 0 or 80 degrees. From () this implies that, a b = 0 From a fact about the magitude we saw i the first sectio we kow that this implies a b = 0 I other words, it wo t be orthogoal to the origial vectors sice we have the zero vector. This does give us aother test for parallel vectors however. Fact If a b = 0 the a ad b will be parallel vectors. Let s also formalize up the fact about the cross product beig orthogoal to the origial vectors. Fact Provided a b 0 the a b is orthogoal to both a ad b. Here are some ice properties about the cross product. Properties If u, v ad w are vectors ad c is a umber the, u v = v u cu v = u cv = c u v u v+ w = u v+ u w u v w = u v w u u u u( v w) = v v v w w w The determiat i the last fact is computed i the same way that the cross product is computed. We will see a example of this computatio shortly. There are a couple of geometric applicatios to the cross product as well. Suppose we have three vectors a, b ad c ad we form the three dimesioal figure show below. 007 Paul Dawkis

323 The area of the parallelogram (two dimesioal frot of this object) is give by, Area = a b ad the volume of the parallelepiped (the whole three dimesioal object) is give by, Volume = a b c Note that the absolute value bars are required sice the quatity could be egative ad volume is t egative. We ca use this volume fact to determie if three vectors lie i the same plae or ot. If three vectors lie i the same plae the the volume of the parallelepiped will be zero. Example Determie if the three vectors a =, 4, 7, b =,, 4 ad c = 0, 9,8 i the same plae or ot. Solutio So, as we oted prior to this example all we eed to do is compute the volume of the parallelepiped formed by these three vectors. If the volume is zero they lie i the same plae ad if the volume is t zero they do t lie i the same plae. lie 007 Paul Dawkis 4

324 4 7 4 a b = 4 ( c) ( 8) ( 4)( 4)( 0) ( 7)( 9) ( 4)( 8) ( 4)( 9) ( 7)( )( 0) = + + = = 0 So, the volume is zero ad so they lie i the same plae. 007 Paul Dawkis 5

325 Three Dimesioal Space Itroductio I this chapter we will start takig a more detailed look at three dimesioal space (-D space or ). This is a very importat topic for Calculus III sice a good portio of Calculus III is doe i three (or higher) dimesioal space. We will be lookig at the equatios of graphs i -D space as well as vector valued fuctios ad how we do calculus with them. We will also be takig a look at a couple of ew coordiate systems for -D space. This is the oly chapter that exists i two places i my otes. Whe I origially wrote these otes all of these topics were covered i Calculus II however, we have sice moved several of them ito Calculus III. So, rather tha split the chapter up I have kept it i the Calculus II otes ad also put a copy i the Calculus III otes. May of the sectios ot covered i Calculus III will be used o occasio there ayway ad so they serve as a quick referece for whe we eed them. Here is a list of topics i this chapter. The -D Coordiate System We will itroduce the cocepts ad otatio for the three dimesioal coordiate system i this sectio. Equatios of Lies I this sectio we will develop the various forms for the equatio of lies i three dimesioal space. Equatios of Plaes Here we will develop the equatio of a plae. Quadric Surfaces I this sectio we will be lookig at some examples of quadric surfaces. Fuctios of Several Variables A quick review of some importat topics about fuctios of several variables. Vector Fuctios We itroduce the cocept of vector fuctios i this sectio. We cocetrate primarily o curves i three dimesioal space. We will however, touch briefly o surfaces as well. Calculus with Vector Fuctios Here we will take a quick look at limits, derivatives, ad itegrals with vector fuctios. Taget, Normal ad Biormal Vectors We will defie the taget, ormal ad biormal vectors i this sectio. Arc Legth with Vector Fuctios I this sectio we will fid the arc legth of a vector fuctio. 007 Paul Dawkis 6

326 Curvature We will determie the curvature of a fuctio i this sectio. Velocity ad Acceleratio I this sectio we will revisit a stadard applicatio of derivatives. We will look at the velocity ad acceleratio of a object whose positio fuctio is give by a vector fuctio. Cylidrical Coordiates We will defie the cylidrical coordiate system i this sectio. The cylidrical coordiate system is a alterate coordiate system for the three dimesioal coordiate system. Spherical Coordiates I this sectio we will defie the spherical coordiate system. The spherical coordiate system is yet aother alterate coordiate system for the three dimesioal coordiate system. 007 Paul Dawkis 7

