CALCULUS II. Paul Dawkins

Transcription

1 CALCULUS II Paul Dawkis

2 Table of Cotets Peface... iii Outlie... v Itegatio Techiques... Itoductio... Itegatio by Pats... Itegals Ivolvig Tig Fuctios... Tig Substitutios... Patial Factios... 4 Itegals Ivolvig Roots... 4 Itegals Ivolvig Quadatics Usig Itegal Tables... 5 Itegatio Stategy Impope Itegals... 6 Compaiso Test fo Impope Itegals Appoximatig Defiite Itegals Applicatios of Itegals... 8 Itoductio... 8 Ac Legth Suface Aea Cete of Mass Hydostatic Pessue ad Foce Pobability Paametic Equatios ad Pola Coodiates Itoductio Paametic Equatios ad Cuves... 0 Tagets with Paametic Equatios... Aea with Paametic Equatios... 8 Ac Legth with Paametic Equatios... Suface Aea with Paametic Equatios... 5 Pola Coodiates... 7 Tagets with Pola Coodiates Aea with Pola Coodiates Ac Legth with Pola Coodiates Suface Aea with Pola Coodiates Ac Legth ad Suface Aea Revisited Sequeces ad Seies... 6 Itoductio... 6 Sequeces... 6 Moe o Sequeces... 7 Seies The Basics Seies Covegece/Divegece Seies Special Seies Itegal Test... 0 Compaiso Test / Limit Compaiso Test... Alteatig Seies Test... 0 Absolute Covegece... 6 Ratio Test... 0 Root Test... 7 Stategy fo Seies Estimatig the Value of a Seies... 4 Powe Seies Powe Seies ad Fuctios... 6 Taylo Seies Paul Dawkis i

3 Applicatios of Seies Biomial Seies Vectos Itoductio Vectos The Basics Vecto Aithmetic... 9 Dot Poduct Coss Poduct Thee Dimesioal Space... 0 Itoductio... 0 The -D Coodiate System... Equatios of Lies... 8 Equatios of Plaes... 4 Quadic Sufaces... 7 Fuctios of Seveal Vaiables... Vecto Fuctios Calculus with Vecto Fuctios Taget, Nomal ad Biomal Vectos... 5 Ac Legth with Vecto Fuctios Cuvatue Velocity ad Acceleatio... 6 Cylidical Coodiates Spheical Coodiates Paul Dawkis ii

4 Peface Hee ae my olie otes fo my Calculus II couse that I teach hee at Lama Uivesity. Despite the fact that these ae my class otes, they should be accessible to ayoe watig to lea Calculus II o eedig a efeshe i some of the topics fom the class. These otes do assume that the eade has a good wokig kowledge of Calculus I topics icludig limits, deivatives ad basic itegatio ad itegatio by substitutio. Calculus II teds to be a vey difficult couse fo may studets. Thee ae may easos fo this. The fist easo is that this couse does equie that you have a vey good wokig kowledge of Calculus I. The Calculus I potio of may of the poblems teds to be skipped ad left to the studet to veify o fill i the details. If you do t have good Calculus I skills, ad you ae costatly gettig stuck o the Calculus I potio of the poblem, you will fid this couse vey difficult to complete. The secod, ad pobably lage, easo may studets have difficulty with Calculus II is that you will be asked to tuly thik i this class. That is ot meat to isult ayoe; it is simply a ackowledgmet that you ca t just memoize a buch of fomulas ad expect to pass the couse as you ca do i may math classes. Thee ae fomulas i this class that you will eed to kow, but they ted to be faily geeal. You will eed to udestad them, how they wok, ad moe impotatly whethe they ca be used o ot. As a example, the fist topic we will look at is Itegatio by Pats. The itegatio by pats fomula is vey easy to emembe. Howeve, just because you ve got it memoized does t mea that you ca use it. You ll eed to be able to look at a itegal ad ealize that itegatio by pats ca be used (which is t always obvious) ad the decide which potios of the itegal coespod to the pats i the fomula (agai, ot always obvious). Fially, may of the poblems i this couse 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 waigs out of the way to my studets who may be hee to get a copy of what happeed o a day that you missed.. Because I wated to make this a faily complete set of otes fo ayoe watig to lea calculus I have icluded some mateial that I do ot usually have time to cove i class ad because this chages fom semeste to semeste it is ot oted hee. You will eed to fid oe of you fellow class mates to see if thee is somethig i these otes that was t coveed i class.. I geeal I ty to wok poblems i class that ae diffeet fom my otes. Howeve, with Calculus II may of the poblems ae difficult to make up o the spu of the momet ad so i this class my class wok will follow these otes faily close as fa as woked poblems go. With that beig said I will, o occasio, wok poblems off the top of my head whe I ca to povide moe examples tha just those i my otes. Also, I ofte 007 Paul Dawkis iii

5 do t have time i class to wok all of the poblems i the otes ad so you will fid that some sectios cotai poblems that wee t woked i class due to time estictios.. Sometimes questios i class will lead dow paths that ae ot coveed hee. I ty to aticipate as may of the questios as possible i witig these up, but the eality is that I ca t aticipate all the questios. Sometimes a vey good questio gets asked i class that leads to isights that I ve ot icluded hee. You should always talk to someoe who was i class o the day you missed ad compae these otes to thei otes ad see what the diffeeces ae. 4. This is somewhat elated to the pevious thee items, but is impotat eough to meit its ow item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Usig these otes as a substitute fo class is liable to get you i touble. As aleady oted ot eveythig i these otes is coveed i class ad ofte mateial o isights ot i these otes is coveed i class. 007 Paul Dawkis iv

6 Outlie Hee is a listig ad bief desciptio of the mateial i this set of otes. Itegatio Techiques Itegatio by Pats Of all the itegatio techiques coveed i this chapte this is pobably the oe that studets ae most likely to u ito dow the oad i othe classes. Itegals Ivolvig Tig Fuctios I this sectio we look at itegatig cetai poducts ad quotiets of tig fuctios. Tig Substitutios Hee we will look usig substitutios ivolvig tig fuctios ad how they ca be used to simplify cetai itegals. Patial Factios We will use patial factios to allow us to do itegals ivolvig some atioal fuctios. Itegals Ivolvig Roots We will take a look at a substitutio that ca, o occasio, be used with itegals ivolvig oots. Itegals Ivolvig Quadatics I this sectio we ae goig to look at some itegals that ivolve quadatics. Usig Itegal Tables Hee we look at usig Itegal Tables as well as elatig ew itegals back to itegals that we aleady kow how to do. Itegatio Stategy We give a geeal set of guidelies fo detemiig how to evaluate a itegal. Impope Itegals We will look at itegals with ifiite itevals of itegatio ad itegals with discotiuous itegads i this sectio. Compaiso Test fo Impope Itegals Hee we will use the Compaiso Test to detemie if impope itegals covege o divege. Appoximatig Defiite Itegals Thee ae may ways to appoximate the value of a defiite itegal. We will look at thee of them i this sectio. Applicatios of Itegals Ac Legth We ll detemie the legth of a cuve i this sectio. Suface Aea I this sectio we ll detemie the suface aea of a solid of evolutio. Cete of Mass Hee we will detemie the cete of mass o cetoid of a thi plate. Hydostatic Pessue ad Foce We ll detemie the hydostatic pessue ad foce o a vetical plate submeged i wate. Pobability Hee we will look at pobability desity fuctios ad computig the mea of a pobability desity fuctio. Paametic Equatios ad Pola Coodiates Paametic Equatios ad Cuves A itoductio to paametic equatios ad paametic cuves (i.e. gaphs of paametic equatios) Tagets with Paametic Equatios Fidig taget lies to paametic cuves. Aea with Paametic Equatios Fidig the aea ude a paametic cuve. 007 Paul Dawkis v

7 Ac Legth with Paametic Equatios Detemiig the legth of a paametic cuve. Suface Aea with Paametic Equatios Hee we will detemie the suface aea of a solid obtaied by otatig a paametic cuve about a axis. Pola Coodiates We ll itoduce pola coodiates i this sectio. We ll look at covetig betwee pola coodiates ad Catesia coodiates as well as some basic gaphs i pola coodiates. Tagets with Pola Coodiates Fidig taget lies of pola cuves. Aea with Pola Coodiates Fidig the aea eclosed by a pola cuve. Ac Legth with Pola Coodiates Detemiig the legth of a pola cuve. Suface Aea with Pola Coodiates Hee we will detemie the suface aea of a solid obtaied by otatig a pola cuve about a axis. Ac Legth ad Suface Aea Revisited I this sectio we will summaize all the ac legth ad suface aea fomulas fom the last two chaptes. Sequeces ad Seies Sequeces We will stat the chapte off with a bief discussio of sequeces. This sectio will focus o the basic temiology ad covegece of sequeces Moe o Sequeces Hee we will take a quick look about mootoic ad bouded sequeces. Seies The Basics I this sectio we will discuss some of the basics of ifiite seies. Seies Covegece/Divegece Most of this chapte will be about the covegece/divegece of a seies so we will give the basic ideas ad defiitios i this sectio. Seies Special Seies We will look at the Geometic Seies, Telescopig Seies, ad Hamoic Seies i this sectio. Itegal Test Usig the Itegal Test to detemie if a seies coveges o diveges. Compaiso Test/Limit Compaiso Test Usig the Compaiso Test ad Limit Compaiso Tests to detemie if a seies coveges o diveges. Alteatig Seies Test Usig the Alteatig Seies Test to detemie if a seies coveges o diveges. Absolute Covegece A bief discussio o absolute covegece ad how it diffes fom covegece. Ratio Test Usig the Ratio Test to detemie if a seies coveges o diveges. Root Test Usig the Root Test to detemie if a seies coveges o diveges. Stategy fo Seies A set of geeal guidelies to use whe decidig which test to use. Estimatig the Value of a Seies Hee we will look at estimatig the value of a ifiite seies. Powe Seies A itoductio to powe seies ad some of the basic cocepts. Powe Seies ad Fuctios I this sectio we will stat lookig at how to fid a powe seies epesetatio of a fuctio. Taylo Seies Hee we will discuss how to fid the Taylo/Maclaui Seies fo a fuctio. Applicatios of Seies I this sectio we will take a quick look at a couple of applicatios of seies. Biomial Seies A bief look at biomial seies. Vectos 007 Paul Dawkis vi

8 Vectos The Basics I this sectio we will itoduce some of the basic cocepts about vectos. Vecto Aithmetic Hee we will give the basic aithmetic opeatios fo vectos. Dot Poduct We will discuss the dot poduct i this sectio as well as a applicatio o two. Coss Poduct I this sectio we ll discuss the coss poduct ad see a quick applicatio. Thee Dimesioal Space This is the oly chapte that exists i two places i my otes. Whe I oigially wote these otes all of these topics wee coveed i Calculus II howeve, we have sice moved seveal of them ito Calculus III. So, athe tha split the chapte up I have kept it i the Calculus II otes ad also put a copy i the Calculus III otes. The -D Coodiate System We will itoduce the cocepts ad otatio fo the thee dimesioal coodiate system i this sectio. Equatios of Lies I this sectio we will develop the vaious foms fo the equatio of lies i thee dimesioal space. Equatios of Plaes Hee we will develop the equatio of a plae. Quadic Sufaces I this sectio we will be lookig at some examples of quadic sufaces. Fuctios of Seveal Vaiables A quick eview of some impotat topics about fuctios of seveal vaiables. Vecto Fuctios We itoduce the cocept of vecto fuctios i this sectio. We cocetate pimaily o cuves i thee dimesioal space. We will howeve, touch biefly o sufaces as well. Calculus with Vecto Fuctios Hee we will take a quick look at limits, deivatives, ad itegals with vecto fuctios. Taget, Nomal ad Biomal Vectos We will defie the taget, omal ad biomal vectos i this sectio. Ac Legth with Vecto Fuctios I this sectio we will fid the ac legth of a vecto fuctio. Cuvatue We will detemie the cuvatue of a fuctio i this sectio. Velocity ad Acceleatio I this sectio we will evisit a stadad applicatio of deivatives. We will look at the velocity ad acceleatio of a object whose positio fuctio is give by a vecto fuctio. Cylidical Coodiates We will defie the cylidical coodiate system i this sectio. The cylidical coodiate system is a alteate coodiate system fo the thee dimesioal coodiate system. Spheical Coodiates I this sectio we will defie the spheical coodiate system. The spheical coodiate system is yet aothe alteate coodiate system fo the thee dimesioal coodiate system. 007 Paul Dawkis vii

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10 Itegatio Techiques Itoductio I this chapte we ae goig to be lookig at vaious itegatio techiques. Thee ae a fai umbe of them ad some will be easie tha othes. The poit of the chapte is to teach you these ew techiques ad so this chapte assumes that you ve got a faily good wokig kowledge of basic itegatio as well as substitutios with itegals. I fact, most itegals ivolvig simple substitutios will ot have ay of the substitutio wok show. It is goig to be assumed that you ca veify the substitutio potio of the itegatio youself. Also, most of the itegals doe i this chapte will be idefiite itegals. It is also assumed that oce you ca do the idefiite itegals you ca also do the defiite itegals ad so to coseve space we cocetate mostly o idefiite itegals. Thee is oe exceptio to this ad that is the Tig Substitutio sectio ad i this case thee ae some subtleties ivolved with defiite itegals that we e goig to have to watch out fo. Outside of that howeve, most sectios will have at most oe defiite itegal example ad some sectios will ot have ay defiite itegal examples. Hee is a list of topics that ae coveed i this chapte. Itegatio by Pats Of all the itegatio techiques coveed i this chapte this is pobably the oe that studets ae most likely to u ito dow the oad i othe classes. Itegals Ivolvig Tig Fuctios I this sectio we look at itegatig cetai poducts ad quotiets of tig fuctios. Tig Substitutios Hee we will look usig substitutios ivolvig tig fuctios ad how they ca be used to simplify cetai itegals. Patial Factios We will use patial factios to allow us to do itegals ivolvig some atioal fuctios. Itegals Ivolvig Roots We will take a look at a substitutio that ca, o occasio, be used with itegals ivolvig oots. Itegals Ivolvig Quadatics I this sectio we ae goig to look at some itegals that ivolve quadatics. Usig Itegal Tables Hee we look at usig Itegal Tables as well as elatig ew itegals back to itegals that we aleady kow how to do. Itegatio Stategy We give a geeal set of guidelies fo detemiig how to evaluate a itegal. 007 Paul Dawkis

11 Impope Itegals We will look at itegals with ifiite itevals of itegatio ad itegals with discotiuous itegads i this sectio. Compaiso Test fo Impope Itegals Hee we will use the Compaiso Test to detemie if impope itegals covege o divege. Appoximatig Defiite Itegals Thee ae may ways to appoximate the value of a defiite itegal. We will look at thee of them i this sectio. 007 Paul Dawkis

12 Itegatio by Pats Let s stat off with this sectio with a couple of itegals that we should aleady be able to do to get us stated. Fist let s take a look at the followig. x x Ú e dx= e + c So, that was simple eough. Now, let s take a look at, x Úx e dx To do this itegal we ll use the followig substitutio. u = x du = xdx fi xdx= du x u u x Úxe dx= du = + c= + c Úe e e Agai, simple eough to do povided you emembe how to do substitutios. By the way make sue that you ca do these kids of substitutios quickly ad easily. Fom this poit o we ae goig to be doig these kids of substitutios i ou head. If you have to stop ad wite these out with evey poblem you will fid that it will take you sigificatly loge to do these poblems. Now, let s look at the itegal that we eally wat to do. 6x Úx e dx 6x If we just had a x by itself o e by itself we could do the itegal easily eough. But, we do t have them by themselves, they ae istead multiplied togethe. Thee is o substitutio that we ca use o this itegal that will allow us to do the itegal. So, at this poit we do t have the kowledge to do this itegal. To do this itegal we will eed to use itegatio by pats so let s deive the itegatio by pats fomula. We ll stat with the poduct ule. fg = f g+ fg Now, itegate both sides of this. fg dx= f g+ fgdx Ú Ú The left side is easy eough to itegate ad we ll split up the ight side of the itegal. fg = f gdx+ fgdx Ú Ú Note that techically we should have had a costat of itegatio show up o the left side afte doig the itegatio. We ca dop it at this poit sice othe costats of itegatio will be showig up dow the oad ad they would just ed up absobig this oe. Fially, ewite the fomula as follows ad we aive at the itegatio by pats fomula. 007 Paul Dawkis

13 Ú Ú fgdx = fg- f gdx This is ot the easiest fomula to use howeve. 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 ae just the stadad Calc I substitutios that hopefully you ae used to by ow. Do t get excited by the fact that we ae usig two substitutios hee. They will wok the same way. Usig these substitutios gives us the fomula that most people thik of as the itegatio by pats fomula. Úudv= uv-ú vdu To use this fomula we will eed to idetify u ad dv, compute du ad v ad the use the fomula. Note as well that computig v is vey easy. All we eed to do is itegate dv. v= Ú dv So, let s take a look at the itegal above that we metioed we wated to do. Example Evaluate the followig itegal. 6 x Ú x e dx Solutio So, o some level, the poblem hee is the x that is i fot of the expoetial. If that was t thee we could do the itegal. Notice as well that i doig itegatio by pats aythig that we choose fo u will be diffeetiated. So, it seems that choosig u = x will be a good choice sice upo diffeetiatig the x will dop out. Now that we ve chose u we kow that dv will be eveythig else that emais. So, hee ae the choices fo u ad dv as well as du ad v. 6 x u = x dv= e dx The itegal is the, du = dx v= Úe dx = e 6 x Û Ú e 6 e Ù ı 6 e x 6 6 6x 6x 6x x dx= - dx 6x 6x = e - e + 6x 6x Oce we have doe the last itegal i the poblem we will add i the costat of itegatio to get ou fial aswe. c 007 Paul Dawkis 4

14 Next, let s take a look at itegatio by pats fo defiite itegals. The itegatio by pats fomula fo defiite itegals is, Itegatio by Pats, Defiite Itegals b udv uv b b = - vdu b Úa a Ú a Note that the uv i the fist tem is just the stadad itegal evaluatio otatio that you should a be familia with at this poit. All we do is evaluate the tem, uv i this case, at b the subtact off the evaluatio of the tem at a. At some level we do t eally eed a fomula hee because we kow that whe doig defiite itegals all we eed to do is do the idefiite itegal ad the do the evaluatio. Let s take a quick look at a defiite itegal usig itegatio by pats. Example Evaluate the followig itegal. 6x Ú xe dx - Solutio This is the same itegal that we looked at i the fist example so we ll use the same u ad dv to get, Ú 6x x 6x 6x xe dx= e - e dx x 6x x = e - e = e + e Sice we eed to be able to do the idefiite itegal i ode to do the defiite itegal ad doig the defiite itegal amouts to othig moe tha evaluatig the idefiite itegal at a couple of poits we will cocetate o doig idefiite itegals i the est of this sectio. I fact, thoughout most of this chapte this will be the case. We will be doig fa moe idefiite itegals tha defiite itegals. Let s take a look at some moe examples. Example Evaluate the followig itegal. Û Ê t ˆ Ù ( t+ 5cos ) Á dt ı Ë 4 Solutio Thee ae two ways to poceed with this example. Fo may, the fist thig that they ty is multiplyig the cosie though the paethesis, splittig up the itegal ad the doig itegatio by pats o the fist itegal. While that is a pefectly acceptable way of doig the poblem it s moe wok tha we eally eed 007 Paul Dawkis 5 Ú

15 to do. Istead of splittig the itegal up let s istead use the followig choices fo u ad dv. Ê t ˆ u = t+ 5 dv= cosá dt Ë 4 Ê t ˆ du = dt v= 4si Á Ë 4 The itegal is the, Û Ê t ˆ Ê t ˆ Û Ê t ˆ Ù( t+ 5cos ) Á dt = 4 ( t+ 5si ) Á -Ùsi Á dt ı Ë4 Ë4 ı Ë4 Ê t ˆ Ê t ˆ = 4 ( t+ 5si ) Á + 48cosÁ + c Ë4 Ë4 Notice that we pulled ay costats out of the itegal whe we used the itegatio by pats fomula. We will usually do this i ode to simplify the itegal a little. Example 4 Evaluate the followig itegal. w si 0w dw Ú Solutio Fo this example we ll use the followig choices fo u ad dv. u = w dv= si 0wdw The itegal is the, du = wdw v=- cos 0w 0 w =- + Ú Ú w si 0wdw cos 0w wcos 0w dw 0 5 I this example, ulike the pevious examples, the ew itegal will also equie itegatio by pats. Fo this secod itegal we will use the followig choices. u = w dv= cos 0wdw So, the itegal becomes, Ú du = dw v= si 0 w 0 w w Á 0 5Ë0 0 Ê w si 0wdw cos 0w si 0w si 0wdw =- + - w Ê w ˆ =- cos( 0w) + Á si( 0w) + cos( 0w) + c 0 5Ë0 00 w w =- cos( 0w) + si( 0w) + cos( 0w) + c Be caeful with the coefficiet o the itegal fo the secod applicatio of itegatio by pats. Sice the itegal is multiplied by 5 we eed to make sue that the esults of actually doig the itegal ae also multiplied by 5. Fogettig to do this is oe of the moe commo mistakes with itegatio by pats poblems. Ú ˆ 007 Paul Dawkis 6

16 As this last example has show us, we will sometimes eed moe tha oe applicatio of itegatio by pats to completely evaluate a itegal. 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 impotat poit about itegatio techiques. Some itegals ca be doe i usig seveal diffeet techiques. That is the case with the itegal i the ext example. Example 5 Evaluate the followig itegal Ú x x+ dx (a) Usig Itegatio by Pats. [Solutio] (b) Usig a stadad Calculus I substitutio. [Solutio] Solutio (a) Evaluate usig Itegatio by Pats. Fist otice that thee ae o tig fuctios o expoetials i this itegal. While a good may itegatio by pats itegals will ivolve tig fuctios ad/o 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 fo u ad dv. u = x dv= x+ dx The itegal is the, Ú du = dx v= x+ x x+ dx= x( x+ ) - Ú x+ dx 5 4 = x( x+ ) - ( x+ ) + c 5 (b) Evaluate Usig a stadad Calculus I substitutio. [Retu to Poblems] Now let s do the itegal 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 fo the quatity ude the squae oot ad oce fo the x i fot of the squae oot. The itegal is the, Ú Ú x x+ dx= u- udu Ú = u -u du 5 = u - u + c 5 5 = ( x+ ) - ( x+ ) + c 5 [Retu to Poblems] 007 Paul Dawkis 7

17 So, we used two diffeet itegatio techiques i this example ad we got two diffeet aswes. The obvious questio the should be : Did we do somethig wog? Actually, we did t do aythig wog. We eed to emembe the followig fact fom Calculus I. If the f x = g x f x = g x + c I othe wods, if two fuctios have the same deivative the they will diffe by o moe tha a costat. So, how does this apply to the above poblem? Fist defie the followig, f x = g x = x x+ The we ca compute f ( x) ad g( x ) by itegatig as follows, = = Ú Ú f x f x dx g x g x dx We ll use itegatio by pats fo the fist itegal ad the substitutio fo the secod itegal. g x should diffe by o moe tha a costat. Let s The accodig to the fact f ( x ) ad veify this ad see if this is the case. We ca veify that they diffe by o moe tha a costat if we take a look at the diffeece of the two ad do a little algebaic maipulatio ad simplificatio. 5 5 Ê 4 ˆ Ê ˆ Á x x+ - x+ - Á x+ - x+ Ë 5 Ë5 Ê 4 ˆ = ( x+ ) Á x- ( x+ ) - ( x+ ) + Ë 5 5 ( x ) ( 0) = + = 0 So, i this case it tus out the two fuctios ae exactly the same fuctio sice the diffeece is zeo. Note that this wo t always happe. Sometimes the diffeece will yield a ozeo costat. Fo a example of this check out the Costat of Itegatio sectio i my Calculus I otes. So just what have we leaed? Fist, thee will, o occasio, be moe tha oe method fo evaluatig a itegal. Secodly, we saw that diffeet methods will ofte lead to diffeet aswes. Last, eve though the aswes ae diffeet it ca be show, sometimes with a lot of wok, that they diffe by o moe tha a costat. Whe we ae faced with a itegal the fist thig that we ll eed to decide is if thee is moe tha oe way to do the itegal. If thee is moe tha oe way we ll the eed to detemie which method we should use. The geeal ule 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 othes fid easiest, but that does t make it the wog method. 007 Paul Dawkis 8

18 Oe of the moe commo mistakes with itegatio by pats is fo people to get too locked ito peceived pattes. Fo istace, all of the pevious examples used the basic patte of takig u to be the polyomial that sat i fot of aothe fuctio ad the lettig dv be the othe fuctio. This will ot always happe so we eed to be caeful ad ot get locked ito ay pattes that we thik we see. Let s take a look at some itegals that do t fit ito the above patte. Example 6 Evaluate the followig itegal. Ú l xdx Solutio So, ulike ay of the othe itegal we ve doe to this poit thee is oly a sigle fuctio i the itegal ad o polyomial sittig i fot of the logaithm. The fist choice of may people hee is to ty ad fit this ito the patte fom above ad make the followig choices fo u ad dv. u = dv= l xdx This leads to a eal poblem howeve sice that meas v must be, v= Ú l xdx I othe wods, we would eed to kow the aswe ahead of time i ode to actually do the poblem. So, this choice simply wo t wok. Also otice that with this choice we d get that du = 0 which also causes poblems ad is aothe easo why this choice will ot wok. Theefoe, if the logaithm 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 The itegal is the, Example 7 Evaluate the followig itegal. du = dx v= x x Ú lxdx= xl x- Û Ù xdx ı x Ú Ú = xl x- dx = xl x- x+ c 5 x x + dx Solutio So, if we agai ty to use the patte fom the fist few examples fo this itegal ou choices fo u ad dv would pobably be the followig. 5 u = x dv= x + dx Howeve, as with the pevious example this wo t wok sice we ca t easily compute v. Ú v= x + dx 007 Paul Dawkis 9

19 This is ot a easy itegal to do. Howeve, otice that if we had a x i the itegal alog with the oot we could vey easily do the itegal with a substitutio. Also otice that we do have a lot of x s floatig aoud i the oigial itegal. So istead of puttig all the x s (outside of the oot) i the u let s split them up as follows. u = x dv= x x + dx du = xdx v= x + 9 We ca ow easily compute v ad afte usig itegatio by pats we get, 5 Úx x + dx= x ( x + ) - ÛÙ x ( x + ) dx 9 ı 5 4 = x ( x + ) - ( x + ) + c 9 45 So, i the pevious two examples we saw cases that did t quite fit ito ay peceived patte that we might have gotte fom the fist couple of examples. This is always somethig that we eed to be o the lookout fo with itegatio by pats. Let s take a look at aothe example that also illustates aothe itegatio techique that sometimes aises out of itegatio by pats poblems. Example 8 Evaluate the followig itegal. q Ú e cosq dq Solutio Okay, to this poit we ve always picked u i such a way that upo diffeetiatig it would make that potio go away o at the vey least put it the itegal ito a fom that would make it easie to deal with. I this case o matte which pat we make u it will eve go away i the diffeetiatio pocess. It does t much matte which we choose to be u so we ll choose i the followig way. Note howeve that we could choose the othe way as well ad we ll get the same esult i the ed. q u = cosq dv= e dq du =- siq dq v= e q The itegal is the, Ú q q q e cosq dq = e cosq + e siq dq Ú So, it looks like we ll do itegatio by pats agai. Hee ae ou choices this time. q u = siq dv= e dq du = cosq dq v= e q The itegal is ow, Ú q q q q e cosq dq = e cosq + e siq - e cosq dq Ú 007 Paul Dawkis 0

20 Now, at this poit it looks like we e just uig i cicles. Howeve, otice that we ow have the same itegal o both sides ad o the ight side it s got a mius sig i fot of it. This meas that we ca add the itegal to both sides to get, Ú e q cosq dq = e q cosq + e q siq All we eed to do ow is divide by ad we e doe. The itegal is, q q q Ú e cosq dq = ( cosq + siq) + c e e Notice that afte dividig by the two we add i the costat of itegatio at that poit. This idea of itegatig util you get the same itegal o both sides of the equal sig ad the simply solvig fo the itegal is kid of ice to emembe. It does t show up all that ofte, but whe it does it may be the oly way to actually do the itegal. We ve got oe moe example to do. As we will see some poblems could equie us to do itegatio by pats umeous times ad thee is a shot had method that will allow us to do multiple applicatios of itegatio by pats quickly ad easily. Example 9 Evaluate the followig itegal. Ú x e dx Solutio We stat off by choosig u ad dv as we always would. Howeve, istead of computig du ad v we put these ito the followig table. We the diffeetiate dow the colum coespodig to u util we hit zeo. I the colum coespodig to dv we itegate oce fo each ety i the fist colum. Thee is also a thid colum which we will explai i a bit ad it always stats with a + ad the alteates sigs as show. 4 x Now, multiply alog the diagoals show i the table. I fot of each poduct put the sig i the thid colum that coespods to the u tem fo that poduct. I this case this would give, 007 Paul Dawkis

21 x x x x x x 4 4 Ê ˆ Ê ˆ Ê ˆ Ê ˆ Ê ˆ Úx e dx= ( x ) Áe - ( 4x ) Á4e + ( x ) Á8e - ( 4x) Á6e + ( 4) Áe Ë Ë Ë Ë Ë x x x x x 4 = x e - 6x e + 96x e - 84xe + 768e + c We ve got the itegal. This is much easie tha witig dow all the vaious u s ad dv s that we d have to do othewise. So, i this sectio we ve see how to do itegatio by pats. I you late math classes this is liable to be oe of the moe fequet itegatio techiques that you ll ecoute. It is impotat to ot get too locked ito pattes that you may thik you ve see. I most cases ay patte that you thik you ve see ca (ad will be) violated at some poit i time. Be caeful! Also, do t foget the shothad method fo multiple applicatios of itegatio by pats poblems. It ca save you a fai amout of wok o occasio. 007 Paul Dawkis

22 Itegals Ivolvig Tig Fuctios I this sectio we ae goig to look at quite a few itegals ivolvig tig fuctios ad some of the techiques we ca use to help us evaluate them. Let s stat off with a itegal that we should aleady be able to do. Ú Ú 5 5 cos si = usig the substitutio = si x xdx u du u x si 6 6 = x+ c This itegal is easy to do with a substitutio because the pesece of the cosie, howeve, what about the followig itegal. Example Evaluate the followig itegal. 5 si xdx Ú Solutio This itegal o loge has the cosie i it that would allow us to use the substitutio that we used above. Theefoe, that substitutio wo t wok ad we ae goig to have to fid aothe way of doig this itegal. Let s fist otice that we could wite the itegal as follows, 5 4 Ú = Ú = Ú si xdx si xsixdx si x si xdx Now ecall the tig idetity, cos x+ si x= fi si x= - cos x With this idetity the itegal ca be witte as, 5 Ú = Ú ( - ) si xdx cos x si xdx ad we ca ow use the substitutio u = cos x. Doig this gives us, Ú Ú Ú u 5 si xdx=- -u du 4 =- - + u du Ê 5ˆ =-Áu- u + u + c Ë 5 =- cos + cos - cos x x x c So, with a little ewitig o the itegad we wee able to educe this to a faily simple substitutio. Notice that we wee able to do the ewite that we did i the pevious example because the expoet o the sie was odd. I these cases all that we eed to do is stip out oe of the sies. 007 Paul Dawkis

23 The expoet o the emaiig sies will the be eve ad we ca easily covet the emaiig 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 stip out a sie, but the emaiig sies would the have a odd expoet ad while we could covet them to cosies the esultig itegal would ofte be eve moe difficult tha the oigial itegal i most cases. Let s take a look at aothe example. Example Evaluate the followig itegal. si xcos Ú 6 Solutio So, i this case we ve got both sies ad cosies i the poblem ad i this case the expoet o the sie is eve while the expoet o the cosie is odd. So, we ca use a simila techique i this itegal. This time we ll stip out a cosie ad covet the est to sies. 6 6 si xcos xdx= si xcos xcos xdx Ú Ú Ú Ú Ú 6 ( ) 6 u u du xdx = si x - si x cosxdx u = si x = = u -u du x x c = si - si + At this poit let s pause fo a secod to summaize what we ve leaed so fa about itegatig powes of sie ad cosie. m Ú si xcos xdx () I this itegal if the expoet o the sies () is odd we ca stip out oe sie, covet the est to cosies usig () ad the use the substitutio u = cos x. Likewise, if the expoet o the cosies (m) is odd we ca stip out oe cosie ad covet the est to sies ad the use the substitutio u = si x. Of couse, if both expoets ae odd the we ca use eithe method. Howeve, i these cases it s usually easie to covet the tem with the smalle expoet. The oe case we have t looked at is what happes if both of the expoets ae eve? I this case the techique we used i the fist couple of examples simply wo t wok ad i fact thee eally is t ay oe set method fo doig these itegals. Each itegal is diffeet ad i some cases thee will be moe tha oe way to do the itegal. With that beig said most, if ot all, of itegals ivolvig poducts of sies ad cosies i which both expoets ae eve ca be doe usig oe o moe of the followig fomulas to ewite the itegad. 007 Paul Dawkis 4

24 cos x= + cos si x= -cos sixcosx= si x ( ( x) ) ( ( x) ) The fist two fomulas ae the stadad half agle fomula fom a tig class witte i a fom that will be moe coveiet fo us to use. The last is the stadad double agle fomula fo sie, agai with a small ewite. Let s take a look at a example. Example Evaluate the followig itegal. si Ú xcos Solutio As oted above thee ae ofte moe tha oe way to do itegals i which both of the expoets ae eve. This itegal is a example of that. Thee ae at least two solutio techiques fo this poblem. We will do both solutios statig with what is pobably the hade of the two, but it s also the oe that may people see fist. Solutio I this solutio we will use the two half agle fomulas above ad just substitute them ito the itegal. Û Êˆ Úsi xcos xdx= Ù ( - cos( x) ) Á ( + cos( x) ) dx ı Ë cos = - ( x) dx 4 Ú So, we still have a itegal that ca t be completely doe, howeve otice that we have maaged to educe the itegal dow to just oe tem causig poblems (a cosie with a eve powe) athe tha two tems causig poblems. I fact to elimiate the emaiig poblem tem all that we eed to do is euse the fist half agle fomula give above. si xcos xdx= Û Ú Ù- ( + cos( 4x) ) dx 4ı = Û Ù - cos ( 4 x) dx 4ı Ê ˆ = Á x- si ( 4 x) + c 4Ë 8 = x- si ( 4 x) + c 8 So, this solutio equied a total of thee tig idetities to complete. xdx 007 Paul Dawkis 5

25 Solutio I this solutio we will use the half agle fomula to help simplify the itegal as follows. Ú si cos si cos x xdx= x x dx ÛÊ ˆ = Ù Á si ( x) ıë si = x dx 4 Now, we use the double agle fomula fo sie to educe to a itegal that we ca do. Ú Ú Ú si xcos xdx= -cos 4x dx 8 = x- si ( 4 x) + c 8 This method equied oly two tig idetities to complete. Notice that the diffeece betwee these two methods is moe oe of messiess. The secod method is ot appeciably easie (othe tha eedig oe less tig idetity) it is just ot as messy ad that will ofte taslate ito a easie pocess. I the pevious example we saw two diffeet solutio methods that gave the same aswe. Note that this will ot always happe. I fact, moe ofte tha ot we will get diffeet aswes. Howeve, as we discussed i the Itegatio by Pats sectio, the two aswes will diffe by o moe tha a costat. I geeal whe we have poducts of sies ad cosies i which both expoets ae eve we will eed to use a seies of half agle ad/o double agle fomulas to educe the itegal ito a fom that we ca itegate. Also, the lage the expoets the moe we ll eed to use these fomulas ad hece the messie the poblem. Sometimes i the pocess of educig itegals i which both expoets ae eve we will u acoss poducts of sie ad cosie i which the agumets ae diffeet. These will equie oe of the followig fomulas to educe the poducts to itegals that we ca do. Ú dx siacosb = Èsi si Î a - b + a + b siasib = Ècos( a b) cos( a b) Î cosacosb = Ècos( a - b) + cos( a + b) Î Let s take a look at a example of oe of these kids of itegals. 007 Paul Dawkis 6

26 Example 4 Evaluate the followig itegal. cos 5x cos 4x dx Ú Solutio This itegal equies the last fomula listed above. Úcos( 5x) cos( 4x) dx= cos( ) cos( 9 ) Ú x + x dx Ê ˆ = Á si( x) + si( 9x) + c Ë 9 Okay, at this poit we ve coveed petty much all the possible cases ivolvig poducts of sies ad cosies. It s ow time to look at itegals that ivolve poducts of secats ad tagets. This time, let s do a little aalysis of the possibilities befoe we just jump ito examples. The geeal itegal will be, m Ú sec xta xdx () The fist thig to otice is that we ca easily covet eve powes of secats to tagets ad eve powes of tagets to secats by usig a fomula simila to (). I fact, the fomula ca be deived fom () 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 e goig to wat to deal with () similaly to how we dealt with (). We ll wat to evetually use oe of the followig substitutios. u = tax du = sec xdx u = secx du = secxta xdx So, if we use the substitutio u = ta x we will eed two secats left fo the substitutio to wok. This meas that if the expoet o the secat () is eve we ca stip two out ad the covet the emaiig secats to tagets usig (4). Next, if we wat to use the substitutio u = sec xwe will eed oe secat ad oe taget left ove i ode 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 itegad we ca stip out oe of the tagets alog with oe of the secats of couse. The taget will the have a eve expoet ad so we ca use (4) to covet the est of the tagets to secats. Note that this method does equie that we have at least oe secat i the itegal as well. If thee ae t ay secats the we ll eed to do somethig diffeet. If the expoet o the secat is eve ad the expoet o the taget is odd the we ca use eithe case. Agai, it will be easie to covet the tem with the smallest expoet. 007 Paul Dawkis 7

27 Let s take a look at a couple of examples. Example 5 Evaluate the followig itegal. sec Ú xta 9 5 Solutio Fist ote that sice the expoet o the secat is t eve we ca t use the substitutio u = ta x. Howeve, the expoet o the taget is odd ad we ve got a secat i the itegal ad so we will be able to use the substitutio u = sec x. This meas stipig out a sigle taget (alog with a secat) ad covetig the emaiig tagets to secats usig (4). xdx Hee s the wok fo this itegal sec xta xdx= sec xta xtaxsec xdx Ú Ú Ú 8 8 u u du 0 8 u u u du 9 = = sec x sec x- taxsecxdx u = sec x Ú Ú = - = - + Example 6 Evaluate the followig itegal. sec sec x sec x sec x c 9 Ú xta 4 6 Solutio So, i this example the expoet o the taget is eve so the substitutio u = sec x wo t wok. The expoet o the secat is eve ad so we ca use the substitutio u = ta x fo this itegal. That meas that we eed to stip out two secats ad covet the est to tagets. Hee is the wok fo this itegal sec xta xdx= sec xta xsec xdx Ú Ú Ú 6 Ú( u ) 8 6 Úu u du xdx 6 = ta x+ ta xsec xdx u = ta x = + = + u du x x c = ta + ta + Both of the pevious examples fit vey icely ito the pattes discussed above ad so wee ot all that difficult to wok. Howeve, thee ae a couple of exceptios to the pattes above ad i these cases thee is o sigle method that will wok fo evey poblem. Each itegal will be diffeet ad may equie diffeet solutio methods i ode to evaluate the itegal. Let s fist take a look at a couple of itegals that have odd expoets o the tagets, but o secats. I these cases we ca t use the substitutio u = sec xsice it equies thee to be at least oe secat i the itegal. 007 Paul Dawkis 8

28 Example 7 Evaluate the followig itegal. ta xdx Ú Solutio To do this itegal all we eed to do is ecall the defiitio of taget i tems of sie ad cosie ad the this itegal is othig moe tha a Calculus I substitutio. si x taxdx= Û Ú Ù dx u = cos x ı cos x =- Û Ù du ı u =- l cosx + c lx= l x - = l cos x + c l sec x + c Example 8 Evaluate the followig itegal. ta xdx Solutio The tick to this oe is do the followig maipulatio of the itegad. ta xdx= taxta xdx Ú Ú Ú Ú Ú = tax sec x- dx ta sec ta = x xdx- xdx We ca ow use the substitutio u = ta x o the fist itegal ad the esults fom the pevious example to o the secod itegal. Ú The itegal is the, Ú xdx= x- x + c ta ta l sec Note that all odd powes of taget (with the exceptio of the fist powe) ca be itegated usig the same method we used i the pevious example. Fo istace, Ú = Ú ( - ) = Ú -Ú 5 ta xdx ta x sec x dx ta xsec xdx ta xdx So, a quick substitutio ( u = ta x) will give us the fist itegal ad the secod itegal will always be the pevious odd powe. 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 wok. It should also be oted that both of the followig two itegals ae itegals that we ll be seeig o occasio i late sectios of this chapte ad i late chaptes. Because of this it would t be a bad idea to make a ote of these esults so you ll have them eady whe you eed them late. 007 Paul Dawkis 9

29 Example 9 Evaluate the followig itegal. Ú sec xdx Solutio This oe is t too bad oce you see what you ve got to do. By itself the itegal ca t be doe. Howeve, if we maipulate the itegad as follows we ca do it. Ú sec xdx= = Û Ù ı Û Ù ı ( + ) secx secx ta x secx+ ta x sec + ta sec dx x x x dx secx+ ta x I this fom we ca do the itegal usig the substitutio u = secx+ ta x. Doig this gives, secxdx= l secx+ ta x + c Ú The idea used i the above example is a ice idea to keep i mid. Multiplyig the umeato ad deomiato of a tem by the same tem above ca, o occasio, put the itegal ito a fom that ca be itegated. Note that this method wo t always wok ad eve whe it does it wo t always be clea what you eed to multiply the umeato ad deomiato by. Howeve, whe it does wok ad you ca figue out what tem you eed it ca geatly simplify the itegal. Hee s the ext example. Example 0 Evaluate the followig itegal. sec xdx Ú Solutio This oe is diffeet fom ay of the othe itegals that we ve doe i this sectio. The fist step to doig this itegal is to pefom itegatio by pats usig the followig choices fo u ad dv. u = secx dv= sec xdx du = secxtaxdx v= ta x Note that usig itegatio by pats o this poblem is ot a obvious choice, but it does wok vey icely hee. Afte doig itegatio by pats we have, sec xdx= secxtax- secxta xdx Ú Ú Now the ew itegal also has a odd expoet o the secat ad a eve expoet o the taget ad so the pevious examples of poducts of secats ad tagets still wo t do us ay good. To do this itegal we ll fist wite the tagets i the itegal i tems of secats. Agai, this is ot ecessaily a obvious choice but it s what we eed to do i this case. sec xdx= secxtax- secx sec x- dx Ú Ú Ú = secxtax- sec xdx+ sec xdx Now, we ca use the esults fom the pevious example to do the secod itegal ad otice that Ú 007 Paul Dawkis 0

30 the fist itegal is exactly the itegal we e beig asked to evaluate with a mius sig i fot. So, add it to both sides to get, sec xdx= secxtax+ l secx+ ta x Fially divide by a two ad we e doe. Ú Ú sec xdx= secxtax+ l secx+ ta x + c Agai, ote that we ve agai used the idea of itegatig the ight side util the oigial itegal shows up ad the movig this to the left side ad dividig by its coefficiet to complete the evaluatio. We fist saw this i the Itegatio by Pats sectio ad oted at the time that this was a ice techique to emembe. Hee is aothe example of this techique. Now that we ve looked at poducts of secats ad tagets let s also ackowledge that because we ca elate cosecats ad cotagets by + cot x= csc x all of the wok that we did fo poducts of secats ad tagets will also wok fo poducts of cosecats ad cotagets. We ll leave it to you to veify that. Thee is oe fial topic to be discussed i this sectio befoe movig o. To this poit we ve looked oly at poducts of sies ad cosies ad poducts of secats ad tagets. Howeve, the methods used to do these itegals 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 itegal. 7 Û si x Ù 4 dx ı cos x Solutio If this wee a poduct of sies ad cosies we would kow what to do. We would stip out a sie (sice the expoet o the sie is odd) ad covet the est of the sies to cosies. The same idea will wok i this case. We ll stip out a sie fom the umeato ad covet the est to cosies as follows, 7 6 Û si x si x dx= Û Ù si xdx 4 Ù 4 ı cos x ı cos x = = Û Ù ı Û Ù ı ( si x) cos 4 ( - cos x) cos x 4 x si xdx si xdx At this poit all we eed to do is use the substitutio u = cos xad we e doe. 007 Paul Dawkis

31 7 Û si x Û Ù dx =-Ù ı cos x ı ( -u ) u 4 4 Ú du = u u u du Ê ˆ =-Á- + + u- u c + Ë u u = cosx cos x c cos x cosx So, ude the ight cicumstaces, we ca use the ideas developed to help us deal with poducts of tig fuctios to deal with quotiets of tig fuctios. The atual questio the, is just what ae the ight cicumstaces? Fist otice that if the quotiet had bee evesed as i this itegal, 4 Û cos x Ù 7 dx ı si x we would t have bee able to stip out a sie. 4 4 Û cos x cos x dx= Û Ù dx 7 Ù 6 ı si x ı si x si x I this case the stipped out sie emais i the deomiato ad it wo t do us ay good fo the substitutio u = cos xsice this substitutio equies a sie i the umeato of the quotiet. Also ote that, while we could covet the sies to cosies, the esultig itegal would still be a faily difficult itegal. So, we ca use the methods we applied to poducts of tig fuctios to quotiets of tig fuctios povided the tem that eeds pats stipped out i is the umeato of the quotiet. 007 Paul Dawkis

32 Tig Substitutios As we have doe i the last couple of sectios, let s stat off with a couple of itegals that we should aleady be able to do with a stadad substitutio. Û x Ú x 5x - 4dx= 5x c Ù dx= 5x - 4+ c 75 ı 5x -4 5 Both of these used the substitutio u = 5x - 4 ad at this poit should be petty easy fo you to do. Howeve, let s take a look at the followig itegal. Example Evaluate the followig itegal. Û Ù ı 5x -4 dx x Solutio I this case the substitutio u = 5x - 4 will ot wok ad so we e goig to have to do somethig diffeet fo this itegal. It would be ice if we could get id of the squae oot somehow. The followig substitutio will do that fo us. x= sec q 5 Do ot woy about whee this came fom at this poit. As we wok the poblem you will see that it woks ad that if we have a simila type of squae oot i the poblem we ca always use a simila substitutio. Befoe we actually do the substitutio howeve let s veify the claim that this will allow us to get id of the squae oot. Ê 4 ˆ Á Ë5 5x - 4 = 5 sec q - 4 = 4sec q - = sec q - To get id of the squae oot all we eed to do is ecall the elatioship, ta q + = sec q fi sec q - = ta q Usig this fact the squae oot becomes, 5-4 = ta = ta x q q Note the pesece of the absolute value bas thee. These ae impotat. Recall that x Thee should always be absolute value bas at this stage. If we kew that taq was always positive o always egative we could elimiate the absolute value bas usig, Ïx if x 0 x =Ì Ó - x if x < 0 = x 007 Paul Dawkis

33 Without limits we wo t be able to detemie if taq is positive o egative, howeve, we will eed to elimiate them i ode to do the itegal. Theefoe, sice we ae doig a idefiite itegal we will assume that taq will be positive ad so we ca dop the absolute value bas. This gives, 5x - 4 = ta So, we wee able to elimiate the squae oot usig this substitutio. Let s ow do the substitutio ad see what we get. I doig the substitutio do t foget that we ll also eed to substitute fo the dx. This is easy eough to get fom the substitutio. x= secq fi dx= secq taq dq 5 5 Usig this substitutio the itegal becomes, Û Ù ı 5x 4 Û taq dx Ù sec ta x ı 5 secq Ë5 - Ê ˆ = Á q q dq Ú = ta q dq With this substitutio we wee able to educe the give itegal to a itegal ivolvig tig fuctios ad we saw how to do these poblems i the pevious sectio. Let s fiish the itegal. q Û 5x -4 Ù dx= q - ı x Ú sec ( q q) dq = ta - + c So, we ve got a aswe fo the itegal. Ufotuately the aswe is t give i x s as it should be. So, we eed to wite ou aswe i tems of x. We ca do this with some ight tiagle tig. Fom ou oigial substitutio we have, 5x hypoteuse secq = = adjacet This gives the followig ight tiagle. Fom this we ca see that, taq = 5x - 4 We ca deal with the q i oe of ay vaiety of ways. Fom ou substitutio we ca see that, 007 Paul Dawkis 4

34 - Ê5x ˆ q = sec Á Ë While this is a pefectly acceptable method of dealig with the q we ca use ay of the possible six ivese tig fuctios ad sice sie ad cosie ae the two tig fuctios most people ae familia with we will usually use the ivese sie o ivese cosie. I this case we ll use the ivese cosie. - Ê ˆ q = cos Á Ë5x So, with all of this the itegal becomes, Û 5x -4 Ê 5x -4 -Ê ˆˆ Ù dx= - cos Á + c ı x Á 5x Ë Ë -Ê ˆ = 5x -4- cos Á + c Ë5x We ow have the aswe back i tems of x. Wow! That was a lot of wok. Most of these wo t take as log to wok howeve. This fist oe eeded lot s of explaatio sice it was the fist oe. The emaiig examples wo t eed quite as much explaatio ad so wo t take as log to wok. Howeve, befoe we move oto moe poblems let s fist addess the issue of defiite itegals ad how the pocess diffes i these cases. Example Evaluate the followig itegal. Û Ù ı x 4 Solutio The limits hee wo t chage the substitutio so that will emai the same. x = sec q 5 Usig this substitutio the squae oot still educes dow to, x - dx 5x - 4 = ta Howeve, ulike the pevious example we ca t just dop the absolute value bas. I this case we ve got limits o the itegal ad so we ca use the limits as well as the substitutio to detemie the age of q that we e i. Oce we ve got that we ca detemie how to dop the absolute value bas. Hee s the limits of q. q 007 Paul Dawkis 5

35 x= fi = sec q fi q = p x= fi = sec q fi q = So, if we ae i the age 4 p 5 x 5 the q is i the age of 0 q ad i this age of q s taget is positive ad so we ca just dop the absolute value bas. Let s do the substitutio. Note that the wok is idetical to the pevious example ad so most of it is left out. We ll pick up at the fial itegal ad the do the substitutio. Û Ù ı x -4 x p sec q 0 dx= - ( q q) = ta - p = - Note that because of the limits we did t eed to esot to a ight tiagle to complete the poblem. Let s take a look at a diffeet set of limits fo this itegal. Example Evaluate the followig itegal. Û Ù ı Ú 5x -4 Solutio Agai, the substitutio ad squae oot ae the same as the fist two examples. sec 5 x= q x - 4 = ta q 5 Let s ext see the limits q fo this poblem. x=- fi - = sec q fi q = p 4 4 p x=- fi - = secq fi q = Note that i detemiig the value of q we used the smallest positive value. Now fo this age of x s we have p q p ad i this age of q taget is egative ad so i this case we ca dop the absolute value bas, but will eed to add i a mius sig upo doig so. I othe wods, x dx 5x - 4 =- ta So, the oly chage this will make i the itegatio pocess is to put a mius sig i fot of the itegal. The itegal is the, q p 0 dq 007 Paul Dawkis 6

36 Û Ù ı x 4 x - p Úp sec q dx=- - ( q q) =- ta - p = - p dq p I the last two examples we saw that we have to be vey caeful with defiite itegals. We eed to make sue that we detemie the limits o q ad whethe o ot this will mea that we ca dop the absolute value bas o if we eed to add i a mius sig whe we dop them. Befoe movig o to the ext example let s get the geeal fom fo the substitutio that we used i the pevious set of examples. a - fi = secq b bx a x Let s wok a ew ad diffeet type of example. Example 4 Evaluate the followig itegal. Û Ù dx 4 ı x 9-x Solutio Now, the squae oot i this poblem looks to be (almost) the same as the pevious oes so let s ty the same type of substitutio ad see if it will wok hee as well. x = secq Usig this substitutio the squae oot becomes, 9- = 9-9sec = - sec = - ta x q q q So usig this substitutio we will ed up with a egative quatity (the taget squaed is always positive of couse) ude the squae oot ad this will be touble. Usig this substitutio will give complex values ad we do t wat that. So, usig secat fo the substitutio wo t wok. Howeve, the followig substitutio (ad diffeetial) will wok. x= siq dx= cosq dq With this substitutio the squae oot is, 9- = - si = cos = cos = cos x q q q q We wee able to dop the absolute value bas because we ae doig a idefiite itegal ad so we ll assume that eveythig is positive. The itegal is ow, 007 Paul Dawkis 7

37 Û Ù ı x Û dx= Ù 4 9 -x ı8si q = Û Ù dq 4 8ı si q csc 4 = q dq 8Ú 4 ( cosq) cosq dq I the pevious sectio we saw how to deal with itegals 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 simila with this. Hee is the itegal. Û Ù ı x dx= csc csc 8Ú q q dq 9 -x = ( cot csc ) cot 8Ú q + q dq u = q =- u du 8Ú + Ê cot ˆ =- Á q + cot q + c 8Ë 4 Now we eed to go back to x s usig a ight tiagle. Hee is the ight tiagle fo this poblem ad tig fuctios fo this poblem. x siq = cotq = 9 - x x The itegal is the, Û Ù ı x Ê Ê ˆ 9-x ˆ 9-x dx=- Á + + c 9- x 8ÁÁ x x Ë Ë 4 ( x ) Hee s the geeal fom fo this type of squae oot x =- - + c 4x 8x 007 Paul Dawkis 8

38 a - fi = siq b a bx x Thee is oe fial case that we eed to look at. The ext itegal will also cotai somethig that we eed to make sue we ca deal with. Example 5 Evaluate the followig itegal. 6 5 Û x Ù ı 0 6 ( x + ) Solutio Fist, otice that thee eally is a squae oot i this poblem eve though it is t explicitly witte out. To see the oot let s ewite thigs a little. dx Ê ˆ ( 6x + ) = Á( 6x + ) = ( 6x + ) Ë This squae oot is ot i the fom we saw i the pevious examples. Hee we will use the substitutio fo this oot. x= taq dx= sec q dq 6 6 With this substitutio the deomiato becomes, 6x + = ta q + = sec q = secq Now, because we have limits we ll eed to covet them to q so we ca detemie how to dop the absolute value bas. x = 0 fi 0= taq fi q = 0 6 p x = fi = ta q fi q = I this age of q secat is positive ad so we ca dop the absolute value bas. Hee is the itegal, Û Ù ı p x Û 7776 ta q Ê ˆ dx= Ù Á sec q ı 0 sec q Ë6 0 ( 6x + ) p 5 4 ta q = Û Ù d q 46656ı secq 0 dq Thee ae seveal ways to poceed fom this poit. Nomally with a odd expoet o the taget we would stip oe of them out ad covet to secats. Howeve, that would equie that we also 007 Paul Dawkis 9

