University of North Dakota

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1 AP E X C II Late Transcendentals University of North Dakota Adapted from AP E X Calculus by Gregory Hartman, Ph.D. Department of Applied Mathema cs Virginia Military Ins tute Revised July 7, 07

2 Contribu ng Authors Troy Siemers, Ph.D. Department of Applied Mathema cs Virginia Military Ins tute Brian Heinold, Ph.D. Department of Mathema cs and Computer Science Mount Saint Mary s University Michael Corral Mathema cs Schoolcra College Paul Dawkins, Ph.D. Department of Mathema cs Lamar University Dimplekumar Chalishajar, Ph.D. Department of Applied Mathema cs Virginia Military Ins tute Editor Jennifer Bowen, Ph.D. Department of Mathema cs and Computer Science The College of Wooster Copyright 05 Gregory Hartman 07 Department of Mathema cs, University of North Dakota This work is licensed under a Crea ve Commons A ribu on-noncommercial 4.0 Interna onal License. Resale and reproduc on restricted.

3 Contents Table of Contents Preface iii vii Calculus I Limits 7. An Introduc on To Limits Epsilon-Delta Defini on of a Limit Finding Limits Analy cally One Sided Limits Limits Involving Infinity Con nuity Deriva ves 77. Instantaneous Rates of Change: The Deriva ve Interpreta ons of the Deriva ve Basic Differen a on Rules The Product and Quo ent Rules The Chain Rule Implicit Differen a on The Graphical Behavior of Func ons Etreme Values The Mean Value Theorem Increasing and Decreasing Func ons Concavity and the Second Deriva ve Curve Sketching iii

4 4 Applica ons of the Deriva ve Related Rates Op miza on Differen als Newton s Method Integra on 9 5. An deriva ves and Indefinite Integra on The Definite Integral Riemann Sums The Fundamental Theorem of Calculus Subs tu on Applica ons of Integra on Area Between Curves Volume by Cross-Sec onal Area; Disk and Washer Methods The Shell Method Work Fluid Forces Calculus II Inverse Func ons and L Hôpital s Rule Inverse Func ons Deriva ves of Inverse Func ons Eponen al and Logarithmic Func ons Hyperbolic Func ons L Hôpital s Rule Techniques of Integra on Integra on by Parts Trigonometric Integrals Trigonometric Subs tu on Par al Frac on Decomposi on Integra on Strategies Improper Integra on Numerical Integra on Sequences and Series Sequences Infinite Series The Integral Test

5 9.4 Comparison Tests Alterna ng Series and Absolute Convergence Ra o and Root Tests Strategy for tes ng series Power Series Taylor Polynomials Taylor Series Curves in the Plane Arc Length and Surface Area Parametric Equa ons Calculus and Parametric Equa ons Introduc on to Polar Coordinates Calculus and Polar Func ons Calculus III 63 Vectors 633. Introduc on to Cartesian Coordinates in Space An Introduc on to Vectors The Dot Product The Cross Product Lines Planes Curvilinear Coordinates Vector Valued Func ons 7. Vector Valued Func ons Calculus and Vector Valued Func ons The Calculus of Mo on Unit Tangent and Normal Vectors The Arc Length Parameter and Curvature Func ons of Several Variables Introduc on to Mul variable Func ons Limits and Con nuity of Mul variable Func ons Par al Deriva ves Differen ability and the Total Differen al The Mul variable Chain Rule Direc onal Deriva ves Tangent Lines, Normal Lines, and Tangent Planes Etreme Values

6 3.9 Lagrange Mul pliers Mul ple Integra on Iterated Integrals and Area Double Integra on and Volume Double Integra on with Polar Coordinates Center of Mass Surface Area Volume Between Surfaces and Triple Integra on Change of Variables in Mul ple Integrals Line and Surface Integrals Line Integrals Proper es of Line Integrals Green s Theorem Surface Integrals and the Divergence Theorem Stokes Theorem Gradient, Divergence, Curl and Laplacian A Solu ons To Selected Problems A. Inde A.5

7 Preface A Note on Using this Tet Thank you for reading this short preface. Allow us to share a few key points about the tet so that you may be er understand what you will find beyond this page. This tet comprises a three volume series on Calculus. The first part covers material taught in many Calculus courses: limits, deriva ves, and the basics of integra on, found in Chapters through 6. The second tet covers material o en taught in Calculus : integra on and its applica ons, along with an introduc on to sequences, series and Taylor Polynomials, found in Chapters 7 through 0. The third tet covers topics common in Calculus 3 or Mul variable Calculus : parametric equa ons, polar coordinates, vector valued func- ons, and func ons of more than one variable, found in Chapters through 5. All three are available separately for free. Prin ng the en re tet as one volume makes for a large, heavy, cumbersome book. One can certainly only print the pages they currently need, but some prefer to have a nice, bound copy of the tet. Therefore this tet has been split into these three manageable parts, each of which can be purchased separately. A result of this spli ng is that some mes material is referenced that is not contained in the present tet. The contet should make it clear whether the missing material comes before or a er the current por on. Downloading the appropriate pdf, or the en re APEX Calculus LT pdf, will give access to these topics. For Students: How to Read this Tet Mathema cs tetbooks have a reputa on for being hard to read. High level mathema cal wri ng o en seeks to say much with few words, and this style o en seeps into tets of lower level topics. This book was wri en with the goal of being easier to read than many other calculus tetbooks, without becoming too verbose. Each chapter and sec on starts with an introduc on of the coming material, hopefully se ng the stage for why you should care, and ends with a look ahead to see how the just-learned material helps address future problems. Addi onally, each chapter includes a sec on zero, which provides a basic review and prac ce problems of pre-calculus skills. Since this content is a pre-requisite for calculus, reviewing and mastering these skills are considered your responsibility. This means that it is your responsibility to seek assistance outside of class from your instructor, a math resource center or other math tutoring available on-campus. A solid understanding of these skills are essen al to your success in solving calculus problems. Please read the tet; it is wri en to eplain the concepts of Calculus. There are numerous eamples to demonstrate the meaning of defini ons, the truth

8 of theorems, and the applica on of mathema cal techniques. When you encounter a sentence you don t understand, read it again. If it s ll doesn t make sense, read on anyway, as some mes confusing sentences are eplained by later sentences. You don t have to read every equa on. The eamples generally show all the steps needed to solve a problem. Some mes reading through each step is helpful; some mes it is confusing. When the steps are illustra ng a new technique, one probably should follow each step closely to learn the new technique. When the steps are showing the mathema cs needed to find a number to be used later, one can usually skip ahead and see how that number is being used, instead of ge ng bogged down in reading how the number was found. Some proofs have been delayed un l later (or omi ed completely). In mathema cs, proving something is always true is etremely important, and entails much more than tes ng to see if it works twice. However, students o en are confused by the details of a proof, or become concerned that they should have been able to construct this proof on their own. To alleviate this poten al problem, we do not include the more difficult proofs in the tet. The interested reader is highly encouraged to find other proofs online or from their instructor. In most cases, one is very capable of understanding what a theorem means and how to apply it without knowing fully why it is true. Work through the eamples. The best way to learn mathema cs is to do it. Reading about it (or watching someone else do it) is a poor subs tute. For this reason, every page has a place for you to put your notes so that you can work out the eamples. That being said, some mes it is useful to watch someone work through an eample. For this reason, this tet also provides links to online videos where someone is working through a similar problem. If you want even more videos, these are generally chosen from Khan Academy: Math Doctor Bob: Just Math Tutorials: (unfortunately, they re not well organized) Some other sites you may want to consider are Larry Green s Calculus Videos: courses/05/videos/videoinde.htm Mathispower4u: Yay Math: (for prerequisite material) All of these sites are completely free (although some will ask you to donate). Here s a sample one: Watch the video: Prac cal Advice for Those Taking College Calculus at Interac ve, 3D Graphics New to Version 3.0 is the addi on of interac ve, 3D graphics in the.pdf version. Nearly all graphs of objects in space can be rotated, shi ed, and zoomed in/out so the reader can be er understand the object illustrated.

9 As of this wri ng, the only pdf viewers that support these 3D graphics are Adobe Reader & Acrobat (and only the versions for PC / Mac / Uni / Linu computers, not tablets or smartphones). To ac vate the interac ve mode, click on the image. Once ac vated, one can click/drag to rotate the object and use the scroll wheel on a mouse to zoom in/out. (A great way to inves gate an image is to first zoom in on the page of the pdf viewer so the graphic itself takes up much of the screen, then zoom inside the graphic itself.) A CTRL-click/drag pans the object le /right or up/down. By right-clicking on the graph one can access a menu of other op ons, such as changing the ligh ng scheme or perspec ve. One can also revert the graph back to its default view. If you wish to deac vate the interac vity, one can right-click and choose the Disable Content op on. Thanks There are many people who deserve recogni on for the important role they have played in the development of this tet. First, I thank Michelle for her support and encouragement, even as this project from work occupied my me and a en on at home. Many thanks to Troy Siemers, whose most important contribu ons etend far beyond the sec ons he wrote or the 7 figures he coded in Asymptote for 3D interac on. He provided incredible support, advice and encouragement for which I am very grateful. My thanks to Brian Heinold and Dimplekumar Chalishajar for their contribu ons and to Jennifer Bowen for reading through so much material and providing great feedback early on. Thanks to Troy, Lee Dewald, Dan Joseph, Meagan Herald, Bill Lowe, John David, Vonda Walsh, Geoff Co, Jessica Liber ni and other faculty of VMI who have given me numerous sugges ons and correc ons based on their eperience with teaching from the tet. (Special thanks to Troy, Lee & Dan for their pa ence in teaching Calc III while I was s ll wri ng the Calc III material.) Thanks to Randy Cone for encouraging his tutors of VMI s Open Math Lab to read through the tet and check the solu ons, and thanks to the tutors for spending their me doing so. A very special thanks to Kris Brown and Paul Janiczek who took this opportunity far above & beyond what I epected, me culously checking every solu on and carefully reading every eample. Their comments have been etraordinarily helpful. I am also thankful for the support provided by Wane Schneiter, who as my Dean provided me with etra me to work on this project. I am blessed to have so many people give of their me to make this book be er. AP E X Affordable Print and Electronic texts AP E X is a consor um of authors who collaborate to produce high quality, low cost tetbooks. The current tetbook wri ng paradigm is facing a poten al revolu on as desktop publishing and electronic formats increase in popularity. However, wri ng a good tetbook is no easy task, as the me requirements alone are substan al. It takes countless hours of work to produce tet, write eamples and eercises, edit and publish. Through collabora on, however, the cost to any individual can be lessened, allowing us to create tets that we freely distribute electronically and sell in printed form for an incredibly low cost. Having said that, nothing is en rely free; someone always bears some cost. This tet cost the authors of this book their me, and that was not enough. APEX Calculus would not eist had not the Virginia Military Ins tute, through a generous Jackson Hope grant, given the lead author significant me away from teaching so he could focus on this tet. Each tet is available as a free.pdf, protected by a Crea ve Commons A ribu on Noncommercial 4.0 copyright. That means you can give the.pdf to

10 anyone you like, print it in any form you like, and even edit the original content and redistribute it. If you do the la er, you must clearly reference this work and you cannot sell your edited work for money. We encourage others to adapt this work to fit their own needs. One might add sec ons that are missing or remove sec ons that your students won t need. The source files can be found at You can learn more at Greg Hartman Crea ng AP E X LT Star ng with the source at faculty at the University of North Dakota made several substan al changes to create AP E X Late Transcendentals. The most obvious change was to rearrange the tet to delay proving the deriva ve of transcendental func ons un l Calculus. UND added Sec ons 7. and 7.3, adapted several sec ons from other resources, created the prerequisite sec ons, included links to videos and Geogebra, and added several eamples and eercises. In the end, every sec on had some changes (some more substan al than others), resul ng in a document that is about 0% longer. The source files can now be found at Etra thanks are due to Michael Corral for allowing us to use por ons of his Vector Calculus, available at (specifically, Sec ons.7, 3.9, and 4.7, and Chapter 5) and to Paul Dawkins for allowing us to use por ons of his online math notes from tutorial.math.lamar.edu/ (specifically, Sec ons 8.5 and 9.7, as well as Area with Parametric Equa ons in Sec- on 0.3). The work on Calculus III was par ally supported by the NDUS OER Ini a ve.

11 Calculus II

12

13 7: I F L H R This chapter completes our differen a on toolkit. The first and most important tool will be how to differen ate inverse func ons. We ll be able to use this to differen ate eponen al and logarithmic func ons, which we stated in Theorem 4 but did not prove. 7. Inverse Func ons We say that two func ons f and g are inverses if g(f()) = for all in the domain of f and f(g()) = for all in the domain of g. A func on can only have an inverse if it is one-to-one, i.e. if we never have f( ) = f( ) for different elements and of the domain. This is equivalent to saying that the graph of the func on passes the horizontal line test. The inverse of f is denoted f, which should not be confused with the func on /f(). Key Idea 7 Inverse Func ons For a one-to one func on f, The domain of f is the range of f; the range of f is the domain of f. f (f()) = for all in the domain of f. f(f ()) = for all in the domain of f. The graph of y = f () is the reflec on across y = of the graph of y = f(). y = f () if and only if f(y) = and y is in the domain of f. Watch the video: Finding the Inverse of a Func on or Showing One Does not Eist, E 3 at

14 Chapter 7 Inverse Func ons and L Hôpital s Rule y (,.5) To determine whether or not f and g are inverses for each other, we check to see whether or not g(f()) = for all in the domain of f,and f(g()) = for all in the domain of g. ( 0.5, 0.375) (0.375, 0.5) (.5, ) Figure 7.: A func on f along with its inverse f. (Note how it does not ma er which func on we refer to as f; the other is f.) Eample Verifying Inverses Determine whether or not the following pairs of func ons are inverses:. f() = 3 + ; g() = 3. f() = 3 + ; g() = /3 S. To check the composi on we plug f() in for in the defini on of g as follows: g(f()) = f() (3 + ) = = = So g(f()) = for all in the domain of f. Likewise, you can check that f(g()) = for all in the domain of g, so f and g are inverses.. If we try to proceed as before, we find that: g(f()) = (f()) /3 = ( 3 + ) /3 This doesn t seem to be the same as the iden ty func on. To verify this, we find a number a in the domain of f and show that g(f(a)) a for that value. Let s try =. Since f() = 3 + =, we find that g(f()) = g() = / Since g(f()), these func ons are not inverses. y (, 4) (, 4) 4 Figure 7.: The func on f() = is not one-to-one. Func ons that are not one-to-one. Unfortunately, not every func on we would like to find an inverse for is one-toone. For eample, the func on f() = is not one-to-one because f( ) = f() = 4. If f is an inverse for f, then f (f( )) = implies that f (4) =. On the other hand, f (f()) =, so f (4) =. We cannot have it both ways if f is a func on, so no such inverse eists. We can find a par al solu on to this dilemma by restric ng the domain of f. There are many possible choices, but tradi onally we restrict the domain to the interval [0, ). The func on f () = is now an inverse for this restricted version of f. 340

15 7. Inverse Func ons The inverse sine func on We consider the func on f() = sin, which is not one-to-one. A piece of the graph of f is in Figure 7.3(a). In order to find an appropriate restric on of the domain of f, we look for consecu ve cri cal points where f takes on its minimum and maimum values. In this case, we use the interval [ π/, π/]. We define the inverse of f on this restricted range by y = sin if and only if sin y = and π/ y π/. The graph is a reflec on of the graph of g across the line y =, as seen in Figure 7.3(b). π π y π π π y sin sin π π (a) π (b) Figure 7.3: (a) A por on of y = sin. (b) A one to one por on of y = sin along with y = sin. The inverse tangent func on Net we consider the func on f() = tan, which is also not one-to-one. A piece of the graph of f is given in Figure 7.4(a). In order to find an interval on which the func on is one-to-one and on which the func on takes on all values in the range, we use an interval between consecu ve ver cal asymptotes. Tradi onally, the interval ( π/, π/) is chosen. Note that we choose the open interval in this case because the func on f is not defined at the endpoints. So we define y = tan if and only if tan y = and π/ < y < π/. The graph of y = tan is shown in Figure 7.4(b). Also note that the ver cal asymptotes of the original func on are reflected to become horizontal asymptotes of the inverse func on. 34

16 Chapter 7 Inverse Func ons and L Hôpital s Rule y π y tan 3π π π π π 3π π tan π π (a) (b) Func on Restricted Domain Figure 7.4: (a) A por on of y = tan. (b) A one to one por on of y = tan along with y = tan. The other inverse trigonometric func ons are defined in a similar fashion. The resul ng domains and ranges are summarized in Figure 7.5. Range Inverse Func on Domain Range sin [ π/, π/] [, ] sin [, ] [ π/, π/] cos [0, π] [, ] cos [, ] [0, π] tan ( π/, π/) (, ) tan (, ) ( π/, π/) csc [ π/, 0) (0, π/] (, ] [, ) csc (, ] [, ) [ π/, 0) (0, π/] sec [0, π/) (π/, π] (, ] [, ) sec (, ] [, ) [0, π/) (π/, π] cot (0, π) (, ) cot (, ) (0, π) Figure 7.5: Domains and ranges of the trigonometric and inverse trigonometric func ons. Some mes, arcsin is used instead of sin. Similar arc func ons are used for the other inverse trigonometric func ons as well. Eample Evalua ng Inverse Trigonometric Func ons Find eact values for the following:. tan () 3. sin (sin(7π/6)). cos(sin ( 3/)) 4. tan(cos (/5)) S. tan () = π/4. cos(sin ( 3/)) = cos(π/3) = / 3. Since 7π/6 is not in the range of the inverse sine func on, we should be careful with this one. sin (sin(7π/6)) = sin ( /) = π/6. 4. Since we don t know the value of cos (/5), we let θ stand for this value. We know that θ is an angle between 0 and π and that cos(θ) = 34

17 7. Inverse Func ons /5. In Figure 7.6, we use this informa on to construct a right triangle with angle θ, where the adjacent side over the hypotenuse must equal /5. Applying the Pythagorean Theorem we find that Finally, we have: y = 5 = 04 = 6. tan(cos (/5)) = tan(θ) = 6. θ 5 Figure 7.6: A right triangle for the situa- on in Eample (4). y 343

18 Eercises 7. Terms and Concepts. T/F: Every func on has an inverse.. In your own words eplain what it means for a func on to be one to one. 3. If (, 0) lies on the graph of y = f(), what can be said about the graph of y = f ()? Problems In Eercises 4 8, verify that the given func ons are inverses. 4. f() = + 6 and g() = 3 5. f() = + 6 +, 3 and g() = 3, 6. f() = + 6 +, 3 and g() = 3,. 7. f() = , 5 and g() =, 0 8. f() = +, and g() = f() In Eercises 9, find a restric on of the domain of the given func on on which the func on will have an inverse. 9. f() = 6 0. g() = 6. r(t) = t 6t + 9. f() = + In Eercises 3 9, find the eact value. 3. tan (0) 4. tan (tan(π/7)) 5. cos(cos ( /5)) 6. sin (sin(8π/3)) 7. sin(tan ()) 8. cos(tan (3/7)) 9. sec(sin ( 3/5)) 344

19 7. Deriva ves of Inverse Func ons 7. Deriva ves of Inverse Func ons In this sec on we will figure out how to differen ate the inverse of a func on. To do so, we recall that if f and g are inverses, then f(g()) = for all in the domain of f. Differen a ng and simplifying yields: f(g()) = f (g())g () = g () = f (g()) assuming f () is nonzero Note that the deriva on above assumes that the func on g is differen able. It is possible to prove that g must be differen able if f is nonzero, but the proof is beyond the scope of this tet. However, assuming this fact we have shown the following: Theorem 46 Deriva ves of Inverse Func ons Let f be differen able and one-to-one on an open interval I, where f () 0 for all in I, let J be the range of f on I, let g be the inverse func on of f, and let f(a) = b for some a in I. Then g is a differen able func on on J, and in par cular, ( f ) (b) = g (b) = f (a) ( f ) () = g () = f (g()) The results of Theorem 46 are not trivial; the nota on may seem confusing at first. Careful considera on, along with eamples, should earn understanding. Watch the video: Deriva ve of an Inverse Func on, E at In the net eample we apply Theorem 46 to the arcsine func on. Eample Finding the deriva ve of an inverse trigonometric func on Let y = sin. Find y using Theorem

20 Chapter 7 Inverse Func ons and L Hôpital s Rule y Figure 7.7: A right triangle defined by y = sin (/) with the length of the third leg found using the Pythagorean Theorem. S Adop ng our previously defined nota on, let g() = sin and f() = sin. Thus f () = cos. Applying Theorem 46, we have g () = f (g()) = cos(sin ). This last epression is not immediately illumina ng. Drawing a figure will help, as shown in Figure 7.7. Recall that the sine func on can be viewed as taking in an angle and returning a ra o of sides of a right triangle, specifically, the ra o opposite over hypotenuse. This means that the arcsine func on takes as input a ra o of sides and returns an angle. The equa on y = sin can be rewri en as y = sin (/); that is, consider a right triangle where the hypotenuse has length and the side opposite of the angle with measure y has length. This means the final side has length, using the Pythagorean Theorem. Therefore cos(sin ) = cos y = / =, resul ng in d ( sin ) = g () = d. π π 4 y = sin π π 4 y ( π 3, 3 ) π 4 π y = sin π 4 π ( 3, π 3 ) Remember that the input of the arcsine func on is a ra o of a side of a right triangle to its hypotenuse; the absolute value of this ra o will be less than. Therefore will be posi ve. In order to make y = sin one-to-one, we restrict its domain to [ π/, π/]; on this domain, the range is [, ]. Therefore the domain of y = sin is [, ] and the range is [ π/, π/]. When = ±, note how the deriva ve of the arcsine func on is undefined; this corresponds to the fact that as ±, the tangent lines to arcsine approach ver cal lines with undefined slopes. In Figure 7.8 we see f() = sin and f () = sin graphed on their respec ve domains. The line tangent to sin at the point (π/3, 3/) has slope cos π/3 = /. The slope of the corresponding point on sin, the point ( 3/, π/3), is ( = = = 3/) 3/4 /4 / =, verifying Theorem 46 yet again: at corresponding points, a func on and its inverse have reciprocal slopes. Using similar techniques, we can find the deriva ves of all the inverse trigonometric func ons a er first restric ng their domains according to Figure 7.5 to allow them to be inver ble. Figure 7.8: Graphs of y = sin and y = sin along with corresponding tangent lines. 346

21 7. Deriva ves of Inverse Func ons Theorem 47 Deriva ves of Inverse Trigonometric Func ons The inverse trigonometric func ons are differen able on all open sets contained in their domains (as listed in Figure 7.5) and their deriva ves are as follows: d (. sin ) d ( = 4. cos ) = d d d (. sec ) = d d ( 5. csc ) = d 3. d ( tan ) = d + 6. d ( cot ) = d + Note how the last three deriva ves are merely the nega ves of the first three, respec vely. Because of this, the first three are used almost eclusively throughout this tet. Eample Finding deriva ves of inverse func ons Find the deriva ves of the following func ons:. f() = cos ( ). g() = sin 3. f() = sin (cos ) S. We use Theorem 47 and the Chain Rule to find: f () = ( ) () = 4. We use Theorem 47 and the Quo ent Rule to compute: ( ) ( ) g (sin ) ( ) () = ( ) = + sin ( ) 3 347

22 Chapter 7 Inverse Func ons and L Hôpital s Rule 3. We apply Theorem 47 and the Chain Rule again to compute: f () = ( sin ) cos = sin sin = sin sin =. Theorem 47 allows us to integrate some func ons that we could not integrate before. For eample, d = sin + C. Combining these formulas with u-subs tu on yields the following: Theorem 48 Let a > Integrals Involving Inverse Trigonometric Func ons a + d = ( a tan a ( a d = sin a a d = a sec ) + C ) + C ( ) + C a We will look at the second part of this theorem. The other parts are similar and are le as eercises. First we note that the integrand involves the number a, but does not eplicitly involve a. We make the assump on that a > 0 in order to simplify what follows. We can rewrite the integral as follows: d a = d a ( (/a) ) = d a (/a) 348

23 7. Deriva ves of Inverse Func ons We net use the subs tu on u = /a and du = d/a to find: d a (/a) = a a u du = du u = sin u + C = sin (/a) + C We conclude this sec on with several eamples. Eample 3 Finding an deriva ves involving inverse func ons Find the following integrals. d sin d 3. d S d = d 0 + = 0 tan (/0) + C. We use the subs tu on u = sin and du = d to find: sin = u du = u + C = ( sin ) + C 3. This does not immediately look like one of the forms in Theorem 48, but we can complete the square in the denominator to see that d = d ( + + ) + 4 = d 4 + ( + ) We now use the subs tu on u = + and du = d to find: d 4 + ( + ) = du 4 + u = tan (u/)+c = tan ( + ) +C. 349

24 Eercises 7. Terms and Concepts. If (, 0) lies on the graph of y = f() and f () = 5, what can be said about y = f ()? Problems In Eercises 6, an inver ble func on f() is given along with a point that lies on its graph. Using Theorem 46, evaluate ( f ) () at the indicated value.. f() = Point= (, 0) Evaluate ( f ) (0) 3. f() = + 4, Point= (3, 7) Evaluate ( f ) (7) 4. f() = sin, π/4 π/4 Point= (π/6, 3/) Evaluate ( f ) ( 3/) 5. f() = Point= (, 8) Evaluate ( f ) (8) 6. f() = +, 0 Point= (, /) Evaluate ( f ) (/) In Eercises 7 5, compute the deriva ve of the given func- on. 7. h(t) = sin (t) 8. f(t) = sec (t) 9. g() = tan () 0. f() = sin. g(t) = sin t cos t. h() = sin cos 3. g() = tan ( ) 4. f() = sec (/) 5. f() = sin(sin ) In Eercises 6 8, compute the deriva ve of the given func- on in two ways: (a) By simplifying first, then taking the deriva ve, and (b) by using the Chain Rule first then simplifying. Verify that the two answers are the same. 6. f() = sin(sin ) 7. f() = tan (tan ) 8. f() = sin(cos ) In Eercises 9 0, find the equa on of the line tangent to the graph of f at the indicated value. 9. f() = sin at = 0. f() = cos () at = 3 4. A regula on hockey goal is 6 feet wide. If a player is skating towards the end line on a line perpendicular to the end line and 0 feet from the imaginary line joining the center of one goal to the center of the other, the angle between the player and the goal first increases and then begins to decrease. In order to maimize this angle, how far from the end line should the player be when they shoot the puck? 0 In Eercises 8, compute the indicated integral / / 3 0 d d sin r r dr d e t 0 e t dt 3 + d ( + ) d θ 6 350

25 7.3 Eponen al and Logarithmic Func ons 7.3 Eponen al and Logarithmic Func ons In this sec on we will define general eponen al and logarithmic func ons and find their deriva ves. General eponen al func ons Consider first the func on f() =. If is ra onal, then we know how to compute. What do we mean by π though? We compute this by first looking at r for ra onal numbers r that are very close to π, then finding a limit. In our case we might compute 3, 3., 3.4, etc. We then define π to be the limit of these numbers. Note that this is actually a different kind of limit than we have dealt with before since we only consider ra onal number close to π, not all real numbers close to π. We will see one way to make this more precise in Chapter 9. Graphically, we can plot the values of for ra onal and get something like the do ed curve in Figure 7.9. In order to define the remaining values, we are connec ng the dots in a way that makes the func on con nuous. It follows from con nuity and the proper es of limits that eponen al func- ons will sa sfy the familiar proper es of eponents (see Sec on.0). This implies that ( ) = ( ) =, y Figure 7.9: The func on for ra onal values of. y so the graph of g() = (/) is the reflec on of f across the y-ais, as in Figure 7.0. We can go through the same process as above for any base a > 0, though we are not usually interested in the constant func on. 0.5 Figure 7.0: The func ons and. Key Idea 8 Proper es of Eponen al Func ons For a > 0 and a the eponen al func on f() = a sa sfies:. a 0 =. lim a = { a > 0 a < 3. a > 0 for all 4. lim a = { 0 a > a < 35

26 Chapter 7 Inverse Func ons and L Hôpital s Rule Deriva ves of eponen al func ons Suppose f() = a for some a > 0. We can use the rules of eponents to find the deriva ve of f: f f( + h) f() () = lim h 0 h a +h a = lim h 0 h a a h a = lim h 0 h a (a h ) = lim h 0 h = a a h lim (since a does not depend on h) h 0 h So we know that f () = a a h lim, but can we say anything about that h 0 h remaining limit? First we note that f a 0+h a 0 a h (0) = lim = lim, h 0 h h 0 h so we have f () = a f a h (0). The actual value of the limit lim depends h 0 h on the base a, but it can be proved that it does eist. We will figure out just what this limit is later, but for now we note that the easiest differen a on formulas a h come from using a base a that makes lim =. This base is the number h 0 h e.788 and the eponen al func on e is called the natural eponen al func on. This leads to the following result. Theorem 49 Deriva ve of Eponen al Func ons For any base a > 0, the eponen al func on f() = a has deriva- ve f () = a f (0). The natural eponen al func on g() = e has deriva ve g () = e. 35

27 7.3 Eponen al and Logarithmic Func ons Watch the video: Deriva ves of Eponen al Func ons at General logarithmic func ons Before reviewing general logarithmic func ons, we ll first remind ourselves of the laws of logarithms. Key Idea 9 Proper es of Logarithms For a,, y > 0 and a, we have. log a (y) = log a + log a y 3. log y = log a y log a, when 5. log a = 0. log a y = log a log a y 4. log a y = y log a 6. log a a = Let us consider the func on f() = a where a. We know that f () = f (0)a, where f (0) is a constant that depends on the base a. Since a > 0 for all, this implies that f () is either always posi ve or always nega ve, depending on the sign of f (0). This in turn implies that f is strictly monotonic, so f is oneto-one. We can now say that f has an inverse. We call this inverse the logarithm with base a, denoted f () = log a. When a = e, this is the natural logarithm func on ln. So we can say that y = log a if and only if a y =. Since the range of the eponen al func on is the set of posi ve real numbers, the domain of the logarithm func on is also the set of posi ve real numbers. Reflec ng the graph of y = a across the line y = we find that (for a > ) the graph of the logarithm looks like Figure 7.. e y e Figure 7.: The func ons y = a and y = log a for a >. 353

28 Chapter 7 Inverse Func ons and L Hôpital s Rule Key Idea 0 Proper es of Logarithmic Func ons For a > 0 and a the logarithmic func on f() = log a sa sfies:. The domain of f() = log a is (0, ) and the range is (, ).. y = log a if and only if a y =. { if a > 3. lim log a = if a < 4. lim 0 + log a = { if a > if a < Using the inverse of the natural eponen al func on, we can determine what the value of f (0) is in the formula (a ) = f (0)a. To do so, we note that a = e ln a since the eponen al and logarithm func ons are inverses. Hence we can write: a = ( e ln a) = e ln a Now since ln a is a constant, we can use the Chain Rule to see that: d d a = d d e ln a = e ln a (ln a) = a ln a Comparing this to our previous result, we can restate our theorem: Theorem 50 Deriva ve of Eponen al Func ons For any base a > 0, the eponen al func on f() = a has deriva ve f () = a ln a. The natural eponen al func on g() = e has deriva- ve g () = e. Change of base In the previous computa on, we found it convenient to rewrite the general eponen al func on in terms of the natural eponen al func on. A related formula allows us to rewrite the general logarithmic func on in terms of the natural 354

29 7.3 Eponen al and Logarithmic Func ons logarithm. To see how this works, suppose that y = log a, then we have: a y = ln(a y ) = ln y ln a = ln y = ln ln a log a = ln ln a. This change of base formula allows us to use facts about the natural logarithm to derive facts about the general logarithm. Deriva ves of logarithmic func ons Since the natural logarithm func on is the inverse of the natural eponen al func on, we can use the formula (f ()) = f (f to find the deriva ve ()) of y = ln. We know that d d e = e, so we get: d d ln = e y = e ln =. Now we can apply the change of base formula to find the deriva ve of a general logarithmic func on: d d log a = d ( ) ln = ( ) d d ln a ln a d ln = ln a. Eample Finding Deriva ves of Logs and Eponen als Find deriva ves of the following func ons.. f() = g() = 3. h() = log 5 S. We apply both the Product and Chain Rules: f () = ( ln 3 ) (4) = ( + 4 ln 3) We apply the Chain Rule: g () = ln () = + ln. 3. Applying the Quo ent Rule: h () = log 5 ( ) ln 5 (log 5 ) = (log 5 )(ln 5) (log 5 ) ln 5 355

30 Chapter 7 Inverse Func ons and L Hôpital s Rule Eample The Deriva ve of the Natural Log Find the deriva ve of the func on y = ln. S We can rewrite our func on as { ln if > 0 y = ln( ) if < 0 Applying the Chain Rule, we see that dy < 0. Hence we have d d ln = d = for > 0, and dy d = = for for 0. An deriva ves Combining these new results, we arrive at the following theorem: Theorem 5 Deriva ves and An deriva ves of Eponen als and Logarithms Given a base a > 0 and a, the following hold: d d e = e d d a = a ln a d d ln = d d log a = ln a 5. e d = e + C 6. a d = a ln a + C d 7. = ln + C Eample 3 Finding An deriva ves Find the following an deriva ves.. 3 d. e 3 d 3. d + 356

31 7.3 Eponen al and Logarithmic Func ons S. Applying our theorem, 3 d = 3 ln 3 + C. We use the subs tu on u = 3, du = 3 d: e 3 d = e u du 3 = 3 eu + C = 3 e3 + C 3. Using the subs tu on u = +, du = d: d + = du u = ln u + C = ln + + C = ln( + ) + C Note that we do not yet have an an deriva ve for the func on f() = ln. We remedy this in Sec on 8. with Eample Logarithmic Differen a on Consider the func on y = ; it is graphed in Figure 7.. It is well defined for > 0 and we might be interested in finding equa ons of lines tangent and normal to its graph. How do we take its deriva ve? The func on is not a power func on: it has a power of, not a constant. It is not an eponen al func on: it has a base of, not a constant. A differen a on technique known as logarithmic differen a on becomes useful here. The basic principle is this: take the natural log of both sides of an equa on y = f(), then use implicit differen a on to find y. We demonstrate this in the following eample. y 4 3 Figure 7.: A plot of y =. 357

32 Chapter 7 Inverse Func ons and L Hôpital s Rule Eample 4 Using Logarithmic Differen a on Given y =, use logarithmic differen a on to find y. S As suggested above, we start by taking the natural log of both sides then applying implicit differen a on. y = ln(y) = ln( ) (apply logarithm rule) ln(y) = ln (now use implicit differen a on) d ( ) ln(y) = d ( ) ln d d y y = ln + y y = ln + y = y ( ln + ) (subs tute y = ) y = ( ln + ). y 4 3 To test our answer, let s use it to find the equa on of the tangent line at =.5. The point on the graph our tangent line must pass through is (.5,.5.5 ) (.5,.837). Using the equa on for y, we find the slope as y =.5.5( ln.5 + ).837(.405).58. Thus the equa on of the tangent line is y =.6833(.5) Figure 7.3 graphs y = along with this tangent line. Figure 7.3: A graph of y = and its tangent line at =

33 Eercises 7.3 Problems In Eercises 4, find the domain of the func on.. f() = e +. f(t) = ln( ) 3. g() = ln( ) 4. f() = log 3 ( + ) In Eercises 5 3, find the deriva ve of the func on. 5. f(t) = e t3 6. g(r) = log 7. f() = log f() = f() = 7 log 7 0. g() = e sin( ln ). h(r) = tan (3 r ). h() = log 0 ( f(t) = ln te t ) 4. Find the two values of n so that the func on y = e n sa sfies the differen al equa on y + y 6y = 0. In Eercises 5 3, evaluate the integral d log 3 d 3 d 8. cos(ln ) d 9. e sin(e ) cos(e ) d log d d t + t dt ( + ) tan d 4. Let f() = and g() =. (a) Since f() = = 4 and g() = = 4, f() = g(). Find a posi ve number c > so that f(c) = g(c). (b) Eplain how you can be sure that there is at least one nega ve number a so that f(a) = g(a). (c) Use the Bisec on Method to es mate the number a accurate to within.05. (d) Assume you were to graph f() and g() on the same graph with unit length equal to inch along both coordinate aes. Approimately how high is the graph of f when = 8? The graph of g? In Eercises 5 3, use logarithmic differen a on to find dy d, then find the equa on of the tangent line at the indicated value. 5. y = ( + ) /, = 6. y = (), = 7. y = +, = 8. y = sin()+, = π/ 9. y = + +, = 30. y = ( + )( + ) ( + 3)( + 4), = 0 3. y = e, = 3. y = (cot ) cos, = π 359

34 Chapter 7 Inverse Func ons and L Hôpital s Rule 7.4 Hyperbolic Func ons y (cos θ,sin θ) A = θ + y = The hyperbolic func ons are func ons that have many applica ons to mathema cs, physics, and engineering. Among many other applica ons, they are used to describe the forma on of satellite rings around planets, to describe the shape of a rope hanging from two points, and have applica on to the theory of special rela vity. This sec on defines the hyperbolic func ons and describes many of their proper es, especially their usefulness to calculus. These func ons are some mes referred to as the hyperbolic trigonometric func ons as there are many connec ons between them and the standard trigonometric func ons. Figure 7.4 demonstrates one such connec on. Just as cosine and sine are used to define points on the circle defined by + y =, the func ons hyperbolic cosine and hyperbolic sine are used to define points on the hyperbola y =. We begin with their defini ons. y y = (cosh θ,sinh θ) A = θ Defini on 7. cosh = e + e Hyperbolic Func ons. sinh = e e 3. tanh = sinh cosh 4. sech = cosh 5. csch = sinh 6. coth = cosh sinh Figure 7.4: Using trigonometric func- ons to define points on a circle and hyperbolic func ons to define points on a hyperbola. The hyperbolic func ons are graphed in Figure 7.5. In the graphs of cosh and sinh, graphs of e / and e / are included with dashed lines. As gets large, cosh and sinh each act like e /; when is a large nega ve number, cosh acts like e / whereas sinh acts like e /. Pronuncia on Note: cosh rhymes with gosh, sinh rhymes with pinch, and tanh rhymes with ranch, No ce the domains of tanh and sech are (, ), whereas both coth and csch have ver cal asymptotes at = 0. Also note the ranges of these func ons, especially tanh : as, both sinh and cosh approach e /, hence tanh approaches. 360

35 7.4 Hyperbolic Func ons 0 y 0 y f() = cosh 5 f() = sinh 0 0 y y 3 f() = coth f() = sech f() = csch f() = tanh 3 Figure 7.5: Graphs of the hyperbolic func ons. Watch the video: Hyperbolic Func ons The Basics at The following eample eplores some of the proper es of these func ons that bear remarkable resemblance to the proper es of their trigonometric counterparts. Eample Eploring proper es of hyperbolic func ons Use Defini on 7 to rewrite the following epressions. 36

36 Chapter 7 Inverse Func ons and L Hôpital s Rule. cosh sinh. tanh + sech 3. cosh sinh ( ) d 4. d cosh ( ) sinh 5. d d 6. d d ( tanh ) S (. e cosh sinh + e ) ( e e = So cosh sinh =. = e + e e + e 4 = 4 4 =. ) e e e + e 4. tanh + sech = sinh cosh + cosh = sinh + cosh Now use iden ty from #. = cosh cosh =. So tanh + sech =. ( 3. e + e ) ( e e ) cosh sinh = = e e 4 = e e = sinh(). 4. Thus cosh sinh = sinh(). d ( ) d cosh = d d ( e + e ) = e e = sinh. So d d( cosh ) = sinh. 36

37 7.4 Hyperbolic Func ons 5. d ( ) d sinh = d d ( e e ) = e + e = cosh. 6. ) So d( d sinh = cosh. ( ) d ( ) d sinh tanh = d d cosh cosh cosh sinh sinh = = cosh = sech. cosh So d d( tanh ) = sech. The following Key Idea summarizes many of the important iden es rela ng to hyperbolic func ons. Each can be verified by referring back to Defini on 7. Key Idea Useful Hyperbolic Func on Proper es Basic Iden es Deriva ves. cosh sinh =. tanh + sech = 3. coth csch = 4. cosh = cosh + sinh 5. sinh = sinh cosh 6. cosh = 7. sinh = cosh + cosh. d d( cosh ) = sinh. d d( sinh ) = cosh 3. d d( tanh ) = sech 4. d d( sech ) = sech tanh 5. d d( csch ) = csch coth 6. d d( coth ) = csch Integrals. cosh d = sinh + C sinh d = cosh + C tanh d = ln(cosh ) + C coth d = ln sinh + C We prac ce using Key Idea. 363

38 Chapter 7 Inverse Func ons and L Hôpital s Rule Eample Deriva ves and integrals of hyperbolic func ons Evaluate the following deriva ves and integrals... d ( ) cosh d sech (7t 3) dt 3. ln 0 cosh d S. Using the Chain Rule directly, we have d d( cosh ) = sinh. Just to demonstrate that it works, let s also use the Basic Iden ty found in Key Idea : cosh = cosh + sinh. d ( ) d ( cosh = cosh + sinh ) = cosh sinh + sinh cosh d d = 4 cosh sinh. Using another Basic Iden ty, we can see that 4 cosh sinh = sinh. We get the same answer either way.. We employ subs tu on, with u = 7t 3 and du = 7dt. Applying Key Idea we have: sech (7t 3) dt = tanh(7t 3) + C ln cosh d = sinh 0 ln 0 = sinh(ln ) sinh 0 = sinh(ln ). We can simplify this last epression as sinh is based on eponen als: sinh(ln ) = eln e ln = / = 3 4. Inverse Hyperbolic Func ons Just as the inverse trigonometric func ons are useful in certain integra ons, the inverse hyperbolic func ons are useful with others. Figure 7.6 shows the restric ons on the domains to make each func on one-to-one and the resul ng domains and ranges of their inverse func ons. Their graphs are shown in Figure

39 7.4 Hyperbolic Func ons Because the hyperbolic func ons are defined in terms of eponen al func- ons, their inverses can be epressed in terms of logarithms as shown in Key Idea. It is o en more convenient to refer to sinh than to ln ( + + ), especially when one is working on theory and does not need to compute actual values. On the other hand, when computa ons are needed, technology is o en helpful but many hand-held calculators lack a convenient sinh bu on. (Often it can be accessed under a menu system, but not conveniently.) In such a situa on, the logarithmic representa on is useful. The reader is not encouraged to memorize these, but rather know they eist and know how to use them when needed. Func on Domain Range Func on Domain Range cosh [0, ) [, ) cosh [, ) [0, ) sinh (, ) (, ) sinh (, ) (, ) tanh (, ) (, ) tanh (, ) (, ) sech [0, ) (0, ] sech (0, ] [0, ) csch (, 0) (0, ) (, 0) (0, ) csch (, 0) (0, ) (, 0) (0, ) coth (, 0) (0, ) (, ) (, ) coth (, ) (, ) (, 0) (0, ) Figure 7.6: Domains and ranges of the hyperbolic and inverse hyperbolic func ons. 0 y y = cosh y = sinh 0 5 y 5 y = cosh y = sinh y y y = coth 3 y = tanh 3 3 y = csch y = sech 3 Figure 7.7: Graphs of the hyperbolic func ons and their inverses. Now let s consider the inverses of the hyperbolic func ons. We begin with the func on f() = sinh. Since f () = cosh > 0 for all real, f is increasing 365

40 Chapter 7 Inverse Func ons and L Hôpital s Rule and must be one-to-one. We proceed as in Sec on 7.: y = e e y = e e (now mul ply by e ) ye = e (a quadra c form ) (e ) ye = 0 (use the quadra c formula) e = y ± 4y 4 e = y ± y + (use the fact that e > 0) e = y + y + = ln(y + y + ) Finally, interchange the variable to find that sinh = ln( + + ). In a similar manner we find that the inverses of the other hyperbolic func ons are given by: Key Idea Logarithmic defini ons of Inverse Hyperbolic Func ons. cosh = ln ( + ) ; 4. sinh = ln ( + + ). tanh = ( ) + ln ; < ( 3. sech + ) = ln ; 0 < 5. coth = ( ) + ln ; > ( ) 6. csch + = ln + ; 0 The following Key Ideas give the deriva ves and integrals rela ng to the inverse hyperbolic func ons. In Key Idea 4, both the inverse hyperbolic and logarithmic func on representa ons of the an deriva ve are given, based on Key Idea. Again, these la er func ons are o en more useful than the former. 366

41 7.4 Hyperbolic Func ons Key Idea 3 Deriva ves Involving Inverse Hyperbolic Func ons. d ( cosh ) = d ; > 4. d ( sech ) = d ; 0 < <. d ( sinh ) = d + 5. d ( csch ) = d + ; 0 3. d ( tanh ) = d ; < 6. d ( coth ) = d ; > Key Idea 4 Integrals Involving Inverse Hyperbolic Func ons ( a d = ) cosh + C; 0 < a < = ln + a a + C ( + a d = ) sinh + C; a > 0 = ln + a + a + C { ( a d = a tanh a) + C < a ( ) = a coth a + C a < a ln a + a + C a d = + a d = ( ) a sech + C; 0 < < a = ( a a ln a + a a csch + C; 0, a > 0 = a a ln a + a + + C ) + C We prac ce using the deriva ve and integral formulas in the following eample. Eample 3 Deriva ves and integrals involving inverse hyperbolic func- ons Evaluate the following. 367

42 Chapter 7 Inverse Func ons and L Hôpital s Rule.. d d [ ( )] 3 cosh 5 d d S. Applying Key Idea 3 with the Chain Rule gives: [ ( )] d 3 cosh = d 5 ( 3 ) Mul plying the numerator and denominator by ( ) gives: of item #3 from Key Idea 4, with a =. Thus d = d. The second integral can be solved with a direct applica on d = d { tanh () + C < = coth () + C < = ln + + C = ln + + C. (7.) 3. This requires a subs tu on, then item # of Key Idea 4 can be applied. Let u = 3, hence du = 3d. We have d = 3 u + 0 du. Note a = 0, hence a = 0. Now apply the integral rule. = 3 sinh ( 3 0 ) + C = 3 3 ln C. 368

43 7.4 Hyperbolic Func ons This sec on covers a lot of ground. New func ons were introduced, along with some of their fundamental iden es, their deriva ves and an deriva ves, their inverses, and the deriva ves and an deriva ves of these inverses. Four Key Ideas were presented, each including quite a bit of informa on. Do not view this sec on as containing a source of informa on to be memorized, but rather as a reference for future problem solving. Key Idea 4 contains perhaps the most useful informa on. Know the integra on forms it helps evaluate and understand how to use the inverse hyperbolic answer and the logarithmic answer. The net sec on takes a brief break from demonstra ng new integra on techniques. It instead demonstrates a technique of evalua ng limits that return indeterminate forms. This technique will be useful in Sec on 8.6, where limits will arise in the evalua on of certain definite integrals. 369