327 The -D Coordiate System We ll start the chapter off with a fairly short discussio itroducig the -D coordiate system ad the covetios that we ll be usig. We will also take a brief look at how the differet coordiate systems ca chage the graph of a equatio. Let s first get some basic otatio out of the way. The -D coordiate system is ofte deoted by. Likewise the -D coordiate system is ofte deoted by ad the -D coordiate system is deoted by. Also, as you might have guessed the a geeral dimesioal coordiate system is ofte deoted by. Next, let s take a quick look at the basic coordiate system. This is the stadard placemet of the axes i this class. It is assumed that oly the positive directios are show by the axes. If we eed the egative axes for ay reaso we will put them i as eeded. Also ote the various poits o this sketch. The poit P is the geeral poit sittig out i -D space. If we start at P ad drop straight dow util we reach a z-coordiate of zero we arrive at the poit Q. We say that Q sits i the xy-plae. The xy-plae correspods to all the poits which have a zero z-coordiate. We ca also start at P ad move i the other two directios as show to get poits i the xz-plae (this is S with a y-coordiate of zero) ad the yz-plae (this is R with a x-coordiate of zero). Collectively, the xy, xz, ad yz-plaes are sometimes called the coordiate plaes. I the remaider of this class you will eed to be able to deal with the various coordiate plaes so make sure that you ca. Also, the poit Q is ofte referred to as the projectio of P i the xy-plae. Likewise, R is the projectio of P i the yz-plae ad S is the projectio of P i the xz-plae. May of the formulas that you are used to workig with i have atural extesios i For istace the distace betwee two poits i is give by,. 007 Paul Dawkis 8

328 While the distace betwee ay two poits i (, ) = ( ) + ( ) d P P x x y y is give by, (, ) = ( ) + ( ) + ( ) d P P x x y y z z Likewise, the geeral equatio for a circle with ceter ( hk, ) ad radius r is give by, ( x h) + ( y k) = r ad the geeral equatio for a sphere with ceter ( hkl,, ) ad radius r is give by, ( x h) + ( y k) + ( z l) = r With that said we do eed to be careful about just traslatig everythig we kow about ito ad assumig that it will work the same way. A good example of this is i graphig to some extet. Cosider the followig example. Example Graph x = i, ad. Solutio I we have a sigle coordiate system ad so x = is a poit i a -D coordiate system. I the equatio vertical lie i a -D coordiate system. x = tells us to graph all the poits that are i the form, y. This is a I the equatio, yz,. If you go back ad look at the coordiate plae poits this is very similar to the coordiates for the yz-plae except this time we have x = istead of x = 0. So, i a -D coordiate system this is a plae that will be parallel to the yz-plae ad pass through the x-axis at x =. Here is the graph of x = i. x = tells us to graph all the poits that are i the form Here is the graph of x = i. 007 Paul Dawkis 9

329 Fially, here is the graph of x = i. Note that we ve preseted this graph i two differet styles. O the left we ve got the traditioal axis system that we re used to seeig ad o the right we ve put the graph i a box. Both views ca be coveiet o occasio to help with perspective ad so we ll ofte do this with D graphs ad sketches. Note that at this poit we ca ow write dow the equatios for each of the coordiate plaes as well usig this idea. z = 0 xy plae y = 0 xz plae x = 0 yz plae Let s take a look at a slightly more geeral example. 007 Paul Dawkis 0

330 Example Graph y = x i ad. Solutio Of course we had to throw out for this example sice there are two variables which meas that we ca t be i a -D space. I this is a lie with slope ad a y itercept of -. However, i this is ot ecessarily a lie. Because we have ot specified a value of z we are forced to let z take ay value. This meas that at ay particular value of z we will get a copy of this lie. So, the graph is the a vertical plae that lies over the lie give by y = x i the xy-plae. Here is the graph i. here is the graph i. Notice that if we look to where the plae itersects the xy-plae we will get the graph of the lie i as oted i the above graph by the red lie through the plae. 007 Paul Dawkis