39 have a secat i the umeato which we do t have. Theefoe, it seems like the best way to do this oe would be to covet the itegad to sies ad cosies. Û Ù ı 6 5 p x 4 ı 0 ( 6x + ) si q dx= Û Ù d q cos q ( -cos q) 007 Paul Dawkis 0 p 4 Û = Ù ı cos q 0 siqdq We ca ow use the substitutio u = cosq ad we might as well covet the limits as well. q = 0 u = cos0= The itegal is the, Û Ù ı p p q = u = cos = x -4 - dx=- u u du Ú ( 6x + ) The geeal fom fo this fial type of squae oot is Ê ˆ =- Á- + + u 46656Ë u u = a + fi = taq b a bx x We have a couple of fial examples to wok i this sectio. Not all tig substitutios will just jump ight out at us. Sometimes we eed to do a little wok o the itegad fist to get it ito the coect fom ad that is the poit of the emaiig examples. Example 6 Evaluate the followig itegal. Û x Ù dx ı x -4x-7 Solutio I this case the quatity ude the oot does t obviously fit ito ay of the cases we looked at above ad i fact is t i the ay of the foms we saw i the pevious examples. Note howeve that if we complete the squae o the quadatic we ca make it look somewhat like the above itegals. Remembe that completig the squae equies a coefficiet of oe i fot of the x. Oce we have that we take half the coefficiet of the x, squae it, ad the add ad subtact it to the

40 quatity. Hee is the completig the squae fo this poblem. Ê 7ˆ Ê 7ˆ Ê 9ˆ Áx -x- = Áx - x+ -- = Á( x-) - = ( x-) -9 Ë Ë Ë So, the oot becomes, x -4x- 7 = x- - 9 This looks like a secat substitutio except we do t just have a x that is squaed. That is okay, it will wok the same way. x- = secq x= + secq dx= secq taq dq Usig this substitutio the oot educes to, -4-7 = = 9sec - 9 = ta = ta = ta x x x q q q q Note we could dop the absolute value bas sice we ae doig a idefiite itegal. Hee is the itegal. Û x Û+ secq Ê ˆ Ù dx= sec ta Ù q q dq x 4x 7 taq Á ı - - ı Ë = Û Ù secq + sec q dq ı = l secq + taq + taq + c Ad hee is the ight tiagle fo this poblem. 4 7 sec x - ta x - q q x - = = The itegal is the, Û x ( x-) x -4x-7 x -4x-7 Ù dx= l c ı x -4x Paul Dawkis

41 Example 7 Evaluate the followig itegal. 4x Ú e + e x dx Solutio This does t look to be aythig like the othe poblems i this sectio. Howeve it is. To see this we fist eed to otice that, = x x e e With this we ca use the followig substitutio. e x = taq e x dx = sec q d q Remembe that to compute the diffeetial all we do is diffeetiate both sides ad the tack o dx o dq oto the appopiate side. With this substitutio the squae oot becomes, x x + e = + e = + ta = sec = sec = sec q q q q Agai, we ca dop the absolute value bas because we ae doig a idefiite itegal. Hee s the itegal. Ú Ú dx x Ú( e ) x x e ( e ) dx Úta ( sec )( sec ) Ú( ) 4 Úu u du e + e dx= e e + e 4x x x x x = + = q q q dq = sec q - sec qsecq taq dq u = secq = - sec q sec q 5 5 = - + Hee is the ight tiagle fo this itegal. x x e + e taq = secq = = + e c x The itegal is the, 5 4x x x x Ú 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 itegals that look othig like the fist few examples ca i fact be tued ito a tig substitutio poblem with a little wok. Befoe leavig this sectio let s summaize all thee cases i oe place. a a -bx fi x= siq b a bx -a fi x= secq b a a + bx fi x= taq b 007 Paul Dawkis

43 Patial Factios I this sectio we ae goig to take a look at itegals of atioal expessios of polyomials ad oce agai let s stat this sectio out with a itegal that we ca aleady do so we ca cotast it with the itegals that we ll be doig i this sectio. Û x- dx= Û Ù du usig u x x 6 ad du Ù = - - = ( x-) dx ı x -x-6 ı u l 6 = x -x- + c So, if the umeato is the deivative of the deomiato (o a costat multiple of the deivative of the deomiato) doig this kid of itegal is faily simple. Howeve, ofte the umeato is t the deivative of the deomiato (o a costat multiple). Fo example, coside the followig itegal. Û x+ Ù dx ı x -x-6 I this case the umeato is defiitely ot the deivative of the deomiato o is it a costat multiple of the deivative of the deomiato. Theefoe, the simple substitutio that we used above wo t wok. Howeve, if we otice that the itegad ca be boke up as follows, x+ 4 = - x -x-6 x- x+ the the itegal is actually quite simple. Û x+ 4 dx= Û Ù Ù - dx ı x -x-6 ı x- x+ = 4l x- - l x+ + c This pocess of takig a atioal expessio ad decomposig it ito simple atioal expessios that we ca add o subtact to get the oigial atioal expessio is called patial factio decompositio. May itegals ivolvig atioal expessios ca be doe if we fist do patial factios o the itegad. So, let s do a quick eview of patial factios. We ll stat with a atioal expessio i the fom, P( x) f ( x) = Q x whee both P(x) ad Q(x) ae polyomials ad the degee of P(x) is smalle tha the degee of Q(x). Recall that the degee of a polyomial is the lagest expoet i the polyomial. Patial factios ca oly be doe if the degee of the umeato is stictly less tha the degee of the deomiato. That is impotat to emembe. So, oce we ve detemied that patial factios ca be doe we facto the deomiato as completely as possible. The fo each facto i the deomiato we ca use the followig table to detemie the tem(s) we pick up i the patial factio decompositio. 007 Paul Dawkis 4

44 Facto i deomiato ax+ b ( ax+ b) k Tem i patial factio decompositio A ax+ b A A A ax+ b ax+ b ax+ b k L, k =,,, k Ax+ B ax + bx+ c ax + bx+ c Ax k + B Ax + B Ax k + Bk ax + bx+ c + + L +, k =,,, K k ax + bx+ c ax + bx+ c ax + bx+ c Notice that the fist ad thid cases ae eally special cases of the secod ad fouth cases espectively. Thee ae seveal methods fo detemiig the coefficiets fo each tem ad we will go ove each of those i the followig examples. Let s stat the examples by doig the itegal above. Example Evaluate the followig itegal. Û x + Ù dx ı x -x-6 Solutio The fist step is to facto the deomiato as much as possible ad get the fom of the patial factio decompositio. Doig this gives, x+ A B = + x- x+ x- x+ The ext step is to actually add the ight side back up. x+ A x+ + B x- = x- x+ x- x+ Now, we eed to choose A ad B so that the umeatos of these two ae equal fo evey x. To do this we ll eed to set the umeatos equal. x+ = A x+ + B x- Note that i most poblems we will go staight fom the geeal fom of the decompositio to this step ad ot bothe with actually addig the tems back up. The oly poit to addig the tems is to get the umeato ad we ca get that without actually witig dow the esults of the additio. At this poit we have oe of two ways to poceed. Oe way will always wok, but is ofte moe wok. The othe, while it wo t always wok, is ofte quicke whe it does wok. I this case both will wok ad so we ll use the quicke way fo this example. We ll take a look at the othe method i a late example. K 007 Paul Dawkis 5

45 What we e goig to do hee is to otice that the umeatos must be equal fo ay x that we would choose to use. I paticula the umeatos must be equal fo x =- ad x =. So, let s plug these i ad see what we get. x=- 5= A 0 + B -5 fi B=- x= 0= A 5 + B 0 fi A= 4 So, by caefully pickig the x s we got the ukow costats to quickly dop out. Note that these ae the values we claimed they would be above. At this poit thee eally is t a whole lot to do othe tha the itegal. Û x+ 4 dx= Û Ù Ù - dx ı x -x-6 ı x- x+ 4 = Û dx- Û Ù Ù dx ı x- ı x+ = 4l x- - l x+ + c Recall that to do this itegal we fist split it up ito two itegals ad the used the substitutios, u = x- v = x+ o the itegals to get the fial aswe. Befoe movig oto the ext example a couple of quick otes ae i ode hee. Fist, may of the itegals i patial factios poblems come dow to the type of itegal see above. Make sue that you ca do those itegals. Thee is also aothe itegal that ofte shows up i these kids of poblems so we may as well give the fomula fo it hee sice we ae aleady o the subject. Û Ù ı x = ta + c + Ë dx - Ê Á ˆ x a a a It will be a example o two befoe we use this so do t foget about it. Now, let s wok some moe examples. Example Evaluate the followig itegal. Û x + 4 Ù dx ıx + 4x -4x Solutio We wo t be puttig as much detail ito this solutio as we did i the pevious example. The fist thig is to facto the deomiato ad get the fom of the patial factio decompositio. x + 4 A B C = + + x x+ x- x x+ x- The ext step is to set umeatos equal. If you eed to actually add the ight side togethe to get 007 Paul Dawkis 6

46 the umeato fo that side the you should do so, howeve, it will defiitely make the poblem quicke if you ca do the additio i you head to get, x + 4= A x+ x- + Bx x- + Cx x+ As with the pevious 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 - fi A=- x=- 8= B( -)(-8) fi B= 40 ʈÊ8ˆ 40 5 x= = CÁ Á fi C = = 9 Ë Ë 6 Note that ulike the fist example most of the coefficiets hee ae factios. That is ot uusual so do t get excited about it whe it happes. Now, let s do the itegal. 5 Û x + 4 dx= Û Ù Ù dx ı x + 4x - 4x ı x x+ x- 5 =- l x + l x+ + l x- + c 6 Agai, as oted above, itegals that geeate atual logaithms ae vey commo i these poblems so make sue you ca do them. Example Evaluate the followig itegal. Û x - 9x+ 5 Ù dx ı( x- 4) ( x + ) Solutio This time the deomiato is aleady factoed so let s just jump ight to the patial factio decompositio. x - 9x+ 5 A B Cx+ D = + + x- 4 x + x- 4 x-4 x + Settig umeatos gives, x x A x x B x Cx D x = I this case we ae t goig to be able to just pick values of x that will give us all the costats. Theefoe, we will eed to wok this the secod (ad ofte loge) way. The fist step is to multiply out the ight side ad collect all the like tems togethe. 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 ae equal. I othe wods we will eed to set the coefficiets of like powes of x equal. This will give a system of equatios that ca be solved. 007 Paul Dawkis 7

47 : + = 0 Ô : = Ô : =-9Ô : = 5 Ô x A C x A B C D x A C D 0 x A B D fi A=, B=- 5, C =-, D= Note that we used x 0 to epeset the costats. Also ote that these systems ca ofte be quite lage ad have a fai amout of wok ivolved i solvig them. The best way to deal with these is to use some fom of compute aided solvig techiques. Now, let s take a look at the itegal. Û Ù ı Û dx= dx Ù - + ı - 4 x + x x x ( x- 4) ( x + ) x ( x-4) Û 5 x = Ù ı x- x + x + ( x-4) 4 5 Ê x ˆ = l l + + ta + x-4 Á Ë dx - x x c I ode to take cae of the thid tem we eeded to split it up ito two sepaate tems. Oce we ve doe this we ca do all the itegals i the poblem. The fist two use the substitutio u = x- 4, the thid uses the substitutio v= x + ad the fouth tem uses the fomula give above fo ivese tagets. Example 4 Evaluate the followig itegal. Û x + 0x + x+ 6 dx Ù ı x- x + 4 Solutio Let s fist get the geeal fom of the patial factio decompositio. x + 0x + x+ 6 A Bx+ C Dx+ E = x- x + 4 x- x + x + 4 Now, set umeatos equal, expad the ight side ad collect like tems. 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 Ô : - = ÔÔ : = 0 fi =, =-, =-, =, = 0 : = Ô Ô : = 6Ô 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 zeo. It happes o occasio. Hee s the itegal. Û x + 0x + x+ 6 Û -x- x dx= + + Ù Ù 4 ı x- x + 4 ı x- x + x + 4 Û x x = Ù ı x- x + x + x Ê xˆ = l -- l ta Á - + Ë x + 4 dx - x x c To this poit we ve oly looked at atioal expessios whee the degee of the umeato was stictly less that the degee of the deomiato. Of couse ot all atioal expessios will fit ito this fom ad so we eed to take a look at a couple of examples whee this is t the case. dx Example 5 Evaluate the followig itegal. 4 Û x - 5x + 6x -8 Ù dx ı x -x Solutio So, i this case the degee of the umeato is 4 ad the degee of the deomiato is. Theefoe, patial factios ca t be doe o this atioal expessio. To fix this up we ll eed to do log divisio o this to get it ito a fom that we ca deal with. Hee is the wok fo that. x- So, fom the log divisio we see that, 4 x -x x - 5x + 6x -8 4 ( x x ) x 6x 8 ( x 6x ) Paul Dawkis 9

49 4 x - 5x + 6x -8 8 = x-- x -x x -x ad the itegal becomes, 4 Û x - 5x + 6x -8 8 dx= Û Ù Ù x- - dx ı x -x ı x -x 8 = x- dx- Û Ú Ù dx ı x -x The fist itegal we ca do easily eough ad the secod itegal is ow i a fom that allows us to do patial factios. So, let s get the geeal fom of the patial factios fo the secod itegad. 8 = A + B + C x x- x x x- Settig umeatos equal gives us, 8= Ax x- + B x- + Cx Now, thee is a vaiatio of the method we used i the fist couple of examples that will wok hee. Thee ae a couple of values of x that will allow us to quickly get two of the thee costats, but thee is o value of x that will just had us the thid. 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 aothe value of x that will be easy to wok with (i.e. it wo t give lage/messy umbes aywhee) ad the we ll use the fact that we also kow the othe two costats to fid the thid. x= 0 8= B - fi B=-6 x= 8= C 9 fi C = x= 8= A - + B - + C =- A+ 4 fi A=- The itegal is the, 4 Û x - 5x + 6x -8 6 dx= x-dx- Û Ù Ù dx ı x -x Ú ı x x x- 6 = x - x+ l x - -l x- + c x I the pevious example thee wee actually two diffeet ways of dealig with the x i the deomiato. Oe is to teat is as a quadatic which would give the followig tem i the decompositio Ax+ B x ad the othe is to teat it as a liea tem i the followig way, x ( x 0) = - which gives the followig two tems i the decompositio, 007 Paul Dawkis 40

50 A B + x x We used the secod way of thikig about it i ou example. Notice howeve that the two will give idetical patial factio decompositios. So, why talk about this? Simple. This will wok fo x, but what about x o x 4? I these cases we eally will eed to use the secod way of thikig about these kids of tems. x Let s take a look at oe moe example. A B C A B C D fi + + x fi x x x x x x x 4 4 Example 6 Evaluate the followig itegal. Û x Ù dx ı x - Solutio I this case the umeato ad deomiato have the same degee. As with the last example we ll eed to do log divisio to get this ito the coect fom. I ll leave the details of that to you to check. Û x Ù dx= Û Ù + dx= dx+ Û dx Ù ı x - ı x - Ú ı x - So, we ll eed to patial factio the secod itegal. Hee s the decompositio. A B = + x- x+ x- x+ Settig umeato equal gives, B( x ) = A x+ + - Pickig value of x gives us the followig coefficiets. x=- = B( -) fi B=- x= = A fi A= The itegal is the, Û x dx= dx+ Û Ù Ù - dx ı x - Ú ı x- x+ = x+ l x-- l x+ + c 007 Paul Dawkis 4

51 Itegals Ivolvig Roots I this sectio we e goig to look at a itegatio techique that ca be useful fo some itegals with oots i them. We ve aleady see some itegals with oots i them. Some ca be doe quickly with a simple Calculus I substitutio ad some ca be doe with tig substitutios. Howeve, ot all itegals with oots will allow us to use oe of these methods. Let s look at a couple of examples to see aothe techique that ca be used o occasio to help with these itegals. Example Evaluate the followig itegal. Û x + Ù dx ı x - Solutio Sometimes whe faced with a itegal that cotais a oot we ca use the followig substitutio to simplify the itegal ito a fom that ca be easily woked with. u = x- So, istead of lettig u be the stuff ude the adical as we ofte did i Calculus I we let u be the whole adical. Now, thee will be a little moe wok hee sice we will also eed to kow what x is so we ca substitute i fo that i the umeato ad so we ca compute the diffeetial, dx. This is easy eough to get howeve. Just solve the substitutio fo x as follows, x= u + dx= u du Usig this substitutio the itegal is ow, Û Ù ı ( u ) = + 5 u Ú udu u udu 5 5 = u + u + c = ( x- ) + ( x- ) + c 5 So, sometimes, whe a itegal cotais the oot g( x ) the substitutio, u = 007 Paul Dawkis 4 g x ca be used to simplify the itegal ito a fom that we ca deal with. Let s take a look at aothe example eal quick. Example Evaluate the followig itegal. Û Ù ı x- x+ 0 dx Solutio We ll do the same thig we did i the pevious example. Hee s the substitutio ad the exta wok we ll eed to do to get x i tems of u.

52 u = x+ 0 x= u - 0 dx= udu With this substitutio the itegal is, Û 4u Ù dx= Û Ù ( u) du = Û Ù du ı x- x+ 0 ıu -0-u ı u -u-0 This itegal ca ow be doe with patial factios. 4 u = A + B u- 5 u+ u- 5 u+ Settig umeatos equal gives, 4u = Au+ + B u- 5 Pickig value of u gives the coefficiets. 8 u =- - 8= B( - 7) B= 7 0 u = 5 0= A( 7) A= 7 The itegal is the, 0 8 Û 7 7 dx= Û Ù Ù + du ı x- x+ 0 ı 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 oots fom the itegal ad put ito a fom that we ca deal with. Note howeve, that this wo t always wok ad sometimes the ew itegal will be just as difficult to do. 007 Paul Dawkis 4

53 Itegals Ivolvig Quadatics To this poit we ve see quite a few itegals that ivolve quadatics. A couple of examples ae, Û x Û -Ê xˆ Ù dx= l x ± a + c dx ta Ù = Á ı x ± a ı x + a a Ëa We also saw that itegals ivolvig with a tig substitutio. bx - a, a - bx ad a + bx could be doe Notice howeve that all of these itegals wee missig a x tem. They all cosist of a quadatic tem ad a costat. Some itegals ivolvig geeal quadatics ae easy eough to do. Fo istace, the followig itegal ca be doe with a quick substitutio. Û x + dx= Û Ù ( 4 4 Ù du u = x + x- du = ( x+ ) dx) ı 4x + x- 4ıu l4 = x + x- + c 4 Some itegals with quadatics ca be doe with patial factios. Fo istace, Û 0x- 6 4 dx= Û Ù 4l 5 l Ù - dx= x+ - x+ + c ıx + 6x+ 5 ı x+ 5 x+ Ufotuately, these methods wo t wok o a lot of itegals. A simple substitutio will oly wok if the umeato is a costat multiple of the deivative of the deomiato ad patial factios will oly wok if the deomiato ca be factoed. This sectio is how to deal with itegals ivolvig quadatics whe the techiques that we ve looked at to this poit simply wo t wok. Back i the Tig Substitutio sectio we saw how to deal with squae oots that had a geeal quadatic i them. Let s take a quick look at aothe oe like that sice the idea ivolved i doig that kid of itegal is exactly what we ae goig to eed fo the othe itegals i this sectio. Example Evaluate the followig itegal. Ú x + 4x+ 5dx Solutio Recall fom the Tig Substitutio sectio that i ode to do a tig substitutio hee we fist eeded to complete the squae o the quadatic. This gives, x x x x x Afte completig the squae the itegal becomes, = = + + Ú Ú x + 4x+ 5dx= x+ + dx 007 Paul Dawkis 44

54 Upo doig this we ca idetify the tig substitutio that we eed. Hee it is, x+ = taq x= taq - dx= sec q dq x + + = ta q + = sec q = secq = secq Recall that sice we ae doig a idefiite itegal we ca dop the absolute value bas. Usig this substitutio the itegal becomes, Ú 4 5 sec x + x+ dx= q d Ú q = ( sec q ta q + l sec q + ta q ) + c We ca fiish the itegal out with the followig ight tiagle. 4 5 ta x + sec x + x + q = q = = x + 4x+ 5 Ú ( ) x + 4x+ 5dx= x+ x + 4x+ 5+ l x+ + x + 4x+ 5 + c So, by completig the squae we wee able to take a itegal that had a geeal quadatic i it ad covet it ito a fom that allowed use a kow itegatio techique. Let s do a quick eview of completig the squae befoe poceedig. Hee is the geeal completig the squae fomula that we ll use. Êbˆ Êbˆ Ê bˆ b x + bx+ c= x + bx+ - + c= x+ + c- Á Á Á Ë Ë Ë 4 This will always take a geeal quadatic ad wite it i tems of a squaed tem ad a costat tem. Recall as well that i ode to do this we must have a coefficiet of oe i fot of the x. If ot we ll eed to facto out the coefficiet befoe completig the squae. I othe wods, Ê ˆ Á b c ax + bx+ c= aáx + x+ Á 44 a a Á complete the Ë squae o this! 007 Paul Dawkis 45

55 Now, let s see how completig the squae ca be used to do itegals that we ae t able to do at this poit. Example Evaluate the followig itegal. Û Ù ı x - x+ dx Solutio Okay, this does t facto so patial factios just wo t wok o this. Likewise, sice the umeato is just we ca t use the substitutio u = x - x+ 8. So, let s see what happes if we complete the squae o the deomiato. Ê ˆ x - x+ = Áx - x+ Ë Ê 9 9 ˆ = Áx - x+ - + Ë 6 6 ÊÊ ˆ 7 ˆ x- + ÁÁ Ë 4 6 Ë With this the itegal is, Û Û Ù dx= ı x - x+ Ù x- + 7 ı 4 6 dx Now this may ot seem like all that geat of a chage. Howeve, otice that we ca ow use the followig substitutio. u = x- du = dx 4 ad the itegal is ow, Û Û Ù dx= Ù du 7 ı x - x+ ı u + We ca ow see that this is a ivese taget! So, usig the fomula fom above we get, Û Ê 4 ˆ - Ê 4u ˆ Ù dx= ta + c ı x - x+ Á Á Ë 7 Ë 7 6 Ê4x- ˆ ta Á 7 Ë 7 - = + Example Evaluate the followig itegal. Û x - Ù dx ı x + 0x+ 8 Solutio This example is a little diffeet fom the pevious oe. I this case we do have a x i the umeato howeve the umeato still is t a multiple of the deivative of the deomiato ad so a simple Calculus I substitutio wo t wok. c 007 Paul Dawkis 46

56 So, let s agai complete the squae o the deomiato ad see what we get, x x x x x = = Upo completig the squae the itegal becomes, Û x- Û x- Ù dx= dx ı x + 0x+ 8 Ù ı x+ 5 + At this poit we ca use the same type of substitutio that we did i the pevious example. The oly eal diffeece is that we ll eed to make sue that we plug the substitutio back ito the umeato as well. u = x+ 5 x= u- 5 dx= du ( u ) Û x- Û -5 - Ù dx= Ù du ı x + 0x+ 8 ı u + u 6 = Û Ù - du ı u + u + 6 Ê u ˆ Á Ë - = l u + - ta + So, i geeal whe dealig with a itegal i the fom, 6 -Ê x+ 5ˆ = l ( x+ 5) + - ta c Á + Ë c Û + B Ù Ax dx ı ax + bx+ c () Hee we ae goig to assume that the deomiato does t facto ad the umeato is t a costat multiple of the deivative of the deomiato. I these cases we complete the squae o the deomiato ad the do a substitutio that will yield a ivese taget ad/o a logaithm depedig o the exact fom of the umeato. Let s ow take a look at a couple of itegals that ae i the same geeal fom as () except the deomiato will also be aised to a powe. I othe wods, let s look at itegals i the fom, Û Ax+ B Ù dx () ı ax + bx+ c Example 4 Evaluate the followig itegal. Û x Ù ı x - 6x+ Solutio Fo the most pat this itegal will wok the same as the pevious two with oe exceptio that will 007 Paul Dawkis 47 dx

57 occu dow the oad. So, let s stat by completig the squae o the quadatic i the deomiato. x x x x x = = - + The itegal is the, Û Ù ı x Û x dx= Ù Î ( x x Ù ) ı È( x ) dx Now, we will use the same substitutio that we ve used to this poit i the pevious two examples. u = x- x = u+ dx = du Û Ù ı x Û dx= Ù u+ ( x - 6x+ ) ı ( u + ) du Û u Û = Ù du+ Ù ı ( u + ) ı ( u + ) Now, hee is whee the diffeeces stat coppig up. The fist itegal ca be doe with the substitutio v= u + ad is t too difficult. The secod itegal howeve, ca t be doe with the substitutio used o the fist itegal ad it is t a ivese taget. It tus out that a tig substitutio will wok icely o the secod itegal ad it will be the same as we did whe we had squae oots i the poblem. u = taq du = sec q dq With these two substitutios the itegals become, Û x dx= Û Û Ù sec Ù dv+ d Ù q q ı x - 6x+ ı v ı ta q + Û sec q =- + dq 4 v Ù ı8ta ( q + ) Û sec q = Ù ( u + ) ı ( sec q) (( x- ) + ) (( x- ) + ) du =- + Û Ù dq ı sec q =- + cos Ú dq q dq 007 Paul Dawkis 48

58 Okay, at this poit we ve got two optios fo the emaiig itegal. We ca eithe use the ideas we leaed i the sectio about itegals ivolvig tig itegals o we could use the followig fomula. cos m q m d q si cos m cos m d m q q - Ú = m Ú q q Let s use this fomula to do the itegal. 4 Úcos q dq = siqcos q + cos 4 4Ú q dq Ê 0 ˆ 0 = siqcos q + Á siqcosq + cos d cos! 4 4 Ú q q q = Ë = siqcos q + siqcosq + q Next, let s use the followig ight tiagle to get this back to x s. u x- x- taq = = siq = cosq = x- + x- + The cosie itegal is the, 4 x- x- -Ê x-ˆ Úcos q dq = + + ta Á - + Ë All told the the oigial itegal is, (( x- ) + ) ( x ) x- x- Ê x-ˆ Ë - = + + ta Á (( x- ) + ) ( x ) 007 Paul Dawkis 49

59 Û Ù ı x dx = ( x x ) 4 ( x ) Ê ˆ Á x- x- -Ê x-ˆ ta (( x ) 8 ) ( x ) 8 Á - + Á - + Ë Ë x- 9 x- 9 Ê x-ˆ 8 64 Á - + Ë - = + + ta + (( x- ) + ) ( x ) It s a log ad messy aswe, but thee it is. Example 5 Evaluate the followig itegal. Û x- Ù ı 4-x-x Solutio As with the othe poblems we ll fist complete the squae o the deomiato. dx 4-x- x =- x + x- 4 =- x + x =- x+ - 5 = 5- x+ The itegal is, Û x- Û x- Ù dx= dx ( 4-x-x Ù ı ) ı È5- ( x+ ) Î Now, let s do the substitutio. u = x+ x = u- dx = du ad the itegal is ow, Û x- Û u-4 Ù dx= du Ù ı 4-x-x ı 5-u Û u Û = Ù du-ù ı 4 ( 5-u ) ı ( 5-u ) I the fist itegal we ll use the substitutio v= 5- u ad i the secod itegal we ll use the followig tig substitutio u = 5siq du = 5cosq dq Usig these substitutios the itegal becomes, du c 007 Paul Dawkis 50

60 Û Ù ı Û dx dv dq x - 4 =- Û ( 5cos ) Ù - Ù q 4-x-x ı v ı 5-5si q 4 5 Û cosq = - Ù v 5 ı si ( - q ) 4 5 cosq = - Û Ù dq 4 v 5 ı cos q 4 5 = - sec v 5 Ú 5 = - ( sec q ta q + l sec q + ta q ) + c v 5 We ll eed the followig ight tiagle to fiish this itegal out. u x+ 5 x+ siq = 5 = 5 secq = taq = 5 - x+ 5 - x+ q dq dq So, goig back to x s the itegal becomes, Û Ê ˆ x- 5 5( x ) 5 x dx Á + + l Ù = c ( 4 x x ) 5 u 5 5 ( x ) ı Á ( x+ ) 5- ( x+ ) Ë 4x- 5 x+ + 5 = + l + c ( x ) ( x ) Ofte the followig fomula is eeded whe usig the tig substitutio that we used i the pevious example. m sec q m m m d q ta sec sec d m q q - Ú = m-ú q q Note that we ll oly eed the two tig substitutios that we used hee. The thid tig substitutio that we used will ot be eeded hee. 007 Paul Dawkis 5

61 Usig Itegal Tables Note : Of all the otes that I ve witte up fo dowload, this is the oe sectio that is tied to the book that we ae cuetly usig hee at Lama Uivesity. I this sectio we discuss usig tables of itegals to help us with some itegals. Howeve, I have t had the time to costuct a table of my ow ad so I will be usig the tables give i Stewat s Calculus, Ealy Tascedetals (6 th editio). As soo as I get aoud to witig my ow table I ll post it olie ad make ay appopiate chages to this sectio. So, with that out of the way let s get o with this sectio. This sectio is etitled Usig Itegal Tables ad we will be usig itegal tables. Howeve, at some level, this is t eally the poit of this sectio. To a cetai extet the eal subject of this sectio is how to take advatage of kow itegals to do itegals that may ot look like aythig the oes that we do kow how to do o ae give i a table of itegals. Fo the most pat we ll be doig this by usig substitutio to put itegals ito a fom that we ca deal with. Howeve, ot all of the itegals will equie a substitutio. Fo some itegals all that we eed to do is a little ewitig of the itegad to get ito a fom that we ca deal with. We ve aleady elated a ew itegal to oe we could deal with least oce. I the last example i the Tig Substitutio sectio we looked at the followig itegal. 4x Ú e + e x dx At fist glace this looks othig like a tig substitutio poblem. Howeve, with the substitutio x u =e we could tu the itegal ito, Ú u + u du which defiitely is a tig substitutio poblem ( u = taq ). We actually did this pocess i a x sigle step by usig e = taq, but the poit is that with a substitutio we wee able to covet a itegal ito a fom that we could deal with. So, let s wok a couple examples usig substitutios ad tables. Example Evaluate the followig itegal. Û Ù ı 7+ 9x dx x Solutio So, the fist thig we should do is go to the tables ad see if thee is aythig i the tables that is close to this. I the tables i Stewat we fid the followig itegal, Û Ù ı a + u a + u du =- + l u+ a + u + c u u This is ealy what we ve got i ou itegal. The oly eal diffeece is that we ve got a 007 Paul Dawkis 5

62 coefficiet i fot of the x ad the fomula does t. This is easily eough dealt with. All we eed to do is the followig maipulatio o the itegad Û 7+ 9 x Û x Û 9 + x Û 9 + x Ù dx = dx dx dx Ù = = x x Ù x Ù ı ı ı ı x So, we ca ow use the fomula with Û Ù ı 7 a = x Ê 9 + x Ê 7 ˆˆ dx = - + l x x c x Á x Á 9 Ë Ë Example Evaluate the followig itegal. Û cos x Ù dx ısix 9six-4 Solutio Goig though ou tables we ae t goig to fid aythig that looks like this i them. Howeve, otice that with the substitutio u = si x we ca ewite the itegal as, Û cosx Û Ù dx= Ù du ısix 9six-4 ı u 9u-4 ad this is i the tables. Û a+ bu - a Ù du = l + c if a> 0 ıu a+ bu a a+ bu + a Ê ˆ = + < -a Á -a Ë ta - a+ bu c if a 0 Notice that this is a fomula that will deped upo the value of a. This will happe o occasio. I ou case we have a =- 4 ad b = 9 so we ll use the secod fomula. Û Ù ı cosx Ê 9u- 4 ˆ dx ta six 9six-4 -(-4) Á -(-4) Ë - = + ta Ê Á Ë 9six-4 ˆ - = + This fial example uses a type of fomula kow as a eductio fomula. c c 007 Paul Dawkis 5

63 Example Evaluate the followig itegal. Û 4 Ê x ˆ Ù cot Á dx ı Ë Solutio x We ll fist eed to use the substitutio u = sice oe of the fomulas i ou tables have that i them. Doig this gives, Û 4Ê x ˆ 4 Ùcot Á dx = cot udu Ë Ú ı To help us with this itegal we ll use the followig fomula Úcot udu = cot u- cot udu - Ú Fomulas like this ae called eductio fomulas. Reductio fomulas geeally do t explicitly give the itegal. Istead they educe the itegal to a easie oe. I fact they ofte educe the itegal to a diffeet vesio of itself! Fo ou itegal we ll use = 4. Û 4Ê cot x ˆ Ê ˆ Ù Á dx = Á- cot u - cot udu Ë Ë Ú ı At this stage we ca eithe euse the eductio fomula with = o use the fomula cot udu =-cot u- u+ c Ú We ll euse the eductio fomula with = so we ca addess somethig that happes o occasio. Û 4Ê xˆ Ê Ê 0 ˆˆ 0 Ùcot Á dx = cot cot cot cot! Á- u -Á- u - Ú udu u = ı Ë Ë Ë cot =- u+ cot u+ du Ú cot =- u+ cot u+ u+ c cot Ê xˆ Ê xˆ =- Á + cot Á + x+ c Ë Ë Do t foget that 0 a =. Ofte people foget that ad the get stuck o the fial itegal! Thee eally was t a lot to this sectio. Just do t foget that sometimes a simple substitutio o ewite of a itegal ca take it fom udoable to doable. 007 Paul Dawkis 54

64 Itegatio Stategy We ve ow see a fai umbe of diffeet itegatio techiques ad so we should pobably pause at this poit ad talk a little bit about a stategy to use fo detemiig the coect techique to use whe faced with a itegal. Thee ae a couple of poits that eed to be made about this stategy. Fist, it is t a had ad fast set of ules fo detemiig the method that should be used. It is eally othig moe tha a geeal set of guidelies that will help us to idetify techiques that may wok. Some itegals ca be doe i moe tha oe way ad so depedig o the path you take though the stategy you may ed up with a diffeet techique tha somebody else who also wet though this stategy. Secod, while the stategy is peseted as a way to idetify the techique that could be used o a itegal also keep i mid that, fo may itegals, it ca also automatically exclude cetai techiques as well. Whe goig though the stategy keep two lists i mid. The fist list is itegatio techiques that simply wo t wok ad the secod list is techiques that look like they might wok. Afte goig though the stategy ad the secod list has oly oe ety the that is the techique to use. If, o the othe had, thee ae moe tha oe possible techique to use we will the have to decide o which is liable to be the best fo us to use. Ufotuately thee is o way to teach which techique is the best as that usually depeds upo the peso ad which techique they fid to be the easiest. Thid, do t foget that may itegals ca be evaluated i multiple ways ad so moe tha oe techique may be used o it. This has aleady bee metioed i each of the pevious poits, but is impotat eough to waat a sepaate metio. Sometimes oe techique will be sigificatly easie tha the othes ad so do t just stop at the fist techique that appeas to wok. Always idetify all possible techiques ad the go back ad detemie which you feel will be the easiest fo you to use. Next, it s etiely possible that you will eed to use moe tha oe method to completely do a itegal. Fo istace a substitutio may lead to usig itegatio by pats o patial factios itegal. Fially, i my class I will accept ay valid itegatio techique as a solutio. As aleady oted thee is ofte moe tha oe way to do a itegal ad just because I fid oe techique to be the easiest does t mea that you will as well. So, i my class, thee is o oe ight way of doig a itegal. You may use ay itegatio techique that I ve taught you i this class o you leaed i Calculus I to evaluate itegals i this class. I othe wods, always take the appoach that you fid to be the easiest. Note that this fial poit is moe geaed towads my class ad it s completely possible that you istucto may ot agee with this ad so be caeful i applyig this poit if you ae t i my class. Okay, let s get o with the stategy. 007 Paul Dawkis 55

65 . Simplify the itegad, if possible. This step is vey impotat i the itegatio pocess. May itegals ca be take fom impossible o vey difficult to vey easy with a little simplificatio o maipulatio. Do t foget basic tig ad algebaic idetities as these ca ofte be used to simplify the itegal. We used this idea whe we wee lookig at itegals ivolvig tig fuctios. Fo example coside the followig itegal. cos xdx This itegal ca t be doe as is howeve, simply by ecallig the idetity, Ú x= ( + ( x) ) cos cos the itegal becomes vey easy to do. Note that this example also shows that simplificatio does ot ecessaily mea that we ll wite the itegad i a simple fom. It oly meas that we ll wite the itegad ito a fom that we ca deal with ad this is ofte loge ad/o messie tha the oigial itegal.. See if a simple substitutio will wok. Look to see if a simple substitutio ca be used istead of the ofte moe complicated methods fom Calculus II. Fo example coside both if the followig itegals. Û x Ù dx x x -dx ı x - Ú The fist itegal ca be doe with patial factios ad the secod could be doe with a tig substitutio. Howeve, both could also be evaluated usig the substitutio u = x - ad the wok ivolved i the substitutio would be sigificatly less tha the wok ivolved i eithe patial factios o tig substitutio. So, always look fo quick, simple substitutios befoe movig o to the moe complicated Calculus II techiques.. Idetify the type of itegal. Note that ay itegal may fall ito moe tha oe of these types. Because of this fact it s usually best to go all the way though the list ad idetify all possible types sice oe may be easie tha the othe ad it s etiely possible that the easie type is listed lowe i the list. a. Is the itegad a atioal expessio (i.e is the itegad a polyomial divided by a polyomial)? If so, the patial factios may wok o the itegal. b. Is the itegad a polyomial times a tig fuctio, expoetial, o logaithm? If so, the itegatio by pats may wok. c. Is the itegad a poduct of sies ad cosies, secat ad tagets, o cosecats ad cotagets? If so, the the topics fom the secod sectio may wok. Likewise, do t foget that some quotiets ivolvig these fuctios ca also be doe usig these techiques. 007 Paul Dawkis 56

66 d. Does the itegad ivolve bx + a, bx - a, o a - bx? If so, the a tig substitutio might wok icely. e. Does the itegad have oots othe tha those listed above i it? If so, the the substitutio u g( x) = might wok. f. Does the itegad have a quadatic i it? If so, the completig the squae o the quadatic might put it ito a fom that we ca deal with. 4. Ca we elate the itegal to a itegal we aleady kow how to do? I othe wods, ca we use a substitutio o maipulatio to wite the itegad ito a fom that does fit ito the foms we ve looked at peviously i this chapte. A typical example hee is the followig itegal. Ú cosx + si xdx This itegal does t obviously fit ito ay of the foms we looked at i this chapte. Howeve, with the substitutio u = si x we ca educe the itegal to the fom, which is a tig substitutio poblem. Ú + u du 5. Do we eed to use multiple techiques? I this step we eed to ask ouselves if it is possible that we ll eed to use multiple techiques. The example i the pevious pat is a good example. Usig a substitutio did t allow us to actually do the itegal. All it did was put the itegal ad put it ito a fom that we could use a diffeet techique o. Do t eve get locked ito the idea that a itegal will oly equie oe step to completely evaluate it. May will equie moe tha oe step. 6. Ty agai. If eveythig that you ve tied to this poit does t wok the go back though the pocess ad ty agai. This time ty a techique that that you did t use the fist time aoud. As oted above this stategy is ot a had ad fast set of ules. It is oly iteded to guide you though the pocess of best detemiig how to do ay give itegal. Note as well that the oly place Calculus II actually aises is i the thid step. Steps, ad 4 ivolve othig moe tha maipulatio of the itegad eithe though diect maipulatio of the itegad o by usig a substitutio. The last two steps ae simply ideas to thik about i goig though this stategy. May studets go though this pocess ad cocetate almost exclusively o Step (afte all this is Calculus II, so it s easy to see why they might do that.) to the exclusio of the othe steps. Oe vey lage cosequece of that exclusio is that ofte a simple maipulatio o substitutio is ovelooked that could make the itegal vey easy to do. Befoe movig o to the ext sectio we should wok a couple of quick poblems illustatig a couple of ot so obvious simplificatios/maipulatios ad a ot so obvious substitutio. 007 Paul Dawkis 57

67 Example Evaluate the followig itegal. Û ta x Ù 4 dx ı sec x Solutio This itegal almost falls ito the fom give i c. It is a quotiet of taget ad secat ad we kow that sometimes we ca use the same methods fo poducts of tagets ad secats o quotiets. The pocess fom that sectio tells us that if we have eve powes of secat to stip two of them off ad covet the est to tagets. That wo t wok hee. We ca split two secats off, but they would be i the deomiato ad they wo t do us ay good thee. Remembe that the poit of splittig them off is so they would be thee fo the substitutio u = ta x. That equies them to be i the umeato. So, that wo t wok ad so we ll have to fid aothe solutio method. Thee ae i fact two solutio methods to this itegal depedig o how you wat to go about it. We ll take a look at both. Solutio I this solutio method we could just covet eveythig to sies ad cosies ad see if that gives us a itegal we ca deal with. Û tax si x 4 dx= Û Ù 4 Ù cos xdx ısec x ı cos x Ú Ú = sixcos xdx u = cos x =- u du cos 4 =- x+ c 4 Note that just covetig to sies ad cosies wo t always wok ad if it does it wo t always wok this icely. Ofte thee will be a lot moe wok that would eed to be doe to complete the itegal. Solutio This solutio method goes back to dealig with secats ad tagets. Let s otice that if we had a secat i the umeato we could just use u = sec x as a substitutio ad it would be a faily quick ad simple substitutio to use. We do t have a secat i the umeato. Howeve we could vey easily get a secat i the umeato simply by multiplyig the umeato ad deomiato by secat. Û tax taxsec x dx= Û Ù sec 4 Ù dx u = x 5 ısec x ı sec x = Û Ù du 5 ı u =- + c 4 4sec x cos 4 =- x+ c Paul Dawkis 58

68 I the pevious example we saw two simplificatios that allowed us to do the itegal. The fist was usig idetities to ewite the itegal ito tems we could deal with ad the secod ivolved multiplyig the umeato ad the deomiato by somethig to agai put the itegal ito tems we could deal with. Usig idetities to ewite a itegal is a impotat simplificatio ad we should ot foget about it. Itegals ca ofte be geatly simplified o at least put ito a fom 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 foget about it. I fact, let s take aothe look at a example i which multiplyig the umeato ad deomiato by somethig will allow us to do a itegal. Example Evaluate the followig itegal. Û Ù ı + si x dx Solutio This is a itegal i which if we just cocetate o the thid step we wo t get aywhee. This itegal does t appea to be ay of the kids of itegals that we woked i this chapte. We ca do the itegal howeve, if we do the followig, Û - si x dx= Û Ù Ù dx ı+ six ı+ six-si x -si x = Û Ù dx ı-si x This does ot appea to have doe aythig fo us. Howeve, if we ow emembe the fist simplificatio we looked at above we will otice that we ca use a idetity to ewite the deomiato. Oce we do that we ca futhe educe the itegal ito somethig we ca deal with. Û - si x dx= Û Ù Ù dx ı+ six ı cos x six = Û Ù - dx ı cos x cosx cos x Ú sec ta sec = x- x xdx = tax- sec x+ c So, we ve see oce agai that multiplyig the umeato ad deomiato by somethig ca put the itegal ito a fom that we ca itegate. Notice as well that this example also showed that simplificatios do ot ecessaily put a itegal ito a simple fom. They oly put the itegal ito a fom that is easie to itegate. Let s ow take a quick look at a example of a substitutio that is ot so obvious. 007 Paul Dawkis 59

69 Example Evaluate the followig itegal. Ú cos x dx Solutio We itoduced this example sayig that the substitutio was ot so obvious. Howeve, this is eally a itegal that falls ito the fom give by e i ou stategy above. Howeve, may people miss that fom ad so do t thik about it. So, let s ty the followig substitutio. u = x x= u dx= udu With this substitutio the itegal becomes, Ú Ú cos x dx= ucosudu This is ow a itegatio by pats itegal. Remembe that ofte we will eed to use moe tha oe techique to completely do the itegal. This is a faily simple itegatio by pats poblem so I ll leave the emaide of the details to you to check. Ú cos x dx= cos x + x si x + c Befoe leavig this sectio we should also poit out that thee ae itegals out thee i the wold that just ca t be doe i tems of fuctios that we kow. Some examples of these ae. -x si ( x) x dx cos( x ) dx Û Úe Ú Ùı dx cos dx x Ú e That does t mea that these itegals ca t be doe at some level. If you go to a compute algeba system such as Maple o Mathematica ad have it do these itegals hee is what it will etu the followig. - x p Ú e dx= ef ( x) p Ê ˆ Ú cos( x ) dx= FeselC x Á p Ë Û Ù ı Ú si x dx = Si x cos ( x ) x x ( e ) dx = Ci( e ) So it appeas that these itegals ca i fact be doe. Howeve this is a little misleadig. Hee ae the defiitios of each of the fuctios give above. Eo Fuctio The Sie Itegal ef = p Ú e x ( x) 0 Si ( x) = Û Ù ı 0 x -t si t dt t dt 007 Paul Dawkis 60

70 The Fesel Cosie Itegal Û Êp ˆ FeselC( x) = Ù cosá t dt ı 0 Ë The Cosie Itegal x cost - Ci( x) = g + l ( x) + Û Ù dt ı 0 t Whee g is the Eule-Mascheoi costat. Note that the fist thee ae simply defied i tems of themselves ad so whe we say we ca itegate them all we ae eally doig is eamig the itegal. The fouth oe is a little diffeet ad yet it is still defied i tems of a itegal that ca t be doe i pactice. It will be possible to itegate evey itegal give i this class, but it is impotat to ote that thee ae itegals that just ca t be doe. We should also ote that afte we look at Seies we will be able to wite dow seies epesetatios of each of the itegals above. x 007 Paul Dawkis 6

71 Impope Itegals I this sectio we eed to take a look at a couple of diffeet kids of itegals. Both of these ae examples of itegals that ae called Impope Itegals. Let s stat with the fist kid of impope itegals that we e goig to take a look at. Ifiite Iteval I this kid of itegals we ae goig to take a look at itegals that i which oe o both of the limits of itegatio ae ifiity. I these cases the iteval of itegatio is said to be ove a ifiite iteval. Let s take a look at a example that will also show us how we ae goig to deal with these itegals. Example Evaluate the followig itegal. Û Ù dx ı x Solutio This is a iocet eough lookig itegal. Howeve, because ifiity is ot a eal umbe we ca t just itegate as omal ad the plug i the ifiity to get a aswe. To see how we e goig to do this itegal let s thik of this as a aea poblem. So istead of askig what the itegal is, let s istead ask what the aea ude f ( x) = o the iteval x [, ) is. We still ae t able to do this, howeve, let s step back a little ad istead ask what the aea ude, t wee t > ad t is fiite. This is a poblem that we ca do. f ( x ) is o the iteval [ ] At t = Û Ù dx =- = - ı x x t Now, we ca get the aea ude f ( x ) o [, ) simply by takig the limit of A t as t goes to ifiity. Ê ˆ A= limat = limá- = tæ tæ Ë t This is the how we will do the itegal itself. t Û dx= lim Û Ù dx Ù t ı x Æ ı x Ê ˆ = lim tæ Á - Ë x Ê ˆ = limá- = tæ Ë t 007 Paul Dawkis 6 t t

72 So, this is how we will deal with these kids of itegals i geeal. We will eplace the ifiity with a vaiable (usually t), do the itegal ad the take the limit of the esult as t goes to ifiity. O a side ote, otice that the aea ude a cuve o a ifiite iteval was ot ifiity as we might have suspected it to be. I fact, it was a supisigly small umbe. Of couse this wo t always be the case, but it is impotat eough to poit out that ot all aeas o a ifiite iteval will yield ifiite aeas. Let s ow get some defiitios out of the way. We will call these itegals coveget if the associated limit exists ad is a fiite umbe (i.e. it s ot plus o mius ifiity) ad diveget if the associated limits eithe does t exist o is (plus o mius) ifiity. Let s ow fomalize up the method fo dealig with ifiite itevals. Thee ae essetially thee cases that we ll eed to look at. t. If Ú x dx exists fo evey t > athe, a f = lim t a tæ a Ú Ú povided the limit exists ad is fiite. b. If f t f x dx f x dx Ú x dx exists fo evey t < b the, b b = lim Ú Ú f x dx f x dx - tæ- povided the limits exists ad is fiite. c. If Ú f ( x) dx ad f - c Ú x dx ae both coveget the, c = + Ú Ú Ú f x dx f x dx f x dx - - Whee c is ay umbe. Note as well that this equies BOTH of the itegals to be coveget i ode fo this itegal to also be coveget. If eithe of the two itegals is diveget the so is this itegal. Let s take a look at a couple moe examples. Example Detemie if the follow itegal is coveget o diveget ad if it s coveget fid its value. Û Ù dx ı x Solutio So, the fist thig we do is covet the itegal to a limit. t Û dx= lim Û Ù Ù dx ı x tæ ı x Now, do the itegal ad the limit. 007 Paul Dawkis 6 t c

73 Û Ù ı dx= liml t x Æ ( x) ( () t l) = lim l - tæ = So, the limit is ifiite ad so the itegal is diveget. If we go back to thikig i tems of aea otice that the aea ude g( x ) = x o the iteval [, ) is ifiite. This is i cotast to the aea ude t f x = which was quite small. Thee eally is t all that much diffeece betwee these two fuctios ad yet thee is a lage diffeece i the aea ude them. We ca actually exted this out to the followig fact. Fact If a > 0 the Û Ù ı is coveget if p > ad diveget if p. dx p a x Oe thig to ote about this fact is that it s i essece sayig that if a itegad goes to zeo fast eough the the itegal will covege. How fast is fast eough? If we use this fact as a guide it looks like itegads that go to zeo faste tha x goes to zeo will pobably covege. Let s take a look at a couple moe examples. Example Detemie if the followig itegal is coveget o diveget. If it is coveget fid its value. 0 Û Ù dx ı- - x Solutio Thee eally is t much to do with these poblems oce you kow how to do them. We ll covet the itegal to a limit/itegal pai, evaluate the itegal ad the the limit. Û Ù ı Û dx= lim Ù dx -x tæ- ı -x = lim - -x tæ- ( t ) = lim tæ- =- + = So, the limit is ifiite ad so this itegal is diveget. t 0 t x 007 Paul Dawkis 64

74 Example 4 Detemie if the followig itegal is coveget o diveget. If it is coveget fid its value. -x Ú xe dx - Solutio I this case we ve got ifiities i both limits ad so we ll eed to split the itegal up ito two sepaate itegals. We ca split the itegal up at ay poit, so let s choose a = 0 sice this will be a coveiet poit fo the evaluate pocess. The itegal is the, 0 -x -x -x Ú Ú Ú - xe dx= xe dx+ xe dx We ve ow got to look at each of the idividual limits. Ú x - x - xe dx= lim xe dx tæ- Ú Ê = lim tæ- Á - e Ë t 0 -x ˆ Ê = lim Á- + e tæ- Ë =- So, the fist itegal is coveget. Note that this does NOT mea that the secod itegal will also be coveget. So, let s take a look at that oe. Ú t -x -x xe dx= lim xe dx Ú 0 tæ 0 Ê = lim tæ Á - e Ë -x ˆ 0 t -t 0 -t ˆ Ê ˆ = limá- e + tæ Ë = This itegal is coveget ad so sice they ae both coveget the itegal we wee actually asked to deal with is also coveget ad its value is, 0 -x -x x x dx x dx - Ú e = x dx 0 - Ú e + =- + = - Ú e 0 Example 5 Detemie if the followig itegal is coveget o diveget. If it is coveget fid its value. si xdx Solutio Fist covet to a limit. Ú - t 007 Paul Dawkis 65

75 t sixdx= lim si xdx - tæ Ú- Ú = lim - tæ ( cos x) - ( cost) = lim cos- tæ This limit does t exist ad so the itegal is diveget. I most examples i a Calculus II class that ae woked ove ifiite itevals the limit eithe exists o is ifiite. Howeve, thee ae limits that do t exist, as the pevious example showed, so do t foget about those. Discotiuous Itegad We ow eed to look at the secod type of impope itegals that we ll be lookig at i this sectio. These ae itegals that have discotiuous itegads. The pocess hee is basically the same with oe o subtle diffeece. Hee ae the geeal cases that we ll look at fo these itegals.. If f ( x ) is cotiuous o the iteval [, ) a b t ab ad ot cotiuous at x t = lim Ú Ú f x dx f x dx - tæb a = b the, povided the limit exists ad is fiite. Note as well that we do eed to use a left had limit hee sice the iteval of itegatio is etiely o the left side of the uppe limit.. If f ( x ) is cotiuous o the iteval (, ] a b ab ad ot cotiuous at x b = lim Ú Ú f x dx f x dx + tæa t = a the, povided the limit exists ad is fiite. I this case we eed to use a ight had limit hee sice the iteval of itegatio is etiely o the ight side of the lowe limit. c b. If f ( x ) is ot cotiuous at x= c whee a< c< b ad Ú ad Ú f ae both coveget the, = + b c b Ú Ú Ú f x dx f x dx f x dx a a c a f x dx c x dx As with the ifiite iteval case this equies BOTH of the itegals to be coveget i ode fo this itegal to also be coveget. If eithe of the two itegals is diveget the so is this itegal. c b 4. If f ( x ) is ot cotiuous at x= aad x= bad if Ú ad f both coveget the, = + b c b Ú Ú Ú a f x dx f x dx f x dx f x dx a a c Ú x dx ae Whee c is ay umbe. Agai, this equies BOTH of the itegals to be coveget i ode fo this itegal to also be coveget. c 007 Paul Dawkis 66

76 Note that the limits i these cases eally do eed to be ight o left haded limits. Sice we will be wokig iside the iteval of itegatio we will eed to make sue that we stay iside that iteval. This meas that we ll use oe-sided limits to make sue we stay iside the iteval. Let s do a couple of examples of these kids of itegals. Example 6 Detemie if the followig itegal is coveget o diveget. If it is coveget fid its value. Û Ù dx ı 0 - x Solutio The poblem poit is the uppe limit so we ae i the fist case above. Û Ù ı dx= Û lim Ù dx -x ı -x - tæ 0 0 t ( x) ( t ) = lim tæ 0 = lim tæ = The limit exists ad is fiite ad so the itegal coveges ad the itegals value is. Example 7 Detemie if the followig itegal is coveget o diveget. If it is coveget fid its value. Û Ù dx ı - x Solutio This itegad is ot cotiuous at x = 0 ad so we ll eed to split the itegal up at that poit. 0 Û dx= Û dx+ Û Ù dx Ù Ù ı x ı x ı x Now we eed to look at each of these itegals ad see if they ae coveget. 0 t Û dx= lim Û Ù dx Ù - ı x tæ0 ı x - - Ê = - Ë lim- tæ0 Á Ê ˆ = lim-á- + tæ0 Ë t 8 =- At this poit we e doe. Oe of the itegals is diveget that meas the itegal that we wee asked to look at is diveget. We do t eve eed to bothe with the secod itegal. t x ˆ t Paul Dawkis 67

77 Befoe leavig this sectio lets ote that we ca also have itegals that ivolve both of these cases. Coside the followig itegal. Example 8 Detemie if the followig itegal is coveget o diveget. If it is coveget fid its value. Û Ù dx ı 0 x Solutio This is a itegal ove a ifiite iteval that also cotais a discotiuous itegad. To do this itegal we ll eed to split it up ito two itegals. We ca split it up aywhee, but pick a value that will be coveiet fo evaluatio puposes. Û dx= Û dx+ Û Ù dx Ù Ù ı x ı x ı x 0 0 I ode fo the itegal i the example to be coveget we will eed BOTH of these to be coveget. If oe o both ae diveget the the whole itegal will also be diveget. We kow that the secod itegal is coveget by the fact give i the ifiite iteval potio above. So, all we eed to do is check the fist itegal. Û dx= lim Û Ù dx Ù + ı x tæ0 ı x 0 Ê ˆ = lim+ tæ0 Á - Ë x Ê ˆ = lim + Á- + tæ0 Ë t = So, the fist itegal is diveget ad so the whole itegal is diveget. t t 007 Paul Dawkis 68