44 Eercises 7.4 Terms and Concepts. In Key Idea, the equa on tanh d = ln(cosh ) + C is given. Why is ln cosh not used i.e., why are absolute values not necessary?. The hyperbolic func ons are used to define points on the right hand por on of the hyperbola y =, as shown in Figure 7.4. How can we use the hyperbolic func ons to define points on the le hand por on of the hyperbola? Problems In Eercises 3 0, verify the given iden ty using Defini on 7, as done in Eample. 3. coth csch = 4. cosh = cosh + sinh 5. cosh cosh + = 6. sinh cosh = d 7. [sech ] = sech tanh d d 8. d [coth ] = csch 9. tanh d = ln(cosh ) + C 0. coth d = ln sinh + C In Eercises, find the deriva ve of the given func on.. f() = cosh. f() = tanh( ) 3. f() = ln(sinh ) 4. f() = sinh cosh 5. f() = sinh cosh 6. f() = sech ( ) 7. f() = sinh (3) 8. f() = cosh ( ) 9. f() = tanh ( + 5) 0. f() = tanh (cos ). f() = cosh (sec ) In Eercises 6, find the equa on of the line tangent to the func on at the given -value.. f() = sinh at = 0 3. f() = cosh at = ln 4. f() = sech at = ln 3 5. f() = sinh at = 0 6. f() = cosh at = In Eercises 7 34, evaluate the given indefinite integral. 7. tanh() d cosh(3 7) d sinh cosh d 9 d 4 4 d + 3 d e e + d sech d (Hint: mul ply by cosh ; set u = sinh.) cosh In Eercises 35 36, evaluate the given definite integral ln ln sinh d cosh d 370

45 7.5 L Hôpital s Rule 7.5 L Hôpital s Rule This sec on is concerned with a technique for evalua ng certain limits that will be useful in later chapters. Our treatment of limits eposed us to 0/0, an indeterminate form. If lim f() = c 0 and lim g() = 0, we do not conclude that lim f()/g() is 0/0; rather, we use c c 0/0 as nota on to describe the fact that both the numerator and denominator approach 0. The epression 0/0 has no numeric value; other work must be done to evaluate the limit. Other indeterminate forms eist; they are: /, 0,, 0 0, and 0. Just as 0/0 does not mean divide 0 by 0, the epression / does not mean divide infinity by infinity. Instead, it means a quan ty is growing without bound and is being divided by another quan ty that is growing without bound. We cannot determine from such a statement what value, if any, results in the limit. Likewise, 0 does not mean mul ply zero by infinity. Instead, it means one quan ty is shrinking to zero, and is being mul plied by a quan ty that is growing without bound. We cannot determine from such a descrip on what the result of such a limit will be. This sec on introduces L Hôpital s Rule, a method of resolving limits that produce the indeterminate forms 0/0 and /. We ll also show how algebraic manipula on can be used to convert other indeterminate epressions into one of these two forms so that our new rule can be applied. Theorem 5 L Hôpital s Rule, Part Let f and g be differen able func ons on an open interval I containing a.. If lim f() = 0, lim g() = 0, and g () 0 ecept possibly at a a = a, then f() lim a g() = lim f () a g ().. If lim a f() = ± and lim a g() = ±, then f() lim a g() = lim f () a g (). A similar statement holds if we just look at the one sided limits lim a and lim a +. 37

46 Chapter 7 Inverse Func ons and L Hôpital s Rule Theorem 53 L Hôpital s Rule, Part Let f and g be differen able func ons on the open interval (c, ) for some value c and g () 0 on (c, ).. If lim f() = 0 and lim g() = 0, then f() lim g() = lim f () g ().. If lim f() = ± and lim g() = ±, then f() lim g() = lim f () g (). Similar statements can be made where approaches. We demonstrate the use of L Hôpital s Rule in the following eamples; we will o en use LHR as an abbrevia on of L Hôpital s Rule. Eample Using L Hôpital s Rule Evaluate the following limits, using L Hôpital s Rule as needed. sin. lim 0. lim lim 0 cos S lim lim e 6. lim 3. This has the indeterminate form 0/0. We proved this limit is in Eample.3.4 using the Squeeze Theorem. Here we use L Hôpital s Rule to show its power. sin by LHR cos lim = lim =. 0 0 While this seems easier than using the Squeeze Theorem to find this limit, we note that applying L Hôpital s Rule here requires us to know the deriva- ve of sin. We originally encountered this limit when we were trying to find that deriva ve. 37

47 7.5 L Hôpital s Rule. This has the indeterminate form 0/ lim by LHR = lim ( + 3) / = This has the indeterminate form 0/0. lim 0 cos by LHR = lim 0 sin. This la er limit also evaluates to the 0/0 indeterminate form. To evaluate it, we apply L Hôpital s Rule again. lim 0 Thus lim 0 cos =. sin by LHR = cos = lim = 0 8 = 0 We cannot use L Hôpital s Rule in this case because the original limit does not return an indeterminate form, so L Hôpital s Rule does not apply. In fact, the inappropriate use of L Hôpital s Rule here would result in the incorrect limit We can evaluate this limit already using Key Idea ; the answer is 3/4. We apply L Hôpital s Rule to demonstrate its applicability. lim by LHR = lim by LHR 6 = lim 8 = 3 4. e 6. lim 3 by LHR = lim e by LHR e = lim 6 by LHR e = lim 6 =. 3 Recall that this means that the limit does not eist; as approaches, the epression e / 3 grows without bound. We can infer from this that e grows faster than 3 ; as gets large, e is far larger than 3. (This has important implica ons in compu ng when considering efficiency of algorithms.) 373

48 Chapter 7 Inverse Func ons and L Hôpital s Rule Indeterminate Forms 0 and L Hôpital s Rule can only be applied to ra os of func ons. When faced with an indeterminate form such as 0 or, we can some mes apply algebra to rewrite the limit so that L Hôpital s Rule can be applied. We demonstrate the general idea in the net eample. Watch the video: L Hop ital s Rule Indeterminate Powers at Eample Applying L Hôpital s Rule to other indeterminate forms Evaluate the following limits.. lim 0 + e /. lim 0 e / 3. lim (ln( + ) ln ) 4. lim ( e ) S. As 0 +, note that 0 and e /. Thus we have the indeterminate form 0. We rewrite the epression e / as e/ ; now, as / 0 +, we get the indeterminate form / to which L Hôpital s Rule can be applied. lim e / e / = lim / by LHR ( / )e / = lim 0 + / = lim e / =. 0 + Interpreta on: e / grows faster than shrinks to zero, meaning their product grows without bound.. As 0, note that 0 and e / e 0. The the limit evaluates to 0 0 which is not an indeterminate form. We conclude then that lim 0 e / =

49 7.5 L Hôpital s Rule 3. This limit ini ally evaluates to the indeterminate form. By applying a logarithmic rule, we can rewrite the limit as lim (ln( + ) ln ) = lim ln ( + As, the argument of the natural logarithm approaches /, to which we can apply L Hôpital s Rule. ). + lim by LHR = lim =. Since implies +, it follows that ( ) + implies ln ln = 0. Thus ( ) + lim (ln( + ) ln ) = lim ln = 0. Interpreta on: since this limit evaluates to 0, it means that for large, there is essen ally no difference between ln( + ) and ln ; their difference is essen ally 0. ( 4. The limit lim e ) ini ally returns the indeterminate form. ( ) We can rewrite the epression by factoring out ; e = e. We need to evaluate how e / behaves as : e lim by LHR e = lim by LHR e = lim =. Thus lim ( e / ) evaluates to ( ), which is not an indeterminate ( form; rather, ( ) evaluates to. We conclude that e ) =. lim Interpreta on: as gets large, the difference between and e grows very large. 375

50 Chapter 7 Inverse Func ons and L Hôpital s Rule Indeterminate Forms 0 0, and 0 When faced with a limit that returns one of the indeterminate forms 0 0,, or 0, it is o en useful to use the natural logarithm to convert to an indeterminate form we already know how to find the limit of, then use the natural eponen al func on find the original limit. This is possible because the natural logarithm and natural eponen al func ons are inverses and because they are both con- nuous. The following Key Idea epresses the concept, which is followed by an eample that demonstrates its use. Key Idea 5 Evalua ng Limits Involving Indeterminate Forms 0 0, and 0 If lim ln ( f() ) = L, then lim f() = lim e ln(f()) = e L. c c c Eample 3 Using L Hôpital s Rule with indeterminate forms involving eponents Evaluate the following limits. (. lim + ). lim. 0 + S. This is equivalent to a special limit given in Theorem 6; these limits have important applica ons in mathema cs and finance. Note that the eponent approaches while the base approaches, leading to the indeterminate form. Let f() = ( + /) ; the problem asks to evaluate lim f(). Let s first evaluate lim ln ( f() ). lim ln ( f() ) ( = lim ln + ) ( = lim ln + ) ( ) ln + = lim / 376

51 7.5 L Hôpital s Rule This produces the indeterminate form 0/0, so we apply L Hôpital s Rule. = lim +/ ( / ) ( / ) = lim + / =. Thus lim ln ( f() ) =. We return to the original limit and apply Key Idea 5. ( lim + ) = lim f() = lim eln(f()) = e = e. This is another way to determine the value of the number e.. This limit leads to the indeterminate form 0 0. Let f() = and consider first lim 0 + ln ( f() ). y lim ln ( f() ) = lim ln ( ) = lim 0 + ln 4 3 This produces the indeterminate form 0( ), so we rewrite it in order to apply L Hôpital s Rule. ln = lim 0 + /. f() = This produces the indeterminate form / so we apply L Hôpital s Rule. Figure 7.8: A graph of f() = supporting the fact that as 0 +, f(). = lim 0 + / / = lim 0 + = 0. Thus lim ln ( f() ) = 0. We return to the original limit and apply Key 0 + Idea 5. lim = lim f() = lim e ln(f()) = e 0 = This result is supported by the graph of f() = given in Figure

52 Eercises 7.5 Terms and Concepts. List the different indeterminate forms described in this sec- on.. T/F: l Hôpital s Rule provides a faster method of compu ng deriva ves. 3. T/F: l Hôpital s Rule states that d [ ] f() = f () d g() g (). 4. Eplain what the indeterminate form means. 5. Fill in the blanks: The Quo ent Rule is applied to f() g() when taking ; l Hôpital s Rule is applied when taking certain. 6. Create (but do not evaluate!) a limit that returns Create a func on f() such that lim f() returns 0 0. Problems In Eercises 8 5, evaluate the given limit. 8. lim lim lim π. lim π/4 sin π sin cos cos() sin(5). lim 0 sin() 3. lim 0 + sin() 4. lim 0 sin(3) 5. lim 0 sin(a) sin(b) 6. lim 0 + e 7. lim 0 + e 8. lim 0 + sin lim e 0. lim e. e lim e. lim e 3. lim 3 4. lim lim ln lim 7. ln( ) lim 8. lim ( ln 9. lim ln 0 + ) 30. lim 0 + ln 3. lim 0 + e/ ( 3. lim 3 ) ( ) 33. lim ln 34. lim e 35. lim 0 + e / 36. lim + )/ 0 +( 37. lim 0 +() 38. lim 0 +(/) 39. lim ) 0 +(sin 40. lim ) +( 4. lim () / 4. lim (/) 43. lim ) +(ln 44. lim ( + ) / 45. lim ( + ) / 46. lim π/ tan cos 47. lim tan sin() π/ ( 48. lim lim 3 + ln ) ( 5 9 ) lim tan(/) (ln ) 3 5. lim 5. lim + ln 378

53 8: T I Chapter 5 introduced the an deriva ve and connected it to signed areas under a curve through the Fundamental Theorem of Calculus. The chapter a er eplored more applica ons of definite integrals than just area. As evalua ng definite integrals will become even important, we will want to find an deriva ves of a variety of func ons. This chapter is devoted to eploring techniques of an differen a on. While not every func on has an an deriva ve in terms of elementary func ons, we can s ll find an deriva ves of a wide variety of func ons. 8. Integra on by Parts Here s a simple integral that we can t yet evaluate: cos d. It s a simple ma er to take the deriva ve of the integrand using the Product Rule, but there is no Product Rule for integrals. However, this sec on introduces Integra on by Parts, a method of integra on that is based on the Product Rule for deriva ves. It will enable us to evaluate this integral. The Product Rule says that if u and v are func ons of, then (uv) = u v+uv. For simplicity, we ve wri en u for u() and v for v(). Suppose we integrate both sides with respect to. This gives (uv) d = (u v + uv ) d. By the Fundamental Theorem of Calculus, the le side integrates to uv. The right side can be broken up into two integrals, and we have uv = u v d + uv d. Solving for the second integral we have uv d = uv u v d. Using differen al nota on, we can write u = du d v = dv d du = u d dv = v d.

54 Chapter 8 Techniques of Integra on Thus, the equa on above can be wri en as follows: u dv = uv v du. This is the Integra on by Parts formula. For reference purposes, we state this in a theorem. Theorem 54 Integra on by Parts Let u and v be differen able func ons of on an interval I containing a and b. Then u dv = uv v du, and applying FTC part we have =b =a u dv = uv b a =b =a v du. Watch the video: Integra on by Parts Definite Integral at Let s try an eample to understand our new technique. Eample Evaluate cos d. Integra ng using Integra on by Parts S The key to Integra on by Parts is to iden fy part of the integrand as u and part as dv. Regular prac ce will help one make good iden- fica ons, and later we will introduce some principles that help. For now, let u = and dv = cos d. It is generally useful to make a small table of these values. u = dv = cos d du =? v =? u = du = d dv = cos d v = sin 380

55 8. Integra on by Parts Right now we only know u and dv as shown on the le ; on the right we fill in the rest of what we need. If u =, then du = d. Since dv = cos d, v is an an deriva ve of cos, so v = sin. Now subs tute all of this into the Integra on by Parts formula, giving cos d = sin sin d. We can then integrate sin to get cos + C and overall our answer is cos d = sin + cos + C. We have two important notes here: () no ce how the an deriva ve contains the product, sin. This product is what makes integra on by parts necessary. And () an differen a ng dv does result in v + C. The intermediate +Cs are all added together and represented by one +C in the final answer. The eample above demonstrates how Integra on by Parts works in general. We try to iden fy u and dv in the integral we are given, and the key is that we usually want to choose u and dv so that du is simpler than u and v is hopefully not too much more complicated than dv. This will mean that the integral on the right side of the Integra on by Parts formula, v du will be simpler to integrate than the original integral u dv. In the eample above, we chose u = and dv = cos d. Then du = d was simpler than u and v = sin is no more complicated than dv. Therefore, instead of integra ng cos d, we could integrate sin d, which we knew how to do. If we had chosen u = cos and dv = d, so that du = sin d and v =, then cos d = ( cos ) sin d. We then need to integrate sin, which is more complicated than our original integral, making this an unproduc ve choice. We now consider another eample. Eample Evaluate e d. Integra ng using Integra on by Parts S No ce that becomes simpler when differen ated and e is unchanged by differen a on or integra on. This suggests that we should let u = and dv = e d: u = dv = e d du =? v =? u = du = d dv = e d v = e 38

56 Chapter 8 Techniques of Integra on The Integra on by Parts formula gives e d = e e d. The integral on the right is simple; our final answer is e d = e e + C. Note again how the an deriva ves contain a product term. Eample 3 Evaluate cos d. Integra ng using Integra on by Parts S Let u = instead of the trigonometric func on, hence dv = cos d. Then du = d and v = sin as shown below. u = dv = cos d u = dv = cos d du =? v =? du = d v = sin The Integra on by Parts formula gives cos d = sin sin d. At this point, the integral on the right is indeed simpler than the one we started with, but to evaluate it, we need to do Integra on by Parts again. Here we choose u = and dv = sin and fill in the rest below. u = dv = sin d du =? v =? u = du = d dv = sin d v = cos This means that ( cos d = sin cos ) cos d. The integral all the way on the right is now something we can evaluate. It evaluates to sin. Then going through and simplifying, being careful to keep all the signs straight, our answer is cos d = sin + cos sin + C. 38

57 8. Integra on by Parts Eample 4 Evaluate e cos d. Integra ng using Integra on by Parts S This is a classic problem. In this par cular eample, one can let u be either cos or e ; we choose u = e and hence dv = cos d. Then du = e d and v = sin as shown below. u = e dv = cos d du =? v =? u = e du = e d dv = cos d v = sin No ce that du is no simpler than u, going against our general rule (but bear with us). The Integra on by Parts formula yields e cos d = e sin e sin d. The integral on the right is not much different than the one we started with, so it seems like we have go en nowhere. Let s keep working and apply Integra on by Parts to the new integral. So what should we use for u and dv this me? We may feel like le ng the trigonometric func on be dv and the eponen al be u was a bad choice last me since we s ll can t integrate the new integral. However, if we let u = sin and dv = e d this me we will reverse what we just did, taking us back to the beginning. So, we let u = e and dv = sin d. This leads us to the following: u = e dv = sin d du =? v =? u = e du = e d dv = sin d v = cos The Integra on by Parts formula then gives: ( ) e cos d = e sin e cos e cos d = e sin + e cos e cos d. It seems we are back right where we started, as the right hand side contains e cos d. But this is actually a good thing. Add e cos d to both sides. This gives e cos d = e sin + e cos 383

58 Chapter 8 Techniques of Integra on Now divide both sides by : e cos d = ( e sin + e cos ). Simplifying a li le and adding the constant of integra on, our answer is thus e cos d = e (sin + cos ) + C. Eample 5 Evaluate ln d. Integra ng using Integra on by Parts: an deriva ve of ln S One may have no ced that we have rules for integra ng the familiar trigonometric func ons and e, but we have not yet given a rule for integra ng ln. That is because ln can t easily be integrated with any of the rules we have learned up to this point. But we can find its an deriva ve by a clever applica on of Integra on by Parts. Set u = ln and dv = d. This is a good strategy to learn as it can help in other situa ons. This determines du = (/) d and v = as shown below. u = ln dv = d du =? v =? u = ln du = / d v = dv = d Pu ng this all together in the Integra on by Parts formula, things work out very nicely: ln d = ln = ln d d = ln + C. Eample 6 Evaluate tan d. Using Integra on by Parts: an deriva ve of tan 384

59 8. Integra on by Parts S The same strategy we used above works here. Let u = tan and dv = d. Then du = /( + ) d and v =. The Integra on by Parts formula gives tan d = tan + d. The integral on the right can be solved by subs tu on. Taking t = +, we get dt = d. The integral then becomes tan d = tan t dt. The integral on the right evaluates to ln t + C, which becomes ln( + ) + C. Therefore, the answer is tan d = tan ln( + ) + C. Since + > 0, we do not need to include the absolute value in the ln( + ) term. Subs tu on Before Integra on When taking deriva ves, it was common to employ mul ple rules (such as using both the Quo ent and the Chain Rules). It should then come as no surprise that some integrals are best evaluated by combining integra on techniques. In par cular, here we illustrate making an unusual subs tu on first before using Integra on by Parts. Eample 7 Evaluate cos(ln ) d. Integra on by Parts a er subs tu on S The integrand contains a composi on of func ons, leading us to think Subs tu on would be beneficial. Le ng u = ln, we have du = / d. This seems problema c, as we do not have a / in the integrand. But consider: du = d du = d. Since u = ln, we can use inverse func ons to solve for = e u. Therefore we have that d = du = e u du. 385

60 Chapter 8 Techniques of Integra on We can thus replace ln with u and d with e u du. Thus we rewrite our integral as cos(ln ) d = e u cos u du. We evaluated this integral in Eample 4. Using the result there, we have: cos(ln ) d = e u cos u du = eu( sin u + cos u ) + C = eln ( sin(ln ) + cos(ln ) ) + C = ( sin(ln ) + cos(ln ) ) + C. Definite Integrals and Integra on By Parts So far we have focused only on evalua ng indefinite integrals. Of course, we can use Integra on by Parts to evaluate definite integrals as well, as Theorem 54 states. We do so in the net eample. Eample 8 Evaluate ln d. Definite integra on using Integra on by Parts S To simplify the integral we let u = ln and dv = d. We then get du = (/) d and v = 3 /3 as shown below. u = ln dv = d du =? v =? u = ln v = 3 /3 du = / d dv = d This may seem counterintui ve since the power on the algebraic factor has 386

61 8. Integra on by Parts increased (v = 3 /3), but as we see this is a wise choice: ln d = 3 3 ln 3 3 = 3 3 ln 3 d = 3 3 ln 3 9 ( ) 3 3 = ln 3 9 ( 8 = 3 ln 8 ) 9 = 8 3 ln 7 9. d ( 3 ln 9 ) In general, Integra on by Parts is useful for integra ng certain products of func ons, like e d or 3 sin d. It is also useful for integrals involving logarithms and inverse trigonometric func ons. As stated before, integra on is generally more difficult than differen a on. We are developing tools for handling a large array of integrals, and eperience will tell us when one tool is preferable/necessary over another. For instance, consider the three similar looking integrals e d, e d and e 3 d. While the first is calculated easily with Integra on by Parts, the second is best approached with Subs tu on. Taking things one step further, the third integral has no answer in terms of elementary func ons, so none of the methods we learn in calculus will get us the eact answer. We will learn how to approimate this integral in Chapter 9 Integra on by Parts is a very useful method, second only to subs tu on. In the following sec ons of this chapter, we con nue to learn other integra on techniques. The net sec on focuses on handling integrals containing trigonometric func ons. 387

62 Eercises 8. Terms and Concepts. T/F: Integra on by Parts is useful in evalua ng integrands that contain products of func ons.. T/F: Integra on by Parts can be thought of as the opposite of the Chain Rule. Problems In Eercises 3 36, evaluate the given indefinite integral sin d e d sin d 3 sin d e d 3 e d e d e sin d. e cos d. e sin(3) d 3. e 5 cos(5) d 4. sin cos d 5. sin d 6. tan () d 7. tan d 8. sin d 9. ln d 0. ( ) ln d. ln( ) d. ln( ) d 3. ln d 4. (ln ) d 5. (ln( + )) d 6. sec d 7. csc d 8. d 9. d 30. sec tan d 3. sec tan d 3. csc cot d 33. cosh d 34. sinh d 35. sinh d 36. tanh d In Eercises 37 4, evaluate the indefinite integral a er first making a subs tu on. 37. sin(ln ) d 38. sin( ) d 39. ln( ) d 40. e d 4. e ln d 4. 3 e d In Eercises 43 5, evaluate the definite integral. Note: the corresponding indefinite integrals appear in Eercises π 0 π/4 π/4 π/ π/ ln 0 sin d e d sin d 3 sin d e d 388

63 e d e d π 0 π/ π/ e sin d e cos d 389

64 Chapter 8 Techniques of Integra on 8. Trigonometric Integrals Trigonometric func ons are useful for describing periodic behavior. This sec on describes several techniques for finding an deriva ves of certain combina ons of trigonometric func ons. Integrals of the form sin m cos n d In learning the technique of Subs tu on, we saw the integral sin cos d in Eample The integra on was not difficult, and one could easily evaluate the indefinite integral by le ng u = sin or by le ng u = cos. This integral is easy since the power of both sine and cosine is. We generalize this integral and consider integrals of the form sin m cos n d, where m, n are nonnega ve integers. Our strategy for evalua ng these integrals is to use the iden ty cos + sin = to convert high powers of one trigonometric func on into the other, leaving a single sine or cosine term in the integrand. We summarize the general technique in the following Key Idea. Watch the video: Trigonometric Integrals Part of 6 at 390

65 8. Trigonometric Integrals Key Idea 6 Integrals Involving Powers of Sine and Cosine Consider sin m cos n d, where m, n are nonnega ve integers.. If m is odd, then m = k + for some integer k. Rewrite sin m = sin k+ = sin k sin = (sin ) k sin = ( cos ) k sin. Then sin m cos n d = ( cos ) k sin cos n d = ( u ) k u n du, where u = cos and du = sin d.. If n is odd, then using subs tu ons similar to that outlined above we have sin m cos n d = u m ( u ) k du, where u = sin and du = cos d. 3. If both m and n are even, use the half angle iden es cos = + cos() and sin = cos() to reduce the degree of the integrand. Epand the result and apply the principles of this Key Idea again. We prac ce applying Key Idea 6 in the net eamples. Eample Evaluate sin 5 cos 8 d. Integra ng powers of sine and cosine S The power of the sine factor is odd, so we rewrite sin 5 as sin 5 = sin 4 sin = (sin ) sin = ( cos ) sin. Our integral is now ( cos ) cos 8 sin d. Let u = cos, hence du = 39

66 Chapter 8 Techniques of Integra on sin d. Making the subs tu on and epanding the integrand gives ( cos ) cos 8 sin d = ( u ) u 8 du ( = u + u 4) u 8 du (u = 8 u 0 + u ) du = 9 u9 + u 3 u3 + C = 9 cos9 + cos 3 cos3 + C. Eample Evaluate sin 5 cos 9 d. Integra ng powers of sine and cosine S The powers of both the sine and cosine factors are odd, therefore we can apply the techniques of Key Idea 6 to either power. We choose to work with the power of the sine factor since that has a smaller eponent. We rewrite sin 5 as sin 5 = sin 4 sin = ( cos ) sin = ( cos + cos 4 ) sin. This lets us rewrite the integral as sin 5 cos 9 d = ( cos + cos 4 ) sin cos 9 d. Subs tu ng and integra ng with u = cos and du = sin d, we have ( cos + cos 4 ) sin cos 9 d ( = u + u 4 )u 9 du = u 9 u + u 3 du = 0 u0 + 6 u 4 u4 + C = 0 cos0 + 6 cos 4 cos4 + C. 39

67 8. Trigonometric Integrals Instead, another approach would be to rewrite cos 9 as cos 9 = cos 8 cos = (cos ) 4 cos = ( sin ) 4 cos = ( 4 sin + 6 sin 4 4 sin 6 + sin 8 ) cos. We rewrite the integral as sin 5 cos 9 d = sin 5 ( 4 sin + 6 sin 4 4 sin 6 + sin 8 ) cos d. Now subs tute and integrate, using u = sin and du = cos d. sin 5 ( 4 sin + 6 sin 4 4 sin 6 + sin 8 ) cos d = u 5 ( 4u + 6u 4 4u 6 + u 8 ) du = (u 5 4u 7 + 6u 9 4u + u 3) du = 6 u6 u u0 3 u + 4 u4 + C = 6 sin6 sin sin0 3 sin + 4 sin4 + C. Technology Note: The work we are doing here can be a bit tedious, but the skills developed (problem solving, algebraic manipula on, etc.) are important. Nowadays problems of this sort are o en solved using a computer algebra system. The powerful program Mathema ca integrates sin 5 cos 9 d as 45 cos() f() = cos(4) cos(6) cos(8) 4096 cos(0) 890 cos() 4576 cos(4) 4688, which clearly has a different form than our second answer in Eample, which is g() = 6 sin6 sin sin0 3 sin + 4 sin4. Figure 8. shows a graph of f and g; they are clearly not equal, but they differ only by a constant: g() = f()+c for some constant C. So we have two different an deriva ves of the same func on, meaning both answers are correct y g() f() Figure 8.: A plot of f() and g() from Eample and the Technology Note. 393

68 Chapter 8 Techniques of Integra on Eample 3 Evaluate sin d. Integra ng powers of sine and cosine S The power of sine is even so we employ a half-angle iden ty, algebra and a u- subs tu on as follows: cos() sin d = d = cos() d = ( ) sin() + C = sin() + C. 4 Eample 4 Evaluate cos 4 sin d. Integra ng powers of sine and cosine S The powers of sine and cosine are both even, so we employ the half angle formulas and algebra as follows. ( ) ( ) + cos() cos() cos 4 sin d = d + cos() + cos () = cos() d 4 ( = + cos() cos () cos 3 () ) d 8 The cos() term is easy to integrate. The cos () term is another trigonometric integral with an even power, requiring the half angle formula again. The cos 3 () term is a cosine func on with an odd power, requiring a subs tu on as done before. We integrate each in turn below. cos() d = sin() + C. + cos(4) cos () d = d = ( + 4 sin(4)) + C. 394

69 8. Trigonometric Integrals Finally, we rewrite cos 3 () as cos 3 () = cos () cos() = ( sin () ) cos(). Le ng u = sin(), we have du = cos() d, hence ( cos 3 () d = sin () ) cos() d = ( u ) du = ( u 3 u3) + C = ( sin() ) 3 sin3 () + C Pu ng all the pieces together, we have cos 4 sin ( d = + cos() cos () cos 3 () ) d 8 = [ + 8 sin() ( + 4 sin(4)) ( sin() )] 3 sin3 () + C = 8[ 8 sin(4) + ] 6 sin3 () + C. The process above was a bit long and tedious, but being able to work a problem such as this from start to finish is important. Integrals of the form tan m sec n d. When evalua ng integrals of the form sin m cos n d, the Pythagorean Theorem allowed us to convert even powers of sine into even powers of cosine, and vise versa. If, for instance, the power of sine was odd, we pulled out one sin and converted the remaining even power of sin into a func on using powers of cos, leading to an easy subs tu on. The same basic strategy applies to integrals of the form tan m sec n d, albeit a bit more nuanced. The following three facts will prove useful: d d (tan ) = sec, d d (sec ) = sec tan, and + tan = sec (the Pythagorean Theorem). 395

70 Chapter 8 Techniques of Integra on If the integrand can be manipulated to separate a sec term with the remaining secant power even, or if a sec tan term can be separated with the remaining tan power even, the Pythagorean Theorem can be employed, leading to a simple subs tu on. This strategy is outlined in the following Key Idea. Key Idea 7 Integrals Involving Powers of Tangent and Secant Consider tan m sec n d, where m, n are nonnega ve integers.. If n is even, then n = k for some integer k. Rewrite sec n as sec n = sec k = sec k sec = ( + tan ) k sec. Then tan m sec n d = tan m ( + tan ) k sec d = u m ( + u ) k du, where u = tan and du = sec d.. If m is odd, then m = k + for some integer k. Rewrite tan m sec n as tan m sec n = tan k+ sec n = tan k sec n sec tan = (sec ) k sec n sec tan. Then tan m sec n d = (sec ) k sec n sec tan d = (u ) k u n du, where u = sec and du = sec tan d. 3. If n is odd and m is even, then m = k for some integer k. Convert tan m to (sec ) k. Epand the new integrand and use Integra on By Parts, with dv = sec d. 4. If m is even and n = 0, rewrite tan m as tan m = tan m tan = tan m (sec ) = tan m sec tan m. So tan m d = tan m sec d } {{ } apply rule # tan m d. }{{} apply rule #4 again The techniques described in items and of Key Idea 7 are rela vely straight- 396

71 8. Trigonometric Integrals forward, but the techniques in items 3 and 4 can be rather tedious. A few eamples will help with these methods. Eample 5 Evaluate tan sec 6 d. Integra ng powers of tangent and secant S Since the power of secant is even, we use rule # from Key Idea 7 and pull out a sec in the integrand. We convert the remaining powers of secant into powers of tangent. tan sec 6 d = tan sec 4 sec d = tan ( + tan ) sec d Now subs tute, with u = tan, with du = sec d. = u ( + u ) du We leave the integra on and subsequent subs tu on to the reader. The final answer is = 3 tan3 + 5 tan5 + 7 tan7 + C. We derived integrals for tangent and secant in Sec on 5.5 and will regularly use them when evalua ng integrals of the form tan m sec n d. As a reminder: tan d = ln sec + C sec d = ln sec + tan + C Eample 6 Evaluate sec 3 d. Integra ng powers of tangent and secant S We apply rule #3 from Key Idea 7 as the power of secant is odd and the power of tangent is even (0 is an even number). We use Integra on by Parts; the rule suggests le ng dv = sec d, meaning that u = sec. 397

72 Chapter 8 Techniques of Integra on u = sec du =? v =? dv = sec d u = sec du = sec tan d dv = sec d v = tan Figure 8.: Se ng up Integra on by Parts. Employing Integra on by Parts, we have sec 3 d = sec }{{} sec }{{ d } u dv = sec tan sec tan d. This new integral also requires applying rule #3 of Key Idea 7: = sec tan sec ( sec ) d = sec tan sec 3 d + sec d = sec tan sec 3 d + ln sec + tan d Note: Remember that in Eample 5.5.8, we found that sec d = ln sec + tan + C In previous applica ons of Integra on by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding sec 3 d to both sides, giving: sec 3 d = sec tan + ln sec + tan sec 3 d = ( ) sec tan + ln sec + tan + C. We give one more eample. Eample 7 Evaluate tan 6 d. Integra ng powers of tangent and secant S We employ rule #4 of Key Idea 7. tan 6 d = tan 4 tan d = tan 4 ( sec ) d = tan 4 sec d tan 4 d 398

73 8. Trigonometric Integrals We integrate the first integral with subs tu on, u = tan and du = sec d; and the second by employing rule #4 again. = u 4 du tan tan d = 5 tan5 tan ( sec ) d = 5 tan5 tan sec d + tan d Again, use subs tu on for the first integral and rule #4 for the second. = 5 tan5 ( 3 tan3 + sec ) d = 5 tan5 3 tan3 + tan + C. Integrals of the form cot m csc n d Not surprisingly, evalua ng integrals of the form cot m csc n d is similar to evalua ng tan m sec n d. The guidelines from Key Idea 7 and the following three facts will be useful: d d (cot ) = csc d (csc ) = csc cot, d and csc = cot + Eample 8 Evaluate cot csc 4 d Integra ng powers of cotangent and cosecant S Since the power of cosecant is even we will let u = cot and save a csc for the resul ng du = csc d. cot csc 4 d = cot csc csc d = cot ( + cot ) csc d = u ( + u ) du. 399

74 Chapter 8 Techniques of Integra on The integra on and subs tu on required to finish this eample are similar to that of previous eamples in this sec on. The result is 3 cot3 5 cot5 + C. Integrals of the form sin(m) sin(n) d, and sin(m) cos(n) d. cos(m) cos(n) d, Func ons that contain products of sines and cosines of differing periods are important in many applica ons including the analysis of sound waves. Integrals of the form sin(m) sin(n) d, cos(m) cos(n) d and sin(m) cos(n) d are best approached by first applying the Product to Sum Formulas of Trigonometry found in the back cover of this tet, namely sin(m) sin(n) = [ cos ( (m n) ) cos ( (m + n) )] cos(m) cos(n) = [ cos ( (m n) ) + cos ( (m + n) )] sin(m) cos(n) = [ sin ( (m n) ) + sin ( (m + n) )] Eample 9 Evaluate Integra ng products of sin(m) and cos(n) sin(5) cos() d. S The applica on of the formula and subsequent integra on are straigh orward: [ ] sin(5) cos() d = sin(3) + sin(7) d = 6 cos(3) cos(7) + C

75 8. Trigonometric Integrals Integra ng other combina ons of trigonometric func ons Combina ons of trigonometric func ons that we have not discussed in this chapter are evaluated by applying algebra, trigonometric iden es and other integra- on strategies to create an equivalent integrand that we can evaluate. To evaluate crazy combina ons, those not readily manipulated into a familiar form, one should use integral tables. A table of common crazy combina ons can be found at the end of this tet. These la er eamples were admi edly long, with repeated applica ons of the same rule. Try to not be overwhelmed by the length of the problem, but rather admire how robust this solu on method is. A trigonometric func on of a high power can be systema cally reduced to trigonometric func ons of lower powers un l all an deriva ves can be computed. The net sec on introduces an integra on technique known as Trigonometric Subs tu on, a clever combina on of Subs tu on and the Pythagorean Theorem. 40

76 Eercises 8. Terms and Concepts. T/F: sin cos d cannot be evaluated using the techniques described in this sec on since both powers of sin and cos are even.. T/F: sin 3 cos 3 d cannot be evaluated using the techniques described in this sec on since both powers of sin and cos are odd. 3. T/F: This sec on addresses how to evaluate indefinite integrals such as sin 5 tan 3 d. Problems In Eercises 4 30, evaluate the indefinite integral. 4. cos d 5. cos 4 d 6. sin 3 cos d 7. sin 3 cos 3 d 8. sin 6 cos 5 d 9. cos tan 3 d 0. sin cos d. sin 3 cos d. sin() cos() d 3. sin(3) sin(7) d 4. sin(π) sin(π) d 5. cos() cos() d 6. ( π ) cos cos(π) d 7. tan d 8. tan sec 4 d 9. tan 3 sec 4 d 0. tan 3 sec d. tan 3 sec 3 d. tan 5 sec 5 d 3. tan 4 d 4. sec 5 d 5. tan sec d 6. tan sec 3 d 7. csc d 8. cot 3 csc 3 d 9. cot 3 d 30. cot 6 csc 4 d In Eercises 3 39, evaluate the definite integral. Note: the corresponding indefinite integrals appear in the previous set π 0 π π π/ π/ π/ 0 π/ π/ π/4 0 π/4 π/4 π π 6 π π 4 sin cos 4 d sin 3 cos d sin cos 7 d sin(5) cos(3) d cos() cos() d tan 4 sec d tan sec 4 d cot d cot 3 d 40

77 8.3 Trigonometric Subs tu on 8.3 Trigonometric Subs tu on In Sec on 5. we defined the definite integral as the signed area under the curve. In that sec on we had not yet learned the Fundamental Theorem of Calculus, so we evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate d = 9π (8.) as we recognized that f() = 9 described the upper half of a circle with radius 3. We have since learned a number of integra on techniques, including Subs tu on and Integra on by Parts, yet we are s ll unable to evaluate the above integral without resor ng to a geometric interpreta on. This sec on introduces Trigonometric Subs tu on, a method of integra on that fills this gap in our integra on skill. This technique works on the same principle as Subs tu on as found in Sec on 5.5, though it can feel backward. In Sec on 5.5, we set u = f(), for some func on f, and replaced f() with u. In this sec on, we will set = f(θ), where f is a trigonometric func on, then replace with f(θ). Watch the video: Trigonometric Subs tu on Eample 3 / Part at We start by demonstra ng this method in evalua ng the integral in Equa on (8.). A er the eample, we will generalize the method and give more eamples. Eample Evaluate 3 3 Using Trigonometric Subs tu on 9 d. S We begin by no ng that 9 sin θ + 9 cos θ = 9, and hence 9 cos θ = 9 9 sin θ. If we let = 3 sin θ, then 9 = 9 9 sin θ = 9 cos θ. Se ng = 3 sin θ gives d = 3 cos θ dθ. We are almost ready to subs tute. We also change our bounds of integra on. The bound = 3 corresponds to θ = π/ (for when θ = π/, = 3 sin θ = 3). Likewise, the bound of 403

78 Chapter 8 Techniques of Integra on = 3 is replaced by the bound θ = π/. Thus d = π/ = = π/ π/ π/ π/ π/ 9 9 sin θ(3 cos θ) dθ 3 9 cos θ cos θ dθ 3 3 cos θ cos θ dθ. On [ π/, π/], cos θ is always posi ve, so we can drop the absolute value bars, then employ a half angle formula: π/ = 9 cos θ dθ π/ π/ 9( ) = + cos(θ) dθ π/ = 9 (θ + ) π/ sin(θ) = 9 π. π/ This matches our answer from before. We now describe in detail Trigonometric Subs tu on. This method ecels when dealing with integrands that contain a, a and + a. The following Key Idea outlines the procedure for each case, followed by more eamples. 404

79 8.3 Trigonometric Subs tu on Key Idea 8 Trigonometric Subs tu on (a) For integrands containing a : Let = a sin θ, for π/ θ π/ and a > 0. On this interval, cos θ 0, so a = a cos θ (b) For integrands containing + a : Let = a tan θ, for π/ < θ < π/ and a > 0. On this interval, sec θ > 0, so + a = a sec θ (c) For integrands containing a : Let = a sec θ, restric ng our work to where a > 0, so /a, and 0 θ < π/. On this interval, tan θ 0, so a = a tan θ Eample Evaluate Using Trigonometric Subs tu on d. 5 + S Using Key Idea 8(b), we recognize a = 5 and set = 5 tan θ. This makes d = 5 sec θ dθ. We will use the fact that 5 + = tan θ = 5 sec θ = 5 sec θ. Subs tu ng, we have: d = tan θ 5 sec θ dθ 5 sec θ = dθ 5 sec θ = sec θ dθ = ln sec θ + tan θ + C. While the integra on steps are over, we are not yet done. The original problem was stated in terms of, whereas our answer is given in terms of θ. We must convert back to. The lengths of the sides of the reference triangle in Figure 8.3 are determined by the Pythagorean Theorem. With = 5 tan θ, we have tan θ = 5 and sec θ = θ Figure 8.3: A reference triangle for Eample 405

80 Chapter 8 Techniques of Integra on This gives 5 + d = ln sec θ + tan θ + C + 5 = ln + + C. 5 5 We can leave this answer as is, or we can use a logarithmic iden ty to simplify it. Note: + 5 ln + + C = ln ( ) + C 5 = ln + ln C 5 = ln C, where the ln ( / 5 ) term is absorbed into the constant C. (In Sec on 7.4 we learned another way of approaching this problem.) Eample 3 Evaluate 4 d. Using Trigonometric Subs tu on S We start by rewri ng the integrand so that it looks like a for some value of a: 4 = = 4 ( ) 4 ( ). So we have a = /, and following Key Idea 8(c), we set = sec θ, and hence 406

81 8.3 Trigonometric Subs tu on d = sec θ tan θ dθ. We now rewrite the integral with these subs tu ons: 4 ( ) d = d = 4 sec θ ( 4 ( = 4 (sec θ ) ) sec θ tan θ dθ ) sec θ tan θ dθ ( ) = 4 tan θ sec θ tan θ dθ = tan θ sec θ dθ = ( ) sec θ sec θ dθ = ( sec 3 θ sec θ ) dθ. We integrated sec 3 θ in Eample 8..6, finding its an deriva ves to be sec 3 θ dθ = ( ) sec θ tan θ + ln sec θ + tan θ + C. Thus 4 d = ( sec 3 θ sec θ ) dθ = ( ( ) ) sec θ tan θ + ln sec θ + tan θ ln sec θ + tan θ + C = (sec θ tan θ ln sec θ + tan θ ) + C. 4 We are not yet done. Our original integral is given in terms of, whereas our final answer, as given, is in terms of θ. We need to rewrite our answer in terms of. With a = /, and = sec θ, we use the Pythagorean Theorem to determine the lengths of the sides of the reference triangle in Figure tan θ = = and sec θ =. 4 θ / /4 Figure 8.4: A reference triangle for Eample 3 407

82 Chapter 8 Techniques of Integra on Therefore, 4 d = ( ) sec θ tan θ ln sec θ + tan θ + C 4 = ( 4 4 ln + 4 = ( ln + 4 = ( 4 4 ln + 4 ) + C ) + C ) + C. θ 4 Eample 4 Using Trigonometric Subs tu on 4 Evaluate d. S We use Key Idea 8(a) with a =, = sin θ, d = cos θ and hence 4 = cos θ. This gives 4 d = = = cos θ 4 sin ( cos θ) dθ θ cot θ dθ (csc θ ) dθ = cot θ θ + C. We need to rewrite our answer in terms of. Using the Pythagorean Theorem we determine the lengths of the sides of the reference triangle in Figure 8.5. We have cot θ = 4 / and θ = sin (/). Thus 4 4 ( d = sin ) + C. Figure 8.5: A reference triangle for Eample 4 Trigonometric Subs tu on can be applied in many situa ons, even those not of the form a, a or + a. In the following eample, we apply it to an integral we already know how to handle. Eample 5 Evaluate Using Trigonometric Subs tu on + d. 408

83 8.3 Trigonometric Subs tu on S We know the answer already as tan + C. We apply Trigonometric Subs tu on here to show that we get the same answer without inherently relying on knowledge of the deriva ve of the arctangent func on. Using Key Idea 8(b), let = tan θ, d = sec θ dθ and note that + = tan θ + = sec θ. Thus + d = = dθ sec θ sec θ dθ = θ + C. Since = tan θ, θ = tan, and we conclude that + d = tan + C. The net eample is similar to the previous one in that it does not involve a square root. It shows how several techniques and iden es can be combined to obtain a solu on. Eample 6 Evaluate Using Trigonometric Subs tu on ( ) d. S We start by comple ng the square, then make the subs tu- on u = + 3, followed by the trigonometric subs tu on of u = tan θ: ( ) d = ( ( + 3) + ) d = (u + ) du. Now make the subs tu on u = tan θ, du = sec θ dθ: = (tan θ + ) sec θ dθ = (sec θ) sec θ dθ = cos θ dθ. Note: Remember the sine and cosine double angle iden es: sin θ = sin θ cos θ cos θ = cos θ sin θ = cos θ = sin θ They are o en needed for wri ng your final answer in terms of. Applying a half angle formula, we have ( = + ) cos(θ) dθ = θ + sin(θ) + C. (8.) 4 409

84 Chapter 8 Techniques of Integra on We need to return to the variable. As u = tan θ, θ = tan u. Using the iden ty sin(θ) = sin θ cos θ and using the reference triangle found in Key Idea 8(b), we have 4 sin(θ) = u u + u + = u u +. Finally, we return to with the subs tu on u = + 3. We start with the epression in Equa on (8.): θ + 4 sin(θ) + C = tan u + u u + + C = + 3 tan ( + 3) + ( ) + C. Sta ng our final result in one line, ( ) d = + 3 tan ( + 3) + ( ) + C. Our last eample returns us to definite integrals, as seen in our first eample. Given a definite integral that can be evaluated using Trigonometric Subs tu on, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of to one in terms of θ, then conver ng back to ) and then evaluate using the original bounds. It is much more straigh orward, though, to change the bounds as we subs tute. Eample 7 Evaluate 5 0 Definite integra on and Trigonometric Subs tu on + 5 d. S Using Key Idea 8(b), we set = 5 tan θ, d = 5 sec θ dθ, and note that + 5 = 5 sec θ. As we subs tute, we change the bounds of integra on. The lower bound of the original integral is = 0. As = 5 tan θ, we solve for θ and find θ = tan (/5). Thus the new lower bound is θ = tan (0) = 0. The original upper bound is = 5, thus the new upper bound is θ = tan (5/5) = π/4. Thus we have 5 π/ d = 5 tan θ 0 5 sec θ 5 sec θ dθ π/4 = 5 tan θ sec θ dθ. 0 40

85 8.3 Trigonometric Subs tu on We encountered this indefinite integral in Eample 3 where we found tan θ sec θ dθ = ( sec θ tan θ ln sec θ + tan θ ). So 5 π/4 0 tan θ sec θ dθ = 5 (sec θ tan θ ln sec θ + tan θ ) = 5 ( ) ln( + ). π/4 0 The net sec on introduces Par al Frac on Decomposi on, which is an algebraic technique that turns complicated frac ons into sums of simpler frac- ons, making integra on easier. 4

86 Eercises 8.3 Terms and Concepts. Trigonometric Subs tu on works on the same principles as Integra on by Subs tu on, though it can feel.. If one uses Trigonometric Subs tu on on an integrand containing 5, then one should set =. 3. Consider the Pythagorean Iden ty sin θ + cos θ =. (a) What iden ty is obtained when both sides are divided by cos θ? (b) Use the new iden ty to simplify 9 tan θ Why does Key Idea 8(a) state that a = a cos θ, and not a cos θ? Problems In Eercises 5, apply Trigonometric Subs tu on to evaluate the indefinite integrals. In Eercises 3 0, evaluate the indefinite integrals. Some may be evaluated without Trigonometric Subs tu on. 3. d 4. 3 d d ( + 9) 3/ 5 0 d ( ) d ( ) 3/ d 5 9. d d d d 4 + d 9 d 6 d 8 + d 3 7 d 5 8 d In Eercises 6, evaluate the definite integrals by making the proper trigonometric subs tu on and changing the bounds of integra on d 6 d + 4 d ( + ) d 9 d d 4