331 Let s take a look at oe more example of the differece betwee graphs i the differet coordiate systems. Example Graph x + y = 4 i ad. Solutio As with the previous example this wo t have a -D graph sice there are two variables. I this is a circle cetered at the origi with radius. I however, as with the previous example, this may or may ot be a circle. Sice we have ot specified z i ay way we must assume that z ca take o ay value. I other words, at ay value of z this equatio must be satisfied ad so at ay value z we have a circle of radius cetered o the z-axis. This meas that we have a cylider of radius cetered o the z-axis. Here are the graphs for this example. Notice that agai, if we look to where the cylider itersects the xy-plae we will agai get the circle from. 007 Paul Dawkis

332 We eed to be careful with the last two examples. It would be temptig to take the results of these ad say that we ca t graph lies or circles i ad yet that does t really make sese. There is o reaso for there to ot be graphs of lies or circles i. Let s thik about the example of the circle. To graph a circle i we would eed to do somethig like x + y = 4 at z = 5. This would be a circle of radius cetered o the z-axis at the level of z = 5. So, as log as we specify a z we will get a circle ad ot a cylider. We will see a easier way to specify circles i a later sectio. We could do the same thig with the lie from the secod example. However, we will be lookig at lies i more geerality i the ext sectio ad so we ll see a better way to deal with lies i there. The poit of the examples i this sectio is to make sure that we are beig careful with graphig equatios ad makig sure that we always remember which coordiate system that we are i. Aother quick poit to make here is that, as we ve see i the above examples, may graphs of equatios i are surfaces. That does t mea that we ca t graph curves i. We ca ad will graph curves i as well as we ll see later i this chapter. 007 Paul Dawkis

333 Equatios of Lies I this sectio we eed to take a look at the equatio of a lie i. As we saw i the previous sectio the equatio y = mx + b does ot describe a lie i, istead it describes a plae. This does t mea however that we ca t write dow a equatio for a lie i -D space. We re just goig to eed a ew way of writig dow the equatio of a curve. So, before we get ito the equatios of lies we first eed to briefly look at vector fuctios. We re goig to take a more i depth look at vector fuctios later. At this poit all that we eed to worry about is otatioal issues ad how they ca be used to give the equatio of a curve. The best way to get a idea of what a vector fuctio is ad what its graph looks like is to look at a example. So, cosider the followig vector fuctio. r t = t, A vector fuctio is a fuctio that takes oe or more variables, oe i this case, ad returs a vector. Note as well that a vector fuctio ca be a fuctio of two or more variables. However, i those cases the graph may o loger be a curve i space. The vector that the fuctio gives ca be a vector i whatever dimesio we eed it to be. I the example above it returs a vector i. Whe we get to the real subject of this sectio, equatios of lies, we ll be usig a vector fuctio that returs a vector i Now, we wat to determie the graph of the vector fuctio above. I order to fid the graph of our fuctio we ll thik of the vector that the vector fuctio returs as a positio vector for poits o the graph. Recall that a positio vector, say v= ab,, is a vector that starts at the origi ad eds at the poit ( ab, ). So, to get the graph of a vector fuctio all we eed to do is plug i some values of the variable ad the plot the poit that correspods to each positio vector we get out of the fuctio ad play coect the dots. Here are some evaluatios for our example. r =, r =, r =, r 5 = 5, So, each of these are positio vectors represetig poits o the graph of our vector fuctio. The poits, (,) (,) (,) ( 5,) are all poits that lie o the graph of our vector fuctio. If we do some more evaluatios ad plot all the poits we get the followig sketch. 007 Paul Dawkis 4

334 I this sketch we ve icluded the positio vector (i gray ad dashed) for several evaluatios as well as the t (above each poit) we used for each evaluatio. It looks like, i this case the graph of the vector equatio is i fact the lie y =. r t t t Here s aother quick example. Here is the graph of = 6cos,si. I this case we get a ellipse. It is importat to ot come away from this sectio with the idea that vector fuctios oly graph out lies. We ll be lookig at lies i this sectio, but the graphs of vector fuctios do ot have to be lies as the example above shows. We ll leave this brief discussio of vector fuctios with aother way to thik of the graph of a vector fuctio. Imagie that a pecil/pe is attached to the ed of the positio vector ad as we icrease the variable the resultig positio vector moves ad as it moves the pecil/pe o the ed sketches out the curve for the vector fuctio. Okay, we ow eed to move ito the actual topic of this sectio. We wat to write dow the equatio of a lie i ad as suggested by the work above we will eed a vector fuctio to do this. To see how we re goig to do this let s thik about what we eed to write dow the 007 Paul Dawkis 5