78 Compaiso Test fo Impope Itegals Now that we ve see how to actually compute impope itegals we eed to addess oe moe topic about them. Ofte we ae t coceed with the actual value of these itegals. Istead we might oly be iteested i whethe the itegal is coveget o diveget. Also, thee will be some itegals that we simply wo t be able to itegate ad yet we would still like to kow if they covege o divege. To deal with this we ve got a test fo covegece o divegece that we ca use to help us aswe the questio of covegece fo a impope itegal. We will give this test oly fo a sub-case of the ifiite iteval itegal, howeve vesios of the test exist fo the othe sub-cases of the ifiite iteval itegals as well as itegals with discotiuous itegads. Compaiso Test f x g x If 0 o the iteval [, ) a the,. If Ú f ( x) dx coveges the so does a Ú g x dx.. If Ú g ( x ) dx diveges the so does f a a a Ú x dx. Note that if you thik i tems of aea the Compaiso Test makes a lot of sese. If f ( x ) is lage tha g( x ) the the aea ude f ( x ) must also be lage tha the aea ude g( x ). So, if the aea ude the lage fuctio is fiite (i.e. f the smalle fuctio must also be fiite (i.e. the smalle fuctio is ifiite (i.e. must also be ifiite (i.e. f a a a Ú x dx coveges) the the aea ude a Ú g x dx coveges). Likewise, if the aea ude Ú g x dx diveges) the the aea ude the lage fuctio Ú x dx diveges). Be caeful ot to misuse this test. If the smalle fuctio coveges thee is o easo to believe that the lage will also covege (afte all ifiity is lage tha a fiite umbe ) ad if the lage fuctio diveges thee is o easo to believe that the smalle fuctio will also divege. Let s wok a couple of example usig the compaiso test. Note that all we ll be able to do is detemie the covegece of the itegal. We wo t be able to detemie the value of the itegals ad so wo t eve bothe with that. 007 Paul Dawkis 69

79 Example Detemie if the followig itegal is coveget o diveget. Û cos x Ù dx ı x Solutio Let s take a secod ad thik about how the Compaiso Test woks. If this itegal is coveget the we ll eed to fid a lage fuctio that also coveges o the same iteval. Likewise, if this itegal is diveget the we ll eed to fid a smalle fuctio that also diveges. So, it seems like it would be ice to have some idea as to whethe the itegal coveges o diveges ahead of time so we will kow whethe we will eed to look fo a lage (ad coveget) fuctio o a smalle (ad diveget) fuctio. To get the guess fo this fuctio let s otice that the umeato is ice ad bouded ad simply wo t get too lage. Theefoe, it seems likely that the deomiato will detemie the covegece/divegece of this itegal ad we kow that Û Ù dx ı x coveges sice p = > by the fact i the pevious sectio. So let s guess that this itegal will covege. So we ow kow that we eed to fid a fuctio that is lage tha cos x x ad also coveges. Makig a factio lage is actually a faily simple pocess. We ca eithe make the umeato lage o we ca make the deomiato smalle. I this case ca t do a lot about the deomiato. Howeve we ca use the fact that 0 cos x to make the umeato lage (i.e. we ll eplace the cosie with somethig we kow to be lage, amely ). So, cos x x x Now, as we ve aleady oted Û Ù dx ı x coveges ad so by the Compaiso Test we kow that must also covege. Û Ù ı cos x dx x Example Detemie if the followig itegal is coveget o diveget. Û Ù dx x ı x + e Solutio Let s fist take a guess about the covegece of this itegal. As oted afte the fact i the last sectio about 007 Paul Dawkis 70

80 Û Ù ı dx p a x if the itegad goes to zeo faste tha x the the itegal will pobably covege. Now, we ve got a expoetial i the deomiato which is appoachig ifiity much faste tha the x ad so it looks like this itegal should pobably covege. So, we eed a lage fuctio that will also covege. I this case we ca t eally make the umeato lage ad so we ll eed to make the deomiato smalle i ode to make the fuctio lage as a whole. We will eed to be caeful howeve. Thee ae two ways to do this ad oly oe, i this case oly oe, of them will wok fo us. Fist, otice that sice the lowe limit of itegatio is we ca say that x > 0 ad we kow that expoetials ae always positive. So, the deomiato is the sum of two positive tems ad if we wee to dop oe of them the deomiato would get smalle. This would i tu make the fuctio lage. The questio the is which oe to dop? Let s fist dop the expoetial. Doig this gives, < x x+ e x This is a poblem howeve, sice Û Ù dx ı x diveges by the fact. We ve got a lage fuctio that is diveget. This does t say aythig about the smalle fuctio. Theefoe, we chose the wog oe to dop. Let s ty it agai ad this time let s dop the x. Also, Ú -x < = e x x x + e e t - x - e dx= lim tæ - x Ú e dx -t - ( e e ) = lim - + = e tæ So, Ú - x dx e is coveget. Theefoe, by the Compaiso test is also coveget. Û Ù ı dx x x + e 007 Paul Dawkis 7

81 Example Detemie if the followig itegal is coveget o diveget. Û Ù dx -x ı x - e Solutio This is vey simila to the pevious example with a couple of vey impotat diffeeces. Fist, otice that the expoetial ow goes to zeo as x iceases istead of gowig lage as it did i the pevious example (because of the egative i the expoet). Also ote that the expoetial is ow subtacted off the x istead of added oto it. The fact that the expoetial goes to zeo meas that this time the x i the deomiato will pobably domiate the tem ad that meas that the itegal pobably diveges. We will theefoe eed to fid a smalle fuctio that also diveges. Makig factios smalle is petty much the same as makig factios lage. I this case we ll eed to eithe make the umeato smalle o the deomiato lage. This is whee the secod chage will come ito play. As befoe we kow that both x ad the expoetial ae positive. Howeve, this time sice we ae subtactig the expoetial fom the x if we wee to dop the expoetial the deomiato will become lage ad so the factio will become smalle. I othe wods, > - x x- e x ad we kow that Û Ù dx ı x diveges ad so by the Compaiso Test we kow that Û Ù dx -x ı x - e must also divege. Example 4 Detemie if the followig itegal is coveget o diveget. 4 Û + si ( x) Ù dx ı x Solutio Fist otice that as with the fist example, the umeato i this fuctio is goig to be bouded sice the sie is eve lage tha. Theefoe, sice the expoet o the deomiato is less tha we ca guess that the itegal will pobably divege. We will eed a smalle fuctio that also diveges. 4 We kow that ( x) 0 si. I paticula, this tem is positive ad so if we dop it fom the umeato the umeato will get smalle. This gives, 4 + si ( x) > x x ad 007 Paul Dawkis 7

82 diveges so by the Compaiso Test also diveges. Û Ù ı dx x 4 si x dx Û + Ù ı x Okay, we ve see a few examples of the Compaiso Test ow. Howeve, most of them woked petty much the same way. All the fuctios wee atioal ad all we did fo most of them was add o subtact somethig fom the umeato o deomiato to get what we wat. Let s take a look at a example that woks a little diffeetly so we do t get too locked ito these ideas. Example 5 Detemie if the followig itegal is coveget o diveget. -x Û e Ù dx ı x Solutio Nomally, the pesece of just a x i the deomiato would lead us to guess diveget fo this itegal. Howeve, the expoetial i the umeato will appoach zeo so fast that istead we ll eed to guess that this itegal coveges. To get a lage fuctio we ll use the fact that we kow fom the limits of itegatio that x >. This meas that if we just eplace the x i the deomiato with (which is always smalle tha x) we will make the deomiato smalle ad so the fuctio will get lage. -x -x e e -x < = e x ad we ca show that - Ú e x dx coveges. I fact, we ve aleady doe this fo a lowe limit of ad chagig that to a wo t chage the covegece of the itegal. Theefoe, by the Compaiso Test -x Û e Ù dx ı x also coveges. We should also eally wok a example that does t ivolve a atioal fuctio sice thee is o easo to assume that we ll always be wokig with atioal fuctios. Example 6 Detemie if the followig itegal is coveget o diveget. - Ú e x dx Solutio We kow that expoetials with egative expoets die dow to zeo vey fast so it makes sese to guess that this itegal will be coveget. We eed a lage fuctio, but this time we do t 007 Paul Dawkis 7

83 have a factio to wok with so we ll eed to do somethig diffeet. -x We ll take advatage of the fact that e is a deceasig fuctio. This meas that -x -x x > x fi e < e I othe wods, plug i a lage umbe ad the fuctio gets smalle. Fom the limits of itegatio we kow that x > ad this meas that if we squae x we will get lage. O, x > x povided x> Note that we ca oly say this sice x >. This wo t be tue if x! We ca ow use the fact -x that e is a deceasig fuctio to get, -x -x -x e < e -x So, e is a lage fuctio tha e ad we kow that - Ú e x dx coveges so by the Compaiso Test we also kow that - Ú e x dx is coveget. 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 lowe limit had bee smalle tha. Example 7 Detemie if the followig itegal is coveget o diveget. -x Û e Ù dx ı x Solutio I this case we ca t just eplace x with i the deomiato ad get a lage fuctio fo all x i the iteval of itegatio as we did i Example 5 above. Remembe that we eed a fuctio (that is also coveget) that is always lage tha the give fuctio.. To see why we ca t just eplace x with plug i x = 4 ito the deomiato ad compae this to what we would have if we plug i x =. -x -x e 4 -x -x e = e < e = 4 So, fo x s i the age x < we wo t get a lage fuctio as equied fo use i the Compaiso Test. Howeve, this is t the poblem it might at fist appea to be. We ca always wite the itegal as follows, 007 Paul Dawkis 74

84 -x -x -x Û e dx= Û e dx+ Û e Ù Ù Ù dx ı x ı x ı x = Û Ù ı -x e x dx We used Maple to get the value of the fist itegal. Now, if the secod itegal coveges it will have a fiite value ad so the sum of two fiite values will also be fiite ad so the oigial itegal will covege. Likewise, if the secod itegal diveges it will eithe be ifiite o ot have a value at all ad addig a fiite umbe oto this will ot all of a sudde make it fiite o exist ad so the oigial itegal will divege. Theefoe, this itegal will covege o divege depedig oly o the covegece of the secod itegal. As we saw i Example 5 the secod itegal does covege ad so the whole itegal must also covege. As we saw i this example, if we eed to, we ca split the itegal up ito oe that does t ivolve ay poblems ad ca be computed ad oe that may cotai poblem what we ca use the Compaiso Test o to detemie its covege. 007 Paul Dawkis 75

85 Appoximatig Defiite Itegals I this chapte we ve spet quite a bit to time o computig the values of itegals. Howeve, ot all itegals ca be computed. A pefect example is the followig defiite itegal. x Ú e dx 0 We ow eed to talk a little bit about estimatig values of defiite itegals. We will look at thee diffeet methods, although oe should aleady be familia to you fom you Calculus I days. We will develop all thee methods fo estimatig Ú b f x dx a by thikig of the itegal as a aea poblem ad usig kow shapes to estimate the aea ude the cuve. Let s get fist develop the methods ad the we ll ty to estimate the itegal show above. Midpoit Rule This is the ule that you should be somewhat familia to you. We will divide the iteval [ ab, ] ito subitevals of equal width, b- a D x = We will deote each of the itevals as follows, [ x0, x], [ x, x], K,[ x-, x] whee x0 = a ad x = b The fo each iteval let * each subiteval with a height of ( i ) * x i be the midpoit of the iteval. We the sketch i ectagles fo f x. Hee is a gaph showig the set up usig = 6. We ca easily fid the aea fo each of these ectagles ad so fo a geeal we get that, b * * * Ú f ( x) dxªd xf ( x) +D xf ( x) + L +Dxf ( x ) a O, upo factoig out a Dx we get the geeal Mid Poit Rule. 007 Paul Dawkis 76

86 b * * * ªD È Ú f x dx x f x f x f x a Î L Tapezoid Rule Fo this ule we will do the same set up as fo the Midpoit Rule. We will beak up the iteval ab, ito subitevals of width, [ ] b-a D x= The o each subiteval we will appoximate the fuctio with a staight lie that is equal to the fuctio values at eithe edpoit of the iteval. Hee is a sketch of this case fo = 6. Each of these objects is a tapezoid (hece the ules ame ) ad as we ca see some of them do a vey good job of appoximatig the actual aea ude the cuve ad othes do t do such a good job. The aea of the tapezoid i the iteval [ x x ], i- i is give by, Dx Ai = ( f ( xi- ) + f ( xi) ) So, if we use subitevals the itegal is appoximately, b Dx Dx Dx Ú f ( x) dxª ( f ( x0) + f ( x) ) + f ( x) + f ( x) + L + f x- a + f x ( ) Upo doig a little simplificatio we aive at the geeal Tapezoid Rule. b Dx Ú f x dx f x0 f x f x f x f x a - Î L ª È Note that all the fuctio evaluatios, with the exceptio of the fist ad last, ae multiplied by. 007 Paul Dawkis 77

87 Simpso s Rule This is the fial method we e goig to take a look at ad i this case we will agai divide up the ab, ito subitevals. Howeve ulike the pevious two methods we eed to equie iteval [ ] that be eve. The easo fo this will be evidet i a bit. The width of each subiteval is, b- a D x = I the Tapezoid Rule we appoximated the cuve with a staight lie. Fo Simpso s Rule we ae goig to appoximate the fuctio with a quadatic ad we e goig to equie that the quadatic agee with thee of the poits fom ou subitevals. Below is a sketch of this usig = 6. Each of the appoximatios is coloed diffeetly so we ca see how they actually wok. Notice that each appoximatio actually coves two of the subitevals. This is the easo fo equiig to be eve. Some of the appoximatios look moe like a lie tha a quadatic, but they eally ae quadatics. Also ote that some of the appoximatios do a bette job tha othes. x x x, x + is, It ca be show that the aea ude the appoximatio o the itevals [, i- i] ad [ i i ] Dx Ai = f xi + f xi + f xi ( ( - ) 4 ( + ) ) If we use subitevals the itegal is the appoximately, b Dx Dx Ú f ( x) dxª ( f ( x0) + 4f ( x) + f ( x) ) + ( f ( x) + 4f ( x) + f ( x4) ) a Dx + L+ f x f x - + f x Upo simplifyig we aive at the geeal Simpso s Rule. ( ( ) ( ) ) b Dx Ú f x dx f x f x f x f x f x f x a Î L ª È Paul Dawkis 78

88 I this case otice that all the fuctio evaluatios at poits with odd subscipts ae multiplied by 4 ad all the fuctio evaluatios at poits with eve subscipts (except fo the fist ad last) ae multiplied by. If you ca emembe this, this a faily easy ule to emembe. Okay, it s time to wok a example ad see how these ules wok. Example Usig = 4 ad all thee ules to appoximate the value of the followig itegal. Ú e x dx 0 Solutio Fist, fo efeece puposes, Maple gives the followig value fo this itegal. x Ú e dx = I each case the width of the subitevals will be, - 0 D x = = 4 ad so the subitevals will be, [ 0, 0.5 ], [ 0.5,, ] [,.5 ], [.5, ] Let s go though each of the methods. Midpoit Rule 0 ( 0.5) Ú e x dx ª = e e e e Remembe that we evaluate at the midpoits of each of the subitevals hee! The Midpoit Rule has a eo of Tapezoid Rule ( 0) () Ú e x dx ª = e e e e e The Tapezoid Rule has a eo of Simpso s Rule ( 0) () Ú e x dx ª = e e e e e The Simpso s Rule has a eo of Noe of the estimatios i the pevious example ae all that good. The best appoximatio i this case is fom the Simpso s Rule ad yet it s still had a eo of almost. To get a bette estimatio we would eed to use a lage. So, fo completeess sake hee ae the estimates fo some lage value of. 007 Paul Dawkis 79

89 Midpoit Tapezoid Simpso s Appox. Eo Appox. Eo Appox. Eo I this case we wee able to detemie the eo fo each estimate because we could get ou hads o the exact value. Ofte this wo t be the case ad so we d ext like to look at eo bouds fo each estimate. These bouds will give the lagest possible eo i the estimate, but it should also be poited out that the actual eo may be sigificatly smalle tha the boud. The boud is oly thee so we ca say that we kow the actual eo will be less tha the boud. So, suppose that f ( x) K ad ( 4 f ) ( x) M fo a x b the if E M, E T, ad E S ae the actual eos fo the Midpoit, Tapezoid 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 Detemie the eo bouds fo the estimatios i the last example. Solutio We aleady kow that = 4, a = 0, ad b = so we just eed to compute K (the lagest value of the secod deivative) ad M (the lagest value of the fouth deivative). This meas that we ll eed the secod ad fouth deivative of f(x). Hee is a gaph of the secod deivative. x = e ( + ) ( 4 ) x 4 = 4e ( ) f x x f x x x Hee is a gaph of the fouth deivative. 007 Paul Dawkis 80

90 So, fom these gaphs it s clea that the lagest value of both of these ae at x =. So, f = fi K = 98 f ( 4 ) = fi M = 56 We ouded to make the computatios simple. Hee ae the bouds fo each ule. E M E T E S ( - ) ( - ) ( - ) = = = I each case we ca see that the eos ae sigificatly smalle tha the actual bouds. 007 Paul Dawkis 8

91 007 Paul Dawkis 8

92 Applicatios of Itegals Itoductio I this sectio we e goig to take a look at some of applicatios of itegatio. It should be oted as well that these applicatios ae peseted hee, as opposed to Calculus I, simply because may of the itegals that aise fom these applicatios ted to equie techiques that we discussed i the pevious chapte. Hee is a list of applicatios that we ll be takig a look at i this chapte. Ac Legth We ll detemie the legth of a cuve i this sectio. Suface Aea I this sectio we ll detemie the suface aea of a solid of evolutio. Cete of Mass Hee we will detemie the cete of mass o cetoid of a thi plate. Hydostatic Pessue ad Foce We ll detemie the hydostatic pessue ad foce o a vetical plate submeged i wate. Pobability Hee we will look at pobability desity fuctios ad computig the mea of a pobability desity fuctio. 007 Paul Dawkis 8

93 Ac Legth I this sectio we ae goig to look at computig the ac legth of a fuctio. Because it s easy eough to deive the fomulas that we ll use i this sectio we will deive oe of them ad leave the othe to you to deive. We wat to detemie the legth of the cotiuous fuctio y = f ( x) o the iteval [, ] ab. Iitially we ll eed to estimate the legth of the cuve. We ll do this by dividig the iteval up ito equal subitevals each of width Dx ad we ll deote the poit o the cuve at each poit by P i. We ca the appoximate the cuve by a seies of staight lies coectig the poits. Hee is a sketch of this situatio fo = 9. Now deote the legth of each of these lie segmets by Pi- Pi ad the legth of the cuve will the be appoximately, Lª Â Pi- Pi i= ad we ca get the exact legth by takig lage ad lage. I othe wods, the exact legth will be, L= limâ Pi- Pi Æ i = Now, let s get a bette gasp o the legth of each of these lie segmets. Fist, o each segmet D y = y - y = f x - f x. We ca the compute diectly the legth of the let s defie lie segmets as follows. i i i- i i- i- i = i - i- + i - i- = D +D i P P x x y y x y By the Mea Value Theoem we kow that o the iteval [ xi, xi] * f ( x ) - f ( x ) = f ( x )( x -x ) i i- i i i- i * ( i ) D y = f x Dx - thee is a poit * x i so that, 007 Paul Dawkis 84

94 Theefoe, the legth ca ow be witte as, * ( i ) P P = x - x + y - y i- i i i- i i- ( i ) * = D x + Èf x Î Dx = + Èf x Î Dx The exact legth of the cuve is the, L= lim P P Â Æ i = Â Æ i = i- * ( i ) = lim + Èf x Î Dx Howeve, usig the defiitio of the defiite itegal, this is othig moe tha, b i L= Û +ÈÎf x dx ı A slightly moe coveiet otatio (i my opiio ayway) is the followig. a b Û Êdy ˆ L= Ù + Á ı Ëdx I a simila fashio we ca also deive a fomula fo x= h( y) o [, ] d Û Êdx ˆ L= Û + Èh ( y) dy = + dy ı Î Ù Á c ı Ëdy c Agai, the secod fom is pobably a little moe coveiet. a d dx cd. This fomula is, Note the diffeece i the deivative ude the squae oot! Do t get too cofused. With oe we diffeetiate with espect to x ad with the othe we diffeetiate with espect to y. Oe way to keep the two staight is to otice that the diffeetial i the deomiato of the deivative will match up with the diffeetial i the itegal. This is oe of the easos why the secod fom is a little moe coveiet. Befoe we wok ay examples we eed to make a small chage i otatio. Istead of havig two fomulas fo the ac legth of a fuctio we ae goig to educe it, i pat, to a sigle fomula. Fom this poit o we ae goig to use the followig fomula fo the legth of the cuve. 007 Paul Dawkis 85

95 Ac Legth Fomula(s) whee, L = Ú ds Êdy ˆ ds = + Á dx if y = f ( x), a x b Ëdx Êdx ˆ ds = + Á dy if x= h( y), c y d Ëdy Note that o limits wee put o the itegal as the limits will deped upo the ds that we e usig. Usig the fist ds will equie x limits of itegatio ad usig the secod ds will equie y limits of itegatio. Thikig of the ac legth fomula as a sigle itegal with diffeet ways to defie ds will be coveiet whe we u acoss ac legths i futue sectios. Also, this ds otatio will be a ice otatio fo the ext sectio as well. Now that we ve deived the ac legth fomula let s wok some examples. Example Detemie the legth of y l( sec x) Solutio p = betwee 0 x. 4 I this case we ll eed to use the fist ds sice the fuctio is i the fom y f ( x) the deivative out of the way. dy sec x ta x Ê ta dy ˆ = = x Á = ta dx sec x Ëdx =. So, let s get Let s also get the oot out of the way sice thee is ofte simplificatio that ca be doe ad thee s o easo to do that iside the itegal. Êdy ˆ ta sec sec sec + Á = + x = x = x = x Ëdx Note that we could dop the absolute value bas hee sice secat is positive i the age give. x The ac legth is the, L= Ú p 4 0 sec xdx = l secx+ ta x = l + p Paul Dawkis 86

96 Example Detemie the legth of x= ( y- ) betwee y 4. Solutio Thee is a vey commo mistake that studets make i poblems of this type. May studets see x= h y ad they immediately decide that it will be too difficult that the fuctio is i the fom to wok with it i that fom so they solve fo y to get the fuctio ito the fom y f ( x) While that ca be doe hee it will lead to a messie itegal fo us to deal with. =. Sometimes it s just easie to wok with fuctios i the fom x h( y) with fuctios i the fom y = f ( x) the you ca wok with fuctios i the fom x h( y) =. I fact, if you ca wok =. Thee eally is t a diffeece betwee the two so do t get excited about fuctios i the fom x= h y. Let s compute the deivative ad the oot. dx Êdx ˆ = ( y-) fi + Á = + y- = y dy Ëdy As you ca see keepig the fuctio i the fom x= h( y) is goig to lead to a vey easy itegal. To see what would happe if we tied to wok with the fuctio i the fom y = f ( x) see the ext example. Let s get the legth. L= = = Ú 4 4 y ydy 4 As oted i the last example we eally do have a choice as to which ds we use. Povided we ca get the fuctio i the fom equied fo a paticula ds we ca use it. Howeve, as also oted above, thee will ofte be a sigificat diffeece i difficulty i the esultig itegals. Let s take a quick look at what would happe i the pevious example if we did put the fuctio ito y = f x. the fom Example Redo the pevious example usig the fuctio i the fom y = f ( x) istead. Solutio I this case the fuctio ad its deivative would be, 007 Paul Dawkis 87

97 - Ê x ˆ dy Ê y x ˆ = Á + = Á Ë dx Ë The oot i the ac legth fomula would the be. x x + x x x Êdy ˆ + Á = + = = Ëdx All the simplificatio wok above was just to put the oot ito a fom that will allow us to do the itegal. Now, befoe we wite dow the itegal we ll also eed to detemie the limits. This paticula ds equies x limits of itegatio ad we ve got y limits. They ae easy eough to get howeve. Sice we kow x as a fuctio of y all we eed to do is plug i the oigial y limits of itegatio ad get the x limits of itegatio. Doig this gives, 0 x ( ) Not easy limits to deal with, but thee they ae. Let s ow wite dow the itegal that will give the legth. Û L= Ù Ù ı 0 x x + dx That s a eally upleasat lookig itegal. It ca be evaluated howeve usig the followig substitutio. Usig this substitutio the itegal becomes, - Êxˆ Êxˆ u = Á + du = Á dx Ë Ë x= 0 fi u = x= fi u = 4 L= Ú 4 = u 4 = So, we got the same aswe as i the pevious example. Although that should t eally be all that supisig sice we wee dealig with the same cuve. udu Paul Dawkis 88

98 Fom a techical stadpoit the itegal i the pevious example was ot that difficult. It was just a Calculus I substitutio. Howeve, fom a pactical stadpoit the itegal was sigificatly moe difficult tha the itegal we evaluated i Example. So, the moal of the stoy hee is that we ca use eithe fomula (povided we ca get the fuctio i the coect fom of couse) howeve oe will ofte be sigificatly easie to actually evaluate. Okay, let s wok oe moe example. Example 4 Detemie the legth of x= y fo 0 x. Assume that y is positive. Solutio We ll use the secod ds fo this oe as the fuctio is aleady i the coect fom fo that oe. Also, the othe ds would agai lead to a paticulaly difficult itegal. The deivative ad oot will the be, dx Êdx ˆ = y fi + Á = + y dy Ëdy Befoe witig dow the legth otice that we wee give x limits ad we will eed y limits fo this ds. With the assumptio that y is positive thee ae easy eough to get. All we eed to do is plug x ito ou equatio ad solve fo y. Doig this gives, 0 y The itegal fo the ac legth is the, Ú L= + 0 y dy This itegal will equie the followig tig substitutio. y = taq dy = sec q dq y = 0 fi 0= taq fi q = 0 The legth is the, p y = fi = taq fi q = 4 + = + ta = sec = sec = sec y q q q q L= Ú p 4 0 sec q dq = sec ta + l sec + ta ( q q q q ) ( l( ) ) = + + The fist couple of examples eded up beig faily simple Calculus I substitutios. Howeve, as this last example had show we ca ed up with tig substitutios as well fo these itegals. p Paul Dawkis 89

99 Suface Aea I this sectio we ae goig to look oce agai at solids of evolutio. We fist looked at them back i Calculus I whe we foud the volume of the solid of evolutio. I this sectio we wat to fid the suface aea of this egio. So, fo the puposes of the deivatio of the fomula, let s look at otatig the cotiuous fuctio y f x ab, about the x-axis. Below is a sketch of a fuctio ad the solid of = i the iteval [ ] evolutio we get by otatig the fuctio about the x-axis. We ca deive a fomula fo the suface aea much as we deived the fomula fo ac legth. We ll stat by dividig the itegal ito equal subitevals of width D x. O each subiteval we will appoximate the fuctio with a staight lie that agees with the fuctio at the edpoits of the each iteval. Hee is a sketch of that fo ou epesetative fuctio usig = 4. Now, otate the appoximatios about the x-axis ad we get the followig solid. 007 Paul Dawkis 90

100 The appoximatio o each iteval gives a distict potio of the solid ad to make this clea each potio is coloed diffeetly. Each of these potios ae called fustums ad we kow how to fid the suface aea of fustums. The suface aea of a fustum is give by, A= p l whee, = ( + ) = adius of ight ed = adius of left ed ad l is the legth of the slat of the fustum. Fo the fustum o the iteval [ x, i- xi] we have, = f ( xi ) = f ( xi -) l = P P ( legth of the lie segmet coectig P ad P ) i- i i i- ad we kow fom the pevious sectio that, ( * ) * [ ] P P = + Èf Î x Dx whee x is some poit i x, x i- i i i i- i Befoe witig dow the fomula fo the suface aea we ae goig to assume that f x is cotiuous we ca the assume that, ad sice ( * ) ( * i ª i ad i- ª i ) f x f x f x f x So, the suface aea of the fustum o the iteval [ x x ], i- i is appoximately, D x is small 007 Paul Dawkis 9

101 + Ê f xi f xi- ˆ Ai = p Á Pi- Pi Ë * * È i ( i ) ª p f x + f x Î Dx The suface aea of the whole solid is the appoximately,  i= * * È i i S ª p f x + Î f x Dx ad we ca get the exact suface aea by takig the limit as goes to ifiity. i * * i ( i ) S = lim p f x + Èf x Dx Æ = Î b = Û p f x +ÈÎf x dx ı a  If we wated to we could also deive a simila fomula fo otatig x= h( y) o [, ] y-axis. This would give the followig fomula. d S = Û p h y +ÈÎh y dy ı c cd about the These ae ot the stadad fomulas howeve. Notice that the oots i both of these fomulas f x ae othig moe tha the two ds s we used i the pevious sectio. Also, we will eplace with y ad h( y ) with x. Doig this gives the followig two fomulas fo the suface aea. Suface Aea Fomulas S = pyds otatio about x-axis whee, Ú Ú S = pxds otatio about y-axis Êdy ˆ ds = + Á dx if y = f ( x), a x b Ëdx Êdx ˆ ds = + Á dy if x= h( y), c y d Ëdy Thee ae a couple of thigs to ote about these fomulas. Fist, otice that the vaiable i the itegal itself is always the opposite vaiable fom the oe we e otatig about. Secod, we ae allowed to use eithe ds i eithe fomula. This meas that thee ae, i some way, fou fomulas hee. We will choose the ds based upo which is the most coveiet fo a give fuctio ad poblem. Now let s wok a couple of examples. 007 Paul Dawkis 9

102 Example Detemie the suface aea of the solid obtaied by otatig - x about the x-axis. y = 9 - x, Solutio The fomula that we ll be usig hee is, S = Ú p yds sice we ae otatig about the x-axis ad we ll use the fist ds i this case because ou fuctio is i the coect fom fo that ds ad we wo t gai aythig by solvig it fo x. Let s fist get the deivative ad the oot take cae of. dy - = ( 9 -x ) (- x) =- dx Hee s the itegal fo the suface aea, x ( 9-x ) Êdyˆ x 9 + Á = + = = Ëdx 9 -x 9 -x 9- x Û S = Ù p y dx ı x Thee is a poblem howeve. The dx meas that we should t have ay y s i the itegal. So, befoe evaluatig the itegal we ll eed to substitute i fo y as well. The suface aea is the, Û S = Ù p 9-x dx ı- = Ú - = 4p 6p dx 9-x Peviously we made the commet that we could use eithe ds i the suface aea fomulas. Let s wok a example i which usig eithe ds wo t ceate itegals that ae too difficult to evaluate ad so we ca check both ds s. Example Detemie the suface aea of the solid obtaied by otatig about the y-axis. Use both ds s to compute the suface aea. y = x, y Solutio Note that we ve bee give the fuctio set up fo the fist ds ad limits that wok fo the secod ds. Solutio This solutio will use the fist ds listed above. We ll stat with the deivative ad oot. 007 Paul Dawkis 9

103 dy = dx x Êdyˆ 9x + 9x + + Á = + = = 4 4 Ëdx 9x 9x x We ll also eed to get ew limits. That is t too bad howeve. All we eed to do is plug i the give y s ito ou equatio ad solve to get that the age of x s is x 8. The itegal fo the suface aea is the, 8 4 Û 9x + S = Ù p x dx Ù ı x 8 4 p = ÛÙ x 9x + dx ı Note that this time we did t eed to substitute i fo the x as we did i the pevious example. I this case we picked up a dx fom the ds ad so we do t eed to do a substitutio fo the x. I fact if we had substituted fo x we would have put y s ito itegal which would have caused poblems. Usig the substitutio the itegal becomes, 4 u = 9x + du = x dx p S = 8 Ú 45 0 p = u udu p Ê ˆ = Á45-0 = Ë Solutio This time we ll use the secod ds. So, we ll fist eed to solve the equatio fo x. We ll also go ahead ad get the deivative ad oot while we e at it. dx x= y = y dy The suface aea is the, Êdx ˆ + Á = + 9y Ëdy Paul Dawkis 94

104 S = px + y 4 dy Ú 9 We used the oigial y limits this time because we picked up a dy fom the ds. Also ote that the pesece of the dy meas that this time, ulike the fist solutio, we ll eed to substitute i fo the x. Doig that gives, 4 4 S = p y + 9y dy u = + 9y Ú p 45 = 8 Ú udu 0 p Ê ˆ = Á45-0 = Ë Note that afte the substitutio the itegal was idetical to the fist solutio ad so the wok was skipped. As this example has show we ca used eithe ds to get the suface aea. It is impotat to poit out as well that with oe ds we had to do a substitutio fo the x ad with the othe we did t. This will always wok out that way. Note as well that i the case of the last example it was just as easy to use eithe ds. That ofte wo t be the case. I may examples oly oe of the ds will be coveiet to wok with so we ll always eed to detemie which ds is liable to be the easiest to wok with befoe statig the poblem. 007 Paul Dawkis 95

105 Cete of Mass I this sectio we ae goig to fid the cete of mass o cetoid of a thi plate with uifom desity. The cete of mass o cetoid of a egio is the poit i which the egio will be pefectly balaced hoizotally if suspeded fom that poit. So, let s suppose that the plate is the egio bouded by the two cuves f ( x ) ad iteval [a,b]. So, we wat to fid the cete of mass of the egio below. g x o the We ll fist eed the mass of this plate. The mass is, M = ( Aea of plate) Ú b a = f x -g x dx Next we ll eed the momets of the egio. Thee ae two momets, deoted by M x ad M y. The momets measue the tedecy of the egio to otate about the x ad y-axis espectively. The momets ae give by, Equatios of Momets ( ÈÎ ÈÎ ) b Mx = Û Ù f x - g x dx ı y Ú a b a ( ) M = x f x -g x dx The coodiates of the cete of mass, ( x, y ), ae the, 007 Paul Dawkis 96

106 Cete of Mass Coodiates b Ú Ú ( - ) M x f x g x dx y a b x = = = x b ( f ( x) g( x) ) dx M AÚ - a f x g x dx a - ( ÈÎ -ÈÎ ) b Û Ù f x g x dx b M a x ı y = = = Û b Ùı ( Èf ( x) -Èg( x) ) dx M f x g x dx A a Î Î Ú a - whee, Ú b a A= f x -g x dx Note that the desity,, of the plate cacels out ad so is t eally eeded. Let s wok a couple of examples. Example Detemie the cete of mass fo the egio bouded by y si( x) È p the iteval Í 0, Î. Solutio Hee is a sketch of the egio with the cete of mass deoted with a dot. =, y = 0 o Let s fist get the aea of the egio. A= Ú p 0 si x dx ( x) =-cos = p 0 Now, the momets (without desity sice it will just dop out) ae, 007 Paul Dawkis 97

107 p p 0 y 0 M = si x dx M = xsi x dx itegatig by pats... x Ú Ú p p p = - cos 4x dx =- xcos x + cos x dx p p p Ê ˆ = Áx- si( 4x) =- xcos x + si x 0 Ë 4 0 p p = = Ú Ú The coodiates of the cete of mass ae the, p p x = = 4 p y p = = 4 Agai, ote that we did t put i the desity sice it will cacel out. Êp p ˆ So, the cete of mass fo this egio is Á, Ë Example Detemie the cete of mass fo the egio bouded by y = x ad y x =. Solutio The two cuves itesect at x= 0 ad x = ad hee is a sketch of the egio with the cete of mass maked with a box. We ll fist get the aea of the egio. 007 Paul Dawkis 98

108 A= x -x dx 0 Ú Ê = Á x - x Ë 4 5 = Now the momets, agai without desity, ae Mx = Û Ù x-x dx ı 0 6 y = Ú ( - ) 0 4 ˆ 4 Ê 7ˆ = x -x dx 0 = Á x - x Ë Ê 5ˆ 5 = Á x - x = Ë5 5 8 = 5 The coodiates of the cete of mass is the, 5 x = = y = = 5 7 Ê ˆ The coodiates of the cete of mass ae the, Á, Ë M x x x dx Ú Paul Dawkis 99

109 Hydostatic Pessue ad Foce I this sectio we ae goig to submege a vetical plate i wate ad we wat to kow the foce that is exeted o the plate due to the pessue of the wate. This foce is ofte called the hydostatic foce. Thee ae two basic fomulas that we ll be usig hee. Fist, if we ae d metes below the suface the the hydostatic pessue is give by, P = gd whee, is the desity of the fluid ad g is the gavitatioal acceleatio. We ae goig to assume that the fluid i questio is wate ad sice we ae goig to be usig the metic system these quatities become, = 000 kg/m g = 9.8 m/s The secod fomula that we eed is the followig. Assume that a costat pessue P is actig o a suface with aea A. The the hydostatic foce that acts o the aea is, F = PA Note that we wo t be able to fid the hydostatic foce o a vetical plate usig this fomula sice the pessue will vay with depth ad hece will ot be costat as equied by this fomula. We will howeve eed this fo ou wok. The best way to see how these poblems wok is to do a example o two. Example Detemie the hydostatic foce o the followig tiagula plate that is submeged i wate as show. Solutio The fist thig to do hee is set up a axis system. So, let s edo the sketch above with the followig axis system added i. 007 Paul Dawkis 00

110 So, we ae goig to oiet the x-axis so that positive x is dowwad, x = 0 coespods to the wate suface ad x = 4 coespods to the depth of the tip of the tiagle. Next we ae beak up the tiagle ito hoizotal stips each of equal width D x ad i each iteval [, * xi- xi] choose ay poit x i. I ode to make the computatios easie we ae goig to make two assumptios about these stips. Fist, we will igoe the fact that the eds ae actually goig to be slated ad assume the stips ae ectagula. If D x is sufficietly small this will ot affect ou computatios much. Secod, we will assume that D x is small eough that the hydostatic pessue o each stip is essetially costat. Below is a epesetative stip. The height of this stip is D x ad the width is a. We ca use simila tiagles to detemie a as follows, a * = fi a = - x * 4 4-x 4 i i 007 Paul Dawkis 0

111 Now, sice we ae assumig the pessue o this stip is costat, the pessue is give by, P = gd = x = 980x * * i i i ad the hydostatic foce o each stip is, * Ê * ˆ * Ê * ˆ Fi = PA i = Pi ad x = xi Á - xi D x= xi Á - xi Dx Ë 4 Ë 4 The appoximate hydostatic foce o the plate is the the sum of the foces o all the stips o, * Ê * ˆ F ª Â 960xi Á- xi Dx i= Ë 4 Takig the limit will get the exact hydostatic foce, * * F = Ê ˆ limâ 960xi Á - xi D x Æ i = Ë 4 Usig the defiitio of the defiite itegal this is othig moe tha, 4 Û F 960 Ê ˆ = x- x dx Ù ı 0 Á Ë 4 The hydostatic foce is the, 4 Û Ê ˆ F = 960 x- x dx Ù ı 0 = N Á Ë 4 Ê ˆ = 960Á x - x Ë Let s take a look at aothe example. Example Fid the hydostatic foce o a cicula plate of adius that is submeged 6 metes i the wate. Solutio Fist, we e goig to assume that the top of the cicula plate is 6 metes ude the wate. Next, we will set up the axis system so that the oigi of the axis system is at the cete of the plate. Settig the axis system up i this way will geatly simplify ou wok. Fially, we will agai split up the plate ito hoizotal stips each of width D y ad we ll * choose a poit y i fom each stip. We ll also assume that the stips ae ectagula agai to help with the computatios. Hee is a sketch of the setup. 007 Paul Dawkis 0

112 The depth below the wate suface of each stip is, * di = 8 - yi ad that i tu gives us the pessue o the stip, P = gd = y * i i i The aea of each stip is, * i A = 4- y D y i The hydostatic foce o each stip is, The total foce o the plate is, * * F = PA = y 4- y D y i i i i i Â Æ i = - ( y) * * ( i ) ( i ) F = lim y 4- y Dy Ú = y dy To do this itegal we ll eed to split it up ito two itegals. F = y dy-960 y 4- y dy Ú Ú Paul Dawkis 0

113 The fist itegal equies the tig substitutio substitutio v= 4 - y. Afte usig these substitutio we get, Ú y = siq ad the secod itegal eeds the p 0 F = cos q dq vdv Ú -p 0 p -p ( q) = 90 + cos dq + 0 Ê = 90Áq + si Ë = 90p ( q) Note that afte the substitutio we kow the secod itegal will be zeo because the uppe ad lowe limit is the same. ˆ p p - Ú 007 Paul Dawkis 04

114 Pobability I this last applicatio of itegals that we ll be lookig at we e goig to look at pobability. Befoe 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 peso, the height of a peso, the amout of time spet waitig i lie, o maybe the lifetime of a battey. Each of these quatities have values that will age ove a iteval of iteges. Because of this these ae called cotiuous adom vaiables. Cotiuous adom vaiables ae ofte epeseted by X. Evey cotiuous adom vaiable, X, has a pobability desity fuctio, f desity fuctios satisfy the followig coditios.. f ( x) 0 fo all x.. Ú f ( x) dx = - x. Pobability Pobability desity fuctios ca be used to detemie the pobability that a cotiuous adom P a X b ad vaiable lies betwee two values, say a ad b. This pobability is deoted by is give by, Let s take a look at a example of this. Example Let f ( x) ( 0 x) b P a X b =Ú f x dx x = - fo 0 x 0 ad f ( x ) = 0 fo all othe values of x Aswe each of the followig questios about this fuctio. f x is a pobability 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 pobability desity fuctio. Fist ote that i the age 0 x 0 is clealy positive ad outside of this age we ve defied it to be zeo. So, to show this is a pobability desity fuctio we ll eed to show that f ( x) dx = - a Ú. 007 Paul Dawkis 05

115 Ú 0 x f ( x) dx= Û Ù ( 0 -x) dx - ı Ê x x ˆ = Á - Ë = Note the chage i limits o the itegal. The fuctio is oly o-zeo i these ages ad so the itegal ca be educed dow to oly the iteval whee the fuctio is ot zeo. [Retu to Poblems] (b) Fid P( X 4) 0 0 I this case we eed to evaluate the followig itegal. 4 x P( X 4) = Û Ù ( 0 -x) dx ı Ê x x ˆ = Á - Ë = So the pobability of X beig betwee ad 4 is 8.658%. 4 [Retu to Poblems] (c) Fid P( x 6) Note that i this case P( x 6) is equivalet to P( 6 X 0) X ca be. So the pobability that X is geate tha o equal to 6 is, 0 x P( X 6) = Û Ù ( 0 -x) dx ı 5000 This pobability is the 66.04% Ê x x ˆ = Á - Ë = sice 0 is the lagest value that 0 6 [Retu to Poblems] Pobability desity fuctios ca also be used to detemie the mea of a cotiuous adom vaiable. The mea is give by, Let s wok oe moe example. m = Ú xf - x dx 007 Paul Dawkis 06

116 Example It has bee detemied that the pobability desity fuctio fo the wait i lie at a coute is give by, Ï0 if t < 0 Ô f () t = Ì t - Ô 0 Ó0.e if t 0 whee t is the umbe of miutes spet waitig i lie. Aswe each of the followig questios about this pobability desity fuctio. (a) Veify that this is i fact a pobability desity fuctio. [Solutio] (b) Detemie the pobability that a peso will wait i lie fo at least 6 miutes.\ [Solutio] (c) Detemie the mea wait i lie. [Solutio] Solutio (a) Veify that this is i fact a pobability desity fuctio. This fuctio is clealy positive o zeo ad so thee s ot much to do hee othe tha compute the itegal. So it is a pobability desity fuctio. Ú t - 0 f () t dt = 0. - Ú e 0 u = lim 0.e uæ Ú 0 Ê = limá-e uæ Ë t - 0 dt t - 0 ˆ u - 0 limá e uæ u 0 dt = Ê - ˆ = Ë (b) Detemie the pobability that a peso will wait i lie fo at least 6 miutes. P x 6. The pobability that we e lookig fo hee is - 0 ( 6) = 0.e P X Ú 6 u = lim 0.e uæ Ú 6 Ê = limá-e uæ Ë t t - 0 dt t - 0 ˆ 6 [Retu to Poblems] 6 u Ê - - ˆ = limáe - e = e = uæ Ë So the pobability that a peso will wait i lie fo moe tha 6 miutes is 54.88%. [Retu to Poblems] u dt 007 Paul Dawkis 07

117 (c) Detemie the mea wait i lie. Hee s the mea wait time. m = = = Ú Ú - () tf t dt t te 0 dt t u - 0 lim 0.t dt itegatig by pats... uæ Ú e 0 Ê = limá- ( t + 0) e uæ Ë t - 0 ˆ = Ê - + ˆ = Ë u - 0 limá0 ( u 0) e 0 uæ u 0 So, it looks like the aveage wait time is 0 miutes. [Retu to Poblems] 007 Paul Dawkis 08

118 Paametic Equatios ad Pola Coodiates Itoductio I this sectio we will be lookig at paametic equatios ad pola coodiates. While the two subjects do t appea to have that much i commo o the suface we will see that seveal of the topics i pola coodiates ca be doe i tems of paametic equatios ad so i that sese they make a good match i this chapte. We will also be lookig at how to do may of the stadad calculus topics such as tagets ad aea i tems of paametic equatios ad pola coodiates. Hee is a list of topics that we ll be coveig i this chapte. Paametic Equatios ad Cuves A itoductio to paametic equatios ad paametic cuves (i.e. gaphs of paametic equatios) Tagets with Paametic Equatios Fidig taget lies to paametic cuves. Aea with Paametic Equatios Fidig the aea ude a paametic cuve. Ac Legth with Paametic Equatios Detemiig the legth of a paametic cuve. Suface Aea with Paametic Equatios Hee we will detemie the suface aea of a solid obtaied by otatig a paametic cuve about a axis. Pola Coodiates We ll itoduce pola coodiates i this sectio. We ll look at covetig betwee pola coodiates ad Catesia coodiates as well as some basic gaphs i pola coodiates. Tagets with Pola Coodiates Fidig taget lies of pola cuves. Aea with Pola Coodiates Fidig the aea eclosed by a pola cuve. Ac Legth with Pola Coodiates Detemiig the legth of a pola cuve. Suface Aea with Pola Coodiates Hee we will detemie the suface aea of a solid obtaied by otatig a pola cuve about a axis. Ac Legth ad Suface Aea Revisited I this sectio we will summaize all the ac legth ad suface aea fomulas fom the last two chaptes. 007 Paul Dawkis 09

119 Paametic Equatios ad Cuves To this poit (i both Calculus I ad Calculus II) we ve looked almost exclusively at fuctios i y f x x= h y ad almost all of the fomulas that we ve developed equie the fom = o that fuctios be i oe of these two foms. The poblem is that ot all cuves o equatios that we d like to look at fall easily ito this fom. Take, fo example, a cicle. It is easy eough to wite dow the equatio of a cicle ceteed at the oigi with adius. x + y = Howeve, we will eve be able to wite the equatio of a cicle dow as a sigle equatio i eithe of the foms above. Sue we ca solve fo x o y as the followig two fomulas show y =± - x x=± - y but thee ae i fact two fuctios i each of these. Each fomula gives a potio of the cicle. ( top) ( ight side) y = - x x= - y ( bottom) ( left side) y =- - x x=- - y Ufotuately we usually ae wokig o the whole cicle, o simply ca t say that we e goig to be wokig oly o oe potio of it. Eve if we ca aow thigs dow to oly oe of these potios the fuctio is still ofte faily upleasat to wok with. Thee ae also a geat may cuves out thee that we ca t eve wite dow as a sigle equatio i tems of oly x ad y. So, to deal with some of these poblems we itoduce paametic y f x x= h y ) we equatios. Istead of defiig y i tems of x ( = ) o x i tems of y ( defie both x ad y i tems of a thid vaiable called a paamete as follows, x= f t y = g t This thid vaiable is usually deoted by t (as we did hee) but does t have to be of couse. Sometimes we will estict the values of t that we ll use ad at othe times we wo t. This will ofte be depedet o the poblem ad just what we ae attemptig to do. Each value of t defies a poit ( xy, ) f ( t), g( t) = that we ca plot. The collectio of poits that we get by lettig t be all possible values is the gaph of the paametic equatios ad is called the paametic cuve. Sketchig a paametic cuve is ot always a easy thig to do. Let s take a look at a example to see oe way of sketchig a paametic cuve. This example will also illustate why this method is usually ot the best. Example Sketch the paametic cuve fo the followig set of paametic equatios. x= t + t y = t- Solutio At this poit ou oly optio fo sketchig a paametic cuve is to pick values of t, plug them ito the paametic equatios ad the plot the poits. So, let s plug i some t s. 007 Paul Dawkis 0

120 t x y The fist questio that should be asked as this poit is, how did we kow to use the values of t that we did, especially the thid choice? Ufotuately thee is o eal aswe to this questio. We simply pick t s util we ae faily cofidet that we ve got a good idea of what the cuve looks like. It is this poblem with pickig good values of t that make this method of sketchig paametic cuves oe of the pooe choices. Sometimes we have o choice, but if we do have a choice we should avoid it. We ll discuss a alteate gaphig method i late examples. We have oe moe idea to discuss befoe we actually sketch the cuve. Paametic cuves have a diectio of motio. The diectio of motio is give by iceasig t. So, whe plottig paametic cuves we also iclude aows that show the diectio of motio. We will ofte give the value of t that gave specific poits o the gaph as well to make it clea the value of t that have that paticula poit. Hee is the sketch of this paametic cuve. So, it looks like we have a paabola that opes to the ight. Befoe we ed this example thee is a somewhat impotat ad subtle poit that we eed to discuss fist. Notice that we made sue to iclude a potio of the sketch to the ight of the poits coespodig to t =- ad t = to idicate that thee ae potios of the sketch thee. Had we simply stopped the sketch at those poits we ae idicatig that thee was o potio of the cuve to the ight of those poits ad thee clealy will be. We just did t compute ay of those poits. This may seem like a uimpotat poit, but as we ll see i the ext example it s moe impotat tha we might thik at this poit. 007 Paul Dawkis

121 Befoe addessig a much easie way to sketch this gaph let s fist addess the issue of limits o the paamete. I the pevious example we did t have ay limits o the paamete. Without limit o the paamete the gaph will cotiue i both diectios as show i the sketch above. We will ofte have limits o the paamete howeve ad this will affect the sketch of the paametic equatios. To see this affect let s look a slight vaiatio of the pevious example. Example Sketch the paametic cuve fo the followig set of paametic equatios. x= t + t y = t- - t Solutio Note that the oly diffeece hee is the pesece of the limits o t. All these limits do is tell us that we ca t take ay value of t outside of this age. Theefoe, the paametic cuve will oly be a potio of the cuve above. Hee is the paametic cuve fo this example. Notice that with this sketch we stated ad stopped the sketch ight o the poits oigiatig fom the ed poits of the age of t s. Cotast this with the sketch i the pevious example whee we had a potio of the sketch to the ight of the stat ad ed poits that we computed. I this case the cuve stats at t =- ad eds at t =, wheeas i the pevious example the cuve did t eally stat at the ight most poits that we computed. We eed to be clea i ou sketches if the cuve stats/eds ight at a poit, o if that poit was simply the fist/last oe that we computed. It is ow time to take a look at a easie method of sketchig this paametic cuve. This method uses the fact that i may, but ot all, cases we ca actually elimiate the paamete fom the paametic equatios ad get a fuctio ivolvig oly x ad y. Thee will be two small poblems with this method, but it will be easy to addess those poblems. It is impotat to ote howeve that we wo t always be able to do this. Just how we elimiate the paamete will deped upo the paametic equatios that we ve got. Let s see how to elimiate the paamete fo the set of paametic equatios that we ve bee wokig with to this poit. 007 Paul Dawkis

122 Example Elimiate the paamete fom the followig set of paametic equatios. x= t + t y = t- Solutio Oe of the easiest ways to elimiate the paamete is to simply solve oe of the equatios fo the paamete (t, i this case) ad substitute that ito the othe equatio. Note that while this may be the easiest to elimiate the paamete, it s usually ot the best way as we ll see soo eough. I this case we ca easily solve y fo t. t = + ( y ) Pluggig this ito the equatio fo x gives, Ê ˆ x y y y y = Á ( + ) + ( + ) = + + Ë 4 4 Sue eough fom ou Algeba kowledge we ca see that this is a paabola that opes to the ight. We wo t bothe with a sketch fo this oe as we ve aleady sketched this oce ad the poit hee was moe to elimiate the paamete ayway. Gettig a sketch of the paametic cuve oce we ve elimiated the paamete is faily simple. All we eed to do is gaph the equatio that we foud by elimiatig the paamete. As oted aleady howeve, thee ae two small poblems with this method. The fist is diectio of motio. The equatio ivolvig oly x ad y will NOT give the diectio of motio of the paametic cuve. This is a easy poblem to fix howeve. All we eed to do is plug i some values of t ito the paametic equatios ad we ca detemie diectio of motio fom that. How may values of t we plug i will deped upo the paametic equatios. I some cases oly two will be equied ad i othes we might eed moe poits. The secod poblem is best illustated i a example as we ll be uig ito this poblem i the emaiig examples. Example 4 Sketch the paametic cuve fo the followig set of paametic equatios. Clealy idicate diectio of motio. x= 5cost y = sit 0 t p Solutio I this case we could elimiate the paamete as we did i the pevious sectio by solvig oe of these fo t ad pluggig this ito the othe. Fo example, -Ê xˆ Ê -Ê xˆˆ t = cos Á fi y = si cos 5 Á Á 5 Ë Ë Ë Ca you see the poblem with doig this? This is defiitely easy to do but we have a geate chace of coectly gaphig the oigial paametic equatios that we do gaphig this! Thee ae may ways to elimiate the paamete fom the paametic equatios ad solvig fo 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 paamete hee? I this case all we eed to do is ecall a vey ice tig idetity ad the equatio of a ellipse. Let s otice that we could do the followig hee. x y 5cos t 4si t + = + = cos t+ si t = Elimiatig the middle steps gives us, ad so it looks like we ve got a ellipse. x y + = 5 4 Befoe poceedig with this example it should be oted that what we did was pobably ot all that obvious to most. Howeve, oce it s bee doe it does clealy wok ad so it s a ice idea that we ca use to elimiate the paamete fom some paametic equatios ivolvig sies ad cosies. It wo t always wok ad sometimes it will take a lot moe maipulatio of thigs that we did hee. A alteate method that we could have used hee was to solve the two paametic equatios fo sie ad cosie as follows, x y cost = si t = 5 The, ecall the tig idetity we used above ad these ew equatio we get, Ê xˆ Ê yˆ x y = t+ t = + = + cos si Á Á Ë5 Ë 5 4 So, the same aswe as the othe method. Which method you use will pobably deped o which you fid easie to use. Both ae pefectly valid ad will get the same esult. Now, let s cotiue o with the example. We ve idetified that the paametic equatios descibe a ellipse, but we ca t just sketch a ellipse ad be doe with it. Recall that all paametic cuves have a diectio of motio. So, we ext eed to detemie the diectio of motio. The equatio of the ellipse tells us othig about the diectio of motio. To get the diectio of motio we ll eed to go back to the paametic equatios ad plug i a few poits. Note as well that i this case we ll eed moe tha two poits to do this. Give ay two poits o a ellipse we could get betwee them by goig eithe clockwise o coute-clockwise about the cicle. So, we ll eed at least thee poits to accuately detemie the diectio of motio. While doig this we should also keep i mid that we ve bee give a age of t s to wok with ad as we saw i Example this may mea that we will oly get a potio of the actual ellipse. So, let s choose t s that will cove the whole age. This will give us the diectio of motio ad eough ifomatio to detemie what potio of the ellipse is i fact taced out. Note that this is the secod poblem alluded to above i elimiatig the paamete. Oce we have elimiated the paamete we ve ot oly elimiated the diectio of motio, but we ve also 007 Paul Dawkis 4