87 8.4 Par al Frac on Decomposi on 8.4 Par al Frac on Decomposi on In this sec on we inves gate the an deriva ves of ra onal func ons. Recall that ra onal func ons are func ons of the form f() = p() q(), where p() and q() are polynomials and q() 0. Such func ons arise in many contets, one of which is the solving of certain fundamental differen al equa ons. We begin with an eample that demonstrates the mo va on behind this sec on. Consider the integral d. We do not have a simple formula for this (if the denominator were +, we would recognize the an deriva ve as being the arctangent func on). It can be solved using Trigonometric Subs tu on, but note how the integral is easy to evaluate once we realize: Thus = / / +. / / d = d + d = ln ln + + C. This sec on teaches how to decompose into / / +. We start with a ra onal func on f() = p(), where p and q do not have q() any common factors. We first consider the degree of p and q. If the deg(p) deg(q) then we use polynomial long division to divide q into p to determine a remainder r() where deg(r) < deg(q). We then write f() = s()+ r() r() and apply par al frac on decomposi on to q() q(). If the deg(p) < deg(q) we can apply par al frac on decomposi on to p() without addi onal work. q() Par al frac on decomposi on is based on an algebraic theorem that guarantees that any polynomial, and hence q, can use real numbers to factor into the product of linear and irreducible quadra cs factors. The following Key Idea states how to decompose a ra onal func on into a sum of ra onal func ons whose denominators are all of lower degree than q. An irreducible quadra c is one that cannot factor into linear terms with real coefficients. 43

88 Chapter 8 Techniques of Integra on Key Idea 9 Par al Frac on Decomposi on Let p() be a ra onal func on, where deg(p) < deg(q). q(). Factor q() : Write q() as the product of its linear and irreducible quadra c factors of the form (a + b) m and (a + b + c) n where m and n are the highest powers of each factor that divide q. Linear Terms: For each linear factor of q() the decomposi on of p() q() will contain the following terms: A (a + b) + A (a + b) + A m (a + b) m Irreducible Quadra c Terms: For each irreducible quadra c factor of q() the decomposi on of p() will contain the following terms: q() B + C (a + b + c) + B + C (a + b + c) + B n + C n (a + b + c) n. Finding the Coefficients A i, B i, and C i : Set p() equal to the sum of its linear and irreducible quadra c terms. q() p() q() = A (a + b) + A m (a + b) m + B + C (a + b + c) + B n + C n (a + b + c) n Mul ply this equa on by the factored form of q() and simplify to clear the denominators. Solve for the coefficients A i, B i, and C i by (a) mul plying out the remaining terms and collec ng like powers of, equa ng the resul ng coefficients and solving the resul ng system of linear equa ons, or (b) subs tu ng in values for that eliminate terms so the simplified equa- on can be solved for a coefficient. 44

89 8.4 Par al Frac on Decomposi on Watch the video: Integra on Using method of Par al Frac ons at The following eamples will demonstrate how to put this Key Idea into prac- ce. In Eample, we focus on the se ng up the decomposi on of a ra onal func on. Eample Decomposing into par al frac ons Decompose f() = ( + 5)( ) 3 ( + + )( without solving + + 7) for the resul ng coefficients. S The denominator is already factored, as both + + and ++7 are irreducible quadra cs. We need to decompose f() properly. Since ( + 5) is a linear factor that divides the denominator, there will be a A + 5 term in the decomposi on. As ( ) 3 divides the denominator, we will have the following terms in the decomposi on: B, C ( ) and D ( ) 3. The + + term in the denominator results in a Finally, the ( + + 7) term results in the terms All together, we have G + H and I + J ( + + 7). E + F + + term. ( + 5)( ) 3 ( + + )( + + 7) = A B + C ( ) + D ( ) 3 + E + F G + H I + J ( + + 7) Solving for the coefficients A, B,, J would be a bit tedious but not hard. In the net eample we demonstrate solving for the coefficients using both methods given in Key Idea 9. 45

90 Chapter 8 Techniques of Integra on Eample Decomposing into par al frac ons Perform the par al frac on decomposi on of. S The denominator can be wri en as the product of two linear factors: = ( )( + ). Thus = A + B +. (8.3) Using the method described in Key Idea 9 (a) to solve for A and B, first mul ply through by = ( )( + ): A( )( + ) B( )( + ) = + + = A( + ) + B( ) (8.4) = A + A + B B = (A + B) + (A B) collect like terms. The net step is key. For clarity s sake, rewrite the equality we have as 0 + = (A + B) + (A B). On the le, the coefficient of the term is 0; on the right, it is (A+B). Since both sides are equal for all values of, we must have that 0 = A + B. Likewise, on the le, we have a constant term of ; on the right, the constant term is (A B). Therefore we have = A B. We have two linear equa ons with two unknowns. This one is easy to solve by hand, leading to A + B = 0 A B = A = / B = /. Thus = / / +. Before solving for A and B using the method described in Key Idea 9 (b), we note that Equa ons (8.3) and (8.4) are not equivalent. Only the second equa on holds for all values of, including = and =, by con nuity of polynomials. Thus, we can choose values for that eliminate terms in the polynomial to solve for A and B. = A( + ) + B( ). 46

91 8.4 Par al Frac on Decomposi on If we choose =, = A(0) + B( ) B =. Net choose = : = A() + B(0) A =. Resul ng in the same decomposi on as above. In Eample 3, we solve for the decomposi on coefficients using the system of linear equa ons (method a). The margin note eplains how to solve using subs tu on (method b). Eample 3 Integra ng using par al frac ons Use par al frac on decomposi on to integrate ( )( + ) d. S Idea 9: We decompose the integrand as follows, as described by Key ( )( + ) = A + B + + C ( + ). (8.5) To solve for A, B and C, we mul ply both sides by ( )( + ) and collect like terms: We have = A( + ) + B( )( + ) + C( ) (8.6) = A + 4A + 4A + B + B B + C C = (A + B) + (4A + B + C) + (4A B C) = (A + B) + (4A + B + C) + (4A B C) leading to the equa ons A + B = 0, 4A + B + C = 0 and 4A B C =. These three equa ons of three unknowns lead to a unique solu on: A = /9, B = /9 and C = /3. Note: Equa ons (8.5) and (8.6) are not equivalent for = and =. However, due to the con nuity of polynomials we can let = to simplify the right hand side to A( + ) = 9A. Since the le hand side is s ll, we have = 9A, so that A = /9. Likewise,when = ; this leads to the equa on = 3C. Thus C = /3. Knowing A and C, we can find the value of B by choosing yet another value of, such as = 0, and solving for B. 47

92 Chapter 8 Techniques of Integra on Thus /9 /9 ( )( + ) d = d + + d + /3 ( + ) d. Each can be integrated with a simple subs tu on with u = or u = +. The end result is ( )( + ) d = 9 ln 9 ln + + 3( + ) + C. Eample 4 Integra ng using par al frac ons 3 Use par al frac on decomposi on to integrate ( 5)( + 3) d. S Key Idea 9 presumes that the degree of the numerator is less than the degree of the denominator. Since this is not the case here, we begin by using polynomial division to reduce the degree of the numerator. We omit the steps, but encourage the reader to verify that = + + ( 5)( + 3) ( 5)( + 3). Using Key Idea 9, we can rewrite the new ra onal func on as: ( 5)( + 3) = A 5 + B + 3 for appropriate values of A and B. Clearing denominators, we have = A( + 3) + B( 5). As in the previous eamples we choose values of to eliminate terms in the polynomial. If we choose = 3, Net choose = 5: 9( 3) + 30 = A(0) + B( 8) B = (5) + 30 = A(8) + B(0) A =

93 8.4 Par al Frac on Decomposi on We can now integrate: 3 ( ( 5)( + 3) d = + + 5/ /8 ) d + 3 = ln ln C. Before the net eample we remind the reader of a ra onal integrand evaluated by trigonometric subs tu on: + a d = ( ) a tan + C. a Eample 5 Integra ng using par al frac ons Use par al frac on decomposi on to evaluate ( + )( ) d. S The degree of the numerator is less than the degree of the denominator so we begin by applying Key Idea 9. We have: ( + )( ) = A + + B + C Now clear the denominators = A( ) + (B + C)( + ). Again, we choose values of to eliminate terms in the polynomial. If we choose =, 30 = 6A + ( B + C)(0) A = 5. Although none of the other terms can be zeroed out, we con nue by le ng A = 5 and subs tu ng helpful values of. Choosing = 0, we no ce 54 = 55 + C C =. Finally, choose = (any value other than and 0 can be used, is easy to work with) 9 = 90 + (B )() B =. 49

94 Chapter 8 Techniques of Integra on Thus 7 ( ) ( + )( ) d = + + d The first term of this new integrand is easy to evaluate; it leads to a 5 ln + term. The second term is not hard, but takes several steps and uses subs tu on techniques. The integrand has a quadra c in the denominator and a linear term in the numerator. This leads us to try subs tu on. Let u = + 6 +, so du = ( + 6) d. The numerator is, not + 6, but we can get a + 6 term in the numerator by adding 0 in the form of = = We can now integrate the first term with subs tu on, leading to a ln term. The final term can be integrated using arctangent. First, complete the square in the denominator: = 7 ( + 3) +. An an deriva ve of the la er term can be found using Key Idea 8 and subs tu on: d = 7 ( ) + 3 tan + C. Let s start at the beginning and put all of the steps together ( + )( ) d ( ) 5 = + + d = + d d d = 5 ln + + ln tan ( + 3 ) + C. As with many other problems in calculus, it is important to remember that one is not epected to see the final answer immediately a er seeing the problem. Rather, given the ini al problem, we break it down into smaller problems that are easier to solve. The final answer is a combina on of the answers of the smaller problems. 40

95 8.4 Par al Frac on Decomposi on Par al Frac on Decomposi on is an important tool when dealing with ra onal func ons. Note that at its heart, it is a technique of algebra, not calculus, as we are rewri ng a frac on in a new form. Regardless, it is very useful in the realm of calculus as it lets us evaluate a certain set of complicated integrals. The net sec on will require the reader to determine an appropriate method for evalua ng a variety of integrals. 4

96 Eercises 8.4 Terms and Concepts. Fill in the blank: Par al Frac on Decomposi on is a method of rewri ng func ons.. T/F: It is some mes necessary to use polynomial division before using Par al Frac on Decomposi on. 3. Decompose without solving for the coefficients, as 3 done in Eample. 4. Decompose 7 without solving for the coefficients, as 9 done in Eample. 5. Decompose 3 without solving for the coefficients, as 7 done in Eample. 6. Decompose + 5 without solving for the coefficients, as done in Eample. Problems In Eercises 7 3, evaluate the indefinite integral d d d ( + 5) d 3 0. d ( + 8) d ( + ) ( )( + 3)(3 ) d (7 + 3)(5 )(3 ) d d d d d d d ( + )(3 + 5 ) d ( 3)( ) d d ( + ) ( + )( + 9) d ( 7)( + + 7) d ( + )( + ) d d 3 8. ( + + 4) d ( 9)( + ) d ( + )( ) d d 3. + d In Eercises 33 36, evaluate the definite integral ( + )( + 3) d (3 + )( + 4) d ( 0)( ) d ( + )( + + ) d 4

97 8.5 Integra on Strategies 8.5 Integra on Strategies We ve now seen a fair number of different integra on techniques and so we should probably pause at this point to talk a li le bit about a strategy to use for determining the correct technique to use when faced with an integral. There are a couple of points that need to be made about this strategy. First, it isn t a hard and fast set of rules for determining the method that should be used. It is really nothing more than a general set of guidelines that will help us to iden fy techniques that may work. Some integrals can be done in more than one way and so depending on the path you take through the strategy you may end up with a different technique than someone else who also went through this strategy. Second, while the strategy is presented as a way to iden fy the technique that could be used on an integral keep in mind that, for many integrals, it can also automa cally eclude certain techniques as well. When going through the strategy keep two lists in mind. The first list is integra on techniques that simply won t work and the second list is techniques that look like they might work. A er going through the strategy, if the second list has only one entry then that is the technique to use. If on the other hand, there is more than one possible technique to use we will have to decide on which is liable to be the best for us to use. Unfortunately there is no way to teach which technique is the best as that usually depends upon the person and which technique they find to be the easiest. Third, don t forget that many integrals can be evaluated in mul ple ways and so more than one technique may be used on it. This has already been men oned in each of the previous points, but is important enough to warrant a separate men on. Some mes one technique will be significantly easier than the others and so don t just stop at the first technique that appears to work. Always iden fy all possible techniques and then go back and determine which you feel will be the easiest for you to use. Net, it s en rely possible that you will need to use more than one method to completely evaluate an integral. For instance a subs tu on may lead to using integra on by parts or par al frac ons integral. 43

98 Chapter 8 Techniques of Integra on Key Idea 30 Guidelines for Choosing an Integra on Strategy. Simplify the integrand, if possible. See if a simple subs tu on will work 3. Iden fy the type of integral 4. Relate the integral to an integral we already know how to do 5. Try mul ple techniques 6. Try again Let s epand on the ideas of the previous Key Idea.. Simplify the integrand, if possible. This step is very important in the integra on process. Many integrals can be taken from very difficult to very easy with a li le simplifica on or manipula on. Don t forget basic trigonometric and algebraic iden es as these can o en be used to simplify the integral. We used this idea when we were looking at integrals involving trigonometric func ons. For eample consider the following integral. cos d the integral can t be done as it is, however by recalling the iden ty, cos = ( + cos ) the integral becomes very easy to do. Note that this eample also shows that simplifica ons does not necessarily mean that we ll write the integrand in a simpler form. It ony means that we ll write the integrand into a form that we can deal with and this is o en longer and/or messier than the original integral.. See if a simple subs tu on will work. Look to see if a simple subs tu on can be used instead of the o en more complicated methods from Calculus II. For eample consider both of the following integrals. d d 44

99 8.5 Integra on Strategies The first integral can be done with the method of par al frac ons and the second could be done with a trigonometric subs tu on. However, both could also be evaluated using the subs tu on u = and the work involved in the subs tu on would be significantly less than the work involved in either par al frac ons or trigonometric subs tu on. So, always look for quick, simple subs tu ons before moving on to the more complicated Calculus II techniques. 3. Iden fy the type of integral. Note that any integral may fall into more than one of these types. Because of this fact it s usually best to go all the way through the list and iden fy all possible types since one may be easier than the other and it s en rely possible that the easier type is listed lower in the list. (a) Is the integrand a ra onal epression (i.e. is the integrand a polynomial divided by a polynomial)? If so then par al frac ons (Sec- on 8.4) may work on the integral. (b) Is the integrand a polynomial mes a trigonometric func on, eponen al, or logarithm? If so, then integra on by parts (Sec on 8.) may work. (c) Is the integrand a product of sines and cosines, secants and tangents, or cosecants and cotangents? If so, then the topics from Sec- on 8. may work. Likewise, don t forget that some quo ents involving these func ons can also be done using these techniques. (d) Does the integrand involve b + a, b a, or a b? If so, then a trigonometric subs tu on (Sec on 8.3) might work nicely. (e) Does the integrand have roots other than those listed above in it? If so then the subs tu on u = n g() might work. (f) Does the integrand have a quadra c in it? If so then comple ng the square on the quadra c might put it into a form that we can deal with. 4. Relate the integral to an integral we already know how to do. In other words, can we use a subs tu on or manipula on to write the integrand into a form that does fit into the forms we ve looked at previously in this chapter. A typical eample is the following integral. cos + sin d This integral doesn t obviously fit into any of the forms we looked at in this chapter. However, with the subs tu on u = sin we can reduce the 45

100 Chapter 8 Techniques of Integra on integral to the form + u d which is a trigonometric subs tu on problem. 5. Try mul ple techniques. In this step we need to ask ourselves if it is possible that we ll need to use mul ple techniques. The eample in the previous part is a good eample. Using a subs tu on didn t allow us to actually do the integral. All it did was put the integral into a form that we could use a different technique on. Don t ever get locked into the idea that an integral will only required one step to completely evaluate it. Many will require more than one step. 6. Try again. If everything that you ve tried to this point doesn t work then go back through the process again. This me try a technique that you didn t use the first me around. As noted above, this strategy is not a hard and fast set of rules. It is only intended to guide you through the process of best determining how to do any given integral. Note as well that the only place Calculus II actually arises is the third step. Steps,, and 4 involve nothing more than manipula on of the integrand either through direct manipula on of the integrand or by using a subs tu on. The last two steps are simply ideas to think about in going through this strategy. Many students go through this process and concentrate almost eclusively on Step 3 (a er all this is Calculus II, so it s easy to see why they might do that...) to the eclusion of the other steps. One very large consequence of that eclusion is that o en a simple manipula on or subs tu on is overlooked that could make the integral very easy to do. Before moving on to the net sec on we will work a couple of eamples illustra ng a couple of not so obvious simplifica ons/manipula ons and a not so obvious subs tu on. Eample Strategies of Integra on Evaluate the integral tan sec 4 d S This integral almost falls into the form given in 3c. It is a quo ent of tangent and secant and we know that some mes we can use the same methods for products of tangents and secants on quo ents. The process from Sec on 8. tells us that if we have even powers of secant to save two of them and convert the rest to tangents. That won t work here. We 46

101 8.5 Integra on Strategies can save two secants, but they would be in the denominator and they won t do us any good here. Remember that the point of saving them is so they could be there for the subs tu on u = tan. That requires them to be in the numerator. So, that won t work. We need to find another solu on method. There are in fact two solu on methods to this integral depending on how you want to go about it. Solu on In this solu on method we could just convert everything to sines and cosines and see if that gives us an integral we can deal with. tan sin sec 4 d = cos cos4 d = sin cos 3 d subs tute u = cos = u 3 du = 4 cos4 + C Note that just conver ng to sines and cosines won t always work and if it does it won t always work this nicely. O en there will be a lot more work that would need to be done to complete the integral. Solu on This solu on method goes back to dealing with secants and tangents. Let s no ce that if we had a secant in the numerator we could just use u = sec as a subs tu on and it would be a fairly quick and simple subs tu on to use. We don t have a secant in the numerator. However we could very easily get a secant in the numerator by mul plying the numerator and denominator by secant (i.e. we mul ply the integrand by ). tan tan sec sec 4 d = sec 5 d subs tute u = sec = u 5 du = 4 sec 4 + C = 4 cos4 + C In the previous eample we saw two simplifica ons that allowed us to evaluate the integral. The first was using iden es to rewrite the integral into terms we could deal with and the second involved mul plying the numerator and denominator by something to again put the integral into terms we could deal with. 47

102 Chapter 8 Techniques of Integra on Using iden es to rewrite an integral is an important simplifica on and we should not forget about it. Integrals can o en be greatly simplified or at least put into a form that can be dealt with by using an iden ty. The second simplifica on is not used as o en, but does show up on occasion so again, it s best to remember it. In fact, let s take another look at an eample in which mul plying the integrand by will allow us to evaluate an integral. Eample Strategy for Integra on Evaluate the integral + sin d S This is an integral which if we just concentrate on the third step we won t get anywhere. This integral doesn t appear to be any of the kinds of integrals that we worked on in this chapter. We can evaluate the integral however, if we do the following, + sin d = sin + sin sin d sin = sin d This does not appear to have done anything for us. However, if we now remember the first simplifica on we looked at above we will no ce that we can use an iden ty to rewrite the denominator. Once we do that we can further manipulate the integrand into something we can evaluate. + sin d = sin cos d = cos sin cos cos d = sec tan sec d = tan sec + C So, we ve just seen once again that mul plying by a helpful form of can put the integral into a form we can integrate. No ce as well that this eample also showed that simplifica ons do not necessarily put an integral into a simpler form. They only put the integrand into a form that is easier to integrate. Let s now take a quick look at an eample of a subs tu on that is not so obvious. 48

103 8.5 Integra on Strategies Eample 3 Strategy for Integra on Evaluate the integral cos d S We introduced this integral by saying that the subs tu on was not so obvious. However, this is really an integral that falls into the form given by 3e in Key Idea 30. Many people miss that form and so don t think about it. So, let s try the following subs tu on. u = = u d = u du With this subs tu on the integral becomes, cos d = u cos u du This is now an integra on by parts. Remember that o en we will need to use more than one technique to completely do the integral. This is a fairly simple integra on by parts problem so we ll leave the remainder of the details for you to check. cos d = (cos + sin ) + C. It will be possible to integrate every integral assigned in this class, but it is important to note that there are integrals that just can t be evaluated. We should also note that a er we look at series in Chapter 9 we will be able to write down a series representa on of many of these types of integrals. 49

104 Eercises 8.5 Problems In Eercises 5, compute the indefinite integral.. sin d. cos 3 sin d ( )( 4 + 3) d 4. tan sec 5 d 5. d ( + 5) 3/ 4 6. d ( ) d 3 8. d e d d. e sin 3 d. cos 3 sin 3 d 3. d d d 3/ + / 6. e sec e d 7. sin 3 d 8. sin 3 cos d 9. e e + d d. sec tan d. csc cot d 3. (8 3 ) /3 d 4. sin d 5. 3 d e 3 + e d d 3 d d tan 5 d etan cos d d cot 6 d 3 5 d ( sech 4) d e 4 d 3 d d 4 tan 7 cos 7 d tan 3 sec d ( 3 + ) cos d 9 4 d ( cot 3) d ( d 4 ) sin d + cos (5 + ) d d ( ) d 3/ ln d 3 d + 3 e ( + ) d 430

105 8.6 Improper Integra on 8.6 Improper Integra on We begin this sec on by considering the following definite integrals: 00 d.5608, d.5698, 0 + 0,000 d No ce how the integrand is /( + ) in each integral (which is sketched in Figure 8.6). As the upper bound gets larger, one would epect the area under the curve would also grow. While the definite integrals do increase in value as the upper bound grows, they are not increasing by much. In fact, consider: b 0 + d = tan b = tan b tan 0 = tan b. 0 As b, tan b π/. Therefore it seems that as the upper bound b grows, b the value of the definite integral d approaches π/ This 0 + should strike the reader as being a bit amazing: even though the curve etends to infinity, it has a finite amount of area underneath it. When we defined the definite integral b a f() d, we made two s pula ons:. The interval over which we integrated, [a, b], was a finite interval, and. The func on f() was con nuous on [a, b] (ensuring that the range of f was finite). 0.5 y In this sec on we consider integrals where one or both of the above condi- ons do not hold. Such integrals are called improper integrals. 5 0 Figure 8.6: Graphing f() = +. 43

106 Chapter 8 Techniques of Integra on Improper Integrals with Infinite Bounds Defini on 8 Improper Integrals with Infinite Bounds. Let f be a con nuous func on on [a, ). For t a let a f() d = lim t t a f() d.. Let f be a con nuous func on on (, b]. For t b let b f() d = b lim t t f() d. 3. Let f be a con nuous func on on (, ). For any real number c (which one doesn t ma er), let f() d = c lim a a f() d + b lim b c f() d. An improper integral is said to converge if its corresponding limit eists; otherwise, it diverges. The improper integral in part 3 converges if and only if both of its limits eist. Watch the video: Improper Integral Infinity in Upper and Lower Limits at Eample Evalua ng improper integrals Evaluate the following improper integrals... d d e d + d 43

107 8.6 Improper Integra on S.. 3. t d = lim t d t = lim t = lim + t t =. A graph of the area defined by this integral is given in Figure 8.7. t d = lim t d = lim ln t = lim t ln(t) =. The limit does not eist, hence the improper integral d diverges. Compare the graphs in Figures 8.7 and 8.8; no ce how the values of f() = / are no ceably larger than those of f() = /. This difference is enough to cause the improper integral to diverge. 0 e d = lim t 0 t t e d 0 = lim t e t = lim t e0 e t =. A graph of the area defined by this integral is given in Figure y f() = 5 0 Figure 8.7: A graph of f() = in Eample part. 0.5 y f() = 5 0 Figure 8.8: A graph of f() = in Eample part. f() = e y 0 5 Figure 8.9: A graph of f() = e in Eample part

108 Chapter 8 Techniques of Integra on f() = y Figure 8.0: A graph of f() = + in Eample part We will need to break this into two improper integrals and choose a value of c as in part 3 of Defini on 8. Any value of c is fine; we choose c = 0. 0 t d = lim d + lim + t t + t 0 + d = lim t tan 0 + lim t t tan t 0 ( = lim tan 0 tan t ) ( + lim tan t tan 0 ) t t ( = 0 π ) ( π ) + 0. = π. A graph of the area defined by this integral is given in Figure 8.0. Sec on 7.5 introduced L Hôpital s Rule, a method of evalua ng limits that return indeterminate forms. It is not uncommon for the limits resul ng from improper integrals to need this rule as demonstrated net. Eample Improper integra on and L Hôpital s Rule ln Evaluate the improper integral d y f() = ln 5 0 Figure 8.: A graph of f() = ln in Eample. S This integral will require the use of Integra on by Parts. Let u = ln and dv = / d. Then ln d = lim t ln d t ( = lim ln t t ) + t d ( = lim ln t ) t ( = lim ln t ) t t t ( ln ). ln t The /t goes to 0, and ln = 0, leaving lim with L Hôpital s Rule. We have: t t ln t lim t t by LHR /t = lim t =

109 8.6 Improper Integra on Thus the improper integral evaluates as: ln d =. Improper Integrals with Infinite Range We have just considered definite integrals where the interval of integra on was infinite. We now consider another type of improper integra on, where the range of the integrand is infinite. Defini on 9 Improper Integra on with Infinite Range Let f() be a con nuous func on on [a, b] ecept at c, a c b, where = c is a ver cal asymptote of f. Define b a f() d = lim t c t a f() d + lim t c + b t f() d. Note that c can be one of the endpoints (a or b). In that case, there is only one limit to consider as part of the defini on. Eample 3 Improper integra on of func ons with infinite range Evaluate the following improper integrals:. 0 d. d. S. A graph of f() = / is given in Figure 8.. No ce that f has a ver cal asymptote at = 0. In some sense, we are trying to compute the area of a region that has no top. Could this have a finite value? 0 y 0 d = lim d t 0 + t = lim t 0 + t = lim t 0 + =. ( t ) 5 f() = 0.5 in E- Figure 8.: A graph of f() = ample

110 Chapter 8 Techniques of Integra on 0 5 y f() = It turns out that the region does have a finite area even though it has no upper bound (strange things can occur in mathema cs when considering the infinite).. The func on f() = / has a ver cal asymptote at = 0, as shown in Figure 8.3, so this integral is an improper integral. Let s eschew using limits for a moment and proceed without recognizing the improper nature of the integral. This leads to: d = = () = Figure 8.3: A graph of f() = in Eample 3. Clearly the area in ques on is above the -ais, yet the area is supposedly nega ve. In this eample we noted the discon nuity of the integrand on [, ] (its improper nature) but con nued anyway to apply the Fundamental Theorem of Calculus. Viola ng the hypothesis of the FTC led us to an incorrect area of. If we now evaluate the integral using Defini- on 9 we will see that the area is unbounded. t d = lim d + lim t 0 t 0 + t = lim t + lim t 0 t 0 + t d = lim t 0 t + + lim t t. Neither limit converges hence the original improper integral diverges. The nonsensical answer we obtained by ignoring the improper nature of the integral is just that: nonsensical. Understanding Convergence and Divergence O en mes we are interested in knowing simply whether or not an improper integral converges, and not necessarily the value of a convergent integral. We provide here several tools that help determine the convergence or divergence of improper integrals without integra ng. Our first tool is knowing the behavior of func ons of the form p. 436

111 8.6 Improper Integra on Eample 4 Improper integra on of / p Determine the values of p for which d converges. p S We begin by integra ng and then evalua ng the limit. t d = lim p t p d = lim t p d (assume p ) t t = lim t p + p+ = lim t p ( t p p). The result of Eample 4 provides an important tool in determining the convergence of other integrals. A similar result is proved in the eercises about improper integrals of the form d. These results are summarized in the 0 p following Key Idea. y f() = p When does this limit converge i.e., when is this limit not? This limit con- verges precisely when the power of b is less than 0: when p < 0 < p. Our analysis shows that if p >, then d converges. When p < p the improper integral diverges; we showed in Eample that when p = the integral also diverges. Figure 8.4 graphs y = / with a dashed line, along with graphs of y = / p, p <, and y = / q, q >. Somehow the dashed line forms a dividing line between convergence and divergence. f() = q p < < q Figure 8.4: Plo ng func ons of the form / p in Eample 4. Key Idea 3 Convergence of Improper Integrals. The improper integral. The improper integral 0 d and p 0 p d. d converges when p > and diverges when p. p d converges when p < and diverges when p. p 437

112 Chapter 8 Techniques of Integra on Note: We used the upper and lower bound of in Key Idea 3 for convenience. It can be replaced by any a where a > 0. A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose convergence is known. We o en use integrands of the form / p in comparisons as their convergence on certain intervals is known. This is described in the following theorem. Theorem 55 Direct Comparison Test for Improper Integrals Let f and g be con nuous on [a, ) where 0 f() g() for all in [a, ).. If. If a a g() d converges, then f() d diverges, then a a f() d converges. g() d diverges. Eample 5 Determining convergence of improper integrals Determine the convergence of the following improper integrals. S. e d. 3 d y f() = 0.5 f() = e 3 4 Figure 8.5: Graphs of f() = e and f() = / in Eample 5.. The func on f() = e does not have an an deriva ve epressible in terms of elementary func ons, so we cannot integrate directly. It is comparable to g() = /, and as demonstrated in Figure 8.5, e < / on [, ). We know from Key Idea 3 that d converges, hence e d also converges.. Note that for large values of, =. We know from Key Idea 3 and the subsequent note that d diverges, so we seek to 3 compare the original integrand to /. It is easy to see that when > 0, we have = >. Taking reciprocals reverses the inequality, giving <. 438

113 8.6 Improper Integra on Using Theorem 55, we conclude that since diverges as well. Figure 8.6 illustrates this. 3 d diverges, 3 d Being able to compare unknown integrals to known integrals is very useful in determining convergence. However, some of our eamples were a li le too nice. For instance, it was convenient that <, but what if the were replaced with a + + 5? That is, what can we say about the con d? We have >, so we cannot vergence of 3 use Theorem 55. In cases like this (and many more) it is useful to employ the following theorem y f() = f() = 4 6 Figure 8.6: Graphs of f() = / and f() = / in Eample 5. Theorem 56 Limit Comparison Test for Improper Integrals Let f and g be con nuous func ons on [a, ) where f() > 0 and g() > 0 for all. If f() lim = L, 0 < L <, g() then a f() d and either both converge or both diverge. a g() d Eample 6 Determining convergence of improper integrals Determine the convergence of d. 3 S As gets large, the quadra c func on will begin to behave much like y =. So we compare to with the Limit Comparison Test: / lim = lim / The immediate evalua on of this limit returns /, an indeterminate form. Using L Hôpital s Rule seems appropriate, but in this situa on, it does not lead 439

114 Chapter 8 Techniques of Integra on 0. y f() = f() = Figure 8.7: Graphing f() = and f() = in Eample to useful results. (We encourage the reader to employ L Hôpital s Rule at least once to verify this.) The trouble is the square root func on. We determine the limit by using a technique we learned in Calculus I: lim Since we know that that = lim = lim = d diverges, by the Limit Comparison Test we know ++5 d also diverges. Figure 8.7 graphs f() = / and f() = /, illustra ng that as gets large, the func ons become indis nguishable. Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. There are versions that apply to improper integrals with an infinite range, but as they are a bit wordy and a li le more difficult to employ, they are omi ed from this tet. This chapter has eplored many integra on techniques. We learned Integra- on by Parts, which reverses the Product Rule of differen a on. We also learned specialized techniques for handling trigonometric and ra onal func ons. All techniques effec vely have this goal in common: rewrite the integrand in a new way so that the integra on step is easier to see and implement. As stated before, integra on is, in general, hard. It is easy to write a func on whose an deriva ve is impossible to write in terms of elementary func ons, and even when a func on does have an an deriva ve epressible by elementary func ons, it may be really hard to discover what it is. The powerful computer algebra system Mathema ca has approimately,000 pages of code dedicated to integra on. Do not let this difficulty discourage you. There is great value in learning integra on techniques, as they allow one to manipulate an integral in ways that can illuminate a concept for greater understanding. There is also great value in understanding the need for good numerical techniques: the Trapezoidal and Simpson s Rules are just the beginning of powerful techniques for approima ng the value of integra on. The net chapter stresses the uses of integra on. We generally do not find an deriva ves for an deriva ve s sake, but rather because they provide the solu on to some type of problem. The following chapter introduces us to several different problems whose solu on is provided by integra on. 440

115 Eercises 8.6 Terms and Concepts. The definite integral was defined with what two s pula- ons? b. If lim f() d eists, then the integral b said to If know that 0 f() d is f() d = 0, and 0 g() f() for all, then we 4. For what values of p will 5. For what values of p will 6. For what values of p will Problems g() d. 0 0 d converge? p d converge? p d converge? p In Eercises 7 37, evaluate the given improper integral e 5 d 3 d 4 d + 9 d d ( ) d + d + 4 d ( ) d ( ) d d d d d d d 3 9 d π 0 π sec d sec d d e d e d e d d e + e ln d ln ln d 0 0 d ln d ln d e sin d e cos d In Eercises 38 47, use the Direct Comparison Test or the Limit Comparison Test to determine whether the given definite integral converges or diverges. Clearly state what test is being used and what func on the integrand is being compared to d e d d e ln d e +3+ d d + sin d + cos d + e d e d 44

116 Chapter 8 Techniques of Integra on 0.5 y y = e 8.7 Numerical Integra on The Fundamental Theorem of Calculus gives a concrete technique for finding the eact value of a definite integral. That technique is based on compu ng an- deriva ves. Despite the power of this theorem, there are s ll situa ons where we must approimate the value of the definite integral instead of finding its eact value. The first situa on we eplore is where we cannot compute an an deriva- ve of the integrand. The second case is when we actually do not know the integrand, but only its value when evaluated at certain points. An elementary func on is any func on that is a combina on of polynomials, n th roots, ra onal, eponen al, logarithmic and trigonometric func ons and their inverses. We can compute the deriva ve of any elementary func on, but there are many elementary func ons of which we cannot compute an an- deriva ve. For eample, the following func ons do not have an deriva ves that we can epress with elementary func ons: 0.5 y y = sin( 3 ) 0.5 e, sin( 3 ) and sin. The simplest way to refer to the an deriva ves of e is to simply write e d. This sec on outlines three common methods of approima ng the value of definite integrals. We describe each as a systema c method of approima ng area under a curve. By approima ng this area accurately, we find an accurate approima on of the corresponding definite integral. We will apply the methods we learn in this sec on to the following definite integrals: y e d, as pictured in Figure 8.8. π π 4 sin( 3 ) d, and 4π 0.5 sin() d, 0.5 y = sin Figure 8.8: Graphically represen ng three definite integrals that cannot be evaluated using an deriva ves. The Le and Right Hand Rule Methods In Sec on 5.3 we addressed the problem of evalua ng definite integrals by approima ng the area under the curve using rectangles. We revisit those ideas here before introducing other methods of approima ng definite integrals. We start with a review of nota on. Let f be a con nuous func on on the interval [a, b]. We wish to approimate b a f() d. We par on [a, b] into n 44

117 8.7 Numerical Integra on equally spaced subintervals, each of length = b a. The endpoints of these n subintervals are labeled as 0 = a, = a +, = a +,..., i = a + i,..., n = b. Sec on 5.3 showed that to use the Le Hand Rule we use the summa on n n f( i ) and to use the Right Hand Rule we use f( i ). We review i= the use of these rules in the contet of eamples. Eample Approimate e 0 spaced subintervals. i= Approima ng definite integrals with rectangles d using the Le and Right Hand Rules with 5 equally S We begin by par oning the interval [0, ] into 5 equally spaced intervals. We have = 0 5 = /5 = 0., so 0 = 0, = 0., = 0.4, 3 = 0.6, 4 = 0.8, and 5 =. Using the Le Hand Rule, we have: y n f( i ) = ( f( 0 ) + f( ) + f( ) + f( 3 ) + f( 4 ) ) i= = ( f(0) + f(0.) + f(0.4) + f(0.6) + f(0.8) ) ( )(0.) Using the Right Hand Rule, we have: 0.5 y = e n f( i ) = ( f( ) + f( ) + f( 3 ) + f( 4 ) + f( 5 ) ) i= = ( f(0.) + f(0.4) + f(0.6) + f(0.8) + f() ) y y = e ( )(0.) Figure 8.9 shows the rectangles used in each method to approimate the definite integral. These graphs show that in this par cular case, the Le Hand Rule is an over approima on and the Right Hand Rule is an under approima- on. To get a be er approima on, we could use more rectangles, as we did in Figure 8.9: Approima ng 0 e d in Eample using (top) the le hand rule and (bo om) the right hand rule. 443

118 Chapter 8 Techniques of Integra on i Eact Appro. sin( 3 i ) 0 π/ π/ π/ π/ π/ π/ π/ π/ π/ π/ π/ Figure 8.0: Table of values used to approimate π π sin( 3 ) d in Eample. y = sin( 3 ) 0.5 y y y = sin( 3 ) Sec on 5.3. We could also average the Le and Right Hand Rule results together, giving = The actual answer, accurate to 4 places a er the decimal, is , showing our average is a good approima on. Eample Approimate π π 4 spaced subintervals. S Approima ng definite integrals with rectangles sin( 3 ) d using the Le and Right Hand Rules with 0 equally We begin by finding : b a n = π/ ( π/4) 0 = 3π It is useful to write out the endpoints of the subintervals in a table; in Figure 8.0, we give the eact values of the endpoints, their decimal approima ons, and decimal approima ons of sin( 3 ) evaluated at these points. Once this table is created, it is straigh orward to approimate the definite integral using the Le and Right Hand Rules. (Note: the table itself is easy to create, especially with a standard spreadsheet program on a computer. The last two columns are all that are needed.) The Le Hand Rule sums the first 0 values of sin( 3 i ) and mul plies the sum by ; the Right Hand Rule sums the last 0 values of sin( 3 i ) and mul plies by. Therefore we have: Le Hand Rule: Right Hand Rule: π π 4 π π 4 sin( 3 ) d (.9)(0.36) = sin( 3 ) d (.7)(0.36) = The average of the Le and Right Hand Rules is The actual answer, accurate to 3 places a er the decimal, is Our approima ons were once again fairly good. The rectangles used in each approima on are shown in Figure 8.. It is clear from the graphs that using more rectangles (and hence, narrower rectangles) should result in a more accurate approima on. Figure 8.: Approima ng π π 4 sin( 3 ) d in Eample using (top) the le hand rule and (bo om) the right hand rule. 444

119 8.7 Numerical Integra on The Trapezoidal Rule In Eample we approimated the value of 0 e d with 5 rectangles of equal width. Figure 8.9 showed the rectangles used in the Le and Right Hand Rules. These graphs clearly show that rectangles do not match the shape of the graph all that well, and that accurate approima ons will only come by using lots of rectangles. Instead of using rectangles to approimate the area, we can instead use trapezoids. In Figure 8., we show the region under f() = e on [0, ] approimated with 5 trapezoids of equal width; the top corners of each trapezoid lies on the graph of f(). It is clear from this figure that these trapezoids more accurately approimate the area under f and hence should give a be er approima on of 0 e d. (In fact, these trapezoids seem to give a great approima on of the area.) 0.5 y y = e Watch the video: The Trapezoid Rule for Approima ng Integrals at Figure 8.: Approima ng 0 e d using 5 trapezoids of equal widths. The formula for the area of a trapezoid is given in Figure 8.3. We approimate 0 e d with these trapezoids in the following eample. Eample 3 Approima ng definite integrals using trapezoids Use 5 trapezoids of equal width to approimate 0 e d. S To compute the areas of the 5 trapezoids in Figure 8., it will again be useful to create a table of values as shown in Figure 8.4. The le most trapezoid has legs of length and 0.96 and a height of 0.. Thus, by our formula, the area of the le most trapezoid is: (0.) = Moving right, the net trapezoid has legs of length 0.96 and 0.85 and a height of 0.. Thus its area is: (0.) = a h b Area = a+b h Figure 8.3: The area of a trapezoid. i e i Figure 8.4: A table of values of e. 445

120 Chapter 8 Techniques of Integra on The sum of the areas of all 5 trapezoids is: (0.) + We approimate (0.) + (0.) (0.) + e d (0.) = There are many things to observe in this eample. Note how each term in the final summa on was mul plied by both / and by = 0.. We can factor these coefficients out, leaving a more concise summa on as: [ ] (0.) (+0.96)+( )+( )+( )+( ). Now no ce that all numbers ecept for the first and the last are added twice. Therefore we can write the summa on even more concisely as 0. [ ] + ( ) This is the heart of the Trapezoidal Rule, wherein a definite integral b a f() d is approimated by using trapezoids of equal widths to approimate the corresponding area under f. Using n equally spaced subintervals with endpoints 0,,, n, we again have = b a n. Thus: Eample 4 b a f() d n i= = = f( i ) + f( i ) n ( f(i ) + f( i ) ) i= [ ] n f( 0 ) + f( i ) + f( n ). i= Using the Trapezoidal Rule Revisit Eample and approimate and 0 equally spaced subintervals. π π 4 sin( 3 ) d using the Trapezoidal Rule 446

121 8.7 Numerical Integra on S We refer back to Figure 8.0 for the table of values of sin( 3 ). Recall that = 3π/ Thus we have: π π 4 sin( 3 ) d 0.36 [ ( ) ] ( 0.03) ( 0.67) = No ce how quickly the Trapezoidal Rule can be implemented once the table of values is created. This is true for all the methods eplored in this sec on; the real work is crea ng a table of i and f( i ) values. Once this is completed, approima ng the definite integral is not difficult. Again, using technology is wise. Spreadsheets can make quick work of these computa ons and make using lots of subintervals easy. Also no ce the approima ons the Trapezoidal Rule gives. It is the average of the approima ons given by the Le and Right Hand Rules! This effec vely renders the Le and Right Hand Rules obsolete. They are useful when first learning about definite integrals, but if a real approima on is needed, one is generally be er off using the Trapezoidal Rule instead of either the Le or Right Hand Rule. We will also show that the Trapezoidal Rule makes using the Midpoint Rule obsolete as well. With much more work, it will turn out that the Midpoint Rule has only a marginal gain in accuracy. But we will include it in our results for the sake of completeness. How can we improve on the Trapezoidal Rule, apart from using more and more trapezoids? The answer is clear once we look back and consider what we have really done so far. The Le Hand Rule is not really about using rectangles to approimate area. Instead, it approimates a func on f with constant func ons on small subintervals and then computes the definite integral of these constant func ons. The Trapezoidal Rule is really approima ng a func on f with a linear func on on a small subinterval, then computes the definite integral of this linear func on. In both of these cases the definite integrals are easy to compute in geometric terms. So we have a progression: we start by approima ng f with a constant func- on and then with a linear func on. What is net? A quadra c func on. By approima ng the curve of a func on with lots of parabolas, we generally get an even be er approima on of the definite integral. We call this process Simpson s Rule, named a er Thomas Simpson (70-76), even though others had used this rule as much as 00 years prior. 447

122 Chapter 8 Techniques of Integra on Simpson s Rule Given one point, we can create a constant func on that goes through that point. Given two points, we can create a linear func on that goes through those points. Given three points, we can create a quadra c func on that goes through those three points (given that no two have the same value). Consider three points (, y ), (, y ) and ( 3, y 3 ) whose values are equally spaced and < < 3. Let f be the quadra c func on that goes through these three points. An eercise will ask you to show that 3 y 3 Figure 8.5: A graph of a func on f and a parabola that approimates it well on [, 3]. 3 f() d = 3 6 ( y + 4y + y 3 ). (8.7) Consider Figure 8.5. A func on f goes through the 3 points shown and the parabola g that also goes through those points is graphed with a dashed line. Using our equa on from above, we know eactly that 3 g() d = 3 ( ) 3 + 4() + = 3. 6 Since g is a good approima on for f on [, 3], we can state that 3 f() d 3. No ce how the interval [, 3] was split into two subintervals as we needed 3 points. Because of this, whenever we use Simpson s Rule, we need to break the interval into an even number of subintervals. In general, to approimate b a f() d using Simpson s Rule, subdivide [a, b] into n subintervals, where n is even and each subinterval has width = (b a)/n. We approimate f with n/ parabolic curves, using Equa on (8.7) to compute the area under these parabolas. Adding up these areas gives the formula: b a f() d 3 [f( 0) + 4f( ) + f( ) + 4f( 3 ) + + f( n ) + 4f( n ) + f( n )]. Note how the coefficients of the terms in the summa on have the pa ern, 4,, 4,, 4,,, 4,. Let s demonstrate Simpson s Rule with a concrete eample. Eample 5 Approimate 0 Using Simpson s Rule e d using Simpson s Rule and 4 equally spaced subintervals. 448

123 8.7 Numerical Integra on S We begin by making a table of values as we have in the past, as shown in Figure 8.6(a). Simpson s Rule states that 0 e d 0.5 [ ] + 4(0.939) + (0.779) + 4(0.570) = Recall in Eample we stated that the correct answer, accurate to 4 places a er the decimal, was Our approima on with Simpson s Rule, with 4 subintervals, is be er than our approima on with the Trapezoidal Rule using 5. Figure 8.6(b) shows f() = e along with its approima ng parabolas, demonstra ng how good our approima on is. The approima ng curves are nearly indis nguishable from the actual func on. y i e i (a) y = e 0.5 Eample 6 Approimate sin( 3 ) d using Simpson s Rule and 0 equally spaced intervals. π π 4 Using Simpson s Rule S Figure 8.7(a) shows the table of values that we used in the past for this problem, shown here again for convenience. Again, = (π/ + π/4)/ Simpson s Rule states that π π 4 sin( 3 ) d 0.36 [ ( 0.466) + 4( 0.65) + ( 0.03) + 3 ] + (0.97) + 4(0.69) + ( 0.67) = Recall that the actual value, accurate to 3 decimal places, is Our approima on is within one /00 th of the correct value. The graph in Figure 8.7(b) shows how closely the parabolas match the shape of the graph. Summary and Error Analysis We summarize the key concepts of this sec on thus far in the following Key Idea (b) Figure 8.6: A table of values to approimate 0 e d in Eample 5, along with a graph of the func on. i sin( 3 i ) (a) y y = sin( 3 ) (b) Figure 8.7: A table of values to approimate π π sin( 3 ) d in Eample 6, along 4 with a graph of the func on. 449