335 equatio of a lie i. I two dimesios we eed the slope (m) ad a poit that was o the lie i order to write dow the equatio. I that is still all that we eed except i this case the slope wo t be a simple umber as it was i two dimesios. I this case we will eed to ackowledge that a lie ca have a three dimesioal slope. So, we eed somethig that will allow us to describe a directio that is potetially i three dimesios. We already have a quatity that will do this for us. Vectors give directios ad ca be three dimesioal objects. So, let s start with the followig iformatio. Suppose that we kow a poit that is o the lie, P0 = ( x0, y0, z0), ad that v= abc,, is some vector that is parallel to the lie. Note, i all likelihood, v will ot be o the lie itself. We oly eed v to be parallel to the lie. Fially, let P= xyz,, be ay poit o the lie. Now, sice our slope is a vector let s also represet the two poits o the lie as vectors. We ll do this with positio vectors. So, let r 0 ad r be the positio vectors for P 0 ad P respectively. Also, for o apparet reaso, let s defie a to be the vector with represetatio PP 0. We ow have the followig sketch with all these poits ad vectors o it. Now, we ve show the parallel vector, v, as a positio vector but it does t eed to be a positio vector. It ca be aywhere, a positio vector, o the lie or off the lie, it just eeds to be parallel to the lie. Next, otice that we ca write r as follows, r = r0 + a If you re ot sure about this go back ad check out the sketch for vector additio i the vector arithmetic sectio. Now, otice that the vectors a ad v are parallel. Therefore there is a umber, t, such that 007 Paul Dawkis 6

336 a= tv We ow have, r= r+ tv= x, y, z + t abc,, This is called the vector form of the equatio of a lie. The oly part of this equatio that is ot kow is the t. Notice that tv will be a vector that lies alog the lie ad it tells us how far from the origial poit that we should move. If t is positive we move away from the origial poit i the directio of v (right i our sketch) ad if t is egative we move away from the origial poit i the opposite directio of v (left i our sketch). As t varies over all possible values we will completely cover the lie. The followig sketch shows this depedece o t of our sketch. There are several other forms of the equatio of a lie. To get the first alterate form let s start with the vector form ad do a slight rewrite. r= x, y, z + t abc,, x, y, z = x + ta, y + tb, z + tc The oly way for two vectors to be equal is for the compoets to be equal. I other words, x = x0 + ta y = y0 + tb z = z + tc 0 This set of equatios is called the parametric form of the equatio of a lie. Notice as well that this is really othig more tha a extesio of the parametric equatios we ve see previously. The oly differece is that we are ow workig i three dimesios istead of two dimesios. 007 Paul Dawkis 7

337 To get a poit o the lie all we do is pick a t ad plug ito either form of the lie. I the vector form of the lie we get a positio vector for the poit ad i the parametric form we get the actual coordiates of the poit. There is oe more form of the lie that we wat to look at. If we assume that a, b, ad c are all o-zero umbers we ca solve each of the equatios i the parametric form of the lie for t. We ca the set all of them equal to each other sice t will be the same umber i each. Doig this gives the followig, x x y y z z = = a b c This is called the symmetric equatios of the lie. If oe of a, b, or c does happe to be zero we ca still write dow the symmetric equatios. To see this let s suppose that b = 0. I this case t will ot exist i the parametric equatio for y ad so we will oly solve the parametric equatios for x ad z for t. We the set those equal ad ackowledge the parametric equatio for y as follows, x x0 z z0 = y = y0 a c Let s take a look at a example. Example Write dow the equatio of the lie that passes through the poits (,, ) ad (, 4, ). Write dow all three forms of the equatio of the lie. Solutio To do this we eed the vector v that will be parallel to the lie. This ca be ay vector as log as it s parallel to the lie. I geeral, v wo t lie o the lie itself. However, i this case it will. All we eed to do is let v be the vector that starts at the secod poit ad eds at the first poit. Sice these two poits are o the lie the vector betwee them will also lie o the lie ad will hece be parallel to the lie. So, v =, 5, 6 Note that the order of the poits was chose to reduce the umber of mius sigs i the vector. We could just have easily goe the other way. Oce we ve got v there really is t aythig else to do. To use the vector form we ll eed a poit o the lie. We ve got two ad so we ca use either oe. We ll use the first poit. Here is the vector form of the lie. r =,, + t, 5,6 = + t, 5 t,+ 6t Oce we have this equatio the other two forms follow. Here are the parametric equatios of the lie. 007 Paul Dawkis 8