124 elimiated ay ifomatio about what potio of the actual gaph is taced out by the paametic equatios. We will always eed to keep i mid that this a potetial poblem whe elimiatig the paamete fom paametic equatios. So, hee is a table of values fo this set of paametic equatios. t x y p 0 p -5 0 p 0 - p 5 0 It looks like we ae movig i a coute-clockwise diectio about the ellipse ad it also looks like we ll make exactly oe complete tace of the ellipse i the age give. Hee is a sketch of the paametic cuve. Let s take a look at aothe example. Example 5 Sketch the paametic cuve fo the followig set of paametic equatios. Clealy idicate diectio of motio. x= 5cos t y = si t 0 t p Solutio Note that the oly diffeece i these paametic equatios is that we eplaced the t with t. We ca elimiate the paamete hee usig eithe of the methods we discussed i the pevious example. I this case we ll do the followig, x y 5cos ( t) 4si ( t) + = + = cos t+ si t = So, we get the same ellipse that we did i the pevious example. Howeve, we also do t get the same paametic cuve i some sese. We saw i the pevious example that we make oe complete tace of the ellipse i the age 0 t p. I this set of paametic cuves we do t have just a t i the tig fuctios howeve. I this set we ve got a t. Whe we have a t we kow that we ll complete a sigle tace whe t = p so to detemie a t 007 Paul Dawkis 5

125 that will complete a sigle tace whe we have a t all we eed to do is, p t = p fi t = So, while we have the same ellipse that we got i the pevious example we ll tace out the cuve exactly oce i the age, p 0 t Sice we ae wokig o the age 0 t p it the looks like the ellipse i this case will be taced out thee times istead of oly oce as we got i the pevious example. Each ellipse will be taced out i the followig ages, p p 4p 4p 0 t t t p The last issue we eed to deal with pio to sketchig is the diectio of motio. By pickig values of t we ca see that the diectio of motio is t chaged i this case. Howeve, because we e goig aoud faste tha befoe we should pobably use a diffeet set this time to make sue we get a accuate idea of the diectio of motio. t x y p 6 0 p -5 0 Hee s the sketch ad ote that it eally is t all that diffeet fom the pevious sketch. The oly diffeece ae the value of t ad the vaious poits we icluded. So, we saw i the last two examples two sets of paametic equatios that i some way gave the same gaph. Yet, because they taced out the gaph a diffeet umbe of times we eally do eed to thik of them as diffeet paametic cuves. This may seem like a diffeece that we do t eed to woy about, but as we will see i late sectios this ca be a vey impotat diffeece. I some of the late sectios we ae goig to eed a cuve that is taced out exactly oce. Let s take a look at a couple moe examples. 007 Paul Dawkis 6

126 Example 6 Sketch the paametic cuve fo the followig set of paametic equatios. Clealy idetify the diectio of motio. If the cuve is taced out moe tha oce give a age of the paamete fo which the cuve will tace out exactly oce. x= si t y = cost Solutio We ca elimiate the paamete much as we did i the pevious two examples. Howeve, we ll eed to ote that the x aleady cotais a si t ad so we wo t eed to squae the x. We will howeve, eed to squae the y as we eed i the pevious two examples. y y x+ = si t+ cos t = fi x= I this case we get a paabola that opes to the left. We will eed to be vey, vey caeful howeve i sketchig this paametic cuve. We will NOT get the whole paabola. A sketch of a paabola, i this fom, will exist fo all possible values of y. We howeve, have defied both x ad y i tems of sie ad cosie ad we kow that the value of these ae limited ad so we wo t get all possible values of x ad y hee. To see what values of x ad y we get let s ote the followig, - fi fi sit 0 si t 0 x - cost fi - cost fi - y So, it is clea fom this that we will oly get the potio of the paabola that is defied by the equatio above. Befoe sketchig let s also get the diectio of motio. Hee ae some poits fo a age of t s. t x y 0 0 p 0 p 0 - p 0 p 0 I the age 0 t p we stat at (0,) ad ed up back at that same poit. Recallig that we must tavel alog the paabola this meas that we must etace ou path to get back to the statig poit. So, it looks like we ve got a paametic cuve that is taced out ove ad ove i both diectios ad we will tace out oce i the age 0 t p. 007 Paul Dawkis 7

127 To this poit we ve see examples that would tace out the complete gaph that we got by elimiatig the paamete if we took a lage eough age of t s. Howeve, i the pevious example we ve ow see that this will ot always be the case. It is moe tha possible to have a set of paametic equatios which will cotiuously tace out just a potio of the cuve. We ca usually detemie if this will happe by lookig fo limits o x ad y that ae imposed up us by the paametic equatio. We will ofte use paametic equatios to descibe the path of a object o paticle. Let s take a look at a example of that. Example 7 The path of a paticle is give by the followig set of paametic equatios. x= cos t y = + cos t Completely descibe the path of this paticle. Do this by sketchig the path, detemiig limits o x ad y ad givig a age of t s fo which the path will be taced out exactly oce (povide it taces out moe tha oce of couse). Solutio Elimiatig the paamete this time will be a little diffeet. We oly have cosies this time ad we ll use that to ou advatage. We ca solve the x equatio fo cosie ad plug that ito the equatio fo y. This gives, cos( ) x Ê x ˆ t = y = + Á = + x Ë 9 This time we ve got a paabola that opes upwad. 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 tace out a potio of the cuve. Hee s a set of evaluatios so we ca detemie a age of t s fo oe tace of the cuve. 007 Paul Dawkis 8

128 t x y 0 p 4 0 p - p 4 0 p So, it looks like the paticle will agai, cotiuously tace out this potio of the cuve ad will make oe tace i the age 0 t p. Hee is a sketch of the paticles path with a few value of t o it. We should give a small waig at this poit. Because of the ideas ivolved i them we cocetated o paametic cuves that etaced potios of the cuve moe tha oce. Do ot howeve, get too locked ito the idea that this will always happe. May, if ot most paametic cuves will oly tace out oce. The fist oe we looked at is a good example of this. That paametic cuve will eve epeat ay potio of itself. Thee is oe fial topic to be discussed i this sectio befoe movig o. So fa we ve stated with paametic equatios ad elimiated the paamete to detemie the paametic cuve. Howeve, thee ae times i which we wat to go the othe way. Give a fuctio o equatio we might wat to wite dow a set of paametic equatios fo it. I these cases we say that we paameteize the fuctio. If we take Examples 4 ad 5 as examples we ca do this fo ellipses (ad hece cicles). Give the ellipse x y + = a b a set of paametic equatios fo it would be, x= acost y = bsi t This set of paametic equatios will tace out the ellipse statig at the poit ( a,0) ad will tace i a coute-clockwise diectio ad will tace out exactly oce i the age 0 t p. This is a faily impotat set of paametic equatios as it used cotiually i some subjects with dealig with ellipses ad/o cicles. 007 Paul Dawkis 9

129 Evey cuve ca be paameteized i moe tha oe way. Ay of the followig will also paameteize the same ellipse. ( w ) si ( w ) ( w ) cos( w ) ( w ) si ( w ) x= acos t y = b t x= asi t y = b t x= acos t y =-b t The pesece of the w will chage the speed that the ellipse otates as we saw i Example 5. Note as well that the last two will tace out ellipse with a clockwise diectio of motio (you might wat to veify this). Also ote that they wo t all stat at the same place (if we thik of t = 0 as the statig poit that is). Thee ae may moe paameteizatios of a ellipse of couse, but you get the idea. It is impotat to emembe that each paameteizatio will tace out the cuve oce with a potetially diffeet age of t s. Each paameteizatio may otate diffeet diectios of motio ad may stat at diffeet poits. You may fid that you eed a paameteizatio of a ellipse that stats at a paticula place ad has a paticula diectio of motio ad so you ow kow that with some wok you ca wite dow a set of paametic equatios that will give you the behavio that you e afte. Now, let s wite dow a couple of othe impotat paameteizatios ad all the commets about diectio of motio, statig poit, ad age of t s fo oe tace (if applicable) ae still tue. Fist, because a cicle is othig moe tha a special case of a ellipse we ca use the paameteizatio of a ellipse to get the paametic equatios fo a cicle ceteed at the oigi of adius as well. Oe possible way to paameteize a cicle is, x= cost y = si t Fially, eve though thee may ot seem to be ay easo to, we ca also paameteize fuctios y f x x= h y. I these cases we paameteize them i the followig way, i the fom = o x= t x= ht () y = f t y = t At this poit it may ot seem all that useful to do a paameteizatio of a fuctio like this, but thee ae may istaces whee it will actually be easie, o it may eve be equied, to wok with the paameteizatio istead of the fuctio itself. Ufotuately, almost all of these istaces occu i a Calculus III couse. 007 Paul Dawkis 0

130 Tagets with Paametic Equatios I this sectio we wat to fid the taget lies to the paametic equatios give by, x= f t y = g t To do this let s fist ecall 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 =, whee = at x= a. Hee the taget Now, otice that if we could figue out how to get the deivative dy fom the paametic dx equatios we could simply euse this fomula sice we will be able to use the paametic equatios to fid the x ad y coodiates of the poit. So, just fo a secod let s suppose that we wee able to elimiate the paamete fom the y = F x. Now, plug the paametic fom ad wite the paametic equatios i the fom paametic equatios i fo x ad y. Yes, it seem silly to elimiate the paamete, the immediately put it back i, but it s what we eed to do i ode to get ou hads o the deivative. Doig this gives, g t = F f t Now, diffeetiate with espect to t ad otice that we ll eed to use the Chai Rule o the ight had side. g t = F f t f t Let s do aothe chage i otatio. We eed to be caeful with ou deivatives hee. Deivatives of the lowe case fuctio ae with espect to t while deivatives of uppe case fuctios ae with espect to x. So, to make sue that we keep this staight let s ewite thigs as follows. dy dx = F ( x) dt dt At this poit we should emid ouselves just what we ae afte. We eeded a fomula fo dy dx o F x that is i tems of the paametic fomulas. Notice howeve that we ca get that fom the above equatio. 007 Paul Dawkis x a dy dy dx = dt, povided 0 dx dx dt dt Notice as well that this will be a fuctio of t ad ot x.

131 As a aside, otice that we could also get the followig fomula with a simila deivatio if we eeded to, Deivative fo Paametic E quatios dx dx dy = dt, povided 0 dy dy dt dt Why would we wat to do this? Well, ecall that i the ac legth sectio of the Applicatios of Itegal sectio we actually eeded this deivative o occasio. So, let s fid a taget lie. Example Fid the taget lie(s) to the paametic cuve give by 5 x= t - 4t y = t at (0,4). Solutio Note that thee is appaetly the potetial fo moe tha oe taget lie hee! We will look ito this moe afte we e doe with the example. The fist thig that we should do is fid the deivative 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 poblem. The deivative is i tems of t ad all we ve got is a x-y coodiate pai. The ext step the is to detemie that value(s) of t which will give this poit. We fid these by pluggig the x ad y values ito the paametic equatios ad solvig fo t. 5 0= t - 4t = t t -4 fi t = 0, ± 4= t fi t =± Ay value of t which appeas i both lists will give the poit. So, sice thee ae 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 e goig to eed to look ito this a little moe. Howeve, befoe we do that let s actually get the taget lies. t = Sice we aleady kow the x ad y-coodiates of the poit all that we eed to do is fid the slope of the taget lie. dy m = =- dx t = Paul Dawkis

132 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 whee we fist a acoss taget lies. A quick gaph of the paametic cuve will explai what is goig o hee. So, the paametic cuve cosses itself! That explais how thee ca be moe tha oe taget lie. Thee is oe taget lie fo each istace that the cuve goes though the poit. The ext topic that we eed to discuss i this sectio is that of hoizotal ad vetical tagets. We ca easily idetify whee these will occu (o at least the t s that will give them) by lookig at the deivative fomula. dy dy = dt dx dx dt Hoizotal tagets will occu whee the deivative is zeo ad that meas that we ll get hoizotal taget at values of t fo which we have, 007 Paul Dawkis

133 Hoizotal Taget fo Paametic Equatios dy dx = 0, povided 0 dt dt Vetical tagets will occu whee the deivative is ot defied ad so we ll get vetical tagets at values of t fo which we have, Vetical Taget fo Paametic Equatios dx dy = 0, povided 0 dt dt Let s take a quick look at a example of this. Example Detemie the x-y coodiates of the poits whee the followig paametic equatios will have hoizotal o vetical tagets. x= t - t y = t - 9 Solutio We ll fist eed the deivatives of the paametic equatios. dx dy = t - = ( t - ) = 6t dt dt Hoizotal Tagets We ll have hoizotal tagets whee, 6t = 0 fi t = 0 Now, this is the value of t which gives the hoizotal tagets ad we wee asked to fid the x-y coodiates of the poit. To get these we just eed to plug t ito the paametic equatios. Theefoe, the oly hoizotal taget will occu at the poit (0,-9). Vetical Tagets I this case we eed to solve, t - = 0 fi t =± The two vetical tagets will occu at the poits (,-6) ad (-,-6). Fo the sake of completeess ad at least patial veificatio hee is the sketch of the paametic cuve. 007 Paul Dawkis 4

134 The fial topic that we eed to discuss i this sectio eally is t elated to taget lies, but does fit i icely with the deivatio of the deivative that we eeded to get the slope of the taget lie. Befoe movig ito the ew topic let s fist emid ouselves of the fomula fo the fist deivative ad i the pocess ewite it slightly. d dy d = ( y) = dt dx dx dx dt Witte i this way we ca see that the fomula actually tells us how to diffeetiate a fuctio y (as a fuctio of t) with espect to x (whe x is also a fuctio of t) whe we ae usig paametic equatios. Now let s move oto the fial topic of this sectio. We would also like to kow how to get the secod deivative of y with espect to x. d y dx ( y ) Gettig a fomula fo this is faily simple if we emembe the ewitte fomula fo the fist deivative above. Secod Deivative fo Paametic Equatios d Êdy ˆ d y d dy Á Ê ˆ dt dx = Ë Á = dx dxëdx dx dt 007 Paul Dawkis 5

135 Note that, d y dx d y dt d x dt Let s wok a quick example. Example Fid the secod deivative fo the followig set of paametic equatios. 5 x= t - 4t y = t Solutio This is the set of paametic equatios that we used i the fist example ad so we aleady have the followig computatios completed. dy dx 4 dy = t = 5t - t = dt dt dx 5t -t We will fist eed the followig, d Ê Á ˆ= = dtë t - t t - t t - t 5 ( t ) t The secod deivative is the, d y dt dx dx d Êdy ˆ Á = Ë 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 deivative? Well, ecall fom you Calculus I class that with the secod deivative we ca detemie whee a cuve is cocave up ad cocave dow. We could do the same thig with paametic equatios if we wated to. 007 Paul Dawkis 6

136 Example 4 Detemie the values of t fo which the paametic cuve give by the followig set of paametic equatios is cocave up ad cocave dow. 7 5 x= - t y = t + t Solutio To compute the secod deivative we ll fist 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 fist deivative above to get some ifomatio about the iceasig/deceasig atue of the cuve as well. I this case it looks like the paametic cuve will be iceasig if t < 0 ad deceasig if t > 0. Now let s move o to the secod deivative. 4 - ( 5 t + 5 t d y ) = = 5t + 5t dx -t 4 It s clea, hopefully, that the secod deivative will oly be zeo at t = 0. Usig this we ca see that the secod deivative will be egative if t < 0 ad positive if t > 0. So the paametic cuve will be cocave dow fo t < 0 ad cocave up fo t > 0. Hee is a sketch of the cuve fo completeess sake. 007 Paul Dawkis 7

137 Aea with Paametic Equatios I this sectio we will fid a fomula fo detemiig the aea ude a paametic cuve give by the paametic equatios, x= f ( t) y = g( t) We will also eed to futhe add i the assumptio that the cuve is taced out exactly oce as t iceases fom a to b. We will do this i much the same way that we foud the fist deivative i the pevious sectio. y = F x o a x b. We will fist ecall how to fid the aea ude A= Ú b a F x dx We will ow thik of the paametic equatio x f ( t) also assume that a = f ( a ) ad b f ( b ) = as a substitutio i the itegal. We will = fo the puposes of this fomula. Thee is actually o easo to assume that this will always be the case ad so we ll give a coespodig fomula late b f a a = f b ). 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 aea fomula above ad makig sue to chage the limits to thei coespodig t values gives us, b A= Ú 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 aive at the fomula that we wat. a Aea Ude Paametic Cuve, Fomula I b A= Ú g() t f () t dt Now, if we should happe to have b= f ( a ) ad a f ( b ) Aea Ude Paametic Cuve, Fomula II a A= Ú g() t f () t dt Let s wok a example. a b = the fomula would be, 007 Paul Dawkis 8

138 Example Detemie the aea ude the paametic cuve give by the followig paametic equatios. x= 6 q - siq y = 6- cosq 0 q p Solutio Fist, otice that we ve switched the paamete to q fo this poblem. This is to make sue that we do t get too locked ito always havig t as the paamete. Now, we could gaph this to veify that the cuve is taced out exactly oce fo the give age if we wated to. We ae goig to be lookig at this cuve i moe detail afte this example so we wo t sketch its gaph hee. Thee eally is t too much to this example othe tha pluggig the paametic equatios ito the fomula. We ll fist eed the deivative of the paametic equatio fo x howeve. dx 6 ( cosq ) dq = - The aea is the, Ú ( q) p A= 6 -cos 0 Ú p = 6 - cos + cos 0 0 q q q p = 6 Û Ù - cosq + cos( q) dq ı = 08p dq Ê ˆ = 6Á q - siq + si( q) Ë 4 d p 0 The paametic cuve (without the limits) we used i the pevious example is called a cycloid. I its geeal fom the cycloid is, ( q siq) ( cosq) x= - y = - The cycloid epesets the followig situatio. Coside a wheel of adius. Let the poit whee the wheel touches the goud iitially be called P. The stat ollig the wheel to the ight. As the wheel olls to the ight tace out the path of the poit P. The path that the poit P taces out is called a cycloid ad is give by the equatios above. I these equatios we ca thik of q as the agle though which the poit P has otated. Hee is a cycloid sketched out with the wheel show at vaious places. The blue dot is the poit P o the wheel that we e usig to tace out the cuve. 007 Paul Dawkis 9

139 Fom this sketch we ca see that oe ach of the cycloid is taced out i the age 0 q p. This makes sese whe you coside that the poit P will be back o the goud afte it has otated though ad agle of p. 007 Paul Dawkis 0

140 Ac Legth with Paametic Equatios I the pevious two sectios we ve looked at a couple of Calculus I topics i tems of paametic equatios. We ow eed to look at a couple of Calculus II topics i tems of paametic equatios. I this sectio we will look at the ac legth of the paametic cuve give by, x= f t y = g t a t b We will also be assumig that the cuve is taced out exactly oce as t iceases fom a to b. We will also eed to assume that the cuve is taced out fom left to ight as t iceases. This is equivalet to sayig, dx 0 fo a t b dt So, let s stat out the deivatio by ecallig the ac legth fomula as we fist deived it i the ac legth sectio of the Applicatios of Itegals chapte. L= Ú ds whee, Êdy ˆ ds = + Á dx if y = f ( x), a x b Ëdx Êdx ˆ ds = + Á dy if x= h( y), c y d Ëdy We will use the fist ds above because we have a ice fomula fo the deivative i tems of the paametic equatios (see the Tagets with Paametic Equatios sectio). To use this we ll also eed to kow that, dx dx= f () t dt = dt dt The ac legth fomula the becomes, b Û dy Û Ê ˆ Êdy ˆ Ù Á dx dt dx L dt Ù Á = dt Ë Ù + Á = + dt dx Ù Á dt Ù Êdx ˆ dt Ù dt Á ıa Ë Ù ı Ë dt This is a paticulaly upleasat fomula. Howeve, if we facto out the deomiato fom the squae oot we aive at, b a 007 Paul Dawkis

141 b Û Êdxˆ Êdyˆ dx L= Ù + dt dx Á Á Ù Ë dt Ë dt dt ıa dt Now, makig use of ou assumptio that the cuve is beig taced out fom left to ight we ca dop the absolute value bas o the deivative which will allow us to cacel the two deivatives that ae outside the squae oot this gives, Ac Legth fo Paametic Equatios b Û Êdxˆ Êdy ˆ L= Ù Á + Á ı Ë dt Ë dt a dt Notice that we could have used the secod fomula fo ds above is we had assumed istead that dy 0 fo a t b dt If we had goe this oute i the deivatio we would have gotte the same fomula. Let s take a look at a example. Example Detemie the legth of the paametic cuve give by the followig paametic equatios. x= si t y = cos t 0 t p Solutio We kow that this is a cicle of adius ceteed at the oigi fom ou pio discussio about gaphig paametic cuves. We also kow fom this discussio that it will be taced out exactly oce i this age. So, we ca use the fomula we deived above. We ll fist eed the followig, dx dy = cos() t =- si () t dt dt The legth is the, () 9cos () p L= 9si t + t dt 0 Ú Ú p 0 p 0 () () = si t + cos t dt = Ú = 6p dt Sice this is a cicle we could have just used the fact that the legth of the cicle is just the cicumfeece of the cicle. This is a ice way, i this case, to veify ou esult. 007 Paul Dawkis

142 Let s take a look at oe possible cosequece of a cuve is taced out moe tha oce ad we ty to fid the legth of the cuve without takig this ito accout. Example Use the ac legth fomula fo the followig paametic equatios. x= si t y = cos t 0 t p Solutio Notice that this is the idetical cicle that we had i the pevious example ad so the legth is still 6p. Howeve, fo the age give we kow tace out the cuve thee times istead oce as equied fo the fomula. Despite that estictio let s use the fomula ayway ad see what happes. I this case the deivatives ae, dx dy = 9cos t =- 9si t dt dt ad the legth fomula gives, () 8cos () p L= Ú 8si t + t dt 0 = Ú p 0 = 8p 9 dt The aswe we got fom the ac legth fomula i this example was times the actual legth. Recallig that we also detemied that this cicle would tace out thee times i the age give, the aswe should make some sese. If we had wated to detemie the legth of the cicle fo this set of paametic equatios we would eed to detemie a age of t fo which this cicle is taced out exactly oce. This is, 0 t p. Usig this age of t we get the followig fo the legth. which is the coect aswe. p L= Ú 8si () t + 8cos () t dt 0 = Ú p 0 = 6p 9 dt Be caeful to ot make the assumptio that this is always what will happe if the cuve is taced out moe tha oce. Just because the cuve taces out times does ot mea that the ac legth fomula will give us times the actual legth of the cuve! Befoe movig o to the ext sectio let s otice that we ca put the ac legth fomula deived i this sectio ito the same fom that we had whe we fist looked at ac legth. The oly diffeece is that we will add i a defiitio fo ds whe we have paametic equatios. The ac legth fomula ca be summaized as, L = Ú ds 007 Paul Dawkis

143 whee, Êdy ˆ ds = + Á dx if y = f ( x), a x b Ëdx Êdx ˆ ds = + Á dy if x= h( y), c y d Ëdy Êdxˆ Êdy ˆ ds = Á + Á dt if x= f () t, y = g() t, a t b Ë dt Ë dt 007 Paul Dawkis 4

144 Suface Aea with Paametic Equatios I this fial sectio of lookig with calculus applicatio with paametic equatios we will take a look at detemiig the suface aea of a egio obtaied by otatig a paametic cuve about the x o y-axis. We will otate the paametic cuve give by, x= f t y = g t a t b about the x o y-axis. We ae goig to assume that the cuve is taced out exactly oce as t iceases fom a to b. At this poit thee actually is t all that much to do. We kow that the suface aea ca be foud by usig oe of the followig two fomulas depedig o the axis of otatio (ecall the Suface Aea sectio of the Applicatios of Itegals chapte). S = pyds otatio about x-axis Ú Ú S = pxds otatio about y-axis All that we eed is a fomula fo ds to use ad fom the pevious sectio we have, Êdxˆ Êdy ˆ ds = Á + Á dt if x= f () t, y = g() t, a t b Ë dt Ë dt which is exactly what we eed. We will eed to be caeful with the x o y that is i the oigial suface aea fomula. Back whe we fist looked at suface aea we saw that sometimes we had to substitute fo the vaiable i the itegal ad at othe times we did t. This was depedet upo the ds that we used. I this case howeve, we will always have to substitute fo the vaiable. The ds that we use fo paametic equatios itoduces a dt ito the itegal ad that meas that eveythig eeds to be i tems of t. Theefoe, we will eed to substitute the appopiate paametic equatio fo x o y depedig o the axis of otatio. Let s take a quick look at a example. Example Detemie the suface aea of the solid obtaied by otatig the followig paametic cuve about the x-axis. p x= cos q y = si q 0 q Solutio We ll fist eed the deivatives of the paametic equatios. dx dy =- cos qsiq = si qcosq dt dt Befoe pluggig ito the suface aea fomula let s get the ds out of the way. 007 Paul Dawkis 5

145 4 4 ds = 9cos qsi q + 9si qcos q dt q q q q = cos si cos + si = cosqsiq Notice that we could dop the absolute value bas sice both sie ad cosie ae positive i this age of q give. Now let s get the suface aea ad do t foget to also plug i fo the y. S = = p yds p Ú si 0 cos si p 6 4 Ú si 0 cos d u si = 6p p q q q dq = p q q q = q = Ú 6p 5 Ú 4 u du Paul Dawkis 6

146 Pola Coodiates Up to this poit we ve dealt exclusively with the Catesia (o Rectagula, o x-y) coodiate system. Howeve, as we will see, this is ot always the easiest coodiate system to wok i. So, i this sectio we will stat lookig at the pola coodiate system. Coodiate systems ae eally othig moe tha a way to defie a poit i space. Fo istace i the Catesia coodiate system at poit is give the coodiates (x,y) ad we use this to defie the poit by statig at the oigi ad the movig x uits hoizotally followed by y uits vetically. This is show i the sketch below. This is ot, howeve, the oly way to defie a poit i two dimesioal space. Istead of movig vetically ad hoizotally fom the oigi to get to the poit we could istead go staight out of the oigi util we hit the poit ad the detemie the agle this lie makes with the positive x- axis. We could the use the distace of the poit fom the oigi ad the amout we eeded to otate fom the positive x-axis as the coodiates of the poit. This is show i the sketch below. Coodiates i this fom ae called pola coodiates. The above discussio may lead oe to thik that must be a positive umbe. Howeve, we also -, p. allow to be egative. Below is a sketch of the two poits 6, p ad Paul Dawkis 7

147 Fom this sketch we ca see that if is positive the poit will be i the same quadat as q. O the othe had if is egative the poit will ed up i the quadat exactly opposite q. Notice as, p 7, p do. The well that the coodiates (- 6 ) descibe the same poit as the coodiates ( 6 ) 7 coodiates, p 6 tells us to otate a agle of 7 6 p fom 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 impotat diffeece betwee Catesia coodiates ad pola coodiates. I Catesia coodiates thee is exactly oe set of coodiates fo ay give poit. With pola coodiates this is t tue. I pola coodiates thee is liteally a ifiite umbe of coodiates fo a give poit. Fo istace, the followig fou poits ae all coodiates fo the same poit. Ê 5 4 5, p ˆ Ê 5, p ˆ Ê 5, p ˆ Ê 5, p ˆ Á = Á - = Á- = Á- - Ë Ë Ë Ë Hee is a sketch of the agles used i these fou sets of coodiates. I the secod coodiate pai we otated i a clock-wise diectio to get to the poit. We should t foget about otatig i the clock-wise diectio. Sometimes it s what we have to do. 007 Paul Dawkis 8

148 The last two coodiate pais use the fact that if we ed up i the opposite quadat fom the poit we ca use a egative to get back to the poit ad of couse thee is both a coute clock-wise ad a clock-wise otatio to get to the agle. These fou poits oly epeset the coodiates of the poit without otatig aoud the system moe tha oce. If we allow the agle to make as may complete otatio about the axis system as we wat the thee ae a ifiite umbe of coodiates fo the same poit. I fact the poit, q ca be epeseted by ay of the followig coodiate pais. ( + ) - + ( + ), q p, q p, whee is ay itege. Next we should talk about the oigi of the coodiate system. I pola coodiates the oigi is ofte called the pole. Because we ae t actually movig away fom the oigi/pole we kow that = 0. Howeve, we ca still otate aoud the system by ay agle we wat ad so the 0,q. coodiates of the oigi/pole ae Now that we ve got a gasp o pola coodiates we eed to thik about covetig betwee the two coodiate systems. Well stat out with the followig sketch emidig us how both coodiate systems wok. Note that we ve got a ight tiagle above ad with that we ca get the followig equatios that will covet pola coodiates ito Catesia coodiates. Pola to Catesia Covesio Fomulas x= cosq y = siq Covetig fom Catesia is almost as easy. Let s fist otice the followig. ( cosq) ( siq) x + y = + = cos q + si q ( cos q si q) = + = This is a vey useful fomula that we should emembe, howeve we ae afte a equatio fo so let s take the squae oot of both sides. This gives, = x + y 007 Paul Dawkis 9

149 Note that techically we should have a plus o mius i fot of the oot sice we kow that ca be eithe positive o egative. We will u with the covetio of positive hee. Gettig a equatio fo q is almost as simple. We ll stat with, y siq = = taq x cosq Takig the ivese taget of both sides gives, - Ê y ˆ q = ta Á Ë x We will eed to be caeful with this because ivese tagets oly etu values i the age - p p < q <. Recall that thee is a secod possible agle ad that the secod agle is give by q + p. Summaizig the gives the followig fomulas fo covetig fom Catesia coodiates to pola coodiates. Catesia to Pola Covesio Fomulas Let s wok a quick example. = x + y = x + y q ta Ê y ˆ Ë x - = Á Example Covet each of the followig poits ito the give coodiate system. Ê p ˆ (a) Á-4, ito Catesia coodiates. [Solutio] Ë (b) (-,-) ito pola coodiates. [Solutio] Solutio (a) Covet Ê p ˆ Á-4, Ë ito Catesia coodiates. This covesio is easy eough. All we eed to do is plug the poits ito the fomulas. Êp ˆ Ê ˆ x=- 4cosÁ =-4Á- = Ë Ë Êp ˆ Ê ˆ y =- 4siÁ =- 4 =- Ë Á Ë So, i Catesia coodiates this poit is (, - ). [Retu to Poblems] 007 Paul Dawkis 40

150 (b) Covet (-,-) ito pola coodiates. Let s fist get. = = Now, let s get q. -Ê-ˆ - p q = ta Á = ta () = Ë- 4 This is ot the coect agle howeve. This value of q is i the fist quadat ad the poit we ve bee give is i the thid quadat. As oted above we ca get the coect agle by addig p oto this. Theefoe, the actual agle is, p 5p q = + p = So, i pola coodiates the poit is ( 4 ), p. Note as well that we could have used the fist q that we got by usig a egative. I this case the poit could also be witte i pola coodiates as (, p 4 ) -. [Retu to Poblems] We ca also use the above fomulas to covet equatios fom oe coodiate system to the othe. Example Covet each of the followig ito a equatio i the give coodiate system. (a) Covet x- 5x = + xy ito pola coodiates. [Solutio] (b) Covet =- 8cosq ito Catesia coodiates. [Solutio] Solutio (a) Covet x- 5x = + xy ito pola coodiates. I this case thee eally is t much to do othe tha pluggig i the fomulas fo x ad y (i.e. the Catesia coodiates) i tems of ad q (i.e. the pola coodiates). ( q) - ( q) = + ( q)( q) cos 5 cos cos si cos - 5 cos = + cos si q q q q (b) Covet =- 8cosq ito Catesia coodiates. [Retu to Poblems] This oe is a little tickie, but ot by much. Fist otice that we could substitute staight fo the. Howeve, thee is o staight substitutio fo the cosie that will give us oly Catesia coodiates. If we had a o the ight alog with the cosie the we could do a diect substitutio. So, if a o the ight side would be coveiet let s put oe thee, just do t foget to put o the left side as well. =- 8cosq We ca ow make some substitutios that will covet this ito Catesia coodiates. x + y =- 8x [Retu to Poblems] 007 Paul Dawkis 4

151 Befoe movig o to the ext subject let s do a little moe wok o the secod pat of the pevious example. The equatio give i the secod pat is actually a faily well kow gaph; it just is t i a fom that most people will quickly ecogize. To idetify it let s take the Catesia coodiate equatio ad do a little eaagig. x + 8x+ y = 0 Now, complete the squae o the x potio of the equatio. x + 8x+ 6+ y = 6 ( x ) So, this was a cicle of adius 4 ad cete (-4,0). This leads us ito the fial topic of this sectio y = 6 Commo Pola Coodiate Gaphs Let s idetify a few of the moe commo gaphs i pola coodiates. We ll also take a look at a couple of special pola gaphs. Lies Some lies have faily simple equatios i pola coodiates.. q = b. We ca see that this is a lie by covetig to Catesia coodiates as follows q = b ta - Ê y Á ˆ= b Ë x y x y = = ta b ( ta b) This is a lie that goes though the oigi ad makes a agle of b with the positive x- axis. O, i othe wods it is a lie though the oigi with slope of ta b.. cosq = a This is easy eough to covet to Catesia coodiates to x= a. So, this is a vetical lie.. siq = b Likewise, this covets to y = b ad so is a hoizotal lie. x 007 Paul Dawkis 4

152 p Example Gaph q =, cosq = 4 ad siq =- o the same axis system. 4 Solutio Thee eally is t too much to this oe othe tha doig the gaph so hee it is. Cicles Let s take a look at the equatios of cicles i pola coodiates.. = a. This equatio is sayig that o matte what agle we ve got the distace fom the oigi must be a. If you thik about it that is exactly the defiitio of a cicle of adius a ceteed at the oigi. So, this is a cicle of adius a ceteed at the oigi. This is also oe of the easos why we might wat to wok i pola coodiates. The equatio of a cicle ceteed at the oigi has a vey ice equatio, ulike the coespodig equatio i Catesia coodiates.. = acosq. We looked at a specific example of oe of these whe we wee covetig equatios to Catesia coodiates. This is a cicle of adius a ad cete ( a,0). Note that a might be egative (as it was i ou example above) ad so the absolute value bas ae equied o the adius. They should ot be used howeve o the cete.. = bsiq. 0,b. This is simila to the pevious oe. It is a cicle of adius b ad cete 4. = acosq + bsiq. This is a combiatio of the pevious two ad by completig the squae twice it ca be show that this is a cicle of adius a + b ad cete ( ab, ). I othe wods, this is the geeal equatio of a cicle that is t ceteed at the oigi. 007 Paul Dawkis 4

153 Example 4 Gaph = 7, = 4cosq, ad =- 7siq o the same axis system. Solutio The fist oe is a cicle of adius 7 ceteed at the oigi. The secod is a cicle of adius ceteed at (,0). The thid is a cicle of adius 7 ceteed at Ê 7 ˆ Á0, - Hee is the gaph of the Ë. thee equatios. Note that it takes a age of 0 q p fo a complete gaph of = a ad it oly takes a age of 0 q p to gaph the othe cicles give hee. Cadioids ad Limacos These ca be boke up ito the followig thee cases.. Cadioids : = a± acosq ad = a± asiq. These have a gaph that is vaguely heat shaped ad always cotai the oigi.. Limacos with a ie loop : = a± bcosq ad = a± bsiq with a< b. These will have a ie loop ad will always cotai the oigi.. Limacos without a ie loop : = a± bcosq ad = a± bsiq with a > b. These do ot have a ie loop ad do ot cotai the oigi. Example 5 Gaph = 5-5siq, = 7-6cosq, ad = + 4cosq. Solutio These will all gaph out oce i the age 0 q p. Hee is a table of values fo each followed by gaphs of each. 007 Paul Dawkis 44

154 q = 5-5siq = 7-6cosq = + 4cosq p 0 7 p 5 - p 0 7 p Paul Dawkis 45

155 Thee is oe fial thig that we eed to do i this sectio. I the thid gaph i the pevious example we had a ie loop. We will, o occasio, eed to kow the value of q fo which the gaph will pass though the oigi. To fid these all we eed to do is set the equatio equal to zeo ad solve as follows, p 4p 0= + 4cosq fi cos q =- fi q =, 007 Paul Dawkis 46

156 Tagets with Pola Coodiates We ow eed to discuss some calculus topics i tems of pola coodiates. We will stat with fidig taget lies to pola cuves. I this case we ae goig to assume that = f q. With the equatio i this fom we ca actually use the the equatio is i the fom equatio fo the deivative dy we deived whe we looked at taget lies with paametic dx equatios. To do this howeve equies us to come up with a set of paametic equatios to epeset the cuve. This is actually petty easy to do. Fom ou wok i the pevious sectio we have the followig set of covesio equatios fo goig fom pola coodiates to Catesia coodiates. x= cosq y = siq Now, we ll use the fact that we e assumig that the equatio is i the fom f ( q ) =. Substitutig this ito these equatios gives the followig set of paametic equatios (with q as the paamete) fo the cuve. cos x= f q q y = f q siq Now, we will eed the followig deivatives. dx dy = f ( q) cosq - f ( q) siq = f ( q) siq + f ( q) cosq dq dq d d = cosq - siq = siq + cosq dq dq The deivative dy dx is the, Deivative with Pola Coodiates dy = dx d siq + cosq dq d cosq - siq dq Note that athe tha tyig to emembe this fomula it would pobably be easie to emembe how we deived it ad just emembe the fomula fo paametic equatios. Let s wok a quick example with this. 007 Paul Dawkis 47

157 Example Detemie the equatio of the taget lie to = + 8siq at Solutio We ll fist eed the followig deivative. The fomula fo the deivative dy dx becomes, dy dx d 8cosq dq = p q =. 6 8cosqsiq + + 8siq cosq 6cosqsiq + cosq = = 8cos q - + 8siq siq 8cos q -siq -8si q The slope of the taget lie is, 4 + dy m = = = dx p q = Now, at q = p 6 we have = 7. We ll eed to get the coespodig x-y coodiates so we ca get the taget lie. Êp ˆ 7 Êp ˆ 7 x= 7cosÁ = y = 7si Á = Ë 6 Ë 6 The taget lie is the, 7 Ê 7 ˆ y = + x- 5 Á Ë Fo the sake of completeess hee is a gaph of the cuve ad the taget lie. 007 Paul Dawkis 48

158 Aea with Pola Coodiates I this sectio we ae goig to look at aeas eclosed by pola cuves. Note as well that we said eclosed by istead of ude as we typically have i these poblems. These poblems wok a little diffeetly i pola coodiates. Hee is a sketch of what the aea that we ll be fidig i this sectio looks like. We ll be lookig fo the shaded aea i the sketch above. The fomula fo fidig this aea is, A= Û Ù ı b a d Notice that we use i the itegal istead of f ( q ) so make sue ad substitute accodigly whe doig the itegal. Let s take a look at a example. Example Detemie the aea of the ie loop of = + 4cosq. Solutio We gaphed this fuctio back whe we fist stated lookig at pola coodiates. Fo this poblem we ll also eed to kow the values of q whee the cuve goes though the oigi. We ca get these by settig the equatio equal to zeo ad solvig. 0= + 4cosq p 4p cos q =- fi q =, Hee is the sketch of this cuve with the ie loop shaded i. q 007 Paul Dawkis 49

159 Ca you see why we eeded to kow the values of q whee the cuve goes though the oigi? These poits defie whee the ie loop stats ad eds ad hece ae also the limits of itegatio i the fomula. So, the aea is the, 4p A= Û Ù + 4cos ı p ( q) 4p = Û Ù ( 4 + 6cos q + 6cos q) dq ı Ú p 4p p 4p p ( ) = + 8cosq + 4+ cos q dq Ú = 6+ 8cosq + 4cos q dq ( 6q 8siq si( q) ) = + + = 4p - 6 =.74 dq 4p p You did follow the wok doe i this itegal did t you? You ll u ito quite a few itegals of tig fuctios i this sectio so if you eed to you should go back to the Itegals Ivolvig Tig Fuctios sectios ad do a quick eview. So, that s how we detemie aeas that ae eclosed by a sigle cuve, but what about situatios like the followig sketch wee we wat to fid the aea betwee two cuves. 007 Paul Dawkis 50

160 I this case we ca use the above fomula to fid the aea eclosed by both ad the the actual aea is the diffeece betwee the two. The fomula fo this is, Let s take a look at a example of this. b A= Û Ù ( - ) dq ı o i a Example Detemie the aea that lies iside = + siq ad outside =. Solutio Hee is a sketch of the egio that we ae afte. To detemie this aea we ll eed to kow that value of q fo which the two cuves itesect. We ca detemie these poits by settig the two equatios ad solvig. 007 Paul Dawkis 5

161 + siq = 7p p si q =- fi q =, 6 6 Hee is a sketch of the figue with these agles added. Note as well hee that we also ackowledged that aothe epesetatio fo the agle 6p is p 6 -. This is impotat fo this poblem. I ode to use the fomula above the aea must be eclosed as we icease fom the smalle to lage agle. So, if we use 7 p 6 to p 6 we will ot eclose the shaded aea, istead we will eclose the bottom most of the thee egios. Howeve if we use the agles - p 6 to 7 p 6 we will eclose the aea that we e afte. So, the aea is the, 7p 6 p - 6 (( q) ) A= Û Ù + si - ı 7p 6 = Û Ù ( 5 + si q + 4si q) dq ı p - 6 7p 6 = Û Ù ( 7 + si q - cos ( q) ) dq ı p - 6 = 7 - cos - si ( q q ( q) ) 4p = + = 4.87 dq 7p 6 p - 6 Let s wok a slight modificatio of the pevious example. 007 Paul Dawkis 5

162 Example Detemie the aea of the egio outside = + siq ad iside =. Solutio This time we e lookig fo the followig egio. So, this is the egio that we get by usig the limits 7 p 6 to p 6. The aea fo this egio is, p 6 7p 6 ( ( q) ) A= Û Ù - + si ı p 6 7p 6 p 6 7p 6 ( 5 si 4si ) = Û Ù - - q - q dq ı = Û Ù (-7- siq + cos( q) ) dq ı = ( 7q cosq si( q) ) 7p = - =.96 dq Notice that fo this aea the oute ad ie fuctio wee opposite! Let s do oe fial modificatio of this example. Example 4 Detemie the aea that is iside both = + siq ad =. Solutio Hee is the sketch fo this example. p 6 7p Paul Dawkis 5

163 We ae ot goig to be able to do this poblem i the same fashio that we did the pevious two. Thee is o set of limits that will allow us to eclose this aea as we icease fom oe to the othe. Remembe that as we icease q the aea we e afte must be eclosed. Howeve, the oly two ages fo q that we ca wok with eclose the aea fom the pevious two examples ad ot this egio. I this case howeve, that is ot a majo poblem. Thee ae two ways to do get the aea i this poblem. We ll take a look at both of them. Solutio I this case let s otice that the cicle is divided up ito two potios ad we e afte the uppe potio. Also otice that we foud the aea of the lowe potio i Example. Theefoe, the aea is, Aea = Aea of Cicle-Aea fom Example = p -.96 = 0.70 Solutio I this case we do petty much the same thig except this time we ll thik of the aea as the othe potio of the limaco tha the potio that we wee dealig with i Example. We ll also eed to actually compute the aea of the limaco i this case. So, the aea usig this appoach is the, 007 Paul Dawkis 54

164 Aea = Aea of Limaco-Aea fom Example p si 4.87 = Û Ù ( + q) dq - ı 0 p = Û Ù ( 9 + si q + 4si q) dq ı 0 p = Û Ù ( + si q - cos ( q) ) dq ı 0 p = ( q - cos ( q) - si ( q) ) = p = 0.70 A slightly loge appoach, but sometimes we ae foced to take this loge appoach. As this last example has show we will ot be able to get all aeas i pola coodiates staight fom a itegal Paul Dawkis 55

165 Ac Legth with Pola Coodiates We ow eed to move ito the Calculus II applicatios of itegals ad how we do them i tems of pola coodiates. I this sectio we ll look at the ac legth of the cuve give by, = f q a q b whee we also assume that the cuve is taced out exactly oce. Just as we did with the taget lies i pola coodiates we ll fist wite the cuve i tems of a set of paametic equatios, x= cosq y = siq cos f = f q q = q siq ad we ca ow use the paametic fomula fo fidig the ac legth. We ll eed the followig deivatives fo these computatios. dx dy = f ( q) cosq - f ( q) siq = f ( q) siq + f ( q) cosq dq dq d d = cosq - siq = siq + cosq dq dq We ll eed the followig fo ou ds. Ê dx ˆ Ê dy ˆ Ê d ˆ Ê d ˆ Á + Á = Á cosq - siq + Á siq + cosq Ëdq Ëdq Ëdq Ëdq Ê d ˆ d = Á cos q - cosqsiq + si q Ëdq dq Ê d ˆ = + Á Ê d ˆ d + Á si q + cosqsiq + cos q Ëdq dq Ê d ˆ = Á Ëdq ( cos q si q) ( cos q si q) Ëdq The ac legth fomula fo pola coodiates is the, whee, L = Ú ds Ê d ˆ ds = + Á dq Ëdq Let s wok a quick example of this. 007 Paul Dawkis 56

166 Example Detemie the legth of = q 0 q. Solutio Okay, let s just jump staight ito the fomula sice this is a faily simple fuctio. L= q + dq We ll eed to use a tig substitutio hee. q = tax dq = sec xdx q = 0 0= tax x= 0 Ú 0 p q = = ta x x= 4 q + = + = = = ta x sec x secx sec x The ac legth is the, L = q + 0 = Ú Ú p 4 0 sec dq xdx = sec ta + l sec + ta ( x x x x) ( l( ) ) = + + p 4 0 Just as a aside befoe we leave this chapte. The pola equatio = q is the equatio of a spial. Hee is a quick sketch of = q fo 0 q 4p. 007 Paul Dawkis 57

167 Suface Aea with Pola Coodiates We will be lookig at suface aea i pola coodiates i this sectio. Note howeve that all we e goig to do is give the fomulas fo the suface aea sice most of these itegals ted to be faily difficult. We wat to fid the suface aea of the egio foud by otatig, = f q a q b about the x o y-axis. As we did i the taget ad ac legth sectios we ll wite the cuve i tems of a set of paametic equatios. x= cosq y = siq cos f = f q q = q siq If we ow use the paametic fomula fo fidig the suface aea we ll get, whee, Ú Ú S = pyds otatio about x-axis S = pxds otatio about y-axis Ê d ˆ ds d f = + Á q = q, a q b Ëdq Note that because we will pick up a dq fom the ds we ll eed to substitute oe of the paametic equatio i fo x o y depedig o the axis of otatio. This will ofte mea that the itegals will be somewhat upleasat. 007 Paul Dawkis 58

168 Ac Legth ad Suface Aea Revisited We wo t be wokig ay examples i this sectio. This sectio is hee solely fo the pupose of summaizig up all the ac legth ad suface aea poblems. Ove the couse of the last two chaptes the topic of ac legth ad suface aea has aise may times ad each time we got a ew fomula out of the mix. Studets ofte get a little ovewhelmed with all the fomulas. Howeve, thee eally ae t as may fomulas as it might seem at fist glace. Thee is exactly oe ac legth fomula ad exactly two suface aea fomulas. These ae, L= Ú Ú Ú ds S = pyds otatio about x-axis S = pxds otatio about y-axis The poblems aise because we have quite a few ds s that we ca use. Agai studets ofte have touble decidig which oe to use. The examples/poblems usually suggest the coect oe to use howeve. Hee is a complete listig of all the ds s that we ve see ad whe they ae used. Êdy ˆ ds = + Á dx if y = f ( x), a x b Ëdx Êdx ˆ ds = + Á dy if x= h( y), c y d Ëdy Êdxˆ Êdy ˆ ds = Á + Á dt if x= f () t, y = g() t, a t b Ë dt Ë dt Ê d ˆ ds d f = + Á q if = q, a q b Ëdq Depedig o the fom of the fuctio we ca quickly tell which ds to use. Thee is oly oe othe thig to woy about i tems of the suface aea fomula. The ds will itoduce a ew diffeetial to the itegal. Befoe itegatig make sue all the vaiables ae i tems of this ew diffeetial. Fo example if we have paametic equatios well use the thid ds ad the we ll eed to make sue ad substitute fo the x o y depedig o which axis we otate about to get eveythig i tems of t. Likewise, if we have a fuctio i the fom x h( y) = the we ll use the secod ds ad if the otatio is about the y-axis we ll eed to substitute fo the x i the itegal. O the othe had if we otate about the x-axis we wo t eed to do a substitutio fo the y. 007 Paul Dawkis 59

169 Keep these ules i mid ad you ll always be able to detemie which fomula to use ad how to coectly do the itegal. 007 Paul Dawkis 60

170 Sequeces ad Seies Itoductio I this chapte we ll be takig a look at sequeces ad (ifiite) seies. Actually, this chapte will deal almost exclusively with seies. Howeve, we also eed to udestad some of the basics of sequeces i ode to popely deal with seies. We will theefoe, sped a little time o sequeces as well. Seies is oe of those topics that may studets do t fid all that useful. To be hoest, may studets will eve see seies outside of thei calculus class. Howeve, seies do play a impotat ole i the field of odiay diffeetial equatios ad without seies lage potios of the field of patial diffeetial equatios would ot be possible. I othe wods, seies is a impotat topic eve if you wo t eve see ay of the applicatios. Most of the applicatios ae beyod the scope of most Calculus couses ad ted to occu i classes that may studets do t take. So, as you go though this mateial keep i mid that these do have applicatios eve if we wo t eally be coveig may of them i this class. Hee is a list of topics i this chapte. Sequeces We will stat the chapte off with a bief discussio of sequeces. This sectio will focus o the basic temiology ad covegece of sequeces Moe o Sequeces Hee we will take a quick look about mootoic ad bouded sequeces. Seies The Basics I this sectio we will discuss some of the basics of ifiite seies. Seies Covegece/Divegece Most of this chapte will be about the covegece/divegece of a seies so we will give the basic ideas ad defiitios i this sectio. Seies Special Seies We will look at the Geometic Seies, Telescopig Seies, ad Hamoic Seies i this sectio. Itegal Test Usig the Itegal Test to detemie if a seies coveges o diveges. Compaiso Test/Limit Compaiso Test Usig the Compaiso Test ad Limit Compaiso Tests to detemie if a seies coveges o diveges. Alteatig Seies Test Usig the Alteatig Seies Test to detemie if a seies coveges o diveges. Absolute Covegece A bief discussio o absolute covegece ad how it diffes fom covegece. 007 Paul Dawkis 6

171 Ratio Test Usig the Ratio Test to detemie if a seies coveges o diveges. Root Test Usig the Root Test to detemie if a seies coveges o diveges. Stategy fo Seies A set of geeal guidelies to use whe decidig which test to use. Estimatig the Value of a Seies Hee we will look at estimatig the value of a ifiite seies. Powe Seies A itoductio to powe seies ad some of the basic cocepts. Powe Seies ad Fuctios I this sectio we will stat lookig at how to fid a powe seies epesetatio of a fuctio. Taylo Seies Hee we will discuss how to fid the Taylo/Maclaui Seies fo a fuctio. Applicatios of Seies I this sectio we will take a quick look at a couple of applicatios of seies. Biomial Seies A bief look at biomial seies. 007 Paul Dawkis 6

172 Sequeces Let s stat off this sectio with a discussio of just what a sequece is. A sequece is othig moe tha a list of umbes witte i a specific ode. The list may o may ot have a ifiite umbe of tems i them although we will be dealig exclusively with ifiite sequeces i this class. Geeal sequece tems ae deoted as follows, a -fist tem a a a -secod tem - + th M tem ( ) st - + tem M Because we will be dealig with ifiite sequeces each tem i the sequece will be followed by aothe tem as oted above. I the otatio above we eed to be vey caeful with the subscipts. The subscipt of + deotes the ext tem i the sequece ad NOT oe plus the th tem! I othe wods, a a + + so be vey caeful whe witig subscipts to make sue that the + does t migate out of the subscipt! This is a easy mistake to make whe you fist stat dealig with this kid of thig. Thee is a vaiety of ways of deotig a sequece. Each of the followig ae equivalet ways of deotig a sequece. {,, K,,, K } { } { } a a a a a a + = I the secod ad thid otatios above a is usually give by a fomula. A couple of otes ae ow i ode about these otatios. Fist, ote the diffeece betwee the secod ad thid otatios above. If the statig poit is ot impotat o is implied i some way by the poblem it is ofte ot witte dow as we did i the thid otatio. Next, we used a statig poit of = i the thid otatio oly so we could wite oe dow. Thee is absolutely o easo to believe that a sequece will stat at =. A sequece will stat whee eve it eeds to stat. Let s take a look at a couple of sequeces. 007 Paul Dawkis 6

173 Example Wite dow the fist few tems of each of the followig sequeces. Solutio Ï + (a) Ì Ó (b) ÏÔ Ì ÔÓ (-) b = (c) { } = + Ô Ô = 0 [Solutio] [Solutio] th, whee b digit of p = [Solutio] Ï + (a) Ì Ó = To get the fist few sequece tems hee all we eed to do is plug i values of ito the fomula give ad we ll get the sequece tems. Ï Ï+ Ô Ô Ì = Ì, {,,,, K Ó = Ô= { 4 { 9 6 { { 5 Ô Ó = = = 4 = 5 Note the iclusio of the at the ed! This is a impotat piece of otatio as it is the oly thig that tells us that the sequece cotiues o ad does t temiate at the last tem. [Retu to Poblems] + ÏÔ( -) Ô (b) Ì ÔÓ Ô = 0 This oe is simila to the fist oe. The mai diffeece is that this sequece does t stat at =. + ÏÔ( -) Ô Ï Ì,,,,, = Ì- - - K ÔÓ Ô Ó = 0 Note that the tems i this sequece alteate i sigs. Sequeces of this kid ae sometimes called alteatig sequeces. [Retu to Poblems] b = (c) { } th, whee b = digit of p This sequece is diffeet fom the fist two i the sese that it does t have a specific fomula fo each tem. Howeve, it does tell us what each tem should be. Each tem should be the th digit of p. So we kow that p = K The sequece is the, {,,4,,5,9,,6,5,,5,K } [Retu to Poblems] 007 Paul Dawkis 64

174 I the fist two pats of the pevious example ote that we wee eally teatig the fomulas as fuctios that ca oly have iteges plugged ito them. O, + + (-) f = g = This is a impotat idea i the study of sequeces (ad seies). Teatig the sequece tems as fuctio evaluatios will allow us to do may thigs with sequeces that could t do othewise. Befoe delvig futhe ito this idea howeve we eed to get a couple moe ideas out of the way. Fist we wat to thik about gaphig a sequece. To gaph the sequece { a } we plot the poits ( a, ) as ages ove all possible values o a gaph. Fo istace, let s gaph the Ï+ sequece Ì Ó =. The fist few poits o the gaph ae, Ê ˆ Ê 4ˆ Ê 5 ˆ Ê 6 ˆ,, Á,, Á,, Á4,, Á5,, Ë 4 Ë 9 Ë 6 Ë 5 K The gaph, fo the fist 0 tems of the sequece, is the, This gaph leads us to a impotat idea about sequeces. Notice that as iceases the sequece tems fom ou sequece tems, i this case, get close ad close to zeo. We the say that zeo is the limit (o sometimes the limitig value) of the sequece ad wite, + lima = lim = 0 Æ Æ This otatio should look familia to you. It is the same otatio we used whe we talked about the limit of a fuctio. I fact, if you ecall, we said ealie that we could thik of sequeces as fuctios i some way ad so this otatio should t be too supisig. Usig the ideas that we developed fo limits of fuctios we ca wite dow the followig wokig defiitio fo limits of sequeces. 007 Paul Dawkis 65