124 Chapter 8 Techniques of Integra on Key Idea 3 Numerical Integra on Let f be a con nuous func on on [a, b], let n be a posi ve integer, and let = b a n. Set 0 = a, = a +,, i = a + i, n = b. Consider Le Hand Rule: Right Hand Rule: Midpoint Rule: Trapezoidal Rule: Simpson s Rule: b a b a b a b a b a b a f() d. [ f() d f( 0 ) + f( ) + + f( n ) ]. [ f() d f( ) + f( ) + + f( n ) ]. [ f() d f ( 0 + ) ( + ) ( n + n )] + f + + f. f() d [ f( 0 ) + f( ) + f( ) + + f( n ) + f( n ) ]. f() d 3 [ f( 0 ) + 4f( ) + f( ) + + 4f( n ) + f( n ) ] (n even). In our eamples, we approimated the value of a definite integral using a given method then compared it to the right answer. This should have raised several ques ons in the reader s mind, such as:. How was the right answer computed?. If the right answer can be found, what is the point of approima ng? 3. If there is value to approima ng, how are we supposed to know if the approima on is any good? These are good ques ons, and their answers are educa onal. In the eamples, the right answer was never computed. Rather, an approima on accurate to a certain number of places a er the decimal was given. In Eample, we do not know the eact answer, but we know it starts with These more accurate approima ons were computed using numerical integra on but with more precision (i.e., more subintervals and the help of a computer). Since the eact answer cannot be found, approima on s ll has its place. How are we to tell if the approima on is any good? Trial and error provides one way. Using technology, make an approima- on with, say, 0, 00, and 00 subintervals. This likely will not take much me at all, and a trend should emerge. If a trend does not emerge, try using yet more subintervals. Keep in mind that trial and error is never foolproof; you might stumble upon a problem in which a trend will not emerge. A second method is to use Error Analysis. While the details are beyond the scope of this tet, there are some formulas that give bounds for how good your 450

125 8.7 Numerical Integra on approima on will be. For instance, the formula might state that the approima on is within 0. of the correct answer. If the approima on is.58, then one knows that the correct answer is between.48 and.68. By using lots of subintervals, one can get an approima on as accurate as one likes. Theorem 57 states what these bounds are. Theorem 57 Error Bounds in Numerical Integra on Suppose that M n is an upper bound on f (n) () on [a, b]. Then a bound for the error of the numerical method of integra on is given by: Method Le /Right Hand Rule Midpoint Rule Trapezoidal Rule Simpson s Rule Error Bound M (b a) n M (b a) 3 4n M (b a) 3 n M 4 (b a) 5 80n 4 There are some key things to note about this theorem.. The larger the interval, the larger the error. This should make sense intui vely.. The error shrinks as more subintervals are used (i.e., as n gets larger). 3. When n doubles, the Le and Right Hand Rules double in accuracy, the Midpoint and Trapezoidal Rules quadruple in accuracy, and Simpson s Rule is 6 mes more accurate. 4. The error in Simpson s Rule has a term rela ng to the 4 th deriva ve of f. Consider a cubic polynomial: its 4 th deriva ve is 0. Therefore, the error in approima ng the definite integral of a cubic polynomial with Simpson s Rule is 0 Simpson s Rule computes the eact answer! We revisit Eamples 3 and 5 and compute the error bounds using Theorem 57 in the following eample. Eample 7 Compu ng error bounds Find the error bounds when approima ng e 0 d using the Trapezoidal Rule and 5 subintervals, and using Simpson s Rule with 4 subintervals. 45

126 Chapter 8 Techniques of Integra on y 0.5 y = e (4 ) Figure 8.8: Graphing f () in Eample 7 to help establish error bounds y y = e ( ) 0.5 Figure 8.9: Graphing f (4) () in Eample 7 to help establish error bounds. S Trapezoidal Rule with n = 5: We start by compu ng the nd deriva ve of f() = e : f () = e (4 ). Figure 8.8 shows a graph of f () on [0, ]. It is clear that the largest value of f, in absolute value, is. Thus we let M = and apply the error formula from Theorem 57. ( 0)3 E T = 5 = Our error es ma on formula states that our approima on of found in Eample 3 is within of the correct answer, hence we know that = e d 0.75 = We had earlier computed the eact answer, correct to 4 decimal places, to be , affirming the validity of Theorem 57. Simpson s Rule with n = 4: We start by compu ng the 4 th deriva ve of f() = e : f (4) () = e ( ). Figure 8.9 shows a graph of f (4) () on [0, ]. It is clear that the largest value of f (4), in absolute value, is. Thus we let M = and apply the error formula from Theorem 57. ( 0)5 E s = = Our error es ma on formula states that our approima on of found in Eample 5 is within of the correct answer, hence we know that = Once again we affirm the validity of Theorem e d = At the beginning of this sec on we men oned two main situa ons where numerical integra on was desirable. We have considered the case where an an deriva ve of the integrand cannot be computed. We now inves gate the situa on where the integrand is not known. This is, in fact, the most widely used applica on of Numerical Integra on methods. Most of the me we observe behavior but do not know the func on that describes it. We instead collect data about the behavior and make approima ons based off of this data. We demonstrate this in an eample. 45

127 8.7 Numerical Integra on Eample 8 Approima ng distance traveled One of the authors drove his daughter home from school while she recorded their speed every 30 seconds. The data is given in Figure Approimate the distance they traveled. S Recall that by integra ng a speed func on we get distance traveled. We have informa on about v(t); we will use Simpson s Rule to approimate v(t) dt. b a The most difficult aspect of this problem is conver ng the given data into the form we need it to be in. The speed is measured in miles per hour, whereas the me is measured in 30 second increments. We need to compute = (b a)/n. Clearly, n = 4. What are a and b? Since we start at me t = 0, we have that a = 0. The final recorded me came a er 4 periods of 30 seconds, which is minutes or /5 of an hour. Thus we have = b a = /5 0 = n 4 0 ; 3 = 360. Thus the distance traveled is approimately: 0. 0 v(t) dt 360 = miles. [ ] f( ) + 4f( ) + f( 3 ) + + 4f( n ) + f( n+ ) [ ] We approimate the author drove 6. miles. (Because we are sure the reader wants to know, the author s odometer recorded the distance as about 6.05 miles.) Time Speed (mph) Figure 8.30: Speed data collected at 30 second intervals for Eample

128 Eercises 8.7 Terms and Concepts. T/F: Simpson s Rule is a method of approima ng an- deriva ves.. What are the two basic situa ons where approima ng the value of a definite integral is necessary? 3. Why are the Le and Right Hand Rules rarely used? 4. Why is the Midpoint Rule rarely used? Problems In Eercises 5, a definite integral is given (a) Approimate the definite integral with the Trapezoidal Rule and n = 4. (b) Approimate the definite integral with Simpson s Rule and n = 4. (c) Find the eact value of the integral. 0 0 π π d 5 d sin d d ( ) d 4 d cos d 9 d In Eercises 3 0, approimate the definite integral with the Trapezoidal Rule and Simpson s Rule, with n = π 0 π/ 0 4 cos ( ) d e d + d sin d cos d ln d sin + d sin + d In Eercises 4, find n such that the error in approima ng the given definite integral is less than when using: (a) the Trapezoidal Rule (b) Simpson s Rule π 0 4 π sin d d cos ( ) d 4 d In Eercises 5 6, a region is given. Find the area of the region using Simpson s Rule: (a) where the measurements are in cen meters, taken in cm increments, and (b) where the measurements are in hundreds of yards, taken in 00 yd increments Let f be the quadra c func on that goes through the points (, y ), ( +, y ) and ( +, y 3). Show that + f()d = (y + 4y + y3)

129 9: S S This chapter introduces sequences and series, important mathema cal construc ons that are useful when solving a large variety of mathema cal problems. The content of this chapter is considerably different from the content of the chapters before it. While the material we learn here definitely falls under the scope of calculus, we will make very li le use of deriva ves or integrals. Limits are etremely important, though, especially limits that involve infinity. One of the problems addressed by this chapter is this: suppose we know informa on about a func on and its deriva ves at a point, such as f() = 3, f () =, f () =, f () = 7, and so on. What can I say about f() itself? Is there any reasonable approima on of the value of f()? The topic of Taylor Series addresses this problem, and allows us to make ecellent approima ons of func ons when limited knowledge of the func on is available. 9. Sequences We commonly refer to a set of events that occur one a er the other as a sequence of events. In mathema cs, we use the word sequence to refer to an ordered set of numbers, i.e., a set of numbers that occur one a er the other. For instance, the numbers, 4, 6, 8,, form a sequence. The order is important; the first number is, the second is 4, etc. It seems natural to seek a formula that describes a given sequence, and o en this can be done. For instance, the sequence above could be described by the func on a(n) = n, for the values of n =,,... (it could also be described by n 4 0n n 48n + 4, to give one of infinitely many other op ons). To find the 0 th term in the sequence, we would compute a(0). This leads us to the following, formal defini on of a sequence. Defini on 30 Sequence A sequence is a func on a(n) whose domain is N. The range of a sequence is the set of all dis nct values of a(n). Nota on: We use N to describe the set of natural numbers, that is, the integers,, 3, The terms of a sequence are the values a(), a(),, which are usually denoted with subscripts as a, a,. A sequence a(n) is o en denoted as {a n }.

130 Chapter 9 Sequences and Series Watch the video: Sequences Eamples showing convergence or divergence at Factorial: The epression 3! refers to the number 3 = 6. In general, n! = n (n ) (n ), where n is a natural number. We define 0! =. While this does not immediately make sense, it makes many mathema cal formulas work properly y y / /4 y a n = 3n n! 3 4 (a) a n = 4 + ( ) n 3 4 (b) n n n Eample Lis ng terms of a sequence List the first four terms of the following sequences. { } { } 3 n ( ). {a n } =. {a n } = {4 + ( ) n n(n+)/ } 3. {a n } = n! n S. a = 3! = 3; a = 3! = 9 ; a 3 = 33 3! = 9 ; a 4 = 34 4! = 7 8 We can plot the terms of a sequence with a sca er plot. The -ais is used for the values of n, and the values of the terms are plo ed on the y-ais. To visualize this sequence, see Figure 9.(a).. a = 4 + ( ) = 3; a = 4 + ( ) = 5; a 3 = 4 + ( ) 3 = 3; a 4 = 4 + ( ) 4 = 5. Note that the range of this sequence is finite, consis ng of only the values 3 and 5. This sequence is plo ed in Figure 9.(b). 3. a = ( )()/ = ; a = ( )(3)/ = 4 a 3 = ( )3(4)/ 3 = 9 a 4 = ( )4(5)/ 4 = 6 ; a 5 = ( )5(6)/ 5 = 5. We gave one etra term to begin to show the pa ern of signs is,, +, +,,,..., due to the fact that the eponent of is a special quadra c. This sequence is plo ed in Figure 9.(c). Eample Determining a formula for a sequence Find the n th term of the following sequences, i.e., find a func on that describes each of the given sequences. a n = ( )n(n+)/ n (c) Figure 9.: Plo ng sequences in Eample. 456

131 9. Sequences., 5, 8,, 4,...., 5, 0, 7, 6, 37,... 3.,,, 6, 4, 0, 70, , 5, 5 8, 5 4, 5 3,... S We should first note that there is never eactly one func on that describes a finite set of numbers as a sequence. There are many sequences that start with, then 5, as our first eample does. We are looking for a simple formula that describes the terms given, knowing there is possibly more than one answer.. Note how each term is 3 more than the previous one. This implies a linear func on would be appropriate: a(n) = a n = 3n+b for some appropriate value of b. As we want a =, we set b =. Thus a n = 3n.. First no ce how the sign changes from term to term. This is most commonly accomplished by mul plying the terms by either ( ) n or ( ) n+. Using ( ) n mul plies the odd terms by ( ); using ( ) n+ mul plies the even terms by ( ). As this sequence has nega ve even terms, we will mul ply by ( ) n+. A er this, we might feel a bit stuck as to how to proceed. At this point, we are just looking for a pa ern of some sort: what do the numbers, 5, 0, 7, etc., have in common? There are many correct answers, but the one that we ll use here is that each is one more than a perfect square. That is, = +, 5 = +, 0 = 3 +, etc. Thus our formula is a n = ( ) n+ (n + ). 3. One who is familiar with the factorial func on will readily recognize these numbers. They are 0!,!,!, 3!, etc. Since our sequences start with n =, we cannot write a n = n!, for this misses the 0! term. Instead, we shi by, and write a n = (n )!. 4. This one may appear difficult, especially as the first two terms are the same, but a li le sleuthing will help. No ce how the terms in the numerator are always mul ples of 5, and the terms in the denominator are always powers of. Does something as simple as a n = 5n n work? When n =, we see that we indeed get 5/ as desired. When n =, we get 0/4 = 5/. Further checking shows that this formula indeed matches the other terms of the sequence. 457

132 Chapter 9 Sequences and Series A common mathema cal endeavor is to create a new mathema cal object (for instance, a sequence) and then apply previously known mathema cs to the new object. We do so here. The fundamental concept of calculus is the limit, so we will inves gate what it means to find the limit of a sequence. Defini on 3 Limit of a Sequence, Convergent, Divergent Let {a n } be a sequence and let L be a real number. Given any ε > 0, if an m can be found such that a n L < ε for all n > m, then we say the limit of {a n }, as n approaches infinity, is L, denoted lim a n = L. n If lim n a n eists, we say the sequence converges; otherwise, the sequence diverges. This defini on states, informally, that if the limit of a sequence is L, then if you go far enough out along the sequence, all subsequent terms will be really close to L. Of course, the terms far enough and really close are subjec ve terms, but hopefully the intent is clear. This defini on is reminiscent of the ε δ proofs of Chapter. In that chapter we developed other tools to evaluate limits apart from the formal defini on; we do so here as well. Theorem 58 Limit of a Sequence Let {a n } be a sequence and let f() be a func on whose domain contains the posi ve real numbers where f(n) = a n for all n in N. If lim f() = L, then lim n a n = L. Theorem 58 allows us, in certain cases, to apply the tools developed in Chapter to limits of sequences. Note two things not stated by the theorem:. If lim f() does not eist, we cannot conclude that lim a n does not eist. n It may, or may not, eist. For instance, we can define a sequence {a n } = {cos(πn)}. Let f() = cos(π). Since the cosine func on oscillates over the real numbers, the limit lim f() does not eist. However, for every posi ve integer n, cos(πn) =, so lim a n =. n 458

133 9. Sequences. If we cannot find a func on f() whose domain contains the posi ve real numbers where f(n) = a n for all n in N, we cannot conclude lim a n does n not eist. It may, or may not, eist. Eample 3 Determining convergence/divergence of a sequence Determine the convergence or divergence of the following sequences. { 3n } { } n + ( ) n. {a n } = n. {a n } = {cos n} 3. {a n } = 000 n S 3 +. Using Key Idea, we can state that lim = 3. (We could 000 have also directly applied L Hôpital s Rule.) Thus the sequence {a n } converges, and its limit is 3. A sca er plot of every 5 values of a n is given in Figure 9. (a). The values of a n vary widely near n = 30, ranging from about 73 to 5, but as n grows, the values approach 3.. The limit lim cos does not eist, as cos oscillates (and takes on every value in [, ] infinitely many mes). Thus we cannot apply Theorem 58. The fact that the cosine func on oscillates strongly hints that cos n, when n is restricted to N, will also oscillate. Figure 9. (b), where the sequence is plo ed, shows that this is true. Because only discrete values of cosine are plo ed, it does not bear strong resemblance to the familiar cosine wave. Based on the graph, we suspect that lim n a n does not eist, but we have not decisively proven it yet. y y a n = 3n n + n 000 (a) a n = cos n n n 3. We cannot actually apply Theorem 58 here, as the func on f() = ( ) / is not well defined. (What does ( ) mean? In actuality, there is an answer, but it involves comple analysis, beyond the scope of this tet.) So for now we say that we cannot determine the limit. (But we will be able to very soon.) By looking at the plot in Figure 9. (c), we would like to conclude that the sequence converges to 0. That is true, but at this point we are unable to decisively say so. y 0.5 (b) n It seems that {( ) n /n} converges to 0 but we lack the formal tool to prove it. The following theorem gives us that tool. Theorem 59 Absolute Value Theorem Let {a n } be a sequence. If lim a n = 0, then lim a n = 0 n n 0.5 a n = ( )n n (c) Figure 9.: Sca er plots of the sequences in Eample

134 Chapter 9 Sequences and Series Proof We know a n a n a n and lim Squeeze Theorem lim n a n = 0. ( a n ) = lim a n = 0. Thus by the n n Eample 4 Determining the convergence/divergence of a sequence Determine the convergence or divergence of the following sequences. { } { ( ) n ( ) n } (n + ). {a n } =. {a n } = n n S. This appeared in Eample 3. We want to apply Theorem 59, so consider the limit of { a n }: lim a n = lim ( ) n n n n = lim n n = 0. Since this limit is 0, we can apply Theorem 59 and state that lim n a n = 0.. Because of the alterna ng nature of this sequence (i.e., every other term ( ) ( + ) is mul plied by ), we cannot simply look at the limit lim. We can try to apply the techniques of Theorem 59: lim a n = lim ( ) n (n + ) n n n n + = lim n n =. y We have concluded that when we ignore the alterna ng sign, the se- quence approaches. This means we cannot apply Theorem 59; it states the the limit must be 0 in order to conclude anything. a n = ( )n (n + ) n n Since we know that the signs of the terms alternate and we know that the limit of a n is, we know that as n approaches infinity, the terms will alternate between values close to and, meaning the sequence diverges. A plot of this sequence is given in Figure 9.3. Figure 9.3: A plot of a sequence in Eample 4, part. 460

135 9. Sequences We con nue our study of the limits of sequences by considering some of the proper es of these limits. Theorem 60 Proper es of the Limits of Sequences Let {a n } and {b n } be sequences such that lim a n = L, lim b n = K, n n and let c be a real number.. lim n (a n ± b n ) = L ± K. lim n (a n b n ) = L K 3. lim n (a n/b n ) = L/K, K 0 4. lim n c a n = c L Eample 5 Applying proper es of limits of sequences Let the following limits be given: lim n a n = 0; lim n b n = e; and lim n c n = 5. Evaluate the following limits.. lim n (a n + b n ). lim n (b n c n ) 3. lim n (000 a n) S We will use Theorem 60 to answer each of these.. Since lim a n = 0 and lim b n = e, we conclude that lim (a n + b n ) = n n n 0 + e = e. So even though we are adding something to each term of the sequence b n, we are adding something so small that the final limit is the same as before.. Since lim b n = e and lim c n = 5, we conclude that lim (b n c n ) = n n n e 5 = 5e. 3. Since lim a n = 0, we have lim 000a n = = 0. It does not n n ma er that we mul ply each term by 000; the sequence s ll approaches 0. (It just takes longer to get close to 0.) 46

136 Chapter 9 Sequences and Series Defini on 3 Geometric Sequence For a constant r, the sequence {r n } is known as a geometric sequence. Theorem 6 Convergence of Geometric Sequences The sequence {r n } is convergent if < r and divergent for all other values of r. Furthermore, { 0 < r < lim n rn = r = Proof We can see from Key Idea 8 and by le ng a = r that { r > lim n rn = 0 0 < r <. We also know that lim n = and lim 0 n = 0. If < r < 0, we know 0 < r < so lim r n = lim r n = 0 and thus by Theorem 59, lim r n = 0. If r, lim r n does not eist. Therefore, the sequence {r n } is convergent if < r and divergent for all other values of r. There is more to learn about sequences than just their limits. We will also study their range and the rela onships terms have with the terms that follow. We start with some defini ons describing proper es of the range. Defini on 33 Bounded and Unbounded Sequences A sequence {a n } is said to be bounded if there eists real numbers m and M such that m < a n < M for all n in N. A sequence {a n } is said to be unbounded if it is not bounded. A sequence {a n } is said to be bounded above if there eists an M such that a n < M for all n in N; it is bounded below if there eists an m such that m < a n for all n in N. 46

137 9. Sequences It follows from this defini on that an unbounded sequence may be bounded above or bounded below; a sequence that is both bounded above and below is simply a bounded sequence. Eample 6 Determining boundedness of sequences Determine the boundedness of the following sequences. { }. {a n } =. {a n } = { n } n y S. The terms of this sequence are always posi ve but are decreasing, so we have 0 < a n < for all n. Thus this sequence is bounded. Figure 9.4(a) illustrates this. / a n = n. The terms of this sequence obviously grow without bound. However, it is also true that these terms are all posi ve, meaning 0 < a n. Thus we can say the sequence is unbounded, but also bounded below. Figure 9.4(b) illustrates this. /4 / (a) n The previous eample produces some interes ng concepts. First, we can recognize that the sequence {/n} converges to 0. This says, informally, that most of the terms of the sequence are really close to 0. This implies that the sequence is bounded, using the following logic. First, most terms are near 0, so we could find some sort of bound on these terms (using Defini on 3, the bound is ε). That leaves a few terms that are not near 0 (i.e., a finite number of terms). A finite list of numbers is always bounded. This logic suggests that if a sequence converges, it must be bounded. This is indeed true, as stated by the following theorem. Theorem 6 Convergent Sequences are Bounded Let {a n } be a convergent sequence. Then {a n } is bounded y a n = n (b) Figure 9.4: A plot of {a n} = {/n} and {a n} = { n } from Eample 6. n In Eample part, we found that lim ( + /) = e. If we consider the sequence {b n } = {( + /n) n }, we see that lim b n = e. Even though n it may be difficult to intui vely grasp the behavior of this sequence, we know immediately that it is bounded. Note: Keep in mind what Theorem 6 does not say. It does not say that bounded sequences must converge, nor does it say that if a sequence does not converge, it is not bounded. 463

138 Chapter 9 Sequences and Series Another interes ng concept to come out of Eample 6 again involves the sequence {/n}. We stated, without proof, that the terms of the sequence were decreasing. That is, that a n+ < a n for all n. (This is easy to show. Clearly n < n +. Taking reciprocals flips the inequality: /n > /(n + ). This is the same as a n > a n+.) Sequences that either steadily increase or decrease are important, so we give this property a name. Defini on 34 Monotonic Sequences. A sequence {a n } is monotonically increasing if a n a n+ for all n, i.e., a a a 3 a n a n+. A sequence {a n } is monotonically decreasing if a n a n+ for all n, i.e., a a a 3 a n a n+ 3. A sequence is monotonic if it is monotonically increasing or monotonically decreasing. Note: It is some mes useful to call a monotonically increasing sequence strictly increasing if a n < a n+ for all n; i.e, we remove the possibility that subsequent terms are equal. A similar statement holds for strictly decreasing. Eample 7 Determining monotonicity Determine the monotonicity of the following sequences. { } n +. {a n } = n { n } +. {a n } = n + { n } 9 3. {a n } = n 0n + 6 { } n 4. {a n } = n! S In each of the following, we will eamine a n+ a n. If a n+ a n 0, we conclude that a n a n+ and hence the sequence is increasing. If a n+ a n 0, we conclude that a n a n+ and the sequence is decreasing. Of course, a sequence need not be monotonic and perhaps neither of the above will apply. We also give a sca er plot of each sequence. These are useful as they suggest a pa ern of monotonicity, but analy c work should be done to confirm a graphical trend. 464

139 9. Sequences. a n+ a n = n + n + n + n (n + )(n) (n + ) = (n + )n = n(n + ) < 0 for all n. y a n = n + n. Since a n+ a n < 0 for all n, we conclude that the sequence is decreasing. a n+ a n = (n + ) + n + n + n + ( (n + ) + ) (n + ) (n + )(n + ) = (n + )(n + ) = n + 4n + (n + )(n + ) > 0 for all n. 5 0 Figure 9.5: A plot of {a n} = { n+ } in Eample n 7(a). 0 y n Since a n+ a n > 0 for all n, we conclude the sequence is increasing We can clearly see in Figure 9.7, where the sequence is plo ed, that it is not monotonic. However, it does seem that a er the first 4 terms it is decreasing. To understand why, perform the same analysis as done before: a n+ a n = (n + ) 9 (n + ) 0(n + ) + 6 n 9 n 0n + 6 = n + n 8 n 8n + 7 n 9 n 0n + 6 = (n + n 8)(n 0n + 6) (n 9)(n 8n + 7) (n 8n + 7)(n 0n + 6) = 0n + 60n 55 (n 8n + 7)(n 0n + 6). a n = n + n Figure 9.6: A plot of {a n} = { n + n+ } in Eample 7(b). 5 0 y n 9 a n = n 0n + 6 n We want to know when this is greater than, or less than, 0. The denominator is always posi ve, therefore we are only concerned with the numerator. Using the quadra c formula, we can determine that 0n + 60n 55 = 0 when n.3, So for n <.3, the sequence is decreasing. Since we are only dealing with the natural numbers, this means that a > a. 5 n 5 0 Figure 9.7: A plot of {a n} = { in Eample 7(c). n 9 } n 0n+6 465

140 Chapter 9 Sequences and Series y a n = n n! Between.3 and 4.87, i.e., for n =, 3 and 4, we have that a n+ > a n and the sequence is increasing. (That is, when n =, 3 and 4, the numerator 0n + 60n + 55 from the frac on above is > 0.) When n > 4.87, i.e, for n 5, we have that 0n + 60n + 55 < 0, hence a n+ a n < 0, so the sequence is decreasing. In short, the sequence is simply not monotonic. However, it is useful to note that for n 5, the sequence is monotonically decreasing. 4. Again, the plot in Figure 9.8 shows that the sequence is not monotonic, but it suggests that it is monotonically decreasing a er the first term. Instead of looking at a n+ a n, this me we ll look at a n /a n+ : 5 0 Figure 9.8: A plot of {a n} = {n /n!} in Eample 7(d). n a n = n (n + )! a n+ n! (n + ) = n n + = n + n + When n =, the above epression is < ; for n, the above epression is >. Thus this sequence is not monotonic, but it is monotonically decreasing a er the first term. Knowing that a sequence is monotonic can be useful. In par cular, if we know that a sequence is bounded and monotonic, we can conclude it converges. Consider, for eample, a sequence that is monotonically decreasing and is bounded below. We know the sequence is always ge ng smaller, but that there is a bound to how small it can become. This is enough to prove that the sequence will converge, as stated in the following theorem. Theorem 63 Bounded Monotonic Sequences are Convergent Let {a n } be a bounded, monotonic sequence. Then {a n } converges; i.e., lim a n eists. n Consider once again the sequence {a n } = {/n}. It is easy to show it is monotonically decreasing and that it is always posi ve (i.e., bounded below by 0). Therefore we can conclude by Theorem 63 that the sequence converges. We already knew this by other means, but in the following sec on this theorem will become very useful. 466

141 9. Sequences Convergence of a sequence does not depend on the first N terms of a sequence. For eample, we could adapt the sequence of the previous paragraph to be, 0, 00, 000, 5, 6, 7, 8, 9, 0,... Because we only changed three of the first 4 terms, we have not affected whether the sequence converges or diverges. Sequences are a great source of mathema cal inquiry. The On-Line Encyclopedia of Integer Sequences ( contains thousands of sequences and their formulae. (As of this wri ng, there are 57,537 sequences in the database.) Perusing this database quickly demonstrates that a single sequence can represent several different real life phenomena. Interes ng as this is, our interest actually lies elsewhere. We are more interested in the sum of a sequence. That is, given a sequence {a n }, we are very interested in a + a + a 3 +. Of course, one might immediately counter with Doesn t this just add up to infinity? Many mes, yes, but there are many important cases where the answer is no. This is the topic of series, which we begin to inves gate in the net sec on. 467

142 Eercises 9. Terms and Concepts. Use your own words to define a sequence.. The domain of a sequence is the numbers. 3. Use your own words to describe the range of a sequence. 4. Describe what it means for a sequence to be bounded. Problems In Eercises 5 8, give the first five terms of the given sequence. { } 4 n 5. {a n} = (n + )! {( 6. {b n} = 3 ) n } } 7. {c n} = { nn+ n + { ( ( ) n ( ) n )} {d n} = 5 In Eercises 9, determine the n th term of the given sequence. 9. 4, 7, 0, 3, 6, , 3, 3 4, 3 8,.... 0, 0, 40, 80, 60,....,,, 6, 4, 0,... In Eercises 3 6, use the following informa on to determine the limit of the given sequences. { } n 0 {a n} = ; lim n n an = {( {b n} = + ) n } ; lim bn = e n n {c n} = {sin(3/n)}; { } n 0 3. {a n} = 7 n 4. {a n} = {3b n a n} { 5. {a n} = sin(3/n) 6. {a n} = ( + n { ( + ) } n n ) n } lim n cn = 0 In Eercises 7 39, determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. { } 7. {a n} = ( ) n n n + { } 4n n {a n} = 3n + { } 4 n 9. {a n} = 5 n { } (n 3)! 0. {a n} = (n + )! { n. {a n} = n }, n n n { } 6 n+3. {a n} = 8 n 3. {a n} = {ln(n)} { } 3n 4. {a n} = n + {( 5. {a n} = + ) n } n { } (n + )! 6. {a n} = (n )! { 7. {a n} = 5 } n { } ( ) n+ 8. {a n} = n { }. n 9. {a n} = n { } n 30. {a n} = n + } 3. {a n} = {( ) n n n } 3. {a n} = { + 9n 8 n { } (n )! 33. {a n} = (n + )! 34. {a n} = {ln(3n + ) ln n} 35. {a n} = {ln(n + 3n + ) ln(n + )} { ( )} 36. {a n} = n sin n { } cos n 37. {a n} = n { } e n + e n 38. {a n} = e n { } ln n 39. {a n} = ln n In Eercises 40 43, determine whether the sequence is bounded, bounded above, bounded below, or none of the above. 40. {a n} = {sin n} { } 4. {a n} = ( ) n 3n n { } 3n 4. {a n} = n 43. {a n} = { n n!} 468

143 In Eercises 44 49, determine whether the sequence is monotonically increasing or decreasing. If it is not, determine if there is an m such that it is monotonic for all n m. { } n 44. {a n} = n + { } n 6n {a n} = n { } 46. {a n} = ( ) n { } n 47. {a n} = n n 3 { ( )} nπ 48. {a n} = cos 49. {a n} = {ne n } 50. Prove Theorem 59; that is, use the defini on of the limit of a sequence to show that if lim an = 0, then lim n n an = Let {a n} and {b n} be sequences such that lim n lim bn = K. n (a) Show that if a n < b n for all n, then L K. (b) Give an eample where L = K. an = L and 5. Prove the Squeeze Theorem for sequences: Let {a n} and {b n} be such that lim n an = L and lim n bn = L, and let {c n} be such that a n c n b n for all n. Then lim n cn = L 469

144 Chapter 9 Sequences and Series 9. Infinite Series Given the sequence {a n } = {/ n } = /, /4, /8,..., consider the following sums: a = / = / a + a = / + /4 = 3/4 a + a + a 3 = / + /4 + /8 = 7/8 a + a + a 3 + a 4 = / + /4 + /8 + /6 = 5/6 Later, we will be able to show that a + a + a a n = n n = n. Let S n be the sum of the first n terms of the sequence {/ n }. From the above, we see that S = /, S = 3/4, and that S n = / n. Now consider the following limit: lim S ( n = lim / n ) =. This limit n n can be interpreted as saying something amazing: the sum of all the terms of the sequence {/ n } is. This eample illustrates some interes ng concepts that we eplore in this sec on. We begin this eplora on with some defini ons. Defini on 35 Infinite Series, n th Par al Sums, Convergence, Divergence Let {a n } be a sequence.. The sum. Let S n = a n is an infinite series (or, simply series). n= n a i ; the sequence {S n } is the sequence of n th par al i= sums of {a n }. 3. If the sequence {S n } converges to L, we say the series a n converges to L, and we write a n = L. n= 4. If the sequence {S n } diverges, the series n= a n diverges. n= 470

145 9. Infinite Series Using our new terminology, we can state that the series / n converges, and / n =. n= n= Watch the video: Finding a Formula for a Par al Sum of a Telescoping Series at We will eplore a variety of series in this sec on. We start with two series that diverge, showing how we might discern divergence. Eample Showing series diverge. Let {a n } = {n }. Show a n diverges. n=. Let {b n } = {( ) n+ }. Show S b n diverges. n=. Consider S n, the n th par al sum. Since lim n S n instruc ve to write S n = a + a + a a n = n n(n + )(n + ) =. by Theorem 35 6 =, we conclude that the series n diverges. It is n= n = for this tells us how the series diverges: it n= grows without bound. A sca er plot of the sequences {a n } and {S n } is given in Figure 9.9. The terms of {a n } are growing, so the terms of the par al sums {S n } are growing even faster, illustra ng that the series diverges y a n 5 0 Figure 9.9: Sca er plots rela ng to the series of Eample part. S n n 47

146 Chapter 9 Sequences and Series. The sequence {b n } starts with,,,,.... Consider some of the par al sums S n of {b n }: y 0.5 S = S = 0 S 3 = S 4 = 0 { n is odd This pa ern repeats; we find that S n = 0 n is even. As {S n} oscillates, repea ng, 0,, 0,..., we conclude that lim S n does not eist, hence n ( ) n+ diverges. n= A sca er plot of the sequence {b n } and the par al sums {S n } is given in Figure 9.0. When n is odd, b n = S n so the marks for b n are drawn oversized to show they coincide n While it is important to recognize when a series diverges, we are generally more interested in the series that converge. In this sec on we will demonstrate a few general techniques for determining convergence; later sec ons will delve deeper into this topic. b n Figure 9.0: Sca er plots rela ng to the series of Eample part. S n Geometric Series One important type of series is a geometric series. Defini on 36 Geometric Series A geometric series is a series of the form ar n = a + ar + ar + ar ar n + Note that the inde starts at n = 0, if the inde starts at n = we have ar n. n= We started this sec on with a geometric series, although we dropped the first term of. One reason geometric series are important is that they have nice convergence proper es. 47

147 9. Infinite Series Theorem 64 Convergence of Geometric Series Consider the geometric series ar n.. If r, the n th par al sum is: S n = a( r n ). r. The series converges if, and only if, r <. When r <, ar n = a r. Proof If r =, then S n = a+a+a+ +a = na. Since lim n S n = ±, the geometric series diverges. If r, we have S n = a + ar + ar + + ar n. Mul ply each term by r and we have rs n = ar + ar + ar 3 + ar n. Subtract these two equa ons and solve for S n. S n rs n = a ar n S n = a( rn ) r From Theorem 6, we know that if < r <, then lim n rn = 0 so lim S n = lim = a( rn ) = a n n r r a r lim n rn = a r. a So when r < the geometric series converges and its sum is r. If either r or r >, the sequence {r n } is divergent by Theorem 6. Thus lim S n does not eist, so the geometric series diverges if r or r > n. 473

148 Chapter 9 Sequences and Series According to Theorem 64, the series n = ( ) = converges as r = /, and n = =. This concurs with our introductory eample; while there we got a sum of, we skipped the first term of /. y Eample Eploring geometric series Check the convergence of the following series. If the series converges, find its sum. ( ) n 3 ( ) n n 4 n= a n S n (a) n S. Since r = 3/4 <, this series converges. By Theorem 64, we have that ( ) n 3 = 4 3/4 = 4. y n However, note the subscript of the summa on in the given series: we are to start with n =. Therefore we subtract off the first two terms, giving: ( ) n 3 = = 9 4. n= This is illustrated in Figure 9.(a). 0.5 y, a n (b) S n. Since r = / <, this series converges, and by Theorem 64, ( ) n = ( /) = 3. The par al sums of this series are plo ed in Figure 9.(b). Note how the par al sums are not purely increasing as some of the terms of the sequence {( /) n } are nega ve. 3. Since r >, the series diverges. (This makes common sense ; we epect the sum to diverge.) This is illustrated in Figure 9.(c). 4 6 a n S n n (c) Figure 9.: Sca er plots rela ng to the series in Eample. 474

149 9. Infinite Series Later sec ons will provide tests by which we can determine whether or not a given series converges. This, in general, is much easier than determining what a given series converges to. There are many cases, though, where the sum can be determined. Eample 3 Telescoping series ( Evaluate the sum n ). n + n= S of this series. It will help to write down some of the first few par al sums S = ( S = ) ( + ) 3 ( S 3 = ) ( + ) 3 ( S 4 = ) ( + 3 ( + ) ( 3 4 ) ) ( + 4 ) 5 = = 3 = 4 = 5 Note how most of the terms in each par al sum subtract out. In general, we see that S n = ( n +. The sequence {S n} converges, as lim S n = lim ) = n n n + (, and so we conclude that n ) =. Par al sums of the series n + n= are plo ed in Figure 9.. The series in Eample 3 is an eample of a telescoping series. Informally, a telescoping series is one in which the par al sums reduce to just a finite number of terms. The par al sum S n did not contain n terms, but rather just two: and /(n + ). When possible, seek a way to write an eplicit formula for the n th par al sum S n. This makes evalua ng the limit lim S n much more approachable. We do so n in the net eample. Eample 4 Evalua ng series Evaluate each of the following infinite series.. n= n + n. ( ) n + ln n n= 0.5 y a n Figure 9.: Sca er plots rela ng to the series of Eample 3. S n n 475

150 Chapter 9 Sequences and Series S. We can decompose the frac on /(n + n) as n + n = n n +. (See Sec on 8.4, Par al Frac on Decomposi on, to recall how this is done, if necessary.) Epressing the terms of {S n } is now more instruc ve: y S = ( 3 S = ) ( + 3 ) 4 ( S 3 = ) ( + 3 ) ( ) 5 ( S 4 = ) ( + 3 ) ( ) 5 ( S 5 = ) ( + 3 ) ( ( + ) ( 4 6 ) ) ( + 5 ) 7 = 3 = = = = We again have a telescoping series. In each par al sum, most of the terms pair up to add to zero and we obtain the formula S n = + n +. Taking limits allows us to determine the convergence of the series: n + ( lim S n = lim + n n n + ) = 3 n +, so n + n = 3. n= a n S n n This is illustrated in Figure We begin by wri ng the first few par al sums of the series: Figure 9.3: Sca er plots rela ng to the series of Eample 4 part. S = ln () ( ) 3 S = ln () + ln ( ) ( ) 3 4 S 3 = ln () + ln + ln 3 ( ) ( 3 4 S 4 = ln () + ln + ln 3 ) + ln ( )

151 9. Infinite Series At first, this does not seem helpful, but recall the logarithmic iden ty: ln + ln y = ln(y). Applying this to S 4 gives: S 4 = ln ()+ln ( ) 3 +ln ( ) 4 +ln 3 We must generalize this for S n. ( ) ( ) 3 n + S n = ln ()+ln + +ln n ( ) ( 5 = ln ) = ln (5). 4 ( = ln 3 n n n + ) n = ln(n+) y We can conclude that {S n } = { ln(n + ) }. This sequence does not converge, as lim S n =. Therefore ln = ; the series ( ) n + n n n= diverges. Note in Figure 9.4 how the sequence of par al sums grows slowly; a er 00 terms, it is not yet over 5. Graphically we may be fooled into thinking the series converges, but our analysis above shows that it does not n a n S n We are learning about a new mathema cal object, the series. As done before, we apply old mathema cs to this new topic. Figure 9.4: Sca er plots rela ng to the series of Eample 4 part. Theorem 65 Proper es of Infinite Series Suppose that a n and b n are convergent series, and that n= n= a n = L, b n = K, and c is a constant. n= n=. Constant Mul ple Rule: c a n = c a n = c L. n= n= ( ). Sum/Difference Rule: an ± b n = a n ± b n = L ± K. n= n= n= Before using this theorem, we will consider the harmonic series n= n. 477

152 Chapter 9 Sequences and Series Eample 5 Divergence of the Harmonic Series Show that the harmonic series n diverges. n= S We will use a proof by contradic on here. Suppose the harmonic series converges to S. That is We then have S = S = = + S This gives us S + S which can never be true, thus our assump on that the harmonic series converges must be false. Therefore, the harmonic series diverges. It may take a while before one is comfortable with this statement, whose truth lies at the heart of the study of infinite series: it is possible that the sum of an infinite list of nonzero numbers is finite. We have seen this repeatedly in this sec on, yet it s ll may take some ge ng used to. As one contemplates the behavior of series, a few facts become clear.. In order to add an infinite list of nonzero numbers and get a finite result, most of those numbers must be very near 0.. If a series diverges, it means that the sum of an infinite list of numbers is not finite (it may approach ± or it may oscillate), and: (a) The series will s ll diverge if the first term is removed. (b) The series will s ll diverge if the first 0 terms are removed. (c) The series will s ll diverge if the first,000,000 terms are removed. (d) The series will s ll diverge if any finite number of terms from anywhere in the series are removed. These concepts are very important and lie at the heart of the net two theorems. 478

153 9. Infinite Series Theorem 66 Convergence of Sequence If the series a n converges, then lim a n = 0. n n= Proof Let S n = a + a + + a n. We have Since n S n = a + a + + a n + a n S n = S n + a n a n = S n S n a n converges, the sequence {S n } converges. Let lim n S n n, n also goes to, so lim n S n = S. We now have lim a n = lim (S n S n ) n n = S. As = lim n S n lim n S n = S S = 0 Theorem 67 Test for Divergence If lim n a n does not eist or lim n a n 0, then the series a n diverges. n= The Test for Divergence follows from Theorem 66. If the series does not diverge, it must converge and therefore lim n a n = 0. Note that the two statements in Theorems 66 and 67 are really the same. In order to converge, the limit of the terms of the sequence must approach 0; if they do not, the series will not converge. Looking back, we can apply this theorem to the series in Eample. In that eample, we had {a n } = {n } and {b n } = {( ) n+ }. lim a n = lim n n n = and lim b n = lim n n ( )n+ which does not eist. 479

154 Chapter 9 Sequences and Series Thus by the Test for Divergence, both series will diverge. Important! This theorem does not state that if lim n a n = 0 then converges. The standard eample of this is the Harmonic Series, as given in Eample 5. The Harmonic Sequence, {/n}, converges to 0; the Harmonic Series, /n, diverges. n= n= a n Theorem 68 Infinite Nature of Series The convergence or divergence remains unchanged by the inser on or dele on of any finite number of terms. That is:. A divergent series will remain divergent with the inser on or dele on of any finite number of terms.. A convergent series will remain convergent with the inser on or dele on of any finite number of terms. (Of course, the sum will likely change.) In other words, when we are only interested in the convergence or divergence of a series, it is safe to ignore the first few billion terms. Eample 6 Removing Terms from the Harmonic Series Consider once more the Harmonic Series which diverges; that is, the parn N n= al sums S N = grow (very, very slowly) without bound. One might think n n= that by removing the large terms of the sequence that perhaps the series will converge. This is simply not the case. For instance, the sum of the first 0 million terms of the Harmonic Series is about 6.7. Removing the first 0 million terms from the Harmonic Series changes the par al sums, effec vely subtracting 6.7 from the sum. However, a sequence that is growing without bound will s ll grow without bound when 6.7 is subtracted from it. The equa on below illustrates this. Even though we have subtracted off the first 0 million terms, this only subtracts a constant off of an epression that is 480

155 9. Infinite Series s ll growing to infinity. Therefore, the modified series is s ll growing to infinity.,000,00 n = lim N N,000,00 n = lim N N n= 0,000,00 n n = lim n= N N n= 6.7 =. n This sec on introduced us to series and defined a few special types of series whose convergence proper es are well known. We know when a geometric series converges or diverges. Most series that we encounter are not one of these types, but we are s ll interested in knowing whether or not they converge. The net three sec ons introduce tests that help us determine whether or not a given series converges. 48

156 Eercises 9. Terms and Concepts. Use your own words to describe how sequences and series are related.. Use your own words to define a par al sum. 3. Given a series a n, describe the two sequences related n= to the series that are important. 4. Use your own words to eplain what a geometric series is. 5. T/F: If {a n} is convergent, then a n is also convergent. Problems In Eercises 6 3, a series n= a n is given. n= (a) Give the first 5 par al sums of the series. (b) Give a graph of the first 5 terms of a n and S n on the same aes. ( ) n 6. n n= n= n cos(πn) n= n n= n= n! 3 n n= n= n= ( 9 0 ( ) n 0 ) n In Eercises 4 9, state whether the given series converges or diverges and provide jus fica on for your conclusion n n= n= 3n n(n + ) 6 n 5 n n n n n= n= n! 0 n 5 n n 5 5 n + n 5 ( n! + ) n n= n= n= n= n= n n + n+ n n 3 n= n= n= ( + ) n n π n 3 n+ 3 n + n n= n= 6 n 3 n(n + ) n In Eercises 30 45, a series is given. (a) Find a formula for S n, the n th par al sum of the series. (b) Determine whether the series converges or diverges. If it converges, state what it converges to n ( ) n n n= n= 5 n e n n= n= n(n + ) 3 n(n + ) 48

157 n= n= n= (n )(n + ) ( ) n ln n + n + n (n + ) ( + ) ( ) + 9 n= n ( ) n sin ( ) ( n(n + ) + 5 ) 4 n n= In Eercises 46 49, find the values of for which the series converges. n n n= ( + 3) n n= n= 4 n n n ( + ) n n= 50. Show the series n= n (n )(n + ) diverges. 483

158 Chapter 9 Sequences and Series 9.3 The Integral Test Knowing whether or not a series converges is very important, especially when we discuss Power Series in Sec on 9.8. Theorem 64 gives criteria for when Geometric series converge and Theorem 67 gives a quick test to determine if a series diverges. There are many important series whose convergence cannot be determined by these theorems, though, so we introduce a set of tests that allow us to handle a broad range of series. We start with the Integral Test. Integral Test We stated in Sec on 9. that a sequence {a n } is a func on a(n) whose domain is N, the set of natural numbers. If we can etend a(n) to have the domain of R, the real numbers, and it is both posi ve and decreasing on [, ), then the convergence of a n is the same as a() d. n= Theorem 69 Integral Test Let a sequence {a n } be defined by a n = a(n), where a(n) is con nuous, posi ve and decreasing on [, ). Then a n converges, if, and only if,. If. If a() d converges. In other words: a() d is convergent, then n= n= a() d is divergent, then n= a n is convergent. a n is divergent. Note: Theorem 69 does not state that the integral and the summa on have the same value. Note that it is not necessary to start the series or the integral at n =. We may use any interval [n, ] on which a(n) is con nuous, posi ve and decreasing. Also the sequence {a n } does not have to be strictly decreasing. It must be ul mately decreasing which means it is decreasing for all n larger than some number N. We can demonstrate the truth of the Integral Test with two simple graphs. In Figure 9.5(a), the height of each rectangle is a(n) = a n for n =,,..., 484

159 9.3 The Integral Test and clearly the rectangles enclose more area than the area under y = a(). Therefore we can conclude that a() d < a n. (9.) n= In Figure 9.5(b), we draw rectangles under y = a() with the Right-Hand rule, star ng with n =. This me, the area of the rectangles is less than the area under y = a(), so a n < a() d. Note how this summa on starts n= with n = ; adding a to both sides lets us rewrite the summa on star ng with n = : a n < a + a() d. (9.) n= n= Combining Equa ons (9.) and (9.), we have a n < a + a() d < a + a n. (9.3) From Equa on (9.3) we can make the following two statements:. If a n diverges, so does a() d (because a n < a + n= n= n= a() d) y y y = a() (a) y = a(). If a n converges, so does n= a() d (because a() d < a n.) Therefore the series and integral either both converge or both diverge. Theorem 68 allows us to etend this theorem to series where a(n) is posi ve and decreasing on [b, ) for some b >. n= (b) Watch the video: Integral Test for Series: Why It Works at Figure 9.5: Illustra ng the truth of the Integral Test. Eample Using the Integral Test ln n Determine the convergence of n. (The terms of the sequence {a n} = n= {ln n/n } and the n th par al sums are given in Figure 9.6.) 485