338 x= + t y = 5t z = + 6t Here is the symmetric form. x y + z = = 5 6 Example Determie if the lie that passes through the poit ( 0,,8) ad is parallel to the lie give by x= 0 + t, y = t ad z = t passes through the xz-plae. If it does give the coordiates of that poit. Solutio To aswer this we will first eed to write dow the equatio of the lie. We kow a poit o the lie ad just eed a parallel vector. We kow that the ew lie must be parallel to the lie give by the parametric equatios i the problem statemet. That meas that ay vector that is parallel to the give lie must also be parallel to the ew lie. Now recall that i the parametric form of the lie the umbers multiplied by t are the compoets of the vector that is parallel to the lie. Therefore, the vector, v =,, is parallel to the give lie ad so must also be parallel to the ew lie. The equatio of ew lie is the, r = 0,,8 + t,, = t, + t,8 t If this lie passes through the xz-plae the we kow that the y-coordiate of that poit must be zero. So, let s set the y compoet of the equatio equal to zero ad see if we ca solve for t. If we ca, this will give the value of t for which the poit will pass through the xz-plae. + t = 0 t = 4 So, the lie does pass through the xz-plae. To get the complete coordiates of the poit all we eed to do is plug t = ito ay of the equatios. We ll use the vector form. 4 r =, +,8 =,0, Recall that this vector is the positio vector for the poit o the lie ad so the coordiates of the poit where the lie will pass through the xz-plae are,0, Paul Dawkis 9

339 Equatios of Plaes I the first sectio of this chapter we saw a couple of equatios of plaes. However, oe of those equatios had three variables i them ad were really extesios of graphs that we could look at i two dimesios. We would like a more geeral equatio for plaes. 0 = Let s also suppose that we have a vector that is orthogoal (perpedicular) to the plae, = abc,,. This P= xyz,, is ay poit i the plae. So, let s start by assumig that we kow a poit that is o the plae, P ( x, y, z ) vector is called the ormal vector. Now, assume that Fially, sice we are goig to be workig with vectors iitially we ll let r 0 ad r be the positio vectors for P 0 ad P respectively. Here is a sketch of all these vectors. Notice that we added i the vector r r 0 which will lie completely i the plae. Also otice that we put the ormal vector o the plae, but there is actually o reaso to expect this to be the case. We put it here to illustrate the poit. It is completely possible that the ormal vector does ot touch the plae i ay way. Now, because is orthogoal to the plae, it s also orthogoal to ay vector that lies i the plae. I particular it s orthogoal to r r 0. Recall from the Dot Product sectio that two orthogoal vectors will have a dot product of zero. I other words, r r = 0 r = r This is called the vector equatio of the plae Paul Dawkis 0

340 A slightly more useful form of the equatios is as follows. Start with the first form of the vector equatio ad write dow a vector for the differece. abc,, xyz,, x, y, z = 0 Now, actually compute the dot product to get, ( ) abc,, x x, y y, z z = a x x + b y y + c z z = This is called the scalar equatio of plae. Ofte this will be writte as, ax + by + cz = d where d = ax0 + by0 + cz0. This secod form is ofte how we are give equatios of plaes. Notice that if we are give the equatio of a plae i this form we ca quickly get a ormal vector for the plae. A ormal vector is, = abc,, Let s work a couple of examples. Example Determie the equatio of the plae that cotais the poits P = (,, 0), Q = (,, 4) ad R = ( 0,, ). Solutio I order to write dow the equatio of plae we eed a poit (we ve got three so we re cool there) ad a ormal vector. We eed to fid a ormal vector. Recall however, that we saw how to do this i the Cross Product sectio. We ca form the followig two vectors from the give poits. PQ =,, 4 PR =,, These two vectors will lie completely i the plae sice we formed them from poits that were i the plae. Notice as well that there are may possible vectors to use here, we just chose two of the possibilities. Now, we kow that the cross product of two vectors will be orthogoal to both of these vectors. Sice both of these are i the plae ay vector that is orthogoal to both of these will also be orthogoal to the plae. Therefore, we ca use the cross product as the ormal vector. i j k i j = PQ PR = 4 = i 8j + 5k The equatio of the plae is the, 007 Paul Dawkis