175 Wokig Defiitio of Limit. We say that lim a = L Æ if we ca make a as close to L as we wat fo all sufficietly lage. I othe wods, the value of the a s appoach L as appoaches ifiity.. We say that lim a = Æ if we ca make a as lage as we wat fo all sufficietly lage. Agai, i othe wods, the value of the a s get lage ad lage without boud as appoaches ifiity.. We say that lim a =- Æ if we ca make a as lage ad egative as we wat fo all sufficietly lage. Agai, i othe wods, the value of the a s ae egative ad get lage ad lage without boud as appoaches ifiity. The wokig defiitios of the vaious sequece limits ae ice i that they help us to visualize what the limit actually is. Just like with limits of fuctios howeve, thee is also a pecise defiitio fo each of these limits. Let s give those befoe poceedig Pecise Defiitio of Limit. We say that lim a = L if fo evey umbe e > 0 thee is a itege N such that Æ. We say that lim a a - L < e wheeve > N Æ. We say that lim a Æ = if fo evey umbe M > 0 thee is a itege N such that a > M wheeve > N =- if fo evey umbe M < 0 thee is a itege N such that a < M wheeve > N We wo t be usig the pecise defiitio ofte, but it will show up occasioally. Note that both defiitios tell us that i ode fo a limit to exist ad have a fiite value all the sequece tems must be gettig close ad close to that fiite value as iceases. Now that we have the defiitios of the limit of sequeces out of the way we have a bit of temiology that we eed to look at. If lim a exists ad is fiite we say that the sequece is Æ Æ coveget. If lim a does t exist o is ifiite we say the sequece diveges. Note that sometimes we will say the sequece diveges to if lim a = ad if lim a =- we will sometimes say that the sequece diveges to -. Æ Æ 007 Paul Dawkis 66

176 Get used to the tems coveget ad diveget as we ll be seeig them quite a bit thoughout this chapte. So just how do we fid the limits of sequeces? Most limits of most sequeces ca be foud usig oe of the followig theoems. Theoem Give the sequece { a } if we have a fuctio f ( x ) such that f = a ad lim the lim a = L Æ f x = L x Æ This theoem 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 eally use this theoem by explicitly witig dow a fuctio. We will moe ofte just teat the limit as if it wee a limit of a fuctio ad take the limit as we always did back i Calculus I whe we wee takig the limits of fuctios. So, ow that we kow that takig the limit of a sequece is ealy idetical to takig the limit of a fuctio we also kow that all the popeties fom the limits of fuctios will also hold. Popeties lim a ± b = lima ± limb. Æ Æ Æ. limca = clim a Æ Æ. lim( ab ) = ( lima)( lim b) Æ Æ Æ 4. a lim a Æ lim =, povided limb 0 Æ b lim b Æ Æ p 5. lima = Èlim a povided a 0 Î Æ Æ p These popeties ca be poved usig Theoem above ad the fuctio limit popeties we saw i Calculus I o we ca pove them diectly usig the pecise defiitio of a limit usig ealy idetical poofs of the fuctio limit popeties. Next, just as we had a Squeeze Theoem fo fuctio limits we also have oe fo sequeces ad it is petty much idetical to the fuctio limit vesio. Squeeze Theoem fo Sequeces a c b fo all > N fo some N ad lima = lim b = L the lim c = L. If Æ Æ Æ 007 Paul Dawkis 67

177 Note that i this theoem the fo all > N fo some N is eally just tellig us that we eed to have a c b fo all sufficietly lage, but if it is t tue fo the fist few that wo t ivalidate the theoem. As we ll see ot all sequeces ca be witte as fuctios that we ca actually take the limit of. This will be especially tue fo sequeces that alteate i sigs. While we ca always wite these sequece tems as a fuctio we simply do t kow how to take the limit of a fuctio like that. The followig theoem will help with some of these sequeces. Theoem If lim a = 0 the lima = 0. Æ Æ Note that i ode fo this theoem to hold the limit MUST be zeo ad it wo t wok fo a sequece whose limit is ot zeo. This theoem is easy eough to pove so let s do that. Poof of Theoem The mai thig to this poof is to ote that, - a a a The ote that, ( a ) lim - =- lim a = 0 Æ Æ We the have ( a ) lim - = lim a = 0 ad so by the Squeeze Theoem we must also have, Æ Æ lima = 0 Æ The ext theoem is a useful theoem givig the covegece/divegece ad value (fo whe it s coveget) of a sequece that aises o occasio. Theoem The sequece { } = 0 coveges if - < ad diveges fo all othe value of. Also, lim Æ Ï0 if - < < = Ì Ó if = Hee is a quick (well ot so quick, but defiitely simple) patial poof of this theoem. Patial Poof of Theoem We ll do this by a seies of cases although the last case will ot be completely pove. Case : > We kow fom Calculus I that lim x xæ = if > ad so by Theoem above we also kow 007 Paul Dawkis 68

178 that lim Æ = ad so the sequece diveges if >. Case : = I this case we have, lim = lim = lim= Æ Æ Æ So, the sequece coveges fo = ad i this case its limit is. Case : 0< < x We kow fom Calculus I that lim = 0 if 0< < ad so by Theoem above we also kow xæ that lim = 0 ad so the sequece coveges if 0< < ad i this case its limit is zeo. Æ Case 4 : = 0 I this case we have, lim = lim0 = lim0= 0 Æ Æ Æ So, the sequece coveges fo = 0 ad i this case its limit is zeo. Case 5 : - < < 0 Fist let s ote that if - < < 0 the 0< < the by Case above we have, lim = lim = 0 Æ Theoem above ow tells us that we must also have, lim = 0 ad so if - < < 0 the sequece coveges ad has a limit of 0. Case 6 : =- I this case the sequece is, { } Æ 007 Paul Dawkis 69 Æ { } {,,,,,,,, } = 0 = - = K = 0 = 0 ad hopefully it is clea that lim Æ - does t exist. Recall that i ode of this limit to exist the tems must be appoachig a sigle value as iceases. I this case howeve the tems just alteate betwee ad - ad so the limit does ot exist. So, the sequece diveges fo =-. Case 7 : <- I this case we e ot goig to go though a complete poof. Let s just see what happes if we let =- fo istace. If we do that the sequece becomes, { } { } {,,4, 8,6,, } = 0 = - = K = 0 = 0 So, if =- we get a sequece of tems whose values alteate i sig ad get lage ad lage

179 ad so lim Æ - does t exist. It does ot settle dow to a sigle value as iceases o do the tems ALL appoach ifiity. So, the sequece diveges fo =-. We could do somethig simila fo ay value of such that <- ad so the sequece diveges fo <-. Let s take a look at a couple of examples of limits of sequeces. Example Detemie if the followig sequeces covege o divege. If the sequece coveges detemie its limit. (a) Ï Ì Ó = Ïe (b) Ì Ó (c) ÏÔ Ì ÔÓ (- ) (d) = Ô Ô = { } = 0 [Solutio] [Solutio] - [Solutio] [Solutio] Solutio Ï - (a) Ì Ó0+ 5 = I this case all we eed to do is ecall the method that was developed i Calculus I to deal with the limits of atioal fuctios. See the Limits At Ifiity, Pat I sectio of my Calculus I otes fo a eview of this if you eed to. To do a limit i this fom all we eed to do is facto fom the umeato ad deomiato the lagest powe of, cacel ad the take the limit. Ê ˆ Á lim = lim Ë = lim = Æ Æ Ê 0 ˆ Æ Á Ë So the sequece coveges ad its limit is 5. [Retu to Poblems] 007 Paul Dawkis 70

180 (b) Ïe Ì Ó = We will eed to be caeful with this oe. We will eed to use L Hospital s Rule o this sequece. The poblem is that L Hospital s Rule oly woks o fuctios ad ot o sequeces. Nomally this would be a poblem, but we ve got Theoem fom above to help us out. Let s defie x f ( x) = e x ad ote that, f = e Theoem says that all we eed to do is take the limit of the fuctio. x x e e e lim = lim = lim = Æ xæ x xæ So, the sequece i this pat diveges (to ). Moe ofte tha ot we just do L Hospital s Rule o the sequece tems without fist covetig to x s sice the wok will be idetical egadless of whethe we use x o. Howeve, we eally should emembe that techically we ca t do the deivatives while dealig with sequece tems. [Retu to Poblems] (c) ÏÔ Ì ÔÓ (- ) Ô Ô = We will also eed to be caeful with this sequece. We might be tempted to just say that the limit of the sequece tems is zeo (ad we d be coect). Howeve, techically we ca t take the limit of sequeces whose tems alteate i sig, because we do t kow how to do limits of fuctios that exhibit that same behavio. Also, we wat to be vey caeful to ot ely too much o ituitio with these poblems. As we will see i the ext sectio, ad i late sectios, ou ituitio ca lead us astay i these poblem if we ae t caeful. So, let s wok this oe by the book. We will eed to use Theoem o this poblem. To this we ll fist eed to compute, (- ) lim = lim = 0 Æ Æ Theefoe, sice the limit of the sequece tems with absolute value bas o them goes to zeo we kow by Theoem that, (-) lim = 0 Æ which also meas that the sequece coveges to a value of zeo. [Retu to Poblems] 007 Paul Dawkis 7

181 { } = 0 (d) (- ) Fo this theoem ote that all we eed to do is ealize that this is the sequece i Theoem above usig =-. So, by Theoem this sequece diveges. [Retu to Poblems] We ow eed to give a waig about misusig Theoem. Theoem oly woks if the limit is zeo. If the limit of the absolute value of the sequece tems is ot zeo the the theoem will ot hold. The last pat of the pevious example is a good example of this (ad i fact this waig the whole easo that pat is thee). Notice that ad yet, lim Æ lim - = lim= Æ Æ - does t eve exist let aloe equal. So, be caeful usig this Theoem. You must always emembe that it oly woks if the limit is zeo. Befoe movig oto the ext sectio we eed to give oe moe theoem that we ll eed fo a poof dow the oad. Theoem 4 Fo the sequece { } lim a = L. Æ a if both lim a Æ = L ad lim a + Æ = L the { a } is coveget ad Poof of Theoem 4 Let e > 0. The sice lim a Æ Likewise, because lim a = L thee is a N > 0 such that if > N we kow that, Æ Now, let N max{ N,N } + a - L < e = L thee is a N > 0 such that if > N we kow that, a - + L < e = + ad let > N. The eithe a = ak fo some k > N o a = a k + fo some k > N ad so i eithe case we have that, a - L < e Theefoe, lim a Æ = L ad so { a } is coveget. 007 Paul Dawkis 7

182 Moe o Sequeces I the pevious sectio we itoduced the cocept of a sequece ad talked about limits of sequeces ad the idea of covegece ad divegece fo a sequece. I this sectio we wat to take a quick look at some ideas ivolvig sequeces. Let s stat off with some temiology ad defiitios. Give ay sequece { a } we have the followig.. We call the sequece iceasig if a < a + fo evey.. We call the sequece deceasig if a > a + fo evey.. If { a } is a iceasig sequece o { } a is a deceasig sequece we call it mootoic. 4. If thee exists a umbe m such that m a fo evey we say the sequece is bouded below. The umbe m is sometimes called a lowe boud fo the sequece. 5. If thee exists a umbe M such that a M fo evey we say the sequece is bouded above. The umbe M is sometimes called a uppe boud fo the sequece. 6. If the sequece is both bouded below ad bouded above we call the sequece bouded. Note that i ode fo a sequece to be iceasig o deceasig it must be iceasig/deceasig fo evey. I othe wods, a sequece that iceases fo thee tems ad the deceases fo the est of the tems is NOT a deceasig sequece! Also ote that a mootoic sequece must always icease o it must always decease. Befoe movig o we should make a quick poit about the bouds fo a sequece that is bouded above ad/o below. We ll make the poit about lowe bouds, but we could just as easily make it about uppe bouds. A sequece is bouded below if we ca fid ay umbe m such that m a fo evey. Note howeve that if we fid oe umbe m to use fo a lowe boud the ay umbe smalle tha m will also be a lowe boud. Also, just because we fid oe lowe boud that does t mea thee wo t be a bette lowe boud fo the sequece tha the oe we foud. I othe wods, thee ae a ifiite umbe of lowe bouds fo a sequece that is bouded below, some will be bette tha othes. I my class all that I m afte will be a lowe boud. I do t ecessaily eed the best lowe boud, just a umbe that will be a lowe boud fo the sequece. Let s take a look at a couple of examples. 007 Paul Dawkis 7

183 Example Detemie if the followig sequeces ae mootoic ad/o bouded. Solutio (a) {- } = 0 { + } = - [Solutio] (a) { } (b) - [Solutio] Ï (c) Ì Ó = 5 = 0 [Solutio] This sequece is a deceasig sequece (ad hece mootoic) because, fo evey. ( ) - >- + Also, sice the sequece tems will be eithe zeo o egative this sequece is bouded above. We ca use ay positive umbe o zeo as the boud, M, howeve, it s stadad to choose the smallest possible boud if we ca ad it s a ice umbe. So, we ll choose M = 0 sice, - 0 fo evey This sequece is ot bouded below howeve sice we ca always get below ay potetial boud by takig lage eough. Theefoe, while the sequece is bouded above it is ot bouded. As a side ote we ca also ote that this sequece diveges (to - if we wat to be specific). [Retu to Poblems] { + } = (b) (- ) The sequece tems i this sequece alteate betwee ad - ad so the sequece is eithe a iceasig sequece o a deceasig sequece. Sice the sequece is eithe a iceasig o deceasig sequece it is ot a mootoic sequece. The sequece is bouded howeve sice it is bouded above by ad bouded below by -. Agai, we ca ote that this sequece is also diveget. [Retu to Poblems] Ï (c) Ì Ó = 5 This sequece is a deceasig sequece (ad hece mootoic) sice, > + The tems i this sequece ae all positive ad so it is bouded below by zeo. Also, sice the 007 Paul Dawkis 74

184 sequece is a deceasig sequece the fist sequece tem will be the lagest ad so we ca see that the sequece will also be bouded above by 5. Theefoe, this sequece is bouded. We ca also take a quick limit ad ote that this sequece coveges ad its limit is zeo. [Retu to Poblems] Now, let s wok a couple moe examples that ae desiged to make sue that we do t get too used to elyig o ou ituitio with these poblems. As we oted i the pevious sectio ou ituitio ca ofte lead us astay with some of the cocepts we ll be lookig at i this chapte. Example Detemie if the followig sequeces ae mootoic ad/o bouded. Solutio Ï (a) Ì Ó + (a) (b) Ï Ì Ó + Ï Ì Ó = = [Solutio] = 0 [Solutio] We ll stat with the bouded pat of this example fist ad the come back ad deal with the iceasig/deceasig questio sice that is whee studets ofte make mistakes with this type of sequece. Fist, is positive ad so the sequece tems ae all positive. The sequece is theefoe bouded below by zeo. Likewise each sequece tem is the quotiet of a umbe divided by a lage umbe ad so is guaateed to be less that oe. The sequece is the bouded above by oe. So, this sequece is bouded. Now let s thik about the mootoic questio. Fist, studets will ofte make the mistake of assumig that because the deomiato is lage the quotiet must be deceasig. This will ot always be the case ad i this case we would be wog. This sequece is iceasig as we ll see. To detemie the iceasig/deceasig atue of this sequece we will eed to esot to Calculus I techiques. Fist coside the followig fuctio ad its deivative. x f ( x) = f ( x) = x+ x+ We ca see that the fist deivative is always positive ad so fom Calculus I we kow that the fuctio must the be a iceasig fuctio. So, how does this help us? Notice that, f = = a + Theefoe because < + ad f x is iceasig we ca also say that, 007 Paul Dawkis 75

185 + a = = f < f + = = a fi a < a I othe wods, the sequece must be iceasig. Note that ow that we kow the sequece is a iceasig sequece we ca get a bette lowe boud fo the sequece. Sice the sequece is iceasig the fist tem i the sequece must be the smallest tem ad so sice we ae statig at = we could also use a lowe boud of fo this sequece. It is impotat to emembe that ay umbe that is always less tha o equal to all the sequece tems ca be a lowe boud. Some ae bette tha othes howeve. A quick limit will also tell us that this sequece coveges with a limit of. Befoe movig o to the ext pat thee is a atual questio that may studets will have at this poit. Why did we use Calculus to detemie the iceasig/deceasig atue of the sequece whe we could have just plugged i a couple of s ad quickly detemied the same thig? The aswe to this questio is the ext pat of this example! [Retu to Poblems] (b) Ï Ì Ó = 0 This is a messy lookig sequece, but it eeds to be i ode to make the poit of this pat. Fist, otice that, as with the pevious pat, the sequece tems ae all positive ad will all be less tha oe (sice the umeato is guaateed to be less tha the deomiato) ad so the sequece is bouded. Now, let s move o to the iceasig/deceasig questio. As with the last poblem, may studets will look at the expoets i the umeato ad deomiato ad detemie based o that that sequece tems must decease. This howeve, is t a deceasig sequece. Let s take a look at the fist few tems to see this. a = ª a = ª a = ª a4 = ª a5 = ª a6 = ª a7 = ª a8 = ª a9 = ª a0 = = The fist 0 tems of this sequece ae all iceasig ad so clealy the sequece ca t be a 007 Paul Dawkis 76

186 deceasig sequece. Recall that a sequece ca oly be deceasig if ALL the tems ae deceasig. Now, we ca t make aothe commo mistake ad assume that because the fist few tems icease the whole sequece must also icease. If we did that we would also be mistake as this is also ot a iceasig sequece. This sequece is eithe deceasig o iceasig. The oly sue way to see this is to do the Calculus I appoach to iceasig/deceasig fuctios. I this case we ll eed the followig fuctio ad its deivative. 4 x -x x f ( x) = f 4 ( x) = 4 x x This fuctio will have the followig thee citical poits, 4 4 x= 0, x= 0000 ª.607, x= ª-.607 Why citical poits? Remembe these ae the oly places whee the fuctio may chage sig! Ou sequece stats at = 0 ad so we ca igoe the thid oe sice it lies outside the values of that we e cosideig. By pluggig i some test values of x we ca quickly detemie that the 4 deivative is positive fo 0< x< 0000 ª.6 ad so the fuctio is iceasig i this age. Likewise, we ca see that the deivative is egative fo will be deceasig i this age. x> ª.6 ad so the fuctio So, ou sequece will be iceasig fo 0 ad deceasig fo. Theefoe the fuctio is ot mootoic. Fially, ote that this sequece will also covege ad has a limit of zeo. [Retu to Poblems] So, as the last example has show we eed to be caeful i makig assumptios about sequeces. Ou ituitio will ofte ot be sufficiet to get the coect aswe ad we ca NEVER make assumptios about a sequece based o the value of the fist few tems. As the last pat has show thee ae sequeces which will icease o decease fo a few tems ad the chage diectio afte that. Note as well that we said fist few tems hee, but it is completely possible fo a sequece to decease fo the fist 0,000 tems ad the stat iceasig fo the emaiig tems. I othe wods, thee is o magical value of fo 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 detemie the iceasig/deceasig atue of a sequece is i sequeces like pat (c) of Example. I this case iceasig oly chaged (i fact iceased) the deomiato ad so we wee able to detemie the behavio of the sequece based o that. 007 Paul Dawkis 77

187 I Example howeve, iceasig iceased both the deomiato ad the umeato. I cases like this thee is o way to detemie which icease will wi out ad cause the sequece tems to icease o decease ad so we eed to esot to Calculus I techiques to aswe the questio. We ll close out this sectio with a ice theoem that we ll use i some of the poofs late i this chapte. Theoem If { a } is bouded ad mootoic the { } a is coveget. Be caeful to ot misuse this theoem. It does ot say that if a sequece is ot bouded ad/o ot mootoic that it is diveget. Example b is a good case i poit. The sequece i that example was ot mootoic but it does covege. Note as well that we ca make seveal vaiats of this theoem. If { a } is bouded above ad iceasig the it coveges ad likewise if { a } is bouded below ad deceasig the it coveges. 007 Paul Dawkis 78

188 Seies The Basics I this sectio we will itoduce the topic that we will be discussig fo the est of this chapte. That topic is ifiite seies. So just what is a ifiite seies? Well, let s stat with a sequece { a } = (ote the = is fo coveiece, it ca be aythig) ad defie the followig, s = a s = a + a s = a + a + a s = a + a + a + a 4 4 M s = a + a + a + a + L+ a = a 4 i i= The s ae called patial sums ad otice that they will fom a sequece, { s } =. Also ecall that the S is used to epeset this summatio ad called a vaiety of ames. The most commo ames ae : seies otatio, summatio otatio, ad sigma otatio. You should have see this otatio, at least biefly, back whe you saw the defiitio of a defiite itegal i Calculus I. If you eed a quick efeshe o summatio otatio see the eview of summatio otatio i my Calculus I otes. Now back to seies. We wat to take a look at the limit of the sequece of patial sums, { } Notatioally we ll defie, lims = limâai = ai Æ Æ i =  i = We will call  ai a ifiite seies ad ote that the seies stats at i = because that is i= whee ou oigial sequece, { } a = s. =, stated. Had ou oigial sequece stated at the ou ifiite seies would also have stated at. The ifiite seies will stat at the same value that the sequece of tems (as opposed to the sequece of patial sums) stats. s If the sequece of patial sums, { } = ifiite seies,  ai i= seies is also called diveget., is coveget ad its limit is fiite the we also call the coveget ad if the sequece of patial sums is diveet the the ifiite Note that sometimes it is coveiet to wite the ifiite seies as,  ai = a+ a + a + L+ a + L i= 007 Paul Dawkis 79

189 We do have to be caeful with this howeve. This implies that a ifiite seies is just a ifiite sum of tems ad as well see i the ext sectio this is ot eally tue. I the ext sectio we e goig to be discussig i geate detail the value of a ifiite seies, povided it has oe of couse as well as the ideas of covegece ad divegece. This sectio is goig to be devoted mostly to otatioal issues as well as makig sue we ca do some basic maipulatios with ifiite seies so we ae eady fo them whe we eed to be able to deal with them i late sectios. Fist, we should ote that i most of this chapte we will efe to ifiite seies as simply seies. If we eve eed to wok with both ifiite ad fiite seies we ll be moe caeful with temiology, but i most sectios we ll be dealig exclusively with ifiite seies ad so we ll just call them seies. Now, i  ai the i is called the idex of summatio o just idex fo shot ad ote that the i= lette we use to epeset the idex does ot matte. So fo example the followig seies ae all the same. The oly diffeece is the lette we ve used fo the idex. = = i + k + +    i= 0 k= 0 = 0 etc. It is impotat to agai ote that the idex will stat at whateve value the sequece of seies tems stats at ad this ca liteally be aythig. So fa we ve used = 0 ad = but the idex could have stated aywhee. I fact, we will usually use  a to epeset a ifiite seies i which the statig poit fo the idex is ot impotat. Whe we dop the iitial value of the idex we ll also dop the ifiity fom the top so do t foget that it is still techically thee. We will be doppig the iitial value of the idex i quite a few facts ad theoems that we ll be seeig thoughout this chapte. I these facts/theoems the statig poit of the seies will ot affect the esult ad so to simplify the otatio ad to avoid givig the impessio that the statig poit is impotat we will dop the idex fom the otatio. Do ot foget howeve, that thee is a statig poit ad that this will be a ifiite seies. Note howeve, that if we do put a iitial value of the idex o a seies i a fact/theoem it is thee because it eally does eed to be thee. Now that some of the otatioal issues ae out of the way we eed to stat thikig about vaious ways that we ca maipulate seies. We ll stat this off with basic aithmetic with ifiite seies as we ll eed to be able to do that o occasio. We have the followig popeties. 007 Paul Dawkis 80

190 Popeties If  a ad  b ae both coveget seies the, 6.  ca, whee c is ay umbe, is also coveget ad  =  ca c a 7. Âa ±  b = k = k is also coveget ad, Âa ± Âb =  ( a ± b). = k = k = k The fist popety is simply tellig us that we ca always facto a multiplicative costat out of a ifiite seies ad agai ecall that if we do t put i a iitial value of the idex that the seies ca stat at ay value. Also ecall that i these cases we wo t put a ifiity at the top eithe. The secod popety says that if we add/subtact seies all we eally eed to do is add/subtact the seies tems. Note as well that i ode to add/subtact seies we eed to make sue that both have the same iitial value of the idex ad the ew seies will also stat at this value. Befoe we move o to a diffeet topic let s discuss multiplicatio of seies biefly. We ll stat both seies at = 0 fo a late fomula ad the ote that, Ê ˆÊ ÁÂa ÁÂb ˆ  ( ab ) Ë = 0 Ë = 0 = 0 To covice youself that this is t tue coside the followig poduct 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 tems, the the x tems, etc. Istead we had to distibute the though the secod polyomial, the distibute the x though the secod polyomial ad fially combie like tems. Multiplyig ifiite seies (eve though we said we ca t thik of a ifiite seies as a ifiite sum) eeds to be doe i the same mae. With multiplicatio we e eally askig us to do the followig, Ê ˆÊ ÁÂa ÁÂb ˆ= ( a0 + a + a + a + L)( b0 + b + b + b + L ) Ë = 0 Ë = 0 To do this multiplicatio we would have to distibute the a 0 though the secod tem, distibute the a though, etc the combie like tems. This is petty much impossible sice both seies have a ifiite set of tems i them, howeve the followig fomula ca be used to detemie the poduct of two seies. Ê ˆÊ ÁÂa ÁÂb ˆ=  c whee c =  ab i - i Ë = 0 Ë = 0 = 0 i= Paul Dawkis 8

191 We also ca t say a lot about the covegece of the poduct. Eve if both of the oigial seies ae coveget it is possible fo the poduct to be diveget. The eality is that multiplicatio of seies is a somewhat difficult pocess ad i geeal is avoided if possible. We will take a bief look at it towads the ed of the chapte whe we ve got moe wok ude ou belt ad we u acoss a situatio whee it might actually be what we wat to do. Util the, do t woy about multiplyig seies. 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 couse. 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 couse does t mea howeve that it is t used ofte i othe classes whee you might u acoss seies. So, we will cove it biefly hee so that you ca say you ve see it. The basic idea behid idex shifts is to stat a seies at a diffeet value fo whateve the easo (ad yes, thee ae legitimate easos fo doig that). Coside the followig seies, + 5 Â = Suppose that fo some easo we wated to stat this seies at = 0, but we did t wat to chage the value of the seies. This meas that we ca t just chage the = to = 0 as this would add i two ew tems to the seies ad thus chagig its value. Pefomig a idex shift is a faily simple pocess to do. We ll stat 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 lowe limit will chage hee. Next, we ca solve this fo to get, = i+ We ca ow completely ewite the seies i tems of the idex i istead of the idex simply by pluggig i ou equatio fo i tems of i. + 5 ( i+ ) + 5 i+ 7 Â = Â = i+ Â i+ = i= 0 i= 0 To fiish the poblem out we ll ecall that the lette we used fo the idex does t matte ad so we ll chage the fial i back ito a to get, Â = Â + = = 0 To covice youselves that these eally ae the same summatio let s wite out the fist couple of tems fo each of them, 007 Paul Dawkis 8

192  = L =  + = L = 0 So, sue eough the two seies do have exactly the same tems. Thee is actually a easie way to do a idex shift. The method give above is the techically coect way of doig a idex shift. Howeve, otice i the above example we deceased the iitial value of the idex by ad all the s i the seies tems iceased by as well. This will always wok i this mae. If we decease the iitial value of the idex by a set amout the all the othe s i the seies tem will icease by the same amout. Likewise, if we icease the iitial value of the idex by a set amout, the all the s i the seies tem will decease by the same amout. Let s do a couple of examples usig this shothad method fo doig idex shifts. Example Pefom the followig idex shifts. - (a) Wite  a as a seies that stats at = 0. = (b) Wite  as a seies that stats at =. + = - Solutio (a) I this case we eed to decease the iitial value by ad so the s (okay the sigle ) i the tem must icease by as well. - ( + ) -  =  =  a a a = = 0 = 0 (b) Fo this poblem we wat to icease the iitial value by ad so all the s i the seies tem must decease 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 moe ofte tha the idex shifts. This fial topic is eally moe about alteate ways to wite seies whe the situatio equies it. Let s stat with the followig seies ad ote that the = statig poit is oly fo coveiece sice we eed to stat the seies somewhee. Âa = a+ a + a+ a4 + a5 + L = 007 Paul Dawkis 8

193 Notice that if we igoe the fist tem the emaiig tems will also be a seies that will stat at = istead of = So, we ca ewite the oigial seies as follows, Â Â a = a + a = = I this example we say that we ve stipped out the fist tem. We could have stipped out moe tems if we wated to. I the followig seies we ve stipped out the fist two tems ad the fist fou tems espectively. Â Â a = a + a + a = = Â a = a + a + a + a + a 4 = = 5 Beig able to stip out tems will, o occasio, simplify ou wok o allow us to euse a pio esult so it s a impotat idea to emembe. Notice that i the secod example above we could have also deoted the fou tems that we stipped out as a fiite seies as follows, Â 4 Â Â Â Â a = a + a + a + a + a = a + a 4 = = 5 = = 5 This is a coveiet otatio whe we ae stippig out a lage umbe of tems o if we eed to stip out a udetemied umbe of tems. I geeal, we ca wite a seies as follows, N Â Â Â a = a + a = = = N+ We ll leave this sectio with a impotat waig about temiology. Do t get sequeces ad seies cofused! A sequece is a list of umbes witte i a specific ode while a ifiite seies is a limit of a sequece of fiite seies ad hece, if it exists will be a sigle value. So, oce agai, a sequece is a list of umbes while a seies is a sigle umbe, povided it makes sese to eve compute the seies. Studets will ofte cofuse the two ad ty to use facts petaiig to oe o the othe. Howeve, sice they ae diffeet beasts this just wo t wok. Thee will be poblems whee we ae usig both sequeces ad seies so we ll always have to emembe that they ae diffeet. 007 Paul Dawkis 84

194 Seies Covegece/Divegece I the pevious sectio we spet some time gettig familia with seies ad we biefly defied covegece ad divegece. Befoe woyig about covegece ad divegece of a seies we wated to make sue that we ve stated to get comfotable with the otatio ivolved i seies ad some of the vaious maipulatios of seies that we will, o occasio, eed to be able to do. As oted i the pevious sectio most of what we wee doig thee wo t be doe much i this chapte. So, it is ow time to stat talkig about the covegece ad divegece of a seies as this will be a topic that we ll be dealig with to oe extet of aothe i almost all of the emaiig sectios of this chapte. So, let s ecap just what a ifiite seies is ad what it meas fo a seies to be coveget o a = diveget. We ll stat with a sequece { } = oly fo the sake of coveiece ad it ca, i fact, be aythig. Next we defie the patial sums of the seies as, s = a s = a + a s = a + a + a s = a + a + a + a ad these fom a ew sequece, { } 4 4 M ad agai ote that we e statig the sequece at s = a + a + a + a + L+ a =Â a 4 i i= s. = A ifiite seies, o just seies hee sice almost evey seies that we ll be lookig at will be a ifiite seies, is the the limit of the patial sums. O, Â i= a = lim s i If the sequece of patial sums is a coveget sequece (i.e. its limit exists ad is fiite) the the Æ seies is also called coveget ad i this case if lim s = s the, Æ Â i= a i = s. Likewise, if the sequece of patial sums is a diveget sequece (i.e. its limit does t exist o is plus o mius ifiity) the the seies is also called diveget. Let s take a look at some seies ad see if we ca detemie if they ae coveget o diveget ad see if we ca detemie the value of ay coveget seies we fid. Example Detemie if the followig seies is coveget o diveget. If it coveges 007 Paul Dawkis 85

195 detemie its value. Solutio To detemie if the seies is coveget we fist eed to get ou hads o a fomula fo the geeal tem i the sequece of patial sums. s  = =  i i= This is a kow seies ad its value ca be show to be, + s =  i = i= Do t woy if you did t kow this fomula (I d be supised if ayoe kew it ) as you wo t be equied to kow it i my couse. So, to detemie if the seies is coveget we will fist eed to see if the sequece of patial sums, ( + ) Ï Ì Ó = is coveget o diveget. That s ot teibly difficult i this case. The limit of the sequece tems is, ( + ) lim = Æ Theefoe, the sequece of patial sums diveges to ad so the seies also diveges. So, as we saw i this example we had to kow a faily obscue fomula i ode to detemie the covegece of this seies. I geeal fidig a fomula fo the geeal tem i the sequece of patial sums is a vey difficult pocess. I fact afte the ext sectio we ll ot be doig much with the patial sums of seies due to the exteme difficulty faced i fidig the geeal fomula. This also meas that we ll ot be doig much wok with the value of seies sice i ode to get the value we ll also eed to kow the geeal fomula fo the patial sums. We will cotiue with a few moe examples howeve, sice this is techically how we detemie covegece ad the value of a seies. Also, the emaiig examples we ll be lookig at i this sectio will lead us to a vey impotat fact about the covegece of seies. So, let s take a look at a couple moe examples. Example Detemie if the followig seies coveges o diveges. If it coveges detemie its sum. 007 Paul Dawkis 86

196 Â = - Solutio This is actually oe of the few seies i which we ae able to detemie a fomula fo the geeal tem i the sequece of patial factios. Howeve, i this sectio we ae moe iteested i the geeal idea of covegece ad divegece ad so we ll put off discussig the pocess fo fidig the fomula util the ext sectio. The geeal fomula fo the patial sums is, s = Â = - - i ad i this case we have, i= Ê ˆ lims = lim Æ Æ Á - - = 4 ( ) Ë + 4 The sequece of patial sums coveges ad so the seies coveges also ad its value is, Â = - 4 = Example Detemie if the followig seies coveges o diveges. If it coveges detemie its sum. Â = 0 (- ) Solutio I this case we eally do t eed a geeal fomula fo the patial sums to detemie the covegece of this seies. Let s just wite dow the fist few patial sums. s 0 etc. = s = - = 0 So, it looks like the sequece of patial sums is, s s = - + = = = 0 { } { } s =,0,,0,,0,,0,, K = 0 ad this sequece diveges sice lim s does t exist. Theefoe, the seies also diveges. Æ Example 4 Detemie if the followig seies coveges o diveges. If it coveges detemie its sum. 007 Paul Dawkis 87

197  - = Solutio Hee is the geeal fomula fo the patial sums fo this seies. Ê ˆ s =  = i- Á - i= Ë Agai, do ot woy about kowig this fomula. This is ot somethig that you ll eve be asked to kow i my class. I this case the limit of the sequece of patial sums is, Ê ˆ lims = lim Á- Æ Æ = Ë The sequece of patial sums is coveget ad so the seies will also be coveget. The value of the seies is,  = - = As we aleady oted, do ot get excited about detemiig the geeal fomula fo the sequece of patial sums. Thee is oly goig to be oe type of seies whee you will eed to detemie this fomula ad the pocess i that case is t too bad. I fact, you aleady kow how to do most of the wok i the pocess as you ll see i the ext sectio. So, we ve detemied the covegece of fou seies ow. Two of the seies coveged ad two diveged. Let s go back ad examie the seies tems fo each of these. Fo each of the seies let s take the limit as goes to ifiity of the seies tems (ot the patial sums!!). lim = Æ (- ) this seies diveged lim = 0 this seies coveged Æ - lim does't exist this seies diveged Æ lim = 0 this seies coveged Æ - Notice that fo the two seies that coveged the seies tem itself was zeo i the limit. This will always be tue fo coveget seies ad leads to the followig theoem. Theoem If  a coveges the lima = 0. Æ Poof Fist let s suppose that the seies stats at =. If it does t the we ca modify thigs 007 Paul Dawkis 88

198 as appopiate below. The the patial sums ae, -  s = a = a + a + a + a + L+ a s = a = a + a + a + a + L + a + a - i 4 - i 4 - i= i= Next, we ca use these two patial sums to wite, a = s - s - Now because we kow that   a is coveget we also kow that the sequece { } s = is also coveget ad that lim s = s fo some fiite value s. Howeve, sice - Æ as Æ we Æ also have lim s - = s. Æ We ow have, lima = lim s - s = lims - lims = s- s = Æ Æ Æ Æ Be caeful to ot misuse this theoem! This theoem gives us a equiemet fo covegece but ot a guaatee of covegece. I othe wods, the covese is NOT tue. If lima = 0 the seies may actually divege! Coside the followig two seies.   = = I both cases the seies tems ae zeo i the limit as goes to ifiity, yet oly the secod seies coveges. The fist seies diveges. It will be a couple of sectios befoe we ca pove this, so at this poit please believe this ad kow that you ll be able to pove the covegece of these two seies i a couple of sectios. Agai, as oted above, all this theoem does is give us a equiemet fo a seies to covege. I ode fo a seies to covege the seies tems must go to zeo i the limit. If the seies tems do ot go to zeo i the limit the thee is o way the seies ca covege sice this would violate the theoem. This leads us to the fist of may tests fo the covegece/divegece of a seies that we ll be seeig i this chapte. Æ Divegece Test If lima 0 the a will divege. Æ Â Agai, do NOT misuse this test. This test oly says that a seies is guaateed to divege if the seies tems do t go to zeo i the limit. If the seies tems do happe to go to zeo the seies may o may ot covege! Agai, ecall the followig two seies, 007 Paul Dawkis 89

199  =  = diveges coveges Oe of the moe commo mistakes that studets make whe they fist get ito seies is to assume that if lima = 0 the  a will covege. Thee is just o way to guaatee this so be caeful! Æ Let s take a quick look at a example of how this test ca be used. Example 5 Detemie if the followig seies is coveget o diveget. 4 -  = 0 0+ Solutio With almost evey seies we ll be lookig at i this chapte the fist thig that we should do is take a look at the seies tems ad see if they go to zeo of ot. If it s clea that the tems do t go to zeo use the Divegece Test ad be doe with the poblem. That s what we ll do hee. 4 - Æ 0+ lim =- 0 The limit of the seies tems is t zeo ad so by the Divegece Test the seies diveges. The divegece test is the fist test of may tests that we will be lookig at ove the couse of the ext seveal sectios. You will eed to keep tack of all these tests, the coditios ude which they ca be used ad thei coclusios all i oe place so you ca quickly efe back to them as you eed to. Next we should talk biefly evisit aithmetic of seies ad covegece/divegece. As we saw i the pevious sectio if  a ad  b ae both coveget seies the so ae  ca ad  = k ( a ± b ). Futhemoe, these seies will have the followig sums o values.  =  Â( ± ) =  ±  ca c a a b a b = k = k = k We ll see a example of this i the ext sectio afte we get a few moe examples ude ou belt. At this poit just emembe that a sum of coveget sequeces is coveget ad multiplyig a coveget sequece by a umbe will ot chage its covegece. We eed to be a little caeful with these facts whe it comes to diveget seies. I the fist case if  a is diveget the  ca will also be diveget (povided c is t zeo of couse) sice multiplyig a seies that is ifiite i value o does t have a value by a fiite value (i.e. c) wo t 007 Paul Dawkis 90

200 chage the fact that the seies has a ifiite o o value. Howeve, it is possible to have both  a ad  b be diveget seies ad yet have  ( a ± b) be a coveget seies. Now, sice the mai topic of this sectio is the covegece of a seies we should metio a stoge type of covegece. A seies  a is said to covege absolutely if  a also coveges. Absolute covegece is stoge tha covegece i the sese that a seies that is absolutely coveget will also be coveget, but a seies that is coveget may o may ot be absolutely coveget. I fact if  a coveges ad  a diveges the seies  a is called coditioally coveget. At this poit we do t eally have the tools at had to popely ivestigate this topic i detail o do we have the tools i had to detemie if a seies is absolutely coveget o ot. So we ll ot say aythig moe about this subject fo a while. Whe we fially have the tools i had to discuss this topic i moe detail we will evisit it. Util the do t woy about it. The idea is metioed hee oly because we wee aleady discussig covegece i this sectio ad it ties ito the last topic that we wat to discuss i this sectio. I the pevious sectio afte we d itoduced the idea of a ifiite seies we commeted o the fact that we should t thik of a ifiite seies as a ifiite sum despite the fact that the otatio we use fo ifiite seies seems to imply that it is a ifiite sum. It s ow time to biefly discuss this. Fist, we eed to itoduce the idea of a eaagemet. A eaagemet of a seies is exactly what it might soud like, it is the same seies with the tems eaaged ito a diffeet ode. = = k Fo example, coside the followig the ifiite seies.  A eaagemet of this seies is,  = a = a + a + a + a + a + a + a + L a = a + a + a + a + a + a + a + L The issue we eed to discuss hee is that fo some seies each of these aagemets of tems ca have a diffeet values despite the fact that they ae usig exactly the same tems. Hee is a example of this. It ca be show that, + ( -) = L = l ()  = Sice this seies coveges 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 9

201 L = l () Now, let s add i a zeo betwee each tem as follows L = l () Note that this wo t chage the value of the seies because the patial sums fo this seies will be the patial sums fo the () except that each tem will be epeated. Repeatig tems i a seies will ot affect its limit howeve ad so both () ad () will be the same. We kow that if two seies covege we ca add them by addig tem by tem ad so add () ad () to get, L = l (4) Now, otice that the tems of (4) ae simply the tems of () eaaged so that each egative tem comes afte two positive tems. The values howeve ae defiitely diffeet despite the fact that the tems ae the same. Note as well that this is ot oe of those ticks that you see occasioally whee you get a cotadictoy esult because of a had to spot math/logic eo. This is a vey eal esult ad we ve ot made ay logic mistakes/eos. Hee is a ice set of facts that gove this idea of whe a eaagemet will lead to a diffeet value of a seies. Facts Give the seiesâ a,. If  a is absolutely coveget ad its value is s the ay eaagemet of  a will also have a value of s.. If  a is coditioally coveget ad is ay eal umbe the thee is a eaagemet of a whose value will be.  Agai, we do ot have the tools i had yet to detemie if a seies is absolutely coveget ad so do t woy about this at this poit. This is hee just to make sue that you udestad that we have to be vey caeful i thikig of a ifiite seies as a ifiite sum. Thee ae times whe we ca (i.e. the seies is absolutely coveget) ad thee ae times whe we ca t (i.e. the seies is coditioally coveget). As a fial ote, the fact above tells us that the seies,  = (-) Paul Dawkis 9

202 must be coditioally coveget sice two eaagemets gave two sepaate values of this seies. Evetually it will be vey simple to show that this seies coditioally coveget. 007 Paul Dawkis 9

203 Seies Special Seies I this sectio we ae goig to take a bief look at thee special seies. Actually, special may ot be the coect tem. All thee have bee amed which makes them special i some way, howeve the mai easo that we e goig to look at two of them i this sectio is that they ae the oly types of seies that we ll be lookig at fo which we will be able to get actual values fo the seies. The thid type is diveget ad so wo t have a value to woy about. I geeal, detemiig the value of a seies is vey difficult ad outside of these two kids of seies that we ll look at i this sectio we will ot be detemiig the value of seies i this chapte. So, let s get stated. Geometic Seies A geometic seies is ay seies that ca be witte i the fom, - Â = o, with a idex shift the geometic seies will ofte be witte as, = 0 a These ae idetical seies ad will have idetical values, povided they covege of couse. If we stat with the fist fom it ca be show that the patial sums ae, a( - ) a a s = = Â The seies will covege povided the patial sums fom a coveget sequece, so let s take the limit of the patial sums. Ê a a ˆ lims = lim Á - Æ Æ Ë - - a a a = lim -lim Æ - Æ - a a = - lim - - Æ Now, fom Theoem fom the Sequeces sectio we kow that the limit above will exist ad be fiite povided - <. Howeve, ote that we ca t let = sice this will give divisio by zeo. Theefoe, this will exist ad be fiite povided - < < ad i this case the limit is zeo ad so we get, a lim s = Æ Paul Dawkis 94

204 Theefoe, a geometic seies will covege if - < <, which is usually witte <, its value is, a = = - - Âa  a = = 0 Note that i usig this fomula we ll eed to make sue that we ae i the coect fom. I othe wods, if the seies stats at = 0 the the expoet o the must be. Likewise if the seies stats at = the the expoet o the must be -. Example Detemie if the followig seies covege o divege. If they covege give the value of the seies. Solutio (a) (a) (b)  9 4 [Solutio] =  = 9 4 (-4)  [Solutio] - = 0 5 This seies does t eally look like a geometic seies. Howeve, otice that both pats of the seies tem ae umbes aised to a powe. This meas that it ca be put ito the fom of a geometic seies. We will just eed to decide which fom is the coect fom. Sice the seies stats at = we will wat the expoets o the umbes to be -. It will be faily easy to get this ito the coect fom. Let s fist ewite thigs slightly. Oe of the s i the expoet has a egative i fot of it ad that ca t be thee i the geometic fom. So, let s fist get id of that ( -) + 4 Â9 4 = Â9 4 =  - 9 = = = Now let s get the coect expoet o each of the umbes. This ca be doe usig simple expoet popeties Â9 4 =  = -  Now, ewite the tem a little. = = = Ê ˆ 9 4 = 6 9 = 44Á 9 Ë Â Â -  = = = 4 So, this is a geometic seies with a = 44 ad = <. Theefoe, sice < we kow the seies will covege ad its value will be, Paul Dawkis 95

205 Â = = = = [Retu to Poblems] (b) ( 4) - Â = Agai, this does t look like a geometic seies, but it ca be put ito the coect fom. I this case the seies stats at = 0 so we ll eed the expoets to be o the tems. Note that this meas we e goig to eed to ewite the expoet o the umeato a little (-4) Ê-64 ˆ = = 5 = 5Á Ë 5 Â Â Â Â - - = 0 = 0 = 0 = 0 64 So, we ve got it ito the coect fom ad we ca see that a = 5 ad =-. Also ote that ad so this seies diveges. 5 [Retu to Poblems] Back i the Seies Basics sectio we talked about stippig out tems fom a seies, but did t eally povide ay examples of how this idea could be used i pactice. We ca ow do some examples. Example Use the esults fom the pevious example to detemie the value of the followig seies. Solutio (a) (a) (b) Â 9 4 [Solutio] = Â 9 4 [Solutio] = Â = I this case we could just ackowledge that this is a geometic seies that stats at = 0 ad so we could put it ito the coect fom ad be doe with it. Howeve, this does povide us with a ice example of how to use the idea of stippig out tems to ou advatage. Let s otice that if we stip out the fist tem fom this seies we aive at, Â Â Â 9 4 = = = 0 = = Fom the pevious example we kow the value of the ew seies that aises hee ad so the value of the seies i this example is, 007 Paul Dawkis 96

206  = = + = 5 5 [Retu to Poblems] (b)  = 9 4 I this case we ca t stip out tems fom the give seies to aive at the seies used i the pevious example. Howeve, we ca stat with the seies used i the pevious example ad stip tems out of it to get the seies i this example. So, let s do that. We will stip out the fist two tems fom the seies we looked at i the pevious example    9 4 = = = = = We ca ow use the value of the seies fom the pevious example to get the value of this seies Â9 4 =  = - 08 = = = 5 5 [Retu to Poblems] Notice that we did t discuss the covegece of eithe of the seies i the above example. Hee s why. Coside the followig seies witte i two sepaate ways (i.e. we stipped out a couple of tems fom it). Let s suppose that we kow  a =   a = a + a + a + a 0 = 0 = is a coveget seies. This meas that it s got a fiite value ad addig thee fiite tems oto this will ot chage that fact. So the value of fiite ad so is coveget. Likewise, suppose that  a = 0 this value we will emai fiite ad aive at the value of so this seies will also be coveget.  a = 0 is also is coveget. I this case if we subtact thee fiite values fom  a =. This is ow a fiite value ad I othe wods, if we have two seies ad they diffe oly by the pesece, o absece, of a fiite umbe of fiite tems they will eithe both be coveget o they will both be diveget. The diffeece of a few tems oe way o the othe will ot chage the covegece of a seies. This is a impotat idea ad we will use it seveal times i the followig sectios to simplify some of the tests that we ll be lookig at. 007 Paul Dawkis 97

207 Telescopig Seies It s ow time to look at the secod of the thee seies i this sectio. I this potio we ae goig to look at a seies that is called a telescopig seies. The ame i this case comes fom what happes with the patial sums ad is best show i a example. Example Detemie if the followig seies coveges o diveges. If it coveges fid its value.  = Solutio We fist eed the patial sums fo this seies. s =  i + i+ Now, let s otice that we ca use patial factios o the seies tem to get, = = - i + i+ i+ i+ i+ i+ i= 0 I ll leave the details of the patial factios to you. By ow you should be faily adept at this sice we spet a fai amout of time doig patial factios back i the Itegatio Techiques chapte. If you eed a efeshe you should go back ad eview that sectio. So, what does this do fo us? Well, let s stat witig out the tems of the geeal patial sum fo this seies usig the patial factio fom. Ê ˆ s = ÂÁ - i= 0 Ëi+ i+ Ê Ê Ê Ê = Á - ˆ+ Á - ˆ+ Á - ˆ+ L + Ë Ë Ë 4 - Ê ˆ Á ˆ+ Á - Ë + Ë + + = - + Notice that evey tem except the fist ad last tem caceled out. This is the oigi of the ame telescopig seies. This also meas that we ca detemie the covegece of this seies by takig the limit of the patial sums. Ê ˆ lims = limá- = Æ Æ Ë + The sequece of patial sums is coveget ad so the seies is coveget ad has a value of  = = I telescopig seies be caeful to ot assume that successive tems will be the oes that cacel. Coside the followig example. 007 Paul Dawkis 98

208 Example 4 Detemie if the followig seies coveges o diveges. If it coveges fid its value. Â = + 4+ Solutio As with the last example we ll leave the patial factios details to you to veify. The patial sums ae, Ê ˆ Ê ˆ s = ÂÁ - i= i i = ÂÁ - Ë + + i= Ëi+ i+ ÈÊ Ê Ê Ê = ÍÁ - ˆ+ Á - ˆ+ Á - ˆ+ L + Á ÍÎË 4 Ë 5 Ë 4 6 Ë È = Í Î + + ˆ Ê ˆ - + Á - + Ë + + I this case istead of successive tems cacelig a tem will cacel with a tem that is fathe dow the list. The ed esult this time is two iitial ad two fial tems ae left. Notice as well that i ode to help with the wok a little we factoed the out of the seies. The limit of the patial sums is, Ê5 ˆ 5 lims = lim Á - - = Æ Æ Ë So, this seies is coveget (because the patial sums fom a coveget sequece) ad its value is, 5 Â = = + 4+ Note that it s ot always obvious if a seies is telescopig o ot util you ty to get the patial sums ad the see if they ae i fact telescopig. Thee is o test that will tell us that we ve got a telescopig seies ight off the bat. Also ote that just because you ca do patial factios o a seies tem does ot mea that the seies will be a telescopig seies. The followig seies, fo example, is ot a telescopig seies despite the fact that we ca patial factio the seies tems. + Ê ˆ = Á Ë+ + Â Â = = I ode fo a seies to be a telescopig we must get tems to cacel ad all of these tems ae positive ad so oe will cacel. Next, we eed to go back ad addess a issue that was fist aised i the pevious sectio. I that sectio we stated that the sum o diffeece of coveget seies was also coveget ad that the pesece of a multiplicative costat would ot affect the covegece of a seies. Now that we have a few moe seies i had let s wok a quick example showig that. 007 Paul Dawkis 99

209 Example 5 Detemie the value of the followig seies. Ê ˆ Â Á = Ë + 4+ Solutio To get the value of this seies all we eed to do is ewite it ad the use the pevious esults. Ê ˆ ÂÁ = Â - Â = Ë + 4+ = + 4+ = = 4Â Â + 4+ = = Ê 5 ˆ 96 = 4Á - Ë 5 86 =- 5 We did t discuss the covegece of this seies because it was the sum of two coveget seies ad that guaateed that the oigial seies would also be coveget. Hamoic Seies This is the thid ad fial seies that we e goig to look at i this sectio. Hee is the hamoic seies. Â = The hamoic seies is diveget ad we ll eed to wait util the ext sectio to show that. This seies is hee because it s got a ame ad so I wated to put it hee with the othe two amed seies that we looked at i this sectio. We e also goig to use the hamoic seies to illustate a couple of ideas about diveget seies that we ve aleady discussed fo coveget seies. We ll do that with the followig example. Example 6 Show that each of the followig seies ae diveget. 5 (a) Â = (b) Â = 4 Solutio 5 (a) Â = To see that this seies is diveget all we eed to do is use the fact that we ca facto a costat out of a seies as follows, 5 Â = 5Â = = 007 Paul Dawkis 00

210 Now,  = is diveget ad so five times this will still ot be a fiite umbe ad so the seies has to be diveget. I othe wods, if we multiply a diveget seies by a costat it will still be diveget. (b)  = 4 I this case we ll stat with the hamoic seies ad stip out the fist thee tems. Ê ˆ  =  fi  = = = 4 = 4 Á - = Ë 6 I this case we ae subtactig a fiite umbe fom a diveget seies. This subtactio will ot chage the divegece of the seies. We will eithe have ifiity mius a fiite umbe, which is still ifiity, o a seies with o value mius a fiite umbe, which will still have o value. Theefoe, this seies is diveget. Just like with coveget seies, addig/subtactig a fiite umbe fom a diveget seies is ot goig to chage the fact the covegece of the seies. So, some geeal ules about the covegece/divegece of a seies ae ow i ode. Multiplyig a seies by a costat will ot chage the covegece/divegece of the seies ad addig o subtactig a costat fom a seies will ot chage the covegece/divegece of the seies. These ae ice ideas to keep i mid. 007 Paul Dawkis 0

211 Itegal Test The last topic that we discussed i the pevious sectio was the hamoic seies. I that discussio we stated that the hamoic seies was a diveget seies. It is ow time to pove that statemet. This poof will also get us stated o the way to ou ext test fo covegece that we ll be lookig at. So, we will be tyig to pove that the hamoic seies, Â = diveges. We ll stat this off by lookig at a appaetly uelated poblem. Let s stat off by askig what the aea ude f ( x) = is o the iteval [, ). Fom the sectio o Impope Itegals we x kow that this is, Û Ù dx = ı x ad so we called this itegal diveget (yes, that s the same tem we e usig hee with seies.). So, just how does that help us to pove that the hamoic seies diveges? Well, ecall that we ca always estimate the aea by beakig up the iteval ito segmets ad the sketchig i ectagles ad usig the sum of the aea all of the ectagles as a estimate of the actual aea. Let s do that fo this poblem as well ad see what we get. We will beak up the iteval ito subitevals of width ad we ll take the fuctio value at the left edpoit as the height of the ectagle. The image below shows the fist few ectagles fo this aea. So, the aea ude the cuve is appoximately, 007 Paul Dawkis 0

212 ʈ ʈ ʈ ʈ ʈ Aª Á () + Á () + Á () + Á () + Á () + L Ë Ë Ë Ë4 Ë5 = L 4 5 = = Now ote a couple of thigs about this appoximatio. Fist, each of the ectagles oveestimates the actual aea ad secodly the fomula fo the aea is exactly the hamoic seies! Puttig these two facts togethe gives the followig, Aª > Û Â Ù dx = ı x Notice that this tells us that we must have,  > fi  = = = = Sice we ca t eally be lage tha ifiity the hamoic seies must also be ifiite i value. I othe wods, the hamoic seies is i fact diveget. So, we ve maaged to elate a seies to a impope itegal that we could compute ad it tus out that the impope itegal ad the seies have exactly the same covegece. Let s see if this will also be tue fo a seies that coveges. Whe discussig the Divegece Test we made the claim that  = coveges. Let s see if we ca do somethig simila to the above pocess to pove this. We will ty to elate this to the aea ude f ( x) = is o the iteval [ ) Impope Itegal sectio we kow that, ad so this itegal coveges. Û Ù ı x dx = x,. Agai, fom the We will oce agai ty to estimate the aea ude this cuve. We will do this i a almost idetical mae as the pevious pat with the exceptio that we will istead of usig the left ed poits fo the height of ou ectagles we will use the ight ed poits. Hee is a sketch of this case, 007 Paul Dawkis 0