160 Chapter 9 Sequences and Series y a n Figure 9.6: Plo ng the sequence and series in Eample. S n n S Figure 9.6 implies that a(n) = (ln n)/n is posi ve and decreasing on [, ). We can determine this analy cally, too. We know a(n) is posi ve as both ln n and n are posi ve on [, ). To determine that a(n) is decreasing, consider a (n) = ( ln n)/n 3, which is nega ve for n. Since a (n) is nega ve, a(n) is decreasing. ln Applying the Integral Test, we test the convergence of d. Integrating this improper integral requires the use of Integra on by Parts, with u = ln and dv = / d. ln d = lim t t ( = lim t ln d ln t t ) + d ( = lim t ln ) t ( = lim t t ln t ). Apply L Hôpital s Rule: t = 0 lim t t = ln Since d converges, so does ln n n. n= 486

161 9.3 The Integral Test p Series Another important type of series is the p-series. Defini on 37 p Series, General p Series. A p series is a series of the form n= n p.. A general p series is a series of the form n= (an + b) p, where a and b are real numbers, and an + b 0 for all n. Like geometric series, one of the nice things about p series is that they have easy to determine convergence proper es. Theorem 70 Convergence of General p Series Assume a and b are real numbers and an + b 0 for all n. A general p series will converge if and only if, p >. (an + b) p n= Proof Consider the integral t d = lim (a + b) p t d; assuming p, (a + b) p (a + b) p d = lim t a( p) (a + t b) p ( = lim (at + b) p (a + b) p). t a( p) 487

162 Chapter 9 Sequences and Series This limit converges if and only if, p >. It is easy to show that the integral also diverges in the case of p =. (This result is similar to the work preceding Key Idea 3.) Therefore converges if, and only if, p >. (an + b) p n= Eample Determining convergence of series Determine the convergence of the following series.. n= n 3. n= n 5. n= ( n 5)3. n= n 4. ( ) n n= n 6. n= n S. This is a p series with p =. By Theorem 70, this series diverges. This series is a famous series, called the Harmonic Series, so named because of its rela onship to harmonics in the study of music and sound.. This is a p series with p =. By Theorem 70, it converges. Note that the theorem does not give a formula by which we can determine what the series converges to; we just know it converges. A famous, unepected result is that this series converges to π /6. 3. This is a p series with p = /; the theorem states that it diverges. 4. This is not a p series; the defini on does not allow for alterna ng signs. Therefore we cannot apply Theorem 70. We will consider this series again in Sec on 9.5. (Another famous result states that this series, the Alternating Harmonic Series, converges to ln.) 5. This is a general p series with p = 3, therefore it converges. 6. This is not a p series, but a geometric series with r = /. It converges. In the net sec on we consider two more convergence tests, both comparison tests. That is, we determine the convergence of one series by comparing it to another series with known convergence. 488

163 Eercises 9.3 Terms and Concepts. In order to apply the Integral Test to a sequence {a n}, the func on a(n) = a n must be, and.. T/F: The Integral Test can be used to determine the sum of a convergent series. Problems In Eercises 3 0, use the Integral Test to determine the convergence of the given series. 3. n n= 4. n 4 n= 5. n n + n= 6. n ln n n= 7. n + n= n= n= n= n(ln n) n n ln n n 3 In Eercises 4, find the value(s) of p for which the series is convergent i= n(ln n) p n( + n ) p i= i= i=3 ln n n p n ln n[ln(ln n)] p 489

164 Chapter 9 Sequences and Series 9.4 Comparison Tests In this sec on we will be comparing a given series with series that we know either converge or diverge. Theorem 7 Direct Comparison Test Let {a n } and {b n } be posi ve sequences where a n b n for all n N, for some N.. If b n converges, then a n converges. n= n= n=. If a n diverges, then b n diverges. n= Note: A sequence {a n } is a posi ve sequence if a n > 0 for all n. Because of Theorem 68, any theorem that relies on a posi ve sequence s ll holds true when a n > 0 for all but a finite number of values of n. Proof First consider the par al sums of each series. S n = n a i and T n = i= Since both series have posi ve terms we know that n i= b i S n S n + a n+ = n n+ a i + a n+ = a i = S n+ i= i= and T n T n + b n+ = n n+ b i + b n+ = b i = T n+ i= Therefore, both of the sequences of par al sums,{s n } and {T n }, are increasing. We also know that because a n b n for all n N that we must have S n T n for all n N. For the first part, assume that b n converges. Since b n 0 we know that n= i= T n = n b i i= i= b i 490

165 9.4 Comparison Tests From above we know that S n T n for all n N so we also have S n i= b i Because b i converges it must have a finite value and {S n } is bounded above. i= We also showed that {S n } is increasing so by Theorem 63 we know {S n } converges and so a n converges. n= For the second part, assume that a n diverges. Because a n 0 we must have n= lim S n =. We also know that for all n, S n T n and so we also know that n lim T n =. Therefore, T n is a divergent sequence and so b n diverges. n i= Watch the video: Direct Comparison Test / Limit Comparison Test for Series Basic Info at Eample Applying the Direct Comparison Test Determine the convergence of 3 n + n. n= S This series is neither a geometric or p-series, but seems related. We predict it will converge, so we look for a series with larger terms that converges. (Note too that the Integral Test seems difficult to apply here.) Since 3 n < 3 n + n, 3 n > 3 n + n for all n. The series 3 n is a convergent geometric series; by Theorem 7, n= 3 n + n converges. n= 49

166 Chapter 9 Sequences and Series Eample Applying the Direct Comparison Test n 3 Determine the convergence of n 4. n= S We know the Harmonic Series diverges, and it seems n n= that the given series is closely related to it, hence we predict it will diverge. We have n 3 n 4 > n3 n 4 = n The Harmonic Series, n, diverges, so we conclude that n 3 n 4 diverges as well. n= for all n. n= The concept of direct comparison is powerful and o en rela vely easy to apply. Prac ce helps one develop the necessary intui on to quickly pick a proper series with which to compare. However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test. Consider n= n 3 n 4. It is very similar to the divergent series given in E- + ample. We suspect that it also diverges, as n n 4 for large n. However, the inequality that we naturally want to use goes the wrong way : since + n 3 n 4 + < n3 n 4 = for all n. The given series has terms less than the terms n of a divergent series, and we cannot conclude anything from this. Fortunately, we can apply another test to the given series to determine its convergence. n3 Limit Comparison Test 49

167 9.4 Comparison Tests Theorem 7 Limit Comparison Test Let {a n } and {b n } be posi ve sequences. a n. If lim = L, where L is a posi ve real number, then n b n b n either both converge or both diverge. n= a n. If lim = 0, then if n b n a n 3. If lim =, then if n b n n= a n and n= b n converges, then so does a n. n= n= b n diverges, then so does a n. n= Proof. We have 0 < L < so we can find two posi ve numbers, m and M such a n that m < L < M. Because L = lim we know that for large enough n n b n the quo ent an b n must be close to L. So there must be a posi ve integer N such that if n > N we also have m < a n < M. Mul ply by b n and we have b n mb n < a n < Mb n for n > N. If b n diverges, then so does mb n. Also since mb n < a n for sufficiently large n, by the Comparison Test n= n= also diverges. Similarly, if b n converges, then so does Mb n. Since a n < Mb n for sufficiently large n, by the Comparison Test n= n= a n also converges. n=. Since lim a n = 0, there is a number N > 0 such that n a n 0 b n < for all n > N a n < b n since a n and b n are posi ve n= a n 493

168 Chapter 9 Sequences and Series Now since b n converges, a n converges by the Comparison Test. n= n= 3. Since lim n a n =, there is a number N > 0 such that n= a n b n > for all n > N a n > b n for all n > N Now since b n diverges, a n diverges by the Comparison Test. n= Theorem 7 is most useful when the convergence of the series from {b n } is known and we are trying to determine the convergence of the series from {a n }. We use the Limit Comparison Test in the net eample to eamine the series n 3 n 4 which mo vated this new test. + n= Eample 3 Applying the Limit Comparison Test n 3 Determine the convergence of n 4 using the Limit Comparison Test. + S Harmonic Sequence n= We compare the terms of n= n : n 3 /(n 4 + ) lim = lim n /n n =. n= n 4 n 4 + = lim n Since the Harmonic Series diverges, we conclude that well. n 3 n 4 to the terms of the + + /n 4 n= n 3 n 4 diverges as + Eample 4 Applying the Limit Comparison Test Determine the convergence of 3 n n n= 494

169 9.4 Comparison Tests S This series is similar to the one in Eample, but now we are considering 3 n n instead of 3 n + n. This difference makes applying the Direct Comparison Test difficult. Instead, we use the Limit Comparison Test and compare with the series 3 n : n= We know as well. /(3 n n ) 3 n lim n /3 n = lim n 3 n n n= L H = L H = lim n L H = (ln 3) 3 n (ln 3) 3 n lim n (ln 3) 3 3 n lim n (ln 3) 3 3 n =. ln 3 3 n ln 3 3 n n 3 n is a convergent geometric series, hence 3 n n converges As men oned before, prac ce helps one develop the intui on to quickly choose a series with which to compare. A general rule of thumb is to pick a series based on the dominant term in the epression of {a n }. It is also helpful to note that factorials dominate eponen als, which dominate algebraic func- ons (e.g., polynomials), which dominate logarithms. In the previous eample, the dominant term of 3 n n was 3n, so we compared the series to. It is 3n n= hard to apply the Limit Comparison Test to series containing factorials, though, as we have not learned how to apply L Hôpital s Rule to n!. n= Eample 5 Applying the Limit Comparison Test n + 3 Determine the convergence of n n +. n= S We naïvely a empt to apply the rule of thumb given above and note that the dominant term in the epression of the series is /n. Knowing that converges, we a empt to apply the Limit Comparison Test: n n= ( n + 3)/(n n + ) n ( n + 3) lim n /n = lim n n n + = (Apply L Hôpital s Rule). 495

170 Chapter 9 Sequences and Series Theorem 7 part (3) only applies when b n diverges; in our case, it converges. Ul mately, our test has not revealed anything about the convergence of our series. The problem is that we chose a poor series with which to compare. Since the numerator and denominator of the terms of the series are both algebraic func ons, we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator. The dominant term of the numerator is n / and the dominant term of the denominator is n. Thus we should compare the terms of the given series to n / /n = /n 3/ : Since the p-series n converges, we conclude that n + 3 3/ n n + converges as well. n= ( n + 3)/(n n + ) n 3/ ( n + 3) lim = lim n /n 3/ n n n + n= = (Apply L Hôpital s Rule). The tests we have encountered so far has required that we analyze series from posi ve sequences (the absolute value of the ra o and the root tests converts the sequence into a posi ve sequence). The net sec on relaes this restric on by considering alterna ng series, where the underlying sequence has terms that alternate between being posi ve and nega ve. n= 496

171 Eercises 9.4 Terms and Concepts. Suppose a n is convergent, and there are sequences {b n} and {c n} such that b n a n c n for all n. What can be said about the series b n and c n? Problems In Eercises 6, use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison n= n= n= n= n= n + 3n 5 4 n + n n ln n n n! + n n In Eercises 7, use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison n= n= n=4 n= n= n 3n n n ln n n 3 n + n n + n In Eercises 4, use the Direct Comparison Test or the Limit Comparison Test to determine the convergence of the given series. State which series is used for comparison n=5 n= n= n n + n + n 3 5 n 0 n + 0n sin ( /n ) n= n= n= n= n= n= n= n= n 5 n + 0 n + 5 n 3 5 n n 4 + n n4 n n n n + 00 n ln n n + 3 n + 7 n= n= + sin n 0 n 5. Given that a n converges, state which of the following n= series converges, may converge, or does not converge. a n (a) n (b) (c) (d) (e) n= a na n+ n= (a n) n= na n n= a n n= In Eercises 6 33, determine the convergence of the given series. State the test used; more than one test may be appropriate n= n= n= n n (n + 5) 3 n! 0 n 497

172 n= n= n= ln n n! 3 n + n n 0n n= n= 3 n n 3 cos(/n) n 498

173 9.5 Alterna ng Series and Absolute Convergence The series convergence tests we have used require that the underlying sequence {a n } be a posi ve sequence. (We can rela this with Theorem 68 and state that there must be an N > 0 such that a n > 0 for all n > N; that is, {a n } is posi ve for all but a finite number of values of n.) In this sec on we eplore series whose summa on includes nega ve terms. We start with a very specific form of series, where the terms of the summa on alternate between being posi ve and nega ve. 9.5 Alterna ng Series and Absolute Convergence Defini on 38 Alterna ng Series Let {b n } be a posi ve sequence. An alterna ng series is a series of either the form ( ) n b n n= or ( ) n+ b n. n= We want to think that an alterna ng sequence {a n } is related to a posi ve sequence {b n } by a n = ( ) n b n. Recall the terms of Harmonic Series come from the Harmonic Sequence {b n } = {/n}. An important alterna ng series is the Alterna ng Harmonic Series: ( ) n+ n = n= Geometric Series can also be alterna ng series when r < 0. For instance, if r = /, the geometric series is ( ) n = Theorem 64 states that geometric series converge when r < and gives the sum: r n =. When r = / as above, we find r ( ) n = ( /) = 3/ = 3. A powerful convergence theorem eists for other alterna ng series that meet a few condi ons. 499

174 Chapter 9 Sequences and Series Theorem 73 Alterna ng Series Test Let {b n } be a posi ve, decreasing sequence where lim b n = 0. Then n converge. ( ) n b n n= and ( ) n+ b n n= The basic idea behind Theorem 73 is illustrated in Figure 9.7. A posi ve, decreasing sequence {b n } is shown along with the par al sums S n = n ( ) i+ b i = b b + b 3 b ( ) n+ b n. i= L 0.5 y b n Figure 9.7: Illustra ng convergence with the Alterna ng Series Test. S n n Because {b n } is decreasing, the amount by which S n bounces up and down decreases. Moreover, the odd terms of S n form a decreasing, bounded sequence, while the even terms of S n form an increasing, bounded sequence. Since bounded, monotonic sequences converge (see Theorem 63) and the terms of {b n } approach 0, we will show below the odd and even terms of S n converge to the same common limit L, the sum of the series. Proof Because {b n } is a decreasing sequence, we have b n b n+ 0. We will consider the even and odd par al sums separately. First consider the even par al sums. S = b b 0 since b b S 4 = b b + b 3 b 4 = S + b 3 b 4 S since b 3 b 4 0 S 6 = S 4 + b 5 b 6 S 4 since b 5 b 6 0. S n = S n + b n b n S n since b n b n 0 We now have 0 S S 4 S 6 S n so {S n } is an increasing sequence. But we can also write S n = b b + b 3 b 4 + b 5 b n + b n b n = b (b b 3 ) (b 4 b 5 ) (b n b n ) b n 500

175 9.5 Alterna ng Series and Absolute Convergence Each term in parentheses is posi ve and b n is posi ve so we have S n b for all n. We now have the sequence of even par al sums, {S n }, is increasing and bounded above so by Theorem 63 {S n } converges. Since we know it converges, we will assume it s limit is L or lim S n = L n Net we determine the limit of the sequence of odd par al sums. lim S n+ = lim (S n + b n+ ) n n = lim n S n + lim n b n+ = L + 0 = L Both the even and odd par al sums converge to s so we have lim S n = L, which n means the series is convergent. Watch the video: Alterna ng Series Another Eample 4 at Eample Applying the Alterna ng Series Test Determine if the Alterna ng Series Test applies to each of the following series.. ( ) n+ n n=. n= ( ) n ln n n 3. n+ sin n ( ) n= n S. This is the Alterna ng Harmonic Series as seen previously. The underlying sequence is {b n } = {/n}, which is posi ve, decreasing, and approaches 0 as n. Therefore we can apply the Alterna ng Series Test and conclude this series converges. While the test does not state what the series converges to, we will see later that ( ) n+ = ln. n n= 50

176 Chapter 9 Sequences and Series. The underlying sequence is {b n } = {ln n/n}. This is posi ve for n and ln n lim n n = lim = 0 (use L Hôpital s Rule). However, the sequence n n is not decreasing for all n. It is straigh orward to compute b 0.347, b 0.366, and b : the sequence is increasing for at least the first terms. We do not immediately conclude that we cannot apply the Alterna ng Series Test. Rather, consider the long term behavior of {b n }. Trea ng b n = b(n) as a con nuous func on of n defined on [, ), we can take its deriva ve: b (n) = ln n n. The deriva ve is nega ve for all n 3 (actually, for all n > e), meaning b(n) = b n is decreasing on [3, ). We can apply the Alterna ng Series Test to the series when we start with n = 3 and conclude that ( ) n ln n converges; adding the terms with n = does not change n n=3 the convergence (i.e., we apply Theorem 68). The important lesson here is that as before, if a series fails to meet the criteria of the Alterna ng Series Test on only a finite number of terms, we can s ll apply the test. 3. The underlying sequence is {b n } = { sin n /n }. This sequence is posi ve and approaches 0 as n. However, it is not a decreasing sequence; the value of sin n oscillates between 0 and as n. We cannot remove a finite number of terms to make {b n } decreasing, therefore we cannot apply the Alterna ng Series Test. Keep in mind that this does not mean we conclude the series diverges; in fact, it does converge. We are just unable to conclude this based on Theorem 73. One of the famous results of mathema cs is that the Harmonic Series, n n= diverges, yet the Alterna ng Harmonic Series, ( ) n+, converges. The n n= no on that alterna ng the signs of the terms in a series can make a series converge leads us to the following defini ons. 50

177 9.5 Alterna ng Series and Absolute Convergence Defini on 39 Absolute and Condi onal Convergence. A series a n converges absolutely if a n converges. n= n=. A series a n converges condi onally if a n converges but n= n= a n diverges. n= Thus we say the Alterna ng Harmonic Series converges condi onally. Eample Determining absolute and condi onal convergence. Determine if the following series converge absolutely, condi onally, or diverge.. ( ) n n + 3 n + n + 5 n= 3. ( ) n 3n 3 5n 0. ( ) n n + n + 5 n=3 n n= Note: In Defini on 39, a n is not n= necessarily an alterna ng series; it just may have some nega ve terms. S. We can show the series n + 3 ( )n n + n + 5 = n + 3 n + n + 5 n= diverges using the Limit Comparison Test, comparing with /n. The series ( ) n n + 3 n converges using the Alterna ng Series + n + 5 n= Test; we conclude it converges condi onally. n=. We can show the series ( )n n + n + 5 = n= n n= n + n + 5 n converges using the Ra o Test. Therefore we conclude ( ) n n + n + 5 n converges absolutely. n= 503

178 Chapter 9 Sequences and Series 3. The series 3n 3 ( )n 5n 0 = 3n 3 5n 0 n=3 diverges using the Test for Divergence, so it does not converge absolutely. The series ( ) n 3n 3 fails the condi ons of the Alterna ng Series 5n 0 n=3 Test as (3n 3)/(5n 0) does not approach 0 as n. We can state further that this series diverges; as n, the series effec vely adds and subtracts 3/5 over and over. This causes the sequence of par al sums to oscillate and not converge. Therefore the series ( ) n 3n 3 5n 0 diverges. n= Knowing that a series converges absolutely allows us to make two important statements, given in the following theorem. The first is that absolute convergence is stronger than regular convergence. That is, just because a n n= converges, we cannot conclude that a n will converge, but knowing a series converges absolutely tells us that n= n=3 a n will converge. n= Theorem 74 Absolute Convergence Theorem Let a n be a series that converges absolutely. n=. a n converges. n=. Let {b n } be any rearrangement of the sequence {a n }. Then b n = a n. n= n= One reason this is important is that our convergence tests all require that the underlying sequence of terms be posi ve. By taking the absolute value of the terms of a series where not all terms are posi ve, we are o en able to apply an 504

179 9.5 Alterna ng Series and Absolute Convergence appropriate test and determine absolute convergence. This, in turn, determines that the series we are given also converges. The second statement relates to rearrangements of series. When dealing with a finite set of numbers, the sum of the numbers does not depend on the order which they are added. (So ++3 = 3++.) One may be surprised to find out that when dealing with an infinite set of numbers, the same statement does not always hold true: some infinite lists of numbers may be rearranged in different orders to achieve different sums. The theorem states that the terms of an absolutely convergent series can be rearranged in any way without affec ng the sum. In Eample, we determined the series in part converges absolutely. Theorem 74 tells us the series converges (which we could also determine using the Alterna ng Series Test). The theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. This implies that perhaps the sum of a condi onally convergent series can change based on the arrangement of terms. Indeed, it can. The Riemann Rearrangement Theorem (named a er Bernhard Riemann) states that any condi onally convergent series can have its terms rearranged so that the sum is any desired value or infinity. Before we consider an eample, we state the following theorem that illustrates how the alterna ng structure of an alterna ng series is a powerful tool when approima ng the sum of a convergent series. Theorem 75 The Alterna ng Series Approima on Theorem Let {b n } be a sequence that sa sfies the hypotheses of the Alterna ng Series Test, and let S n and L be the n th par al sums and sum, respec- vely, of either ( ) n b n or ( ) n+ b n. Then n= n=. S n L < b n+, and. L is between S n and S n+. Part of Theorem 75 states that the n th par al sum of a convergent alterna ng series will be within b n+ of its total sum. Consider the alterna ng series we looked at before the statement of the theorem,. ( ) n+ Since b 4 = / , we know that S 3 is within of the total sum. Moreover, Part of the theorem states that since S and S , we know the sum L lies between 0.80 and One use of this is n= n 505

180 Chapter 9 Sequences and Series the knowledge that S 4 is accurate to two places a er the decimal. Some alterna ng series converge slowly. In Eample we determined the n+ ln n series ( ) converged. With n = 00, we find ln n/n , n n= meaning that S is accurate to one, maybe two, places a er the decimal. Since S , we know the sum L is L Eample 3 Approima ng the sum of convergent alterna ng series Approimate the sum of the following series, accurate to within ( ) n+ n 3. n= n+ ln n ( ) n. n= S. Using Theorem 75, we want to find n where /n 3 < 0.00: n = 000 n n n 0. Let L be the sum of this series. By Part of the theorem, S 9 L < b 0 = /000. We can compute S 9 = 0.906, which our theorem states is within 0.00 of the total sum. We can use Part of the theorem to obtain an even more accurate result. As we know the 0 th term of the series is /000, we can easily compute S 0 = Part of the theorem states that L is between S 9 and S 0, so < L < We want to find n where ln(n)/n < We start by solving ln(n)/n = 0.00 for n. This cannot be solved algebraically, so we will use Newton s Method to approimate a solu on. Let f() = ln()/ 0.00; we want to know where f() = 0. We make a guess that must be large, so our ini al guess will be = 000. Recall how Newton s Method works: given an approimate solu on n, our net approima on n+ is given by n+ = n f( n) f ( n ). 506

181 9.5 Alterna ng Series and Absolute Convergence We find f () = ( ln() ) /. This gives = 000 = 000. ln(000)/ ( ln(000) ) /000 Using a computer, we find that Newton s Method seems to converge to a solu on = 98.0 a er 8 itera ons. Taking the net integer higher, we have n = 99, where ln(99)/99 = < Again using a computer, we find S 98 = Part of the theorem states that this is within 0.00 of the actual sum L. Already knowing the 9,9 th term, we can compute S 99 = , meaning < L < No ce how the first series converged quite quickly, where we needed only 0 terms to reach the desired accuracy, whereas the second series took over 9,000 terms. We now consider the Alterna ng Harmonic Series once more. We have stated that ( ) n+ n = = ln, 7 n= (see Eample ). Consider the rearrangement where every posi ve term is followed by two nega ve terms: (Convince yourself that these are eactly the same numbers as appear in the Alterna ng Harmonic Series, just in a different order.) Now group some terms and simplify: ( ) ( ) ( ) 0 + = = ( ) + = ln. By rearranging the terms of the series, we have arrived at a different sum. (One could try to argue that the Alterna ng Harmonic Series does not actually 507

182 Chapter 9 Sequences and Series converge to ln, because rearranging the terms of the series shouldn t change the sum. However, the Alterna ng Series Test proves this series converges to L, for some number L, and if the rearrangement does not change the sum, then L = L/, implying L = 0. But the Alterna ng Series Approima on Theorem quickly shows that L > 0. The only conclusion is that the rearrangement did change the sum.) This is an incredible result. We men oned earlier that the Integral Test did not work well with series containing factorial terms. The net sec on introduces the Ra o Test, which does handle such series well. We also introduce the Root Test, which is good for series where each term is raised to a power. 508

183 Eercises 9.5 Terms and Concepts. Why is sin n not an alterna ng series? n=. A series ( ) n a n converges when {a n} is, n= and lim n an =. 3. Give an eample of a series where a n converges but a n does not. 4. The sum of a convergent series can be changed by rearranging the order of its terms. Problems In Eercises 5 0, an alterna ng series a n is given. n=i (a) Determine if the series converges or diverges. (b) Determine if a n converges or diverges. (c) If a n converges, determine if the convergence is condi onal or absolute. ( ) n+ n= n= n ( ) n+ n! ( ) n n + 5 3n 5 ( ) n n n= n ( ) n+ 3n + 5 n 3n + n= ( ) n ln n + ( ) n n ln n n= n= ( ) n (n ) cos ( πn ) n= 4. sin ( (n + /)π ) n ln n n= 5. ( ) n ( e) n ( ) n n n! ( ) n n n= ( ) n n ( 000) n n= n! Let S n be the n th par al sum of a series. In Eercises 4, a convergent alterna ng series is given and a value of n. Compute S n and S n+ and use these values to find bounds on the sum of the series n= ( ) n ln(n + ), n = 5 ( ) n+, n = 4 n 4 n= ( ) n, n! n = 6 ( ) n, n = 9 In Eercises 5 8, a convergent alterna ng series is given along with its sum and a value of ε. Use Theorem 75 to find n such that the n th par al sum of the series is within ε of the sum of the series ( ) n+ n= n 4 ( ) n n! = 7π4 70, ε = 0.00 = e, ε = ( ) n n + = π 4, ε = 0.00 ( ) n (n)! = cos, ε =

184 Chapter 9 Sequences and Series 9.6 Ra o and Root Tests Theorem 67 states that if a series n= a n converges, then lim n a n = 0. That is, the terms of {a n } must get very small. Not only must the terms approach 0, they must approach 0 fast enough : while lim /n = 0, the Harmonic Series n diverges as the terms of {/n} do not approach 0 fast enough. n n= The comparison tests of Sec on 9.4 determine convergence by comparing terms of a series to terms of another series whose convergence is known. This sec on introduces the Ra o and Root Tests, which determine convergence by analyzing the terms of a series to see if they approach 0 fast enough. Ra o Test Theorem 76 Ra o Test Let {a n } be a sequence where lim a n+ n a n = L.. If L <, then a n converges. n=. If L > or L =, then a n diverges. n= 3. If L =, the Ra o Test is inconclusive. The principle of the Ra o Test is this: if lim a n+ n a n = L <, then for large n, each term of {a n } is significantly smaller than its previous term which is enough to ensure convergence. A full proof can be found at math.lamar.edu/classes/calcii/ratiotest.asp. Watch the video: Using the Ra o Test to Determine if a Series Converges # at 50

185 9.6 Ra o and Root Tests Eample Applying the Ra o Test Use the Ra o Test to determine the convergence of the following series:. n= n n!. n= 3 n n 3 3. n= n +. S n. n! : n+ /(n + )! n+ n! lim n= n n = lim /n! n n (n + )! = lim n n + = Since the limit is 0 <, by the Ra o Test n= 3 n n 3 : n= n n! converges. 3 n+ /(n + ) 3 3 n+ n 3 lim n 3 n /n 3 = lim n 3 n (n + ) 3 3n 3 = lim n (n + ) 3 = 3. Since the limit is 3 >, by the Ra o Test n= n + : n= 3 n n 3 diverges. / ( (n + ) + ) n + lim n /(n = lim + ) n (n + ) + =. Since the limit is, the Ra o Test is inconclusive. We can easily show this series converges using the Direct or Limit Comparison Tests, with each comparing to the series n. n= The Ra o Test is not effec ve when the terms of a series only contain algebraic func ons (e.g., polynomials). It is most effec ve when the terms contain some factorials or eponen als. The previous eample also reinforces our developing intui on: factorials dominate eponen als, which dominate algebraic func ons, which dominate logarithmic func ons. In Part of the eample, 5

186 Chapter 9 Sequences and Series the factorial in the denominator dominated the eponen al in the numerator, causing the series to converge. In Part, the eponen al in the numerator dominated the algebraic func on in the denominator, causing the series to diverge. While we have used factorials in previous sec ons, we have not eplored them closely and one is likely to not yet have a strong intui ve sense for how they behave. The following eample gives more prac ce with factorials. Eample Applying the Ra o Test n!n! Determine the convergence of (n)!. n= S Before we begin, be sure to note the difference between (n)! and n!. When n = 4, the former is 8! = = 40, 30, whereas the la er is (4 3 ) = 48. Applying the Ra o Test: (n + )!(n + )!/ ( (n + ) )! (n + )!(n + )!(n)! lim = lim n n!n!/(n)! n n!n!(n + )! No ng that (n + )! = (n + ) (n + ) (n)!, we have (n + )(n + ) = lim n (n + )(n + ) = /4. Since the limit is /4 <, by the Ra o Test we conclude n= n!n! (n)! converges. Root Test The final test we introduce is the Root Test, which works par cularly well on series where each term is raised to a power, and does not work well with terms containing factorials. 5

187 9.6 Ra o and Root Tests Theorem 77 Root Test Let {a n } be a sequence where lim a n /n = L. n. If L <, then a n converges. n=. If L > or L =, then a n diverges. n= 3. If L =, the Root Test is inconclusive. Eample 3 Applying the Root Test Determine the convergence of the following series using the Root Test:. n= ( ) n 3n +. 5n n= n 4 (ln n) n 3. n= n n. S (( ) n ) /n 3n + 3n +. lim = lim n 5n n 5n = 3 5. Since the limit is less than, we conclude the series converges. Note: it is difficult to apply the Ra o Test to this series. ( ) n 4 /n ( ) n /n 4. lim n (ln n) n = lim n ln n. As n grows, the numerator approaches (apply L Hôpital s Rule) and the denominator grows to infinity. Thus lim n ( ) n /n 4 = 0. ln n Since the limit is less than, we conclude the series converges. 3. lim n ( n n ) /n = lim ( ) n n /n =. Since this is greater than, we conclude the series diverges. 53

188 Chapter 9 Sequences and Series We end here our study of tests to determine convergence. The net sec- on of this tet provides strategies for tes ng series, while the back of the book contains a table summarizing the tests that one may find useful. While series are worthy of study in and of themselves, our ul mate goal within calculus is the study of Power Series, which we will consider in Sec on 9.8. We will use power series to create func ons where the output is the result of an infinite summa on. 54

189 Eercises 9.6 Terms and Concepts. The Ra o Test is not effec ve when the terms of a sequence only contain func ons.. The Ra o Test is most effec ve when the terms of a sequence contains and/or func ons. 3. What three convergence tests do not work well with terms containing factorials? 4. The Root Test works par cularly well on series where each term is to a. Problems In Eercises 5 6, determine the convergence of the given series using the Ra o Test. If the Ra o Test is inconclusive, state so and determine convergence with another test n= n= n= n n! 5 n 3n 4 n n!0 n (n)! 5 n + n 4 7 n + n n 3n n 7 n 3 ( 3 n 5 n= n= n= ) n n n n= n! (5n) e n n! n= n= e /n n 3 In Eercises 7 6, determine the convergence of the given series using the Root Test. If the Root Test is inconclusive, state so and determine convergence with another test. ( ) n n n + n= ( ).9n n n 3 8. n + n n= n= n= n= n= n n 3 n n n 3 n n n+ 4 n+7 7 n ( ) n n n n + n ( n ) n n n= n= n= n= ( ln n ) n n ( ln n ) n 55

190 Chapter 9 Sequences and Series 9.7 Strategy for tes ng series We have now covered all of the tests for determining the convergence or divergence of a series, which we summarize here. Because more than one test may apply to a given series, you should always go completely through the guidelines and iden fy all possible tests that you can use. Once you ve done this, you can iden fy the test that will be the easiest for you to use.. With a quick glance does it look like the series terms don t converge to zero in the limit, i.e. does lim a n 0? If so, use the Test for Divergence. n Note that you should only use the Test for Divergence if a quick glance suggests that the series terms may not converge to zero in the limit.. Is the series a p-series ( n p ) or a geometric series ( ar n )? If so, use the fact that p-series will converge only if p > and a geometric series will only converge if r <. Remember as well that o en some algebraic manipula on is required to get a geometric series into the correct form. 3. Is the series similar to a p-series or a geometric series? If so, try the Comparison Test. 4. Is the series a ra onal epression involving only polynomials or polynomials under radicals? If so, try the Comparison test or the Limit Comparison Test. Remember however, that in order to use the Comparison Test and the Limit Comparison Test the series terms all need to be posi ve. 5. Is the series of the form ( ) n a n? If so, then the Alterna ng Series Test may work. 6. Does the series contain factorials or constants raised to powers involving n? If so, then the Ra o Test may work. Note that if the series term contains a factorial then the only test that we have that will work is the Ra o Test. 7. Can the series terms be wri en in the form a n = (b n ) n? If so, then the Root Test may work. 8. If a n = f(n) for some posi ve, decreasing func on and to evaluate then the Integral Test may work. a f() d is easy Again, remember that these are only a set of guidelines and not a set of hard and fast rules to use when trying to determine the best test to use on a series. If more that one test can be used, try to use the test that will be the easiest for you to use. These guidelines are also summarized in a table in the back of the book. We now consider several eamples. 56

191 9.7 Strategy for tes ng series Eample Tes ng Series Determine whether the given series converges absolutely, converges condi onally, or diverges.. n= ( ) n n n + 3. n= n 3n 4n n + 3. n= e n (n + 3)! S. We see that this series is alterna ng so we use the alterna ng series test. The underlying sequence is {a n } = { n n +3 } which is posi ve and decreasing since a n (n) = 3 n (n +3) < 0 for n. We also see lim n n + 3 = 0 ( ) n n so by the Alterna ng Series Test n converges. We now determine if it converges absolutely. Consider the sequence ( ) n n n + 3 = + 3 n= n= n n + 3. We compare this series to n n = which is a divergent p-series. We also have so by the Comparison n n= n= n= n test, n=. lim n n n + 3 diverges. Therefore, n + 3 > n n = n ( ) n n n + 3 n= converges condi onally. n 3n 4n n + = 4 so by the Test for Divergence n= diverges. n 3n 4n n + 3. We see the factorial and use the Ra o Test. All terms of the series are 57

192 Chapter 9 Sequences and Series posi ve so we consider a n+ lim = lim n a n n e n+ (n+4)! e n (n+3)! e n+ (n + 3)! = lim n e n (n + 4)! = lim n = lim n e e n (n + 3)! e n (n + 4)(n + 3)! e n + 4 = 0 < e n So by the Ra o Test, converges. Because all of the series (n + 3)! n= terms are posi ve it converges absolutely. 58

193 Eercises 9.7 Problems In Eercises 37, determine whether the given series converges absolutely, converges condi onally, or diverges.. 3 n(n + )(n + 4) n= ( 3 n= n= 3 n+ n5 n n e n n= n= ) n n! ln(n + ) (n + 4)( ) n n= n= n= n= n= n= n + 4 n e n n e ( ) n 4 n sin( 4πn 3 ) n 4π/3 3 n n! (n + )! ( ) n n n + n= n= n= cos n n n 5n 3 + n 3 ( ) n n + n 4 + n= n= n= n= (3n) n n 3n e n (n )! ln ( ) n ln n n n= n= n= n= n= n= n= n= n= n= n ( ) n ( ) n (n + 5) 3 n! ( 0) n ln n n! ( 3) n + n ( ) n (n ) 0n + 5 ( 3) n n 3 cos(/n) n ( ) n (n + 4n ) n 3 + 4n 3n + 7 n 4 ( 4) n n= n= n= n= n= n= n= n= n= n! n ( 3) n + n ( ) n n n + 4n + ( 3) n n n n!n!n! (3n)! ( ) n ln n ( ) n n + n + n 3 ( ln n ) n ( n ) n + 59

194 Chapter 9 Sequences and Series 9.8 Power Series So far, our study of series has eamined the ques on of Is the sum of these infinite terms finite?, i.e., Does the series converge? We now approach series from a different perspec ve: as a func on. Given a value of, we evaluate f() by finding the sum of a par cular series that depends on (assuming the series converges). We start this new approach to series with a defini on. Defini on 40 Power Series Let {a n } be a sequence, let be a variable, and let c be a real number.. The power series in is the series a n n = a 0 + a + a + a The power series in centered at c is the series a n ( c) n = a 0 + a ( c) + a ( c) + a 3 ( c) 3 + Eample Eamples of power series Write out the first five terms of the following power series:. n. n+ ( + )n ( ) n n= 3. n+ ( π)n ( ). (n)! S. One of the conven ons we adopt is that 0 = regardless of the value of. Therefore n = This is a geometric series in.. This series is centered at c =. Note how this series starts with n =. We could rewrite this series star ng at n = 0 with the understanding that 50

195 9.8 Power Series a 0 = 0, and hence the first term is 0. n+ ( + )n ( ) n n= ( + ) ( + )3 ( + )4 ( + )5 = (+) This series is centered at c = π. Recall that 0! =. n+ ( π)n ( ) (n)! = + ( π) ( π)4 4 + ( π)6 6! ( π)8 8! We introduced power series as a type of func on, where a value of is given and the sum of a series is returned. Of course, not every series converges. For instance, in part of Eample, we recognized the series n as a geometric series in. Theorem 64 states that this series converges only when <. This raises the ques on: For what values of will a given power series converge?, which leads us to a theorem and defini on. Theorem 78 Convergence of Power Series Let a power series a n ( c) n be given. Then one of the following is true:. The series converges only at = c.. There is an R > 0 such that the series converges for all in (c R, c + R) and diverges for all < c R and > c + R. 3. The series converges for all. The value of R is important when understanding a power series, hence it is given a name in the following defini on. Also, note that part of Theorem 78 makes a statement about the interval (c R, c + R), but the not the endpoints of that interval. A series may or may not converge at these endpoints. 5

196 Chapter 9 Sequences and Series Defini on 4 Radius and Interval of Convergence. The number R given in Theorem 78 is the radius of convergence of a given series. When a series converges for only = c, we say the radius of convergence is 0, i.e., R = 0. When a series converges for all, we say the series has an infinite radius of convergence, i.e., R =.. The interval of convergence is the set of all values of for which the series converges. To find the values of for which a given series converges, we will use the convergence tests we studied previously (especially the Ra o Test). However, the tests all required that the terms of a series be posi ve. The following theorem gives us a work around to this problem. Theorem 79 The Radius of Convergence of a Series and Absolute Convergence The series a n ( c) n and a n ( c) n have the same radius of convergence R. Theorem 79 allows us to find the radius of convergence R of a series by applying the Ra o Test (or any applicable test) to the absolute value of the terms of the series. We prac ce this in the following eample. Watch the video: Power Series Finding the Interval of Convergence at J-0 5

197 9.8 Power Series Eample Determining the radius and interval of convergence. Find the radius and interval of convergence for each of the following series: n n+ n.. ( ) 3. n ( 3) n 4. n! n n! n S n=. We apply the Ra o Test to the series lim n n+ /(n + )! n /n! n n! : = lim n+ n! n n (n + )! = lim n n + = 0 for all. The Ra o Test shows us that regardless of the choice of, the series converges. Therefore the radius of convergence is R =, and the interval of convergence is (, ).. We apply the Ra o Test to the series lim n n+ /(n + ) n /n n ( )n+ n = n= = lim n = lim n =. n+ n n n + n n + The Ra o Test states a series converges if the limit of a n+ /a n = L <. We found the limit above to be ; therefore, the power series converges when <, or when is in (, ). Thus the radius of convergence is R =. To determine the interval of convergence, we need to check the endpoints of (, ). When =, we have the opposite of the Harmonic Series: n+ ( )n ( ) n+ ( ) = n n n= = n= n= =. n n= n n : 53

198 Chapter 9 Sequences and Series The series diverges when =. n+ ()n When =, we have the series ( ), which is the Alterna ng n n= Harmonic Series, which converges. Therefore the interval of convergence is (, ]. 3. We apply the Ra o Test to the series lim n n+ ( 3) n+ n ( 3) n n ( 3) n : = lim n n+ n = lim ( 3). n ( 3)n+ ( 3) n According to the Ra o Test, the series converges when ( 3) < = 3 < /. The series is centered at 3, and must be within / of 3 in order for the series to converge. Therefore the radius of convergence is R = /, and we know that the series converges absolutely for all in (3 /, 3 + /) = (.5, 3.5). We check for convergence at the endpoints to find the interval of convergence. When =.5, we have: n (.5 3) n = n ( /) n = ( ) n, which diverges. A similar process shows that the series also diverges at = 3.5. Therefore the interval of convergence is (.5, 3.5). 4. We apply the Ra o Test to lim n n! n : (n + )! n+ n! n = lim (n + ) n = for all, ecept = 0. The Ra o Test shows that the series diverges for all ecept = 0. Therefore the radius of convergence is R = 0. 54

199 9.8 Power Series Power Series as Func ons We can use a power series to define a func on: f() = a n n where the domain of f is a subset of the interval of convergence of the power series. One can apply calculus techniques to such func ons; in par cular, we can find deriva ves and an deriva ves. Theorem 80 Deriva ves and Indefinite Integrals of Power Series Func ons Let f() = a n ( c) n be a func on defined by a power series, with radius of convergence R.. f() is con nuous and differen able on (c R, c + R).. f () = 3. a n n ( c) n, with radius of convergence R. n= f() d = C + A few notes about Theorem 80: ( c) n+ a n, with radius of convergence R. n +. The theorem states that differen a on and integra on do not change the radius of convergence. It does not state anything about the interval of convergence. They are not always the same.. No ce how the summa on for f () starts with n =. This is because the constant term a 0 of f() goes to Differen a on and integra on are simply calculated term by term using previous rules of integra on and differen a on. Eample 3 Let f() = convergence. Deriva ves and indefinite integrals of power series n. Find the following along with their respec ve intervals of. f () and. F() = f() d 55

200 Chapter 9 Sequences and Series We find the deriva ve and indefinite integral of f(), follow- S ing Theorem 80.. f() = = f () = = In Eample, we recognized that n n= n n n is a geometric series in. We know that such a geometric series converges when < ; that is, the interval of convergence is (, ). To determine the interval of convergence of f (), we consider the endpoints of (, ). When = we have f ( ) = n( ) n which diverges by the Test for Divergence and when = we have n= f () = n which also diverges by the Test for Divergence. Therefore, the interval of convergence of f () is (, ).. f() = = n F() = n= f() d = C = C + n+ n + = C + To find the interval of convergence of F(), we again consider the endpoints of (, ). When = we have F( ) = C + ( ) n The value of C is irrelevant; no ce that the rest of the series is an Alterna ng Series that whose terms converge to 0. By the Alterna ng Series n= n n= n n 56

201 9.8 Power Series Test, this series converges. (In fact, we can recognize that the terms of the series a er C are the opposite of the Alterna ng Harmonic Series. We can thus say that F( ) = C ln.) F() = C + No ce that this summa on is C + the Harmonic Series, which diverges. Since F converges for = and diverges for =, the interval of convergence of F() is [, ). n= n The previous eample showed how to take the deriva ve and indefinite integral of a power series without mo va on for why we care about such opera- ons. We may care for the sheer mathema cal enjoyment that we can, which is mo va on enough for many. However, we would be remiss to not recognize that we can learn a great deal from taking deriva ves and indefinite integrals. Recall that f() = n in Eample 3 is a geometric series. According to Theorem 64, this series converges to /( ) when <. Thus we can say f() = n =, on (, ). Integra ng the power series, (as done in Eample 3,) we find F() = C + while integra ng the func on f() = /( ) gives Equa ng Equa ons (9.4) and (9.5), we have F() = C + n+ n +, (9.4) F() = ln + C. (9.5) n+ n + = ln + C. Le ng = 0, we have F(0) = C = C. This implies that we can drop the constants and conclude n+ = ln. n + 57

202 Chapter 9 Sequences and Series n+ We established in Eample 3 that the series converges at = ; n + subs tu ng = on both sides of the above equality gives = ln. 5 On the le we have the opposite of the Alterna ng Harmonic Series; on the right, we have ln. We conclude that = ln. 4 In Eample 9.5. of Sec on 9.5 we said the Alterna ng Harmonic Series converges to ln, but did not show why this was the case. The work above shows how we conclude that the Alterna ng Harmonic Series Converges to ln. We use this type of analysis in the net eample. Eample 4 Let f() = Analyzing power series func ons f() d, and use these to analyze the behavior of f(). n n!. Find f () and S We start by making two notes: first, in Eample, we found the interval of convergence of this power series is (, ). Second, we will find it useful later to have a few terms of the series wri en out: n n! = (9.6) We now find the deriva ve: f () = n n n! n= n = (n )! = + +! +. n= Since the series starts at n = and each term refers to (n ), we can re-inde the series star ng with n = 0: = = f(). n n! 58

203 9.8 Power Series We found the deriva ve of f() is f(). The only func ons for which this is true are of the form y = ce for some constant c. As f(0) = (see Equa on (9.6)), c must be. Therefore we conclude that for all. We can also find f() d: f() = f() d = C + = C + n n! = e n+ n!(n + ) n+ (n + )! We write out a few terms of this last series: n+ C + (n + )! = C The integral of f() differs from f() only by a constant, again indica ng that f() = e. Eample 4 and the work following Eample 3 established rela onships between a power series func on and regular func ons that we have dealt with in the past. In general, given a power series func on, it is difficult (if not impossible) to epress the func on in terms of elementary func ons. We chose eamples where things worked out nicely. Representa ons of Func ons with Power Series It can be difficult to recognize an elementary func on by its power series epansion. It is far easier to start with a known func on, epressed in terms of elementary func ons, and represent it as a power series func on. One may wonder why we would bother doing so, as the la er func on probably seems more complicated. Let s start off with a series we already know how to do, although when we first ran across this series we didn t think of it as a power series nor did we acknowledge that it represented a func on. Recall that the geometric series is ar n = a r provided r <. 59

204 Chapter 9 Sequences and Series We also know that if r the series diverges. Now, if we take a = and r = this becomes, n = provided < (9.7) Turning this around we can see that we can represent the func on with the power series f() = (9.8) n provided <. (9.9) This provision is important. We can clearly plug any number other than = into the func on, however, we will only get a convergent power series if <. This means the equality in Equa on (9.7) will only hold if <. For any other value of the equality won t hold. Note as well that we can also use this to acknowledge that the radius of convergence of this power series is R = and the interval of convergence is <. This idea of convergence is important here. We will be represen ng many func ons as power series and it will be important to recognize that the representa ons will o en only be valid for a range of s and that there may be values of that we can plug into the func on that we can t plug into the power series representa on. In this sec on we are going to concentrate on represen ng func ons with power series where the func on can be related back to a geometric series. In this way we will hopefully become familiar with some of the kinds of manipula ons that we will some mes need when working with power series. We will see in Sec on 9.0 that this strategy is useful for integra ng func ons that don t have elementary deriva ves. Eample 5 Finding a Power Series Find a power series representa on for g() = and determine its interval + 3 of convergence. S We want to relate this func on back to Equa on (9.8). This is actually easier than it might look. Recall that the in Equa on (9.8) is simply a variable and can represent anything. So, a quick rewrite of g() gives, g() = ( 3 ) 530