341 ( x ) ( y ) ( z ) = 0 x 8y+ 5z = 8 We used P for the poit, but could have used ay of the three poits. Example Determie if the plae give by x+ z = 0 ad the lie give by r = 5, t,0 + 4t are orthogoal, parallel or either. Solutio This is ot as difficult a problem as it may at first appear to be. We ca pick off a vector that is ormal to the plae. This is =, 0,. We ca also get a vector that is parallel to the lie. This is v = 0,, 4. Now, if these two vectors are parallel the the lie ad the plae will be orthogoal. If you thik about it this makes some sese. If ad v are parallel, the v is orthogoal to the plae, but v is also parallel to the lie. So, if the two vectors are parallel the lie ad plae will be orthogoal. Let s check this. i j k i j v = 0 0 = i + 4j + k So, the vectors are t parallel ad so the plae ad the lie are ot orthogoal. Now, let s check to see if the plae ad lie are parallel. If the lie is parallel to the plae the ay vector parallel to the lie will be orthogoal to the ormal vector of the plae. I other words, if ad v are orthogoal the the lie ad the plae will be parallel. Let s check this. v= = 8 0 The two vectors are t orthogoal ad so the lie ad plae are t parallel. So, the lie ad the plae are either orthogoal or parallel. 007 Paul Dawkis

342 Quadric Surfaces I the previous two sectios we ve looked at lies ad plaes i three dimesios (or ) ad while these are used quite heavily at times i a Calculus class there are may other surfaces that are also used fairly regularly ad so we eed to take a look at those. I this sectio we are goig to be lookig at quadric surfaces. Quadric surfaces are the graphs of ay equatio that ca be put ito the geeral form Ax + By + Cz + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0 where A,, J are costats. There is o way that we ca possibly list all of them, but there are some stadard equatios so here is a list of some of the more commo quadric surfaces. Ellipsoid Here is the geeral equatio of a ellipsoid. x y z + + = a b c Here is a sketch of a typical ellipsoid. If a = b= c the we will have a sphere. Notice that we oly gave the equatio for the ellipsoid that has bee cetered o the origi. Clearly ellipsoids do t have to be cetered o the origi. However, i order to make the discussio i this sectio a little easier we have chose to cocetrate o surfaces that are cetered o the origi i oe way or aother. Coe Here is the geeral equatio of a coe. Here is a sketch of a typical coe. x y z + = a b c 007 Paul Dawkis

343 Note that this is the equatio of a coe that will ope alog the z-axis. To get the equatio of a coe that opes alog oe of the other axes all we eed to do is make a slight modificatio of the equatio. This will be the case for the rest of the surfaces that we ll be lookig at i this sectio as well. I the case of a coe the variable that sits by itself o oe side of the equal sig will determie the axis that the coe opes up alog. For istace, a coe that opes up alog the x-axis will have the equatio, y z x + = b c a For most of the followig surfaces we will ot give the other possible formulas. We will however ackowledge how each formula eeds to be chaged to get a chage of orietatio for the surface. Cylider Here is the geeral equatio of a cylider. x a y b + = This is a cylider whose cross sectio is a ellipse. If a= b we have a cylider whose cross sectio is a circle. We ll be dealig with those kids of cyliders more tha the geeral form so the equatio of a cylider with a circular cross sectio is, x + y = r Here is a sketch of typical cylider with a ellipse cross sectio. 007 Paul Dawkis 4

344 The cylider will be cetered o the axis correspodig to the variable that does ot appear i the equatio. Be careful to ot cofuse this with a circle. I two dimesios it is a circle, but i three dimesios it is a cylider. Hyperboloid of Oe Sheet Here is the equatio of a hyperboloid of oe sheet. x y z + = a b c Here is a sketch of a typical hyperboloid of oe sheet. The variable with the egative i frot of it will give the axis alog which the graph is cetered. 007 Paul Dawkis 5