213 I this case the aea estimatio is, Aª Ê Á ˆ () + Ê Á ˆ () + Ê Á ˆ () + Ê Á ˆ () + L Ë Ë Ë4 Ë5 = L 4 5 This time, ulike the fist case, the aea will be a udeestimatio of the actual aea ad the estimatio is ot quite the seies that we ae wokig with. Notice howeve that the oly diffeece is that we e missig the fist tem. This meas we ca do the followig, = < + Û Â L dx Ù = + = = ı x Aea Estimatio O, puttig all this togethe we see that,  = < With the hamoic seies this was all that we eeded to say that the seies was diveget. With this seies howeve, this is t quite eough. Fo istace -< ad if the seies did have a value of - the it would be diveget (whe we wat coveget). So, let s do a little moe wok. Fist, let s otice that all the seies tems ae positive (that s impotat) ad that the patial sums ae, s = i Because the tems ae all positive we kow that the patial sums must be a iceasig sequece. I othe wods, + s =  < s  = + i i i= i= i= 007 Paul Dawkis 04

214 I s + we ae addig a sigle positive tem oto s ad so must get lage. Theefoe, the patial sums fom a iceasig (ad hece mootoic) sequece. Also ote that, sice the tems ae all positive, we ca say, s =  <  < fi s < i= i = ad so the sequece of patial sums is a bouded sequece. I the secod sectio o Sequeces we gave a theoem that stated that a bouded ad mootoic sequece was guaateed to be coveget. This meas that the sequece of patial sums is a coveget sequece. So, who caes ight? Well ecall that this meas that the seies must the also be coveget! So, oce agai we wee able to elate a seies to a impope itegal (that we could compute) ad the seies ad the itegal had the same covegece. We wet though a fai amout of wok i both of these examples to detemie the covegece of the two seies. Luckily fo us we do t eed to do all this wok evey time. The ideas i these two examples ca be summaized i the followig test. Itegal Test Suppose that f ( x ) is a cotiuous, positive ad deceasig fuctio o the iteval [, ) f that = a the, Ú x dx is coveget so is  a k. = k. If f Ú x dx is diveget so is  a. k = k. If f A fomal poof of this test ca be foud at the ed of this sectio. k ad Thee ae a couple of thigs to ote about the itegal test. Fist, the lowe limit o the impope itegal must be the same value that stats the seies. Secod, the fuctio does ot actually eed to be deceasig ad positive eveywhee i the iteval. All that s eally equied is that evetually the fuctio is deceasig ad positive. I othe wods, it is okay if the fuctio (ad hece seies tems) iceases o is egative fo a while, but evetually the fuctio (seies tems) must decease ad be positive fo all tems. To see why this is tue let s suppose that the seies tems icease ad o ae egative i the age k N ad the decease ad ae positive fo N +. I this case the seies ca be witte as, N    a = a + a = k = k = N+ Now, the fist seies is othig moe tha a fiite sum (o matte how lage N is) of fiite tems ad so will be fiite. So the oigial seies will be coveget/diveget oly if the secod ifiite seies o the ight is coveget/diveget ad the test ca be doe o the secod seies as it satisfies the coditios of the test. 007 Paul Dawkis 05

215 A simila agumet ca be made usig the impope itegal as well. The equiemet i the test that the fuctio/seies be deceasig ad positive eveywhee i the age is equied fo the poof. I pactice howeve, we oly eed to make sue that the fuctio/seies is evetually a deceasig ad positive fuctio/seies. Also ote that whe computig the itegal i the test we do t actually eed to stip out the iceasig/egative potio sice the pesece of a small age o which the fuctio is iceasig/egative will ot chage the itegal fom coveget to diveget o fom diveget to coveget. Thee is oe moe vey impotat poit that must be made about this test. This test does NOT give the value of a seies. It will oly give the covegece/divegece of the seies. That s it. No value. We ca use the above seies as a pefect example of this. All that the test gave us was that,  < = So, we got a uppe boud o the value of the seies, but ot a actual value fo the seies. I fact, fom this poit o we will ot be askig fo the value of a seies we will oly be askig whethe a seies coveges o diveges. I a late sectio we look at estimatig values of seies, but eve i that sectio still wo t actually be gettig values of seies. Just fo the sake of completeess the value of this seies is kow. p  = = < 6 Let s wok a couple of examples. = Example Detemie if the followig seies is coveget o diveget.  = l Solutio I this case the fuctio we ll use is, f ( x) = xl x This fuctio is clealy positive ad if we make x lage the deomiato will get lage ad so the fuctio is also deceasig. Theefoe, all we eed to do is detemie the covegece of the followig itegal. t Û dx= lim Û Ù Ù dx u = l x ı xlx tæ ı xl x t ( ( x) ) ( ( t) l( l) ) = lim l l tæ = lim l l - tæ = The itegal is diveget ad so the seies is also diveget by the Itegal Test. 007 Paul Dawkis 06

216 Example Detemie if the followig seies is coveget o diveget. -  e = 0 Solutio The fuctio that we ll use i this example is, f ( x) = xe -x This fuctio is always positive o the iteval that we e lookig at. Now we eed to check that the fuctio is deceasig. It is ot clea that this fuctio will always be deceasig o the iteval give. We ca use ou Calculus I kowledge to help us howeve. The deivative of this fuctio is, -x e ( ) f x = - x This fuctio has two citical poits (which will tell us whee the deivative chages sig) at x =±. Sice we ae statig at = 0 we ca igoe the egative citical poit. Pickig a couple of test poits we ca see that the fuctio i iceasig o the iteval È0, Î ad it is deceasig o È Î, ). Theefoe, evetually the fuctio will be deceasig ad that s all that s equied fo us to use the Itegal Test. Ú t -x -x Ú xe dx= lim xe dx u =-x 0 tæ 0 Ê = lim tæ Á - e Ë -x ˆ 0 -t ˆ Ê = limá - e = tæ Ë The itegal is coveget ad so the seies must also be coveget by the Itegal Test. We ca use the Itegal test to get the followig fact/test fo some seies. Fact ( The p seies Test) If k > 0 the  coveges if p > ad diveges if p k p. = Sometimes the seies i this fact ae called p-seies ad so this fact is sometimes called the p- seies test. This fact follows diectly fom the Itegal Test ad a simila fact we say i the Impope Itegal sectio. This fact says that the itegal, Û Ù dx p ı k x coveges if p > ad diveges if p. t 007 Paul Dawkis 07

217 Usig the p-seies test makes it vey easy to detemie the covegece of some seies. Example Detemie if the followig seies ae coveget o diveget. (a)  7 = 4 (b)  = Solutio (a) I this case p = 7> ad so by this fact the seies is coveget. (b) Fo this seies p = ad so the seies is diveget by the fact. The last thig that we ll do i this sectio is give a quick poof of the Itegal Test. We ve essetially doe the poof aleady at the begiig of the sectio whe we wee motivatig the Itegal Test, but let s go though it fomally fo a geeal fuctio. Poof of Itegal Test Fist, fo the sake of the poof we ll be wokig with the seies  a =. The oigial test statemet was fo a seies that stated at a geeal = k ad while the poof ca be doe fo that it will be easie if we assume that the seies stats at =. Aothe way of dealig with the = k is we could do a idex shift ad stat the seies at = ad the do the Itegal Test. Eithe way povig the test fo = will be sufficiet. Let s stat off ad estimate the aea ude the cuve o the iteval [, ] ad we ll udeestimate the aea by takig ectagles of width oe ad whose height is the ight edpoit. This gives the followig figue. Now, ote that, f = a f ( ) = a L f = a 007 Paul Dawkis 08

218 The appoximate aea is the, Aª f + f ( ) + L+ f = a + a + L a ad we kow that this udeestimates the actual aea so, Â i= i a = a + a + L a < f x dx Now, let s suppose that Ú f ( x) dx is coveget ad so f Also, because f ( x ) is positive we kow that, Ú < f x dx f x dx Ú Ú Ú x dx must have a fiite value. This i tu meas that, Â i < Ú < Ú i= a f x dx f x dx Ou seies stats at = so this is t quite what we eed. Howeve, that s easy eough to deal with. Â Â Ú 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 patial sums, ae bouded above by M. s = Â ai i= Next, because the tems ae positive we also kow that, ad so the sequece { } = of patial sums { } + Â Â s s + a = a + a = a = s fi s s + i + i + + i= i= s s coveges ad hece ou seies = Â a = is also a iceasig sequece. So, we ow kow that the sequece is coveget. So, the fist pat of the test is pove. The secod pat is somewhat easie. This time let s ove estimate the aea ude the cuve by usig the left edpoits of iteval fo the height of the ectagles as show below. 007 Paul Dawkis 09

219 I this case the aea is appoximately, Aª f + f + L+ f - = a + a + L a - Sice we kow this oveestimates the aea we also the kow that, - Â - i - i= - s = a = a + a + L a > f x dx Now, suppose that Ú f ( x) dx is diveget. I this case this meas that f Ú Ú x dx Æ as Æ because f ( x) 0. Howeve, because - Æ as Æ we also kow that - Ú f x dx Æ. - Theefoe, sice s > - Ú f x dx we kow that as Æ we must have s - Æ. This i tu tells us that s Æ as Æ. s So, we ow kow that the sequece of patial sums, { } = Â a = is a diveget seies., is a diveget sequece ad so It is impotat to ote befoe leavig this sectio that i ode to use the Itegal Test the seies tems MUST evetually be deceasig ad positive. If they ae ot the the test does t wok. Also emembe that the test oly detemies the covegece of a seies ad does NOT give the value of the seies. 007 Paul Dawkis 0

220 Compaiso Test / Limit Compaiso Test I the pevious sectio we saw how to elate a seies to a impope itegal to detemie the covegece of a seies. While the itegal test is a ice test, it does foce us to do impope itegals which ae t always easy ad i some cases may be impossible to detemie the covegece of. Fo istace coside the followig seies.  0 = + I ode to use the Itegal Test we would have to itegate Û Ù dx ı 0 x + x ad I m ot eve sue if it s possible to do this itegal. Nicely eough fo us thee is aothe test that we ca use o this seies that will be much easie to use. Fist, let s ote that the seies tems ae positive. As with the Itegal Test that will be impotat i this sectio. Next let s ote that we must have x > 0 sice we ae itegatig o the iteval 0 x <. Likewise, egadless of the value of x we will always have x > 0. So, if we dop the x fom the deomiato the deomiato will get smalle ad hece the whole factio will get lage. So, + < Now, is a geometic seies ad we kow that sice will be,  0 = =  = = = 0 - < the seies will covege ad its value Now, if we go back to ou oigial seies ad wite dow the patial sums we get, s =  i + i i= Sice all the tems ae positive addig a ew tem will oly make the umbe lage ad so the sequece of patial sums must be a iceasig sequece. + s =  < i  = s i + + i + i The sice, i= i= 007 Paul Dawkis

221 + < ad because the tems i these two sequeces ae positive we ca also say that, s =  < i  < i  = fi s < + i i= i= = Theefoe, the sequece of patial sums is also a bouded sequece. The fom the secod sectio o sequeces we kow that a mootoic ad bouded sequece is also coveget. So, the sequece of patial sums of ou seies is a coveget sequece. This meas that the seies itself,  0 = + is also coveget. So, what did we do hee? We foud a seies whose tems wee always lage tha the oigial seies tems ad this ew seies was also coveget. The sice the oigial seies tems wee positive (vey impotat) this meat that the oigial seies was also coveget. To show that a seies (with oly positive tems) was diveget we could go though a simila agumet ad fid a ew diveget seies whose tems ae always smalle tha the oigial seies. I this case the oigial seies would have to take a value lage tha the ew seies. Howeve, sice the ew seies is diveget its value will be ifiite. This meas that the oigial seies must also be ifiite ad hece diveget. We ca summaize all this i the followig test. Compaiso Test Suppose that we have two seies  a ad  b with, 0 a b fo all ad a b fo all. The,. If  b is coveget the so is  a.. If  a is diveget the so is  b. I othe wods, we have two seies of positive tems ad the tems of oe of the seies is always lage tha the tems of the othe seies. The if the lage seies is coveget the smalle seies must also be coveget. Likewise, if the smalle seies is diveget the the lage seies must also be diveget. Note as well that i ode to apply this test we eed both seies to stat at the same place. A fomal poof of this test is at the ed of this sectio. Do ot misuse this test. Just because the smalle of the two seies coveges does ot say aythig about the lage seies. The lage seies may still divege. Likewise, just because we kow that the lage of two seies diveges we ca t say that the smalle seies will also divege! Be vey caeful i usig this test 007 Paul Dawkis

222 Recall that we had a simila test fo impope itegals back whe we wee lookig at itegatio techiques. So, if you could use the compaiso test fo impope itegals you ca use the compaiso test fo seies as they ae petty much the same idea. Note as well that the equiemet that a, b 0 ad a b eally oly eed to be tue evetually. I othe wods, if a couple of the fist tems ae egative o a b fo a couple of the fist few tems we e okay. As log as we evetually each a poit whee a, b 0 ad a b fo all sufficietly lage the test will wok. To see why this is tue let s suppose that the seies stat at = k ad that the coditios of the test ae oly tue fo fo N + ad fo k N at least oe of the coditios is ot tue. If we the look at  a (the same thig could be doe fo  b ) we get, N    a = a + a = k = k = N+ The fist seies is othig moe tha a fiite sum (o matte how lage N is) of fiite tems ad so will be fiite. So the oigial seies will be coveget/diveget oly if the secod ifiite seies o the ight is coveget/diveget ad the test ca be doe o the secod seies as it satisfies the coditios of the test. Let s take a look at some examples. Example Detemie if the followig seies is coveget o diveget.  = - cos Solutio Sice the cosie tem i the deomiato does t get too lage we ca assume that the seies tems will behave like, = which, as a seies, will divege. So, fom this we ca guess that the seies will pobably divege ad so we ll eed to fid a smalle seies that will also divege. Recall that fom the compaiso test with impope itegals that we detemied that we ca make a factio smalle by eithe makig the umeato smalle o the deomiato lage. I this case the two tems i the deomiato ae both positive. So, if we dop the cosie tem we will i fact be makig the deomiato lage sice we will o loge be subtactig off a positive quatity. Theefoe, The, sice > = - cos 007 Paul Dawkis

223 Â = diveges (it s hamoic o the p-seies test) by the Compaiso Test ou oigial seies must also divege. Example Detemie if the followig seies coveges o diveges. + Â 4 = + 5 Solutio I this case the + ad the +5 do t eally add aythig to the seies ad so the seies tems should behave petty much like = 4 which will covege as a seies. Theefoe, we ca guess that the oigial seies will covege ad we will eed to fid a lage seies which also coveges. This meas that we ll eithe have to make the umeato lage o the deomiato smalle. We ca make the deomiato smalle by doppig the +5. Doig this gives, + + < At this poit, otice that we ca t dop the + fom the umeato sice this would make the tem smalle ad that s ot what we wat. Howeve, this is actually all the futhe that we eed to go. Let s take a look at the followig seies. + Â 4 = Â 4 + Â 4 = = = = Â + Â 4 = = As show, we ca wite the seies as a sum of two seies ad both of these seies ae coveget by the p-seies test. Theefoe, sice each of these seies ae coveget we kow that the sum, + Â 4 = is also a coveget seies. Recall that the sum of two coveget seies will also be coveget. Now, sice the tems of this seies ae lage tha the tems of the oigial seies we kow that the oigial seies must also be coveget by the Compaiso Test. The compaiso test is a ice test that allows us to do poblems that eithe we could t have doe with the itegal test o at the best would have bee vey difficult to do with the itegal test. That does t mea that it does t have poblems of its ow. Coside the followig seies. Â 0 = Paul Dawkis 4

224 This is ot much diffeet fom the fist seies that we looked at. The oigial seies coveged because the gets vey lage vey fast ad will be sigificatly lage tha the. Theefoe, the does t eally affect the covegece of the seies i that case. The fact that we ae ow subtactig the off ow istead of addig the o eally should t chage the covegece. We ca say this because the gets vey lage vey fast ad the fact that we e subtactig off wo t eally chage the size of this tem fo all sufficietly lage value of. So, we would expect this seies to covege. Howeve, the compaiso test wo t wok with this seies. To use the compaiso test o this seies we would eed to fid a lage seies that we could easily detemie the covegece of. I this case we ca t do what we did with the oigial seies. If we dop the we will make the deomiato lage (sice the was subtacted off) ad so the factio will get smalle ad just like whe we looked at the compaiso test fo impope itegals kowig that the smalle of two seies coveges does ot mea that the lage of the two will also covege. So, we will eed somethig else to do help us detemie the covegece of this seies. The followig vaiat of the compaiso test will allow us to detemie the covegece of this seies. Limit Compaiso Test Suppose that we have two seies  a ad  b with a, b 0 fo all. Defie, a c = lim Æ b If c is positive (i.e. c > 0 ) ad is fiite (i.e. c <) the eithe both seies covege o both seies divege. The poof of this test is at the ed of this sectio. Note that it does t eally matte which seies tem is i the umeato fo this test, we could just have easily defied c as, b c = lim Æ a ad we would get the same esults. To see why this is, coside the followig two defiitios. a b c= lim c = lim Æ b Æ a Stat with the fist defiitio ad ewite it as follows, the take the limit. a c= lim = lim = = Æ b Æ b b lim c a Æ a I othe wods, 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 esults fom the test so do t woy about which seies tems should be i the umeato ad which should be i the deomiato. Choose this to make the limit easy to compute. 007 Paul Dawkis 5

225 Also, this eally is a compaiso test i some ways. If c is positive ad fiite this is sayig that both of the seies tems will behave i geeally the same fashio ad so we ca expect the seies themselves to also behave i a simila fashio. If c = 0 o c = we ca t say this ad so the test fails to give ay ifomatio. The limit i this test will ofte be witte as, c= lim a Æ b sice ofte both tems will be factios ad this will make the limit easie to deal with. Let s see how this test woks. Example Detemie if the followig seies coveges o diveges.  0 = - Solutio To use the limit compaiso test we eed to fid a secod seies that we ca detemie the covegece of easily ad has what we assume is the same covegece as the give seies. O top of that we will eed to choose the ew seies i such a way as to give us a easy limit to compute fo c. We ve aleady guessed that this seies coveges ad sice it s vaguely geometic let s use  0 = as the secod seies. We kow that this seies coveges ad thee is a chace that sice both seies have the i it the limit wo t be too bad. Hee s the limit. - c = lim Æ = lim- Æ Now, we ll eed to use L Hospital s Rule o the secod tem i ode to actually evaluate this limit. c = -lim Æ l = So, c is positive ad fiite so by the Compaiso Test both seies must covege sice coveges.  0 = 007 Paul Dawkis 6

226 Example 4 Detemie if the followig seies coveges o diveges. 4 + Â 7 = + Solutio Factios ivolvig oly polyomials o polyomials ude adicals will behave i the same way as the lagest powe of will behave i the limit. So, the tems i this seies should behave as, = = 7 7 ad as a seies this will divege by the p-seies test. I fact, this would make a ice choice fo ou secod seies i the limit compaiso test so let s use it lim = lim Æ 7 Æ + 7Ê ˆ Á+ 4 Ë 7 Ê ˆ Á4 + = lim Ë Æ = = 4 = c So, c is positive ad fiite ad so both limits will divege sice diveges. Â = Fially, to see why we eed to c must be positive ad fiite (i.e. c 0 ad c ) coside the followig two seies. Â Â = = The fist diveges ad the secod coveges. Now compute each of the followig limits. lim g = lim= lim g = lim = 0 Æ Æ Æ Æ I the fist case the limit fom the limit compaiso test yields c = ad i the secod case the limit yields c = 0. Clealy, both seies do ot have the same covegece. Note howeve, that just because we get 0 c = o c = does t mea that the seies will have the opposite covegece. To see this coside the seies, 007 Paul Dawkis 7

227   = = Both of these seies covege ad hee ae the two possible limits that the limit compaiso test uses. lim g = lim = 0 lim g = lim = Æ Æ Æ Æ So, eve though both seies had the same covegece we got both c = 0 ad c =. The poit of all of this is to emid us that if we get c = 0 o c = fom the limit compaiso test we will kow that we have chose the secod seies icoectly ad we ll eed to fid a diffeet choice i ode to get ay ifomatio about the covegece of the seies. We ll close out this sectio with poofs of the two tests. Poof of Compaiso Test The test statemet did ot specify whee each seies should stat. We oly eed to equie that they stat at the same place so to help with the poof we ll assume that the seies stat at =. If the seies do t stat at = the poof ca be edoe i exactly the same mae o you could use a idex shift to stat the seies at = ad the this poof will apply. We ll stat off with the patial sums of each seies.  s = a t = b i i i= i= Let s otice a couple of ice facts about these two patial sums. Fist, because a, b 0 we kow that,  + s s + a = a + a = a = s fi s s + i + i + + i= i=   +  t t + b = b + b = b = t fi t t + i + i + + i= i= So, both patial sums fom iceasig sequeces. Also, because a b fo all we kow that we must have s t fo all. With these pelimiay facts out of the way we ca poceed with the poof of the test itself.  Let s stat out by assumig that  b is a coveget seies. Sice 0 = t = Âbi  bi i= i= b we kow that, 007 Paul Dawkis 8

228 Howeve, we also have established that s t fo all ad so fo all we also have, s  bi i= Fially sice  b is a coveget seies it must have a fiite value ad so the patial sums, s = ae bouded above. Theefoe, fom the secod sectio o sequeces we kow that a mootoic ad bouded sequece is also coveget ad so { s } is a coveget ad so =  a is = coveget. Next, let s assume that  a is diveget. Because a 0 we the kow that we must have = s Æ as Æ. Howeve, we also kow that fo all we have s t ad theefoe we also kow that t Æ as Æ. t is a diveget sequece as so =  b is diveget. = So, { } Poof of Limit Compaiso Test Because 0 < c < we ca fid two positive ad fiite umbes, m ad M, such that m< c< M. a a Now, because lim c = we kow that fo lage eough the quotiet must be close to c Æ b b ad so thee must be a positive itege N such that if > N we also have, a m< < M b Multiplyig though by b gives, povided > N. mb < a < Mb Now, if  b diveges the so does  mb ad so sice mb < a fo all sufficietly lage by the Compaiso Test  a also diveges. Likewise, if  b coveges the so does  Mb ad sice a < Mb fo all sufficietly lage by the Compaiso Test  a also coveges. 007 Paul Dawkis 9

229 Alteatig Seies Test The last two tests that we looked at fo seies covegece have equied that all the tems i the seies be positive. Of couse thee ae may seies out thee that have egative tems i them ad so we ow eed to stat lookig at tests fo these kids of seies. The test that we ae goig to look ito i this sectio will be a test fo alteatig seies. A alteatig seies is ay seies, a, fo which the seies tems ca be witte i oe of the followig two foms.  a = - b b 0 + a = - b b 0 Thee ae may othe ways to deal with the alteatig sig, but they ca all be witte as oe of the two foms above. Fo istace, + (- ) = (-) (- ) = (-) (- ) = (-) (- ) = (-) Thee ae of couse may othes, but they all follow the same basic patte of educig to oe of the fist two foms give. If you should happe to u ito a diffeet fom tha the fist two, do t woy about covetig it to oe of those foms, just be awae that it ca be ad so the test fom this sectio ca be used. Alteatig Seies Test Suppose that we have a seies fo all. The if,  a ad eithe a = - b o a + = - b whee b 0. limb = 0 ad, Æ. { b } is a deceasig sequece the seies  a is coveget. A poof of this test is at the ed of the sectio. Thee ae a couple of thigs to ote about this test. Fist, ulike the Itegal Test ad the Compaiso/Limit Compaiso Test, this test will oly tell us whe a seies coveges ad ot if a seies will divege. Secodly, i the secod coditio all that we eed to equie is that the seies tems, b will be evetually deceasig. It is possible fo the fist few tems of a seies to icease ad still have the test be valid. All that is equied is that evetually we will have b b + fo all afte some poit. To see why this is coside the followig seies, 007 Paul Dawkis 0

230 { } Let s suppose that fo N deceasig. The seies ca the be witte as, Â (-) b = b is ot deceasig ad that fo N N Â - b = Â - b + Â (-) b = = = N+ b is + { } The fist seies is a fiite sum (o matte how lage N is) of fiite tems ad so we ca compute its value ad it will be fiite. The covegece of the seies will deped solely o the covegece of the secod (ifiite) seies. If the secod seies has a fiite value the the sum of two fiite values is also fiite ad so the oigial seies will covege to a fiite value. O the othe had if the secod seies is diveget eithe because its value is ifiite o it does t have a value the addig a fiite umbe oto this will ot chage that fact ad so the oigial seies will be diveget. The poit of all this is that we do t eed to equie that the seies tems be deceasig fo all. We oly eed to equie that the seies tems will evetually be deceasig sice we ca always stip out the fist few tems that ae t actually deceasig ad look oly at the tems that ae actually deceasig. Note that, i pactice, we do t actually stip out the tems that ae t deceasig. All we do is check that evetually the seies tems ae deceasig ad the apply the test. Let s wok a couple of examples. Example Detemie if the followig seies is coveget o diveget. Solutio Fist, idetify the b fo the test. (- ) (- ) + Â = + + Â = Â (- ) b = = = Now, all that we eed to do is u though the two coditios i the test. limb = lim = 0 Æ Æ b = > = b+ + Both coditios ae met ad so by the Alteatig Seies Test the seies must covege. The seies fom the pevious example is sometimes called the Alteatig Hamoic Seies. Also, the (- ) + could be Alteatig Hamoic Seies. 007 Paul Dawkis - o ay othe fom of alteatig sig ad we d still call it a

231 I the pevious example it was easy to see that the seies tems deceased sice iceasig oly iceased the deomiato fo the tem ad hece made the tem smalle. I geeal howeve we will eed to esot to Calculus I techiques to pove the seies tems decease. We ll see a example of this i a bit. Example Detemie if the followig seies is coveget o diveget. Solutio Fist, idetify the b fo the test. (-) (-) Â = + 5 = (-) fi b = Â Â = = Let s check the coditios. limb = lim = 0 Æ Æ + 5 So, the fist coditio is t met ad so thee is o easo to check the secod. Sice this coditio is t met we ll eed to use aothe test to check covegece. I these cases whee the fist coditio is t met it is usually best to use the divegece test. ( ) Ë ( lim ) - Ê ˆ lim = lim - lim Á Æ + 5 Æ Æ + 5 = - Æ = lim - does't exist Æ This limit does t exist ad so by the Divegece Test this seies diveges. Example Detemie if the followig seies is coveget o diveget. - (-) Â = Solutio Notice that i this case the expoet o the - is t o +. That wo t chage how the test woks howeve so we wo t woy about that. I this case we have, b = + 4 so let s check the coditios. The fist is easy eough to check. limb = lim = 0 Æ Æ Paul Dawkis

232 The secod coditio equies some wok howeve. It is ot immediately clea that these tems will decease. Iceasig to + will icease both the umeato ad the deomiato. Iceasig the umeato says the tem should also icease while iceasig the deomiato says that the tem should decease. Sice it s ot clea which of these will wi out we will eed to esot to Calculus I techiques to show that the tems decease. Let s stat with the followig fuctio ad its deivative. x 4-x f ( x) = f ( x) = x+ 4 x x+ 4 Now, thee ae thee citical poits fo this fuctio, x=- 4, x= 0, ad x= 4. The fist is outside the boud of ou seies so we wo t eed to woy about that oe. Usig the test poits, f () = f ( 5) = ad so we ca see that the fuctio i iceasig o 0 x 4 ad deceasig o x 4. f = b we kow as well that the b ae also iceasig o 0 4 ad Theefoe, sice deceasig o 4. The b ae the evetually deceasig ad so the secod coditio is met. Both coditios ae met ad so by the Alteatig Seies Test the seies must be covegig. As the pevious example has show, we sometimes eed to do a fai amout of wok to show that the tems ae deceasig. Do ot just make the assumptio that the tems will be deceasig ad let it go at that. Let s do oe moe example just to make a poit. Example 4 Detemie if the followig seies is coveget o diveget. cos p  = Solutio The poit of this poblem is eally just to ackowledge that it is i fact a alteatig seies. To see this we eed to ackowledge that, ad so the seies is eally, cos ( p) (- ) = = ( p ) = (- ) cos = fi b =   Checkig the two coditio gives, limb = lim = 0 Æ Æ 007 Paul Dawkis

233 b = > = b + + The two coditios of the test ae met ad so by the Alteatig Seies Test the seies is coveget. It should be poited out that the ewite we did i pevious example oly woks because is a itege ad because of the pesece of the p. Without the p we could t do this ad if was t guaateed to be a itege we could t do this. Let s close this sectio out with a poof of the Alteatig Seies Test. Poof of Alteatig Seies Test Without loss of geeality we ca assume that the seies stats at =. If ot we could modify the poof below to meet the ew statig place o we could do a idex shift to get the seies to stat at =. Fist, otice that because the tems of the sequece ae deceasig fo ay two successive tems we ca say, b -b + 0 Now, let s take a look at the eve patial sums. s = b -b 0 s = b - b + b - b = s + b -b s because b -b s = s + b -b s because b -b M s = s + b -b s because b -b So, { s } is a iceasig sequece. Next, we ca also wite the geeal tem as, s = b - b + b - b + b + L- b + b -b L = b - b -b - b - b + - b -b -b Each of the quatities i paethesis ae positive ad by assumptio we kow that b is also positive. So, this tells us that s b fo all. We ow kow that { s } is a iceasig sequece that is bouded above ad so we kow that it must also covege. So, let s assume that its limit is s o, lim s = s Æ Next, we ca quickly detemie the limit of the sequece of odd patial sums, { } s +, as follows, 007 Paul Dawkis 4

234 lims = lim s + b = lims + limb = s+ 0= s Æ Æ Æ Æ So, we ow kow that both { s } ad { s + } ae coveget sequeces ad they both have the same limit ad so we also kow that { s } is a coveget sequece with a limit of s. This i tu tells us that  a is coveget. 007 Paul Dawkis 5

235 Absolute Covegece Whe we fist talked about seies covegece we biefly metioed a stoge type of covegece but did t do aythig with it because we did t have ay tools at ou disposal that we could use to wok poblems ivolvig it. We ow have some of those tools so it s ow time to talk about absolute covegece i detail. Fist, let s go back ove the defiitio of absolute covegece. Defiitio A seies  a is called absolutely coveget if  a is coveget. If  a is coveget ad  a is diveget we call the seies coditioally coveget. We also have the followig fact about absolute covegece. Fact If  a is absolutely coveget the it is also coveget. Poof Fist otice that say, a is eithe a o it is - a depedig o its sig. This meas that we ca the 0 a + a a Now, sice we ae assumig that  a is coveget the  a is also coveget sice we ca just facto the out of the seies ad times a fiite value will still be fiite. This howeve allows us to use the Compaiso Test to say that  a + a is also a coveget seies. Fially, we ca wite,    a = a + a - a ad so  a is the diffeece of two coveget seies ad so is also coveget. This fact is oe of the ways i which absolute covegece is a stoge type of covegece. Seies that ae absolutely coveget ae guaateed to be coveget. Howeve, seies that ae coveget may o may ot be absolutely coveget. Let s take a quick look at a couple of examples of absolute covegece. Example Detemie if each of the followig seies ae absolute coveget, coditioally coveget o diveget. 007 Paul Dawkis 6

236 Solutio (a) = (a) (b) = (- ) Â [Solutio] (- ) + Â [Solutio] = si (c) (- ) Â Â [Solutio] = This is the alteatig hamoic seies ad we saw i the last sectio that it is a coveget seies so we do t eed to check that hee. So, let s see if it is a absolutely coveget seies. To do this we ll eed to check the covegece of. (- ) Â = Â = = This is the hamoic seies ad we kow fom the itegal test sectio that it is diveget. Theefoe, this seies is ot absolutely coveget. It is howeve coditioally coveget sice the seies itself does covege. [Retu to Poblems] (b) (- ) + Â = I this case let s just check absolute covegece fist sice if it s absolutely coveget we wo t eed to bothe checkig covegece as we will get that fo fee. (- ) + = Â Â = = This seies is coveget by the p-seies test ad so the seies is absolute coveget. Note that this does say as well that it s a coveget seies. [Retu to Poblems] si (c) Â = I this pat we eed to be a little caeful. Fist, this is NOT a alteatig seies ad so we ca t use ay tools fom that sectio. What we ll do hee is check fo absolute covegece fist agai sice that will also give covegece. This meas that we eed to check the covegece of the followig seies. 007 Paul Dawkis 7

237 si =   si = = To do this we ll eed to ote that - si fi si ad so we have, si Now we kow that  = coveges by the p-seies test ad so by the Compaiso Test we also kow that si coveges.  = Theefoe the oigial seies is absolutely coveget (ad hece coveget). [Retu to Poblems] Let s close this sectio off by ecappig a topic we saw ealie. Whe we fist discussed the covegece of seies i detail we oted that we ca t thik of seies as a ifiite sum because some seies ca have diffeet sums if we eaage thei tems. I fact, we gave two eaagemets of a Alteatig Hamoic seies that gave two diffeet values. We closed that sectio off with the followig fact, Facts Give the seiesâ a,. If  a is absolutely coveget ad its value is s the ay eaagemet of  a will also have a value of s. 4. If  a is coditioally coveget ad is ay eal umbe the thee is a eaagemet of  a whose value will be. Now that we ve got the tools ude ou belt to detemie absolute ad coditioal covegece we ca make a few moe commets about this. Fist, as we showed above i Example a a Alteatig Hamoic is coditioally coveget ad so o matte what value we chose thee is some eaagemet of tems that will give that value. Note as well that this fact does ot tell us what that eaagemet must be oly that it does exist. Next, we showed i Example b that, 007 Paul Dawkis 8

238 (- ) +  = is absolutely coveget ad so o matte how we eaage the tems of this seies we ll always get the same value. I fact, it ca be show that the value of this seies is, + (-) p  =- = 007 Paul Dawkis 9

239 Ratio Test I this sectio we ae goig to take a look at a test that we ca use to see if a seies is absolutely coveget o ot. Recall that if a seies is absolutely coveget the we will also kow that it s coveget ad so we will ofte use it to simply detemie the covegece of a seies. Befoe poceedig with the test let s do a quick emide of factoials. This test will be paticulaly useful fo seies that cotai factoials (ad we will see some i the applicatios) so let s make sue we ca deal with them befoe we u ito them i a example. If is a itege such that 0 the factoial is defied as, L! = - - if 0! = by defiitio Let s compute a couple eal quick.! = () () () ()! = =! = = 6 4! = 4 = 4 5! = 54 = 0 I the last computatio above, otice that we could ewite the factoial i a couple of diffeet ways. Fo istace, 5! = 54 ( ) = 54! 444 4! () 5! = 54 = 54! 44 I geeal we ca always stip out tems fom a factoial as follows.! = - - L -k - k+ L L L(! )! = - - -k - k+! = - - -k -k- We will eed to do this o occasio so do t foget about it. Also, whe dealig with factoials we eed to be vey caeful with paethesis. Fo istace,!! as we ca see if we wite each of the followig factoials out. ( )! = ( -)( -) L( )! = È( -)( -) L( )() Î 007 Paul Dawkis 0

240 Agai, we will u acoss factoials with paethesis so do t dop them. This is ofte oe of the moe commo mistakes that studets make whe the fist u acoss factoials. Okay, we ae ow eady fo the test. Ratio Test Suppose we have the seies  a. Defie, a lim + L = Æ a The,. if L < the seies is absolutely coveget (ad hece coveget).. if L > the seies is diveget.. if L = the seies may be diveget, coditioally coveget, o absolutely coveget. A poof of this test is at the ed of the sectio. Notice that i the case of L = the atio test is petty much wothless ad we would eed to esot to a diffeet test to detemie the covegece of the seies. Also, the absolute value bas i the defiitio of L ae absolutely equied. If they ae ot thee it will be possible fo us to get the icoect aswe. Let s take a look at some examples. Example Detemie if the followig seies is coveget o diveget. (-0)  + = 4 ( + ) Solutio With this fist example let s be a little caeful ad make sue that we have eveythig dow coectly. Hee ae the seies tems a. (-0) + ( + ) a = 4 Recall that to compute a + all that we eed to do is substitute + fo all the s i a. Now, to defie L we will use, a ( ) ( + + ) ( + ) = = ( ) L= lim a + Æ a sice this will be a little easie whe dealig with factios as we ve got hee. So, 007 Paul Dawkis

241 + ( ) ( ) ( + ) lim Æ ( ) lim L= Æ = = lim 6 Æ + 0 = < 6 So, L< ad so by the Ratio Test the seies coveges absolutely ad hece will covege. As see i the pevious example thee is usually a lot of cacelig that will happe i these. Make sue that you do this cacelig. If you do t do this kid of cacelig it ca make the limit faily difficult. Example Detemie if the followig seies is coveget o diveget.! Â 0 5 = Solutio Now that we ve woked oe i detail we wo t go ito quite the detail with the est of these. Hee is the limit. ( +! ) 5 ( +! ) L= lim = lim Æ + 5! Æ 5! I ode to do this limit we will eed to elimiate the factoials. We simply ca t do the limit with the factoials i it. To elimiate the factoials we will ecall fom ou discussio o factoials above that we ca always stip out tems fom a factoial. If we do that with the umeato (i this case because it s the lage of the two) we get, ( + )! L= lim Æ 5! at which poit we ca cacel the! fo the umeato a deomiato to get, ( + ) L= lim => Æ 5 So, by the Ratio Test this seies diveges. Example Detemie if the followig seies is coveget o diveget. Â = -! Solutio I this case be caeful i dealig with the factoials. 007 Paul Dawkis

242 ( ) ( ) ( + - ) ( + ) ( - ) ( + ) ( ) + -! L= lim Æ!! = lim Æ! ( + ) ( - ) ( + )( - ) ( + ) ( + )( )! = lim Æ! = lim Æ = 0< So, by the Ratio Test this seies coveges absolutely ad so coveges. Example 4 Detemie if the followig seies is coveget o diveget. 9 Â = - Solutio Do ot mistake this fo a geometic seies. The i the deomiato meas that this is t a geometic seies. So, let s compute the limit. L= lim Æ (-) + (- ) ( + ) 9 9 = lim Æ (- )( + ) 9 = lim Æ + 9 = > Theefoe, by the Ratio Test this seies is diveget. I the pevious example the absolute value bas wee equied to get the coect aswe. If we 9 had t used them we would have gotte L=- < which would have implied a coveget seies! Now, let s take a look at a couple of examples to see what happes whe we get L=. Recall that the atio test will ot tell us aythig about the covegece of these seies. I both of these examples we will fist veify that we get L= ad the use othe tests to detemie the covegece Paul Dawkis

243 Example 5 Detemie if the followig seies is coveget o diveget. Solutio Let s fist get L. (-) Â = (- ) + + Æ ( ) ( ) L= lim = lim = Æ So, as implied ealie we get L = which meas the atio test is o good fo detemiig the covegece of this seies. We will eed to esot to aothe test fo this seies. This seies is a alteatig seies ad so let s check the two coditios fom that test. limb = lim = 0 Æ Æ + b = > = b The two coditios ae met ad so by the Alteatig Seies Test this seies is coveget. We ll leave it to you to veify this seies is also absolutely coveget. Example 6 Detemie if the followig seies is coveget o diveget. + Â = Solutio Hee s the limit ( + )( + 7) L= lim = lim = Æ Æ Agai, the atio test tells us othig hee. We ca howeve, quickly use the divegece test o this. I fact that pobably should have bee ou fist choice o this oe ayway. + lim = 0 Æ + 7 By the Divegece Test this seies is diveget. So, as we saw i the pevious two examples if we get L = fom the atio test the seies ca be eithe coveget o diveget. Thee is oe moe thig that we should ote about the atio test befoe we move oto the ext sectio. The last seies was a polyomial divided by a polyomial ad we saw that we got L = fom the atio test. This will always happe with atioal expessio ivolvig oly polyomials o polyomials ude adicals. So, i the futue it is t eve woth it to ty the atio test o these kids of poblems sice we ow kow that we will get L =. 007 Paul Dawkis 4

244 Also, i the secod to last example we saw a example of a alteatig seies i which the positive tem was a atioal expessio ivolvig polyomials ad agai we will always get L = i these cases. Let s close the sectio out with a poof of the Ratio Test. Poof of Ratio Test Fist ote that we ca assume without loss of geeality that the seies will stat at = as we ve doe fo all ou seies test poofs. Let s stat off the poof hee by assumig that L < ad we ll eed to show that  a is absolutely coveget. To do this let s fist ote that because L < thee is some umbe such that L< <. Now, ecall that, a L = lim Æ a ad because we also have chose such that L< thee is some N such that if N we will have, a+ < fi a+ < a a + Next, coside the followig, So, fo k =,,, K we have look at the followig seies. a N+ < a a < a < a N N+ N+ N a < a < a N+ N+ N M a < a < a k N+ k N+ k- N k an+ k < an. Just why is this impotat? Well we ca ow  an k= 0 k This is a geometic seies ad because 0< < we i fact kow that it is a coveget seies. k Also because an+ k < an by the Compaiso test the seies is coveget. Howeve sice,  a =  a N+ k = N+ k= N    a = a + a = = = N+ 007 Paul Dawkis 5

245 we kow that  a = is also coveget sice the fist tem o the ight is a fiite sum of fiite tems ad hece fiite. Theefoe  a is absolutely coveget. = Next, we eed to assume that L > ad we ll eed to show that that, a L = lim Æ a +  a is diveget. Recallig ad because L > we kow that thee must be some N such that if N we will have, a+ > fi a+ > a a Howeve, if a+ > a fo all N the we kow that, lim a 0 Æ because the tems ae gettig lage ad guaateed to ot be egative. This i tu meas that, lima 0 Theefoe, by the Divegece Test Æ Â a is diveget. Fially, we eed to assume that L = ad show that we could get a seies that has ay of the thee possibilities. To do this we just eed a seies fo each case. We ll leave the details of checkig to you but all thee of the followig seies have L = ad each oe exhibits oe of the possibilities.  absolutely coveget = (-)  =  = coditioally coveget diveget 007 Paul Dawkis 6

246 Root Test This is the last test fo seies covegece that we e goig to be lookig at. As with the Ratio Test this test will also tell whethe a seies is absolutely coveget o ot athe tha simple covegece. Root Test Suppose that we have the seies  a. Defie, L= lim a = lim a Æ The, 4. if L < the seies is absolutely coveget (ad hece coveget). 5. if L > the seies is diveget. 6. if L = the seies may be diveget, coditioally coveget, o absolutely coveget. A poof of this test is at the ed of the sectio. As with the atio test, if we get L = the oot test will tell us othig ad we ll eed to use aothe test to detemie the covegece of the seies. Also ote that if L = i the Ratio Test the the Root Test will also give L =. Æ We will also eed the followig fact i some of these poblems. Fact lim = Æ Let s take a look at a couple of examples. Example Detemie if the followig seies is coveget o diveget.  + = Solutio Thee eally is t much to these poblems othe tha computig the limit ad the usig the oot test. Hee is the limit fo this poblem. So, by the Root Test this seies is diveget. L Æ + Æ + lim lim = = = => Example Detemie if the followig seies is coveget o diveget. Solutio Agai, thee is t too much to this seies. Ê5-  Á = 0 Ë Paul Dawkis 7 ˆ

247 Ê5- ˆ 5- - L= lim Á lim = = = < Æ 7 Æ Ë Theefoe, by the Root Test this seies coveges absolutely ad hece coveges. Note that we had to keep the absolute value bas o the factio util we d take the limit to get the sig coect. Example Detemie if the followig seies is coveget o diveget. - Solutio Hee s the limit fo this seies. (- )  = L= lim = lim = = > Æ Æ Afte usig the fact fom above we ca see that the Root Test tells us that this seies is diveget. Poof of Root Test Fist ote that we ca assume without loss of geeality that the seies will stat at = as we ve doe fo all ou seies test poofs. Also ote that this poof is vey similia to the poof of the Ratio Test. Let s stat off the poof hee by assumig that L< ad we ll eed to show that  a is absolutely coveget. To do this let s fist ote that because L< thee is some umbe such that L< <. Now, ecall that, L= lim a = lim a Æ ad because we also have chose such that L< thee is some N such that if N we will have, Now the seies Æ a < fi a <  = 0 is a geometic seies ad because 0< < we i fact kow that it is a coveget seies. Also because a < N by the Compaiso test the seies  a = N 007 Paul Dawkis 8

248 is coveget. Howeve sice, N-    a = a + a = = = N we kow that  a is also coveget sice the fist tem o the ight is a fiite sum of fiite = tems ad hece fiite. Theefoe  a is absolutely coveget. = Next, we eed to assume that L> ad we ll eed to show that a is diveget. Recallig that, L= lim a = lim a Æ ad because L> we kow that thee must be some N such that if N we will have, a Æ Â > fi a > = Howeve, if a > fo all N the we kow that, lim a 0 Æ This i tu meas that, lima 0 Æ Theefoe, by the Divegece Test  a is diveget. Fially, we eed to assume that L= ad show that we could get a seies that has ay of the thee possibilities. To do this we just eed a seies fo each case. We ll leave the details of checkig to you but all thee of the followig seies have L= ad each oe exhibits oe of the possibilities.  absolutely coveget = (-)  =  = coditioally coveget diveget 007 Paul Dawkis 9

249 Stategy fo Seies Now that we ve got all of ou tests out of the way it s time to thik about ogaizig all of them ito a geeal set of guidelies to help us detemie the covegece of a seies. Note that these ae a geeal set of guidelies ad because some seies ca have moe tha oe test applied to them we will get a diffeet esult depedig o the path that we take though this set of guidelies. I fact, because moe tha oe test may apply, you should always go completely though the guidelies ad idetify all possible tests that ca be used o a give seies. Oce this has bee doe you ca idetify the test that you feel will be the easiest fo you to use. With that said hee is the set of guidelies fo detemiig the covegece of a seies.. With a quick glace does it look like the seies tems do t covege to zeo i the limit, i.e. does lima 0? If so, use the Divegece Test. Note that you should oly do the Æ divegece test if a quick glace suggests that the seies tems may ot covege to zeo i the limit. -. Is the seies a p-seies (  ) o a geometic seies ( a p  o  a )? If so use = 0 = the fact that p-seies will oly covege if p > ad a geometic seies will oly covege if <. Remembe as well that ofte some algebaic maipulatio is equied to get a geometic seies ito the coect fom.. Is the seies simila to a p-seies o a geometic seies? If so, ty the Compaiso Test. 4. Is the seies a atioal expessio ivolvig oly polyomials o polyomials ude adicals (i.e. a factio ivolvig oly polyomials o polyomials ude adicals)? If so, ty the Compaiso Test ad/o the Limit Compaiso Test. Remembe howeve, that i ode to use the Compaiso Test ad the Limit Compaiso Test the seies tems all eed to be positive. 5. Does the seies cotai factoials o costats aised to powes ivolvig? If so, the the Ratio Test may wok. Note that if the seies tem cotais a factoial the the oly test that we ve got that will wok is the Ratio Test. 6. Ca the seies tems be witte i the fom a = (- ) b o a the Alteatig Seies Test may wok. 7. Ca the seies tems be witte i the fom 8. wok. a + = - b? If so, the = b? If so, the the Root Test may If a = f fo some positive, deceasig fuctio ad f the the Itegal Test may wok. Ú x dx is easy to evaluate a 007 Paul Dawkis 40

250 Agai, emembe that these ae oly a set of guidelies ad ot a set of had ad fast ules to use whe tyig to detemie the best test to use o a seies. If moe tha oe test ca be used ty to use the test that will be the easiest fo you to use ad emembe that what is easy fo someoe else may ot be easy fo you! Also just so we ca put all the tests ito oe place hee is a quick listig of all the test that we ve got. Divegece Test If lima 0 the a will divege Æ Â Itegal Test Suppose that f ( x) is a positive, deceasig fuctio o the iteval [ k,) ad that f = a the, Ú x dx is coveget so is  a k. = k. If f Ú x dx is diveget so is  a. k = k. If f Compaiso Test Suppose that we have two seies  a ad  b with a, b 0 fo all ad a b fo all. The,. If  b is coveget the so is  a.. If  a is diveget the so is  b. Limit Compaiso Test Suppose that we have two seies  a ad  b with, 0 a b fo all. Defie, a c = lim Æ b If c is positive (i.e. c > 0 ) ad is fiite (i.e. c <) the eithe both seies covege o both seies divege. Alteatig Seies Test Suppose that we have a seies fo all. The if,  a ad eithe a = (- ) b o a + = - b whee b 0. limb = 0 ad, Æ. { b } is evetually a deceasig sequece the seies  a is coveget 007 Paul Dawkis 4

251 Ratio Test Suppose we have the seies  a. Defie, a L = lim Æ a The,. if L < the seies is absolutely coveget (ad hece coveget).. if L > the seies is diveget.. if L = the seies may be diveget, coditioally coveget, o absolutely coveget. Root Test Suppose that we have the seies  a. Defie, + L= lim a = lim a Æ The,. if L < the seies is absolutely coveget (ad hece coveget).. if L > the seies is diveget.. if L = the seies may be diveget, coditioally coveget, o absolutely coveget. Æ 007 Paul Dawkis 4

252 Estimatig the Value of a Seies We have ow spet quite a few sectios detemiig the covegece of a seies, howeve, with the exceptio of geometic ad telescopig seies, we have ot talked about fidig the value of a seies. This is usually a vey difficult thig to do ad we still ae t goig to talk about how to fid the value of a seies. What we will do is talk about how to estimate the value of a seies. Ofte that is all that you eed to kow. Befoe we get ito how to estimate the value of a seies let s emid ouselves how seies covegece woks. It does t make ay sese to talk about the value of a seies that does t covege ad so we will be assumig that the seies we e wokig with coveges. Also, as we ll see the mai method of estimatig the value of seies will come out of this discussio. So, let s stat with the seiesâ a (the statig poit is ot impotat, but we eed a statig = poit to do the wok) ad let s suppose that the seies coveges to s. Recall that this meas that if we get the patial sums, s = Â ai the they will fom a coveget sequece ad its limit is s. I othe wods, lim s = s Æ Now, just what does this mea fo us? Well, sice this limit coveges it meas that we ca make the patial sums, s, as close to s as we wat simply by takig lage eough. I othe wods, if we take lage eough the we ca say that, i= s ª s This is oe method of estimatig the value of a seies. We ca just take a patial sum ad use that as a estimatio of the value of the seies. Thee ae ow two questios that we should ask about this. Fist, how good is the estimatio? If we do t have a idea of how good the estimatio is the it eally does t do all that much fo us as a estimatio. Secodly, is thee ay way to make the estimate bette? Sometimes we ca use this as a statig poit ad make the estimatio bette. We wo t always be able to do this, but if we ca that will be ice. So, let s stat with a geeal discussio about the detemiig how good the estimatio is. Let s fist stat with the full seies ad stip out the fist tems. Âa = Âa + Â a () i i i i= i= i= + Note that we coveted ove to a idex of i i ode to make the otatio cosistet with pio otatio. Recall that we ca use ay lette fo the idex ad it wo t chage the value. 007 Paul Dawkis 4

253 Now, otice that the fist seies (the tems that we ve stipped out) is othig moe tha the patial sum s. The secod seies o the ight (the oe statig at i = + ) is called the emaide ad deoted by R. Fially let s ackowledge that we also kow the value of the seies sice we ae assumig it s coveget. Takig this otatio ito accout we ca ewite () as, s = s + R We ca solve this fo the emaide to get, R = s- s So, the emaide tells us the diffeece, o eo, betwee the exact value of the seies ad the value of the patial sum that we ae usig as the estimatio of the value of the seies. Of couse we ca t get ou hads o the actual value of the emaide because we do t have the actual value of the seies. Howeve, we ca use some of the tests that we ve got fo covegece to get a petty good estimate of the emaide povided we make some assumptios about the seies. Oce we ve got a estimate o the value of the emaide we ll also have a idea o just how good a job the patial sum does of estimatig the actual value of the seies. Thee ae seveal tests that will allow us to get estimates of the emaide. We ll go though each oe sepaately. Itegal Test Recall that i this case we will eed to assume that the seies tems ae all positive ad will evetually be deceasig. We deived the itegal test by usig the fact that the seies could be f = a. We ca do thought of as a estimatio of the aea ude the cuve of f ( x ) whee somethig simila with the emaide. Fist, let s ecall that the emaide is, Â R = a = a + a + a + a + L i i= + Now, if we stat at x= +, take ectagles of width ad use the left edpoit as the height of +, as show i the the ectagle we ca estimate the aea ude f ( x ) o the iteval [ ) sketch below. 007 Paul Dawkis 44

254 We ca see that the emaide, R, is exactly this aea estimatio ad it will ove estimate the exact aea. So, we have the followig iequality. R Ú f x dx () + Next, we could also estimate the aea by statig at x=, takig ectagles of width agai ad the usig the ight edpoit as the height of the ectagle. This will give a estimatio of the,. This is show i the followig sketch. aea ude f ( x ) o the iteval [ ) Agai, we ca see that the emaide, R, is agai this estimatio ad i this case it will udeestimate the aea. This leads to the followig iequality, Combiig () ad () gives, R Ú f x dx () Ú+ Ú f x dx R f x dx So, povided we ca do these itegals we ca get both a uppe ad lowe boud o the emaide. This will i tu give us a uppe boud ad a lowe boud o just how good the patial sum, s, is as a estimatio of the actual value of the seies. I this case we ca also use these esults to get a bette estimate fo the actual value of the seies as well. Fist, we ll stat with the fact that Now, if we use () we get, s = s + R = + +Ú + s s R s f x dx 007 Paul Dawkis 45

255 Likewise if we use () we get, = + +Ú s s R s f x dx Puttig these two togethe gives us, Ú+ Ú (4) s + f x dx s s + f x dx This gives a uppe ad a lowe boud o the actual value of the seies. We could the use as a estimate of the actual value of the seies the aveage of the uppe ad lowe boud. Let s wok a example with this. Example Usig = 5 to estimate the value of Â. = Solutio Fist, fo compaiso puposes, we ll ote that the actual value of this seies is kow to be, p  = = = Usig = 5 let s fist get the patial sum. 5 s5 =  = i i= Note that this is close to the actual value i some sese, but is t eally all that close eithe. Now, let s compute the itegals. These ae faily simple itegals so we ll leave it to you to veify the values. Û dx= Û Ù dx Ù = ı x 5 ı x Pluggig these ito (4) gives us, s s Both the uppe ad lowe boud ae ow vey close to the actual value ad if we take the aveage of the two we get the followig estimate of the actual value. That is petty da close to the actual value. s ª Paul Dawkis 46

256 So, that is how we ca use the Itegal Test to estimate the value of a seies. Let s move o to the ext test. Compaiso Test I this case, ulike with the itegal test, we may o may ot be able to get a idea of how good a paticula patial sum will be as a estimate of the exact value of the seies. Much of this will deped o how the compaiso test is used. Fist, let s emid ouselves o how the compaiso test actually woks. Give a seies let s assume that we ve used the compaiso test to show that it s coveget. Theefoe, we foud a secod seies  b that coveged ad a b fo all. What we wat to do is detemie how good of a job the patial sum, will do i estimatig the actual value of the seies s =  ai this. Let s actually wite dow the emaide fo both seies. i=    R = a T = b i i i= + i= + a Â. Agai, we will use the emaide to do a Now, sice a b we also kow that R T Whe usig the compaiso test it is ofte the case that the b ae faily ice tems ad that we might actually be able to get a idea o the size of T. Fo istace, if ou secod seies is a p- seies we ca use the esults fom above to get a uppe boud o T as follows, Ú whee + R T g x dx g = b Also, if the secod seies is a geometic seies the we will be able to compute T exactly. If we ae uable to get a idea of the size of T the usig the compaiso 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 Â. = Solutio To do this we ll fist eed to go though the compaiso test so we ca get the secod seies. So, ad ʈ = Á Ë 007 Paul Dawkis 47