205 9.8 Power Series and so the 3 holds the same place as the in Equa on (9.8). Therefore, all we need to do is replace the in Equa on (9.9) and we ve got a power series representa on for g(). g() = ( 3 ) n provided 3 < No ce that we replaced both the in the power series and in the interval of convergence. All we need to do now is a li le simplifica on. g() = ( ) n 3n provided < So, in this case the interval of convergence is the same as the original power series. This usually won t happen. More o en than not the new interval of convergence will be different from the original interval of convergence. Eample 6 Finding a Power Series Find a power series representa on for h() = of convergence. and determine its interval + 3 S This func on is similar to the previous func on, however the numerator is different. Since Equa on (9.8) doesn t have an in the numerator it appears that we can t relate this func on back to that. However, now that we ve worked the first eample this one is actually very simple since we can use the result of the answer from that eample. To see how to do this let s first rewrite the func on a li le. h() = + 3. Now, from the first eample we ve already got a power series for the second term so let s use that to write the func on as, h() = ( ) n 3n provided < No ce that the presence of s outside of the series will NOT affect its convergence and so the interval of convergence remains the same. The last step is to bring the coefficient into the series and we ll be done. When we do this make sure and combine the s as well. We typically only want a single in a power series. h() = ( ) n 3n+ provided <. 53

206 Chapter 9 Sequences and Series As we saw in the previous eample we can o en use previous results to help us out. This is an important idea to remember as it can o en greatly simplify our work. Eample 7 Finding a Power Series Find a power series representa on for f() = of convergence. and determine its interval 5 S So again, we have an in the numerator. As with the last eample factor out and we have f() =. If we had a power series 5 representa on for g() = we could get a power series representa on for 5 f(). We need the number in the denominator to be a one so we rewrite the denominator. g() = 5 5 Now all we need to do to get a power series representa on is to replace the in Equa on (9.9) with. Doing this gives 5 g() = 5 Now simplify the series. ( ) n provided <. 5 5 g() = 5 = n 5 n n 5 n+ The interval of convergence for this series is < 5 < < 5 5 We now have a power series representa on for g() but we need to find a power series representa on for the original func on. All we need to do for this 53

207 9.8 Power Series is to mul ply the power series representa ve for g() by and we ll have it. f() = 5 n = 5 n+ = n+ 5 n+ The interval of convergence doesn t change and so it will be < 5. We now consider several eamples where differen a on and integra on of power series from Theorem 80 are used to write the power series for a func on. Eample 8 Differen a ng a Power Series Find a power series representa on for g() = and determine its radius ( ) of convergence. S We know that ( ) = d d ( ). Since we have a power series representa on for, we can differen ate that power series to get a power series representa on for g(). g() = ( ) = d d ( = d ) n d = n= n n Since the original power series had a radius of convergence of R = the deriva ve, and hence g(), will also have a radius of convergence of R =. 533

208 Chapter 9 Sequences and Series Eample 9 Integra ng a Power Series Find a power series representa on for h() = ln(5 ) and determine its radius of convergence. S In this case we need the fact that d = ln(5 ). 5 Recall that we found a power series representa on for now have ln(5 ) = 5 d n = d where < 5 5n+ n+ = C (n + )5 n+ where < 5 in Eample 7. We 5 We can find the constant of integra on, C, by subs tu ng in a value of. A good choice is = 0 as the series is usually easy to evaluate there. So, the final answer is, ln(5 0) = C ln(5 0) = C ln(5 ) = ln(5) 0 n+ (n + )5 n+ n+ (n + )5 n+, and the radius of convergence is 5. No ce that = 5 allows for convergence so the interval of convergence is [ 5, 5). 534

209 Eercises 9.8 Terms and Concepts. We adopt the conven on that 0 =, regardless of the value of.. What is the difference between the radius of convergence and the interval of convergence? 3. If the radius of convergence of a n n is 5, what is the radius of convergence of n a n n? n= 4. If the radius of convergence of radius of convergence of Problems a n n is 5, what is the ( ) n a n n? In Eercises 5 8, write out the sum of the first 5 terms of the given power series. 5. n n n= n n n! n ( ) n (n)! n In Eercises 9 8, a power series is given. (a) Find the radius of convergence. (b) Find the interval of convergence. ( ) n+ 9. n n! n n ( ) n ( 3) n n= n ( + 4) n n n n! ( ) n ( 5) n 0 n 5 n ( ) n ( ) n n n n n 3 n n 3 n ( 5)n n! ( ) n n!( 0) n n n ( + ) n n= n= n 3 ( ) n n! 0 ( ) n n (3 ) n n3 n n= n 5 n n 5 n= n (ln n) n n= ( ) n n+ (n + )! n= In Eercises 9 33, write the following func ons as a power series and give the radius of convergence. 9. f() = f() = f() = f() = f() = (a) Use differen a on to find a power series representa on for f() =. What is the radius of ( + ) convergence? (b) Use part (a) to find a power series for f() = ( + ) 3. (c) Use part (b) to find a power series for f() = ( + )

210 In Eercises 35 4, find a power series representa on for the func on and determine the radius of convergence. 35. f() = ln(3 ) 36. f() = ( + 9) ( ) f() = ln 38. f() = tan 39. f() = tan ( 3 ) 40. f() = + ( ) ( 4. f() = ) 3 536

211 9.9 Taylor Polynomials 9.9 Taylor Polynomials Consider a func on y = f() and a point ( c, f(c) ). The deriva ve, f (c), gives the instantaneous rate of change of f at = c. Of all lines that pass through the point ( c, f(c) ), the line that best approimates f at this point is the tangent line; that is, the line whose slope (rate of change) is f (c). In Figure 9.8, we see a func on y = f() graphed. The table below the graph shows that f(0) = and f (0) = ; therefore, the tangent line to f at = 0 is p () = ( 0)+ = +. The tangent line is also given in the figure. Note that near = 0, p () f(); that is, the tangent line approimates f well. One shortcoming of this approima on is that the tangent line only matches the slope of f; it does not, for instance, match the concavity of f. We can find a polynomial, p (), that does match the concavity without much difficulty, though. The table in Figure 9.8 gives the following informa on: f(0) = f (0) = f (0) =. Therefore, we want our polynomial p () to have these same proper es. That is, we need p (0) = p (0) = p (0) =. This is simply an ini al value problem. We can solve this using the techniques first described in Sec on 5.. To keep p () as simple as possible, we ll assume that not only p (0) =, but that p () =. That is, the second deriva- ve of p is constant. If p () =, then p () = + C for some constant C. Since we have determined that p (0) =, we find that C = and so p () = +. Finally, we can compute p () = ++C. Using our ini al values, we know p (0) = so C =. We conclude that p () = + +. This func on is plo ed with f in Figure 9.9. We can repeat this approima on process by crea ng polynomials of higher degree that match more of the deriva ves of f at = 0. In general, a polynomial of degree n can be created to match the first n deriva ves of f. Figure 9.9 also shows p 4 () = 4 / 3 /6+ ++, whose first four deriva ves at 0 match those of f. (Using the table in Figure 9.8, start with p (4) 4 () = and solve the related ini al value problem.) As we use more and more deriva ves, our polynomial approima on to f gets be er and be er. In this eample, the interval on which the approima on is good gets bigger and bigger. Figure 9.0 shows p 3 (); we can visually affirm that this polynomial approimates f very well on [, 3]. The polynomial p 3 () is fairly complicated: y = f() y = p () f(0) = f (0) = f (0) = 5 y f (0) = f (4) (0) = f (5) (0) = 9 Figure 9.8: Plo ng y = f() and a table of deriva ves of f evaluated at 0. y = p () 4 4 y = p 4() 5 5 y Figure 9.9: Plo ng f, p, and p 4. 5 y 4 4 y = p 3() 5 Figure 9.0: Plo ng f and p

212 Chapter 9 Sequences and Series The polynomials we have created are eamples of Taylor polynomials, named a er the Bri sh mathema cian Brook Taylor who made important discoveries about such func ons. While we created the above Taylor polynomials by solving ini al value problems, it can be shown that Taylor polynomials follow a general pa ern that make their forma on much more direct. This is described in the following defini on. Defini on 4 Taylor Polynomial, Maclaurin Polynomial Let f be a func on whose first n deriva ves eist at = c.. The Taylor polynomial of degree n of f at = c is p n () = f(c) + f (c)( c) + f (c)! n f (k) (c) = ( c) k. k! k=0 ( c) + f (c) 3! ( c) f (n) (c) ( c) n n!. A special case of the Taylor polynomial is the Maclaurin polynomial, where c = 0. That is, the Maclaurin polynomial of degree n of f is p n () = f(0) + f (0) + f (0)! n f (k) (0) = k. k! k=0 + f (0) 3! f (n) (0) n n! Note: The summa ons in this defini- on use the conven on that 0 = even when = 0 and that f (0) = f. They also use the defini on that 0! =. Generally, we order the terms of a polynomial to have decreasing degrees, and that is how we began this sec on. This defini on, and the rest of this chapter, reverses this order to reflect the greater importance of the lower degree terms in the polynomials that we will be finding. Watch the video: Taylor Polynomial to Approimate a Func on, E 3 at 538

213 9.9 Taylor Polynomials We will prac ce crea ng Taylor and Maclaurin polynomials in the following eamples. Eample Finding and using Maclaurin polynomials. Find the n th Maclaurin polynomial for f() = e.. Use p 5 () to approimate the value of e. S. We start with crea ng a table of the deriva ves of e evaluated at = 0. In this par cular case, this is rela vely simple, as shown in Figure 9.. By the defini on of the Maclaurin series, we have p n () = n k=0 f (k) (0) k = k!. Using our answer from part, we have n k=0 k! k. p 5 () = To approimate the value of e, note that e = e = f() p 5 (). It is very straigh orward to evaluate p 5 (): p 5 () = = This is an error of about 0.006, or 0.06%. A plot of f() = e and p 5 () is given in Figure 9.. f() = e f(0) = f () = e f (0) = f () = e f (0) =.. f (n) () = e f (n) (0) = Figure 9.: The deriva ves of f() = e evaluated at = y Eample Finding and using Taylor polynomials. Find the n th Taylor polynomial of y = ln at =. y = p 5() Figure 9.: A plot of f() = e and its 5 th degree Maclaurin polynomial p 5().. Use p 6 () to approimate the value of ln Use p 6 () to approimate the value of ln. 539

214 Chapter 9 Sequences and Series S f() = ln f() = 0 f () = / f () = f () = / f () = f () = / 3 f () = f (4) () = 6/ 4 f (4) () = 6. f (n) () = f (n) () = ( ) n+ (n )! n ( ) n+ (n )! Figure 9.3: Deriva ves of ln evaluated at =. y. y = ln 3. We begin by crea ng a table of deriva ves of ln evaluated at =. While this is not as straigh orward as it was in the previous eample, a pa ern does emerge, as shown in Figure 9.3. Using Defini on 4, we have p n () = n k=0 f (k) (c) ( c) k = k!. We can compute p 6 () using our work above: n k= k ( )k. p 6 () = ( ) ( ) + 3 ( )3 4 ( )4 + 5 ( )5 6 ( )6. Since p 6 () approimates ln well near =, we approimate ln.5 p 6 (.5): p 6 (.5) = (.5 ) (.5 ) + (.5 )3 3 = (.5 )4 + 5 (.5 )5 (.5 )6 6 4 y = p 6() This is a good approima on as a calculator shows that ln Figure 9.4 plots y = ln with y = p 6 (). We can see that ln.5 p 6 (.5). Figure 9.4: A plot of y = ln and its 6 th degree Taylor polynomial at =. 3. We approimate ln with p 6 (): p 6 () = ( ) ( ) + ( )3 3 4 ( )4 + 5 ( )5 ( )6 6 y y = ln = = This approima on is not terribly impressive: a hand held calculator shows 3 y = p 0() 4 Figure 9.5: A plot of y = ln and its 0 th degree Taylor polynomial at =. 540

215 9.9 Taylor Polynomials that ln The graph in Figure 9.4 shows that p 6 () provides less accurate approima ons of ln as gets close to 0 or. Surprisingly enough, even the 0 th degree Taylor polynomial fails to approimate ln for >, as shown in Figure 9.5. We ll soon discuss why this is. Taylor polynomials are used to approimate func ons f() in mainly two situa ons:. When f() is known, but perhaps hard to compute directly. For instance, we can define y = cos as either the ra o of sides of a right triangle ( adjacent over hypotenuse ) or with the unit circle. However, neither of these provides a convenient way of compu ng cos. A Taylor polynomial of sufficiently high degree can provide a reasonable method of compu ng such values using only opera ons usually hard wired into a computer (+,, and ).. When f() is not known, but informa on about its deriva ves is known. This occurs more o en than one might think, especially in the study of differen al equa ons. In both situa ons, a cri cal piece of informa on to have is How good is my approima on? If we use a Taylor polynomial to compute cos, how do we know how accurate the approima on is? We had the same problem when studying Numerical Integra on. Theorem 57 provided bounds on the error when using, say, Simpson s Rule to approimate a definite integral. These bounds allowed us to determine that, for eample, using 0 subintervals provided an approima on within ±.0 of the eact value. The following theorem gives similar bounds for Taylor (and hence Maclaurin) polynomials. Note: Even though Taylor polynomials could be used in calculators and computers to calculate values of trigonometric func ons, in prac ce they generally aren t. Other more efficient and accurate methods have been developed, such as the CORDIC algorithm. 54

216 Chapter 9 Sequences and Series Theorem 8 Taylor s Theorem. Let f be a func on whose (n+) th deriva ve eists on an interval I and let c be in I. Then, for each in I, there eists z between and c such that f() = n k=0 f (k) ( c) k! where R n () = f (n+) (z ) (n + )! ( c)n+.. R n () ma f (n+) (z) (n + )! c n+. + R n (), The first part of Taylor s Theorem states that f() = p n () + R n (), where p n () is the n th order Taylor polynomial and R n () is the remainder, or error, in the Taylor approima on. The second part gives bounds on how big that error can be. If the (n + ) th deriva ve is large, the error may be large; if is far from c, the error may also be large. However, the (n + )! term in the denominator tends to ensure that the error gets smaller as n increases. The following eample computes error es mates for the approima ons of ln.5 and ln made in Eample. Eample 3 Finding error bounds of a Taylor polynomial Use Theorem 8 to find error bounds when approima ng ln.5 and ln with p 6 (), the Taylor polynomial of degree 6 of f() = ln at =, as calculated in Eample. S. We start with the approima on of ln.5 with p 6 (.5). The theorem references an open interval I that contains both and c. The smaller the interval we use the be er; it will give us a more accurate (and smaller) approima on of the error. We let I = (0.9,.6), as this interval contains both c = and =.5. The theorem references ma f (n+) (z). In our situa on, this is asking How big can the 7 th deriva ve of y = ln be on the interval (0.9,.6)? The seventh deriva ve is y = 6!/ 7. The largest value it a ains on I is 54

217 9.9 Taylor Polynomials about 506. Thus we can bound the error as: R 6 (.5) ma f (7) (z) 7! We computed p 6 (.5) = ; using a calculator, we find ln , so the actual error is about (or 0.%), which is less than our bound of This affirms Taylor s Theorem; the theorem states that our approima on would be within about thousandths of the actual value, whereas the approima on was actually closer.. We again find an interval I that contains both c = and = ; we choose I = (0.9,.). The maimum value of the seventh deriva ve of f on this interval is again about 506 (as the largest values come near = 0.9). Thus R 6 () ma f (7) (z) 7! This bound is not as nearly as good as before. Using the degree 6 Taylor polynomial at = will bring us within 0.3 of the correct answer. As p 6 () , our error es mate guarantees that the actual value of ln is somewhere between and These bounds are not par cularly useful. In reality, our approima on was only off by about 0.07 (or %). However, we are approima ng ostensibly because we do not know the real answer. In order to be assured that we have a good approima on, we would have to resort to using a polynomial of higher degree. We prac ce again. This me, we use Taylor s theorem to find n that guarantees our approima on is within a certain amount. Eample 4 Finding sufficiently accurate Taylor polynomials Find n such that the n th Taylor polynomial of f() = cos at = 0 approimates cos to within 0.00 of the actual answer. What is p n ()? 543

218 Chapter 9 Sequences and Series S Following Taylor s theorem, we need bounds on the size of the deriva ves of f() = cos. In the case of this trigonometric func on, this is easy. All deriva ves of cosine are ± sin or ± cos. In all cases, these func ons are never greater than in absolute value. We want the error to be less than To find the appropriate n, consider the following inequali es: ma f (n+) (z) 0 n (n + )! (n + )! n We find an n that sa sfies this last inequality with trial and error. When n = 8, 8+ we have (8 + )! 0.004; when n = 9, we have 9+ (9 + )! < Thus we want to approimate cos with p 9 (). f() = cos f(0) = f () = sin f (0) = 0 f () = cos f (0) = f () = sin f (0) = 0 f (4) () = cos f (4) (0) = f (5) () = sin f (5) (0) = 0 f (6) () = cos f (6) (0) = f (7) () = sin f (7) (0) = 0 f (8) () = cos f (8) (0) = f (9) () = sin f (9) (0) = 0 Figure 9.6: A table of the deriva ves of f() = cos evaluated at = 0. y y = p 8() f() = cos We now set out to compute p 9 (). We again need a table of the deriva ves of f() = cos evaluated at = 0. A table of these values is given in Figure 9.6. No ce how the deriva ves, evaluated at = 0, follow a certain pa ern. All the odd powers of in the Taylor polynomial will disappear as their coefficient is 0. While our error bounds state that we need p 9 (), our work shows that this will be the same as p 8 (). Since we are forming our polynomial at = 0, we are crea ng a Maclaurin polynomial, and: p 8 () = 8 k=0 We finally approimate cos : f (k) (0) k = k!! + 4! 4 6! 6 + 8! 8 cos p 8 () = Our error bound guarantee that this approima on is within 0.00 of the correct answer. Technology shows us that our approima on is actually within about (or 0.07%) of the correct answer. Figure 9.7 shows a graph of y = p 8 () and y = cos. Note how well the two func ons agree on about ( π, π). Eample 5 Finding and using Taylor polynomials. Find the degree 4 Taylor polynomial, p 4 (), for f() = at = 4.. Use p 4 () to approimate Find bounds on the error when approima ng 3 with p 4 (3). Figure 9.7: A graph of f() = cos and its degree 8 Maclaurin polynomial. 544

219 9.9 Taylor Polynomials S. We begin by evalua ng the deriva ves of f at = 4. This is done in Figure 9.8. These values allow us to form the Taylor polynomial p 4 (): p 4 () = + 4 ( 4) + /3! ( 4) + 3/56 3! ( 4) 3 + 5/048 ( 4) 4. 4!. As p 4 () near = 4, we approimate 3 with p 4 (3) = To find a bound on the error, we need an open interval that contains = 3 and = 4. We set I = (.9, 4.). The largest value the fi h deriva ve of f() = takes on this interval is near =.9, at about Thus f() = f(4) = f () = f (4) = 4 f () = f (4) = 4 3/ 3 f () = 3 f (4) = 3 8 5/ 56 f (4) () = 5 f (4) (4) = 5 6 7/ 048 Figure 9.8: A table of the deriva ves of f() = evaluated at = 4. R 4 (3) ! This shows our approima on is accurate to at least the first places a er the decimal. It turns out that our approima on has an error of , or 0.004%. A graph of f() = and p 4 () is given in Figure 9.9. Note how the two func ons are nearly indis nguishable on (, 7). y 3 y = y = p 4() Most of this chapter has been devoted to the study of infinite series. This sec on has stepped aside from this study, focusing instead on finite summa on of terms. In the net sec on, we will combine power series and Taylor polynomials into Taylor Series, where we represent a func on with an infinite series. 5 0 Figure 9.9: A graph of f() = and its degree 4 Taylor polynomial at =

220 Eercises 9.9 Terms and Concepts. What is the difference between a Taylor polynomial and a Maclaurin polynomial?. T/F: In general, p n() approimates f() be er and be er as n gets larger. 3. For some func on f(), the Maclaurin polynomial of degree 4 is p 4() = What is p ()? 4. For some func on f(), the Maclaurin polynomial of degree 4 is p 4() = What is f (0)? Problems In Eercises 5, find the Maclaurin polynomial of degree n for the given func on. 5. f() = e, n = 3 6. f() = sin, n = 8 7. f() = e, n = 5 8. f() = tan, n = 6 9. f() = e, n = 4 0. f() =, n = 4. f() = +, n = 4. f() = +, n = 7 In Eercises 3 0, find the Taylor polynomial of degree n, at = c, for the given func on. 3. f() =, n = 4, c = 4. f() = ln( + ), n = 4, c = 5. f() = cos, n = 6, c = π/4 6. f() = sin, n = 5, c = π/6 7. f() =, n = 5, c = 8. f() =, n = 8, c = 9. f() =, n = 3, c = + 0. f() = cos, n =, c = π In Eercises 4, approimate the func on value with the indicated Taylor polynomial and give approimate bounds on the error.. Approimate sin 0. with the Maclaurin polynomial of degree 3.. Approimate cos with the Maclaurin polynomial of degree Approimate 0 with the Taylor polynomial of degree centered at = Approimate ln.5 with the Taylor polynomial of degree 3 centered at =. Eercises 5 8 ask for an n to be found such that p n() approimates f() within a certain bound of accuracy. 5. Find n such that the Maclaurin polynomial of degree n of f() = e approimates e within of the actual value. 6. Find n such that the Taylor polynomial of degree n of f() =, centered at = 4, approimates 3 within of the actual value. 7. Find n such that the Maclaurin polynomial of degree n of f() = cos approimates cos π/3 within of the actual value. 8. Find n such that the Maclaurin polynomial of degree n of f() = sin approimates cos π within of the actual value. In Eercises 9 33, find the n th term of the indicated Taylor polynomial. 9. Find a formula for the n th term of the Maclaurin polynomial for f() = e. 30. Find a formula for the n th term of the Maclaurin polynomial for f() = cos. 3. Find a formula for the n th term of the Maclaurin polynomial for f() =. 3. Find a formula for the n th term of the Maclaurin polynomial for f() = Find a formula for the n th term of the Taylor polynomial for f() = ln. In Eercises 34 36, approimate the solu on to the given differen al equa on with a degree 4 Maclaurin polynomial. 34. y = y, y(0) = 35. y = 5y, y(0) = y = y, y(0) = 546

221 9.0 Taylor Series 9.0 Taylor Series In Sec on 9.8, we showed how certain func ons can be represented by a power series func on. In Sec on 9.9, we showed how we can approimate func ons with polynomials, given that enough deriva ve informa on is available. In this sec on we combine these concepts: if a func on f() is infinitely differen able, we show how to represent it with a power series func on. Defini on 43 Taylor and Maclaurin Series Let f() have deriva ves of all orders at = c.. The Taylor Series of f(), centered at c is f (n) (c) ( c) n. n!. Se ng c = 0 gives the Maclaurin Series of f(): f (n) (0) n. n! Watch the video: Taylor and Maclaurin Series Eample at The difference between a Taylor polynomial and a Taylor series is the former is a polynomial, containing only a finite number of terms, whereas the la er is a series, a summa on of an infinite set of terms. When crea ng the Taylor polynomial of degree n for a func on f() at = c, we needed to evaluate f, and the first n deriva ves of f, at = c. When crea ng the Taylor series of f, we need to find a pa ern that describes the n th deriva ve of f at = c. We demonstrate this in the net two eamples. Eample The Maclaurin series of f() = cos Find the Maclaurin series of f() = cos. 547

222 Chapter 9 Sequences and Series f() = cos f(0) = f () = sin f (0) = 0 f () = cos f (0) = f () = sin f (0) = 0 f (4) () = cos f (4) (0) = f (5) () = sin f (5) (0) = 0 f (6) () = cos f (6) (0) = f (7) () = sin f (7) (0) = 0 f (8) () = cos f (8) (0) = f (9) () = sin f (9) (0) = 0 Figure 9.30: A table of the deriva ves of f() = cos evaluated at = 0. S In Eample we found the 8 th degree Maclaurin polynomial of cos. In doing so, we created the table shown in Figure No ce how f (n) (0) = 0 when n is odd, f (n) (0) = when n is divisible by 4, and f (n) (0) = when n is even but not divisible by 4. Thus the Maclaurin series of cos is + 4 4! 6 6! + 8 8! We can go further and write this as a summa on. Since we only need the terms where the power of is even, we write the power series in terms of n : ( ) n n (n)!. Eample The Taylor series of f() = ln at = Find the Taylor series of f() = ln centered at =. S Figure 9.3 shows the n th deriva ve of ln evaluated at = for n = 0,..., 5, along with an epression for the n th term: f (n) () = ( ) n+ (n )! for n. f() = ln f() = 0 f () = / f () = f () = / f () = f () = / 3 f () = f (4) () = 6/ 4 f (4) () = 6 f (5) () = 4/ 5 f (5) () = 4. f (n) () = f (n) () = ( ) n+ (n )! n ( ) n+ (n )! Figure 9.3: Deriva ves of ln evaluated at =.. Remember that this is what dis nguishes Taylor series from Taylor polynomials; we are very interested in finding a pa ern for the n th term, not just finding a finite set of coefficients for a polynomial. Since f() = ln = 0, we skip the first term and start the summa on with n =, giving the Taylor series for ln, centered at =, as ( ) n+ (n )! n! ( )n n+ ( )n = ( ). n n= It is important to note that Defini on 43 defines a Taylor series given a func- on f(); however, we cannot yet state that f() is equal to its Taylor series. We will find that most of the me they are equal, but we need to consider the condi ons that allow us to conclude this. Theorem 8 states that the error between a func on f() and its n th degree Taylor polynomial p n () is R n (), where R n () ma f (n+) (z) (n + )! n= c n+. If R n () goes to 0 for each in an interval I as n approaches infinity, we conclude that the func on is equal to its Taylor series epansion. 548

223 9.0 Taylor Series Theorem 8 Func on and Taylor Series Equality Let f() have deriva ves of all orders at = c, let R n () be as stated in Theorem 8, and let I be an interval on which the Taylor series of f() converges. If lim R n() = 0 for all in I, then n f() = f (n) (c) ( c) n on I. n! We demonstrate the use of this theorem in an eample. Eample 3 Establishing equality of a func on and its Taylor series Show that f() = cos is equal to its Maclaurin series, as found in Eample, for all. S bounded by Given a value, the magnitude of the error term R n () is R n () ma f (n+) (z) (n + )! n+. Since all deriva ves of cos are ± sin or ± cos, whose magnitudes are bounded by, we can state R n () (n + )! n+ which implies n+ n+ (n + )! R n() n+ (n + )!. (9.0) For any, lim = 0. Applying the Squeeze Theorem to Equa on (9.0), n (n + )! we conclude that lim R n() = 0 for all, and hence n cos = ( ) n n (n)! for all. It is natural to assume that a func on is equal to its Taylor series on the series interval of convergence, but this is not the case. In order to properly establish equality, one must use Theorem 8. This is a bit disappoin ng, as we developed beau ful techniques for determining the interval of convergence of a power series, and proving that R n () 0 can be cumbersome as it deals with high order deriva ves of the func on. 549

224 Chapter 9 Sequences and Series There is good news. A func on f() that is equal to its Taylor series, centered at any point the domain of f(), is said to be an analy c func on, and most, if not all, func ons that we encounter within this course are analy c func ons. Generally speaking, any func on that one creates with elementary func ons (polynomials, eponen als, trigonometric func ons, etc.) that is not piecewise defined is probably analy c. For most func ons, we assume the func on is equal to its Taylor series on the series interval of convergence and only use Theorem 8 when we suspect something may not work as epected. We develop the Taylor series for one more important func on, then give a table of the Taylor series for a number of common func ons. Eample 4 The Binomial Series Find the Maclaurin series of f() = ( + ) k, k 0. S When k is a posi ve integer, the Maclaurin series is finite. For instance, when k = 4, we have f() = ( + ) 4 = The coefficients of when k is a posi ve integer are known as the binomial coefficients, giving the series we are developing its name. When k = /, we have f() = +. Knowing a series representa on of this func on would give a useful way of approima ng.3, for instance. To develop the Maclaurin series for f() = ( + ) k for any value of k 0, we consider the deriva ves of f evaluated at = 0: f() = ( + ) k f(0) = f () = k( + ) k f (0) = k f () = k(k )( + ) k f (0) = k(k ) f () = k(k )(k )( + ) k 3 f (0) = k(k )(k ).. f (n) () = k(k ) (k (n ) ) ( + ) k n f (n) (0) = k(k ) (k (n ) ) Thus the Maclaurin series for f() = ( + ) k is + k + k(k )! + k(k )(k ) 3! + + k(k ) (k (n ) ) n! + It is important to determine the interval of convergence of this series. With a n = k(k ) (k (n ) ) n, n! 550

225 9.0 Taylor Series we apply the Ra o Test: a n+ lim = lim n a n n = lim n =. n+ ( ) k (n ) n! n k(k ) (k n) (n+)! k(k ) k n n The series converges absolutely when the limit of the Ra o Test is less than ; therefore, we have absolute convergence when <. While outside the scope of this tet, the interval of convergence depends on the value of k. When k > 0, the interval of convergence is [, ]. When < k < 0, the interval of convergence is [, ). If k, the interval of convergence is (, ). We learned that Taylor polynomials offer a way of approima ng a difficult to compute func on with a polynomial. Taylor series offer a way of eactly represen ng a func on with a series. One probably can see the use of a good approima on; is there any use of represen ng a func on eactly as a series? While we should not overlook the mathema cal beauty of Taylor series (which is reason enough to study them), there are prac cal uses as well. They provide a valuable tool for solving a variety of problems, including problems rela ng to integra on and differen al equa ons. In Key Idea 33 (on the following page) we give a table of the Maclaurin series of a number of common func ons. We then give a theorem about the algebra of power series, that is, how we can combine power series to create power series of new func ons. This allows us to find the Taylor series of func ons like f() = e cos by knowing the Taylor series of e and cos. Before we inves gate combining func ons, consider the Taylor series for the arctangent func on (see Key Idea 33). Knowing that tan () = π/4, we can use this series to approimate the value of π: π 4 = tan () = π = 4 ( ) Unfortunately, this par cular epansion of π converges very slowly. The first 00 terms approimate π as 3.359, which is not par cularly good. 55

226 Chapter 9 Sequences and Series Key Idea 33 Func on and Series e n = n! sin = Important Maclaurin Series Epansions ( ) n n+ (n + )! cos = ( ) n n (n)! ln( + ) = ( ) n= n+ n n First Few Terms Interval of Convergence + +! (, ) 3! 3 3! + 5 5! 7 + (, ) 7!! + 4 4! 6 + (, ) 6! + 3 (, ] 3 = n (, ) ( + ) k k(k ) (k (n ) ) k(k ) (, ) k = n + k + + [, ) < k < 0 n!! [, ] 0 < k tan = ( ) n n+ 3 n [, ] 7 55

227 9.0 Taylor Series Theorem 83 Algebra of Power Series Let f() = a n n and g() = b n n converge absolutely for < R, and let h() be con nuous.. f() ± g() = (a n ± b n ) n for < R. ( ) ( ). f()g() = a n n b n n = ( ) a0 b n + a b n + + a n b 0 n for < R. 3. f ( h() ) ( ) n = a n h() for h() < R. Eample 5 Combining Taylor series Write out the first 3 terms of the Maclaurin Series for f() = e cos using Key Idea 33 and Theorem 83. S Key Idea 33 informs us that e = + +! and cos = 3!! + 4 4! +. Applying Theorem 83, we find that ) ) e cos = ( + + (! + 3 3! +! + 4 4! +. Distribute the right hand epression across the le : = ( +! + 4 4! )! + 4 4! + ) (! + 4 4! ! (! + 4 4! + + ( ) + ) 4! +! + 4 (! + 4 4! + ) 553

228 Chapter 9 Sequences and Series Distribute again and collect like terms. = While this process is a bit tedious, it is much faster than evalua ng all the necessary deriva ves of e cos and compu ng the Taylor series directly. Because the series for e and cos both converge on (, ), so does the series epansion for e cos. Eample 6 Crea ng new Taylor series Use Theorem 83 to create the Taylor series for y = sin( ) centered at = 0 and a series for y = ln( ) centered at c =. S Given that sin = ( ) n n+ (n + )! = 3 3! + 5 5! 7 7! +, we simply subs tute for in the series, giving sin( ) = ( ) n ( ) n+ (n + )! = ( ) n 4n+ (n + )! = 6 3! + 0 5! 4 7!. Since the Taylor series for sin has an infinite radius of convergence, so does the Taylor series for sin( ). The Taylor epansion for ln given in Key Idea 33 is centered at =, so we will center the series for ln( ) at = as well. With ln = n+ ( )n ( ) n n= = ( ) ( ) + ( )3 3, Note: In Eample 6, one could create a series for ln( ) by simply recognizing that ln( ) = ln( / ) = / ln, and hence mul plying the Taylor series for ln by /. This eample was chosen to demonstrate other aspects of series, such as the fact that the interval of convergence changes. we subs tute for to obtain ln( ) = n= ( ) n+ ( ) n n = ( ) ( ) + ( ) 3 3. While this is not strictly a power series because of the, it is a series that allows us to study the func on ln( ). Since the interval of convergence of ln is (0, ], and the range of on (0, 4] is (0, ], the interval of convergence of this series epansion of ln( ) is (0, 4]. 554

229 9.0 Taylor Series Eample 7 Using Taylor series to evaluate definite integrals Use the Taylor series of e to evaluate 0 e d. S We learned, when studying Numerical Integra on, that e does not have an an deriva ve epressible in terms of elementary func ons. This means any definite integral of this func on must have its value approimated, and not computed eactly. We can quickly write out the Taylor series for e using the Taylor series of e : e n = n! = + +! + 3 3! + and so e = = ( ) n n! ( ) n n n! = + 4! 6 3! +. We use Theorem 80 to integrate: e d = C ! 7 7 3! + + n+ ( )n (n + )n! + This is the an deriva ve of e ; while we can write it out as a series, we cannot write it out in terms of elementary func ons. We can evaluate the definite integral 0 e d using this an deriva ve; subs tu ng and 0 for and subtrac ng gives 0 e d = 3 + 5! 7 3! + 9 4!. Summing the 5 terms shown above give the approima on of Since this is an alterna ng series, we can use the Alterna ng Series Approima on Theorem, (Theorem 75), to determine how accurate this approima on is. The net term of the series is /( 5!) Thus we know our approima on is within of the actual value of the integral. This is arguably much less work than using Simpson s Rule to approimate the value of the integral. 555

230 Chapter 9 Sequences and Series Another advantage to using Taylor series instead of Simpson s Rule is for making subsequent approima ons. We found in Eample that the error in using Simpson s Rule for e 0 d with four intervals was If we wanted to decrease that error, we would need to use more intervals, essen ally star ng the problem over. Using a Taylor series, if we wanted a more accurate approima on, we can just subtract the net term /( 5!) to get an approima on of , with an error of at most /(3 6!) Finding a pa ern in the coefficients that match the series epansion of a known func on, such as those shown in Key Idea 33, can be difficult. What if the coefficients are given in their reduced form; how could we s ll recover the func on? Suppose that all we know is that a 0 =, a =, a =, a 3 = 4 3, a 4 = 3. Defini on 43 states that each term of the Taylor epansion of a func on includes an n!. This allows us to say that a = = b!, a 3 = 4 3 = b 3 3!, and a 4 = 3 = b 4 4! for some values b, b 3 and b 4. Solving for these values, we see that b = 4, b 3 = 8 and b 4 = 6. That is, we are recovering the pa ern b n = n, allowing us to write f() = a n n b n = n! n = + + 4! + 8 3! ! 4 + From here it is easier to recognize that the series is describing an eponen al func on. This chapter introduced sequences, which are ordered lists of numbers, followed by series, wherein we add up the terms of a sequence. We quickly saw that such sums do not always add up to infinity, but rather converge. We studied tests for convergence, then ended the chapter with a formal way of defining func ons based on series. Such series defined func ons are a valuable tool in solving a number of different problems throughout science and engineering. Coming in the net chapters are new ways of defining curves in the plane apart from using func ons of the form y = f(). Curves created by these new methods can be beau ful, useful, and important. 556

231 Eercises 9.0 Terms and Concepts. What is the difference between a Taylor polynomial and a Taylor series?. What theorem must we use to show that a func on is equal to its Taylor series? Problems Key Idea 33 gives the n th term of the Taylor series of common func ons. In Eercises 3 6, verify the formula given in the Key Idea by finding the first few terms of the Taylor series of the given func on and iden fying a pa ern. 3. f() = e ; c = 0 4. f() = sin ; c = 0 5. f() = /( ); c = 0 6. f() = tan ; c = 0 5. f() = ln( + ) (show equality only on (0, )). 6. f() = /( ) (show equality only on (, 0)) In Eercises 7 0, use the Taylor series given in Key Idea 33 to verify the given iden ty. 7. cos( ) = cos 8. sin( ) = sin d ( ) 9. d sin = cos ( ) d 0. d cos = sin In Eercises 4, write out the first 5 terms of the Binomial series with the given k-value.. k = /. k = / 3. k = /3 4. k = 4 In Eercises 7, find a formula for the n th term of the Taylor series of f(), centered at c, by finding the coefficients of the first few powers of and looking for a pa ern. (The formulas for several of these are found in Key Idea 33; show work verifying these formula.) 7. f() = cos ; c = π/ 8. f() = /; c = 9. f() = e ; c = 0 0. f() = ln( + ); c = 0. f() = /( + ); c =. f() = sin ; c = π/4 In Eercises 3 6, show that the Taylor series for f(), as given in Key Idea 33, is equal to f() by applying Theorem 8; that is, show lim Rn() = 0. n 3. f() = e 4. f() = sin In Eercises 5 30, use the Taylor series given in Key Idea 33 to create the Taylor series of the given func ons. 5. f() = cos ( ) 6. f() = e 7. f() = sin ( + 3 ) 8. f() = tan ( / ) 9. f() = e sin (only find the first non-zero 4 terms) 30. f() = ( + ) / cos (only find the first non-zero 4 terms) In Eercises 3 3, approimate the value of the given definite integral by using the first 4 nonzero terms of the integrand s Taylor series π 0 π /4 0 sin ( ) d cos ( ) d 557

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233 0.0 Chapter Prerequisites Conic Sec ons 0.0 Chapter Prerequisites Conic Sec ons The material in this sec on provides a basic review of and prac ce problems for pre-calculus skills essen al to your success in Calculus. You should take me to review this sec on and work the suggested problems (checking your answers against those in the back of the book). Since this content is a pre-requisite for Calculus, reviewing and mastering these skills are considered your responsibility. This means that minimal, and in some cases no, class me will be devoted to this sec on. When you iden fy areas that you need help with we strongly urge you to seek assistance outside of class from your instructor or other student tutoring service. The ancient Greeks recognized that interes ng shapes can be formed by intersec ng a plane with a double napped cone (i.e., two iden cal cones placed p to p as shown in the following figures). As these shapes are formed as sec ons of conics, they have earned the official name conic sec ons. The three most interes ng conic sec ons are given in the top row of Figure 0.. They are the parabola, the ellipse (which includes circles) and the hyperbola. In each of these cases, the plane does not intersect the ps of the cones (usually taken to be the origin). Parabola Ellipse Circle Hyperbola Point Line Crossed Lines Figure 0.: Conic Sec ons When the plane does contain the origin, three degenerate cones can be formed as shown the bo om row of Figure 0.: a point, a line, and crossed lines. We focus here on the nondegenerate cases. While the above geometric constructs define the conics in an intui ve, visual way, these constructs are not very helpful when trying to analyze the shapes 559

234 Chapter 0 Curves in the Plane algebraically or consider them as the graph of a func on. It can be shown that all conics can be defined by the general second degree equa on A + By + Cy + D + Ey + F = 0. While this algebraic defini on has its uses, most find another geometric perspec ve of the conics more beneficial. Each nondegenerate conic can be defined as the locus, or set, of points that sa sfy a certain distance property. These distance proper es can be used to generate an algebraic formula, allowing us to study each conic as the graph of a func on. Parabolas Defini on 44 Parabola A parabola is the locus of all points equidistant from a point (called a focus) and a line (called the directri) that does not contain the focus. Directri Ais of Symmetry Focus } p } Verte p d (, y) d Figure 0. illustrates this defini on. The point halfway between the focus and the directri is the verte. The line through the focus, perpendicular to the directri, is the ais of symmetry, as the por on of the parabola on one side of this line is the mirror image of the por on on the opposite side. The geometric defini on of the parabola and distance formula can be used to derive the quadra c func on whose graph is a parabola with verte at the origin. Figure 0.: Illustra ng the defini on of the parabola and establishing an algebraic formula. y = 4p. Applying transforma ons of func ons we get the following standard form of the parabola. 560

235 0.0 Chapter Prerequisites Conic Sec ons Key Idea 34 General Equa on of a Parabola. Ver cal Ais of Symmetry: The equa on of the parabola with verte at (h, k), directri y = k p, and focus at (h, k + p) in standard form is y = 4p ( h) + k.. Horizontal Ais of Symmetry: The equa on of the parabola with verte at (h, k), directri = h p, and focus at (h + p, k) in standard form is = 4p (y k) + h. Note: p is not necessarily a posi ve number. Eample Finding the equa on of a parabola Give the equa on of the parabola with focus at (, ) and directri at y = 3. S The verte is located halfway between the focus and directri, so (h, k) = (,.5). This gives p = 0.5. Using Key Idea 34 we have the equa on of the parabola as y = 4( 0.5) ( ) +.5 = ( ) +.5. The parabola is sketched in Figure 0.3. y 5 Ellipses 4 6 Defini on 45 Ellipse An ellipse is the locus of all points whose sum of distances from two fied points, each a focus of the ellipse, is constant. Figure 0.3: The parabola described in Eample. An easy way to visualize this construc on of an ellipse is to pin both ends of a string to a board. The pins become the foci. Holding a pencil ght against the 56

236 Chapter 0 Curves in the Plane d d + d = constant d string places the pencil on the ellipse; the sum of distances from the pencil to the pins is constant: the length of the string. See Figure 0.4. As shown in Figure 0.5, the values of a and b have meaning. In general, the two foci of an ellipse lie on the major ais of the ellipse, and the midpoint of the segment joining the two foci is the center. The major ais intersects the ellipse at two points, each of which is a verte. The line segment through the center and perpendicular to the major ais is the minor ais. The constant sum of distances that defines the ellipse is the length of the major ais, i.e., a. Allowing for the shi ing of the ellipse gives the following standard equa ons. Figure 0.4: Illustra ng the construc on of an ellipse with pins, pencil and string. Ver ces b } {{ } a } {{ } c Foci Key Idea 35 Standard Equa on of the Ellipse The equa on of an ellipse centered at (h, k) with major ais of length a and minor ais of length b in standard form is:. Horizontal major ais:. Ver cal major ais: ( h) (y k) a + b =. ( h) (y k) b + a =. The foci lie along the major ais, c units from the center, where c = a b. Major ais Minor ais Figure 0.5: Labeling the significant features of an ellipse y Figure 0.6: The ellipse used in Eample. Eample Finding the equa on of an ellipse Find the general equa on of the ellipse graphed in Figure 0.6. S The center is located at ( 3, ). The distance from the center to a verte is 5 units, hence a = 5. The minor ais seems to have length 4, so b =. Thus the equa on of the ellipse is ( + 3) + 4 (y ) 5 =. Eample 3 Graphing an ellipse Graph the ellipse defined by 4 + 9y 8 36y = 4. S It is simple to graph an ellipse once it is in standard form. In order to put the given equa on in standard form, we must complete the square with both the and y terms. We first rewrite the equa on by regrouping: 4 + 9y 8 36y = 4 (4 8) + (9y 36y) = 4. 56

237 0.0 Chapter Prerequisites Conic Sec ons Now we complete the squares. (4 8) + (9y 36y) = 4 4( ) + 9(y 4y) = 4 4( + ) + 9(y 4y + 4 4) = 4 4 ( ( ) ) + 9 ( (y ) 4 ) = 4 4( ) 4 + 9(y ) 36 = 4 4( ) + 9(y ) = 36 ( ) + 9 (y ) 4 =. We see the center of the ellipse is at (, ). We have a = 3 and b = ; the major ais is horizontal, so the ver ces are located at (, ) and (4, ). We find c = 9 4 = 5.4. The foci are located along the major ais, approimately.4 units from the center, at ( ±.4, ). This is all graphed in Figure 0.7. Hyperbolas The defini on of a hyperbola is very similar to the defini on of an ellipse; we essen ally just change the word sum to difference. Defini on 46 Hyperbola A hyperbola is the locus of all points where the absolute value of the difference of distances from two fied points, each a focus of the hyperbola, is constant. 4 3 y 3 4 Figure 0.7: Graphing the ellipse in Eample 3. We do not have a convenient way of visualizing the construc on of a hyperbola as we did for the ellipse. The geometric defini on does allow us to find an algebraic epression that describes it. It will be useful to define some terms first. The two foci lie on the transverse ais of the hyperbola; the midpoint of the line segment joining the foci is the center of the hyperbola. The transverse ais intersects the hyperbola at two points, each a verte of the hyperbola. The line through the center and perpendicular to the transverse ais is the conjugate ais. This is illustrated in Figure 0.8. It is easy to show that the constant difference of distances used in the defini on of the hyperbola is the distance between the ver ces, i.e., a. Conjugate ais a c {}}{{}}{ Transverse ais Ver ces Foci Figure 0.8: Labeling the significant features of a hyperbola. 563

238 Chapter 0 Curves in the Plane Key Idea 36 Standard Equa on of a Hyperbola The equa on of a hyperbola centered at (h, k) in standard form is: y. Horizontal Transverse Ais:. Ver cal Transverse Ais: ( h) (y k) a b =. (y k) ( h) a b =. The ver ces are located a units from the center and the foci are located c units from the center, where c = a + b. 5 5 Figure 0.9: Graphing the hyperbola 9 y = along with its asymptotes, y = ±/3. y k + b k k b Graphing Hyperbolas Consider the hyperbola 9 y =. Solving for y, we find y = ± /9. As grows large, the part of the equa on for y becomes less significant and y ± /9 = ±/3. That is, as gets large, the graph of the hyperbola looks very much like the lines y = ±/3. These lines are asymptotes of the hyperbola, as shown in Figure 0.9. This is a valuable tool in sketching. Given the equa on of a hyperbola in general form, draw a rectangle centered at (h, k) with sides of length a parallel to the transverse ais and sides of length b parallel to the conjugate ais. (See Figure 0.0 for an eample with a horizontal transverse ais.) The diagonals of the rectangle lie on the asymptotes. These lines pass through (h, k). When the transverse ais is horizontal, the slopes are ±b/a; when the transverse ais is ver cal, their slopes are ±a/b. This gives equa ons: Horizontal Transverse Ais Ver cal Transverse Ais h a h h + a y = ± b a ( h) + k y = ±a ( h) + k. b Figure 0.0: Using the asymptotes of a hyperbola as a graphing aid. 0 5 y Eample 4 Graphing a hyperbola (y ) Sketch the hyperbola given by 5 ( ) 4 =. S The hyperbola is centered at (, ); a = 5 and b =. In Figure 0. we draw the prescribed rectangle centered at (, ) along with the asymptotes defined by its diagonals. The hyperbola has a ver cal transverse Figure 0.: Graphing the hyperbola in Eample

239 0.0 Chapter Prerequisites Conic Sec ons ais, so the ver ces are located at (, 7) and (, 3). This is enough to make a good sketch. We also find the loca on of the foci: as c = a + b, we have c = Thus the foci are located at (, ± 5.4) as shown in the figure. Eample 5 Graphing a hyperbola Sketch the hyperbola given by 9 y + y = 0. S We must complete the square to put the equa on in general form. (We recognize this as a hyperbola since it is a general quadra c equa on and the and y terms have opposite signs.) y 9 y + y = 0 9 (y y) = 0 9 (y y + ) = 0 9 ( (y ) ) = 0 9 (y ) = 9 (y ) 9 We see the hyperbola is centered at (0, ), with a horizontal transverse ais, where a = and b = 3. The appropriate rectangle is sketched in Figure 0. along with the asymptotes of the hyperbola. The ver ces are located at (±, ). We have c = 0 3., so the foci are located at (±3., ) as shown in Figure 0.. = 0 Figure 0.: Graphing the hyperbola in Eample 5. This chapter eplores curves in the plane, in par cular curves that cannot be described by func ons of the form y = f(). In this sec on, we learned of ellipses and hyperbolas that are defined implicitly, not eplicitly. In the following sec ons, we will learn completely new ways of describing curves in the plane, using parametric equa ons and polar coordinates, then study these curves using calculus techniques. 565

240 Eercises 0.0 Problems y In Eercises 8, find the equa on of the parabola defined by the given informa on. Sketch the parabola.. Focus: (3, ); directri: y = 7.. Focus: (, 4); directri: y = 3. Focus: (, 5); directri: = 3 4. Focus: (/4, 0); directri: = /4 5. Focus: (, ); verte: (, ) 6. Focus: ( 3, 0); verte: (0, 0) y 5 7. Verte: (0, 0); directri: y = /6 8. Verte: (, 3); directri: = In Eercises 9 0, sketch the ellipse defined by the given equa- on. Label the center, foci and ver ces. 5 ( ) (y ) 9. + = (y + 3) = In Eercises, find the equa on of the ellipse shown in the graph. 9. y 6 4 y 5. 4 y 6. 4 y In Eercises 3 6, write the equa on of the given ellipse in standard form y 8y = y = y y + 6 = y 4 4y + 4 = 0 In Eercises 7 0, find the equa on of the hyperbola shown in the graph In Eercises, sketch the hyperbola defined by the given equa on. Label the center.. ( ) 6. (y 4) (y + ) 9 ( + ) 5 = = In Eercises 3 6, write the equa on of the hyperbola in standard form y = 4. 3 y + y = y + 40y = (4y )(4y + ) =

241 0: C P We have eplored func ons of the form y = f() closely throughout this tet. We have eplored their limits, their deriva ves and their an deriva ves; we have learned to iden fy key features of their graphs, such as rela ve maima and minima, inflec on points and asymptotes; we have found equa ons of their tangent lines, the areas between por ons of their graphs and the -ais, and the volumes of solids generated by revolving por ons of their graphs about a horizontal or ver cal ais. Despite all this, the graphs created by func ons of the form y = f() are limited. Since each -value can correspond to only y-value, common shapes like circles cannot be fully described by a func on in this form. Fi ngly, the ver cal line test ecludes ver cal lines from being func ons of, even though these lines are important in mathema cs. In this chapter we ll eplore new ways of drawing curves in the plane. We ll s ll work within the framework of func ons, as an input will s ll only correspond to one output. However, our new techniques of drawing curves will render the ver cal line test pointless, and allow us to create important and beau ful new curves. Once these curves are defined, we ll apply the concepts of calculus to them, con nuing to find equa ons of tangent lines and the areas of enclosed regions. One aspect that we ll be interested in is how long is this curve? Before we eplore that idea for these new ways to draw curves, we ll start by eploring how long a curve is when we ve go en it from a regular y = f() func on. 0. Arc Length and Surface Area In previous sec ons we have used integra on to answer the following ques ons: y. Given a region, what is its area?. Given a solid, what is its volume? In this sec on, we address a related ques on: Given a curve, what is its length? This is o en referred to as arc length. Consider the graph of y = sin on [0, π] given in Figure 0.3 (a). How long is this curve? That is, if we were to use a piece of string to eactly match the shape of this curve, how long would the string be? As we have done in the past, we start by approima ng; later, we will refine our answer using limits to get an eact solu on. y π 4 π (a) 3π 4 π π 4 π 3π 4 π (b) Figure 0.3: Graphing y = sin on [0, π] and approima ng the curve with line segments.