345 Hyperboloid of Two Sheets Here is the equatio of a hyperboloid of two sheets. x y z + = a b c Here is a sketch of a typical hyperboloid of two sheets. The variable with the positive i frot of it will give the axis alog which the graph is cetered. Notice that the oly differece betwee the hyperboloid of oe sheet ad the hyperboloid of two sheets is the sigs i frot of the variables. They are exactly the opposite sigs. Elliptic Paraboloid Here is the equatio of a elliptic paraboloid. x y z + = a b c As with cyliders this has a cross sectio of a ellipse ad if a = b it will have a cross sectio of a circle. Whe we deal with these we ll geerally be dealig with the kid that have a circle for a cross sectio. Here is a sketch of a typical elliptic paraboloid. 007 Paul Dawkis 6

346 I this case the variable that is t squared determies the axis upo which the paraboloid opes up. Also, the sig of c will determie the directio that the paraboloid opes. If c is positive the it opes up ad if c is egative the it opes dow. Hyperbolic Paraboloid Here is the equatio of a hyperbolic paraboloid. x y z = a b c Here is a sketch of a typical hyperbolic paraboloid. These graphs are vaguely saddle shaped ad as with the elliptic paraboloid the sig of c will determie the directio i which the surface opes up. The graph above is show for c positive. 007 Paul Dawkis 7

347 With the both of the types of paraboloids discussed above the surface ca be easily moved up or dow by addig/subtractig a costat from the left side. For istace z = x y + 6 is a elliptic paraboloid that opes dowward (be careful, the - is o the x ad y istead of the z) ad starts at z = 6 istead of z = 0. Here are a couple of quick sketches of this surface. Note that we ve give two forms of the sketch here. The sketch o the left has the stadard set of axes but it is difficult to see the umbers o the axis. The sketch o the right has bee boxed ad this makes it easier to see the umbers to give a sese of perspective to the sketch. I most sketches that actually ivolve umbers o the axis system we will give both sketches to help get a feel for what the sketch looks like. 007 Paul Dawkis 8

348 Fuctios of Several Variables I this sectio we wat to go over some of the basic ideas about fuctios of more tha oe variable. First, remember that graphs of fuctios of two variables, z f( xy, ) dimesioal space. For example here is the graph of = are surfaces i three z = x + y 4. This is a elliptic parabaloid ad is a example of a quadric surface. We saw several of these i the previous sectio. We will be seeig quadric surfaces fairly regularly later o i Calculus III. Aother commo graph that we ll be seeig quite a bit i this course is the graph of a plae. We have a covetio for graphig plaes that will make them a little easier to graph ad hopefully visualize. Recall that the equatio of a plae is give by ax + by + cz = d or if we solve this for z we ca write it i terms of fuctio otatio. This gives, f ( x, y) = Ax + By + D To graph a plae we will geerally fid the itersectio poits with the three axes ad the graph the triagle that coects those three poits. This triagle will be a portio of the plae ad it will give us a fairly decet idea o what the plae itself should look like. For example let s graph the plae give by, f xy, = x 4y 007 Paul Dawkis 9

349 For purposes of graphig this it would probably be easier to write this as, z = x 4y x+ 4y+ z = Now, each of the itersectio poits with the three mai coordiate axes is defied by the fact that two of the coordiates are zero. For istace, the itersectio with the z-axis is defied by x= y = 0. So, the three itersectio poits are, Here is the graph of the plae. x axis : 4,0,0 y axis : 0,, 0 z axis : 0,0, Now, to exted this out, graphs of fuctios of the form w f( xyz,, ) = would be four dimesioal surfaces. Of course we ca t graph them, but it does t hurt to poit this out. We ext wat to talk about the domais of fuctios of more tha oe variable. Recall that domais of fuctios of a sigle variable, y = f ( x), cosisted of all the values of x that we could plug ito the fuctio ad get back a real umber. Now, if we thik about it, this meas that the domai of a fuctio of a sigle variable is a iterval (or itervals) of values from the umber lie, or oe dimesioal space. The domai of fuctios of two variables, z= f( xy, ), are regios from two dimesioal space ad cosist of all the coordiate pairs, ( xy, ), that we could plug ito the fuctio ad get back a real umber. 007 Paul Dawkis 40