257 Ê ˆ  Á = 0 Ë is a geometic seies ad coveges because = <. Now that we ve gotte ou secod seies let s get the estimate. 5 s5 =  = So, how good is it? Well we kow that, = 0 Ê ˆ R5 T5 =  Á = 6Ë will be a uppe boud fo the eo betwee the actual value ad the estimate. Sice ou secod seies is a geometic seies we ca compute this diectly as follows. 5 Ê ˆ Ê ˆ Ê ˆ Á = Á + Á Ë Ë Ë Â Â Â = 0 = 0 = 6 The seies o the left is i the stadad fom ad so we ca compute that diectly. The fist seies o the ight has a fiite umbe of tems ad so ca be computed exactly ad the secod seies o the ight is the oe that we d like to have the value fo. Doig the wok gives, 5 Ê ˆ Ê ˆ Ê ˆ Á = Á - Á Ë Ë Ë =    = 6 = 0 = 0 = So, accodig to this if we use s ª as a estimate of the actual value we will be off fom the exact value by o moe tha ad that s ot too bad. I this case it ca be show that  = = ad so we ca see that the actual eo i ou estimatio is, Eo = Actual- Estimate= = Note that i this case the estimate of the eo is actually faily close (ad i fact exactly the same) as the actual eo. This will ot always happe ad so we should t expect that to happe i all cases. The eo estimate above is simply the uppe boud o the eo ad the actual eo will ofte be less tha this value. 007 Paul Dawkis 48

258 Befoe movig o to the fial pat of this sectio let s agai ote that we will oly be able to detemie how good the estimate is usig the compaiso test if we ca easily get ou hads o the emaide of the secod tem. The eality is that we wo t always be able to do this. Alteatig Seies Test Both of the methods that we ve looked at so fa have equied the seies to cotai oly positive tems. If we allow seies to have egative tems i it the pocess is usually moe difficult. Howeve, with that said thee is oe case whee it is t too bad. That is the case of a alteatig seies. Oce agai we will stat off with a coveget seies a = (- )   b which i this case happes to be a alteatig seies, so we kow that b 0 fo all. Also ote that we could have ay powe o the - we just used fo the sake of coveiece. We wat to kow how good of a estimatio of the actual seies value will the patial sum, s, be. As with the pio cases we kow that the emaide, R, will be the eo i the estimatio ad so if we ca get a hadle o that we ll kow appoximately how good the estimatio is. Fom the poof of the Alteatig Seies Test we ca see that s will lie betwee s ad s + fo ay ad so, s- s s - s = b Theefoe, + + R = s- s b + We eeded absolute value bas because we wo t kow ahead of time if the estimatio is lage o smalle tha the actual value ad we kow that the b s ae positive. Let s take a look at a example. Example Usig = 5 to estimate the value of (- ) Â. = Solutio This is a alteatig seies ad is does covege. I this case the exact value is kow ad so fo compaiso puposes, (-) p  = =- = Now, the estimatio is, s 5  (-) = = = Fom the fact above we kow that 007 Paul Dawkis 49

259 R5 = s- s5 b6 = = So, ou estimatio will have a eo of o moe tha I this case the exact value is kow ad so the actual eo is, R5 = s- s5 = I the pevious example the estimatio had oly half the estimated eo. It will ofte be the case the actual eo will be less tha the estimated eo. Remembe that this is oly a uppe boud fo the actual eo. Ratio Test This will be the fial case that we e goig to look at fo estimatig seies values ad we ae goig to have to put a couple of faily stiget estictio o the seies tems i ode to do the wok. Oe of the mai estictios we e goig to make is to assume that the seies tems ae positive. We ll also be addig o aothe estictio i a bit. I this case we ve used the atio test to show that ad foud that L <. Â a L = lim Æ a a + is coveget. To do this we computed As with the pevious cases we ae goig to use the emaide, R, to detemie how good of a estimatio of the actual value the patial sum, s, is. To get a estimate of the emaide let s fist defie the followig sequece, a+ = a We ow have two possible cases.. If { } is a deceasig sequece ad < the,. If { } is a iceasig sequece the, R R + a a+ - L Poof 007 Paul Dawkis 50

260 Both pats will eed the followig wok so we ll do it fist. We ll stat with the emaide.  R = a = a + a + a + a + L i i= + a Ê a a a = + Á a+ a+ a+ Ë ˆ L Next we eed to do a little wok o a couple of these tems. Ê a a a a 4 a a ˆ R = a+ Á L Ë a+ a+ a+ a+ a+ a+ a Ê Ë a a a a a a = + Á a+ a+ a+ a+ a+ a+ Now use the defiitio of to wite this as, R = a L Okay ow let s do the poof ˆ L Fo the fist pat we ae assumig that { } is deceasig ad so we ca estimate the emaide as, ( L) R = a ( L ) a k + + k= 0 = a  Fially, the seies hee is a geometic seies ad because + < we kow that it coveges ad we ca compute its value. So, a+ R = - Fo the secod pat we ae assumig that { } is iceasig ad we kow that, ad so we kow that + a lim lim Æ + = = Æ a < L fo all. The emaide ca the be estimated as, ( L) R = a ( L ) a + L+ L + L + k + L k= 0 = a  L 007 Paul Dawkis 5

261 This is a geometic seies ad sice we ae assumig that ou oigial seies coveges we also kow that L < ad so the geometic seies above coveges ad we ca compute its value. So, a+ R = - L Note that thee ae some estictios o the sequece { } ad at least oe of its tems i ode to use these fomulas. If the estictios ae t met the the fomulas ca t be used. Let s take a look at a example of this. Example 4 Usig = 5 to estimate the value of Â. 0 = Solutio Fist, let s use the atio test to veify that this is a coveget seies. + + L= lim = lim = < Æ + Æ So, it is coveget. Now let s get the estimate. 5 s = = Â 0 = To detemie a estimate o the emaide, ad hece the eo, let s fist get the sequece { }. + + Ê ˆ = = = + Á + Ë The last ewite was just to simplify some of the computatios a little. Now, otice that, Ê ˆ f ( x) = Á+ f ( x) =- < 0 Ë x x f = ad so, this sequece is deceasig. Sice this fuctio is always deceasig ad Also ote that ( ) = + <. Theefoe we ca use the fist case fom the fact above to get, 6 6 R a = = So, it looks like ou estimate is pobably quite good. I this case the exact value is kow. Â = = 0 4 ad so we ca compute the actual eo. R5 = s- s5 = This is less tha the uppe boud, but ulike i the pevious example this actual eo is quite close to the uppe boud. 007 Paul Dawkis 5

262 I the last two examples we ve see that the uppe boud computatios o the eo ca sometimes be quite close to the actual eo ad at othe times they ca be off by quite a bit. Thee is usually o way of kowig ahead of time which it will be ad without the exact value i had thee will eve be a way of detemiig which it will be. Notice that this method did equie the seies tems to be positive, but that does t mea that we ca t deal with atio test seies if they have egative tems. Ofte seies that we used atio test o ae also alteatig seies ad so if that is the case we ca always esot to the pevious mateial to get a uppe boud o the eo i the estimatio, eve if we did t use the alteatig seies test to show covegece. Note howeve that if the seies does have egative tems, but does t happe to be a alteatig seies the we ca t use ay of the methods discussed i this sectio to get a uppe boud o the eo. 007 Paul Dawkis 5

263 Powe Seies We ve spet quite a bit of time talkig about seies ow ad with oly a couple of exceptios we ve spet most of that time talkig about how to detemie if a seies will covege o ot. It s ow time to stat lookig at some specific kids of seies ad we ll evetually each the poit whee we ca talk about a couple of applicatio of seies. I this sectio we ae goig to stat talkig about powe seies. A powe seies about a, o just powe seies, is ay seies that ca be witte i the fom, Â c ( x- a) = 0 whee a ad c ae umbes. The c s ae ofte called the coefficiets of the seies. The fist thig to otice about a powe seies is that it is a fuctio of x. That is diffeet fom ay othe kid of seies that we ve looked at to this poit. I all the pio sectios we ve oly allowed umbes i the seies ad ow we ae allowig vaiables to be i the seies as well. This will ot chage how thigs wok howeve. Eveythig that we kow about seies still holds. I the discussio of powe seies covegece is still a majo questio that we ll be dealig with. The diffeece is that the covegece of the seies will ow deped upo the value of x that we put ito the seies. A powe seies may covege fo some values of x ad ot fo othe values of x. Befoe we get too fa ito powe seies thee is some temiology that we eed to get out of the way. Fist, as we will see i ou examples, we will be able to show that thee is a umbe R so that the powe seies will covege fo, x- a < Rad will divege fo x- a > R. This umbe is called the adius of covegece fo the seies. Note that the seies may o may ot covege if x- a = R. What happes at these poits will ot chage the adius of covegece. Secodly, the iteval of all x s, icludig the ed poits if eed be, fo which the powe seies coveges is called the iteval of covegece of the seies. These two cocepts ae faily closely tied togethe. If we kow that the adius of covegece of a powe seies is R the we have the followig. a- R< x< a+ R x< a- R ad x> a+ R powe seies coveges powe seies diveges The iteval of covegece must the cotai the iteval a- R< x< a+ R sice we kow that the powe seies will covege fo these values. We also kow that the iteval of covegece ca t cotai x s i the ages x< a- R ad x > a+ R sice we kow the powe seies diveges fo these value of x. Theefoe, to completely idetify the iteval of covegece all that we have to do is detemie if the powe seies will covege fo x = a- R o x = a+ R. If the powe seies coveges fo oe o both of these values the we ll eed to iclude those i the iteval of covegece. 007 Paul Dawkis 54

264 Befoe gettig ito some examples let s take a quick look at the covegece of a powe seies fo the case of x= a. I this case the powe seies becomes, 0 Â Â Â Â c a- a = c 0 = c 0 + c 0 = c + 0= c + 0= c = 0 = 0 = = ad so the powe seies coveges. Note that we had to stip out the fist tem sice it was the oly o-zeo tem i the seies. It is impotat to ote that o matte what else is happeig i the powe seies we ae guaateed to get covegece fo x= a. The seies may ot covege fo ay othe value of x, but it will always covege fo x= a. Let s wok some examples. We ll put quite a bit of detail ito the fist example ad the ot put quite as much detail i the emaiig examples. Example Detemie the adius of covegece ad iteval of covegece fo the followig powe seies. (-) Â = 4 x ( + ) Solutio Okay, we kow that this powe seies will covege fo x =-, but that s it at this poit. To detemie the emaide of the x s fo which we ll get covegece we ca use ay of the tests that we ve discussed to this poit. Afte applicatio of the test that we choose to wok with we will aive at coditio(s) o x that we ca use to detemie which values of x fo which the powe seies will covege ad which values of x fo which the powe seies will divege. Fom this we ca get the adius of covegece ad most of the iteval of covegece (with the possible exceptio of the edpoits. With all that said, the best tests to use hee ae almost always the atio o oot test. Most of the powe seies that we ll be lookig at ae set up fo oe o the othe. I this case we ll use the atio test. L= lim Æ = lim Æ (- ) ( + )( x+ ) ( )( x ) (- ) ( x+ ) + 4 Befoe goig ay fathe with the limit let s otice that sice x is ot depedet o the limit ad so it ca be factoed out of the limit. Notice as well that i doig this we ll eed to keep the absolute value bas o it sice we eed to make sue eveythig stays positive ad x could well be a value that will make thigs egative. The limit is the, + L= x+ lim Æ 4 = x + 4 So, the atio test tells us that if L < the seies will covege, if L > the seies will divege, 007 Paul Dawkis 55

265 ad if L = we do t kow what will happe. So, we have, x+ < 4 fi x+ < 4 seies coveges x+ > 4 fi x+ > 4 seies diveges We ll deal with the L = case i a bit. Notice that we ow have the adius of covegece fo this powe seies. These ae exactly the coditios equied fo the adius of covegece. The adius of covegece fo this powe seies is R = 4. Now, let s get the iteval of covegece. We ll get most (if ot all) of the iteval by solvig the fist iequality fom above. - 4< x + < 4-7< x < So, most of the iteval of validity is give by - 7< x <. All we eed to do is detemie if the powe seies will covege o divege at the edpoits of this iteval. Note that these values of x will coespod to the value of x that will give L =. The way to detemie covegece at these poits is to simply plug them ito the oigial powe seies ad see if the seies coveges o diveges usig ay test ecessay. x =- 7 : I this case the seies is, = - - Â (- 4) = 4 Â - = 4 = 4 Â( ) ( ) ( ) ( ) ( ) = = = - = = Â This seies is diveget by the Divegece Test sice lim = 0. x = : I this case the seies is, (-) 4 Æ Â Â ( 4) = (-) = = This seies is also diveget by the Divegece Test sice lim Æ - does t exist. So, i this case the powe seies will ot covege fo eithe edpoit. The iteval of covegece is the, - 7< x < 007 Paul Dawkis 56

266 I the pevious example the powe seies did t covege fo eithe ed poit of the iteval. Sometimes that will happe, but do t always expect that to happe. The powe seies could covege at eithe both of the ed poits o oly oe of the ed poits. Example Detemie the adius of covegece ad iteval of covegece fo the followig powe seies. Â ( 4x - 8) = Solutio Let s jump ight ito the atio test. L = lim Æ ( x - ) ( x- ) 4 8 = lim Æ + = 4x-8lim Æ + = 4x ( x ) So we will get the followig covegece/divegece ifomatio fom this. 4x- 8 < seies coveges 4x- 8 > seies diveges We eed to be caeful hee i detemiig the iteval of covegece. The iteval of covegece equies x- a < Rad x- a > R. I othe wods, we eed to facto a 4 out of the absolute value bas i ode to get the coect adius of covegece. Doig this gives, 8 x- < fi x- < 8 seies coveges 8 x- > fi x- > 8 seies diveges So, the adius of covegece fo this powe seies is R = 8. Now, let s fid the iteval of covegece. Agai, we ll fist solve the iequality that gives covegece above. - < x - < < x < 8 8 Now check the ed poits. 007 Paul Dawkis 57

267 5 x = : 8 The seies hee is,  Ê5 ˆ Ê ˆ Á - 8 =  Á- Ë Ë = = (-) =  = (-) =  = This is the alteatig hamoic seies ad we kow that it coveges. 7 x = : 8 The seies hee is,  Ê7 ˆ ʈ Á - 8 =  Á Ë Ë = = = =  =  = This is the hamoic seies ad we kow that it diveges. So, the powe seies coveges fo oe of the ed poits, but ot the othe. This will ofte happe so do t get excited about it whe it does. The iteval of covegece fo this powe seies is the, 5 7 x < 8 8 We ow eed to take a look at a couple of special cases with adius ad itevals of covegece. Example Detemie the adius of covegece ad iteval of covegece fo the followig powe seies. = 0 ( x+ )! Solutio We ll stat this example with the atio test as we have fo the pevious oes.  007 Paul Dawkis 58

268 L= lim Æ = lim Æ + ( +! )( x+ )! ( x+ ) ( + )! ( x+ )! ( ) = x+ lim + Æ At this poit we eed to be caeful. The limit is ifiite, but thee is that tem with the x s i fot of the limit. We ll have L=> povided x -. So, this powe seies will oly covege if x=-. If you thik about it we actually aleady kew that howeve. Fom ou iitial discussio we kow that evey powe seies will covege fo x= a ad i this case coefficiet of the x must be a oe!. a=-. Remembe that we get a fom ( x a) -, ad otice the I this case we say the adius of covegece is R= 0 ad the iteval of covegece is x=-, ad yes we eally did mea iteval of covegece eve though it s oly a poit. Example 4 Detemie the adius of covegece ad iteval of covegece fo the followig powe seies. ( x-6) Â = Solutio I this example the oot test seems moe appopiate. So, L= lim Æ ( x- 6) x- 6 = lim Æ = x-6lim Æ = 0 So, sice L= 0< egadless of the value of x this powe seies will covege fo evey x. I these cases we say that the adius of covegece is R= ad iteval of covegece is -< x<. So, let s summaize the last two examples. If the powe seies oly coveges fo x= a the the adius of covegece is R= 0 ad the iteval of covegece is x= a. Likewise if the powe seies coveges fo evey x the adius of covegece is R= ad iteval of covegece is -< x<. 007 Paul Dawkis 59

269 Let s wok oe moe example. Example 5 Detemie the adius of covegece ad iteval of covegece fo the followig powe seies. x  = - Solutio Fist otice that a = 0 i this poblem. That s ot eally impotat to the poblem, but it s woth poitig out so people do t get excited about it. The impotat diffeece i this poblem is the expoet o the x. I this case it is athe tha the stadad. As we will see some powe seies will have expoets othe tha a ad so we still eed to be able to deal with these kids of poblems. This oe seems set up fo the oot test agai so let s use that. x L = lim Æ (- ) x = lim Æ - x = So, we will get covegece if x < fi x < The adius of covegece is NOT howeve. The adius of covegece equies a expoet of o the x. Theefoe, x x Be caeful with the absolute value bas! I this case it looks like the adius of covegece is R =. Notice that we did t bothe to put dow the iequality fo divegece this time. The iequality fo divegece is just the iteval fo covegece that the test gives with the iequality switched ad geeally is t eeded. We will usually skip that pat. Now let s get the iteval of covegece. Fist fom the iequality we get, - < x < Now check the ed poits. x =- : Hee the powe seies is, < < 007 Paul Dawkis 60

270 (- ) Calculus II Â = = ((- ) ) = - - = Â ( ) Â = ( -) ( ) Â = = - This seies is diveget by the Divegece Test sice lim Æ - does t exist. x = : Because we e squaig the x this seies will be the same as the pevious step. which is diveget. ( ) Â Â (-) = = = - The iteval of covegece is the, - < x < 007 Paul Dawkis 6

271 Powe Seies ad Fuctios We opeed the last sectio by sayig that we wee goig to stat thikig about applicatios of seies ad the pomptly spet the sectio talkig about covegece agai. It s ow time to actually stat with the applicatios of seies. With this sectio we will stat talkig about how to epeset fuctios with powe seies. The atual questio of why we might wat to do this will be asweed i a couple of sectios oce we actually lea how to do this. Let s stat off with oe that we aleady kow how to do, although whe we fist a acoss this seies we did t thik of it as a powe seies o did we ackowledge that it epeseted a fuctio. Recall that the geometic seies is  = 0 a a = povided < - Do t foget as well that if the seies diveges. Now, if we take a = ad = x this becomes,  x = - x povided x < () = 0 Tuig this aoud we ca see that we ca epeset the fuctio f ( x) = - x () with the powe seies  x povided x < () = 0 This povisio is impotat. We ca clealy plug ay umbe othe tha x = ito the fuctio, howeve, we will oly get a coveget powe seies if x <. This meas the equality i () will oly hold if x <. Fo ay othe value of x the equality wo t hold. Note as well that we ca also use this to ackowledge that the adius of covegece of this powe seies is R = ad the iteval of covegece is x <. This idea of covegece is impotat hee. We will be epesetig may fuctios as powe seies ad it will be impotat to ecogize that the epesetatios will ofte oly be valid fo a age of x s ad that thee may be values of x that we ca plug ito the fuctio that we ca t plug ito the powe seies epesetatio. I this sectio we ae goig to cocetate o epesetig fuctios with powe seies whee the fuctios ca be elated back to (). 007 Paul Dawkis 6

272 I this way we will hopefully become familia with some of the kids of maipulatios that we will sometimes eed to do whe wokig with powe seies. So, let s jump ito a couple of examples. Example Fid a powe seies epesetatio fo the followig fuctio ad detemie its iteval of covegece. g( x) = + x Solutio What we eed to do hee is to elate this fuctio back to (). This is actually easie tha it might look. Recall that the x i () is simply a vaiable ad ca epeset aythig. So, a quick ewite of g( x ) gives, g x = - -x ad so the - x i g( x ) holds the same place as the x i (). Theefoe, all we eed to do is eplace the x i () ad we ve got a powe seies epesetatio fo g( x ).  = 0 g x = -x povided - x < Notice that we eplaced both the x i the powe seies ad i the iteval of covegece. All we eed to do ow is a little simplificatio. povided  g x = - x x < fi x < = 0 So, i this case the iteval of covegece is the same as the oigial powe seies. This usually wo t happe. Moe ofte tha ot the ew iteval of covegece will be diffeet fom the oigial iteval of covegece. Example Fid a powe seies epesetatio fo the followig fuctio ad detemie its iteval of covegece. x h( x) = + x Solutio This fuctio is simila to the pevious fuctio. The diffeece is the umeato ad at fist glace that looks to be a impotat diffeece. Sice () does t have a x i the umeato it appeas that we ca t elate this fuctio back to that. Howeve, ow that we ve woked the fist example this oe is actually vey simple sice we ca use the esult of the aswe fom that example. To see how to do this let s fist ewite the fuctio a little. h( x) = x + x 007 Paul Dawkis 6

273 Now, fom the fist example we ve aleady got a powe seies fo the secod tem so let s use that to wite the fuctio as, povided  h x = x - x x < = 0 Notice that the pesece of x s outside of the seies will NOT affect its covegece ad so the iteval of covegece emais the same. The last step is to big the coefficiet ito the seies ad we ll be doe. Whe we do this make sue ad combie the x s as well. We typically oly wat a sigle x i a powe seies.  + h x = - x povided x < = 0 As we saw i the pevious example we ca ofte use pevious esults to help us out. This is a impotat idea to emembe as it ca ofte geatly simplify ou wok. Example Fid a powe seies epesetatio fo the followig fuctio ad detemie its iteval of covegece. x f ( x) = 5 -x Solutio So, agai, we ve got a x i the umeato. So, as with the last example let s facto that out ad see what we ve got left. f ( x) = x 5-x If we had a powe seies epesetatio fo g( x) = 5 -x f x. we could get a powe seies epesetatio fo So, let s fid oe. We ll fist otice that i ode to use (4) we ll eed the umbe i the deomiato to be a oe. That s easy eough to get. g( x) = 5 x - 5 Now all we eed to do to get a powe seies epesetatio is to eplace the x i () with x 5. Doig this gives, g x Now let s do a little simplificatio o the seies. Ê xˆ x =  Á povided < 5 = 0Ë Paul Dawkis 64

274 g x x = Â 5 5 = = 0 = 0 x 5 The iteval of covegece fo this seies is, x < fi x < 5 5 fi x < 5 Okay, this was the wok fo the powe seies epesetatio fo g( x ) let s ow fid a powe seies epesetatio fo the oigial fuctio. All we eed to do fo this is to multiply the powe seies epesetatio fo g( x ) by x ad we ll have it. Â = 0 + f ( x) = x 5-x x = xâ5 = + + Â = 0 The iteval of covegece does t chage ad so it will be x < 5. So, hopefully we ow have a idea o how to fid the powe seies epesetatio fo some fuctios. Admittedly all of the fuctios could be elated back to () but it s a stat. We ow eed to look at some futhe maipulatio of powe seies that we will eed to do o occasio. We eed to discuss diffeetiatio ad itegatio of powe seies. Let s stat with diffeetiatio of the powe seies, Â 0 ( ) = 0 f x = c x- a = c + c x- a + c x- a + c x- a + L Now, we kow that if we diffeetiate a fiite sum of tems all we eed to do is diffeetiate each of the tems ad the add them back up. With ifiite sums thee ae some subtleties ivolved that we eed to be caeful with, but ae somewhat beyod the scope of this couse. Nicely eough fo us howeve, it is kow that if the powe seies epesetatio of f ( x ) has a adius of covegece of R > 0 the the tem by tem diffeetiatio of the powe seies will also have a adius of covegece of R ad (moe impotatly) will i fact be the powe seies f x povided we stay withi the adius of covegece. epesetatio of Agai, we should make the poit that if we ae t dealig with a powe seies the we may o may ot be able to diffeetiate each tem of the seies to get the deivative of the seies. x Paul Dawkis 65

275 So, what all this meas fo is that, d f x = Âc x- a = c+ c x- a + c x- a + L = Âc x-a dx = 0 = - Note the iitial value of this seies. It has bee chaged fom = 0 to =. This is a ackowledgemet of the fact that the deivative of the fist tem is zeo ad hece is t i the deivative. Notice howeve, that sice the =0 tem of the above seies is also zeo, we could stat the seies at = 0 if it was equied fo a paticula poblem. I geeal howeve, this wo t be doe i this class. We ca ow fid fomulas fo highe ode deivatives as well ow. - etc.  f x = - c x-a = f x = - - c x-a  ( - ) = Oce agai, otice that the iitial value of chages with each diffeetiatio i ode to ackowledge that a tem fom the oigial seies diffeetiated to zeo. Let s ow biefly talk about itegatio. Just as with the diffeetiatio, whe we ve got a ifiite seies we eed to be caeful about just itegatio tem by tem. Much like with deivatives it tus out that as log as we e wokig with powe seies we ca just itegate the tems of the seies to get the itegal of the seies itself. I othe wods, Ú Û f ( x) dx= Ù Â c( x-a) dx ı = 0 = ÂÚc( x-a) dx = 0 + ( x-a) = C+ Âc = 0 + Notice that we pick up a costat of itegatio, C, that is outside the seies hee. Let s summaize the diffeetiatio ad itegatio ideas befoe movig o to a example o two. Fact  has a adius of covegece 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 adius of covegece of R. Ú Â 007 Paul Dawkis 66

276 Now, let s see how we ca use these facts to geeate some moe powe seies epesetatios of fuctios. Example 4 Fid a powe seies epesetatio fo the followig fuctio ad detemie its iteval of covegece. g( x) = - x Solutio To do this poblem let s otice that ( - x) d Ê ˆ = Á dxë- x The sice we ve got a powe seies epesetatio fo - x all that we ll eed to do is diffeetiate that powe seies to get a powe seies epesetatio fo g x. g x = ( - x) - = d Ê ˆ = Á dxë- x d Ê ˆ = x dx ÁÂ Ë = 0 = The sice the oigial powe seies had a adius of covegece of R = the deivative, ad hece g(x), will also have a adius of covegece of R =. Example 5 Fid a powe seies epesetatio fo the followig fuctio ad detemie its iteval of covegece. h x = l 5- x Solutio I this case we eed to otice that  x Û Ù dx=-l 5- ı 5- x ad the ecall that we have a powe seies epesetatio fo 5 - x ( x) Remembe we foud a epesetatio fo this i Example. So, 007 Paul Dawkis 67

277 l( 5- x) =- Û Ù dx ı 5-x Û =-Ù = C - x 5 dx  + ı = 0 + x  + = 0 5 ( + ) We ca fid the costat of itegatio, C, by pluggig i a value of x. A good choice is x = 0 sice that will make the seies easy to evaluate. + 0 l( 5-0) = C- l 5 = C = 0 So, the fial aswe is, l 5- = l 5 - ( x)  = 0 + x 5 ( + ) + Note that it is okay to have the costat sittig outside of the seies like this. I fact, thee is o way to big it ito the seies so do t get excited about it. 007 Paul Dawkis 68

278 Taylo Seies I the pevious sectio we stated lookig at witig dow a powe seies epesetatio of a fuctio. The poblem with the appoach i that sectio is that eveythig came dow to eedig to be able to elate the fuctio i some way to - x ad while thee ae may fuctios out thee that ca be elated to this fuctio thee ae may moe that simply ca t be elated to this. So, without takig aythig away fom the pocess we looked at i the pevious sectio, we eed to do is come up with a moe geeal method fo witig a powe seies epesetatio fo a fuctio. So, fo the time beig, let s make two assumptios. Fist, let s assume that the fuctio f ( x ) does i fact have a powe seies epesetatio about x = a, Â 0 4( 4 ) = 0 f x = c x- a = c + c x- a + c x- a + c x- a + c x- a + L Next, we will eed to assume that the fuctio, f ( x ), has deivatives of evey ode ad that we ca i fact fid them all. Now that we ve assumed that a powe seies epesetatio exists we eed to detemie what the coefficiets, c, ae. This is easie tha it might at fist appea to be. Let s fist just evaluate eveythig at x= a. This gives, = c0 f a So, all the tems except the fist ae zeo ad we ow kow what c 0 is. Ufotuately, thee is t ay othe value of x that we ca plug ito the fuctio that will allow us to quickly fid ay of the othe coefficiets. Howeve, if we take the deivative of the fuctio (ad its powe seies) the plug i x= a we get, ad we ow kow c. 4 4 f x = c + c x- a + c x- a + c x- a + L f a = c Lets cotiue with this idea ad fid the secod deivative. So, it looks like, 4 4 f x = c + c x- a + c x- a + L f a = c Usig the thid deivative gives, c f = ( a) 007 Paul Dawkis 69

279 4 f x = c + c x- a + L 4 f a = c fi c = f ( a) Usig the fouth deivative gives, 4 f x = 4 c + 54 c x-a L 4 5 ( 4 ) ( a) ( 4 ) f f ( a) = 4 c4 fi c4 = 4 Hopefully by this time you ve see the patte hee. It looks like, i geeal, we ve got the followig fomula fo the coefficiets. ( f ) ( a) c =! This eve woks fo =0 if you ecall that 0! = ad defie ( 0 ) f x = f x. So, povided a powe seies epesetatio fo the fuctio f ( x ) about x= a exists the Taylo Seies fo f ( x ) about x= ais, Taylo Seies = 0 ( ) ( a) f f ( x) = Â x-a! f a f a = f ( a) + f ( a)( x- a) + x- a + x- a + L!! If we use a = 0, so we talkig about the Taylo Seies about x = 0, we call the seies a f x o, Maclaui Seies fo Maclaui Seies = 0 ( ) ( 0) f f ( x) = Â x! ( 0) f ( 0) f f ( 0) f ( 0) x x x!! = L Befoe wokig ay examples of Taylo Seies we fist eed to addess the assumptio that a Taylo Seies will i fact exist fo a give fuctio. Let s stat out with some otatio ad defiitios that we ll eed. To detemie a coditio that must be tue i ode fo a Taylo seies to exist fo a fuctio let s f x as, fist defie the th degee Taylo polyomial of 007 Paul Dawkis 70

280 i= 0 i! ( i ) ( a) f T x = x-a  Note that this eally is a polyomial of degee at most! If we wee to wite out the sum without the summatio otatio this would clealy be a th degee polyomial. We ll see a ice applicatio of Taylo polyomials i the ext sectio. Notice as well that fo the full Taylo Seies, f  = 0! ( ) ( a) ( x-a ) the th degee Taylo polyomial is just the patial sum fo the seies. Next, the emaide is defied to be, R x = f x - T x So, the emaide is eally just the eo betwee the fuctio f ( x ) ad the th degee Taylo polyomial fo a give. With this defiitio ote that we ca the wite the fuctio as, f x = T x + R x We ow have the followig Theoem. Theoem Suppose that f ( x) T ( x) R ( x) fo x- a < R the, o x- a < R. = +. The if, I geeal showig that R ( x) ( x) limr = 0 Æ = 0! ( ) ( a) f f x = x-a  lim = 0 is a somewhat difficult pocess ad so we will be Æ assumig that this ca be doe fo some R i all of the examples that we ll be lookig at. Now let s look at some examples. x Example Fid the Taylo Seies fo f ( x ) =e about x= 0. Solutio This is actually oe of the easie Taylo Seies that we ll be asked to compute. To fid the Taylo Seies fo a fuctio we will eed to detemie a geeal fomula fo f a. This is oe of the i 007 Paul Dawkis 7

281 few fuctios whee this is easy to do ight fom the stat. To get a fomula fo ad so, ( ) ( 0) f all we eed to do is ecogize that, f ( ) x = e = 0,,,, f x ( ) 0 K 0 = e = = 0,,,, K x Theefoe, the Taylo seies fo f ( x ) = e about x=0 is, e x x = Â x = Â!! = 0 = 0 -x Example Fid the Taylo Seies fo f ( x) = e about x = 0. Solutio Thee ae two ways to do this poblem. Both ae faily simple, howeve oe of them equies sigificatly less wok. We ll wok both solutios sice the loge oe has some ice ideas that we ll see i othe examples. Solutio As with the fist example we ll eed to get a fomula fo f ( ) ( 0). Howeve, ulike the fist oe we ve got a little moe wok to do. Let s fist take some deivatives 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 = -x f x =- e f 0 =- M ( ) -x f x = - e f 0 = - = 0,,, M Afte a couple of computatios we wee able to get geeal fomulas fo both f ( ) ( 0). We ofte wo t be able to get a geeal fomula fo f ( ) f ( ) x ad x so do t get too excited about gettig that fomula. Also, as we will see it wo t always be easy to get a geeal fomula fo ( f ) ( a ). So, i this case we ve got geeal fomulas so all we eed to do is plug these ito the Taylo Seies fomula ad be doe with the poblem. e -x = Â (-) = 0! x 007 Paul Dawkis 7

282 Solutio The pevious solutio was t too bad ad we ofte have to do thigs i that mae. Howeve, i this case thee is a much shote solutio method. I the pevious sectio we used seies that we ve aleady foud to help us fid a ew seies. Let s do the same thig with this oe. We x aleady kow a Taylo Seies fo e about x = 0 ad i this case the oly diffeece is we ve got a -x i the expoet istead of just a x. So, all we eed to do is eplace the x i the Taylo Seies that we foud i the fist example with -x. e -x (-x) (-) x = Â = Â!! = 0 = 0 This is a much shote method of aivig at the same aswe so do t foget about usig peviously computed seies whee possible (ad allowed of couse). 4 x Example Fid the Taylo Seies fo f ( x) x - = e about x = 0. Solutio Fo this example we will take advatage of the fact that we aleady have a Taylo Seies fo about x = 0. I this example, ulike the pevious example, doig this diectly would be sigificatly loge ad moe difficult. (- x ) Â 4 - x 4 x e = x = x = = 0! = 0 4 (-) Â = 0 (-) Â To this poit we ve oly looked at Taylo Seies about x = 0 (also kow as Maclaui Seies) so let s take a look at a Taylo Seies that is t about x = 0. Also, we ll pick o the expoetial fuctio oe moe time sice it makes some of the wok easie. This will be the fial Taylo Seies fo expoetials i this sectio. -x Example 4 Fid the Taylo Seies fo f ( x) Solutio Fidig a geeal fomula fo The Taylo Seies is the, ( ) ( 4) ( ) e! f - is faily simple.! x x + 4 = e about x =- 4. ( ) 4 -x f x = - f - 4 = - e x e 007 Paul Dawkis 7

283 e -x = 0 (-) 4 e = Â +! ( x 4) Okay, we ow eed to wok some examples that do t ivolve the expoetial fuctio sice these will ted to equie a little moe wok. Example 5 Fid the Taylo Seies fo f ( x) cos( x) = about x = 0. Solutio Fist we ll eed to take some deivatives of the fuctio ad evaluate them at x= f x = cosx f 0 = () () f x =- six f 0 = 0 f x =- cosx f 0 =- f x = six f 0 = f x = cosx f 0 = 5 5 f x =- six f 0 = f x =- cosx f 0 =- M I this example, ulike the pevious oes, thee is ot a easy fomula fo eithe the geeal deivative o the evaluatio of the deivative. Howeve, thee is a clea patte to the evaluatios. So, let s plug what we ve got ito the Taylo seies ad see what we get, cos x= ( f ) ( 0 ) Â = 0! x M 4 5 f 0 f 0 f 0 4 f 0 5 = f ( 0) + f ( 0) x+ x + x + x + x + L!! 4! 5! 4 6 = { + { 0 - x + { 0 + x + { 0 - x + L {! { 4! { 6! = 0 = = = 5 = = 4 = 6 So, we oly pick up tems with eve powes o the x s. This does t eally help us to get a geeal fomula fo the Taylo Seies. Howeve, let s dop the zeoes ad eumbe the tems as follows to see what we ca get. 4 6 cosx= { - x + x - x +L {! { 4! { 6! = 0 = = = By eumbeig the tems as we did we ca actually come up with a geeal fomula fo the Taylo Seies ad hee it is, 007 Paul Dawkis 74

284 cos x =  (-) x = 0 ( )! This idea of eumbeig the seies tems as we did i the pevious example is t used all that ofte, but occasioally is vey useful. Thee is oe moe seies whee we eed to do it so let s take a look at that so we ca get oe moe example dow of eumbeig seies tems. Example 6 Fid the Taylo Seies fo f ( x) si ( x) = about x = 0. Solutio As with the last example we ll stat off i the same mae. 0 0 f x = six f 0 = 0 () () f x = cosx f 0 = f x =- six f 0 = 0 f x =- cosx f 0 =- 4 4 f x = six f 0 = f x = cosx f 0 = 6 6 f x =- six f 0 = 0 M So, we get a simila patte fo this oe. Let s plug the umbes ito the Taylo Seies. ( ) ( 0 ) f si x=  x = 0! x x x x!! 5! 7! 5 7 = I this case we oly get tems that have a odd expoet o x ad as with the last poblem oce we igoe the zeo tems thee is a clea patte ad fomula. So eumbeig the tems as we did i the pevious example we get the followig Taylo Seies. si x = + (-) x  = 0 ( +! ) We eally eed to wok aothe example o two i which is t about x = 0. Example 7 Fid the Taylo Seies fo f ( x) l ( x) Solutio Hee ae the fist few deivatives ad the evaluatios. M L = about x =. 007 Paul Dawkis 75

285 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 M M ( ) ( ) ( ) + + ( ) - -! - -! f ( x) = f = =,,, K x Note that while we got a geeal fomula hee it does t wok fo = 0. This will happe o occasio so do t woy about it whe it does. I ode to plug this ito the Taylo Seies fomula we ll eed to stip out the = 0 tem fist. = 0 ( ) ( x ) f l( x) = Â -! = ( ) ( x ) f = f + Â -! = l + -! ( x ) = l + - = + (-) ( -! ) Â ( x ) = + (-) Â Notice that we simplified the factoials i this case. You should always simplify them if thee ae moe tha oe ad it s possible to simplify them. Also, do ot get excited about the tem sittig i fot of the seies. Sometimes we eed to do that whe we ca t get a geeal fomula that will hold fo all values of. 007 Paul Dawkis 76

286 = about x=-. x Solutio Agai, hee ae the deivatives ad evaluatios. ( 0 ) ( 0 ) f x = f (- ) = = x - Example 8 Fid the Taylo Seies fo f ( x) ( ) ( ) 007 Paul Dawkis 77 () () f ( x) =- f (- ) =- = x (-) 4 (-) 5 (-) ( ) f ( x) = f - = = 4 x ( ) 4 4 f ( x) =- f - =- = 4 5 x M ( ) ( ) + (-) ( ) - +! - +! f ( x) = f - = = ( +! + ) x Notice that all the egatives sigs will cacel out i the evaluatio. Also, this fomula will wok fo all, ulike the pevious example. Hee is the Taylo Seies fo this fuctio. f = Â + x! M ( ) (-) ( x ) = 0 ( +! ) Â ( x ) = 0 = +! Â( )( x ) = 0 = + + Now, let s wok oe of the easie examples i this sectio. The poblem fo most studets is that it may ot appea to be that easy (o maybe it will appea to be too easy) at fist glace. Example 9 Fid the Taylo Seies fo f x = x - 0x + 6 about x =. Solutio Hee ae the deivatives fo this poblem. 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

287 This Taylo seies will temiate afte =. This will always happe whe we ae fidig the Taylo Seies of a polyomial. Hee is the Taylo Seies fo this oe. ( ) f ( ) x - 0x + 6=  ( x- )! = 0 ( x ) ( x ) ( x ) ( ) f ( ) f = f ( ) + f ( )( x- ) + x- + x- + 0!! = Whe fidig the Taylo Seies of a polyomial we do t do ay simplificatio of the ight had side. We leave it like it is. I fact, if we wee to multiply eveythig out we just get back to the oigial polyomial! While it s ot appaet that witig the Taylo Seies fo a polyomial is useful thee ae times whee this eeds to be doe. The poblem is that they ae beyod the scope of this couse ad so ae t coveed hee. Fo example, thee is oe applicatio to seies i the field of Diffeetial Equatios whee this eeds to be doe o occasio. So, we ve see quite a few examples of Taylo Seies to this poit ad i all of them we wee able to fid geeal fomulas fo the seies. This wo t always be the case. To see a example of oe that does t have a geeal fomula check out the last example i the ext sectio. Befoe leavig this sectio thee ae thee impotat Taylo Seies that we ve deived i this sectio that we should summaize up i oe place. I my class I will assume that you kow these fomulas fom this poit o. x e = cos x = si x = x  = 0! (-) x  = 0 ( )! + (-) x  = 0 ( +! ) 007 Paul Dawkis 78

288 Applicatios of Seies Now, that we kow how to epeset fuctio as powe seies we ca ow talk about at least a couple of applicatios of seies. Thee ae i fact may applicatios of seies, ufotuately most of them ae beyod the scope of this couse. Oe applicatio of powe seies (with the occasioal use of Taylo Seies) is i the field of Odiay Diffeetial Equatios whe fidig Seies Solutios to Diffeetial Equatios. If you ae iteested i seeig how that woks you ca check out that chapte of my Diffeetial Equatios otes. Aothe applicatio of seies aises i the study of Patial Diffeetial Equatios. Oe of the moe commoly used methods i that subject makes use of Fouie Seies. May of the applicatios of seies, especially those i the diffeetial equatios fields, ely o the fact that fuctios ca be epeseted as a seies. I these applicatios it is vey difficult, if ot impossible, to fid the fuctio itself. Howeve, thee ae methods of detemiig the seies epesetatio fo the ukow fuctio. While the diffeetial equatios applicatios ae beyod the scope of this couse thee ae some applicatios fom a Calculus settig that we ca look at. Example Detemie a Taylo Seies about x = 0 fo the followig itegal. Û Ù ı si x dx x Solutio To do this we will fist eed to fid a Taylo Seies about x = 0 fo the itegad. This howeve is t teibly difficult. We aleady have a Taylo Seies fo sie about x = 0 so we ll just use that as follows, + six (-) x (-) x = Â = Â x x +! +! We ca ow do the poblem. Û si x Û Ù dx = Ù ı x = 0 = 0 Â (-) x ı = 0( + ) C Â = 0 = + dx! + (-) x ( + )( + )! So, while we ca t itegate this fuctio i tems of kow fuctios we ca come up with a seies epesetatio fo the itegal. This idea of deivig a seies epesetatio fo a fuctio istead of tyig to fid the fuctio itself is used quite ofte i seveal fields. I fact, thee ae some fields whee this is oe of the mai ideas used ad without this idea it would be vey difficult to accomplish aythig i those fields. 007 Paul Dawkis 79

289 Aothe applicatio of seies is t eally a applicatio of ifiite seies. It s moe a applicatio of patial sums. I fact, we ve aleady see this applicatio i use oce i this chapte. I the Estimatig the Value of a Seies we used a patial sum to estimate the value of a seies. We ca do the same thig with powe seies ad seies epesetatios of fuctios. The mai diffeece is that we will ow be usig the patial sum to appoximate a fuctio istead of a sigle value. We will look at Taylo seies fo ou examples, but we could just as easily use ay seies epesetatio hee. Recall that the th degee Taylo Polyomial of f(x) is give by, ( i f ) ( a) i T( x) = Â ( x-a ) i Let s take a look at example of this. i= 0! Example Fo the fuctio f ( x) = cos( x) plot the fuctio as well as T( x ), T4 T8( x ) o the same gaph fo the iteval [-4,4]. x, ad Solutio Hee is the geeal fomula fo the Taylo polyomials fo cosie. T ( x) =Â i= 0 i (-) ( i) x! The thee Taylo polyomials that we ve got ae the, x T( x) = - 4 x x T4( x) = x x x x T8( x) = Hee is the gaph of these thee Taylo polyomials as well as the gaph of cosie. i 007 Paul Dawkis 80

290 As we ca see fom this gaph as we icease the degee of the Taylo polyomial it stats to look T x the oly diffeece is moe ad moe like the fuctio itself. I fact by the time we get to ight at the eds. The highe the degee of the Taylo polyomial the bette it appoximates the fuctio. Also the lage the iteval the highe degee Taylo polyomial we eed to get a good appoximatio fo the whole iteval. Befoe movig o let s otice wite dow a couple moe Taylo polyomials fom the pevious example. Notice that because the Taylo seies fo cosie does t cotai ay tems with odd powes o x we get the followig Taylo polyomials. T T T 5 9 ( x) ( x) ( x) x = - 4 x x = x x x x = These ae idetical to those used i the example. Sometimes this will happe although that was ot eally the poit of this. The poit is to otice that the th degee Taylo polyomial may actually have a degee that is less tha. It will eve be moe tha, but it ca be less tha. The fial example i this sectio eally is t a applicatio of seies ad pobably beloged i the pevious sectio. Howeve, the pevious sectio was gettig too log so the example is i this sectio. This is a example of how to multiply seies togethe ad while this is t a applicatio of seies it is somethig that does have to be doe o occasio i the applicatios. So, i that sese it does belog i this sectio Paul Dawkis 8

291 f x = e x about x Example Fid the fist thee o-zeo tems i the Taylo Seies fo cos x = 0. Solutio Befoe we stat let s ackowledge that the easiest way to do this poblem is to simply compute the fist -4 deivatives, evaluate them at x = 0, plug ito the fomula ad we d be doe. Howeve, as we oted pio to this example we wat to use this example to illustate how we multiply seies. We will make use of the fact that we ve got Taylo Seies fo each of these so we ca use them i this poblem. x Ê x ˆÊ (-) x ˆ e cos x = Á Á = 0! Á Â Ë = 0 ( )! Ë We e ot goig to completely multiply out these seies. We e goig to eough of the multiplicatio to get a aswe. The poblem statemet says that we wat the fist thee o-zeo tems. That o-zeo bit is impotat as it is possible that some of the tems will be zeo. If oe of the tems ae zeo this would mea that the fist thee o-zeo tems would be the costat tem, x tem, ad x tem. Howeve, because some might be zeo let s assume that if we get all the tems up though x 4 we ll have eough to get the aswe. If we ve assumed wog it will be vey easy to fix so do t woy about that. Now, let s wite dow the fist few tems of each seies ad we ll stop at the x 4 tem i each. 4 4 Ê cos x x x ˆÊ x x x ˆ e x= Á + x L Á L Ë 6 4 Ë 4 Note that we do eed to ackowledge that these seies do t stop. That s the pupose of the +L at the ed of each. Just fo a secod howeve, let s suppose that each of these did stop ad ask ouselves how we would multiply each out. If this wee the case we would take evey tem i the secod ad multiply by evey tem i the fist. I othe wods, we would fist multiply evey tem i the secod seies by, the evey tem i the secod seies by x, the by x etc. By stoppig each seies at x 4 we have ow guaateed that we ll get all tems that have a expoet of 4 o less. Do you see why? Each of the tems that we eglected to wite dow have a expoet of at least 5 ad so multiplyig by o ay powe of x will esult i a tem with a expoet that is at a miimum 5. Theefoe, oe of the eglected tems will cotibute tems with a expoet of 4 o less ad so wee t eeded. So, let s stat the multiplicatio pocess. 007 Paul Dawkis 8

292 4 4 Ê ˆÊ ˆ x x x x x x e cosx= Á+ x L Á- + + L Ë 6 4 Ë x x x x x x x = L+ x- + + L L Secod Seies Secod Seies x Secod Seies x 6 Secod Seies x x x x x x x L L+ L Secod Seies x 4 Now, collect like tems igoig eveythig with a expoet of 5 o moe sice we wo t have all those tems ad do t wat them eithe. Doig this gives, x Ê ˆ Ê ˆ Ê ˆ 4 e cosx= + x+ Á- + x + Á- + x + Á - + x + L Ë Ë 6 Ë x x = + x- - + L 6 Thee we go. It looks like we ove guessed ad eded up with fou o-zeo tems, but that s okay. If we had ude guessed ad it tued out that we eeded tems with x 5 i them all we would eed to do at this poit is go back ad add i those tems to the oigial seies ad do a couple quick multiplicatios. I othe wods, thee is o easo to completely edo all the wok. 007 Paul Dawkis 8

293 Biomial Seies I this fial sectio of this chapte we ae goig to look at aothe seies epesetatio fo a fuctio. Befoe we do this let s fist ecall the followig theoem. Biomial Theoem If is ay positive itege the, whee, ʈ -i i a+ b = ÂÁ a b i= 0 Ëi ( -) = a + a b+ a b + L+ ab + b! ( -)( -) L( - + ) Ê i Á ˆ= i,,, i = K Ë i! Ê Á ˆ= Ë0 This is useful fo expadig ( a+ b) fo lage whe staight fowad multiplicatio would t be easy to do. Let s take a quick look at a example. Example Use the Biomial Theoem to expad ( x - ) 4 Solutio Thee eally is t much to do othe tha pluggig ito the theoem. 4 4 Ê4ˆ 4-i i ( x- ) = ÂÁ ( x) (-) i= 0Ëi Ê4ˆ Ê4ˆ Ê4ˆ Ê4ˆ Ê4ˆ = Á 0 x + Á x - + Á x - + Á x - + Á 4 - Ë Ë Ë Ë Ë = ( x) + 4 ( x)( - ) + ( x ) ( - ) + 4 ( x )( - ) + ( - ) 4 = 6x - 96x + 6x - 6x Now, the Biomial Theoem equied that be a positive itege. Thee is a extesio to this howeve that allows fo ay umbe at all. 007 Paul Dawkis 84

294 Biomial Seies If k is ay umbe ad x < the, whee, k Êk ˆ + = ÂÁ x = 0Ë k( k-) k( k-)( k-) kx x x!! ( x) = ( -)( -) L( - + ) Ê k k k k k Á ˆ=,,, = K Ë! Êk Á ˆ= Ë0 So, simila to the biomial theoem except that it s a ifiite seies ad we must have x < i ode to get covegece. Let s check out a example of this. Example Wite dow the fist fou tems i the biomial seies fo 9 - x Solutio So, i this case k = ad we ll eed to ewite the tem a little to put it ito the fom equied. Ê xˆ Ê Ê xˆˆ 9- x = Á - = Á Á 9 Ë Ë Ë The fist fou tems i the biomial seies is the, Ê Ê x ˆˆ 9- x = Á + Á - 9 Ë Ë Ê ˆÊ x ˆ = ÂÁ - = 0 Á Ë Ë 9 (- ) (- )(- ) È ÊˆÊ xˆ Ê xˆ Ê xˆ = Í + Á Á- + Á- + Á- + L ÍÎ Ë Ë 9 Ë 9 6 Ë 9 x x x = L L 007 Paul Dawkis 85

295 Vectos Itoductio This is a faily shot chapte. We will be takig a bief look at vectos ad some of thei popeties. We will eed some of this mateial i the ext chapte ad those of you headig o towads Calculus III will use a fai amout of this thee as well. Hee is a list of topics i this chapte. Vectos The Basics I this sectio we will itoduce some of the basic cocepts about vectos. Vecto Aithmetic Hee we will give the basic aithmetic opeatios fo vectos. Dot Poduct We will discuss the dot poduct i this sectio as well as a applicatio o two. Coss Poduct I this sectio we ll discuss the coss poduct ad see a quick applicatio. 007 Paul Dawkis 86

296 Vectos The Basics Let s stat this sectio off with a quick discussio o what vectos ae used fo. Vectos ae used to epeset quatities that have both a magitude ad a diectio. Good examples of quatities that ca be epeseted by vectos ae foce ad velocity. Both of these have a diectio ad a magitude. Let s coside foce fo a secod. A foce of say 5 Newtos that is applied i a paticula diectio ca be applied at ay poit i space. I othe wods, the poit whee we apply the foce does ot chage the foce itself. Foces ae idepedet of the poit of applicatio. To defie a foce all we eed to kow is the magitude of the foce ad the diectio that the foce is applied i. The same idea holds moe geeally with vectos. Vectos oly impat magitude ad diectio. They do t impat ay ifomatio about whee the quatity is applied. This is a impotat idea to always emembe i the study of vectos. I a gaphical sese vectos ae epeseted by diected lie segmets. The legth of the lie segmet is the magitude of the vecto ad the diectio of the lie segmet is the diectio of the vecto. Howeve, because vectos do t impat ay ifomatio about whee the quatity is applied ay diected lie segmet with the same legth ad diectio will epeset the same vecto. Coside the sketch below. Each of the diected lie segmets i the sketch epesets the same vecto. I each case the vecto stats at a specific poit the moves uits to the left ad 5 uits up. The otatio that we ll use fo this vecto is, v = -,5 ad each of the diected lie segmets i the sketch ae called epesetatios of the vecto. Be caeful to distiguish vecto otatio, -,5, fom the otatio we use to epeset coodiates of poits, (,5) -. The vecto deotes a magitude ad a diectio of a quatity while the poit deotes a locatio i space. So do t mix the otatios up! 007 Paul Dawkis 87

297 A epesetatio of the vecto v = a, a uuu AB, fom the poit A= ( xy, ) to the poit B ( x a, y a ) the vecto v = a, a, a i two dimesioal space is ay diected lie segmet, = + +. Likewise a epesetatio of uuu i thee dimesioal space is ay diected lie segmet, AB, fom the poit A= ( xyz,, ) to the poit B ( x a, y a, z a ) = Note that thee is vey little diffeece betwee the two dimesioal ad thee dimesioal fomulas above. To get fom the thee dimesioal fomula to the two dimesioal fomula all we did is take out the thid compoet/coodiate. Because of this most of the fomulas hee ae give oly i thei thee dimesioal vesio. If we eed them i thei two dimesioal fom we ca easily modify the thee dimesioal fom. Thee is oe epesetatio of a vecto that is special i some way. The epesetatio of the vecto v = a, a, a that stats at the poit ( 0,0,0) B= a, a, a A = ad eds at the poit ( ) is called the positio vecto of the poit ( a, a, a ). So, whe we talk about positio vectos we ae specifyig the iitial ad fial poit of the vecto. Positio vectos ae useful if we eve eed to epeset a poit as a vecto. As we ll see thee ae times i which we defiitely ae goig to wat to epeset poits as vectos. I fact, we e goig to u ito topics that ca oly be doe if we epeset poits as vectos. Next we eed to discuss biefly how to geeate a vecto give the iitial ad fial poits of the epesetatio. Give the two poits A= ( a, a, a) ad B= ( b, b, b) the vecto with the uuu epesetatio AB is, v = b -a, b -a, b -a Note that we have to be vey caeful with diectio hee. The vecto above is the vecto that stats uuu at A ad eds at B. The vecto that stats at B ad eds at A, i.e. with epesetatio BA is, w= a -b, a -b, a -b These two vectos ae diffeet ad so we do eed to always pay attetio to what poit is the statig poit ad what poit is the edig poit. Whe detemiig the vecto betwee two poits we always subtact the iitial poit fom the temial poit. Example Give the vecto fo each of the followig. (a) The vecto fom (, - 7,0) to (, -,- 5). (b) The vecto fom (, -,- 5) to(, - 7,0). (c) The positio vecto fo (- 90,4) Solutio (a) Remembe that to costuct this vecto we subtact coodiates of the statig poit fom the edig poit. 007 Paul Dawkis 88