242 Chapter 0 Curves in the Plane y i y i y y i i i Figure 0.4: Zooming in on the i th subinterval [ i, i] of a par on of [a, b]. i The length of straight line segments is easy to compute using the Distance Formula. We can approimate the length of the given curve by approima ng the curve with straight lines and measuring their lengths. In Figure 0.3 (b), the curve y = sin has been approimated with 4 line segments (the interval [0, π] has been divided into 4 equally lengthed subintervals). It is clear that these four line segments approimate y = sin very well on the first and last subinterval, though not so well in the middle. Regardless, the sum of the lengths of the line segments is 3.79, so we approimate the arc length of y = sin on [0, π] to be In general, we can approimate the arc length of y = f() on [a, b] in the following manner. Let a = 0 < <... < n < n = b be a par on of [a, b] into n subintervals. Let i represent the length of the i th subinterval [ i, i ]. Figure 0.4 zooms in on the i th subinterval where y = f() is approimated by a straight line segment. The dashed lines show that we can view this line segment as the hypotenuse of a right triangle whose sides have length i and y i. Using the Pythagorean Theorem, the length of this line segment is ( i ) + ( y i ). Summing over all subintervals gives an arc length approima on L n ( i ) + ( y i ). i= As it is wri en, this is not a Riemann Sum. While we could conclude that taking a limit as the subinterval length goes to zero gives the eact arc length, we would not be able to compute the answer with a definite integral. We need first to do a li le algebra. In the above epression factor out a i term: n ( i ) + ( y i ) = i= n i= ( i ) ( + ( y i) ( i ) Now pull the ( i ) term out of the square root: n L + ( y i) ( i ) i. i= This is nearly a Riemann Sum. Consider the ( y i ) /( i ) term. The epression y i / i measures the change in y/change in, that is, the rise over run of f on the i th subinterval. The Mean Value Theorem of Differen a on (Theorem 4) states that there is a c i in the i th subinterval where f (c i ) = y i / i. Thus we can rewrite our above epression as: n L + [f (c i )] i. i= ). 568

243 0. Arc Length and Surface Area This is a Riemann Sum. As long as f is con nuous on [a, b], we can invoke Theorem 36 and conclude L = b a + [f ()] d. Key Idea 37 Arc Length Let f be differen able on an open interval containing [a, b], where f is also con nuous on [a, b]. Then the arc length of f from = a to = b is b L = + [f ()] d. a Watch the video: Arc Length at As the integrand contains a square root, it is o en difficult to use the formula in Key Idea 37 to find the length eactly. When eact answers are difficult to come by, we resort to using numerical methods of approima ng definite integrals. The following eamples will demonstrate this. Eample Finding arc length Find the arc length of f() = 3/ from = 0 to = 4. y S A graph of f is given in Figure 0.5. We begin by finding 4 Figure 0.5: A graph of f() = 3/ from Eample. 569

244 Chapter 0 Curves in the Plane f () = 3 /. Using the formula, we find the arc length L as 4 ( ) 3 L = + 0 / d = = = = d ( ) / d ( ) 3/ 4 ( 0 3/ 0 ) units. Eample Finding arc length Find the arc length of f() = 8 ln from = to =. 0.5 y 3 Figure 0.6: A graph of f() = 8 ln from Eample. S A graph of f is given in Figure 0.6; the por on of the curve measured in this problem is in bold. This func on was chosen specifically because the resul ng integral can be evaluated eactly. We begin by finding f () = /4 /. The arc length is ( L = + 4 ) d = d = d ( = 4 + ) d ( = 4 + ) d ( ) = 8 + ln = 3 + ln units

245 0. Arc Length and Surface Area The previous eamples found the arc length eactly through careful choice of the func ons. In general, eact answers are much more difficult to come by and numerical approima ons are necessary. Eample 3 Approima ng arc length numerically Find the length of the sine curve from = 0 to = π. S This is somewhat of a mathema cal curiosity; in Eample we found the area under one hump of the sine curve is square units; now we are measuring its arc length. The setup is straigh orward: f() = sin and f () = cos. Thus π L = + cos d. 0 This integral cannot be evaluated in terms of elementary func ons so we will approimate it with Simpson s Method with n = 4. Figure 0.7 gives + cos evaluated at 5 evenly spaced points in [0, π]. Simpson s Rule then states that π + cos d π 0 ( ) + 4 3/ + () + 4 3/ Using a computer with n = 00 the approima on is L 3.80; our approima on with n = 4 is quite good. Our approima on of 3.79 from the beginning of this sec on isn t as close. + cos 0 π/4 3/ π/ 3π/4 3/ π Figure 0.7: A table of values of y = + cos to evaluate a definite integral in Eample 3. Surface Area of Solids of Revolu on We have already seen how a curve y = f() on [a, b] can be revolved around an ais to form a solid. Instead of compu ng its volume, we now consider its surface area. We begin as we have in the previous sec ons: we par on the interval [a, b] with n subintervals, where the i th subinterval is [ i, i+ ]. On each subinterval, we can approimate the curve y = f() with a straight line that connects f( i ) and f( i+ ) as shown in Figure 0.8(a). Revolving this line segment about the -ais creates part of a cone (called a frustum of a cone) as shown in Figure 0.8(b). The surface area of a frustum of a cone is A = πr avg L, where r avg is the average of R and R. The length is given by L; we use the material just covered by arc length to state that (a) L + [f (c i )] i (b) Figure 0.8: Establishing the formula for surface area. 57

246 Chapter 0 Curves in the Plane for some c i in the i th subinterval. The radii are just the func on evaluated at the endpoints of the interval: f( i ) and f( i ). Thus the surface area of this sample frustum of the cone is approimately π f( i ) + f( i ) + [f (c i )] i. Since f is a con nuous func on, the Intermediate Value Theorem states there is some d i in [ i, i ] such that f(d i ) = f( i ) + f( i ) ; we can use this to rewrite the above equa on as πf(d i ) + [f (c i )] i. Summing over all the subintervals we get the total surface area to be approimately n Surface Area πf(d i ) + [f (c i )] i, i= which is a Riemann Sum. Taking the limit as the subinterval lengths go to zero gives us the eact surface area, given in the upcoming Key Idea. If instead we revolve y = f() about the y-ais, the radii of the resul ng frustum are i and i ; their average value is simply the midpoint of the interval. In the limit, this midpoint is just. This gives the second part of Key Idea 38. Key Idea 38 Surface Area of a Solid of Revolu on Let f be differen able on an open interval containing [a, b] where f is also con nuous on [a, b].. The surface area of the solid formed by revolving the graph of y = f(), where f() 0, about the -ais is Surface Area = π b a f() + [f ()] d.. The surface area of the solid formed by revolving the graph of y = f() about the y-ais, where a, b 0, is Surface Area = π b a + [f ()] d. 57

247 0. Arc Length and Surface Area Eample 4 Finding surface area of a solid of revolu on Find the surface area of the solid formed by revolving y = sin on [0, π] around the -ais, as shown in Figure 0.9. S The setup turns out to be easier than the resul ng integral. Using Key Idea 38, we have the surface area SA is: π SA = π sin + cos d 0 = π + u du subs tute u = cos π/4 = π sec 3 θ dθ subs tute u = tan θ π/4 = π (sec θ tan θ + ln sec θ + tan θ ) π/4 π/4 by Eample 8..6 ( ( ) ( = π + ln + ( ))) + ln ( = π ( )) + + ln ( ( )) = π + ln + units ra onalize the denominator. It is interes ng to see that the surface area of a solid, whose shape is defined by a trigonometric func on, involves both a square root and a natural logarithm. Figure 0.9: Revolving y = sin on [0, π] about the -ais. Eample 5 Finding surface area of a solid of revolu on Find the surface area of the solid formed by revolving the curve y = on [0, ] about:. the -ais. the y-ais. S. The solid formed by revolving y = around the -ais is graphed in Figure 0.0(a). Like the integral in Eample 4, this integral is easier to setup than to actually integrate. While it is possible to use a trigonometric subs tu on to evaluate this integral, it is significantly more difficult than a solu on employing the hyperbolic sine: SA = π 0 + () d. = π ( (8 3 + ) ) sinh () 0 = π ( 8 ) 5 sinh units. 3 (a) (b) Figure 0.0: The solids used in Eample

248 Chapter 0 Curves in the Plane. Since we are revolving around the y-ais, the radius of the solid is not f() but rather. Thus the integral to compute the surface area is: SA = π () d = π u du subs tute u = = π u3/ = π ( 5 ) 5 units. 6 The solid formed by revolving y = about the y-ais is graphed in Figure 0.0 (b). Our final eample is a famous mathema cal parado. Eample 6 The surface area and volume of Gabriel s Horn Consider the solid formed by revolving y = / about the -ais on [, ). Find the volume and surface area of this solid. (This shape, as graphed in Figure 0., is known as Gabriel s Horn since it looks like a very long horn that only a supernatural person, such as an angel, could play.) Figure 0.: A graph of Gabriel s Horn. S We have: To compute the volume it is natural to use the Disk Method. V = π = lim b π = lim b π d b d ( ) b ( = lim π ) b b = π units 3. Gabriel s Horn has a finite volume of π cubic units. Since we have already seen that regions with infinite length can have a finite area, this is not too difficult to accept. 574

249 0. Arc Length and Surface Area We now consider its surface area. The integral is straigh orward to setup: SA = π + /4 d. Integra ng this epression is not trivial. We can, however, compare it to other improper integrals. Since < + / 4 on [, ), we can state that π d < π + /4 d. By Key Idea 3, the improper integral on the le diverges. Since the integral on the right is larger, we conclude it also diverges, meaning Gabriel s Horn has infinite surface area. Hence the parado : we can fill Gabriel s Horn with a finite amount of paint, but since it has infinite surface area, we can never paint it. Somehow this parado is striking when we think about it in terms of volume and area. However, we have seen a similar parado before, as referenced above. We know that the area under the curve y = / on [, ) is finite, yet the shape has an infinite perimeter. Strange things can occur when we deal with the infinite. 575

250 Eercises 0. Terms and Concepts. T/F: The integral formula for compu ng Arc Length was found by first approima ng arc length with straight line segments.. T/F: The integral formula for compu ng Arc Length includes a square root, meaning the integra on is probably easy. Problems In Eercises 3, find the arc length of the func on on the given interval. 3. f() = on [0, ]. 4. f() = 8 on [, ]. 5. f() = 3 3/ / on [0, ]. 6. f() = 3 + on [, 4]. 7. f() = 3/ 6 on [0, 9]. 8. f() = cosh on [ ln, ln ]. 9. f() = ( e + e ) on [0, ln 5]. 0. f() = 5 + on [., ]. 53. f() = ln ( sin ) on [π/6, π/].. f() = ln ( cos ) on [0, π/4]. In Eercises 3 0, set up the integral to compute the arc length of the func on on the given interval. Do not evaluate the integral. 3. f() = on [0, ]. 4. f() = 0 on [0, ]. 5. f() = on [0, ]. 6. f() = ln on [, e]. 7. f() = on [, ]. (Note: this describes the top half of a circle with radius.) 8. f() = /9 on [ 3, 3]. (Note: this describes the top half of an ellipse with a major ais of length 6 and a minor ais of length.) 9. f() = on [, ]. 0. f() = sec on [ π/4, π/4]. In Eercises 8, use Simpson s Rule, with n = 4, to approimate the arc length of the func on on the given interval. Note: these are the same problems as in Eercises f() = on [0, ].. f() = 0 on [0, ]. 3. f() = on [0, ]. (Note: f () is not defined at = 0.) 4. f() = ln on [, e]. 5. f() = on [, ]. (Note: f () is not defined at the endpoints.) 6. f() = /9 on [ 3, 3]. (Note: f () is not defined at the endpoints.) 7. f() = on [, ]. 8. f() = sec on [ π/4, π/4]. In Eercises 9 33, find the surface area of the described solid of revolu on. 9. The solid formed by revolving y = on [0, ] about the -ais. 30. The solid formed by revolving y = on [0, ] about the y-ais. 3. The solid formed by revolving y = 3 on [0, ] about the -ais. 3. The solid formed by revolving y = on [0, ] about the -ais. 33. The sphere formed by revolving y = on [, ] about the -ais. 576

251 0. Parametric Equa ons 0. Parametric Equa ons We are familiar with sketching shapes, such as parabolas, by following this basic procedure: Choose Use a func on f to find y ( y = f() ) Plot point (, y) In the rectangular coordinate system, the rectangular equa on y = f() works well for some shapes like a parabola with a ver cal ais of symmetry, but in precalculus and the review of conic sec ons in Sec on 0.0, we encountered several shapes that could not be sketched in this manner. (To plot an ellipse using the above procedure, we need to plot the top and bo om separately.) In this sec on we introduce a new sketching procedure: Use a func on f to find ( = f(t) ) Choose t Use a func on g to find y ( y = g(t) ) Plot point (, y) Here, and y are found separately but then plo ed together. This leads us to a defini on. Defini on 47 Parametric Equa ons and Curves Let f and g be con nuous func ons on an interval I. The graph of the ( parametric ) ( equa ons ) = f(t) and y = g(t) is the set of all points, y = f(t), g(t) in the Cartesian plane, as the parameter t varies over I. A curve is a graph along with the parametric equa ons that define it. 577

252 Chapter 0 Curves in the Plane Watch the video: Parametric Equa ons Some basic ques ons at y 4 t = 0 t y (a) t = t = t = 4 (b) t = Figure 0.: A table of values of the parametric equa ons in Eample along with a sketch of their graph y t y 0 π/4 / + / π/ 0 3π/4 / / π 0 (a) t = 0 t = π/ t = π/4 t = 3π/4 t = π (b) Figure 0.3: A table of values of the parametric equa ons in Eample along with a sketch of their graph. This is a formal defini on of the word curve. When a curve lies in a plane (such as the Cartesian plane), it is o en referred to as a plane curve. Eamples will help us understand the concepts introduced in the defini on. Eample Plo ng parametric func ons Plot the graph of the parametric equa ons = t, y = t + for t in [, ]. S We plot the graphs of parametric equa ons in much the same manner as we plo ed graphs of func ons like y = f(): we make a table of values, plot points, then connect these points with a reasonable looking curve. Figure 0.(a) shows such a table of values; note how we have 3 columns. The points (, y) from the table are plo ed in Figure 0.(b). The points have been connected with a smooth curve. Each point has been labeled with its corresponding t-value. These values, along with the two arrows along the curve, are used to indicate the orienta on of the graph. This informa on describes the path of a par cle traveling along the curve. We o en use the le er t as the parameter as we o en regard t as representing me. Certainly there are many contets in which the parameter is not me, but it can be helpful to think in terms of me as one makes sense of parametric plots and their orienta on (for instance, At me t = 0 the posi on is (, ) and at me t = 3 the posi on is (5, ). ). Eample Plo ng parametric func ons Sketch the graph of the parametric equa ons = cos t, y = cos t + for t in [0, π]. S We again start by making a table of values in Figure 0.3(a), then plot the points (, y) on the Cartesian plane in Figure 0.3(b). The curves in Eamples and are por ons of the same parabola (y ) + =. While the parabola is the same, the curves are different. In Eample, if we let t vary over all real numbers, we d obtain the en re parabola. In this eample, le ng t vary over all real numbers would s ll produce the same graph; this por on of the parabola would be traced, and re traced, infinitely o en. The orienta on shown in Figure 0.3 shows the orienta on on [0, π], but this orienta on is reversed on [π, π]. 578

253 0. Parametric Equa ons Conver ng between rectangular and parametric equa ons It is some mes useful to transform rectangular form equa ons (i.e., y = f()) into parametric form equa ons, and vice versa. Conver ng from rectangular to parametric can be very simple: given y = f(), the parametric equa ons = t, y = f(t) produce the same graph. As an eample, given y = 6, the parametric equa ons = t, y = t t 6 produce the same parabola. However, other parameteriza ons can be used. The following eample demonstrates one possible alterna ve. Eample 3 Conver ng from rectangular to parametric Find parametric equa ons for f() = 6. S Solu on : For any choice for we can determine the corresponding y by subs tu on. If we choose = t then y = (t ) (t ) 6 = t 3t 4. Thus f() can be represented by the parametric equa ons = t y = t 3t 4. On the graph of this parameteriza on (Figure 0.4(a)) the points have been labeled with the corresponding t-values and arrows indicate the path of a par cle traveling on this curve. The par cle would move from the upper le, down to the verte at (.5, 5.75) and then up to the right. Solu on : If we choose = 3 t then y = (3 t) (3 t) 6 = t 5t. Thus f() can also be represented by the parametric equa ons = 3 t y = t 5t. On the graph of this parameteriza on (Figure 0.4(b)) the points have been labeled with the corresponding t values and arrows indicate the path of a par cle traveling on this curve. The par cle would move down from the upper right, to the verte at (.5, 5.75) and then up to the le. Solu on 3: We can also parameterize any y = f() by se ng t = dy d. That is, t = a corresponds to the point on the graph whose tangent line has a slope a. Compu ng dy d, f () = we set t =. Solving for we find = t+ and by subs tu on y = 4 t 5 4. Thus f() can be represented by the parametric equa ons = t + y = 4 t 5 4. The graph of this parameteriza on is shown in Figure 0.4(c). To find the point where the tangent line has a slope of 0, we set t = 0. This gives us the point (.5, 5.75) which is the verte of f(). t = t = 5 t = t = 5 y (a) y t = 3 t = t = 3 5 (b) y t = t = t = (c) t = 3 t = 5 Figure 0.4: The equa on f() = 6 with different parameteriza ons. 579

254 Chapter 0 Curves in the Plane Eample 4 Conver ng from rectangular to parametric Find parametric equa ons for the circle + y = 4. S We will present three different approaches: ( ) ( ) y Solu on : Consider the equivalent equa on + = and the Pythagorean Iden ty, sin t + cos t =. We set cos t = and sin t = y, which gives = cos t and y = sin t. To trace the circle once, we must have 0 t π. Note that when t = 0 a par cle tracing the curve would be at the point (, 0) and would move in a counterclockwise direc on. Solu on : Another parameteriza on of the same circle would be = sin t and y = cos t for 0 t π. When t = 0 a par cle would be at the point (0, ) and would move in a clockwise direc on. Solu on 3: We could let = sin t and y = cos t for 0 t π. Also note that we could use = cos t and y = sin t for 0 t π. As we have shown in the previous eamples, there are many different ways to parameterize any given curve. We some mes choose the parameter to accurately model physical behavior. Eample 5 Conver ng from rectangular to parametric ( ) (y + 3) Find a parameteriza on that traces the ellipse at the point (, 3) in a clockwise direc on. = star ng S Applying the Pythagorean Iden ty, cos t + sin t =, we set cos ( ) t = and sin (y + 3) t =. Solving these equa ons for and 9 4 y we set = 3 cos t + and y = sin t 3 for 0 t π. Eample 6 Conver ng from rectangular to parametric ( ) (y 3) Find a parameteriza on for the hyperbola 9 4 =. S We use an alterna ve form of the Pythagorean Iden ty, sec t tan t =. We let sec ( ) t = and tan (y 3) t =. Solving these equa- 9 4 ons for and y we have = 3 sec t + and y = tan t + 3 for 0 t π. 580

255 0. Parametric Equa ons Eample 7 Conver ng from rectangular to parametric An object is fired from a height of 0 and lands 6 seconds later, 9 away. Assuming ideal projec le mo on, the height, in feet, of the object can be described by h() = /64 + 3, where is the distance in feet from the ini al loca on. (Thus h(0) = h(9) = 0.) Find parametric equa ons = f(t), y = g(t) for the path of the projec le where is the horizontal distance the object has traveled at me t (in seconds) and y is the height at me t. S Physics tells us that the horizontal mo on of the projec le is linear; that is, the horizontal speed of the projec le is constant. Since the object travels 9 in 6s, we deduce that the object is moving horizontally at a rate of 3 /s, giving the equa on = 3t. As y = /64 + 3, we find y = 6t + 96t. We can quickly verify that y = 3 /s, the accelera on due to gravity, and that the projec le reaches its maimum at t = 3, halfway along its path. These parametric equa ons make certain determina ons about the object s loca on easy: seconds into the flight the object is at the point ( (), y() ) = ( 64, 8 ). That is, it has traveled horizontally 64 and is at a height of 8, as shown in Figure y t = It is some mes necessary to convert given parametric equa ons into rectangular form. This can be decidedly more difficult, as some simple looking parametric equa ons can have very complicated rectangular equa ons. This conversion is o en referred to as elimina ng the parameter, as we are looking for a rela onship between and y that does not involve the parameter t. 50 = 3t y = 6t + 96t Figure 0.5: Graphing projec le mo on in Eample 7. Eample 8 Elimina ng the parameter Find a rectangular equa on for the curve described by = t + and y = t t +. S There is not a set way to eliminate a parameter. One method is to solve for t in one equa on and then subs tute that value in the second. We use that technique here, then show a second, simpler method. Star ng with = /(t + ), solve for t: t = ± /. Subs tute this 58

256 Chapter 0 Curves in the Plane value for t in the equa on for y: y = y = t + y = t t + Figure 0.6: Graphing parametric and rectangular equa ons for a graph in Eample 8. y = t t + = / / + = / / ( ) = =. Thus y =. One may have recognized this earlier by manipula ng the equa on for y: y = t t + = t + =. This is a shortcut that is very specific to this problem; some mes shortcuts eist and are worth looking for. We should be careful to limit the domain of the func on y =. The parametric equa ons limit to values in (0, ], thus to produce the same graph we should limit the domain of y = to the same. The graphs of these func ons are given in Figure 0.6. The por on of the graph defined by the parametric equa ons is given in a thick line; the graph defined by y = with unrestricted domain is given in a thin line. Eample 9 Elimina ng the parameter Eliminate the parameter in = 4 cos t + 3, y = sin t + S We should not try to solve for t in this situa on as the resul ng algebra/trig would be messy. Rather, we solve for cos t and sin t in each equa on, respec vely. This gives cos t = 3 4 and sin t = y. The Pythagorean Theorem gives cos t + sin t =, so: cos t + sin t = ( ) ( ) 3 y + = 4 ( 3) + 6 (y ) 4 = 58

257 0. Parametric Equa ons This final equa on should look familiar it is the equa on of an ellipse. Figure 0.7 plots the parametric equa ons, demonstra ng that the graph is indeed of an ellipse with a horizontal major ais and center at (3, ). y Graphs of Parametric Equa ons These eamples begin to illustrate the powerful nature of parametric equa ons. Their graphs are far more diverse than the graphs of func ons produced by y = f() func ons. One nice feature of parametric equa ons is that their graphs are easy to shi. While this is not too difficult in the y = f() contet, the resul ng func on can look rather messy. (Plus, to shi to the right by two, we replace with, which is counterintui ve.) The following eample demonstrates this. Eample 0 Shi ing the graph of parametric func ons Sketch the graph of the parametric equa ons = t + t, y = t t. Find new parametric equa ons that shi this graph to the right 3 units and down. S The graph of the parametric equa ons is given in Figure 0.8 (a). It is a parabola with an ais of symmetry along the line y = ; the verte is at (0, 0). It should be noted that finding the verte is not a trivial ma er and not something you will be asked to do in this tet. In order to shi the graph to the right 3 units, we need to increase the - value by 3 for every point. The straigh orward way to accomplish this is simply to add 3 to the func on defining : = t + t + 3. To shi the graph down by units, we wish to decrease each y-value by, so we subtract from the func on defining y: y = t t. Thus our parametric equa ons for the shi ed graph are = t + t + 3, y = t t. This is graphed in Figure 0.8 (b). No ce how the verte is now at (3, ). Because the - and y-values of a graph are determined independently, the graphs of parametric func ons o en possess features not seen on y = f() type graphs. The net eample demonstrates how such graphs can arrive at the same point more than once. Eample Graphs that cross themselves Plot the parametric func ons = t 3 5t + 3t + and y = t t + 3 and determine the t-values where the graph crosses itself Figure 0.7: Graphing the parametric equa ons = 4 cos t + 3, y = sin t + in Eample y y = t + t y = t t = t + t + 3 y = t t (a) (b) Figure 0.8: Illustra ng how to shi graphs in Eample

258 Chapter 0 Curves in the Plane 0 5 y 5 = t 3 5t + 3t + y = t t Figure 0.9: A graph of the parametric equa ons from Eample. S Using the methods developed in this sec on, we again plot points and graph the parametric equa ons as shown in Figure 0.9. It appears that the graph crosses itself at the point (, 6), but we ll need to analy cally determine this. We are looking for two different values, say, s and t, where (s) = (t) and y(s) = y(t). That is, the -values are the same precisely when the y-values are the same. This gives us a system of equa ons with unknowns: s 3 5s + 3s + = t 3 5t + 3t + s s + 3 = t t + 3 Solving this system is not trivial but involves only algebra. Using the quadra c formula, one can solve for t in the second equa on and find that t = ± s s +. This can be subs tuted into the first equa on, revealing that the graph crosses itself at t = and t = 3. We confirm our result by compu ng ( ) = (3) = and y( ) = y(3) = 6. We now present a small gallery of interes ng and famous curves along with parametric equa ons that produce them. 584

259 0. Parametric Equa ons y 6 y 6 y 4 4 Astroid = cos 3 t y = sin 3 t π Cycloid = r(t sin t) y = r( cos t) π Witch of Agnesi = at y = a/( + t ) y y 5 5 y Hypotrochoid = cos(t) + 5 cos(t/3) y = sin(t) 5 sin(t/3) Epicycloid = 4 cos(t) cos(4t) y = 4 sin(t) sin(4t) Folium of Descartes = 3at/( + t 3 ) y = 3at /( + t 3 ) One might note a feature shared by three of these graphs: sharp corners, or cusps. We have seen graphs with cusps before and determined that such func ons are not differen able at these points. This leads us to a defini on. Defini on 48 Smooth A curve C defined by = f(t), y = g(t) is smooth on an interval I if f and g are con nuous on I and not simultaneously 0 (ecept possibly at the endpoints of I). A curve is piecewise smooth on I if I can be par oned into subintervals where C is smooth on each subinterval. Consider the astroid, given by = cos 3 t, y = sin 3 t. Taking deriva ves, we have: = 3 cos t sin t and y = 3 sin t cos t. 585

260 Chapter 0 Curves in the Plane It is clear that each is 0 when t = 0, π/, π,.... Thus the astroid is not smooth at these points, corresponding to the cusps seen in the figure. However, by restric ng the domain of the astroid to all reals ecept t = kπ for k Z we have a piecewise smooth curve. We demonstrate this once more. Eample Determine where a curve is not smooth Let a curve C be defined by the parametric equa ons = t 3 t + 7 and y = t 4t + 8. Determine the points, if any, where it is not smooth. 8 y S We begin by taking deriva ves. = 3t, y = t Figure 0.30: Graphing the curve in Eample ; note it is not smooth at (, 4). We set each equal to 0: = 0 3t = 0 t = ± y = 0 t 4 = 0 t = We consider only the value of t = since both and y must be 0. Thus C is not smooth at t =, corresponding to the point (, 4). The curve is graphed in Figure 0.30, illustra ng the cusp at (, 4). If a curve is not smooth at t = t 0, it means that (t 0 ) = y (t 0 ) = 0 as defined. This, in turn, means that rate of change of (and y) is 0; that is, at that instant, neither nor y is changing. If the parametric equa ons describe the path of some object, this means the object is at rest at t 0. An object at rest can make a sharp change in direc on, whereas moving objects tend to change direc on in a smooth fashion. Eample 3 The Cycloid A well-known parametric curve is the cycloid. Fi r, and let = r(t sin t), y = r( cos t). This represents the path traced out by a point on a wheel of radius r as starts rolling to the right. We can think of t as the angle through which the point has rotated. t = 0 t = π 3 t = 4π 3 t = π t = 8π 3 t = 0π 3 t = 4π Figure 0.3: A cycloid traced through two revolu ons. 586

261 0. Parametric Equa ons Figure 0.3 shows a cycloid sketched out with the wheel shown at various places. The dot on the rim is the point on the wheel that we re using to trace out the curve. From this sketch we can see that one arch of the cycloid is traced out in the range 0 t π. This makes sense when you consider that the point will be back on the ground a er it has rotated through an angle of π. One should be careful to note that a sharp corner does not have to occur when a curve is not smooth. For instance, one can verify that = t 3, y = t 6 produce the familiar y = parabola. However, in this parametriza on, the curve is not smooth. A par cle traveling along the parabola according to the given parametric equa ons comes to rest at t = 0, though no sharp point is created. Our previous eperience with cusps taught us that a func on was not differen able at a cusp. This can lead us to wonder about deriva ves in the contet of parametric equa ons and the applica on of other calculus concepts. Given a curve defined parametrically, how do we find the slopes of tangent lines? Can we determine concavity? We eplore these concepts and more in the net sec- on. 587

262 Eercises 0. Terms and Concepts. T/F: When sketching the graph of parametric equa ons, the and y values are found separately, then plo ed together.. The direc on in which a graph is moving is called the of the graph. 3. An equa on wri en as y = f() is wri en in form. 4. Create parametric equa ons = f(t), y = g(t) and sketch their graph. Eplain any interes ng features of your graph based on the func ons f and g. Problems In Eercises 5 8, sketch the graph of the given parametric equa ons by hand, making a table of points to plot. Be sure to indicate the orienta on of the graph. 5. = t + t, y = t, 3 t 3 6. =, y = 5 sin t, π/ t π/ 7. = t, y =, t 8. = t 3 t + 3, y = t +, t In Eercises 9 7, sketch the graph of the given parametric equa ons; using a graphing u lity is advisable. Be sure to indicate the orienta on of the graph. 9. = t 3 t, y = t, t 3 0. = /t, y = sin t, 0 < t 0. = 3 cos t, y = 5 sin t, 0 t π. = 3 cos t +, y = 5 sin t + 3, 0 t π 3. = cos t, y = cos(t), 0 t π 4. = cos t, y = sin(t), 0 t π 5. = sec t, y = 3 tan t, π/ < t < π/ 6. = cos t+ cos(8t), 4 y = sin t+ sin(8t), 4 0 t π 7. = cos t+ sin(8t), 4 y = sin t+ cos(8t), 4 0 t π In Eercises 8 9, four sets of parametric equa ons are given. Describe how their graphs are similar and different. Be sure to discuss orienta on and ranges (a) = t y = t, < t < (b) = sin t y = sin t, < t < (c) = e t y = e t, < t < (d) = t y = t, < t < (a) = cos t y = sin t, 0 t π (b) = cos(t ) y = sin(t ), 0 t π (c) = cos(/t) y = sin(/t), 0 < t < (d) = cos(cos t) y = sin(cos t), 0 t π In Eercises 0, find a parameteriza on for the curve. 0. y = y = 5. ( + 9) + (y 4) = 49 In Eercises 3 6, find a parametric equa on and a parameter interval. 3. The line segment with endpoints (, 3) and (4, ) 4. The line segment with endpoints (, 3) and (3, ) 5. The le half of the parabola y = + 6. The lower half of the parabola = y In Eercises 7 30, find parametric equa ons for the given rectangular equa on using the parameter t = dy. Verify that d at t =, the point on the graph has a tangent line with slope of. 7. y = y = e 9. y = sin on [0, π] 30. y = on [0, ) 3. Find parametric equa ons and a parameter interval for the mo on of a par cle that starts at (, 0) and traces the circle + y = (a) once clockwise (c) twice clockwise (b) once (d) twice counter-clockwise counter-clockwise 3. Find parametric equa ons and a parameter interval for the mo on of a par cle that starts at (a, 0) and traces the ellipse + y = a b (a) once clockwise (c) twice clockwise (b) once (d) twice counter-clockwise counter-clockwise In Eercises 33 4, find parametric equa ons that describe the given situa on. 33. A projec le is fired from a height of 0, landing 6 away in 4s. 34. A projec le is fired from a height of 0, landing 00 away in 4s. 35. A projec le is fired from a height of 0, landing 00 away in 0s. 36. A circle of radius, centered at the origin, that is traced clockwise once on [0, π]. 37. A circle of radius 3, centered at (, ), that is traced once counter clockwise on [0, ]. 38. An ellipse centered at (, 3) with ver cal major ais of length 6 and minor ais of length. 39. An ellipse with foci at (±, 0) and ver ces at (±5, 0). 40. A hyperbola with foci at (5, 3) and (, 3), and with ver ces at (, 3) and (3, 3). 4. A hyperbola with ver ces at (0, ±6) and asymptotes y = ±3. 588

263 In Eercises 4 5, eliminate the parameter in the given parametric equa ons. 4. = t + 5, y = 3t = sec t, y = tan t 44. = 4 sin t +, y = 3 cos t 45. = t, y = t = t +, y = 3t + 5 t = e t, y = e 3t = ln t, y = t 49. = cot t, y = csc t 50. = cosh t, y = sinh t 5. = cos(t), y = sin t In Eercises 5 55, eliminate the parameter in the given parametric equa ons. Describe the curve defined by the parametric equa ons based on its rectangular form. 53. = r cos t, y = r sin t 54. = a cos t + h, y = b sin t + k 55. = a sec t + h, y = b tan t + k In Eercises 56 59, find the values of t where the graph of the parametric equa ons crosses itself. 56. = t 3 t + 3, y = t = t 3 4t + t + 7, y = t t 58. = cos t, y = sin(t) on [0, π] 59. = cos t cos(3t), y = sin t cos(3t) on [0, π] In Eercises 60 63, find the value(s) of t where the curve defined by the parametric equa ons is not smooth. 60. = t 3 + t t, y = t + t = t 4t, y = t 3 t 4t 6. = cos t, y = cos t 63. = cos t cos(t), y = sin t sin(t) 5. = at + 0, y = bt + y 0 589

264 Chapter 0 Curves in the Plane 0.3 Calculus and Parametric Equa ons The previous sec on defined curves based on parametric equa ons. In this sec- on we ll employ the techniques of calculus to study these curves. We are s ll interested in lines tangent to points on a curve. They describe how the y-values are changing with respect to the -values, they are useful in making approima ons, and they indicate instantaneous direc on of travel. The slope of the tangent line is s ll dy d, and the Chain Rule allows us to calculate this in the contet of parametric equa ons. If = f(t) and y = g(t), the Chain Rule states that Solving for dy d, we get dy dy dt = dy d d dt. d = dy/dt d/dt = g (t) f (t), provided that f (t) 0. This is important so we label it a Key Idea. Key Idea 39 Finding dy d with Parametric Equa ons. Let = f(t) and y = g(t), where f and g are differen able on some open interval I and f (t) 0 on I. Then dy d = dy/dt d/dt = g (t) f (t). We use this to define the tangent line. Defini on 49 Tangent Lines Let a curve C be parameterized by = f(t) and y = g(t), where f and g are differen able func ons on some interval I containing t = t 0. The tangent line to C at t = t 0 is the line through (f(t 0 ), g(t 0 )) with slope m = g (t 0) f (t, provided f 0) (t 0 ) 0. It is possible for parametric curves to have horizontal and ver cal tangents. As epected a horizontal tangent occurs whenever dy d = 0 or when dy dt = 0 (provided d dy dt 0). Similarly, a ver cal tangent occurs whenever d is undefined or when d dy dt = 0 (provided dt 0). 590

265 0.3 Calculus and Parametric Equa ons Defini on 50 Normal Lines The normal line to a curve C at a point P is the line through P and perpendicular to the tangent line at P. For t = t 0 the normal line is the line through (f(t 0 ), g(t 0 )) with slope m = f (t 0) g (t, provided 0) g (t 0 ) 0. As with the tangent line we note that it is possible for a normal line to be ver cal or horizontal. A horizontal normal line occurs whenever dy d is undefined or when d dt = 0 (provided dy dt 0). Similarly, a ver cal normal line occurs whenever dy dy d = 0 or when dt = 0 (provided d dt 0). In other words, if the curve C has a ver cal tangent at (f(t 0 ), g(t 0 )) the normal line will be horizontal and if the tangent is horizontal the normal line will be a ver cal line. Watch the video: Deriva ves of Parametric Func ons at Eample Tangent and Normal Lines to Curves Let = 5t 6t + 4 and y = t + 6t, and let C be the curve defined by these equa ons.. Find the equa ons of the tangent and normal lines to C at t = 3.. Find where C has ver cal and horizontal tangent lines. S. We start by compu ng f (t) = 0t 6 and g (t) = t + 6. Thus dy d = t + 6 0t 6. Make note of something that might seem unusual: dy d is a func on of t, not. Just as points on the curve are found in terms of t, so are the slopes of the tangent lines. The point on C at t = 3 is (3, 6). The slope of the tangent line is m = / and the slope of the normal line is m =. Thus, 59

266 Chapter 0 Curves in the Plane the equa on of the tangent line is y = ( 3) + 6, and the equa on of the normal line is y = ( 3) y Figure 0.3: Graphing tangent and normal lines in Eample. This is illustrated in Figure To find where C has a horizontal tangent line, we set dy d = 0 and solve for t. In this case, this amounts to se ng g (t) = 0 and solving for t (and making sure that f (t) 0). g (t) = 0 t + 6 = 0 t = 3. The point on C corresponding to t = 3 is (67, 0); the tangent line at that point is horizontal (hence with equa on y = 0). To find where C has a ver cal tangent line, we find where it has a horizontal normal line, and set f (t) g (t) = 0. This amounts to se ng f (t) = 0 and solving for t (and making sure that g (t) 0). f (t) = 0 0t 6 = 0 t = 0.6. The point on C corresponding to t = 0.6 is (.,.96). The tangent line at that point is =.. The points where the tangent lines are ver cal and horizontal are indicated on the graph in Figure 0.3. Eample Tangent and Normal Lines to a Circle. Find where the unit circle, defined by = cos t and y = sin t on [0, π], has ver cal and horizontal tangent lines.. Find the equa on of the normal line at t = t 0. S. We compute the deriva ve following Key Idea 39: dy d = g (t) f (t) = cos t sin t. The deriva ve is 0 when cos t = 0; that is, when t = π/, 3π/. These are the points (0, ) and (0, ) on the circle. The normal line is horizontal (and hence, the tangent line is ver cal) when sin t = 0; that is, when t = 0, π, π, corresponding to the points (, 0) and (0, ) on the circle. These results should make intui ve sense. 59

267 0.3 Calculus and Parametric Equa ons. The slope of the normal line at t = t 0 is m = sin t 0 cos t 0 = tan t 0. This normal line goes through the point (cos t 0, sin t 0 ), giving the line y = sin t 0 cos t 0 ( cos t 0 ) + sin t 0 = (tan t 0 ), as long as cos t 0 0. It is an important fact to recognize that the normal lines to a circle pass through its center, as illustrated in Figure Stated in another way, any line that passes through the center of a circle intersects the circle at right angles. y Eample 3 Tangent lines when dy d is not defined Find the equa on of the tangent line to the astroid = cos 3 t, y = sin 3 t at t = 0, shown in Figure S We start by finding (t) and y (t): (t) = 3 sin t cos t, y (t) = 3 cos t sin t. Note that both of these are 0 at t = 0; the curve is not smooth at t = 0 forming a cusp on the graph. Evalua ng dy d at this point returns the indeterminate form of 0/0. We can, however, eamine the slopes of tangent lines near t = 0, and take the limit as t 0. y (t) lim t 0 (t) = lim 3 cos t sin t t 0 3 sin t cos t = lim sin t t 0 cos t = 0. (We can reduce as t 0.) We have accomplished something significant. When the deriva ve dy d returns an indeterminate form at t = t 0, we can define its value by se ng it to be lim d, if that limit eists. This allows us to find slopes of tangent lines at cusps, which can be very beneficial. We found the slope of the tangent line at t = 0 to be 0; therefore the tangent line is y = 0, the -ais. t t0 dy Figure 0.33: Illustra ng how a circle s normal lines pass through its center. y Figure 0.34: A graph of an astroid. 593

268 Chapter 0 Curves in the Plane Concavity We con nue to analyze curves in the plane by considering their concavity; that is, we are interested in d y d, the second deriva ve of y with respect to. To find this, we need to find the deriva ve of dy d with respect to ; that is, d y d = d d [ ] dy, d but recall that dy d is a func on of t, not, making this computa on not straightforward. To make the upcoming nota on a bit simpler, let h(t) = dy d. We want d dh d [h(t)]; that is, we want d. We again appeal to the Chain Rule. Note: dh dt = dh d d dt dh d = dh/dt d/dt. In words, to find d y dy, we first take the deriva ve of d d then divide by (t). We restate this as a Key Idea. with respect to t, Key Idea 40 Finding d y d with Parametric Equa ons Let = f(t) and y = g(t) be twice differen able func ons on an open interval I, where f (t) 0 on I. Then [ ] [ ] d dy d dy d y dt d dt d d = = d f. (t) dt Eamples will help us understand this Key Idea. Eample 4 Concavity of Plane Curves Let = 5t 6t+4 and y = t +6t as in Eample. Determine the t-intervals on which the graph is concave up/down. S Concavity is determined by the second deriva ve of y with respect to, d y d, so we compute that here following Key Idea

269 0.3 Calculus and Parametric Equa ons In Eample, we found dy d = t + 6 0t 6 and f (t) = 0t 6. So: d y d = d dt [ t+6 0t 6 ] 0t 6 = 7 (0t 6) 0t 6 7 = (0t 6) 3 9 = (5t 3) 3 The graph of the parametric func ons is concave up when d y d > 0 and concave down when d y d < 0. We determine the intervals when the second deriva- ve is greater/less than 0 by first finding when it is 0 or undefined. 9 d As the numerator of is never 0, y (5t 3) 3 d 0 for all t. It is undefined when 5t 3 = 0; that is, when t = 3/5. Following the work established in Sec on 3.4, we look at values of t greater or less than 3/5 on a number line: 3 5 f + f CU CD Reviewing Eample, we see that when t = 3/5 = 0.6, the graph of the parametric equa ons has a ver cal tangent line. This point is also a point of inflec on for the graph, illustrated in Figure y t > 3/5; concave down t < 3/5; concave up Figure 0.35: Graphing the parametric equa ons in Eample 4 to demonstrate concavity. Eample 5 Concavity of Plane Curves Find the points of inflec on of the graph of the parametric equa ons = t, y = sin t, for 0 t 6. S We need to compute dy d and d y d. dy d = y (t) (t) = [ d d y dy ] d = dt d (t) cos t /( t) = t cos t. = cos t/ t t sin t /( t) = cos t 4t sin t. 595