350 Example Determie the domai of each of the followig. f xy, = x+ y [Solutio] (a) (b) f( xy, ) = x+ y [Solutio] (c) f( xy, ) l ( 9 x 9y ) = [Solutio] Solutio (a) I this case we kow that we ca t take the square root of a egative umber so this meas that we must require, x+ y 0 Here is a sketch of the graph of this regio. [Retur to Problems] (b) This fuctio is differet from the fuctio i the previous part. Here we must require that, x 0 ad y 0 ad they really do eed to be separate iequalities. There is oe for each square root i the fuctio. Here is the sketch of this regio. [Retur to Problems] (c) I this fial part we kow that we ca t take the logarithm of a egative umber or zero. Therefore we eed to require that, 007 Paul Dawkis 4

351 x 9 x 9y > 0 + y < 9 ad upo rearragig we see that we eed to stay iterior to a ellipse for this fuctio. Here is a sketch of this regio. Note that domais of fuctios of three variables, w f( xyz,, ) dimesioal space. [Retur to Problems] =, will be regios i three Example Determie the domai of the followig fuctio, f( xyz,, ) = x + y + z 6 Solutio I this case we have to deal with the square root ad divisio by zero issues. These will require, x + y + z 6 > 0 x + y + z > 6 So, the domai for this fuctio is the set of poits that lies completely outside a sphere of radius 4 cetered at the origi. The ext topic that we should look at is that of level curves or cotour curves. The level curves z= f xy, are two dimesioal curves we get by settig z = k, where k is ay of the fuctio umber. So the equatios of the level curves are f( xy, ) = k. Note that sometimes the equatio will be i the form f( xyz,, ) = 0 ad i these cases the equatios of the level curves are f( xyk,, ) = 0. You ve probably see level curves (or cotour curves, whatever you wat to call them) before. If you ve ever see the elevatio map for a piece of lad, this is othig more tha the cotour curves for the fuctio that gives the elevatio of the lad i that area. Of course, we probably do t have the fuctio that gives the elevatio, but we ca at least graph the cotour curves. Let s do a quick example of this. 007 Paul Dawkis 4

352 Example Idetify the level curves of f xy, = x + y. Sketch a few of them. Solutio First, for the sake of practice, let s idetify what this surface give by f( xy, ) is. To do this let s rewrite it as, z = x + y Now, this equatio is ot listed i the Quadric Surfaces sectio, but if we square both sides we get, z = x + y ad this is listed i that sectio. So, we have a coe, or at least a portio of a coe. Sice we kow that square roots will oly retur positive umbers, it looks like we ve oly got the upper half of a coe. Note that this was ot required for this problem. It was doe for the practice of idetifyig the surface ad this may come i hady dow the road. Now o to the real problem. The level curves (or cotour curves) for this surface are give by the equatio are foud by substitutig z = k. I the case of our example this is, k = x + y x + y = k where k is ay umber. So, i this case, the level curves are circles of radius k with ceter at the origi. We ca graph these i oe of two ways. We ca either graph them o the surface itself or we ca graph them i a two dimesioal axis system. Here is each graph for some values of k. 007 Paul Dawkis 4

353 Note that we ca thik of cotours i terms of the itersectio of the surface that is give by z= f( xy, ) ad the plae z = k. The cotour will represet the itersectio of the surface ad the plae. For fuctios of the form f(,, ) of level surfaces are give by f( xyz,, ) = kwhere k is ay umber. xyz we will occasioally look at level surfaces. The equatios The fial topic i this sectio is that of traces. I some ways these are similar to cotours. As oted above we ca thik of cotours as the itersectio of the surface give by z= f( xy, ) ad the plae z = k. Traces of surfaces are curves that represet the itersectio of the surface ad the plae give by x= a or y = b. Let s take a quick look at a example of traces. Example 4 Sketch the traces of f( xy, ) = 0 4x y for the plae x = ad y =. Solutio We ll start with x =. We ca get a equatio for the trace by pluggig x = ito the equatio. Doig this gives, z = f, y = 0 4 y z = 6 y ad this will be graphed i the plae give by x =. Below are two graphs. The graph o the left is a graph showig the itersectio of the surface ad the plae give by x =. O the right is a graph of the surface ad the trace that we are after i this part. 007 Paul Dawkis 44

354 For y = we will do pretty much the same thig that we did with the first part. Here is the equatio of the trace, z = f ( x,) = 0 4x z = 6 4x ad here are the sketches for this case. 007 Paul Dawkis 45