298 (b) Same thig hee. -, --(-7, ) -5-0 = -,4,- 5 -, -7-(-,0 ) -(- 5) =, - 4,5 Notice that the oly diffeece betwee the fist two is the sigs ae all opposite. This diffeece is impotat as it is this diffeece that tells us that the two vectos poit i opposite diectios. (c) Not much to this oe othe tha ackowledgig that the positio vecto of a poit is othig moe tha a vecto with the poits coodiates as its compoets. - 90,4 We ow eed to stat discussig some of the basic cocepts that we will u ito o occasio. Magitude The magitude, o legth, of the vecto v = a, a, a is give by, v = a + a + a Example Detemie the magitude of each of the followig vectos. (a) a =, -5,0 (b) u =,- 5 5 (c) w = 0,0 (d) i =,0,0 Solutio Thee is t too much to these othe tha plug ito the fomula. (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 squae all the compoets the oly way we ca get zeo out of the fomula was fo the compoets to be zeo i the fist place. Uit Vecto Ay vecto with magitude of, i.e. u =, is called a uit vecto. 007 Paul Dawkis 89

299 Example Which of the vectos fom Example ae uit vectos? Solutio Both the secod ad fouth vectos had a legth of ad so they ae the oly uit vectos fom the fist example. Zeo Vecto The vecto w = 0,0 that we saw i the fist example is called a zeo vecto sice its compoets ae all zeo. Zeo vectos ae ofte deoted by 0. Be caeful to distiguish 0 (the umbe) fom 0 (the vecto). The umbe 0 deote the oigi i space, while the vecto 0 deotes a vecto that has o magitude o diectio. Stadad Basis Vectos The fouth vecto fom the secod example, i =,0,0, is called a stadad basis vecto. I thee dimesioal space thee ae thee stadad basis vectos, i =,0,0 j = 0,,0 k = 0,0, I two dimesioal space thee ae two stadad basis vectos, i =,0 j = 0, Note that stadad basis vectos ae also uit vectos. Waig We ae petty much doe with this sectio howeve, befoe poceedig to the ext sectio we should poit out that vectos ae ot esticted to two dimesioal o thee dimesioal space. Vectos ca exist i geeal -dimesioal space. The geeal otatio fo a -dimesioal vecto is, v = a, a, a, K, a ad each of the a i s ae called compoets of the vecto. Because we will be wokig almost exclusively with two ad thee dimesioal vectos i this couse most of the fomulas will be give fo the two ad/o thee dimesioal cases. Howeve, most of the cocepts/fomulas will wok with geeal vectos ad the fomulas ae easily (ad atually) modified fo geeal -dimesioal vectos. Also, because it is easie to visualize thigs i two dimesios most of the figues elated to vectos will be two dimesioal figues. So, we eed to be caeful to ot get too locked ito the two o thee dimesioal cases fom ou discussios i this chapte. We will be wokig i these dimesios eithe because it s easie to visualize the situatio o because physical estictios of the poblems will efoce a dimesio upo us. 007 Paul Dawkis 90

300 Vecto Aithmetic I this sectio we eed to have a bief discussio of vecto aithmetic. We ll stat with additio of two vectos. So, give the vectos a = a, a, a ad b = b, b, b the additio of the two vectos is give by the followig fomula. a+ b = a + b, a + b, a + b The followig figue gives the geometic itepetatio of the additio of two vectos. This is sometimes called the paallelogam law o tiagle law. Computatioally, subtactio is vey simila. Give the vectos a = a, a, a b = b, b, b the diffeece of the two vectos is give by, a- b = a -b, a -b, a -b ad Hee is the geometic itepetatio of the diffeece of two vectos. 007 Paul Dawkis 9

301 It is a little hade to see this geometic itepetatio. To help see this let s istead thik of subtactio as the additio of a ad -b. Fist, as we ll see i a bit -b is the same vecto as b with opposite sigs o all the compoets. I othe wods, a + - b. b. Hee is the vecto set up fo -b goes i the opposite diectio as + - As we ca see fom this figue we ca move the vecto epesetig a ( b) we ve got i the fist figue showig the diffeece of the two vectos. to the positio Note that we ca t add o subtact two vectos uless they a have the same umbe of compoets. If they do t have the same umbe of compoets the additio ad subtactio ca t be doe. The ext aithmetic opeatio that we wat to look at is scala multiplicatio. Give the vecto a = a, a, a ad ay umbe c the scala multiplicatio is, ca = ca, ca, ca So, we multiply all the compoets by the costat c. To see the geometic itepetatio of scala multiplicatio let s take a look at a example. Example Fo the vecto a =,4 the same axis system. compute a, a ad -a. Gaph all fou vectos o Solutio Hee ae the thee scala multiplicatios. a = 6, a =, - a = -4,-8 Hee is the gaph fo each of these vectos. 007 Paul Dawkis 9

302 I the pevious example we ca see that if c is positive all scala multiplicatio will do is stetch (if c > ) o shik (if c < ) the oigial vecto, but it wo t chage the diectio. Likewise, if c is egative scala multiplicatio will switch the diectio so that the vecto will poit i exactly the opposite diectio ad it will agai stetch o shik the magitude of the vecto depedig upo the size of c. Thee ae seveal ice applicatios of scala multiplicatio that we should ow take a look at. The fist is paallel vectos. This is a cocept that we will see quite a bit ove the ext couple of sectios. Two vectos ae paallel if they have the same diectio o ae i exactly opposite diectios. Now, ecall agai the geometic itepetatio of scala multiplicatio. Whe we pefomed scala multiplicatio we geeated ew vectos that wee paallel to the oigial vectos (ad each othe fo that matte). So, let s suppose that a ad b ae paallel vectos. If they ae paallel the thee must be a umbe c so that, a = cb So, two vectos ae paallel if oe is a scala multiple of the othe. Example Detemie if the sets of vectos ae paallel o ot. (a) a =, - 4,, b = -6,,- (b) a = 4,0, b =, -9 Solutio (a) These two vectos ae paallel sice b =-a (b) These two vectos ae t paallel. This ca be see by oticig that ( ) 0 = 5-9. I othe wods we ca t make a be a scala multiple of b. The ext applicatio is best see i a example. 4 = ad yet 007 Paul Dawkis 9

303 Example Fid a uit vecto that poits i the same diectio as w = -5,,. Solutio Okay, what we e askig fo is a ew paallel vecto (poits i the same diectio) that happes to be a uit vecto. We ca do this with a scala multiplicatio sice all scala multiplicatio does is chage the legth of the oigial vecto (alog with possibly flippig the diectio to the opposite diectio). Hee s what we ll do. Fist let s detemie the magitude of w. w = = 0 Now, let s fom the followig ew vecto, 5 u = w= - 5,, = -,, w The claim is that this is a uit vecto. That s easy eough to check u = + + = = This vecto also poits i the same diectio as w sice it is oly a scala multiple of w ad we used a positive multiple. So, i geeal, give a vecto w, u = as w. w will be a uit vecto that poits i the same diectio w Stadad Basis Vectos Revisited I the pevious sectio we itoduced the idea of stadad basis vectos without eally discussig why they wee impotat. We ca ow do that. Let s stat with the vecto a = a, a, a We ca use the additio of vectos to beak this up as follows, a = a, a, a = a,0,0 + 0, a,0 + 0,0, a Usig scala multiplicatio we ca futhe ewite the vecto as, a = a,0,0 + 0, a,0 + 0,0, a = a,0,0 + a 0,,0 + a 0,0, Fially, otice that these thee ew vectos ae simply the thee stadad basis vectos fo thee dimesioal space. a, a, a = ai + a j + ak 007 Paul Dawkis 94

304 So, we ca take ay vecto ad wite it i tems of the stadad basis vectos. Fom this poit o we will use the two otatios itechageably so make sue that you ca deal with both otatios. Example 4 If a =, -9, ad w=- i + 8k compute a -w. Solutio I ode to do the poblem we ll covet to oe otatio ad the pefom the idicated opeatios. a- w=, -9, - -,0,8 = 6, -8, - -,0,4 = 9, -8,- We will leave this sectio with some basic popeties of vecto aithmetic. Popeties If v, w ad u ae vectos (each with the same umbe of compoets) ad a ad b ae two umbes the we have the followig popeties. 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 poofs of these ae petty much just computatio poofs so we ll pove oe of them ad leave the othes to you to pove. Poof of a( v+ w) = av+ aw We ll stat with the two vectos, v = v, v, K, v ad w= w, w, K, w ad yes we did mea fo these to each have compoets. The theoem woks fo geeal vectos so we may as well do the poof fo geeal vectos. Now, as oted above this is petty much just a computatioal poof. What that meas is that we ll compute the left side ad the do some basic aithmetic o the esult to show that we ca make the left side look like the ight side. Hee is the wok. a v+ w = a v, v, K, v + w, w, K, w ( ) = a v + w, v + w, K, v + w,, K, = a v + w a v + w a v + w = av + aw, av + aw, K, av + aw = av, av, K, av + aw, aw, K, aw = a v, v, K, v + a w, w, K, w = av+ aw 007 Paul Dawkis 95

305 Dot Poduct The ext topic fo discussio is that of the dot poduct. Let s jump ight ito the defiitio of the dot poduct. Give the two vectos a = a, a, a ad b = b, b, b the dot poduct is, ab g = ab + ab + ab () Sometimes the dot poduct is called the scala poduct. The dot poduct is also a example of a ie poduct ad so o occasio you may hea it called a ie poduct. Example Compute the dot poduct fo each of the followig. (a) v = 5i - 8 j, w= i + j (b) a = 0,, - 7, b =,, Solutio Not much to do with these othe tha use the fomula. (a) vw= g 5-6=- (b) ab= g = Hee ae some popeties of the dot poduct. Popeties ug v w uv g uw g cv gw vg cw c vw g vw g = wv g vg0= 0 vv g = v If vv g = 0 the v = 0 ( + ) = + = = The poofs of these popeties ae mostly computatioal poofs ad so we e oly goig to do a couple of them ad leave the est to you to pove. Poof of u g( v + w ) = uv g + uw g We ll stat with the thee vectos, u = u, u, K, u, v = v, v, K, v ad w= w, w, K, w ad yes we did mea fo these to each have compoets. The theoem woks fo geeal vectos so we may as well do the poof fo geeal vectos. Now, as oted above this is petty much just a computatioal poof. What that meas is that we ll compute the left side ad the do some basic aithmetic o the esult to show that we ca make the left side look like the ight side. Hee is the wok. 007 Paul Dawkis 96

306 ug v w u u K u g v v K v w w K w ( + ) =,,, (,,, +,,, ) = u, u, K, u g v + w, v + w, K, v + w,, K, = u v + w u v + w u v + w = uv + uw, uv + uw, K, uv + uw = uv, uv, K, uv + uw, uw, K, uw = u, u, K, u g v, v, K, v + u, u, K, u g w, w, K, w = uv g + uw g Poof of : If vv= g 0 the v = 0 This is a petty simple poof. Let s stat with v = v, v, K, v ad compute the dot poduct. vv g = v, v, K, v g v, v, K, v = 0 = v + v + L + v Now, sice we kow vi 0 fo all i the the oly way fo this sum to be zeo is to i fact have v = 0. This i tu howeve meas that we must have v = 0 ad so we must have had v = 0. i i Thee is also a ice geometic itepetatio to the dot poduct. Fist suppose that q is the agle betwee a ad b such that 0 q p as show i the image below. We ca the have the followig theoem. 007 Paul Dawkis 97

307 Theoem ab g = a b cosq () Poof Let s give a modified vesio of the sketch above. The thee vectos above fom the tiagle AOB ad ote that the legth of each side is othig moe tha the magitude of the vecto fomig that side. The Law of Cosies tells us that, a- b = a + b - a b cosq Also usig the popeties of dot poducts we ca wite the left side as, a- b = ( a-b) g( a-b) = aa g -ab g - ba g + bb g = a - ab g + b Ou oigial equatio is the, a- b = a + b - a b cosq a - ab g + b = a + b - a b cosq - ab g =- a b cosq ab g = a b cosq 007 Paul Dawkis 98

308 The fomula fom this theoem is ofte used ot to compute a dot poduct but istead to fid the agle betwee two vectos. Note as well that while the sketch of the two vectos i the poof is fo two dimesioal vectos the theoem is valid fo vectos of ay dimesio (as log as they have the same dimesio of couse). Let s see a example of this. Example Detemie the agle betwee a =, -4,- ad b = 0,5,. Solutio We will eed the dot poduct as well as the magitudes of each vecto. ab g =- a = 6 b = 9 The agle is the, ab g - cosq = = = a b 6 9 q = - = - cos adias=4.4 degees The dot poduct gives us a vey ice method fo detemiig if two vectos ae pepedicula ad it will give aothe method fo detemiig whe two vectos ae paallel. Note as well that ofte we will use the tem othogoal i place of pepedicula. Now, if two vectos ae othogoal the we kow that the agle betwee them is 90 degees. Fom () this tells us that if two vectos ae othogoal the, ab= g 0 Likewise, if two vectos ae paallel the the agle betwee them is eithe 0 degees (poitig i the same diectio) o 80 degees (poitig i the opposite diectio). Oce agai usig () this would mea that oe of the followig would have to be tue. ab g = a b = ab g =- a b = ( q 0 ) OR ( q 80 ) Example Detemie if the followig vectos ae paallel, othogoal, o eithe. (a) a = 6, -,-, b =,5, (b) u = i - j, v =- i + j 4 Solutio (a) Fist get the dot poduct to see if they ae othogoal. ab= g -0- = 0 The two vectos ae othogoal. (b) Agai, let s get the dot poduct fist. 007 Paul Dawkis 99

309 5 uv=- g - =- 4 4 So, they ae t othogoal. Let s get the magitudes ad see if they ae paallel. 5 5 u = 5 v = = 6 4 Now, otice that, 5 5 uv 5 Ê ˆ g =- =- =- u v 4 Á 4 Ë So, the two vectos ae paallel. Thee ae seveal ice applicatios of the dot poduct as well that we should look at. Pojectios The best way to udestad pojectios is to see a couple of sketches. So, give two vectos a ad b we wat to detemie the pojectio of b oto a. The pojectio is deoted by poj a b. Hee ae a couple of sketches illustatig the pojectio. So, to get the pojectio of b oto a we dop staight dow fom the ed of b util we hit (ad fom a ight agle) with the lie that is paallel to a. The pojectio is the the vecto that is paallel to a, stats at the same poit both of the oigial vectos stated at ad eds whee the dashed lie hits the lie paallel to a. Thee is a ice fomula fo fidig the pojectio of b oto a. Hee it is, g poj a b = a ab a Note that we also eed to be vey caeful with otatio hee. The pojectio of a oto b is give by 007 Paul Dawkis 00

310 g poj b a = b ab b We ca see that this will be a totally diffeet vecto. This vecto is paallel to b, while poj a b is paallel to a. So, be caeful with otatio ad make sue you ae fidig the coect pojectio. Hee s a example. Example 4 Detemie the pojectio of b =,, - oto a =,0, -. Solutio We eed the dot poduct ad the magitude of a. ab g = 4 a = 5 The pojectio is the, poj a ab g b = a a 4 =,0, =,0,- 5 5 Fo compaiso puposes let s do it the othe way aoud as well. Example 5 Detemie the pojectio of a =,0, - otob =,, -. Solutio We eed the dot poduct ad the magitude of b. ab g = 4 b = 6 The pojectio is the, poj b a = ab g b b 4 =,, =,,- As we ca see fom the pevious two examples the two pojectios ae diffeet so be caeful. 007 Paul Dawkis 0

311 Diectio Cosies This applicatio of the dot poduct equies that we be i thee dimesioal space ulike all the othe applicatios we ve looked at to this poit. Let s stat with a vecto, a, i thee dimesioal space. This vecto will fom agles with the x- axis (a ), the y-axis (b ), ad the z-axis (g ). These agles ae called diectio agles ad the cosies of these agles ae called diectio cosies. Hee is a sketch of a vecto ad the diectio agles. The fomulas fo the diectio cosies ae, ai g a agj a ak g a a = = b = = g = = a a a a a a cos cos cos whee i, j ad k ae the stadad basis vectos. Let s veify the fist dot poduct above. We ll leave the est to you to veify. ai g = a, a, a g,0,0 = a Hee ae a couple of ice facts about the diectio cosies.. The vecto u = cos a,cos b,cosg is a uit vecto. cos a + cos b + cos g =.. a = a cos a,cos b,cosg Let s do a quick example ivolvig diectio cosies. 007 Paul Dawkis 0

312 Example 6 Detemie the diectio cosies ad diectio agles fo a =,, -4. Solutio We will eed the magitude of the vecto. a = = The diectio cosies ad agles ae the, cosa = a =.9 adias= 64. degees cosb = b =.5 adias= degees cosg = -4 g =.6 adias= degees 007 Paul Dawkis 0

313 Coss Poduct I this fial sectio of this chapte we will look at the coss poduct of two vectos. We should ote that the coss poduct equies both of the vectos to be thee dimesioal vectos. Also, befoe gettig ito how to compute these we should poit out a majo diffeece betwee dot poducts ad coss poducts. The esult of a dot poduct is a umbe ad the esult of a coss poduct is a vecto! Be caeful ot to cofuse the two. So, let s stat with the two vectos a = a, a, a give by the fomula, b = b, b, b ad a b = ab -ab, ab -ab, ab -ab the the coss poduct is This is ot a easy fomula to emembe. Thee ae two ways to deive this fomula. Both of them use the fact that the coss poduct is eally the detemiat of a x matix. If you do t kow what this is that is do t woy about it. You do t eed to kow aythig about matices o detemiats to use eithe of the methods. The otatio fo the detemiat is as follows, i j k a b = a a a b b b The fist ow is the stadad basis vectos ad must appea i the ode give hee. The secod ow is the compoets of a ad the thid ow is the compoets of b. Now, let s take a look at the diffeet methods fo gettig the fomula. The fist method uses the Method of Cofactos. If you do t kow the method of cofactos that is fie, the esult is all that we eed. Hee is the fomula. whee, a a a a a a a b = i - j + k b b b b b b a b ad bc c d = - This fomula is ot as difficult to emembe as it might at fist appea to be. Fist, the tems alteate i sig ad otice that the x is missig the colum below the stadad basis vecto that multiplies it as well as the ow of stadad basis vectos. The secod method is slightly easie; howeve, may textbooks do t cove this method as it will oly wok o x detemiats. This method says to take the detemiat as listed above ad the copy the fist two colums oto the ed as show below. 007 Paul Dawkis 04

314 i j k i j a b = a a a a a b b b b b We ow have thee diagoals that move fom left to ight ad thee diagoals that move fom ight to left. We multiply alog each diagoal ad add those that move fom left to ight ad subtact those that move fom ight to left. This is best see i a example. We ll also use this example to illustate a fact about coss poducts. Example If a =,, - ad b = -,4, (a) a b (b) b a Solutio (a) Hee is the computatio fo this oe. i j k i j a b = - compute each of the followig = i + j k - j -i - -k - = 5i + j + k (b) Ad hee is the computatio fo this oe. i j k i j b a =- 4-4 ()() ( ) ( 4) () ( 4) ()( ) - = i - + j + k - - j - - -i -k =-5i - j -k ( 4) ()( ) ( )() ( ) ()() ( 4) Notice that switchig the ode of the vectos i the coss poduct simply chaged all the sigs i the esult. Note as well that this meas that the two coss poducts will poit i exactly opposite diectios sice they oly diffe by a sig. We ll fomalize up this fact shotly whe we list seveal facts. Thee is also a geometic itepetatio of the coss poduct. Fist we will let q be the agle betwee the two vectos a ad b ad assume that 0 q p, the we have the followig fact, a b = a b siq () ad the followig figue. 007 Paul Dawkis 05

315 Thee should be a atual questio at this poit. How did we kow that the coss poduct poited i the diectio that we ve give it hee? Fist, as this figue, implies the coss poduct is othogoal to both of the oigial vectos. 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 upwad diectio (i this case) by the ight had ule. This says that if we take ou ight had, stat at a ad otate ou figes towads b ou thumb will poit i the diectio of the coss poduct. Theefoe, if we d sketched i b a above we would have gotte a vecto i the dowwad diectio. Example A plae is defied by ay thee poits that ae i the plae. If a plae cotais the,0,0,, R =, -, fid a vecto that is othogoal to the plae. poits P =, Q = ad Solutio The oe way that we kow to get a othogoal vecto is to take a coss poduct. So, if we could fid two vectos that we kew wee i the plae ad took the coss poduct of these two vectos we kow that the coss poduct would be othogoal to both the vectos. Howeve, sice both the vectos ae i the plae the coss poduct would the also be othogoal to the plae. So, we eed two vectos that ae i the plae. This is whee the poits come ito the poblem. Sice all thee poits lie i the plae ay vecto betwee them must also be i the plae. Thee ae may ways to get two vectos betwee these poits. We will use the followig two, uuu PQ = -,-0,- 0 = 0,, uuu PR = -, --0,- 0 =, -, The coss poduct of these two vectos will be othogoal to the plae. So, let s fid the coss poduct. 007 Paul Dawkis 06

316 uuu uuu PQ PR = i j k i j = 4i + j -k So, the vecto 4i + j -k will be othogoal to the plae cotaiig the thee poits. Now, let s addess the oe time whee the coss poduct will ot be othogoal to the oigial vectos. If the two vectos, a ad b, ae paallel the the agle betwee them is eithe 0 o 80 degees. Fom () this implies that, a b = 0 Fom a fact about the magitude we saw i the fist sectio we kow that this implies a b = 0 I othe wods, it wo t be othogoal to the oigial vectos sice we have the zeo vecto. This does give us aothe test fo paallel vectos howeve. Fact If a b = 0 the a ad b will be paallel vectos. Let s also fomalize up the fact about the coss poduct beig othogoal to the oigial vectos. Fact Povided a b 0 the a b is othogoal to both a ad b. Hee ae some ice popeties about the coss poduct. Popeties If u, v ad w ae vectos ad c is a umbe the, u v =- v u cu v = u cv = cu v u v+ w = u v+ u w ug v w = u v gw u u u ug v w = v v v w w w The detemiat i the last fact is computed i the same way that the coss poduct is computed. We will see a example of this computatio shotly. Thee ae a couple of geometic applicatios to the coss poduct as well. Suppose we have thee vectos a, b ad c ad we fom the thee dimesioal figue show below. 007 Paul Dawkis 07

317 The aea of the paallelogam (two dimesioal fot of this object) is give by, Aea = a b ad the volume of the paallelpiped (the whole thee dimesioal object) is give by, Volume = ag b c Note that the absolute value bas ae equied sice the quatity could be egative ad volume is t egative. We ca use this volume fact to detemie if thee vectos lie i the same plae o ot. If thee vectos lie i the same plae the the volume of the paallelpiped will be zeo. Example Detemie if the thee vectos a =,4, -7 i the same plae o ot., b =, -,4 ad c = 0, -9,8 Solutio So, as we oted pio to this example all we eed to do is compute the volume of the paallelpiped fomed by these thee vectos. If the volume is zeo they lie i the same plae ad if the volume is t zeo they do t lie i the same plae. lie 007 Paul Dawkis 08

318 4-7 4 ag b = ( c) ()( )( 8) ( 4)( 4)( 0) ( 7)( 9) ( 4)( 8) -()( 4)( -9)-(-7)(-)( 0) = = = 0 So, the volume is zeo ad so they lie i the same plae. 007 Paul Dawkis 09

319 Thee Dimesioal Space Itoductio I this chapte we will stat takig a moe detailed look at thee dimesioal space (-D space o ). This is a vey impotat topic fo Calculus III sice a good potio of Calculus III is doe i thee (o highe) dimesioal space. We will be lookig at the equatios of gaphs i -D space as well as vecto valued fuctios ad how we do calculus with them. We will also be takig a look at a couple of ew coodiate systems fo -D space. This is the oly chapte that exists i two places i my otes. Whe I oigially wote these otes all of these topics wee coveed i Calculus II howeve, we have sice moved seveal of them ito Calculus III. So, athe tha split the chapte 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 coveed i Calculus III will be used o occasio thee ayway ad so they seve as a quick efeece fo whe we eed them. Hee is a list of topics i this chapte. The -D Coodiate System We will itoduce the cocepts ad otatio fo the thee dimesioal coodiate system i this sectio. Equatios of Lies I this sectio we will develop the vaious foms fo the equatio of lies i thee dimesioal space. Equatios of Plaes Hee we will develop the equatio of a plae. Quadic Sufaces I this sectio we will be lookig at some examples of quadic sufaces. Fuctios of Seveal Vaiables A quick eview of some impotat topics about fuctios of seveal vaiables. Vecto Fuctios We itoduce the cocept of vecto fuctios i this sectio. We cocetate pimaily o cuves i thee dimesioal space. We will howeve, touch biefly o sufaces as well. Calculus with Vecto Fuctios Hee we will take a quick look at limits, deivatives, ad itegals with vecto fuctios. Taget, Nomal ad Biomal Vectos We will defie the taget, omal ad biomal vectos i this sectio. Ac Legth with Vecto Fuctios I this sectio we will fid the ac legth of a vecto fuctio. 007 Paul Dawkis 0

320 Cuvatue We will detemie the cuvatue of a fuctio i this sectio. Velocity ad Acceleatio I this sectio we will evisit a stadad applicatio of deivatives. We will look at the velocity ad acceleatio of a object whose positio fuctio is give by a vecto fuctio. Cylidical Coodiates We will defie the cylidical coodiate system i this sectio. The cylidical coodiate system is a alteate coodiate system fo the thee dimesioal coodiate system. Spheical Coodiates I this sectio we will defie the spheical coodiate system. The spheical coodiate system is yet aothe alteate coodiate system fo the thee dimesioal coodiate system. 007 Paul Dawkis

321 The -D Coodiate System We ll stat the chapte off with a faily shot discussio itoducig the -D coodiate system ad the covetios that we ll be usig. We will also take a bief look at how the diffeet coodiate systems ca chage the gaph of a equatio. Let s fist get some basic otatio out of the way. The -D coodiate system is ofte deoted by. Likewise the -D coodiate system is ofte deoted by ad the -D coodiate system is deoted by. Also, as you might have guessed the a geeal dimesioal coodiate system is ofte deoted by. Next, let s take a quick look at the basic coodiate system. This is the stadad placemet of the axes i this class. It is assumed that oly the positive diectios ae show by the axes. If we eed the egative axis fo ay easo we will put them i as eeded. Also ote the vaious poits o this sketch. The poit P is the geeal poit sittig out i -D space. If we stat at P ad dop staight dow util we each a z-coodiate of zeo we aive at the poit Q. We say that Q sits i the xy-plae. The xy-plae coespods to all the poits which have a zeo z-coodiate. We ca also stat at P ad move i the othe two diectios as show to get poits i the xz-plae (this is S with a y-coodiate of zeo) ad the yz-plae (this is R with a x-coodiate of zeo). Collectively, the xy, xz, ad yz-plaes ae sometimes called the coodiate plaes. I the emaide of this class you will eed to be able to deal with the vaious coodiate plaes so make sue that you ca. Also, the poit Q is ofte efeed to as the pojectio of P i the xy-plae. Likewise, R is the pojectio of P i the yz-plae ad S is the pojectio of P i the xz-plae. May of the fomulas that you ae used to wokig with i have atual extesios i Fo istace the distace betwee two poits i is give by,. 007 Paul Dawkis

322 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 geeal equatio fo a cicle with cete ( hk, ) ad adius is give by, ( x- h) + ( y- k) = ad the geeal equatio fo a sphee with cete ( hkl,, ) ad adius is give by, ( x- h) + ( y- k) + ( z- l) = With that said we do eed to be caeful about just taslatig eveythig we kow about ito ad assumig that it will wok the same way. A good example of this is i gaphig to some extet. Coside the followig example. Example Gaph x = i, ad. Solutio I we have a sigle coodiate system ad so x = is a poit i a -D coodiate system. I the equatio x = tells us to gaph all the poits that ae i the fom (, y ). This is a vetical lie i a -D coodiate system. I the equatio x = tells us to gaph all the poits that ae i the fom (,, ) yz. If you go back ad look at the coodiate plae poits this is vey simila to the coodiates fo the yz-plae except this time we have x = istead of x = 0. So, i a -D coodiate system this is a plae that will be paallel to the yz-plae ad pass though the x-axis at x =. Hee is the gaph of x = i. Hee is the gaph of x = i. 007 Paul Dawkis

323 Fially, hee is the gaph of x = i. Note that we ve peseted this gaph i two diffeet styles. O the left we ve got the taditioal axis system ad we e used to seeig ad o the ight we ve put the gaph i a box. Both views ca be coveiet o occasio to help with pespective ad so we ll ofte do this with D gaphs ad sketches. Note that at this poit we ca ow wite dow the equatios fo each of the coodiate 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 moe geeal example. 007 Paul Dawkis 4

324 Example Gaph y = x- i ad. Solutio Of couse we had to thow out fo this example sice thee ae two vaiables which meas that we ca t be i a -D space. I this is a lie with slope ad a y itecept of -. Howeve, i this is ot ecessaily a lie. Because we have ot specified a value of z we ae foced to let z take ay value. This meas that at ay paticula value of z we will get a copy of this lie. So, the gaph is the a vetical plae that lies ove the lie give by y = x- i the xy-plae. Hee is the gaph i. hee is the gaph i. Notice that if we look to whee the plae itesects the xy-plae we will get the gaph of the lie i as oted i the above gaph by the ed lie though the plae. 007 Paul Dawkis 5

325 Let s take a look at oe moe example of the diffeece betwee gaphs i the diffeet coodiate systems. Example Gaph x + y = 4 i ad. Solutio As with the pevious example this wo t have a -D gaph sice thee ae two vaiables. I this is a cicle ceteed at the oigi with adius. I howeve, as with the pevious example, this may o may ot be a cicle. Sice we have ot specified z i ay way we must assume that z ca take o ay value. I othe wods, at ay value of z this equatio must be satisfied ad so at ay value z we have a cicle of adius ceteed o the z-axis. This meas that we have a cylide of adius ceteed o the z-axis. Hee ae the gaphs fo this example. Notice that agai, if we look to whee the cylide itesects the xy-plae we will agai get the cicle fom. 007 Paul Dawkis 6

326 We eed to be caeful with the last two examples. It would be temptig to take the esults of these ad say that we ca t gaph lies o cicles i ad yet that does t eally make sese. Thee is o easo fo thee to ot be gaphs lies o cicles i. Let s thik about the example of the cicle. To gaph a cicle i we would eed to do somethig like x + y = 4 at z = 5. This would be a cicle of adius ceteed o the z-axis at the level of z = 5. So, as log as we specify a z we will get a cicle ad ot a cylide. We will see a easie way to specify cicles i a late sectio. We could do the same thig with the lie fom the secod example. Howeve, we will be lookig at lie i moe geeality i the ext sectio ad so we ll see a bette way to deal with lies i thee. The poit of the examples i this sectio is to make sue that we ae beig caeful with gaphig equatios ad makig sue that we always emembe which coodiate system that we ae i. Aothe quick poit to make hee is that, as we ve see i the above examples, may gaphs of equatios i ae sufaces. That does t mea that we ca t gaph cuves i. We ca ad will gaph cuves i as well as we ll see late i this chapte. 007 Paul Dawkis 7

327 Equatios of Lies I this sectio we eed to take a look at the equatio of a lie i. As we saw i the pevious sectio the equatio y = mx+ b does ot descibe a lie i, istead it descibes a plae. This does t mea howeve that we ca t wite dow a equatio fo a lie i -D space. We e just goig to eed a ew way of witig dow the equatio of a cuve. So, befoe we get ito the equatios of lies we fist eed to biefly look at vecto fuctios. We e goig to take a moe i depth look at vecto fuctios late. At this poit all that we eed to woy about is otatioal issues ad how they ca be used to give the equatio of a cuve. The best way to get a idea of what a vecto fuctio is ad what its gaph looks like is to look at a example. So, coside the followig vecto fuctio. t = t, A vecto fuctio is a fuctio that takes oe o moe vaiables, oe i this case, ad etus a vecto. Note as well that a vecto fuctio ca be a fuctio of two o moe vaiables. Howeve, i those cases the gaph may o loge be a cuve i space. The vecto that the fuctio gives ca be a vecto i whateve dimesio we eed it to be. I the example above it etus a vecto i. Whe we get to the eal subject of this sectio, equatios of lies, we ll be usig a vecto fuctio that etus a vecto i Now, we wat to detemie the gaph of the vecto fuctio above. I ode to fid the gaph of ou fuctio we ll thik of the vecto that the vecto fuctio etus as a positio vecto fo poits o the gaph. Recall that a positio vecto, say v = ab,, is a vecto that stats at the oigi ad eds at the poit ( ab, ). So, to get the gaph of a vecto fuctio all we eed to do is plug i some values of the vaiable ad the plot the poit that coespods to each positio vecto we get out of the fuctio ad play coect the dots. Hee ae some evaluatios fo ou example. - = -, - = -, =, 5 = 5, So, each of these ae positio vectos epesetig poits o the gaph of ou vecto fuctio. The poits, -, -,, 5, ae all poits that lie o the gaph of ou vecto fuctio. If we do some moe evaluatios ad plot all the poits we get the followig sketch. 007 Paul Dawkis 8

328 I this sketch we ve icluded the positio vecto (i gay ad dashed) fo seveal evaluatios as well as the t (above each poit) we used fo each evaluatio. It looks like, i this case the gaph of the vecto equatio is i fact the lie y =. t t t Hee s aothe quick example. Hee is the gaph of = 6cos,si. I this case we get a ellipse. It is impotat to ot come away fom this sectio with the idea that vecto fuctios oly gaph out lies. We ll be lookig at lies i this sectio, but the gaphs of vecto fuctio do ot have to be lies as the example above shows. We ll leave this bief discussio of vecto fuctio with aothe way to thik of the gaph of a vecto fuctio. Imagie that a pecil/pe is attached to the ed of the positio vecto ad as we icease the vaiable the esultig positio vecto moves ad as it moves the pecil/pe o the ed sketches out the cuve fo the vecto fuctio. Okay, we ow eed to move ito the actual topic of this sectio. We wat to wite dow the equatio of a lie i ad as suggested by the wok above we will eed a vecto fuctio to do this. To see how we e goig to do this let s thik about what we eed to wite dow the 007 Paul Dawkis 9

329 equatio of a lie i. I two dimesios we eed the slope (m) ad a poit that was o the lie i ode to wite dow the equatio. I that is still all that we eed except i this case the slope wo t be a simple umbe as it was i two dimesios. I this case we will eed to ackowledge that a lie ca have a thee dimesioal slope. So, we eed somethig that will allow us to descibe a diectio that is potetially i thee dimesios. We aleady have a quatity that will do this fo us. Vectos give diectios ad ca be thee dimesioal objects. So, let s stat with the followig ifomatio. Suppose that we kow a poit that is o the lie, P = x, y, z, ad that v = abc,, is some vecto that is paallel to the lie. Note, i all likelihood, v will ot be o the lie itself. We oly eed v to be paallel to the lie. Fially, let P= xyz,, be ay poit o the lie. Now, sice ou slope is a vecto let s also epeset the two poits o the lie as vectos. We ll do this with positio vectos. So, let u 0 ad be the positio vectos fo P 0 ad P espectively. Also, fo o appaet easo, let s defie a uuu to be the vecto with epesetatio PP 0. We ow have the followig sketch with all these poits ad vectos o it. Now, we ve show the paallel vecto, v, as a positio vecto but it does t eed to be a positio vecto. It ca be aywhee, a positio vecto, o the lie o off the lie, it just eeds to be paallel to the lie. Next, otice that we ca wite as follows, u = 0 + a If you e ot sue about this go back ad check out the sketch fo vecto additio i the vecto aithmetic sectio. Now, otice that the vectos a ad v ae paallel. Theefoe thee is a umbe, t, such that 007 Paul Dawkis 0

330 a = tv We ow have, u = + tv = x, y, z + t abc,, This is called the vecto fom of the equatio of a lie. The oly pat of this equatio that is ot kow is the t. Notice that tv will be a vecto that lies alog the lie ad it tells us how fa fom the oigial poit that we should move. If t is positive we move away fom the oigial poit i the diectio of v (ight i ou sketch) ad if t is egative we move away fom the oigial poit i the opposite diectio of v (left i ou sketch). As t vaies ove all possible values we will completely cove the lie. The followig sketch shows this depedece o t of ou sketch. Thee ae seveal othe foms of the equatio of a lie. To get the fist alteate fom let s stat with the vecto fom ad do a slight ewite. = x, y, z + t abc,, xyz,, = x + ta, y + tb, z + tc The oly way fo two vectos to be equal is fo the compoets to be equal. I othe wods, x= x + ta 0 y = y + tb 0 0 z = z + tc This set of equatios is called the paametic fom of the equatio of a lie. Notice as well that this is eally othig moe tha a extesio of the paametic equatios we ve see peviously. The oly diffeece is that we ae ow wokig i thee dimesios istead of two dimesios. 007 Paul Dawkis

331 To get a poit o the lie all we do is pick a t ad plug ito eithe fom of the lie. I the vecto fom of the lie we get a positio vecto fo the poit ad i the paametic fom we get the actual coodiates of the poit. Thee is oe moe fom of the lie that we wat to look at. If we assume that a, b, ad c ae all o-zeo umbes we ca solve each of the equatios i the paametic fom of the lie fo t. We ca the set all of them equal to each othe sice t will be the same umbe i each. Doig this gives the followig, x-x y- y z-z = = a b c This is called the symmetic equatios of the lie. If oe of a, b, o c does happe to be zeo we ca still wite dow the symmetic equatios. To see this let s suppose that b = 0. I this case t will ot exist i the paametic equatio fo y ad so we will oly solve the paametic equatios fo x ad z fo t. We the set those equal ad ackowledge the paametic equatio fo y as follows, x-x0 z-z0 = y = y0 a c Let s take a look at a example. Example Wite dow the equatio of the lie that passes though the poits (, -,) ad (,4, - ). Wite dow all thee foms of the equatio of the lie. Solutio To do this we eed the vecto v that will be paallel to the lie. This ca be ay vecto as log as it s paallel to the lie. I geeal, v wo t lie o the lie itself. Howeve, i this case it will. All we eed to do is let v be the vecto that stats at the secod poit ad eds at the fist poit. Sice these two poits ae o the lie the vecto betwee them will also lie o the lie ad will hece be paallel to the lie. So, v =, -5,6 Note that the ode of the poits was chose to educe the umbe of mius sigs i the vecto. We could just have easily goe the othe way. Oce we ve got v thee eally is t aythig else to do. To use the vecto fom we ll eed a poit o the lie. We ve got two ad so we ca use eithe oe. We ll use the fist poit. Hee is the vecto fom of the lie. =, -, + t, - 5,6 = + t, -- 5, t + 6t Oce we have this equatio the othe two foms follow. Hee ae the paametic equatios of the lie. 007 Paul Dawkis

332 x= + t y =--5t z = + 6t Hee is the symmetic fom. x - y + z - = = -5 6 Example Detemie if the lie that passes though the poit ( 0, -,8) ad is paallel to the lie give by x = 0+ t, y = t ad z =-- t passes though the xz-plae. If it does give the coodiates of that poit. Solutio To aswe this we will fist eed to wite dow the equatio of the lie. We kow a poit o the lie ad just eed a paallel vecto. We kow that the ew lie must be paallel to the lie give by the paametic equatios i the poblem statemet. That meas that ay vecto that is paallel to the give lie must also be paallel to the ew lie. Now ecall that i the paametic fom of the lie the umbes multiplied by t ae the compoets of the vecto that is paallel to the lie. Theefoe, the vecto, v =,, - is paallel to the give lie ad so must also be paallel to the ew lie. The equatio of ew lie is the, = 0, -,8 + t,, - =, t - + t,8-t If this lie passes though the xz-plae the we kow that the y-coodiate of that poit must be zeo. So, let s set the y compoet of the equatio equal to zeo ad see if we ca solve fo t. If we ca, this will give the value of t fo which the poit will pass though the xz-plae. - + t = 0 fi t = 4 So, the lie does pass though the xz-plae. To get the complete coodiates of the poit all we eed to do is plug t = ito ay of the equatios. We ll use the vecto fom. 4 ʈ ʈ = Á, - + Á,8 - =,0, Ë4 Ë Recall that this vecto is the positio vecto fo the poit o the lie ad so the coodiates of the Ê ˆ poit whee the lie will pass though the xz-plae ae Á,0, Ë Paul Dawkis

333 Equatios of Plaes I the fist sectio of this chapte we saw a couple of equatios of plaes. Howeve, oe of those equatios had thee vaiables i them ad wee eally extesios of gaphs that we could look at i two dimesios. We would like a moe geeal equatio fo plaes. So, let s stat by assumig that we kow a poit that is o the plae, P ( x, y, z ) 0 = Let s also suppose that we have a vecto that is othogoal (pepedicula) to the plae, = abc,,. This vecto is called the omal vecto. Now, assume that P ( xyz,, ) = is ay poit i the plae. Fially, sice we ae goig to be wokig with vectos iitially we ll let u 0 ad be the positio vectos fo P 0 ad P espectively. Hee is a sketch of all these vectos. Notice that we added i the vecto u - 0 which will lie completely i the plae. Also otice that we put the omal vecto o the plae, but thee is actually o easo to expect this to be the case. We put it hee to illustate the poit. It is completely possible that the omal vecto does ot touch the plae i ay way. Now, because is othogoal to the plae, it s also othogoal to ay vecto that lies i the plae. I paticula it s othogoal to u - 0. Recall fom the Dot Poduct sectio that two othogoal vectos will have a dot poduct of zeo. I othe wods, u u g - = 0 fi g = g This is called the vecto equatio of the plae Paul Dawkis 4

334 A slightly moe useful fom of the equatios is as follows. Stat with the fist fom of the vecto equatio ad wite dow a vecto fo the diffeece. abc,, g xyz,, - x, y, z = 0 Now, actually compute the dot poduct to get, ( ) abc,, g x-x, y- y, z- z = a x- x + b y- y + c z- z = This is called the scala equatio of plae. Ofte this will be witte as, ax+ by+ cz = d whee d = ax0 + by0 + cz0. This secod fom is ofte how we ae give equatios of plaes. Notice that if we ae give the equatio of a plae i this fom we ca quickly get a omal vecto fo the plae. A omal vecto is, = abc,, Let s wok a couple of examples. Example Detemie the equatio of the plae that cotais the poits P = (, -,0), Q = (,,4) ad R = ( 0, -,). Solutio I ode to wite dow the equatio of plae we eed a poit (we ve got thee so we e cool thee) ad a omal vecto. We eed to fid a omal vecto. Recall howeve, that we saw how to do this i the Coss Poduct sectio. We ca fom the followig two vectos fom the give poits. uuu uuu PQ=,,4 PR = -,, These two vectos will lie completely i the plae sice we fomed them fom poits that wee i the plae. Notice as well that thee ae may possible vectos to use hee, we just chose two of the possibilities. Now, we kow that the coss poduct of two vectos will be othogoal to both of these vectos. Sice both of these ae i the plae ay vecto that is othogoal to both of these will also be othogoal to the plae. Theefoe, we ca use the coss poduct as the omal vecto. i j k i j uuu uuu = PQ PR= 4 = i - 8j + 5k - - The equatio of the plae is the, 007 Paul Dawkis 5

335 ( x ) ( y ) ( z ) = 0 x- 8y+ 5z = 8 We used P fo the poit, but could have used ay of the thee poits. Example Detemie if the plae give by - x+ z = 0 ad the lie give by = 5, - t,0+ 4t ae othogoal, paallel o eithe. Solutio This is ot as difficult a poblem as it may at fist appea to be. We ca pick off a vecto that is omal to the plae. This is = -,0,. We ca also get a vecto that is paallel to the lie. This is v = 0, -,4. Now, if these two vectos ae paallel the the lie ad the plae will be othogoal. If you thik about it this makes some sese. If ad v ae paallel, the v is othogoal to the plae, but v is also paallel to the lie. So, if the two vectos ae paallel the lie ad plae will be othogoal. Let s check this. i j k i j v =- 0-0 = i + 4j + k So, the vectos ae t paallel ad so the plae ad the lie ae ot othogoal. Now, let s check to see if the plae ad lie ae paallel. If the lie is paallel to the plae the ay vecto paallel to the lie will be othogoal to the omal vecto of the plae. I othe wods, if ad v ae othogoal the the lie ad the plae will be paallel. Let s check this. v= g = 8 0 The two vectos ae t othogoal ad so the lie ad plae ae t paallel. So, the lie ad the plae ae eithe othogoal o paallel. 007 Paul Dawkis 6

336 Quadic Sufaces I the pevious two sectios we ve looked at lies ad plaes i thee dimesios (o ) ad while these ae used quite heavily at times i a Calculus class thee ae may othe sufaces that ae also used faily egulaly ad so we eed to take a look at those. I this sectio we ae goig to be lookig at quadic sufaces. Quadic sufaces ae the gaphs of ay equatio that ca be put ito the geeal fom Ax + By + Cz + Dxy+ Exz+ Fyz+ Gx+ Hy+ Iz+ J = 0 whee A,, J ae costats. Thee is o way that we ca possibly list all of them, but thee ae some stadad equatios so hee is a list of some of the moe commo quadic sufaces. Ellipsoid Hee is the geeal equatio of a ellipsoid. x y z + + = a b c Hee is a sketch of a typical ellipsoid. If a = b= c the we will have a sphee. Notice that we oly gave the equatio fo the ellipsoid that has bee ceteed o the oigi. Clealy ellipsoids do t have to be ceteed o the oigi. Howeve, i ode to make the discussio i this sectio a little easie we have chose to cocetate o sufaces that ae ceteed o the oigi i oe way o aothe. Coe Hee is the geeal equatio of a coe. Hee is a sketch of a typical coe. x y z + = a b c 007 Paul Dawkis 7

337 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 othe axes all we eed to do is make a slight modificatio of the equatio. This will be the case fo the est of the sufaces that we ll be lookig at i this sectio as well. I the case of a coe the vaiable that sits by itself o oe side of the equal sig will detemie the axis that the coe opes up alog. Fo istace, a coe that opes up alog the x-axis will have the equatio, y z x + = b c a Fo most of the followig sufaces we will ot give the othe possible fomulas. We will howeve ackowledge how each fomula eeds to be chaged to get a chage of oietatio fo the suface. Cylide Hee is the geeal equatio of a cylide. x a y b + = This is a cylide whose coss sectio is a ellipse. If a = b we have a cylide whose coss sectio is a cicle. We ll be dealig with those kids of cylides moe tha the geeal fom so the equatio of a cylide with a cicula coss sectio is, x + y = Hee is a sketch of typical cylide with a ellipse coss sectio. 007 Paul Dawkis 8

338 The cylide will be ceteed o the axis coespodig to the vaiable that does ot appea i the equatio. Be caeful to ot cofuse this with a cicle. I two dimesios it is a cicle, but i thee dimesios it is a cylide. Hypeboloid of Oe Sheet Hee is the equatio of a hypeboloid of oe sheet. x y z + - = a b c Hee is a sketch of a typical hypeboloid of oe sheet. The vaiable with the egative i fot of it will give the axis alog which the gaph is ceteed. 007 Paul Dawkis 9

339 Hypeboloid of Two Sheets Hee is the equatio of a hypeboloid of two sheets. x y z = a b c Hee is a sketch of a typical hypeboloid of two sheets. The vaiable with the positive i fot of it will give the axis alog which the gaph is ceteed. Notice that the oly diffeece betwee the hypeboloid of oe sheet ad the hypeboloid of two sheets is the sigs i fot of the vaiables. They ae exactly the opposite sigs. Elliptic Paaboloid Hee is the equatio of a elliptic paaboloid. x y z + = a b c As with cylides this has a coss sectio of a ellipse ad if a = b it will have a coss sectio of a cicle. Whe we deal with these we ll geeally be dealig with the kid that have a cicle fo a coss sectio. Hee is a sketch of a typical elliptic paaboloid. 007 Paul Dawkis 0

340 I this case the vaiable that is t squaed detemies the axis upo which the paaboloid opes up. Also, the sig of c will detemie the diectio that the paaboloid opes. If c is positive the it opes up ad if c is egative the it opes dow. Hypebolic Paaboloid Hee is the equatio of a hypebolic paaboloid. x y z - = a b c Hee is a sketch of a typical hypebolic paaboloid. These gaphs ae vaguely saddle shaped ad as with the elliptic paaoloid the sig of c will detemie the diectio i which the suface opes up. The gaph above is show fo c positive. 007 Paul Dawkis

341 With the both of the types of paaboloids discussed above the suface ca be easily moved up o dow by addig/subtactig a costat fom the left side. Fo istace z =-x - y + 6 is a elliptic paaboloid that opes dowwad (be caeful, the - is o the x ad y istead of the z) ad stats at z = 6 istead of z = 0. Hee is a couple of quick sketches of this suface. Note that we ve give two foms of the sketch hee. The sketch o the ight has the stadad set of axes but it is difficult to see the umbes o the axis. The sketch o the left has bee boxed ad this makes it easie to see the umbes to give a sese of pespective to the sketch. I most sketches that actually ivolve umbes o the axis system we will give both sketches to help get a feel fo what the sketch looks like. 007 Paul Dawkis

342 Fuctios of Seveal Vaiables I this sectio we wat to go ove some of the basic ideas about fuctios of moe tha oe vaiable. Fist, emembe that gaphs of fuctios of two vaiables, z f ( xy, ) dimesioal space. Fo example hee is the gaph of = ae sufaces i thee z x y = This is a elliptic paabaloid ad is a example of a quadic suface. We saw seveal of these i the pevious sectio. We will be seeig quadic sufaces faily egulaly late o i Calculus III. Aothe commo gaph that we ll be seeig quite a bit i this couse is the gaph of a plae. We have a covetio fo gaphig plaes that will make them a little easie to gaph ad hopefully visualize. Recall that the equatio of a plae is give by ax+ by+ cz = d o if we solve this fo z we ca wite it i tems of fuctio otatio. This gives, f ( xy, ) = Ax+ By+ D To gaph a plae we will geeally fid the itesectio poits with the thee axes ad the gaph the tiagle that coects those thee poits. This tiagle will be a potio of the plae ad it will give us a faily decet idea o what the plae itself should look like. Fo example let s gaph the plae give by, f xy, = -x- 4y 007 Paul Dawkis

343 Fo puposes of gaphig this it would pobably be easie to wite this as, z = -x-4y fi x+ 4y+ z = Now, each of the itesectio poits with the thee mai coodiate axes is defied by the fact that two of the coodiates ae zeo. Fo istace, the itesectio with the z-axis is defied by x= y = 0. So, the thee itesectio poits ae, Hee is the gaph of the plae. x - axis : 4,0,0 y - axis : 0,,0 z - axis : 0,0, Now, to exted this out, gaphs of fuctios of the fom w f ( xyz,, ) = would be fou dimesioal sufaces. Of couse we ca t gaph them, but it does t hut to poit this out. We ext wat to talk about the domais of fuctios of moe tha oe vaiable. Recall that y = f x, cosisted of all the values of x that we domais of fuctios of a sigle vaiable, could plug ito the fuctio ad get back a eal umbe. Now, if we thik about it, this meas that the domai of a fuctio of a sigle vaiable is a iteval (o itevals) of values fom the umbe lie, o oe dimesioal space. The domai of fuctios of two vaiables, z = f ( xy, ), ae egios fom two dimesioal space ad cosist of all the coodiate pais, ( xy, ), that we could plug ito the fuctio ad get back a eal umbe. 007 Paul Dawkis 4

344 Example Detemie 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 squae oot of a egative umbe so this meas that we must equie, x+ y 0 Hee is a sketch of the gaph of this egio. [Retu to Poblems] (b) This fuctio is diffeet fom the fuctio i the pevious pat. Hee we must equie that, x 0 ad y 0 ad they eally do eed to be sepaate iequalities. Thee is oe fo each squae oot i the fuctio. Hee is the sketch of this egio. [Retu to Poblems] (c) I this fial pat we kow that we ca t take the logaithm of a egative umbe o zeo. Theefoe we eed to equie that, 007 Paul Dawkis 5

345 x 9-x - 9y > 0 fi + y < 9 ad upo eaagig we see that we eed to stay iteio to a ellipse fo this fuctio. Hee is a sketch of this egio. Note that domais of fuctios of thee vaiables, w f ( xyz,, ) dimesioal space. Example Detemie the domai of the followig fuctio, f ( xyz,, ) = x + y + z -6 [Retu to Poblems] =, will be egios i thee Solutio I this case we have to deal with the squae oot ad divisio by zeo issues. These will equie, x + y + z - 6> 0 fi x + y + z > 6 So, the domai fo this fuctio is the set of poits that lies completely outside a sphee of adius 4 ceteed at the oigi. The ext topic that we should look at is that of level cuves o cotou cuves. The level cuves z = f xy, ae two dimesioal cuves we get by settig z = k, whee k is ay of the fuctio umbe. So the equatios of the level cuves ae f ( xy, ) = k. Note that sometimes the equatio will be i the fom f ( xyz,, ) = 0 ad i these cases the equatios of the level cuves ae f ( xyk,, ) = 0. You ve pobably see level cuves (o cotou cuves, whateve you wat to call them) befoe. If you ve eve see the elevatio map fo a piece of lad, this is othig moe tha the cotou cuves fo the fuctio that gives the elevatio of the lad i that aea. Of couse, we pobably do t have the fuctio that gives the elevatio, but we ca at least gaph the cotou cuves. Let s do a quick example of this. 007 Paul Dawkis 6

346 Example Idetify the level cuves of Solutio f xy, = x + y. Sketch a few of them. Fist, fo the sake of pactice, let s idetify what this suface give by f (, ) let s ewite it as, xy is. To do this z = x + y Now, this equatio is ot listed i the Quadic Sufaces sectio, but if we squae both sides we get, z = x + y ad this is listed i that sectio. So, we have a coe, o at least a potio of a coe. Sice we kow that squae oots will oly etu positive umbes, it looks like we ve oly got the uppe half of a coe. Note that this was ot equied fo this poblem. It was doe fo the pactice of idetifyig the suface ad this may come i hady dow the oad. Now o to the eal poblem. The level cuves (o cotou cuves) fo this suface ae give by the equatio ae foud by substitutig z = k. I the case of ou example this is, k = x + y fi x + y = k whee k is ay umbe. So, i this case, the level cuves ae cicles of adius k with cete at the oigi. We ca gaph these i oe of two ways. We ca eithe gaph them o the suface itself o we ca gaph them i a two dimesioal axis system. Hee is each gaph fo some values of k. 007 Paul Dawkis 7

347 Note that we ca thik of cotous i tems of the itesectio of the suface that is give by z = f xy, ad the plae z = k. The cotou will epeset the itesectio of the suface ad the plae. Fo fuctios of the fom f (,, ) of level sufaces ae give by f ( xyz,, ) = k whee k is ay umbe. xyz we will occasioally look at level sufaces. The equatios The fial topic i this sectio is that of taces. I some ways these ae simila to cotous. As z = f xy, ad oted above we ca thik of cotous as the itesectio of the suface give by the plae z = k. Taces of sufaces ae cuves that epeset the itesectio of the suface ad the plae give by x= a o y = b. Let s take a quick look at a example of taces. Example 4 Sketch the taces of f xy, = 0-4x - y fo the plae x = ad y =. Solutio We ll stat with x =. We ca get a equatio fo the tace by pluggig x = ito the equatio. Doig this gives, (), z = f y = - - y fi z = - y ad this will be gaphed i the plae give by x =. Below ae two gaphs. The gaph o the left is a gaph showig the itesectio of the suface ad the plae give by x =. O the ight is a gaph of the suface ad the tace that we ae afte i this pat. 007 Paul Dawkis 8

348 Fo y = we will do petty much the same thig that we did with the fist pat. Hee is the equatio of the tace, ad hee ae the sketches fo this case. z = f x, = 0-4x - fi z = 6-4x 007 Paul Dawkis 9