270 Chapter 0 Curves in the Plane 50 y The possible points of inflec on are found by se ng d y d = 0. This is not trivial, as equa ons that mi polynomials and trigonometric func ons generally do not have nice solu ons. In Figure 0.36(a) we see a plot of the second deriva ve. It shows that it has zeros at approimately t = 0.5, 3.5, 6.5, 9.5,.5 and 6. These approima- ons are not very good, made only by looking at the graph. Newton s Method provides more accurate approima ons. Accurate to decimal places, we have: t t = 0.65, 3.9, 6.36, 9.48,.6 and y y = cos t 4t sin t (a) 3 4 The corresponding points have been plo ed on the graph of the parametric equa ons in Figure 0.36(b). Note how most occur near the -ais, but not eactly on the ais. Area with Parametric Equa ons We will now find a formula for determining the area under a parametric curve given by the parametric equa ons = f(t) y = g(t). (b) Figure 0.36: In (a), a graph of d y d, showing where it is approimately 0. In (b), graph of the parametric equa ons in Eample 5 along with the points of inflec- on. We will also need to further add in the assump on that the curve is traced out eactly once as t increases from α to β. First, recall how to find the area under y = F() on a b: A = b a F() d. Now think of the parametric equa on = f(t) as a subs tu on in the integral, assuming that a = f(α) and b = f(β) for the purposes of this formula. (There is actually no reason to assume that this will always be the case and so we ll give a corresponding formula later if it s the opposite case (b = f(α) and a = f(β)).) In order to subs tute, we ll need d = f (t) dt. Plugging this into the area formula above and making sure to change the limits to their corresponding t values gives us A = β α F(f(t))f (t) dt. Since we don t know what F() is, we ll use the fact that and arrive at the formula that we want. y = F() = F(f(t)) = g(t) 596

271 0.3 Calculus and Parametric Equa ons Key Idea 4 Area Under a Parametric Curve The area under the parametric curve given by = f(t), y = g(t), for f(α) = a < < b = f(β) is A = β α g(t)f (t) dt. On the other hand, if we should happen to have b = f(α) and a = f(β), then the formula would be Let s work an eample. A = α β g(t)f (t) dt. Eample 6 Finding the area under a parametric curve Determine the area under the cycloid given by the parametric equa ons = 6(θ sin θ) y = 6( cos θ) 0 θ π. S First, no ce that we ve switched the parameter to θ for this problem. This is to make sure that we don t get too locked into always having t as the parameter. Now, we could graph this to verify that the curve is traced out eactly once for the given range if we wanted to. There really isn t too much to this eample other than plugging the parametric equa ons into the formula. We ll first need the deriva ve of the parametric equa on for however. d = 6( cos θ). dθ The area is then A = π = 36 = 36 = 36 0 π 36( cos θ) dθ 0 π 0 = 08π. cos θ + cos θ dθ 3 cos θ + cos(θ) dθ [ 3 θ sin θ + 4 sin(θ) ] π 0 597

272 Chapter 0 Curves in the Plane Arc Length We con nue our study of the features of the graphs of parametric equa ons by compu ng their arc length. Recall in Sec on 0. we found the arc length of the graph of a func on, from = a to = b, to be L = b a + ( ) dy d. d We can use this equa on and convert it to the parametric equa on contet. Le ng = f(t) and y = g(t), we know that dy d = g (t)/f (t). It will also be useful to calculate the differen al of : d = f (t)dt dt = f (t) d. Star ng with the arc length formula above, consider: b ( ) dy L = + d a d b = + [g (t)] [f (t)] d = = a b a t t [f (t)] + [g (t)] [f (t)] + [g (t)] dt. f (t) d }{{} =dt Factor out the [f (t)] Note the new bounds (no longer bounds, but t bounds). They are found by finding t and t such that a = f(t ) and b = f(t ). This formula is important, so we restate it as a theorem. Theorem 84 Arc Length of Parametric Curves Let = f(t) and y = g(t) be parametric equa ons with f and g con- nuous on some open interval I containing t and t on which the graph traces itself only once. The arc length of the graph, from t = t to t = t, is t L = [f (t)] + [g (t)] dt. t 598

273 0.3 Calculus and Parametric Equa ons As before, these integrals are o en not easy to compute. We start with a simple eample, then give another where we approimate the solu on. Eample 7 Arc Length of a Circle Find the arc length of the circle parametrized by = 3 cos t, y = 3 sin t on [0, 3π/]. S By direct applica on of Theorem 84, we have L = 3π/ Apply the Pythagorean Theorem. = 0 3π/ 0 = 3t 3π/ 0 ( 3 sin t) + (3 cos t) dt. 3 dt = 9π/. This should make sense; we know from geometry that the circumference of a circle with radius 3 is 6π; since we are finding the arc length of 3/4 of a circle, the arc length is 3/4 6π = 9π/. Eample 8 Arc Length of a Parametric Curve The graph of the parametric equa ons = t(t ), y = t crosses itself as shown in Figure 0.37, forming a teardrop. Find the arc length of the teardrop. S We can see by the parameteriza ons of and y that when t = ±, = 0 and y = 0. This means we ll integrate from t = to t =. Applying Theorem 84, we have L = (3t ) + (t) dt = 9t4 t + dt. Unfortunately, the integrand does not have an an deriva ve epressible by elementary func ons. We turn to numerical integra on to approimate its value. Using 4 subintervals, Simpson s Rule approimates the value of the integral as Using a computer, more subintervals are easy to employ, and n = 0 gives a value of Increasing n shows that this value is stable and a good approima on of the actual value. y Figure 0.37: A graph of the parametric equa ons in Eample 8, where the arc length of the teardrop is calculated. t 599

274 Chapter 0 Curves in the Plane Surface Area of a Solid of Revolu on Related to the formula for finding arc length is the formula for finding surface area. We can adapt the formula found in Key Idea 38 from Sec on 0. in a similar way as done to produce the formula for arc length done before. Key Idea 4 Surface Area of a Solid of Revolu on Consider the graph of the parametric equa ons = f(t) and y = g(t), where f and g are con nuous on an open interval I containing t and t on which the graph does not cross itself.. The surface area of the solid formed by revolving the graph about the -ais is (where g(t) 0 on [t, t ]): t Surface Area = π g(t) [f (t)] + [g (t)] dt. t. The surface area of the solid formed by revolving the graph about the y-ais is (where f(t) 0 on [t, t ]): t Surface Area = π f(t) [f (t)] + [g (t)] dt. t Eample 9 Surface Area of a Solid of Revolu on Consider the teardrop shape formed by the parametric equa ons = t(t ), y = t as seen in Eample 8. Find the surface area if this shape is rotated about the -ais, as shown in Figure Figure 0.38: Rota ng a teardrop shape about the -ais in Eample 9. S The teardrop shape is formed between t = and t =. Using Key Idea 4, we see we need for g(t) 0 on [, ], and this is not the case. To fi this, we simplify replace g(t) with g(t), which flips the whole graph about the -ais (and does not change the surface area of the resul ng solid). The surface area is: Area S = π = π ( t ) (3t ) + (t) dt ( t ) 9t 4 t + dt. Once again we arrive at an integral that we cannot compute in terms of elementary func ons. Using Simpson s Rule with n = 0, we find the area to be 600

275 0.3 Calculus and Parametric Equa ons S = Using larger values of n shows this is accurate to places a er the decimal. A er defining a new way of crea ng curves in the plane, in this sec on we have applied calculus techniques to the parametric equa on defining these curves to study their proper es. In the net sec on, we define another way of forming curves in the plane. To do so, we create a new coordinate system, called polar coordinates, that iden fies points in the plane in a manner different than from measuring distances from the y- and - aes. 60

276 Eercises 0.3 Terms and Concepts. T/F: Given parametric equa ons = f(t) and y = g(t), dy = f (t)/g (t), as long as g (t) 0. d. Given parametric equa ons = f(t) and y = g(t), the deriva ve dy as given in Key Idea 39 is a func on of d? 3. T/F: Given parametric equa ons ( = ) f(t) and y = g(t), to find d y, one simply computes d dy. d dt d 4. T/F: If dy = 0 at t = t0, then the normal line to the curve d at t = t 0 is a ver cal line. Problems In Eercises 5, parametric equa ons for a curve are given. (a) Find dy d. (b) Find the equa ons of the tangent and normal line(s) at the point(s) given. (c) Sketch the graph of the parametric func ons along with the found tangent and normal lines. 5. = t, y = t ; t = 6. = t, y = 5t + ; t = 4 7. = t t, y = t + t; t = 8. = t, y = t 3 t; t = 0 and t = 9. = sec t, y = tan t on ( π/, π/); t = π/4 0. = cos t, y = sin(t) on [0, π]; t = π/4. = cos t sin(t), y = sin t sin(t) on [0, π]; t = 3π/4. = e t/0 cos t, y = e t/0 sin t; t = π/ In Eercises 3 0, find t-values where the curve defined by the given parametric equa ons has a horizontal tangent line. Note: these are the same equa ons as in Eercises = t, y = t 4. = t, y = 5t + 5. = t t, y = t + t 6. = t, y = t 3 t 7. = sec t, y = tan t on ( π/, π/) 8. = cos t, y = sin(t) on [0, π] 9. = cos t sin(t), y = sin t sin(t) on [0, π] 0. = e t/0 cos t, y = e t/0 sin t In Eercises 4, find t = t 0 where the graph of the given dy parametric equa ons is not smooth, then find lim t t0 d.. = t +, y = t3. = t 3 + 7t 6t + 3, y = t 3 5t + 8t 3. = t 3 3t + 3t, y = t t + 4. = cos t, y = sin t In Eercises 5 3, parametric equa ons for a curve are given. Find d y, then determine the intervals on which the graph of d the curve is concave up/down. Note: these are the same equa- ons as in Eercises = t, y = t 6. = t, y = 5t + 7. = t t, y = t + t 8. = t, y = t 3 t 9. = sec t, y = tan t on ( π/, π/) 30. = cos t, y = sin(t) on [0, π] 3. = cos t sin(t), y = sin t sin(t) on [ π/, π/] 3. = e t/0 cos t, y = e t/0 sin t In Eercises 33 40, find the arc length of the graph of the parametric equa ons on the given interval(s). 33. = 3 sin(t), y = 3 cos(t) on [0, π] 34. = e t/0 cos t, y = e t/0 sin t on [0, π] and [π, 4π] 35. = 5t +, y = 3t on [, ] 36. = t 3/, y = 3t on [0, ] 37. = cos t, y = sin t on [0, π] 38. = + 3t, y = 4 + t 3 on [0, ] 39. = t, + t y = ln( + t) on [0, ] 40. = e t t, y = 4e t/ on [ 8, 3] In Eercises 4 44, numerically approimate the given arc length. 4. Approimate the arc length of one petal of the rose curve = cos t cos(t), y = sin t cos(t) using Simpson s Rule and n = Approimate the arc length of the bow e curve = cos t, y = sin(t) using Simpson s Rule and n = Approimate the arc length of the parabola = t t, y = t + t on [, ] using Simpson s Rule and n = A common approimate of the circumference of an ellipse a + b given by = a cos t, y = b sin t is C π. Use this formula to approimate the circumference of = 5 cos t, y = 3 sin t and compare this to the approima- on given by Simpson s Rule and n = 6. In Eercises 45 5, a solid of revolu on is described. Find or approimate its surface area as specified. 45. Find the surface area of the sphere formed by rota ng the circle = cos t, y = sin t about: (a) the -ais and (b) the y-ais. 46. Find the surface area of the torus (or donut ) formed by rota ng the circle = cos t +, y = sin t about the y-ais. 60

277 47. Find the surface area of the solid formed by rota ng the curve = a cos 3 θ, y = a sin 3 θ on [0, π/] about the ais 48. Find the surface area of the solid formed by rota ng the curve = t 3, y = t on [0, ] about the ais 49. Find the surface area of the solid formed by rota ng the curve = 3t, y = t 3 on [0, 5] about the y ais 50. Approimate the surface area of the solid formed by rotating the upper right half of the bow e curve = cos t, y = sin(t) on [0, π/] about the -ais, using Simpson s Rule and n = Approimate the surface area of the solid formed by rota ng the one petal of the rose curve = cos t cos(t), y = sin t cos(t) on [0, π/4] about the -ais, using Simpson s Rule and n =

278 Chapter 0 Curves in the Plane 0.4 Introduc on to Polar Coordinates We are generally introduced to the idea of graphing curves by rela ng -values to y-values through a func on f. That is, we set y = f(), and plot lots of point pairs (, y) to get a good no on of how the curve looks. This method is useful but has limita ons, not least of which is that curves that fail the ver cal line test cannot be graphed without using mul ple func ons. The previous two sec ons introduced and studied a new way of plo ng points in the, y-plane. Using parametric equa ons, and y values are computed independently and then plo ed together. This method allows us to graph an etraordinary range of curves. This sec on introduces yet another way to plot points in the plane: using polar coordinates. Polar Coordinates O r (r, θ) θ ini al ray Start with a point O in the plane called the pole (we will always iden fy this point with the origin). From the pole, draw a ray, called the ini al ray (we will always draw this ray horizontally, iden fying it with the posi ve -ais). A point P in the plane is determined by the distance r that P is from O, and the angle θ formed between the ini al ray and the segment OP (measured counter-clockwise). We record the distance and angle as an ordered pair (r, θ). Illustra ng polar coordi- Figure 0.39: nates. Watch the video: Polar Coordinates The Basics at Prac ce will make this process more clear. Eample Plo ng Polar Coordinates Plot the following polar coordinates: A(, π/4) B(.5, π) C(, π/3) D(, π/4) B A O 3 S To aid in the drawing, a polar grid is provided at the bo om of this page. To place the point A, go out unit along the ini al ray (pu ng you on the inner circle shown on the grid), then rotate counter-clockwise π/4 radians (or 45 ). Alternately, one can consider the rota on first: think about the ray from O that forms an angle of π/4 with the ini al ray, then move out unit along this ray (again placing you on the inner circle of the grid). D C Figure 0.40: Plo ng polar points in Eample. O 3 604

279 0.4 Introduc on to Polar Coordinates To plot B, go out.5 units along the ini al ray and rotate π radians (80 ). To plot C, go out units along the ini al ray then rotate clockwise π/3 radians, as the angle given is nega ve. To plot D, move along the ini al ray units in other words, back up unit, then rotate counter-clockwise by π/4. The results are given in Figure Consider the following two points: A(, π) and B(, 0). To locate A, go out unit on the ini al ray then rotate π radians; to locate B, go out units on the ini al ray and don t rotate. One should see that A and B are located at the same point in the plane. We can also consider C(, 3π), or D(, π); all four of these points share the same loca on. This ability to iden fy a point in the plane with mul ple polar coordinates is both a blessing and a curse. We will see that it is beneficial as we can plot beau ful func ons that intersect themselves (much like we saw with parametric func ons). The unfortunate part of this is that it can be difficult to determine when this happens. We ll eplore this more later in this sec on. Polar to Rectangular Conversion It is useful to recognize both the rectangular (or, Cartesian) coordinates of a point in the plane and its polar coordinates. Figure 0.4 shows a point P in the plane with rectangular coordinates (, y) and polar coordinates (r, θ). Using trigonometry, we can make the iden es given in the following Key Idea. Key Idea 43 Conver ng Between Rectangular and Polar Coordinates Given the polar point P(r, θ), the rectangular coordinates are determined by = r cos θ y = r sin θ. Given the rectangular coordinates (, y), the polar coordinates are determined by r = + y tan θ = y. O r θ Figure 0.4: Conver ng between rectangular and polar coordinates. P y Eample Conver ng Between Polar and Rectangular Coordinates. Convert the polar coordinates A(, π/3) and B(, 5π/4) to rectangular coordinates.. Convert the rectangular coordinates (, ) and (, ) to polar coordinates. 605

280 Chapter 0 Curves in the Plane A(, π 3 ) O (a) B(, 5π 4 ) S. (a) We start with A(, π/3). Using Key Idea 43, we have = cos(π/3) = y = sin(π/3) = 3. So the rectangular coordinates are (, 3) (,.73). (b) The polar point B(, 5π/4) is converted to rectangular with: = cos(5π/4) = / y = sin(5π/4) = /. So the rectangular coordinates are ( /, /) (0.707, 0.707). These points are plo ed in Figure 0.4 (a). The rectangular coordinate system is drawn lightly under the polar coordinate system so that the rela onship between the two can be seen.. (a) To convert the rectangular point (, ) to polar coordinates, we use the Key Idea to form the following two equa ons: + = r tan θ =. The first equa on tells us that r = 5. Using the inverse tangent func on, we find tan θ = θ = tan. radians Thus polar coordinates of (, ) are ( 5,.). (b) To convert (, ) to polar coordinates, we form the equa ons ( ) + = r tan θ =. (, ) π 4 (0, 0) (b) (, ) 3π 4. Figure 0.4: Plo ng rectangular and polar points in Eample. Thus r =. We need to be careful in compu ng θ: using the inverse tangent func on, we have tan θ = θ = tan ( ) = π/4. This is not the angle we desire. The range of tan is ( π/, π/); that is, it returns angles that lie in the st and 4 th quadrants. To find loca ons in the nd and 3 rd quadrants, add π to the result of tan. So π + ( π/4) puts the angle at 3π/4. Thus the polar point is (, 3π/4). An alternate method is to use the angle θ given by arctangent, but change the sign of r. Thus we could also refer to (, ) as (, π/4). These points are plo ed in Figure 0.4 (b). The polar system is drawn lightly under the rectangular grid with rays to demonstrate the angles used. 606

281 0.4 Introduc on to Polar Coordinates Polar Func ons and Polar Graphs Defining a new coordinate system allows us to create a new kind of func on, a polar func on. Rectangular coordinates lent themselves well to crea ng func- ons that related and y, such as y =. Polar coordinates allow us to create func ons that relate r and θ. Normally these func ons look like r = f(θ), although we can create func ons of the form θ = f(r). The following eamples introduce us to this concept. Eample 3 Introduc on to Graphing Polar Func ons Describe the graphs of the following polar func ons.. r =.5. θ = π/4 S. The equa on r =.5 describes all points that are.5 units from the pole; as the angle is not specified, any θ is allowable. All points.5 units from the pole describes a circle of radius.5. We can consider the rectangular equivalent of this equa on; using r = +y, we see that.5 = +y, which we recognize as the equa on of a circle centered at (0, 0) with radius.5. This is sketched in Figure The equa on θ = π/4 describes all points such that the line through them and the pole make an angle of π/4 with the ini al ray. As the radius r is not specified, it can be any value (even nega ve). Thus θ = π/4 describes the line through the pole that makes an angle of π/4 = 45 with the ini al ray. We can again consider the rectangular equivalent of this equa on. Combine tan θ = y/ and θ = π/4: tan π/4 = y/ tan π/4 = y y =. This graph is also plo ed in Figure The basic rectangular equa ons of the form = h and y = k create ver cal and horizontal lines, respec vely; the basic polar equa ons r = h and θ = α create circles and lines through the pole, respec vely. With this as a founda on, we can create more complicated polar func ons of the form r = f(θ). The input is an angle; the output is a length, how far in the direc on of the angle to go out. We sketch these func ons much like we sketch rectangular and parametric func ons: we plot lots of points and connect the dots with curves. We demonstrate this in the following eample. Figure 0.43: plots. θ = π 4 r =.5 O Plo ng standard polar 607

282 Chapter 0 Curves in the Plane Eample 4 Sketching Polar Func ons Sketch the polar func on r = + cos θ on [0, π] by plo ng points. π 3π/4 5π/4 θ r = + cos θ 0 π/6 + 3/ π/4 + / π/3 3/ π/ π/3 / 3π/4 / 5π/6 3/ π 0 7π/6 3/ 5π/4 / 4π/3 / 3π/ 5π/3 3/ 7π/4 + / π/6 + 3/ π/ π/4 O 3π/ 7π/4 Figure 0.44: Graph of the polar func on in Eample 4 by plo ng points. S A common ques on when sketching curves by plo ng points is Which points should I plot? With rectangular equa ons, we o en chose easy values integers, then added more if needed. When plo ng polar equa ons, start with the common angles mul ples of π/6 and π/4. Figure 0.44 gives a table of just a few values of θ in [0, π]. Consider the point (, 0) determined by the first line of the table. The angle is 0 radians we do not rotate from the ini al ray then we go out units from the pole. When θ = π/6, r = + 3/; so rotate by π/6 radians and go out + 3/ units. Eample 5 Sketching Polar Func ons Sketch the polar func on r = cos(θ) on [0, π] by plo ng points. S We start by making a table of cos(θ) evaluated at common angles θ, as shown in Figure These points are then plo ed in Figure This par cular graph moves around quite a bit and one can easily forget which points should be connected to each other. To help us with this, we numbered each point in the table and on the graph. Pt. θ cos(θ) Pt. θ cos(θ) 0 0 7π/6 0.5 π/ π/4 0 3 π/4 0 4π/ π/ π/ 5 π/ 4 5π/ π/ π/ π/4 0 6 π/ π/ π 9 π Figure 0.46: Tables of points for plo ng a polar curve. This plot is an eample of a rose curve It is some mes desirable to refer to a graph via a polar equa on, and other mes by a rectangular equa on. Therefore it is necessary to be able to convert between polar and rectangular func ons, which we prac ce in the following eample. We will make frequent use of the iden es found in Key Idea Figure 0.45: Polar plots from Eample

283 0.4 Introduc on to Polar Coordinates Eample 6 Conver ng between rectangular and polar equa ons. Convert from rectangular to polar.. y =. y = Convert from polar to rectangular. 3. r = sin θ cos θ 4. r = cos θ S. Replace y with r sin θ and replace with r cos θ, giving: y = r sin θ = r cos θ sin θ cos θ = r We have found that r = sin θ/ cos θ = tan θ sec θ. The domain of this polar func on is ( π/, π/); plot a few points to see how the familiar parabola is traced out by the polar equa on.. We again replace and y using the standard iden es and work to solve for r: y = r cos θ r sin θ = r = cos θ sin θ r = cos θ sin θ This func on is valid only when the product of cos θ sin θ is posi ve. This occurs in the first and third quadrants, meaning the domain of this polar func on is (0, π/) (π, 3π/). We can rewrite the original rectangular equa on y = as y = /. This is graphed in Figure 0.47; note how it only eists in the first and third quadrants. 3. There is no set way to convert from polar to rectangular; in general, we look to form the products r cos θ and r sin θ, and then replace these with and y, respec vely. We start in this problem by mul plying both sides y Figure 0.47: Graphing y = from Eample

284 Chapter 0 Curves in the Plane by sin θ cos θ: r = sin θ cos θ r(sin θ cos θ) = r sin θ r cos θ =. Now replace with y and : y = y = +. The original polar equa on, r = /(sin θ cos θ) does not easily reveal that its graph is simply a line. However, our conversion shows that it is. The upcoming gallery of polar curves gives the general equa ons of lines in polar form. 4. By mul plying both sides by r, we obtain both an r term and an r cos θ term, which we replace with + y and, respec vely. r = cos θ r = r cos θ + y =. We recognize this as a circle; by comple ng the square we can find its radius and center. + y = 0 ( ) + y =. The circle is centered at (, 0) and has radius. The upcoming gallery of polar curves gives the equa ons of some circles in polar form; circles with arbitrary centers have a complicated polar equa on that we do not consider here. Some curves have very simple polar equa ons but rather complicated rectangular ones. For instance, the equa on r = + cos θ describes a cardioid (a shape important to the sensi vity of microphones, among other things; one is graphed in the gallery in the Limaçon sec on). It s rectangular form is not nearly as simple; it is the implicit equa on 4 + y 4 + y y 3 y = 0. The conversion is not hard, but takes several steps, and is le as an eercise. 60

285 0.4 Introduc on to Polar Coordinates Gallery of Polar Curves There are a number of basic and classic polar curves, famous for their beauty and/or applicability to the sciences. This sec on ends with a small gallery of some of these graphs. We encourage the reader to understand how these graphs are formed, and to inves gate with technology other types of polar func ons. Lines Through the origin: Horizontal line: Ver cal line: Not through origin: θ = α r = a csc θ r = a sec θ b r = sin θ m cos θ α { a a {}}{ } b slope = m 6

286 Circles Sprial Centered on origin: ( a ) + y = a 4 + (y a ) = a 4 Archimedean spiral r = a r = a cos θ r = a sin θ r = θ a a {}}{ a {}}{ Limaçons Symmetric about -ais: r = a ± b cos θ; Symmetric about y-ais: r = a ± b sin θ; a, b > 0 With inner loop: Cardioid: Dimpled: Conve: a b < a b = < a b < a b > Rose Curves Symmetric about -ais: r = a cos(nθ); Symmetric about y-ais: r = a sin(nθ) Curve contains n petals when n is even and n petals when n is odd. r = a cos(θ) r = a sin(θ) r = a cos(3θ) r = a sin(3θ) Special Curves Rose curves Lemniscate: Eight Curve: r = a sin(θ/5) r = a sin(θ/5) r = a cos(θ) r = a sec 4 θ cos(θ)

287 0.4 Introduc on to Polar Coordinates Earlier we discussed how each point in the plane does not have a unique representa on in polar form. This can be a good thing, as it allows for the beau ful and interes ng curves seen in the preceding gallery. However, it can also be a bad thing, as it can be difficult to determine where two curves intersect. Eample 7 Finding points of intersec on with polar curves Determine where the graphs of the polar equa ons r = +3 cos θ and r = cos θ intersect. S As technology is generally readily available, it is usually a good idea to start with a graph. We have graphed the two func ons in Figure 0.48(a); to be er discern the intersec on points, part (b) of the figure zooms in around the origin. We start by se ng the two func ons equal to each other and solving for θ: + 3 cos θ = cos θ cos θ = cos θ = θ = π 3, 4π 3. (There are, of course, infinite solu ons to the equa on cos θ = /; as the limaçon is traced out once on [0, π], we restrict our solu ons to this interval.) We need to analyze this solu on. When θ = π/3 we obtain the point of intersec on that lies in the 4 th quadrant. When θ = 4π/3, we get the point of intersec on that lies in the nd quadrant. There is more to say about this second intersec on point, however. The circle defined by r = cos θ is traced out once on [0, π], meaning that this point of intersec on occurs while tracing out the circle a second me. It seems strange to pass by the point once and then recognize it as a point of intersec on only when arriving there a second me. The first me the circle arrives at this point is when θ = π/3. It is key to understand that these two points are the same: (cos π/3, π/3) and (cos 4π/3, 4π/3). To summarize what we have done so far, we have found two points of intersec on: when θ = π/3 and when θ = 4π/3. When referencing the circle r = cos θ, the la er point is be er referenced as when θ = π/3. There is yet another point of intersec on: the pole (or, the origin). We did not recognize this intersec on point using our work above as each graph arrives at the pole at a different θ value. A graph intersects the pole when r = 0. Considering the circle r = cos θ, r = 0 when θ = π/ (and odd mul ples thereof, as the circle is repeatedly traced). π/ 0.5 (a) π/ (b) Figure 0.48: Graphs to help determine the points of intersec on of the polar func ons given in Eample

288 Chapter 0 Curves in the Plane The limaçon intersects the pole when + 3 cos θ = 0; this occurs when cos θ = /3, or for θ = cos ( /3). This is a nonstandard angle, approimately θ =.906 radians The limaçon intersects the pole twice in [0, π]; the other angle at which the limaçon is at the pole is the reflec on of the first angle across the -ais. That is, θ = If all one is concerned with is the (, y) coordinates at which the graphs intersect, much of the above work is etraneous. We know they intersect at (0, 0); we might not care at what θ value. Likewise, using θ = π/3 and θ = 4π/3 can give us the needed rectangular coordinates. However, in the net sec on we apply calculus concepts to polar func ons. When compu ng the area of a region bounded by polar curves, understanding the nuances of the points of intersec on becomes important. 64

289 Eercises 0.4 Terms and Concepts. In your own words, describe how to plot the polar point P(r, θ).. T/F: When plo ng a point with polar coordinate P(r, θ), r must be posi ve. 3. T/F: Every point in the Cartesian plane can be represented by a polar coordinate. 4. T/F: Every point in the Cartesian plane can be represented uniquely by a polar coordinate. Problems 5. Plot the points with the given polar coordinates. (a) A(, 0) (b) B(, π) (c) C(, π/) (d) D(, π/4) 6. Plot the points with the given polar coordinates. (a) A(, 3π) (b) B(, π) (c) C(, ) (d) D(/, 5π/6) 7. For each of the given points give two sets of polar coordinates that iden fy it, where 0 θ π. C D A O 3 B 8. For each of the given points give two sets of polar coordinates that iden fy it, where π θ π. D C A O 3 B 9. Convert the polar coordinates A and B to rectangular, and the rectangular coordinates C and D to polar. (a) A(, π/4) (b) B(, π/4) (c) C(, ) (d) D(, ) 0. Convert the polar coordinates A and B to rectangular, and the rectangular coordinates C and D to polar. (a) A(3, π) (b) B(, π/3) (c) C(0, 4) (d) D(, 3) In Eercises 3, graph the polar func on on the given interval.. r =, 0 θ π/. θ = π/6, r 3. r = cos θ, [0, π] 4. r = + sin θ, [0, π] 5. r = sin θ, [0, π] 6. r = sin θ, [0, π] 7. r = + sin θ, [0, π] 8. r = cos(θ), [0, π] 9. r = sin(3θ), [0, π] 0. r = cos(θ/3), [0, 3π]. r = cos(θ/3), [0, 6π]. r = θ/, [0, 4π] 3. r = 3 sin(θ), [0, π] 4. r = 4 sin(θ), [0, π] 5. r = cos(θ), [0, π] 6. r = 3 cos(θ), [0, π] 7. r = cos θ sin θ, [0, π] 8. r = θ (π/), [ π, π] 3 9. r =, 5 sin θ cos θ [0, π] 30. r =, 3 cos θ sin θ [0, π] 3. r = 3 sec θ, ( π/, π/) 3. r = 3 csc θ, (0, π) In Eercises 33 43, convert the polar equa on to a rectangular equa on. 33. r = cos θ 34. r = 4 sin θ 35. r = 3 sin(θ) 36. r = 3 cos(θ) 37. r = cos θ + sin θ r = 5 sin θ cos θ 39. r = 3 cos θ 40. r = 4 sin θ 4. r = tan θ 4. r = 43. θ = π 6 In Eercises 44 5, convert the rectangular equa on to a polar equa on. 44. y = 45. y = = y = = y 65

290 49. y = y = 7 5. ( + ) + y = In Eercises 5 59, find the points of intersec on of the polar graphs. 5. r = sin(θ) and r = cos θ on [0, π] 53. r = cos(θ) and r = cos θ on [0, π] 54. r = cos θ and r = sin θ on [0, π] 55. r = sin θ and r = sin θ on [0, π] 56. r = sin(3θ) and r = cos(3θ) on [0, π] 57. r = 3 cos θ and r = + cos θ on [ π, π] 58. r = and r = sin(θ) on [0, π] 59. r = cos θ and r = + sin θ on [0, π] 60. Pick a integer value for ( n, where n, 3, and use technology to plot r = sin m ) n θ for three different integer values of m. Sketch these and determine a minimal interval on which the en re graph is shown. 6. Create your own polar func on, r = f(θ) and sketch it. Describe why the graph looks as it does. 66

291 0.5 Calculus and Polar Func ons 0.5 Calculus and Polar Func ons The previous sec on defined polar coordinates, leading to polar func ons. We inves gated plo ng these func ons and solving a fundamental ques on about their graphs, namely, where do two polar graphs intersect? We now turn our a en on to answering other ques ons, whose solu ons require the use of calculus. A basis for much of what is done in this sec on is the ability to turn a polar func on r = f(θ) into a set of parametric equa ons. Using the iden es = r cos θ and y = r sin θ, we can create the parametric equa ons = f(θ) cos θ, y = f(θ) sin θ and apply the concepts of Sec on 0.3. Polar Func ons and dy d We are interested in the lines tangent to a given graph, regardless of whether that graph is produced by rectangular, parametric, or polar equa ons. In each of these contets, the slope of the tangent line is dy d. Given r = f(θ), we are generally not concerned with r = f (θ); that describes how fast r changes with respect to θ. Instead, we will use = f(θ) cos θ, y = f(θ) sin θ to compute dy d. Using Key Idea 39 we have dy d = dy / d dθ dθ. Each of the two deriva ves on the right hand side of the equality requires the use of the Product Rule. We state the important result as a Key Idea. Key Idea 44 Finding dy d with Polar Func ons Let r = f(θ) be a polar func on. With = f(θ) cos θ and y = f(θ) sin θ, dy dy d = dθ d dθ = f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ. Watch the video: The Slope of Tangent Lines to Polar Curves at 67

292 Chapter 0 Curves in the Plane Eample Finding dy d with polar func ons. Consider the limaçon r = + sin θ on [0, π].. Find the rectangular equa ons of the tangent and normal lines to the graph at θ = π/4.. Find where the graph has ver cal and horizontal tangent lines. S. We start by compu ng dy d. With f (θ) = cos θ, we have dy cos θ sin θ + cos θ( + sin θ) = d cos θ sin θ( + sin θ) cos θ(4 sin θ + ) = (cos θ sin θ) sin θ. When θ = π/4, dy d = (this requires a bit of simplifica on). In rectangular coordinates, the point on the graph at θ = π/4 is ( + /, + /). Thus the rectangular equa on of the line tangent to the limaçon at θ = π/4 is y = ( ) ( ( + /) ) + + / The limaçon and the tangent line are graphed in Figure The normal line has the opposite reciprocal slope as the tangent line, so its equa on is y To find the horizontal lines of tangency, we find where dy d = 0 (when the denominator does not equal 0); thus we find where the numerator of our equa on for dy d is 0. cos θ(4 sin θ + ) = 0 cos θ = 0 or 4 sin θ + = 0. On [0, π], cos θ = 0 when θ = π/, 3π/. Se ng 4 sin θ + = 0 gives θ = sin ( /4) 0.57 = We want the results in [0, π]; we also recognize there are two solu ons, one in the 3 rd quadrant and one in the 4 th. Using reference angles, we have our two solu ons as θ = 3.39 and 6.03 radians. The four points we obtained where the limaçon has a horizontal tangent line are given in Figure 0.49 with black filled dots. 68

293 0.5 Calculus and Polar Func ons To find the ver cal lines of tangency, we determine where dy d is undefined by se ng the denominator of dy d = 0 (when the numerator does not equal 0). (cos θ sin θ) sin θ = 0. Convert the cos θ term to sin θ: ( sin θ sin θ) sin θ = 0 4 sin θ + sin θ = 0. Recognize this as a quadra c in the variable sin θ. Using the quadra c formula, we have sin θ = ± We solve sin θ = and sin θ = 33 8 : sin θ = θ = sin ( ) sin θ = 33 8 θ = sin ( 33 8 θ θ.0030 In each of the solu ons above, we only get one of the possible two solu ons as sin only returns solu ons in [ π/, π/], the 4 th and st quadrants. Again using reference angles, we have: and sin θ = sin θ = 33 8 θ ,.5067 radians θ 4.446, 5.80 radians. These points are also shown in Figure 0.49 with white filled dots. ) 3 π/ 0 Figure 0.49: The limaçon in Eample with its tangent line at θ = π/4 and points of ver cal and horizontal tangency. When the graph of the polar func on r = f(θ) intersects the pole, it means that f(α) = 0 for some angle α. Making this subs tu on in the formula for dy d given in Key Idea 44 we see dy d = f (α) sin α + f(α) cos α f (α) cos α + f(α) sin α = sin α = tan α. cos α 69

294 Chapter 0 Curves in the Plane This equa on makes an interes ng point. It tells us the slope of the tangent line at the pole is tan α; some of our previous work (see, for instance, Eample 0.4.3) shows us that the line through the pole with slope tan α has polar equa on θ = α. Thus when a polar graph touches the pole at θ = α, the equa on of the tangent line at the pole is θ = α π/ Figure 0.50: Graphing the tangent lines at the pole in Eample. 0 Eample Finding tangent lines at the pole Let r = + sin θ, a limaçon. Find the equa ons of the lines tangent to the graph at the pole. S We need to know when r = 0. + sin θ = 0 sin θ = / θ = 7π 6, π 6. Thus the equa ons of the tangent lines, in polar, are θ = 7π/6 and θ = π/6. In rectangular form, the tangent lines are y = tan(7π/6) = 3 and y = tan(π/6) = 3. The full limaçon can be seen in Figure 0.49; we zoom in on the tangent lines in Figure Note: Recall that the area of a sector of a circle with radius r subtended by an angle θ is A = θr. θ r Area When using rectangular coordinates, the equa ons = h and y = k defined ver cal and horizontal lines, respec vely, and combina ons of these lines create rectangles (hence the name rectangular coordinates ). It is then somewhat natural to use rectangles to approimate area as we did when learning about the definite integral. When using polar coordinates, the equa ons θ = α and r = c form lines through the origin and circles centered at the origin, respec vely, and combina ons of these curves form sectors of circles. It is then somewhat natural to calculate the area of regions defined by polar func ons by first approima ng with sectors of circles. Consider Figure 0.5 (a) where a region defined by r = f(θ) on [α, β] is given. (Note how the sides of the region are the lines θ = α and θ = β, whereas in rectangular coordinates the sides of regions were o en the ver cal lines = a and = b.) Par on the interval [α, β] into n equally spaced subintervals as α = θ 0 < θ <... < θ n = β. The radian of each subinterval is θ = (β α)/n, represen ng a small change in angle. The area of the region defined by the i th 60

295 0.5 Calculus and Polar Func ons subinterval [θ i, θ i ] can be approimated with a sector of a circle with radius f(c i ), for some c i in [θ i, θ i ]. The area of this sector is [f(c i)] θ. This is shown in part (b) of the figure, where [α, β] has been divided into 4 subintervals. We approimate the area of the whole region by summing the areas of all sectors: Area n i= [f(c i)] θ. This is a Riemann sum. By taking the limit of the sum as n, we find the eact area of the region in the form of a definite integral. 0.5 π/ θ = β r = f(θ) Theorem 85 Area of a Polar Region Let f be con nuous and non-nega ve on [α, β], where 0 β α π. The area A of the region bounded by the curve r = f(θ) and the lines θ = α and θ = β is π/ θ = α 0.5 (a) 0 A = β α [f(θ)] dθ = β α r dθ θ = β r = f(θ) The theorem states that 0 β α π. This ensures that region does not overlap itself, which would give a result that does not correspond directly to the area. Eample 3 Area of a polar region Find the area of the circle defined by r = cos θ. (Recall this circle has radius /.) 0.5 θ = α 0.5 (b) 0 S This is a direct applica on of Theorem 85. The circle is traced out on [0, π], leading to the integral Area = π cos θ dθ 0 = π + cos(θ) dθ 0 = π ( θ + 4 sin(θ)) 0 = π 4. Of course, we already knew the area of a circle with radius /. We did this eample to demonstrate that the area formula is correct. Figure 0.5: Compu ng the area of a polar region. Note: Eample 3 requires the use of the integral cos θ dθ. This is handled well by using the half angle formula as found in the back of this tet. Due to the nature of the area formula, integra ng cos θ and sin θ is required o en. We offer here these indefinite integrals as a me saving measure. cos θ dθ = θ + 4 sin(θ) + C sin θ dθ = θ 4 sin(θ) + C 6

296 Chapter 0 Curves in the Plane Eample 4 Area of a polar region Find the area of the cardioid r = +cos θ bound between θ = π/6 and θ = π/3, as shown in Figure 0.5. π/ θ = π/3 θ = π/6 0 S This is again a direct applica on of Theorem 85. Area = π/3 π/6 π/3 ( + cos θ) dθ = ( + cos θ + cos θ) dθ π/6 = [θ + sin θ + θ + 4 ] π/3 sin(θ) π/6 Figure 0.5: Finding the area of the shaded region of a cardioid in Eample 4. = 8( π ). Area Between Curves 0.5 π/ r = f (θ) θ = β θ = α r = f (θ) 0.5 Figure 0.53: Illustra ng area bound between two polar curves. 0 Our study of area in the contet of rectangular func ons led naturally to finding area bounded between curves. We consider the same in the contet of polar func ons. Consider the shaded region shown in Figure We can find the area of this region by compu ng the area bounded by r = f (θ) and subtrac ng the area bounded by r = f (θ) on [α, β]. Thus Area = β α r dθ β α r dθ = β α ( r r ) dθ. Key Idea 45 Area Between Polar Curves The area A of the region bounded by r = f (θ) and r = f (θ), θ = α and θ = β, where f (θ) f (θ) on [α, β], is A = β α [f (θ)] [f (θ)] dθ = β α ( r r ) dθ. π/ Eample 5 Area between polar curves Find the area bounded between the curves r = + cos θ and r = 3 cos θ, as shown in Figure Figure 0.54: Finding the area between polar curves in Eample 5. 6

297 0.5 Calculus and Polar Func ons S We need to find the points of intersec on between these two func ons. Se ng them equal to each other, we find: + cos θ = 3 cos θ cos θ = / θ = ±π/3 Thus we integrate ( (3 cos θ) ( + cos θ) ) on [ π/3, π/3]. Area = = π/3 π/3 π/3 π/3 ( (3 cos θ) ( + cos θ) ) dθ ( 8 cos θ cos θ ) dθ = ( ) π/3 sin(θ) sin θ + 3θ π/3 = π. Amazingly enough, the area between these curves has a nice value. Eample 6 Area defined by polar curves Find the area bounded between the polar curves r = and r = cos(θ), as shown in Figure 0.55 (a). π/ S We need to find the point of intersec on between the two curves. Se ng the two func ons equal to each other, we have 0 cos(θ) = cos(θ) = θ = π/3 θ = π/6. In part (b) of the figure, we zoom in on the region and note that it is not really bounded between two polar curves, but rather by two polar curves, along with θ = 0. The dashed line breaks the region into its component parts. Below the dashed line, the region is defined by r =, θ = 0 and θ = π/6. (Note: the dashed line lies on the line θ = π/6.) Above the dashed line the region is bounded by r = cos(θ) and θ = π/6. Since we have two separate regions, we find the area using two separate integrals. Call the area below the dashed line A and the area above the dashed line A. They are determined by the following integrals: π/ 0.5 (a) A = π/6 0 () dθ A = π/4 π/6 ( cos(θ) ) dθ. 0.5 (b) 0 Figure 0.55: Graphing the region bounded by the func ons in Eample 6. 63

298 Chapter 0 Curves in the Plane (The upper bound of the integral compu ng A is π/4 as r = cos(θ) is at the pole when θ = π/4.) We omit the integra on details and let the reader verify that A = π/ and A = π/ 3/8; the total area is A = π/6 3/8. Arc Length As we have already considered the arc length of curves defined by rectangular and parametric equa ons, we now consider it in the contet of polar equa ons. Recall that the arc length L of the graph defined by the parametric equa ons = f(t), y = g(t) on [a, b] is L = b a b [f (t)] + [g (t)] dt = [ (t)] + [y (t)] dt. (0.) Now consider the polar func on r = f(θ). We again use the iden es = f(θ) cos θ and y = f(θ) sin θ to create parametric equa ons based on the polar func on. We compute (θ) and y (θ) as done before when compu ng dy d, then apply Equa on (0.). The epression [ (θ)] + [y (θ)] can be simplified a great deal; we leave this as an eercise and state that This leads us to the arc length formula. [ (θ)] + [y (θ)] = [f (θ)] + [f(θ)]. a Key Idea 46 Arc Length of Polar Curves Let r = f(θ) be a polar func on with f con nuous on an open interval I containing [α, β], on which the graph traces itself only once. The arc length L of the graph on [α, β] is L = β α β [f (θ)] + [f(θ)] dθ = (r ) + r dθ. α Eample 7 Arc Length of Polar Curves Find the arc length of the cardioid r = + cos θ. 64

299 0.5 Calculus and Polar Func ons S With r = + cos θ, we have r = sin θ. The cardioid is traced out once on [0, π], giving us our bounds of integra on. Applying Key Idea 46 we have L = = = = = = = π 0 π 0 π 0 π 0 π 0 π 0 π 0 ( sin θ) + ( + cos θ) dθ sin θ + ( + cos θ + cos θ) dθ + cos θ dθ + cos θ cos θ cos θ dθ 4 4 cos θ cos θ dθ cos θ cos θ dθ sin θ cos θ dθ Since the sin θ > 0 on [0, π] and sin θ < 0 on [π, π] we separate the integral into two parts π 0 sin θ π dθ cos θ π sin θ cos θ dθ Using the symmetry of the cardioid and u-subs tu on (u = cos θ) we simplify the integra on to π sin θ L = 4 dθ cos θ = u du 4 = 4u / = 8. 0 Eample 8 Arc length of a limaçon Find the arc length of the limaçon r = + sin t. 65

300 Chapter 0 Curves in the Plane S With r = + sin t, we have r = cos t. The limaçon is traced out once on [0, π], giving us our bounds of integra on. Applying Key Idea 46, we have π/ 3 0 L = = = π 0 π 0 π ( cos θ) + ( + sin θ) dθ 4 cos θ + 4 sin θ + 4 sin θ + dθ 4 sin θ + 5 dθ The final integral cannot be solved in terms of elementary func ons, so we resorted to a numerical approima on. (Simpson s Rule, with n = 4, approimates the value with Using n = gives the value above, which is accurate to 4 places a er the decimal.) Figure 0.56: The limaçon in Eample 8 whose arc length is measured. Surface Area The formula for arc length leads us to a formula for surface area. The following Key Idea is based on Key Idea 4. Key Idea 47 Surface Area of a Solid of Revolu on Consider the graph of the polar equa on r = f(θ), where f is con nuous on an open interval containing [α, β] on which the graph does not cross itself.. The surface area of the solid formed by revolving the graph about the ini al ray (θ = 0) is: Surface Area = π β α f(θ) sin θ [f (θ)] + [f(θ)] dθ.. The surface area of the solid formed by revolving the graph about the line θ = π/ is: Surface Area = π β α f(θ) cos θ [f (θ)] + [f(θ)] dθ. 66

301 0.5 Calculus and Polar Func ons Eample 9 Surface area determined by a polar curve Find the surface area formed by revolving one petal of the rose curve r = cos(θ) about its central ais (see Figure 0.57). S We choose, as implied by the figure, to revolve the por on of the curve that lies on [0, π/4] about the ini al ray. Using Key Idea 47 and the fact that f (θ) = sin(θ), we have Surface Area = π π/ cos(θ) sin(θ) ( sin(θ) ) + ( cos(θ) ) dθ π/ 0 The integral is another that cannot be evaluated in terms of elementary func- ons. Simpson s Rule, with n = 4, approimates the value at This chapter has been about curves in the plane. While there is great mathema cs to be discovered in the two dimensions of a plane, we live in a three dimensional world and hence we should also look to do mathema cs in 3D that is, in space. The net chapter begins our eplora on into space by introducing the topic of vectors, which are incredibly useful and powerful mathema cal objects. (a) (b) Figure 0.57: Finding the surface area of a rose curve petal that is revolved around its central ais. 67